MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ and $Y$ be smooth projective varieties, say over $\mathbb C$. Fixing a point $y\in Y$, we obtain a smooth, closed subvariety $X\times\{y\}$ of $X\times Y$, which in turn corresponds to a point $P_y$ on the Hilbert scheme $\mathcal{Hilb}(X\times Y)$.

What technology can I use to decide whether $P_y\in \mathcal{Hilb}(X\times Y)$ is smooth or not?

share|cite|improve this question
up vote 6 down vote accepted

The tangent space to the Hilbert scheme at point $P \in Hilb$ corresponding to a subvariety $Z$ is $H^0(Z,N_Z)$, where $N_Z$ is the normal bundle, and the obstruction space is $H^1(Z,N_Z)$. In your case $N_Z = T_yY\otimes O_X$ is a trivial vector bundle, so the tangent space is $T_y\otimes H^0(X,O_X)$. If $X$ is connected, it is just $T_yY$ and it is clear that all obstructions vanish (since there is a family of deformations with the same tangent space). So, in this case $P_y$ is a smooth point of the Hilbert scheme.

share|cite|improve this answer

First, rigidity lemma: let $X\times Y \leftarrow Z\to T$ be a flat family of closed subschemes of $X\times Y$, with the fiber $Z_0$ over $0\in T$ equal to $X\times {y}$. Then the composition $Z\to X\times Y \to Y$ contracts $Z_0$ to a point, therefore it has to contract nearby fibers by the Rigidity Lemma.

I omitted a considerable amount of details here, but it should follow that the connected component of the Hilbert scheme corresponding to $P_y$ is isomorphic to $Y$, hence smooth.

A good reference for the Rigidity Lemma(s) is Mumford's book on abelian varieties, chapters 2 and 3.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.