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1 Matrix Lie Groups

1.1 De?nition of a Matrix Lie Group
We begin with a very important class of groups, the general linear groups. The groups we will study in this book will all be subgroups (of a certain sort) of one of the general linear groups. This chapter makes use of various standard results from linear algebra that are summarized in Appendix B. This chapter also assumes basic facts and de?nitions from the theory of abstract groups; the necessary information is provided in Appendix A. De?nition 1.1. The general linear group over the real numbers, denoted GL(n; R), is the group of all n × n invertible matrices with real entries. The general linear group over the complex numbers, denoted GL(n; C), is the group of all n × n invertible matrices with complex entries. The general linear groups are indeed groups under the operation of matrix multiplication: The product of two invertible matrices is invertible, the identity matrix is an identity for the group, an invertible matrix has (by de?nition) an inverse, and matrix multiplication is associative. De?nition 1.2. Let Mn (C) denote the space of all n×n matrices with complex entries. De?nition 1.3. Let Am be a sequence of complex matrices in Mn (C). We say that Am converges to a matrix A if each entry of Am converges (as m → ∞) to the corresponding entry of A (i.e., if (Am )kl converges to Akl for all 1 ≤ k, l ≤ n). De?nition 1.4. A matrix Lie group is any subgroup G of GL(n; C) with the following property: If Am is any sequence of matrices in G, and Am converges to some matrix A then either A ∈ G, or A is not invertible. The condition on G amounts to saying that G is a closed subset of GL(n; C). (This does not necessarily mean that G is closed in Mn (C).) Thus, De?nition

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1 Matrix Lie Groups

1.4 is equivalent to saying that a matrix Lie group is a closed subgroup of GL(n; C). The condition that G be a closed subgroup, as opposed to merely a subgroup, should be regarded as a technicality, in that most of the interesting subgroups of GL(n; C) have this property. (Most of the matrix Lie groups G we will consider have the stronger property that if Am is any sequence of matrices in G, and Am converges to some matrix A, then A ∈ G (i.e., that G is closed in Mn (C)).) 1.1.1 Counterexamples An example of a subgroup of GL(n; C) which is not closed (and hence is not a matrix Lie group) is the set of all n × n invertible matrices all of whose entries are real and rational. This is in fact a subgroup of GL(n; C), but not a closed subgroup. That is, one can (easily) have a sequence of invertible matrices with rational entries converging to an invertible matrix with some irrational entries. (In fact, every real invertible matrix is the limit of some sequence of invertible matrices with rational entries.) Another example of a group of matrices which is not a matrix Lie group is the following subgroup of GL(2; C). Let a be an irrational real number and let eit 0 G= t∈R . 0 eita Clearly, G is a subgroup of GL(2, C). Because a is irrational, the matrix ?I is not in G, since to make eit equal to ?1, we must take t to be an odd integer multiple of π , in which case ta cannot be an odd integer multiple of π . On the other hand (Exercise 1), by taking t = (2n + 1)π for a suitably chosen integer n, we can make ta arbitrarily close to an odd integer multiple of π . Hence, we can ?nd a sequence of matrices in G which converges to ?I , and so G is not a matrix Lie group. See Exercise 1 and Exercise 18 for more information.

1.2 Examples of Matrix Lie Groups
Mastering the subject of Lie groups involves not only learning the general theory but also familiarizing oneself with examples. In this section, we introduce some of the most important examples of (matrix) Lie groups. 1.2.1 The general linear groups GL(n; R) and GL(n; C) The general linear groups (over R or C) are themselves matrix Lie groups. Of course, GL(n; C) is a subgroup of itself. Furthermore, if Am is a sequence of matrices in GL(n; C) and Am converges to A, then by the de?nition of GL(n; C), either A is in GL(n; C), or A is not invertible.

1.2 Examples of Matrix Lie Groups

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Moreover, GL(n; R) is a subgroup of GL(n; C), and if Am ∈ GL(n; R) and Am converges to A, then the entries of A are real. Thus, either A is not invertible or A ∈ GL(n; R). 1.2.2 The special linear groups SL(n; R) and SL(n; C) The special linear group (over R or C) is the group of n × n invertible matrices (with real or complex entries) having determinant one. Both of these are subgroups of GL(n; C). Furthermore, if An is a sequence of matrices with determinant one and An converges to A, then A also has determinant one, because the determinant is a continuous function. Thus, SL(n; R) and SL (n; C) are matrix Lie groups. 1.2.3 The orthogonal and special orthogonal groups, O(n) and SO(n) An n × n real matrix A is said to be orthogonal if the column vectors that make up A are orthonormal, that is, if
n

Alj Alk = δjk ,
l=1

1 ≤ j, k ≤ n.

(Here δjk is the Kronecker delta, equal to 1 if j = k and equal to zero if j = k.) Equivalently, A is orthogonal if it preserves the inner product, namely if x, y = Ax, Ay for all vectors x, y in Rn . ( Angled brackets denote the usual inner product on Rn , x, y = k xk yk .) Still another equivalent de?nition is that A is orthogonal if Atr A = I , i.e., if Atr = A?1 . (Here, Atr is the transpose of A, (Atr )kl = Alk .) See Exercise 2. Since det Atr = det A, we see that if A is orthogonal, then det(Atr A) = 2 (det A) = det I = 1. Hence, det A = ±1, for all orthogonal matrices A. This formula tells us in particular that every orthogonal matrix must be invertible. However, if A is an orthogonal matrix, then A?1 x, A?1 y = A(A?1 x), A(A?1 y ) = x, y . Thus, the inverse of an orthogonal matrix is orthogonal. Furthermore, the product of two orthogonal matrices is orthogonal, since if A and B both preserve inner products, then so does AB . Thus, the set of orthogonal matrices forms a group. The set of all n × n real orthogonal matrices is the orthogonal group O(n), and it is a subgroup of GL(n; C). The limit of a sequence of orthogonal matrices is orthogonal, because the relation Atr A = I is preserved under taking limits. Thus, O(n) is a matrix Lie group. The set of n × n orthogonal matrices with determinant one is the special orthogonal group SO(n). Clearly, this is a subgroup of O(n), and hence of

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1 Matrix Lie Groups

GL(n; C). Moreover, both orthogonality and the property of having determinant one are preserved under limits, and so SO(n) is a matrix Lie group. Since elements of O(n) already have determinant ±1, SO(n) is “half” of O(n). Geometrically, elements of O(n) are either rotations or combinations of rotations and re?ections. The elements of SO(n) are just the rotations. See also Exercise 6. 1.2.4 The unitary and special unitary groups, U(n) and SU(n) An n × n complex matrix A is said to be unitary if the column vectors of A are orthonormal, that is, if
n

Alj Alk = δjk .
l=1

Equivalently, A is unitary if it preserves the inner product, namely if x, y = Ax, Ay for all vectors x, y in Cn . (Angled brackets here denote the inner product on Cn , x, y = k xk yk . We will adopt the convention of putting the complex conjugate on the left.) Still another equivalent de?nition is that A is unitary if A? A = I , i.e., if A? = A?1 . (Here, A? is the adjoint of A, (A? )jk = Akj .) See Exercise 3. Since det A? = det A, we see that if A is unitary, then det(A? A) = 2 |det A| = det I = 1. Hence, |det A| = 1, for all unitary matrices A. This, in particular, shows that every unitary matrix is invertible. The same argument as for the orthogonal group shows that the set of unitary matrices forms a group. The set of all n × n unitary matrices is the unitary group U(n), and it is a subgroup of GL(n; C). The limit of unitary matrices is unitary, so U(n) is a matrix Lie group. The set of unitary matrices with determinant one is the special unitary group SU(n). It is easy to check that SU(n) is a matrix Lie group. Note that a unitary matrix can have determinant eiθ for any θ, and so SU(n) is a smaller subset of U(n) than SO(n) is of O(n). (Speci?cally, SO(n) has the same dimension as O(n), whereas SU(n) has dimension one less than that of U(n).) See also Exercise 8. 1.2.5 The complex orthogonal groups, O(n; C) and SO(n; C) Consider the bilinear form (·, ·) on Cn de?ned by (x, y ) = k xk yk . This form is not an inner product (Section B.6) because, for example, it is symmetric rather than conjugate-symmetric. The set of all n × n complex matrices A which preserve this form (i.e., such that (Ax, Ay ) = (x, y ) for all x, y ∈ Cn ) is the complex orthogonal group O(n; C), and it is a subgroup of GL(n; C). Repeating the arguments for the case of SO(n) and O(n) (but now permitting complex entries), we ?nd that an n × n complex matrix A is in O(n; C) if and

1.2 Examples of Matrix Lie Groups

7

only if Atr A = I , that O(n; C) is a matrix Lie group, and that det A = ±1 for all A in O(n; C). Note that O(n; C) is not the same as the unitary group U(n). The group SO(n; C) is de?ned to be the set of all A in O(n; C) with det A = 1 and it is also a matrix Lie group. 1.2.6 The generalized orthogonal and Lorentz groups Let n and k be positive integers, and consider Rn+k . De?ne a symmetric bilinear form [·, ·]n,k on Rn+k by the formula [x, y ]n,k = x1 y1 + · · · + xn yn ? xn+1 yn+1 ? · · · ? xn+k yn+k (1.1)

The set of (n + k ) × (n + k ) real matrices A which preserve this form (i.e., such that [Ax, Ay ]n,k = [x, y ]n,k for all x, y ∈ Rn+k ) is the generalized orthogonal group O(n; k ). It is a subgroup of GL(n + k ; R) and a matrix Lie group (Exercise 4). If A is an (n + k ) × (n + k ) real matrix, let A(i) denote the ith column vector of A, that is, ? ? A1,i ? . ? A(i) = ? . . ?. An+k,i Then, A is in O(n; k ) if and only if the following conditions are satis?ed: A(l) , A(j ) A(l) , A(l) A(l) , A(l)
n,k n,k n,k

= 0 = 1 = ?1

l = j, 1 ≤ l ≤ n, n + 1 ≤ l ≤ n + k.

(1.2)

Let g denote the (n + k ) × (n + k ) diagonal matrix with ones in the ?rst n diagonal entries and minus ones in the last k diagonal entries. Then, A is in O(n; k ) if and only if Atr gA = g (Exercise 4). Taking the determinant of this equation gives (det A)2 det g = det g , or (det A)2 = 1. Thus, for any A in O(n; k ), det A = ±1. Of particular interest in physics is the Lorentz group O(3; 1). See also Exercise 7. 1.2.7 The symplectic groups Sp(n; R), Sp(n; C), and Sp(n) The special and general linear groups, the orthogonal and unitary groups, and the symplectic groups (which will be de?ned momentarily) make up the classical groups. Of the classical groups, the symplectic groups have the most confusing de?nition, partly because there are three sets of them (Sp(n; R), Sp(n; C), and Sp(n)) and partly because they involve skew-symmetric bilinear forms rather than the more familiar symmetric bilinear forms. To further

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confuse matters, the notation for referring to these groups is not consistent from author to author. Consider the skew-symmetric bilinear form B on R2n de?ned as follows:
n

B [x, y ] =
k=1

xk yn+k ? xn+k yk .

(1.3)

The set of all 2n × 2n matrices A which preserve B (i.e., such that B [Ax, Ay ] = B [x, y ] for all x, y ∈ R2n ) is the real symplectic group Sp(n; R), and it is a subgroup of GL(2n; R). It is not di?cult to check that this is a matrix Lie group (Exercise 5). This group arises naturally in the study of classical mechanics. If J is the 2n × 2n matrix J= 0 I ?I 0 ,

then B [x, y ] = x, Jy , and it is possible to check that a 2n×2n real matrix A is in Sp(n; R) if and only if Atr JA = J . (See Exercise 5.) Taking the determinant 2 2 of this identity gives (det A) det J = det J , or (det A) = 1. This shows that det A = ±1, for all A ∈ Sp(n; R). In fact, det A = 1 for all A ∈ Sp(n; R), although this is not obvious. One can de?ne a bilinear form on C2n by the same formula (1.3). (This form involves no complex conjugates.) The set of 2n × 2n complex matrices which preserve this form is the complex symplectic group Sp(n; C). A 2n × 2n complex matrix A is in Sp(n; C) if and only if Atr JA = J . (Note: This condition involves Atr , not A? .) This relation shows that det A = ±1, for all A ∈ Sp(n; C). In fact, det A = 1, for all A ∈ Sp(n; C). Finally, we have the compact symplectic group Sp(n) de?ned as Sp(n) = Sp (n; C) ∩ U(2n). See also Exercise 9. For more information and a proof that det A = 1 for all A ∈ Sp(n; C), see Section 9.4 of Miller (1972). What we call Sp (n; C) Miller calls Sp(n), and what we call Sp(n), Miller calls USp(n). 1.2.8 The Heisenberg group H The set of all 3 × 3 real matrices A of the form ? ? 1ab A = ?0 1 c?, 001

(1.4)

where a, b, and c are arbitrary real numbers, is the Heisenberg group. It is easy to check that the product of two matrices of the form (1.4) is again of that form, and, clearly, the identity matrix is of the form (1.4). Furthermore, direct computation shows that if A is as in (1.4), then

1.2 Examples of Matrix Lie Groups

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A?1

? 1 ?a ac ? b = ? 0 1 ?c ? . 0 0 1

?

Thus, H is a subgroup of GL(3; R). Clearly, the limit of matrices of the form (1.4) is again of that form, and so H is a matrix Lie group. The reason for the name “Heisenberg group” is that the Lie algebra of H gives a realization of the Heisenberg commutation relations of quantum mechanics. (See especially Chapter 4, Exercise 8.) See also Exercise 10. 1.2.9 The groups R? , C? , S 1 , R, and Rn Several important groups which are not naturally groups of matrices can (and will in these notes) be thought of as such. The group R? of non-zero real numbers under multiplication is isomorphic to GL(1; R). Thus, we will regard R? as a matrix Lie group. Similarly, the group C? of nonzero complex numbers under multiplication is isomorphic to GL(1; C), and the group S 1 of complex numbers with absolute value one is isomorphic to U(1). The group R under addition is isomorphic to GL(1; R)+ (1 × 1 real matrices with positive determinant) via the map x → [ex ]. The group Rn (with vector addition) is isomorphic to the group of diagonal real matrices with positive diagonal entries, via the map ? ? x 0 e 1 ? ? .. (x1 , . . . , xn ) → ? ?. . 0 exn

1.2.10 The Euclidean and Poincar? e groups E(n) and P(n; 1) The Euclidean group E(n) is, by de?nition, the group of all one-to-one, onto, distance-preserving maps of Rn to itself, that is, maps f : Rn → Rn such that d(f (x), f (y )) = d(x, y ) for all x, y ∈ Rn . Here, d is the usual distance on Rn : d(x, y ) = |x ? y |. Note that we do not assume anything about the structure of f besides the above properties. In particular, f need not be linear. The orthogonal group O(n) is a subgroup of E(n) and is the group of all linear distance-preserving maps of Rn to itself. For x ∈ Rn , de?ne the translation by x, denoted Tx , by Tx (y ) = x + y. The set of translations is also a subgroup of E(n). Proposition 1.5. Every element T of E(n) can be written uniquely as an orthogonal linear transformation followed by a translation, that is, in the form

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T = Tx R with x ∈ Rn and R ∈ O(n). We will not prove this. The key step is to prove that every one-to-one, onto, distance-preserving map of Rn to itself which ?xes the origin must be linear. We will write an element T = Tx R of E(n) as a pair {x, R}. Note that for y ∈ Rn , {x, R} y = Ry + x and that {x1 , R1 }{x2 , R2 }y = R1 (R2 y + x2 ) + x1 = R1 R2 y + (x1 + R1 x2 ). Thus, the product operation for E(n) is the following: {x1 , R1 }{x2 , R2 } = {x1 + R1 x2 , R1 R2 }. The inverse of an element of E(n) is given by {x, R}?1 = {?R?1 x, R?1 }. As already noted, E(n) is not a subgroup of GL(n; R), since translations are not linear maps. However, E(n) is isomorphic to a subgroup of GL(n + 1; R), via the map which associates to {x, R} ∈ E(n) the following matrix: ? ? x1 ? ? . . ? R . ? (1.6) ? ?. ? xn ? 0 ··· 0 1 This map is clearly one-to-one, and direct computation shows that multiplication of elements of the form (1.6) follows the multiplication rule in (1.5), so that this map is a homomorphism. Thus, E(n) is isomorphic to the group of all matrices of the form (1.6) (with R ∈ O(n)). The limit of things of the form (1.6) is again of that form, and so we have expressed the Euclidean group E(n) as a matrix Lie group. We similarly de?ne the Poincar? e group P(n; 1) to be the group of all transformations of Rn+1 of the form T = Tx A with x ∈ Rn+1 and A ∈ O(n; 1). This is the group of a?ne transformations of Rn+1 which preserve the Lorentz “distance” dL (x, y ) = (x1 ? y1 )2 + · · · + (xn ? yn )2 ? (xn+1 ? yn+1 )2 . (An a?ne transformation is one of the form x → Ax + b, where A is a linear transformation and b is constant.) The group product is the obvious analog of the product (1.5) for the Euclidean group. (1.5)

1.3 Compactness

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The Poincar? e group P(n; 1) is isomorphic to the group of (n + 2) × (n + 2) matrices of the form ? ? x1 ? ? . . ? A . ? (1.7) ? ? ? xn+1 ? 0 ··· 0 1 with A ∈ O(n; 1). The set of matrices of the form (1.7) is a matrix Lie group.

1.3 Compactness
De?nition 1.6. A matrix Lie group G is said to be compact if the following two conditions are satis?ed: 1. If Am is any sequence of matrices in G, and Am converges to a matrix A, then A is in G. 2. There exists a constant C such that for all A ∈ G, |Aij | ≤ C for all 1 ≤ i, j ≤ n. This is not the usual topological de?nition of compactness. However, the 2 set Mn (C) of all n × n complex matrices can be thought of as Cn . The above 2 de?nition says that G is compact if it is a closed, bounded subset of Cn . It is n2 a standard theorem from elementary analysis that a subset of C is compact if and only if it is closed and bounded. All of our examples of matrix Lie groups except GL(n; R) and GL(n; C) have property (1). Thus, it is the boundedness condition (2) that is most important. 1.3.1 Examples of compact groups The groups O(n) and SO(n) are compact. Property (1) is satis?ed because the limit of orthogonal matrices is orthogonal and the limit of matrices with determinant one has determinant one. Property (2) is satis?ed because if A is orthogonal, then the column vectors of A have norm one, and hence |Akl | ≤ 1, for all 1 ≤ k, l ≤ n. A similar argument shows that U(n), SU(n), and Sp(n) are compact. (This includes the unit circle, S 1 ? = U(1).) 1.3.2 Examples of noncompact groups All of the other examples given of matrix Lie groups are noncompact. The groups GL(n; R) and GL(n; C) violate property (1), since a limit of invertible matrices may be noninvertible. The groups SL(n; R) and SL (n; C) violate (2), (except in the trivial case n = 1) since

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? Am ? ? ? =? ? ?

m
1 m

? 1 .. . 1 ? ? ? ? ? ?

has determinant one, no matter how large m is. The following groups also violate (2), and hence are noncompact: O(n; C) and SO(n; C); O(n; k ) and SO(n; k ) (n ≥ 1, k ≥ 1); the Heisenberg group H ; Sp(n; R) and Sp(n; C); E(n) and P(n; 1); R and Rn ; R? and C? . It is left to the reader to provide examples to show that this is the case.

1.4 Connectedness
De?nition 1.7. A matrix Lie group G is said to be connected if given any two matrices A and B in G, there exists a continuous path A(t), a ≤ t ≤ b, lying in G with A(a) = A and A(b) = B . This property is what is called path-connected in topology, which is not (in general) the same as connected. However, it is a fact (not particularly obvious at the moment) that a matrix Lie group is connected if and only if it is path-connected. So, in a slight abuse of terminology, we shall continue to refer to the above property as connectedness. (See Section 1.8.) A matrix Lie group G which is not connected can be decomposed (uniquely) as a union of several pieces, called components, such that two elements of the same component can be joined by a continuous path, but two elements of di?erent components cannot. Proposition 1.8. If G is a matrix Lie group, then the component of G containing the identity is a subgroup of G. Proof. Saying that A and B are both in the component containing the identity means that there exist continuous paths A(t) and B (t) with A(0) = B (0) = I , A(1) = A, and B (1) = B . Then, A(t)B (t) is a continuous path starting at I and ending at AB . Thus, the product of two elements of the identity component is again in the identity component. Furthermore, A(t)?1 is a continuous path starting at I and ending at A?1 , and so the inverse of any element of the identity component is again in the identity component. Thus, the identity component is a subgroup. Note that because matrix multiplication and matrix inversion are continuous on GL(n; C), it follows that if A(t) and B (t) are continuous, then so are A(t)B (t) and A(t)?1 . The continuity of the matrix product is obvious. The continuity of the inverse follows from the formula for the inverse in terms of cofactors; this formula is continuous as long as we remain in the set of invertible matrices where the determinant in the denominator is nonzero.

1.4 Connectedness

13

Proposition 1.9. The group GL(n; C) is connected for all n ≥ 1. Proof. Consider ?rst the case n = 1. A 1 × 1 invertible complex matrix A is of the form A = [λ] with λ in C? , the set of nonzero complex numbers. Given any two nonzero complex numbers, we can easily ?nd a continuous path which connects them and does not pass through zero. For the case n ≥ 2, we will show that any element of GL(n; C) can be connected to the identity by a continuous path lying in GL(n; C). Then, any two elements A and B of GL(n; C) can be connected by a path going from A to the identity and then from the identity to B. We make use of the result that every matrix is similar to an upper triangular matrix (Theorem B.7). That is, given any n × n complex matrix A, there exists an invertible n × n complex matrix C such that A = CBC ?1 where B is upper triangular: ? ? B=? λ1 .. 0 . λn ? ? ? ?.

If we now assume that A is invertible, then all the λi ’s must be nonzero, since det A = det B = λ1 · · · λn . Let B (t) be obtained by multiplying the part of B above the diagonal by (1 ? t), for 0 ≤ t ≤ 1, and let A(t) = CB (t)C ?1 . Then, A(t) is a continuous path which starts at A and ends at CDC ?1 , where D is the diagonal matrix ? ? λ1 0 ? ? .. D=? ?. . 0 λn This path lies in GL(n; C) since det A(t) = λ1 · · · λn = det A for all t. Now, as in the case n = 1, we can de?ne λi (t), which connects each λi to 1 in C? as t goes from 1 to 2. Then, we can de?ne A(t) on the interval 1 ≤ t ≤ 2 by ? ? λ1 (t) 0 ? ? ?1 .. A(t) = C ? ?C . . 0 λn (t) This is a continuous path which starts at CDC ?1 when t = 1 and ends at I (= CIC ?1 ) when t = 2. Since the λk (t)’s are always nonzero, A(t) lies in GL(n; C). We see, then, that every matrix A in GL(n; C) can be connected to the identity by a continuous path lying in GL(n; C). An alternative proof of this result is given in Exercise 12.

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Proposition 1.10. The group SL(n; C) is connected for all n ≥ 1. Proof. The proof is almost the same as for GL(n; C), except that we must be careful to preserve the condition det A = 1. Let A be an arbitrary element of SL(n; C). The case n = 1 is trivial, so we assume n ≥ 2. We can de?ne A(t) as before for 0 ≤ t ≤ 1, with A(0) = A and A(1) = CDC ?1 , since det A(t) = det A = 1. Now, de?ne λk (t) as before for 1 ≤ k ≤ n ? 1 and de?ne ?1 λn (t) to be [λ1 (t) · · · λn?1 (t)] . (Note that since λ1 · · · λn = 1, λn (1) = λn .) This allows us to connect A to the identity while staying within SL(n; C). Proposition 1.11. The groups U(n) and SU(n) are connected, for all n ≥ 1. Proof. By a standard result of linear algebra (Theorem B.3), every unitary matrix has an orthonormal basis of eigenvectors, with eigenvalues of the form eiθ . It follows that every unitary matrix U can be written as ? iθ ? e 1 0 ? ? ?1 .. U = U1 ? (1.8) ? U1 . 0 eiθn with U1 unitary and θi ∈ R. Conversely, as is easily checked, every matrix of the form (1.8) is unitary. Now, de?ne ? i(1?t)θ ? 1 e 0 ? ? ?1 .. U (t) = U1 ? ? U1 . . i(1?t)θn 0 e As t ranges from 0 to 1, this de?nes a continuous path in U(n) joining U to I . Thus, any two elements U and V of U(n) can be connected to each other by a continuous path that runs from U to I and then from I to V. A slight modi?cation of this argument, as in the proof of Proposition 1.10, shows that SU(n) is connected. Proposition 1.12. The group GL(n; R) is not connected, but has two components. These are GL(n; R)+ , the set of n × n real matrices with positive determinant, and GL(n; R)? , the set of n × n real matrices with negative determinant. Proof. GL(n; R) cannot be connected, for if det A > 0 and det B < 0, then any continuous path connecting A to B would have to include a matrix with determinant zero and hence pass outside of GL(n; R). The proof that GL(n; R)+ is connected is sketched in Exercise 15. Once GL(n; R)+ is known to be connected, it is not di?cult to see that GL(n; R)? is also connected. Let C be any matrix with negative determinant and take A and B in GL(n; R)? . Then, C ?1 A and C ?1 B are in GL(n; R)+ and can be joined by a continuous path D(t) in GL(n; R)+ . However, then, CD(t) is a continuous path joining A and B in GL(n; R)? .

1.5 Simple Connectedness

15

The following table lists some matrix Lie groups, indicates whether or not the group is connected, and gives the number of components: Group Connected? Components GL(n; C) yes 1 SL(n; C) yes 1 GL(n; R) no 2 SL(n; R) yes 1 O(n) no 2 SO(n) yes 1 U(n) yes 1 SU(n) yes 1 O(n; 1) no 4 SO(n; 1) no 2 Heisenberg yes 1 E(n) no 2 P(n; 1) no 4 Proofs of some of these results are given in Exercises 7, 13, 14, and 15.

1.5 Simple Connectedness
De?nition 1.13. A matrix Lie group G is said to be simply connected if it is connected and, in addition, every loop in G can be shrunk continuously to a point in G. More precisely, assume that G is connected. Then, G is simply connected if given any continuous path A(t), 0 ≤ t ≤ 1, lying in G with A(0) = A(1), there exists a continuous function A(s, t), 0 ≤ s, t ≤ 1, taking values in G and having the following properties: (1) A(s, 0) = A(s, 1) for all s, (2) A(0, t) = A(t), and (3) A(1, t) = A(1, 0) for all t. One should think of A(t) as a loop and A(s, t) as a family of loops, parameterized by the variable s which shrinks A(t) to a point. Condition 1 says that for each value of the parameter s, we have a loop; condition 2 says that when s = 0 the loop is the speci?ed loop A(t); and condition 3 says that when s = 1 our loop is a point. Proposition 1.14. The group SU(2) is simply connected. Proof. Exercise 8 shows that SU(2) may be thought of (topologically) as the three-dimensional sphere S 3 sitting inside R4 . It is well known that S 3 is simply connected. The condition of simple connectedness is extremely important. One of our most important theorems will be that if G is simply connected, then there is a

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natural one-to-one correspondence between the representations of G and the representations of its Lie algebra. For any path-connected topological space, one can de?ne an object called the fundamental group. See Appendix E for more information. A topological space is simply connected if and only if the fundamental group is the trivial group {1}. I now provide the following tables of fundamental groups, ?rst for compact groups and then for noncompact groups. See Appendix E for the methods of proof. Here, SOe (n; 1) denotes the identity component of SO(n; 1) (since one de?nes the fundamental group only for connected groups). In each entry, the result is understood to apply for all n ≥ 1 unless otherwise stated. Group Simply SO(2) SO(n) (n ≥ 3) U(n) SU(n) Sp(n) connected? Fundamental group no Z no Z/2 no Z yes { 1} yes { 1} connected? Fundamental group no same as SO(n) no Z no same as SO(n) yes {1} no same as SO(n) yes {1} no same as SO(n) no Z yes {1}

Group Simply GL(n; R)+ (n ≥ 2) GL(n; C) SL(n; R) (n ≥ 2) SL(n; C) SO(n; C) SOe (1; 1) SOe (n; 1) (n ≥ 2) Sp(n; R) Sp(n; C)

We conclude this section with a discussion of the case of SO(3). If v is a unit vector in R3 , let Rv,θ be the element of SO(3) consisting of a “right-handed” rotation by angle θ in the plane perpendicular to v. Here, right-handed means that if one places the thumb of one’s right hand in the v -direction, the rotation is in the direction that one’s ?ngers curl. To say this more mathematically, let v ⊥ denote the plane perpendicular to v and let us choose an orthonormal basis (u1 , u2 ) for v ⊥ in such a way that the basis (u1 , u2 , v ) for R3 has the same orientation as the standard basis (e1 , e2 , e3 ). (This means that the linear map taking (u1 , u2 , v ) to (e1 , e2 , e3 ) has positive determinant.) We then use the basis (u1 , u2 ) to identify v ⊥ with R2 , and the rotation is then in the counterclockwise direction in R2 . It is easily seen that R?v,θ is the same as Rv,?θ . It is also not hard to show (Exercise 16) that every element of SO(3) can be expressed as Rv,θ , for some v and θ with ?π ≤ θ ≤ π. Furthermore, we can arrange that 0 ≤ θ ≤ π by replacing v with ?v if necessary. Now let B denote the closed ball of radius π in R3 and consider the map Φ : B → SO(3) given by

1.6 Homomorphisms and Isomorphisms

17

Φ(u) = Ru ?, Φ(0) = I.

u

,

u = 0,

Here, u ? = u/ u is the unit vector in the u-direction. The map Φ is continuous, even at I, since Rv,θ approaches the identity as θ approaches zero, regardless of how v is behaving. The discussion in the preceding paragraph shows that Φ maps B onto R3 . The map Φ is almost injective, but not quite. Since Rv,π = R?v,π , antipodal points on the boundary of B (i.e., pairs of points of the form (u, ?u) with u = π ) map to the same element of SO(3). This means that SO(3) can be identi?ed (homeomorphically) with B/?, where ? denotes identi?cation of antipodal points on the boundary. It is known that B/? is not simply connected. Speci?cally, consider the loop in B/? that begins at some vector u of length π and goes in a straight line through the origin until it reaches ?u. (Since u and ?u are identi?ed, this is a loop in B/?.) It can be shown that this loop cannot be shrunk continuously to a point in B/?. This, then, shows that SO(3) is not simply connected. In fact, B/? is homeomorphic to the manifold RP3 (real projective space of dimension 3) which has fundamental group Z/2.

1.6 Homomorphisms and Isomorphisms
De?nition 1.15. Let G and H be matrix Lie groups. A map Φ from G to H is called a Lie group homomorphism if (1) Φ is a group homomorphism and (2) Φ is continuous. If, in addition, Φ is one-to-one and onto and the inverse map Φ?1 is continuous, then Φ is called a Lie group isomorphism. The condition that Φ be continuous should be regarded as a technicality, in that it is very di?cult to give an example of a group homomorphism between two matrix Lie groups which is not continuous. In fact, if G = R and H = C? , then any group homomorphism from G to H which is even measurable (a very weak condition) must be continuous. (See Exercise 17 in Chapter 9 of Rudin (1987).) Note that the inverse of a Lie group isomorphism is continuous (by de?nition) and a group homomorphism (by elementary group theory), and thus a Lie group isomorphism. If G and H are matrix Lie groups and there exists a Lie group isomorphism from G to H , then G and H are said to be isomorphic, and we write G ? = H . Two matrix Lie groups which are isomorphic should be thought of as being essentially the same group. The simplest interesting example of a Lie group homomorphism is the determinant, which is a homomorphism of GL(n; C) into C? . Another simple example is the map Φ : R → SO(2) given by Φ(θ) = cos θ ? sin θ sin θ cos θ .

18

1 Matrix Lie Groups

This map is clearly continuous, and calculation (using standard trigonometric identities) shows that it is a homomorphism. (Compare Exercise 6.) 1.6.1 Example: SU(2) and SO(3) A very important topic for us will be the relationship between the groups SU(2) and SO(3). This example is designed to show that SU(2) and SO(3) are almost (but not quite!) isomorphic. Speci?cally, there exists a Lie group homomorphism Φ which maps SU(2) onto SO(3) and which is two -to-one. We now describe this map. Consider the space V of all 2 × 2 complex matrices which are self-adjoint (i.e., A? = A) and have trace zero. This is a three-dimensional real vector space with the following basis: A1 = 01 10 ; A2 = 0 i ?i 0 ; A3 = 1 0 0 ?1 .

We may de?ne an inner product (Section B.6 of Appendix B) on V by the formula 1 A, B = trace(AB ). 2 (Except for the factor of 1 2 , this is simply the restriction to V of the Hilbert– Schmidt inner product described in Section B.6.) Direct computation shows that {A1 , A2 , A3 } is an orthonormal basis for V . Having chosen an orthonormal basis for V , we can identify V with R3 . Now, suppose that U is an element of SU(2) and A is an element of V, and consider UAU ?1 . Then (Section B.5), trace(UAU ?1 ) = trace(A) = 0 and (UAU ?1 )? = (U ?1 )?AU ? = UAU ?1 , and so UAU ?1 is again in V. Furthermore, for a ?xed U, the map A → UAU ?1 is linear in A. Thus for each U ∈ SU(2), we can de?ne a linear map ΦU of V to itself by the formula ΦU (A) = UAU ?1 .
?1 ?1 Note that U1 U2 AU2 U1 = (U1 U2 )A(U1 U2 )?1 , and so ΦU1 U2 = ΦU1 ΦU2 . Moreover, given U ∈ SU(2) and A, B ∈ V , we have

ΦU (A), ΦU (B ) =

1 1 trace(UAU ?1 UBU ?1 ) = trace(AB ) = A, B . 2 2

Thus, ΦU is an orthogonal transformation of V . Once we identify V with R3 (using the above orthonormal basis), then we may think of ΦU as an element of O(3). Since ΦU1 U2 = ΦU1 ΦU2 , we see that Φ (i.e., the map U → ΦU ) is a homomorphism of SU(2) into O(3). It is easy to see that Φ is continuous and, thus, a Lie group homomorphism. Recall that every element of O(3) has determinant ±1. Now, SU(2) is connected (Exercise

1.7 The Polar Decomposition for SL(n; R) and SL(n; C)

19

8), Φ is continuous, and ΦI is equal to I, which has determinant one. It follows that Φ must actually map SU(2) into the identity component of O(3), namely SO(3). The map U → ΦU is not one-to-one, since for any U ∈ SU(2), ΦU = Φ?U . (Observe that if U is in SU(2), then so is ?U .) It is possible to show that ΦU is a two-to-one map of SU(2) onto SO(3). (The least obvious part of this assertion is that Φ maps onto SO(3). This will be easy to prove once we have introduced the concept of the Lie algebra and proved Theorem 2.21.) The signi?cance of this homomorphism is that SO(3) is not simply connected, but SU(2) is. The map Φ allows us to relate problems on the non-simply-connected group SO(3) to problems on the simply-connected group SU(2).

1.7 The Polar Decomposition for SL(n; R) and SL(n; C)
In this section, we consider the polar decompositions for SL(n; R) and SL(n; C). These decompositions can be used to prove the connectedness of SL(n; R) and SL(n; C) and to show that the fundamental groups of SL (n; R) and SL (n; C) are the same as those of SO(n) and SU(n), respectively (Appendix E). These decompositions are supposed to be analogous to the unique decomposition of a nonzero complex number z as z = up, with |u| = 1 and p real and positive. A real symmetric matrix P is said to be positive if x, P x > 0 for all nonzero vectors x ∈ Rn . (Symmetric means that P tr = P.) Equivalently, a symmetric matrix is positive if all of its eigenvalues are positive. Given a symmetric positive matrix P, there exists an orthogonal matrix R such that P = RDR?1 , where D is diagonal with positive diagonal entries λ1 , . . . , λn . (If we choose an orthonormal basis v1 , . . . , vn of eigenvectors for P, then R is the matrix whose columns are v1 , . . . , vn .) We can then construct a square root of P as P 1/2 = RD1/2 R?1 , where D1/2 is the diagonal matrix whose (positive) diagonal entries are 1 /2 1 /2 λ1 , . . . , λn . Then, P 1/2 is also symmetric and positive. It can be shown 1 /2 that P is the unique positive symmetric matrix whose square is P (Exercise 21). We now prove the following result. Proposition 1.16. Given A in SL(n; R), there exists a unique pair (R, P ) such that R ∈ SO(n), P is real, symmetric, and positive, and A = RP. The matrix P satis?es det P = 1. Proof. If there were such a pair, then we would have Atr A = P R?1 RP = P 2 . Now, Atr A is symmetric (check!) and positive, since x, Atr Ax = Ax, Ax > 0, where Ax = 0 because A is invertible. Let us then de?ne P by

20

1 Matrix Lie Groups

P = (Atr A)1/2 , so that P is real, symmetric, and positive. Since we want A = RP, we must set R = AP ?1 = A((Atr A)1/2 )?1 . We check that R is orthogonal: RRtr = A((Atr A)1/2 )?1 ((Atr A)1/2 )?1 Atr = A(Atr A)?1 Atr = I. This shows that R is in O(n). To check that R is in SO(n), we note that 1 = det A = det R det P. Since P is positive, we have det P > 0. This means that we cannot have det R = ?1, so we must have det R = 1. It follows that det P = 1 as well. We have now established the existence of a pair (R, P ) with the desired properties. To establish the uniqueness of the pair, we recall that if such a pair exists, then we must have P 2 = Atr A. However, we have remarked earlier that a real, positive, symmetric matrix has a unique real, positive, symmetric square root, so P is unique. It follows that R = AP ?1 is also unique. If P is a self-adjoint complex matrix (i.e., P ? = P ), then we say P is positive if x, P x > 0 for all nonzero vectors x in Cn . An argument similar to the one above establishes the following polar decomposition for SL(n; C). Proposition 1.17. Given A in SL(n; C), there exists a unique pair (U, P ) with U ∈ SU(n), P self-adjoint and positive, and A = U P. The matrix P satis?es det P = 1. It is left to the reader to work out the appropriate polar decompositions for the groups GL(n; R), GL(n; R)+ , and GL(n; C).

1.8 Lie Groups
As explained in this section and in Appendix C, a Lie group is something that is simultaneously a smooth manifold and a group. As the terminology suggests, every matrix Lie group is a Lie group. (This is not at all obvious from the de?nition of a matrix Lie group, but it is true nevertheless, as we will prove in the next chapter.) The reverse is not true: Not every Lie group is isomorphic to a matrix Lie group. Nevertheless, I have restricted attention in this book to matrix Lie groups for several reasons. First, not everyone who wants to learn about Lie groups is familiar with manifold theory. Second, even for someone familiar with manifolds, the de?nitions of the Lie algebra and exponential mapping for a general Lie group are substantially more complicated and abstract than in the matrix case. Third, most of the interesting examples of Lie groups are matrix Lie groups. Fourth, the results we will prove for matrix Lie groups (e.g., about the relationship between Lie group homomorphisms and Lie algebra homomorphisms) continue to hold for general Lie groups. Indeed,

1.8 Lie Groups

21

the proofs of these results are much the same as in the general case, except that one can get started more quickly in the matrix case. Although in the long run the manifold approach to Lie groups is unquestionably the right one, the matrix approach allows one to get into the meat of Lie group theory with minimal preparation. This section gives a very brief account of the manifold approach to Lie groups. Appendix C gives more information, and complete accounts can be found in standard textbooks such as those by Br¨ ocker and tom Dieck (1985), Varadarajan (1974), and Warner (1983). Appendix C gives two examples of Lie groups that cannot be represented as matrix Lie groups and also discusses two important constructions (covering groups and quotient groups) which can be performed for general Lie groups but not for matrix Lie groups. De?nition 1.18. A Lie group is a di?erentiable manifold G which is also a group and such that the group product G×G→G and the inverse map g → g ?1 are di?erentiable. A manifold is an object that looks locally like a piece of Rn . An example would be a torus, the two-dimensional surface of a “doughnut” in R3 , which looks locally (but not globally) like R2 . For a precise de?nition, see Appendix C. Example. As an example, let G = R × R × S 1 = (x, y, u)|x ∈ R, y ∈ R, u ∈ S 1 ? C and de?ne the group product G × G → G by (x1 , y1 , u1 ) · (x2 , y2 , u2 ) = (x1 + x2 , y1 + y2 , eix1 y2 u1 u2 ). Let us ?rst check that this operation makes G into a group. It is not obvious but easily checked that this operation is associative; the product of three elements with either grouping is (x1 + x2 + x3 , y1 + y2 + y3 , ei(x1 y2 +x1 y3 +x2 y3 ) u1 u2 u3 ). There is an identity element in G, namely e = (0, 0, 1) and each element (x, y, u) has an inverse given by (?x, ?y, eixy u?1 ). Thus, G is, in fact, a group. Furthermore, both the group product and the map that sends each element to its inverse are clearly smooth, and so G is a Lie group. Note that there is nothing about matrices in the way we have de?ned G; that is, G is not given to us as a matrix group. We may still ask whether G is isomorphic to some matrix Lie group, but even this is not true. As shown in Appendix C, there is no continuous, injective homomorphism of G into any GL(n; C). Thus, this example shows that not every Lie group is

22

1 Matrix Lie Groups

a matrix Lie group. Nevertheless, G is closely related to a matrix Lie group, namely the Heisenberg group. The reader is invited to try to work out what the relationship is before consulting the appendix. Now let us think about the question of whether every matrix Lie group is a Lie group. This is certainly not obvious, since nothing in our de?nition of a matrix Lie group says anything about its being a manifold. (Indeed, the whole point of considering matrix Lie groups is that one can de?ne and study them without having to go through manifold theory ?rst!) Nevertheless, it is true that every matrix Lie group is a Lie group, and it would be a particularly misleading choice of terminology if this were not so. Theorem 1.19. Every matrix Lie group is a smooth embedded submanifold of Mn (C) and is thus a Lie group. The proof of this theorem makes use of the notion of the Lie algebra of a matrix Lie group and is given in Chapter 2. Let us think ?rst about the case of GL(n; C). This is an open subset of the space Mn (C) and thus a manifold of (real) dimension 2n2 . The matrix product is certainly a smooth map of Mn (C) to itself, and the map that sends a matrix to its inverse is smooth on GL(n; C), by the formula for the inverse in terms of the classical adjoint. Thus, GL(n; C) itself is a Lie group. If G ? GL(n; C) is a matrix Lie group, then we will prove in Chapter 2 that G is a smooth embedded submanifold of GL(n; C). (See Corollary 2.33 to Theorem 2.27.) The matrix product and inverse will be restrictions of smooth maps to smooth submanifolds and, thus, will be smooth. This will show, then, that G is also a Lie group. It is customary to call a map Φ between two Lie groups a Lie group homomorphism if Φ is a group homomorphism and Φ is smooth, whereas we have (in De?nition 1.15) required only that Φ be continuous. However, the following proposition shows that our de?nition is equivalent to the more standard one. Proposition 1.20. Let G and H be Lie groups and let Φ be a group homomorphism from G to H . If Φ is continuous, it is also smooth. Thus, group homomorphisms from G to H come in only two varieties: the very bad ones (discontinuous) and the very good ones (smooth). There simply are not any intermediate ones. (See, for example, Exercise 19.) We will prove this in the next chapter (for the case of matrix Lie groups). See Corollary 2.34 to Theorem 2.27. In light of Theorem 1.19, every matrix Lie group is a (smooth) manifold. As such, a matrix Lie group is automatically locally path-connected. It follows that a matrix Lie group is path-connected if and only if it is connected. (See the remarks following De?nition 1.7.)

1.9 Exercises

23

1.9 Exercises
1. Let a be an irrational real number and let G be the following subgroup of GL(2; C): eit 0 t∈R . G= 0 eita Show that G= eit 0 0 eis t, s ∈ R ,

where G denotes the closure of the set G inside the space of 2 × 2 matrices. Assume the following result: The set of numbers of the form e2πina , n ∈ Z, is dense in S 1 . Note : The group G can be thought of as the torus S 1 × S 1 , which, in turn, can be thought of as [0, 2π ] × [0, 2π ], with the ends of the intervals identi?ed. The set G ? [0, 2π ] × [0, 2π ] is called an irrational line. Drawing a picture of this set should make it plausible that G is dense in [0, 2π ] × [0, 2π ]. 2. Orthogonal groups . Let ·, · denote the standard inner product on Rn : x, y = k xk yk . Show that a matrix A preserves this inner product if and only if the column vectors of A are orthonormal. Show that for any n × n real matrix B , Bx, y = x, B tr y , where (B tr )kl = Blk . Using this, show that a matrix A preserves the inner product on Rn if and only if Atr A = I . Note : A similar analysis applies to the complex orthogonal groups O(n; C) and SO(n; C). 3. Unitary groups . Let ·, · denote the standard inner product on Cn : x, y = k xk yk . Following Exercise 2, show that Ax, Ay = x, y for all x, y ∈ Cn if and only if A? A = I and that this holds if and only if the columns of A are orthonormal. Here, (A? )kl = Alk . 4. Generalized orthogonal groups. Let [·, ·]n,k be the symmetric bilinear form on Rn+k de?ned in (1.1). Let g be the (n + k ) × (n + k ) diagonal matrix with ?rst n diagonal entries equal to one and last k diagonal entries equal to minus one: In 0 . g= 0 ?Ik Show that for all x, y ∈ Rn+k , [x, y ]n,k = x, gy . Show that a (n + k ) × (n + k ) real matrix A is in O(n; k ) if and only if Atr gA = g . Show that O(n; k ) and SO(n; k ) are subgroups of GL(n + k ; R) and are matrix Lie groups.

24

1 Matrix Lie Groups

5. Symplectic groups. Let B [x, y ] be the skew-symmetric bilinear form on n R2n given by B [x, y ] = k=1 (xk yn+k ? xn+k yk ). Let J be the 2n × 2n matrix 0I J= . ?I 0 Show that for all x, y ∈ R2n , B [x, y ] = x, Jy Show that a 2n × 2n matrix A is in Sp(n; R) if and only if Atr JA = J . Show that Sp(n; R) is a subgroup of GL(2n; R) and a matrix Lie group. Note : A similar analysis applies to Sp(n; C). 6. The groups O(2) and SO(2). Show that the matrix cos θ ? sin θ sin θ cos θ is in SO(2) and that cos θ ? sin θ sin θ cos θ cos φ ? sin φ sin φ cos φ = cos(θ + φ) ? sin(θ + φ) sin(θ + φ) cos(θ + φ) .

Show that every element A of O(2) is of one of the two forms: A= cos θ ? sin θ sin θ cos θ or A = cos θ sin θ sin θ ? cos θ .

(Note that if A is of the ?rst form, then det A = 1, and if A is of the second form, then det A = ?1.) Hint : Recall that for A to be in O(2), the columns of A must be orthonormal. 7. The groups O(1; 1) and SO(1; 1). Show that the matrix cosh t sinh t sinh t cosh t is in SO(1; 1) and that cosh t sinh t sinh t cosh t cosh s sinh s sinh s cosh s = cosh(t + s) sinh(t + s) sinh(t + s) cosh(t + s) .

Show that every element of O(1; 1) can be written in one of the four forms: cosh t sinh t sinh t cosh t cosh t ? sinh t sinh t ? cosh t ; ; ? cosh t sinh t sinh t ? cosh t ? cosh t ? sinh t sinh t cosh t ; .

1.9 Exercises

25

(Note that since cosh t is always positive, there is no overlap among the four cases. Note also that matrices of the ?rst two forms have determinant one and matrices of the last two forms have determinant minus one.) Hint : Use condition (1.2). 8. The group SU(2). Show that if α and β are arbitrary complex numbers 2 2 satisfying |α| + |β | = 1, then the matrix A= α ?β β α

9. 10. 11.

12.

13.

is in SU(2). Show that every A ∈ SU(2) can be expressed in this form 2 2 for a unique pair (α, β ) satisfying |α| + |β | = 1. (Thus, SU(2) can be thought of as the three-dimensional sphere S 3 sitting inside C2 = R4 . In particular, this shows that SU(2) is simply connected.) The groups Sp(1; R), Sp(1; C), and Sp(1). Show that Sp(1; R) = SL(2; R), Sp(1; C) = SL(2; C), and Sp(1) = SU(2). The Heisenberg group. Determine the center Z (H ) of the Heisenberg group H . Show that the quotient group H/Z (H ) is abelian. A subset E of a matrix Lie group G is called discrete if for each A in E there is a neighborhood U of A in G such that U contains no point in E except for A. Suppose that G is a connected matrix Lie group and N is a discrete normal subgroup of G. Show that N is contained in the center of G. This problem gives an alternative proof of Proposition 1.9, namely that GL(n; C) is connected. Suppose A and B are invertible n × n matrices. Show that there are only ?nitely many complex numbers λ for which det (λA + (1 ? λ)B ) = 0. Show that there exists a continuous path A(t) of the form A(t) = λ(t)A + (1 ? λ(t))B connecting A to B and such that A(t) lies in GL(n; C). Here, λ(t) is a continuous path in the plane with λ(0) = 0 and λ(1) = 1. Connectedness of SO(n). Show that SO(n) is connected, using the following outline. For the case n = 1, there is nothing to show, since a 1 × 1 matrix with determinant one must be [1]. Assume, then, that n ≥ 2. Let e1 denote the unit vector with entries 1, 0, . . . , 0 in Rn . Given any unit vector v ∈ Rn , show that there exists a continuous path R(t) in SO(n) with R(0) = I and R(1)v = e1 . (Thus, any unit vector can be “continuously rotated” to e1 .) Now, show that any element R of SO(n) can be connected to a blockdiagonal matrix of the form 1 R1 with R1 ∈ SO(n ? 1) and proceed by induction.

26

1 Matrix Lie Groups

14. The connectedness of SL(n; R). Using the polar decomposition of SL(n; R) (Proposition 1.16) and the connectedness of SO(n) (Exercise 13), show that SL(n; R) is connected. Hint : Recall that if P is a real, symmetric matrix, then there exists a real, ?1 orthogonal matrix R1 such that P = R1 DR1 , where D is diagonal. + 15. The connectedness of GL(n; R) . Using the connectedness of SL(n; R) (Exercise 14) show that GL(n; R)+ is connected. 16. If R is an element of SO(3), show that R must have an eigenvector with eigenvalue 1. Hint : Since SO(3) ? SU(3), every (real or complex) eigenvalue of R must have absolute value 1. 17. Show that the set of translations is a normal subgroup of the Euclidean group E(n). Show that the quotient group E(n)/(translations) is isomorphic to O(n). (Assume Proposition 1.5.) 18. Let a be an irrational real number. Show that the set of numbers of the form e2πina , n ∈ Z, is dense in S 1 . (See Problem 1.) 19. Show that every continuous homomorphism Φ from R to S 1 is of the form Φ(x) = eiax for some a ∈ R. (This shows in particular that every such homomorphism is smooth.) 20. Suppose G ? GL(n1 ; C) and H ? GL(n2 ; C) are matrix Lie groups and that Φ : G → H is a Lie group homomorphism. Then, the image of G under Φ is a subgroup of H and thus of GL(n2 ; C). Is the image of G under Φ necessarily a matrix Lie group? Prove or give a counter-example. 21. Suppose P is a real, positive, symmetric matrix with determinant one. Show that there is a unique real, positive, symmetric matrix Q whose square is P. Hint : The existence of Q was discussed in Section 1.7. To prove uniqueness, consider two real, positive, symmetric square roots Q1 and Q2 of P and show that the eigenspaces of both Q1 and Q2 coincide with the eigenspaces of P.

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