Equivariant de Rham cohomology: theory and applications

Oliver Goertsches; Leopold Zoller

doi:10.1007/s40863-019-00129-4

São Paulo Journal of Mathematical Sciences

December 2019, Volume 13, Issue 2, pp 539–596 | Cite as

Equivariant de Rham cohomology: theory and applications

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Oliver Goertsches
Leopold Zoller

Survey

First Online: 11 April 2019

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Abstract

This is a survey on the equivariant cohomology of Lie group actions on manifolds, from the point of view of de Rham theory. Emphasis is put on the notion of equivariant formality, as well as on applications to ordinary cohomology and to fixed points.

Keywords

Lie group actions Equivariant Cohomology Cartan model Equivariant formality Fixed points

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Notes

Acknowledgements

Parts of this paper stem from the first named author’s lectures at the University of Hamburg in 2012, and at the Philipps University of Marburg in 2018. We would like to thank the participants of these courses for their interest in the topic and their valuable comments. We are grateful to Michèle Vergne for several remarks on a previous version of this paper. We are especially indebted to Jeffrey Carlson for several enlightening discussions, as well as for a very thorough reading of a previous version and numerous suggestions that improved the presentation of this paper. The second named author is supported by the German Academic Scholarship foundation.

Appendix A. Spectral sequences and the module structure on equivariant cohomology

We present the basics of the spectral sequence of a filtration and apply them to the Cartan model of equivariant cohomology. By also paying attention to the multiplicative structure on spectral sequences, this tool allows us to derive some fundamental properties of the $S(\mathfrak {g}^*)^G$-module structure on $H^*_G(M)$: it is finitely generated and its rank agrees with that of the final page of the spectral sequence associated to a certain filtration. Also, we use spectral sequences to prove the torus case of Remark 5.5. Finally we give an example where $E_\infty $ and $H^*_G(M)$ are not isomorphic as $S(\mathfrak {g}^*)^G$-modules, a point which is in several places unclear in the literature.

Before we start, we want to point out that the goal here is not to give a complete introduction to spectral sequences but rather to provide the reader with all the algebraic background that is needed for our (and many other topological) applications. In particular, we avoid the finer details of convergence by restricting to first-quadrant spectral sequences. For an in-depth introduction we recommend, e.g., Chapter 5 of [79].

Basic definitions

Let R be a commutative ring. When applying algebraic results to equivariant cohomology we will always take $R={\mathbb {R}}$.

Definition A.1

A (cohomology) spectral sequence is a sequence $\{(E_r,d_r)\}_{r\ge 0}$ of bigraded R-modules $E_r=\bigoplus _{p,q\in {\mathbb {Z}}}E^{p,q}_r$ with R-linear differentials

$$\begin{aligned} d_r^{p,q}:E_r^{p,q}\rightarrow E^{p+r,q-r+1}_r \end{aligned}$$

satisfying $d_r\circ d_r=0$ and isomorphisms

$$\begin{aligned} E_{r+1}^{p,q} \cong \frac{\ker d_r^{p,q}}{{\text {im}}\, d_r^{p-r,q+r-1}}. \end{aligned}$$

A spectral sequence is often compared to a book, where for turning the rth page $E_r$ one takes cohomology to arrive at the next page $E_{r+1}\cong H^*(E_r,d_r)$. The advantage of spectral sequences is that they can be used to approximate the cohomology of a cochain complex by breaking down the transition $(C^*,d)\rightsquigarrow H^*(C^*,d)$ into smaller steps. Let us now make this idea precise by defining a suitable notion of convergence.

A first-quadrant spectral sequence is a spectral sequence $(E_r,d_r)$ where $E_r^{p,q}=0$ whenever $p<0$ or $q<0$. Note that if we fix a bidegree (p, q) and start turning through the pages, the differentials $d_r^{p,q}$ (resp. $d_r^{p-r,q+r-1}$) eventually leave (resp. come from outside) the first-quadrant and thus are trivial. This implies that $E_r^{p,q}\cong E_l^{p,q}$ for all $l\ge r$. This stable value is denoted by $E_\infty ^{p,q}$ and the the bigraded R-module $E_\infty $ is called the final page of the spectral sequence. If for some r we have $d_i=0$ for $i\ge r$, or equivalently $E_r=E_\infty $, we say that the spectral sequence collapses at $E_r$. While we will solely be interested in first-quadrant spectral sequences, the definition of $E_\infty $ is not limited to this special case and makes sense whenever the pointwise limit exists.

Definition A.2

A filtration of a (graded) R-module H is a sequence of (graded) submodules

$$\begin{aligned} \ldots \subset F^pH\subset F^{p-1}H\subset \ldots \end{aligned}$$

We say that the spectral sequence $(E_r,d_r)$converges to a graded module $H^*$ if there is a filtration of $H^*$ such that in any degree n we have

$$\begin{aligned} 0=F^sH^n\subset \ldots \subset F^pH^n\subset F^{p-1}H^n\subset \ldots \subset F^tH^n=H^n \end{aligned}$$

for some $s,t\in {\mathbb {Z}}$ and $E_\infty ^{p,q}\cong F^pH^{p+q}/F^{p+1}H^{p+q}$.

Note that when working with ${\mathbb {R}}$-coefficients (or over any field) there is a highly non-canonical isomorphism of vector spaces

$$\begin{aligned} H^n=\bigoplus _p F^pH^n/ F^{p+1}H^n=\bigoplus _{p+q=n}E_\infty ^{p,q}. \end{aligned}$$

In particular $H^*\cong E_\infty $ as graded vector spaces when we consider $E_\infty ^{p,q}$ to be of degree $p+q$.

Spectral sequence of a filtration

As hinted at above, the usefulness of spectral sequences stems from the fact that they can be used to break the process of taking cohomology down into several steps. Consider, e.g., the Cartan model $C_G(M)=(S(\mathfrak {g}^*)\otimes \Omega (M))^G$ with its differential $d_G=1\otimes d+\delta $ where $\delta $ is the component which raises the degree in $S(\mathfrak {g}^*)$ and d is just the differential on $\Omega (M)$. Algebraically speaking, $C_G(M)$ is a huge and complicated object, but its cohomology under the differential $1\otimes d$ is much smaller (see Proposition A.8 below). Consequently, when analysing $H_G(M)$, it can be helpful to take cohomology with respect to $1\otimes d$ first, and then worry about the rest of $d_G$. This process of singling out the $1\otimes d$ component is achieved via a suitable filtration and the associated spectral sequence.

Definition A.3

A filtration of a cochain complex (C, d) of R-modules is a family

$$\begin{aligned} \ldots \subset F^pC\subset F^{p-1}C\subset \ldots \end{aligned}$$

of subcomplexes of C. The filtration is said to be canonically bounded if $F^0C=C$ and $F^{n+1}C^n=0$.

Remark A.4

A filtration of a complex (C, d) induces a filtration $F^*H^*(C,d)$ of $H^*(C,d)$, where $F^pH^n(C,d)$ is the image of the map $H^n(F^pC,d)\rightarrow H^n(C,d)$.

Theorem A.5

Let (C, d) be a cochain complex and $F^*C$ a canonically bounded filtration. Then the construction below gives rise to a first-quadrant spectral sequence $(E_r,d_r)$ converging to $H^*(C,d)$. More precisely we have

$$\begin{aligned} E^{p,q}_\infty \cong F^{p}H^{p+q}(C,d)/F^{p+1}H^{p+q}(C,d), \end{aligned}$$

where $F^pH^n(C,d)$ is defined as above.

In the construction we, for the moment, forget about the cohomological degree and focus purely on the filtration degree. The second component of the bidegree will be added in the end. We start by setting

$$\begin{aligned} E_0^{p}=F^pC/F^{p+1}C. \end{aligned}$$

This carries a differential induced by d and $E_0=\bigoplus _p E_0^p$ is known as the associated graded chain complex. Its cohomology $E_1$ is a first approximation of the cohomology of (C, d), where cocycles are represented by elements whose filtration degree increases under the differential. Note that there is a subquotient of $E_0$ that is a much better approximation of the cohomology, namely $E_\infty =\bigoplus _p E_\infty ^p$ where

$$\begin{aligned} E_\infty ^p=\frac{\ker d\cap F^pC+F^{p+1}C}{{\text {im}}d\cap F^pC+F^{p+1}C}. \end{aligned}$$

To interpolate between the two we introduce the approximate cycles

$$\begin{aligned} A_r^p=\{x\in F^pC~|~dx\in F^{p+r}C\} \end{aligned}$$

whose filtration degree increases by r under the differential. Now set

$$\begin{aligned} E_r^p= \frac{A^p_r+F^{p+1}C}{d(A^{p-r+1}_{r-1})+F^{p+1}C} \cong \frac{A_r^{p}}{d\left( A_{r-1}^{p-r+1}\right) +A_{r-1}^{p+1}}. \end{aligned}$$

The usefulness of these interpolations stems from the fact that $E_{r+1}$ can be computed from $E_r$: by definition d induces a map $d_r:E_r^p\rightarrow E_r^{p+r}$ and one can identify $E_{r+1}$ with $H(E_r,d_r)$ (see [79, Theorems 5.4.1] for details).

The bigrading in the spectral sequence arises from additionally considering the grading on C. We want the latter to correspond to the total degree of the bigrading so we set $A^{p,q}_r=A_r^p\cap C^{p+q}$, which naturally induces a bigrading on $E_r$. Explicitly we have

$$\begin{aligned} E_r^{p,q}=\frac{A_r^{p,q}}{d\left( A_{r-1}^{p-r+1,q+r-2}\right) +A_{r-1}^{p+1,q-1}}. \end{aligned}$$

Since $d_r$ raises the total degree by one and the filtration degree by r, it is of bidegree $(r,-r+1)$. To construct the isomorphism $E^{p,q}_\infty \cong F^{p}H^{p+q}(C,d)/F^{p+1}H^{p+q}(C,d)$, note that d vanishes on $A^{p,q}_r$ for $r>q+1$ because the filtration is canonically bounded. Thus $E_r^{p,q}$ is represented by cocycles from $\ker (d)\cap F^pC^{p+q}$. The isomorphism is then defined by just mapping those cocycles onto their image in $F^{p}H^{p+q}(C,d)/F^{p+1}H^{p+q}(C,d)$. For further details like well-definedness of the last map we again refer to [79, Theorems 5.4.1 and 5.5.1].

The spectral sequence of the Cartan model

From now on let G be a compact, connected group acting on a manifold M. Recall from the definitions in Sect. 4 that the Cartan model $C_G(M)\subset S(\mathfrak {g}^*)\otimes \Omega ^*(M)$ inherits a bigrading via

$$\begin{aligned} \left( S(\mathfrak {g}^*)\otimes \Omega ^*(M)\right) ^{p,q}= S^\frac{p}{2}(\mathfrak {g}^*)\otimes \Omega ^q(M), \end{aligned}$$

whenever p is even and $C_G^{p,q}(M)=0$ when p is odd. In particular, $S(\mathfrak {g}^*)$ is concentrated in even degrees when considered as the subalgebra $C_G^{*,0}$. We also assign a total degree via $C_G^n(M)=\bigoplus _{p+q=n}C^{p,q}_G(M)$. The Cartan differential is $d_G=1\otimes d+\delta $ with d just the regular differential in $\Omega ^*(M)$ and $(\delta \omega )(X)=-i_{\overline{X}}(\omega (X))$. Note that $1\otimes d$ and $\delta $ are themselves differentials of bidegree (0, 1) and $(2,-1)$.

Remark A.6

Doing a suitable degree shift one can achieve that the bidegrees of the differentials are (0, 1) and (1, 0). With this grading $C_G(M)$ becomes a double complex in the classical sense and the spectral sequence we construct below is (up to degree shifts) the spectral sequence associated to this double complex (c.f. [53]). As the degree shift will not simplify our presentation of the material and the original bigrading is more in line with the topological conventions, we decide to stick to the original one.

In what follows we will write C instead of $C_G(M)$. The filtration we consider on C is defined by

$$\begin{aligned} F^pC:= C^{\ge p,*}=\bigoplus _{l\ge p,q\ge 0}C^{l,q}. \end{aligned}$$

It is canonically bounded as

$$\begin{aligned} F^pC^n=\bigoplus _{l=p}^nC^{l,n-l}. \end{aligned}$$

The differential $d_G$ restricts to the $F^pC$, so this is indeed a filtration by subcomplexes and we have an associated spectral sequence to which we just refer as the spectral sequence of C. Let us now explicitly compute the first pages.

We have $E_0^{p,q}=F^pC^{p+q}/F^{p+1}C^{p+q}$, which is canonically isomorphic to $C^{p,q}$ via the projection onto this summand. The differential $d_0:E^{p,q}_0\rightarrow E^{p,q+1}_0$ is just the one induced by $d_G$ on the quotient. The composition with the isomorphisms

$$\begin{aligned} C^{p,q}\cong F^pC^{p+q}/F^{p+1}C^{p+q}\xrightarrow {d_G} F^pC^{p+q+1}/F^{p+1}C^{p+q+1}\cong C^{p,q+1} \end{aligned}$$

is precisely the its bidegree (0, 1) component $1\otimes d$. Thus we see that $(E_0,d_0)$ is isomorphic to $(C,1\otimes d)$ as a cochain complex.

Remark A.7

The following observation will become relevant when discussing multiplicative aspects in Sect. A.4. In fact the above isomorphism $(C,1\otimes d)\cong (E_0,d_0)$ is one of commutative differential graded algebras (cdga, see Sect. A.4) with respect to the product

$$\begin{aligned} F^pC/F^{p+1}C\otimes F^qC/F^{q+1}C\rightarrow F^{p+q}C/F^{p+q+1}C \end{aligned}$$

on $E_0$ which is induced by multiplication in C. The cohomology of a cdga is naturally a commutative graded algebra. Morphisms between cdgas, i.e. multiplicative maps that respect the grading and commute with the differential, induce multiplicative maps in cohomology. The isomorphism in the following proposition is of this form and hence respects the algebra structure.

Proposition A.8

If G is a compact, connected Lie group acting on a compact differentiable manifold, then the $E_1$-term in the spectral sequence associated to the Cartan complex is

$$\begin{aligned} E_1 \cong S(\mathfrak {g}^*)^G\otimes H^{*}(M). \end{aligned}$$

Proof

We just need to compute the cohomology of $(E_0,d_0)$. Consider the inclusion of complexes

$$\begin{aligned} (C,1\otimes d) = ((S(\mathfrak {g}^*)\otimes \Omega (M))^G,1\otimes d) \longrightarrow (S(\mathfrak {g}^*)\otimes \Omega (M),1\otimes d). \end{aligned}$$

(13.1)

With regards to Remark A.7 note that it is an inclusion of cdgas. We obtain the induced map on cohomology

$$\begin{aligned} i:H^*(C,1\otimes d) \longrightarrow S(\mathfrak {g}^*)\otimes H^*(M). \end{aligned}$$

Let us show first that it is injective. Assume that $\omega \in C$ is such that $\omega =(1\otimes d)(\sigma )$ for some $\sigma \in S(\mathfrak {g}^*)\otimes \Omega (M)$. As $\omega $ is G-invariant and $1\otimes d$ commutes with the diagonal G-action on $S(\mathfrak {g}^*)\otimes \Omega (M)$, we have $(1\otimes d)(g^*\sigma )=\omega $ for all $g\in G$. But then also

$$\begin{aligned} (1\otimes d)\left( \int _G g^*\sigma \, dg\right) = \int _G(1\otimes d)g^*\sigma \, dg = \int _G \omega \, dg = \omega . \end{aligned}$$

Because $\int _G g^*\sigma \, dg\in C$, it follows that $[\omega ]=0\in H^*(C,1\otimes d)$.

We next claim that the map i takes values in $S(\mathfrak {g}^*)^G\otimes H^*(M)$, which means that for every $[\omega ]$ on the left hand side, the element $i[\omega ]$ is G-invariant when considered as a polynomial function with values in $H^*(M)$. For $g\in G$ the diffeomorphism $g^{-1}:M\rightarrow M$ is homotopic to the identity, because G is connected. Then, for any $X\in \mathfrak {g}$ we have $[\omega ({\text {Ad}}_gX)] = [(g^{-1})^*\omega (X)] = [\omega (X)]$.

Finally we show that $i:H^*(C,1\otimes d)\rightarrow S(\mathfrak {g}^*)^G\otimes H^*(M)$ is surjective. For this we precompose (13.1) with the inclusion

$$\begin{aligned} (S(\mathfrak {g}^*)^G\otimes \Omega (M)^G,1\otimes d) \longrightarrow (C,1\otimes d). \end{aligned}$$

In cohomology we obtain the composition

$$\begin{aligned} S(\mathfrak {g}^*)^G\otimes H^*(\Omega (M)^G,d)\longrightarrow H^*(C,1\otimes d)\overset{i}{\longrightarrow }S(\mathfrak {g}^*)^G\otimes H^*(M) \end{aligned}$$

which, by Theorem 2.2, is an isomorphism. Thus i is surjective. $\square $

Remark A.9

Note that the proof is simpler in case of a torus action: in this case the coadjoint action on $S(\mathfrak {t}^*)$ is trivial, so the isomorphism $E_1 = S(\mathfrak {t}^*)\otimes H^{*}(M)$ follows directly from Theorem 2.2.

Corollary A.10

If the cohomology of M is concentrated in even degrees, i.e., $H^n(M)=0$ whenever n is odd, then the spectral sequence of the Cartan model degenerates at the $E_1$-term.

Proof

Under the hypothesis we know that $E_1^{p,q}$ vanishes whenever p or q is odd. Thus $d_1$ vanishes for degree reasons. The same argument applies to all subsequent pages. $\square $

Remark A.11

The differential $d_r$ on $E_r$ vanishes whenever $r\ge 1$ is odd, because $S(\mathfrak {g}^*)^G$ is concentrated in even degrees. In particular, the spectral sequence collapses at $E_1$ if and only if it collapses at $E_2$.

Example A.12

Consider the diagonal action of $S^1\subset {\mathbb {C}}$ on the unit sphere $S^{2n+1}\subset {\mathbb {C}}^{n+1}$. The Weyl-invariant polynomials are just ${\mathbb {R}}[u]$, where u is the dual of some generator X of the Lie algebra of $S^1$. The $E_1$ term of the spectral sequence is isomorphic to ${\mathbb {R}}[u]\otimes H^*(S^{2n+1})$, so it consists just of two copies of ${\mathbb {R}}[u]$, embedded as $E_1^{*,0}$ and $E_1^{*,2n+1}$. A differential can only be nonzero if it maps from the $(2n+1)^{\mathrm {st}}$ row to the $0^{\mathrm {th}}$ row. Consequently we have $d_r=0$ for $1\le r\le 2n+1$ and $E_1\cong E_{2n+2}$. By the same reasoning we have $d_r=0$ for $r\ge 2n+3$ and $E_{2n+3}=E_\infty $. All that remains to understand is what the differential $d_{2n+2}$ does on $E_{2n+2}$:

Often spectral sequence arguments can work entirely without knowing the explicit definition of the differentials if one adds an extra ingredient. In this case for example, we know by Theorem 5.2 that $E_\infty $ is the cohomology of a 2n-dimensional manifold and vanishes in degrees above 2n. This knowledge implies that no elements of greater (total) degree must survive the transition from $E_{2n+2}$ to $E_{2n+3}$. Consequently $d_{2n+2}:\smash {E_{2n+2}^{p,2n+1}}\rightarrow \smash {E^{p+2n+2,0}_{2n+2}}$ has to be an isomorphism for every $p\ge 0$. All that remains on the page $E_{2n+3}=E_\infty $ is therefore ${\mathbb {R}}[u]/(u^{n+1})$ in the 0th row. We have shown that $H^*({\mathbb {C}}P^n)\cong H_{S^1}(S^{2n+1})\cong {\mathbb {R}}[u]/(u^{n+1})$ as graded vector spaces. With the help of the discussion of the ${\mathbb {R}}[u]$-module and algebra structures from the subsequent sections, one can deduce that this isomorphism is actually one of ${\mathbb {R}}[u]$-algebras. However, this is false in general and only holds because in the example, $E_\infty $ is concentrated in a single row, implying there is only one step in the filtration of $H_{S^1}(S^{2n+1})$.

Finally, let us examine explicitly the generator of $\smash {E_{2n+2}^{0,2n+1}}\cong H^{2n+1}(S^{2n+1})$. Let $\omega _0$ be a $S^1$-invariant volume form on $S^{2n+1}$. Other than suggested by the isomorphism, $\omega _0$ does not represent a generator of $\smash {E_{2n+2}^{0,2n+1}}$ because $d_{S^1}\omega _0=ui_{\overline{X}}\omega _0$ has filtration degree 2. So $\omega _0$ is not an element of $\smash {A^{0,2n+1}_{2n+2}}$. However, we find a form $\omega _1$ such that $i_{\overline{X}}(\omega _0)=d\omega _1$ because $H^{2n}(S^{2n+1})=0$. Now $d_G(\omega _0+u\omega _1)=u^2i_{\overline{X}}\omega _1$ lies in filtration degree 4. Inductively we construct a zigzag $\omega =\omega _0+\cdots +u^n\omega _n$ such that $d_G\omega $ is a multiple of $u^{n+1}$. So $\omega $ lies in $\smash {A^{0,2n+1}_{2n+2}}$ and induces an element of $\smash {E_{2n+2}^{0,2n+1}}$. Using the fact that the bidegree-$(0,2n+1)$ component of $\omega $, which is precisely $\omega _0$, does not lie in the the projection ${\text {im}}d$ of ${\text {im}}d_G$ to the $(0,2n+1)$ component, we conclude that $\omega $ descends to a generator.

Multiplicative structure

Definition A.13

A gradedR-algebra is an R-algebra $A=\bigoplus _{k\in {\mathbb {Z}}} A^k$ (where $A^k$ are R-modules) such that the multiplication map respects the grading, i.e., $A^p\cdot A^q\subset A^{p+q}$. It is called commutative if $xy=(-1)^{|x||y|}yx$ for homogeneous elements x, y of degrees |x|, |y|. If $d:A\rightarrow A$ is an R-linear map which raises the degree by 1 and satisfies $d^2=0$ as well as the graded Leibniz rule

$$\begin{aligned} d(xy)=dx\cdot y+(-1)^{|x|}x\cdot dy, \end{aligned}$$

we call (A, d) a commutative differential graded algebra (cdga). A filtration $F^*A$ of A (as a graded R-module) is called multiplicative if $F^pA\cdot F^lA\subset F^{p+l}A$.

Remark A.14

The cohomology $H^*(A,d)$ of any cdga (A, d) inherits an algebra structure which turns it into a commutative graded algebra. If $F^*A$ is a multiplicative filtration of (A, d) by subcomplexes, then the induced filtration on $H^*(A,d)$ (see Remark A.4) is multiplicative with respect to the induced algebra structure. In this case we have well defined product maps

$$\begin{aligned} \frac{F^pH^n}{F^{p+1}H^n}\otimes \frac{F^lH^m}{F^{l+1}H^m}\longrightarrow \frac{F^{p+l}H^{n+m}}{F^{p+l+1}H^{n+m}}, \end{aligned}$$

where we write $H^k$ for $H^k(A,d)$.

Example A.15

The differential forms $(\Omega (M),d)$ and the Cartan model $(C_G(M),d_G)$ are cdgas with the total degree which is the sum of both components of the bidegree. The filtration of the Cartan model as defined in the previous section is a multiplicative filtration.

We have seen that for a suitably filtered complex (C, d) the last page of the associated spectral sequence carries information on $H^*(C,d)$ and the two are even abstractly isomorphic as vector spaces if we use field coefficients. It is natural to ask if in case of a cdga (A, d), $E_\infty $ carries information on the algebra structure on $H^*(A,d)$. While we cannot expect to have $E_\infty \cong H^*(A,d)$ as algebras, the algebra structure does indeed leave its mark on $E_\infty $ in the following manner.

Theorem A.16

Let (A, d) be a cdga with a canonically bounded multiplicative filtration $F^*A$. Then the spectral sequence from Theorem A.5 carries a multiplicative structure, i.e., for any r there exist multiplication maps $\mu _r:E^{p,q}_r\otimes E^{s,t}_r\rightarrow E_r^{p+s,q+t}$ with the following properties:

$(E_r,d_r)$ is a cdga with respect to the total degree of the bigrading.
The multiplication $\mu _{r+1}$ is induced by $\mu _r$ under the isomorphism $E_{r+1}\cong H(E_{r},d_{r})$.

In particular we get an induced multiplication on $E_\infty $. Under the isomorphism

$$\begin{aligned} E_\infty ^{p,q}= F^pH^{p+q}(A,d)/F^{p+1}H^{p+q}(A,d), \end{aligned}$$

this product coincides with the one described in Remark A.14.

Details of the proof are given e.g. in [65, Sect. 2.3]. Let us just quickly demystify the products $\mu _r$ by giving their definition: in the explicit construction of $E_r^{p,q}$ from Sect. A.2 one easily checks that multiplication in A restricts to $A_r^{p,q}\otimes A_r^{s,t}\rightarrow A_r^{p+s,q+t}$ and that this descends to quotients inducing the map $\mu _r:E_r^{p,q}\otimes E_r^{s,t}\rightarrow E_r^{p+s,q+t}$ from the above theorem. Finally we want to draw the reader’s attention to Remark A.7, where we argue that

$$\begin{aligned} E_1\cong S(\mathfrak {g^*})^G\otimes H^*(M) \end{aligned}$$

as algebras.

On the module structure of the equivariant cohomology

One of the interesting features of equivariant cohomology is that it is not only an algebra over ${\mathbb {R}}$ but over $S(\mathfrak {g}^*)^G$. As we have seen, multiplicative structures carry over to the spectral sequence, so we can use the latter to analyse the $S(\mathfrak {g}^*)^G$-module structure on $H^*_G(M)$.

As the differential $d_G$ of the Cartan model vanishes on $S(\mathfrak {g}^*)^G\otimes {\mathbb {R}}$, we have $S^p(\mathfrak {g}^*)^G \subset A^{2p,0}_r$ for all r. The degreewise projection onto $E^{2p,0}_r$ yields a map

$$\begin{aligned} S(\mathfrak {g}^*)^G\rightarrow E_r \end{aligned}$$

whose image is the zeroth row $E_r^{*,0}$. On the page $E_1\cong S(\mathfrak {g}^*)^G\otimes H^*(M)$ (see Proposition A.8) it is just the inclusion of $S(\mathfrak {g}^*)^G\otimes {{\mathbb {R}}}$. Note that we also obtain an induced map $S(\mathfrak {g}^*)^G\rightarrow E_\infty $. These maps are easily checked to be morphisms of algebras. Thus, the $E_r$ carry the structure of $S(\mathfrak {g}^*)^G$-modules.

For degree reasons the differentials $d_r$ vanish on $E_r^{*,0}$ for $r\ge 1$ so by the Leibniz rule we have $d_r(fx)=fd_r(x)$ for any $f\in S(\mathfrak {g}^*)^G,x\in E_r$. The module structure on $E_{r+1}$ is just the one that $H(E_r,d_r)$ inherits from the differential graded $S(\mathfrak {g}^*)^G$-module $(E_r,d_r)$.

Lemma A.17

Let $x_1,\ldots ,x_k\in E_\infty $ be homogeneous elements (with respect to the bigrading) that generate $E_\infty $ as an $S(\mathfrak {g}^*)^G$-module. Choose representatives $y_1,\ldots ,y_k\in H^*_G(M)$ via the isomorphisms

$$\begin{aligned} E_\infty ^{p,q}\cong F^pH^{p+q}_G(M)/F^{p+1}H^{p+q}_G(M). \end{aligned}$$

Then the $y_i$ generate $H^*_G(M)$ as an $S(\mathfrak {g}^*)^G$-module.

Proof

Let $c_0\in H^l_G(M)$ be any element. It is contained in some $F^pH_G^l(M)$, so we may consider its image $\overline{c_0}\in E_\infty ^{p,l-p}$. We find elements $f_1,\ldots ,f_k\in S(\mathfrak {g}^*)^G$ such that

$$\begin{aligned} \overline{c_0}=\sum f_ix_i. \end{aligned}$$

Recall that the multiplication in $E_\infty $ respects the bigrading. We may therefore choose the $f_i$ in such a way that they are homogeneous and if $x_i\in E_\infty ^{p-m,l-p}$, we have $|f_i|=m$ (in the grading inherited from the Cartan model) or $f_i=0$. This ensures that $\sum _i f_iy_i$ lies in $F^pH^l_G(M)$. Now by the description of the multiplicative structure on $E_\infty $ from Theorem A.16 one verifies that $\sum _i f_iy_i$ projects to $\overline{c_0}$ in $E_\infty ^{p,l-p}$. In particular

$$\begin{aligned} c_1=c_0-\sum f_iy_i \end{aligned}$$

projects to 0 and thus lies in $F^{p+1}H^l_G(M)$. Now we repeat this process for $c_1$ until eventually $c_{l-p+1}\in F^{l+1}H^l_G(M)=0$. We have written $c_0$ as a linear combination of the $y_i$. $\square $

The following proposition applies in particular to compact manifolds. The proof is taken from [3, Proposition 3.10.1]

Proposition A.18

If $\dim H^*(M)<\infty $, then $H^*_G(M)$ is finitely generated as an $S(\mathfrak {g}^*)^G$-module.

Proof

By Lemma A.17, it suffices to show that $E_\infty $ is finitely generated. We have seen that $E_1$ is the free module $S(\mathfrak {g}^*)^G\otimes H^*(M)$. The cohomology $H^*(M)$ is finite-dimensional and in particular $E_1$ is finitely generated as an $S(\mathfrak {g}^*)^G$-module. The ring $S(\mathfrak {g}^*)^G$ is is a polynomial ring (see Sect. 3). In particular it is Noetherian, which implies that submodules and quotients of finitely generated $S(\mathfrak {g}^*)^G$-modules are again finitely generated, see [8, Proposition 6.5]. Thus if $E_r$ is finitely generated, the same is true for $E_{r+1}=H(E_r,d_r)$: the differential respects the module structure so the cohomology is a quotient of the submodule $\ker d_r$. As the spectral sequence collapses after a finite number of pages (at most $\dim M$), we conclude that $E_\infty $ is finitely generated. $\square $

Note that, since $S(\mathfrak {g}^*)^G$ is concentrated in even degrees, the module structure preserves even and odd degree elements. With regard to the resulting decomposition we have the following

Corollary A.19

If $\dim H^*(M)<\infty $, then the ranks of the $S(\mathfrak {g}^*)^G$-modules $E^{{\text {even}}}_\infty $ (resp. $E_\infty ^{{\text {odd}}}$) and $H^{{\text {even}}}_G(M)$ (resp. $H^{{\text {odd}}}_G(M)$) coincide.

Proof

For a finitely generated graded module M over the polynomial ring $S(\mathfrak {g}^*)^G$, the rank is encoded in its Hilbert-Poincaré series $H_M(t)=\sum _i \dim ( M^i)\, t^i$: the latter takes the form $f(t)\prod _{i=1}^r(1-t^{k_i})^{-1}$ for some $f\in {\mathbb {Z}}[t]$, where r is the number of variables of $S(\mathfrak {g}^*)^G$ and the $k_i$ are their degrees [8, Theorem 11.1]. The rank is then precisely f(1) (check this for a free module first and then deduce it for general M via a free resolution). As we have already seen, $E_\infty $ and $H^*_G(M)$ are isomorphic as graded vector spaces, so the claim follows. $\square $

Remark A.20

In the corollary above, it is tempting to argue that a basis of a free submodule in $H^*_G(M)$ projects down to the basis of a free submodule of $E_\infty $. However this is false in general.

Naturality and the comparison theorem

We briefly discuss maps between spectral sequences and the important comparison theorem. The latter enables us to prove Remark 5.5 in case G and H are tori. Also, a construction made in said proof is needed in the next and final section.

Definition A.21

A morphism of spectral sequences $(E_r,d_r)\rightarrow (E_r',d_r')$ is a family of morphisms $f_r:E_r\rightarrow E'_r$, defined for large r, that preserve the bigrading, commute with the differentials, and have the property that $f_{r+1}$ is the map induced by $f_r$ in cohomology.

In particular, if $E_\infty $ is defined, we obtain a map $f_\infty :E_\infty \rightarrow E'_\infty $. Morphisms of spectral sequences associated to filtrations arise naturally via filtration-preserving maps: suppose (C, d) and $(C',d')$ are canonically bounded filtered cochain complexes and $f:C\rightarrow C'$ is a filtration-preserving chain map. Then f maps $A_r^{p,q}$ (see the construction in Sect. A.2) to ${A'}_r^{p,q}$ and induces maps $f_r:E_r\rightarrow E_r'$ for $r\ge 0$. One checks directly via the definitions that this is a morphism of spectral sequences. For proofs of this and the theorem below we refer to [79, Theorem 5.5.11].

Theorem A.22

(comparison theorem) If, in the above setting, one of the $f_r$ is an isomorphism, then so are all subsequent $f_r$ and f induces an isomorphism in cohomology.

To illustrate the usefulness of the above theorem, we prove Remark 5.5 in the case of tori:

Proposition A.23

Let a torus $T=T'\times T''$ act on M in such a way that the restricted action of the $T''$-factor is free. Then there is a map $C_{T'}(M/T'')\rightarrow C_{T}(M)$ of cdgas inducing an isomorphism in cohomology.

Proof

It suffices to prove the proposition in case $T=T'\times S^1$. Then the general case follows by induction. Consider now an action of $T=T'\times S^1$ on M with the $S^1$ factor acting freely. Via the above product decomposition the Lie algebra of T decomposes as $\mathfrak {t}\oplus \mathfrak {t}_1$. In Corollary 5.3 it was proved that $\Omega (M/S^1)\cong \Omega _{\mathrm {bas}~S^1}(M)\rightarrow C_{S^1}(M)$ induces an isomorphism on cohomology. Note that if we restrict this map to $\Omega (M/S^1)^{T'}$, it will take values in $S(\mathfrak {t}_1^*)\otimes \Omega ^{T}(M)$. We want to argue that in the diagram

Open image in new window

the map $\psi _1$ induces an isomorphism in cohomology. By Theorem 2.2 and Corollary 5.3 (applied to the proved $S^1$ case) we know that $\psi _2$ and $\psi _4$ induce isomorphisms. Consequently, if we show that $\psi _3$ induces an isomorphism, the same will hold for $\psi _1$.

Filter both complexes, $S(\mathfrak {t}_1^*)\otimes \Omega ^{T}(M)$ and $C_{S^1}(M)$, by the degree of $S(\mathfrak {t}_1^*)$ as we did for the construction of the spectral sequence for $C_{S^1}(M)$ (see Sect. A.3). As $\psi _3$ is $S(\mathfrak {t}_1^*)$-linear it respects the filtration and induces a morphism of spectral sequences. As argued before, the 0th pages of the spectral sequences are isomorphic to the respective filtered complexes $ S(\mathfrak {t}_1^*)\otimes \Omega ^{T}(M)$ and $C_{S^1}(M)$ and one quickly checks that the map between the 0th pages is just $\psi _3$. On both 0th pages, the differential $d_0$ is $1\otimes d$, with d the exterior derivative on $\Omega (M)$. The inclusion $\Omega (M)^T\rightarrow \Omega (M)$ factors through $\Omega (M)^{S^1}\rightarrow \Omega (M)$ and both induce isomorphisms in cohomology by Theorem 2.2. Consequently the inclusion $i:\Omega (M)^{T}\rightarrow \Omega (M)^{S^1}$ induces an isomorphism as well and we deduce that $\psi _3={{\text {id}}_{S(\mathfrak {t}_1^*)}}{\otimes }\ { i}$ induces an isomorphism on $E_1=H(E_0,d_0)$. Now by the Comparison Theorem A.22, $\psi _3$ induces an isomorphism in cohomology.

The final step is to show that the map

$$\begin{aligned} \varphi :C_{T'}(M/S^1)=S(\mathfrak {t}^*)\otimes \Omega (M/S^1)^{T'} \longrightarrow S(\mathfrak {t}^*)\otimes \left( S(\mathfrak {t}_1^*)\otimes \Omega (M)^{T}\right) =C_{T}(M) \end{aligned}$$

defined as ${\text {id}}_{S(\mathfrak {t}_l^*)}\otimes \psi _1$ induces an isomorphism in cohomology. To see this one proceeds analogously to before: Filter both complexes by the degree of $S(\mathfrak {t}^*)$. Then the 0th pages will be isomorphic to $C_{T'}(M/S^1)$ and $C_{T}(M)$ (the bigrading on the latter is not the usual one!) and $\varphi $ induces a morphism of spectral sequences which on $E_0$ is just $\varphi $ itself. The differentials $d_0$ are $1\otimes d$ and $1\otimes d_{S^1}$. In particular $\varphi $ induces an isomorphism on the cohomology $E_1$ because $\psi _1$ does so on the right tensor factor. Another application of Theorem A.22 yields the result. $\square $

A counterexample

In [76] it was shown that under certain topological conditions, e.g. for compact manifolds, the equivariant cohomology of a $S^1$-action and the final page of the spectral sequence are isomorphic as $S(\mathfrak {t}^*)$-modules. For tori of greater dimension this is no longer true. We construct here a $T^2$-action on a compact manifold such that the final page of the spectral sequence associated to the Cartan model is not isomorphic as a graded $S(\mathfrak {t}^*)$-module to the equivariant cohomology.

Consider the standard action of the diagonal maximal torus $T^3$ of $\mathrm {SU}(4)$ by left multiplication, where we identify (s, t, u) with the diagonal matrix with entries $(stu,\overline{s},\overline{t},\overline{u})$. The maximal diagonal torus of $\mathrm {SU}(2)$ is a circle and together they yield a product action of $T^4=S^1\times T^3$ on $\mathrm {SU}(2)\times \mathrm {SU}(4)$. We pull back this action along the homomorphism $T^3\rightarrow T^4,~(s,t,u)\mapsto (s,s,t,u)$. Now we take the quotient of the first circle factor of $T^3$ and consider the action of the middle and right circle factors to obtain an action of $T^2$ on the space

$$\begin{aligned} M:=(\mathrm {SU}(2)\times \mathrm {SU}(4))/S^1. \end{aligned}$$

This action has the desired properties as we will now show. In what follows the Lie algebra of the r-torus will be denoted $\mathfrak {t}_r$.

As it is our goal to show that $H_{T^2}^*(M)$ and $E_\infty $ are not isomorphic let us begin by pointing out the structural difference in the two modules.

Claim

In $E_\infty $ there exists a nontrivial degree 2 element which is torsion with respect to some linear polynomial in $S(\mathfrak {t}_2^*)$. The same does not hold for $H^*_{T^2}(M)$.

To analyse $H^*_{T^2}(M)$ we will use that it is isomorphic to $H^*_{T^3}(N)$, where $N=\mathrm {SU}(2)\times \mathrm {SU}(4)$ with the aforementioned $T^3$-action. The isomorphism is induced by the cdga morphism

$$\begin{aligned} \varphi :C_{T^2}(M)=S(\mathfrak {t}_2^*)\otimes \Omega (M)^{T^2}\longrightarrow S(\mathfrak {t}_2^*)\otimes \left( S(\mathfrak {t}_1^*)\otimes \Omega (N)^{T^3}\right) =C_{T^3}(N) \end{aligned}$$

which was constructed in the proof of Proposition A.23, where we decompose $\mathfrak {t}_3$ as $\mathfrak {t}_2\oplus \mathfrak {t}_1$ in such a way that $\mathfrak {t}_1$ corresponds to the subcircle of $T^3$ such that $M=N/S^1$. In the proof we also argued that $\varphi $ induces an isomorphism between the $E_\infty $-term of the spectral sequence of $C_{T^2}(M)$ and the final page $E_\infty '$ of the spectral sequence obtained by filtering $C_{T^3}(N)$ by the degree of $S(\mathfrak {t}^*_2)$. This allows us to work with the latter spectral sequence when analysing the $E_\infty $-term. Note that under the isomorphisms $H_{T^2}(M)\cong H_{T^3}(N)$ and $E_\infty \cong E_\infty '$, the $S(\mathfrak {t}_2^*)$-module structure on the left side corresponds to the pullback of the $S(\mathfrak {t}_3^*)$-module structure on the right side along the inclusion $S(\mathfrak {t}_2^*)\rightarrow S(\mathfrak {t}^*_3)$.

Now let $X,Y,Z\in \mathfrak {t}_3^*$ be the dual basis of the standard basis of $\mathfrak {t}_3$, with X in the $\mathfrak {t}_1^*$ summand of the decomposition $\mathfrak {t}_3^*=\mathfrak {t}_2^*\oplus \mathfrak {t}_1^*$.

Lemma A.24

The map $S(\mathfrak {t}^*_3)\rightarrow H_{T^3}^*(N)$ is injective in degrees up to 3 and its kernel in degree 4 is generated by $X^2$ and $X^2+XY+Y^2+YZ+Z^2 +ZX$.

Proof

Let $(E_r,d_r)$ denote the spectral sequence of $C_{T^3}(N)$. The map $S^p(\mathfrak {t}_3^*)\rightarrow H_{T^3}^{2p}(N)$ factors as

$$\begin{aligned} S^p(\mathfrak {t}_3^*)\rightarrow E_\infty ^{2p,0}\cong F^{2p}H^{2p}_{T^3}(N)\subset H^{2p}_{T^3}(N), \end{aligned}$$

where we have used that $F^{2p+1}H^{2p}_{T^3}(N)=0$ (see the definition of the isomorphism at the end of Sect. A.2). In particular the kernels of $S(\mathfrak {t}_3^*)\rightarrow E_\infty $ and $S(\mathfrak {t}_3^*)\rightarrow H_{T^3}^*(N)$ coincide.

We have $E_1=S(\mathfrak {t}^*_3)\otimes H^*(\mathrm {SU}(2)\times \mathrm {SU}(4))$. By the Künneth formula, $H^*(\mathrm {SU}(2)\times \mathrm {SU}(4))$ is trivial in degrees 1 and 2. For degree reasons, no elements in $E_1^{2,0}$ can be hit by a differential, and thus they live to infinity. This shows injectivity. Elements in $E_1^{4,0}$ live to $E_3^{4,0}$, where they can potentially be hit by $d_3:E_3^{0,3}\rightarrow E_3^{4,0}\cong H^3(\mathrm {SU}(2)\times \mathrm {SU}(4))$. This is the only nonzero differential entering this position and thus the kernel in degree 4 corresponds to the image of $d_3$ in $E_3^{4,0}$. In particular it is at most 2-dimensional because $\dim H^3(\mathrm {SU}(2)\times \mathrm {SU}(4))=2$. It remains to show that the polynomials from the lemma actually lie in the kernel in which case they will span it.

Recall that the $T^3$-action is defined as a pullback of the product $T^4$-action on N along a homomorphism which on Lie algebras is given by $i:\mathfrak {t}_3\rightarrow \mathfrak {t}_4,~(x,y,z)\mapsto (x,x,y,z)$ where we use the standard bases. We have a commutative diagram

Open image in new window

induced by the pullback map $i^*:C_{T^4}(N)\rightarrow C_{T^3}(N)$. Let W, X, Y, Z denote the dual basis of the standard basis of $\mathfrak {t}_4$, where W corresponds to the circle factor acting on $\mathrm {SU}(2)$ and X, Y, Z correspond to the maximal torus of $\mathrm {SU}(4)$. Note that N is actually a Lie group and that the $T^4$-action is the action of a maximal torus of N. By Remark 10.5, the kernel of $S(\mathfrak {t}_4^*)\rightarrow H_{T^4}^*(N)$ consists of the Weyl-invariant polynomials which in (cohomological) degree 4 are $p_1=W^2$ and $p_2=X^2+XY+Y^2+YZ+Z^2+ZX$. Hence the elements $i^*(p_1),i^*(p_2)$ lie in the kernel of $S(\mathfrak {t}_3^*)\rightarrow H_{T^3}^*(N)$. They are precisely the polynomials from the lemma because $i^*$ maps W to X and X, Y, Z to themselves. $\square $

As we see from the spectral sequence of $C_{T^3}(N)$, the elements X, Y, Z induce a basis of $H^2_{T^3}(N)$. No element of the degree-4 part of $\ker (S(\mathfrak {t}^*_3)\rightarrow H_{T^3}^*(N))$ is divisible by a linear polynomial from $S(\mathfrak {t}^*_2)$. Indeed, for an element of the form $aY+bZ$ to divide a nonzero element of the form $cX^2+d(X^2+X(Y+Z)+Y^2+YZ+Z^2)$ it is certainly necessary that $c=-d$ and $a=b$. But $Y+Z$ does not divide $X(Y+Z)+Y^2+YZ+Z^2$. This proves the claim that no nonzero element of $H_{T^2}^2(M)$ is sent to 0 by multiplication with a linear polynomial from $S(\mathfrak {t}^*_2)$.

On the contrary, consider the element $\overline{X}\in {E'}_\infty ^{0,2}$ induced by X in the spectral sequence obtained by filtering $C_{T^3}(N)$ by the degree in ${\mathbb {R}}[Y,Z]$ (recall that $E_\infty '$ is isomorphic to the final page associated to $C_{T^2}(M)$). By the lemma, $X(Y+Z)+Y^2+YZ+Z^2$ is a coboundary. But this shows that $X(Y+Z)$ is a coboundary up to elements of filtration degree 4 and therefore becomes trivial in ${E'}_\infty ^{2,2}$. Thus $\overline{X}(Y+Z)=0$. We have shown that $E_\infty $ and $H^*_{T^2}(M)$ are not isomorphic as graded modules.

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Oliver Goertsches
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Leopold Zoller
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1.Fachbereich Mathematik und InformatikPhilipps Universität MarburgMarburgGermany

Equivariant de Rham cohomology: theory and applications

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