Cohomology ring of $\mathbb R P^\infty$ with $\mathbb Z_{2k}$ coefficients

Question

Let $k$ be a positive integer. I am trying to show that as rings, $H^*(\mathbb RP^\infty ; \mathbb Z_{2k}) \cong \mathbb Z_{2k}[a,b]/(2a , 2b , a^2 - kb)$. This is exercise 3.2.5 in Hatcher. The hint is to "Use the coefficient map $\mathbb Z_{2k} \rightarrow\mathbb Z_2$ and the proof of theorem 3.12". I tried to adapt the proof of theorem 3.12 (the computation of $H^*(\mathbb RP^\infty, \mathbb Z_{2k})$) but was unable to do so: the proof basically shows that cup products of a generator of $H^1$ generate all of the higher cohomology groups. In the $\mathbb Z_{2k}$ case the cohomology ring doesn't have such a simple description, so I couldn't find a way to make an inductive proof work. Does anyone have a proof?

Answer 1

Here's an answer which uses the unique non-trivial homomorphism $\phi:\mathbb{Z}_{2k}\rightarrow \mathbb{Z}_2$ and the Bockstein homomorphism associated to the short exact sequence $$0\rightarrow \mathbb{Z}_{2k}\rightarrow \mathbb{Z}_{(2k)^2}\rightarrow \mathbb{Z}_{2k}\rightarrow 0.$$ In fact, just using $\phi$ we can determine the ring structure when $k$ is odd, and make partial progress when $k$ is even. The rest of the ring structure when $k$ is even will follow using the Bockstein.

So, let's work with the group homomorphism $\phi:\mathbb{Z}_{2k}\rightarrow \mathbb{Z}_2$ which sends $1$ to $1$. Note that this is, in fact, a ring homomorphism. Then $\phi$ will induce a ring map $H^\ast(\phi):H^\ast(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\rightarrow H^\ast(\mathbb{R}P^\infty, \mathbb{Z}_2).$ Let's compute what $\phi$ does in cohomology.

To begin with, we use the usual CW structure on $\mathbb{R}P^\infty$ which has a unique cell in each dimension. The corresponding chain complexes with $\mathbb{Z}_{2k}$ coefficients and $\mathbb{Z}_2$ coefficients fit into a commutative diagram

$$\begin{array} 00 & {\longrightarrow} & \mathbb{Z}_{2k} & \stackrel{\times 0}\longrightarrow & \mathbb{Z}_{2k} & \stackrel{\times 2}\longrightarrow & \mathbb{Z}_{2k} & \stackrel{\times 0} \longrightarrow & \mathbb{Z}_{2k} & \stackrel{\times 2}\longrightarrow... \\ \downarrow & & \downarrow{\phi_0} & & \downarrow{\phi_1} & & \downarrow{\phi_2} & & \downarrow{\phi_3} \\ 0 & {\longrightarrow} & \mathbb{Z}_2 & \stackrel{\times 0}\longrightarrow &\mathbb{Z}_2 & \stackrel{\times 0}\longrightarrow &\mathbb{Z}_2 & \stackrel{\times 0}\longrightarrow & \mathbb{Z}_2 & \stackrel{\times 0}\longrightarrow ... \end{array}$$

In this diagram, all the $\phi_i$ are really just copies $\phi$, but the subscripts will allow me to talk about specific points in the diagram.

Note that apart from $H^0$, it follows from this chain complex that $H^i(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\cong \mathbb{Z}_2$ no matter the value of $k$. Thus, for $i>0$, we will identify $H^i(\phi_i)$ as a self map of $\mathbb{Z}_2$. I will use $H(\phi_i)$ to denote the induced self-map of $\mathbb{Z}_2$.

Proposition 1: For $i>0$ even, $H(\phi_i)$ is the identity. For $i$ odd, $H(\phi_i)$ is the identity if $k$ is odd, and is the $0$-map if $k$ is even.

Proof: We begin with the case where $i>0$ is even. Then, on the chain complex level, the kernel of the differential $d:\mathbb{Z}_{2k}\xrightarrow{\times 0} \mathbb{Z}_{2k}$ is, of course, everything. But the image of the previous differential consists of all the even elements. Hence, the cohomology $\mathbb{Z}_2$ is generated by $[1]$, and so $H(\phi_i)([1]) = [\phi(1)] = [1]$, which tells us that $H(\phi^i)$ is the identity.

We now work on the case where $i$ is odd. Returning to the chain complex level, the kernel of $d:\mathbb{Z}_{2k}\xrightarrow{\times 2}\mathbb{Z}_{2k}$ is a $\mathbb{Z}_2$ generated by $[k]$. The image of the prior differential is trivial, so $[k]$ generates the cohomology.

Then $H(\phi^i)([k]) = [\phi(k)]$. Since $\phi(k) = \begin{cases}0 & k\text{ even}\\ 1 & k\text{ odd}\end{cases},$ the result follows. $\square$

When $k$ is odd, this means that $H(\phi)$ injects $H^\ast(\mathbb{R}P^\infty;\mathbb{Z}_{2k})$ into $H^\ast(\mathbb{R}P^\infty; \mathbb{Z}_2)$, which allows us to determine the ring structure.

Set $H^\ast(\mathbb{R}P^\infty;\mathbb{Z}_2)\cong \mathbb{Z}_2[\gamma]$.

Proposition 2: Suppose $k$ is odd, then $$H^\ast(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\cong \mathbb{Z}_{2k}[\alpha,\beta]/(2\alpha, 2\beta, \alpha^2-k\beta).$$

Proof: We know that $H^i(\mathbb{R}P^\infty; \mathbb{Z}_{2k})\cong \mathbb{Z}_2$ for any $i>0$. Let $\alpha_i$ denote a generator of $H^i(\mathbb{R}P^\infty;\mathbb{Z}_{2k})$. Of course, $2\alpha_i = 0$ for all $i$.

We claim that $\alpha_1^i = \alpha_i$. To see this, note that Proposition 1 implies that $H^\ast(\phi)(\alpha_i) = \gamma^i$. Then $H^\ast(\phi)(\alpha_1^i) = (H^\ast(\phi)(\alpha_1))^i = \gamma_i = H^\ast(\phi)(\alpha_i)$. Since $k$ is odd, $H^\ast(\phi)$ is injective, so we conclude that $\alpha_1^i = \alpha_i$.

This tells us that $H^\ast(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\cong \mathbb{Z}_{2k}[\alpha_1]/2\alpha_1$, which is isomorphic to the ring given in the statement of the Proposition. $\square$

This leaves the case where $k$ is even. As before, $H^i(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\cong \mathbb{Z}_2$ for any $i>0$. Again, let $\alpha_i$ denote a generator of $H^i$.

Proposition 3: We must have $\alpha_1^2 = 0$ and $\alpha_2^i\neq 0$ for any $i\geq 1$.

Proof: We begin proving that $\alpha_1^2 = 0$. Assume this is false. Then $\alpha_1^2 = \alpha_2$, so $H^\ast(\phi)(\alpha_1^2) = H^\ast(\phi)( \alpha_2) = \alpha^2\neq 0$, which implies that $H^\ast(\phi)(\alpha_1)^2\neq 0$, which implies that $H^\ast(\phi)(\alpha_1)\neq 0$. This contradicts Proposition 1.

Now, we prove that $\alpha_2^i\neq 0$. From Proposition $1$, $H^\ast(\phi)(\alpha_2^i) = H^\ast(\phi)(\alpha_2)^i = (\alpha^2)^i = \alpha^{2i}\neq 0$, so $H^\ast(\phi)(\alpha_2^i)\neq 0$. $\square$

Thus, to finish, we need only show that $\alpha_1\alpha_2^i\neq 0$ for all $i$. As far as I can tell, there is no direct proof just using $\phi$; we actually need something else. That something else is Bockstein $\beta:H^i(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\rightarrow H^{i+1}(\mathbb{R}P^\infty;\mathbb{Z}_{2k})$.

Proposition 4: For any $i\geq 0$, $\alpha_1\alpha_2^i\neq 0$.

Proof: Since $\beta$ is a homomorphism, to show that $\alpha_1\alpha_2^i\neq 0$, it is sufficient to show that $\beta(\alpha_1\alpha_2^i) \neq 0$. To do this calculation we will use the fact that $\beta$ is a (graded) derivation: $\beta(\alpha_1 \alpha_2^i) = \beta(\alpha_1)\alpha_2^i \pm \alpha_1\beta(\alpha_2^i)$.

To compute $\beta$, we use the definition. Namely, from the short exact sequence at the top of this answer, we get a long exact sequence in cohomology groups. Starting with $0 = H^{-1}(\mathbb{R}P^\infty;\mathbb{Z}_{2k}) \rightarrow H^0(\mathbb{R}P^\infty;\mathbb{Z}_{2k})\cong \mathbb{Z}_{2k}$, we get $$0\rightarrow \mathbb{Z}_{2k}\rightarrow \mathbb{Z}_{(2k)^2}\rightarrow \mathbb{Z}_{2k}\xrightarrow{\beta} \mathbb{Z}_2\rightarrow \mathbb{Z}_2\rightarrow \mathbb{Z}_2\xrightarrow{\beta} \mathbb{Z}_2\rightarrow \mathbb{Z}_2\rightarrow \mathbb{Z}_2\xrightarrow{\beta} \mathbb{Z}_2 ...$$

Starting from the left, exactness clearly forces the first $\beta$ to be trivial, which then forces the next one to be an isomorphism (that is $\beta(\alpha_1) = \alpha_2$), which then forces the next $\beta$ to be trivial (that is, $\beta(\alpha_2) = 0$.)

Using the derivation property, it follows easily that $\beta(\alpha_2^i) = 0$. Thus, we find that $$\beta(\alpha_1\alpha_2^i) = \beta(\alpha_1)\alpha_2^i \pm \alpha_1\beta(\alpha_2^i) = \alpha_2^{i+1}.$$ From Proposition 3, we know that $\alpha_2^{i+1}\neq 0$, which proves that $\beta(\alpha_1\alpha_2^i)\neq 0$ as claimed. $\square$

Answer 2

Let's use the proof of the theorem as Hatcher suggested. It is easy to see that in the diagram below the map $H^1(P^n) \to H^1(P^1)$ is a multiplication by $k$ ( injectivity and $H^1(P^1) = Z_{2k}$). The same applies to the two downwards maps on the right

Proceeding as in the Hatcher's proof, we conclude that $H^1(P^n,P^n-P^1) \to H^1(R^n,R^n - R^1)$ has to be a multiplication by $k$ in order to make the diagram commute:

Therefore in the first cohomology the left downwards map is a multiplication by $k$.

So $a\cup a = kb$

Answer 3

I realize this question is old but, since nobody ever answered, I decided to post a solution.

Consider the short exact coefficient sequence $$0\to \mathbb{Z}_2\overset{\times k}{\to} \mathbb{Z}_{2k}\to \mathbb{Z}_{2}\to 0\;.$$ This induces a long exact Bockstien sequence $$... \to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_2)\to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to H^*(\mathbb{R} P^{\infty};\mathbb{Z}_2)\overset{\beta}{\to} H^{*+1}(\mathbb{R} P^{\infty};\mathbb{Z}_2)\to...$$ Recall that $$H^*(\mathbb{R}P^{\infty};\mathbb{Z}_{2})\simeq \mathbb{Z}_2[t]$$ This gives the exact segment $$\mathbb{Z}_2(t^n)\to H^n(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to \mathbb{Z}_2(t^n) \to \mathbb{Z}_2(t^{n+1})\;.$$ Since $\beta$ satisfies the Leibniz rule and we are working over $\mathbb{Z}_2$, we have $\beta(t^{n})=0$ is $n$ is even and $t^{n+1}$ if $n$ is odd. This gives two possibilities. In the odd case, we have an iso $$0\to \mathbb{Z}_2(t^{2n+1})\to H^{2n+1}(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to 0\;$$ and in the even case, we have the iso $$0\to H^{2n}(\mathbb{R} P^{\infty};\mathbb{Z}_{2k})\to \mathbb{Z}_2(t^{2n}) \to 0\;.$$ By the first isomorphism we conclude that $H^1$ is generated by a single element $a=kt$ satisfying $2a=0$. From the second isomorphism, we conclude that $H^2$ is generated by a single element $b$, satisfying $2b=0$. Since $b$ mod $k$ must be $t^2$, we have $b=kt^2$ and $kb=k^2t^2=a^2$. The full ring structure can be deduced from these two cases.