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## Five proofs that the Euler characteristic of a closed orientable surface is even

Let $\Sigma_g$ be a closed orientable surface of genus $g$. (Below we will occasionally write $\Sigma$, omitting the genus.) Then its Euler characteristic $\chi(\Sigma_g) = 2 - 2g$ is even. In this post we will give five proofs of this fact that do not use the fact that we can directly compute the Euler characteristic to be $2 - 2g$, roughly in increasing order of sophistication. Along the way we’ll end up encountering or proving more general results that have other interesting applications.

Proof 1: Poincaré duality

A corollary of Poincaré duality is that if $X$ is a closed orientable manifold of dimension $n$, then the Betti numbers $b_k = \dim H^k(X, \mathbb{Q})$ satisfy $b_k = b_{n-k}$. When $n$ is odd, this implies that the Euler characteristic

$\displaystyle \chi(X) = \sum_{k=0}^n (-1)^k b_k$

is equal to zero, since $(-1)^k b_k + (-1)^{n-k} b_{n-k} = 0$. In fact slightly more is true.

Proposition: Let $X$ be a closed manifold of dimension $n$, not necessarily orientable. If $n$ is odd, then $\chi(X) = 0$. If $n$ is even and $X$ is a boundary, then $\chi(X) \equiv 0 \bmod 2$.

Proof. When $\dim X$ is odd, let $\widetilde{X}$ be the orientable double cover of $X$, so that $\chi(\widetilde{X}) = 2 \chi(X)$. By Poincaré duality, $\chi(\widetilde{X}) = 0$, so the same is true for $\chi(X)$. Alternatively, because the Euler characteristic can also be calculated using the cohomology over $\mathbb{F}_2$, we can also use Poincaré duality over $\mathbb{F}_2$, which holds for all closed manifolds since all closed manifolds have fundamental classes over $\mathbb{F}_2$.

When $\dim X$ is even and $X = \partial Y$ is the boundary of a compact manifold $Y$, let $Z = Y \sqcup_X Y$ be the manifold obtained from two copies of $Y$ by gluing along their common boundary. Then $Z$ is a closed odd-dimensional manifold, hence $\chi(Z) = 0$. But

$\displaystyle \chi(Z) = 2 \chi(Y) - \chi(X)$

(e.g. by an application of Mayer-Vietoris), from which it follows that $\chi(X) \equiv \chi(Z) \equiv 0 \bmod 2$. $\Box$

Corollary: The Euler characteristic of $\Sigma_g$ is even.

Proof. $\Sigma_g$ is the boundary of the solid $g$-holed torus. $\Box$

Corollary: No product of the even-dimensional real projective spaces $\mathbb{RP}^{2k}, k \ge 1$ is a boundary.

Proof. Since $\chi(S^{2k}) = 2$ and $\mathbb{RP}^{2k}$ is double covered by $S^{2k}$, we have $\chi(\mathbb{RP}^{2k}) = 1$, hence any product of even-dimensional real projective spaces also has Euler characteristic $1$, which in particular is odd. $\Box$

Corollary: The Euler characteristic $\bmod 2$ is a cobordism invariant.

Proof. Let $X, Y$ be two closed manifolds which are cobordant, so that there exists a closed manifold $Z$ such that $\partial Z = X \sqcup Y$. Then $\chi(\partial Z) = \chi(X) + \chi(Y) \equiv 0 \bmod 2$, hence $\chi(X) \equiv \chi(Y) \bmod 2$. $\Box$

In addition to satisfying $\chi(\partial Y) \equiv 0 \bmod 2$, the Euler characteristic also satisfies $\chi(X \times Y) = \chi(X) \chi(Y)$ (e.g. by the Künneth theorem). It follows that the Euler characteristic $\bmod 2$ is a genus of unoriented manifolds, or equivalently that it defines a ring homomorphism

$\displaystyle \chi : MO_{\bullet}(\text{pt}) \to \mathbb{F}_2$

where $MO_{\bullet}(\text{pt})$ is the unoriented cobordism ring and $MO$ is the Thom spectrum for unoriented cobordism. This is arguably the simplest example of a genus.

Warning. The Euler characteristic itself is not a genus because it is not a cobordism invariant. For example, $\Sigma_2$ is a boundary, hence cobordant to the empty manifold, but $\chi(\Sigma_2) = -2$. There is an integer-valued genus lifting the Euler characteristic $\bmod 2$ on oriented manifolds, although it is not the Euler characteristic but the signature

$\displaystyle \sigma : MSO_{\bullet}(\text{pt}) \to \mathbb{Z}$

where $MSO_{\bullet}(\text{pt})$ is the oriented cobordism ring and $MSO$ is the Thom spectrum for oriented cobordism.

Proof 2: Poincaré duality again

Let $X$ be a closed oriented manifold of even dimension $2k$. Then the cup product defines a pairing

$\displaystyle H^k(X, \mathbb{Q}) \times H^k(X, \mathbb{Q}) \to H^{2k}(X, \mathbb{Q}) \cong \mathbb{Q}$

on middle cohomology which is nondegenerate by Poincaré duality, symmetric if $k$ is even, and skew-symmetric if $k$ is odd. Previously we used this pairing when $k = 2$ and over $\mathbb{Z}$ to understand 4-manifolds. When $k$ is odd we can say the following.

Proposition: With hypotheses as above (in particular, $k$ odd), the Betti number $b_k$ is even.

Proof. On $H^k(X, \mathbb{Q})$ the cup product pairing is a symplectic form, and symplectic vector spaces are even-dimensional. (This follows from the fact that by induction on the dimension, every symplectic vector space $(V, \omega)$ has a symplectic basis, namely a basis $x_1, \dots x_k, y_1, \dots y_k$ such that $\omega(x_i, y_j) = \delta_{ij}$ and $\omega(x_i, x_j) = \omega(y_i, y_j) = 0$. This is a pointwise form of Darboux’s theorem.) $\Box$

Corollary: The Euler characteristic of a closed orientable manifold $X$ of dimension $2 \bmod 4$ is even. In particular, the Euler characteristic of $\Sigma$ is even.

Proof. As above, let $\dim X = 2k$. In the sum

$\displaystyle \chi(X) \equiv \sum_{i=0}^{2k} b_i \bmod 2$

every term $b_i$ is canceled by the corresponding term $b_{2k-i} = b_i$ by Poincaré duality, except for the middle term $b_k$, which we now know is even. $\Box$

Remark. Although this proof also uses Poincaré duality and has the same conclusion as the previous proof, it proves a genuinely different fact about manifolds: on the one hand, it only applies to manifolds of dimension $2 \bmod 4$ and requires orientability over $\mathbb{Z}$ and not just over $\mathbb{F}_2$, but on the other hand it applies in principle to manifolds which are not boundaries.

Going back to the particular case of surfaces $\Sigma_g$, we can even write down a fairly explicit choice of symplectic basis for $H^1(\Sigma_g, \mathbb{Q})$ as follows: thinking of $\Sigma_g$ as a $g$-holed torus, hence equivalently as the connected sum of $g$ tori, we can write down the usual basis $a_k, b_k$ of the first homology of the $k^{th}$ torus. Together these give the standard choice of generators $a_1, b_2, \dots a_g, b_g$ of the fundamental group $\pi_1(\Sigma_g)$, as well as of the first homology $H_1(\Sigma_g, \mathbb{Z})$, and their Poincaré duals in $H^1(\Sigma_g, \mathbb{Q})$ form the symplectic basis we want by the standard relationship between intersections and cup products.

The symplectic structure on $H^1(\Sigma, \mathbb{Q})$ is a shadow of a more general construction of symplectic structures on character varieties $\text{Hom}(\pi_1(\Sigma), G)/G$ of surfaces; these are moduli spaces of flat $G$-bundles with connection on $\Sigma$. The connection is that $H^1(\Sigma, \mathbb{R})$ is the tangent space at the identity of the moduli space $\text{Hom}(\pi_1(\Sigma), U(1))$ of flat (unitary, complex) line bundles on $\Sigma$. These moduli spaces are what classical Chern-Simons theory assigns to $\Sigma$, and applying geometric quantization to these moduli spaces is one way to rigorously construct quantum Chern-Simons theory.

Proof 3: characteristic classes (and Poincaré duality)

For a closed surface $\Sigma$, the Euler characteristic $\bmod 2$ is equivalently the Stiefel-Whitney number $w_2 [\Sigma]$, where $w_2 \in H^2(\Sigma, \mathbb{F}_2)$ is the second Stiefel-Whitney class and $[\Sigma] \in H_2(\Sigma, \mathbb{F}_2)$ is the $\mathbb{F}_2$-fundamental class, which, as above, exists whether or not $\Sigma$ is orientable. In general, the top Stiefel-Whitney class $w_n$ of an $n$-dimensional real vector bundle is its $\bmod 2$ Euler class.

Proof 1 showed that this Stiefel-Whitney number is a cobordism invariant; in fact every Stiefel-Whitney number is a cobordism invariant, although we will not use this. In any case, to show that the Euler characteristic of $\Sigma$ is even when $\Sigma$ is orientable it suffices to show that $w_2 = 0$.

Proposition: Let $\Sigma$ be a closed orientable surface. Then $w_2 = 0$.

Proof. We will again appeal to the relationship between the Stiefel-Whitney classes and the Wu classes $\nu_k \in H^k(-, \mathbb{F}_2)$. Since $\Sigma$ is orientable, $\nu_1 = w_1 = 0$, so $\nu_2 = w_2$, where $\nu_2$ represents the second Steenrod square $\text{Sq}^2 : H^0(\Sigma, \mathbb{F}_2) \to H^2(\Sigma, \mathbb{F}_2)$ in the sense that

$\displaystyle x \nu_2 = \text{Sq}^2(x), x \in H^0(\Sigma, \mathbb{F}_2)$

where $x \nu_2$ denotes the cup product of $x$ and $\nu_2$. But $\text{Sq}^k$ vanishes on classes of degree less than $k$, so $\text{Sq}^2(x) = 0$ above, hence (by Poincaré duality $\bmod 2$) $\nu_2 = 0$ as well. $\Box$

Corollary: The Euler characteristic of $\Sigma$ is even.

Corollary: $\Sigma$ admits a spin structure.

Remark. Atiyah observed that spin structures on $\Sigma$ turn out to be equivalent to theta characteristics, after picking a complex structure. See Akhil Mathew’s blog post on this topic for more.

So we’ve shown that all of the Stiefel-Whitney classes of $\Sigma_g$ vanish. It follows that all of the Stiefel-Whitney numbers of $\Sigma_g$ vanish, and this is known to be a necessary and sufficient criterion for $\Sigma_g$ to be a boundary, a fact which we used in Proof 1. Essentially the same argument shows that all of the Stiefel-Whitney classes of a closed orientable $3$-manifold vanish, so all of the Stiefel-Whitney numbers vanish, and we get the less trivial fact that all closed orientable $3$-manifolds are boundaries. We also get that they all admit spin structures.

In the next two proofs we’ll finally stop using Poincaré duality, but now we’ll start using the fact that $\Sigma$ admits not only an orientation but a complex structure.

Proof 4: the Hodge decomposition

Any compact orientable surface $\Sigma$ can be given the structure of a compact Riemann surface, and so in particular the structure of a compact Kähler manifold, with Kähler metric inherited from any embedding into $\mathbb{CP}^n$ with the Fubini-Study metric. For any compact Kähler manifold $X$, its complex cohomology $H^k(X, \mathbb{C})$ has a Hodge decomposition

$\displaystyle H^k(X, \mathbb{C}) \cong \bigoplus_{p+q=k} H^{p, q}(X)$

where $H^{p, q}(X)$ is equivalently either the subspace of $H^k(X, \mathbb{C})$ represented by complex differential forms of type $(p, q)$ or the Dolbeault cohomology group

$\displaystyle H^{p, q}(X) \cong H^q(X, \Omega^p)$.

Here $\Omega^p$ is the sheaf of holomorphic $p$-forms and the cohomology being taken is sheaf cohomology. Moreover, since $H^k(X, \mathbb{C}) \cong H^k(X, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C}$, the LHS has a notion of complex conjugate, hence we can define the complex conjugate of a subspace, and with respect to this complex structure we have Hodge symmetry: $\overline{H^{p, q}(X)} = H^{q, p}(X)$. This implies the following.

Proposition: Let $X$ be a compact Kähler manifold (e.g. a smooth projective algebraic variety over $\mathbb{C}$). If $k$ is odd, then the Betti number $b_k$ is even.

Proof. Let $h^{p, q} = \dim H^{p, q}(X)$ be the Hodge number of $X$. The Hodge decomposition implies that

$\displaystyle b_k = \sum_{p+q=k} h^{p, q}$

and Hodge symmetry implies that $h^{p, q} = h^{q, p}$. When $k$ is odd, every term $h^{p, q}$ in the above sum is paired with a different term $h^{q, p}$ equal to it, hence $b_k \equiv 0 \bmod 2$ as desired. $\Box$

Corollary: The Euler characteristic of $\Sigma$ is even.

Proof. As before, we have $\chi(\Sigma) = b_0 - b_1 + b_2 = 2 - b_1$, and $b_1$ is even by the above. $\Box$

Corollary: Let $G$ be a finitely presented group. If $G$ has a finite index subgroup $H$ such that the first Betti number

$\displaystyle b_1(H) = \dim H^1(H, \mathbb{Q}) = \dim \text{Hom}(H, \mathbb{Q})$

of $H$ is odd, then $G$ cannot be the fundamental group of a compact Kähler manifold, and in particular cannot be the fundamental group of a smooth projective complex variety.

Fundamental groups of compact Kähler manifolds are called Kähler groups; see these two blog posts by Danny Calegari for more.

Proof. Since a finite cover of a compact Kähler manifold is naturally a compact Kähler manifold, if $G$ is a Kähler group then so are all of its finite index subgroups; taking the contrapositive, if any of the finite index subgroups of $G$ are not Kähler, then neither is $G$. If $X$ is any space with $\pi_1(X) = H$, then $b_1(X) = b_1(H)$, hence the former is odd iff the latter is. It follows that if $H$ is the fundamental group of a compact Kähler manifold then $b_1(H)$ is even; taking the contrapositive, we get the desired result. $\Box$

Example. The free abelian groups $\mathbb{Z}^{2k+1}$ of odd rank have first Betti number $2k + 1$ and hence are not Kähler groups. On the other hand, the free abelian groups $\mathbb{Z}^{2k}$ of even rank are the fundamental groups of complex tori $\mathbb{C}^k / \Gamma$ (e.g. products of elliptic curves).

Example. The free groups $F_{2k+1}$ of odd rank have first Betti number $2k + 1$ and hence are not Kähler groups. The free groups of even rank $F_{2k}$ turn out to have free groups of odd rank as finite index subgroups and hence are also not Kähler.

To see this, first note that if $F_n$ is any free group, then $F_n$ admits finite index subgroups of every possible index because it is possible to write down surjections from $F_n$ into finite groups of every possible size (e.g. cyclic groups). Second, by the standard topological argument every finite index subgroup of $F_n$ is again free because every finite cover of the wedge of $n$ circles is a graph and hence homotopy equivalent to a wedge of circles; moreover, by the multiplicativity of Euler characteristics under coverings, if $F_m$ is an index $k$ subgroup of $F_n$ then

$\displaystyle \chi(F_m) = 1 - m = k \chi(F_n) = k(1 - n)$

and hence $F_n$ has subgroups of index $k$ and first Betti number

$\displaystyle b_1(F_m) = m = 1 + k(n - 1)$

for all $k \in \mathbb{N}$. This is odd whenever $k$ is even, and in particular when $k = 2$. More explicitly, if $F_n$ is free on generators $g_1, \dots g_n$, then

$\displaystyle F_n \ni g_i \mapsto 1 \in \mathbb{Z}_2$

is a surjection onto a finite group of order $2$, and hence its kernel must be free on $1 + 2(n - 1) = 2n - 1$ generators. One possible choice of generators is

$g_1^2, g_2^2, \dots g_n^2, g_1 g_2, g_1 g_3, \dots g_1 g_n$.

Corollary: The fundamental groups $\pi_1(\Sigma_g)$ of compact Riemann surfaces are not free.

There is a great MO question on the topic of why $\pi_1(\Sigma_g)$ is not free in which this argument is given in the comments. As it happens, that MO question loosely inspired this post.

Above, instead of using Hodge symmetry, we can also do the following. In the particular case of surfaces $\Sigma_g$, we in fact have $b_1 = 2g = h^{0, 1} + h^{1, 0}$, hence the two interesting Hodge numbers are

$\displaystyle h^{0, 1} = h^{1, 0} = g$.

In terms of Dolbeault cohomology, this gives

$\displaystyle \dim H^1(\Sigma, \Omega^0) = \dim H^0(\Sigma, \Omega^1) = g$.

Here $\Omega^0$ is the sheaf of holomorphic $0$-forms, or equivalently the structure sheaf $\mathcal{O}_{\Sigma}$ of holomorphic functions.

The identity $\dim H^0(\Sigma, \Omega^1) = g$ gives us one possible definition of the genus of a compact Riemann surface, namely the dimension of the space of holomorphic forms. In general, if $X$ is a complex manifold we can define its geometric genus to be the Hodge number $h^{n, 0} = \dim H^0(X, \Omega_X^n)$, where $\Omega_X^n$ is the canonical bundle, hence the dimension of the space of top forms.

The identity $\dim H^1(\Sigma, \Omega^0) = g$ can be thought of in terms of Hodge symmetry, but it can also be thought of in terms of Serre duality. On the Dolbeault cohomology groups of a compact complex manifold $X$ of complex dimension $n$, Serre duality gives an identification

$\displaystyle H^q(X, \Omega^p) \cong H^{n-q}(X, \Omega^{n-p})^{\ast}$

and hence $h^{p, q} = h^{n-p, n-q}$, which is a different symmetry of the Hodge numbers than Hodge symmetry gives. When $X$ is Kähler, in terms of the Hodge decomposition Serre duality refines Poincaré duality, which only gives

$\displaystyle H^k(X, \mathbb{C}) \cong H^{n-k}(X, \mathbb{C})^{\ast}$.

In particular, we have

$\displaystyle H^1(\Sigma, \Omega^0) \cong H^0(\Sigma, \Omega^1)^{\ast}$

which gives a second proof, independent of Hodge symmetry but still depending on the Hodge decomposition, that $b_1$ is even.

Moreover, since Serre duality is a refinement of Poincaré duality we conclude that $H^1(\Sigma, \mathbb{C})$ is, as a symplectic vector space (as in Proof 2), isomorphic (possibly up to a scalar) to $V \oplus V^{\ast}$ with its standard symplectic structure

$\displaystyle \omega(v_1 \oplus f_1, v_2 \oplus f_2) = f_2(v_1) - f_1(v_2)$

where $V$ is either $H^1(\Sigma, \Omega^0)$ or $H^0(\Sigma, \Omega^1)$. Hence a complex structure on $\Sigma$ equips the symplectic vector space $H^1(\Sigma, \mathbb{C})$ with a Lagrangian subspace.

Digression: the Riemann-Roch theorem

The motivation for the fifth proof starts from the observation that one way to write down the Riemann-Roch theorem for compact Riemann surfaces $\Sigma$ is

$\displaystyle \ell(D) - \ell(K - D) = \deg D + \frac{\chi(\Sigma)}{2}$.

If we can write down a proof of the Riemann-Roch theorem with the genus $g$ appearing directly in this form, in terms of half the Euler characteristic, as opposed to the other ways the genus can appear in a formula involving Riemann surfaces (e.g. as the dimension of the space of holomorphic forms), then since all of the other terms are manifestly integers we would get a proof that $\chi(\Sigma)$ is even.

Here is a proof which does not accomplish this. Let $\mathcal{O}(D)$ denote the line bundle associated to the divisor $D$. Then

$\ell(D) = \dim H^0(\Sigma, \mathcal{O}(D))$

and

$\ell(K - D) = \dim H^0(\Sigma, \mathcal{O}(D)^{\ast} \otimes \Omega)$

since $K$ is the divisor corresponding to the canonical bundle $\Omega$ and $\mathcal{O}(D)^{\ast} \cong \mathcal{O}(-D)$. Now Serre duality gives

$\displaystyle H^1(\Sigma, \mathcal{O}(D)) \cong H^0(\Sigma, \mathcal{O}(D)^{\ast} \otimes \Omega)^{\ast}$

and hence we can rewrite the LHS as an Euler characteristic

$\displaystyle \chi(\mathcal{O}(D)) = \dim H^0(\Sigma, \mathcal{O}(D)) - \dim H^1(\Sigma, \mathcal{O}(D))$

where we are using that the cohomology of sheaves on $\Sigma$ vanishes above its complex dimension, namely $1$. This lets us rewrite Riemann-Roch in the form

$\displaystyle \chi(\mathcal{O}(D)) = \deg D + \frac{\chi(\Sigma)}{2}$.

Let $D = \sum n_p p$ and let $p \in \Sigma$ be a point, so that the meromorphic functions in $H^0(\Sigma, \mathcal{O}(D))$ can have poles of order at most $n_p$ at $p$. Then there is an evaluation map

$\displaystyle \mathcal{O}(D) \to \mathbb{C}_p$

given by taking the coefficient of $z^{-n_p}$ where $z$ is a local coordinate at $p$; here $\mathbb{C}_p$ denotes the skyscraper sheaf supported at $p$ with stalk $\mathbb{C}$. The kernel of this evaluation map consists of functions in $\mathcal{O}(D)$ which have poles of order at most $n_p - 1$ at $p$, which are precisely the sections of the sheaf $\mathcal{O}(D - p)$. Hence we have a short exact sequence of sheaves

$\displaystyle 0 \to \mathcal{O}(D - p) \to \mathcal{O}(D) \to \mathbb{C}_p \to 0$.

Since the Euler characteristic of sheaf cohomology is additive in short exact sequences, it follows that

$\displaystyle \chi(\mathcal{O}(D)) = \chi(\mathcal{O}(D - p)) + \chi(\mathbb{C}_p)$.

Since $H^0(\Sigma, \mathbb{C}_p) \cong \mathbb{C}$ and, being a skyscraper sheaf, $\mathbb{C}_p$ has no higher sheaf cohomology, we have $\chi(\mathbb{C}_p) = 1$, hence

$\displaystyle \chi(\mathcal{O}(D)) = \chi(\mathcal{O}(D - p)) + 1$.

Noting that we also have $\deg D = \deg (D - p) + 1$, by adding and removing points suitably we conclude that if $D_1, D_2$ are any two divisors, then

$\displaystyle \chi(\mathcal{O}(D_1)) - \chi(\mathcal{O}(D_2)) = \deg D_1 - \deg D_2$

or equivalently that there is a constant $c$ such that

$\displaystyle \chi(\mathcal{O}(D)) = \deg D + c$

for all divisors $D$. To determine $c$ it suffices to determine the Euler characteristic of any of the sheaves $\mathcal{O}(D)$, which we can do with a second application of Serre duality: for $D = 0$, so that $\mathcal{O}(D) \cong \mathcal{O}_{\Sigma}$ is the structure sheaf, we have

$\displaystyle \dim H^0(\Sigma, \mathcal{O}_{\Sigma}) = 1$

since the holomorphic functions on a compact Riemann surface are constant, and

$\displaystyle \dim H^1(\Sigma, \mathcal{O}_{\Sigma}) = \dim H^0(\Sigma, \Omega^1) = g$

by Serre duality and the definition of $g$ in terms of holomorphic forms. Hence

$\displaystyle \chi(\mathcal{O}_{\Sigma}) = 1 - g$

from which it follows that $c = 1 - g$. This proves Riemann-Roch, but $1 - g$ appears as the holomorphic Euler characteristic of $\Sigma$ rather than as half the topological Euler characteristic like we wanted. The two can be related using the Hodge decomposition, which shows more generally that for $X$ a compact Kähler manifold of complex dimension $n$,

$\displaystyle \chi(X) = \sum_{k=0}^{2n} (-1)^k \dim H^k(X, \mathbb{C}) = \sum_{k=0}^{2n} (-1)^k b_k$

can be written in terms of Hodge numbers as

$\displaystyle \sum_{k=0}^{2n} (-1)^k \sum_{p+q=k} h^{p, q} = \sum_{p=0}^n (-1)^p \sum_{q=0}^n (-1)^q h^{p, q}$

which we can further rewrite as an alternating sum of Euler characteristics

$\displaystyle \sum_{p=0}^n (-1)^p \sum_{q=0}^n (-1)^q \dim H^q(X, \Omega^p) = \sum_{p=0}^n (-1)^p \chi(\Omega^p).$

Abstractly this identity reflects the fact that the sheaves $\Omega^p$ together form a resolution of the constant sheaf $\mathbb{C}$, just as in the case of smooth differential forms on a smooth manifold. However, in the smooth case, the sheaves of smooth differential forms do not themselves have any higher sheaf cohomology, whereas in the complex case, the sheaves of holomorphic differential forms do in general have higher cohomology. This resolution also exists on any complex manifold, not necessarily compact or Kähler. It gives rise to the Hodge-to-de Rham (or Frölicher) spectral sequence in general, and the existence of the Hodge decomposition reflects the fact that on compact Kähler manifolds this spectral sequence degenerates.

Returning to the case of a compact Riemann surface $\Sigma$, we get that

$\displaystyle \chi(\Sigma) = \chi(\Omega^0) - \chi(\Omega^1)$

but by Serre duality $\chi(\Omega^1) = - \chi(\Omega^0)$, hence

$\displaystyle \chi(\Sigma) = \chi(\Omega^0) + \chi(\Omega^0) = 2 \chi(\mathcal{O}_{\Sigma})$.

Hence the topological Euler characteristic of $\Sigma$ is twice its holomorphic Euler characteristic. This argument not only shows that the topological Euler characteristic is even but gives an interpretation of the number obtained by dividing it by $2$.

But we used the Hodge decomposition and Serre duality already, so let’s do something else.

Proof 5: the Hirzebruch-Riemann-Roch theorem

The Riemann-Roch theorem has the following more general form. Let $V$ be a holomorphic vector bundle on a compact complex manifold $X$ of complex dimension $n$. Let

$\displaystyle \chi(V) = \sum_{k=0}^n (-1)^k \dim H^k(X, V)$

denote the Euler characteristic of the sheaf of holomorphic sections of $V$, as we did above for line bundles. Let $\text{ch}(V)$ denote the Chern character of $V$, which is defined via the splitting principle as

$\displaystyle \text{ch}(L_1 \oplus \dots \oplus L_k) = \sum_{i=1}^k e^{c_1(L_k)} \in H^{\bullet}(X, \mathbb{Q})$

for a direct sum of complex line bundles. $\text{ch}(V)$ can be written in terms of the Chern classes $c_n(V)$ using the fact that the total Chern class

$\displaystyle H^{\bullet}(X, \mathbb{Z}) \ni c(V) = \sum_{k=0}^n c_k(V), c_k(V) \in H^{2k}(X, \mathbb{Z})$

can be defined via the splitting principle as

$\displaystyle c(L_1 \oplus \dots \oplus L_k) = \prod_{i=1}^k (1 + c_1(L_i))$.

Equivalently, $c_n$ is the $n^{th}$ elementary symmetric function in the Chern roots $\alpha_i = c_1(L_i)$. Expanding out the definition of $\text{ch}(V)$ gives power symmetric functions of the Chern roots which we can write as a polynomial in the elementary symmetric functions, e.g. using Newton’s identities, hence as a polynomial in the Chern classes. The first three terms are

$\displaystyle \text{ch}(V) = \dim V + c_1(V) + \frac{c_2(V) - c_1(V)^2}{2} + \dots$.

Similarly, let $\text{td}(X)$ denote the Todd class of (the tangent bundle of) $X$, which is defined via the splitting principle as

$\displaystyle \text{td}(L_1 \oplus \dots \oplus L_k) = \prod_{i=1}^k \frac{c_1(L_i)}{1 - e^{-c_1(L_i)}} \in H^{\bullet}(X, \mathbb{Q})$

for a direct sum of complex line bundles. Again we can use symmetric function identities to write $\text{td}(X)$ in terms of the Chern classes $c_n(X)$ of (the tangent bundle of) $X$. The first three terms are

$\displaystyle \text{td}(X) = 1 + \frac{c_1(X)}{2} + \frac{c_2(X) + c_1(X)^2}{12} + \dots$.

Finally, suppose that

$\displaystyle H^{\bullet}(X, \mathbb{Q}) \ni \alpha = \sum_{k=0}^{2n} \alpha_k, \alpha_k \in H^k(X, \mathbb{Q})$

is a (mixed) cohomology class, and let

$\displaystyle \int_X \alpha = \alpha [X] = \alpha_{2n}[X] \in \mathbb{Q}$

denote the pairing of the degree $2n$ part of $\alpha$ with the fundamental class $[X] \in H_{2n}(X, \mathbb{Q})$.

Theorem (Hirzebruch-Riemann-Roch): With hypotheses as above, the Euler characteristic $\chi(V)$ satisfies

$\displaystyle \chi(V) = \int_X \text{ch}(V) \text{td}(X)$.

We’ll make no attempt to prove this, but here are some notable features of this theorem.

First, 1) $\text{ch}(V)$ only depends on the isomorphism class of $V$ as a topological, rather than holomorphic, complex vector bundle, 2) $\text{td}(X)$ only depends on the isomorphism class of the tangent bundle of $X$ as a topological complex vector bundle, and 3) $\int_X$ only depends on the orientation of $X$ coming from the complex structure on its tangent bundle. In other words, the RHS consists of topological, rather than holomorphic, data. This reflects the way the Hirzebruch-Riemann-Roch theorem is a special case of the Atiyah-Singer index theorem.

In addition, the RHS is a rational linear combination of certain characteristic numbers, hence is a priori rational, but Hirzebruch-Riemann-Roch tells us that it is in fact an integer. This implies divisibility relations which substantially generalize the divisibility relation we’re looking for, namely that $2 | \chi(\Sigma)$.

Corollary (Riemann-Roch): Let $L = \mathcal{O}(D)$ be a holomorphic line bundle on a compact Riemann surface $\Sigma$. Then

$\displaystyle \chi(L) = \left( c_1(L) + \frac{c_1(\Sigma)}{2} \right) [\Sigma] = \deg D + \frac{\chi(\Sigma)}{2}$.

In particular, the holomorphic Euler characteristic satisfies $\chi(\mathcal{O}_{\Sigma}) = \frac{\chi(\Sigma)}{2}$.

Proof. In general, the top Chern class $c_n$ of an $n$-dimensional complex vector bundle is its Euler class $e$. In particular, $c_1(\Sigma)$ is the Euler class $e(\Sigma)$, hence $e(\Sigma) [\Sigma] = \chi(\Sigma)$.

It remains to show that $c_1(L) [\Sigma] = \deg D$. Morally speaking this is because if $D = \sum n_p p$ then $c_1(L) \in H^2(\Sigma, \mathbb{Z})$ is Poincaré dual to $\sum n_p p \in H_0(\Sigma, \mathbb{Z})$, which is morally the vanishing locus of a generic section of $L$. But I am not sure how to make this precise easily. An unsatisfying proof that gets the job done is to use the same additivity argument involving skyscraper sheaves as in the previous proof of Riemann-Roch to conclude that $c_1(L) [\Sigma] = \deg D + c$ for some constant $c$ and then to note that, since $\mathcal{O}(0)$ is topologically the trivial line bundle, $c_1(\mathcal{O}(0)) [\Sigma] = 0$, hence $c = 0$. $\Box$

Corollary: The holomorphic Euler characteristic $\chi(\mathcal{O}_X)$ is equal to the Todd genus of $X$:

$\displaystyle \chi(\mathcal{O}_X) = \int_X \text{td}(X)$.

Proof. The underlying topological line bundle of the structure sheaf $\mathcal{O}_X$ is the trivial line bundle, and hence has trivial Chern character. $\Box$

In particular, $\chi(\mathcal{O}_X)$ only depends on the Chern numbers of $X$. These are known to be complex cobordism invariants, and in fact the Todd genus is a genus: it gives a ring homomorphism

$\displaystyle \text{td} : MU_{\bullet}(\text{pt}) \to \mathbb{Z}$

where $MU_{\bullet}(\text{pt})$ is the complex cobordism ring and $MU$ is the Thom spectrum for complex cobordism.

In the next dimension up (complex dimension $2$, real dimension $4$), the Hirzebruch-Riemann-Roch theorem gives the following divisibility relation.

Corollary (Noether’s formula): The holomorphic Euler characteristic of a compact complex surface $X$ satisfies

$\displaystyle \chi(\mathcal{O}_X) = \left( \frac{c_2(X) + c_1(X)^2}{12} \right) [X]$.

In particular, the RHS is an integer.

Corollary: If $X$ is a compact complex surface with $c_1 = 0$ (in particular if $X$ is Calabi-Yau; the converse holds if $X$ is Kähler), then $c_2(X) [X] = \chi(X) \equiv 0 \bmod 12$.

Examples include the hypersurface of degree $4$ in $\mathbb{CP}^3$ (as we saw previously), and more generally any K3 surface, with Euler characteristic $24$.

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