Finding the Dynkin basis of the weight space.

Question

I'm trying to understand how Dynkin diagrams work but I'm struggling quite hard.

For example I want to find the Dynkin basis of the weight space. I know that the weight space is the dual space of the root space and that if we take as basis of the root space $\mathcal{B} = \{2\alpha^{(i)}/(\alpha^{(i)}, \alpha^{(i)})\}$ where $\alpha^{(i)}$ are the simple roots. Now the weight space consists of linear maps $\Lambda_{(i)}$ with $\Lambda_{(i)}\left(\frac{2\alpha^{(j)}}{(\alpha^{(j)}, \alpha^{(j)})}\right) = \delta^j_i.$ These are the fundamental weights and form the Dynkin basis of the weight space: $\mathcal{B}^* = \{\Lambda_{(i)} \vert i = 1, \dots, r\}$ where $r$ is the rank of the algebra.

Now I know we can relate the Dynkin diagrams to the Dynkin basis, but I don't really udnerstand how. For example, I want to do this for $G_2$, but I don't know where to start. It seems too much work to first compute all the simple roots...

Answer 1

I am not completely sure where exactly you have difficulties. Maybe an example can help.

How Cartan matrices and root systems can be retrieved from the Dynkin diagram on the example of $G_2\,.$

The Dynkin diagram tells us that $\alpha \prec \beta$ and $\langle \alpha,\beta \rangle \langle \beta ,\alpha \rangle=3\,.$ The cosine formula tells us, that the angle they enclose is $30°$ but this doesn't matter here. Since the only ways to get an integer product of three are $3\cdot 1 = (-3)\cdot (-1)=3$ we may assume w.l.o.g. and the sign in the theorem of root systems in mind, that $\langle \alpha,\beta\rangle = -1$ and $\langle \beta,\alpha\rangle=-3\,.$ This produces the Cartan matrix $$ G_2\, : \,\begin{bmatrix}2&-1\\-3&2\end{bmatrix} $$ Next we calculate by linearity in the first argument \begin{align*} \alpha - \langle \alpha,\beta\rangle\cdot \beta &= \alpha+\beta \\ \beta - \langle \beta,\alpha \rangle \cdot \alpha &= 3\alpha+\beta \\ (\alpha+\beta)-\langle \alpha+\beta,\alpha\rangle\cdot \alpha &=2\alpha+\beta \\ (3\alpha+\beta)-\langle3\alpha+\beta,\beta \rangle\cdot \beta&=3\alpha+2\beta \end{align*} From the decomposition formula $$ \mathfrak{g} = \mathfrak{h}\oplus \sum_{\alpha \in\Phi^+} \mathbb{F}E_\alpha \oplus \sum_{\alpha \in\Phi^-} \mathbb{F}E_\alpha $$ we get with a two dimensional Cartan subalgebra $\mathfrak{h}=\operatorname{span}\{\,H_\alpha,H_\beta\,\}$ the roots \begin{align*} \Phi^+ &= \{\,\alpha,\beta,\alpha+\beta,2\alpha+\beta,3\alpha+\beta,3\alpha+2\beta\,\}\\ \Phi^- &= \{\,-\alpha,-\beta,-\alpha-\beta,-2\alpha-\beta,-3\alpha-\beta,-3\alpha-2\beta\,\}\\ \end{align*} and $$ G_2 = \operatorname{span}\{\,H_\alpha,H_\beta\,\} \oplus \sum_{\gamma \in \Phi^+\,\cup\, \Phi^-} \mathbb{F}\cdot E_\gamma $$

If you want to create the weight spaces of arbitrary finite-dimensional representations other than the adjoint, you could look e.g. at the proof of theorem 7.2 for the other simple case $\mathfrak{sl}(2)$ in Humphreys's book about Lie algebras and their representations, GTM 9. But I am sure that this example can be found easily on other sources.

Answer 2

So there appears to be a little confusion here. I've not heard the term "Dynkin basis" before but you seem to want the fundamental weight basis. But this is directly deductible from the simple root basis which the diagram gives you. Note this is not a basis of the "weight space" but the "weight lattice".

So as you have noted: To the simple roots we have a natural set of dual elements, the simple coroots. These are the elements $\alpha_i^\vee \in \mathfrak{h}$ such that $\alpha_j(\alpha_i^\vee) = 2\frac{(\alpha_j,\alpha_i)}{(\alpha_i,\alpha_i)}$. Then the fundamental weights are dual to the simple coroots so $\omega_j(\alpha_i^\vee) = \delta^{i,j}$.

In particular we note that they are in 1-to-1 correspondence with the simple roots and they both form bases of $\mathfrak{h}^*$. Indeed we know exactly what the change of basis matrix is: It is the Cartan matrix corresponding to the diagram. You can deduce this from the relations we have just stated.