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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMay 3rd 2010
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   Author: Urs
   Format: MarkdownItexedited [[Weil algebra]] a bit. More to come.

   edited Weil algebra a bit. More to come.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2010
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   Author: Urs
   Format: MarkdownItexadded to [[Weil algebra]] a section <a href="http://ncatlab.org/nlab/show/Weil+algebra#Properties">Properties</a> with a discussion of the free property of the Weil algebra.

   added to Weil algebra a section Properties with a discussion of the free property of the Weil algebra.

3. • CommentRowNumber3.
   • CommentAuthordomenico_fiorenza
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
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   Author: domenico_fiorenza
   Format: MarkdownItexI'm not sure I'm following your argument: it seems you're saying that a dgca morphism $f:W(\mathfrak{g}) \to A$ is the same thing as a linear map $\mathfrak{g}^*\to U(A)$ which moreover satisfies the compatibility with differentials resticted to $\mathfrak{g}^*$, i.e. that argument seems to prove that $Hom_{dgca}(W(\mathfrak{g}),A)\to Hom_{grVect}(\mathfrak{g},U(A))$ is injective, not bijective. Where am I lost?

   I’m not sure I’m following your argument: it seems you’re saying that a dgca morphism $f:W(\mathfrak{g}) \to A$ is the same thing as a linear map $\mathfrak{g}^*\to U(A)$ which moreover satisfies the compatibility with differentials resticted to $\mathfrak{g}^*$, i.e. that argument seems to prove that $Hom_{dgca}(W(\mathfrak{g}),A)\to Hom_{grVect}(\mathfrak{g},U(A))$ is injective, not bijective. Where am I lost?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
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   Author: Urs
   Format: MarkdownItexThe argument I gave is supposed to show that once you fix the underlying $\mathfrak{g}^* \to U(A)$, there is a unique morphism $W(\mathfrak{g}) \to A$ extending that. See, fix any images $f : \mathfrak{g}^* \to A$ of the unshifted generators $t \in \mathfrak{g}^*$. Then you can solve uniquely for the images of the shifted generators $\sigma t$ by using that in order to have a dg-homomorphism we need $$ d_A f (t) = f (d_{W} t ) = f (d_{CE} t) + f (\sigma t) $$ so that we find that $f$ has to act on the shifted generators as $$ f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,. $$ So then the only remaining question is if this unique way to solve $[d,f] = 0$ on the unshifted generators also solves it on the shifted generators. This I typed out the computation for. Please let me know if this helps.

   The argument I gave is supposed to show that once you fix the underlying $\mathfrak{g}^* \to U(A)$, there is a unique morphism $W(\mathfrak{g}) \to A$ extending that.

   See, fix any images $f : \mathfrak{g}^* \to A$ of the unshifted generators $t \in \mathfrak{g}^*$.

   Then you can solve uniquely for the images of the shifted generators $\sigma t$ by using that in order to have a dg-homomorphism we need

   $$d_A f (t) = f (d_{W} t ) = f (d_{CE} t) + f (\sigma t)$$

   so that we find that $f$ has to act on the shifted generators as

   $$f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,.$$

   So then the only remaining question is if this unique way to solve $[d,f] = 0$ on the unshifted generators also solves it on the shifted generators. This I typed out the computation for.

   Please let me know if this helps.

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexwhat I don't see is how you can have $d_A f(t)=f(d_W t)$ without imposing this condition on $f:\mathfrak{g}^*\to U(A)$. Only thing I'm able to see is that if you have this, then the unique ectension of $f$ to shifted generators is a differential on the whole $W(\mathfrak{g})$.

   what I don’t see is how you can have $d_A f(t)=f(d_W t)$ without imposing this condition on $f:\mathfrak{g}^*\to U(A)$. Only thing I’m able to see is that if you have this, then the unique ectension of $f$ to shifted generators is a differential on the whole $W(\mathfrak{g})$.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexWe must be misunderstanding each other on some simple point. > what I don't see is how you can have $d_A f(t)=f(d_W t)$ without imposing this condition on $f:\mathfrak{g}^*\to U(A)$. Which condition do you mean? For me $f:\mathfrak{g}^*\to U(A)$ means the images of the unshifted generators. Fix any such, without any condition. Then the equation $d_A f(t) = f(d_W t)$ has a unique solution encoded by a map $f : \mathfrak{g}^*[1] \to U(A)$: these shifted generators must be mapped as $$ f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,. $$ You see, the Weil algebra is precisely built such that it introduces precisely one new generator for each condition in the Chevalley-Eilenberg algebra, precisely such that all these conditions get unique soliutions: the new generators pick up precisely the failure o the old generators to satisfy their original conditions.

   We must be misunderstanding each other on some simple point.

   what I don’t see is how you can have $d_A f(t)=f(d_W t)$ without imposing this condition on $f:\mathfrak{g}^*\to U(A)$.

   Which condition do you mean? For me $f:\mathfrak{g}^*\to U(A)$ means the images of the unshifted generators. Fix any such, without any condition. Then the equation $d_A f(t) = f(d_W t)$ has a unique solution encoded by a map $f : \mathfrak{g}^*[1] \to U(A)$: these shifted generators must be mapped as

   $$f : \sigma t \mapsto d_A (f(t)) - f(d_{CE} t) \,.$$

   You see, the Weil algebra is precisely built such that it introduces precisely one new generator for each condition in the Chevalley-Eilenberg algebra, precisely such that all these conditions get unique soliutions: the new generators pick up precisely the failure o the old generators to satisfy their original conditions.

7. • CommentRowNumber7.
   • CommentAuthordomenico_fiorenza
   • CommentTimeAug 16th 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexAh, ok! now that's perfectly clear! do you mind if I rewrite the proof on the nLab here and there? ;)

   Ah, ok! now that’s perfectly clear! do you mind if I rewrite the proof on the nLab here and there? ;)

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeAug 16th 2010
   • PermaLink
   Author: Urs
   Format: Markdown> do you mind if I rewrite the proof on the nLab here and there? ;) Please do.

   do you mind if I rewrite the proof on the nLab here and there? ;)

   Please do.

9. • CommentRowNumber9.
   • CommentAuthordomenico_fiorenza
   • CommentTimeAug 16th 2010
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexDone. Please have a look. I've left also the old proof for the moment. One day we should take care of fixing the grading..

   Done. Please have a look. I’ve left also the old proof for the moment. One day we should take care of fixing the grading..

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeAug 16th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownThanks, looks good. I removed the redundant old paragraph now. > One day we should take care of fixing the grading. There are some comments on that at [[Chevalley-Eilenberg algebra]], at least.

    Thanks, looks good. I removed the redundant old paragraph now.

    One day we should take care of fixing the grading.

    There are some comments on that at Chevalley-Eilenberg algebra, at least.

11. • CommentRowNumber11.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexMinor edits. Added a line to explain in which sense $\sigma$ is extended to a derivation.

    Minor edits. Added a line to explain in which sense $\sigma$ is extended to a derivation.

12. • CommentRowNumber12.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexAdded a one line construction of the canonical morphism $W(\mathfrak{g})\to CE(\mathfrak{g})$ :)

    Added a one line construction of the canonical morphism $W(\mathfrak{g})\to CE(\mathfrak{g})$ :)

13. • CommentRowNumber13.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
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    Author: domenico_fiorenza
    Format: MarkdownItexHow do we see that the cohomology of $(W(\mathfrak{g}),d_{W(\mathfrak{g})})$ is trivial?

    How do we see that the cohomology of $(W(\mathfrak{g}),d_{W(\mathfrak{g})})$ is trivial?

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 17th 2010
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    Author: Todd_Trimble
    Format: MarkdownItexI am probably off base here since I am only glancing at this discussion from afar, but the construction of $W(\mathfrak{g})$ does remind me of taking the exponential (free commutative monoid) of the mapping cone of an identity morphism, which has trivial cohomology on general grounds.

    I am probably off base here since I am only glancing at this discussion from afar, but the construction of $W(\mathfrak{g})$ does remind me of taking the exponential (free commutative monoid) of the mapping cone of an identity morphism, which has trivial cohomology on general grounds.

15. • CommentRowNumber15.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexIndeed that is what it should be! but I'm unable to switch between the general grounds definition of $W(\mathfrak{g} $ and the explicit one :(

    Indeed that is what it should be! but I’m unable to switch between the general grounds definition of $W(\mathfrak{g}$ and the explicit one :(

16. • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeAug 17th 2010
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    Author: Todd_Trimble
    Format: MarkdownItexOr rather, I guess the differential on $W(\mathfrak{g})$ is some twisted version of this. Anyway, what is in the back of my mind is that there is a satisfying bar construction story to be told (or fleshed out more) here which would shed light on all this.

    Or rather, I guess the differential on $W(\mathfrak{g})$ is some twisted version of this. Anyway, what is in the back of my mind is that there is a satisfying bar construction story to be told (or fleshed out more) here which would shed light on all this.

17. • CommentRowNumber17.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
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    Author: Urs
    Format: MarkdownItexYes, it comes from the mapping cone on the identity. An explicit way to see the vanishing of the cohomology is this: Let $F(\mathfrak{g})$ be the free dg-algebra on the graded vector space $\mathfrak{g}^*$, i.e. $F(\mathfrak{g}) = \wedge^\bullet (\mathfrak{g}^* \oplus \mathfrak{g}^*[1])$ with differential this shift isomorphism extended as a derivation $d_{F} = \sigma$. This manifestly has trivial cohomology. Then notice that there is a dg-algebra isomorphism $$f : F(\mathfrak{g}) \to W(\mathfrak{g}) $$ which is the identity on the unshifted genertors, and which on the shifted generators is $ f : \sigma (t) \mapsto d_W t = d_{CE} t + \sigma (t) $.

    Yes, it comes from the mapping cone on the identity.

    An explicit way to see the vanishing of the cohomology is this:

    Let $F(\mathfrak{g})$ be the free dg-algebra on the graded vector space $\mathfrak{g}^*$, i.e. $F(\mathfrak{g}) = \wedge^\bullet (\mathfrak{g}^* \oplus \mathfrak{g}^*[1])$ with differential this shift isomorphism extended as a derivation $d_{F} = \sigma$.

    This manifestly has trivial cohomology.

    Then notice that there is a dg-algebra isomorphism

    $$f : F(\mathfrak{g}) \to W(\mathfrak{g})$$

    which is the identity on the unshifted genertors, and which on the shifted generators is

    $f : \sigma (t) \mapsto d_W t = d_{CE} t + \sigma (t)$.

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
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    Author: Urs
    Format: MarkdownItexI have added that remark to the Properties-section at [[Weil algebra]].

    I have added that remark to the Properties-section at Weil algebra.

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
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    Author: Urs
    Format: MarkdownItexAlso added a remark in the section on invariant polynomials on how the dg--algebraic definition of invariant polynomial implies that these are indeed invariant under the ad-action of the $\infty$-Lie algebra.

    Also added a remark in the section on invariant polynomials on how the dg–algebraic definition of invariant polynomial implies that these are indeed invariant under the ad-action of the $\infty$-Lie algebra.

20. • CommentRowNumber20.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
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    Author: domenico_fiorenza
    Format: MarkdownItexConcerning $F(\mathfrak{g})$ vs. $W(\mathfrak{g})$, I'd rather use the following half-line argument: given an $L_\infty$ algebra $\mathfrak{g}$, let $\mathfrak{g}_null$ be the undelying graded vector spce, which we think of as an $L_\infty$ algebra with trivial opeartions. Then the Weil algebra of $\mathfrak{g}_{null}$ is $W(\mathfrak{g}_{null})=F(\mathfrak{g})$, and the freeness property of Weil algebras tells that the identity morphism of graded vector spaces $\mathfrak{g}_{null}=\mathfrak{g}$ induces an isomorphism of differential graded commutative algebras $F(\mathfrak{g})\cong W(\mathfrak{g})$.

    Concerning $F(\mathfrak{g})$ vs. $W(\mathfrak{g})$, I’d rather use the following half-line argument: given an $L_\infty$ algebra $\mathfrak{g}$, let $\mathfrak{g}_null$ be the undelying graded vector spce, which we think of as an $L_\infty$ algebra with trivial opeartions. Then the Weil algebra of $\mathfrak{g}_{null}$ is $W(\mathfrak{g}_{null})=F(\mathfrak{g})$, and the freeness property of Weil algebras tells that the identity morphism of graded vector spaces $\mathfrak{g}_{null}=\mathfrak{g}$ induces an isomorphism of differential graded commutative algebras $F(\mathfrak{g})\cong W(\mathfrak{g})$.

21. • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
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    Author: Urs
    Format: MarkdownItexSure, I would think this is a good extra remark to be added.

    Sure, I would think this is a good extra remark to be added.

22. • CommentRowNumber22.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexby the way, to prove that $F(\mathfrak{g})$ has trivial cohomology, I would add a line saying that $\sigma^{-1}:\mathfrak{g}^*[1]\to \mathfrak{g}^*$ extends to a degree -1 derivation $K$ of $F(\mathfrak{g})$. Since $[d_F,K]=2 Id$, multiplication by 2 is homotopic to zero, and that's all. It's completely trivial and standard, but being so short maybe it's worth adding to the nLab page.

    by the way, to prove that $F(\mathfrak{g})$ has trivial cohomology, I would add a line saying that $\sigma^{-1}:\mathfrak{g}^*[1]\to \mathfrak{g}^*$ extends to a degree -1 derivation $K$ of $F(\mathfrak{g})$. Since $[d_F,K]=2 Id$, multiplication by 2 is homotopic to zero, and that’s all. It’s completely trivial and standard, but being so short maybe it’s worth adding to the nLab page.

23. • CommentRowNumber23.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
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    Author: Urs
    Format: MarkdownItexYou want me to do it?

    You want me to do it?

24. • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexWait a sec, it's just $[d_F, K] = Id$. (Sufficient to check this on generators.)

    Wait a sec, it’s just $[d_F, K] = Id$.

    (Sufficient to check this on generators.)

25. • CommentRowNumber25.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexoh, yes, sure! I had just used $\sigma \sigma^{-1}+\sigma^{-1}\sigma$ without thinking that half of this expression is actually zero on generators (which half depending on which generators one considers) :) I can make the additions in a couple of hours, if you can do them before, please do. Otherwise I'll do them.

    oh, yes, sure! I had just used $\sigma \sigma^{-1}+\sigma^{-1}\sigma$ without thinking that half of this expression is actually zero on generators (which half depending on which generators one considers) :)

    I can make the additions in a couple of hours, if you can do them before, please do. Otherwise I’ll do them.

26. • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
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    Author: Urs
    Format: MarkdownItex> I had just used [...] Yeah, I thought that's what you did. :-) > I can make the additions in a couple of hours, if you can do them before, please do. Okay, I did. See the last few paragraphs of the properties section. I also added a remark on what it means that $W(\mathfrak{g})$ is contractible, lest any reader is led to conclude that we are writing a long entry on just the point.

    I had just used […]

    Yeah, I thought that’s what you did. :-)

    I can make the additions in a couple of hours, if you can do them before, please do.

    Okay, I did. See the last few paragraphs of the properties section. I also added a remark on what it means that $W(\mathfrak{g})$ is contractible, lest any reader is led to conclude that we are writing a long entry on just the point.

27. • CommentRowNumber27.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
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    Author: domenico_fiorenza
    Format: MarkdownItexFine. I'm not convinced of the "elementary way" to see that Weil algebra has trivial cohomology. Namely the space $d_W(W(\mathfrak{g})$ of $d_W$-exact elements does not seems to me to be contained in $\wedge^\bullet(\mathfrak{g}^*[1])$, so a fortiori $d_W$-closed elements do not necessaily belong to that space. Maybe there's an argument which just looks at closed elements, but the "identity is homotopic to zero" seems so simple and neat to me that I would leave just that. Now I'm looking at invariant polynomials, but I have to think a while on them.

    Fine. I’m not convinced of the “elementary way” to see that Weil algebra has trivial cohomology. Namely the space $d_W(W(\mathfrak{g})$ of $d_W$-exact elements does not seems to me to be contained in $\wedge^\bullet(\mathfrak{g}^*[1])$, so a fortiori $d_W$-closed elements do not necessaily belong to that space. Maybe there’s an argument which just looks at closed elements, but the “identity is homotopic to zero” seems so simple and neat to me that I would leave just that.

    Now I’m looking at invariant polynomials, but I have to think a while on them.

28. • CommentRowNumber28.
    • CommentAuthorUrs
    • CommentTimeAug 17th 2010
    • PermaLink
    Author: Urs
    Format: MarkdownItexYou are right about the "elementary way". I wrote this for $F(\mathfrak{g})$, not for $W(\mathfrak{g})$, but anyway.

    You are right about the “elementary way”. I wrote this for $F(\mathfrak{g})$, not for $W(\mathfrak{g})$, but anyway.

29. • CommentRowNumber29.
    • CommentAuthordomenico_fiorenza
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
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    Author: domenico_fiorenza
    Format: MarkdownItexyes, I meant $F(\mathfrak{g})$, but I messed up with notations :) I just wanted to point out that the differential of a product of elements in $\mathfrak{g}^*$ should not be an element of $\wedge^\bullet(\mathfrak{g}^*[1])$.

    yes, I meant $F(\mathfrak{g})$, but I messed up with notations :) I just wanted to point out that the differential of a product of elements in $\mathfrak{g}^*$ should not be an element of $\wedge^\bullet(\mathfrak{g}^*[1])$.

30. • CommentRowNumber30.
    • CommentAuthorUrs
    • CommentTimeFeb 9th 2011
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    Author: Urs
    Format: MarkdownItexWhile on the plane to Paris, I have worked on [[Weil algebra]] a little. * Have split the definition section in two pieces, first the easier for $L_\infty$ algebras, then the genera one for $L_\infty$-algebroids; * Gave full detail in the version of $L_\infty$-algebroids of how it works over a Fermat theory and with Kähler differentials in first shifted degree. (Now at Charles-de Gaulle, about to board to Lisbon)

    While on the plane to Paris, I have worked on Weil algebra a little.

    • Have split the definition section in two pieces, first the easier for $L_\infty$ algebras, then the genera one for $L_\infty$-algebroids;

    • Gave full detail in the version of $L_\infty$-algebroids of how it works over a Fermat theory and with Kähler differentials in first shifted degree.

    (Now at Charles-de Gaulle, about to board to Lisbon)

31. • CommentRowNumber31.
    • CommentAuthorjim_stasheff
    • CommentTimeJul 7th 2021
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    Author: jim_stasheff
    Format: TextIn re:[[CE algebra]] and [[Weil algebra]]. Add something about the restriction to finite conditions can be surpassed by using the early definitions Instead of the graded symmetric algebra on $g^*$, use Hom(the graded symmetric coalgebra on g, R) The algebroid versions use e.g. C^infty(M). What about $\Omega^*(M)\otimes g$?

    In re:CE algebra and Weil algebra. Add something about the restriction to finite conditions can be surpassed by using the early definitions
    Instead of the graded symmetric algebra on $g^*$, use Hom(the graded symmetric coalgebra on g, R)
    
    The algebroid versions use e.g. C^infty(M). What about $\Omega^*(M)\otimes g$?