Connected subgroups of SU(2) and SU(3)

Question

I am reading Lie groups, Lie Algebras, and Representations: An Introduction (2nd edition) by Brian C. Hall and I am unable to do the problem 12 in chapter 5. It says

Show that every connected Lie subgroup of SU(2) is closed. Show that this is not the case for SU(3).

Also I am wondering if the Lie algebra of SU(2) has any two-dimensional subalgebras.

Is there a systematic way of listing all the Lie subalgebras of the Lie algebras of SU(2) and SU(3) ? What about Lie subgroups ?

Answer 1

Since I will be invoking theorem 5.20 from Hall's book several times, I will cite it here.

Theorem 5.20. Let $G$ be a matrix Lie group with Lie algebra $\mathfrak{g}$ and let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Then there exists a unique connected Lie subgroup $H$ of $G$ with Lie algebra $\mathfrak{h}$. Namely, $H=\left\{e^{X_{1}} e^{X_{2}} \cdots e^{X_{N}} \mid X_{1}, \ldots, X_{N} \in \mathfrak{h}\right\}$.

We want to show that every Lie connected subgroup of $\operatorname{SU}(2)$ is closed, but these subgroups are in one-to-one correspondence with the subalgebras of $\operatorname{su}(2)$ (follows after theorem 5.20 and Hall's book definition of 'connected Lie subgroup', definition 5.19 on his book). We will be therefore studying who are these subalgebras. Let's write before the Lie algebra $\operatorname{su}(2)$ in a more familiar way. It turns out $\operatorname{su}(2)$ is isomorphic to the Lie algebra $(\mathbb{R}^3,\times)$, where $\times$ is the cross product, since if we pick the following basis for $\operatorname{su}(2)$,

$$ \left\{E_{1}=\frac{1}{2}\left(\begin{array}{rr} i & 0 \\ 0 & -i \end{array}\right) , \quad E_{2}=\frac{1}{2}\left(\begin{array}{ll} 0 & i \\ i & 0 \end{array}\right) , \quad E_{3}=\frac{1}{2}\left(\begin{array}{lr} 0 & -1 \\ 1 & 0 \end{array}\right) \right\}, $$

then the linear isomorphism $\operatorname{su}(2)\to \mathbb{R}^3$ given by $E_j\mapsto e_j$ (where $\{e_1,e_2,e_3\}$ is the standard basis of $\mathbb{R}^3$) can be verified to be a Lie algebra homomorphism (and thus, it is a Lie algebra isomorphism).

Let's classify the Lie subalgebras of $\operatorname{su}(2)$ by its dimension. We will be using theorem 5.20.

• Dim 3. It's $\operatorname{su}(2)$ himself.
• Dim 2. There are no Lie subalgebras of $\operatorname{su}(2)$ of dimension 2. Why not? Use the isomorphism $\operatorname{su}(2)\to \mathbb{R}^3$ and pick any two linearly independent vectors from $\mathbb{R}^3$. Using the right-hand rule, can you tell me who is the Lie subalgebra of $(\mathbb{R}^3,\times)$ generated by these two vectors?
• Dim 1. They are the ones of the form $\operatorname{span}_\mathbb{R}(v)$, where $v\in\operatorname{su}(2)$ is any non-zero vector. We now observe that $\operatorname{span}_\mathbb{R}(v)$ is a commutative Lie algebra (all 1-dimensional Lie algebras are) and that it is a maximal Lie subalgebra. The result is then given by Hall's Proposition 5.24.

Proposition 5.24 Suppose $G\subset\operatorname{GL}(n;\mathbb{C})$ is a matrix Lie group with Lie algebra $\mathfrak{g}$ and that $\mathfrak{h}$ is commutative and $\mathfrak{h}$ is a maximal commutative subalgebra of $\mathfrak{g}$, meaning that $\mathfrak{h}$ is commutative and $\mathfrak{h}$ is not contained in any larger commutative subalgebra of $\mathfrak{g}$. Then the connected Lie subgroup $H$ of $G$ with Lie algebra $\mathfrak{h}$ is closed.

If you would like a more elementary approach for tackling dim 1 which doesn't rely on Proposition 5.24, you can also use Hall's exercise 6 from chapter 2:

Exercise 6. Show that every $2\times 2$ matrix $X$ with $\operatorname{trace}(X)=0$ satisfies $$ X^{2}=-\operatorname{det}(X) I$$ If $X$ is $2\times 2$ with trace zero, show by direct calculation using the power series for the exponential that $$ \tag{1}\label{ec}e^{X}=\cos (\sqrt{\operatorname{det} X}) I+\frac{\sin \sqrt{\operatorname{det} X}}{\sqrt{\operatorname{det} X}} X, $$ where $\sqrt{\det X}$ is either of the two (possibly complex) square roots of $\det X$.

Note: The value of the coefficient of $X$ in \eqref{ec} is to be interpreted as $1$ when $\det X=0$, in accordance with the limit $\lim_{θ\to 0}\sin θ/θ=1$.

• Dim 0. The trivial Lie subalgebra $\{0\}\subset\operatorname{su}(2)$ gives rise to the trivial Lie connected subgroup $\{I\}\subset\operatorname{SU}(2)$, which is closed.

That makes it for $\operatorname{SU}(2)$. We now go after $\operatorname{SU}(3)$. We will use AnonymousCoward's answer hint, but I will phrase it in more understandable way for someone who has only read the five first chapters of Hall's book. The following comes from the beginning of section 5.9 of Hall's book. Using notations of theorem 5.20, we have that if $G=\operatorname{U}(1)\times \operatorname{U}(1)$ and $$ \mathfrak{h}=\left\{\left(\begin{array}{cc} it & 0 \\ 0 & i t a \end{array}\right) : t \in \mathbb{R}\right\}, $$ where $a$ is an irrational real number, then $$ \tag{2}\label{ec2} H=\left\{\left(\begin{array}{cc} e^{i t} & 0 \\ 0 & e^{i t a} \end{array}\right) : t \in \mathbb{R}\right\}, $$ where the group product of matrix Lie groups is realized as a matrix Lie group using exercise 5 from chapter 3,

Exercise 5. If $G_1\subset\operatorname{GL}(n_1;\mathbb{C})$ and $G_2\subset\operatorname{GL}(n_2;\mathbb{C})$ are matrix Lie groups and $G_1\times G_2$ is their direct product (regarded as a subgroup of $\operatorname{GL}(n_1+n_2;\mathbb{C})$ in the obvious way), show that the Lie algebra of $G_1\times G_2$ is isomorphic to the Lie algebra direct sum $\mathfrak{g}_1\oplus\mathfrak{g}_2$.

In exercise 10 from chapter 1 it is proven that the $H$ from \eqref{ec2} isn't closed. In fact, in that exercise it is shown that $H$ is dense in $G$ (I have also found this on MSE). A subset of $\operatorname{U}(1)\times \operatorname{U}(1)=S^1\times S^1$ which has the form of $H$ is called an irrational line on the torus.

It follows that any matrix Lie group which contains a matrix Lie group which is isomorphic to $\operatorname{U}(1)\times \operatorname{U}(1)$ also contains a connected Lie subgroup which isn't closed. That's what we will do with $\operatorname{SU}(3)$. By theorem 5.20, the Lie subalgebra of $\operatorname{su}(3)$ generated by the matrices $$ \left( \begin{array}{ccc} i & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -i \end{array} \right),\quad \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & -i \end{array} \right), $$ (which belong to $\operatorname{su}(3)$) gives rise to the connected Lie subgroup of $\operatorname{SU}(3)$ $$ H=\left\{ \left( \begin{array}{ccc} e^{it} & 0 & 0 \\ 0 & e^{is} & 0 \\ 0 & 0 & e^{-i(t+s)} \end{array} \right) : t,s\in\mathbb{R} \right\} = \left\{ \left( \begin{array}{ccc} z & 0 & 0 \\ 0 & w & 0 \\ 0 & 0 & \overline{zw} \end{array} \right) : z,w\in \operatorname{U}(1) \right\}. $$ The map $$ \begin{align*} H&\longrightarrow \operatorname{U}(1)\times \operatorname{U}(1)\\ \left( \begin{array}{ccc} z & 0 & 0 \\ 0 & w & 0 \\ 0 & 0 & \overline{zw} \end{array} \right) &\longmapsto (z,w) \end{align*} $$ is a Lie group isomorphism, so $H$ is a torus. Thus $H$ contains an irrational line and therefore so does $\operatorname{SU}(3)$.