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I am (self-)learning how to calculate cohomology rings using the Serre spectral sequence, but I am having troubles in understanding the relation among the multiplication in $E_\infty$ and the multiplication in the cohomology ring. I am working on these two examples.


First example. Want to compute $H^*(U(2); \mathbb{Z})$. There is the fibration $S^1 \to U(2) \to S^3$. The $E_\infty$ page of the associated spectral sequence looks like

\begin{array}{|c|c|c|c|c|} \hline 1 & \mathbb{Z}[y] & 0 & 0 & \mathbb{Z}[xy] \\ \hline 0 & \mathbb{Z}[1] & 0 & 0 & \mathbb{Z}[x]\\ \hline & 0 & 1 & 2 & 3\\ \hline \end{array}

Then $E_\infty \cong \mathbb{Z}[\alpha_1] \oplus \mathbb{Z}[\alpha_3] \oplus \mathbb{Z} [\alpha_4]$ as additive structure. The multiplication table is given by

\begin{array}{|c|c|c|c|} \hline & \alpha_1 & \alpha_3 & \alpha_4 \\ \hline \alpha_1 & 0 & \alpha_4 & 0 \\ \hline \alpha_3 & -\alpha_4 & 0 & 0 \\ \hline \alpha_4 & 0 & 0 & 0 \\ \hline \end{array}

Because there is at most one group on each diagonal, $H^*(U(2); \mathbb{Z}) \cong \mathbb{Z}[\alpha_1] \oplus \mathbb{Z}[\alpha_3] \oplus \mathbb{Z} [\alpha_4]$ as additive structure. By anti-commutativity, the multiplication table is given by

\begin{array}{|c|c|c|c|} \hline & \alpha_1 & \alpha_3 & \alpha_4 \\ \hline \alpha_1 & 0 & * & 0 \\ \hline \alpha_3 & -* & 0 & 0 \\ \hline \alpha_4 & 0 & 0 & 0 \\ \hline \end{array}


Second example. Want to compute $H^*(RP^3; \mathbb{Z}_2)$. There is the fibration $S^1 \to RP^3 \to S^2$. The $E_\infty$ page of the associated spectral sequence looks like

\begin{array}{|c|c|c|c|} \hline 1 & \mathbb{Z}_2[y] & 0 & \mathbb{Z}_2[xy] \\ \hline 0 & \mathbb{Z}_2[1] & 0 & \mathbb{Z}_2[x] \\ \hline & 0 & 1 & 2 \\ \hline \end{array}

Then $E_\infty \cong \mathbb{Z}_2[\alpha_1] \oplus \mathbb{Z}_2[\alpha_2] \oplus \mathbb{Z}_2 [\alpha_3]$ as additive structure. The multiplication table is given by

\begin{array}{|c|c|c|c|} \hline & \alpha_1 & \alpha_2 & \alpha_3 \\ \hline \alpha_1 & 0 & \alpha_3 & 0 \\ \hline \alpha_2 & \alpha_3 & 0 & 0 \\ \hline \alpha_3 & 0 & 0 & 0 \\ \hline \end{array}

Because there is at most one group on each diagonal, $H^*(RP^3; \mathbb{Z}_2) \cong \mathbb{Z}_2[\alpha_1] \oplus \mathbb{Z}_2[\alpha_2] \oplus \mathbb{Z}_2 [\alpha_3]$ as additive structure. By anti-commutativity, the multiplication table is given by

\begin{array}{|c|c|c|c|} \hline & \alpha_1 & \alpha_2 & \alpha_3 \\ \hline \alpha_1 & * & ** & 0 \\ \hline \alpha_2 & ** & 0 & 0 \\ \hline \alpha_3 & 0 & 0 & 0 \\ \hline \end{array}


My questions are

  1. How can I prove that the multiplication table in the first example is correct, i.e. that $ * = \alpha_4$ as in the multiplication table of $E_\infty$?
  2. How can I prove that the multiplication table in the second example is incorrect, i.e. that $* \neq 0$, but $* = \alpha_2$?
  3. How can I prove that the other information in the second example is correct, i.e. that $** = \alpha_3$?
  4. Which is the difference between these two examples? In both, the groups in the $E_\infty$ page are free and one for diagonal!
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For your first example, the spectral sequences degenerates at the $E_2$ page so the multiplicative structure of the $E_\infty $ page is the tensor product of the cohomologies of $S^1$ and of $S^3$. You are hoping to show that this is the multiplicative structure of the cohomology of the total space. You lift $x$ to $\alpha_1$ and $y$ to $\alpha_2$. Since $xy = x \otimes y$, we have that $\alpha_1 \alpha_2$ is a lift of $x \otimes y$.

The possible ambiguity is that the groups on the diagonals are filtration quotients, and so the lifts vary by elements of higher filtration. However, since the diagonals have no other groups besides the trivial group, we are in highest filtration already.

Your second example illustrates that, in general, the extension problem must be solved by outside information. You do have lifting issues because the multiplication on the $E_\infty$ page takes you to 0 which is in lowest filtration, and there are elements in filtration above it. The best approach would be to just calculate it via triangulations.

To address your fourth question, the difference was in the first we could formally address extension issues because we knew how extension issues came about (having elements in higher filtration), and we could either observe no such elements existed or use the fact that the cup product is anticommutative and we are torsion free. In the second we lose both of these advantages.

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  • $\begingroup$ Thank you for your answer, I think I have understood now. I want to ask to you just one more thing: how do you use $S^0 \to S^3 \to RP^3$? Don't you get cohomology with $\mathbb{Z}_2 \times \mathbb{Z}_2$ coefficients? $\endgroup$ – Marco Nervo Apr 20 '20 at 17:05
  • $\begingroup$ @MarcoFrancescoNervo Ah, apologies. There are multiple issues with it. First, the base isn't simply connected, and what I envisioned can't happen. The spectral sequence fibration $S^1 \rightarrow S^{2n-1} \rightarrow\mathbb{C}P^n$ can be used to deduce the multiplicative structure of $H^* {\mathbb{C}P^n)$, but not in the real case. Thankfully, there are other methods to find the multiplication on real projective space. $\endgroup$ – Connor Malin Apr 20 '20 at 17:27
  • $\begingroup$ I have really hoped that what you say would have worked! I know that this works with CPn. I know that it is possible to find out the cohomology ring of RPn via Poincaré duality and also via an ad-hoc analysis of the cellular structure. I dislike both: do you know a better one? $\endgroup$ – Marco Nervo Apr 20 '20 at 18:03
  • $\begingroup$ @MarcoFrancescoNervo Those are probably the simplest. If you can construct spaces that you know have an element in $H^1$ have nontrivial n-fold cup products, this is probably enough since $\mathbb{R}P^\infty$ represents first cohomology. However, I don't know how to do this without just pointing out the $\mathbb{R}P^n$ are examples of such spaces. $\endgroup$ – Connor Malin Apr 20 '20 at 18:46

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