I am a Physicist, so probably my question could sound trivial or even stupid to most of you. I apologize in advance. I'm working with a set of quantum-mechanical operators, which realize a 10-dimensional Lie algebra. Starting from the defining commutators between the various operators, is there a way to find a matrix realization of this algebra?

First of all: is there a way to determine the dimension of the square matrices that must be employed?

Can you please suggest me a way, or an "algorithm" to write these 10 matrices?


The statement that this always CAN be done is called 'Ado's theorem'. One way to get an answer is to look for proofs of this theorem online. This blog post by Terrence Tao looks useful (but I didn't read it all): https://terrytao.wordpress.com/2011/05/10/ados-theorem/.

EDITTED IN: A good start (see also the blogpost) is to first try and find the center of the Lie algebra, so all linear combinations of your operators that have the property of commuting with all the operators. If your center is only the zero operator, things are very simple. And if it is one dimensional things are not that bad either.

FURTHER EDIT AFTER READING COMMENT: There is one thing you can always do but only gives the desired result when the center is zero (because it will always represent operators in the center with the zero-matrix whether it is justified or not) and that is taking the adjoint representation. Here is an algorithm:

The vector space the matrices act on is just the Lie algebra itself (so it is ten dimensional). The vector $[1, 0, 0, 0, 0, 0, 0, 0, 0, 0]$ corresponds to the first operator on your list, the vector $[0, 1, 0, 0, 0, 0, 0, 0, 0, 0]$ to the second etc. (When I write lying vectors I mean standing ones. They are just more annoying to type.) I will call the operators in order $X_1, \ldots, X_{10}$.

Now to create a matrix for an operator $X$ in the Lie algebra we remember that the first column of any matrix $A$ is just the vector you get when letting $A$ act on the first basisvector $[1, 0, 0, 0, 0, 0, 0, 0, 0, 0]$, you get the second column of $A$ by seeing what happens when you let $A$ act on $[0, 1, 0, 0, 0, 0, 0, 0, 0, 0]$ etc.

Now what we want the matrix assigned to operator $X$ to do is send $[1, 0, 0, 0, 0, 0, 0, 0, 0, 0]$ to the vector corresponding to $[X, X_1]$, send $[0, 1, 0, 0, 0, 0, 0, 0, 0, 0]$ to the vector corresponding to $[X, X_2]$ etc.

More explicit: if you calculate that $[X, X_1] = 3 X_2 + 5 X_3 + 10 X_{10}$ the first column of the matrix you look for is $[0, 3, 5, 0, 0, 0, 0, 0, 0, 10]$. Etc.

At the end of the day every operator has then both a matrix AND a vector attached to it, but once you have all the matrices you can (if you want) forget the vectors. From the description it is not completely clear that the matrices associated to the operators satisfy the same commutation relations as the operators themselves, but you can check that by hand. (Or prove it in the general case. Terrence Tao states that this 'follows from the Jacobi Identity', which is the identity $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$. The Jacobi identity is part of the definition of an abstract Lie algebra but more relevantly, it is automatically true for every Lie algebra where the elements are linear operators and the Lie bracket is the commutator bracket.)

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  • $\begingroup$ There are no linear combinations of operators of mine that commute with all the others. So the center of the Algebra is zero?! Can you please expand "things are very simple" and explain what I should do? As I said, I've not an algebraic / group-theoretic background, so I would appreciate a more detailed help. Thanks a lot in advance. $\endgroup$ – AndreaPaco Nov 18 '16 at 11:37
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    $\begingroup$ Also when the center is zero there is a finite list of Lie algebras your Lie algebra could be isomorphic to. By 'could' I mean it could not be isomorphic to any other so it must be in the list, only perhaps presented differently. The point is that a centerless Lie algebra is semisimple and a semisimple lie algebra is a direct sum of simple ones. If you look at the list of simple Lie algberas in wikipedia you can find all the ones with dimension $\leq 10$ and see which ones can add up to give dimension 10 $\endgroup$ – Vincent Nov 18 '16 at 12:09
  • $\begingroup$ There is actually one simple one of actual dimension 10: it is so(5) which is isomorphic to sp(4) (B_2 and C_2 in root system nomenclatura). If your Lie algebra actually turns out to be so(5) it has a representation in anti-symmetric 5 by 5 matrices, half as big as the ones in the adjoint representation. This is a common theme. The adjoint is never the smallest rep, except in the E_8 case. $\endgroup$ – Vincent Nov 18 '16 at 12:12

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