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I'm trying to find the root space decomposition of a lie algebra wrt a toral subalgebra h. Both a matrix lie algebras. I'm confused about how do I find the linear forms $\lambda \in \mathfrak{h}^*$ on $\mathfrak{h}$.

$\mathfrak{g}_{\lambda } {\text{:=} \{ X\in \mathfrak{g}|\text{ad}_H(X).=\lambda (H)X \forall H\in \mathfrak{h} \} } $

So the above formula for the root space is simple enough, but how do I find all the $\lambda$ such that $\mathfrak{g}_{\lambda }$ is nonzero?

In my example, $\mathfrak{h}= \{ \left( \begin{array}{cccc} 0 & u & 0 & 0 \\ -u & 0 & 0 & 0 \\ 0 & 0 & 0 & v \\ 0 & 0 & -v & 0 \end{array} \right) | u,v \in \mathbb{C} \}$ and $\mathfrak{g} = \mathfrak{so_4}\mathbb{C}$ but I'm looking for more general help on how to find all the elements of $\mathfrak{h}^*$ that give nonzero root spaces.

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up vote 2 down vote accepted

I'm not sure if there's a better way, but one way is to write a basis of $\mathfrak{g}$ and a basis ${ h_1, h_2}$for $\mathfrak{h}$ and then create the matrix for each $Ad_{h_i}$ in the basis of $\mathfrak{g}$.

Then, one solves the usual eigenvalue problem for both of the $Ad_{h_i}$. Since $Ad_{h_1}$ and $Ad_{h_2}$ commute, you will end up finding they have a common eigenspaces. Then, reading off the eigenvalues for the common eigenspaces and using linearity gives the $\lambda$ functionals.

I don't have my copy of Fulton and Harris's Representation Theory book handy, but they might even work this out as an example. At the very least, they'd work out some example, which may give you better tools to work with than the one I've described.


Here is a start on working it out. Let $h_1 = E_{12}$, the matrix with a 1 in the $1,2$ slot an a zero else. Let $h_2 = E_{34}$.

Consider the matrix $X = \begin{bmatrix} 0 & 0 & a & b \\ 0 & 0 & c & d \\ -a & -b & 0 & 0 \\ -b & -d & 0 & 0 \end{bmatrix}$ in $\mathfrak{g}$. (I've set the entries corresponding to $\mathfrak{h}$ equal to 0 because we know what $Ad_{h_i}$ does to those). To simplify notation, I'll only describe the top right 2 x 2 block at all times. The top left and bottom right 2 x 2 blocks are always 0s, while the bottom left 2 x 2 block is always whatever is necessary to guarantee I'm looking at an element of $\mathfrak{g}$.

Computing, one finds $$ Ad_{h_1}X = \begin{bmatrix} c & d \\ -a & -b\end{bmatrix}$$ (meaning, by the above convention, that the bottom left is what it needs to be to make $Ad_{h_1}X$ antisymmetric and the top left and bottom right are both 0s).

In order for $X$ to be an eigenvector of $Ad_{h_i}$ with eigenvalue $\eta$, we must have $$\begin{align} c &=& \eta a \\ d &=& \eta b\\ -a &=& \eta c\\ -b &=& \eta d\end{align}$$ simultaneously holding.

This implies that $\eta = \pm i$. When $\eta = i$, the eigenvectors are $\begin{bmatrix} 1 & 0 \\ i & 0\end{bmatrix}$ and $\begin{bmatrix} 0 & 1 \\ 0 & i \end{bmatrix}$. Let $V = E_i$ be the eigenspace to $Ad_{h_1}$ corresponding to the eigenvalue $i$ (so $V$ is just the span of these two $X$s).

I claim that $Ad_{h_2}:V\rightarrow V$. In fact, this is general: if $A$ and $B$ commute, then $B$ maps any eigenvector of $A$ to a new eigenvector of $A$ with the same eigenvalue. For, if $Ax = \lambda x$, then $ABx = BAx = B\lambda x = \lambda B x$.

We now wish to find eigenvalues of $Ad_{h_2}|_V$.

Let $Y = \begin{bmatrix} a & b \\ ai & bi\end{bmatrix}$ be an arbitrary element of $V$. Then $Ad_{h_2}Y = \begin{bmatrix} b & -a\\ bi & - ai\end{bmatrix}$.

We see that $Y$ is an eigenvector of $Ad_{h_2}$ with eigenvalue $\mu$ iff $$\begin{align} b &=& \mu a \\ -a &=& \mu b\\ bi &=& \mu a i \\ -ai &=& \mu bi\end{align}$$

Again, this implies that $\mu = \pm i$. Choosing $\mu = i$, we see that $Y = \begin{bmatrix} 1 & i \\ i & -1\end{bmatrix}$ is both in $V$ and is an eigenvector for $Ad_{h_2}$.

This implies that $span\{Y\}$ is an eigenspace for both $Ad_{h_1}$ and $Ad_{h_2}$, both with eigenvalue $i$. Thus, if $\lambda(h_1) = \lambda(h_2) = i$ and we extend $\lambda$ to all of $\mathfrak{h}$ by linearity, then $Y\in \mathfrak{g}_\lambda$.

From here, you find the other 3 eigenspaces by considering the cases $\eta = \pm i$ and $\mu = \pm i$ separately. (I've just done the case where we choose plus signs for both).

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Okay by letting – Zeophlite Feb 14 '11 at 16:28
@Jason, Okay by letting h1 = {u=1,v=0} and h2 = {u=0, v=1}, then taking ad_hi of a generic element of g I get two matrices. They both have the same eigenvalues, $\pm \frac{\sqrt{-b^2-c^2-d^2-e^2\pm \sqrt{\left((c+d)^2+(b-e)^2\right) \left((c-d)^2+(b+e)^2\right)}}}{\sqrt{2}}$ , with a,b,c,d,e,f the entries in my matrix for g. What do you mean by reading them off and using linearity to get the lambda functionals – Zeophlite Feb 14 '11 at 16:35
You still need to split things into eigenspaces. This should lead to writing $\mathfrak{g}$ as a direct sum of vector spaces, each of which is an eigenspace for both $Ad_{h_1}$ and $Ad_{h_2}$. – Jason DeVito Feb 14 '11 at 18:02
Then, if, say $v\in E$ where $E$ is an eigenspace, you'd have $Ad_{h_i}v = \lambda_i v$ for some $\lambda_i$. This gives $\lambda:\mathfrak{h}\rightarrow \mathbb{C}$ where $\lambda(h_i) = \lambda_i$, and then you extend by linearity. – Jason DeVito Feb 14 '11 at 18:03
@Jason, But I have 4 different eigenvalues and only two basis elements for h – Zeophlite Feb 14 '11 at 21:03

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