Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 64 / 5

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

Identify ${\mathbb{R}}^4$ with the quaternions, and use the identification of unit quaternions with $SU(2)$ proved in the root Solution. The actionpreserves the Euclidean norm by multiplicativity of quaternion norms. It is a homomorphism $SU(2)\times SU(2)\to SO(4)$: its domain is connected, so the orthogonal image has determinant $+1$.

An element in the kernel fixes $v=1$, giving $q_L=q_R=q$. Fixing every other $v$ means $q$ commutes with all quaternions, so $q$ is real; since it has unit norm, $q=\pm 1$. Hence the kernel is the diagonal subgroup $\{(1,1),(-1,-1)\}$.

The differential acts as $v\mapsto av-vb$ for imaginary quaternions $a,b$. If it vanishes, $v=1$ gives $a=b$, and commutation with all $v$ makes $a$ real and imaginary, hence zero. The differential is injective, and both Lie algebras have dimension six. The image is therefore an open subgroup of connected $SO(4)$, hence all of $SO(4)$. Thuswhere ${\mathbb{Z}}_2$ acts diagonally, not separately on the two factors.

Identify a Minkowski vector $(t,x,y,z)$ with the Hermitian matrix representation of Minkowski four-vectorsFor $A\in SL(2,\mathbb{C})$, the real-linear action $X\mapsto AX{A}^{†}$ preserves Hermiticity and the determinant, so it preserves the Lorentz quadratic form. It also preserves the cone of positive-definite Hermitian matrices, hence the future timelike cone. The group $SL(2,\mathbb{C})$ is connected, by polar decomposition of an invertible complex matrix into $SU(2)$ and positive determinant-one Hermitian matrices. Thus its image lies in the proper orthochronous Lorentz group.

If $AX{A}^{†}=X$ for every Hermitian $X$, taking $X=I$ makes $A$ unitary; the remaining equations then say it commutes with every Hermitian matrix. Their real span is all Hermitian matrices and their complex span is all complex matrices, so $A$ is scalar. The determinant condition leaves precisely $A=\pm I$. A discrete kernel gives an injective differential, and the real dimensions of both groups are six. The image is therefore open and equals the connected Lorentz group:The quotient does not include the disconnected time-reversing component of $SO(3,1)$.

The space of real $2\times 2$ matrices has a determinant form of signature $(2,2)$: writeThe action $(A,B):X\mapsto AX{B}^{-1}$ of $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ preserves this determinant. Each factor is connected by polar decomposition of an invertible real matrix: its orthogonal factor lies in connected $SO(2)$ and its positive factor is an exponential of a symmetric traceless matrix. Thus the image is contained in $S{O}_0(2,2)$.

If the action is trivial, $X=I$ gives $A=B$, and fixing every $X$ makes $A$ a scalar matrix. Determinant one gives $A=B=\pm I$. The kernel is therefore the diagonal ${\mathbb{Z}}_2$. Its discreteness makes the differential injective; domain and target both have dimension six. An open subgroup of connected $S{O}_0(2,2)$ is the whole group, soAgain the kernel is diagonal, and the target is the identity component.

The imaginary quaternions form Euclidean ${\mathbb{R}}^3$. Conjugation by a unit quaternion acts byIt preserves the imaginary subspace and its norm, and continuity from $q=1$ gives determinant $+1$. The kernel consists of quaternions commuting with all imaginary quaternions, hence is $\{\pm 1\}$.

For a unit imaginary quaternion $n$, take $q=\cos (\theta /2)+n\sin (\theta /2)$. If $v$ is perpendicular to $n$, quaternion multiplication giveswhile the component parallel to $n$ is unchanged. This produces the rotation through angle $\theta$ about axis $n$, and every element of $SO(3)$ has such an axis-angle description. The map is surjective, yieldingThe half-angle also explains why $q$ and $-q$ give the same rotation.

Use the real three-dimensional space of traceless matricesThus $-\det X$ is a quadratic form of signature $(2,1)$. The adjoint action $X\mapsto AX{A}^{-1}$ of $SL(2,\mathbb{R})$ preserves trace and determinant. Connectedness puts its image in $S{O}_0(2,1)$.

A kernel element commutes with every traceless matrix and with the identity, hence with every real matrix; it is therefore scalar. Determinant one leaves $\pm I$. The derivative is injective and both groups have dimension three, so the image is open and equals the identity component. Consequently