# Irreducibility and weights of a representation

For some reason I can't get a good hold of those topics (I'm reading Brian C. Hall's Lie Groups, Lie algebras and Representations. So it's matrices only). I'll try to narrow it a bit more:

Irreducibility - I've found only one example of proof showing that representation is irreducible and it was straightforward from the definition: that it doesn't contain non-trivial, invariant subspaces. Is there any other, not too advanced way? Using roots/weights?

Weights - Only one example was given in the book, for the space of $sl(3,\mathbb{C})$. For this space, a special basis was chosen with $H_1 = diag(1,-1,0)$ and $H_2 = diag(0,1,-1)$ and then the weights are $\mu = (m_1,m_2) \in \mathbb{C}$ such that $$\pi(H_1)v = m_1v,$$ $$\pi(H_2)v=m_2v.$$ How do I find a "good" basis in general? The trivial basis seems not to work in that case.

And one last question, about the highest weight - I have found two, possibly equivalent definitions and I'm not sure which one to use:

1. (Lecture) All $v$ such that $\pi(z)v = v$ and $z \in Z(G)$, when $Z(G)$ is the triangular matrices with ones along the diagonal in $GL_n$.

2. (Book) By comparison of the weights using positive roots of a representation - $\mu_1 > \mu_2$ iff $\mu_1 - \mu_2 = a\alpha_1 + b\alpha_2$, when $\alpha_1,\alpha_2$ are the positive roots and $a,b \geq 0$.

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What do you mean by "trivial basis"? –  Eric O. Korman Mar 2 '13 at 15:49
Not sure if it's the right definition, but what I meant is that each base element is a matrix with 1 in one cell only. –  Pavel Mar 2 '13 at 15:58
That won't give a basis for $sl(3,\mathbb C)$ since that is the Lie algebra of matrices with zero trace. In general, to talk about weights one must first start with a Cartan subalgebra, i.e. a maximal abelian subalgebra. For many algebras, this will be the diagonal matrices. For $sl(n,\mathbb C)$, the simplest basis for it's diagonal matrices are those matrices with two entries along the diagonal, one being 1 and the other -1. –  Eric O. Korman Mar 2 '13 at 16:08
Yes, my bad, this basis is not the right one for my example. So the $H_1,H_2$ chosen above, are the elements of the Cartan subalgebra of $sl(3,\mathbb{C})$ (and the only ones)? –  Pavel Mar 2 '13 at 16:42

B) Note that the weights of a representation, considered as elements of the dual $\mathfrak{h}^*$ of the Cartan subalgebra, do not depend on the basis of $\mathfrak{h}$. However, if you want to draw them in a weight diagram, you have to pick a basis for $\mathfrak{h}$, which chooses the dual basis for $\mathfrak{h}^*$. For example, the Cartan subalgebra of $\mathfrak{sl}(3)$ consists of all 3x3 traceless matrices, so a possible basis is $$\begin{pmatrix}1 & 0 & 0\\0 & -1 & 0\\ 0 & 0 & 0 \end{pmatrix},\quad \begin{pmatrix}1 & 0 & 0\\0 & 0 & 0\\ 0 & 0 & -1 \end{pmatrix}$$ There's nothing like a good or bad basis, although with some choices (when the Killing form is the identity), the weight diagrams will look more symmetric.
C) A highest weight vector is one annihilated by all the positive roots (considered as elements of the Lie algebra). This is just your definition (1) translated from Lie group to Lie algebra setting. To explain your definition (2) from this point of view, note that if a positive root $\alpha$ does not annihilate a given vector with weight $\beta$, the resulting vector will have weight $\alpha+\beta$, leading to a greater value of $\mu$. Thus a vector which maximizes $\mu$ must be annihilated by all positive roots. Note that in a highest weight representation (with a unique highest weight vector), the converse is also true.
If you know exactly how each element of the Lie algebra acts in your representation, then you also know all the weights. For example in $\mathfrak{sl}(3)$, if you parametrize an element of the Cartan subalgebra as $a\alpha + b\beta$, where $\alpha$, $\beta$ are the two matrices above, the weights of the fundamental representation will be $a+b$, $-a$ and $-b$. –  Dalimil Mazáč Mar 11 '13 at 21:15