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LINEAR LIE GROUPS 1. the General Linear Group Recall That M(N,R)

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LINEAR LIE GROUPS 1. the General Linear Group Recall That M(N,R) LECTURE 7: LINEAR LIE GROUPS 1. The general linear group 2 Recall that M(n; R), the set of all n × n real matrices, is diffeomorphic to Rn . Definition 1.1. A linear Lie group, or matrix Lie group, is a submanifold of M(n; R) which is also a Lie group, with group structure the matrix multiplication. Let's begin with the \largest" linear Lie group, the general linear group GL(n; R) = fX 2 M(n; R) j det X 6= 0g: Since the determinant map is continuous, GL(n; R) is open in M(n; R) and thus a sub- manifold. 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) = fX 2 M(n; R) j det X > 0g and GL−(n; R) = fX 2 M(n; R) j det X < 0g: 2 The fact that GL(n; R) is an open subset of M(n; R) ' Rn also implies that the Lie algebra of GL(n; R), as the tangent space at e = In, is the set M(n; R) itself, i.e. gl(n; R) = fA j A is an n × n real matrixg: To figure out the Lie bracket operation, we take a matrix A = (Aij)n×n 2 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-invariant 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 2 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 commutator [A; B] = AB − BA: Given any A 2 gl(n; R), we can define the matrix exponential A2 A3 An eA = I + A + + + ··· + + ··· : n 2! 3! n! 1 2 LECTURE 7: LINEAR LIE GROUPS 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 2 GL(n; R). So e d tA is a one-parameter subgroup of GL(n; R). Since dt jt=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 2 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 subgroups 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 inclusion map ιH : H,! G is an immersion (and therefore a Lie group homomorphism). Suppose H is a Lie subgroup of G, and h be the Lie algebra of H. Since ι : H,! G is an immersion and is a Lie group homomorphism, dιH : h ! g is injective and is a Lie algebra homomorphism. So we can think of h as a Lie subalgebra (= a linear subspace 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 = fX 2 g j expG(tX) 2 H for all t 2 Rg: Now we are ready to study other linear Lie groups. By definition they are Lie subgroups of GL(n; R). Example. (The special linear group) The special linear group is defined as SL(n; R) = fX 2 GL(n; R) : det X = 1g: 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 2 SL(n; R) if and only if Tr(A) = 0. We conclude sl(n; R) = fA 2 gl(n; R) j Tr(A) = 0g: Example. (The orthogonal group) Next let's consider the orthogonal group T O(n) = fX 2 GL(n; R): X X = Ing: 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 2 o(n) if and only if AT = −A. So T o(n) = fA 2 gl(n; R) j A + A = 0g; 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 component of identity is the called the special orthogonal groups T SO(n) = fX 2 GL(n; R): X X = In; det X = 1g = O(n) \ SL(n; R): Its Lie algebra so(n) is the same as o(n). Example. (The symplectic group) The symplectic group is by definition T Sp(2n; R) = fX 2 GL(2n; R): X JX = Jg; 0 I where Let J = n . It is a Lie group of dimension n(2n + 1) with Lie algebra −In 0 T sp(2n; R) = fA 2 gl(2n; R) j JA + A J = 0g A1 A2 T T T = fA = j Ai 2 M(n; R);A1 = −A4 ;A2 = A2 ;A3 = A3 g: 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 bilinear form on Rn. Consider the set of all invertible n × n matrices that preserves β, n GLβ(n; R) = fX 2 GL(n; R) j β(Xu; Xv) = β(u; v)for all u; v 2 R g: In matrix form, there is a matrix B such that β(u; v) = uT Bv. Then T GLβ(n; R) = fX 2 GL(n; R) j X BX = Bg: 4 LECTURE 7: LINEAR LIE GROUPS Lemma 2.3. GLβ(n; R) is a linear Lie group with Lie algebra T glβ(n; R) = fA 2 gl(n; R) j A B + BA = 0g: 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 2 glβ(n; R) if and only if e 2 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) = fX 2 GL(n; R) j X I(p; n − p)X = I(p; n − p)g; where I(p; n − p) = diag(Ip; −In−p). Its Lie algebra is T o(p; n − p) = fX 2 gl(n; R) j A I(p; n − p) + I(p; n − p)A = 0:g: Example. (The unitary group) All the previous examples generalize to complex matrices. In particular, the unitary groups T U(n) = fX 2 GL(n; C): X X = Ing is a Lie subgroup of GL(n; C) with Lie algebra T u(n) = fA 2 gl(n; C) j A + A = 0g; the space of skew-Hermitian matrices.
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