Let $G = GL\_n$ over a field $F$, and let $\gamma \in G(F)$ be a semisimple element. The characteristic polynomial $c\_\gamma(t)$ of $\gamma$ encodes a fair bit of information about $\gamma$. Importantly (for me at least), $c\_\gamma(t)$ decomposes into linear factors (over $F$) if and only if $\gamma$ lives in a maximal split torus, and the splitting field $E\_\gamma$ of $c\_\gamma$ is the minimal field extension $E$ such that $\gamma$ lives in an $E$-split torus.


Is there an appropriate generalization to other split groups? For instance, is there a function $f\_\gamma$ (or set of functions) in $k[G]^G$ such that the statement `$\gamma$ lives in an $F$-split maximal torus' is equivalent to some statement about the algebraic properties of $f\_\gamma$?


More generally, if $G$ is quasi-split over $F$, is there a function $f\_\gamma$ in $k[G]^G$ whose properties determine if $\gamma$ lives in a Borel defined over $F$? (This is a generalization of the previous, since if $G$ is split then $\gamma$ lies in a maximal split torus if and only if it lives in a Borel).


Just for the record, my first thought was to let $f\_\gamma$ be the characteristic polynomial of $Ad(\gamma)$ acting on the Lie algebra $\mathfrak{g}$. This fails even over $GL\_2$, where $F$ is any field not having a square root of $-1$. Then $\gamma = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ has $f\_\gamma(t) = (t-1)^2(t+1)^2$, but $\gamma$ cannot be diagonalized over $F$.



Is there a generalization of the "characteristic polynomial" to other split/quasi-split algebraic groups?
There is a problem you will face in generality. I don't think such a minimal field extension exists, and there isn't a nice invariant. 


Let $V$ be any vector space with a quadratic form, and let $-V$ be the same vector space with minus that quadratic form. Then $V \oplus -V$ is a vector space with a split quadratic form. So its orthogonal group is split. The element of that orthogonal group that acts as $1$ on $V$ and $-1$ on $-V$ is part of a split maximal torus if and only if its centralizer, $O(V) \times O(V)$, has a split maximal torus, which happens if and only if $V$ is split. There is no simple invariant that tells you whether or not a quadratic form is split over an arbitrary field.


The problem is that this is not really an invariant of the geometric conjugacy class - there can be two elements that are conjugate over an algebraically closed field that are not conjugate over the base field. Indeed, the geometric conjugacy class in $O\_n$ is just determined by the characteristic polynomial, and this has the same characteristic polynomial as an element that is part of a split torus.


If you're interested only in conjugacy over an algebraically closed field, it's sufficient to take the characteristic polynomial of any faithful representation - if that factors completely into linear factors, then your element lives in a split torus.


To prove this, take any split torus $T \subseteq G$. Your element must be conjugate to some element of $T$ over an algebraically closed field. Let's show that element lies in the base field. It's sufficient to show that every character of that torus sends it to an element of the base field. But because the representation is faithful, the characters of the torus that appear as eigenfunctions of the representation generate all the characters. Because the eigenvalues are in the base field, your element will be as well.


