Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question
    
First compute the cohomology groups using Mayer-Vietoris. See here: math.stackexchange.com/questions/31287/… –  AlexE Jan 14 '12 at 10:46

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.