Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 4

One form of the Jacobi triple product is, for $0<|q|<1$ and $z\ne 0$,Both sides converge locally uniformly on the punctured $z$ plane. Here is a finite-product proof which also determines the multiplicative constant.

Put $Q=q^2$ and $(Q;Q{)}_m={\prod }_{j=1}^m(1-Q^j)$, with value one at $m=0$. Define the Gaussian binomial coefficient ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q=(Q;Q{)}_m/[(Q;Q{)}_k(Q;Q{)}_{m-k}]$. The recurrence ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q={\left[\genfrac{}{}{0ex}{}{m-1}{k}\right]}_Q+Q^{m-k}{\left[\genfrac{}{}{0ex}{}{m-1}{k-1}\right]}_Q$ proves by induction thatRewrite the two finite factors asThe finite identity, with $n=k-N$, makes this ${\sum }_{n=-N}^N{z}^n{q}^{n^2}{\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q$. Multiply by $(Q;Q{)}_N$. For each fixed integer $n$, the resulting coefficient tends to one, because $(Q;Q{)}_m$ tends to a nonzero limit and ${\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q\to 1/(Q;Q{)}_{\infty }$.

All finite $(Q;Q{)}_m$ are uniformly bounded above and away from zero. Thus these coefficients have one uniform bound, and ${\sum }_{n\in \mathbb{Z}}|q{|}^{n^2}|z{|}^n$ is uniformly summable on compact annuli. Dominated convergence passes the limit through the finite sums and proves the boxed formula. At $q=0$ the formula follows by its limiting value one. The frequently used alternative form isIt follows by the substitution $q^2=t$, $z=u/q$, and is the version used for the charge-counting identity in Question 8.

Write $p:U(H)\to PU(H)$ for the scalar quotient. Local strongly continuous sections exist: near a fixed projective unitary, fix its phase by requiring one nonzero matrix coefficient, taken relative to a fixed representative, to be positive real. Thus the pullback $E=\{(z,V):p(V)=\rho (z)\}$ is a locally trivial central extension of the circle by the circle. It is compact, since finitely many closed arcs inside such trivializing neighbourhoods cover the compact base and each pullback is a product with the compact fibre.

Any two lifts commute up to a scalar. Their commutator defines a continuous alternating bicharacter $c(z,w)$ on the base: the scalar commutator is independent of phase choices and multiplicative in each argument. If $z$ is an irrational rotation, its powers are dense and $c(z,z^m)=1$ for every integer $m$. Continuity makes $c(z,w)=1$ for every $w$. Such $z$ are themselves dense, so all the lifts commute, and $E$ is abelian.

The tautological representation $R(z,V)=V$ is strongly continuous. Choose a unit vector $\xi$ and average its rank-one orthogonal projection over Haar measure on $E$:The integrand is trace-norm continuous, so $K$ is positive compact with trace one, and commutes with $R(E)$. A positive eigenvalue has a nonzero finite-dimensional eigenspace invariant under $E$. Commuting unitary matrices on that space have a common unit eigenvector $\eta$. Therefore $R(e)\eta =\chi (e)\eta$ for a continuous character $\chi$ of $E$.

On the central fibre, $R(1,\lambda I)=\lambda I$, so $\chi (1,\lambda I)=\lambda$. Consequentlyis independent of that lift. It is a homomorphism and strongly continuous, because the continuous map on $E$ descends through its quotient to the base. It satisfies $p(U_z)=\rho (z)$. This proves circle projective representations lift to unitary representations using an actual central character, rather than assuming a global choice of phases.

First, all lifts of the real one-parameter projective group commute. Its scalar commutator $c(s,t)$ is again a continuous alternating bicharacter. For fixed nonzero $t$ and an integer $n>0$, a lift of $t$ is a phase multiple of the $n$th power of a lift of $t/n$; hence $c(t,t/n)=1$ and $c(t,mt/n)=1$ for every integer $m$. Rational multiples of $t$ are dense in $\mathbb{R}$, so continuity makes $c(t,s)=1$ for every $s$.

Choose a unitary $U$ representing $\rho (1)$ and the allowed self-adjoint logarithm $A\in U^{\prime \prime }$ with $U=e^{iA}$. Every lift of $\rho (t)$ commutes with $U$, hence with $A$ by the double commutant theorem for star-algebras condition. Thereforeis a continuous projective homomorphism, with $\tilde{\rho }(1)=1$. It is periodic of period one and descends to $\mathbb{R}/\mathbb{Z}$. Part (b) lifts it to a continuous unitary homomorphism $L$ of that circle. Each $L(t)$ is a lift of $\tilde{\rho }(t)$ and commutes with $A$, since a representative has the form of an original lift times $e^{-itA}$. Thusis a strongly continuous unitary homomorphism with projective class $\rho (t)$. This gives the required lift of the whole real group, not just unrelated unitary representatives at each time.