# Two approaches to compute the signature of a Kaehler manifold

Given a compact Kaehler manifold $M$ of complex dimension $2n$, there are essentially two ways to compute its signature $\sigma(M)$, i.e. the index of the intersection form on $H_{2n}(M,\mathbb{R})$:

1.by Hodge index theorem $\sigma(M)=\sum_{p,q}(-1)^p h^{p,q}$, here $h^{p,q}$ stands for the Hodge numbers.

2.by Hirzebruch signature theorem $\sigma(M)=L[M]$, here $L[M]$ stands for the $L$-genus, i.e. the characteristic number of the top $L$-class.This approach is more general since it works on any $4k$ dimensional real manifolds.

My questions are

1.Since these two approaches rest on different levels of cohomology theory, how are they interrelated?

2.Of course, one possible way to answer Question 1 is to generalize both by the Hirzebruch-Riemann-Roch on Kaehler manifolds, a point already mentioned in Hirzebruch's Neue topologische Methoden. However, I am wondering if someone could relate these two approaches on a more fundamental level.

To be precise,

Is there a formula to express the Chern numbers/Pontryagin numbers out of the Hodge numbers on a compact Kaehler manifold $M$ of complex dimension $n$? Surely it is the case for $c_n[M]$ interpreted as the Euler characteristic number.

Or, does anyone know such counterexamples that two Kaehler manifolds(notably, Kaehler surfaces, I guess) have the same Hodge numbers but different Chern numbers?

Many thanks!

• For KE in dimension 2 I know:By $\chi(M)$, $Sign(M)$ and $\mathcal A(M)$ we denote the Euler-Poincare characteristic, the signature and the arithmetic genus of $M$. Let $M$ be a compact complex manifold of complex dimension $2$. If $M$ admits a Kahler metric $\omega$,then$$2\chi(M) + 3 Sign(M) = 8 \mathcal A(M) + Sign(M) = 12 \mathcal A(M) - \chi(M) \leq 2 Vol(M) a^2,$$with equality if and only if $c_1(M)=a[\omega]$. So if first chern class be proportional to kahler form (like Kahler-Einstein manifolds), then by this formula we can compute the signature of a kahler manifold in dimension 2.
– user21574
Oct 23 '17 at 20:29
• Let $M$ be a compact Kaehler manifold of complex dimension 2, then $$48\pi^2Sign(M)=-\int_M(|R|^2-2|\rho|^2)dv$$ where $|R|$ denotes the length of the curvature tensor $R$ of $M$ and $\rho$ is the Ricci tensor. See link.springer.com/article/10.1007/BF00147936
– user21574
Oct 23 '17 at 20:35
• The two ways of computing the signature are related via the Hirzebruch $\chi_y$ genus. More precisely, $\chi_1(M)$ can be identified with $\sigma(M)$ via the L genus (the second approach), and in the Kähler case, $\chi_1(M)$ can also be written as in the Hodge Index Theorem. Using the $\chi_y$ genus, one also obtains a formula for the signature of a complex manifold (not necessarily Kähler) in terms of Hodge numbers, see here. For more details on Hirzebruch's $\chi_y$ genus, see this note. Aug 24 '18 at 14:56

About your last question, a recent theorem of Kotschick-Schreieder (see http://arxiv.org/abs/1202.2676 page 2) says that a linear combination of Hodge numbers equals a linear combination of Chern numbers for all projective manifolds (modulo the usual Kähler symmetries) iff it is a linear combination of the numbers $\chi_p=\sum_p (-1)^q h^{p,q}$.

Similarly, a linear combination of Hodge numbers equals a linear combination of Pontryagin numbers iff it is multiple of the signature.

This shows that, apart for the signature and the $\chi_p$'s, there is no universal formula to express Chern or Pontryagin numbers purely in terms of Hodge numbers. So the Hirzebruch signature formula is really an isolated phenomenon, in this sense.

• Exactly what I wondered about. Fascinating! Apr 9 '12 at 1:00

"counterexamples that two Kaehler manifolds have the same Hodge numbers but different Chern numbers?"

As you explained above, Chern numbers of surfaces can be expressed in terms of the Euler number and the first Pontrjagin number, so you need dimension at least 3 for a counterexample.

In dimension 3, consider a projective space and a smooth quadric threefold. These two have the same Hodge numbers, same $c_3 = 4$ (Euler number), same $c_1 c_2 = 24$ (by Todd's theorem), but distinct degrees $c_1^3$: for the projective space it equals $64$ and for quadric it is $54$.

• This counterexample is related to the fact that the Coxeter group $B_2$ has a symmetry that doesn't lift to the Lie group. Namely, $\mathbb{P}^3$ and a quadric $3$-fold in $\mathbb{P}^4$ both are homogeneous spaces for the Lie group $SP(4) \cong PSO(5)$. The combinatorial data indexing them is a choice of vertex of the Dynkin diagram, and these choices of vertex are switched by the Coxeter symmetry. The Betti numbers can be described in terms of the Coxeter group, so they must match, but the Chern numbers cannot so they don't have to. (The cohomology rings also don't match.) Mar 8 '19 at 16:18
• Sorry, $SO(5) \cong PSp(4)$, not vice versa. Mar 8 '19 at 16:44
• David, thanks for your comment. I have not considered this example from your perspective. Geometrically you could say that complete flag variety F for $B_2$ has two structures of $\mathbb{P}^1$-bundle -- one over $\mathbb{P}^3$ and another over Q. The same diagram in case of $G_2$ is used in arxiv.org/abs/1606.04210 to construct annihilations of $\mathbb{A}^1$ in the Grothendieck ring of varieties - just consider a "Cayley hypersurface" H in F that intersects with generic fibers of both projections by a single point. Apr 1 '19 at 22:51
• Thanks for showing me the $G_2$ reference! I remember being confused by Borisov's paper in the past, and this one is a useful alternate perspective. Apr 3 '19 at 17:54