Calculating relative root systems

Question

Let $\mathbf{G}$ be a connected semisimple algebraic group defined over a field $k$. Let $T$ be a maximal torus of $\mathbf{G}$ defined over $k$, and let $S \subset T$ be a maximal $k$-split torus. Let $\Phi = \Phi(T,\mathbf{G})$ be the absolute root system of $\mathbf{G}$, and let $\Phi_{\text{rel}} = \Phi(S,\mathbf{G})$ be the relative root system.

Let $T^{\ast}$ and $S^{\ast}$ be the groups of characters of $S$ and $T$. The restriction map $T^{\ast} \rightarrow S^{\ast}$ restricts to a surjection $\Phi \sqcup \{0\} \rightarrow \Phi_{\text{rel}} \sqcup \{0\}$. We can view $\Phi$ and $\Phi_{\text{rel}}$ as subsets of $T^{\ast} \otimes \mathbb{R}$ and $S^{\ast} \otimes \mathbb{R}$. The root system structures on $\Phi$ and $\Phi_{\text{rel}}$ come from Euclidean inner products on these Euclidean spaces that are invariant under the actions of the Weyl group and the relative Weyl group.

Let's say that I am working with $k$-forms of a semisimple algebraic group that I understand well. I can therefore compute the Euclidean inner product on $T^{\ast} \otimes \mathbb{R}$ as well as the kernel $K$ of the restriction map $T^{\ast} \rightarrow S^{\ast}$. This is enough to determine $\Phi_{\text{rel}}$ as a subset of $S^{\ast} \otimes \mathbb{R}$. But what I don't know how to do is compute the Euclidean inner product on $S^{\ast} \otimes \mathbb{R}$, which is the last thing I need to actually determine the root system $\Phi_{\text{rel}}$.

Another way of saying what I want to be able to compute is the inner product between elements of $\Phi_{\text{rel}}$. My guess is that what you do is as follows. Consider elements $r,r' \in \Phi_{\text{rel}}$. There are then unique lifts $\tilde{r},\tilde{r}' \in T^{\ast} \otimes \mathbb{R}$ that are orthogonal to the kernel $K$, and I suspect that the inner product of $r$ and $r'$ equals the inner product of $\tilde{r}$ and $\tilde{r}'$.

Question: Is my guess correct, and is there a source that spells it out like this?

I apologize for two things:

1. I have never really worked with algebraic groups over fields that are not algebraically closed before, so since I am self-taught in that area I might be writing things in non-standard ways.

2. I know that instead of $\Phi_{\text{rel}}$ I should write $\Phi$ with the field $k$ as a left subscript. I know several ways to do this in LaTeX, but they all seem to break the in-line LaTeX processor used by MathOverflow.

Answer 1

Since $W(G, S)(k) = (N_G(S)/C_G(S))(k) = N_G(S)(k)/C_G(S)(k)$ is finite, there is what Borel, in Linear algebraic groups, §14.7, calls an admissible scalar product on $S^* \otimes_{\mathbb Z} \mathbb R$, obtained simply by averaging any random scalar (i.e., inner) product over its $W(G, S)(k)$-orbit. Then his Theorem 21.6 says that any such choice of admissible scalar product makes $\Phi_\text{rel}$ a root system, with Weyl group $W(G, S)(k)$. Per your comment and my response, if you wish to do this without knowing $W(G, S)(k)$ directly, then Borel’s Corollary 21.4 says that you can just work with the subgroup $W(G, T, S)(k^\text{sep})$ of $W(G, T)(k^\text{sep})$ that stabilises $S$.

You can typeset a left subscript in a fairly naïve way, using $_k\Phi$ _k\Phi. This does not work well if it's not at the beginning of a formula, as in $\Phi(G, S) = _k\Phi$ \Phi(G, S) = _k\Phi. I prefer to shut this off with surrounding braces, like $\Phi(G, S) = {_k\Phi}$ \Phi(G, S) = {_k\Phi}, but many people just use an extra atom, like $\Phi(G, S) = {}_k\Phi$ \Phi(G, S) = {}_k\Phi. (A full solution might be something like \newcommand\lsub[2]{\vphantom{#2}_{#1}#2}, and then $\newcommand\lsub[2]{\vphantom{#2}_{#1}#2}\lsub k\Phi$ \lsub k\Phi. I suspect that packages like leftidx do something like that, but cleverer.)