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Let $G$ be a connected algebraic group over an algebraically closed field. I'm trying to understand the phrase "the subvariety of semisimple elements in $G$ which are not regular." This tacitly implies that the conditions of semisimplicity and regularity must be open or closed.

For reference, an element in $g\in G$ is semisimple if $g=su$ is its Jordan decomposition and $u$ is the identity element of $G$. Also, an element is regular if the dimension of its centralizer is equal to the dimension of a maximal torus in $G$.

I believe I can prove that regularity is an open condition for $G=\mathrm{GL}_n$, as a matrix $A$ is regular if and only if $\{I,A,A^2,\ldots,A^{n-1}\}$ is a linearly independent set. How can I see that regularity is an open condition for arbitrary $G$?

Concerning semisimplicity, I read on pg. $99$ of Humphrey's "Linear Algebraic Groups" that the set of semisimple elements "$G_s$ is rarely a closed subset of $G$." In light of the phrase I'm trying to understand, are there conditions on $G$ which make semisimplicity a closed or open condition?

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  • $\begingroup$ I think semisimple is more likely to be an open condition, just from the GL case. $\endgroup$ – Jack Schmidt May 8 '14 at 16:47
  • $\begingroup$ Yeah, regular semisimple are dense open in connected semisimple. Still looking for general connected. $\endgroup$ – Jack Schmidt May 8 '14 at 16:52
  • $\begingroup$ Humphrey's Linear Algebraic Groups only says it contains an open dense set, but doesn't specify whether it is open itself. Borel's Linear Algebraic Groups page 160 says the regular semisimple elements form an open dense set all the time (but I think say there are no unipotent regular elements). $\endgroup$ – Jack Schmidt May 8 '14 at 17:13
  • $\begingroup$ math.stackexchange.com/a/412110/583 $\endgroup$ – Jack Schmidt May 8 '14 at 17:14
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    $\begingroup$ I am not sure if this is entirely correct but here's an alternative approach. Any affine algebraic group over an algebraically closed field has a finite dimensional faithful representation inside its coordinate ring. This representation gives an isomorphism from the group to a subgroup of some $GL(V)$ which preserves semisimplicity because a element of an affine group is semisimple if and only if it acts semisimply on its coordinate ring. Hence, the set of semisimple elements inside the group $G$ is the intersection of $G$ with the set of semisimple elements of $GL(V)$ which is hence open. $\endgroup$ – Siddharth Venkatesh May 8 '14 at 18:32
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The locus of regular elements is open for arbitrary G. To see this, use the fact that in a group scheme, the dimension of the fibre is a semicontinuous function on the base. (Apply this to the centraliser scheme with base G where the fibre over g is the centraliser of g).

If G=GLn it is easy to see that the semisimple elements are dense. The identity element lies in the closure of the set of unipotent elements so the semisimple locus is not open. The image of GxT in G under the map (g,t)-->gtg-1 is the semisimple locus so Chevalley's theorem implies that the semisimple locus is a constructible set.

For a general G, embed G into GLn and use the above result to see the semisimple locus is constructible for a general linear algebraic group. This is the best we can say in general (as the above GLn example shows). Only for very special groups (e.g. unipotent groups or tori) is the set of semisimple elements closed.

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