## Equivariant DeRham Cohomology

My lectures on equivariant cohomology are spinning a bit out of control.  The questions and lively discussion, while always welcome, have stretched what was meant to be a hand-waving tour through the basics into a three week mini-course (at least, I hope its only three weeks).  I’m starting to feel a bit sheepish, since the result I’m trying to get to might not really merit a full month-long preamble.

Last time, I talked about how to define the equivariant cohomology of a space $M$ in terms of the cohomology of some big infinite-dimensional space $M\times E/G$.  This is good on a conceptual level, but unless $E$ is particularly nice, we will have a bitch of a time computing the cohomology of anything.  What we need is a more effective model for the cohomology of $M\times E/G$.

The idea is to start by pining for the existance of a nice de Rham complex $\Omega(E)$ on $E$.  We’ll say “Oh, if only it $\Omega(E)$ existed, it would look like this, and this…”.  Since $E$ was only defined by its properties (contractibilty and a free action), this amounts to listing what properties a differential graded algebra should have to correspond to those of $E$.  Such DGAs will be called ‘locally-free, acyclic $G^*$-algebras’.

From there, its a three step process.  First, show that every such DGA computes the same cohomology.  Second, show that there is an (almost) universal locally-free, acyclic $G^*$-algebra called the ‘Weil algebra’, which is simple enough in structure to make computations effective.  Third, show that there exists any such DGA which correctly computes the equivariant cohomology (this last step should probably be first, but it isn’t very exciting).

Oh, before I forget, I’m imposing some assumptions on the Lie group $G$: it is compact, connected and finite-dimensional.

$G^*$ Algebras

We start with a basic question.  If $G$ acts on some manifold $M$, what structure does this induce on $\Omega(M)$?  Most obviously, it induces a $G$ action.  We can also take any element $v$ of the Lie algebra $g$ and look at the vector field $\xi_v$ corresponding to the infinitesmal action.  We can then do vector field-y things to $\Omega(M)$; specifically, we can contract forms and we can take their Lie derivative

However, there are relationships between all these actions.  Lie derivative and contraction are linked by the Cartan Formula:

$L_v(\omega)=d\iota_v(\omega)+\iota_vd(\omega)$

while the Lie derivative is required to be the infinitesmal version of the group action:

$L_v(\omega)=\frac{d}{dt}exp(vt)(\omega)|_{t=0}$

Let us define a $\mathbb{Z}$-graded Lie superalgebra $\tilde{g}=g_{-1}\oplus g_0\oplus \mathbb{C}\{d\}_1$, where the subscripts denote the grading.  This algebra is supposed to be capturing the information of a Lie derivative in degree zero, and contraction in degree -1.  As such, we define the bracket to capture some standard facts: $dL_v-L_vd=0$, $L_v\iota_w-\iota_wL_v=\iota_{[v,w]}$, and $d\iota_v+\iota_vd=L_v$ (remember, this is a super commutator).  Note that the group $G$ acts on this Lie superalgebra, by the adjoint action on the first two pieces, and fixing $d$.

We now have what we need to define a ‘full’ group action on a DGA.  A $G^*$-algebra is a (super)commutative DGA $A$, together with:

1) a $G$ action on $A$.

2) a $G$-equivariant $\tilde{g}$ action on $A$, such that the action of $g_0$ coincides with the infinitesmal action of $G$.

The obvious example is $\Omega(M)$, for any manifold $M$ on which $G$ acts; we think of any $G^*$-algebra as an ‘abstract manifold’ on which $G$ acts.

We need to be able to extract the information about the cohomology of the quotient from these complexes.  Therefore, we define the basic subcomplex of a $G^*$-algebra to be the subcomplex consisting of $G$-invariant elements which are killed by contraction.  In the case of $\Omega(M)$ with a free action, this subcomplex is the image of $\Omega(M/G)$ under the pullback map.  The ‘basic cohomology’ $H_{bas}$ of a $G^*$-algebra is the cohomology of the basic subcomplex.

What might $\Omega(E)$ look like?

Now that we know what a group action on a space means in terms of the algebra of forms, lets try to translate the properties ‘contractible’ and ‘free action’ into this new language, to see what $\Omega(E)$ might look like.

Contractiblity is easy.  Whitehead’s theorem, together with Hurewicz’s Lemma, tells us that this is equivalent to the complex of differential forms being acyclic.  That is, $H^i(\Omega(M))=0$ if $i>0$ and $H^0(\Omega(M))=\mathbb{C}$.

Free actions are impossible.  This is because differential forms keep track of local information, so if our group action is locally free, but not free, it will be extremely hard to detect this.  Think, for instance, $S^1$ acting on itself by $\theta\cdot \theta'=2\theta+\theta'$.  We will have to reconcile ourself to seeing when an action is locally free.

Ok, so what does locally free look like?  This means that for each $v\in g$, the corresponding vector field $\xi_v$ vanishes nowhere.  Therefore, we can find a 1-form $\theta$ such that $\theta(\xi_v)=1$, the constant function.  Heck, we can do one better; for any element $f$ in $g^*$ (the linear dual), we can find a 1-form $\theta_f$ such that $\theta_f(\xi_v)=f(v)$, again a constant function.  It should also be clear that if we can do this for every $f$, then the action is necessarily locally free.

So for an abstract $G^*$-algebra $A$, if we can find a map $\theta:g^*\rightarrow A^1$ such that $\iota_v(\theta_f)=f(v)$, we will call $A$ locally free.  Note that we are not distinguishing a $\theta$, just guarenteeing one exists.

The map $\theta$ is not $G$-equivariant, but because we can always average over $G$ (its compact), we can find a new $\theta$ which is equivariant.  If $\theta$ is equivariant, it is usually called a connection, and rightfully so.  Giving a map from $g^*\rightarrow A^1$ is the same as picking an element in $g\otimes A^1$, so if $A=\Omega^1(M)$, these are Lie algebra valued 1-forms.  The eager reader can verify that, in the case of a truly free action, this element will always be a connection.

So, in summation, if $\Omega(E)$ were to exist, it would be a locally free, acyclic $G^*$-algebra.  Even more so, if $E'$ is were any space with a $G$-action such that $\Omega(E')$ were a locally free, acyclic $G^*$-algebra, then $E'$ is necessarily a contractible space on which $G$ acts locally freely.  Therefore, modulo the difference between free and locally free, we could have used $E'$ in the definition of equivariant cohomology.

The reason we didn’t have to pick a specific choice of $E$ in the definition of equivariant cohomology is because any two contractible spaces on which $G$ acts freely are weakly homotopic.  We can hope that we will have a similar result here, even with the locally freeness.  As it happens, this is true!

Theorem.  Let $A$ be a locally free, acyclic $G^*$-algebra.  Then there is another locally free, acyclic $G^*$-algebra $W$, together with a $G^*$-algebra map $W\rightarrow A$ which induce an isomorphism on each basic cohomology group (this is the DGA version of a homotopy equivalence).

Furthermore, if $D$ is any other $G^*$-algebra, then the induced map $D\otimes W\rightarrow D\otimes A$ also induces an isomorphism on basic cohomology.

It should be noted that the maps are not canonical (they depend on a particular choice of connection), and so the final isomorphism is non-canonical.  However, surprisingly, there is a canonical choice of $W$, independant even of $A$!

The Weil Algebra

The claim now is as follows.  There exists a universal $G^*$-algebra with connection called $W$ (which is acyclic).  Remember that ‘with connection’ means with a distinguished choice of $\theta$ which is equivariant.

Universality here means that, given any other $G^*$-algebra with connection $A$, there is a unique map $W\rightarrow A$.  We will construct $W$ from this principle in the most naive way.

We start with $\theta:g^*\rightarrow A^1$, a $G$-equivariant map.  Well, we want a universal guy, which means we want it to be free as possible.  So we start with a copy of $g^*$ in degree 1, and since degree one things anti-commute with each other, we know it’ll generate a copy of $\Lambda(g^*)$.  Furthermore, we know we can take $d$ of anything in the image of $\theta$, so this will give us a copy of $g^*$ in degree two.  Since degree two things commute with each other, we know it’ll generate a copy of $S(g^*)$.

So what do we have?  We have $\Lambda(g^*)\otimes S(g^*)$, with a boundary map that takes generators of $\Lambda(g^*)$ to generators of $S(g^*)$, and kills generators of $S(g^*)$

We want to give this the structure of a $G^*$-algebra next.  The $G$ action is required to be the coadjoint one, since $\theta$ is $G$-equivariant.  The infinitesmal version of this action tells us that $L_v(\theta_f)=ad_v^*(\theta_f)$, were $ad_v^*$ is precomposition by the usual $ad_v$ action.  That is, $ad_v^*(\theta_f)[w]=f([v,w])$.

Finally, we need contraction.  Half of it is done for us, since we know that $\iota_v(\theta_f)=f(v)$, so the contraction action on $\Lambda(g^*)$ is generated by evalution $g\otimes g^*\rightarrow \mathbb{C}$.  Then,

$\iota_vd(\theta_f)=L_v(\theta_f)-d\iota_v(\theta_f)=ad_v^*(\theta_f)-d(f(v))=ad_v^*(\theta_v)$

Thus, we have made $W:=\Lambda(g^*)\otimes S(g^*)$ into a $G^*$-algebra, which has a connection by construction.  To show acyclicity, consider the map which sends a generator $d(\theta_f)$ of $S(g^*)$ to $\theta_f$ in $\Lambda(g^*)$ and kills $\Lambda(g^*)$.  This is a chain homotopy between the identity map and projection onto degree 0.

Therefore, $W$ is the universal $G^*$-algebra with connection.  If we replace ‘with connection’ with ‘locally free’, then $W$ only has a map which is unique up to chain homotopy, which is why I called it ‘almost universal’.

So what does this have to do with $H^\bullet_G(M)$ again?

The final piece of the puzzle is to show that there is any locally free, acyclic $G^*$-algebra which computes the equivariant cohomology.  I have glossed over this detail because I find it the least interesting; much like the construction of spaces like $E$, its more technical then it is worth.

The rough idea is to present $E$ as a direct limit of nested, finite-dimensional manifolds $E_i$ with free G-action.  Then, we define $\tilde{\Omega}(E)$ to be the indirect limit of the $\Omega(E_i)$‘s.  Since each of the steps in the limit were locally free $G^*$-algebras, so is $\tilde{\Omega}(E)$.  If we were smart enough to pick the $E_i$ so that $H^j(E_i)=0$ for $i>>j$, then the complex $\tilde{\Omega}(E)$ will also be acyclic.  We then just need to fiddle with complexes to get:

Theorem. $H^\bullet(M\times E/G)=H_{bas}^\bullet(\Omega(M)\otimes \tilde{\Omega}(E))$.

Of course, $\tilde{\Omega}(E)$ is about as miserable to work with as $E$ was, so we use the simple replacement algebra we just constructed:

Theorem. $H^\bullet _{bas}(\Omega(M)\otimes \tilde{\Omega(E)})=H^\bullet_{bas}(\Omega(M)\otimes W)$.

As long as $M$ is a fairly tractible space, this complex is very reasonable, and is called the ‘Weil Model’ of the equivariant cohomology.  However, there is an even simpler model one can adopt, called the ‘Cartan model’, which I probably won’t talk about.

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