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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexThere is IMO some confusion in the article [[exterior algebra]], in the examples section "Over a super vector space". Very generally, for a cocomplete symmetric monoidal closed $Vect$-enriched category $V$ (these hypotheses can be weakened), we have two distinct constructions: (1) The symmetric algebra $S(X) = \sum_{n \geq 0} X^{\otimes n}/S_n$. <i>This</i> is the free commutative monoid on $X$. (2) The exterior algebra $\Lambda X = \sum_{n \geq 0} sgn(n) \otimes_{S_n} X^{\otimes n}$. These are different things, although there are of course connections between them. Particularly, for $V =$ super vector spaces, we have the fundamental relation $$(\Lambda V)[1] \cong S(V[1])$$ where the $[1]$ indicates the degree 1 shift (mod 2). Now, in the article, there is the following statement > For $V$ a [[super vector space]], the exterior algebra $\Lambda V$ is often called the **Grassmann algebra** over $V$. This is the $\Lambda V$ or $\wedge^\bullet V$ is the [[free object|free]] [[supercommutative algebra|graded commutative]] [[superalgebra]] on $V$. This says in other words that the exterior algebra $\Lambda X$ is the free graded commutative algebra on $X$. Which is wrong, or at least confused; to make a correct statement, one has to take into account the two degree 1 shifts: $\Lambda X = S(X[1])[-1]$. Matters become potentially even more confusing when the grading is not just mod 2, but graded by $\mathbb{Z}$ (or $\mathbb{N}$). This relates to what I was trying to say yesterday about the definition of CE-algebra. You might say that the underlying graded object is a Grassmann algebra on $\mathfrak{g}$, but in order to do that you have to undergo contortions. You can't really say Grassmann algebra on $\mathfrak{g}$ if $\mathfrak{g}$ is regarded as concentrated in degree 0, because strictly speaking that Grassmann algebra = exterior algebra would again be concentrated in degree 0, according to the definition given in [[exterior algebra]]. Nor can you really say Grassmann algebra = exterior algebra on the degree 1 shift $\mathfrak{g}[1]$, because by what was said above, the exterior algebra construction on *that* gives, in degree $n$, the object of commuting monomials $\mathfrak{g}^{\otimes n}/S_n$! No, what you really want to say in order to be coherent is that we are taking the free graded commutative algebra on $\mathfrak{g}[1]$ (and then remark that the degree $n$ component is $\Lambda^n \mathfrak{g}$). This is a conceptually simpler way of thinking about it IMO, because one can invoke some general abstract nonsense, to the effect that in any (cocomplete symmetric monoidal closed $Vect$-enriched) $V$, if $S(X)$ is the free commutative monoid, then any map $X \to S(X)$ extends uniquely to a derivation $S(X) \to S(X)$ (and similarly for the free monoid $T(X)$ in place of $S(X)$). This is what I was getting at when I had written yesterday at [[Chevalley-Eilenberg algebra]].

   There is IMO some confusion in the article exterior algebra, in the examples section “Over a super vector space”.

   Very generally, for a cocomplete symmetric monoidal closed $Vect$-enriched category $V$ (these hypotheses can be weakened), we have two distinct constructions:

   (1) The symmetric algebra $S(X) = \sum_{n \geq 0} X^{\otimes n}/S_n$. This is the free commutative monoid on $X$.

   (2) The exterior algebra $\Lambda X = \sum_{n \geq 0} sgn(n) \otimes_{S_n} X^{\otimes n}$.

   These are different things, although there are of course connections between them. Particularly, for $V =$ super vector spaces, we have the fundamental relation

   $$(\Lambda V)[1] \cong S(V[1])$$

   where the $[1]$ indicates the degree 1 shift (mod 2).

   Now, in the article, there is the following statement

   For $V$ a super vector space, the exterior algebra $\Lambda V$ is often called the Grassmann algebra over $V$. This is the $\Lambda V$ or $\wedge^\bullet V$ is the free graded commutative superalgebra on $V$.

   This says in other words that the exterior algebra $\Lambda X$ is the free graded commutative algebra on $X$. Which is wrong, or at least confused; to make a correct statement, one has to take into account the two degree 1 shifts: $\Lambda X = S(X[1])[-1]$. Matters become potentially even more confusing when the grading is not just mod 2, but graded by $\mathbb{Z}$ (or $\mathbb{N}$).

   This relates to what I was trying to say yesterday about the definition of CE-algebra. You might say that the underlying graded object is a Grassmann algebra on $\mathfrak{g}$, but in order to do that you have to undergo contortions. You can’t really say Grassmann algebra on $\mathfrak{g}$ if $\mathfrak{g}$ is regarded as concentrated in degree 0, because strictly speaking that Grassmann algebra = exterior algebra would again be concentrated in degree 0, according to the definition given in exterior algebra. Nor can you really say Grassmann algebra = exterior algebra on the degree 1 shift $\mathfrak{g}[1]$, because by what was said above, the exterior algebra construction on that gives, in degree $n$, the object of commuting monomials $\mathfrak{g}^{\otimes n}/S_n$!

   No, what you really want to say in order to be coherent is that we are taking the free graded commutative algebra on $\mathfrak{g}[1]$ (and then remark that the degree $n$ component is $\Lambda^n \mathfrak{g}$). This is a conceptually simpler way of thinking about it IMO, because one can invoke some general abstract nonsense, to the effect that in any (cocomplete symmetric monoidal closed $Vect$-enriched) $V$, if $S(X)$ is the free commutative monoid, then any map $X \to S(X)$ extends uniquely to a derivation $S(X) \to S(X)$ (and similarly for the free monoid $T(X)$ in place of $S(X)$). This is what I was getting at when I had written yesterday at Chevalley-Eilenberg algebra.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJul 22nd 2010
   • (edited Jul 22nd 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThe degree shift conventions are an endless source of amusement in this business. > This says in other words that the exterior algebra $\wedge X$ is the free graded commutative algebra on $X$. Which is wrong, To me the symbol $\wedge$ means shift up by one and apply $Sym$. That fits nicely with the ordinary convention that people think of a Lie algebra as having a vector space in degree 0 underlying it. Then $\wedge^\bullet \mathfrak{g}^*$ is the correct thing that underlies the CE-algebra. And it is $Sym(\mathfrak{g}^*[1])$ not $Sym(\mathfrak{g}^*[1])[-1]$. Why do you mean to shift down again? We want $$ \wedge^\bullet \mathfrak{g}^* = k \oplus \mathfrak{g}^* \oplus \mathfrak{g}^* \wedge \mathfrak{g}^* \oplus \cdots $$ with the $n$th power in degree $n$. This is arranged to nicely fit with the common (though unfortunately not universal) convention that people like to take their Lie $n$-algebra $\mathfrakg{$ to be concentrated in degree 0, -1, -2, ... -$(n-1)$. Then $\mathfrak{g}^* $ is concentrated in degree 0 to $(n-1)$. And $\wedge^\bullet \mathfrak{g}^*$ is the correct algebra underlying its CE-algebra.

   The degree shift conventions are an endless source of amusement in this business.

   This says in other words that the exterior algebra $\wedge X$ is the free graded commutative algebra on $X$. Which is wrong,

   To me the symbol $\wedge$ means shift up by one and apply $Sym$.

   That fits nicely with the ordinary convention that people think of a Lie algebra as having a vector space in degree 0 underlying it. Then $\wedge^\bullet \mathfrak{g}^*$ is the correct thing that underlies the CE-algebra.

   And it is $Sym(\mathfrak{g}^*[1])$ not $Sym(\mathfrak{g}^*[1])[-1]$. Why do you mean to shift down again? We want

   $$\wedge^\bullet \mathfrak{g}^* = k \oplus \mathfrak{g}^* \oplus \mathfrak{g}^* \wedge \mathfrak{g}^* \oplus \cdots$$

   with the $n$th power in degree $n$.

   This is arranged to nicely fit with the common (though unfortunately not universal) convention that people like to take their Lie $n$-algebra to be concentrated in degree 0, -1, -2, … -$(n-1)$. Then $\mathfrak{g}^*$ is concentrated in degree 0 to $(n-1)$. And $\wedge^\bullet \mathfrak{g}^*$ is the correct algebra underlying its CE-algebra.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexA reference example that is good to remain consistent with is the archetypical one: the de Rham complex on $X$ This is written $$ \wedge^\bullet_{C^\infty(X)} \Gamma(T^* X) $$ and has to come out as $$ \cdots = Sym_{C^\infty(X)}( \Gamma(T^* X)[1]) $$ (in fact many/most people would write $[-1]$ instead of $[1]$ for the up-shift.)

   A reference example that is good to remain consistent with is the archetypical one: the de Rham complex on $X$

   This is written

   $$\wedge^\bullet_{C^\infty(X)} \Gamma(T^* X)$$

   and has to come out as

   $$\cdots = Sym_{C^\infty(X)}( \Gamma(T^* X)[1])$$

   (in fact many/most people would write $[-1]$ instead of $[1]$ for the up-shift.)

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexUrs, I think maybe you missed my point (probably, paradoxically, because all this is obvious to you anyway). In the article [[Chevalley-Eilenberg algebra]], the term "Grassmann algebra" is hyperlinked to exterior algebra. The reader who follows this link is then given the definition of exterior algebra which says essentially $$\Lambda(X) = \sum_{n \geq 0} sgn(n) \otimes_{S_n} X^{\otimes n}$$ Unless I missed something, it does not say in the definition section > the symbol $\wedge$ means shift up by one and apply $Sym$. which, exactly, was my whole point: the Grassmann algebra as graded algebra is *this*. (And that's what we should be saying, but it's said very confusingly at [[exterior algebra]]. Look again at the quotation again in my first comment.) It is the result of applying the symmetric algebra functor to an appropriately graded object. It is *not* the result of applying the exterior algebra functor, *as defined above*, to some object in the symmetric monoidal category of graded vector spaces. For if you apply that definition of exterior algebra *literally* to an object in degree 0, you still get an object in degree 0. If you apply the exterior algebra functor to an object in degree 1, you get instead the symmetric algebra with its standard grading. See my point? You understand this perfectly well, but the neophyte attempting to follow links could quite easily become very confused, because the statements in the article are garbled IMO. > And it is $Sym(\mathfrak{g}^*[1])$ not $Sym(\mathfrak{g}^*[1])[-1]$. Why do you mean to shift down again? Oh, sorry; you're right. There was a reason I did that: you do that if you want the super (or graded) Grassmann algebra to be a *monad* on super (or graded) vector spaces (where morphisms are by definition degree 0 maps). The way I view it is like this: we are *conjugating* the symmetric algebra monad by the degree 1 (mod 2) shift functor to get another monad. You don't get a monad if you just shift and apply $Sym$, because then you've shifted degrees (e.g., the unit of the "monad" would be a degree 1 map).

   Urs, I think maybe you missed my point (probably, paradoxically, because all this is obvious to you anyway).

   In the article Chevalley-Eilenberg algebra, the term “Grassmann algebra” is hyperlinked to exterior algebra. The reader who follows this link is then given the definition of exterior algebra which says essentially

   $$\Lambda(X) = \sum_{n \geq 0} sgn(n) \otimes_{S_n} X^{\otimes n}$$

   Unless I missed something, it does not say in the definition section

   the symbol $\wedge$ means shift up by one and apply $Sym$.

   which, exactly, was my whole point: the Grassmann algebra as graded algebra is this. (And that’s what we should be saying, but it’s said very confusingly at exterior algebra. Look again at the quotation again in my first comment.) It is the result of applying the symmetric algebra functor to an appropriately graded object. It is not the result of applying the exterior algebra functor, as defined above, to some object in the symmetric monoidal category of graded vector spaces. For if you apply that definition of exterior algebra literally to an object in degree 0, you still get an object in degree 0. If you apply the exterior algebra functor to an object in degree 1, you get instead the symmetric algebra with its standard grading.

   See my point?

   You understand this perfectly well, but the neophyte attempting to follow links could quite easily become very confused, because the statements in the article are garbled IMO.

   And it is $Sym(\mathfrak{g}^*[1])$ not $Sym(\mathfrak{g}^*[1])[-1]$. Why do you mean to shift down again?

   Oh, sorry; you’re right. There was a reason I did that: you do that if you want the super (or graded) Grassmann algebra to be a monad on super (or graded) vector spaces (where morphisms are by definition degree 0 maps). The way I view it is like this: we are conjugating the symmetric algebra monad by the degree 1 (mod 2) shift functor to get another monad. You don’t get a monad if you just shift and apply $Sym$, because then you’ve shifted degrees (e.g., the unit of the “monad” would be a degree 1 map).

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexOkay, here is an attempt to clarify the CE-entry locally, first: i added a section <a href="http://ncatlab.org/nlab/show/Chevalley-Eilenberg+algebra#GradingConvention">Grading conventions</a>

   Okay, here is an attempt to clarify the CE-entry locally, first: i added a section Grading conventions

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexExcellent; thanks Urs! Looks good.

   Excellent; thanks Urs! Looks good.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexAll right, thanks for pushing me to improve this CE entry (and fill it with content in the first place).

   All right, thanks for pushing me to improve this CE entry (and fill it with content in the first place).

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeJul 22nd 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexI ought to state my feeling that CE-algebra fits in a general family of cobar constructions; I'm not sure how much this has been brought out at the Lab (maybe under $L_\infty$-algebra?). This sort of thing has been on my mind lately. I am returning to thinking about bar and cobar constructions for operads, Koszul duality for quadratic operads, cyclic operads, and the like, where there are quite a few parallels between the theory of operads and the theory of algebras. But I also want to get back to understanding your theory of universal connections on classifying bundles. I may have asked you this before, but can one use such a universal connection to construct an explicit homotopy inverse to the canonical map $G \to \Omega B G$ for $G$ a Lie group? (We know on general grounds that the map is a weak homotopy equivalence, hence a homotopy equivalence, but it would be nice to have an explicit homotopy inverse.)

   I ought to state my feeling that CE-algebra fits in a general family of cobar constructions; I’m not sure how much this has been brought out at the Lab (maybe under $L_\infty$-algebra?). This sort of thing has been on my mind lately. I am returning to thinking about bar and cobar constructions for operads, Koszul duality for quadratic operads, cyclic operads, and the like, where there are quite a few parallels between the theory of operads and the theory of algebras.

   But I also want to get back to understanding your theory of universal connections on classifying bundles.

   I may have asked you this before, but can one use such a universal connection to construct an explicit homotopy inverse to the canonical map $G \to \Omega B G$ for $G$ a Lie group? (We know on general grounds that the map is a weak homotopy equivalence, hence a homotopy equivalence, but it would be nice to have an explicit homotopy inverse.)

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeJul 23rd 2010
   • (edited Jul 23rd 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItex> I may have asked you this before, but can one use such a universal connection to construct an explicit homotopy inverse to the canonical map It's right, you asked me that before. The thing here is that this universal connection that i am talking about is constructed in a sheaf topos, not in $Top$ or $Diff$. So it has the familiar tautological flavor to it and does not seem to readily yield explicit topological constructions such as you are after. At least not without further thinking. In fact, if you look at the consstruction, it boils down to something very tautological indeed. The interesting point (as far as I see it) is not so much that it exists, but how it fits into a grander scheme of things. You see, what I am currently still trying to work out better is this: I have a model for what a general $\infty$-connection on a general principal $\infty$-bundle is. I know from a bunch of examples that this model is correct. But only for parts of this model do I have a first-principles abstract nonsense _derivation_ that explains _why_ this must be the right answer. What I do have fully clarified, I think, is the case where the coefficient object $\mathbf{B}G$ is once deloopable. In this "semi-abelian" case there is, I can show, a general-abstract theory for obstruction classes to extending $\mathbf{B}G$-cocycles to _flat_ differential cocycles and when one turns the crank on this nice bit ob abstract $(\infty,1)$-category theory, out drops semi-abelian differential cohomology, such as Deligne cohomology, describing abelian $n$-gerbes with connection. Okay, so that much is a done deal. Now the next observation is this: I take the explicit _model_ (read: model category-model) that I proved present the above-mentioned abstract theory for semi-abelian differential cohomology in terms of 1-categorical data. Looking at this model with the eyes of the articles I wrote with Stasheff and Sati, there is an evident generalization of the _model_ to the fully general non-necessarily semi-abelian case. And looking at examples, this evident generalization does the right thing. Okay. So back then I was happy with this. But today I can't accept this. I have now a model of which I don't fully know what it models. There must be a general-abstract mechanism at work in the background, which is a tad more complicated than that for the semi-abelian case wich I do understand. So I am trying to find this general abstract mechanism. The stuff that wrote about universal connections on principal 1-bundles is something I came up with in this process. It is an observation of how the simple familiar notion of a connection on a principal bundle may be reformulated as a bunch of almost-fully-abstract nonsense. For instance, I finally figured out the fully abstract way to say "Maurer-Cartan form". I mean, in such generality that it works in any $(\infty,1)$-topos. I don't even need to have a notion of "differential form" to say this. But it reduces to that case in the $(\infty,1)$-topos of $\infty$-Lie groupoids. When I finally had this, I saw that this provides a step in understanding the abstract procedure behind my model for nonabelian differential cohomology. In a pretty precise sense, this goes two thirds of the way! The remaining third is that open question which I tried to ask for help about in another thread. What I wrote about the universal connection on a 1-bundle in an $\infty$-stack topos is just a proof-of-principle sort of that shows how this abstract machinery really does reproduce the familiar notions. The construciton does not really say anything new about these familiar notions, and likely does not help much with questions such as you were asking. What it does accomplish, for me, is to point the way for generalizing this to differential cohomology beyond the familiar notions.

   I may have asked you this before, but can one use such a universal connection to construct an explicit homotopy inverse to the canonical map

   It’s right, you asked me that before. The thing here is that this universal connection that i am talking about is constructed in a sheaf topos, not in $Top$ or $Diff$. So it has the familiar tautological flavor to it and does not seem to readily yield explicit topological constructions such as you are after. At least not without further thinking.

   In fact, if you look at the consstruction, it boils down to something very tautological indeed. The interesting point (as far as I see it) is not so much that it exists, but how it fits into a grander scheme of things.

   You see, what I am currently still trying to work out better is this: I have a model for what a general $\infty$-connection on a general principal $\infty$-bundle is. I know from a bunch of examples that this model is correct. But only for parts of this model do I have a first-principles abstract nonsense derivation that explains why this must be the right answer.

   What I do have fully clarified, I think, is the case where the coefficient object $\mathbf{B}G$ is once deloopable. In this “semi-abelian” case there is, I can show, a general-abstract theory for obstruction classes to extending $\mathbf{B}G$-cocycles to flat differential cocycles and when one turns the crank on this nice bit ob abstract $(\infty,1)$-category theory, out drops semi-abelian differential cohomology, such as Deligne cohomology, describing abelian $n$-gerbes with connection.

   Okay, so that much is a done deal. Now the next observation is this: I take the explicit model (read: model category-model) that I proved present the above-mentioned abstract theory for semi-abelian differential cohomology in terms of 1-categorical data. Looking at this model with the eyes of the articles I wrote with Stasheff and Sati, there is an evident generalization of the model to the fully general non-necessarily semi-abelian case. And looking at examples, this evident generalization does the right thing.

   Okay. So back then I was happy with this. But today I can’t accept this. I have now a model of which I don’t fully know what it models. There must be a general-abstract mechanism at work in the background, which is a tad more complicated than that for the semi-abelian case wich I do understand.

   So I am trying to find this general abstract mechanism. The stuff that wrote about universal connections on principal 1-bundles is something I came up with in this process. It is an observation of how the simple familiar notion of a connection on a principal bundle may be reformulated as a bunch of almost-fully-abstract nonsense.

   For instance, I finally figured out the fully abstract way to say “Maurer-Cartan form”. I mean, in such generality that it works in any $(\infty,1)$-topos. I don’t even need to have a notion of “differential form” to say this. But it reduces to that case in the $(\infty,1)$-topos of $\infty$-Lie groupoids.

   When I finally had this, I saw that this provides a step in understanding the abstract procedure behind my model for nonabelian differential cohomology. In a pretty precise sense, this goes two thirds of the way! The remaining third is that open question which I tried to ask for help about in another thread.

   What I wrote about the universal connection on a 1-bundle in an $\infty$-stack topos is just a proof-of-principle sort of that shows how this abstract machinery really does reproduce the familiar notions. The construciton does not really say anything new about these familiar notions, and likely does not help much with questions such as you were asking. What it does accomplish, for me, is to point the way for generalizing this to differential cohomology beyond the familiar notions.