Let $G$ be a finite group and $V$ be an integral representation of $G$, i.e. a free abelian group of finite rank with $G$-action. Now consider the symmetric power $Sym(V)$ of $V$ over $\mathbb{Z}$, which has again a $G$-action by the functoriality of $Sym$. While for the tensor algebra instead, one could compute its group cohomology via a Künneth theorem [edit: as Thorsten Ekedahl pointed out, this computes only the cohomology of $G\times G$], I have not found a similar theorem for symmetric powers in the literature. So my question is:

Is there a general procedure for computing the cohomology groups $H^i(G; Sym(V))$ (with $H^i(G; V)$ as input)?

It is not clear for me how to do this in a conceptual way even for simple examples like the two-dimensional indecomposable representation of the cyclic group $C_3$. With other words, the example I am especially interested in is to compute the $C_3 = \langle t|t^3 =1\rangle$-cohomology of $\mathbb{Z}[x,y]$ with $tx = y$ and $ty = -x - y$. Of course, it would be enough to have the decomposition in indecomposables. This computation is related to an alternative derivation of the cohomology of the sheaves $\omega^k$ on the moduli stack of elliptic curves (at the prime $3$).

  • $\begingroup$ But isn't Sym(V) a direct factor of the tensor product as a G-representation? So you may just compute the cohomology of the tensor product, and then apply the projector. $\endgroup$
    – Xandi Tuni
    Oct 24, 2011 at 16:21
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    $\begingroup$ You have to pass to $\mathbb Q$ to get the projector. $\endgroup$ Oct 24, 2011 at 17:19
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    $\begingroup$ You can not in any obvious way compute even the cohomology of the tensor power. The Künneth theorem is about the action of $G^n$ on $V^{\otimes n}$ not about the action of $G$. $\endgroup$ Oct 24, 2011 at 17:22
  • $\begingroup$ Lennart -- I've added an example with a free module. $\endgroup$
    – algori
    Oct 26, 2011 at 17:52
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    $\begingroup$ If $V$ is the rank two module, there is a short exact sequence $0\to\mathbb Z\to\mathbb ZC_3\to V\to 0$, which provides a filtration on $S^N(\mathbb ZC_3)$ whose subquotients are exterior powers of $V$. From this one gets a spectral sequence going from $E_1=H^p(C_3,S^qV)$ to $H^\bullet(C_3,S^N(\mathbb ZC_3)$. Maybe you can use this? $\endgroup$ Oct 26, 2011 at 23:21

2 Answers 2


This is a sketch on how to get along with the computation of $H^*(C_3;\mathbb{Z}[x,y])$.

It's well-known that $H^{2*}(C_3;\mathbb{Z}[x,y]) = \frac{\mathbb{Z}[x,y]^{C_3}}{(1+t+t^2)\mathbb{Z}[x,y]}$.

Let $\zeta \in \mathbb{C}$ be a primitive 3th root of unity, $R := \mathbb{Z}[\zeta,1/3]$ and set $$u := -\zeta^2 x+y \hspace{30pt} v := -\zeta x+y$$ ($u,v$ are just the eigenvectors of the linear representation). Extend the action of $C_3$ to $R \otimes_\mathbb{Z} \mathbb{Z}[x,y] = R[x,y] = R[u,v]$. Then $$t \cdot u = \zeta u \hspace{30pt} t \cdot v = \zeta^2 v.$$ Let $u^iv^j$ be invariant under the action of $C_3$. This is equivalent to $\zeta^{i+2j} = 1$, i.e. $i - j \equiv 0(3)$. It follows that $R[u,v]^{C_3}$ is the free $R[u^3,v^3]$-module with generators $1, uv, (uv)^2$.

Moreover, $(1+t+t^2)u^iv^j = (1 + \zeta^{i+2j} + \zeta^{2(i+2j)})u^iv^j$. Thus $(1+t+t^2)u^iv^j = 0$ if $i - j$ is not congruent mod 3 and $(1+t+t^2)u^iv^j = 3u^iv^j$ if $i-j \equiv 0(3)$. This shows $(1+t+t^2)R[u,v] = 3R[u,v]^{C_3}$.

Hence $H^{2*}(C_3;R[u,v])$ is the free $\mathbb{F}_3[\zeta][u^3,v^3]$-module with generators $1, uv, (uv)^2$.

Now let's translate the result back into $\mathbb{Z}[x,y]$-coefficients using $$\mathbb{Z}[x,y]^{C_3} = \mathbb{Z}[x,y] \cap R[u,v]^{C_3}.$$

An easy computation shows (miscalculations are of course possible) $$c := uv = x^2+xy + y^2$$ $$a := u^3 + v^3 = -2 x^3 +3xy^2-3x^2 y + 2y^3$$ $$b := \zeta v^3 + \zeta^2 u^3 = x^3 + 3xy^2+6x^2 y -y^3$$ lie in $\mathbb{Z}[x,y]^{C_3}$ and with some more work (for example by comparing ranks with $R[u,v]^{C_3}$) one can show that $\mathbb{Z}[x,y]^{C_3}$ is the free $\mathbb{Z}[a,b]$-module with generators $1, c, c^2$.

Hence one concludes that $H^{2*}(C_3;\mathbb{Z}[x,y])$ is the free $\mathbb{F}_3[a,b]$-module with generators $1, c, c^2$.

The computation of $H^{2*+1}(C_3;\mathbb{Z}[x,y])$ is similar (if you should get trouble just let me know).

Edit: The computation of $H^{2\ast}(C_3;\mathbb{Z}[x,y])$ turns out to be more interesting as it seemed at a first glance. As I want to save the result for myself, the following exposition is somewhat more extensive.

Step 1: Computation of $\mathbb{Z}[x,y]^{C_3}$

From $a-2b = 9(x^2y + xy^2)$ it follows that

$$\bar{a} := x^2y + xy^2 \in \mathbb{Z}[x,y]^{C_3}$$ $$\bar{b} := b-3\bar{a} = x^3 + 3x^2y - y^3 \in \mathbb{Z}[x,y]^{C_3}$$

should be good canditates for a hsop of $\mathbb{Z}[x,y]^{C_3}$. To simplify notation, set $$a := x^2y + xy^2, \quad b := x^3 + 3x^2y - y^3.$$

Using Singular, I found

$$x^6 = (-3x^3+5x^2 y +2xy^2-2y^3)a +(x^3-2xy^2)b \in (a,b)$$ $$y^6 = (2x^2y+5xy^2-2y^3)a - (2xy^2+y^3)b \in (a,b)$$

Thus $\mathbb{Z}[x,y]/ (a,b)$ vanishes (at least) in degrees > 25. This is equivalent to $\mathbb{Z}[x,y]$ being finitely generated as a module over $\mathbb{Z}[a,b]$.

The action of the Galois group $G(\mathbb{Q}(\zeta)/\mathbb{Q})$ on $\mathbb{Q}(\zeta)$ induces an action of $G(\mathbb{Q}(\zeta)/\mathbb{Q})$ on $\mathbb{Q}(\zeta)[x,y]$ via operation on the coefficients.

Let $f \in \mathbb{Z}[x,y]^{C_3}$ be homogeneous. We already know $R[x,y]^{C_3}=R [a,b] (1,c,c^2)$. Thus we may write $f = \sum_{i,j}q_{ij}a^i b^j c^k$ ($0 \le k \le 2$) with $q_{ij} \in \mathbb{Q}(\zeta)$. Since $f$ and $a^i b^j c^k$ have integral coefficients, they are invariant under the action of $G(\mathbb{Q}(\zeta)/\mathbb{Q})$. This implies that the $q_{ij}$ are also invariant, i.e. $q_{ij} \in \mathbb{Q}$. By multiplying with the lcm $l$ of the denominators of the $q_{ij}$ we then obtain a relation $$lf = \sum_{i,j}m_{ij} a^i b^j c^k$$ with coprime integers $m_{ij}$. Suppose that $l \neq 1$. Let $p$ be a prime-divisor of $l$. By reducing mod p and dividing out $c^k \neq 0$, we obtain the relation $$0 = \sum_{i,j}m_{ij} a^i b^j$$ in $\mathbb{Z}/p [a,b]$. Since the $m_{ij}$ are coprime, not all are zero in $\mathbb{F}_p$. Thus $a, b$ are algebraic dependent over $\mathbb{F}_p$ and there is an epimorphism from a polynomial ring $\mathbb{F}_p[A,B] \to \mathbb{F}_p[a,b]$ those kernel contains a non-zero homogeneous polynomial of positive degree, say $g$. Thus the Krull-dimension satisfies $1 = \dim \mathbb{F}_p[A,B]/(g) \ge \dim \mathbb{F}_p[a,b]$.

Since $\mathbb{Z}[x,y]$ is finitely generated as module over $\mathbb{Z}[a,b]$ it follows that $\mathbb{F}_p[x,y]$ is finitely generated as module over $\mathbb{F}_p[a,b]$. Thus $\mathbb{F}_p[x,y]$ is an integral extension of $\mathbb{F}_p[a,b]$. In particular, they have the same Krull-dimension, that equals 2, contadicting $\dim \mathbb{F}_p[a,b] \le 1$.

Therefore one obtains $l = 1$, i.e. $q_{ij} \in \mathbb{Z}$ and so $f \in \mathbb{Z} [a,b] (1,c,c^2)$. This shows $\mathbb{Z}[x,y]^{C_3} \le$ $\mathbb{Z} [a,b] (1,c,c^2)$ and thus $\mathbb{Z}[x,y]^{C_3} = \mathbb{Z} [a,b] (1,c,c^2)$.

Remark: In order to show $l=1$, one may wonder, if it isn't sufficient to argue that, because $a,b$ are algebraic independent over $\mathbb{Z}$, $\mathbb{Z}[a,b]$ is isomorphic to a polynomial ring $\mathbb{Z}[A,B]$ and hence $\mathbb{F}_3[a,b] \cong \mathbb{F}_3[A,B]$. But this doesn't work as the example $a=3x$ shows.

Step 2: Computation of $H^{2\ast}$

Let $N$ be the norm map, i.e. multiplication with $1+t+t^2$. $N$ is linear over $\mathbb{Z}[x,y]^{C_3}$ und thus over $\mathbb{Z}[a,b]$. Because of $c=N(c-x^2)$ and $c^2=N(c^2-x^4)$, $H^{2*}$ is the quotient of $\mathbb{Z}[a,b]$ by the ideal $I$ generated by norm's of degree $\equiv 0(3)$.

Since $x^6,y^6 \in (a,b)$, $\mathbb{Z}[x,y]$ is generated as $\mathbb{Z}[a,b]$-modul by $x^iy^j$, $0 \le i,j \le 5$. Thus $I$ is generated by $N(x^iy^j)$ where $i+j \in \lbrace 0,3,6 \rbrace$. Using Singular, I found $I=(3,b) \trianglelefteq \mathbb{Z}[a,b]$, yielding finally

$$H^0(C_3;\mathbb{Z}[x,y]) =\mathbb{Z} [a,b] (1,c,c^2)$$ $$H^{2i}(C_3;\mathbb{Z}[x,y]) = \mathbb{F}_{3}[a] ,\quad i> 0.$$

Simplifications in the arguments are welcome.

  • $\begingroup$ Thank you very much for your answer, although it seems to be partially flawed. 1) $R[x,y] \neq R[u,v]$ since $\zeta-\zeta^2$ (the determinant of your base change) is not a unit in $R$, but only after tensorizing with $\mathbb{Z}[\frac{1}{3}]$. This seems to be no problem for your determination of the invariant elements, but $(1+t+t^2)\mathbb{Z}[x,y]$ is bigger. $\endgroup$ Oct 28, 2011 at 13:20
  • $\begingroup$ 2) In the determination of $\mathbb{Z}[x,y]^{C_3}$ you seem to forget the element $u^3v^3$. I think, only $3u^3v^3$ is in your module. This can be seen to be all elements without really working since one can use that $\mathbb{Z}[x,y]^{C_3}$ are the Galois invariant elements of $R[x,y]^{C_3}$ and then use the elementary symmetric polynomials. $\endgroup$ Oct 28, 2011 at 13:21
  • $\begingroup$ Thanks for your notice. I didn't forget $u^3v^3$. The actual problem was that in my hasty computation $a,b$ weren't of minimal length in the lattice. $\bar{a}, \bar{b}$ above do the trick. If you don't mind, you could post the final result for the cohomology groups. I would be interested. $\endgroup$
    – Ralph
    Nov 1, 2011 at 0:20
  • $\begingroup$ Very nice, thank you. I think, one could simplify the calculations slightly if one does the calculations in your edit only over $\mathbb{F}_3$, which should be enough. $\endgroup$ Dec 14, 2011 at 12:49

The answer is no. Take $G=\mathbb{Z}/2$ and $V=\mathbb{Z}/3$ with the nontrivial action of $G$. Then for even $n$ the module $Sym^n(V)$ is trivial, so $H^0(G,Sym^n(V))=\mathbb{Z}/3\neq 0$. For odd $n$ we have $Sym^n(V)=V$ and every $H^i(G,Sym^n(V))=0$.

One way to see this is this. Take $H=\mathbb{Z}$ and a surjective group map $f:H\to G$. Then $V$ becomes an $H$-module. Moreover, $V$ is trivial as a $\ker f$-module and $G$ acts as multiplication by $-1$ on $H^*(\ker f,V)$. So both rows of the Serre-Hochschild spectral sequence of $f$ will be identical, and if one is non-zero, so would be the other. On the other hand we have $H^*(H,V)=0$ and so both rows of the spectral sequence, which are just the cohomology $H^*(G,V)$, are zero.

(This is just the simplest possible example, but there are many more. In general there is no way to reduce the cohomology of tensor products/symmetric/alternating powers of a sheaf to the cohomology of the sheaf itself.)

[upd: slightly modifying the above construction one can get an example of a free module $G$-module $V$ with $G$ cyclic such that $H^*(G,V)=0$ and $H^0(G,Sym^2(V))\neq 0$.

Namely, take $V$ to be the $\mathbb{Z}$-module $\mathbb{Z}^2$ on which 1 acts as $A=\begin{pmatrix}0&1\\-1&1\end{pmatrix}$. Notice that the order of $A$ is 6, so we $V$ is a $G=\mathbb{Z}/6$-module. Notice also that $I-A\in SL_2(\mathbb{Z})$, so $H^*(\mathbb{Z},V)=0$.

Now consider the Serre-Hochschild spectral sequence as above. Both the rows of it contain $H^*(G,V)$ and so both must be 0 since $H^*(\mathbb{Z},V)=0$. On the other hand it is not too difficult to construct an invariant in $Sym^2(V)$: pass to the dual and then construct an invariant quadratic polynomial.

Using Serre-Hochschild for this may be an overkill. It is possible that one can do all this using just the standard cochain complexes.]

  • $\begingroup$ This is a nice example, although no counterexample to my original question since I wanted $V$ to be free abelian. An alternative derivation of your result might use simply the short exact sequence $0\to \mathbb{Z}^\sigma \to \mathbb{Z}^\sigma \to V \to 0$ and the resulting long exact sequence in cohomology. $\endgroup$ Oct 26, 2011 at 15:06
  • $\begingroup$ Lennart -- sorry, I must have misread the question. I think there must be a counterexample with a free $V$ as well. $\endgroup$
    – algori
    Oct 26, 2011 at 15:16
  • $\begingroup$ Could whoever downvoted this please give their reasons? $\endgroup$
    – algori
    Nov 1, 2011 at 12:50

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