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Let $G$ be an algebraic group, $B$ a Borel subgroup.

We have a one to one correspondence between the points of the variety of Borel subgroups $\mathfrak{B}$ and the points of the variety $G/B$, for any chosen $B\in\mathfrak{B}$.

We send $B'\in \mathfrak{B}$ to the unique fixed point $xB$ of $G/B$.

Question 1: This fixed point comes about sice $x^{-1}B'x=B$ means $$B'x=xB\implies B'xB=xBB\implies B'xB=xB$$ So that the action of $B'$ on $xB$ gives us $xB$, so that from this, indeed $B'$ has atleast one fixed point from the conjugacy theorem.Showing it is unique is fine.

Question 2: Under this one to one correspondence, we have an action of $G$ on $\mathfrak{B}$ by $g\cdot B=gBg^{-1}$ and this gives us, over that map, an action of $g\cdot xB=gxB$ (left multiplication).

Let $H$ be a subgroup of $G$, then the fixed points of the action of $H$ (restricted from $G$) on $G/B$, should correspond to the set of all Borels in $\mathfrak{B}$ containing $H$. How to show that this is the case?

I would have some fixed point $xB$ so that $HxB=xB$ right, so this would give us the corresponding Borel: $hxBh^{-1}x^{-1}$, why does this contain $H$? If it did it would reduce to $hxBx^{-1}=xBx^{-1}$, so this Borel contains $H$?

alggeo
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1 Answers1

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Let $G$ be an algberaic group over some field $k$. Let $B$ be a Borel subgroup of $G$. I'll write, as you do, $\mathfrak{B}$ as the set of Borel subgroups of $G$.

There is an action $G\times \mathfrak{B}\rightarrow \mathfrak{B}$ given by conjugation: $(g, B')\mapsto gB'g^{-1}$. Since $B$ is equal its normalizer in $G$, we find the isotropy group at $B\in\mathfrak{B}$ is equal $B$. This induces the isomorphism $G/B\cong \mathfrak{B}$ (a noncanonical isomorphism since we had to choose a point to obtain it). The choice didn't matter, since any other choice differs by a conjugation (which is an isomorphism; this uses that the conjugation on Borel subgroups is transitive).

What we've done is fixed a map $G\cong G\times \{B\}\rightarrow \mathfrak{B}$ which passes through the quotient $G/G_B$ to induce an isomorphism. A more explicit definition for the isomorphism is then $gB\mapsto gBg^{-1}$; hence also the inverse is given by $B'=gBg^{-1}\mapsto gB$.

Now I'll answer your question. If $S\subset G(k)$ is some set of rational points, we can find a bijection between those Borel subgroups containing $S$ in their rational points, and the fixed points for the action of $S$ on $\mathfrak{B}$. To do this, write $\mathfrak{B}^S$, the fixed set of the action by all elements of $S$, as the intersection $\bigcap_{s\in S} \mathfrak{B}^s$.

Each $\mathfrak{B}^s$ is the scheme fixed by the action $\mathfrak{B}\cong\{s\}\times\mathfrak{B}\rightarrow \mathfrak{B}$, or the equalizer of this and the identity map $\mathfrak{B}\rightarrow \mathfrak{B}$. By the definition we had $(s,B)\mapsto sBs^{-1}$ which equals $B$ only if $s\in N_G(B)(k)$ which, as mentioned above, is equal $B(k)$ since Borels equal their normalizer.

Eoin
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