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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 8th 2011
   • (edited Jun 8th 2011)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexLet $\mathfrak{g}$ be an $L_\infty$-algebra (over a characteristic zero field $k$). then, if $\mathfrak{g}$ is finite-dimensional in each degree, the $L_\infty$-algebra $inn(\mathfrak{g})$ can be defined as the $L_\infty$-algebra whose Chevalley-Eilenberg algebra is the [[Weil algebra]] $W(\mathfrak{g})$ of $\mathfrak{g}$. The Weil algebra has a remarkable freeness property: $$ Hom_{dgca}(W(\mathfrak{g}),\Omega^\bullet)=Hom_{dgVect}(\mathfrak{g}^*[-1],\Omega^\bullet)=\Omega^1(\mathfrak{g}), $$ where $\Omega^\bullet$ is an arbitrary dgca and $\Omega^i(\mathfrak{g})$ denotes the vector space of degree $i$ elements of $\Omega^\bullet\otimes\mathfrak{g}$. The identification $W(\mathfrak{g})=CE(inn{\mathfrak{g})$ can then be used to define _curvature_ of $\mathfrak{g}$-connections, and to write down their Bianchi identities. Indeed $$ Hom_{dgca}(CE(inn(\mathfrak{g})),\Omega^\bullet)=Maurer-Cartan\, elements\, in\, \Omega^\bullet(inn(\mathfrak{g})) $$ and the underlying graded vector space of $inn(\mathfrak{g})$ is $\mathfrak{g}\oplus\mathfrak{g}[1]$, so that the identification $W(\mathfrak{g})=CE(inn{\mathfrak{g})$ together with the freeness property of $W(\mathfrak{g})$ gives the following: for any $A\in \Omega^1(\mathfrak{g})$ there exists a unique $F_A\in \Omega^2(\mathfrak{g})$ such that the pair $(A,F_A)$ satisfies the Maurer-Cartan equation in $\Omega^\bullet(inn(\mathfrak{g})$. The element $F_A$ is the curvature of $A$ and the Maurer-Cartan equation expresses both the relation between $A$ and $F_A$ and the Bianchi identities. To write these equations in a fully explicit form in a way that allows making contact with classical equations from differential geometry, it would be convenient to have the $L_\infty$-algebra structure on $inn(\mathfrak{g})$ spelled out in terms of lots of brackets (in terms of the lots of brackets defining the $L_\infty$-algebra sturcuture on $\mathfrak{g}$). Do we have these brackets already spelled out somewhere?

   Let $\mathfrak{g}$ be an $L_\infty$-algebra (over a characteristic zero field $k$). then, if $\mathfrak{g}$ is finite-dimensional in each degree, the $L_\infty$-algebra $inn(\mathfrak{g})$ can be defined as the $L_\infty$-algebra whose Chevalley-Eilenberg algebra is the Weil algebra $W(\mathfrak{g})$ of $\mathfrak{g}$. The Weil algebra has a remarkable freeness property:

   $$Hom_{dgca}(W(\mathfrak{g}),\Omega^\bullet)=Hom_{dgVect}(\mathfrak{g}^*[-1],\Omega^\bullet)=\Omega^1(\mathfrak{g}),$$

   where $\Omega^\bullet$ is an arbitrary dgca and $\Omega^i(\mathfrak{g})$ denotes the vector space of degree $i$ elements of $\Omega^\bullet\otimes\mathfrak{g}$. The identification can then be used to define curvature of $\mathfrak{g}$-connections, and to write down their Bianchi identities. Indeed

   $$Hom_{dgca}(CE(inn(\mathfrak{g})),\Omega^\bullet)=Maurer-Cartan\, elements\, in\, \Omega^\bullet(inn(\mathfrak{g}))$$

   and the underlying graded vector space of $inn(\mathfrak{g})$ is $\mathfrak{g}\oplus\mathfrak{g}[1]$, so that the identification together with the freeness property of $W(\mathfrak{g})$ gives the following:

   for any $A\in \Omega^1(\mathfrak{g})$ there exists a unique $F_A\in \Omega^2(\mathfrak{g})$ such that the pair $(A,F_A)$ satisfies the Maurer-Cartan equation in $\Omega^\bullet(inn(\mathfrak{g})$. The element $F_A$ is the curvature of $A$ and the Maurer-Cartan equation expresses both the relation between $A$ and $F_A$ and the Bianchi identities.

   To write these equations in a fully explicit form in a way that allows making contact with classical equations from differential geometry, it would be convenient to have the $L_\infty$-algebra structure on $inn(\mathfrak{g})$ spelled out in terms of lots of brackets (in terms of the lots of brackets defining the $L_\infty$-algebra sturcuture on $\mathfrak{g}$). Do we have these brackets already spelled out somewhere?

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJun 8th 2011
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   Author: Urs
   Format: MarkdownItexHi Domenico, no, I don't think we have written out these $L_\infty$-brakcets already. (By the way, after it says "Indeed" in your above message, the intended equation does not display. I think maybe you need whitespace around the double dollar signs, or something like that.)

   Hi Domenico,

   no, I don’t think we have written out these $L_\infty$-brakcets already.

   (By the way, after it says “Indeed” in your above message, the intended equation does not display. I think maybe you need whitespace around the double dollar signs, or something like that.)

3. • CommentRowNumber3.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 8th 2011
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItex(The first `inn` is followed by a `{` without a matching closing one. I suspect that it ought to be a `(` instead.)

   (The first inn is followed by a { without a matching closing one. I suspect that it ought to be a ( instead.)

4. • CommentRowNumber4.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 8th 2011
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexHi Urs, > no, I don't think we have written out these $L_\infty$-brakcets already. ok, then I'll try to work them out. as you can imagine I'm interested in this in connection with local differential form data for string connections > (The first `inn` is followed by a `{`... thanks Andrew, I've now edited above

   Hi Urs,

   no, I don’t think we have written out these $L_\infty$-brakcets already.

   ok, then I’ll try to work them out. as you can imagine I’m interested in this in connection with local differential form data for string connections

   (The first inn is followed by a {…

   thanks Andrew, I’ve now edited above

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 9th 2011
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexIn the end (if I didn't get lost in combinatorics), that was not so hard. In terms of a lot of brackets, the defining equation of curvature and the Bianchi identity (which are the expression I was after, rather than the brackets themeselves) read (up to some sign that I have almost surely missed) $$ F_A=d_\Omega A+\sum_{n=1}^\infty\frac{1}{n!}[A,A,\dots,A]_n $$ and $$ d_\Omega F_A=\sum_{n=1}^\infty\frac{1}{(n-1)!}[A,A,\dots,A,F_A]_n $$ where $A\in \Omega^1(\mathfrak{g})$, and $[\,,\dots,\,]_n$ are the brackets on the $L_\infty$-algebra $\mathfrak{g}$; the curvature $F_A$ is an element in $\Omega^2(\mathfrak{g})$. In particular, when $\mathfrak{g}$ is a Lie algebra, we recover the usual formulae $$ F_A=d_\Omega A+\frac{1}{2}[A,A] $$ and $$ d_\Omega F_A=[A,F_A] $$ Another example is the Lie 3-algebra $\mathfrak{g}=(b\mathbb{R}\to\mathfrak{string})$ described at [[differential string structure]]. Here the connection $A$ will have three components: a $\mathfrak{so}$-valued 1-form $\omega$, an $\mathbb{R}$-valued 2-form $B$ and an $\mathbb{R}$-valued 3-form $C$ (taking values in another copy of $\mathbb{R}$!! to explicitly read this in the following formulae, we will write $(B,0)$ and $(0,C)$ when confusion is possible). Also the curvature $F_A$ will have 3-components: a $\mathfrak{so}$-valued 2-form $F$, an $\mathbb{R}$-valued 3-form $H$ and an $\mathbb{R}$-valued 4-form $G$. The defining equation for $F_A$ then gives $$ F+H+G=d_\omega \omega+d_\omega B+d_\Omega C + [\omega]+[B]+[C]+\frac{1}{2}[\omega,\omega]+[\omega,B]+[\omega,C]+\frac{1}{2}[B,B]+[B,C]+\frac{1}{2}[C,C]+ $$ $$ +\frac{1}{6}[\omega,\omega,\omega]+\frac{1}{2}[\omega,\omega,B]+\frac{1}{2}[\omega,\omega,C]+\frac{1}{2}[\omega,B,B]+[\omega,B,C]+\frac{1}{2}[\omega,C,C] $$ $$ +\frac{1}{6}[B,B,B]+\frac{1}{2}[B,B,C]+\frac{1}{2}[B,B,C]+\frac{1}{6}[C,C,C], $$ since the brackets $[\,,\dots,\,]_n$ of $(b\mathbb{R}\to\mathfrak{string})$ are trivial for $n\geq 4$. Recalling the definition of the brackets $[\,,\dots,\,]_n$ of $(b\mathbb{R}\to\mathfrak{string})$ for $n=1,2,3$, the above equation reduces to $$ F+H+G=d_\omega \omega+d_\omega B+d_\Omega C + (C,0)+\frac{1}{2}[\omega,\omega]+\frac{1}{6}\mu(\omega,\omega,\omega) $$ and so, looking at the various components, $$ F=d_\Omega \omega+\frac{1}{2}[\omega,\omega]; \qquad H=d_\Omega B+C+\frac{1}{6}\mu(\omega,\omega,\omega); \qquad G=d_\Omega C $$ Similarly, the Bianchi identities are $$ d_\Omega (F+H+G)=(G,0)+[\omega,F]+\frac{1}{2}\mu(\omega,\omega,F) $$ i.e., $$ d_\Omega F=[\omega,F];\qquad d_\Omega H=G+\frac{1}{2}\mu(\omega,\omega,F);\qquad d_\Omega G=0. $$ These are best written in terms of a [[Chern-Simons element]] giving the transgression between the 3-cocycle $\mu$ and the Killing form of $\mathfrak{so}$. This will be done in the following post.

   In the end (if I didn’t get lost in combinatorics), that was not so hard. In terms of a lot of brackets, the defining equation of curvature and the Bianchi identity (which are the expression I was after, rather than the brackets themeselves) read (up to some sign that I have almost surely missed)

   $$F_A=d_\Omega A+\sum_{n=1}^\infty\frac{1}{n!}[A,A,\dots,A]_n$$

   and

   $$d_\Omega F_A=\sum_{n=1}^\infty\frac{1}{(n-1)!}[A,A,\dots,A,F_A]_n$$

   where $A\in \Omega^1(\mathfrak{g})$, and $[\,,\dots,\,]_n$ are the brackets on the $L_\infty$-algebra $\mathfrak{g}$; the curvature $F_A$ is an element in $\Omega^2(\mathfrak{g})$.

   In particular, when $\mathfrak{g}$ is a Lie algebra, we recover the usual formulae

   $$F_A=d_\Omega A+\frac{1}{2}[A,A]$$

   and

   $$d_\Omega F_A=[A,F_A]$$

   Another example is the Lie 3-algebra $\mathfrak{g}=(b\mathbb{R}\to\mathfrak{string})$ described at differential string structure. Here the connection $A$ will have three components: a $\mathfrak{so}$-valued 1-form $\omega$, an $\mathbb{R}$-valued 2-form $B$ and an $\mathbb{R}$-valued 3-form $C$ (taking values in another copy of $\mathbb{R}$!! to explicitly read this in the following formulae, we will write $(B,0)$ and $(0,C)$ when confusion is possible). Also the curvature $F_A$ will have 3-components: a $\mathfrak{so}$-valued 2-form $F$, an $\mathbb{R}$-valued 3-form $H$ and an $\mathbb{R}$-valued 4-form $G$. The defining equation for $F_A$ then gives

   $$F+H+G=d_\omega \omega+d_\omega B+d_\Omega C + [\omega]+[B]+[C]+\frac{1}{2}[\omega,\omega]+[\omega,B]+[\omega,C]+\frac{1}{2}[B,B]+[B,C]+\frac{1}{2}[C,C]+$$ $$+\frac{1}{6}[\omega,\omega,\omega]+\frac{1}{2}[\omega,\omega,B]+\frac{1}{2}[\omega,\omega,C]+\frac{1}{2}[\omega,B,B]+[\omega,B,C]+\frac{1}{2}[\omega,C,C]$$ $$+\frac{1}{6}[B,B,B]+\frac{1}{2}[B,B,C]+\frac{1}{2}[B,B,C]+\frac{1}{6}[C,C,C],$$

   since the brackets $[\,,\dots,\,]_n$ of $(b\mathbb{R}\to\mathfrak{string})$ are trivial for $n\geq 4$. Recalling the definition of the brackets $[\,,\dots,\,]_n$ of $(b\mathbb{R}\to\mathfrak{string})$ for $n=1,2,3$, the above equation reduces to

   $$F+H+G=d_\omega \omega+d_\omega B+d_\Omega C + (C,0)+\frac{1}{2}[\omega,\omega]+\frac{1}{6}\mu(\omega,\omega,\omega)$$

   and so, looking at the various components,

   $$F=d_\Omega \omega+\frac{1}{2}[\omega,\omega]; \qquad H=d_\Omega B+C+\frac{1}{6}\mu(\omega,\omega,\omega); \qquad G=d_\Omega C$$

   Similarly, the Bianchi identities are

   $$d_\Omega (F+H+G)=(G,0)+[\omega,F]+\frac{1}{2}\mu(\omega,\omega,F)$$

   i.e.,

   $$d_\Omega F=[\omega,F];\qquad d_\Omega H=G+\frac{1}{2}\mu(\omega,\omega,F);\qquad d_\Omega G=0.$$

   These are best written in terms of a Chern-Simons element giving the transgression between the 3-cocycle $\mu$ and the Killing form of $\mathfrak{so}$. This will be done in the following post.

6. • CommentRowNumber6.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 9th 2011
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   Author: domenico_fiorenza
   Format: MarkdownItexRecalling that on $\mathfrak{so}$ $$ \mu(x,y,z)=\langle x,[y,z]\rangle + \langle y,[z,x]\rangle + \langle z,[x,y]\rangle $$ and using $[\omega,F]=[F,\omega]$ since both $\omega$ and $F$ have degree 1 in $\Omega^\bullet(inn(\mathfrak{g}))$, we have $$ \frac{1}{2}\mu(\omega,\omega,F)=\langle \omega,[\omega,F]\rangle+\frac{1}{2}\langle F,[\omega,\omega]\rangle=\langle \omega,d_\Omega F\rangle+\langle F,\frac{1}{2}[\omega,\omega]\rangle=d_\Omega\langle \omega,F\rangle+\langle F,F\rangle $$ Thus we can rewrite equation $$ d_\Omega H=G+\frac{1}{2}\mu(\omega,\omega,F) $$ as $$ d_\Omega(H-\langle \omega,F\rangle)=G+\langle F,F\rangle $$ It is therefore convenient to set $\tilde{H}=H-\langle \omega,F\rangle$. In terms of this new field, the Bianchi identity reads $$ d_\Omega \tilde{H}=G+\langle F,F\rangle $$ while the defining equation of the curvature component $H$ is translated into $$ \tilde{H}=d_\Omega B+C+\frac{1}{6}\langle\omega,[\omega,\omega]\rangle+\langle \omega,d_\Omega\omega\rangle+\frac{1}{2}\langle \omega,[\omega,\omega]\rangle= d_\Omega B+C+\left(\langle \omega,d_\Omega \omega\rangle+\frac{2}{3}\langle \omega,[\omega,\omega]\rangle\right)=d_\Omega B+C+\mathbf{cs}(\omega) $$ (there are a few signs and coefficients to be fixed)

   Recalling that on $\mathfrak{so}$

   $$\mu(x,y,z)=\langle x,[y,z]\rangle + \langle y,[z,x]\rangle + \langle z,[x,y]\rangle$$

   and using $[\omega,F]=[F,\omega]$ since both $\omega$ and $F$ have degree 1 in $\Omega^\bullet(inn(\mathfrak{g}))$, we have

   $$\frac{1}{2}\mu(\omega,\omega,F)=\langle \omega,[\omega,F]\rangle+\frac{1}{2}\langle F,[\omega,\omega]\rangle=\langle \omega,d_\Omega F\rangle+\langle F,\frac{1}{2}[\omega,\omega]\rangle=d_\Omega\langle \omega,F\rangle+\langle F,F\rangle$$

   Thus we can rewrite equation

   $$d_\Omega H=G+\frac{1}{2}\mu(\omega,\omega,F)$$

   as

   $$d_\Omega(H-\langle \omega,F\rangle)=G+\langle F,F\rangle$$

   It is therefore convenient to set $\tilde{H}=H-\langle \omega,F\rangle$. In terms of this new field, the Bianchi identity reads

   $$d_\Omega \tilde{H}=G+\langle F,F\rangle$$

   while the defining equation of the curvature component $H$ is translated into

   $$\tilde{H}=d_\Omega B+C+\frac{1}{6}\langle\omega,[\omega,\omega]\rangle+\langle \omega,d_\Omega\omega\rangle+\frac{1}{2}\langle \omega,[\omega,\omega]\rangle= d_\Omega B+C+\left(\langle \omega,d_\Omega \omega\rangle+\frac{2}{3}\langle \omega,[\omega,\omega]\rangle\right)=d_\Omega B+C+\mathbf{cs}(\omega)$$

   (there are a few signs and coefficients to be fixed)

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJun 9th 2011
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   Author: Urs
   Format: MarkdownItexThanks, looks good. Are you putting this into the entries?

   Thanks, looks good. Are you putting this into the entries?

8. • CommentRowNumber8.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 9th 2011
   • (edited Jun 9th 2011)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexYes, as soon as I'm able to fix signs and factorials :) (hopefully this won't take long) A large part of what I wrote above should go into [[differential string structure]]. but there's a difference with the approach you are currently having there, so let us discuss this here before I change the entry: it seems to me that the final form of the equations one gets (the ones with $\tilde{H}$) is interesting since it reproduces formulae from physics literature, whereas the formulae from the math machinery would be those with $H$ (in the notations of the above posts). In your description you switch to $\tilde{H}$ at an early stage, introducing the algebra $\tilde{W}$, which is isomorphic to the Weil algebra one considers here. In the descrition I'm sketching above, on the contrary, I'm following the math machinery till the end, and only then I'm manipulating the formulae. But to adopt this point of view I will also have to rewrite completely the proof of Proposition 9 there (it is now written in terms of the "modified" Weil algebra). I think I can do this: I'll write my proof here in a series of posts as I'm able to work on that.

   Yes, as soon as I’m able to fix signs and factorials :) (hopefully this won’t take long)

   A large part of what I wrote above should go into differential string structure. but there’s a difference with the approach you are currently having there, so let us discuss this here before I change the entry: it seems to me that the final form of the equations one gets (the ones with $\tilde{H}$) is interesting since it reproduces formulae from physics literature, whereas the formulae from the math machinery would be those with $H$ (in the notations of the above posts). In your description you switch to $\tilde{H}$ at an early stage, introducing the algebra $\tilde{W}$, which is isomorphic to the Weil algebra one considers here. In the descrition I’m sketching above, on the contrary, I’m following the math machinery till the end, and only then I’m manipulating the formulae. But to adopt this point of view I will also have to rewrite completely the proof of Proposition 9 there (it is now written in terms of the “modified” Weil algebra). I think I can do this: I’ll write my proof here in a series of posts as I’m able to work on that.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeJun 10th 2011
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   Author: Urs
   Format: MarkdownItexI see. I used the isomorphism to the modified Weil algebra as a tool to take care of the horizontality constraint on the curvature invariants. That isomorphism "shifts the curvatures into horizontal form". So I'd imagine if you don't use this, you get more involved constraints on your forms? Or what do you do?

   I see. I used the isomorphism to the modified Weil algebra as a tool to take care of the horizontality constraint on the curvature invariants. That isomorphism “shifts the curvatures into horizontal form”.

   So I’d imagine if you don’t use this, you get more involved constraints on your forms? Or what do you do?

10. • CommentRowNumber10.
    • CommentAuthorjim_stasheff
    • CommentTimeJun 10th 2011
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    Author: jim_stasheff
    Format: TextDomenico: F A=d ΩA+∑ n=1 ∞1n![A,A,…,A] n and d ΩF A=∑ n=1 ∞1(n−1)![A,A,…,A,F A] n Notice the first corresponds to the infty-MC eqn - as one would expect I thought you were looking for some formula special to inn, but these are indeed the universal ones. As to modified Weil alg, where is that? how does it relate to Cartan versus Weil?

    Domenico:
    F A=d ΩA+∑ n=1 ∞1n![A,A,…,A] n
    
    and
    
    
    d ΩF A=∑ n=1 ∞1(n−1)![A,A,…,A,F A] n
    
    Notice the first corresponds to the infty-MC eqn - as one would expect
    
    I thought you were looking for some formula special to inn, but these are indeed the universal ones.
    
    As to modified Weil alg, where is that? how does it relate to Cartan versus Weil?
11. • CommentRowNumber11.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 10th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexUrs: >So I'd imagine if you don't use this, you get more involved constraints on your forms? Or what do you do? I'm just going to hide the role of the shift in the computations, not to make another algebra come explicitly into play. Jim: > I thought you were looking for some formula special to inn, but these are indeed the universal ones. well, these formulas are indeed special to inn: they are the _single_ MC equation in $inn(\mathfrak{g})$; then, splitting $inn(\mathfrak{g})$ into two copies of $\mathfrak{g}$ (one of which is shifted) one gets the _two_ equations above. >As to modified Weil alg, where is that? how does it relate to Cartan versus Weil? That is some auxiliary dgca currently appearing in [[differential string structure]]. I'm now working at rewriting the proofs there in a way that does not involve this auxiliary algebra explicitly.

    Urs:

    So I’d imagine if you don’t use this, you get more involved constraints on your forms? Or what do you do?

    I’m just going to hide the role of the shift in the computations, not to make another algebra come explicitly into play.

    Jim:

    I thought you were looking for some formula special to inn, but these are indeed the universal ones.

    well, these formulas are indeed special to inn: they are the single MC equation in $inn(\mathfrak{g})$; then, splitting $inn(\mathfrak{g})$ into two copies of $\mathfrak{g}$ (one of which is shifted) one gets the two equations above.

    As to modified Weil alg, where is that? how does it relate to Cartan versus Weil?

    That is some auxiliary dgca currently appearing in differential string structure. I’m now working at rewriting the proofs there in a way that does not involve this auxiliary algebra explicitly.

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeJun 10th 2011
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    Author: Urs
    Format: MarkdownItex> I'm just going to hide the role of the shift in the computations, not to make another algebra come explicitly into play. Oh, I see, that makes sense. My use of these modified Weil algebras is maybe historical: the way I originally found the relation of twisted string bundles to Chern-Simons gerbes was by observing that there are these "Chern-Simons L-oo algebras", as I originally called them. So me, personally, I found that theory of "twisted differential string structures" from the fact that these modified Weil algebras exist. But I'll accept if you say that that is not actually the best way to discuss the theory.

    I’m just going to hide the role of the shift in the computations, not to make another algebra come explicitly into play.

    Oh, I see, that makes sense. My use of these modified Weil algebras is maybe historical: the way I originally found the relation of twisted string bundles to Chern-Simons gerbes was by observing that there are these “Chern-Simons L-oo algebras”, as I originally called them. So me, personally, I found that theory of “twisted differential string structures” from the fact that these modified Weil algebras exist. But I’ll accept if you say that that is not actually the best way to discuss the theory.

13. • CommentRowNumber13.
    • CommentAuthorjim_stasheff
    • CommentTimeJun 11th 2011
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    Author: jim_stasheff
    Format: TextDomenico: well, these formulas are indeed special to inn: they are the single MC equation in inn(

    Domenico:
    well, these formulas are indeed special to inn: they are the single MC equation in
    inn(
14. • CommentRowNumber14.
    • CommentAuthorjim_stasheff
    • CommentTimeJun 11th 2011
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    Author: jim_stasheff
    Format: TextThe rest of my comment disappeared? I had overlooked the Omega in your formulas. So the emphasis for inn is on the splitting? how/why did this ? originate?

    The rest of my comment disappeared?
    I had overlooked the Omega in your formulas.
    So the emphasis for inn is on the splitting?
    how/why did this ? originate?
15. • CommentRowNumber15.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 11th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> The rest of my comment disappeared? apparently.. :) > So the emphasis for inn is on the splitting? how/why did this ? originate? what I'm stressing here is just the "two-sided" nature of the Weil algebra: on one side it is a free object, so a morphism out of the Weil algebra is just a degree 1 $\mathfrak{g}$-valued differential form: the local connection form; on the other side, the Weil algebra of $\mathfrak{g}$ is the Chevalley-Eilenberg algebra of $inn(\mathfrak{g})$, so morphisms out of it are solutions of the MC equation. this means that each connection $A$ comes with a unique degree 2 $\mathfrak{g}$-valued form $F_A$, such that the pair $(A,F_A)$ is a solution of the MC equation for $inn(\mathfrak{g})$-valued differential forms. this single MC equation is _both_ the defining equation of the curvature $F_A$ and the Bianchi identity.

    The rest of my comment disappeared?

    apparently.. :)

    So the emphasis for inn is on the splitting? how/why did this ? originate?

    what I’m stressing here is just the “two-sided” nature of the Weil algebra: on one side it is a free object, so a morphism out of the Weil algebra is just a degree 1 $\mathfrak{g}$-valued differential form: the local connection form; on the other side, the Weil algebra of $\mathfrak{g}$ is the Chevalley-Eilenberg algebra of $inn(\mathfrak{g})$, so morphisms out of it are solutions of the MC equation. this means that each connection $A$ comes with a unique degree 2 $\mathfrak{g}$-valued form $F_A$, such that the pair $(A,F_A)$ is a solution of the MC equation for $inn(\mathfrak{g})$-valued differential forms. this single MC equation is both the defining equation of the curvature $F_A$ and the Bianchi identity.

16. • CommentRowNumber16.
    • CommentAuthorDavidRoberts
    • CommentTimeJun 12th 2011
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    Author: DavidRoberts
    Format: MarkdownItexJim >So the emphasis for inn is on the splitting? >how/why did this ? originate? If you recall, back in the dim dark past, my [paper with Urs](http://golem.ph.utexas.edu/category/2007/07/the_inner_automorphism_3group.html) (link is to cafe discussion and Urs' explanation of this), we describe what is called there $INN(G)$ - a 2-group model for the total space of the universal $G$-bundle for $G$ an ordinary group. Ditto for a strict 2-group $\mathcal{G}$, we get a strict 3-group $INN(\mathcal{G})$ which is a model for the universal $\mathcal{G}$-2-bundle. This construction is essentially the mapping cone of the identity, calculated in various ways (e.g. through crossed modules and their higher companions). Now around this time is was discovered (by whom I cannot recall) that if we are talking about Lie groups here, then connections given by parallel transport functors with values in the '$INN$' version gave non-flat connections (whereas before, cf Baez-Schreiber paper on 2-connections, such things gave connections satisfying a flatness conditions). There's some subtlety I skipping in the last sentence, but I hope you get the gist of it. Anyway, the differential form version of a connection taking values in the Lie ($n$-)algebra of $INN$ of a(n $n$-)group shows the same phenomenon. It is this way you get all the connection data from e.g. Breen and Messing's paper on the geometry of nonabelian gerbes, with no flatness condition ('fake flatness'). So the story is: take a Lie $n$-group $G$, take $INN$ of it, which is a groupal model of the (total space of the) universal bundle, then apply the Lie functor, to get a Lie $n$-algebra. Connections with values in this are connections on a $G$-bundle. Or, as we have a commuting square laying around, take $G$, apply the Lie functor, then apply the $L_\infty$-algebra version of $INN$, called $inn$ (which at some level is the mapping cone of the identity), and you end up with essentially the same thing.

    Jim

    So the emphasis for inn is on the splitting?

    how/why did this ? originate?

    If you recall, back in the dim dark past, my paper with Urs (link is to cafe discussion and Urs’ explanation of this), we describe what is called there $INN(G)$ - a 2-group model for the total space of the universal $G$-bundle for $G$ an ordinary group. Ditto for a strict 2-group $\mathcal{G}$, we get a strict 3-group $INN(\mathcal{G})$ which is a model for the universal $\mathcal{G}$-2-bundle. This construction is essentially the mapping cone of the identity, calculated in various ways (e.g. through crossed modules and their higher companions).

    Now around this time is was discovered (by whom I cannot recall) that if we are talking about Lie groups here, then connections given by parallel transport functors with values in the ’$INN$’ version gave non-flat connections (whereas before, cf Baez-Schreiber paper on 2-connections, such things gave connections satisfying a flatness conditions). There’s some subtlety I skipping in the last sentence, but I hope you get the gist of it.

    Anyway, the differential form version of a connection taking values in the Lie ($n$-)algebra of $INN$ of a(n $n$-)group shows the same phenomenon. It is this way you get all the connection data from e.g. Breen and Messing’s paper on the geometry of nonabelian gerbes, with no flatness condition (’fake flatness’).

    So the story is: take a Lie $n$-group $G$, take $INN$ of it, which is a groupal model of the (total space of the) universal bundle, then apply the Lie functor, to get a Lie $n$-algebra. Connections with values in this are connections on a $G$-bundle. Or, as we have a commuting square laying around, take $G$, apply the Lie functor, then apply the $L_\infty$-algebra version of $INN$, called $inn$ (which at some level is the mapping cone of the identity), and you end up with essentially the same thing.

17. • CommentRowNumber17.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 14th 2011
    • (edited Jun 14th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexI've worked a bit on Proposition 7 in [[differential string structure]]. It seems to me that what's going on there can be thought as follows: in the notations of post 5 above, a $k$-simplex in $\exp_\Delta(b\mathbb{R}\to\mathfrak{string})$ is the datum of a triple $(\omega,B,C)$ with $\omega\in\Omega^1(\Delta^k;\mathfrak{so})\otimes C^\infty(U;\mathbb{R})$, $B\in\Omega^2(\Delta^k;\mathbb{R})\otimes C^\infty(U;\mathbb{R})$ and $C\in\Omega^2(\Delta^k;\mathbb{R})\otimes C^\infty(U;\mathbb{R})$, subject to the constrains $$ d_{\Omega^\bullet(\Delta^k)} \omega+\frac{1}{2}[\omega,\omega]=0; \qquad d_{\Omega^\bullet(\Delta^k)} B+C+\frac{1}{6}\mu (\omega,\omega,\omega)=0;\qquad d_{\Omega^\bullet(\Delta^k)} C=0 $$ The second equation defines $C$: $$ C=-d_{\Omega^\bullet(\Delta^k)} B-\frac{1}{6}\mu(\omega,\omega,\omega) $$ (there are signs to be fixed here, I guess), so the third equation becomes $$ d_{\Omega^\bullet(\Delta^k)}\mu(\omega,\omega,\omega)=0 $$ which is indeed automatically satisfied by the first equation and by the fact that $\mu$ is a cocycle. Hence the data of a $k$-simplex in $\exp_\Delta(b\mathbb{R}\to\mathfrak{string})$ boil down to a pair $(\omega,B)$, where $\omega$ is $k$-simplex in $\exp_\Delta(\mathfrak{so})$ and $B$ is an arbitrary element in $\Omega^2(\Delta^k;\mathbb{R})\otimes C^\infty(U;\mathbb{R})$. But such an element is precisely a $k$-simplex in $\exp_\Delta(inn(b\mathbb{R}))$, so what we are saying is that there is a natural isomorphism of simplicial presheaves $$ \exp_\Delta(b\mathbb{R}\to\mathfrak{string})\cong \exp_\Delta(\mathfrak{so})\times \exp_\Delta(inn(b\mathbb{R})) $$ and the fact that the natural morphism $\mathfrak{so}\to (b\mathbb{R}\to\mathfrak{string})$ induces a weak equivalence of simplicial presheaves reduces to the fact that $\exp_\Delta(inn(b\mathbb{R}))$ is contractible, which is easily verified (the argument at [[differential string structure]] saying that $B$ can be rescaled smoothly to zero does precisely this job). This accounts for the first part of Proposition 7, the one concerning the weak equivalence. Now going to work on the second part, the one stating that $\exp_\Delta(b\mathbb{R}\to \mathfrak{string}))\to \exp_\Delta(b^2\mathbb{R})$ is a fibration.

    I’ve worked a bit on Proposition 7 in differential string structure. It seems to me that what’s going on there can be thought as follows: in the notations of post 5 above, a $k$-simplex in $\exp_\Delta(b\mathbb{R}\to\mathfrak{string})$ is the datum of a triple $(\omega,B,C)$ with $\omega\in\Omega^1(\Delta^k;\mathfrak{so})\otimes C^\infty(U;\mathbb{R})$, $B\in\Omega^2(\Delta^k;\mathbb{R})\otimes C^\infty(U;\mathbb{R})$ and $C\in\Omega^2(\Delta^k;\mathbb{R})\otimes C^\infty(U;\mathbb{R})$, subject to the constrains

    $$d_{\Omega^\bullet(\Delta^k)} \omega+\frac{1}{2}[\omega,\omega]=0; \qquad d_{\Omega^\bullet(\Delta^k)} B+C+\frac{1}{6}\mu (\omega,\omega,\omega)=0;\qquad d_{\Omega^\bullet(\Delta^k)} C=0$$

    The second equation defines $C$:

    $$C=-d_{\Omega^\bullet(\Delta^k)} B-\frac{1}{6}\mu(\omega,\omega,\omega)$$

    (there are signs to be fixed here, I guess), so the third equation becomes

    $$d_{\Omega^\bullet(\Delta^k)}\mu(\omega,\omega,\omega)=0$$

    which is indeed automatically satisfied by the first equation and by the fact that $\mu$ is a cocycle. Hence the data of a $k$-simplex in $\exp_\Delta(b\mathbb{R}\to\mathfrak{string})$ boil down to a pair $(\omega,B)$, where $\omega$ is $k$-simplex in $\exp_\Delta(\mathfrak{so})$ and $B$ is an arbitrary element in $\Omega^2(\Delta^k;\mathbb{R})\otimes C^\infty(U;\mathbb{R})$. But such an element is precisely a $k$-simplex in $\exp_\Delta(inn(b\mathbb{R}))$, so what we are saying is that there is a natural isomorphism of simplicial presheaves

    $$\exp_\Delta(b\mathbb{R}\to\mathfrak{string})\cong \exp_\Delta(\mathfrak{so})\times \exp_\Delta(inn(b\mathbb{R}))$$

    and the fact that the natural morphism $\mathfrak{so}\to (b\mathbb{R}\to\mathfrak{string})$ induces a weak equivalence of simplicial presheaves reduces to the fact that $\exp_\Delta(inn(b\mathbb{R}))$ is contractible, which is easily verified (the argument at differential string structure saying that $B$ can be rescaled smoothly to zero does precisely this job).

    This accounts for the first part of Proposition 7, the one concerning the weak equivalence. Now going to work on the second part, the one stating that $\exp_\Delta(b\mathbb{R}\to \mathfrak{string}))\to \exp_\Delta(b^2\mathbb{R})$ is a fibration.

18. • CommentRowNumber18.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 14th 2011
    • (edited Jun 14th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexwe are faced with the lifting problem $$ \array{ \Lambda[k]_i &\to& \exp_\Delta(b\mathbb{R} \to \mathfrak{string}) \\ \downarrow && \downarrow \\ \Delta[k] &\to& \exp_\Delta(b^2\mathbb{R}) } $$ where the projection map on the right is $(\omega,B,C)\mapsto C$. So the lifting problem is: given a 3-form $C$ on $\Delta^k$ (depending smoothly on a parameter $u\in U$), and data $(\omega,B)$ on a horn such that $\pm d B\pm\frac{1}{6}\mu(\omega,\omega,\omega)=C\vert_{\Lambda^i[k]}$, can we extend $(\omega,B)$ to $(\tilde{\omega},\tilde{B})$ on the whole $\Delta[k]$ in such a way that $\pm d\tilde{B}\pm\frac{1}{6}\mu(\tilde{\omega},\tilde{\omega},\tilde{\omega})=C$? The contraction $\pi:\Delta[k]\to \Lambda^i[k]$ induces a dgca morphism $\Omega^\bullet(\Lambda^i[k];\mathbb{R})\otimes C^\infty(U;\mathbb{R})\to\Omega^\bullet(\Delta[k];\mathbb{R})\otimes C^\infty(U;\mathbb{R})$ and so $\pi^*\mu$ is a $k$-simplex in $\exp_\Delta(\mathfrak{so})$ extending $\mu$. The lifting problem thus reduces to finding a 2-form $\tilde{B}$ on the $k$-simplex such that $d\tilde{B}=C\pm\frac{1}{6}\mu(\tilde{\omega},\tilde{\omega},\tilde{\omega})$ and $\tilde{B}\vert_{\Lambda^i[3]}=B$. Write $\tilde{B}=B'\pm\pi^*B$ and $C'=C-\pi^*(C\vert_{\Lambda^i[k]}}$ Then the lifting problem is translated into $$ d B'=C'; \qquad B'\vert_{\Lambda^i[k]}=0 $$ And since $C'$ is closed on the $k$-simplex and satisfies $C'\vert_{\Lambda^i[k]}=0$, such a $B'$ exists (pick any solution $B''$ of $d B''=C'$; then notice that $d (B''\vert_{\Lambda^i[k]})=(d B'')\vert_{\Lambda^i[k]})=0$ and so $B'=B''-\pi^*(B''\vert_{\Lambda^i[k]})$ is a solution with the given boundary condition)

    we are faced with the lifting problem

    $$\array{ \Lambda[k]_i &\to& \exp_\Delta(b\mathbb{R} \to \mathfrak{string}) \\ \downarrow && \downarrow \\ \Delta[k] &\to& \exp_\Delta(b^2\mathbb{R}) }$$

    where the projection map on the right is $(\omega,B,C)\mapsto C$. So the lifting problem is: given a 3-form $C$ on $\Delta^k$ (depending smoothly on a parameter $u\in U$), and data $(\omega,B)$ on a horn such that $\pm d B\pm\frac{1}{6}\mu(\omega,\omega,\omega)=C\vert_{\Lambda^i[k]}$, can we extend $(\omega,B)$ to $(\tilde{\omega},\tilde{B})$ on the whole $\Delta[k]$ in such a way that $\pm d\tilde{B}\pm\frac{1}{6}\mu(\tilde{\omega},\tilde{\omega},\tilde{\omega})=C$?

    The contraction $\pi:\Delta[k]\to \Lambda^i[k]$ induces a dgca morphism $\Omega^\bullet(\Lambda^i[k];\mathbb{R})\otimes C^\infty(U;\mathbb{R})\to\Omega^\bullet(\Delta[k];\mathbb{R})\otimes C^\infty(U;\mathbb{R})$ and so $\pi^*\mu$ is a $k$-simplex in $\exp_\Delta(\mathfrak{so})$ extending $\mu$. The lifting problem thus reduces to finding a 2-form $\tilde{B}$ on the $k$-simplex such that $d\tilde{B}=C\pm\frac{1}{6}\mu(\tilde{\omega},\tilde{\omega},\tilde{\omega})$ and $\tilde{B}\vert_{\Lambda^i[3]}=B$.

    Write $\tilde{B}=B'\pm\pi^*B$ and Then the lifting problem is translated into

    $$d B'=C'; \qquad B'\vert_{\Lambda^i[k]}=0$$

    And since $C'=C-\pi^*(C\vert_{\Lambda^i[k]}}C'$ is closed on the $k$-simplex and satisfies $C'\vert_{\Lambda^i[k]}=0$, such a $B'$ exists (pick any solution $B''$ of $d B''=C'$; then notice that $d (B''\vert_{\Lambda^i[k]})=(d B'')\vert_{\Lambda^i[k]})=0$ and so $B'=B''-\pi^*(B''\vert_{\Lambda^i[k]})$ is a solution with the given boundary condition)

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeJun 14th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Domenico, this looks good, yes. I think maybe one reason why the discussion at the nLab entry looks more complicated at times is that there it is also discussed that one can find the lift with the required sitting instants. Isn't that something one would have to add to the above discussion, too, around the point where $B'$ appears? Maybe not, maybe I am looking at it in too complicated a way.

    Hi Domenico,

    this looks good, yes. I think maybe one reason why the discussion at the nLab entry looks more complicated at times is that there it is also discussed that one can find the lift with the required sitting instants.

    Isn’t that something one would have to add to the above discussion, too, around the point where $B'$ appears? Maybe not, maybe I am looking at it in too complicated a way.

20. • CommentRowNumber20.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 14th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, I haven't looked at this in detail, but preserving sitting instants should be that story we once discussed (and which led to the current definition of forms with sitting intants): the complex of differential forms with sitting intants on a simplex is acyclic. so each closed form with sitting instants has a primitive with sitting instants. this works for the horn, too, since the projection on the opposite face of the simplex is a diffeomorphism (except on the edges, where this is cured by the sitting instants condition).

    Hi Urs,

    I haven’t looked at this in detail, but preserving sitting instants should be that story we once discussed (and which led to the current definition of forms with sitting intants): the complex of differential forms with sitting intants on a simplex is acyclic. so each closed form with sitting instants has a primitive with sitting instants. this works for the horn, too, since the projection on the opposite face of the simplex is a diffeomorphism (except on the edges, where this is cured by the sitting instants condition).

21. • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeJun 14th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Domenico, okay, I see, right, so we should state the re-occuring manipulations concerning sitting instants separately and then just refer to them, not to clutter the proofs. But let's see, what do we actually know? Here is something I know: For $\omega$ a closed $k \geq n$-form on the $n$-disk with sitting instants, there is an $(n-1)$-form $\lambda$ with sitting instants such that $d \lambda = \omega$. Notice the constraint on the degrees. This ensures that $\omega$ in fact _vanishes_ in a neighbourhood of the boundary, instead of being just constant on some non-vanishing value. Using this in the standard formula for the Poincare lemma, one gets a $\lambda$ that is constant perpendicular to the boundary: since the Poincare lemma formula adds up radial contributions of $\omgea$, and since that vanishes near the boundary, we don't add further contributions there. The same argument fails for $\omega$ of degree $\lt n$. That's okay for the kind of arguments at [[differential string structure]], but maybe it means that we need to be careful with saying that forms with sitting instants on a disk are acyclic under the de Rham differential. I think.

    Hi Domenico,

    okay, I see, right, so we should state the re-occuring manipulations concerning sitting instants separately and then just refer to them, not to clutter the proofs.

    But let’s see, what do we actually know? Here is something I know:

    For $\omega$ a closed $k \geq n$-form on the $n$-disk with sitting instants, there is an $(n-1)$-form $\lambda$ with sitting instants such that $d \lambda = \omega$.

    Notice the constraint on the degrees. This ensures that $\omega$ in fact vanishes in a neighbourhood of the boundary, instead of being just constant on some non-vanishing value.

    Using this in the standard formula for the Poincare lemma, one gets a $\lambda$ that is constant perpendicular to the boundary: since the Poincare lemma formula adds up radial contributions of $\omgea$, and since that vanishes near the boundary, we don’t add further contributions there.

    The same argument fails for $\omega$ of degree $\lt n$. That’s okay for the kind of arguments at differential string structure, but maybe it means that we need to be careful with saying that forms with sitting instants on a disk are acyclic under the de Rham differential. I think.

22. • CommentRowNumber22.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 14th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> it means that we need to be careful with saying that forms with sitting instants on a disk are acyclic under the de Rham differential. I think. right, but acyclicity is a fundamental property for what we do: it means we are replacing teh base ield $\mathbb{R}$ with an equivalent dgca. notice for instance that in Withney-Sullivan-Hinich-Getzler the dgca of polynomial differential forms is considered, and that is acyclic. so whichever definition of sitting instants we work with, we should ensure that acyclicity holds. I have a vague memory that when we discussed this issue a time ago we had managed to convince ourself that forms with sitting instants were an acyclic complex. I'll try to recover our argument (and to work in detail what happens for $k$ simplices with $k\le 2$, o have some concrete computation at hand).

    it means that we need to be careful with saying that forms with sitting instants on a disk are acyclic under the de Rham differential. I think.

    right, but acyclicity is a fundamental property for what we do: it means we are replacing teh base ield $\mathbb{R}$ with an equivalent dgca. notice for instance that in Withney-Sullivan-Hinich-Getzler the dgca of polynomial differential forms is considered, and that is acyclic. so whichever definition of sitting instants we work with, we should ensure that acyclicity holds. I have a vague memory that when we discussed this issue a time ago we had managed to convince ourself that forms with sitting instants were an acyclic complex. I’ll try to recover our argument (and to work in detail what happens for $k$ simplices with $k\le 2$, o have some concrete computation at hand).

23. • CommentRowNumber23.
    • CommentAuthorUrs
    • CommentTimeJun 14th 2011
    • (edited Jun 14th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItexWait, we do need a suitably acyclic simplicial complex, but not necessarily acylcity via the _de Rham differential_ . I don't think that we ever need that $\Omega^\bullet_si(\Delta^n)$ is acyclic. In fact this complex never appears in our constructions. What we do need is that $\Omega^n_{cl}(\Delta^\bullet)$ is $\simeq \mathbf{B}\mathbb{R}^n$ (and checking that involves the kind of argument I mentioned above).

    Wait, we do need a suitably acyclic simplicial complex, but not necessarily acylcity via the de Rham differential .

    I don’t think that we ever need that $\Omega^\bullet_si(\Delta^n)$ is acyclic. In fact this complex never appears in our constructions. What we do need is that $\Omega^n_{cl}(\Delta^\bullet)$ is $\simeq \mathbf{B}\mathbb{R}^n$ (and checking that involves the kind of argument I mentioned above).

24. • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeJun 14th 2011
    • (edited Jun 14th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItex.

    .

25. • CommentRowNumber25.
    • CommentAuthorjim_stasheff
    • CommentTimeJun 15th 2011
    • PermaLink
    Author: jim_stasheff
    Format: Text#24 was blank? by all means, discussion of the acyclicity should be a separate section, to be referred to where needed

    #24 was blank?
    
    by all means, discussion of the acyclicity should be a separate section, to be referred to where needed
26. • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeJun 15th 2011
    • (edited Jun 15th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> 24 was blank? Yes, sorry, I had added another remark, but then realized that it was at best not helpful. So I removed it.

    24 was blank?

    Yes, sorry, I had added another remark, but then realized that it was at best not helpful. So I removed it.

27. • CommentRowNumber27.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 15th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexLet me assume acyclicity for the moment (I guess one should be able to prove it using Poincare' lemma by choosing a clever retraction of the simplex, along a vector field which is orthogonla to the boundary and with a single sink in the interior of the simplex; I've just asked for a confirmation of this argument on MO). And let me try to see if I'm able to give a proof of Proposition 9 in [[differential string structures]] along the lines of posts 17 and 18. to begin with, let me recall the definition of $\exp_\Delta(\mathfrak{g})_{conn}$: its $k$-simplices over a Cartesian space $U$ are $\mathfrak{g}$-valued connection forms on $U\times\Delta^k$ whose curvature forms are of type $(p,0)$ with $p\gt0$, with respect to the type decomposition of forms on $\U\times\Delta^k$ induced by the product structure. So, for $\mathfrak{g}=(b\mathbb{R}\to\mathfrak{string}$, our data are triples $(\omega,B,C)$ as above, subject to the constrains: $d_\Omega \omega+\frac{1}{2}[\omega,\omega]$ is of type $(2,0)$; $d_\Omega B+C+\frac{1}{6}\mu(\omega,\omega,\omega)$ is of type $(3,0)$ $d_\Omega C$ is of type $(4,0)$. (math is not displayed, I don't understand why. maybe a problem on nforum. I will try back later)

    Let me assume acyclicity for the moment (I guess one should be able to prove it using Poincare’ lemma by choosing a clever retraction of the simplex, along a vector field which is orthogonla to the boundary and with a single sink in the interior of the simplex; I’ve just asked for a confirmation of this argument on MO). And let me try to see if I’m able to give a proof of Proposition 9 in differential string structures along the lines of posts 17 and 18.

    to begin with, let me recall the definition of $\exp_\Delta(\mathfrak{g})_{conn}$: its $k$-simplices over a Cartesian space $U$ are $\mathfrak{g}$-valued connection forms on $U\times\Delta^k$ whose curvature forms are of type $(p,0)$ with $p\gt0$, with respect to the type decomposition of forms on $\U\times\Delta^k$ induced by the product structure.

    So, for $\mathfrak{g}=(b\mathbb{R}\to\mathfrak{string}$, our data are triples $(\omega,B,C)$ as above, subject to the constrains:

    $d_\Omega \omega+\frac{1}{2}[\omega,\omega]$ is of type $(2,0)$;

    $d_\Omega B+C+\frac{1}{6}\mu(\omega,\omega,\omega)$ is of type $(3,0)$

    $d_\Omega C$ is of type $(4,0)$.

    (math is not displayed, I don’t understand why. maybe a problem on nforum. I will try back later)

28. • CommentRowNumber28.
    • CommentAuthorDavidRoberts
    • CommentTimeJun 16th 2011
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItexHere's the maths from #27: $d_\Omega \omega+\frac{1}{2}[\omega,\omega]$ is of type $(2,0)$; $d_\Omega B+C+\frac{1}{6}\mu(\omega,\omega,\omega)$ is of type $(3,0)$ $d_\Omega C$ is of type $(4,0)$. (Domenico, if you need to do this put 4 spaces at the beginning of each new paragraph and the input will display as-is)

    Here’s the maths from #27:

    $d_\Omega \omega+\frac{1}{2}[\omega,\omega]$ is of type $(2,0)$; 
    
    $d_\Omega B+C+\frac{1}{6}\mu(\omega,\omega,\omega)$ is of type $(3,0)$
    
    $d_\Omega C$ is of type $(4,0)$.
    

    (Domenico, if you need to do this put 4 spaces at the beginning of each new paragraph and the input will display as-is)

29. • CommentRowNumber29.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 16th 2011
    • (edited Jun 16th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex(now I've been able to have math displayed by introducing a few spaces within the formulas (look at the source). there must be something weird happening in the background) Thanks to a suggestion by Tom Goodwillie on MO (<a href="http://mathoverflow.net/questions/67861/differential-forms-on-the-simplex-which-are-constant-towards-the-boundary">here</a>), I finally have a proof of the acyclicity of the complex of differential forms with sitting instants on the simplex. Given this, and given the trick for having math displayed, let me go on with post 27. Just rewriting the last part of it for the moment, will continue later. to begin with, let me recall the definition of $\exp_\Delta (\mathfrak{g})_{conn}$: its $k$-simplices over a Cartesian space $U$ are $\mathfrak{g}$-valued connection forms on $U\times\Delta^k$ whose curvature forms are of type $(p, 0)$ with $p \gt 0$, with respect to the type decomposition of forms on $U \times\Delta^k$ induced by the product structure. So, for $\mathfrak{g}=(b\mathbb{R}\to\mathfrak{string}$, our data are triples $(\omega,B,C)$ as above, subject to the constrains: $d_\Omega \omega +\frac{1}{2} [\omega,\omega]$ is of type $(2, 0)$; $d_\Omega B +C+\frac{1}{6}\mu(\omega, \omega,\omega)$ is of type $(3, 0)$ $ d_\Omega C$ is of type $(4, 0)$.

    (now I’ve been able to have math displayed by introducing a few spaces within the formulas (look at the source). there must be something weird happening in the background)

    Thanks to a suggestion by Tom Goodwillie on MO (here), I finally have a proof of the acyclicity of the complex of differential forms with sitting instants on the simplex. Given this, and given the trick for having math displayed, let me go on with post 27. Just rewriting the last part of it for the moment, will continue later.

    to begin with, let me recall the definition of $\exp_\Delta (\mathfrak{g})_{conn}$: its $k$-simplices over a Cartesian space $U$ are $\mathfrak{g}$-valued connection forms on $U\times\Delta^k$ whose curvature forms are of type $(p, 0)$ with $p \gt 0$, with respect to the type decomposition of forms on $U \times\Delta^k$ induced by the product structure.

    So, for $\mathfrak{g}=(b\mathbb{R}\to\mathfrak{string}$, our data are triples $(\omega,B,C)$ as above, subject to the constrains:

    $d_\Omega \omega +\frac{1}{2} [\omega,\omega]$ is of type $(2, 0)$;

    $d_\Omega B +C+\frac{1}{6}\mu(\omega, \omega,\omega)$ is of type $(3, 0)$

    $d_\Omega C$ is of type $(4, 0)$.

30. • CommentRowNumber30.
    • CommentAuthorUrs
    • CommentTimeJun 16th 2011
    • (edited Jun 16th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> I finally have a proof of the acyclicity of the complex of differential forms with sitting instants on the simplex Great. Thanks for looking into this. That allows to simplify the look of these proofs considerably, simply by always pointing to this fact. Concerning the remainder of your message: somehow it seems that you break off before you have ended with a statement. I am not sure what kind of reply you might be looking for, if any. Could you clarify?

    I finally have a proof of the acyclicity of the complex of differential forms with sitting instants on the simplex

    Great. Thanks for looking into this. That allows to simplify the look of these proofs considerably, simply by always pointing to this fact.

    Concerning the remainder of your message: somehow it seems that you break off before you have ended with a statement. I am not sure what kind of reply you might be looking for, if any. Could you clarify?

31. • CommentRowNumber31.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 16th 2011
    • (edited Jun 16th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> Concerning the remainder of your message: somehow it seems that you break off before you have ended with a statement. I am not sure what kind of reply you might be looking for, if any. Could you clarify? I was just displaying the math from post 27 there. I'll now go on with that: let us write $\eta_{p,q}$ for the $(p,q)$ component of a differential form $\eta$. then $\omega=\omega_{0,1}+\omega_{1,0}$, $B=B_{0,2}+B_{1,1}+B_{2,0}$ and $C=C_{0,3}+C_{1,2}+C_{2,1}+C_{3,0}$. the curvature constrains then read $$ d_\Delta \omega_{0,1} +\frac{1}{2}[\omega_{0,1},\omega_{0,1}]=0\qquad ( the\, connection\, is \, flat\, in\, the\, vertical\, direction) $$ $$ d_\Delta\omega_{1,0} +d_U\omega_{0,1}+[\omega_{0,1},\omega_{1,0}]=0 $$ $$ C_{0,3}=\pm d_\Delta B_{0,2}\pm \frac{1}{6}[\omega_{0,1},\omega_{0,1},\omega_{0,1}] \qquad ( this\, is\, the\, vertical\, equation) $$ $$ C_{2,1}=\pm d_\Delta B_{2,0} \pm d_U B_{1,1}\pm\frac{1}{2} \mu(\omega_{0,1},\omega_{1,0},\omega_{1,0}) $$ $$ C_{1,2}=\pm d_\Delta B_{1,1} \pm d_U B_{0,2}\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},\omega_{1,0}) $$ $$ d_\Delta C_{0,3}=0 \qquad (the\, vertical\, equation) $$ $$ d_\Delta C_{1,2}+d_U C_{0,3}=0 $$ $$ d_\Delta C_{2,1}+d_U C_{1,2}=0 $$ $$ d_\Delta C_{3,0}+d_U C_{2,1}=0 $$ Thus one sees that there are no constrains on $B$ and that $C_{0,3}$, $C_{1,2}$ and $C_{2,1}$ are determined by $\omega$ and $B$. furthermore the equation $d_\Delta C_{0,3}=0$ is, as we have seen, automatically staisfied. I'll now have to check what happens to the last three equations above.

    Concerning the remainder of your message: somehow it seems that you break off before you have ended with a statement. I am not sure what kind of reply you might be looking for, if any. Could you clarify?

    I was just displaying the math from post 27 there. I’ll now go on with that:

    let us write $\eta_{p,q}$ for the $(p,q)$ component of a differential form $\eta$. then $\omega=\omega_{0,1}+\omega_{1,0}$, $B=B_{0,2}+B_{1,1}+B_{2,0}$ and $C=C_{0,3}+C_{1,2}+C_{2,1}+C_{3,0}$. the curvature constrains then read

    $$d_\Delta \omega_{0,1} +\frac{1}{2}[\omega_{0,1},\omega_{0,1}]=0\qquad ( the\, connection\, is \, flat\, in\, the\, vertical\, direction)$$ $$d_\Delta\omega_{1,0} +d_U\omega_{0,1}+[\omega_{0,1},\omega_{1,0}]=0$$ $$C_{0,3}=\pm d_\Delta B_{0,2}\pm \frac{1}{6}[\omega_{0,1},\omega_{0,1},\omega_{0,1}] \qquad ( this\, is\, the\, vertical\, equation)$$ $$C_{2,1}=\pm d_\Delta B_{2,0} \pm d_U B_{1,1}\pm\frac{1}{2} \mu(\omega_{0,1},\omega_{1,0},\omega_{1,0})$$ $$C_{1,2}=\pm d_\Delta B_{1,1} \pm d_U B_{0,2}\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},\omega_{1,0})$$ $$d_\Delta C_{0,3}=0 \qquad (the\, vertical\, equation)$$ $$d_\Delta C_{1,2}+d_U C_{0,3}=0$$ $$d_\Delta C_{2,1}+d_U C_{1,2}=0$$ $$d_\Delta C_{3,0}+d_U C_{2,1}=0$$

    Thus one sees that there are no constrains on $B$ and that $C_{0,3}$, $C_{1,2}$ and $C_{2,1}$ are determined by $\omega$ and $B$. furthermore the equation $d_\Delta C_{0,3}=0$ is, as we have seen, automatically staisfied. I’ll now have to check what happens to the last three equations above.

32. • CommentRowNumber32.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 16th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexusing the definitions of $C_{0,3}$ and $C_{1,2}$, the equation $d_\Delta C_{1,2}+d_U C_{0,3}=0$ is $$ \pm \frac{1}{2}\mu([\omega_{0,1},\omega_{0,1}],\omega_{0,1},\omega_{1,0})\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},d_\Delta \omega_{1,0}) \pm \frac{1}{2}\mu(d_U\omega_{0,1},\omega_{0,1},\omega_{0,1})= $$ $$ \pm \frac{1}{2}\mu([\omega_{0,1},\omega_{0,1}],\omega_{0,1},\omega_{1,0})\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},d_\Delta \omega_{1,0}+d_U\omega_{0,1})= $$ $$ \pm \frac{1}{2}\mu([\omega_{0,1},\omega_{0,1}],\omega_{0,1},\omega_{1,0})\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},[\omega_{0,1},\omega_{1,0}])=0 $$ since $\mu$ is a cocycle. similarly I expect also the equation $ d_\Delta C_{2,1}+d_U C_{1,2}=0$ to be automatically satisfied. let me assume for a moment this is true (there must be some better argument than the brute force computation above). then a $k$-simplex in $\exp_\Delta(b\mathbb{R}\to \mathfrak{string})_{conn}$ would be a triple $(\omega,B,C_{3,0})$ with $$ d_\Delta \omega_{0,1} +\frac{1}{2}[\omega_{0,1},\omega_{0,1}]=0;\qquad d_\Delta\omega_{1,0} +d_U\omega_{0,1}+[\omega_{0,1},\omega_{1,0}]=0 $$ and with a third equation relating $d_\Delta C_{3,0}$ with $\omega$ and $B$. The first two equations are nice: tehy say that $\omega$ is a $k$-simplex in $\exp_\Delta(\mathfrak{so})_{conn}$. the third equation, however, looks a bit awkward. this puzzled me for a while, since I finally understood it has to be so! namely, $\mathfrak{g}\mapsto \exp_\Delta(\mathfrak{g})$ is a functor on $L_\infty$-algebras, but $\mathfrak{g}\mapsto \exp_\Delta(\mathfrak{g})_{conn}$ is not! the datum of a morphism of $L_\infty$-algebras $\mathfrak{g}\to \mathfrak{h}$ does not suffice to give a morphism of simplicial presheaves $\exp_\Delta(\mathfrak{g})_{conn}\to \exp_\Delta(\mathfrak{h})_{conn}$. rather $\mathfrak{g}\mapsto \exp_\Delta(\mathfrak{g})_{conn}$ is a functor defined on the category whose morphisms are triples $$ \array{ CE(\mathfrak{g})&\leftarrow & CE(\mathfrak{h})\\ \uparrow &&\uparrow\\ W(\mathfrak{g})&\leftarrow & W(\mathfrak{h})\\ \uparrow &&\uparrow\\ inv(\mathfrak{g})&\leftarrow & inv(\mathfrak{h})\\ } $$ For $\mathfrak{g}=\mathfrak{so}$ and $\mathfrak{h}=(b\mathbb{R}\to\mathfrak{string})$ we are a priori given the morphism $CE(\mathfrak{g})\leftarrow CE(\mathfrak{h})$. I strongly suspect that a "Chern-Simons" morphism $W(\mathfrak{g})\leftarrow W(\mathfrak{h})$ can actually be given in the case at hand, and that it will be precisely this morphism to involve and justify Urs' modified equations. The problem they solve is (I guess): how d we define a dgca morphism $$ W(\mathfrak{so})\leftarrow W(b\mathbb{R}\to \mathfrak{string}) $$ ?

    using the definitions of $C_{0,3}$ and $C_{1,2}$, the equation $d_\Delta C_{1,2}+d_U C_{0,3}=0$ is

    $$\pm \frac{1}{2}\mu([\omega_{0,1},\omega_{0,1}],\omega_{0,1},\omega_{1,0})\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},d_\Delta \omega_{1,0}) \pm \frac{1}{2}\mu(d_U\omega_{0,1},\omega_{0,1},\omega_{0,1})=$$ $$\pm \frac{1}{2}\mu([\omega_{0,1},\omega_{0,1}],\omega_{0,1},\omega_{1,0})\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},d_\Delta \omega_{1,0}+d_U\omega_{0,1})=$$ $$\pm \frac{1}{2}\mu([\omega_{0,1},\omega_{0,1}],\omega_{0,1},\omega_{1,0})\pm \frac{1}{2}\mu(\omega_{0,1},\omega_{0,1},[\omega_{0,1},\omega_{1,0}])=0$$

    since $\mu$ is a cocycle. similarly I expect also the equation $d_\Delta C_{2,1}+d_U C_{1,2}=0$ to be automatically satisfied. let me assume for a moment this is true (there must be some better argument than the brute force computation above). then a $k$-simplex in $\exp_\Delta(b\mathbb{R}\to \mathfrak{string})_{conn}$ would be a triple $(\omega,B,C_{3,0})$ with

    $$d_\Delta \omega_{0,1} +\frac{1}{2}[\omega_{0,1},\omega_{0,1}]=0;\qquad d_\Delta\omega_{1,0} +d_U\omega_{0,1}+[\omega_{0,1},\omega_{1,0}]=0$$

    and with a third equation relating $d_\Delta C_{3,0}$ with $\omega$ and $B$. The first two equations are nice: tehy say that $\omega$ is a $k$-simplex in $\exp_\Delta(\mathfrak{so})_{conn}$. the third equation, however, looks a bit awkward. this puzzled me for a while, since I finally understood it has to be so!

    namely, $\mathfrak{g}\mapsto \exp_\Delta(\mathfrak{g})$ is a functor on $L_\infty$-algebras, but $\mathfrak{g}\mapsto \exp_\Delta(\mathfrak{g})_{conn}$ is not! the datum of a morphism of $L_\infty$-algebras $\mathfrak{g}\to \mathfrak{h}$ does not suffice to give a morphism of simplicial presheaves $\exp_\Delta(\mathfrak{g})_{conn}\to \exp_\Delta(\mathfrak{h})_{conn}$. rather $\mathfrak{g}\mapsto \exp_\Delta(\mathfrak{g})_{conn}$ is a functor defined on the category whose morphisms are triples

    $$\array{ CE(\mathfrak{g})&\leftarrow & CE(\mathfrak{h})\\ \uparrow &&\uparrow\\ W(\mathfrak{g})&\leftarrow & W(\mathfrak{h})\\ \uparrow &&\uparrow\\ inv(\mathfrak{g})&\leftarrow & inv(\mathfrak{h})\\ }$$

    For $\mathfrak{g}=\mathfrak{so}$ and $\mathfrak{h}=(b\mathbb{R}\to\mathfrak{string})$ we are a priori given the morphism $CE(\mathfrak{g})\leftarrow CE(\mathfrak{h})$.

    I strongly suspect that a "Chern-Simons" morphism $W(\mathfrak{g})\leftarrow W(\mathfrak{h})$ can actually be given in the case at hand, and that it will be precisely this morphism to involve and justify Urs’ modified equations. The problem they solve is (I guess): how d we define a dgca morphism

    $$W(\mathfrak{so})\leftarrow W(b\mathbb{R}\to \mathfrak{string})$$

    ?

33. • CommentRowNumber33.
    • CommentAuthorUrs
    • CommentTimeJun 16th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Domenico, thanks for all this. So this is the equations needed to consider if one works not with the modified Weil algebra, right? It seems at this point kind of striking that working with the modified Weil algebra gives a much simpler computation. Why not just use that? The point of it is somehow that by making one simple transformation on the coeffients once and for all, all these complicated equations are taken care of at once. Or am I overlooking something? You say something about justifying the modification. It is justified by organizing these kind of computations in one stroke, i think. Let me know if you think I am wrong about that. Concerning the question at the very end: a morphism out of the Weil algebra is given by a morphism out of the generating complex. So we get a morphism simply from the evident underlying projection. Or is this not what you are after here?

    Hi Domenico,

    thanks for all this. So this is the equations needed to consider if one works not with the modified Weil algebra, right? It seems at this point kind of striking that working with the modified Weil algebra gives a much simpler computation. Why not just use that? The point of it is somehow that by making one simple transformation on the coeffients once and for all, all these complicated equations are taken care of at once.

    Or am I overlooking something? You say something about justifying the modification. It is justified by organizing these kind of computations in one stroke, i think. Let me know if you think I am wrong about that.

    Concerning the question at the very end: a morphism out of the Weil algebra is given by a morphism out of the generating complex. So we get a morphism simply from the evident underlying projection. Or is this not what you are after here?

34. • CommentRowNumber34.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 17th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> Concerning the question at the very end: a morphism out of the Weil algebra is given by a morphism out of the generating complex. So we get a morphism simply from the evident underlying projection. Or is this not what you are after here? Yes, sure. This is the freeness property of the Weil algebra. Right, it is not this what I had in mind. My problem (maybe it's trivial, I havent checked, yet) is we have to define a morphism between the Weil algebras that is compatible with the morphism of CE-algebras, and that preserves invariant polynomials. My suspect (I may be wrong) is that one should see the modification of the Weil algebra come in already at this level, to write in a nice way this morphism. The fact that subsequent computations simplify would then be a consequence of this.

    Concerning the question at the very end: a morphism out of the Weil algebra is given by a morphism out of the generating complex. So we get a morphism simply from the evident underlying projection. Or is this not what you are after here?

    Yes, sure. This is the freeness property of the Weil algebra. Right, it is not this what I had in mind. My problem (maybe it’s trivial, I havent checked, yet) is we have to define a morphism between the Weil algebras that is compatible with the morphism of CE-algebras, and that preserves invariant polynomials.

    My suspect (I may be wrong) is that one should see the modification of the Weil algebra come in already at this level, to write in a nice way this morphism. The fact that subsequent computations simplify would then be a consequence of this.

35. • CommentRowNumber35.
    • CommentAuthorUrs
    • CommentTimeJun 17th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexAh, now I understand. Yes, I think that' right: the isomorphism from the ordinary to the modified Weil algebra is precisely that which collects the original shifted generators together to make them into invariant polynomials, so that it is then easy to see which combinations of these to send through subsequent maps.

    Ah, now I understand. Yes, I think that’ right: the isomorphism from the ordinary to the modified Weil algebra is precisely that which collects the original shifted generators together to make them into invariant polynomials, so that it is then easy to see which combinations of these to send through subsequent maps.

36. • CommentRowNumber36.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 17th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexso, here's the compact story: i) the cocycle $\mu: CE(b^2\mathbb{R})\to CE(\mathfrak{so})$ factors as $$ CE(b^2\mathbb{R})\to CE(b\mathbb{R}\to\mathfrak{string})\to CE(\mathfrak{so}), $$ where the first morphism is given on the generators by $c\mapsto c$ and the second one by $c\mapsto \mu; \quad b\mapsto 0; t^a\mapsto t^a$ (see [[differential string structure]] for these notations). This factorization induces a factorization of $\exp_\Delta(b^2\mathbb{R})\to\exp_\Delta(\mathfrak{so})$ in $$ \exp_\Delta(b^2\mathbb{R})\stackrel{\sim}{\to}\exp_\Delta(b\mathbb{R}\to\mathfrak{string})\twoheadrightarrow \exp_\Delta(\mathfrak{so}) $$ ii) the commutative diagram $$ \array{ CE(b^2\mathbb{R})& \to & CE(\mathfrak{so})\\ \uparrow & & \uparrow\\ W(b^2\mathbb{R})& \to & W(\mathfrak{so})\\ \uparrow & & \uparrow\\ inv(b^2\mathbb{R})& \to & inv(\mathfrak{so})\\ } $$ factors as $$ \array{ CE(b^2\mathbb{R})& \to & CE(b\mathbb{R}\to\mathfrak{string})& \to & CE(\mathfrak{so})\\ \uparrow & & \uparrow& & \uparrow\\ W(b^2\mathbb{R})& \to & W(b\mathbb{R}\to\mathfrak{string})& \to & W(\mathfrak{so})\\ \uparrow & & \uparrow& & \uparrow\\ inv(b^2\mathbb{R})& \to & inv(b\mathbb{R}\to\mathfrak{string})& \to & inv(\mathfrak{so})\\ } $$ Here the upper morphisms are those describes above; before writing the middle morphisms on the generators (which I hope to be able to do soon) let me note that the commutativity of the left bottom corner $$ \array{ W(b^2\mathbb{R})& \to & W(b\mathbb{R}\to\mathfrak{string})\\ \uparrow & & \uparrow\\ inv(b^2\mathbb{R})& \to & inv(b\mathbb{R}\to\mathfrak{string})\\ } $$ forces the generator $g$ to be mapped to an invariant polynomial in $W(b\mathbb{R}\to\mathfrak{string})$.

    so, here’s the compact story:

    i) the cocycle $\mu: CE(b^2\mathbb{R})\to CE(\mathfrak{so})$ factors as

    $$CE(b^2\mathbb{R})\to CE(b\mathbb{R}\to\mathfrak{string})\to CE(\mathfrak{so}),$$

    where the first morphism is given on the generators by $c\mapsto c$ and the second one by $c\mapsto \mu; \quad b\mapsto 0; t^a\mapsto t^a$ (see differential string structure for these notations). This factorization induces a factorization of $\exp_\Delta(b^2\mathbb{R})\to\exp_\Delta(\mathfrak{so})$ in

    $$\exp_\Delta(b^2\mathbb{R})\stackrel{\sim}{\to}\exp_\Delta(b\mathbb{R}\to\mathfrak{string})\twoheadrightarrow \exp_\Delta(\mathfrak{so})$$

    ii) the commutative diagram

    $$\array{ CE(b^2\mathbb{R})& \to & CE(\mathfrak{so})\\ \uparrow & & \uparrow\\ W(b^2\mathbb{R})& \to & W(\mathfrak{so})\\ \uparrow & & \uparrow\\ inv(b^2\mathbb{R})& \to & inv(\mathfrak{so})\\ }$$

    factors as

    $$\array{ CE(b^2\mathbb{R})& \to & CE(b\mathbb{R}\to\mathfrak{string})& \to & CE(\mathfrak{so})\\ \uparrow & & \uparrow& & \uparrow\\ W(b^2\mathbb{R})& \to & W(b\mathbb{R}\to\mathfrak{string})& \to & W(\mathfrak{so})\\ \uparrow & & \uparrow& & \uparrow\\ inv(b^2\mathbb{R})& \to & inv(b\mathbb{R}\to\mathfrak{string})& \to & inv(\mathfrak{so})\\ }$$

    Here the upper morphisms are those describes above; before writing the middle morphisms on the generators (which I hope to be able to do soon) let me note that the commutativity of the left bottom corner

    $$\array{ W(b^2\mathbb{R})& \to & W(b\mathbb{R}\to\mathfrak{string})\\ \uparrow & & \uparrow\\ inv(b^2\mathbb{R})& \to & inv(b\mathbb{R}\to\mathfrak{string})\\ }$$

    forces the generator $g$ to be mapped to an invariant polynomial in $W(b\mathbb{R}\to\mathfrak{string})$.

37. • CommentRowNumber37.
    • CommentAuthorUrs
    • CommentTimeJun 17th 2011
    • (edited Jun 17th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI agree, that's a good way to put it. Now next I would simply consider one more diagram factor: The isomorphism $W(b \mathbb{R} \to \mathfrak{string}) \stackrel{\simeq}{\to} \tilde W(b \mathbb{R} \to \mathfrak{string})$ induces another diagram $$ \array{ CE(b\mathbb{R} \to \mathfrak{string}) &\stackrel{Id}{\to}& CE(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ \tilde W(b\mathbb{R} \to \mathfrak{string}) &\stackrel{\simeq}{\to}& W(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ inv(b\mathbb{R} \to \mathfrak{string}) &\stackrel{Id}{\to}& inv(b\mathbb{R} \to \mathfrak{string}) } $$ and moreover there is, by construction of $\tilde W$, an easy diagram $$ \array{ CE(b^2 \mathbb{R}) &\stackrel{}{\to}& CE(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ W(b^2 \mathbb{R}) &\stackrel{}{\to}& \tilde W(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ inv(b^2 \mathbb{R}) &\stackrel{}{\to}& inv(b\mathbb{R} \to \mathfrak{string}) } \,. $$ Pasting this all together gives the desired factorization.

    I agree, that’s a good way to put it. Now next I would simply consider one more diagram factor: The isomorphism $W(b \mathbb{R} \to \mathfrak{string}) \stackrel{\simeq}{\to} \tilde W(b \mathbb{R} \to \mathfrak{string})$ induces another diagram

    $$\array{ CE(b\mathbb{R} \to \mathfrak{string}) &\stackrel{Id}{\to}& CE(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ \tilde W(b\mathbb{R} \to \mathfrak{string}) &\stackrel{\simeq}{\to}& W(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ inv(b\mathbb{R} \to \mathfrak{string}) &\stackrel{Id}{\to}& inv(b\mathbb{R} \to \mathfrak{string}) }$$

    and moreover there is, by construction of $\tilde W$, an easy diagram

    $$\array{ CE(b^2 \mathbb{R}) &\stackrel{}{\to}& CE(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ W(b^2 \mathbb{R}) &\stackrel{}{\to}& \tilde W(b\mathbb{R} \to \mathfrak{string}) \\ \uparrow && \uparrow \\ inv(b^2 \mathbb{R}) &\stackrel{}{\to}& inv(b\mathbb{R} \to \mathfrak{string}) } \,.$$

    Pasting this all together gives the desired factorization.

38. • CommentRowNumber38.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 17th 2011
    • (edited Jun 17th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexThat's it! What I'm suggesting is: once we've used the auxiliary $\tilde{W}$ to find the seeked morphisms, let us write directly the composite morphisms and omit $\tilde{W}$. by the way, Urs, if you have already these computations, can you post them here? (otherwise I can work them out as soon as I have a minute for this). Have you also the explicit morphism $$ \array{ CE(b\mathbb{R} \to \mathfrak{string}) &\to& CE(\mathfrak{so}) \\ \uparrow && \uparrow \\ W(b\mathbb{R} \to \mathfrak{string}) &\to& W(\mathfrak{so}) \\ \uparrow && \uparrow \\ inv(b\mathbb{R} \to \mathfrak{string}) &\to& inv(\mathfrak{so}) } $$ ?

    That’s it! What I’m suggesting is: once we’ve used the auxiliary $\tilde{W}$ to find the seeked morphisms, let us write directly the composite morphisms and omit $\tilde{W}$. by the way, Urs, if you have already these computations, can you post them here? (otherwise I can work them out as soon as I have a minute for this). Have you also the explicit morphism

    $$\array{ CE(b\mathbb{R} \to \mathfrak{string}) &\to& CE(\mathfrak{so}) \\ \uparrow && \uparrow \\ W(b\mathbb{R} \to \mathfrak{string}) &\to& W(\mathfrak{so}) \\ \uparrow && \uparrow \\ inv(b\mathbb{R} \to \mathfrak{string}) &\to& inv(\mathfrak{so}) }$$

    ?

39. • CommentRowNumber39.
    • CommentAuthorUrs
    • CommentTimeJun 17th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Domenico, what I have are the computations that you have seen in the entry. I am thinking by that extra factorization that I mentioned, these are sufficient to give all the rest automatically. So in my message above there is one double-square diagram with an isomorphism in the middle. That is automatic: the remaining morphism are just those of the original double-square composed with the inverse of that iso. Then there is the double-square which I called "the easy diagram". Since that involves $\tilde W$ it is really evident: with the morphism all send generators of some name to the generator of the same name. But, sure, I can write that out. Probably not tonight, though.

    Hi Domenico,

    what I have are the computations that you have seen in the entry. I am thinking by that extra factorization that I mentioned, these are sufficient to give all the rest automatically.

    So in my message above there is one double-square diagram with an isomorphism in the middle. That is automatic: the remaining morphism are just those of the original double-square composed with the inverse of that iso.

    Then there is the double-square which I called “the easy diagram”. Since that involves $\tilde W$ it is really evident: with the morphism all send generators of some name to the generator of the same name.

    But, sure, I can write that out. Probably not tonight, though.

40. • CommentRowNumber40.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 17th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex> I am thinking by that extra factorization that I mentioned, these are sufficient to give all the rest automatically. That's what I'm sure of, too. That's why I wrote that as soon as I have a minute I will write this computaton out. I was just hoping you had it already explicited. if not, no hurry to do it now.

    I am thinking by that extra factorization that I mentioned, these are sufficient to give all the rest automatically.

    That’s what I’m sure of, too. That’s why I wrote that as soon as I have a minute I will write this computaton out. I was just hoping you had it already explicited. if not, no hurry to do it now.

41. • CommentRowNumber41.
    • CommentAuthorUrs
    • CommentTimeJun 18th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexokay, I am now in the process of typing this expanded proof up: at a new entry: [[differential string structure -- proofs]]

    okay, I am now in the process of typing this expanded proof up: at a new entry: differential string structure – proofs

42. • CommentRowNumber42.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 18th 2011
    • (edited Jun 18th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexhow are the $(\dots)$ morphisms defined? (I mean are you in the process of typing these or is this left for a later time?)

    how are the $(\dots)$ morphisms defined? (I mean are you in the process of typing these or is this left for a later time?)

43. • CommentRowNumber43.
    • CommentAuthorUrs
    • CommentTimeJun 18th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexRight, that's what I am in the process of typing up. I'll tell you when I am done. I just thought I'd already announce that I have started working on it.

    Right, that’s what I am in the process of typing up. I’ll tell you when I am done. I just thought I’d already announce that I have started working on it.

44. • CommentRowNumber44.
    • CommentAuthorUrs
    • CommentTimeJun 18th 2011
    • (edited Jun 18th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItex(20 min later...) Okay, I think I am now close to something complete, even if in need of more polishing/expansion. Please have another look at [[differential string structure -- proofs]]

    (20 min later…)

    Okay, I think I am now close to something complete, even if in need of more polishing/expansion.

    Please have another look at differential string structure – proofs

45. • CommentRowNumber45.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 18th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexok, fine. in the meanwhile, let me notice that the isomorphism $CE(b\mathbb{R}\to\mathfrak{string})\cong CE(\mathfrak{so})\otimes\CE(inn(b\mathbb{R}))$ precisely gives natural isomorphisms $Hom_{dgca}(CE(b\mathbb{R}\to\mathfrak{string}),\Omega^\bullet(U\times\Delta^k)_{vert})\cong Hom_{dgca}(CE(\mathfrak{so}),\Omega^\bullet(U\times\Delta^k)_{vert})\times Hom_{dgca}(CE(inn(b\mathbb{R})),\Omega^\bullet(U\times\Delta^k)_{vert})$, thus the isomorphism $$ \exp_\Delta(b\mathbb{R}\to\mathfrak{string})\cong \exp_\Delta(\mathfrak{so})\times\exp_\Delta(inn(b\mathbb{R})) $$

    ok, fine. in the meanwhile, let me notice that the isomorphism $CE(b\mathbb{R}\to\mathfrak{string})\cong CE(\mathfrak{so})\otimes\CE(inn(b\mathbb{R}))$ precisely gives natural isomorphisms $Hom_{dgca}(CE(b\mathbb{R}\to\mathfrak{string}),\Omega^\bullet(U\times\Delta^k)_{vert})\cong Hom_{dgca}(CE(\mathfrak{so}),\Omega^\bullet(U\times\Delta^k)_{vert})\times Hom_{dgca}(CE(inn(b\mathbb{R})),\Omega^\bullet(U\times\Delta^k)_{vert})$, thus the isomorphism

    $$\exp_\Delta(b\mathbb{R}\to\mathfrak{string})\cong \exp_\Delta(\mathfrak{so})\times\exp_\Delta(inn(b\mathbb{R}))$$
46. • CommentRowNumber46.
    • CommentAuthorUrs
    • CommentTimeJun 18th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItexOkay, good. Maybe we should put all these little technical points now into the new Proofs-entry, where they don't obstruct the readability of the main entry. We should in fact move more of the long proofs from there. I'll have to go offline in a few minutes and will come online again later this evening.

    Okay, good. Maybe we should put all these little technical points now into the new Proofs-entry, where they don’t obstruct the readability of the main entry. We should in fact move more of the long proofs from there.

    I’ll have to go offline in a few minutes and will come online again later this evening.

47. • CommentRowNumber47.
    • CommentAuthorUrs
    • CommentTimeJun 18th 2011
    • (edited Jun 18th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Domenico, I have now fine-tuned the discussion a bit further. One point I changed is this: it's actually not quite true that we get a factorization by double square diagrams that is a weak equivalence on the left if we use $inv(b \mathbb{R} \to \mathfrak{string})$, rather, it is true if we use $\tilde inv(b \mathbb{R} \to \mathfrak{string})$. That's another way of saying what the isomorphism $W(b \mathbb{R} \to \mathfrak{string}) \simeq \tilde W(b \mathbb{R} \to \mathfrak{string})$ is about: it is the one that makes the quasi-isomorphism $CE(\mathfrak{so}) \simeq CE(b \mathbb{R} \to \mathfrak{string})$ descend to invariant polynomials: for the invariant polynomials of $(b\mathbb{R} \to \mathfrak{string})$ are those of $\mathfrak{so}$ with the generator $g$ adjoined. But those of the twiddled version are those of $\mathfrak{so}$ with generators $g$ and $\tilde h$ adjoined and quotiented by the relation $d \tilde h = g- \langle - , -\rangle$. That makes the superfluous $g$ drop out in cohomology again. I suppose this means that one should in principle eventually try to think about an actual model structure on the category whose morphisms are such $CE \leftarrow W \leftarrow inv$-double squares.

    Hi Domenico,

    I have now fine-tuned the discussion a bit further.

    One point I changed is this: it’s actually not quite true that we get a factorization by double square diagrams that is a weak equivalence on the left if we use $inv(b \mathbb{R} \to \mathfrak{string})$, rather, it is true if we use $\tilde inv(b \mathbb{R} \to \mathfrak{string})$.

    That’s another way of saying what the isomorphism $W(b \mathbb{R} \to \mathfrak{string}) \simeq \tilde W(b \mathbb{R} \to \mathfrak{string})$ is about: it is the one that makes the quasi-isomorphism $CE(\mathfrak{so}) \simeq CE(b \mathbb{R} \to \mathfrak{string})$ descend to invariant polynomials:

    for the invariant polynomials of $(b\mathbb{R} \to \mathfrak{string})$ are those of $\mathfrak{so}$ with the generator $g$ adjoined. But those of the twiddled version are those of $\mathfrak{so}$ with generators $g$ and $\tilde h$ adjoined and quotiented by the relation $d \tilde h = g- \langle - , -\rangle$. That makes the superfluous $g$ drop out in cohomology again.

    I suppose this means that one should in principle eventually try to think about an actual model structure on the category whose morphisms are such $CE \leftarrow W \leftarrow inv$-double squares.

48. • CommentRowNumber48.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 18th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexthere's something unclear to me about the role of $\tilde{inv}$: written thsi way the second from the left column is not (at least directly) related to $\exp_{\Delta}(b\mathbb{R}\to \mathfrak{string})_{conn}$, so it is unclear to me how the big diagram decribes a factorization $$ \exp_{\Delta}(\mathfrak{so})_{conn}\to \exp_{\Delta}(b\mathbb{R}\to \mathfrak{string})_{conn}\to \exp_{\Delta}(b^2\mathbb{R})_{conn} $$

    there’s something unclear to me about the role of $\tilde{inv}$: written thsi way the second from the left column is not (at least directly) related to $\exp_{\Delta}(b\mathbb{R}\to \mathfrak{string})_{conn}$, so it is unclear to me how the big diagram decribes a factorization

    $$\exp_{\Delta}(\mathfrak{so})_{conn}\to \exp_{\Delta}(b\mathbb{R}\to \mathfrak{string})_{conn}\to \exp_{\Delta}(b^2\mathbb{R})_{conn}$$
49. • CommentRowNumber49.
    • CommentAuthorUrs
    • CommentTimeJun 18th 2011
    • (edited Jun 18th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItexRight, as I said, the middle term uses the twiddled invariant polynomials instead.

    Right, as I said, the middle term uses the twiddled invariant polynomials instead.

50. • CommentRowNumber50.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 18th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexmmm.. so you're syng that we are writing a factorization of $\exp_{\Delta}(\mathfrak{so})_{conn}\to \exp_{\Delta}(b^2\mathbb{R})_{conn}$ of the form $$ \exp_{\Delta}(\mathfrak{so})_{conn}\stackrel{\sim}{\to} P\twoheadrightarrow \exp_{\Delta}(b^2\mathbb{R})_{conn} $$ where $P$ is some simplicial presheaf which, however, is not of the form $\exp_\Delta(\mathfrak{h})$ for an $L_\infty$-algebra $\mathfrak{h}$? clearly this could be, but would make the whole construction much less nice that hoped. but most of all, I'm confused by the fact that if one writes the composition of the two leftmost morphisms, so to have a diagram in which $\tilde{W}$ does not appear, the assignment $\tilde{h}\mapsto h+(cs-\mu)$ seems to disappear and to be hidden in the only non explicit arrow in the diagram (the bottom vertical one in the middle): $$ \array{ CE(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu } \right) }{\leftarrow} & CE(b \mathbb{R} \to \mathfrak{string}) &\stackrel{ \left( \array{ c \mapsto c } \right) }{\leftarrow} & CE(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{i^*_{\mathfrak{so}}} && \uparrow^\mathrlap{i^*_{(b\mathbb{R} \to \mathfrak{string})}} && \uparrow^\mathrlap{i^*_{b^2 \mathbb{R}}} \\ W(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu \\ r^a \mapsto r^a \\ h \mapsto 0 \\ g \mapsto \langle-,-\rangle } \right) }{\leftarrow} & W(b \mathbb{R} \to \mathfrak{string}) & \stackrel{ \left( \array{ c \mapsto c \\ g \mapsto g } \right) }{\leftarrow} & W(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{p^*_{\mathfrak{so}}} && \uparrow^\mathrlap{} && \uparrow^\mathrlap{p^*_{b^2 \mathbb{R}}} \\ inv(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ \tilde h \mapsto 0 \\ g \mapsto \langle -,-\rangle \\ \langle \cdots \rangle \mapsto \langle \cdots \rangle } \right) }{\leftarrow} & \tilde inv(b \mathbb{R} \to \mathfrak{string}) & \stackrel{ \left( \array{ g \mapsto g } \right) }{\leftarrow} & inv(b^2 \mathbb{R}) } \,. $$ on the other hand, the arrows at the inv-level should not be defined on the generators of the inv's but rather should be just the restrictions of the maps at the W-level. in other words, once the maps at the W-lever are defined one need only to check whether the inv-subspaces are preserved. inv-elements are defined by two conditions: being $d_W$-closed and being polynomials in the shifted generators. the first condition is always satisfied, dince the maps at the W-level are dgca maps; so one only needs to check whether shifted geerators are mapped to polynomial in the shifted generators. therefore I'm not sure I can see which is the problem with this diagram here below: $$ \array{ CE(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu } \right) }{\leftarrow} & CE(b \mathbb{R} \to \mathfrak{string}) &\stackrel{ \left( \array{ c \mapsto c } \right) }{\leftarrow} & CE(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{i^*_{\mathfrak{so}}} && \uparrow^\mathrlap{i^*_{(b\mathbb{R} \to \mathfrak{string})}} && \uparrow^\mathrlap{i^*_{b^2 \mathbb{R}}} \\ W(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu \\ r^a \mapsto r^a \\ h \mapsto 0 \\ g \mapsto \langle-,-\rangle } \right) }{\leftarrow} & W(b \mathbb{R} \to \mathfrak{string}) & \stackrel{ \left( \array{ c \mapsto c \\ g \mapsto g } \right) }{\leftarrow} & W(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{p^*_{\mathfrak{so}}} && \uparrow^\mathrlap{p^*_{b\mathbb{R}\to\mathfrak{string}}} && \uparrow^\mathrlap{p^*_{b^2 \mathbb{R}}} \\ inv(\mathfrak{so}) & \leftarrow & inv(b \mathbb{R} \to \mathfrak{string}) & \leftarrow & inv(b^2 \mathbb{R}) } \,. $$

    mmm.. so you’re syng that we are writing a factorization of $\exp_{\Delta}(\mathfrak{so})_{conn}\to \exp_{\Delta}(b^2\mathbb{R})_{conn}$ of the form

    $$\exp_{\Delta}(\mathfrak{so})_{conn}\stackrel{\sim}{\to} P\twoheadrightarrow \exp_{\Delta}(b^2\mathbb{R})_{conn}$$

    where $P$ is some simplicial presheaf which, however, is not of the form $\exp_\Delta(\mathfrak{h})$ for an $L_\infty$-algebra $\mathfrak{h}$? clearly this could be, but would make the whole construction much less nice that hoped. but most of all, I’m confused by the fact that if one writes the composition of the two leftmost morphisms, so to have a diagram in which $\tilde{W}$ does not appear, the assignment $\tilde{h}\mapsto h+(cs-\mu)$ seems to disappear and to be hidden in the only non explicit arrow in the diagram (the bottom vertical one in the middle):

    $$\array{ CE(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu } \right) }{\leftarrow} & CE(b \mathbb{R} \to \mathfrak{string}) &\stackrel{ \left( \array{ c \mapsto c } \right) }{\leftarrow} & CE(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{i^*_{\mathfrak{so}}} && \uparrow^\mathrlap{i^*_{(b\mathbb{R} \to \mathfrak{string})}} && \uparrow^\mathrlap{i^*_{b^2 \mathbb{R}}} \\ W(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu \\ r^a \mapsto r^a \\ h \mapsto 0 \\ g \mapsto \langle-,-\rangle } \right) }{\leftarrow} & W(b \mathbb{R} \to \mathfrak{string}) & \stackrel{ \left( \array{ c \mapsto c \\ g \mapsto g } \right) }{\leftarrow} & W(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{p^*_{\mathfrak{so}}} && \uparrow^\mathrlap{} && \uparrow^\mathrlap{p^*_{b^2 \mathbb{R}}} \\ inv(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ \tilde h \mapsto 0 \\ g \mapsto \langle -,-\rangle \\ \langle \cdots \rangle \mapsto \langle \cdots \rangle } \right) }{\leftarrow} & \tilde inv(b \mathbb{R} \to \mathfrak{string}) & \stackrel{ \left( \array{ g \mapsto g } \right) }{\leftarrow} & inv(b^2 \mathbb{R}) } \,.$$

    on the other hand, the arrows at the inv-level should not be defined on the generators of the inv’s but rather should be just the restrictions of the maps at the W-level. in other words, once the maps at the W-lever are defined one need only to check whether the inv-subspaces are preserved. inv-elements are defined by two conditions: being $d_W$-closed and being polynomials in the shifted generators. the first condition is always satisfied, dince the maps at the W-level are dgca maps; so one only needs to check whether shifted geerators are mapped to polynomial in the shifted generators. therefore I’m not sure I can see which is the problem with this diagram here below:

    $$\array{ CE(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu } \right) }{\leftarrow} & CE(b \mathbb{R} \to \mathfrak{string}) &\stackrel{ \left( \array{ c \mapsto c } \right) }{\leftarrow} & CE(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{i^*_{\mathfrak{so}}} && \uparrow^\mathrlap{i^*_{(b\mathbb{R} \to \mathfrak{string})}} && \uparrow^\mathrlap{i^*_{b^2 \mathbb{R}}} \\ W(\mathfrak{so}) & \underoverset{\simeq}{ \left( \array{ t^a \mapsto t^a \\ b \mapsto 0 \\ c \mapsto \mu \\ r^a \mapsto r^a \\ h \mapsto 0 \\ g \mapsto \langle-,-\rangle } \right) }{\leftarrow} & W(b \mathbb{R} \to \mathfrak{string}) & \stackrel{ \left( \array{ c \mapsto c \\ g \mapsto g } \right) }{\leftarrow} & W(b^2 \mathbb{R}) \\ \uparrow^\mathrlap{p^*_{\mathfrak{so}}} && \uparrow^\mathrlap{p^*_{b\mathbb{R}\to\mathfrak{string}}} && \uparrow^\mathrlap{p^*_{b^2 \mathbb{R}}} \\ inv(\mathfrak{so}) & \leftarrow & inv(b \mathbb{R} \to \mathfrak{string}) & \leftarrow & inv(b^2 \mathbb{R}) } \,.$$
51. • CommentRowNumber51.
    • CommentAuthorUrs
    • CommentTimeJun 19th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItex> mmm.. so you're saying [...] Yes, right. We are just finding a means to factor that morphism as a weak equivalence followed by a fibration. The definition of invariant polynomials that we have used does not send all weak equivalences to quasi-isomorphisms, so I am fixing that by hand here. I agree that this smells like there should be a better theory, but for the moment it just does the job. One should remember how the deifnition of the invariant polynomials arises in the first place: after finding the simplicial model for $\mathbf{B}^n U(1)_{conn}$ these just happen to do the job of extending the simplicial models of morphisms $\mathbf{B}G \to \mathbf{B}^n U(1)$. Concerning the problem with the other factorization: yes, you could factor like this. But the problem to be solved is that when integrating via $exp(-)_{conn}$ we are supposed to keep the invariant polynmoials horizontal, and solving that conditon requires forming certain linear combinations of generators. This forming of linear combinations is what the isomorphism to $\tilde W$ takes care of.

    mmm.. so you’re saying […]

    Yes, right. We are just finding a means to factor that morphism as a weak equivalence followed by a fibration. The definition of invariant polynomials that we have used does not send all weak equivalences to quasi-isomorphisms, so I am fixing that by hand here. I agree that this smells like there should be a better theory, but for the moment it just does the job. One should remember how the deifnition of the invariant polynomials arises in the first place: after finding the simplicial model for $\mathbf{B}^n U(1)_{conn}$ these just happen to do the job of extending the simplicial models of morphisms $\mathbf{B}G \to \mathbf{B}^n U(1)$.

    Concerning the problem with the other factorization: yes, you could factor like this. But the problem to be solved is that when integrating via $exp(-)_{conn}$ we are supposed to keep the invariant polynmoials horizontal, and solving that conditon requires forming certain linear combinations of generators. This forming of linear combinations is what the isomorphism to $\tilde W$ takes care of.

52. • CommentRowNumber52.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 19th 2011
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItex>we are supposed to keep the invariant polynmoials horizontal ah, ok, now I see it! you're saying that $exp(-)_{CW}$ is functorial on the triples, but $exp(-)_{conn}$ is not, so this needs be fixed at hand, am I right? if so I agree $\tilde{W}$ is crucial, but we should state this more clearly (i.e., we should stress the fact that ${}_{conn}$ pich some nice representatives, but ${}_CW$ is the real (functorial) thing. I'm sure we're already saying this, but maybe it can be stressed, it seems important)

    we are supposed to keep the invariant polynmoials horizontal

    ah, ok, now I see it! you’re saying that $exp(-)_{CW}$ is functorial on the triples, but $exp(-)_{conn}$ is not, so this needs be fixed at hand, am I right? if so I agree $\tilde{W}$ is crucial, but we should state this more clearly (i.e., we should stress the fact that ${}_{conn}$ pich some nice representatives, but ${}_CW$ is the real (functorial) thing. I’m sure we’re already saying this, but maybe it can be stressed, it seems important)

53. • CommentRowNumber53.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJun 19th 2011
    • (edited Jun 19th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexon second thought, can't we trade "strange algebras with nice morphisms" for "nice algebras with strange morphisms"? as soon as the algebras involved are isomorphic this should be possible, just by spelling out the compositions of the "nice morphisms" with the isomorphism between the nice algebra and the stange one. I'll think on this.

    on second thought, can’t we trade “strange algebras with nice morphisms” for “nice algebras with strange morphisms”? as soon as the algebras involved are isomorphic this should be possible, just by spelling out the compositions of the “nice morphisms” with the isomorphism between the nice algebra and the stange one. I’ll think on this.

54. • CommentRowNumber54.
    • CommentAuthordomenico_fiorenza
    • CommentTimeSep 8th 2011
    • (edited Sep 8th 2011)
    • PermaLink
    Author: domenico_fiorenza
    Format: MarkdownItexHi Urs, at [[differential string structure -- proofs]] shouldn't it be $c\mapsto cs$ and $h\mapsto \mu-cs$ in the last diagram (on the left, in the middle row of the diagram)? this way the composite morphism $W(b^2\mathbb{R})\to W(\mathfrak{so})$ would be $c\mapsto cs$ and $g \mapsto \langle-,-\rangle$, which seems to be correct, since $d_W cs= \langle-,-\rangle$, while the compatibility with the upper row is given by $i^* (cs)= \mu$. this would also say that the morphism $\tilde{W}(b\mathbb{R}\to \mathfrak{string})\to W(\mathfrak{so})$ is $$ t^a\mapsto t^a;\qquad b\mapsto 0;\qquad c\mapsto cs; $$ $$ r^a\mapsto r^a;\qquad \tilde{h}\mapsto 0; \qquad g\mapsto \langle-,-\rangle $$

    Hi Urs,

    at differential string structure – proofs shouldn’t it be $c\mapsto cs$ and $h\mapsto \mu-cs$ in the last diagram (on the left, in the middle row of the diagram)?

    this way the composite morphism $W(b^2\mathbb{R})\to W(\mathfrak{so})$ would be $c\mapsto cs$ and $g \mapsto \langle-,-\rangle$, which seems to be correct, since $d_W cs= \langle-,-\rangle$, while the compatibility with the upper row is given by $i^* (cs)= \mu$.

    this would also say that the morphism $\tilde{W}(b\mathbb{R}\to \mathfrak{string})\to W(\mathfrak{so})$ is

    $$t^a\mapsto t^a;\qquad b\mapsto 0;\qquad c\mapsto cs;$$ $$r^a\mapsto r^a;\qquad \tilde{h}\mapsto 0; \qquad g\mapsto \langle-,-\rangle$$
55. • CommentRowNumber55.
    • CommentAuthorUrs
    • CommentTimeSep 8th 2011
    • (edited Sep 8th 2011)
    • PermaLink
    Author: Urs
    Format: MarkdownItexYes, of course. I think that's a copy-and-paste typo from the CE-row. I'll change it.

    Yes, of course. I think that’s a copy-and-paste typo from the CE-row. I’ll change it.