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.
 A: 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.
