10
$\begingroup$

I was reading Luna's paper Toute variété magnifique est sphérique and stumbled on a few facts about Bialynicki-Birula decompositions and fixed points that I don't understand.

Here is the setup. Let $G$ be a connected reductive group over an algebraically closed field $k$ (of characteristic 0, though I'm not certain if that matters), and fix a Borel subgroup $B \subset G$ and a maximal torus $T \subset B$. Denote by $B^-$ the opposite Borel group to $B$ containing $T$. Let $X$ be an irreducible, normal, complete $G$-variety, and suppose that $X$ has finitely many $G$-orbits. Given any one-parameter subgroup $\lambda: \mathbb{G}_m \to T$ and any $y \in X^T$, we write $$X(\lambda,y) = \{x \in X\ |\ \lim_{t \to 0} \lambda(t)x = y\}.$$

Here are the claims Luna makes that I don't understand:

(1) The fixed point set $X^T$ is finite.

(2) We are mainly interested in the case where $\lambda$ is in the Weyl chamber of $B$ (i.e.\ where $\langle \lambda, \alpha \rangle > 0$ for all positive roots $\alpha$), so that $X(\lambda,y)$ is $B$-stable. Luna states that for a "sufficiently general" such $\lambda$, we will have $X^{\mathbb{G}_m} = X^T$, where $\mathbb{G}_m$ acts on $X$ via its image under $\lambda$. He also states that in this case, the $X(\lambda,y)$ for various $y \in X^T$ form the Bialynicki-Birula decomposition of $X$.

(3) With $\lambda$ satisfying the conditions in (2), if $X(\lambda,y)$ is open, then $y$ is fixed by the opposite Borel subgroup $B^-$. (Luna doesn't say anything about this, but I'm also curious: is it true that if $y$ is fixed by $B^-$, then $X(\lambda,y)$ is open?)

All of these statements seem pretty reasonable to me, and I've worked them out in the case where $X = \mathbb{P}(V)$, $G = \mathrm{SL}(V)$, and $B$ (resp. $T$) is the subgroup of upper triangular (resp. diagonal) matrices. In this case, everything is clear using projective coordinates, but I don't know how to make these types of arguments without appealing to coordinates like that. Any proofs (or references to proofs) would be much appreciated!

$\endgroup$
2
$\begingroup$

I think all of these should be easy enough to resolve. First note that (1) is a triviality from your assumption that $G$ has finitely many orbits on $X$, because a maximal torus of $G$ can only have finitely many fixed points on $G/H$.

Now recall that by a beautiful result of Sumihiro (in the case $G$ is an connected linear algebraic group) given a normal $G$-variety $X$ we and an orbit $Y \subset X$ we can find a $G$-stable opens $Y \subset U \subset X$ such that $U$ is isomorphic to a $G$-stable locally closed subset of $\mathbb{P}(\rho)$ for $\rho$ a finite dimensional representation of $G$, a well-known corollary allows us to let $U$ be affine when $T$ is a split $k$-torus. Since all of your questions are about the local structure of orbits in $X$ unless I am misreading you, I will just assume $X \subset \mathbb{P}(\rho)$ is locally closed for $(V, \rho)$ some fixed representation from now on.

For (2), $T$ acts on $V$ as such $V \cong \oplus_i V_{\chi_i}$ for $\chi_i$ various distinct characters of $T$, then let $\lambda$ be sufficiently general in the sense that $\langle \lambda, \chi_i - \chi_j \rangle \neq 0$ for $i \neq j$. Then the eigenvalues $\langle \lambda, \chi_i \rangle$ are distinct so $\lambda$ and $T$ induce the same eigendecomposition of $V$ and thus a point $x \in \mathbb{P}(V)$ is fixed by one iff it is fixed by the other. To me the definition you gave is the definition of the B-B decomposition, so you will have to elaborate on what definition you are working with if you want me to show that this induces the B-B decomposition.

For (3) lets take $y \in X^T$ such that $X(\lambda, y) = U$ is the big cell, and take $\lambda$ as above to be a regular cocharacter of $T$ wrt $B$. We know that for any $b \in B^-$ we have that $\text{lim}_{t \to 0} \lambda(t)^{-1}b\lambda(t) \in T$ by explicit computation of the BB decomposition for $G$ (alternatively if you use the dynamic approach to parabolics this is by definition).

Further considering $A: B^- \to X$ the action map $A(b) = b\cdot y$ we know that there is an open subscheme $V$ of $B^-$ such that $V \cdot y \subset U$. But for $b \in B^-$ $\text{lim}_{t \to 0} \lambda(t)^{-1} b \cdot y = \text{lim}_{t \to 0} b^{\lambda(t)^{-1}} \lambda(t)^{-1} \cdot y = y$, since $y$ is torus-stable. But then for $b \in V$ we have that $b \cdot y$ has both limits $\text{lim}_{t \to 0} \lambda(t) \cdot b \cdot y$ and $\text{lim}_{t \to \infty} \lambda(t) \cdot b \cdot y$ defined, which defines a map $\mathbb{P}^1 \to X$, the image of this map lies inside of $B^- \cdot y$, which is affine because $B^-$ is solvable, therefore it is constant. Thus $b\cdot y$ is a fixed point for $\lambda$, thus by $(2)$ it is a fixed point for $T$, for any $b \in V$. Because the stabilizer of $y$ is closed and $V$ is dense in $B^-$ we are done, $B^- \cdot y = y$.

Hope this helps, let me know if anything is unclear.

$\endgroup$
3
  • $\begingroup$ Thanks so much for your reply! This is really helpful to me. Two minor questions: (1) When you say that $B^- \cdot y$ is affine, are you using that $y$ is $T$-stable and that orbits of unipotent groups are affine, or is there some other result like this? (2) I'm not familiar with the statement that $T$ has finitely many fixed points on $G/H$. I flipped through Borel's book on algebraic groups but couldn't find it. Can you explain more and/or provide a reference? $\endgroup$ – Michael Christianson Apr 29 '20 at 5:05
  • $\begingroup$ Hi Michael, it’s a general result that solvable affine linear algebraic groups have affine orbits, the result is just predicated on the fact that it’s true for unipotents and for tori, and one can do induction by dimension. But you are right that in this case it’s simpler. $\endgroup$ – Sempliner Apr 29 '20 at 6:34
  • $\begingroup$ So if T has a fixed point on G/H then there is some g in G such that T is conjugate by g to a subgroup of H, whence a maximal torus of H. As all maximal tori of H are conjugate we may as well assume g sends T to a specific maximal torus inside of H. But for T, T’ maximal tori of G, the set of g in G which send T to T’ is clearly a torsor for the normalizer of either torus, whence finiteness modulo H. $\endgroup$ – Sempliner Apr 29 '20 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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