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 ?

  • $\begingroup$ Do you need to think about the Lie algebras at all to do the problem? $\endgroup$ Commented Oct 11, 2013 at 23:40
  • $\begingroup$ @StefanSmith May be it is not necessary, but I was thinking to argue like the following : If there is no 2-dimensional subalgebra, then there is no two dimensional Lie subgroup. Three dimensional subalgebra would be the Lie algebra of SU(2) itself and the associated Lie group would be SU(2). One dimensional subalgebras are generated by a single element and may be easier to handle. But may be I am thinking along wrong lines, I would welcome any solution not necessarily along these lines. $\endgroup$
    – user90041
    Commented Oct 12, 2013 at 4:13
  • $\begingroup$ I am also working through Hall's book, though I haven't looked at it in a while. I have been doing the problems, and haven't made it to this problem yet. What is the "dimension" of a matrix Lie group? Is it the number of degrees of freedom? The Lie algebra is a real vector space (not necessarily closed under complex multiplication), so when you refer to the dimension of a matrix Lie group, are you using complex or real scalars? The Lie algebra of SU(2) is the set of 2-by-2 skew-Hermitian matrices with zero trace. That gives you five restrictions, so the dimension of ... $\endgroup$ Commented Oct 13, 2013 at 16:11
  • $\begingroup$ su(2) (the Lie algebra of SU(2)) appears to be three (as a real vector space). I am not an expert, so I may be wrong. $\endgroup$ Commented Oct 13, 2013 at 16:12
  • 1
    $\begingroup$ @StefanSmith Dimension of a matrix Lie group is the dimension of the underlying manifold which is same as the vector space dimension of its Lie algebra (as a real vector space). (Sorry, if this is not the standard terminology.) In my understanding too the dimension of su(2) (the Lie algebra of SU(2)) is three as a real vector space. $\endgroup$
    – user90041
    Commented Oct 13, 2013 at 19:19

2 Answers 2


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 $\operatorname{SU}(3)$.

  • 1
    $\begingroup$ I know it is quite some time since the answer is posted. There is another way to approach $\dim \mathfrak{h} = 1$ without using the result of Chapter 2. Proposition 5.24 is more helpful. Since any subalgebra of dimension $1$ is commutative, and $\mathfrak{su}(2)$ has no other commutative subalgebras (other than those of dimension $1$), we can invoke Proposition 5.24 to directly say that the induced connected Lie subgroup is closed. $\endgroup$ Commented May 13, 2021 at 6:21
  • $\begingroup$ @AniruddhaDeshmukh I have just added your easier argument to the answer. Thanks! $\endgroup$ Commented May 18, 2021 at 15:17

Hint: Consider the torus $T = \mathbb{R}^2/\mathbb{Z}^2$. Let $\alpha \in \mathbb{R}\setminus \mathbb{Q}$ an irrational number and define the subgroup $$H = \left\{ [x,\alpha x] \in \mathbb{R}^2/\mathbb{Z}^2 \colon\, x \in \mathbb{R} \right\}.$$ You may easily verify that this is a connected Lie subgroup in the sense defined in Hall (Definition 5.19, page 129, 2nd edition).

Remark: This subgroup is the image of the injective immersion $$ \psi:\mathbb{R} \rightarrow T,\, x\mapsto [x,\alpha x]$$ which is a group homomorphism, but $\psi$ is not an embedding since it is not a homeomorphism onto its image, $\psi(\mathbb{R}) = H$ (observe also that $\psi$ is not proper).

Claim (that you may want to check for yourself): The subgroup $H$ is not closed; in fact, the closure of $H$ is the entire torus $T$.

It follows immediately from this claim that if $G$ is a Lie group of rank at least 2, then $G$ has connected Lie subgroups that are not closed.

Is such a subgroup possible if the rank of $G$ is less than 2?


  • $\begingroup$ I am having trouble trying to find a torus of dimension 2 inside $\operatorname{SU}(3)$. Also, in Hall's book, maximal tori aren't introduced until chapter 11, and the exercise OP quotes is from chapter 5, so maybe there could be an easier way. I have only read myself until chapter 5, so I don't know either about maximal tori theory for Lie groups. How do you find such a torus inside $\operatorname{SU}(3)$? $\endgroup$ Commented Apr 22, 2021 at 16:58
  • $\begingroup$ @ElíasGuisado: Consider the collection of diagonal matrices in $SU(3)$. $\endgroup$ Commented Apr 23, 2021 at 16:43
  • $\begingroup$ @JasonDeVito yes, I found it and I posted the answer, thanks $\endgroup$ Commented Apr 24, 2021 at 16:46
  • $\begingroup$ @JasonDeVito: Your short answer here, is another answer to this post. $\endgroup$
    – C.F.G
    Commented May 4, 2021 at 0:52

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