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• CommentRowNumber1.
• CommentAuthoramg
• CommentTimeNov 13th 2016
• (edited Nov 16th 2016)

The cohomology of a Lie algebra $g$ is by definition that of its Chevalley–Eilenberg complex, $CE^*(g)$. It’s fairly straightforward to see that this cdga can also be obtained as the $G$-invariant differential forms on $G$, for any Lie group $G$ integrating $g$:

$CE^*(g) \cong \Omega^*(G)^G$

(where the $G$-action on forms is given by pullback along the multiplication map).

It occurred to me that this should be expressible in the language of smooth sets. First of all we have $\Omega^k(G) = hom_{SmthSet}(G,\Omega^k)$, and the $G$-action is now given by precomposition. This means that instead of taking $G$-invariants of the hom-set we could just map out of $G/G = pt$. But this clearly doesn’t recover $CE^k(g)$.

For a moment I thought that maybe the issue is that I’m forgetting the cohesiveness of the group $G$ here: so really, I should be looking at an internal hom $[G,\Omega^k]$, and this carries a $G$-action which is witnessed by having it as the fiber of some object (its (homotopy) quotient) over $\mathbf{B}G$. After all, in the previous paragraph I was asking a smooth group to act on a discrete set, which should be a no-no.

But this would come from looking at internal hom in ${\mathbf H}_{/{\mathbf B}G}$ of $G/G$ into $\Omega^k \times {\mathbf B}G$ (which classifies the trivial action of $G$ on $\Omega^k$). And so once again we see $G/G$ appearing, which even speaking cohesively is just the terminal object. So this doesn’t fix the problem either.

Does anyone see what I’m doing wrong here?

• CommentRowNumber2.
• CommentAuthorDavid_Corfield
• CommentTimeNov 14th 2016
• (edited Nov 14th 2016)

That sounds a lot like what confused me once about Urs’s treatment of general covariance. One needs to get internal hom in a symmetric context right:

$[\Sigma//Diff(\Sigma),\; \mathbf{Fields}] \simeq [\Sigma,\; \mathbf{Fields}]//Diff(\Sigma) \,.$

Here to read the above equivalence as a theorem, we have to read the left hand side, as it should, be “in the context of $Diff(\Sigma)$-actions”

• CommentRowNumber3.
• CommentAuthorDmitri Pavlov
• CommentTimeNov 14th 2016

Freed and Hopkins do precisely this (i.e., recover the Chevalley-Eilenberg complex in terms of smooth sets) in their paper “Chern-Weil forms and abstract homotopy theory”.

The key idea is to map out of the stack B_∇(G) of G-bundles with connection instead of just BG.

• CommentRowNumber4.
• CommentAuthorDavid_Corfield
• CommentTimeNov 14th 2016
• (edited Nov 14th 2016)

Here I am struggling my way out of confusion.

And Mike on a related post:

in the monoidal category of objects with $G$-action, whose morphisms are $G$-equivariant maps, the internal hom $[X,Y]$ consists of all maps from $X$ to $Y$ (not necessarily $G$-equivariant), with $G$ acting by conjugation (so that the fixed points of the action are the $G$-equivariant maps). This means that categories enriched over $G$-spaces can be confusing if you’re not used to it: they have hom-spaces on which $G$ acts, and hence an “underlying category” in a naive sense whose morphisms are the points of these hom-spaces, but this is not the “underlying category” in the formal sense of enriched category theory — that consists of the fixed points of the $G$-action, which one should think of as the equivariant maps.

• CommentRowNumber5.
• CommentAuthoramg
• CommentTimeNov 14th 2016
• (edited Nov 14th 2016)

@David #2: Unfortunately I don’t think I’m screwing this up. Given a base object $B \in C$ and over-objects $(Y \rightarrow B),(Z \rightarrow B) \in C_{/B}$, the internal hom in $C_{/B}$ if it exists will be given on a test object $(f : K \rightarrow B) \in C_{/B}$ by $hom_{/K}(f^*Y,f^*Z)$. From here, the space of “equivariant” maps (the unenriched hom) is the space of global sections, i.e. the value on the terminal object $(id : B \rightarrow B)$, on which the internal-hom evaluates as the space of maps over $B$. Again, the over-object representing the $G$-action on $G$ is just $(pt \rightarrow \mathbf{B}G)$, and so I’m getting

$hom_{\mathbf{H}_{/\mathbf{B}G}}(pt , \Omega^k \times {\mathbf{B}} G) \simeq \hom_{\mathbf{H}}(pt , \Omega^k ) .$

@David #4: I haven’t yet read through your linked thread, but in light of the previous paragraph (which of course could itself be wrong) does it explain what I’m doing wrong?

@Dmitri #3: Thanks. Yes, I’m aware of that paper. Here I’m just trying to understand a baby example. There, they “bake the connections into the pie”, so to speak.

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeNov 14th 2016

Hmm, interesting. I wonder if the problem is that “invariance” is not supposed to be internal? We have a smooth group $G$, and its underlying discrete group $\flat G$, and maybe what you want to look at is the $\flat G$-invariants of $Hom(G,\Omega^k)$ (or maybe $\flat Hom(G,\Omega^k)$)?

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 14th 2016
• (edited Nov 14th 2016)

Hey Aaron, that’s a good question.

(And sorry for my slow reactions, here and by email. On top of everything else, today my computer failed me and stole me half od the day thereby.)

I think the issue here is that of need of “differential concretification”. The internal hom of a manifold into differential forms is not in fact the correct moduli space of forms on that manifold, only its “concretification” is. (You saw that discussed in dcct when you caught typos for me).

In particular the inclusion of vertical forms (not depending on the parameter space) into the internal hom is not stable under pullback along parameterized group elements. I think this is where your argument breaks down.

• CommentRowNumber8.
• CommentAuthorMarc Hoyois
• CommentTimeNov 14th 2016

@amg This probably amounts to what Mike and Urs are saying, but you should be mapping out of the smooth set $G/\flat G$, which is not representable. I think it classifies flat connections on the trivial principal $G$-bundle.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeNov 14th 2016

Urs, can you give a pointer to the appropriate place in dcct?

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeNov 15th 2016

On the nLab it’s here.

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeNov 15th 2016

Thanks! Do you think we also need to be quotienting by $\flat G$ instead of $G$, or does concretification allow us to keep the quotient by $G$? Or hmm, maybe $G$ itself doesn’t even act on the concretification?

• CommentRowNumber12.
• CommentAuthoramg
• CommentTimeNov 16th 2016

Hi everyone, thanks for these helpful comments.

Marc’s suggestion seems promising, since $CE^*(\mathfrak{g})$ is closely related to $\flat_{dR} \mathbf{B}G := fib(\flat \mathbf{B} G \rightarrow \mathbf{B}G)$.

Urs, thanks for the reminder about concretification – originally I figured I could ignore it when just computing the ordinary hom-space $hom(-,\Omega^*)$, but then I forgot to take it back into account when passing to the internal hom.

@Mike #11: I’d guess you could concretify in $\mathbf{H}_{/\mathbf{B}G}$, which would retain the $G$-action? We’d also want to check that concretification commutes with the forgetful functor $\mathbf{H}_{/\mathbf{B}G} \rightarrow \mathbf{H}$, though, which may not be immediate since I believe that’s only a left adjoint (whereas concretification is a limit of right adjoints).