Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am trying to compute $\chi(\mathbb{C}\mathrm{P}^2)$ using only elementary techniques from differential topology and this is proving to be trickier than I thought. I am aware of the usual proof for this result, which uses the cellular decomposition of $\mathbb{C}\mathrm{P}^2$ to get $\chi(\mathbb{C}\mathrm{P}^2) = 3$, but I would like to find a proof of this result that relies on concepts like indices of isolated zeros on a vector field. So for the purposes of this question, I would like to utilize the following definition of the Euler characteristic: For a closed orientable manifold $M$ we define $\chi(M) = \sum_i \mathrm{Ind}_{d_i} \mathrm{v}$ where $\mathrm{v}$ is a vector field on $M$ with isolated zeros.

In my first attempt at this problem I thought about finding a vector field on $\tilde{\mathrm{v}}$ on $S^5$, and then using the identification $\mathbb{C}\mathrm{P}^2 \cong S^5/\mathrm{U}(1)$, seeing if $\tilde{\mathrm{v}}$ descended to a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros that lent itself to computing the Euler characteristic. I had difficulty making this work out, so I am unsure if this is a good approach to tackling the problem. Any insights?

share|improve this question
6  
Try the gradient of the function $[a,b,c] \mapsto (|a|^2 + 2|b|^2 + 3|c^2|) / (|a|^2 + |b|^2 + |c|^2)$. Its critical points are at the coordinate points, with all indices $1$ (and Morse indices $0,2,4$). –  Allen Knutson May 20 '12 at 3:05
3  
Dear Joshua, If you are the same person who posted this question on Math.SE, please note that it is standard etiquette not to post on both sites simultaneously. Also, MO is not a place to ask homework questions. (If it is actually just coincidence that the question appeared on both sites, please accept my apologies.) Regards, –  Emerton May 20 '12 at 5:08

3 Answers 3

up vote 3 down vote accepted

There is a canonical way to construct holomorphic vector fields on $\mathbb{C}P^2,$ and that way is described in Zoladek's "Monodromy Group", page 335. If you read the description, it will be pretty clear what the index is (note that if the vector field is holomorphic, it is given by at most quadratic polynomials, otherwise there are poles. There is a large literature [much of it by Iliashenko] studying such fields and the associated foliations.

share|improve this answer

Take a $3 \times 3$ complex diagonal matrix $A$ with distinct nonzero diagonal entries. The 1-parameter subgroup $exp(At)$ acts on $CP^2$; the fixed points are the lines in $C^3$ containing eigenvectors of $A$. There are $3$ of them and the derivative of the action is a vector field with $3$ zeroes. As the vector field is holomorphic, the index at each zero is $+1$.

share|improve this answer

By the way, a very cool (to my mind) way of computing the Euler characteristic of $\mathbb{C}P^n$ is to treat it as the $n$-fold symmetric product of $\mathbb{C}P^1 = \mathbb{S}^2$ with itself. Then, it is apparently a result of MacDonald (of Atiyah and M fame) that the Euler characteristic of an $n$-fold symmetric product of a space $X$ with itself equals $\binom{\chi(X)+n-1}{\chi(X) - 1}.$`

EDIT Following @Qiaochu's penetrating remark, I looked up MacDonald's paper (I was citing from secondary sources before, shame on me). The result is:

The $k$-th Betti number of the $n$th symmetric power of $X$ is the coefficient of $x^k t^n$ in $\prod_i (1- (-x)^i t)^{- (-1)^i (-B_i)}.$ where the $B_i$ are the betti numbers of $X$. So, for Euler characteristic you evaluate this product at $x=-1,$ and find the coefficient of $t^n.$

share|improve this answer
    
I can't make sense of this expression if $\chi(X) \le 0$. Do you mean ${\chi(X) + n-1 \choose n}$? –  Qiaochu Yuan May 21 '12 at 0:07
    
No, I don't mean that (that would not work for complex projective spaces...), but your point is very well taken. I will go back to the original sources, check it out, and report on the findings... –  Igor Rivin May 21 '12 at 1:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.