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!
Best Answer
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).