# Differences in computing $H^*(U(2); \mathbb{Z})$ and $H^*(RP^3; \mathbb{Z_2})$ using the Serre spectral sequence

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!

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 $$x \otimes y$$.
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.
• 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? – Marco Nervo Apr 20 '20 at 17:05
• @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. – Connor Malin Apr 20 '20 at 17:27
• @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. – Connor Malin Apr 20 '20 at 18:46