Show: $\phi: \mathbb{R}^3 \rightarrow \mathcal{su}(2)$, $h \mapsto h \cdot \sigma$ is an isometric isomorphism

Question

I found this problem and need some help. It is given:

$$ \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$

$$ \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$

$$ \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$

These are the Pauli matrices.

Furthermore I have:

$$h \in \mathbb{R}^3$$ $$h \sigma := h_1 \sigma_1 + h_2 \sigma_2 + h_3 \sigma_3$$

and

$$\mathcal{su}(2) := \lbrace A \in M(2,\mathbb{C}) | A = h \cdot \sigma, h \in \mathbb{R}^3\rbrace$$

is a vector space.

Now I have a scalar product given by

$$\langle \cdot , \cdot \rangle_{\mathcal{su}(2)}: \mathcal{su}(2) \times \mathcal{su}(2) \rightarrow \mathbb{R} $$ $$\langle A, B\rangle_{\mathcal{su}(2)} = \frac{1}{2} trace (AB)$$

I got this and now I want to show, that

$$\phi: \mathbb{R}^3 \rightarrow \mathcal{su}(2)$$ $$h \mapsto h \cdot \sigma$$

is an isometric isomorphism

An isometric isomorphism has to be bijectiv, continous, the inverse has to be continous and the norm must be retained.

But how can I show it? I need some help!

Answer 1

If you define $su(2)$ like this, in other words, the span of those Pauli matrices, you get the vector space of all hermitian matrices with trace $0$. (It is more common to define $su(2)$ as the set of all anti-hermitian matrices with zero trace, but let's skip this for the moment.) The map $\phi$ is clearly linear and its inverse $\phi^{-1}$ is $$\begin{pmatrix}a & b+ci \\ b-ci & -a\end{pmatrix}\mapsto (b,-c,a),$$ so $\phi$ is a linear isomorphism. It is an isometry because the three Pauli matrices form an orthonormal basis under the given scalar product (check that $(1/2) Tr(\sigma_i\,\sigma_j)=\delta_{ij}$) and if a linear map maps an orthonormal basis $\{e_1, e_2, e_3\}$ to an orthonormal basis $\{\sigma_1, \sigma_2, \sigma_3\}$, it is an isometry.