# Wednesday, February 24 Last time: \[ \chi(C, L) &= h^0(C, L) - h^1(C, L) \\ &= h^0(C, L) - h^0(C, L ^{-1} \tensor K_C) \\ &= \deg L + 1 -g ,\] which is determined by purely topological information. We can generalize this to arbitrary ranks of the bundle and arbitrary dimensions of manifold: :::{.theorem title="Hirzebruch-Riemann-Roch (HRR) Formula"} Let $X$ be a compact complex manifold and let \( \bundle{E} \to X \) be a holomorphic vector bundle. Then \[ \chi( \bundle{E} ) = \int_C \ch( \bundle{E} ) \td(X) .\] The constituents here: - The **Chern character**, summed over $R$ the *Chern roots*, which is in mixed cohomological degree. \[ \ch( \bundle{E} ) \da \sum_{x_i \in R} e^{x_i} = \ch_0( \bundle{E} ) + \ch_1( \bundle{E} ) + \cdots + \ch_i( \bundle{E} ) \in H^{2i}(X; \QQ) .\] - The **Todd class**, defined as \[ \td( F) \da \prod_{x_i \in R} {x_i \over 1 - e^{-x_i} } \] where $\td(X) \da \td(TX)$ is viewed as a complex vector bundle, which is again in mixed cohomological degree. ::: :::{.remark} Note that integrating over cohomology classes in mixed degree is just equal to the integral over the top degree terms. Applying this to $X = C$ a curve and \( \bundle{E} \da \OO \), we obtain \[ \chi(C, \OO) = \int_C \ch( \OO) \td(C) .\] We have - $\ch(\OO) = e^{c_1(\OO)} = e^0 = 1$ - $\td(C) \da \td(TC) = c_1(TC) / (1- e^{ - c_1(TC) } )$, whose Taylor coefficients are the Bernoulli numbers. We can expand $x/(1 -e^{-x}) = 1 + (x/2) + (x^2/12) - x^4(720) + \cdots$, and since terms above degree 2 vanish, we have \[ \cdots &= \int_C 1 + \qty{ 1 + {c_1(TC) \over 2} } \\ &= \int_C \qty{c_1(TC) \over 2 }\\ &= {1\over 2} \chi_\Top(C) && \text{Chern-Gauss-Bonnet} \\ &= {2-2g \over 2} \\ &= 1-g .\] We thus obtain \[ \chi(C, L) &= \int_C \ch(L) \td(C) \\ &= \int_C (1 + c_1(L) ) \qty {1 + {c_1(L) \over 2} }\\ &= \int_C c_1(L) + {c_1(TC) \over 2} \\ &= \deg L + 1-g .\] ::: :::{.remark} Note that this is a better definition of genus than the previous one, which was just the correction term in Riemann-Roch. Here we can define it as $g \da h^1/2$. ::: :::{.exercise title="?"} Try to state and prove a Riemann-Roch formula for vector bundles on curves. ::: :::{.proposition title="Formula for Euler characteristic of a line bundle on a complex surface"} Let $S$ be a compact complex surface, i.e. $S\in \Mfd_\CC^2$. An example might be $C\cross D$ for $C,D$ two complex curves, or $\CP^2$. Let $L\to S$ be a holomorphic vector bundle. Then \[ \chi(L) = \chi(\OO_S) + {1\over 2} \qty{ L^2 - L \cdot K} .\] Note that $L^2 \da \int_S c_1(L) c_1(L)$ is just shorthand for taking the intersection of $L$ with itself. Recall that $K \da \Omega_S^2$ is the space of holomorphic top forms. ::: :::{.proof title="?"} Let $x_1, x_2$ be the Chern roots of $TS$. By HRR, we have \[ \chi(L) &= \int_S \ch(L) \td(S) \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{ {x_1 \over 1 - e^{-x_1} } {x_2 \over 1-e^{-x_2}} }\\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + {x_1 \over 2} + {x_1^2 \over 12} }\qty{ 1 + {x_2 \over 2} + {x_2^2\over 12}} \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + {x_1 + x_2 \over 2} + {x_1^2 + x_2^2 + 3x_1 x_2 \over 12} } \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + {c_1(x_1, x_2) \over 2} + {c_1(x_1, x_2)^2 + c_2(x_1, x_2) \over 12 } } \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + { c_1(T) \over 2} + {c_1(T)^2 + c_2(T) \over 2 } } \\ &= \int_S {c_1(L)^2 \over 2} + {c_1(L) c_1(T) \over 2} + {c_1(T)^2 \over 2} + {c_2(T) \over 12} \quad \text{Take deg 4} \\ &= \int_S \qty{ c_1(L)^2 + c_1(L) c_1(T) \over 2} + \chi(\OO_S) \quad \text{HRR on last two terms} .\] where we've applied HRR to $\OO_S$. It remains to show that $c_1(T) = -c_1(K)$. We have \[ K = \Omega_S^2 = \Wedgepower^2 T\dual .\] Note that $\Wedgepower^{\text{top}} \bundle{E} \da \det( \bundle{E} )$ for any bundle $\bundle{E}$ since this is a 1-dimensional bundle. We have $c_1(T) = -c_1(T\dual)$ since the Chern roots of $T\dual$ are $-x_1, -x_2$. So it suffices to show $c_1(T\dual) = c_1(K)$, but there is a general result that $c_1(\bundle{E}) = c_1( \det \bundle{E} )$. This uses the splitting principle $\bundle{E} = \bigoplus_{i=1}^r L_i$ with $x_i = c_1(L_i)$. We have $c_1(\bundle{E}) = \sum x_i$ and $\det\bundle{E} = \bigotimes_{i=1}^r L_i$, so $\sum x_i = c_1(L_1\tensor \cdots \tensor L_r)$. ::: :::{.remark} We want to use the following formula: \[ \chi(S, L) = \chi(\OO_S) = {1\over 2}(L^2 - L\cdot K) .\] This requires knowing $\chi(\OO_S)$. Applying HRR yields \[ \chi(\OO_S) &= \int_S {c_1(T)^2 + c_2(T) \over 12}\\ &= \int_S { (-c_1(K))^2 + c_2(T) \over 12}\\ &= {K^2 + \displaystyle\int_S c_2(T) \over 12} ,\] so we just need to understand $\int_S c_2(T)$. But for $n=\rank \bundle{E}$, $c_n( \bundle{E} )$ (the top Chern class) is the fundamental class of a zero locus of a section of $\bundle{E}$. Note that $S \in \Mfd_\RR^4$ is oriented, so $\int_S c_2(T)$ is the signed number of zeros of a smooth vector field. ![image_2021-02-25-20-42-49](figures/image_2021-02-25-20-42-49.png) \todo[inline]{Check.} Looking at the tangent bundle of the surface, the local sign of an intersection will be the number of incoming directions $\mod 2$, i.e. the index of the critical point. Then the signed number of zeros here yields $1-6+1 = -4 = \chi_{\Top}(C)$. More generally, we have \[ \chi_{\Top}(M^n) = \int_C c_{n}(TM) ,\] the **Chern-Gauss-Bonnet** formula. We can thus write \[ \chi(\OO_S) = {K^2 + \chi_\Top(S) \over 12 } .\] :::