Let $H$ be the punctual Hilbert scheme of $3$ points in $\mathbb A^3$, over the complex numbers. Then $H$ can be described in the following equivalent ways:

  • set-theoretically, $H$ is the set of subschemes $Z\subset \mathbb A^3$ of length $3$ concentrated at the origin;
  • $H=\textrm{Hilb}^3(\textrm{Spec }R)$, with $R=\mathbb C[x,y,z]/(x,y,z)^3$;
  • $H=p^{-1}(3[P])$, where $p:\textrm{Hilb}^3(\mathbb A^3)\to \textrm{Sym}^3(\mathbb A^3)$ is the Hilbert-Chow morphism and $P\in\mathbb A^3$ is a closed point.

I am struggling to find the dimension of $H$ and I would like some advise on how to compute it.

On the one hand, there are only two non-isomorphic structures on a triple point in $\mathbb A^3$, but on the other hand the dimension of $H$ seems to be bigger than the expected dimension $3\cdot\dim R=0$. I am not able to compute the dimension of the fiber over $3[P]$, as it lies in the diagonal and I do not really know what to expect.

Thanks for any help.


I am sure there are references online for these kinds of computations: for instance, you could look at the papers of Iarrobino, who has written a lot about these schemes. Anyway, here is one way to calculate what you want; there may be a better one.

Let $A= \mathbf C[x,y,z]$.

We want to find ideals $I$ supported at $0$ (so contained in $\mathfrak m = (x,y,z)$) such that $A/I$ is a vector space of dimension 3. The first observation is that $I$ must contain $\mathfrak m^3$, for otherwise we can easily find 4 linearly independent elements in the quotient. So we are really looking for ideals $\tilde{I}$ contained in $\mathfrak m / \mathfrak m^3$, and the condition that $\operatorname{dim} A/I =3 $ translates to $\operatorname{dim} \mathfrak m/\tilde{I} =2 $.

Let $L$ be the vector subspace in $\mathfrak m / \mathfrak m^3$ spanned by the linear forms. Since $L$ has has dimension 3, it must have nonzero intersection with $\tilde{I}$, otherwise the quotient $\mathfrak m/\tilde{I}$ would have dimension at least 3. On the other hand, we cannot have $L \subset \tilde{I}$, since $L$ generates $\mathfrak m$ as an ideal.

So $\operatorname{dim}( L \cap \tilde{I}) = 1 \text{ or } 2$.

If it is 1, then the dimensions tell us $\tilde{I}$ must contain all of $\mathfrak m^2 / \mathfrak m^3$, so $\tilde{I}$ is determined by its intersection with $L$.

If it is 2, then (exercise) we calculate that $(\tilde{I} \cap L)^2$ has dimension 5, hence again by dimensions is all of the degree-2 part of $\tilde{I}$; so again in this case, $\tilde{I}$ is determined by its intersection with $L$.

So the upshot of all this is that the possible choices of the ideal $I$ are exactly the possible choices of a linear subspace of dimension 1 or 2 in the 3-dimensional vector space $L$. So the Hilbert scheme is isomorphic to $G(1,3) \amalg G(2,3) = \mathbf P^2 \amalg \mathbf P^2$.

By the way, note that $PGL(3)$ acts transitively on each component, giving the two possible scheme structure on a triple point that you mention.

  • $\begingroup$ Dear @Brenin, if there's something you find lacking or unclear in my answer, please let me know and I'll try to fix it. $\endgroup$
    – user64687
    Feb 8 '14 at 20:41
  • $\begingroup$ Your answer is very clear, I just need time to digest it. (I confess I get a bit lost in the commutative algebra calculations.) I was wondering if the answer $G(1,3)\coprod G(2,3)$ can be thought of as reflecting the fact that a triple point is either the degeneration of three points on a smooth curve (this is the choice of an $\mathbb A^1\subset \mathbb A^3$, i.e. $G(1,3)$), or the degeneration of two points on the two components of a reducible nodal curve, onto the node (this it the choice of two distinct lines, i.e. $G(2,3)$). $\endgroup$
    – Brenin
    Feb 9 '14 at 11:56
  • $\begingroup$ Dear @Brenin, yes that interpretation seems correct to me. Notice though that my 3-dimensional vector space $L$ is (more or less) the cotangent space, rather than the tangent space. So the $G(1,3)$ you describe is the $G(2,3)$ in my answer, and vice-versa. $\endgroup$
    – user64687
    Feb 10 '14 at 13:17

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