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:

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

  • $\begingroup$ WP. Your handwritten formula does not illustrate the TeX'd formula preceding it. It might be more intuitive for you to express the group results with algebra entities throughout. The adjoint rep is real. $\endgroup$ Nov 27, 2021 at 1:24

2 Answers 2

  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.


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.

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 $U(R) = e^{i\tilde A}$.

  • $\begingroup$ Really helpful. To ensure I'm following completely, these are the lines of algebra omitted, right? $(R\tilde{L}_k)_\times = (R^\mu_\nu \delta^\nu_k)_\times = (R^\mu_k)_\times = (\epsilon_{n\mu m}R^\mu_k) = (-\epsilon_{\mu n m}R^\mu_k) = L_\mu R^\mu_k$ $\endgroup$ Nov 28, 2021 at 22:35
  • $\begingroup$ In your explanation involving $\vec{L}$, the rotation operator $U(R)$ acts on the wavefunction. The manipulation of the expectation value gives us a perspective where instead the rotation may be viewed as an adjoint action on the $L_j$ operator itself. Then the intuitive step is in connecting both perspectives to the fact that we know the rotation of the wavefunctions should induce a rotation on $\langle L_k \rangle$. Would you agree with how I've phrased this? $\endgroup$ Nov 28, 2021 at 22:59
  • $\begingroup$ @samporterbridges Yes, I think that looks right! $\endgroup$
    – J. Murray
    Nov 29, 2021 at 17:37
  • $\begingroup$ The explanation you've given using $A_\times$, and the isomorphism between $\mathbf{R}^3$ and $3\times3$ anitsymmetric matrices appears to me to be highly specific to the case of the defining representation of $SO(3)$. However, the statement turns out to be general for unitary representations of $SO(3)$ of any dimension, acting in this adjoint fashion on $\mathfrak{so}(3)$ (or equivalently $\mathfrak{su}(2)$) in a representation of the same dimension. Right? It seems that it must be given the discussion of $\langle L_j\rangle$. $\endgroup$ Dec 1, 2021 at 3:02
  • $\begingroup$ Furthermore, the same statement holds for representations of $SU(2)$ acting again on $\mathfrak{so}(3)$ (or equivalently $\mathfrak{su}(2)$). This form, using $U \in SU(2)$, is what is written in my original question and in the question I linked to. Your answer doesn't obviously connect to the $SU(2)$ case, correct? $\endgroup$ Dec 1, 2021 at 3:05

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