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Is there a procedure for finding the basis of a Lie algebra generated by a (known) finite set $A_{1}, \ldots, A_{K}$ of skew-hermitian operators on an Hermitian space $\mathbb{C}^{K}$? The Lie bracket is defined as usual: $$ [A, B] = AB - BA. $$ My main concern is that the Lie brackets of pairs of $A_{k}$'s are generally not in the span of the $A_{k}$'s. So, I am looking for a systematic way to generate all the brackets that are needed to span the entire Lie algebra generated by the $A_{k}$'s.

I have found out about the Dynkin diagrams and root systems, which determine the Lie algebra in the classification, but I am not yet understanding enough about them to determine whether they can be used to devise the procedure described in the question. Any guidance?

avs
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    A basis of the Lie algebra by definition is just a basis of the underlying vector space. So this is linear algebra. – Dietrich Burde Dec 06 '20 at 14:26
  • Can't the Lie bracket of two matrices from among the given $A_{k}$'s be a matrix that is not in the span of the given $A_{k}$'s? If so, my main question is, how to organize the process of forming, systematically, the Lie brackets of the $A_{k}$'s, and then their Lie brackets, and when to stop, knowing that I know have enough of them to generate the entire Lie algebra? – avs Dec 06 '20 at 17:21
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    Dynkin diagrams and root systems are helpful only for semisimple, or maybe reductive, Lie algebras. There are tons of other Lie algebras, in particular solvable ones whose classification is "wild" and will not be helpful here. – Torsten Schoeneberg Dec 06 '20 at 18:19
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    Do I understand the question correctly: Given a finite set of matrices in $M_K(\mathbb C)$, what is the Lie algebra generated by them? – Torsten Schoeneberg Dec 06 '20 at 18:30
  • Yes, essentially, that is it. – avs Dec 06 '20 at 21:13

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Since the Lie algebra is also a vector space, you can choose some basis and treat your matrices $A_i$ as vectors in that basis (for example, if you just flatten a matrix into a column vector, that will be okay). If you arrange these columns into a matrix, its rank will be equal to the dimension of the vector space spanned by the vectors.

To find the basis of the Lie algebra, you can calculate commutators, add them to the collection of vectors, and stop when it's clear that the rank will no longer increase. More specifically, I'd do it as follows:

  1. Calculate all commutators of the "first generation": $[A_i, A_j]$. Turn them into column vectors and append them to the matrix of vectorized matrices, then calculate its rank.

  2. Calculate the vectors of the "second generation" by calculating all commutators between the vectors of all preceding generations (perhaps you can skip calculating commutators in the zeroth generation because you already have at this point). Again, vectorize the new matrices, stack all columns so far into a matrix and calculate its rank.

  3. Continue until the rank stops increasing with adding new generations. This means that any commutator is now included in the Lie algebra.

If the rank of the final matrix is $r$, then any $r$ linearly independent columns can serve as a basis for your Lie algebra.

One more remark: to my understanding, the rank of a matrix is numerically calculated by calculating its SVD and counting nonzero singular values. If your matrices have singular values comparable to machine precision, there might be spurious results.

Alexey Uvarov
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