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Let $X$ be a smooth variety with an action of $\mathbb{C}^*.$ One has the so-called Bialynicki-Birula decomposition of $X$ given by stable manifolds: $$X=\bigcup_N X^+(N),$$ where $N$ varies in the set of connected components of the fixed point locus $X^{\mathbb{C}^*}.$ Rougly speaking, $X^+(N)$ is related to the positive part of the weight decomposition of $T_p X$, for $p\in N$ (see Theorem 4.1 in [2] for a more precise definition).

Define the following relation on the set of connected components of $X^{\mathbb{C}^*}$:

Relation 1: $N_1 \prec N_2 \Leftrightarrow \mbox{ the boundary of } X^+(N_1) \mbox{ intersects } X^+(N_2).$

Question 1: Is this relation antisymmetric?

I know that the answer to this question is affermative in the following case I shall explain in the following.

First, I need to recall some Morse theory. Let $M$ be a compact Riemannian Manifold and $f\colon M\rightarrow \mathbb{R}$ a smooth function.

Definition. ([1]) Let $N \subset M$ be a connected submanifold of $M.$ It shall be called a non-degenerate critical manifold for $f$ if and only if

  1. the differential $df$ of $f$ is identically zero along $N$,
  2. the Hessian $H_N f$ of $f$ is non-degenerate on the normal bundle $\nu(N)$ of $N.$

Assume that the critical set of $f$ is a union of non-degenerate critical manifolds. In this case, we shall call $f$ Morse function. To any non-degenerate critical manifold $N$ one can associate the so-called stable manifold $M^+(N)$, that is, the set of points of $M$ belonging on trajectories of the gradient flow of $f$ converging to $N.$ One has the so-called Morse decomposition $$M=\bigcup_N M^+(N).$$

In the paper [1], Atiyah and Bott define a relation on the set of critical manifolds in the following way:

Relation 2: $N_1< N_2 \Leftrightarrow \mbox{ the boundary of } M^+(N_1 ) \mbox{ intersects } M^+(N_2 ).$

Moreover, they state that $N_1 < N_2 \Rightarrow f(N_1 ) < f(N_2 )$, hence Relation 2 is antisymmetric.

Let $X$ be a compact Kahler manifold endowed with an action of a compact holomorphic one-dimensional torus $T.$ One can define a Morse function $f$ associated to the torus action.

Facts (Chapter 3 in [3]): The non-degenerate critical manifolds coincide with the connected components of $X^{T}.$ Moreover, the stable manifolds of the Morse decomposition associated to $f$ coincide with the stable manifolds of the Bialynicki-Birula decomposition associated to the $T$-action.

Thus, in this case Relation 1 coincides with Relation 2, hence Relation 1 is antisymmetric.

Question 2: Do we really need Morse theory to prove the antisymmetric property of Relation 1?


[1] Atiyah, Bott. The Yang-Mills equations over Riemann surfaces. Philosophical Transactions of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 308, No. 1505 (Mar. 17, 1983), pp. 523-615.

[2] Bialynicki-Birula. Some theorems on actions of algebraic groups. The Annals of Mathematics, Second Series, Vol. 98, No. 3 (Nov., 1973), pp. 480-497.

[3] Nakajima. Lectures on Hilbert schemes of points on surfaces, American Math. Society, Providence RI, 1999.

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up vote 2 down vote accepted

If $X$ is projective, the Bialynicki-Birula decomposition is filterable; this means that there is a filtration $Z_1 \subseteq Z_2 \subseteq \cdots \subseteq Z_r = X$ by closed subsets, such each difference $Z_i \smallsetminus Z_{i-1}$ is a piece of the Bialynicki-Birula decomposition (Białynicki-Birula, Some properties of the decompositions of algebraic varieties determined by actions of a torus. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 24 (1976), no. 9, 667–674). This implies that your relation is antisymmetric.

[Edit] When $X$ is only quasi-projective, the Białynicki-Birula decomposition does not exist in general, even if $X^{{\mathbb C}^*}$ is proper. It does exists when $\lim_{t \to 0}t \cdot x$ exists for any $x \in X$. I would guess that in this case it is still filterable, and that you can prove this by choosing a smooth equivariant compactification.

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If $X$ is only quasi-projective with $X^{\mathbb{C}^*}$ proper, is still my relation antisymmetric? – Francesco Apr 21 '12 at 9:39

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