Finding Cartan Subalgebras for Matrix Algebras

[leo.]

Homework Statement

Homework 6.2.30 (Quantum Field Theory for Mathematicians by Robin Ticciati): Find Cartan subalgebras of $\mathfrak{u}(n), \mathfrak{su}(n),\mathfrak{so}(n)$ and $\mathfrak{so}(1,3)$. Show that these subalgebras have dimensions $n,n-1,[n/2]$ (the greatest integer less than $n/2$), and $2$ respectively.

Relevant Equations

Definition of a Cartan subalgebra: A Cartan subalgebra of a Lie algebra $\mathfrak{g}$ is a maximal commutative Lie subalgebra of $\mathfrak{g}$.

This is one problem from Robin Ticciati's Quantum Field Theory for Mathematicians essentially asking us to find Cartan subalgebras for the matrix algebras $\mathfrak{u}(n), \mathfrak{su}(n),\mathfrak{so}(n)$ and $\mathfrak{so}(1,3)$. The only thing he gives is the definition of a Cartan subalgebra so I guess we should work from there.

The cases $\mathfrak{u}(n)$ and $\mathfrak{su}(n)$ I think I got correctly. Let's consider $\mathfrak{u}(n)$, it is the Lie algebra of the unitary group $U(n)$ and so its elements are anti-hermitian matrices. Now let $\{X_a\}\subset \mathfrak{u}(n)$ be a set of linearly independent commuting matrices. Since the matrices are anti-hermitian we can diagonalize them simultaneously. This means from $\{X_a\}$ we can build $\{X_a'\}$ a set of linearly independent diagonal matrices. The maximal set cardinality seems intuitively to be the number of diagonal entries and we can take the set $$X_1 = \operatorname{diag}(i,0,\dots, 0),X_2 = \operatorname{diag}(0,i,\dots, 0),\ \dots \ ,X_n = \operatorname{diag}(0,\dots, 0, i).$$ So I guess for $\mathfrak{u}(n)$ the subalgebra generated by $\{X_a\}$ is a Cartan subalgebra and its dimension is $n$.

For $\mathfrak{su}(n)$ everything goes the same, but now the matrices must have zero trace. A way to ensure this is to take as basis elements $$X_1 = \operatorname{diag}(i,-i,\dots, 0),X_2 = \operatorname{diag}(0,i,-i,\dots, 0),\ \dots \ , X_{n-1} = \operatorname{diag}(0,\dots, i,-i)$$
and clearly we have just one less generator. So we get $\mathfrak{su}(n)$ a Cartan subalgebra of dimension $n-1$.

Now for $\mathfrak{so}(n)$ I have no idea what to do. The thing is that for $\mathfrak{u}(n),\mathfrak{su}(n)$ since the matrices were anti-hermitian, commuting sets are in a sense equivalent to sets of simultaneously diagonal matrices. So we could look for sets of diagonal matrices which are much simpler. For $\mathfrak{so}(n)$ this is not the case.

So how can I proceed to solve this? Is my solution for $\mathfrak{u}(n),\mathfrak{su}(n)$ correct? Can the argument be improved? And what about $\mathfrak{so}(n)$? How do we proceed to find a Cartan subalgebra without resorting to diagonalization?

 

[fresh_42]

Homework Statement: Homework 6.2.30 (Quantum Field Theory for Mathematicians by Robin Ticciati): Find Cartan subalgebras of $\mathfrak{u}(n), \mathfrak{su}(n),\mathfrak{so}(n)$ and $\mathfrak{so}(1,3)$. Show that these subalgebras have dimensions $n,n-1,[n/2]$ (the greatest integer less than $n/2$), and $2$ respectively.
Homework Equations: Definition of a Cartan subalgebra: A Cartan subalgebra of a Lie algebra $\mathfrak{g}$ is a maximal commutative Lie subalgebra of $\mathfrak{g}$.

This is one problem from Robin Ticciati's Quantum Field Theory for Mathematicians essentially asking us to find Cartan subalgebras for the matrix algebras $\mathfrak{u}(n), \mathfrak{su}(n),\mathfrak{so}(n)$ and $\mathfrak{so}(1,3)$. The only thing he gives is the definition of a Cartan subalgebra so I guess we should work from there.

The cases $\mathfrak{u}(n)$ and $\mathfrak{su}(n)$ I think I got correctly. Let's consider $\mathfrak{u}(n)$, it is the Lie algebra of the unitary group $U(n)$ and so its elements are anti-hermitian matrices. Now let $\{X_a\}\subset \mathfrak{u}(n)$ be a set of linearly independent commuting matrices. Since the matrices are anti-hermitian we can diagonalize them simultaneously. This means from $\{X_a\}$ we can build $\{X_a'\}$ a set of linearly independent diagonal matrices. The maximal set cardinality seems intuitively to be the number of diagonal entries and we can take the set $$X_1 = \operatorname{diag}(i,0,\dots, 0),X_2 = \operatorname{diag}(0,i,\dots, 0),\ \dots \ ,X_n = \operatorname{diag}(0,\dots, 0, i).$$ So I guess for $\mathfrak{u}(n)$ the subalgebra generated by $\{X_a\}$ is a Cartan subalgebra and its dimension is $n$.

Unitary means $U^\dagger U=1$ which becomes $A^\dagger + A = 0$ for the Lie algebra. This means skew symmetric matrices. $\mathfrak{u}(n)$ is of dimension $\dfrac{n(n-1)}{2}$. The CSA is of dimension $n-1$ (I think), but still a subalgebra, so the main diagonal has to be zero.

For $\mathfrak{su}(n)$ everything goes the same, but now the matrices must have zero trace.

Yes. It is basically the same CSA as $\det U=\pm 1$ becomes the only condition $\operatorname{tr}A=0$. Hence it is the same Lie algebra, although not the same Lie group.

A way to ensure this is to take as basis elements $$X_1 = \operatorname{diag}(i,-i,\dots, 0),X_2 = \operatorname{diag}(0,i,-i,\dots, 0),\ \dots \ , X_{n-1} = \operatorname{diag}(0,\dots, i,-i)$$
and clearly we have just one less generator. So we get $\mathfrak{su}(n)$ a Cartan subalgebra of dimension $n-1$.

Now for $\mathfrak{so}(n)$ I have no idea what to do. The thing is that for $\mathfrak{u}(n),\mathfrak{su}(n)$ since the matrices were anti-hermitian, commuting sets are in a sense equivalent to sets of simultaneously diagonal matrices. So we could look for sets of diagonal matrices which are much simpler. For $\mathfrak{so}(n)$ this is not the case.

So how can I proceed to solve this? Is my solution for $\mathfrak{u}(n),\mathfrak{su}(n)$ correct? Can the argument be improved? And what about $\mathfrak{so}(n)$? How do we proceed to find a Cartan subalgebra without resorting to diagonalization?

If you want to read more about Lie algebras, see
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-basics/#toggle-id-1