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The simple root $\alpha_i$ basis is not an orthonormal basis, as can be seen from the Cartan matrix, which encodes how much they aren't orthonormal.

For simplicity, let's assume a simply-laced Lie algebra where the simple simple roots are self-dual.

In the Dynkin basis we write each weight with the components with respect to the fundamental weights $\Lambda_i$. These are defined by their property

$$ \Lambda_i(\alpha_j) = \delta_{ij} . $$

Does this mean that the Dynkin basis is an orthonormal basis?

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No. Look at $\mathfrak{sl}_3$. The simple roots are $\alpha_1,\alpha_2$ with $$(\alpha_1,\alpha_1)=2\;\;\;\mbox{and}\;\;\;(\alpha_1,\alpha_2)=-1.$$

Write $\Lambda_1=a\alpha_1+b\alpha_2$ and solve the system $$\begin{cases}(\Lambda_1,\alpha_1)=1\\(\Lambda_1,\alpha_2)=0\end{cases}$$ to get $\Lambda_1=\frac{2}{3}\alpha_1+\frac{1}{3}\alpha_2$ (similarly, $\Lambda_2=\frac{1}{3}\alpha_1+\frac{2}{3}\alpha_2$).

Now compute that $(\Lambda_1,\Lambda_1)=\frac{2}{3}$ and $(\Lambda_1,\Lambda_2)=\frac{1}{3}$.

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  • $\begingroup$ Thx a lot! The same calculation for $su(3)$ can be found here books.google.de/… $\endgroup$ – JakobH Aug 26 '15 at 9:18

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