# Definition The [[special unitary group in 2 dimensions]] $SU(2)$ is the double cover of the [[special orthogonal group in 3 dimensions]] $SO(3)$. This means that there is a two-to-one (double) and onto (cover) [[group homomorphism|homomorphism]] from the former to the latter. Moreover, the [[Lie algebra|Lie algebras]] $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ are [[Lie Algebra Isomorphism|isomorphic]], $\mathfrak{su}(2) \cong \mathfrak{so}(3)$. # Derivation ## Preliminaries Recall that the [[Lie algebra]] of $SU(2)$, denoted $\mathfrak{su}(2)$, is the vector space of all $2\times2$ traceless anti-[[Hermitian Operator|Hermitian matrices]], as discussed [[Special Unitary Group in 2 Dimensions#Infinitesimal Generators and Lie Algebra|here]]. An arbitrary element $X \in \mathfrak{su}(2)$ can be written as: $ X = x S_x + y S_y + z S_z $ where $ S_\alpha \equiv -\frac{i}{2} \sigma_\alpha $ and $\sigma_\alpha$ are the 3 [[Pauli matrices]]. Thus, in this basis $\mathcal{B} = \{S_x, S_y, S_z\}$, the column vector corresponding to $X $ is: $ [X] = \begin{pmatrix} x \\ y \\ z \end{pmatrix} $ and observe $ \det(X) = \frac{1}{4}(x^2 + y^2 + z^2) = \frac{1}{4} \braket{[X], [X]} = \frac{1}{4} ||[X]||^2 $ so that the determinant of $X \in \mathfrak{su}(2)$ is proportional to the [[Euclidean norm]] of $[X] \in \mathbb{R}^3$. ## Action of $SU(2)$ Now let $A \in SU(2)$ and $X \in \mathfrak{su}(2)$ $ \begin{align} A &= \left(\begin{array}{rr} \alpha & \beta \\ -\beta^* & \alpha^* \end{array}\right) \quad \alpha, \beta \in \mathbb{C}, \quad|\alpha|^{2}+|\beta|^{2}=1\\ X &= \frac{1}{2} \begin{pmatrix} -i z & -y - i x \\ y - i x & i z \end{pmatrix} \quad x,y,z \in \mathbb{R} \end{align} \tag{1} $ Computing $A X A^\dagger$ using Mathematica, it is trivial to see that it is anti-Hermitian and traceless, i.e., $A X A^\dagger \in \mathfrak{su}(2)$. Note also that: $ A (X + Y) A^\dagger = A X A^\dagger + A Y A^\dagger $ so that the map: $ \rho: A \mapsto A X A^\dagger $ is actually a [[linear operator]] on $\mathfrak{su}(2)$. It can be represented in the basis $\mathcal{B} = \{S_x, S_y, S_z\}$ by a $3\times3$ matrix which we'll denote by $\rho(A)$ $ [A X A^\dagger] = \rho(A) [X] $ also observe that: $ ||\rho(A)[X]||^2 = ||[A X A^\dagger]||^2 = 4 \det (A X A^\dagger) = 4 \det(X) = ||[X]||^2 $ so that: $ \braket{\rho(A)[X]|\rho(A)[X]} = \braket{[X]|[X]} $ and $\rho(A)$ preserves the norm of $[X]$ and is thus an [[Isometry Group|isometry]]. Therefore, since we are dealing with a [[Euclidean norm]] (i.e. an [[inner product]] on a real [[vector space]]), we conclude that $\rho(A) \in O(3)$. ## Explicit form for $\rho(A)$ Using the forms of $A$ and $X$ mentioned [[#Action of SU 2|above]], we have: $ A X A^\dagger = \left( \begin{array}{cc} -\frac{1}{2} i \left(\alpha ^* (\alpha z+\beta (x+i y))+\beta ^* (\alpha x-i \alpha y-\beta z)\right) & \frac{1}{2} i \left(x \left(\beta ^2-\alpha ^2\right)+i y \left(\alpha ^2+\beta ^2\right)+2 \alpha \beta z\right) \\ \frac{1}{2} \left(\left(\alpha ^*\right)^2 (y-i x)+\left(\beta ^*\right)^2 (y+i x)+2 i z \alpha ^* \beta ^*\right) & \frac{1}{2} i \left(\alpha ^* (\alpha z+\beta (x+i y))+\beta ^* (\alpha x-i \alpha y-\beta z)\right) \\ \end{array} \right) $ Now let $\alpha = a + ib$ and $\beta = c + id$ $ \frac{1}{2} \left( \begin{array}{cc} -i z_a & -y_a -i x_a \\ +y_a - i x_a & +i z_a \\ \end{array} \right) $ Where we have defined $ \begin{align} x_a &= a^2 x+2 a b y-2 a c z-b^2 x+2 b d z-c^2 x+2 c d y+d^2 x\\ y_a &= a^2 y-2 a b x+2 a d z-b^2 y+2 b c z+c^2 y+2 c d x-d^2 y, \\ z_a &= z \left(a^2+b^2-c^2-d^2\right)+2 (a c x-a d y+b c y+b d x), \end{align} $ so that we can write a column vector corresponding to $A X A^\dagger$ in the $\mathcal{B} = \{S_x, S_y, S_z\}$ basis as: $ [A X A^\dagger] = \begin{pmatrix} x_a \\y_a\\z_a \end{pmatrix} $ Now we can work on finding $\rho(A)$, which we have previously defined by: $ \rho(A) [X] = [A X A^\dagger] $ Solving this equation via Mathematica, we compute the matrix elements of $\rho(A)$: $ \begin{align} \rho_{11} &= a^2 - b^2 - c^2 + d^2,\\ \rho_{12} &= 2(ab + cd),\\ \rho_{13} &= 2(-ac + bd)\\ \rho_{21} &= 2 (c d-a b)\\ \rho_{22} &= a^2-b^2+c^2-d^2\\ \rho_{23} &= 2 (a d+b c)\\ \rho_{31} &= 2 (a c+ b d) \\ \rho_{32} &= 2 (b c- a d)\\ \rho_{33} &= \left(a^2+b^2-c^2-d^2\right) \end{align} $ Parameterizing this as: $ \begin{align} a &= \cos\theta/2 ,\\ b &= - n_z \sin\theta/2,\\ c &= - n_y \sin\theta/2,\\ d &= - n_x \sin\theta/2\\ \end{align} $ we get: $ \left(\begin{array}{ccc} n_{x}^{2}(1-\cos \theta)+\cos \theta & n_{x} n_{y}(1-\cos \theta)-n_{z} \sin \theta & n_{x} n_{z}(1-\cos \theta)+n_{y} \sin \theta \\ n_{y} n_{x}(1-\cos \theta)+n_{z} \sin \theta & n_{y}^{2}(1-\cos \theta)+\cos \theta & n_{y} n_{z}(1-\cos \theta)-n_{x} \sin \theta \\ n_{z} n_{x}(1-\cos \theta)-n_{y} \sin \theta & n_{z} n_{y}(1-\cos \theta)+n_{x} \sin \theta & n_{z}^{2}(1-\cos \theta)+\cos \theta \end{array}\right) $ where is one of the [[Special Orthogonal Group in 3 Dimensions#Elements|standard forms]] of SO(3) matrices, proving that $\rho(A) \in SO(3)$. ## The Homomorphism So far, we have a map $ \begin{align} \rho: SU(2) &\rightarrow SO(3)\\ A &\mapsto \rho(A) \end{align} $ Recall the action of $A \in SU(2)$ on $X \in \mathfrak{su}(2)$: $ X \mapsto A X A^\dagger $ and the action of $\rho(A) \in SO(3)$ on $X$ represented in the $\mathcal{B} = \{S_x, S_y, S_z\}$ basis: $ [X] \mapsto \rho(A) [X] $ Now observe: $ \rho(AB)[X] = [ABX(AB)^\dagger] = [ABXB^\dagger A^\dagger] = \rho(A)[BXB^\dagger] = \rho(A)\rho(B)[X] $ and we conclude: $ \rho(AB) = \rho(A)\rho(B), $ i.e., $ \begin{align} \rho: SU(2) &\rightarrow SO(3)\\ A &\mapsto \rho(A) \end{align} $ is a homomorphism! In fact, it is a homomorphism from $SU(2)$ [[Types of Maps|onto]] $SO(3)$, but it is *not* one-to-one, i.e., it is not a [[group isomorphism]]. The [[Kernel of a Homomorphism|kernel]] of $\rho$ is $\{I,-I\}$. Note that, from the definition of $\rho$, $\rho(A) = \rho(-A) \forall A \in SU(2)$. Thus, for every rotation $R \in SO(3)$, there exists *two* matrices in $SU(2)$ which [[map]] to $R$ under $\rho$. In conclusion, we say $SU(2)$ is the *double cover* of $SO(3)$, since there is a two-to-one (double) and onto (cover) homomorphism from the former to the latter. # The Induced Lie Algebra Homomorphism We have a *continuous* homomorphism $\rho: SU(2) \rightarrow SO(3)$ defined by the equation $ [A X A^\dagger] = \rho(A) [X] $ where $A \in SU(2)$ and $X \in \mathfrak{su}(2)$ in the $\mathcal{B} = \{S_x, S_y, S_z\}$ [[basis]]. We thus have an [[Induced Lie Algebra Homomorphism]] $\phi: \mathfrak{su}(2) \rightarrow \mathfrak{so}(3)$ defined by: $ \phi(Y) = \frac{d}{dt} \rho(e^{tY})\bigg|_{t=0} $ Let's work this out. Plugging in $A = e^{tS_i}$ for $i \in \{x,y,z\}$, and an arbitrary $X$ from Eq. $(1)$ we get: $ \begin{align} e^{tS_x}X (e^{tS_x})^\dagger &= \frac{1}{2} \left( \begin{array}{cc} -i [y \sin (t)+z \cos (t)] & -[y \cos (t)-z \sin (t)]-i x \\ [y \cos (t)-z \sin (t)]-i x & i [y \sin (t)+z \cos (t)] \\ \end{array} \right)\\ e^{tS_y}X (e^{tS_y})^\dagger &= \frac{1}{2} \left( \begin{array}{cc} -i [z \cos (t)-x \sin (t)] & -y-i[x \cos (t)+ z \sin (t)] \\ y-i[x \cos (t)+ z \sin (t)] & i [z \cos (t)-x \sin (t)] \\ \end{array} \right)\\ e^{tS_z}X (e^{tS_z})^\dagger &= \frac{1}{2} \left( \begin{array}{cc} -i z & -[x \sin (t)+y \cos (t)]-i[x \cos (t)-y \sin (t)] \\ [x \sin (t)+y \cos (t)]-i[x \cos (t)-y \sin (t)] &i z \\ \end{array} \right) \end{align} $ So we get: $ \begin{align} [e^{tS_x}X (e^{tS_x})^\dagger] &= \begin{pmatrix} x\\ y \cos (t)-z \sin (t)\\ y \sin (t)+z \cos (t) \end{pmatrix}\\ [e^{tS_y}X (e^{tS_y})^\dagger] &= \begin{pmatrix} x \cos(t) + z \sin(t)\\ y\\ z \cos(t) - x \sin(t) \end{pmatrix}\\ [e^{tS_z}X (e^{tS_z})^\dagger] &= \begin{pmatrix} x \cos(t) - y \sin(t) \\ x \sin(t) + y \cos(t) \\ z \end{pmatrix}\\ \end{align} $ Using the definition of $\rho$ $ [e^{tS_i}X (e^{tS_i})^\dagger] = \rho(e^{tS_i}) [X] $ we can solve for $\rho(e^{tS_i})$ $ \begin{align} \rho(e^{tS_x}) &= \begin{pmatrix} 1 & 0 & 0\\ 0 & \cos(t) & -\sin(t)\\ 0 & \sin(t) & \cos(t)\\ \end{pmatrix}\\ \rho(e^{tS_y}) &= \begin{pmatrix} \cos(t) & 0 & \sin(t)\\ 0 & 1 & 0\\ -\sin(t) & 0 & \cos(t) \end{pmatrix}\\ \rho(e^{tS_z}) &= \begin{pmatrix} \cos(t) & -\sin(t) & 0 \\ \sin(t) & \cos(t) & 0 \\ 0 & 0 & 1 \end{pmatrix} \end{align} $ Note that these are the rotation matrices about the $x,y,z$ axes, an unsurprising result considering the homomorphism between $SU(2)$ and $SO(3)$. Now we can finally compute: $ \begin{aligned} &\phi(S_x) = \frac{d}{dt} \rho(e^{t S_x})\bigg|_{t=0} = \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{array}\right) = L_x\\ &\phi(S_y) = \frac{d}{dt} \rho(e^{t S_y})\bigg|_{t=0} =\left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{array}\right) = L_y\\ &\phi(S_z) = \frac{d}{dt} \rho(e^{t S_z})\bigg|_{t=0} =\left(\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) = L_z. \end{aligned} $ This proves that $\mathfrak{su}(2) \cong \mathfrak{so}(3)$. Since we have proved $\phi(S_i) = L_i$, and the $S_i$ and $L_i$ have the same [[Structure Constants of a Lie Algebra|structure constants]], then the [[Lie Algebra Homomorphism|homomorphism]] $\phi$ is actually a [[types of maps|bijection]], making it an isomorphism. # The Adjoint Homomorphism Note that the homomorphism $ \begin{align} \rho: SU(2) &\rightarrow SO(3)\\ A & \mapsto \rho(A) \end{align} $ where $ \begin{align} \rho(A) [X] &= [A X A^{\dagger}]\\ \rho(A) [X] &= [\text{Ad}_A X] \end{align} $ Recalling that the [[Ad Homomorphism 1]] maps $G$ to $\text{Isom}(\mathfrak{g})$ where $\text{Isom}(\mathfrak{g})$ is the [[Isometry group]] of the [[vector space]] $\mathfrak{g}$, and this identification is due to $\text{Ad}_\cdot$ preserving the [[Killing Form]]. $ \text{Ad}: G \rightarrow \text{Isom}(\mathfrak{g}) $ Let's set $G = SU(2)$ and $\mathfrak{g} = \mathfrak{su}(2)$, then we can prove that the [[Killing Form]] is positive definite in the basis $\mathcal{B} = \{S_x, S_y, S_z\}$ (shown below) and we can conclude that $\text{Ad}_A \in \text{Isom}(\mathfrak{su}(2)) \cong O(3)$. Thus, we have $\text{Ad}: SU(2) \rightarrow O(3)$. Now we need to prove that this map is actually to $SO(3)$, i.e., the [[identity component]] of $O(3)$ and not the entirety of $O(3)$. ==I don't quite know how to do that, honestly.== After carrying out this calculation, we can conclude that $\rho(A)$ and $\text{Ad}_A$ are identical, and thus the [[Group Homomorphism|homomorphism]] between $SU(2)$ and $SO(3)$ is nothing but the [[Ad Homomorphism 1]]. Finally, corresponding to the [[Ad Homomorphism 1]] is a [[Lie algebra homomoprhism]], which of course is nothing but the [[Ad Homomorphism 1]]: $\text{ad}: \mathfrak{su}(2) \rightarrow \mathfrak{so}(3)$. This must be equal to the $\phi$ isomorphism discussed above, thus, we get: $ [\text{ad}_{S_i}] = \phi(S_i) = L_i $ This can be verified explicitly using the commutation relations. Finally, let's prove the [[Killing Form]] is positive defining the $\{S_i\}$ basis. We have: $ [\text{ad}_{S_i}] = L_i $ using the identity: $ \begin{align} (L_iL_j)_{kl} &= \sum_m (L_i)_{km} (L_j)_{ml}\\ &= \sum_m \epsilon_{ikm} \epsilon_{jml}\\ &= -\sum_m \epsilon_{mik} \epsilon_{mjl}\\ &= -\delta_{ij} \delta_{kl} + \delta_{il} \delta_{kj}\\ \end{align} $ we can show that: $ \begin{align} K(S_i, S_j) &= -\text{tr}(\text{ad}_{S_i} \text{ad}_{S_j})\\ &= -\text{tr}(L_i L_j)\\ &= \sum_{kl}\left(\delta_{ij} \delta_{kl} - \delta_{il} \delta_{kj}\right)\\ &= 3 \delta _{i,j} -\delta _{1,i} \delta _{1,j}-\delta _{2,i} \delta _{2,j}-\delta _{3,i} \delta _{3,j} &= 2 \delta _{i,j} \end{align} $ So we end up with: $ [K] = 2 I $ which is positive-definite. $\square$