I just realized that Hodge numbers can be defined for every $\mathbb C$-variety, not only the smooth proper ones. At least we can define them using the Grothendieck ring $K_0(\text{Var}/\mathbb C)$ since it is generated by smooth projective varieties. However, in this way, the geometric meaning is not so clear to me. If we use the original definition via $H^q(\Omega^p)$, the $\Omega^1$ is not necessarily a vector bundle, and I am not sure if we can still do the wedge product. So, my first question is:

1) Is there some good interpretation of Hodge numbers of singular varieties?

and I also would like to know that

2) is the Kunneth formular also holds in the singular case?


3) if $X$ is a smooth hypersurface of degree $d$ in $\mathbb P^n$ and $X'$ be a singular one, is it true that the Hodge numbers of $X$ and $X'$ coincide?

I guess 2) is true since we only need if $X_1\sim X_2$ and $Y_1 \sim Y_2$ then we have $X_1 \times Y_1 \sim X_2 \times Y_2$. But I didn't see how to show this in detail. For 3) I guess it is not true, but still I cannot give a reason.

I think there should be some reference for this, but I did not find it. Any help would be appreciated. Thanks!

  • $\begingroup$ A small remark "at least we can define them using the Grothendieck ring". It is non-trivial to see that the Hodge polynomial is compatible with the relations of the Grothendieck ring (in fact the usual one is not, one has to take the $E$-polynomial described in my answer, which coincide with the usual polynomial in the smooth projective case). The "deep" reason is Bittner's theorem, stating that $K_0$ is generated by smooth projective varieties, with relations $[X'] - [E] = [X] - [Z]$ where $X'$ is the blow-up of $X$ along $Z$ and $E$ the exceptional divisor. $\endgroup$ Dec 15, 2019 at 15:02

1 Answer 1


1) This polynomial correspond to the so-called "$E$-polynomial". It is computed in term of compactly-supported mixed Hodge structure. This is claimed in thes notes. The precise formulas for the $E$ polynomial is given (see here) by $E(u,v) = P(u,v,-1)$ where $P(u,v,t)$ is the "compactly-supported mixed Hodge polynomial" defined by $P(u,v,t) = \sum_{i,j} \dim Gr^i_FGr^{i+j}_W H_c^k(X, \Bbb C)u^iv^jt^k$. Here $W$ is the weight filtration and $F$ is the Hodge filtration. A really helpful reference for mixed Hodge structure is here.

Just a word about mixed Hodge structure : the quick idea is to "resolve" varieties by smooth, projective varieties to define Hodge structure. By resolving and compactifying our varieties, we can obtain have long exact sequence (for example the Gysin sequence associated to $X = X' \backslash D$, where $D$ is a divisor) where each $H^k(X; \Bbb C)$ term is squeezed between two terms that have a Hodge structure on it, and use it to define Hodge structure on $H^k(X; \Bbb C)$ !

2) Sure, it's a ring morphism from $\mathrm K_0(\mathrm{Var}/\Bbb C)$. That's a very strong condition !

3) No, because already the ordinary Poincaré polynomial do not coincide (for example, adding a node typically makes the Euler characteristic drop).

EDIT (regarding 3) : I'm not sure you can use directly the Grothendieck ring of varieties to study family of hypersurfaces in $\Bbb P^n$, because hypersurfaces of the same degree are not isomorphic in general, so they are a fiber bundle in the $C^{\infty}$-category but not the algebraic category. However maybe these methods can still be used, I would be interested to see anything in this direction.

  • $\begingroup$ Many thanks! Your answer is always so helpful and I can learn a lot from it. $\endgroup$
    – Akatsuki
    Dec 15, 2019 at 20:53
  • $\begingroup$ Concerning the last paragraph: I am trying to use this to study some special isotrivial families of hypersurfaces, say, if we have a (flat) family of hypersurfaces $S \to T$ such that over a dense open set $U\subset T$ the family is a fiber bundle with fiber some smooth hypersurface $F$, I am wondering if the Hodge polynomial of $S$ is simply the same as the Hodge polynomial of $T\times F$. I have computed some examples and they indeed coincided, so I guess this might be true. If you are interested in this, I would be very appreciated if you can remark something. $\endgroup$
    – Akatsuki
    Dec 15, 2019 at 20:53
  • $\begingroup$ @Akatsuki : with pleasure ! If you want, you can send me an e-mail. I'm not sure I can say anything useful but I know that explaining a problem to someone else can be very helpful :-) Intuitively, I would say that it depends on the topology of the locus over $T \backslash U$. Typically, if it's smooth I believe it should be good. Anyway feel free to send me a message ! $\endgroup$ Dec 15, 2019 at 21:00
  • $\begingroup$ Thank you for this! I will send you a detailed version tomorrow. $\endgroup$
    – Akatsuki
    Dec 15, 2019 at 23:19

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