"Nice" way to compute the signature of a toric manifold?

Question

Is there a "nice" way to compute the signature of a smooth toric manifold of even complex dimension in terms of the moment polytope? By signature I mean in the sense of topology (see https://en.wikipedia.org/wiki/Signature_(topology)).

If one goes through all the machinery it is clear that the signature is encoded. I am asking if there is a some way you can "see" this quantity from the moment polytope, quickly and effortlessly. For example the topological Euler characteristic is the number of vertices. The signature won't be as nice as this, certainly, but that is the spirit of what I am looking for (i.e. the less machinery the better). In complex dimension 2 the signature is also a function of the number of vertices by the minimal model program, but that seems to be a stroke of fortune owing to the low dimension.

Answer 1

The Hodge index theorem for smooth projective varieties (or compact Kähler manifolds) says that the signature is $$\sum_{p,q} (-1)^p h^{p,q}(X)$$

For toric varieties, $h^{p,q}=0$ unless $p=q$, and equals the Betti number $h^{2p}(X)$, so the signature is $\sum_p (-1)^p h^{2p}(X)$.

The Betti number $h^{2p}$ is equal to $\sum_{i=p}^n (-1)^{i-p} (-1)^{i-p} \binom{i}{p} N_i$ where $N_i$ is the number of $i$-dimensional cells in the polytope.

So the signature is $$ \sum_p(-1)^p \sum_{i=p}^n (-1)^{i-p} (-1)^{i-p} \binom{i}{p} N_i = \sum_i (-1)^i \sum_{o=0}^i \binom{i}{p} N_i = \sum_i (-1)^i2^i N_i = \sum_i (-2)^i N_i.$$

Thus, for a two-dimensional variety, this is the number of vertices minus twice the number of edges plus four times the number of faces. For a polygon, the number of vertices is the number of edges, and the number of faces is one, so this is four minus the number of vertices, which should be your formula.

Similarly, in higher dimensions we can use the Dehn-Somerville equations for simplicial polyhedra to eliminate half of the variables, as smooth toric varieties have simplicial moment polytopes, as pointed out by David Speyer in the comments.