Lie Groups and Differentiable Manifolds course of the University of Oxford (C3.1)

Lie Groups and Differentiable Manifolds, Mathematics Parts C and B; Wednesday 28 May 2008, 9.30am–12.30pm. The complete four-page source and both sections are privately transcribed from original Part C bundle pages 117–120, supplied through the past-papers disclosure. No exact worked-answer advertisement was found in the bounded availability check. Representations are finite-dimensional and continuous; manifolds have no boundary unless stated otherwise.

For a Matrix Lie group, $T_{g}G$ consists of derivatives ${\gamma }^{\prime }(0)$ of differentiable curves in $G$ with $\gamma (0)=g$, viewed inside $M_n(\mathbb{C})$ as a real vector space. At the identity, multiplying two such curves adds their derivatives, and reparametrizing a curve multiplies its derivative by any real scalar. Thus $L(G)=T_{I}G$ is a real vector subspace.

If $Y\in L(G)$ and $g\in G$, conjugating a curve with derivative $Y$ proves $gY{g}^{-1}\in L(G)$. Let $g(s)$ have $g(0)=I$ and $g^{\prime }(0)=X\in L(G)$. Differentiating the curve $g(s)Yg(s{)}^{-1}$ inside the finite-dimensional, hence closed, vector subspace $L(G)$ givesThe matrix commutator is real bilinear, antisymmetric, and satisfies the Jacobi identity by associativity of matrix multiplication. This proves that $L(G)$ is a real Lie algebra.

The exponential-map theorem for a linear Lie group says $\exp X\in G$ whenever $X\in L(G)$. Applying it to every real multiple of $X$ gives $\exp (tX)\in G$ for every real $t$. Conversely that curve has derivative $X$ at zero, so $X\in L(G)$. Hence

Here the orthogonal group is the real group $O_n(\mathbb{R})$. Differentiating $g(t{)}^{T}g(t)=I$ at zero gives $X^T+X=0$. Conversely a real skew-symmetric matrix satisfies $(e^{tX}{)}^T{e}^{tX}=e^{-tX}{e}^{tX}=I$. ThereforeFor $n\ge 2$ this is not a complex Lie subalgebra of $M_n(\mathbb{C})$: multiplying a nonzero real member by $i$ leaves this real subspace. For $n=1$ the zero algebra is the trivial complex Lie algebra as well. The complex orthogonal group would be a different group from the real orthogonal group used here.

Differentiating the affine block form givesConversely the exponential of such a block matrix has bottom row $(0,1)$ and invertible upper-left block $e^{tB}$, so it remains in the affine group. Its commutator corresponds toThe space is closed under multiplication by $i$ and its bracket is complex bilinear. Thus it is a complex Lie algebra of dimension $n^2+n$, and its real dimension is $2n^2+2n$.

The matrix exponential series differentiates term by term on bounded intervals and gives $(e^{tX}{)}^{\prime }=X{e}^{tX}$ and $e^{0X}=I$, proving existence. If $a$ is another differentiable solution, thenIts initial value is $I$, so $e^{-tX}a(t)=I$, giving $a(t)=e^{tX}$ uniquely.

The exponential satisfies the initial conditions and $e^{(s+t)X}=e^{sX}{e}^{tX}$ because the two matrix exponents commute. Conversely, differentiating $a(s+t)=a(s)a(t)$ with respect to $s$ at zero gives $a^{\prime }(t)=Xa(t)$. Even if only differentiability at zero was initially specified, the identity $a(t+h)=a(h)a(t)$ supplies differentiability at every $t$. The uniqueness result just proved yields $\boxed{a(t)=e^{tX}}$.

For fixed $X\in L(G)$, set $b(t)=\varphi (e^{tX})$. The homomorphism property gives $b(s+t)=b(s)b(t)$ and $b(0)=I$. Its initial derivative is $L(\varphi )(X)=D{\varphi }_I(X)$. The one-parameter subgroup uniqueness theorem givesFor the Adjoint representation, ${\mathrm{Ad}}_{g}Y=gY{g}^{-1}$. ThusThe differential $L(\varphi )$ is a real linear map of Lie algebras; the bracket identity follows by differentiating the compatibility of conjugation with $\varphi$.

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.

With the convention applicable also to disconnected compact groups, the Weyl group is $W=N_G(T)/T$, where the normalizer consists of elements taking $T$ to itself by conjugation. Put $\mathfrak{t}=L(T)$ and $\Lambda =\ker (\exp :\mathfrak{t}\to T)$. This is a full lattice with $T\cong \mathfrak{t}/\Lambda$.

If $\psi$ is a torus automorphism, exponential compatibility gives $\psi (\exp X)=\exp (D{\psi }_{I}X)$. Therefore $D{\psi }_I$ sends $\Lambda$ into itself; applying the inverse proves it sends the lattice onto itself. In a lattice basis this derivative is an element of $G{L}_r(\mathbb{Z})$.

Conjugation by a connected subgroup $Q\subset N_G(T)$ induces a continuous family of these integer matrices. Since $G{L}_r(\mathbb{Z})$ is discrete, this family is constant and equals its value at the identity. Its action on $\mathfrak{t}$ is therefore trivial, and surjectivity of the torus exponential shows that $Q$ centralizes $T$.

Apply this to the identity component $N_G(T{)}^0$. It lies in $G^0$ and centralizes $T$. The allowed maximal-abelian property of $T$ in $G^0$ gives $N_G(T{)}^0\subseteq T$, while the connected group $T$ lies in that identity component. Thus $N_G(T{)}^0=T$. The normalizer is a closed compact Lie subgroup, so its component quotient is discrete and compact, hence finite. Consequently $W$ is finite and discrete. For disconnected $G$, its action on the lattice need not be faithful; the component argument is essential.

The distinction is recorded in Weyl group of a disconnected compact Lie group.

A finite discrete group has trivial identity component, so its only torus is $T=\{1\}$. Thus $N_G(T)=G$ and, under the definition $N_G(T)/T$,The alternative quotient by the centralizer would give a different answer for a disconnected group; it is not the convention used here.

A maximal torus is $T=S{O}_2$, which is normal of index two in $O_2$. ThereforeA reflection represents its nontrivial element and conjugates a rotation of angle $\theta$ to the rotation of angle $-\theta$.

Use the given torus of independent rotations in $n$ mutually orthogonal two-planes and a fixed final axis. A normalizer element permutes the two-plane rotation characters, and on each plane either preserves or reverses its orientation. Its induced action therefore permutes the $n$ angle coordinates and changes their signs independently.

Every such signed permutation is realized. Permutations of two-dimensional blocks have determinant one. A reflection in any chosen plane reverses that angle; reflecting the remaining one-dimensional axis as necessary restores total determinant one. The kernel consists precisely of independent plane rotations: commuting with a generic torus element forces preservation of its distinct rotation planes, and the determinant condition then fixes the final axis. Thus the kernel is $T$ andThis is the signed permutation group, the Weyl group of type $B_n$.

As usual here, representations are finite-dimensional and continuous. Start with any positive definite Hermitian form $h_0$ and normalized Haar measure $\mu$ on the compact group. DefineThis is Hermitian, sesquilinear, and positive definite: for $v\ne 0$ its continuous integrand $h_0(gv,gv)$ is strictly positive. Bi-invariance of Haar measure shows $h(av,aw)=h(v,w)$ for every $a\in G$.

If $U\subset V$ is invariant, then so is its orthogonal complement: for $v\perp U$ and $u\in U$, $h(av,u)=h(v,a^{-1}u)=0$. Hence $V=U\oplus U^{\perp }$ as a representation. Induct on dimension, splitting whenever a nontrivial proper invariant subspace exists. This proves every such representation is a direct sum of irreducible representations.

Schur orthogonality for irreducible complex characters of a compact group, with normalized Haar measure, isEquivalent representations clearly have equal characters, since trace is invariant under conjugation. Conversely complete reducibility gives ${\chi }_V={\sum }_{\alpha }{m}_{\alpha }{\chi }_{\alpha }$, where the nonnegative integer multiplicity is recovered by $m_{\alpha }=⟨{\chi }_V,{\chi }_{\alpha }⟩$. Equal characters thus give equal multiplicities of every irreducible constituent, hence equivalent representations.

The usual Weyl character formula requires a compact connected Lie group. Choose a maximal torus $T$, a positive root system ${\Phi }^{+}$, Weyl group $W$, and $\rho =\frac{1}{2}{\sum }_{\alpha \in {\Phi }^{+}}\alpha$. Irreducible representations are indexed by dominant integral highest weights $\lambda$ belonging to the character lattice of the actual group $T$. For $X\in L(T)$ such that $\exp X$ is regular,Weights are written as linear functionals on the complexified torus algebra; their values on the compact real torus algebra are imaginary. The denominator is $e^{\rho (X)}{\prod }_{\alpha >0}(1-e^{-\alpha (X)})$. If $\rho$ is not itself a character of $T$, the individual expressions are interpreted on its covering algebra; their quotient is single-valued and descends to $T$. The apparent singularities at nonregular elements are removable, and conjugacy into a maximal torus determines the character on the whole connected group.

The formula explicitly determines all irreducible characters from their highest weights, and its limit at the identity gives the dimension formulaCharacter multiplication and orthogonality then compute tensor-product multiplicities. For $S{U}_2$, highest weights are integers $m\ge 0$, and at $\mathrm{diag}(z,z^{-1})$,Connectedness cannot simply be omitted from this version. For a nontrivial finite group the maximal torus is trivial, yet characters can differ away from the identity. The root data on that torus alone cannot supply the characters of all disconnected compact groups.

One definition of $T_{x}M$ is the space of derivations at $x$: linear maps $v:C^{\infty }(M)\to \mathbb{R}$ satisfying $v(fh)=f(x)v(h)+h(x)v(f)$. In local coordinates $u^1,\dots ,u^n$, each has the unique form $v={\sum }_i{v}^i\partial /\partial u^i{|}_x$, identifying $T_{x}M$ with ${\mathbb{R}}^n$. Equivalently it is the space of velocities of curves through $x$, identifying curves with the same coordinate derivative.

For a chart $\varphi :U\to {\mathbb{R}}^n$, give $TM{|}_U$ coordinates $(x,v)\mapsto (\varphi (x),D{\varphi }_{x}v)$. If $F=\psi \circ {\varphi }^{-1}$ is a base transition, the induced transition isThis is a smooth bijection with smooth inverse coming from the reverse chart. The resulting bundle atlas makes $TM$ a smooth manifold of dimension $2n$. These local product charts also give the usual Hausdorff and second-countable topology, inherited from a countable base atlas and its fibre coordinates.

For $n\ge 1$, identify nonzero vectors in each fibre up to nonzero real scalar multiplication. Locally this gives $P(TM){|}_U\cong U\times {\mathbb{RP}}^{n-1}$. In a projective chart where $v^j\ne 0$, use the ratios $v^i/v^j$ for $i\ne j$. Under a base change, the fibre line maps by $[v]\mapsto [D{F}_{u}v]$; its projective coordinates are ratios of smooth linear expressions with nonzero denominator on the overlap. This defines a smooth manifold of dimension $n+(n-1)=2n-1$.

For the sphere, a point is a pair $(x,L)$ with $|x|=1$ and an unoriented line $L\subset x^{\perp }$. Choose a unit $v\in L$. The triple $(x,v,x\times v)$ is a positively oriented orthonormal frame. Given a second pair, the unique orientation-preserving linear map between chosen frames lies in $S{O}_3$ and carries the first pair to the second. Thus the action is transitive.

A rotation stabilizing $(x,L)$ fixes $x$, so rotates its tangent plane. To preserve the unoriented line it must act by $v\mapsto v$ or $v\mapsto -v$, giving exactly the identity and a half-turn about $x$. Hence the stabilizer has order two, and $P(T{S}^2)\cong S{O}_3/C_2$.

This is an example of the projectivized tangent bundle.

In a coordinate chart, define the exterior derivative byThe chain rule makes this independent of coordinates. It is real linear, increases form degree by one, agrees with the differential on functions, satisfies $d^2=0$, and obeys the graded Leibniz ruleIt also commutes with pullback by a smooth map. These properties characterize it locally and hence globally.

de Rham cohomology is the quotient of closed forms by exact forms. If $d\alpha =d\beta =0$, Leibniz gives $d(\alpha \wedge \beta )=0$. Changing $\alpha$ to $\alpha +d\eta$ changes the product by $d(\eta \wedge \beta )$; changing $\beta$ to $\beta +d\xi$ changes it by $(-1{)}^{r}d(\alpha \wedge \xi )$. Thusis independent of representatives. This gives the graded-commutative de Rham cohomology ring.

The nonvanishing top-degree form ${\omega }^m$ is a volume form. Declare an ordered tangent basis positive when its value under ${\omega }^m$ is positive. Smooth nonvanishing ensures these choices agree on overlapping oriented charts and vary continuously. Hence a symplectic manifold is orientable; this is its symplectic orientation.

Use the standard convention that the compact manifold is nonempty and has no boundary. Orient it by ${\omega }^m$. Since $d\omega =0$, every power ${\omega }^k$ is closed. If ${\omega }^k=d\eta$ for some $1\le k\le m$, thenagain by Leibniz and closedness. Stokes theorem would give ${\int }_M{\omega }^m=0$. But this integral is strictly positive in the symplectic orientation, a contradiction. ThereforeThe boundaryless hypothesis matters: a compact disk has a symplectic area form but no nonzero ordinary second cohomology class.

Let the two oriented surfaces carry positive area forms $\alpha$ and $\beta$, obtainable from oriented Riemannian metrics. A top-degree form on a surface is automatically closed. On their product setIt is closed, andis a nowhere-vanishing four-form; the two squared terms vanish because each comes from a two-dimensional factor. Thus the product is symplectic, with no compactness assumption needed.

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.

A Riemannian metric is a smooth positive definite symmetric bilinear form $g_q$ on each tangent space. The inverse form on the cotangent space defines the kinetic HamiltonianIt is smooth and satisfies $H(q,ap)=a^{2}H(q,p)$. With the canonical symplectic form $\Omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, define ${\iota }_{X_H}\Omega =dH$. This gives Hamilton's equationsThus $p_i=g_{ij}{\dot{q}}^j$. Differentiating the inverse metric yields ${\dot{p}}_i=\frac{1}{2}{\partial }_i{g}_{jk}{\dot{q}}^j{\dot{q}}^k$. Differentiating $p_i=g_{ij}{\dot{q}}^j$ and substituting givesTherefore the projections of the Hamiltonian integral curves are affinely parametrized geodesics, and lifting a geodesic by its metric covector gives a Hamiltonian integral curve.

For the Killing vector field define the fibrewise linear function $f(q,p)=p(Y_q)=p_i{Y}^i(q)$. Along the lifted geodesic it equals $g(\dot{q},Y)$. Metric compatibility and the geodesic equation giveThus $f$ is conserved. Equivalently the cotangent lift of the isometry flow preserves $H$, and its generator has Hamiltonian $f$, so their Poisson bracket vanishes.

For the final two metrics write $s=\ell (x)+m(y)$, restricting to an open set where the printed tensors are smooth nondegenerate metrics. Their inverse forms giveAlong the $H_1$ flow, $H_1$ is constant, whileHenceThis is the Liouville metric geodesic integral, a quadratic conserved quantity. It does not require a Killing vector field; it arises from the separated metric coefficients. When both printed metrics are Riemannian, their positivity in particular requires $s>0$, $m>0$ and $\ell <0$; the calculation also applies wherever the corresponding inverse tensors are defined.