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# Lecture 13: October 27, 2022 ## Last time: * The group $SO(3)$ of rotations in $\mathbb{R}^3$. * We showed that $SO(3)$ can be parametrized by Euler angles $(\alpha, \beta, \gamma)$. $$ R(\alpha, \beta, \gamma) = R_z(\alpha) R_x(\beta) R_z(\gamma) $$ * We also discussed the pa...

# Lecture 13: October 27, 2022 ## Last time: * The group $SO(3)$ of rotations in $\mathbb{R}^3$. * We showed that $SO(3)$ can be parametrized by Euler angles $(\alpha, \beta, \gamma)$. $$ R(\alpha, \beta, \gamma) = R_z(\alpha) R_x(\beta) R_z(\gamma) $$ * We also discussed the parametrization of $SO(3)$ by axis and angle $(\hat{n}, \theta)$. ## Today: * More on $SO(3)$. * The group $SU(2)$. * The Lie algebra $\mathfrak{su}(2)$. * The Lie algebra $\mathfrak{so}(3)$. * The map $\varphi: SU(2) \rightarrow SO(3)$. * The group $SO(3)$ is not simply connected. ### The exponential map Let $A$ be a $3 \times 3$ real matrix. Define $$ e^A = \sum_{n=0}^{\infty} \frac{A^n}{n!} $$ **Claim:** If $A$ is antisymmetric (i.e. $A^T = -A$), then $e^A \in SO(3)$. **Proof:** Check that $e^A$ is orthogonal and has determinant 1. $$ (e^A)^T = \left( \sum_{n=0}^{\infty} \frac{A^n}{n!} \right)^T = \sum_{n=0}^{\infty} \frac{(A^T)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-A)^n}{n!} $$ Also, $$ e^{-A} = \sum_{n=0}^{\infty} \frac{(-A)^n}{n!} $$ Therefore, $(e^A)^T = e^{-A}$. We want to show that $(e^A)^T e^A = \mathbb{I}$. $(e^A)^T e^A = e^{-A} e^A = e^{-A + A} = e^0 = \mathbb{I}$. To show that $\det(e^A) = 1$, notice that $\det(e^A) = e^{\text{Tr}(A)}$. So we need to show that $\text{Tr}(A) = 0$. Since $A^T = -A$, we have $A_{ii} = -A_{ii}$, which implies that $A_{ii} = 0$. Therefore, $\text{Tr}(A) = \sum_{i=1}^3 A_{ii} = 0$. ### The group $SU(2)$ $SU(2)$ is the group of $2 \times 2$ unitary matrices with determinant 1. A matrix $U$ is unitary means that $U U^{\dagger} = \mathbb{I}$, where $U^{\dagger} = (U^T)^*$. Let $U = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $a, b, c, d \in \mathbb{C}$. Then $U^{\dagger} = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$. The condition $U U^{\dagger} = \mathbb{I}$ implies: * $|a|^2 + |b|^2 = 1$ * $|c|^2 + |d|^2 = 1$ * $a c^* + b d^* = 0$ The condition $\det(U) = 1$ implies $ad - bc = 1$. **Claim:** $c = -b^*$, $d = a^*$ **Proof:** $a c^* + b d^* = 0 \implies a c^* = -b d^* \implies c^* = -\frac{b d^*}{a}$. Then $c = -\frac{b^* d}{a^*}$. $ad - bc = 1 \implies ad + b b^* \frac{d}{a^*} = 1 \implies d(a + \frac{b b^*}{a^*}) = 1 \implies d(\frac{a a^* + b b^*}{a^*}) = 1 \implies d \frac{1}{a^*} = 1 \implies d = a^*$. Then, since $c = -\frac{b^* d}{a^*}$, we have $c = -\frac{b^* a^*}{a^*} = -b^*$. Therefore, $U = \begin{pmatrix} a & b \\ -b^* & a^* \end{pmatrix}$, with $|a|^2 + |b|^2 = 1$. Let $a = x_0 + i x_3$ and $b = x_2 + i x_1$, where $x_0, x_1, x_2, x_3 \in \mathbb{R}$. Then the condition $|a|^2 + |b|^2 = 1$ implies $x_0^2 + x_1^2 + x_2^2 + x_3^2 = 1$. We can therefore parametrize $SU(2)$ by the 3-sphere $S^3$.