Borel subalgebras fixed by the adjoint action of a semisimple element Let $s$ be a semisimple (diagonalisable, and in this case we will take $s$ to be a diagonal matrix) element in $SL_n(\mathbb{C})$.
I want to describe explicitly what $\mathcal{B}^s = \left\lbrace \mathfrak{b} : s \cdot \mathfrak{b} \cdot s^{-1} \subset \mathfrak{b} \right\rbrace$ is.
I know there is the property that a Borel subalgebra $\mathfrak{b}$ is fixed by $s$ if and only if $s \in B$, where $B$ is the corresponding Borel subgroup, and I understand the bijection $G/B \cong \mathcal B$ between the homogeneous space $G/B$ and the flag variety.
Also I know that Borel subalgebras in this setting are in correspondence with stabilisers of flags in $\mathbb{C}^n$, so intuitively we would ask for flags which are also fixed by the action of $s$.
All of these concepts I have seen before but I am struggling to wrap my head around how to synthesise them to describe $\mathcal{B}^s$.
I have tried working through a small example, $n=2$.
Here, $s = \operatorname{diag}[q,q^{-1}]$ for $q \in \mathbb{C}^\times$, and I am quite sure that $\mathcal{B}^s$ consists only of the two Borel subalgebras of upper and lower triangular matrices in $\mathfrak{sl}_2(\mathbb{C})$.
But this is only an observation of a pattern from some small calculations: I find a line in $\mathbb{C}^2$, compute its Borel subalgebra (given by matrices $\begin{pmatrix} a & b \\ c & -a \end{pmatrix}$ with some linear relation among $a, b, c$), and then show that these matrices aren't fixed by conjugation by $s$.
However this is obviously not exhaustive, and I don't know how well this argument translates to larger $n$.
Clarification on this part would be greatly appreciated - I cannot find such an exercise/example in any of the usual Lie theory textbooks like Serre, Humphreys, Fulton-Harris, Goodman-Wallach, Springer, nor Steinberg.
 A: Inherently, this answer is going to depend on $s$. In your example you look at this in terms of diagonal elements. What does this mean geometrically? Well if $s=\mathrm{diag}[a_1,\dots,a_k]$ with $a_1\cdots a_k = 1$ then We have taken some decomposition $\mathbb{C}^n = L_1 \oplus \cdots \oplus L_n $ and $s$ acts as $a_i$ on $L_i$. Clearly $s$ is going to preserve any flag built out of the $L_i$ for example $L_1 \leq L_1 \oplus L_2 \leq \cdots \leq L_1 \oplus \cdots L_{n-1} \leq \mathbb{C}^n$. There are as many of these as there permutations of $n$ elements i.e. $n!$.
Is this all of them? Well this is where it's going to depend on $s$. If all the $a_i$ are distinct then yes we are done. In this case, I would call $s$ a regular element. In fact its centraliser is precisely a maximal torus and this defines a root system. Then the Borel subalgebras it preserves are the direct sum of the corresponding Cartan subalgebra and the positive root spaces (for each possible choice of positive roots). It's not too hard to check that there are $n!$ choices of which roots are positive (Same as the number of Weyl chambers or as the distinct choices of simple roots).
However, if $s$ has eigenvalues with dimension greater than one (i.e. some of the $a_i$ are equal) it's not quite so simple. Consider the case $a_1 = a_2$ but the rest are distinct. Since $L_1 \oplus L_2$ is an eigenspace for $s$ it preserves any line contained in it. So $s$ preserves flags of the form $L \leq L_1 \oplus L_2 \leq \cdots \leq L_1 \oplus \cdots L_{n-1} \leq \mathbb{C}^n$ for any $L$ as an example. Indeed rearranging all the $L_i$'s around this I see a disjoint union of $(n-1)!$ copies of $\mathbb{P}^1$.
Generalising this further $s$ clearly preserves all subflags on its eigenspaces and the $\mathcal{B}^s$ looks like a product of smaller flag varieties and a discrete set of size $k!$ where $k$ is the number of distinct eigenvalues of s.
As an extreme example if $s$ is the identity it preserves all flags automatically and so $\mathcal{B}^s$ would be the whole flag variety.
