# Euler characteristic of complex projective plane

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How to compute $\chi(\mathbb{C}\mathrm{P}^2)$?

This problem is from a class on differential topology, so we have defined the Euler characteristic as the sum of the indices of isolated zeros on a non-vanishing vector field. Off the top of my head, I cannot think of any theorems which really help with this computation, so I am thinking I need to do this by finding a sufficiently nice vector field on $\mathbb{C}\mathrm{P}^2$ and the just calculating the indices of the isolated zeros by hand. Could someone help get me started with this?

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quasar987
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You know that CP^2 is the quotient of S^3 by S^1. Do you know of any nonvanishing vector field on S^3? It is easy to construct 3 of them. Check wheter any of them descends on CP^2 to a nonvanishing vector field.

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You know that CP^2 is the quotient of S^3 by S^1. Do you know of any nonvanishing vector field on S^3? It is easy to construct 3 of them. Check wheter any of them descends on CP^2 to a nonvanishing vector field.
I thought the identification was $\mathbb{C}\mathrm{P}^2 \cong S^5/S^1$, but maybe I am remembering wrong. In any case, this does help me find a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros.

quasar987
Homework Helper
Gold Member
You're remembering right. But yeah, in any case, my idea was just this: there are "natural" nonvanishing vector field on S^n for n odd. Check whether by any chance any of them descends on CP² to one of the kind that helps you to compute the Euler caracteristic.

mathwonk
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CP^2 is obtained by attaching a single 4-cell to CP^1. Does that help?

mathwonk
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Oh you have a vector field definition of the euler characteristic. ok. think of a triangulation of the spherical earth say as an icosahedron. then poke in the center of every face, so it becomes a valley, and raise up every vertex so it becomes a mountain peak. Then the edges sag in the middle as paths joining the mountain peaks.

now the gradient vector field for the height above sea level function has zeroes at each valley, each peak, and each mountain "pass" between two peaks. Moreover the valleys are local minima, the peaks are loical maxima, and the passes are saddle points. Thus the index of the vector field = V-E+F = #vertices - #edges + #faces. the usual euler characteristic.

eventually it can be deduced that also the euler characteristic equals the alternating sum of ( (-1)^k times) the number of k cells needed to construct the space as either a simplicial complex or a cell complex. Since the complex projective space is made by adding to real 4 space a complex projective line at infinity, it is constructed from adding a 4 cell to a 2-sphere.

I.e. it takes one 0-cell, one 2-cell, and one 4 cell. Since these are all even dimensional cells, so the answer is 3.

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mathwonk
Homework Helper
here's another computation via a useful theorem, the Riemann Roch theorem for surfaces. If D is any divisor, i.e. curve on the plane, then

chi(D) - chi(O) = (1/2)(D.[D-K]) where e is the euler characteristic of the surface,

chi is the alternating sum of the dimensions of the cohomology groups, K is the divisor of a differential form, and " . " means intersection number,.

In this case a differential form like dz has a pole only at infinity, and there of order three, so K = -3H where H is the class of a line. Moreover if we take D = H, then H is an ample divisor so chi(H) = # independent linear polynomials.

Moreover, chi(O) = (1/12)(K^2 + e), intersection number.

So we get 3 - (1/12)(9+e) = (1/2)(H.4H) = 2, or 3-2 = 1 = (1/12)(9 + e). So e = 3.

I hope I got this right.

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mathwonk