3
$\begingroup$

I strongly suspect that there is a connection between the Hilbert function $H_M(d)$ for a graded $K[x_0,...,x_n]$-module $M$ and the Euler characteristic of a topological space. For the obvious connection we have that $$H_M(d)=\sum_i (-1)^i H_{F_i}(d)$$ if we have the graded free resolution $$\cdots \to F_1 \to F_0 \to M \to 0$$ of $M$.
The Euler characteristic of a topological space $X$ can be (glossing over details) computed as the alternating sum of different dimensional holes, or if we triangulate and take simplicial homology as $$\chi(X) =\sum_i (-1)^i \dim H_i(C_{\bullet}(X))$$ where $H_i(C_{\bullet}(X))$ is the $i$-th homology group of the complex of $n$-chains of $X$.

Now of course these two look very similar. But why do they look similar? I am looking for a connection for example on the level of algebraic geometry, something along the lines of:
Let $R:=K[x_0,...,x_n]$, then $R$ is the homogeneous coordinate ring of projective space $\mathbb{P}^n$. If we take a homogeneous ideal $I \subset R$ then $R/I$ is a graded $R$-module and thus we can calculate the above Hilbert function $H_{R/I}$. But at the same time $R/I$ is the homogeneous coordinate ring of the projective variety defined by the vanishing locus $V(I)$. Now $V(I)$ lives inside of the projective space $\mathbb{P}^n$ and has a topological structure from the Zariski topology or maybe even from an induced topology of $K$ (?) Could it be that $V(I)$ has as a topological space a Euler characteristic equal to the Hilbert function of $R/I$ at some degrees?

Somehow I think that there must lie a connection. But maybe I am mistaken. Anyway I would be glad to hear if there are also other possible connections.

$\endgroup$

1 Answer 1

2
$\begingroup$

There is a general definition of the Euler characteristic of a chain complex; one way to say it is the following. If $0 \to V_0 \to V_1 \to \dots \to V_n \to 0$ is a chain complex of finite-dimensional vector spaces then its Euler characteristic can be defined as the alternating sum of dimensions

$$\chi(V) = \sum_{i=0}^n (-1)^i \dim V_i$$

and if the $V_i$ are furthermore graded then we can take the graded dimension instead, etc. The general fact, which is a kind of "inclusion-exclusion" argument, is that the Euler characteristic is also given by the alternating sum of dimensions

$$\chi(V) = \sum_{i=0}^n (-1)^i \dim H_i(V)$$

of the homology of $V$. Roughly speaking, $\chi(V)$ is a homological version of "the dimension of $V$," and is in some sense the most basic invariant of $V$.

The connection to the Euler characteristic of a topological space $X$ is that we can take $V$ to be any of several different chain complexes whose homology computes the homology of $X$. In general we can take singular chains, which is infinite-dimensional, so we can't discuss the alternating sum of dimensions and can only discuss the alternating sum of homologies. But, for example, if $X$ is a finite CW complex then we can take $V$ to be cellular chains, which is finite-dimensional, of dimension $c_i$ in degree $i$ where $c_i$ is the number of $i$-cells. This implies that the Euler characteristic of $X$ can be computed as an alternating sum over the number of cells,

$$\chi(X) = \sum_{i=0}^n (-1)^i c_i$$

which is a generalization of Euler's polyhedral formula; the fact that this is also the alternating sum of the homologies implies that the alternating sum of the number of cells is a homotopy invariant. Roughly speaking $\chi(X)$ is a homotopy-theoretic version of cardinality, and is also in some sense the most basic invariant of $X$ (from the point of view of homotopy theory).

The basic connection to the Hilbert function is ultimately not that interesting as far as I can tell, it just boils down to the definition of a resolution. A free resolution is a chain complex of free modules $F$ such that $H_0(F) \cong M$ and the higher homologies vanish. This means its Euler characteristic is $\chi(F) = \dim H_0(F) = \dim M$ and the same is true for graded dimensions, etc. This is again a kind of "inclusion-exclusion" argument, and it's a useful computational tool but it doesn't by itself imply any kind of deep relationship to the topological Euler characteristic, which involves some other unrelated chain complex, with interesting homologies in higher degree but no extra grading.

However there is a different connection to the Hilbert function which is less direct, and which passes through the Hirzebruch-Riemann-Roch theorem. Namely, the Hilbert function of, say, a smooth projective variety $X \subset \mathbb{CP}^n$ is eventually equal to a polynomial, the Hilbert polynomial $HP(d)$. The HRR theorem implies that the Hilbert polynomial is given by a sum

$$HP(d) = \chi(X, \mathcal{O}(d)) = \int_X \text{ch}(\mathcal{O}(d)) \text{td}(X)$$

where

  • $\chi$ here is now the Euler characteristic of a coherent sheaf, which is defined as the alternating sum over the dimensions of its sheaf cohomology; for large $d$ the higher cohomology vanishes (by Serre vanishing) and this is the same as $\dim H^0(X, \mathcal{O}(d))$, which is the dimension of homogeneous polynomials on $X$ of degree $d$, and this is why we get the Hilbert polynomial.
  • $\mathcal{O}(d)$ is the line bundle on $\mathbb{P}^n$ whose sections are the homogeneous polynomials of degree $d$, pulled back to $X$.
  • $\text{ch}$ and $\text{td}$ are the Chern character and Todd class, and
  • $\int_X$ refers to pairing a cohomology class with the fundamental class of $X$.

This means the Hilbert polynomial contains information about the Chern classes of $X$. And one of these Chern classes, namely the top Chern class $c_{n-1}$, does in fact know the topological Euler characteristic of $X$! However, this information is mixed up with information about the Chern classes of $\mathcal{O}(d)$, so it's not as simple as saying that the Hilbert polynomial computes $\chi(X)$. The Todd class also mixes things up a lot, so I don't even know if we get that the Hilbert polynomial determines $\chi(X)$ in general.

That was a lot of abstraction so let's get more concrete. Suppose $X \subset \mathbb{CP}^2$ is a smooth projective plane curve over $\mathbb{C}$, such as an elliptic curve in Weierstrass form. In this case HRR specializes to the Riemann-Roch theorem, and we get that the Hilbert polynomial of $X$ is

$$HP(d) = d - g + 1.$$

So the Hilbert polynomial knows the genus $g$ of $X$, which is the same information as $\chi(X) = 2 - 2g$. But they're not the same number, and the degree $d$ is in there too (which is not an invariant of $X$ and depends on the embedding into projective space). Note, interestingly, that the constant term

$$HP(0) = \chi(X, \mathcal{O}_X) = 1 - g$$

of the Hilbert polynomial is exactly half $\chi(X)$; here $\chi(X, \mathcal{O}_X)$ is not $\chi(X)$ but the "coherent Euler characteristic"

$$\chi(X, \mathcal{O}_X) = \sum (-1)^i \dim H^i(X, \mathcal{O}_X).$$

This is also not the value $H(0)$ of the Hilbert function at $0$; that would just be $\dim H^0(X, \mathcal{O}_X) = 1$. This does not directly generalize to higher dimensions but it's not a coincidence and can be explained using Hodge theory; for some more details see the fifth proof here.

$\endgroup$
2
  • $\begingroup$ I am truly delighted by this answer. I greatly appreciate your effort. There are a lot of words and theorems which I do not understand yet, but I will read into them and I am sure that this will open up a nice big picture. Already the first few paragraphs have cleared up many thoughts about the Euler characteristic and its ubiquity. $\endgroup$ Commented Aug 5, 2024 at 6:53
  • $\begingroup$ @Flynn: yes, there's a lot of great classical stuff to understand here, and I can't claim to be an expert in most of it either, I've just done a couple of relevant calculations. If you're curious you can find some of them here: qchu.wordpress.com/2014/10/14/…. $\endgroup$ Commented Aug 5, 2024 at 7:09

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.