Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 57 / 6 / a

Identify ${\mathbb{R}}^3$ with the traceless Hermitian matrices $X=x\cdot \sigma$, where ${\sigma }_i$ are the Pauli matrices. Then $\det X=-|x{|}^2$. For $U\in SU(2)$ the action $X\mapsto UX{U}^{†}$ preserves trace, Hermiticity and determinant, so gives an orthogonal transformation of ${\mathbb{R}}^3$. Since $SU(2)$ is connected, the determinant of this transformation is $+1$.

An element in the kernel commutes with every ${\sigma }_i$, hence is scalar. Unitarity and determinant one restrict it to $\pm I$. Every three-dimensional rotation is a rotation through some angle $\theta$ around a unit axis $n$. It is obtained byIndeed multiplication with ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$ gives the Rodrigues rotation formula for $UX{U}^{†}$. This proves surjectivity, not just equality of Lie-algebra dimensions. ThereforeA full $2\pi$ rotation lifts from $I$ to $-I$, explaining the double cover and the Spin group interpretation.

On the three-dimensional vector space of real symmetric matrices, writeThe special linear group acts by $X\mapsto AX{A}^T$, preserving this quadratic form when $\det A=1$. A kernel element satisfies $A{A}^T=I$ and commutes with every symmetric matrix, so it is scalar and equals $\pm I$. Its derivative is injective: if $aX+X{a}^T=0$ for every symmetric $X$, taking $X=I$ makes $a$ skew-symmetric, and commuting with diagonal matrices then makes $a=0$.

Both real Lie group dimensions are three. The derivative is consequently an isomorphism, so the image is an open subgroup. The source $SL(2,\mathbb{R})$ is connected: polar decomposition deforms it onto $SO(2)$, with the positive symmetric determinant-one factor contractible. Its image is thus the full identity component $S{O}_0(2,1)$, since a connected group has no proper open subgroup. This is the symmetric-matrix double cover of SO0(2,1).

For $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, direct multiplication gives $A^{T}JA=(\det A)J$. Hence preserving the two-dimensional symplectic form is exactly the determinant-one condition, soThe subscript zero is necessary. The full $SO(2,1)$ also contains a component reversing time orientation, which cannot be the image of the connected source.

Use the four-dimensional real vector space of Hermitian matricesFor $A\in SL(2,\mathbb{C})$, $X\mapsto AX{A}^{†}$ preserves determinant, hence defines a Lorentz transformation. Positive definite $X$ stays positive definite, so future timelike vectors remain future timelike. Connectedness also fixes the orientation, giving an image in $S{O}_0(3,1)$.

If the action is trivial, $X=I$ makes $A$ unitary. The remaining condition says it commutes with every Hermitian matrix, forcing $A=\lambda I$ with ${\lambda }^2=1$. Infinitesimally $aX+X{a}^{†}=0$ first makes $a$ anti-Hermitian and then scalar; tracelessness forces $a=0$. Both real Lie-algebra dimensions are six, so the image is open. Polar decomposition shows $SL(2,\mathbb{C})$ is connected, with connected unitary factor $SU(2)$ and a contractible positive Hermitian determinant-one factor. The image is therefore the full proper orthochronous Lorentz group.This is the Hermitian-matrix double cover of SO0(3,1). The full determinant-one Lorentz group also contains transformations reversing time orientation, so the identity-component qualification cannot be omitted.

Let $V={\mathbb{R}}^4$ have symplectic basis $(e_1,f_1,e_2,f_2)$. Fix $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$ and define a symmetric bilinear form $Q$ on ${\Lambda }^{2}V$ by $u\wedge v=Q(u,v)\mathrm{vol}$. The pairings between complementary basis bivectors split this six-dimensional space into three hyperbolic planes, so $Q$ has signature $(3,3)$.

The symplectic group preserves volume and the bivector $\Omega =e_1\wedge f_1+e_2\wedge f_2$, obtained from the inverse symplectic form, up to its fixed sign convention. Since $Q(\Omega ,\Omega )=2$, its orthogonal complement $W={\Omega }^{\perp }$ has dimension five and signature $(2,3)$. On $W$ use $-Q$, which has signature $(3,2)$. This constructs the exterior-square double cover of SO0(3,2).

If an element acts trivially on $W$, it also fixes $\Omega$, so its action on the entire exterior square is the identity. Every decomposable bivector then shows that its two-plane is preserved. A line is the intersection of two such planes; hence every line is preserved, and the matrix is scalar. Its exterior-square action forces its scalar square to equal one, giving precisely $\pm I$.

The derivative is injective by the same infinitesimal argument: zero action on ${\Lambda }^{2}V$ forces an infinitesimal scalar, whose induced scalar there is twice its value and therefore zero. The symplectic Lie algebra has dimension $2^2+2(2\cdot 3/2)=10$, from blocks $\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with symmetric $B,C$. The orthogonal Lie algebra in five dimensions also has dimension ten. Thus the image is open. Real symplectic polar decomposition has connected compact factor $U(2)$ and a contractible positive symplectic factor, so $Sp(4,\mathbb{R})$ is connected and its image is exactly the identity component:As in the preceding noncompact cases, this does not identify the quotient with the disconnected full $SO(3,2)$.