Chern–Weil Forms and Abstract Homotopy TheoryThe statement

# 1 The statement

The main theorem of the Freed–Hopkins paper Chern–Weil forms and abstract homotopy theory is that Chern–Weil forms are the only natural way to get a differential form from a principal $G$-bundle.

Theorems along these lines are of interest historically. It is an important ingredient in the heat kernel proof of the Atiyah–Singer index theorem. Essentially, the idea of the proof is to use the heat equation to show that there is some formula for the index of a vector bundle in terms of the derivatives of the metric, and then by invariant theory, this must be given by the Chern–Weil forms we know and love. One then computes this for sufficiently many examples to figure out exactly which characteristic class it is, as Hirzebruch originally did for his signature formula.

To state the theorem, we work in the category $\operatorname{Shv}(\mathsf{Man}, \mathcal{S})$. For the purposes of this theorem, it actually suffices to work with sheaves of groupoids, i.e. $\operatorname{Shv}(\mathsf{Man}, \tau _{\leq 1} \mathcal{S})$. This only requires $2$-category theory instead of $\infty$-category theory. However, working with $\infty$-categories presents no additional difficulty, and is what we shall do.

We now introduce the main characters of the story.

Example 1

Any $M \in \mathsf{Man}$ defines a representable (discrete) sheaf, which we denote by $M$ again.

Example 2

Any sheaf of sets on $\mathsf{Man}$ is in particular sheaf of (discrete) spaces. Thus, for $p \geq 0$, we have a discrete sheaf

$\Omega ^p \in \operatorname{Shv}(\mathsf{Man}, \mathcal{S}).$

This is in fact a sheaf of vector spaces, and moreover, there are linear natural transformations $\mathrm{d}: \Omega ^p \to \Omega ^{p + 1}$. Thus, we get a sheaf of chain complexes $\Omega ^{\bullet }$, and

$[M, \Omega ^\bullet ] = \Omega ^\bullet (M).$

In general, for any sheaf $\mathcal{F}$, we can think of $\Omega ^\bullet (\mathcal{F}) \equiv [\mathcal{F}, \Omega ^{\bullet }]$ as the de Rham complex of $\mathcal{F}$.

From now on, fix $G$ a Lie group.

Example 3

For $M \in \mathsf{Man}$, define $B_\nabla G(M)$ to be the groupoid of principal $G$-bundles on $M$ with connection and isomorphisms, which we think of as a 1-truncated space. This defines $B_\nabla G \in \operatorname{Shv}(\mathsf{Man}, \mathcal{S})$.

The main theorem is
Theorem 4

The Chern–Weil homomorphism induces an isomorphism

$(\operatorname{Sym}^\bullet \mathfrak {g}^*)^G \overset {\sim }{\to } \Omega ^\bullet (B_\nabla G).$

To prove the theorem, we consider the universal principal $G$-bundle $E_\nabla G \to B_\nabla G$. The point is that $E_\nabla G$ admits a much more explicit description, and then we use $B_\nabla G = E_\nabla G /\! /G$ to understand $B_\nabla G$ itself.

$E_\nabla G$ can be described explicitly as follows:

Example 5

Define $E_\nabla G (M)$ to be the groupoid of trivialized $G$-bundles on $M$ with connection. Equivalently, this is the groupoid of connections on the trivial $G$-bundle $M \times G \to G$. So $E_\nabla G \cong \Omega ^1 \otimes \mathfrak {g}$.

There is then a natural map $E_\nabla G(M) \to B_\nabla G(M)$, which one can easily check is the universal principal $G$-bundle. Our next claim is that $B_\nabla G(M) = E_\nabla G(M) /\! /G$, which is clear once we know what the latter is.

Definition 6

Let $\mathcal{F} \in \operatorname{Shv}(\mathsf{Man}, \mathcal{S})$, and let $\alpha : G \times \mathcal{F} \to \mathcal{F}$ be an action by $G$. Explicitly, for each $M \in \mathsf{Man}$, there is a group action

$\operatorname{Hom}_{\mathsf{Man}}(M, G) \times \mathcal{F}(M) \to \mathcal{F}(M)$

where $\operatorname{Hom}_{\mathsf{Man}}(M, G)$ is given the pointwise group structure. We can then define the action groupoid

$(\mathcal{F}/\! /G)_\bullet = G^{\times \bullet } \times \mathcal{F} \in \operatorname{Shv}(\mathsf{Man}, \mathcal{S})^{\Delta ^\mathrm{op}}.$

The homotopy quotient of $\mathcal{F}$ by $G$ is then

$\mathcal{F}/\! /G = |(\mathcal{F}/\! /G)_\bullet |.$

Note that this geometric realization is taken in the category $\operatorname{Shv}(\mathsf{Man}, \mathcal{S})$. To compute this, one takes the geometric realization in the category of presheaves, then sheafify.

We then see that $B_\nabla G = E_\nabla G /\! /G$. Explicitly, the action of the gauge group can be described as follows — given $g: M \to G$ and $\alpha \in E_\nabla G(M) = \Omega ^1(M; \mathfrak {g})$, we have

$g \cdot \alpha = g^* \theta + \operatorname{Ad}_{g^{-1}} \alpha .$

Remark 7

Formally, to prove that $B_\nabla G = E_\nabla G /\! /G$, we first form the quotient of $E_\nabla G$ by $G$ in the category of presheaves. Since $E_\nabla G$ is discrete, this is given by (the nerve of) the action groupoid of the $G$-action on $E_\nabla G$. This gives the presheaf of trivial principal $G$-bundles. To show that the sheafification is $B_\nabla G$, observe that there is a natural map from this presheaf to $B_\nabla G$, and it is an equivalence on stalks since all principal $G$-bundles on contractible spaces are trivial. So it induces an isomorphism after sheafification.

Our proof then naturally breaks into two steps. First, we compute $\Omega ^\bullet (E_\nabla G)$, and then we need to know how to compute $\Omega ^\bullet (\mathcal{F}/\! /G)$ from $\Omega ^\bullet (\mathcal{F})$ for any discrete sheaf $\mathcal{F}$.

We first do the second part.

Lemma 8

Let $\mathcal{F} \in \operatorname{Shv}(\mathsf{Man}, \mathcal{S})$ be a discrete sheaf with a $G$-action $\alpha : G \times \mathcal{F} \to \mathcal{F}$. Then $\Omega ^\bullet (\mathcal{F} /\! /G)$ is the subcomplex of $\Omega ^\bullet (\mathcal{F})$ consisting of the $\omega$ such that

1. $\alpha ^* \omega |_{\{ g\} \times \mathcal{F}} = \omega$ for all $g \in G$; and

2. $\iota _\xi \omega = 0$ for all $\xi \in \mathfrak {g}$.

The first condition says $\omega$ should be $G$-invariant, and the second condition says $\omega$ is suitably “horizontal”.

Remark 9

Let us explain what we mean by $\iota _\xi \omega$. In general, for $M$ a manifold and $X$ is a vector field on $M$, we can define $\iota _X: \Omega ^p(M \times N) \to \Omega ^{p - 1}(M \times N)$ for all manifolds $N$. Then by left Kan extension, this induces a map $\iota _X: \Omega ^p(M \times \mathcal{F}) \to \Omega ^{p - 1}(M \times \mathcal{F})$ for all $\mathcal{F} \in \operatorname{Shv}(\mathsf{Man}, \mathcal{S})$.

Now if $\mathcal{F}$ has a $G$-action and $\xi \in \mathfrak {g}$, then $\xi$ induces an invariant vector field on $G$, which we also call $\xi$. We then define $\iota _\xi : \Omega ^p(\mathcal{F}) \to \Omega ^{p - 1}(\mathcal{F})$ by the following composition where the last map is induced by the inclusion.

This gives us a very explicit method to compute the natural transformation $\iota _\xi \omega$ for $\omega \in \Omega ^p(\mathcal{F})$ and $\xi \in \mathfrak {g}$. Given a test manifold $M$ and $\phi \in \mathcal{F}(M)$, which we think of as a natural transformation $\phi : M \to \mathcal{F}$, we form the composiite This defines a differential form $\eta \in \Omega ^p(G \times M)$. Then we have

$(\iota _\xi \omega )_M(\phi ) = \iota _{\xi } \eta |_{\{ e\} \times M}.$

Proof
We have

$\Omega ^p (\mathcal{F} /\! /G) = \Omega ^p(|(\mathcal{F}/\! /G)_\bullet |) = \operatorname{Tot}(\Omega ^p((\mathcal{F}/\! /G)_\bullet )).$

Since $(\mathcal{F}/\! /\mathcal{G})_\bullet$ is a simplicial discrete sheaf, its totalization can be computed by where $\mathrm{pr}: G \times \mathcal{F} \to \mathcal{F}$ is the projection.

To prove the lemma, we have to show that $\mathrm{pr}^* \omega = \alpha ^* \omega$ iff the conditions in the lemma are satisfied. This follows from the more general claim below with $\eta = \alpha ^* \omega - \mathrm{pr}^* \omega$.

Claim

Let $M$ be a manifold and $\mathcal{F}$ a sheaf. Then $\eta \in \Omega ^p(M \times \mathcal{F})$ is zero iff

1. $\eta |_{\{ x\} \times \mathcal{F}} = 0$ for all $x \in M$

2. $\iota _X \eta = 0$ for any vector field $X$ on $M$.

The conditions (1) and (1') match up exactly. Unwrapping the definition of $\iota _\xi$ and noting that $\iota _X \mathrm{pr}^* \omega = 0$ always, the only difference between (2) and (2') is that in (2), we only test on invariant vector fields on $G$, instead of all vector fields, and we only check the result is zero after restricting to a fiber $\{ e\} \times \mathcal{F}$. The former is not an issue because the condition $C^\infty (G)$-linear and the invariant vector fields span as a $C^\infty (G)$-module. The latter also doesn't matter because we have assumed that $\alpha ^* \omega$ is invariant.

To prove the claim, if $\mathcal{F}$ were a manifold, this is automatic, since the first condition says $\eta$ vanishes on vectors in the $N$ direction while the second says it vanishes on vectors in the $M$ direction.

If $\mathcal{F}$ were an arbitrary sheaf, we know $\eta$ is zero when pulled back along any map $(1 \times \phi ): M \times N \to M \times \mathcal{F}$ where $N$ is a manifold, by naturality of the conditions. But since $M \times \mathcal{F}$ is a colimit of such maps, $\eta$ must already be zero on $M \times \mathcal{F}$.

Proof

Now it remains to describe $\Omega ^\bullet (E_\nabla G) = \Omega ^\bullet (\Omega ^1 \otimes \mathfrak {g})$. More generally, for any vector space $V$, we can calculate $\Omega ^\bullet (\Omega ^1 \otimes V)$. We first state the result in the special case where $V = \mathbb {R}$.

Theorem 10
$\Omega ^p(\Omega ^1) \cong \mathbb {R}\text{ for all }p \geq 0.$

For $p = 2q$, it sends $\omega$ to $(\mathrm{d}\omega )^q$. For $p = 2q + 1$, it sends $\omega$ to $\omega \wedge (\mathrm{d}\omega )^q$.

The general case is no harder to prove, and the result is described in terms of the Koszul complex.

Definition 11

Let $V$ be a vector space. The Koszul complex $\operatorname{Kos}^\bullet V$ is a differential graded algebra whose underlying algebra is

$\operatorname{Kos}^\bullet V = {\textstyle \bigwedge }^\bullet V \otimes \operatorname{Sym}^\bullet V.$

For $v \in V$, we write $v$ for the corresponding element in ${\textstyle \bigwedge }^1 V$, and $\tilde{v}$ for the corresponding element in $\operatorname{Sym}^1 V$. We set $|v| = 1$ and $|\tilde{v}| = 2$. The differential is then

$\mathrm{d}(v) = \tilde{v},\quad d(\tilde{v}) = 0.$

Theorem 12

For any vector space $V$, we have an isomorphism of differential graded algebras

$\eta : \operatorname{Kos}^\bullet V^* \overset {\sim }{\to } \Omega ^\bullet (\Omega ^1 \otimes V).$

In particular,

$\Omega ^\bullet (E_\nabla G) = \operatorname{Kos}^\bullet \mathfrak {g}^*.$

Explicitly, for $\ell \in V^* = {\textstyle \bigwedge }^1 V^*$, the element $\eta (\ell ) \in \Omega ^1(\Omega ^1 \otimes V)$ is defined by

$\eta (\ell )(\alpha \otimes v) = \langle v, \ell \rangle \, \alpha$

for $\alpha \in \Omega ^1$ and $v \in V$. This is then extended to a map of differential graded algebras.

In other words, the theorem says every natural transformation

$\omega _M: \Omega ^1(M; V) \to \Omega ^p(M)$

is (uniquely) a linear combination of transformations of the form

$\sum \alpha _i \otimes v_i \mapsto \sum _{I, J} M_{I, J}(v_{i_1}, \ldots , v_{i_k}, v_{j_1}, \ldots , v_{j_\ell })\, \alpha _{i_1} \wedge \cdots \wedge \alpha _{i_k} \wedge \mathrm{d}\alpha _{j_1} \wedge \cdots \wedge \mathrm{d}\alpha _{j_\ell }$

where $M_{I, J}$ is anti-symmetric in the first $k$ variables and symmetric in the last $\ell$.

Using this, we conclude

Theorem 13

The Chern–Weil homomorphism gives an isomorphism

$(\operatorname{Sym}^\bullet \mathfrak {g}^*)^G \overset {\sim }{\to } \Omega ^\bullet (B_\nabla G),$

and the differential on $\Omega ^\bullet (B_\nabla G)$ is zero.

Note that this $\operatorname{Sym}^\bullet \mathfrak {g}^*$ is different from that appearing in the Koszul complex.

Proof
We apply the criteria in Lemma 8. The first condition is the $G$-invariance condition, and translates to the $(\cdots )^G$ part of the statement. So we have to check that the forms satisfying the second condition are isomorphic to $\operatorname{Sym}^\bullet \mathfrak {g}^*$.

To do so, we have to compute the action of $\iota _\xi$ on $E_\nabla G$ following the recipe in Remark 9. Fix $\omega \in \Omega ^p(E_\nabla G)$ and $\xi \in \mathfrak {g}$.

Let $\phi : M \to E_\nabla G$ be a trivial principal $G$-bundle with connection $A \in \Omega ^1(M; \mathfrak {g})$. The induced principal $G$-bundle on $G \times M$ under the action then has connection $\theta + \operatorname{Ad}_{g^{-1}} A$. So by definition,

$(\iota _\xi \omega )_M(A) = \left.\iota _\xi \left(\omega (\theta + \operatorname{Ad}_{g^{-1}} A)\right)\right|_{\{ e\} \times M}.$

To compute the action on $\operatorname{Kos}^\bullet \mathfrak {g}^*$, it suffices to compute it on ${\textstyle \bigwedge }^1 \mathfrak {g}^*$ and $\operatorname{Sym}^1 \mathfrak {g}^*$.

1. If $\lambda \in \mathfrak {g}^* = {\textstyle \bigwedge }^1 \mathfrak {g}^*$, then $\lambda (A) = \langle A, \lambda \rangle$, and

$\iota _\xi \langle \theta + \operatorname{Ad}_{g^{-1}} A, \lambda \rangle = \langle \iota _\xi \theta + \iota _\xi \operatorname{Ad}_{g^{-1}} A, \lambda \rangle .$

We know $\iota _\xi \theta = \xi$, and $\iota _\xi \operatorname{Ad}_{g^{-1}} A = 0$ since $\operatorname{Ad}_{g^{-1}} A$ vanishes on all vectors in the $G$ direction. So we know

$\iota _\xi \lambda = \langle \xi , \lambda \rangle \in {\textstyle \bigwedge }^0 \mathfrak {g}^*.$
2. Next, $\tilde{\lambda }(A) = \langle \mathrm{d}A, \lambda \rangle$. We compute

$\iota _\xi \langle \mathrm{d}(\theta + \operatorname{Ad}_{g^{-1}}A), \lambda \rangle |_{\{ e\} \times M} = \left.\iota _\xi \left\langle -\frac{1}{2}[\theta , \theta ] + \operatorname{Ad}_{\mathrm{d}g^{-1}} \wedge A + \operatorname{Ad}_{g^{-1}} \mathrm{d}A, \lambda \right\rangle \right|_{\{ e\} \times M} = \langle -\operatorname{Ad}_\xi A, \lambda \rangle = \langle A, -\operatorname{Ad}_\xi ^* \lambda \rangle .$

So

$\iota _\xi \tilde{\lambda } = -\operatorname{Ad}^*_\xi \lambda \in {\textstyle \bigwedge }^1 \mathfrak {g}^*.$

First observe that in ${\textstyle \bigwedge }^\bullet \mathfrak {g}^*$, the only elements killed by $\iota _\xi$ are those in ${\textstyle \bigwedge }^0 \mathfrak {g}^* \cong \mathbb {R}$. To take care of the $\operatorname{Sym}$ part, set

$\Omega _\lambda = \tilde{\lambda } + \frac{1}{2}[\lambda , \lambda ].$

Since $\tilde{\lambda }(A) = \langle \mathrm{d}A, \lambda \rangle$, we see that $\Omega _\lambda (A) = \langle \Omega _A, \lambda \rangle$, where $\Omega _A$ is the curvature, and one calculates $\iota _\xi \Omega _\lambda = 0$. By a change of basis, we can identify

$\operatorname{Kos}^\bullet \mathfrak {g}^* \cong {\textstyle \bigwedge }^\bullet \mathfrak {g}^* \otimes \operatorname{Sym}^\bullet \langle \Omega _\lambda : \lambda \in \mathfrak {g}^*\rangle ,$

and $\iota _\xi$ vanishes on the second factor entirely. So we are done.

Proof

More generally, the same proof shows that

Theorem 14

If $M$ is a smooth manifold, the de Rham complex of $M \times (\Omega ^1 \otimes V)$ is $\Omega (M; \operatorname{Kos}V^*)^\bullet$ (the total complex of $\Omega ^\bullet (M; \operatorname{Kos}^\bullet V^*)$).

In particular, if $M$ has a $G$-action, then $M \times E_\nabla G /\! /G$ is exactly the Cartan model for equivariant de Rham cohomology.

This would follow immediately if we had a result that says $\Omega ^\bullet (M \times \mathcal{F}) \cong \Omega ^\bullet (M) \hat{\otimes } \Omega ^\bullet (\mathcal{F})$, and since $\Omega ^\bullet (E_\nabla G)$ is finite dimensional, the completed tensor product is the usual tensor product.