The Degree of the Tangent Bundle for a Riemann Surface

A well known result states that the degree of the tangent bundle $TX$ of a Riemann Surface $X$ of genus $g$ is exactly $2-2g$.

In my mind the genus is intuitively the number of "handles" of the surface, and precisely it is the dimension of the $\mathbb{C}$-vector space $\Omega_1(X)$ of holomorphic 1-forms on $X$. Googling around I noticed that the number $2-2g$ is the Euler characteristic of the surface.

Question 1 (Answered in the comments) I checked the relation for some small-genus surfaces by finding an explicit triangulation, but how can I prove it?

Then it comes the definition of degree of a line bundle $L$ over $X$. I can view it in two different, but equivalent, ways. On one hand it is the degree of the associated divisor on $X$ as in Hartshorne, on the other hand it is the "weighted" sum of zeroes of a general section, as in page 16 of those lecture notes. It is clear why the two definitions are the same and are independent from the choice of the general section.

The unique way I see for computing the degree of a line bundle is to find a "smart" section, but I failed to find one in this case of the tangent bundle.

Question 2 There are other effective ways to compute it? Otherwise what could be a good choice for a section?

I tried to keep the question as more general as possible because I´m more interested in a clarification of the notions involved than in a short proof of the statement of the title. Also hints or suggestions are more than welcome!

Thank you for your time!

• For Question1: First you need to prove that the Euler characteristic is a toplogical invariant (in particular does not depend on the chosen triangulation). Then, calculate the Euler characteristic of a torus (you said you did it). Finally, note that if $S_1,S_2$ are two surfaces and you remove a maximal simplex $\sigma_1,\sigma_2$ of each of them and then you identify $\partial \sigma_1$ with $\partial \sigma_2$, you obtain a surface of bigger genus. Then you can prove it by induction – user17786 Jan 25 '12 at 14:20
• @User17786, First thank you for the suggestion! I have some questions on it. By topological invariant you mean that the Euler characteristic is homotopically invariant, am I right? I did steps 2 and 3 as you suggested, but in order to proceed by induction I think I need a statement like "two surfaces with the same genus are homotopically equivalent". Unfortunately I guess this is not true unless we assume something like the compactness of our surfaces. Does it hold a statement like the previous one? Does it require compactness or other additional suppositions? Thank you again! – Giovanni De Gaetano Jan 25 '12 at 14:39
• Looking for a proof of the invariance by homotopy of the Euler characteristic I found the very useful paragraph 8 of Chapter I of Massey´s "A Basic Course in Algebraic Topology". It answers completely Question 1, once given the suggestion of User17786. In specific two compact surfaces with the same Euler characteristic are homotopically equivalent and the non-compact case is non-sense since the Euler characteristic is not defined in that case. – Giovanni De Gaetano Jan 25 '12 at 14:55
• If you want a nice section, you can put your genus $g$ surface in $R^3$ as a ladder (the holes are all above each other). Than the gradient flow of the height function is a very nice section of the tangent bundle. It has two zero's at each hole, plus the top and the bottom of the surface. – Thomas Rot Apr 21 '15 at 11:13

First let us get rid of non-compact Riemann surfaces: every holomorphic vector bundle of any rank on such a surface is holomorphically trivial and this seals the fate of its tangent bundle: it is trivial.

From now on, we assume the Riemann surface $X$ is compact.
Here are three useful ways to compute its genus $g$ (which is equivalent to computing the degree of its tangent bundle because of the formula $deg (T_X)=2-2g$ you mentioned) :

I) If the Riemann surface $X\subset \mathbb P^2( \mathbb C)$ is the smooth zero locus of a homogeneous polynomial $f(x_0,x_1,x_2)\in \mathbb C[x_0,x_1,x_2]$ of degree $d$ whose partial derivatives $\frac{\partial f}{\partial x_i}(x_0,x_1,x_2)$ don't vanish simultaneously on $X$ then the genus $g$ of $X$ is $$g= \frac{(d-1)(d-2)}{2}$$

II) Suppose the Riemann surface $X\subset \mathbb P^3( \mathbb C)$ is the complete intersection of two surfaces given by the homogeneous polynomials$f(x_0,x_1,x_2,x_3)$ and $g(x_0,x_1,x_2,x_3)$ of degrees $d,e$ whose gradients are linearly independent along $X$. Then $X$ has genus $$g=1+ \frac{de(d+e-4)}{2}$$ Edit (March 2017)
This formula can be obtained by remembering that the genus of X is equal to its arithmetic genus $p_a=1-P_X(0)$, where $P_X(t)$ is the Hilbert polynomial of X. For the calculation of that polynomial, see Hassett page 210. See also Hartshorne, Exercise 7.2.(d). page 54.

III) If $\phi:X\to Y$ is a ramified covering of degree $d$ between two compact Riemann surfaces, with $r$ ramification points (counted with multiplicities), then their genera are related by the Riemann-Hurwitz formula : $$g(X)= 1+ \frac{r}{2} +d\cdot(g(Y)+1)$$

Bibliography
The astonishing result that on a non-compact Riemann surface every holomorphic vector bundle of any rank is holomorphically trivial is proved, with all necessary prerequisites, in the fundamental reference O.Forster, Lectures on Riemann Surfaces , Theorem 30.3.
(Note that if you replace "holomorphic" by "algebraic" the result becomes completely false: already non trivial algebraic line bundles are plentiful on non-complete non-rational algebraic curves)
The Riemann-Hurwitz formula is proved on page 140 of the same book.

For calculations of group actions on a Riemann Surfaces and the associated ramified covering, in relation to Riemann-Hurwitz, I recommend R.Miranda, Algebraic Curves and Riemann Surfaces , Ch.III, §3 .

Edit: An example
As an illustration of II), take the quadrics $x_0x_2-x_1^2=0$ and $x_0x_1+x_1x_2-x_3^2=0$ in $\mathbb P^3(\mathbb C)$ . Their intersection in $\mathbb P^3(\mathbb C)$ is a smooth curve of genus $1$ and the last chapter of Harris's Algebraic Geometry will show you that it is isomorphic to the curve $y^2z-x^3-xz^2=0$ in $\mathbb P^2(\mathbb C)$, which indeed has genus $1$ by I).

• Thank you for your answer! Do you have a reference for the first statement (regarding non-compact surfaces)? For the computation of the genus, the Riemann Surfaces in my setup arose as quotient of the (compactified) upper half plane by a Fuchsian group. It seems to me that the only criteria which could work in this case is the third one, but I don´t know how to use it. Any suggestions? – Giovanni De Gaetano Jan 26 '12 at 15:59
• Dear @Student73, I have added a bibliography addressing your requests. I think Miranda's book will interest you: it doesn't mention Fuchsian groups, but the quotients he computes might serve as a warm-up to your problem. – Georges Elencwajg Jan 26 '12 at 19:51
• @Student: I think Milne's free online course on modular functions is what you want about Fuchsian functions – Georges Elencwajg Jan 26 '12 at 22:39
• Dear Georges, I was reading your excellent answer and I have a question : could you provide a reference how to get your formula II) ? Many thanks in advance ! – user171326 Mar 5 '17 at 13:24
• Dear @N.H. I have added a little bibliographic edit for formula II). – Georges Elencwajg Mar 5 '17 at 19:14