Cohomology ring $H^*(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)$.

I am interested in computing the cohomology ring $H^*(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)$. Here # is the connected sum. Using a suggestion here on my earlier post, I computed the additive structure as $$H^i(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)=\begin{cases} \mathbb{Z}_2 & \mbox{if } i=0 \\ \mathbb{Z}_2 \oplus \mathbb{Z}_2 & \mbox{if } i=1 \\ \mathbb{Z}_2 \oplus \mathbb{Z}_2 & \mbox{if } i=2 \\ \mathbb{Z}_2 & \mbox{if } i=3 \\ \end{cases}$$

But I am having hard time in computing the ring structure. I actually want to compute $H^*(\mathbb{R}P^n \# \mathbb{R}P^n; \mathbb{Z}_2)$ for odd $n$ and think that the case $n=3$ should help me get the general case. Ths case $n=1$ is trivial as $\mathbb{R}P^1 \# \mathbb{R}P^1= \mathbb{S}^1$.

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you can just edit earlier post instead of posting updated version as a separate question – Grigory M Jan 20 '12 at 9:15
My sincere apology. I am a new user and would be careful from now on. – kelly Jan 20 '12 at 10:28

This is a ellaboration of the above answer by Juan S. Further ellaborations are welcome.

Let $M,N$ be compact $d$-manifolds, let $B$ be a small ball along which we connect sum, and let $u\in H^p(M\# N), v\in H^q(M\# N)$ with $p,q\neq 0,d$, and -as you already have calculated the module structure of $H^*(M\# N)$- such that $u,v$ come from certain elements $u^*,v^*$ in $H^*(M)$ or $H^*(N)$. Moreover, as none of them has dimension $d$, we can assume $u$ (resp. $v$) has a representative $\bar{u}$ (resp. $\bar{v}$) supported in $M\backslash B$ or $N\backslash B$. Then you have the following cases:

1) $u^*,v^*\in H^*(M)$. Then the product of $\bar{u}$ and $\bar{v}$ has support contained in $M$, and it defines a representative $\overline{uv}$ which is in fact a extension by zero of a representative of $u^*\cup v^*$, so $u\cup v= w$, where $w$ comes from $u^*\cap v^*$.

2) $u^*,v^*\in H^*(N)$. The same as above.

3) $u^*\in H^*(M), v^*\in H^*(N)$. Then, the supports of $\bar{u}$ and $\bar{v}$ are dsjoint, hence $u\cup v=0$.

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Thanks. After working out your suggestions, I found that $H^*(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2) \cong \mathbb{Z}_2[a,b]/\langle a^4,ab, a^3-b^3 \rangle$, where $a,b$ are elements of degree one each. Please correct me if I am wrong. – kelly Jan 20 '12 at 16:19
You're right. If I were you, I would also write for myself more formally the reasoning above. – user17786 Jan 20 '12 at 16:48

Do you know what the ring structure of $H^*(\mathbb{R}P^3)$ is?

Then the cohomology ring of a connected sum is the direct sum of the cohomology rings modulo identification of the 0 and $n$-th cohomology groups (Here is the precise statement). I guess you can kind of see this since you have worked out the additive structure anyhow.

Note that in your case, since you are using $\mathbb{Z}/2\mathbb{Z}$ coefficients, everything is orientable.

A reference for the result in the orientable case is Bredon - 'Geometric Topology'. This is Problem 1 on page 358.

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Could you please give some reference to what you are suggesting. I mean some paper or textbook and not a online wiki which has no reference. – kelly Jan 20 '12 at 6:44
You have misquoted the wiki, which says : the cohomology ring of the connected sum is the sum of the cohomology rings, modulo some quotienting at the zeroth, $(n-1)^{th}$ and $n^{th}$ stages. "Some quotienting" is not what I would call a precise statement. – Georges Elencwajg Jan 20 '12 at 8:24
Well, it also says "In particular, if both are compact connected orientable manifolds, the cohomology ring of the connected sum is the connected sum of the cohomology rings, modulo identification of the cohomology groups at the zeroth and n-th stage." – user17786 Jan 20 '12 at 12:03
@Georges: I agree. However in this case, since we are working with $\mathbb{Z}/2$ coefficients, the statement that user17786 quotes is enough for this problem – Juan S Jan 20 '12 at 23:17
@Kelly: I have updated with a reference – Juan S Jan 20 '12 at 23:26