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Let $L$ be semisimple Lie algebra over $\mathbb{C}$ of finite dimension.

Let $H$ be a maximal toral (abelian) subalgebra of $L$.

Then set of semi-simple elements $\{ ad_h:L\rightarrow L : h\in H\}$ is simultaneously diagonalizable.

So $L$ is direct sum of subspaces $L_a:=\{ x\in L : ad_h(x)=a(h)x\}$, where $h\mapsto a(h)$ is in $Hom(H,\mathbb{C}$).

Problem: I confused in thinking what $L_a$ is, in explicit words? Which one of the following is correct way to think about $L_a$?

(1) $L_a$ is the representation of $H$, where $H$ acts via $ad$ as one dimensional representation.

(2) $L_a$ is the one dimensional representation of $H$ via $ad$.

Especially, when $H$ is set of scalar matrices in $\mathfrak{gl}(n,\mathbb{C})$ then what $L_a$ should be taken?

Please help me to clarify the definition of $L_a$ which should we intuitively think.

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  • $\begingroup$ Since $A = \{ad_h : h \in H\}$ is simultaneously diagonalizable, $L$ has a basis of common eigenvectors for $A$. Let $x$ be one of these eigenvectors. We know $ad_{h_1}(x) = \lambda_1 x$ and $ad_{h_2}(x) = \lambda_2 x$, but in general $\lambda_1 \neq \lambda_2$ for $h_1 \neq h_2$. However there is a linear functional $\alpha : H \to \mathbb C$ such that $ad_h(x) = \alpha(h)x$ for all $h \in H$. Perhaps this will clarify the definition of $L_\alpha$. $\endgroup$ Commented Jan 13, 2017 at 5:23
  • $\begingroup$ Let $\langle v_i\rangle$, $i=1,2,...,n$ be the $n$-eigenspaces from which we obtain basis for diagonalization. Suppose, for example, $ad_h(v_1)=a(h)v_1$ and also $ad_h(v_2)=a(h)v_2$. Then in $L_a$, we should take both $v_1,v_2$ or just one of them? (I read somewhere in online notes that $L_a$'s are one-dimensional; without going into their proof, that statement confused.) $\endgroup$
    – Beginner
    Commented Jan 13, 2017 at 5:29
  • $\begingroup$ I'm not sure I understand what you're asking. It is true that the $L_\alpha$'s are one dimensional, but this is not obvious from the definition. But if you believe this, you can think of $L_\alpha$ as being generated by one of the common eigenvectors for $\{ad_h : H \in H\}$. $\endgroup$ Commented Jan 13, 2017 at 5:33
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    $\begingroup$ @Beginner The problem with your example is that $H$ is not maximal. Take all diagonal matrices instead. Then it should also be clear how to decompose $L$. I also weren't aware of the statement that all the $L_a$ are one-dimensional. Does anybody have a reference on this? $\endgroup$
    – Maik Pickl
    Commented Jan 13, 2017 at 14:15
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    $\begingroup$ @MaikPickl Humphrey's "Introduction to Lie Algebras and Representation Theory," or any other intro Lie algebra book will have this proof. $\endgroup$ Commented Jan 16, 2017 at 4:51

1 Answer 1

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$\mathfrak{gl}(n, \Bbb{C})$ is not semisimple, but let us look at $L = \mathfrak{sl}(n, \Bbb{C})$. Let's say $n=4$ to have a very concrete example.

Choose $H$ to be the diagonal matrices, a general element being

$$h = \pmatrix{h_1 & 0 & 0 & 0\\ 0 & h_2 & 0 & 0\\ 0 & 0 & h_3 & 0\\ 0 & 0 & 0 & h_4}.$$

(With the extra condition $\sum_1^4 h_i = 0$ to ensure we are in $\mathfrak{sl}(4,\mathbb{C})$, which will not make any actual difference in what follows; see remark at the very end.)

Now let us see what happens if we let this operate (via $ad$) on a completely arbitrary element of $L$, say, $$x = \pmatrix{d_1 & e_{12} & e_{13} & e_{14}\\ f_{21} & d_2 & e_{23} & e_{24}\\ f_{31} & f_{32} & d_3 & e_{34}\\ f_{41} & f_{42} & f_{43} & d_4}.$$ We have:

(*)$\qquad ad(h) (x) = [h,x] = \pmatrix{0 & (h_1-h_2)\,e_{12} & (h_1-h_3)\,e_{13} & (h_1-h_4)\,e_{14}\\ (h_2-h_1)\,f_{21} & 0 & (h_2-h_3)\,e_{23} & (h_2-h_4)\,e_{24}\\ (h_3-h_1)\,f_{31} & (h_3-h_2)\,f_{32} & 0 & (h_3-h_4)\,e_{34}\\ (h_4-h_1)\,f_{41} & (h_4-h_2)\,f_{42} & (h_4-h_3)\,f_{43} & 0}.$

I claim that this, in this concrete example, is nothing else than the famous $$(**) \qquad L = H\oplus \bigoplus_{\alpha\in \Phi} L_\alpha$$ where $\Phi \subset Hom(H,\Bbb{C})$ is the root system, and the $L_\alpha$ are the root spaces you ask about. This is a decomposition of $L$ as $H$-representation (where $H$ operates via $ad$); $L_\alpha$ is the subspace on which $H$ operates via the character $\alpha$. (It turns out that such an $L_\alpha$ is always one-dimensional, but that is a non-trivial result in general.)

Let's check how equation (*) "is" equation (**) applied to our concrete example. The roots are linear forms (characters, one-dimensional representations) of $H$. In this case, they are all of the form $$\alpha\left(\pmatrix{h_1 & 0 & 0 & 0\\ 0 & h_2 & 0 & 0\\ 0 & 0 & h_3 & 0\\ 0 & 0 & 0 & h_4}\right) = h_i-h_j$$ for certain $1\le i\neq j\le 4$.

Actually, let's call $\alpha_1$ the one sending $h$ to $h_1-h_2$. You will notice in (*) that there is exactly one matrix entry which gets multiplied by $h_1-h_2$ when the whole matrix is acted on by $h$: The one at position 12. Well, that means the root space to that root $\alpha_1: h\mapsto h_1-h_2$ is $$L_{\alpha_1} = \pmatrix{0 & * & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0}.$$

Similarly, if $\alpha_2$ is the one that sends $h$ to $h_2-h_3$, we have

$$L_{\alpha_2} = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & * & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0}$$

and $\alpha_3:h \mapsto h_3-h_4$ is the one that rules over $$L_{\alpha_3} = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & *\\ 0 & 0 & 0 & 0}$$

Now I chose these three for a purpose; namely, all the others can be written as combinations of them. For example, what about the $\alpha$ that sends $h$ to $h_1-h_4$, the one that operates on the upper right corner? Well, that is $\alpha_1+\alpha_2+\alpha_3$ (because it sends $h$ to $(h_1-h_2)+(h_2-h_3)+(h_3-h_4) = h_1-h_4$). Or the one that operates on the entry $f_{31}$? That is $-\alpha_1-\alpha_2$, which sends $h$ to $h_3-h_1$.

We can rewrite our whole equation (*) as:

$$ad(h) (x) = \pmatrix{0 & \alpha_1(h)\,e_{12} & (\alpha_1+\alpha_2)(h)\,e_{13} & (\alpha_1+\alpha_2+\alpha_3)(h)\,e_{14}\\ -\alpha_1(h)\,f_{21} & 0 & \alpha_2(h)\,e_{23} & (\alpha_2+\alpha_3)(h)\,e_{24}\\ (-\alpha_1-\alpha_2)(h)\,f_{31} & -\alpha_2(h)\,f_{32} & 0 & \alpha_3(h)\,e_{34}\\ (-\alpha_1-\alpha_2-\alpha_3)(h)\,f_{41} & (-\alpha_2-\alpha_3)(h)\,f_{42} & -\alpha_3(h)\,f_{43} & 0}.$$

(This might look a bit awkward and not really simplifying in this example, but it is something that generalises and is of great importance to all root systems: $(\alpha_1, \alpha_2, \alpha_3)$ is a basis or set of simple roots of our $\Phi$.)

It turns out that in our case, $\Phi$ consists of the twelve roots $$\{\pm \alpha_1, \quad \pm\alpha_2, \quad \pm \alpha_3, \quad \pm(\alpha_1+\alpha_2), \quad \pm(\alpha_2+\alpha_3), \quad \pm(\alpha_1+\alpha_2+\alpha_3\}.$$ And you can explicitly see those root spaces and that they are one-dimensional. Just one more example: $$L_{-\alpha_2-\alpha_3} = \pmatrix{0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & * & 0 & 0}$$ etc.

Their sum is $$\bigoplus_{\alpha\in \Phi} L_\alpha = \pmatrix{0 & * & * & *\\ * & 0 & * & *\\ * & * & 0 & *\\ * & * & * & 0}$$ and this should clarify equation ($**$). Finally, the $H$ in ($**$) is sometimes also written as $L_0$, and that is totally consistent, because in (*) you see how on the diagonal, $H$ actually acts via the constant eigenvalue zero ("$\alpha_0 = 0$"), and $$L_0 = H = \pmatrix{* & 0 & 0 & 0\\ 0 & * & 0 & 0\\ 0 & 0 & * & 0\\ 0 & 0 & 0 & *}$$ is the only space in the decomposition which is not one-dimensional.

All this works for $\mathfrak{gl}(4,\mathbb{C})$ as well. The extra condition in $\mathfrak{sl}(4, \Bbb{C})$ that all elements $h$ satisfy $\sum h_i=0$ basically "mods out" the centre of $\mathfrak{gl}(4) =$ the scalar matrices $\pmatrix{h & 0 & 0 & 0\\ 0 & h & 0 & 0\\ 0 & 0 & h & 0\\ 0 & 0 & 0 & h}$ and thus ensures that for every root $\alpha$, there are elements in $H$ on which $\alpha$ is not zero. The fact that the roots span the whole dual space $H^* = Hom(H, \Bbb{C})$ is useful in the abstract theory, so here $H$ should better be $3$-dimensional.

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