Pontryagin classes of $\mathbb{CP}^{2n}\#\mathbb{CP}^{2n}$

Question

I am trying to compute the Pontryagin classes of $\mathbb{CP}^{2n}\#\mathbb{CP}^{2n}$. But I have no clue how to do this. I tried to use Mayer-Vietoris and I think $$ p_k(\mathbb{CP}^{2n}\#\mathbb{CP}^{2n}) = {2n+1\choose k}(x^{2} +y^2)^k $$ for $k < n$ and $x,y$ are two generators of $H^2(\mathbb{CP}^{2n}\#\mathbb{CP}^{2n}$). But I dont see how to compute the top Pontryagin class.

Edit 1: I changed the formula slightly.

Edit 2: For $k<n$ we have an isomorphism from the Mayer Vietoris sequence $$H^k(\mathbb{CP}^{2n}\#\mathbb{CP}^{2n}) \cong H^k(\mathbb{CP}^{2n}\setminus D) \oplus H^k(\mathbb{CP}^{2n}\setminus D),$$ where $D$ is a small open $4n$--Disk. We can choose $D$ such that there is a $\mathbb{CP}^{2n-1} \subset\mathbb{CP}^{2n}\setminus D$ which is a strong deformation retract of $\mathbb{CP}^{2n}\setminus D$ (I don't know if this is necessary). Let $i \colon \mathbb{CP}^{2n}\setminus D \to \mathbb{CP}^{2n}\# \mathbb{CP}^{2n}$ be the inclusion map and $\tau$ the tangent bundle of $\mathbb{CP}^{2n}\# \mathbb{CP}^{2n}$. We (should) have that $i^*(\tau)$ is the tangent bundle of $\mathbb{CP}^{2n}$ restricted to $\mathbb{CP}^{2n}\setminus D$. Hence for $k<n$ $$ p_k(i^*(\tau)) ={2n+1\choose k}x^{2k}, $$ where $x \in H^2(\mathbb{CP}^{2n})$ is an appropriate generator.

Answer 1

You can put cell structures on $M, N,$ and $M \# N$ so that $(M \# N)_{n-1} \cong M_{n-1} \vee N_{n-1}$, and so that this isomorphism respects (the restrictions of) tangent bundles, where on the last space we've glued the bundles together at the basepoints by some isomorphism between $T_p M$ and $T_{p'} N$. As a corollary, you get an isomorphism (below top degree) $H^k(M \# N) \cong H^k(M) \oplus H^k(N)$, and this isomorphism respects characteristic classes. This answers your question for $k < n/4$.

In top degree, remember that Pontryagin numbers (and hence the top Pontryagin class in dimension $n = 4k$) are invariant under oriented bordism, and $M \# N$ is oriented bordant to $M \sqcup N$, where $p_k(M \sqcup N) = p_k(M) + p_k(N)$. So $$p_n(\Bbb{CP}^{2n} \# \Bbb{CP}^{2n}) = 2\binom{2n+1}{n}.$$