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Let $\frak{g}$ be a finite-dimensional complex nilpotent Lie algebra. Given $\xi\in\frak{g}$, what is known about the intersection of $im(ad_{\xi})$ (the image of $ad_{\xi}:\frak{g}\rightarrow\frak{g}$) and $z(\frak{g})$ (the centre of $\frak{g}$)? In particular, under what conditions is it true that the intersection is non-trivial?

Alternatively, under what conditions will $im(ad_{\xi})$ contain a 1-dimensional ideal?

Thanks

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3 Answers

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Whenever $ad_{\xi}$ is an endomorphism of $\mathfrak{g}$ whose corresponding partition $\pi: 1^{s_{1}}2^{s_{2}} \cdots \;$ of $\dim \mathfrak{g}$ is such that $s_{1} =0$, then we have $im\; ad_{\xi} \cap Z(\mathfrak{g}) \neq \{0\}$.

As $\mathfrak{g}$ is nilpotent, $ad_{\xi}$ acts as a nilpotent endomomorphism of $\mathfrak{g}$, for every $\xi \in \mathfrak{g}$, and we can associate to $ad_{\xi}$ a partition $\pi(\xi)$ of $n=\dim \mathfrak{g}$ as follows: the Jordan form of $ad_{\xi}$ is

$$J=\begin{bmatrix} J_{1}\\\ & \ddots \\\ && J_{r}\end{bmatrix}$$

where $J_{i}$ is an $n_{i}\times n_{i}$ $0$-Jordan block and $n_{1}\geq \ldots \geq n_{r} >0$. Then, $n_{1} +\ldots +n_{r}=n$ is a partition of $n$, which we will denote $\pi(\xi): 1^{s_{1}}2^{s_{2}} \cdots $, so that $$1s_{1}+2s_{2}+\ldots =n$$ is the partition of $n$ determined by $J$.

Let $\xi \in \mathfrak{g}$ and let $\pi(\xi) : 1^{s_{1}}2^{s_{2}} \cdots $ be the partition associated to $ad_{\xi}$ and suppose that $(x_{1},\ldots,x_{n})$ is a Jordan basis of $ad_{\xi}$, so that the matrix of $ad_{\xi}$ is in Jordan form. Thus, $x_{1},x_{n_{1}+1},\ldots, x_{n_{r-1}+1}$ is a basis of $$Z_{\mathfrak{g}}(\xi)=\ker ad_{\xi}.$$

As $\mathfrak{g}$ is nilpotent we must have $Z(\mathfrak{g}) \neq \{0\}$. Let $w \in Z(\mathfrak{g})$ be nonzero. Then, $w \in Z_{\mathfrak{g}}(\xi)$ and so $w= c_{1}x_{1}+\ldots + c_{r} x_{n_{r-1}+1}$.

Assume that $s_{1}=0$ in the partition $\pi$ (so that $\pi$ contains no parts equal to $1$). Thus, $x_{2},\ldots, x_{n_{r-1}+2} \notin Z_{\mathfrak{g}}(\xi)$ and $x=c_{1}x_{2}+\ldots +c_{r}x_{n_{r-1}+2}$ is such that $[\xi,x] =w\in Z(\mathfrak{g})$.

While this condition is sufficient it is not necessary: for example, consider the nilpotent Lie algebra $\mathfrak{n}_{3}$ of strictly upper triangular $3\times 3$ complex matrices. Then,

$$\xi = \begin{bmatrix} 0&1&0\\\0&0&1\\\0&0&0\end{bmatrix}$$

is such that $im \; ad_{\xi}=Z(\mathfrak{n}_{3})$, while $\pi(\xi): 12$.

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Of course, the easiest case is when $\mathfrak{g}$ has nilpotence class 2 (that is, $[[\mathfrak{g},\mathfrak{g}],\mathfrak{g}]=0$). Under this assumption one has trivially that $im(ad_{\xi})⊆Z(\mathfrak{g})$, and $im(ad_{\xi}$) contains a 1-dimensional ideal whenever $\xi$ is not in $Z(\mathfrak{g})$.

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For generic $\xi$, the answer is yes.

Denote the lower central series as $$ \mathfrak{g}_0 = \mathfrak{g} \supset \mathfrak{g}_1 = [\mathfrak{g},\mathfrak{g}_0] \supset \mathfrak{g}_2 = [\mathfrak{g},\mathfrak{g}_1]\supset\ldots $$

Let $r$ be minimal such that $\mathfrak{g}_r\neq \{0\}$, $\mathfrak{g}_{r+1} = \{0\}$, and $s$ minimal such that $\textrm{ad}_\xi^s \neq 0$, and $\textrm{ad}_\xi^{s+1} = 0$. Suppose $\xi$ is such that $\textrm{Im}(\textrm{ad}_\xi)\cap Z(\mathfrak{g}) = \{0\}$, and also that it has maximal $s$ with respect to this property. We must have $s<r$, since $\mathfrak{g}_r \subset Z(\mathfrak{g})$.

Now for any other $x\in\mathfrak{g}$, view $\textrm{ad}_x^{s+1} = 0 $ as a system of polynomials on $\mathfrak{g}$. It's nontrivial since $s<r$, and since $\xi$ is a solution.

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