Cup products of connected sum

Question

Hej,

I am interested in the cohomology ring of the connected sum $M \# N$ of two oriented manifolds $M$ and $N$ in terms of the corresponding cohomology rings of $M$ and $N$.

Mayer-Vietoris shows that in dimensions $0<k<n$ the additive structure is given by

$H^k(M \# N)=H^k(M)\oplus H^k(N)$

via the induced maps of the inclusions. And because this isomorphisms are given by induced maps the cup product translates into componentwise product whenever this is possible, i.e.:

$$([\omega_1],[\eta_1])\cup([\omega_2],[\eta_2])=([\omega_1\cup\omega_2],[\eta_1\cup\eta_2])$$

for $([\omega_1],[\eta_1])\in H^i(M)\oplus H^i(N),([\omega_2],[\eta_2])\in H^j(M)\oplus H^j(N)$ AND $i+j<n$.

But how is cup product given if $i+j=n$?

Thanks a lot for your answers!

Answer 1

The natural description is like this. There are augmentations $\epsilon_X:H^*(X)\to\mathbb{Z}$ (for $X\in\{M,N\}$) and orientation classes $u_X\in H^d(X)$. Put $$R'=\{(a,b)\in H^*(M)\times H^*(N):\epsilon_M(a)=\epsilon_N(b)\} = H^*(M\vee N)$$ and $R=R'/(u_M,-u_N)$. Then one can use the cofibration $$ S^{d-1} \to M \# N \to M\vee N $$ to identify $H^*(M\#N)$ canonically with $R$. Thus, if $i+j=d$ with $i,j>0$ then the product of $H^i(M)$ with $H^j(N)$ in $H^d(M\# N)$ is zero, but the product of $H^i(M)$ with $H^j(M)$ is the same as in the original manifold $M$.