Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 1

Put $\Lambda (t)=I+tX+O(t^2)$ in ${\Lambda }^T\eta \Lambda =\eta$. The coefficient of $t$ is $X^T\eta +\eta X$, so it must vanish. For $X=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ this condition isHence $a=d=0$ and $b=c$. Thus the Lie algebra is one-dimensional with basis

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Write $X(\phi )=\phi K$. Since $K^2=I$, the Exponential map of a Lie group givesThe hyperbolic addition formulas give $\Lambda (\phi )\Lambda (\psi )=\Lambda (\phi +\psi )$ and $\Lambda (\phi {)}^{-1}=\Lambda (-\phi )$, so these matrices form a subgroup of the One-dimensional Lorentz group. It is Abelian because addition in $\mathbb{R}$ is commutative. It is noncompact because $\cosh \phi$ is unbounded, equivalently because the subgroup is homeomorphic to $\mathbb{R}$.

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For the null coordinates in two-dimensional Minkowski spacetime, direct substitution givesThus a boost dilates one null direction and contracts the other by the reciprocal factor. It preservesThe invariant curves are therefore the level sets $x^2-y^2=c$: the branches of hyperbolas for $c\ne 0$, together with the two null lines when $c=0$. Each connected branch is preserved by the identity component.

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Solving ${\Lambda }^T\eta \Lambda =\eta$ with $\det \Lambda =1$ shows that every element is either $\Lambda (\phi )$ or $-\Lambda (\phi )$. HenceEach set is connected, but the sign of the $(1,1)$ entry cannot change continuously because its absolute value is at least one. Thus $SO(1,1)$ has two connected components. The matrix $-I$ lies in the component disjoint from the identity.

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For $\phi \ne \pm 1$, the Cayley transform isWriting its diagonal and off-diagonal entries as $a,b$, one has $a^2-b^2=1$, so $C^T\eta C=\eta$ and $\det C=1$. For $|\phi |<1$, put $\phi =\tanh (t/2)$ to obtain $C(\phi K)=\Lambda (t)$, so this interval covers the identity component. For $|\phi |>1$ the image lies in the other component and covers it except for $-I$, approached only as $|\phi |\to \infty$. The Exponential map of a Lie group reaches only the identity component, whereas the Cayley transform also reaches nonidentity-component elements but omits $-I$ and is undefined at $\phi =\pm 1$.

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