Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 1 (Lie Groups)

Identify the Lie algebra $\mathfrak{g}$ with the tangent space $T_{e}G$. For $X\in \mathfrak{g}$, its left-invariant vector field is $X^L(g)=(d{L}_g{)}_{e}X$. Let ${\gamma }_X$ be its integral curve with ${\gamma }_X(0)=e$. Uniqueness of integral curves and left invariance give ${\gamma }_X(s+t)={\gamma }_X(s){\gamma }_X(t)$ wherever initially defined. Repetition of a local curve extends it to all real times. Thus ${\gamma }_X$ is the unique one-parameter subgroup with derivative $X$ at zero, and the Exponential map of a Lie group is defined bySmooth dependence of solutions of differential equations on their initial data and parameters makes this map smooth.

The derivative at zero is particularly simple:It is the identity linear map from $\mathfrak{g}$ to $T_{e}G$. The inverse function theorem therefore proves that the Exponential map of a Lie group is a local diffeomorphism at zero, producing a local exponential chart. This proves the requested local assertion.

There is a necessary qualification to the other assertion: the image of every one-parameter subgroup lies in the identity component of a Lie group $G^{\circ }$. The correct statement without a connectedness hypothesis isIndeed, if $\exp$ is a group homomorphism, its image is an abelian subgroup. It contains an identity neighborhood by the local result. An open subgroup of a connected group is the entire group: every coset is open, so its complement is also open. Thus $\exp (\mathfrak{g})=G^{\circ }$, proving that this component is abelian.

Conversely, if $G^{\circ }$ is abelian, then $\exp (tX)\exp (tY)$ is a one-parameter subgroup: its group law follows by commuting the factors. Its derivative at zero is $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$ and, at $t=1$, additivity. This proves the additivity of the Lie exponential map criterion. For connected $G$ it is exactly the stated equivalence with an abelian group. For disconnected $G$ the unrestricted claim is false: $S_3\times \mathbb{R}$ is a nonabelian Lie group, but its exponential is the homomorphism $x\mapsto (e,x)$. The original PDF does not impose connectedness in that sentence, so that qualification cannot be omitted.

Now suppose $G$ is connected and abelian, of dimension $n$. The exponential is a surjective group homomorphism, and its kernel $\Lambda$ is discrete by local injectivity at zero. The induced mapis a bijective Lie group homomorphism and a local diffeomorphism, hence a Lie group isomorphism. To identify the quotient, take linearly independent elements ${\lambda }_1,\dots ,{\lambda }_b\in \Lambda$ spanning the real linear span of $\Lambda$, and let ${\Lambda }_0={\sum }_i\mathbb{Z}{\lambda }_i$. Every coset of ${\Lambda }_0$ in $\Lambda$ has a representative in the bounded fundamental parallelepiped of these vectors. A discrete additive subgroup has finite intersection with a compact set: otherwise differences of arbitrarily close elements would approach zero, contradicting discreteness at zero. Thus $\Lambda /{\Lambda }_0$ is finite.

It follows that $\Lambda$ is a finitely generated torsion-free abelian group of rank $b$, hence has a $\mathbb{Z}$-basis of $b$ elements. These basis elements also span its $b$-dimensional real span and are linearly independent over $\mathbb{R}$. Extending them to a real basis identifies $\Lambda$ with ${\mathbb{Z}}^b\times \{0{\}}^{n-b}$. Consequently the classification of connected abelian Lie groups isThe compact circle factors record the periods of the one-parameter subgroups.

The Lie algebra of SL2R is the space of traceless real two-by-two matrices. For $A$ in this space, the Cayley-Hamilton theorem gives $A^2=cI$, where $c=-\det A$. Expanding the matrix exponential, if $c=s^2>0$ thenIf $c=0$, then $\exp A=I+A$ and its trace is $2$. If $c=-s^2<0$, thenIn every case $\mathrm{tr}(\exp A)\ge -2$. Buthas determinant one and trace $-5/2$. Thus $g\in S{L}_2(\mathbb{R})$ is not an exponential, and the exponential map is not surjective.

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.

For a compact Lie group $G$, the complex representation ring of a compact group $R(G)$ is the Grothendieck group of finite-dimensional continuous complex group representations, with the relations $[M\oplus N]=[M]+[N]$. Multiplication is $[M][N]=[M\otimes N]$, and the unit is the trivial one-dimensional representation. The ring of class functions iswith pointwise addition and multiplication. The character of $M$ is ${\chi }_M(g)=\mathrm{tr}{\rho }_M(g)$. The trace identities for direct sums and tensor products makea well-defined ring homomorphism $\chi :R(G)\to c\ell (G)$.

Normalize Haar measure by ${\int }_{G}dg=1$. Averaging a positive definite Hermitian inner product giveswhich is positive definite and $G$-invariant. This proves unitarization of a compact-group representation. In a unitary representation the orthogonal complement of an invariant subspace is invariant. Induction on dimension therefore gives complete reducibility of compact-group representations. Hence every element of $R(G)$ is a finite integral linear combination of irreducible classes.

We establish the needed character orthogonality for compact groups carefully. For irreducible unitary representations $V,W$, let $G$ act on ${\mathrm{Hom}}_{\mathbb{C}}(W,V)$ byIts average $P={\int }_G(g\cdot )dg$ is a projection onto ${\mathrm{Hom}}_G(W,V)$: averaging makes every image invariant, and it fixes every intertwiner. The trace of this action is ${\chi }_V(g)\overline{{\chi }_W(g)}$, since the inverse of a unitary matrix has conjugate trace. Taking the trace of the projection givesFor completeness, Schur's lemma follows here from invariance of the kernel and image of an intertwiner. A nonzero intertwiner between irreducibles is an isomorphism. An endomorphism of an irreducible complex representation has an eigenvalue $\lambda$; the noninvertible intertwiner $A-\lambda I$ must vanish. Thus the last dimension is one for isomorphic irreducibles and zero for inequivalent ones.

If a virtual character ${\sum }_V{n}_V{\chi }_V$ vanishes, integrating it against $\overline{{\chi }_W}$ gives $n_W=0$ for every irreducible $W$. The irreducible multiplicities therefore distinguish the classes, andThis also proves that two finite-dimensional representations with the same character are isomorphic.

For SU(2), the irreducibles are $V_k={\mathrm{Sym}}^k({\mathbb{C}}^2)$, $k\ge 0$, of dimension $k+1$. One can see this classification through the usual ${\mathfrak{sl}}_2$ operators $H,E,F$: a highest-weight vector $v$ has $Hv=kv$, $Ev=0$, and the relations $[H,E]=2E$, $[H,F]=-2F$, $[E,F]=H$ give, by induction,The torus weights are integers. If $F^{r}v\ne 0$ but $F^{r+1}v=0$, the identity at $j=r+1$ forces $k=r\ge 0$. The vectors $v,Fv,\dots ,F^{k}v$ span an invariant irreducible module, which in an irreducible representation is the whole space. This is the symmetric-power module. Conversely its successive monomial weight vectors are linked by $E,F$ with nonzero coefficients, proving its irreducibility. Thus the classification of finite-dimensional representations of SU2 gives every irreducible, rather than just a list of examples.

On the maximal torus $T=\{\mathrm{diag}(z,z^{-1}):|z|=1\}$, their characters areWith $x={\chi }_1=z+z^{-1}$, multiplication gives ${\chi }_{k+1}=x{\chi }_k-{\chi }_{k-1}$, starting with ${\chi }_0=1$. Hence each ${\chi }_k$ is a monic integral polynomial of degree $k$ in $x$. For example ${\chi }_2=x^2-1$ and ${\chi }_3=x^3-2x$. They form a basis over $\mathbb{Z}$, soThe character map realizes this ring as the finite integral polynomials in $z+z^{-1}$, interpreted as continuous class functions. Its image is not the whole infinite-dimensional ring of continuous class functions; injectivity is the assertion being proved.

We prove uniform approximation, not just approximation in $L^2$. The analytic facts used are: a continuous square-integrable kernel gives a compact integral operator on $L^2$; the spectral theorem for compact self-adjoint operators decomposes the closure of its range into its finite-dimensional nonzero eigenspaces; and continuous functions are dense in $L^2$ for normalized Haar measure on a compact Lie group. Normalized Haar measure on a compact group is both left- and right-invariant and is preserved by inversion.

Choose a continuous nonnegative function $k$ of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with $k(g^{-1})=k(g)$. Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the convolution operatorIts kernel is continuous on $G\times G$, so $T$ is compact. The symmetry of $k$ makes it self-adjoint. It commutes with the left translations $(L_{g}h)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.

By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show $Th$ is continuous for every $h\in L^2(G)$. Moreover, rewriting $y=x{z}^{-1}$ gives $(Tf)(x)=\int f(x{z}^{-1})k(z)dz$. Uniform continuity on the compact group therefore makes $\Vert Tf-f{\Vert }_{\infty }$ arbitrarily small when the support of $k$ is small. Also $\Vert T{\Vert }_{\infty \to \infty }\le 1$ because $k\ge 0$ and has integral one.

Fix $\epsilon >0$ and take $k$ so $\Vert Tf-f{\Vert }_{\infty }<\epsilon /3$. ThenLet $P_N$ be the orthogonal projection onto increasing finite sums of the nonzero eigenspaces of $T$. Since $Tf$ lies in the closure of the range, the spectral theorem gives $P_{N}Tf\to Tf$ in $L^2$. Applying the smoothing estimate givesThus some $h=T{P}_{N}Tf$ satisfies $\Vert h-f{\Vert }_{\infty }<\epsilon$.

Each nonzero eigenspace is finite dimensional, invariant under $L_g$, and consists of continuous functions: if $Tv=\lambda v$ with $\lambda \ne 0$, then $v=Tv/\lambda$. The finite sum $M$ containing $h$ therefore carries the finite-dimensional unitary representation $\rho (g)=L_g{|}_M$. It is continuous, since translating each of its finitely many continuous basis functions depends continuously on $g$ in the uniform and hence the $L^2$ norm.

Let $\ell$ be evaluation at the identity on $M$, and let the dual representation on $M^{\ast }$ be $\theta (g)\ell =\ell \circ \rho (g^{-1})$. For the linear functional $L_h$ on $M^{\ast }$ defined by $L_h(u)=u(h)$,The rank-one endomorphism $\alpha =\ell \otimes L_h$ of $M^{\ast }$ satisfies $\mathrm{tr}(\alpha \theta (g))=L_h(\theta (g)\ell )$. ThereforeThis is the requested Peter-Weyl theorem, proved via the convolution proof of uniform Peter-Weyl approximation. The extra convolution after the spectral truncation is what upgrades the $L^2$ estimate to a uniform one.

To deduce a faithful finite-dimensional representation, first note that these representations separate points. For $g\ne e$, choose a continuous function taking different values at $g$ and $e$, and approximate it closely enough by a trace coefficient to retain that difference. The representation appearing in that trace must satisfy $\theta (g)\ne I$.

There is an identity neighborhood $U$ containing no nontrivial subgroup. Here is the Lie groups have no small subgroups argument: choose an injective exponential chart on a Lie-algebra ball of radius $r$, and take $U=\exp (B_{\delta })$ with $0<\delta <r/2$. If a subgroup contained a nonidentity $\exp X$ in $U$, choose the first integer $m$ with $m\Vert X\Vert \ge \delta$. Then $\delta \le m\Vert X\Vert <2\delta <r$, so its power $\exp (mX)$ lies outside $U$, a contradiction. For a zero-dimensional group, take $U=\{e\}$.

For every $g\notin U$, choose a representation nontrivial at $g$. The sets on which these representations are not the identity are open and cover the compact set $G\setminus U$. A finite subcover gives finitely many representations, whose direct sum has kernel contained in $U$. A subgroup contained in $U$ is trivial, so this direct sum is faithful. Averaging its Hermitian form makes it unitary. Thus the finite faithful representation from point-separating representations yieldsUsing the no-small-subgroups neighborhood is essential; $G\setminus \{e\}$ alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.

A maximal torus in a compact connected Lie group is a compact connected abelian Lie subgroup maximal among such subgroups. By the connected abelian classification it is isomorphic to a product of circles. Such subgroups exist: choose one of largest dimension among compact connected abelian subgroups; a proper inclusion of connected Lie subgroups would increase dimension. Write $\mathfrak{t}$ for its Lie algebra.

Here are the main steps of the conjugacy assertion, with the critical arguments supplied. Average an inner product on $\mathfrak{g}$ over the adjoint action to obtain an $\mathrm{Ad}(G)$-invariant positive definite inner product. It defines a bi-invariant Riemannian metric. Compactness makes this metric complete, and the Hopf-Rinow theorem joins the identity to any $g$ by a geodesic. For a bi-invariant metric the Levi-Civita connection on left-invariant fields is ${\nabla }_{X}Y=[X,Y]/2$, so the geodesic starting with $X$ is $\exp (tX)$. Consequently every $g$ has the form $\exp X$.

The Lie-algebra centralizer of $\mathfrak{t}$ is exactly $\mathfrak{t}$. If $X$ commutes with $\mathfrak{t}$, its one-parameter subgroup commutes with $T$. The closure of the subgroup generated by $T$ and $\exp (\mathbb{R}X)$ is compact, connected and abelian, so it is a torus; maximality forces it to equal $T$, giving $X\in \mathfrak{t}$. The torus action on ${\mathfrak{g}}_{\mathbb{C}}$ decomposes into finitely many character weight spaces. The zero-weight space is ${\mathfrak{t}}_{\mathbb{C}}$. Choose $H\in \mathfrak{t}$ outside the finitely many kernels of the nonzero weight differentials. Then the centralizer of $H$ is exactly $\mathfrak{t}$.

For an arbitrary $X\in \mathfrak{g}$, the continuous function $g\mapsto ⟨{\mathrm{Ad}}_{g}X,H⟩$ attains a maximum. Put $Y={\mathrm{Ad}}_{g_0}X$ at a maximizer. Differentiating along $\exp (sZ)g_0$ givesHence $[Y,H]=0$, so $Y\in \mathfrak{t}$. Exponentiating gives $g_0(\exp X)g_0^{-1}=\exp Y\in T$. This adjoint-orbit criterion for maximal-torus conjugacy proves

Restriction of representations is a ring homomorphism $\mathrm{Res}:R(G)\to R(T)$. If a virtual representation restricts to zero, its character vanishes on $T$. Since characters are class functions and every element is conjugate into $T$, it vanishes on all of $G$. The injectivity of the character map proved in Question 3 then givesThe Weyl group is $W=N_G(T)/T$. For $n\in N_G(T)$, the operator $\rho (n)$ intertwines the restriction of $\rho$ with its twist by $t\mapsto nt{n}^{-1}$. Therefore every restricted representation, and every virtual difference of them, is fixed by $W$:

For the special orthogonal group $G=SO(2m+1)$, takewhere $R_{\theta }$ is a real planar rotation. It is a maximal torus: its Lie-algebra centralizer consists of the independent infinitesimal rotations in those planes. Its character lattice is ${\mathbb{Z}}^m$, so, writing $z_i=e^{i{\theta }_i}$,Indeed commuting unitary torus matrices diagonalize simultaneously into one-dimensional characters, and the characters of the circle are exactly the integral powers.

The normalizer permutes the real rotation planes and can reverse each angle independently. Permutations of whole two-dimensional blocks have determinant one. A reflection in one rotation plane reverses its angle and has determinant minus one; combining it with reflection of the remaining fixed line gives an element of $SO(2m+1)$. Thus all independent inversions occur. Conversely a normalizer must permute the distinct nontrivial torus character pairs $\{\pm e_i\}$ and preserve the fixed line, so there are no additional actions. The kernel of this action is exactly $T$, giving

It follows that the restriction identifies $R(SO(2m+1))$ with a subring of the signed-permutation-invariant Laurent polynomials. In this case we can prove equality. Invariance under individual inversions first givesFor one variable, every invariant is an integral combination of $1$ and $z^k+z^{-k}$, and these satisfy the recurrence $s_{k+1}=x{s}_k-s_{k-1}$. Applying this one coordinate at a time proves the assertion over the other Laurent coefficient rings. Permutation invariance and the Fundamental theorem of symmetric polynomials then givewhere $e_k$ are the elementary symmetric polynomials.

Let $V$ be the complexification of the defining $(2m+1)$-dimensional real representation. Its torus weights are $1,z_i,z_i^{-1}$. The characters of its exterior powers obeyFor $k\le m$, choosing $j$ linear terms and $r$ quadratic terms shows that the coefficient $c_k={\chi }_{{\Lambda }^{k}V}$ isIts leading term is $e_k$ with coefficient one; all other terms use $e_j$ with $j<k$. By induction $e_k$ lies in the restriction image, because $c_k$ is the character of a genuine representation and all smaller $e_j$ already lie in the image. Hence every invariant polynomial lies there, proving the representation ring of an odd-dimensional special orthogonal group descriptionThe triangular change of generators is integral, so $[{\Lambda }^{1}V],\dots ,[{\Lambda }^{m}V]$ are algebraically independent polynomial generators. For $m=1$, $c_1=1+z+z^{-1}$ and this reduces to $R(SO(3))=\mathbb{Z}[V_3]$, with the three-dimensional defining representation as generator. These are genuine representations of the orthogonal group, not the half-integral spin weights of its simply connected double cover.