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Could someone tell me what is the cohomology algebra $H^*(\mathbb{R}P^n \# \mathbb{R}P^n; \mathbb{Z}_2)$ and how to compute it. Here $\#$ is the connected sum.


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First compute the cohomology groups using Mayer-Vietoris. See here:… – AlexE Jan 14 '12 at 10:46

Given two compact connected differential manifolds $M_1$ and $M_2$ of dimension $n$, the reduced homology with $\mathbb Z$ coefficients of their connected sum is given by
$$\tilde H_i(M_1 \#M_2) =\tilde H_i(M_1) \oplus \tilde H_i(\#M_2) \quad \text {for}\;\; i\lt n-1$$
$$ H_{n-1}(M_1 \#M_2) = H_{n-1}(M_1) \oplus \ H_{n-1}(\#M_2)$$ if one of $M_1, M_2 $at least is orientable
$$H_n(M_1 \#M_2) = \mathbb Z \; \text {or } \;0$$ according as both $M_1, M_2$ are orientable or not.
This results from Mayer-Vietoris.
You should then calculate cohomology with coefficients in $\mathbb Z/2$ by the universal coefficient theorem.
Warning: if both $M_1,M_2$ are non-orientable (in your case if your projective space $\mathbb P^n$ is even-dimensional), I am not sure that $M_1 \#M_2$ is even well defined, i.e. is independent of the choices made in the construction of the connected sum.

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Thanks. Actually I want to compute $H^*(\mathbb{R}P^n \# \mathbb{R}P^n; \mathbb{Z}_2)$ when $n$ is odd. I am interested in knowing the algebra structure and computing the product seems tricky. – kelly Jan 14 '12 at 11:12
For $n=3$, I computed that $$H^i(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)=\begin{cases} \mathbb{Z}_2 & \mbox{if } i=0\\ \mathbb{Z}_2 \oplus \mathbb{Z}_2 & \mbox{if } i=1 \\ \mathbb{Z}_2 \oplus \mathbb{Z}_2 & \mbox{if } i=2 \\ \mathbb{Z}_2 & \mbox{if } i=3.\\ \end{cases}$$ But I am not able to determine the product. Could someone help me. – kelly Jan 19 '12 at 13:34

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