Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 20 / 1

For $X\in \mathfrak{g}=T_{e}G$, let $X^L$ be the left-invariant vector field with value $X$ at the identity. Its integral curve through the identity is the unique one-parameter subgroup ${\gamma }_X$ with ${\gamma }_X^{\prime }(0)=X$. It exists for all real time: local existence followed by translation and the identity ${\gamma }_X(t+s)={\gamma }_X(t){\gamma }_X(s)$ extends it indefinitely. The Exponential map of a Lie group isIt is smooth and $d_0\exp ={\mathrm{id}}_{\mathfrak{g}}$, because differentiating $t\mapsto \exp (tX)$ at zero gives $X$. The inverse function theorem therefore makes it a local diffeomorphism at zero.

If $G$ is abelian, the product $t\mapsto \exp (tX)\exp (tY)$ is a one-parameter subgroup with initial tangent $X+Y$. Uniqueness gives $\exp (t(X+Y))=\exp (tX)\exp (tY)$, henceConversely, if this additive homomorphism identity holds, all $\exp (tX)$ and $\exp (sY)$ commute. Differentiating their adjoint action shows $[X,Y]=0$. Alternatively, their commuting images contain an identity neighbourhood; that neighbourhood generates the identity component of a Lie group, so the identity component is abelian. Conversely, if the Lie algebra is abelian, its left-invariant vector fields commute, so their flows commute. Hence all one-parameter subgroups commute; their product with initial tangent $X+Y$ is the one-parameter subgroup for $X+Y$. The Lie exponential map is therefore additive, and its image generates an abelian identity component. Thus the stated equivalence holds for connected $G$. Without connectedness the exact assertion is that the Lie exponential map is a homomorphism precisely when $G^{\circ }$ is abelian. The converse implication to all of $G$ is false: in $O(2)$ the Lie algebra consists of $tJ$, whereRotations satisfy $R_{t+s}=R_t{R}_s$, but a reflection $S$ satisfies $S{R}_t{S}^{-1}=R_{-t}$, so the full orthogonal group is not abelian.

Now take a connected abelian Lie group of dimension $n$. Its Lie exponential map is a homomorphism whose image contains an identity neighbourhood, hence an open subgroup. A connected group has no proper open subgroup, so this map is onto. Its kernel $\Lambda$ is discrete, since it is injective on a neighbourhood of zero. The induced map is a bijective local diffeomorphism and therefore a Lie group isomorphismHere is the needed classification of $\Lambda$. Let $r$ be the dimension of its real span. Choose ${\lambda }_1,\dots ,{\lambda }_r\in \Lambda$ that form a real basis of that span. Their integer span ${\Lambda }_0$ has finite index in $\Lambda$: subtract integer multiples of the ${\lambda }_i$ to put each coset representative in a compact parallelepiped. Only finitely many points of $\Lambda$ lie there, since discreteness at zero gives a uniform positive separation between distinct points. Thus $\Lambda$ is finitely generated and torsion-free, and the structure theorem for finitely generated abelian groups makes it free abelian of rank $r$. Its integer basis is a real basis of the span. A linear change of coordinates consequently sends $\Lambda$ to ${\mathbb{Z}}^r\times \{0\}$ in ${\mathbb{R}}^n$.

This proves the classification of connected abelian Lie groups:The isomorphism depends on the choice of lattice basis and complementary vector space. Compactness is equivalent to $s=0$.

A complex analytic Lie group is a complex manifold whose multiplication and inversion are holomorphic maps between complex manifolds. Its identity tangent space is a complex Lie algebra, and the Adjoint representation of a Lie groupis holomorphic: differentiating the holomorphic conjugation map with respect to its second variable gives holomorphically varying matrix entries.

Every scalar holomorphic function on a compact connected complex manifold is constant. Indeed its modulus attains a maximum; the maximum modulus principle, applied in complex coordinate charts and along complex lines, makes it locally constant there, and connectedness propagates that constancy. Apply this to each entry of $\mathrm{Ad}$. Since ${\mathrm{Ad}}_e=I$, it follows that ${\mathrm{Ad}}_g=I$ for every $g$. Differentiation givesThus $\mathfrak{g}$ is abelian, and the connected group $G$ is abelian by the identity-neighbourhood argument in part (i). This proves that compact connected complex Lie groups are complex tori.

If $n={\dim }_{\mathbb{C}}G$, identify $\mathfrak{g}$ with ${\mathbb{C}}^n$. The Lie exponential map is holomorphic, additive, onto, and a local biholomorphism, by the holomorphic left-invariant differential equation and its identity derivative. Its kernel $B$ is a discrete subgroup of the additive space ${\mathbb{C}}^n$. The induced map yieldsas complex Lie groups, not merely as real groups. Compactness further forces $B$ to be a full-rank real lattice. If its real span $W$ were proper, projection would give a surjective continuous map ${\mathbb{C}}^n/B\to {\mathbb{R}}^{2n}/W$, contradicting compactness. The discrete-subgroup argument in part (i) therefore gives $B\cong {\mathbb{Z}}^{2n}$. The quotient is a complex torus; it need not split into one-dimensional complex tori.