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It is known from either Morse theory or Bialynicki-Birula decomposition that the fixed points of a ${\mathbb{C}}^*$ action on a smooth algebraic variety over $\mathbb{C}$ determine a cell decomposition of this variety (cells correspond to fixed points by sending each point to a fixed point by the ${\mathbb{C}}^*$ action). Such cell decomposition of the Hilbert scheme of points on the plane $({\mathbb{C}}^2)^{[n]}$ is described in several sources, notably the original papers by G. Ellingsrud and S.A. Stromme and Chapter 5 of the book about Hilbert schemes of points on surfaces by H.Nakajima.

It follows that the fixed points of the torus action (and hence the cells) on $({\mathbb{C}}^2)^{[n]}$ are indexed by partitions of $n$. However, it is hard for me to infer from the above references what is the combinatorial order of these cells (i.e. which cells lie on the boundary of which other cells) -- this should induce a certain order on the partitions of $n$. Is there a reference which would state what this order is explicitly?

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Fixed. On MO asterisks get interpreted as italics, so you should use \ast. –  Qiaochu Yuan Jan 22 '11 at 1:40
    
Try this: J. Briancon: Description de $Hilb^n C\{x, y\}$, Invent. Math. 41 (1977), no. 1, 45–89. –  Sasha Jan 22 '11 at 4:21
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3 Answers

Another reference for Hilbert schemes is chapter 18 of Miller and Sturmfels' book Combinatorial Commutative Algebra, although I don't think they address your question specifically.

It's hard to imagine that the partial order on cells could be anything other than refinement of partitions -- merging two parts into a bigger part corresponds to merging two points of a configuration, thus getting something in a less generic cell. But I am not up on what the exact cell decomposition is in terms of torus-fixed points, so I can't swear to this. Certainly, the partial order must have a unique maximal element, because the Hilbert scheme is irreducible for $n=2$.

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The order is dominance (or anti-dominance, depending on one's conventions) order on partitions. I first learned about such things from Nakajima's paper here http://arxiv.org/pdf/alg-geom/9610021 (see section 4).

Edit: To be more specific, the punchline is on page 13. See the second displayed equation there, together with (4.13) and Proposition 4.14. Nakajima works with $Hilb_n(X)$ where $X$ is the total space of a line bundle on $\mathbb{P}^1$, so for what you want take the line bundle to be trivial and consider the open subset of $Hilb_n(X)$ supported on $\mathbb{C}^2$.

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It might be useful to see quickly why Jeremy's answer, although a very reasonable guess, is wrong. Consider the two partitions $(2,2)$ and $(3,1)$. In refinement order, neither one is greater than the other.

Consider the subscheme of $\mathbb{P}^1 \times \mathbb{C}^2$ cut out by $$x^2=u xy + v y^2=y^3=0$$ where $(x,y)$ are the coordinates on $\mathbb{C}^2$ and $(u: v)$ are the homogenous coordinates on $\mathbb{P}^1$. Every fiber over $\mathbb{P}^1$ has length $4$, so this is a flat family, and we get a map $\mathbb{P}^1 \to \mathrm{Hilb}_4(\mathbb{C}^2)$.

The image of this map is a torus invariant curve in the Hilbert scheme. Its two torus fixed points are the monomial ideals $\langle x^2, y^2 \rangle$ and $\langle x^2, xy, y^3 \rangle$, corresponding to the partitions $(2,2)$ and $(3,1)$.

So, in any Bialynicki-Birula decomposition, one of these partitions must dominate the other.

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