# Lie Algebra-Roots and Dynkin basis

• December 29th 2010, 06:44 AM
ppyvabw
Lie Algebra-Roots and Dynkin basis
Given a Lie algebra I understand all the stuff about the Cartan-Weyl basis, the roots and stuff, so for example, for A2, the roots are

$\alpha=(1,0), \ \ \beta=\frac{1}{2}(-1,\sqrt{3}),\ \ \gamma=\frac{1}{2}(1,\sqrt{3})$

with simple roots $+\alpha$ and $+\beta$. The scalar products are

$(\alpha,\alpha)=1,\ \ (\beta,\beta)=\frac{-1}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}=1$ and $(\alpha,\beta)=-\frac{1}{2}$ using the definition of the scalar product in this basis from which one calculates the Cartan matrix and angles between roots blah blah blah...

I am reading 'Symmetries, Lie Algebras and Representations: A Graduate Course for Physicists' By Jürgen Fuchs, Christoph Schweigert and I do not understand the Dynkin basis. It says the quadratic form matrix is given by

$G_{ij}=(\Lambda_{(i)},\Lambda_{(j)})$ and $G^{ij}=(\check{\alpha}^{(i)},\check{\alpha}^{(j)}) =\frac{2}{(\alpha^{(i)},\alpha^{(i)})}A_{ij}$

where $A^{ij}$ is the cartan matrix, $\check{\alpha}$ is the co-root $\check{\alpha}=\frac{2\alpha}{(\alpha,\alpha)}$ and $\Lambda$ is the weight in the dual space with $\Lambda_{(i)}(\check{\alpha}^{(j)})=\delta_i^j$. From this, they claim $(\check{\alpha}^{(j)})_i=(\Lambda_{(i)})^j=\delta_ i^j$ and the Dynkin basis is

$(\alpha^{(i)})^j=A^{ij}$.

How does one calculate the matrix elements $G_{ij}$? It seems a recursive definition of $G_{ij}$ because it is written in terms of the inner product but to define the inner product in this basis one needs to already know $G_{ij}$

Does my confusion make sense to anyone and could someone please clear this up for me?
• December 30th 2010, 06:17 PM
TheArtofSymmetry
Quote:

Originally Posted by ppyvabw
Given a Lie algebra I understand all the stuff about the Cartan-Weyl basis, the roots and stuff, so for example, for A2, the roots are

$\alpha=(1,0), \ \ \beta=\frac{1}{2}(-1,\sqrt{3}),\ \ \gamma=\frac{1}{2}(1,\sqrt{3})$

with simple roots $+\alpha$ and $+\beta$. The scalar products are

$(\alpha,\alpha)=1,\ \ (\beta,\beta)=\frac{-1}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}=1$ and $(\alpha,\beta)=-\frac{1}{2}$ using the definition of the scalar product in this basis from which one calculates the Cartan matrix and angles between roots blah blah blah...

I am reading 'Symmetries, Lie Algebras and Representations: A Graduate Course for Physicists' By Jürgen Fuchs, Christoph Schweigert and I do not understand the Dynkin basis. It says the quadratic form matrix is given by

$G_{ij}=(\Lambda_{(i)},\Lambda_{(j)})$ and $G^{ij}=(\check{\alpha}^{(i)},\check{\alpha}^{(j)}) =\frac{2}{(\alpha^{(i)},\alpha^{(i)})}A_{ij}$

where $A^{ij}$ is the cartan matrix, $\check{\alpha}$ is the co-root $\check{\alpha}=\frac{2\alpha}{(\alpha,\alpha)}$ and $\Lambda$ is the weight in the dual space with $\Lambda_{(i)}(\check{\alpha}^{(j)})=\delta_i^j$. From this, they claim $(\check{\alpha}^{(j)})_i=(\Lambda_{(i)})^j=\delta_ i^j$ and the Dynkin basis is

$(\alpha^{(i)})^j=A^{ij}$.

How does one calculate the matrix elements $G_{ij}$? It seems a recursive definition of $G_{ij}$ because it is written in terms of the inner product but to define the inner product in this basis one needs to already know $G_{ij}$

Does my confusion make sense to anyone and could someone please clear this up for me?

Let $\Delta={\alpha_1, \ldots, \alpha_r}$ be a base consisting of positive simple roots. The fundamental weights with respect to $\Delta$ are the elements $\mu_1, \ldots, \mu_r$ satisfying

$2\dfrac{<\mu_k, \alpha_m>}{<\alpha_m, \alpha_m>}=\delta_{km}, k, m=1,\ldots, r$.

Then, the kth fundamental weight is

$\mu_k=\dfrac{1}{2}\mu\dfrac{<\alpha_k, \alpha_k>}{<\mu, \alpha_k>}$, where $\mu$ is an integral element.

I didn't study your book, but I think p38 shows the relevant example. I think $G_{ij}$ needs fundamental weights which can be obtained from the above formula. My knowledge of this area is limited, so hope this is a right direction:)