Past exam of the lie Groups and Differentiable Manifolds course of the University of Oxford / 2008 / A (Lie Groups)

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.