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The example I'm working off of is $A_2$, so in this case I've determined that the dimension of the Lie algebra is 8, with the dimension of the Cartan subalgebra H being 2. Denoting the basis vectors of the subalgebra as $h_1$ and $h_2$, and the corresponding roots of H* as $\lambda_i$ where i goes from 1 to 6. A defining feature of the Cartan subalgebra is that

$$[h_i, e_j] = ad(h_i)e_j = \lambda_j(h_i)e_j$$

where $e_j$ denotes the other Lie algebra vectors. Therefore the structure constants should be

$$C^i_{ji} = \lambda_j(h_i)$$ (no summation)

But I have no clue what these numbers actually are or how to calculate them.

The lecture series I have been following, https://youtu.be/G9uVcit_VwY?t=4986 seemingly assigns the structure constants the coordinates of a basis vector in R^2, but I'm not sure why. In principle the two fundamental roots have to be 120 degrees apart, so any arbitrary two roots could work as long as it fulfills that condition, meaning you could get an infinite number of different structure constants for $C^i_{ji}$.

In addition, I'm not exactly sure how $h_1$ and $h_2$ are chosen. If $\pi_1$ and $\pi_2$ are the fundamental roots, ie the basis of H*, then the standard choice would be to choose the h's such that

$$\pi_j(h_i) = \delta_{ij}$$

but as far as I understand, this choice is mostly for convenience and not strictly necessary, given that this is a subset of the problem for finding the lambda's, what is the exact process for choosing this basis? Is there a certain degree of freedom we have when choosing?

Kevin Guo
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    See this post, or see this post. The Dynkin diagram gives the Cartan matrix, and this gives you the structure constants by the Serre relations. – Dietrich Burde Sep 24 '22 at 18:13
  • I see, so in the first post they picked $\pi_1$ to have coords (1, 0) and $\pi_2$ to have $\frac{-1}{2},\frac{\sqrt(3)}{2}$, and then chose $h_1$ and $h_2$ such that the structure constants equaled those coords. The only thing I'm confused about about is if you instead took $\pi = (\frac{1}{\sqrt(2)},\frac{1}{\sqrt(2)})$ and then had $\pi_2$ to be rotated 120 degrees from there. Obviously the structure constants would be different, so would this constitute a different Lie algebra? I can't imagine it should, but I don't know why Lie algebra is independent of the coords chosen for the $\pi$s – Kevin Guo Sep 24 '22 at 21:06
  • Does this post help? The "rotated" Lie algebra then is isomorphic. – Dietrich Burde Sep 25 '22 at 08:56
  • I may be misunderstanding something, but if two Lie algebras are isomorphic, then shouldn't their structure constants be the same? I get that the root systems would be functionally the same(same angles between fundamental roots), but wouldn't a rotated root system lead to different structure constants? – Kevin Guo Sep 25 '22 at 14:02
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    No, take the $2$-dimensional nonabelian Lie algebra with basis $(x,y)$. All Lie brackets $[x,y]=rx+sy$ are isomorphic for all $r,s\neq 0$. So isomorphic Lie algebras can have different structure constants. – Dietrich Burde Sep 25 '22 at 14:55
  • I see, my intution is that if two Lie algebras are essentially rotated by some arbitrary degree, then they are isomorphic, and this rotation can originate from the rotation of the root system, but I am having trouble showing that. How would you formally prove this isomorphism? – Kevin Guo Sep 26 '22 at 10:00
  • Nathan Jacobson's "Lie Algebras" gives a proof for an isomorphism theorem on page 127 that seems to more or less describe what Dietrich is saying, but the proof shows something about the "multiplication table" of the two Lie algebras to be the same, and I can't seem to find if "multiplication table" is an old word for structure constant – Kevin Guo Sep 26 '22 at 20:34
  • An important thing here is that structure constants are dependent on your basis so changing your basis will automatically change your constants. There may be a nice basis to choose (e.g one following a root space decomposition) but even rescaling the basis vectors will change the constants so they are not fixed in any way. – Callum Sep 27 '22 at 06:19

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