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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.

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    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,2012-05-20

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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.