1. The general linear

2 Recall that M(n, R), the of all n × n real matrices, is diffeomorphic to Rn . Definition 1.1. A linear , or Lie group, is a of M(n, R) which is also a Lie group, with group structure the matrix .

Let’s begin with the “largest” linear Lie group, the general GL(n, R) = {X ∈ M(n, R) | det X 6= 0}. Since the map is continuous, GL(n, R) is open in M(n, R) and thus a sub- . Moreover, GL(n, R) is closed under the group multiplication and inversion operations, so it is a Lie group. Obviously GL(n, R) is an n2-dimensional noncom- pact Lie group, and it is not connected. In fact, it consists of exactly two connected components, GL+(n, R) = {X ∈ M(n, R) | det X > 0} and GL−(n, R) = {X ∈ M(n, R) | det X < 0}. 2 The fact that GL(n, R) is an open of M(n, R) ' Rn also implies that the Lie of GL(n, R), as the tangent at e = In, is the set M(n, R) itself, i.e. gl(n, R) = {A | A is an n × n real matrix}.

To figure out the Lie operation, we take a matrix A = (Aij)n×n ∈ g, and take the global coordinate system by (Xij). Then the corresponding tangent vector P ∂ at TIn GL(n, R) is Aij ∂Xij , and the corresponding left- vector on G at the ij P ik ∂ matrix X = (X ) is X Akj ∂Xij . It follows that the Lie bracket [A, B] between matrices A, B ∈ g is the matrix corresponding to X ∂ X ∂  X ∂ X ∂ XikA , XpqB = XikA B − XpqB A kj ∂Xij qr ∂Xpr kj jr ∂Xir qr rj ∂Xpj X ∂ = Xik (A B − B A ) . kr rj kr rj ∂Xij In other words, the Lie bracket operation on g is the matrix [A, B] = AB − BA.

Given any A ∈ gl(n, R), we can define the A2 A3 An eA = I + A + + + ··· + + ··· . n 2! 3! n!


It is easy to check that the series converges, and esAetA = e(s+t)A. 0A tA −1 −tA tA tA Notice that e = In, we have (e ) = e . In particular, e ∈ GL(n, R). So e d tA is a one-parameter of GL(n, R). Since dt |t=0e = A, we conclude that the exponential map exp : gl(n, R) → GL(n, R) is A2 A3 exp(A) = eA = I + A + + + ··· . n 2! 3! Note that exp is not surjective, not even surjective to GL+(n, R). Remark. Alternatively, one can first show that etA is the one-parameter subgroup of GL(n, R) generated by A ∈ gl(n, R), and thus the Lie bracket structure on gl(n, R) is

d d tA sB −tA d tA −tA [A, B] = (e e e ) = (e Be ) = AB − BA. dt t=0 ds s=0 dt t=0

2. Lie of GL(n, R) Before we continue to study other linear Lie groups, we need some general results on Lie subgroups. More details and proofs of these results will be discussed later. Definition 2.1. A subgroup H of a Lie group G is called a Lie subgroup if it is a Lie group (with respect to the induced group operation), and the ιH : H,→ G is an (and therefore a Lie group ). Suppose H is a Lie subgroup of G, and h be the of H. Since ι : H,→ G is an immersion and is a Lie , dιH : h → g is injective and is a Lie algebra homomorphism. So we can think of h as a Lie (= a that is closed under the Lie bracket) of g. Note that a one-parameter subgroup of H is automatically a one-parameter of G (with initial vector in TeH), so the exponential map expH : h → H is exactly the restriction of expG : g → G onto h. The following theorem, which we will prove later, is very useful in determine the Lie algebra of a Lie subgroup. Theorem 2.2. Suppose H is a Lie subgroup of G. Then as a Lie subalgebra of g,

h = {X ∈ g | expG(tX) ∈ H for all t ∈ R}. Now we are ready to study other linear Lie groups. By definition they are Lie subgroups of GL(n, R). Example. (The ) The special linear group is defined as SL(n, R) = {X ∈ GL(n, R) : det X = 1}. It is easy see that SL(n, R) is a subgroup of GL(n, R). In PSET 1 we have seen that SL(n, R) is an n2 − 1 dimensional submanifold of GL(n, R). It follows that SL(n, R) is a (connected non-compact) Lie subgroup of GL(n, R). LECTURE 7: LINEAR LIE GROUPS 3

To determine its Lie algebra sl(n, R), we notice that det eA = eTr(A). So for an n × n matrix A, eA ∈ SL(n, R) if and only if Tr(A) = 0. We conclude sl(n, R) = {A ∈ gl(n, R) | Tr(A) = 0}. Example. (The ) Next let’s consider the orthogonal group T O(n) = {X ∈ GL(n, R): X X = In}. n(n−1) This is another subgroup of GL(n, R). In PSET 1 we have seen that O(n) is an 2 n(n−1) dimensional submanifold of GL(n, R). So O(n) is an 2 dimensional Lie subgroup of GL(n, R). Note that O(n) is compact. To figure out its Lie algebra o(n), we note that (eA)T = eAT , so

tA T tA tAT −tA (e ) e = In ⇐⇒ e = e For all t. Since the exponential map is locally bijective, we conclude that A ∈ o(n) if and only if AT = −A. So T o(n) = {A ∈ gl(n, R) | A + A = 0}, which is the space of n × n skew-symmetric matrices. Notice that O(n) is not connected. It consists of two connected components, and the connected of identity is the called the special orthogonal groups T SO(n) = {X ∈ GL(n, R): X X = In, det X = 1} = O(n) ∩ SL(n, R). Its Lie algebra so(n) is the same as o(n). Example. (The ) The symplectic group is by definition T Sp(2n, R) = {X ∈ GL(2n, R): X JX = J},  0 I  where Let J = n . It is a Lie group of n(2n + 1) with Lie algebra −In 0 T sp(2n, R) = {A ∈ gl(2n, R) | JA + A J = 0}   A1 A2 T T T = {A = | Ai ∈ M(n, R),A1 = −A4 ,A2 = A2 ,A3 = A3 }. A3 A4 This, as well as the orthogonal group in the previous example, are special cases of the following more general example. Let β : Rn × Rn → R be a on Rn. Consider the set of all invertible n × n matrices that preserves β, n GLβ(n, R) = {X ∈ GL(n, R) | β(Xu, Xv) = β(u, v)for all u, v ∈ R }. In matrix form, there is a matrix B such that β(u, v) = uT Bv. Then T GLβ(n, R) = {X ∈ GL(n, R) | X BX = B}. 4 LECTURE 7: LINEAR LIE GROUPS

Lemma 2.3. GLβ(n, R) is a linear Lie group with Lie algebra T glβ(n, R) = {A ∈ gl(n, R) | A B + BA = 0}.

Proof. One can easily check that GLβ(n, R) is a subgroup of GL(n, R), and it is topo- logically a closed subset. According to the Cartan’s closed subgroup theorem that we will prove later, it is a Lie subgroup. tA To describe its Lie algebra, notice that A ∈ glβ(n, R) if and only if e ∈ GLβ(n, R), i.e. etAT BetA = B. By taking t derivative at t = 0, we get AT B + BA = 0. Conversely, if AT B + BA = 0, i.e. tAT B = B(−tA), one can easily derive by definition that tAT −tA tAT tA e B = Be , i.e. e Be = B. This completes the proof.  n Notice that in the case B = In (β = the standard inner product on R ), we get the orthogonal group O(n), and in the case B = J(β = the standard symplectic form on R2n), we get the symplectic group above. If we take β to be the standard inner product of signature (p, n − p), p n X X β(x, y) = xiyi − xiyi, i=1 i=p+1 then we will get the indefinite orthogonal group O(p, n − p), T O(p, n − p) = {X ∈ GL(n, R) | X I(p, n − p)X = I(p, n − p)}, where I(p, n − p) = diag(Ip, −In−p). Its Lie algebra is T o(p, n − p) = {X ∈ gl(n, R) | A I(p, n − p) + I(p, n − p)A = 0.}. Example. (The ) All the previous examples generalize to complex matrices. In particular, the unitary groups T U(n) = {X ∈ GL(n, C): X X = In} is a Lie subgroup of GL(n, C) with Lie algebra T u(n) = {A ∈ gl(n, C) | A + A = 0}, the space of skew-Hermitian matrices. Also the special unitary groups SU(n) = U(n) ∩ SL(n, C) has Lie algebra T su(n) = {A ∈ M(n, C) | A + A = 0, Tr(A) = 0}. Also we have the compact symplectic group Sp(n) = U(2n) ∩ Sp(2n, C). This is a real compact Lie group of dimension n(2n + 1).