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Let $G = \mathrm{SL}_n$ (say); let $K$ be a field. Let $g$ be a regular semisimple element of $G(K)$, and $\mathrm{Cl}_g$ its conjugacy class, considered as an algebraic variety. Then $\mathrm{Cl}_g$ is a variety of codimension $n-1$.

For which $h_1,\dotsc,h_n\in G(K)$ is the intersection $$\mathrm{Cl}_g \cap h_1 \mathrm{Cl}_g \cap \dotsc \cap h_n \mathrm{Cl}_g$$ zero-dimensional?

Notes:

(1) It is easy to show that this is the case for $(h_1,\dotsc,h_n)$ generic, i.e., for all such tuples outside a subvariety of $G^n$ of positive codimension.

(2) In the case $G=\mathrm{SL}_2$, it is easy to show that the intersection is zero-dimensional whenever $I$, $h_1$ and $h_2$ are distinct as elements of $\mathrm{SL}_2(K)/\{\pm 1\}$, unless both $h_1$ and $h_2$ are elements of a unitary subgroup stable under conjugation by $g$. (I'll sketch the reasoning in the comments.)

(3) Any hope to get something nicer than (1) will of course have to rely on the fact that $\mathrm{Cl}_g$ is a conjugacy class, as opposed to a garden-variety variety. Can character theory be useful here?

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  • $\begingroup$ Let $n=2$. Then the conjugacy class of a regular semisimple element is given by the equation $\textrm{tr}(g) = c$, where $c\ne \pm 2$. The intersection above is thus the set of $g$ satisfying $\textrm{tr}(g) = \textrm{tr}(h_1^{-1} g) = \textrm{tr}(h_2^{-1} g) = c$. This is an intersection of three planes; the only way for it not to be a line or the empty set would be for $h_2^{-1}$ to equal $r h_1^{-1} + (1-r) I$ for some $r$, and it is easy to see that this cannot be the case for $h_1^{-1}$ semisimple and $h_2^{-1}$ distinct from $I$ and $h_1^{-1}$. $\endgroup$
    – H A Helfgott
    Commented Apr 18 at 7:02
  • $\begingroup$ From that it's a short walk to conclusion (b) in the above. (I should really have said $\textrm{tr}(x) = c$, the set of $x$ satisfying $\textrm{tr}(x) = \textrm{tr}(h_1^{-1} x) = \textrm{\tr}(h^{-2} x)$, etc.) $\endgroup$
    – H A Helfgott
    Commented Apr 18 at 7:22
  • $\begingroup$ Caveat: the same reasoning shows that there is a wider class of pairs $(h_1,h_2)$ for which the three varieties $\textrm{Cl}_g$, $h_1 \textrm{Cl}_g$, $h_2 \textrm{Cl}_g$ do not necessarily intersect transversally. For a lack of transversality, it is enough that $g^{-1}$, $I$, $h_1^{1}$ and $h_2^{-1}$ be linearly dependent. $\endgroup$
    – H A Helfgott
    Commented Apr 18 at 7:26
  • $\begingroup$ I shoukd add that this question addresses a special case of mathoverflow.net/questions/469174/… $\endgroup$
    – H A Helfgott
    Commented Apr 18 at 9:40

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