Timeline for semisimple conjugacy classes over general bases
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 18, 2013 at 8:26 | comment | added | A Stasinski | @Margaux: You are right, I should have said that the semisimple conjugacy classes in $G(R)$ are in bijection with those of $G(k)$. The semisimple elements in $T(R)K$ are the conjugates of elements in the reductive part. | |
| Oct 17, 2013 at 21:08 | comment | added | Marguax | @A Stasinki: Your conclusion is incorrect since the torus is not normal. (I was expecting you were making this mistake, but wanted to see your argument to be sure.) As a simple analogue, consider the semidirect product $\mathbf{G}_m \ltimes \mathbf{G}_a$ with the usual scaling action (say all over a field). This has tons of semisimple elements not coming from the left factor. | |
| Oct 17, 2013 at 16:46 | comment | added | prochet | As the result is true over a field you can ask the question for artinian ring if you look at elements in G(k[[t]]) to see which statements generalise | |
| Oct 17, 2013 at 15:05 | comment | added | A Stasinski | @Margaux: The kernel $K$ of $G(R)\rightarrow G(k)$ lies in the unipotent radical of $G(R)$. Hence any maximal torus of $G(R)$ lies inside $T(R)K$, where $T$ is a maximal torus of $G$. But the only semisimple elements in $T(R)K$ are the ones in the reductive part of $T(R)$, which is isomorphic to $T(k)$ (e.g. via the Teichmuller section $T(k)\rightarrow T(R)$). The group $G(R)$ has a canonical structure of lin. alg. gp. over $k$ so we have a natural notion of semisimple elements. If these are not appropriate for the present purposes perhaps one should not consider semisimple elements. | |
| Oct 16, 2013 at 16:38 | comment | added | Marguax | @A Stasinski: Why do you claim there is such a bijection of semisimple elements? If one applies conjugation by $G(R)$ on itself then such a claim seems unlikely (though you don't say what your bijection is). In addition, it's hard to know if your proposed definition of "semisimple" is the appropriate one for present purposes; prochet probably should explain the motivation for the question in order to determine what is a suitable definition. | |
| Oct 16, 2013 at 10:09 | comment | added | A Stasinski | To state the question precisely you need to define the Steinberg map over $R$. If this can be done the answer to the question should be analogous to the case over $k$. | |
| Oct 16, 2013 at 9:41 | comment | added | A Stasinski | Suppose that $R$ is local Artinian with residue field $k$. Then the Greenberg functor identifies $G(R)$ with a linear algebraic group over $k$. We therefore have a natural notion of semisimple elements in $G(R)$. A maximal torus in $G(R)$ is then the reductive part of $T(R)$ where $T$ is a maximal torus in $G$, so the semisimple elements of $G(R)$ are in bijection with those of $G(k)$. | |
| Oct 16, 2013 at 9:34 | comment | added | prochet | There is a stronger notion of semisimplicity regarding to a faithful representation $\rho:G\rightarrow GL_{n}$ that say that $g\in G(R)$ is strongly semisimple if $R[\rho(g)]$ is a finite étale algebra locally a direct factor of $\mathfrak{gl}_{n}(R)$, this definition avoids your example I presume. | |
| Oct 16, 2013 at 9:31 | comment | added | prochet | I don't know for your question on $R$. Probably we need a stronger definition of semisimple. | |
| Oct 16, 2013 at 6:58 | comment | added | Marguax | I don't think it is true that infinitesimal multiplicative type subgroups with non-cyclic Cartier dual necessarily lie in a subtorus of $G$ (even over an algebraically closed field), so by your definition the "universal point" of such a subgroup scheme would not be semisimple. Not that there's anything "wrong" with that, as it depends on your motivation (not given), but are you sure that this phenomenon doesn't give you pause about your definition? Do you know if the weaker condition of admitting such a $T$ over a finite flat local extension of $R$ is equivalent to have one over $R$? | |
| Oct 16, 2013 at 6:50 | comment | added | prochet | $g \in G(R)$ is semisimple, if there exists a torus T over R, such that there is a morphism from $T\rightarrow G$ over $R$ which contains $g$ in its image. | |
| Oct 15, 2013 at 22:12 | comment | added | Marguax | Define "semisimple" for an $R$-valued point. | |
| Oct 15, 2013 at 21:53 | history | asked | prochet | CC BY-SA 3.0 |