The corresponding simply connected group is the group of unipotent up-per triangular 3 by 3 matrices: the exponential map is a bijection b ecausethe exponential and logarithm maps are polynomials. The center of thesimply connected group is R, and there is an outer automorphism thatacts by recaling the center, so there are two possible groups with this Liealgebra, one simple connected, and one with center S1. The simply con-nected one can be represented as upper triangular unipotent matrices, andis the group associated with one of the 8 Thurston geometries (the nil-manifolds: take a quotient of the group by a discrete subgroup, such as thesubgroup of matrices with integer coefficients). The other group can berepresented as the group of transformations of L2(R) generated by trans-lations and multiplication by eixy. These satisfy the Weyl commutationrelations. The center is multiplication by constants of absolute value 1.This group has no faithful finite dimensional representations: in any finitedimensional representation the center must act trivially. One way to seethis is to observe that any element of the center of a characteristic 0 Lie al-gebra in the derived algebra must act nilpotently in any finite dimensionalrepresentation (chop the representation up into generalized eigenspaces,and then look at the trace on any generalized eigenspace. The trace mustbe zero as the element is in the derived algebra, so the eigenvalue mustbe zero.) But the only way a nilpotent element can generate a compactgroup is if it acts trivially. There are several variations and generalizationsof these groups. There is a Heisenberg group of dimension 2n + 1 for anypositive integer n associated to a symplectic form of dimension 2n. Wecan also define Heisenberg groups over finite fields in a similar way.Exercise 97 Show that over a finite field of prime order p for p odd, everyelement of the Heisenberg group has order 1 or p, and the exponential mapis a bijection. What happens over the field of order 2?The universal enveloping algebra of the Heisenberg algebra becomes thering of polynomials in x and d/dx if we take a quotient by identifing thecenter of the Heisenberg algebra with the real numbers. This gives a rep-resentation of the Lie algebra on the ring of polynomials, with the centeracting as scalars. The center of this Lie algebra cannot make up its mindwhether it is semisimple or nilpotent: in finite dimensional representationsit acts nilpotently, but in the infinite dimensional representations we havedescribed its acts semisimply.• A is not nilpotent and not semisimple. Both eigenvalues must be thesame, and we can normalize A so they are both 1. So we can assume A is1 10 1. (Bianchi type IV)Exercise 98 Show that there is a unique connected Lie group with thisLie algebra, and represent it by 3 by 3 upper triangular matrices. Findthe conjugacy classes of this group that are in the 2-dimensional derivedsubalgebra, and sketch a picture of them, paying careful attention to whathappens near the origin (the answer may be slightly stranger than youexpect).44• A is semisimple, nonzero, with one eigenvector zero. The Lie algebra is theproduct of the 1-dimensional abelian Lie algebra with the 2-dimensionalnon-abelian Lie algebra (Bianchi type III). There are two correspondingLie groups.• A is semisimple, nonzero, with real non-zero eigenvectors (Bianchi typeVI if the eigenvalues are distinct, type V if they are the same). Herewe get an uncountable infinite family of distinct Lie algebras, as we canchange the smallest eigenvalue to anything we want, but then the secondis determined. There is only one connected group for each of these Liealgebras. If the eigenvalues have sum 0 the Lie algebra has an e xtrasymmetry (Bianchi type V I0) This is the Lie algebra of isometries of 2-dimensional Minkowski space. This also appears as the group of one of the8 Thurston geometries, giving the solv manifolds. For example one cantake a quotient of this group by a cocompact discrete subgroup. Some ofthese manifolds are the mapping cyliner of an Anosov map of the 2-torus(given by an integral matrix A with distinct real nonzero eigenvectorswhose product is 1).Exercise 99 Find an example of an Anosov map. Show how to constructa cocompact discrete subgroup of the Bianchi group V I0from any Anosovmap.Exercise 100 Show that the outer automorphism group of this connectedLie group is dihedral of order 8. (Some elements correspond to time re-versal, parity reversal, and changing the sign of the metric of Minkowskispace.)When the eigenvalues are the same the group consists of translations anddilations of the plane.• A is semisimple, nonzero, with non-real eigenvectors. Bianchi type VII.Again we get an infinite family of Lie algebras. The simply connectedgroup has trivial center except for the following special case (Bianchi typeV II0): this is the one with imaginary eigenvalues, and is the Lie algebraof isometries of the plane. It has an extra symmetry. There is an obviousconnected group with this Lie algebra: we can take orientation-preservingisometries of the plane. However this group is not simply connected, as ithas homotopy type the circle, so we can also take its universal cover, orthe cover of any order 1, 2, 3, . . .. We came across this group earlier as asolvable connected Lie group whose exponential map is not surjective.Exercise 101 Show that the real Lie algebras of type V I0and V II0arenot isomorphic, but become isomophic when tensored with the complexnumbers.The remaining cases are where G is not solvable, in which case it must be sim-ple as all groups of smaller dimension are solvable. (Similarly the non-solvablefinite group of smallest order is necessarily simple.) We will postpone the classi-fication of these as this will be easier when we have developed more theory, andjust state the result. There are 2 possible Lie algebras, su(2) (Bianchi type IX)45and sl2(R) (Bianchi type VIII). The first has simply connected group SU(2)with center of order 2, so we get two possible Lie groups (one is SO3(R)). Forthe other there are two obvious groups SL2(R) and the quotient by its centerP SL2(R). However there are infinitely many other groups because SL2(R) isnot simply connected: its fundamental group is Z so we can take its universalcover (which also has fundamental group Z) and quotient out by any subgroupof Z. These covers have no faithful finite dimensional representations. Thedouble cover of SL2(Z) appears