Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 1 / 2 / b

Let $E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, so $A(t)=I+t{E}_{12}$ and $E_{12}^2=0$. Multiplication givesThe nilpotence also gives $(I+t{E}_{12}{)}^k=I+kt{E}_{12}$ for every nonnegative integer $k$, since all higher binomial terms vanish. Hence, for each positive integer $m$,These are unipotent matrices in SL2R. The integer is positive, since the displayed diagonal matrix is undefined at $m=0$.

Put $U=\varphi (A(t))$. Applying the group homomorphism $\varphi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its eigenvalues is unchanged by taking their $m^2$th powers.

Take $m=2$. The fourth-power operation permutes this finite multiset. For each eigenvalue $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne 0$, so $z^{4^r-1}=1$: every eigenvalue is a root of unity. Choose a positive integer $d$ divisible by all their orders. Every eigenvalue of $U^{d^2}$ is then one. Because its eigenvalue multiset is also the original one, we obtainThis is square-power rigidity of a finite nonzero spectrum. Finiteness of the dimension is essential to the permutation-cycle argument.

Let $N$ be the normal closure of the upper unipotents $U(t)=A(t)$. Conjugation by $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesThus $N$ also contains every lower unipotent $L(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. We prove explicitly that these elementary unipotent generators of SL2R generate the full group.

For $a\ne 0$, multiplication givesSo all determinant-one diagonal matrices lie in $N$. If $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has $a\ne 0$ and $ad-bc=1$, thenbecause its lower-right entry is $(1+bc)/a=d$. If $a=0$, then $c\ne 0$ and left multiplication by $U(1)$ makes the upper-left entry equal to $c$. The resulting matrix lies in $N$ by the preceding factorization, and $U(1{)}^{-1}\in N$ restores $g$. Therefore

Every unitary matrix is unitarily diagonalizable. Part (ii) therefore proves the stronger matrix statement $\varphi (A(t))=I$, not merely a statement about its eigenvalues. The kernel of a group homomorphism is a normal subgroup, and it contains every upper unipotent. By part (iii) their normal closure is the entire group. ThusThis proves the triviality of finite-dimensional unitary representations of SL2R. In fact the argument used only the algebraic homomorphism property and finite-dimensional unitarity; it did not need continuity.