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Given a fiber bundle $(E,B,p,F)$ with path connected base $B$ and fiber $F$, both closed smooth manifolds of finite dimensions. The second page $E_2^{p,q}$ of the Leray-Serre spectral sequence over $\mathbb{Z}_2$ is give by $H^p(B;\mathcal{H}^q(F;\mathbb{Z}_2))$, where $\mathcal{H}^q(F;\mathbb{Z}_2)$ is the local system (sheaf) of $\mathbb{Z}_2$-vector spaces on $B$ given by $\mathcal{H}^q(F;\mathbb{Z}_2)|_b=H^q(p^{-1}(b);\mathbb{Z}_2)$ for any $b\in B$. In the case this local system is trivial, namely, $\pi_1(B)$ acts trivially on it, we have $$E_2^{p,q}=H^p(B;\mathcal{H}^q(F;\mathbb{Z}_2))=H^p(B;\mathbb{Z}_2)\otimes H^q(F;\mathbb{Z}_2).$$

But if the local system is nontrivial, is there still any relation between two vector spaces $E_2^{p,q}$ and $H^p(B;\mathbb{Z}_2)\otimes H^q(F;\mathbb{Z}_2)$? More generally, is it true that the bi-graded $\mathbb{Z}_2$-algebra $E_2^{*,*}$ is isomorphic as bi-graded $\mathbb{Z}_2$-algebras to $\left(H^*(B;\mathbb{Z}_2)\otimes H^*(F;\mathbb{Z}_2)\right)/I$ for some ideal $I$?

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    $\begingroup$ The answer to your second question is no. In fact $H^*(B;\mathbb{Z}/2)\otimes H^*(F;\mathbb{Z}/2)$ could be much smaller than $H^*(B; \mathcal{H}^*(F;\mathbb{Z}/2))$. $\endgroup$ – Denis Nardin Dec 30 '16 at 10:38
  • $\begingroup$ @Denis Nardin Thanks. Do you have some example for this situation. It is hard to imagine that $E_2^{*,*}$ has generators that cannot be expressed as a polynomial in generators of $H^*(B;\mathbb{Z}_2)$ and $H^*(F;\mathbb{Z}_2)$. $\endgroup$ – Rami Dec 30 '16 at 13:19
  • $\begingroup$ This is not quite finite dimensional, but one can take $B=BZ/2$, $F=X^2$, $E=EZ/2\times _{Z/2}\times X \times X$. Then $H^*(B;/2)\otimes H^*(F;Z/2)$ is {\it larger} than the $E_2$ term of the Serre spectral sequence. $\endgroup$ – user43326 Dec 30 '16 at 14:22
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Let $\pi$ be the fundamental group of $B$ and $R$ be whatever ring of coefficients you'd like. Suppose $B$ admits a universal cover $\tilde{B}$. Then there is a spectral sequence $\text{Ext}^{*,*}_{R[\pi]}(H_*(\tilde{B}), H^*(F))\Rightarrow H^*(B; \underline{H^*(F)})$. With field coefficients, say $k$, you can rephrase this as saying that the derived tensor product $H^*(\tilde{B}) \otimes^{\mathbb{L}}_{k[\pi]} H^*(F)$ serves as an upper bound for the cohomology $H^*(B; \underline{H^*(F)})$.

Now, there is certainly a map from $H^*(B) \otimes^{\mathbb{L}}_{k[\pi]} H^*(F)$. If the order of $\pi$ is finite and invertible in $k$ then we even get a map from $H^*(B) \otimes_{k[\pi]} H^*(F) = H^*(B) \otimes (H^*(F)/\pi)$ to $H^*(B; \underline{H^*(F)})$. Unfortunately, even in this case I see no reason for that map to be surjective in general.

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Let me first explain why your intuition is wrong, then let me provide a counterexample. You claim that "It is hard to imagine that $E_2^{*,*}$ has generators that cannot be expressed as a polynomial in generators of $H^*(B;\mathbb Z_2)$ and $H^∗(F;\mathbb Z_2)$". But it is not so hard to imagine, as the generators of $H^∗(F;\mathbb Z_2)$ in general do not correspond to any elements of $E_2^{*,*}$ at all. If they are not there, obviously elements that are there do not correspond to polynomials in them.

There are maps, but in the other direction, from $H^0(B;\mathcal H^q(F, \mathbb Z_2))$ to $H^q(F,\mathbb Z_2)$, but those don't help you.

A counterexample is provided by any space where $H^0(B;\mathcal H^1(F, \mathbb Z_2))$ vanishes but $H^1(B;\mathcal H^1(F, \mathbb Z_2))$ is nonzero. Then there is no such ideal $I$, because it would have to contain all of $H^1( F;\mathbb Z_2)$ but not all of $H^1(B;\mathbb Z_2) \otimes H^1(F,\mathbb Z_2)$.

We can do this using a suitable $T^2$-bundle on a Riemann surface $S$of genus $g \geq 2$. (If you used $\mathbb Q$-coefficients, an $S^1$-bundle would suffice.) Indeed, we can choose the torus bundle so the monodromy representation of $\pi_1(S)$ on $H^1(T_2,\mathbb Z)$ (and hence $H^1(T_2,\mathbb Z_2)$) factors through a cyclic subgroup of order $3$ acting by an order $3$ element of $GL_2(\mathbb Z)$. So $\mathcal H^1(F,\mathbb Z_2)$, as a sheaf, comes from a nontrivial irreducible two-dimensional representation of $\pi_1$. Then there are no global sctions of $\mathcal H^1(F,\mathbb Z_2)$, so $H^0(S, \mathcal H^1(F,\mathbb Z_2)) =0$, but by the Euler characteristic formula

$$\dim H^0(S, \mathcal H^1(F,\mathbb Z_2)) - \dim H^1(S, \mathcal H^1(F,\mathbb Z_2)) + \dim H^2(S, \mathcal H^1(F,\mathbb Z_2)) =2 (2-2g)$$

we can see that $H^1(S, \mathcal H^1(F,\mathbb Z_2)) $ is nonvanishing.

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