Notation. For any Lie group G, we set G 0 to be the connected component of the identity.

Transcription

1 Notation. For any Lie group G, we set G 0 to be the connected component of the identity. Problem 1 Prove that GL(n, R) is homotopic to O(n, R). (Hint: Gram-Schmidt Orthogonalization.) Here is a sequence of n homotopies whose composition will be a deformation retract onto O(n, R). Let M(a 1,, a n ) be the matrix whose ith column is a i. Suppose the starting matrix is M(v 1,, v n ). The jth homotopy will fix all columns except the jth, and at time t [0, 1] it will take it to (v j t[(v 1, v j )v 1 (v j 1, v j )v j 1 ]) exp ( t log v j (v 1, v j )v 1 (v j 1, v j )v j 1 ) We never take a log of zero due to linear independence of the columns, and so we also always stay in GL(n, R) as the span of the columns never changes. Assuming that v 1,, v j 1 were orthonormal, this sends v j to a vector which is of unit length and orthogonal to the rest. So an induction on j shows that after performing these n homotopies in order, we end up with an element of O(n, R). Also, these homotopies never moved an element of O(n, R), so this was a deformation retract, and hence, a homotopy equivalence. This basically is following Gram-Schmidt process and noting it gives the required deformation retract. Problem 2 A flag in C n is a sequence of subspaces {0} = V 0 V 1...V n = C n where dimv i = i for each i = 0, 1,, n. Let F n (C) be the set of all flags in C n. Show that F n (C) = GL(n, C)/B(n, C), where B(n, C) denotes the group of upper triangular matrices (this subgroup is also called the Borel subgroup of GL(n, C), hence the notation B(n, C)). Find the dimension of F n (C) over C. Show, moreover, that F n = U(n)/T (n), where U(n) = {g GL(n, C) g t ḡ = I n } is the unitary group in n-variables and T(n) is the n n diagonal unitary matrices. This, in particular, shows that F n (C) is a compact complex manifold. First note that U(n) acts transitively on the flags: given flags V i and V i, Graham-Schmidt orthogonalization allows us to inductively choose orthonormal (under a fixed Hermitian inner product) bases {v 1,, v i } and {v 1,, v i } for V i and V i respectively, and the unique M U(n) sending v i v i satisfies M(V i) = V i. In particular GL(n, C) also acts transitively, so in both cases flags correspond bijectively to cosets with respect to the stabilizer of any fixed flag. If we fix a basis e 1,, e n and write matrices in this basis, then the stabilizer of the flag V i = span{e 1,, e i } in GL(n, C) is B(n, C). Hence the stabilizer in U(n) will be B(n, C) U(n). The upper triangular unitary matrices must be diagonal, because if the ith column were the first column with an off-diagonal piece, say v i = a 1 e a i e i, then for j < i we have v j being a multiple of e j, so (v j, v i ) = 0 would automatically force all the a j for j < i to be zero, contradicting our assumption (of not being diagonal already). In this way, F n (C) = GL(n, C)/ B(n, C) = U(n)/ T(n). As a complex manifold, the dimension must then be dim GL(n, C) dim B(n, C) = n 2 n(n + 1)/2 = n(n 1)/2 (as GL(n, C) is an open subset of C n2 and B(n, C) is an open subset of C n(n+1)/2 ). Also, we see that the flag space is compact because U(n) is compact and the quotient topology will also make the quotient be compact. Problem 3 Let SO(3, R) = {g GL(3, R) g t g = I 3 } and so(3, R) denote its Lie algebra. Let J x = , J y = , J z = be a basis for so(3, R). The standard action of SO(3, R) on R 3 then induces an action of so(3, R) on R 3 via vector fields. By abuse of notation, let J x, J y, and J z denote the corresponding vector fields on R 3. Let, sph := J 2 x + J 2 y + J 2 z where we are considering J x, J y, and J z as differential operators. Then, sph is a second degree differential operator on R 3. 1

2 (a) Denote the standard coordinates on R 3 by (x, y, z) and the corresponding coordinates on the tangent space by x, y and z. Write sph in terms of these coordinates. (b) Show that sph gives a well-defined differential operator on S 2 = { x R 3 x }. (i.e. You should show that if sph (f) S 2 depends only on f S 2.) (c) Show that for g SO(3, R) one has sph (gf) = g sph f. i.e. sph is rotationally invariant. Computations yield exp(tj x ) = cos t 0 sin t cos t sin t 0 0 cos t sin t, exp(tj y ) = 0 1 0, exp(tj z ) = sin t cos t 0 0 sin t cos t sin t 0 cos t so taking derivatives of exp(tj α )(x, y, z) for α {x, y, z} at t = 0 identifies J α with the following vector fields. J x : z y + y z J y : x z + z x J z : y x + x y, so if we use the notation cyc for the cyclic sum in the variables x, y, z we get sph = J 2 x + J 2 y + J 2 z = ( z y + y z ) 2 cyc = cyc (z 2 2 y + y 2 2 z 2yz y z y y z z ) = x 2 ( 2 y + 2 z) + y 2 ( 2 x + 2 z) + z 2 ( 2 x + 2 y) 2xy x y 2yz y z 2zx z x 2x x 2y y 2z z. To show that sph is a well-defined differential operator on S 2, we show this instead for J x. Then by symmetry the same holds for J y, J z, and hence for sph. To do so, we check that the vector field J x always lies in the tangent bundle of S 2 by checking that it is orthogonal to the radial direction. As J x = z y + y z and the radial direction is x x + y y + z z, we compute the dot product to be x 0 z y + y z = 0, and get the orthogonality. So the vector fields J x, J y, J z do lie in the tangent bundle, and hence sph only depends on f S 2. To show that sph is rotationaly invariant, it is enough to show that it commutes with the lie algebra action (because SO(3) compact connected). Let s check this for J x. We have J x (J 2 x + J 2 y + J 2 z ) = J 3 x + J y J x J y + J z J x J z J x = (J 2 x + J 2 y + J 2 z )J x. The J y, J z invariance follows by symmetry. Problem 4 Show that a connected, compact, complex Lie group is commutative. (Hint: You can use the fact that Ad : G gl(g) is complex analytic.) As mentioned in the hint, the map Ad : G gl(g) is complex analytic. Thinking of gl(g) as matrices, the image Ad(G) in each matrix entry is bounded by compactness, hence, as G connected, this is constant by Liouville s theorem. Since Ad(I) is the identity matrix, it follows that Ad is the constant map to the identity, so for any g G it follows that Ad(g) : g g is the identity map. Letting V be an open subset of the identity in G such that exp is a homeomorphism, and setting U = V g 1 V g, we see that if C g : G G is the conjugation action x gxg 1, then the restriction of C g to U takes U to gug 1 V, so C g = exp 1 Ad(g) exp on U is the identity. By connectivity of G we know that any neighborhood of the identity generates G, so (as C g is a homomorphism) C g must be the identity on all of G. This holds for any g G, so G is abelian. 2

3 Problem 5 Give an example of a connected Lie group G having non-conjugate maximal tori. How many conjugacy classes of tori are there in G of your example? Let G = SL(2, R). This is a connected Lie group, but it is not compact. Let T be any positive dimensional torus, and let t be its Lie algebra, which is a Lie subalgebra of sl(2, R). As T is a torus, exp : t T is a surjective homomorphism. Suppose X, Y t. Then exp(tx), exp(sy ) T and hence they commute. We know that d 2 /dtds of the function exp(sy ) exp(tx) exp( sy ) exp( tx) will be [X, Y ], and as we know exp(sy ) exp(tx) exp( sy ) exp( tx) = 1 we conclude [X, Y ] = 0. Using the standard basis E, F, H of sl(2, R) with relations [H, E] = 2E, [H, F ] = 2F, [E, F ] = H, a computation reveals that if X = x E E + x F F + x H H and Y = y E E + y F F + y H H, then [X, Y ] = 0 implies x E y F = x F y E and similarly 2 other relations, so [x E : x F : x H ] = [y E : y F : y H ] P(sl(2, R)). If X 0, this implies Y is a multiple of X. Therefore t is at most one dimensional. As we assumed T has positive dimension, we conclude that t is exactly one dimensional. In particular these are maximal (positive dimension implies one dimensional implies maximal). For two maximal tori T 1 and T 2 with corresponding Lie algebras t i, if there were some g G such that Ad(g)t 1 = t 2, then it would follow from exponentiation that gt 1 g 1 = T 2. So we first consider SL(2, R)-conjugacy classes in sl(2, R). This is the same as GL(2, R)-conjugacy classes, as we divide out by the determinant in any change of basis. Let X be any nonzero element in sl(2, R). Then the minimal polynomial has to be of degree exactly 2, as degree 1 would imply X is a multiple of the identity, which implies it is 0 (as the trace is 0). The roots of the characteristic polynomial (eigenvalues) are then either (i) both 0, (ii) a pair (λ, λ) for λ R, or (iii) a pair (iθ, iθ) for θ R. As the characteristic polynomial is always degree 2, it is in particular also the minimal polynomial, so via rational canonical form we conclude that X is conjugate to one of the following: {( ) ( ) ( )} λ 0 0 θ,, 0 λ θ 0 In particular, we only have to consider t {RE, RH, R(F E)}, so we only have to consider the following tori (computed by considering exp(xe), exp(xh), exp(x(f E))): {( ) } {( ) } {( ) } 1 x e x 0 cos(x) sin(x) : x R e x : x R : x R sin(x) cos(x) So by the previous computations these are all the tori up to conjugation and in fact are all different. Problem 6 Let G be a compact, connected Lie group and g G such that the closure of the subgroup generated by g is a torus. Show that C G (g) is connected. Give a counterexample to the statement when the closure of the group generated by g is not a torus.(hint: Think of finite subgroups of compact Lie groups.) We first note that for any g G, C G (g) is a closed subgroup as it is cut out by the closed condition hgh 1 g 1 = 1, and hence it is a closed Lie subgroup. (We also use this in the next problem.) Since g is a topological generator for a torus S (not necessarily maximal), an element of G centralizes g if and only if it centralizes all of S. Hence C G (g) = C G (S), and it suffices to check that C G (S) is connected. Suppose x C G (S). Then S C G (x), and connectedness of S produces S C G (x) 0. As G is compact and connected, the exponential map is surjective and there exists X g with exp(x) = x, so the path γ : [0, 1] G given by t exp(tx) satisfies γ(0) = 1, γ(1) = x, and γ(t) C G (x), so it follows that x C G (x) 0. Now we have S, {x} C G (x) 0. As C G (x) 0 is compact connected we can find a maximal torus T x in C G (x) 0 containing S, and we can find y C G (x) 0 such that yxy 1 T x. As y commutes with x, this means x T x. Therefore, S, {x} T x, and that gives x T x C G (S). Since we did this for arbitrary x C G (S), it follows that C G (S) = x CG (S)T x, and hence C G (S) is a union of tori. All tori are connected and contain the identity, so their union is connected. Therefore C G (S) = C G (g) is connected. 3

4 For a counterexample, when g does not generate a torus, pick G = SO(3) and consider the rotation about the x-axis by an angle of π: g = Then an easy computation shows that 0 0 C G (g) = 0 SO(3) 0 (the * entries can be anything as long as the result is in SO(3)). Therefore the upper left * is forced to be ±1, and the bottom 2 2 block has determinant ±1 (depending on what the upper left * is), and is in O(2). Therefore, we have an isomorphism of Lie groups O(2) C G (g) given by det(a) 0 0 A O(2) 0 0 A However O(2) is disconnected since it has two connected compoents, and hence the same is true for C G (g). In this case g did not generate a torus, but it had order 2. Problem 7 Let G be a compact, connected Lie group, T a maximal torus in G. Show that if t 1, t 2 T are conjugate in G, then they already are conjugate in N G (T ). (i.e. Show that ( g G s.t. gt 1 g 1 = t 2 ) ( n N G (T ) s.t. nt 1 n 1 = t 2 ).) As T is abelian and contains t 1, t 2 we have T C G (t 1 ), C G (t 2 ). Furthermore, T being connected implies T C G (t 2 ) 0. Pick g G such that gt 1 g 1 = t 2, so that T C G (t 1 ) = gt g 1 gc G (t 1 )g 1 = C G (t 2 ), and connectedness of gt g 1 then gives gt g 1 C G (t 2 ) 0. Hence T, gt g 1 are two tori in C G (t 2 ) 0. Moreover, they are maximal as they were maximal in G. Using conjugacy of maximal tori for C G (t 2 ) 0, find h C G (t 2 ) 0 such that gt g 1 = ht h 1, so that h 1 g N G (T ). Set n = h 1 g, and this element will conjugate t 1 to t 2 : nt 1 n 1 = h 1 gt 1 g 1 h = h 1 t 2 h = t 2. Problem 8 Let G = U(n) (the unitary group in n-variables), T G the diagonal Torus. Using the previous question show that if g G has distinct eigenvalues then there are exactly n! different elements t T that g can be conjugated into. What if g has an eigenvalue repeated exactly 2 times and the rest are distinct? What if it has one eigenvalue repeated m 1 -times, the second one m 2 -times,..., the rth one m r -times, such that m 1 + m m r = n? From the previous problem, we can find G-conjugacy classes by just looking at N G (T )-conjugacy classes. Moreover, since T is abelian, we may as well look at N G (T )/T representatives. We can see that N G (T ) consisted of monomial matrices, so representatives for N G (T )/T can be chosen to be permutation matrices. Therefore two elements in T are conjugate if and only if their diagonal entries can be permuted to each other. Now this is a combinatorial problem; it suffices to find the number of distinct ways of ordering m i copies of λ i. There are a total of n! ways if they were all distinct, but we have to divide by m i! for each i in order to avoid overcounting. So the answer is n! m 1! m r!. In particular, if exactly one eigenvalue was repeated twice and the rest were distinct, it would be n!/2 ways. Problem 9 Classify (up to conjugacy) all subgroups of H U(4) such that H = C G (g) for some g U(4). 4

5 Nothing is special about 4, so set 4 = n. We may freely conjugate g into the diagonal maximal torus, as this does not change the conjugacy class of C G (g). If h C G (g), then h preserves the eigenspaces of g, so it will be block diagonal with respect to the eigenspace decomposition of g. This is indeed the only condition, so if the eigenvalues of g were λ i with multiplicity m i (so that m i = n), then C G (g) will be block diagonal with block sizes m i. Each of the blocks also has to be unitary, so C G (g) = i U(m i). Now we come back to n = 4. For notational convenience, let T (k) abbreviate the k-fold product of U(1) (the notation comes from this being the maximal torus in U(k)). For n = 4, all centralizers are conjugate to one of the following: U(4), U(3) T (1), U(2) U(2), U(2) T (2), T (4). Their respective dimensions are 4 2 = 16, = 10, = 8, = 6, and 4, so none of these are conjugate to one another. 5