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S Apr 30 at 9:05 history bounty ended CommunityBot
S Apr 30 at 9:05 history notice removed CommunityBot
S Apr 22 at 7:48 history bounty started H A Helfgott
S Apr 22 at 7:48 history notice added H A Helfgott Canonical answer required
Apr 18 at 9:40 comment added H A Helfgott I shoukd add that this question addresses a special case of mathoverflow.net/questions/469174/…
Apr 18 at 7:26 comment added H A Helfgott 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.
Apr 18 at 7:22 comment added H A Helfgott 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.)
Apr 18 at 7:02 comment added H A Helfgott 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 $h_1^{-1}$.
Apr 18 at 7:01 history edited H A Helfgott CC BY-SA 4.0
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Apr 18 at 6:47 history asked H A Helfgott CC BY-SA 4.0