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Is there an efficient way of finding commutation relations for a Lie algebra?

For $\mathfrak{su}(2)$ with the Pauli matrices multiplied by $-\frac i2$ we get only three non-trivial commutation relations, but for larger Lie algebras there can be many, many times more.

How does one efficiently find them all? Perhaps they fall into classes of some sort?


I.e. The basis of the three-dimensional Lie algebra $\mathfrak{sl}(2,\Bbb F)$ is given by:

$$e=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad h=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\quad f=\begin{bmatrix}0&0\\1&0\end{bmatrix}$$

The corresponding nontrivial commutation relations are:

$$[h,e]=2e,\quad [h,f]=-2f,\quad [e,f]=h$$

Is there a faster way of finding these, than computing the bracket for basis elements and seeing if they are related to another basis element?

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  • $\begingroup$ @DietrichBurde I didn't know how to answer that, so I editted the question with a clarification for what I meant to say $\endgroup$ – So many hats Aug 20 '15 at 0:27
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    $\begingroup$ OK, then the answer might be helpful for you (hopefully, I mean). $\endgroup$ – Dietrich Burde Aug 20 '15 at 14:38
  • $\begingroup$ I would like to suggest you Maple packages LieAlgebra and DifferentialGeometry for automatically calculating commutation relations. $\endgroup$ – IgotiT Jun 19 '16 at 13:47
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In general, given a subalgebra $\mathfrak{h}$ of $\mathbb{gl}_n(K)$ consisiting of matrices, one needs to compute all commutators of a basis $f_1,\ldots ,f_m$ of the vector space $\mathfrak{h}$. Using the canonical basis $E_{ij}$ of $\mathbb{gl}_n(K)$, there are formulas available for the commutators $[E_{ij},E_{kl}]$, i.e., $$ [E_{jk},E_{lm}]=\delta_{kl}E_{jm}-\delta_{jm}E_{lk}. $$ For example, in the case of $\mathbb{sl}_2(K)$ we can choose a basis $e=E_{12}$, $f=E_{21}$ and $h=E_{11}-E_{22}$ and apply the formula. Then we obtain the commutation relations without further computation. Applying more theory we can almost guess the commutator relations, e.g., since $ad(h)$ acts diagonally, we already know that $[h,e]=\lambda e$, $[h,f]=\mu f$ for some scalars $\lambda,\mu$.

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  • $\begingroup$ I like the answer, but I'll leave it open until I am fully happy with (my understanding of) it. Thanks $\endgroup$ – So many hats Aug 21 '15 at 15:39
  • $\begingroup$ The last part can be explained in more detail, but requires then some theory for semisimple Lie algebras. For the commutation relations one can use the Serre relations, too. $\endgroup$ – Dietrich Burde Aug 22 '15 at 17:39

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