Bialynicki-Birula decompositions and fixed points

Question

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!

Answer 1

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.