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Let us consider the su(4) Lie algebra which is with 15 orthogonal generators and is of rank 3. Suppose I know the fundamental representation $F_{1-15}$ (which are 4-by-4 traceless Hermitian matrices) of these 15 generators, and according to which I can also obtain the adjoint representation $A_{1-15}$ of these generators by the commutators ($A_{1-15}$ are 15-by-15 purely-imaginary and Hermitian matrices).

Here are the questions:

(1) Given a certain Hermitian matrix $B = \sum_i a_i A_i$ with $a_i$ being real numbers, whether I can always find some other two Hermitian matrices, say B' and B'', such that $\{B,B',B''\}$ construct a Cartan subalgebra of su(4)? If so, how to find out the explicit forms of B' and B''?

(2) One step further, given two Hermitian matrices $B$ and $B'$ which do not commute, i.e. $[B,B']\neq 0$, whether there must exist a Hermitian matrix $B''$ such that $\{B,B',B''\}$ can construct a subalgebra of su(4)? If so, how to find out the minimal subalgebra, and what is the explicit form of $B''$?

These two questions were written in the adjoint representation, but it seems that they do not rely on certain representations. Namely, e.g., I can replace all $A_i$ by $F_i$ in both questions.

For the question (2), I can raise some special cases that make this work. For example, if we consider the two matrices $B = \sigma^x\otimes I$ and $B' = \sigma^y\otimes I$ with $\sigma^{x,y,z}$ being the Pauli 2-by-2 matrices and $I$ being the 2-by-2 identity. And obviously, we can take $B'' = \sigma^z\otimes I$ such that $\{B,B',B''\}$ form an su(2) subalgebra in the su(4).

However, I have no idea about the more general case. I want to try this proof by first putting $B$ in certain Cartan subalgebra, and then demonstrating that $B$ and $B'$ cannot generate all the simple roots.

Dietrich Burde
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Lee Chan
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  • Re 2, https://link.springer.com/article/10.1007/s00209-008-0397-3 shows in particular that in every simple real Lie algebra there are two elements $B_1,B_2$ which generate the entire Lie algebra. Does that answer that question? – Torsten Schoeneberg Dec 14 '20 at 17:17
  • Your answer helps a lot. Starting from the paper you cited, I went through some other papers and now I can come with some temperary answers about the question (2). For any semi-Lie algebra over a field of characteristic 0, there must be two elements that can generate the entire algebra. Therefore, the answer of the question (2) is no. About this question, I'd also like to recommend the paper 'on everywhere dense imbedding of free groups in Lie groups' by Kuranishi Masatake. – Lee Chan Dec 15 '20 at 08:15
  • sorry, I mean semi-simple Lie algebra in the previous comment. – Lee Chan Dec 15 '20 at 08:48

1 Answers1

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For question (1):

A subalgebra of a (semisimple) Lie algebra is called toral if it is abelian and all its elements are $ad$-semisimple.

Lemma 1: If $\mathfrak g$ is a Lie algebra over a field over characteristic $0$, then a subalgebra $\mathfrak h$ of $\mathfrak g$ is a Cartan subalgebra if and only if it is maximal toral (i.e. it is toral, and no subalgebra which properly contains $\mathfrak h$ is toral).

Proof: See Equivalence of Two Cartan Subalgebra Definitions in Semi-Simple Lie Algebra and If X commutes with all elements of the Cartan subalgebra, then X is in the Cartan Subalgebra?. Confer also Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?.

Fact 2 (alluded to in https://math.stackexchange.com/a/3820346/96384): For $n \ge 2$, every element of the Lie algebra $\mathfrak{su}(n)$ is $ad$-semisimple. Equivalently, the only nilpotent element of $\mathfrak{su}(n)$ is zero. Equivalently, the only $ad$-diagonalisable element of $\mathfrak{su}(n)$ is zero.

(Try to show this fact and/or the equivalences by hand. Such Lie algebras are called "anisotropic". For the ground field $\mathbb R$, this is the same as what is commonly called "compact". An analogous statement on the Lie group level is in Compact semisimple Lie groups contain no nontrivial unipotent elements? )

Conclusion: For every element $0\neq x \in \mathfrak{su}(4)$ there exist elements $x', x''$ such that $x, x', x''$ span a Cartan subalgebra of $\mathfrak{su}(4)$.

Proof: The one-dimensional subalgebra spanned by $x$ is toral because of Fact 2. Every toral subalgebra is contained in a maximal toral subalgebra, i.e. (by Lemma 1) a Cartan subalgebra. We know that Cartan subalgebras of $\mathfrak{su}(4)$ are three-dimensional.

This translates to your statement for a representation which consist of whatever matrices.

I am not entirely sure what you mean in question 2, but as said in a comment, it is known that every simple Lie algebra can be generated by two elements. In particular, for each simple Lie algebra contained in $\mathfrak{su}(4)$ (of which there are plenty, of various dimensions, including of course the entire $\mathfrak{su}(4)$), one can find two elements $x,x' \in \mathfrak{su}(4)$ which generate that subalgebra (and one could then just choose a third $x''$ contained in that one). If one can even pick a third element, there are obviously many more subalgebras one could generate. I find it hard to imagine any subalgebra of $\mathfrak{su}_4$ for which one could not find three elements which generate it.

But then again, as said I am not sure if that is what you ask. In particular I don't understand what you mean by finding "the minimal subalgebra" and "the explicit form" of the third element?

Added in response to comment: In principle, it is straightforward to find elements $x', x''$ to a given $x$ as above: Because as said every non-zero element of our LA is semisimple, just compute the centralizer of $x$, take one element $x'$ from that which is not a scalar multiple of $x$. Next, compute the centralizer of $x'$, take the intersection with the centralizer of $x$ you computed before; by general theory, this intersection still contains elements linearly independent from the space spanned by $x$ and $x'$; take one of them and call it $x''$. (I am not saying the computations of centralizers are easy, but they are doable.)

Maybe you should clarify for yourself here that starting with one element $x$, you can certainly find elements $x', x''$ such that they form a basis of a CSA, but in general this will be far from unique. Even in the "most unique" case of $x$ being a regular element, i.e. it is contained in a unique CSA, if you have found such $x', x''$, you can obviously replace them by any linear combination of $x, x', x''$ as long as the resulting space is still three-dimensional.

This is e.g. the case for $B = diag(0,i,-i,0)$. (Also note that I use standard math notation, for me $\mathfrak{su}(4)$ consists of traceless skew-Hermitian matrices. To get physics notation, divide everything by $i$.) Here the only choices for $B'$ and $B''$ are two linearly independent elements of the space $\{ diag(ia, -ia,ib,-ib) : a,b \in \mathbb R \}$.

But e.g. if your first element is $B=diag(i,-i,0,0)$, you can also choose $B' = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0}$, a non-diagonal matrix, and then e.g. $B'' = diag(0,2i,-i,-i)$. (Some centralizers are easily computed.)

Torsten Schoeneberg
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  • Your answer is adequet for both questions, and now I think I am entirely clear about the question (2).

    One minor question about the answer of the question (1). Now, I know that given a certain element x, there must be two other elements x' and x'' such that {x,x',x''} span a Cartan subalgebra of su(4). Now, in certain basis, I am with particular matrix representation of the x, how can I find the matrix representation of x' and x''?

     – Lee Chan Dec 15 '20 at 08:54
  • See added paragraphs responding to your comment. If you are satisfied with the answer, you might want to accept it. – Torsten Schoeneberg Dec 17 '20 at 18:13