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1

I'll sketch how you do it. In a general Lie group $\mathfrak{G}$ setting, the mapping $\mathrm{Ad}$ is clearly a homomorphism, since the action of the image of $\gamma_1\,\gamma_2$ under $\mathrm{Ad}$ is $$X\mapsto \gamma_1\,\gamma_2\,X\,(\gamma_1\,\gamma_2)^{-1} = \gamma_1\,\gamma_2\,X\,\gamma_2^{-1}\,\gamma_1^{-1} = ... 1 A more physical construction: Let R_3(\theta, \bf{n}) be the matrix of a rotation of angle \theta around axis \bf{n} in \mathbb{R}^3. Then if \hat{J}_i, i=1,2,3 are corresponding SO(3) generators,$$ \left[\hat{J}_i, \hat{J}_j \right] = i \epsilon_{ijk} \hat{J}_k $$we have$$ R_3(\theta, {\bf n} ) = exp\left(-i\;\theta \;n^i \hat{J}_i\right) ...

1

Given an element $\phi$ of $SU(2)$, let the first row of $\phi$ be $(P,Q)$ where $P$ and $Q$ are complex numbers. Let $q(\phi)$ be the quaternion $P+Qj$. Now for any $(x,y,z)\in {\mathbb R}^3$, consider the quaternion $q(\phi)(xi+yj+zk)q(\phi)^{-1} = (ai+bj+ck)$. The map $(x,y,z)\mapsto (a,b,c)$ is a rotation of ${\mathbb R}^3$ and hence an element of ...

0

I interpret the question as follows. Let $SU(3)$ be the group of complex $3\times 3$ matrices with $U^\dagger U=I$ and $\det U=1$ acting as linear operators in $\mathbb C^3$. Is there a subspace $M \subset \mathbb C^3$ with $M \neq \{0\}$, $M \neq \mathbb C^3$ such that $U(M) \subset M$ for every $U \in SU(3)$? In other words, is the natural action of ...

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An algorithm that solves for $\alpha$, $\beta$, and $\gamma$ for any given proper 3×3 rotation matrix constitutes a constructive proof. R_z (\alpha) R_y (\beta) R_z (\gamma) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \beta ...

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