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edited Weil algebra a bit. More to come.
added to Weil algebra a section Properties with a discussion of the free property of the Weil algebra.
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?
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.
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 $W(\mathfrak{g})$.
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.
Ah, ok! now that’s perfectly clear! do you mind if I rewrite the proof on the nLab here and there? ;)
do you mind if I rewrite the proof on the nLab here and there? ;)
Please do.
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..
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.
Minor edits. Added a line to explain in which sense $\sigma$ is extended to a derivation.
Added a one line construction of the canonical morphism $W(\mathfrak{g})\to CE(\mathfrak{g})$ :)
How do we see that the cohomology of $(W(\mathfrak{g}),d_{W(\mathfrak{g})})$ is trivial?
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.
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 :(
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.
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
$f : \sigma (t) \mapsto d_W t = d_{CE} t + \sigma (t)$.
I have added that remark to the Properties-section at Weil 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.
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 $F(\mathfrak{g})\cong W(\mathfrak{g})$.
Sure, I would think this is a good extra remark to be added.
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.
You want me to do it?
Wait a sec, it’s just $[d_F, K] = Id$.
(Sufficient to check this on generators.)
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.
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.
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.
You are right about the “elementary way”. I wrote this for $F(\mathfrak{g})$, not for $W(\mathfrak{g})$, but anyway.
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 $\wedge^\bullet(\mathfrak{g}^*[1])$.
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)
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