Why does the adjoint action of $SU(2)$ on its own Lie algebra implement $SO(3)$?

Question

This is a follow-up on this question. Is it obvious that the adjoint action, $Ad(g)$ of $SU(2)$ on its own Lie algebra implements $SO(3)$? I understand that the adjoint representation for a group is defined by the homomorphism, \begin{align} Ad: g\in G \rightarrow Ad(g) \in GL(\mathfrak{g}) \end{align} where $\mathfrak{g}$ is the Lie algebra of the group $G$. Here $Ad(g)$ acts on $X\in \mathfrak{g}$ by \begin{align} (Ad(g))(X) = gXg^{-1} \end{align} Now consider the formula quoted in the answer to the linked question, \begin{align} U(R)\sigma_i U^\dagger(R) = \sigma_jR^j_i \end{align} where $U(R)$ is an $SU(2)$ matrix, $\sigma_i$ is an $SU(2)$ generator (in other words an element of the $\mathfrak{su}(2)$ Lie algebra for the same dimension as is $U(R)$), and $R^j_i$ is an $SO(3)$ rotation matrix.

How general is this formula? In the comments to the answer of the linked question, the following formula is quoted (I've adjusted it to my notation here), \begin{align} Ad(g): X_i \rightarrow gX_ig^{-1} = X_j [Ad(g)]^j_{\,i} \end{align} and I think this gets to the crux of my confusion. The RHS of this equation is some sort of rotation of a vector whose components are themselves the group generators (Lie algebra basis elements). But the LHS is a conjugation? I'm uncomfortable with the idea that these are equivalent! Here's my picture of how I understand what's happening:

As I said at the start, is this an unsurprising fact?

Answer 1

1. In physics the Lie algebra $su(2)$ is often identified with the space of $2\times2 $ traceless Hermitian matrices.

2. There is a bijective isometry from the Euclidean 3D space $(\mathbb{R}^3,||\cdot||^2)$ to the space of $2\times2 $ traceless Hermitian matrices $(su(2),-\det(\cdot))$, $$\mathbb{R}^3 ~\cong ~ su(2) ~:=~\{\sigma\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid \sigma^{\dagger}=\sigma\wedge {\rm tr}(\sigma)=0 \} ~=~ {\rm span}_{\mathbb{R}} \{\sigma_k \mid k=1,2,3\}, $$ $$\mathbb{R}^3~\ni~\vec{x}~=~(x^1,x^2,x^3) \quad\mapsto \quad\sigma~=~\sum_{k=1}^3x^k\sigma_k~\in~ su(2), $$ $$ ||\vec{x}||^2 ~=~\sum_{k=1}^3(x^k)^2 ~=~-\det(\sigma) .\tag{1}$$

3. There is an adjoint group action ${\rm Ad}: SU(2)\times su(2) \to su(2)$ given by $$g\quad \mapsto\quad{\rm Ad}(g)\sigma~:= ~g\sigma g^{-1}~=~g\sigma g^{\dagger}, \qquad g\in SU(2),\qquad\sigma\in su(2), \tag{2}$$ which is length preserving, i.e. $g$ is an orthogonal transformation. In other words, there is a Lie group homomorphism
   $${\rm Ad}: SU(2) \quad\to\quad O(su(2))~\cong~ O(3) .\tag{3}$$

4. Since $SU(2)$ is connected and the map ${\rm Ad}$ is continuous, ${\rm Ad}$ actually maps into the connected component $SO(3)$. In fact, one may show that $SU(2)$ is a double-cover of $SO(3)$.

5. For more details and interesting connections to quaternions, see e.g. this Phys.SE post.

Answer 2

The RHS of this equation is some sort of rotation of a vector whose components are themselves the group generators (Lie algebra basis elements). But the LHS is a conjugation? I'm uncomfortable with the idea that these are equivalent!

Recall that a Lie algebra $\mathfrak g$ is in particular a vector space, which can be equipped with some basis $\{T_k\}$. If $[\mathrm{Ad}(g)](T_k)$ is indeed another element of $\mathfrak{g}$, then we must be able to expand it as $ T_j c^j_k$ for some coefficients $c^j_{\ \ k}$. In that sense, we could always write $$\big[\mathrm{Ad}(g)\big](T_k)= \big[\mathrm{Ad}(g)\big]^j_{\ \ k}T_j$$ where $\big[\mathrm{Ad}(g)\big]^j_{\ \ k}$ is the $(j,k)$-component of the linear map $\mathrm{Ad}(g)$. At this point it remains only to compute those components in some chosen basis.

In the specific case of $\mathrm{SO}(3)$, there's a nice isomorphism between the set of vectors in $\mathbb R^3$ and the antisymmetric $3\times 3$ matrices given by $$A = \pmatrix{A_1\\A_2\\A_3} \leftrightarrow \pmatrix{0 &-A_3& A_2 \\ A_3&0&-A_1\\-A_2&A_1&0} \equiv A_\times$$ where the notation is chosen because for any vector $V\in \mathbb R^3$, $A_\times(V) = A \times V$. This is very useful here, because we note that if $R\in \mathrm SO(3)$, we have that for any vectors $V,W\in \mathbb R^3$ $$R(V \times W) = R(V) \times R(W)$$ This is ordinarily expressed as "the cross product of two vectors behaves like a vector under proper rotations." But this implies that

$$(RA)_\times V = (RA) \times V = (RA)\times (RR^\mathrm T V) = R\big(A \times (R^{\mathrm T} V)\big) = (RA_\times R^\mathrm T) V$$ $$\iff (RA)_\times = RA_\times R^\mathrm T$$ However, the right-hand side is precisely how $A_\times$ (understood as an element of $\mathfrak{so}(3)$) transforms under the adjoint action of $R$. As a result, we have that

$$\big[\mathrm{Ad}(R)\big](A_\times) = (RA)_\times$$ If we choose the standard basis $(L_k)_{\ell m} = -\epsilon_{k\ell m}$ for $\mathfrak{so}(3)$, then the vector $\tilde L_k$ corresponding to $L_k$ has components $(\tilde L_k)^i = \delta^i_{\ \ k}$, and so $$\big[\mathrm{Ad}(R)\big](L_k) := R L_k R^\mathrm T = (R\tilde L_k)_\times \overset{\star}{=} R^\mu_{\ \ k} L_\mu$$ where the $\star$ denotes the omission of a few fairly straightforward lines of algebra.

How general is this formula?

Not particularly general. $\mathrm{SO}(3)$ is a special case in which the components of $\mathrm{Ad}(R)$ work out to be numerically equal to the components of $R$ itself; this is not typical.

Perhaps more intuitively than the formal tricks employed above, the fact that the $L_k$'s transform in this way under the adjoint action of $\mathrm{SO}(3)$ is equivalent to the statement that $\vec L \equiv (L_1,L_2,L_3)$ is a vector operator with the property that under rotations induced by the unitary operator $U(R)$, we should have that $$\langle L_j\rangle \equiv \langle \psi, L_j \psi \rangle \mapsto \langle U(R)^\dagger\psi, L_j U(R)^\dagger\psi\rangle = \langle U(R) L_j U(R)^\dagger\rangle \overset{!}{=} R^k_{\ \ j} \langle L_k\rangle$$

Similar expressions exist for tensor operators of higher rank.

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As a final note, I spoke in this about the action of $\mathrm{SO}(3)$ on $\mathfrak{so}(3)$, not $\mathfrak{su}(2)$; however, it is not difficult to show that these two Lie algebras are isomorphic, with the linear isomorphism $L_i \leftrightarrow \frac{1}{2}\sigma_i$. We ordinarily do not distinguish between them for this reason; $\mathrm{Ad}(R)$ can be understood as a linear map $\mathfrak{su}(2)\rightarrow\mathfrak{su}(2)$ with the same components as above, i.e. $$\big[\mathrm{Ad}(R)\big](\sigma_k) = R^j_{\ \ k} \sigma_j$$ Along similar lines, because $\mathrm{SO}(3)$ is compact and connected, we can write any $R$ as $e^A$ for some $A\in \mathfrak{so}(3)$. Mapping this $A = A^\mu L_\mu \mapsto \frac{1}{2}A^\mu \sigma_\mu = \tilde A\in \mathfrak{su}(2)$, we exponentiate to obtain the unitary matrices $U(R) = e^{i\tilde A}$.