Lecture 32: One-Parameter Subgroups (Cont.)

Lecture 32: More About One-Parameter Subgroups One-Parameter Subgroups, Continued 32 32.1 Review Essentially, we want to restrict the image of φ to lie in G, in order to understand G better. Guiding Question How can we characterize which matrices A defne one-parameter subgroups satisfying this property? As a motivation, the real numbers with addition, (R, +), is essentially the simplest one-dimensional group, and here the notion of a one-parameter maps the real numbers into other groups, allowing us to study more complicated groups using the additive structure of the real numbers. Let’s see an example we covered last time. Example 32.1 (Unitary Matrices) Let G = Un ⊂ GLn (C). Last time, we showed that the condition that etA ∈ Un for all t ∈ R is equivalent to requiring that A∗ = −Aa ; that is, if and only if A is skew-Hermitian, the one-parameter subgroup described by A maps to only unitary matrices. a Note 32.2 that it is not required for A to be in GLn /invertible; it simply has to be some n×n matrix. Examples! Another example is given by upper triangular matrices. Example 32.2 (Upper Triangular Matrices) Let   1 G = 0  0 be the real upper triangular matrices. Then the set of matrices   0 0  0  ⋆ ⋆  1 ⋆ ≤ GL3 (R)  0 1 the corresponding A for which etA ∈ G for all t ∈ R make up  ⋆ ⋆  0 ⋆ ⊆ Mat3×3 (R).  0 0 The one-parameter group is a homomorphism φ(t) = etA , for some matrix A; in particular, A is φ′ (0). So fnding what A looks like, given G, is done by taking the derivative at t = 0. The image of φ lies in G, and since the 1s down the diagonal are not dependent on t, while the upper right entries could be nonzero,     1 ⋆ ⋆ 0 ⋆ ⋆ d 0 1 ⋆ = 0 0 ⋆ . A= dt t=0 0 0 1 0 0 0 It is also necessary to show the other direction that if A is of such a form, then the associated  one-parameter    0 ⋆ ⋆   0 ⋆ ⋆  group φ = etA will actually have its image lie in G. If A ∈ 0 0 ⋆ , then Ak = 0 0 ⋆ for     0 0 0 0 0 0 P 1 k k ≥ 1. Then, the exponential is a sum k! A , so       1 0 0 0 ⋆ ⋆ 1 ⋆ ⋆ eA = 0 1 0 + 0 0 ⋆ + · · · = 0 1 ⋆ , 0 0 1 0 0 0 0 0 1 and therefore etA ∈ G. So in fact etA ∈ G is equivalent to A being upper triangular. Now, let’s consider the orthogonal matrices. Lecture 32: More About One-Parameter Subgroups Example 32.3 Consider the orthogonal matrices On ⊂ GLn (R). For a one-parameter subgroup of On , the matrix A must satisfy (etA )T = (etA )−1 for all t. This is equivalent to T etA = e−tA T d for all t.a Taking the derivative dt , we get AT etA = −Ae−tA , and evaluating at t = 0 gives that the possible A are the ones satisfying {AT = −A}.b T Conversely, if AT = −A, then etA = e−tA , which is equivalent to (etA )T = (etA )−1 , and so etA ∈ On when AT = −A. So these are the correct matrices A. T out the exponential as a sum gives (etA )T = et(A ) , and it is clear that the inverse of etA is e−tA . is also possible to get this simply by the fact that On is Un ∩ GLn (R), and so A must satisfy the same property as for the unitary matrices, that A∗ = −A, and for real matrices this condition is the same as AT = −A. a Writing b It 32.3 The Special Linear Group SLn (C) In order to study SLn (C), which is in some sense the frst subgroup of matrices that we ever studied, an important identity about the matrix exponential must be established. Guiding Question What about SLn (C)? Lemma 32.4 For any A ∈ Matn×n (C), det eA = etrace(A) . This property is clearly true for  diagonal  matrices, and from there we hope that this is true for other matrices λ1 0 as well. For example, for A = , 0 λ2  λ  e 1 0 eA = , 0 eλ2 and det(eA ) = eλ1 eλ2 = eλ1 +λ2 = etrace(A) . Trying to steamroll through the proof of the lemma becomes very difcult, but fortunately the exponential, determinant, and trace all behave well with respect to conjugation. Proof. We have det(P AP −1 ) = det(A), trace(P AP −1 ) = trace(A), eP AP −1 = P eA P −1 . Thus, if the lemma is true for a matrix conjugate to A, it is true for A, and so only one representative from each conjugacy class needs to be considered. Then, without loss of generality, we can assume  that A is inJordan λ1 · · · ⋆  ..  to be .. canonical form, and the proof follows identically to the diagonal case. We take A =  ... . .   k λ1  upper triangular and in Jordan form. Then Ak =  ... 0 P det(eA ) = e ··· .. . ··· λi  λ ⋆ e1  .. A ...  , and so e =  . λkn 0  = etrace(A) . 0 ··· .. . ··· · · · λn  ⋆ . ..  , so enλ Lecture 32: More About One-Parameter Subgroups Now we can take a look at SLn (C). Example 32.5 Consider A ∈ Matn×n (C) such that etA ∈ SLn (C) for all t ∈ R. Then etA ∈ SLn (C), and so det(etA ) = 1 for all t. Therefore, by Lemma 32.4, etrace (tA) = 1, which is equivalent to stating that trace(A) ∈ 2πiZ for all t ∈ R, which is possible only if the trace of A is 0. So the one-parameter groups in SLn (C) are etA where A ∈ Matn×n (C) is traceless.a a The trace is 0. These conditions on A can obviously be combined for diferent groups G. The one-parameter groups in SUn correspond with the matrices A such that A∗ = −A and trace(A) = 0. In particular, the one-parameter groups in SU2 consist of etA where the 2×2 matrix A is skew-symmetric and has zero trace. Example 32.6 (SU2 ) α β For A = , these conditions end up givinga that γ δ  A= ix1 −x2 + x3 x2 + ix3 −ix1   = x1 i 0     0 0 1 0 + x2 + x3 −i −1 0 i  i . 0 These are the matrices that showed up when we studied the equator! We have A = x1 I + x2 J + x3 K, and so A = c⃗v for ⃗v ∈ E ⊂ SU2 .b Then etA = etc⃗v = cos(tc)I + sin(tc)⃗v c ∈ Long⃗v d . Essentially, there is a one-parameter subgroup of SLn for each ⃗v for some vector ⃗v on the equator, where the coefcient c in A = c⃗v determines the speed at which the subgroup is swept out. The image of etA simply corresponds to the longitude given by ⃗v . a Actually going through the process is slightly tedious and not that informative; try it for yourself if you want to! q 2 there is no requirement that x1 + x22 + x23 = 1, A does not actually have to be on the equator, but it is just some multiple c of a vector ⃗v on the equator. When c = 1, A is actually on the equator, by the characterization we gave in a previous lecture. c This is given by simply writing out the expansion of the matrix exponential as a sum and collecting the terms into a Taylor series for cosine and a Taylor series for sine; since ⃗v is on the equator, ⃗v 2 = −1, and so the result is essentially the same as the result that eiθ = cos θ + i sin θ. d This is the characterization given in a previous lecture. b Because 32.4 Tangent Vectors So far, several examples have been shown, but we would like to see what else we can say more generally about one-parameter subgroups. To do so, we will introduce some new tools. Guiding Question What else can we say about the setsa of matrices A defning one-parameter groups? a In fact, they will be vector spaces Since they are given by derivatives at t = 0, corresponding to the identity, the matrices A are "tangent vectors" to G ≤ GLn (R) at the identity I. Intuitively, a tangent vector at some point is a vector lying in the tangent plane. Lecture 32: More About One-Parameter Subgroups There are three approaches to rigorously defning tangent vectors. 1. The approach taken so far is that a tangent vector will be given by a matrix A ∈ Matn×n (R) such that the corresponding one-parameter group etA ∈ G for all t ∈ R.102 2. The second approach is more general, and builds on the idea that a one-parameter group is the trajectory of a particle moving in G in a specifc way defned by φ. Instead of taking a path that happens to be a homomorphism, take any diferentiable path103 from some interval f : (−ε, ε) −→ GLn (R) such that f (0) = I and f (t) ∈ (G) for all t. Then a tangent vector is simply the velocity at time t = 0, f ′ (0) ∈ Matn×n (R).104 It is not obvious, but it turns out that the frst defnition is equivalent to the second defnition, giving the same subsets of matrices A as tangent vectors. For the frst defnition, the advantage is that each tangent vector corresponds to only one path through the identity, so there is a bijection. The second approach gives lots of paths through the identity that have a given velocity vector, but it turns out that it is easier to use it to show that the set of tangent vectors actually forms a vector space, called the tangent space.105 3. Suppose G is defned by polynomial constraints on the matrix entries.106 For example, the constraints could be that it is upper triangular; then aij = 0 for all i > j is a bunch of polynomial conditions on the matrix entries. Orthogonality can also be phrased as a polynomial condition. Then, when working with polynomials, the derivative can be mimicked without actually needing to know analysis. We work with an object R[ε] := R + Rε where ε2 = 0.107 This allows us to defne a derivative without actually taking any limits. For example, for f (x) = x2 + 2x, evaluating f on x + ε ∈ R[ε] will give f (x + ε) = (x + ε)2 + 2(x + ε) = x2 + 2xε + ε2 + 2x + 2ε = x2 + 2xε + 2x + 2ε = (x2 + 2x) + (2x + 2)ε. (x) So with this funky multiplication, any terms of order more than two in ε disappear, and we get f (x+ε)−f = ε ′ f (x), even though we haven’t actually defned the derivative from an analysis perspective. The upshot is that to fnd A ∈ Matn×n (R), we simply look at A with the property that In + εA, the identity matrix perturbed by A, satisfes the same system of equations defning G, but in the sense of the funky multiplication of R[ε], where ε2 = 0. We’ll explain this more on Friday. 102 Every matrix A will defne some path in GLn ; if the entire path lives in G, then we can think of A as a tangent vector. matrix-valued function 104 The idea is that all the paths through the identity will give velocity vectors (with lots of redundancy), which we will consider as tangent vectors. 105 There is actually more structure on it, which we will talk about on Friday. 106 This approach is not necessary for this class, but it is fun, so we will do it. 107 Here ε is not some number in R, so it is not actually true that ε must be 0; it is simply a formal construct where we impose the √ condition that ε2 = 0. It is similar to the defnition of i = −1; there is no such real number, so we simply defne some number satisfying this property. In the same way, we simply defne R[ε] to be R + Rε for some object ε satisfying ε2 = 0. 103 Any