Showing that a $SU(2)$-Matrix is in $\mathfrak{su}(2)$.

Question

Let $I \subset \mathbb{R}$ and $\phi : I \to SU(2)$ be a differentiable mapping. I want to show that $U := \phi'\phi^{-1}$ is an element of $\mathfrak{su}(2)$, that is $U + U^{*} = 0$ and $\operatorname{tr}(U) = 0$.

I know that $SU(2)$-Matrices are of the form $A=\begin{pmatrix}z_1&z_2\\-\overline{z_2}&\overline{z_1}\end{pmatrix}$, where $z_1,z_2 \in \mathbb{C}$. My attempt was to just calculate the product $\phi'\phi^{-1}$ by hand and show that it has the desired properties. I got the following:

$U = \phi'\phi^{-1} = \begin{pmatrix} z_1^{'}\overline{z_1} + z_2^{'}\overline{z_2} & -z_1^{'}z_2 + z_1z_2^{'} \\-\overline{z_1}\overline{z_2^{'}} + \overline{z_2}\overline{z_1^{'}} &\overline{z_2^{'}}z_2 + \overline{z_1^{'}}z_1 \end{pmatrix}$.

Unfortunately, there seems to be no obvious way to deduce anything helpful from this calculation. Is there maybe a better way to approach this problem? Any help is appreciated!

Answer 1

The Lie algebra $\mathfrak{g}$ of a Lie group $G$ can be thought of as the space of tangent vectors at the identity element $1_G$ of $G$ to paths passing through $1_G$. Given some $t_0 \in I$, the path $t \mapsto \phi(t) \cdot \phi(t_0)^{-1}$ is a path through the point $(t_0, 1_G)$ whose derivative at $t_0$ is $\phi(t_0)' \cdot \phi(t_0)^{-1}$, which therefore lies in the Lie algebra. This works for any Lie group, including $\mathrm{SU}(2)$.

If you don't want to appeal directly to facts about Lie groups, the following is all you need.

• The tangent space to the identity of $\mathrm{SU}(2)$ can be thought of as the space of matrices that arise as derivatives at the identity of paths passing through the identity.
• The matrix exponential map $\exp \colon \operatorname{Mat}_{2 \times 2}(\mathbb{C}) \to \mathrm{GL}_2(\mathbb{C})$ which sends a matrix $X$ to $\sum_{n \geq 0} X^n/n!$ is a smooth map with the property that $\left.\frac{d}{dt} \exp(t \cdot X)\right|_{t = 0} = X$ for any $X \in \operatorname{Mat}_{2 \times 2}(\mathbb{C})$.
• Consider the paths of the form $t \mapsto \exp(t \cdot X)$ for $X \in \operatorname{Mat}_{2 \times 2}(\mathbb{C})$. Check that such a path has image in $\mathrm{SU}(2)$ if and only if $X \in \mathfrak{su}(2)$.
• The derivative at $t = 0$ of the path $t \mapsto \exp(t \cdot X)$ is $X$, so the tangent space to the identity of $\mathrm{SU}(2)$ contains all of $\mathfrak{su}(2)$. But the dimension of $\mathrm{SU}(2)$ as a manifold is equal to the dimension of $\mathfrak{su}(2)$ as a vector space, so in fact, the tangent space to the identity of $\mathrm{SU}(2)$ equals $\mathfrak{su}(2)$.
• It follows that every derivative at the identity of a path through the identity lies in the tangent space and hence in $\mathfrak{su}(2)$.