⇤Math diary
May 9, 2024: Riemann roch
I’ve passed my quals (yayyy), so I’m back in math writing mode. Im working on a problem about using geometric quantization to access subtle invariants of symplectic manifolds. I have a big question mark around the metaplectic correction – I think this is the key to unlocking other mysteries for me. For example, why does the half sum of positive roots appear everywhere in representation theory? How can spinors help us in geometric quantization?
Riemann-Roch
The Riemann Roch formula is primary tool for computing cohomology of line bundles on Riemann surfaces. If $L$ is a line bundle on a Riemann surface $\Sigma$ with genus $g$, then
\[h^0(L) - h^1(L) = d - g+1\]Let’s prove Riemann-Roch! First, a holomorphic line bundle $L$ on $\Sigma$ can be obtained by twisting the structure sheaf $\mathcal{O}$ of $\Sigma$ at the divisor of $L$. Choose some meromorphic section of $L$ , and let $P$ be the divisor of poles and $Z$ the divisor of zeros. Then, the divisor associated to $L$ is $P-Z$, which means $L = \mathcal{O}(P)\otimes\mathcal{O}(-Z)$. We have the short exact sequences of sheafs
\[\begin{align} \mathcal{O}(-P) &\to \mathcal{O} \to \mathcal{O}_P \\ \mathcal{O}(-Z) &\to \mathcal{O} \to \mathcal{O}_Z \end{align}\]where $\mathcal{O}_P,\mathcal{O}_Z$ is the skyscraper sheaf at the divisor $P,Z$. Essentially, they record a complex number for each point on the divisor. For a section of $\mathcal{O}$ (a holomorphic function), we get a section of $\mathcal{O}_P$ by simply evaluating the function at each point along the divisor. We call the map $\mathcal{O} \to \mathcal{O}_P$ restriction. The kernel of restriction is exactly the holomorphic functions with zeros along $P$, which is encoded in the sheaf $\mathcal{O}(-P)$.
To get our line bundle to appear, we tensor the top and bottom by $\mathcal{O}(P)$:
\[\begin{align} \mathcal{O} &\to \mathcal{O}(P) \to \mathcal{O}_P \otimes \mathcal{O}(P)\\ \mathcal{O}(P-Z) \cong L &\to \mathcal{O}(P) \to \mathcal{O}_Z \otimes \mathcal{O}(P) \end{align}\]Next we take the holomorphic Euler characteristic. For a sheaf $F$, we define
\[\chi(F) = h^0(F) - h^1(F) + h^2(F) - \dots\]The Euler characteristic is useful because it is additive: for any short exact sequence of sheaves, we get an identity on their euler characteristic:
\[\begin{align} F &\to G \to H \\ \chi(F) - & \chi(G) +\chi(H) = 0 \end{align}\]This is not mysterious, if you believe that short exact sequences induce a long exact sequence on homology. the alternating sum of the dimensions of the long exact sequence must vanish. Rearranging the terms, we get the identity above.
Applied to the short exact sequences defining the line bundle $L$, we get
\[\begin{align} \chi(L) &= \chi(\mathcal{O}_Z\otimes \mathcal{O}(P)) - \chi(\mathcal{O}(P))\\ \chi(\mathcal{O}) &= \chi(\mathcal{O}_P\otimes \mathcal{O}(P)) - \chi(\mathcal{O}(P)) \end{align}\]Rearranging once again, we get the formula for the Euler characteristic of the line bundle
\[\chi(L) = \chi(O) + \chi(\mathcal{O}_Z\otimes \mathcal{O}(P)) - \chi(\mathcal{O}_P\otimes \mathcal{O}(P))\]This form of Riemann-Roch is very general. It applies for line bundles on any any dimensional manifold. We get explicit forms for Riemann-Roch by evaluating the terms on the right hand side. For Riemann surfaces, this is a quick application of Hodge theory. On surfaces, things are somewhat more involved (we need the adjunction formula and the Noether formula), but we can still be explicit. But let’s focus on Riemann surfaces.
Riemann surfaces are dimension 1, so the zero/pole locus of a section must have dimension 0. These subvarieties $Z$ and $P$ are just collections of points, and the sheafs $\mathcal{O}_P$ are skyscraper sheaves. First, note that the skyscraper sheaf can’t see the twisting of the line bundle, so $\mathcal{O}_Z\otimes \mathcal{O}(P) = \mathcal{O}_Z$. More conceptually, $\mathcal{O}_Z\otimes \mathcal{O}(P)$ is the line bundle $\mathcal{O}(P)$ restricted to the subvariety $Z$. This is a zero dimensional subvariety, so all line bundles are necessarily trivial. On the zero dimensional variety $Z$, all cohomology groups $H^i$ for $i\geq 1$ necessarily vanish. So, the Euler characteristic of $\mathcal{O}_Z$ computes exactly the dimension of the space of global sections of $\mathcal{O}_Z$. We get one complex number per point in $Z$, so $\chi(\mathcal{O}_Z) = h^0(\mathcal{O}_Z) = \vert Z \vert$, the number of points in $Z$. Likewise, $\chi(\mathcal{O}_P\otimes \mathcal{O}(P)) = h^0(\mathcal{O}_P) = \vert P\vert$. All together,
\[\chi(L) = \chi(O) + (\vert P\vert - \vert Z\vert )\]The number $\vert P\vert-\vert Z\vert$ is the degree of the line bundle $L$, which agrees with its first Chern class $c_1(L)$. Next we compute the euler charecteristic of the structure sheaf on a Riemann surface $\Sigma$. Because $\Sigma$ is 1 dimensional, all cohomology in degree more than 2 vanishes. So, there are only two things we need to compute:
$h^0(\Sigma,\mathcal{O}) = 1$
Since $\Sigma$ is compact, every global holomorphic function must be constant. Constant functions span $H^0(\Sigma,\mathcal{O})$, so this is a 1 dimensional space.
$h^1(\Sigma,\mathcal{O})=g$
This is a result of hodge theory. By the Dolbeaut theorem, we know $h^1 = h^{0,1}$ By the hodge theorem, we can represent these classes by harmonic forms. These are persevered by complex conjugation, so the $1,0$ cohomology and the $0,1$ cohomology groups agree. By the hodge decomposition, \(2h^{0,1} = h^{0,1} + h^{1,0}= h^{1}_{\mathrm{dR}}\) By the de-Rahm theorem, $h^{1}_{\mathrm{d}} = \dim H^1(\Sigma, \mathbb{Z})= 2g$. Therefore, $h^1 = g$.
Putting the pieces together, $\chi(\mathcal{O}) = 1-g$, so
\[\boxed{\chi(L) = (1-g) + d}\]Generally, computing $h^0$ of a line bundle is easier than computing $h^1$. To make Riemann-Roch practically useful, we need a way to compute $h^1(L)$. The answer comes from Serre duality:
\[h^1(L) = h^0(K\otimes L^*)\]Here $K$ is the canonical bundle, which on the Riemann surface coincides with the holomorphic cotangent bundle $(T^*)^{(1,0)}\Sigma$. Once again, this follows from hodge theory. A section of $h^1(L)$ is an $L$-valued $(0,1)$ form. Taking the hodge dual, we get a $L^*$-valued $(1,0)$ form. Finally, an $L^*$-valued $(1,0)$ form is a holomorphic section of the line bundle $L^* \otimes K$. This gives a formula for the left hand side of Riemann-Roch:
\[\chi(L) = h^0(L) - h^0(K\otimes L^*)\]Can we simplify things?
Everywhere above we get these annoying shifts showing up. In Serre duality, we can’t simply dualize the line bunde, we need to shift by the canonical bundle. For the Riemann-Roch formula, we have to carry around a factor of $\chi(\mathcal{O}) = 1-g$. There are two canonically defined bundles attached to the Riemann surface: The structure sheaf, and the canonical bundle. What if we started somewhere halfway in between?
Let $\theta$ be a line bundle such that $\theta^2=K$, A “square root” of the canonical bundle. Equivalently, $\theta = K \otimes \theta^*$. Such a bundle is called a theta characteristic (hence the symbol). We can phrase the equations above relative to $\theta$. First, Serre duality:
\[h^1(L\otimes \theta) = h^0((L\otimes \theta)^* \otimes K) = h^0(L^*\otimes \theta)\]Serre duality inverts line bundles relative to $\theta$. Now to Riemann-Roch. First, the canonical bundle $K$ has degree $2g-2$. We can see this from Riemann-Roch plus Serre duality. Riemann roch says $\text{deg} (K) = \chi(K) - \chi(\mathcal{O})$. Then, Serre duality says
\[\chi(K) = h^0(K)-h^1(K) = h^1(\mathcal{O}) - h^0(\mathcal{O}) = -\chi(\mathcal{O})\]putting these together, we get $\text{deg}(K)=-2\chi(\mathcal{O}) = 2g-2$. Therefore, since $\theta$ has half the degree of $K$, $\deg{\theta} = g-1$. This cancels out the factor of $1-g$ in Riemann Roch:
\[\chi(L\otimes \theta) = 1-g+\text{deg}(L\otimes \theta) = \text{deg}(L)\]Soon I want to explain why $\theta$ plays such a distinguished role. Short answer is, $\theta$ defines a spin structure, and the sections of $L\otimes \theta$ are harmonic spinors. More in next week!