8
$\begingroup$

The following sentence is quoted from the paper ON THE COHOMOLOGY OF SPLIT EXTENSIONS by D. J. BENSON AND M. FESHBACH:

In general, the differentials in the Lyndon-Hochschild-Serre spectral sequence for a group extension are difficult to understand. In the case of a central extension, the $E_2$ page is easy to calculate and the transgressions may be computed using the fact that they commute with the Steenrod operations on the base and fiber.

May I know what is the complete statement of the result "the transgressions commute with the Steenrod operations on the base and fiber" and which reference can I refer to for it? And why this makes calculation easy? Can anyone give some examples illustrating that?

$\endgroup$

1 Answer 1

8
$\begingroup$

The statement in question refers to the Kudo-Serre transgression theorem: If $E_r$ is, for example, the LHS spectral sequence of a group extension and $x \in E_{2k+1}^{0,2k}$ is transgressive with $d_{2k+1}(x)=y$, then $$d_i(x^p)=0\;\;(i=2k+1,...,2kp)\quad \text{and }\;\; d_{2kp+1}(x^p)=P^ky$$ where $P^k$ denotes the Steenrod power operation (a textbook reference is McCleary: A User's guide to spectral sequences, Theorem 6.14).

I'll give an example that is related to the Benson-Feshbach paper: Let $p$ be a prime and let $$1 \to C \to G \to Q \to 1$$ be a central extension with $C \cong (\mathbb{Z}/p)^n$ elementray abelian. If $k$ is a field of char. $p$ (with trivial $G$-action), than $Q$ acts trivially on $H^\ast(C;k)$ and by the universal coefficient theorem, the LHS spectral sequence satisfies $$E_2^{ij} \cong H^i(Q;k) \otimes_k H^j(C;k).$$ Moreover, $E_2 \cong H^\ast(Q;k) \otimes_k H^\ast(C;k)$ is an isomorphism of $k$-algebras (up to a sign). This is what is ment by B-F when they say the $E_2$-page is easy to compute.

The cohomology of $C$ is given by $H^\ast(C;k) = k[x_1,...,x_n,y_1,...,y_n]$ with $\deg(x_i)=1, \deg(y_j)=2$ and the relations $x_i^2=y_i$ if $p=2$ and $x_i^2=0$ for $p$ odd. Note that $y_i = \beta(x_i)$ where $\beta$ is the Bockstein homomorphism.

In terms of the spectral sequence, we have $x_i \in E_2^{0,1}$. Hence $d_2(x_i) =: z_i \in E_2^{2,0}=H^2(Q;k)$, i.e. $x_i$ is transgressive. Now Kudo's theorem says that $d_2, \beta$ commute, i.e.
$$d_2(y_i) = d_2(\beta(x_i))=\beta(d_2(x_i))=\beta(z_i) \in H^3(Q;k).$$ Hence $y_i$ is again transgressive and by Kudo's theorem $d_5(y_i^2)=P^1\beta(z_i).$ Continuing this way, one obtains $$d_{2p^k+1}(y_i^{p^k})=P^{p^{k-1}}P^{p^{k-2}}\cdots P^1\beta(z_i).$$ As a result, the differentials on the fiber $E_\ast^{0,\ast}$ are completely determined by the values of $z_i \in H^2(Q;k)$ (the base) under the power operations.

$\endgroup$
11
  • $\begingroup$ Thank you @Ralph for your great answer!! Just to clarify a doubt: In Theorem 6.14 of McCleary's book, the condition that the extension is central is not explicitly mentioned. In your example for the extension $1\to G\to Q\to1$ (I think you mean $1\to C\to G\to Q\to1$, right?), where did you actually use the "central" condition? $\endgroup$
    – Minghui
    May 1, 2012 at 22:25
  • $\begingroup$ @Minghui: To say $Q$ acts trivially on the cohomology of $C$. $\endgroup$
    – Steve D
    May 1, 2012 at 22:31
  • $\begingroup$ Also, in your example $k$ is assumed to be a field of char. $p$, while in Theorem 6.14, the coefficient field is required to be $\mathbb{F}_p$. Did you mean that Theorem 6.14 also holds for any field of char. $p$? $\endgroup$
    – Minghui
    May 1, 2012 at 22:33
  • $\begingroup$ Thank you @Steve D! May I ask why the extension being central implies that $Q$ acts trivially on the cohomology of $C$? $\endgroup$
    – Minghui
    May 1, 2012 at 22:34
  • 1
    $\begingroup$ Kudo's theorem holds in general (i.e. not only for central extensions and not only for the LHS spectral sequence). Centrality is used here to ensure that $x_i \in E_2^{0,1}=H^1(C;k)^Q$. Kudo's theorem is simplest to apply, if you can start with transgressive elements. $\endgroup$
    – Ralph
    May 1, 2012 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.