Let $X$ be a projective (or affine) variety over $\mathbb{C}$ defined by some homogenous ideal $I = (f_1,\ldots,f_n)$. How can we interpret the Euler characteristic of $X$ other than as just an invariant to distinguish non isomorphic objects? What I mean is what could it tell us if anything about either $X$ or the polynomials defining $I$ if $\chi(X) > 0$? How about if $\chi(X) = 0$ or $\chi(X) < 0$? How about in the case of $X$ being a hypersurface cut out by the single (homogeneous) polynomial $f$. What can we deduce about $f$ if $\chi(X) > 0, \chi(X) = 0,$ or $\chi(X) < 0$. I've been trying to relate conditions on the Euler characteristic to anything explicit about the polynomials defining $X$ but haven't had any success so anything that could be said would be interesting and helpful.

4$\begingroup$ The Euler characteristic is only a topological invariant, so it doesn't seem to tell too much of the scheme structure of $X.$ For a hypersurface, it only tells info about $f$ up to its degree: if the projective hypersurfaces defined by $f=0$ and $g=0$ are both smooth, and $\deg f=\deg g,$ then they are diffeomorphic and hence have the same Euler char. Even the degree of $f$ cannot in general be determined: think of a projective line and a conic in a plane. $\endgroup$ – shenghao Sep 8 '11 at 10:33

3$\begingroup$ While this doesn't answer your question about projective varieties, it is a notable fact that if a complex scheme $X$ is decomposed as a disjoint union of other schemes, then $\chi$ is additive. (Not true for arbitrary decompositions of topological spaces!) So you can think of $\chi$ as a characteristic zero analogue of "counting points over ${\mathbb F}_p$". $\endgroup$ – Allen Knutson Sep 8 '11 at 13:35
Hi Dori! I think that you will find Paolo Aluffi's paper "Computing characteristic classes of projective schemes" useful, if not for the exact formulas, at least for the algorithm.
Among other things it proves the version of the formula in J.C. Ottem's answer for singular hypersurfaces. If $f\in \mathbb C[x_0,\dots,x_n]$ is a nonconstant homogeneous polynomial, let $X=\mathcal Z(f)$ and $\Sigma = \mathcal Z_{\mathbb P^n}(\partial_0 f,\dots,\partial_n f)$. If we denote by $g_0,g_1,\dots,g_{n1}$ the degrees of the gradient morphism $$\mathbb P^n\backslash \Sigma\to \mathbb P^n, x=(x_0,\dots,x_n)\to (\partial_0f(x),\dots,\partial_nf(x))$$ then we have $$\chi(X)=n+\sum g_i (1)^{n+1i}.$$
As far as restricting to graph hypersurfaces, I would say look at the recent papers of Aluffi and Marcolli but you know more than me about these matters so... :)
Added: A toy application of this is in the case of graphs $\Gamma_n$ which have two vertices and $n$ parallel edges, so called banana graphs. the formula above gives $$\chi(X_{\Gamma_n})=n+(1)^n$$ so the Euler characteristic determines the graph. The motivic generalization is done in "Feynman motives of banana graphs". The motivic viewpoint also shows that one shouldn't expect a nice relation to hold for all graphs, since graph hypersurfaces are generators for the Grothendieck ring (details and actual statement are here).

$\begingroup$ Thanks Gjergji! That's exactly the type of result I was looking for. I hadn't seen that paper or that result in any of their recent papers on graph hypersurfaces. $\endgroup$ – Dori Bejleri Sep 9 '11 at 7:57
This is unlikely unless $X$ is say, a hypersurface. The reason is that the Euler characteristic is invariant of the projective embedding (it is a topological invariant), while the equations defining a variety are certainly not. For example, it is known that any projective variety can be defined by determinental quadratic equations in some large projective space.
EDIT: Now, if $X\subset \mathbb{P}^n$ is a smooth hypersurface of degree $d$, then we have $$ \chi(X)=((1d)^{n+1}1)d^{1}+n+1. $$Hence when $d\ge 3$ or $n$ is odd, one finds that knowing the Euler characteristic would at least give you the degree of the polynomial (the ambiguous cases $d=1,2$ with $n$ even, are easy to distinguish). More generally, when $X$ is a complete intersection, I think there is also some explicit polynomial expressing $\chi(X)$ in terms of the degrees of the generators, but I doubt you can extract any useful information from this.

$\begingroup$ Hi John, I think you forgot to divide by $d$ the first part of the formula for the Euler characteristic. $\endgroup$ – M P Sep 8 '11 at 10:33

1$\begingroup$ Also, note that you can reconstruct the degree of $X$ if you know its dimension, with the ambiguity that odd dimensional quadrics have the same cohomology groups (not ring!) as projective space. I think that, for fixed dimension, this is the only ambiguity. $\endgroup$ – M P Sep 8 '11 at 10:41

$\begingroup$ Yes, you are right (and it is indeed the only ambiguous case). $\endgroup$ – J.C. Ottem Sep 8 '11 at 11:07

1$\begingroup$ This is quite wellknown, and goes back to Mumfords "Varieties defined by quadratic equations", Questions on algebraic varieties, C.I.M.E. Varenna, 1969 , Cremonese (1970) pp. 29–100. To see why, just take any embedding $X\to \mathbb{P}^n$ and take sufficiently many veronese remembeddings of $\mathbb{P}^n$ into some large projective space $\mathbb{P}^N$. In this $\mathbb{P}^N$ the ideal defining $X$ will be defined by linear polynomials (coming from the original embedding) and quadratic binomials defining the veronese variety. $\endgroup$ – J.C. Ottem Sep 8 '11 at 17:30

1$\begingroup$ Thanks! Something like that formula is actually extremely useful for me. The hypersurfaces I'm working with are cut out by polynomials that have a very specific form, where each variable appears at most once in every monomial (specifically the Kerchoff polynomial of a graph) so knowing the degree from χ will still be very informative. Is there any generalization of this formula to singular hypersurfaces? Also @MP does your comment refer to using the expression for $\Chi$ that J. C. Ottem gave to deduce the degree or some other technique and if so what? $\endgroup$ – Dori Bejleri Sep 8 '11 at 18:09