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Let $(e, h, f)$ be an $\operatorname{SL}_2$-triple in $\mathfrak g$. My understanding is that $e + C_{\mathfrak g}(f)$ is called a Kostant section only in case $e$ is regular; but I don't impose this restriction. (EDIT: It seems that, in general, it's called a Slodowy slice.) Given such a datum, there is a unique decomposition (*) $\mathfrak g = [e, \mathfrak g] + C_{\mathfrak g}(e, f) + C_{\mathfrak g}(f)(< 0)$, where “$< 0$” in the last term refers to the grading coming from the action of $h$.

Now suppose that we are given only a nilpotent element $e \in \mathfrak g$, and an arbitrary element $X \in \mathfrak g$. Can we complete $e$ to an $\operatorname{sl}_2$-triple $(e, h, f)$ so that, in the decomposition $X_1 + X_2 + X_3$ of $X$ according to (*), we have that $X_2$, or at least its semisimple part, commutes with $X_3$? The obvious answer is “No, because, in certain characteristic, you can't complete $e$ to an $\operatorname{sl}_2$-triple at all.”  To avoid such trivialities, assume the characteristic is as large as necessary.

As a very mild hint that the answer is ‘yes’, the weak form of the question works whenever $e$ is distinguished (since then the semisimple part of $X_2$ is central in $\mathfrak g$); and, by explicit calculation, it works when $\mathfrak g = \mathfrak{gl}_3$ and $e$ is the $(1, 2)$-nilpotent, as long as $p > 3$. Because of the first hint, I'd like to make an appeal to Bala–Carter(–Pommerening–Premet …) theory, but I can't figure out how.

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up vote 6 down vote accepted

In some cases the answer to the weaker version of the question (involving the semisimple part of $X_2$) is YES. This will happen if $C_g(e)$ is self-dual which is the case, for instance, when $g=gl_N$, $N=nm$, and $e$ has $n$ Jordan blocks of size $m$. To see this, one can use the fact that all maximal toral subalgebras of $C_g(e)$ are conjugate.

However, in general the answer to the question is NO as the following example shows. Let $g=sp_4$ and $e$ a long root vector in $g$. Then the action of $ad\ h$ on $g$ gives us a $\mathbb{Z}$-grading $g=g(-2)\oplus g(-1)\oplus g(0)\oplus g(1)\oplus g(2)$ with $g(-2)=\mathbb{C}f$, $g(2)=\mathbb{C}e$, $g(0)\cong gl(2)$, $g(-1)$ the standard $2$-dimensional module for $g(0)$, and $g(1)\cong g(-1)^*$ as $g(0)$-modules. We also have that $C_g(f)= g(0)'\oplus g(-1)\oplus g(-2)$ and $C_g(e,f)=g(0)'$. Let {$x,H,y$} be an $sl_2$-triple in $g(0)'$ which spans $g(0)'\cong sl_2$ and let {$v_1,v_{-1}$} and {$w_1,w_{-1}$} be eigenbases for $ad\ H$ of $g(1)$ and $g(-1)$, respectively, so that $[H,v_1]=v_1$, $[H,w_{-1}]=-w_{-1}$, etc.

We choose $X=X_1+X_2+X_3$ such that $X_1=0$, $X_2=H+y$ and $X_3=w_{1}$. Then $X_2$ is semisimple and $[X_2,X_3]=w_1+w_{-1}$. We may adjust $X$ effectively only by applying automorphisms of the form $Ad\ g$ with $g$ in the unipotent radical of the centraliser of $e$ in $Sp_4$. Then $Ad\ g=\exp(a\ ad\ e)\exp(b\ ad\ v_1)\exp(c\ ad\ v_{-1})$ for some $a,b,c\in\mathbb{C}$. Such an adjustment is not going to change $X_3=w_1$, but it will replace $X_2=H+y$ by $H+y+b[v_1,w_1]+c[v_{-1},w_1]=H+y+\lambda x+\mu H$ for some $\lambda,\mu\in\mathbb{C}$. (It will change $X_1$ as well, but this is not important.) Since $[x,w_1]=0$ and $[y,w_1]=w_{-1}$, this is not going to help us to improve $X$ the way we wanted.

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A beautifully comprehensive answer. Thank you! –  L Spice Sep 24 '12 at 15:20
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