Let $X$ be a compact Kahler manifold. We denote it's topological Euler characteristic by $\chi$ and it's holomorphic Euler characteristic by \begin{align*} \chi(\mathcal{O})=\sum\limits_i (-1)^i\text{ dim}H^i(X;\mathcal{O}) \end{align*} Here $\mathcal{O}$ denotes the sheaf of holomorphic functions and $H^i$s are corresponding Cech cohomologies. I know that for a compact Riemann surface $X,\chi=2-2g$ and we also have \begin{align*} &\chi(\mathcal{O})=\sum\limits_i (-1)^i\text{ dim}H^i(X;\mathcal{O})=\text{ dim}H^0(X;\mathcal{O})-\text{ dim} H^1(X;\mathcal{O})=1-g\\ &\Rightarrow \chi=2\chi(\mathcal{O}) \end{align*} Is there any such relation which is true in general for any compact Kahler manifold or may be a projective hypersurface?
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1Not an answer, but if $X$ is a smooth projective hypersurface of degree $d$ and dimension $n$, its topological Euler characteristic is $\frac{1}{d}[(1-d)^{n+2} - 1 + (n+2)d]$. If $X$ is Fano, hwever, (i.e., $d \leq n$ in the case of a smooth projective hypersurface), the holomorphic Euler characteristic is $1$. – Andrew D. Hwang May 24 '21 at 15:37
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@AndrewD.Hwang can you please give a reference for the genus formula? Thanks. – Partha May 24 '21 at 16:01
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One place is Proposition 10.7 of A beginner's guide to holomorphic manifolds (page 113). – Andrew D. Hwang May 24 '21 at 16:50
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The topological Euler characteristic $\chi$ and the Euler number of the trivial line bundle $\chi(\mathcal O_X)$ do not seem to be strictly related by a simple equation as in the case of a curve. But there are indeed "some" relations between them.
First, the topological Euler characteristic is more related to Euler classe of the tangent bundle. In fact, for an n-dimension compact complex manifold $X$, the Poincaré-Hopf theorem shows that $$ \chi = \int_Xe(T_X)= \int_Xc_n(T_X), $$ where $e(T_X)$ is the Euler class of $T_X$.
On the other hand, we have the Hirzebruch-Riemann-Roch theorem which gives us a relation between the Euler number of the trivial line bundle and "topological informations" : $$ \chi(\mathcal O_X)=\int_X ch(\mathcal O_X)Td(T_X) $$ where $ch(\mathcal O_X$ is the Chern character of $O_X$ and $Td(T_X)$ is the Todd class of $T_X$. In developing $ch(\mathcal O_X)$ and $Td(T_X)$ explicitly, we find $$ \chi(\mathcal O_X)=\int_X Td_n(T_X) $$ where $Td_n(T_X)$ is the top degree direct summand of $Td(T_X)$ (in $H^{2n}(X)$). In general, the expression of $Td_n(T_X)$ can be really complicated. I list below the expression of $Td_n(T_X)$ when $n$ is relatively small :
- $n=1, Td_1(T_X)=c_1(T_X)/2$;
- $n=2, Td_2(T_X)=\frac{c_1(T_X)^2+c_2(T_X)}{12}$;
- $n=3, Td_3(T_X)=\frac{c_1(T_X)c_2(T_X)}{24}$;
- $n=4, Td_4(T_X)=\frac{-c_1^4+4c_1^2c_2+c_1c_3+3c_2^2-c_4}{720}$;
- ...
Hence, in general, $\chi(\mathcal O_X)$ and $\chi$ do not have some explicit easy relation: the former often contains information of the Chern classes of $T_X$ of all degrees but the latter is roughly the top degree Chern class of $T_X$.
But we do have some interesting relations if we know other conditions. For example, when the dimension $n=1$, since $Td(T_X)=c_1(T_X)/2$, we see that $\chi=2\chi(\mathcal O_X)$, as you noticed in the description of the question.
When the dimension $n=2$, if we know what $c_1(T_X)(=-\deg(K_X))$ is, then we have the simple relation $$ \chi(\mathcal O_X)=\chi/12+\deg(K_X)^2/12 $$ where $K_X$ is the canonical bundle of X. For example, if $X$ is a K3 surface, or a two-dimension complex torus, we see that $\chi(\mathcal O_X)=\chi/12$.
When the dimension $n=3$, things seem interesting : it seems there are no relations between $\chi(\mathcal O_X)$ and $\chi$, at all, since in this case, $\chi(\mathcal O_X)=\int_X\frac{c_1(T_X)c_2(T_X)}{24}$ whereas $\chi=\int_Xc_3(T_X)$.
Pène Papin
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