I started learning about signature of a $4k$-manifold and one of the most common example is the signature of $\mathbb{C}P^{2n}$. The only reference I found is tom Dieck's Algebraic Topology.

Even though the "structure" of the proof (i.e. the idea behind it) seems clear there are some obscure passage in it, first of all, given that strange map $p \colon (\mathbb{C}P^1)^n\to \mathbb{C}P^{2n}$, realising what is the behaviour of $p_*$ or $p^*$ for me it's completely impossible.

So I thought about it and realised that maybe there is an easier (but surely sloppier) argument: Recall that $H^*(\mathbb{C}P^{2n};\mathbb{Z}) \cong \mathbb{Z}[\alpha]/(\alpha^{n+1}), |\alpha|=2$. Due to the fact that $H^{4n-1}(\mathbb{C}P^{2n};\mathbb{Z})=0$, then $H^{4n}(\mathbb{C}P^{2n};\mathbb{Z})\cong \hom_{\mathbb{Z}}(H_{4n}(\mathbb{C}P^{2n}); \mathbb{Z})$, so we can choose the generator of $H^{4n}(\mathbb{C}P^{2n};\mathbb{Z})$ as the dual of $[\mathbb{C}P^{2n}]$, and we can check wether $(\alpha^{n}\smile \alpha^{n})\frown [\mathbb{C}P^{2n}]$ is $1$ or $-1$, i.e. we check if $\alpha^{2n}:=\alpha^{n}\smile \alpha^{n}$ is the dual of the fundamental class or not. in the first case, $\sigma(\mathbb{C}P^{2n})=1$, in the other case $\sigma(\mathbb{C}P^{2n})=-1$. So it's just a matter of wether the element $\alpha$ "induces" the chosen orientation or not.

I'm aware that this reasoning is somewhat sloppy, BUT I really didn't find anything else, and I can't prove the passage tom Dieck does with the induced map $p_*,p^*$ in (co)-homology.

So I'm here asking if someone has a better proof of this fact OR can explain why (according to tom Dieck), $p􏰙^*(\alpha) = \alpha_1 + \cdots + \alpha_n$, where $\alpha_i$ is the dual of the fundamental class of the $i$-th copy of $\mathbb{C}P^1$ in $(\mathbb{C}P^1)^n$.


1 Answer 1


Let me explain why $p^*(\alpha)=\alpha_1+\dots+\alpha_n$. We have $H_2((\mathbb{C}P^1)^n)=\mathbb{Z}^n$, with basis $(i_k)_*(x)$, where $i_k:\mathbb{C}P^1\to (\mathbb{C}P^1)^n$ is the inclusion of the $k$th factor and $x\in H_2(\mathbb{C}P^1)$ is the fundamental class. Explicitly, choosing $[0,1]$ as a basepoint for $\mathbb{C}P^1$, the map $i_k$ is defined by $i_k([a,b])=([0,1],\dots,[0,1],[a,b],[0,1],\dots,[0,1])$, where $[a,b]$ is in the $k$th coordinate. The classes $\alpha_k\in H^2((\mathbb{C}P^1)^n)$ are then just the dual basis to the classes $(i_k)_*(x)$. Now the composition $pi_k:\mathbb{C}P^1\to\mathbb{C}P^n$ can be explicitly computed to be given by the formula $pi_k([a,b])=[0,0,\dots,0,a,b]$ for each $k$. That is, $pi_k$ is just the standard inclusion of $\mathbb{C}P^1$ into $\mathbb{C}P^n$. In particular, this means that $p_*(i_k)_*(x)$ is the same for every $k$ and is a generator of $H_2(\mathbb{C}P^n)=\mathbb{Z}$. The class $\alpha\in H^2(\mathbb{C}P^n)$ then satisfies $\alpha(p_*(i_k)_*(x))=\pm 1$; let's suppose $\alpha(p_*(i_k)_*(x))=1$ (if not, we could just change which generator of $H^2(\mathbb{C}P^n)$ we're calling $\alpha$).

We now have $\alpha(p_*(i_k)_*(x))=1$ for each $k$, or equivalently, $p^*(\alpha)((i_k)_*(x))=1$ for each $k$. Since the classes $\alpha_k$ are the dual basis to the classes $(i_k)_*(x)$, this means that $$p^*(\alpha)=\sum_k p^*(\alpha)((i_k)_*(x))\alpha_k=\sum_k\alpha_k.$$

In short, the map $p$ might be hard to understand, but its compositions with the inclusions $i_k$ turn out to be very easy to understand, and this is all you need to compute $p^*(\alpha)$.

  • $\begingroup$ wow, you really solve every doubts I could have with this map. Thanks! As a side-note, do you know where this map $p$ comes from? It kind of reminds me the Veronese embedding, even though it is different (not even an embedding I fear). Is it "famous"? or it is something build exactly for this proof? $\endgroup$
    – Luigi M
    Oct 11, 2015 at 20:13
  • $\begingroup$ The map $p$ is well-known, though I don't know of any particular name for it. It is the quotient map for the group action of $\Sigma_n$ permuting the coordinates of $(\mathbb{C}P^1)^n$, and thus identifies $\mathbb{C}P^n$ with the symmetric power $Sym^n(\mathbb{C}P^1)$. $\endgroup$ Oct 11, 2015 at 20:23

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