Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 1

The matrices are rotations about the third coordinate axis. Direct multiplication givesand $g(\theta {)}^{\mathsf{T}}g(\theta )=I$. Thus they form a one-parameter subgroup of the orthogonal group $O(3)$, isomorphic to the circle group.

If $X$ belongs to the Special orthogonal Lie algebra, which is also the Lie algebra of $O(3)$, then $e^{tX}\in O(3)$ near $t=0$. Differentiatingat zero gives $X^{\mathsf{T}}+X=0$. Conversely, the matrix exponential of every real skew-symmetric matrix is orthogonal. Hence

For $A\in O(3)$ and $X^{\mathsf{T}}=-X$,so ${\mathrm{Ad}}_A$ maps $\mathfrak{o}(3)$ into itself. Moreover,It also sends the identity to the identity linear map, so $A\mapsto {\mathrm{Ad}}_A$ is the Adjoint representation of a Lie group.

For a smooth Lie-group representation $D:G\to GL(V)$, define its derived representationDifferentiation makes $dD$ linear. Applying $D$ to the group commutator curveand taking the mixed derivative at $(0,0)$ givesThus $dD:\mathfrak{g}\to \mathfrak{gl}(V)$ is a Lie algebra representation.

Differentiate ${\mathrm{Ad}}_{e^{tX}}Y=e^{tX}Y{e}^{-tX}$ at zero. This gives the Adjoint representation of a Lie algebraThe Jacobi identity is exactlyso this is a Lie-algebra representation on $\mathfrak{o}(3)$.

The Exponential map of a Lie group sends $X\in \mathfrak{o}(3)$ to $e^X\in SO(3)$ and is a local diffeomorphism at zero. Every element of $SO(3)$ is an exponential of a skew-symmetric matrix, but no determinant-$-1$ element of $O(3)$ is an exponential because $\det e^X=e^{\mathrm{tr}X}=1$.

A representation $d:\mathfrak{o}(3)\to \mathfrak{gl}(V)$ always integrates uniquely to the simply connected covering group $\mathrm{Spin}(3)\simeq SU(2)$. It descends to $SO(3)$ exactly when the nontrivial element in the kernel of $SU(2)\to SO(3)$ acts trivially. Extending it further to disconnected $O(3)$ requires an additional parity operator compatible with conjugation by a reflection. Thus the Lie-algebra representation alone need not define a representation of all of $O(3)$.