# Compute $\chi(\mathbb{C}\mathrm{P}^2)$.

I need to compute $\chi(\mathbb{C}\mathrm{P}^2)$ using techniques from differential topology. I cannot think of any theorems that are particularly useful for this computation, so I think that I will have to find a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros and compute the index of the vector field about these zeros. My first idea to find a vector field with isolated zeros was to recall the diffeomorphism $\mathbb{C}\mathrm{P}^2 \cong S^5/\sim$ where $(z^1,z^2,z^3) \sim (u^1,u^2,u^3)$ if and only if there exists $w \in S^1$ with $(z^1,z^2,z^3) = (wu^1,wu^2,wu^3)$. Then it would suffice to find a vector field on $S^5$ which descends to a vector field on $S^5/\sim$ with isolated zeros. However, I have had some difficulties making this approach work, so I was hoping that someone could help me out here.

Edit 1: I should add that while I am limited to the tools of differential topology for this problem, I do not have to follow the outline I have thus far; that is, I can find a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros and compute its Euler characteristic from there in anyway (the vector field does not have to come from $S^5$).

Edit 2: I am also not limited to computing the Euler characteristic directly from a vector field with isolated zeros. For example, I can use things like the Gauss mapping too.

-
Just to be sure: the standard (easy) technique to compute $\chi(\mathbb{C} P^n)$ is to note that it has a cell decomposition with exactly one cell in each even dimension $0 \leq 2k \leq 2n$. Then you get $\chi = n+1$ without breaking a sweat. But I presume that this method is not "differential topological" enough, so is off limits? – Pete L. Clark May 20 '12 at 0:10
That is correct. The definition of Euler characteristic that we are working with is the sum of the indices of isolated zeros on a vector field and we are not allowed to use things like cell decompositions, homology groups, etc. – jgensler May 20 '12 at 0:27
OK, I thought so. Sorry to be of such little help (I don't have a differential topological answer to offer you at the moment), but perhaps my comment will forestall other answers not of the sort you're looking for. Good luck! – Pete L. Clark May 20 '12 at 0:29
I don't have time right now to work out an answer to give hints, so let me just make some hopefully helpful off-the-cuff comments. Your initial approach would also be my initial approach, so I think you're on the right track. I'd also try working in projective coordinates or using the cell structure by defining a vector field $X$ on $e^0$ and then looking at obstructions (in $\pi_k$) to extending $X$ over $e^k$ without introducing zeros. – Neal May 20 '12 at 1:45
Dear jgensler, I am guessing that you are the Joshua who posted this same question on MO. Just so you know, it is a matter of ettiquette not to post the same question on MO as here, and also, MO is not intended for homework questions. (If it's just coincidence that the same question appeared on both sites, please accept my apologies.) Regards, – Matt E May 20 '12 at 5:06

Here's a way to, fairly explicitly, get your hands on a vector field on $\mathbb{C}P^2$ with isolated zeroes. It extends nicely to all $\mathbb{C}P^n$s in the following sense: First, it has isolated $0$s at precisely the points of the form $[0:0:...:1:0:...:0]$. Second, on $\mathbb{C}P^1 = S^2$, it's the vector field given by spinning the $S^2$ about an axis and taking the velocity vector field. Third, the vector field given on $\mathbb{C}P^n$ is an extension of the same "nice" one given to an appropriate $\mathbb{C}P^k\subseteq \mathbb{C}P^n$ (given by the first $k+1$ homogeneous coordinates).
Consider the $S^1$ action on $\mathbb{C}P^n$ given by $$z*[z_0:z_1:z_2:...:z_n] = [z_0:zz_1:z^2z_2:...:z^nz_n],$$ where we think of $z\in S^1$ as a unit complex number. To be clear, the power of $z$ on he $k$th homogeneous coordinate is $k$.
Lets figure out the fixed points. First, it's obvious that each of the points where single homogenous coordinate is $1$ and all others are $0$ is a fixed point. So lets show these are the only ones by contradiction.
Assume we have a fixed point with at least 2 nonzero coordinates, $z_i$ and $z_j$. Then, not bothering to write the other coordinates, we get $[z^i z_i: z^j z_j] = [z_i,z_j]$. Equivalently, $[z^{i-j} \frac{z_i}{z_j}:1] = [\frac{z_i}{z_j}:1]$ so $z^{i-j}\frac{z_i}{z_j} = \frac{z_i}{z_j}$. Since $z_i\neq 0$, this implies $z^{i-j} = 1$. But since $i\neq j$, this isn't true of all $z\in S^1$, giving a contradiction.
Finally, to make this into a vector field $X$, take the velocity vector field of this action. More specifically, define $X(p) = \frac{d}{dt}|_{t=0} e^{it}p$. Then, we'll have $X(p) = 0$ iff $p$ is a fixed point of the action, so $X(p)$ has isolated $0$s.