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Let $X$ be a projective variety over $\mathbb{C}$. Is there a way to define some number $\tilde{\chi}(X)\in \mathbb{Z}$ satisfying both of the following two properties?

$\boldsymbol{(1)} \;$ When $X$ is smooth, $\tilde{\chi}(X)=\chi(X)$, where $\chi(X)$ is the usual Euler number of $X$ (as a topological space).

$\boldsymbol{(2)} \;$ $\tilde{\chi}(X)$ is invariant under deformation, i.e. if we have a family of variety over a connected base $B$, then any two fibers have the same $\tilde{\chi}$.

If I consider an affine variety (still over $\mathbb{C}$) instead of a projective variety, then what is the answer to the corresponding question?

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    $\begingroup$ (1) The answer is yes, because complex projective varieties are triangulable, so it sufficies to take the Euler characteristic of the underlying topological space $X(\mathbb{C})$. See mathoverflow.net/questions/35156/… (2) This definition of course is not preserved by arbitrary deformatons: take a family of elliptic curves (any of them is diffeomorphic to $S^1 \times S^1$) that degenerates to a cuspidal curve (which is diffeomorphic to $S^2$). $\endgroup$
    – Francesco Polizzi
    Commented Sep 19, 2016 at 5:55
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    $\begingroup$ Homeomorphic to $S^2$... :) $\endgroup$
    – diverietti
    Commented Sep 19, 2016 at 9:48
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    $\begingroup$ I think a more down-to-earth question would be: suppose a projective variety admits a smoothing (= a flat deformation to a smooth variety). Is the Euler number of the general fiber independent of the smoothing? $\endgroup$
    – abx
    Commented Sep 19, 2016 at 11:57
  • $\begingroup$ Sorry. I did not state the problem clearly enough. See the revised version.I agree that the question @abx asked is a more down-to-earth version. $\endgroup$
    – user44651
    Commented Sep 19, 2016 at 12:32
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    $\begingroup$ @diverietti: yes, of course I wanted to say homeomorphic. Thanks :-) $\endgroup$
    – Francesco Polizzi
    Commented Sep 19, 2016 at 13:23

1 Answer 1

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The following example shows that the answer to abx's down-to-earth question

Is the Euler number of the general fiber independent of the smoothing?

is in general no.

Take $X \subset \mathbb{P}^5$, the cone over a rational normal curve $C_4 \subset \mathbb{P}^4$. It is well-known that $X$ admits two differents smoothings: a $\mathbb{Q}$-Gorenstein smoothing whose general fibre isomorphic to $\mathbb{P}^2$ and a non-$\mathbb{Q}$-Gorenstein smoothing whose general fibre is isomorophic to $\mathbb{P}^1 \times \mathbb{P}^1$, see for instance this MathOverflow question and the corresponding answers.

In the first case, the topological Euler number of the smooth fibre is $3$, whereas in the second case it is $4$.

So, at least for varieties that are not $\mathbb{Q}$-Gorenstein, the topological Euler number of the general fibre actually depends on the smoothing we are choosing.

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  • $\begingroup$ OK! I see. That implies the invariant $\tilde{\chi}$ does not exist. Somehow it violate my (possible wrong) intuition that some kind of local structure (e.g. the structural sheaf) for a singular variety will be enough to recover the Euler number of a "slight perturbation" of it (i.e. a smoothing). $\endgroup$
    – user44651
    Commented Sep 19, 2016 at 16:18
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    $\begingroup$ Well, the Euler characteristic of a slight perturbation can be recovered from the sheaf of nearby cycles of the perturbation (as its Euler characteristic), so the intuition is not really wrong. However, different perturbations might have different nearby cycles sheaves... $\endgroup$
    – t3suji
    Commented Sep 19, 2016 at 18:26
  • $\begingroup$ In the following paper: math.ubc.ca/~behrend/microlocal.pdf, the author defined a certain "weighted Euler characteristic" for the moduli space of stable sheaves over Kalabi-Yau 3-folds. It seems very closed to what I want. $\endgroup$
    – user44651
    Commented Oct 2, 2016 at 2:40

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