Mrthfx

As the question title asks for, how do others visualize the Riemann-Roch theorem with complex analysis or geometric topology considerations? That is all Riemann would have had back in the day, and he reminds me of Thurston in the sense he just drew a picture and then the proof popped out. Nowadays most formulations of Riemann-Roch are couched in algebraic language and do not invoke any geometric intuition for the complex plane, so I am asking here. Bonus points for pictures.

I'll try to give this a shot, but I'm coming from an algebraic geometry perspective. In the case of curves, it's not so clear to me what the line between algebraic geometry and complex geometry is, so I'm not sure if what I'm saying is what you're looking for. Anyways, let's assume that our curve $C$ is not hyperelliptic because I want to use the canonical embedding because embeddings are easier to visualize for me. The Riemann-Roch theorem says for a divisor $D$ on $C$ that
$$
h^0(D)=1-g+deg(D)+h^0(K-D)
$$
We know that $H^0(K-D)$ is measuring sections of the canonical divisor $K$ that vanish on $D$. But in the canonical embedding, $H^0(C,K)=H^0(mathbb{P}^{g-1},mathcal{O}(1))$. Hence, sections of the canonical divisor are just linear homogeneous polynomials on a projective space. Therefore, $H^0(K-D)$ is just hyperplanes in the canonical embedding that contain $D$. Now consider the linear span of $D$, which is some projective space $mathbb{P}^kcong text{Span}(D)$. The linear polynomials that vanish on $D$ are the same as the linear polynomials vanishing on its span, $mathbb{P}^{k}$, and we can compute the number of these. There are $g-1-k$ of them. Plugging back into Riemann-Roch, we get
$$
h^0(D)=deg D-dim text{Span}(D)
$$
I think this statement is more amenable to visualizing because the right hand side just asks you to place a bunch of points on a curve in projective space, and then take their linear span. I have no idea if this is how Riemann thought about it, but it's certainly geometric. If you want to read more, I learned about this from Geometry of Algebraic Curves Volume 1 by Arbarello, Cornalba, Griffiths, and Harris.

Here is my "visual" coming from the complex (and functional) analysis perspective, due to a novel proof by Taubes. For a holomorphic line bundle $E$ over a genus $g$ surface $C$, the RR theorem states that $ind_mathbb{C}(barpartial)=c_1(E)+1-g$.

Because the index of a Fredholm operator over a closed manifold does not change if we perturb by a 0th order (compact) operator, $ind_mathbb{R}(barpartial)=ind_mathbb{R}(barpartial+rB)$ for every $rinmathbb{R}_+$ and generic section $BinGamma(E^2otimes T^{0,1}C)$. Here $B$ acts as a complex anti-linear map in order to define the perturbation of $barpartial:Gamma(E)toGamma(Eotimes T^{0,1}C)$.

Now we can visualize the (co)kernel of our perturbed operator as $rtoinfty$, because the elements of the (co)kernel localize/concentrate around the zeros of $B$. Thus $ind_mathbb{R}(barpartial)=# B^{-1}(0)$.

Now we can visualize and appeal to the "geometry" of characteristic classes and sections of bundles: $# B^{-1}(0)=c_1(E^2otimes T^{0,1}C)$, which equals $2c_1(E)+chi(C)$.