Chapter X. Matrix Lie Groups.

Transcription

1 Notes c F.P. Greenleaf and S. Marques LAII-s16-matrixgps.tex version 7/28/2016 Chapter X. Matrix Lie Groups. X.1. Matrix Groups and Implicit Function Theorem. The rank of a linear operator T : V W is dim(range(t)) = dim(v ) dim(ker(t)). If X, Y are bases in V, W the rank of T can be determined from the matrix A = [T] Y,X as follows. A k k submatrix is obtained by designating k rows and k columns and extracting from A the k k array where these meet. To describe the outcome we must specify the row indices I = {i 1 < < i k } and column indices J = {j 1 < < j k }, and then we might indicate how the submatrix was constructed by writing A IJ. Note that A itself is not necessarily square; it is n m if dim(v ) = m, dim(w) = n Lemma. Given a nonzero n m matrix A its rank rk(a) is equal to k max = max{k N : A has a nonsingular k k submatrix A IJ } Proof: Obviously rk(a) k max : if A IJ is nonsingular its columns {C j 1,, C j k } are truncated versions of the corresponding columns {C j1,, C jk } of A, which forces the latter to be linearly independent. Hence J k max rk(a). We also have rk(a) k max, for if rk(a) = k there is some set of column indices with J = k such that {C j : i J} are linearly independent. If B is the n k matrix [C j1 ; ; C jk ], it is well known that row rank(b) = column rank(b), so we can find a set I of row indices with I = J = k such that the rows {R i (B) : i I} are linearly independent. The rows in the n k matrix B = C j1 ; ; C jk are truncated versions of the corresponding rows in A, and those with row indices in J are precisely the rows of the k k submatrix A IJ. Obviously, this submatrix is nonsingular, so k max k = rk(a IJ ) = rk(a). Note that various choices I, J of row and column indices may yield nonsingular square submatrices A IJ of maximal size. Smooth Mappings and their Differentials. Now consider a mapping y = φ(x) = (φ 1 (x),..., φ n (x)) from F m F n (F = R or C, but mostly R in our discussion). We say that φ is a C map (or smooth map) if the scalar components φ k (x) have continuous partial derivatives of all orders. Figure A square k k submatrix A IJ is extracted from an n m matrix by specifying row indices I = {i 1 < < i k } and column indices J = {j 1 < < j k }. The rank rk(a) is equal to k if A IJ is nonsingular and all nonsingular square submatrices have size r k 75

2 The Jacobian matrix for φ at base point p is the n m matrix y 1 x 1 (p) (dφ) p = y n x 1 (p) y 1 x m (p) y n x m (p) whose entries are smooth scalar-valued functions of x F m. We will be concerned with the rank rk(dφ) x of the Jacobian matrix at and near various base point. We assign a linear operator, the differential of φ at p, n m (dφ) p : F m F n such that (dφ) p (v) = (dφ) p col(v 1,..., v m ) at each base point in F m where φ is smooth. The operator (dφ) p is the unique linear operator F m F n that closely approximates the behavior of the (nonlinear) map φ : F m F n near p, in the sense that φ = φ(p + x) φ(p) = (dφ) p ( x) + E( x), in which the error term E( x) becomes very small compared to x for small increments away from the base point p: (48) Error Estimate: E( x) x 0 in F n as x 0 in F m. As a function of the base point p F m, the linear operator (matrix) (dφ) p is a C map from F m into the matrix space M(n m, F). We define the rank of φ at p to be the rank r = rk(dφ) p of its Jacobian matrix. As above, we have rk(dφ) p = r dim(range(dφ) p ) m dim(ker(dφ) p ) = m r there are r row indices I = {i 1 < < i r } and column indices J = {j 1 < < j r }, such that 1. The submatrix (dφ p ) IJ is non singular, and 2. No larger square submatrix (with k > r) can be nonsingular, so r r is the maximum size of any nonsingular square submatrix. Note the following points: 1. Various choices of I, J may yield nonsingular submatrices (dφ) IJ of maximal size r r. The valid choices of I, J may also vary with the base point p. 2. For fixed choices of indices I, J the entries in the r r matrix (dφ x ) IJ vary smoothly with x, and so does the determinant det(dφ) x, so if the determinant is nonzero at p it must also be nonzero for all x near p. Hence for fixed choice of I, J we have rk(dφ x ) IJ r for all x near p if rk(dφ p ) IJ = r. Now let r max be the largest value rk(dφ x ) achieves on F m. If rk(dφ) p = r max it follows that rk(dφ) x = r max (constant rank) on some open neighborhood of p in F m. Quite often, as when φ : F m F n has scalar components φ = (φ 1 (x),..., φ n (x)) that are polynomials in x = (x 1,, x m ), this open set is dense in F m and its complement has Lebesgue measure zero i.e. maximal (constant) rank is achieved at almost all points in F m. 76

3 Figure The projection maps π J, π J and direct sum decomposition F m = F J F J associated with a partition of column indices [1, m] = J J, J J =. Near any base point p M this splits the variables in x = (x 1,..., x m) into two groups so x = (x J, x J ) with x J = (x k1,..., x r ) and x J = (x l1,..., x ls ), where r = J, s = J, and r+s = m = dim(f m ). In our narrative such partitions of variables arise in discussing the rank r = rk(dφ) p of the the n m Jacobian matrix [ f i / x j ] of a differentiable map φ : F m F n at points on a typical level set L(φ = q). 3. For any choice of indices I, J with I = J = k min{m, n} define J = [1, m] J and let F J, F J F m be the subspaces F J = R-span{e i : i J}, F J = R-span{e i : i J }, where {e i } is the standard basis in F m. Then F m splits as a direct sum F J F J, and this decomposition determines projections onto these subspaces. π J : F m F J and π J : F m F J Note: By composing φ with translations in F m and F n we can assume φ maps the origin in F m to the origin in F n. This will not change rk(dφ p ), but greatly simplifies the notation. The following exercise shows that this maneuver does not affect the Jacobian matrices or their determinants Exercise. If p R m (a) Consider a translation operator y = φ(x) = (x 1 +p 1,...,x m +p m ) from R m R m. Prove that (dφ) p = I m m at every base point. (b) Given smooth maps R m φ R n ψ R k and base points p R m, q = φ(p) R n, explain why the differential of a composite map ψ φ : R m R k is the matrix product of their differentials d(ψ φ) p = (dψ) φ(p) (dφ) p Smooth Hypersurfaces and the Implicit Function Theorem. The Implicit Function Theorem (IFT) concerns itself with level sets L(φ = q) = {x F m : φ(x) = q} F m, on which a smooth mapping φ : F m F n has constant (vector) value φ(x) = q F n. In essence, the IFT says that if p lies in a level set L(φ = q), and if (dφ) x has constant rank = r at and near p, then the locus L(φ = q) can be described locally as a smooth surface of dimension m r in F m. That is to say, near p the level set coincides with the graph Γ = {(x, f(x) : x F m r } F m = F m r F r 77

4 Figure Level sets for the map f : R 2 R with f(x, y) = z 2 1 2, identifying z = x+iy with (x, y) R 2. If c < 0 the level set L c = L(f = c) is empty; when c = 0 it consists of three isolated points z = 1, 0,+1 ; and for c > 0 the locus is usually a smooth curve (perhaps with more than one connected component, as when 0 < c < 1). But when c = 1 the locus has a singularity at the origin. It cannot be represented near the origin as the graph of any smooth function y = h(x) or x = g(y). The origin is a branch point for the locus. of a smooth map f : F m r F r. The idea is illustrated in the following example (see also Figure 10.2.) The map y = f(x) is the implicit function of the IFT Example. Define φ(z) = z ifor z C and regard it as a map R 2 R by identifying z = x+iy C with x = (x, y) R 2. Then φ becomes a 4 th degree polynomial in x and y, φ(x, y) = x 4 + 2x 2 y 2 + y 4 2x 2 + 2y The level sets L c = L(φ = c) are empty if c < 0; reduce to the isolated points { 1, 0, +1} if c = 0; and for c > 0 are smooth curves (sometimes with more than one connected component if 0 < c < 1). However there is one exception. When c = 1 the locus L(φ = 1), shown in Figure 10.3, has a singularity at the origin. Near z = 0 + i0 it cannot be described locally as the graph of a smooth function y = h(x) or x = g(y). The 1 2 Jacobian matrix Jφ(z) = [ φ/ x(z), φ/ y(z)] = (dφ) z has rk(dφ) z 1 (constant) throughout R 2 except at z = 0, z = 1 and z = +1 on the real axis (the critical points where both partial derivatives of f are zero) Exercise. Let φ : R 2 R be the function in Example 1.3 (a) Verify that the Jacobian matrix (dφ) x = [ φ x, φ y ] has rank zero (both components = 0) if and only if x = ( 1, 0), (0, 0), or (+1, 0) in R 2, by solving the system of equations φ x (x) = 0 φ y (x) = 0 (b) At which points x is one of the derivatives φ/ x and φ/ y zero, while the other is nonzero? Draw pictures of the sets S 1 = {x : φ/ x = 0 and φ/ y 0} S 2 = {x : φ/ x 0 and φ/ y = 0} 78

5 (c) Verify that rk(dφ) x = 1 at all points of S 1 and S 2 identified in (b). How are these points related to the pattern of level curves shown in Figure 10.3? Note: The point sets in (b) are infinite. In discussing the IFT we will discover that a level set L c = L(φ = c) in Example 1.3 can be described as the graph of a smooth function y = f(x) near any point p on the locus where φ/ x(p) 0, and similarly we can write x = g(y) if φ/ y(p) 0. In Example 1.3, at least one of these condition is satisfied at every base point, except the origin x = (0, 0), which lies on the locus L(φ = 1), and the points x = ( 1, 0) and (1, 0) which make up the degenerate locus L(φ = 0). Consequently, for c 0 or 1 the non-empty level curves L c = L(φ = c) can be described locally as smooth curves (the graphs of smooth functions y = f(x) or x = g(y)). Furthermore, L c can be described both ways (with y = f(x) or with x = g(y)) near most points p L c, but at a few points only one such description is possible these are the points on the curves in Figure 10.3 at which L c has either a horizontal or vertical tangent line. We will apply the IFT to show that the classical matrix groups O(n), SO(n), U(n), SO(n, C), etc. are actually smooth hypersurfaces in matrix space M(n, F) F n2, and hence have well-defined dimensions, tangent spaces, etc. If φ : F m F n is a C map and if rk(dφ) x r (constant) on some open neighborhood of p in F m, the IFT asserts that the level set S p = {x F m : φ(x) = φ(p)} passing through p can be described near p as a smooth hypersurface of dimension m r in F m. For simplicity we state the result taking F = R, though it remains true almost verbatim for F = C Theorem (Implicit Function Theorem). Let φ : R m R n be a C map defined near p R m and let M = L(φ = φ(p)) be the level set containing p. Assume rk(dφ) x r (constant) for x near p in R m. Given index sets I = {i 1 < < i r } [1, n], J = {j 1 < < j r } [1, m] such that the square submatrix [(dφ) p ] IJ is non singular, let J = [1, m] J and let π J,π J be the projections of R m onto R J, R J associated with the decomposition R m = R J R J, in which J = r, J = m r. Then there is an open rectangular neighborhood B 1 B 2 of p in R m = R J R J such that 1. On the relatively open neighborhood U p = (B 1 B 2 ) M of p in M, the restriction π J Up : U p B 1 of the linear projection π J : R m R J = R m r is a bicontinuous bijection between the open set U p M and the open set B 1 R m r. It assigns unique Euclidean coordinates x = (x 1,, x m r ) to every point in U p. 2. The inverse map Ψ = (π J Up ) 1 : B 1 U p R m is a C map from the open set B 1 R J into all of R m, and maps B 1 onto the relatively open neighborhood U p of p in the level set M. Then the map f : R r R n obtained by following Ψ with the horizontal projection π I shown in Figure 10.4 f = π I Ψ : B 1 B 2 R n is a C map. Furthermore Ψ is the graph map for the smooth function f because Ψ(x) = (π J (Ψ(x)), π I (Ψ(x))) = (x, π I Ψ(x)) = (x, f(x)) for x B 1. In particular the open neighborhood U p in M is the graph of the C map f : R J R I. Conclusion: Near p the locus M = L(φ = φ(p)) passing through p looks like part of a 79

6 Figure A diagram showing the players in the Implicit Function Theorem. Here S p is the level set passing through p for a C map φ : R m R n. By splitting coordinates in R m into two groups we get a decomposition R m = R J R J and associated projections π J, π J from F m to F J, F J such that (i) the restriction π J Up becomes a bijective bicontinuous map to an open set B 1 in R J for a suitably chosen open neighborhood U p of p in S p, and (ii) the inverse Ψ = (π J Up ) 1 : B 1 U p is a C map from the open set B 1 R J into the entire Euclidean space R m in which the level set S p lives. smooth hypersurface in R m of dimension k = m n. The situation described in the IFT is shown in Figure A rough general principle is at work here. If φ : R m R is a scalar valued C function we often find that the solution set L(φ = c), c R, is a smooth hypersurface of dimension m 1. A level set of a vector valued map φ : R m R n with φ = (φ 1,...,φ n ) is the intersection of the solution sets for a system of scalar constraint equations φ 1 (x) = c 1,..., φ n (x) = c n The solution set tends to lose one degree of freedom for each imposed constraint, so the outcome is usually a smooth hypersurface in R m of dimension m n, but that is not always the case and the point of the IFT is to make clear when it is true. This principle also suggests why it is often natural to restrict attention to the case m n, in which maximal rank means rk(dφ) p = n. If the number of constraints n exceeds the dimension m of the space R m in which the level set lives, the locus may be degenerate with solutions at all, or it may reduce to a set of isolated points in R m. The Maximal Rank Case: As a particular example, if φ maps F m F n and m n, the maximum possible value for the rank of (dφ p ) is n. If this maximal rank is achieved at some base point p F m, φ will automatically have the same (maximal) rank at all points x near p in F m. The maximal rank case is often encountered, but the IFT is proved in the more general constant rank case, in which we do not assume m n, or that the constant rank is the maximum possible rank of (dφ) p on all of F m Exercise. Consider the C map φ : R 4 R 2 given by y = f(x) = φ(x 1, x 2, x 3, x 4 ) = (y 1 (x), y 2 (x)) = (x x 2 2, x 2 3 x x 1 x 4 ) (a) Show that the locus M = L(φ = q) can be described as a smooth two-dimensional hypersurface in R 4 near p = (1, 2, 1, 3), at which q = φ(p) = (5, 5). Identify all pairs of variables x i, x j, (1 i < j 4) that can be used to smoothly parametrize this hypersurface near p 80

7 (b) Is this locus smooth near all of its points? Hint: In (b), start by showing x 4 0 for every point x M, so you can assume x 4 0 in calculations involving points on this locus, even if you cannot draw a picture. In answering (b) you will have to compute rk(a IJ ) for various square submatrices of the 2 4 Jacobian matrix [ y i / x j ], which has variable coefficients. Do this using symbolic row operations Exercise. Consider the C scalar-valued function f : R 2 R, φ(x, y) = x 3 y + 2e xy (a) Find all critical points, where both partial derivatives φ/ x and φ/ y are zero. At a critical point p there is no way to represent the level set S p = L(φ = φ(p)) passing through p as the graph of a smooth function y = f(x) or x = g(y). (b) Locate all points p where one of the partial derivatives is zero but the other is not (two cases to consider). Find the value of φ at each such base point to determine which level sets L(φ = c), c R, contain such points. (c) The locus M = L(φ = 2) obviously contains the horizontal and vertical axes. Prove that there are no other points on this locus. (Thus the origin is a singularity for the locus M, and there are no others.) Hint: In (c): Quadrant-by-quadrant, what is the sign of φ/ y off of the x- and y-axes? 1.8. Exercise. Let y = φ(x) = (y 1 (x), y 2 (x)) be a C map from R 3 R 2 such that (i) φ(p) = q = (0, 0) at base point p = (3, 1, 2) ( ) (ii) At p the Jacobian matrix is [ y i / x j ] = Answer the following questions without knowing anything more about φ. (a) Can the level set M = L(φ = q) be described near p = (3, 1, 2) as a smooth hypersurface in R 3? Of What dimension k? (b) Which of the variables x 1, x 2, x 3 can be legitimately be used to parametrize M near p = (3, 1, 2) as the graph of a smooth map f : R k R 3? List all valid choices of the parametrizing variables x i1,..., x ik Exercise. If f : F r F s is a C map defined on open set B F r, its graph Γ = {(x, f(x)) F r F s : x B} is the range of the graph map F(x) = (x, f(x)) from F r F r+s Show that the graph map is C for x B F r and that rk(df) x r (constant) for all x B. Smooth Submanifolds in F m. A space M is locally Euclidean of dimension d if it can be covered by a family of charts {(x α, U α ) : α in some index set I}, where x α : U α V α R d is a bicontinuous map from an open subset U α M to an open set V α = x α (U α ) in the Euclidean space R d. The chart maps x α assign locally defined Euclidean coordinates x α (u) = (x (α) 1 (u),..., x(α) (u)) k for u U α. Thus M looks locally like Euclidean coordinate space R d. Where the domains of two charts (x α, U α ), (x β, U β ) overlap we have the situation shown in Figure The intersection U α U β is an open set in M, the images N α = x α (U α U β ), N β = x β (U α U β ) are open sets in coordinate space R k, and we have 81

8 between two charts (x α, U α), (x β, U β ) and their (shaded) domains of definition in R r are shown. Both shaded domains N α, N β in R r correspond to the intersection U α U β of the chart domains, which is an open set in the locally Euclidean space M. Figure The coordinate transition maps x α x 1 β and x β x 1 α induced coordinate transition maps that tell us how the coordinates x = x α (u) and y = x β (u) assigned to u M by the chart maps are related. These transition maps x = (x α x 1 β )(y) y = (x β x 1 α )(x) from N β N α from N α N β are bicontinuous bijections between the open sets N α and N β in R k Definition. A locally Euclidean space M is a smooth manifold of dimension dim(m) = d if the charts (U α, x α ) that cover M map it into R d and are C -related, so the coordinate transition maps are C between the open sets N α, N β R d that correspond to the (open) intersection U α U β of chart domains in M. This allows us to make sense of smooth manifolds without requiring that they be embedded in some surrounding Euclidean space. IT is the starting point for modern differential geometry. Once we make M a C manifold by introducing C related covering charts, we can begin to do Calculus on M. The following concepts now make sense: 1. Given any chart (x α, U α ) on M, a scalar function f : M F on M becomes a function of local chart coordinates if we write y = F(x) = (f x 1 α )(x), which is defined on the open set V α = x α (U α ) in coordinate space F m. We say that f is a C function on M if y = F(x) has continuous partial derivatives of all orders, for each of the covering charts that determine the manifold structure of M. 2. A map φ : M N is a C mapping between manifolds M and N of dimensions m and n if it becomes a C map from F m F n when described in local coordinates on M and N. Thus if (x α, U α ) and (y β, U β ) are charts on M and N respectively, the composite y = Φ(x) = y β φ xα 1(x) is a C map from F m F n wherever it is well-defined. 3. A parametric curve in M is any continuous map y = γ(t) from some interval [a, b] R into M. It is a C -curve in M if it becomes a C vector-valued map x = (y 1 (t),..., y m (t)) = x α γ(t) for t [a, b] 82

9 for every chart (x α, U α ) on M Definition. Suppose M h N f R are C maps between C manifolds. In terms of the preceding definitions, explain why the composite f h : M R is a C map wherever it is well-defined. Hint: If M, N, R are Euclidean coordinate spaces F m, F n, F r this follows by the Chain Rule of multivariate Calculus. Smooth Manifolds and the IFT. One consequence of the IFT is this: If φ is a C map R m R n and if rk(dφ) x r (constant) near every point in a level set M = L(φ = q), q R n, we can use the IFT to create a family of C -related charts (x α, U α ) that cover M. The resulting standard C structure makes M into a smooth r-dimensional manifold. The crucial fact that the charts are C -related follows directly from the way the standard charts are constructed (see Proposition 1.13 below) Constructing Standard Charts on a Level Set M. The IFT and the constant rank condition allow us to construct a chart (x α, U α ) about a typical base point p M. 1. Write φ as v = φ(u) in terms of the standard coordinates u = (u 1,..., u m ) and v = (v 1,...,v n ) in R m and R n. By Lemma 1.1 and the constant rank condition we can, for each u M, choose row and column indices I [ 1, n] and J [ 1, m] with I = J = r = rk(dφ) u such that the square submatrices [ v i / u j (u)] IJ are nonsingular for u near p in R m. By Lemma 1.1 this cannot be done for any larger square submatrix. 2. Using the column indices determined in Step 1, let J = [ 1, m] J, split R m = R J R J and let π J,π J be the projection maps from R m to R J or R J. By the IFT there is a rectangular open neighborhood B 1 B 2 of p in F m such that the projection π J : (B 1 B 2 ) B 1 maps the relatively open neighborhood U p = (B 1 B 2 ) M in M onto the open set B 1 R J = R m r. To get a chart (x α, U α ) that imposes Euclidean coordinates on M near p we take U α = U p and bijective chart map x α = (π J Up ) : U p B 1 (an open set in R m r ). The charts (x α, U α ) obviously cover M owing to constancy of rk(dφ) near every point in M. 3. The inverse map Ψ = (π J Up ) 1 : B 1 U p M R m is C from B 1 R J into all of R m, and its range is precisely the chart domain U p. We now show that charts created this way, perhaps about different base points, are always C -related wherever the chart domains overlap Proposition. If φ : R m R n is a C map and M = L(φ = q) a level set such that rk(dφ) x r (constant) on an open neighborhood of every point in M, then all standard charts on M obtained by the preceding construction are C related where they overlap. This determines the standard C structure on M. The dimension of the resulting C manifold is k = m r. Proof: Consider two standard charts (x α, U α ) and (x β, U β ) about a typical point p in U α U β. The chart (x α, U α ) is determined by a partition of column indices [1, m] = J (α) J(α) and a choice of row indices I(α) [1, n] with J(α) = r and J (α) = m r, such that [(dφ) p ] IJ is nonsingular. In the notation of the IFT we then have U α = (B α 1 B α 2 ) M x α = (π J (α) Uα ) and V α = x α (U α ) = B α 1 R m r 83

10 The chart map is the restriction to U α of a linear projection map R m R J (α) = R m r, Its inverse Ψ = x 1 α x α = (π J (α) Uα ) : U α B α 1 Rm r. is the graph map Ψ α = (π J (α) Uα ) 1 : V α U α, which is actually a C map from B α 1 into all of Rm. The chart (x β, U β ) corresponds to some other choice of column and row indices [1, m] = J (β) J(β) and I(β) [1, n], and a corresponding rectangular open neighborhood B β 1 Bβ 2 of p in Rm. The chart map on U β = (B β 1 Bβ 2) M is just the restriction to U β of a linear projection π J (β), and by the IFT its inverse is a C map from B β 1 into all of Rm. Therefore the coordinate transition map x β x 1 α is the composite of a linear map and a C map R m r = R J (α) = π J (β) (π J (α) Uα ) 1 = π J (β) Ψ α x 1 α R m x β R J (β) = R m r and is certainly C. Likewise for the transition map in the reverse direction. The preceding proof is burdened by the complicated notation needed to label all the players. Here is a shorter proof that emphasizes the intuition behind the proof. Alternative Proof of Proposition 1.13: Suppose p is any point in U α U β and x 0 = x α (p),y 0 = x β (p) in R m r. To show y = x β x 1 α (x) is C near x 0 we observe that Near x 0 the chart map xα 1 coincides with the map (π J (α) Uα ) 1, which by the IFT is a C map from an open set in R m r into all of R m that sends x 0 p, and whose range is contained in M. Near p the chart map x β coincides with the globally defined linear projection map π J (β), which is certainly C. Therefore the transition map x β x 1 α is the composite of a linear map and a C map R m r = R J (α) x 1 α R m x β R J (β) = R m r and is C. Likewise for the transition map in the reverse direction Example. Let φ : R 3 R 1 with φ(x) = x 2 3 x2 1 x2 2. At any p = (x 1, x 2, x 3 ) the 1 3 Jacobian matrix (dφ) p = [ φ x 1,, φ x 3 ] = [ 2x 1, 2x 2, 2x 3 ] is just the classical gradient vector φ(p). The rank rk(dφ) x is constant 1 unless all three entries are zero, which happens only at the origin p = (0, 0, 0). The level set M 0 = L(φ = 0) is the double cone shown in Figure 10.6(b). This two-dimensional hypersurface has a singularity at the origin in R 3, where it fails to be locally Euclidean. Thus the locus L(φ = 0) cannot be made into a smooth manifold by covering it with suitably defined coordinate charts. All other level sets M c (c 0) are smooth twodimensional manifolds; a few of these level surfaces are shown in Figure 10.6(a). 84

11 Figure In (a) we show some level sets L c = L(φ = c) for the map φ(x 1, x 2, x 3 ) = x 2 3 x2 2 x2 1 from R3 R 1. For c = 0 the level set where x 2 1 x2 2 x2 3 = 0, shown in (b), is a double cone with a singularity at the origin, where it fails to be locally Euclidean. All other level sets are smooth two-dimensional hypersurfaces in R 3, but the geometry of L c changes as we pass from c < 0 to c > 0. For c < 0, we get a single connected surface; for c > 0 there are two isolated pieces, both smooth. Consider the possible charts we might impose near the point p = (1, 1, 3) on the particular level set M = L(φ = 1). Entries in (dφ) p (dφ) p = [ 2x 1, 2x 2, 2x 3 ] = ( 2 2, 2 2, 2 3) at p, are all nonzero near p, so we have constant rank rk(dφ) x 1 near p, ]and may apply the IFT to define standard charts about p. Each nonzero entry in (dφ) p corresponds to a nonsingular 1 1 submatrix, so several legitimate groupings of variables are available to parametrize M near p: (49) I = {1}, J = {2, 3} or I = {2}, J = {1, 3} or I = {3}, J = {1, 3} Thus L(φ = 1) can be described as the graph in R 3 of various smooth functions x k = f k (x i, x j ) by solving 1 = φ(x) = x 2 3 x 2 2 x 2 1 for one variable in terms of the other two. 1. x 1 = f 1 (x 2, x 3 ) = + x 2 3 x2 2 1 near (1, 3) in the (x 2, x 3 )-plane. 2. x 2 = f 2 (x 1, x 3 ) = + x 2 3 x2 1 1 near (1, 3) in the (x 1, x 3 )-plane. 3. x 3 = f 3 (x 1, x 2 ) = (x x2 2 ) near (1, 1) in the (x 1, x 2 )-plane. The coordinate transition map (x 1, x 3 ) = y β x 1 α (x 2, x 3 ) can be computed directly by writing x 1 = f 1 (x 2, x 3 ) to get (x 1, x 3 ) in terms of (x 2, x 3 ). The resulting transition map (x 1, x 3 ) = Φ(x 2, x 3 ) = y β x 1 α (x 2, x 3 ) = (x 1, x 3 ) x1=f 1(x 2,x 3) = ( + x 2 3 x2 2 1, x 3) is clearly a C map from (x 2, x 3 ) to (x 1, x 3 ). So is its inverse Exercise. In Example 1.14, 85

12 (a) Compute the inverse (x 2, x 3 ) = Φ 1 (x 1, x 3 ). (b) One of the valid splittings [1, 3] = J J of column indices listed in (49) is J = {1, 2} J = {3}. Find an explicit formula for the corresponding projection map (x 1, x 2 ) = π J (x 1, x 2, x 3 ) that assigns Euclidean coordinates to points x = (x 1, x 2, x 3 ) on M near p = (1, 1, 3). (c) Give an explicit formula for the inverse (x 1, x 2, x 3 ) = (π J R J ) 1 (x 1, x 3 ) of the projection map in (b) Exercise. Consider the points (a) p = (1, 0, 2) (b) p = (0, 0, 1) on the two-dimensional hypersurface M = L(φ = 1) of Example In each case determine all pairs of coordinates x J = (x i, x j ) that give a legitimate parametrization of M near the prescribed base point p Exercise. Verify that the unit sphere S 2 = L(φ = 1) for φ(x) = x x2 2 + x2 3 is a C manifold in R 3 by showing that rk(dφ) x 1 near every point x S Exercise. Describe a set of standard charts covering the unit sphere S 2 = L(φ = 1) where φ(x) = x x2 2 + x2 3, taking for your chart domains the relatively open hemispheres (boundary circles excluded) U + k = {x S2 : x k > 0} U k = {x S2 : x k < 0} for k = 1, 2, 3. All six hemispheres are required to fully cover S 2. The chart maps x ± k : U ± k R 2 project points x U ± k onto the open unit disc x x 2 3 < 1 in the (x 2, x 3 )-plane when k = 1; project U 2 ± onto the open disc in the (x 1, x 3 )-plane when k = 2; and project onto the disc in the (x 1, x 2 )-plane when k = 3. (a) Give explicit formulas for the chart maps on the particular domains U + 1 and U 3. (b) Compute the coordinate transition maps in both directions for these two charts, noting that they have the form (x i, x j ) = x α (x) = x α (x 1, x 2, x 3 ) for x M. Note: These are examples of standard charts on the level set L(φ = 1) Exercise (Stereographic Projection). Let H + be the two-dimensional hyperplane in R 3 that is tangent to the unit sphere M = S 2 at its north pole N = (0, 0, +1), and consider the punctured sphere U α = S 2 {S} obtained by deleting the south pole S = (0, 0, 1) from the sphere. Each point u U α determines a unique straight line in R 3 that passes through S and the point u; continuing along this line, we will meet the hyperplane H + in a unique point with coordinates (x(u), y(u), +1). The resulting bijection Φ + : U α H + is an example of stereographic projection. Dropping the redundant coordinate entry 1 we obtain the stereographic projection map x α : U α R 2, x α (u) = (x 1 (u), x 2 (u)) R 2 which is is bicontinuous from the open subset U + α S2 onto all of coordinate space R 2. Similarly we may stereographically project the punctured sphere U β = S 2 {N} onto the hyperplane H tangent to the sphere at the south pole S = (0, 0, 1), to define a second chart map x β (v) = (x, y ) R 2 for v U β. 86

13 (a) Give an explicit formula for the stereographic projection map (x 1, x 2 ) = x α (u) = x α (u 1, u 2, u 3 ). Note carefully that Φ + maps triples u with u 2 1+u 2 2 +u 3 3 = 1 to pairs (x 1, x 2 ) R 2. (b) Compute the coordinate transition map v = x β x 1 α (u) and its inverse, and check that it is C where defined. Stereographic projection allows us to cover the sphere S 2 with just two C -related charts, the minimum number possible since it is well known that S 2 cannot be mapped bicontinuously to the plane R 2. But for many purposes the covering with hemispheres leads to simpler computations. Note: These charts are not of the standard form in Proposition 1.12 but they are C - related to all standard charts (which by 1.13 are C -related to each other). Hint: In (a) use similar triangles, and rotational symmetry of the problem. X.2. Matrix Lie Groups. 1 The classical groups are the level sets of certain polynomial maps F m F n, except for the general linear group GL = GL(n, F) = {A M(n, F) : det(a) 0}, which is an open subset in matrix space M(n, F) F n2. This is a smooth manifold and it is covered by a single chart with chart domain U α = GL and chart map x α = the identity map of GL(n, F) F n2, which we shall write as x α (A) = (A 11,..., A in ; A 21,...,A 2n ;... ; A n1,..., A nn ) in what follows. Obviously dim F (GL) = n 2 since GL is an open set in M(n, F). All other classical groups are closed lower-dimensional subsets in GL(n, F) and in M(n, F) = F n Definition. A smooth manifold G is an abstract Lie group if 1. It is a group under some product operation P : G G G and under the inversion map J : G G that sends x x The product operation and inverse operation are both C maps. In particular if (U α, x α ), (U β, y β ) are coordinate charts the product operation becomes a C map from F m F m F m when expressed in these coordinates. Thus if x U α, y U β, and (U γ, z γ ) is a chart containing z = P(x, y) = x y, the composite map z γ P (x 1 α y 1 β ) : Fm F m F n is a C map. Similarly, if z U α and z 1 U β, x β J x 1 α : Fm F m is a C map The dimension d = dim F (G) is the dimension of the charts that cover G. If F = C, we regard F R 2 and view G as a real manifold of dimension dim R (G) = 2 dim C (G), with charts x α : U α R 2d. The general theory of Lie groups has become a vast subject. To keep things simple we restrict attention to matrix Lie groups, subsets G M(n, F) such that G is: (i) A group under matrix multiplication, as in Definition 2.1. (ii) A C manifold (smooth hypersurface) in matrix space, as in Definition That s pronounced Lee Groups. Sophus Lie was a Norwegian mathematician who pioneered the study of these structures toward the end of the 1800s. Esoteric concepts then, they are ubiquitous in modern physics and differential geometry. 87

14 The simplest example of a Lie group is G = (R n, +) with structure given by the single identity chart (U α, x α ) = (R n, id). Clearly, the (+) operation is a C map R n R n R n, as is the inverse map J(x) = x on R n. But this in not a matrix Lie group because its elements are not matrices. The general linear group GL(n, F) is a (noncommutive) matrix Lie group of dim F (G) = n 2. The group operation, expressed in chart coordinates x α (A) = (A 11,..., A 1n ;... ; A n1,..., A nn ) F n2, is a polynomial map of F n2 F n2 with and inversion is the rational map (AB) ij = A 1 = n A ik B kj, k=1 1 det(a) Cof(A)t This involves the transpose of the cofactor matrix Cof(A), whose entries are polynomials in the entries of A; det(a) is also a polynomial in the entries of A. Other matrix Lie groups are level sets in matrix space for various C (polynomial, actually) maps φ : M(n, F) F k with k n 2 = dim F M(n, F). We will soon indicate why they are all smooth manifolds. To see that they are also Lie groups we prove: 2.2. Theorem. Suppose M is the level set through p for some C map φ : M(n, F) F k, and that rk(dφ) x r (constant) on some open neighborhood of each p M. If M is also a subgroup of GL(n, F) under matrix multiplication, then M is a Lie group in the standard C structure it inherits as a smooth submanifold in matrix space M(n, F) F n2. Proof: If a, b M, let (U α, x α ), (U β, x β ) be standard charts about a, b and let (U γ, z γ ) be a chart about c = ab = P(a, b). By the IFT x γ is the restriction to U γ of a linear projection π J (γ) from F n2 to an open set in F d = F J (γ). The product map P(a, b) = a b is defined and C on all of matrix space M(n, F) M(n, F) M(n, F); and in fact it is a polynomial map. To verify that it is a C map on the manifold G we must show that z γ P (x 1 α y 1) β : Fd F d F d = (π J (γ) Uγ ) P (x 1 α y 1) β : Fd F d F d is C, where d = dim F (G) = n 2 r and r = rk(dφ), by breaking this composite into steps F d2 = F d F d x 1 α y 1 β z P γ=π J (γ) U α U β Uγ F d The matrix product operation P : F n2 F n2 U γ is defined and C on all of M(n, F). In particular points near (a, b) map to points near a b in U γ ; furthermore P(U α U β ) G because G is a group. The map z γ : U γ F d is the restriction to U γ G of a globally defined linear projection map π J (γ) : F n2 F J (γ) = F d, so it is the restriction to U γ of a globally C map. Clearly then, z γ P (x 1 α y 1 β ) = π P J (x 1 (γ) α y 1 β ) : Fd2 F d is C too. Similarly, the inversion map J : G G is C when expressed in standard chart coordinates because J : GL(n, F) GL(n, F) is a rational function of matrix coordinates. 88

15 The first example where we really need the Implicit Function Theorem is 2.3. Example. The special linear group G = SL(n, F) has dimension n 2 1 if F = C. But if we regard F = R 2 and SL(n, C) M(n, C) = R 2n2, then dim R (G) = 2n 2 2. On the other hand, if F = R we have dim R SL(n, R) = n 2 1. (Geometrically, we have SL(n, R) = SL(n, C) M(n, R)+i0 when we identify M(n, C) = M(n, R)+ 1M(n, R).) We will restrict attention is the case F = R. Discussion: SL(n, R) is the level set L(φ = 1) where φ : R n2 = M(n, R) R is φ(a) = det(a). To see that SL is a smooth manifold in matrix space we must show that the 1 n 2 Jacobian matrix has rk(dφ) X = 1 near every X SL(n, R). If we write coordinates of A as (a 11,..., a 1n ;...;a n1,..., a nn ) R n2, then φ = det(a) = σ S n sgn(σ) a 1,σ(1)... a n,σ(n), and by the product formula for derivatives we get φ = n sgn(σ) [ a 1,σ(1)... a j,σ(j)... a n,σ(n) ] a k,l a σ j=1 k,l = sgn(σ) a 1,σ(1)... â k,l... a n,σ(n) σ:σ(k)=l (Here we use the standard math notation b 1 b i b n = j i b j.) Indeed, a j,σ(j) / a k,l = 0 unless (j, σ(j)) = (k, l) which happens if and only if j = k and σ(k) = l, in which case it is equal to 1. Therefore we have a kl φ a k,l = σ:σ(k)=l sgn(σ) a 1,σ(1)... a k,l... a n,σ(n) But the 1 n 2 matrix [ φ/ a kl ] has rank = 1 (maximal rank) unless all entries are zero, in which case we get det(a) = ( ) + ( ) ( ) σ(1)=1 = σ(1)=2 σ S n sgn(σ) a 1,σ(1) a n,σ(n) = 0 σ(1)=n That contradicts the hypothesis det(a) = 1, and cannot occur. Thus (dφ) X has maximal rank (= 1) at each point in SL(n, R), and in fact at every point in GL(n, R) Example. The (real) orthogonal groups O(n) and SO(n), for which F = R, both have dimension 1 2 (n2 n). Discussion: Both are closed subgroups of M(n, R). On O(n), det(a) can only achieve values ±1 (and 1 is actually achieved); the value is +1 precisely on the subgroup SO(n). The full orthogonal group O(n) consists of two disjoint closed cosets, the subgroup SO(n) and the coset J SO(n), where J is any orthogonal matrix with det(j) = 1 such as J = diag(1,, 1, 1). The coset J SO(n) is the set of orientation-reversing orthogonal maps on R n, which is not a subgroup in O(n), while SO(n) is the group of invertible orientation-preserving maps in O(n). To show O(n) is smooth manifold recall that A O(n) A t A = I the rows R i are an ON basis in R n, so (R i, R j ) = δ ij (Kronecker delta) for 1 i, j n. Eliminating redundant identities in this list of n 2 identities by requiring i j, we define the map φ : R n2 = M(n, R) R (n 2 +n)/2 = R d 89

16 given by φ(a) = ((R 1, R 2 ),..., (R 1, R n ); (R 2, R 2 ),..., (R 2, R n ); ; (R n, R n )). Then O(n) is the level set L(φ = x) where x = (1, 0,, 0; ; 1, 0 ; 1) We must show rk(dφ) A = d for all A M(n, R) near an arbitrary point p O(n). The idea is clearly revealed by the case n = 3, where φ(a) = ( 3 i=1 a 2 1i, i a 1i a 2i, i a 1i a 3i ; i Writing x = (a 11, a 12, a 13 ; a 22, a 23 ; a 33 ) we have [ φ i a ij ] = a 2 2i, i a 2i a 3i ; i a 2 3i) 2a 11 2a 12 2a a 21 a 22 a 23 a 11 a 12 a a 31 a 32 a a 11 a 12 a a 21 2a 22 2a a 31 a 32 a 33 a 21 a 22 a a 31 2a 32 2a 333 Recall that row rank and column rank of any matrix are equal. Symbolic row/column operations show that the row rank rk(dφ) A of this matrix is 6 = 1 2 (n2 +n), so dim R O(3) = dim R SO(3) = 9 6 = 3 (see next exercise for hints on this calculation) Exercise. Verify that the row rank of the matrix above is 6, at every A O(3). Note: The row rank is equal to the number of linearly independent rows. But rk(dφ) A is always less than or equal to 6, so the rank must in fact be 6 (constant) in a neighborhood of every A O(3). Hint: The original rows and columns of A are independent so we can transform A I 3 3 by column operations. If we scale the first row of (dφ) A by 1 the matrix A appears as 2 a 3 3 submatrix in the upper left corner, with zeros below it. What happens if you apply the same column operations that worked for A to the full Jacobian matrix? Etc. (In later steps you may have to use both row and column operations.) All the classical groups mentioned in the previous section can be shown to be smooth closed manifolds over F = R; we will not do that here. However, if G M(n, C) we should identify C = R 2 and M(n, C) = R 2n2 in discussing G as a real submanifold. For example SU(2) is M(2, C) but dim R SU(2) = 3, so there is no way complex coordinate charts x α : U α C n can be introduced into SU(2) (or SU(n) for that matter). The classical groups are generally viewed as smooth hypersurfaces in R n2 or R 2n2, even if they are defined as subsets of M(n, C). However, a few actually are complex manifolds. For instance, dim C GL(n, C) = n 2 and dim C SL(n, C) = n 2 1, although these groups can also be regarded as real manifolds with dim R = 2n 2 or 2n 2 2. Other examples of intrinsically complex matrix Lie groups are O(n, C) and SO(n, C) which have dim C = 1 2 (n2 n) Exercise. Prove that the matrix group G = U(2) = {A M(2, C) : A A = I} is a smooth hypersurface of dimension dim R (G) = 4 when we identify M(2, C) = C 4 = R 8 via the correspondence that sends A U(2), ( ) z11 z A = 12 with z z 21 z ij = x ij + 1y ij, (x ij, y ij R),

17 to the real 8-tuple x = (x 11,...,x 44 ; y 11,...,y 44 ) in R 8 (see next exercise for further comments). Hint: Write the identities implied by A A = I 2 2 as a system of several scalar equations in the variables x R 8, deleting any that are redundant. Define a C map φ : R 8 R k has constant rank on and near U(2). (What is the appropriate value of k in this situation?) Note: SU(2) cannot be a smooth complex manifold even though it is carved out of complex matrix space M(2, C), because dim R SU(2) = 4 1 = 3. Observe that dim R SO(3) = 3 is also the real dimension of SU(2). This coincidence is no accident because there is a two-to-one homomorphism mapping SU(2) onto the Euclidean rotation group SO(3). This double covering of SO(3) has profound implications in quantum mechanics. In the previous exercise you were asked to show that the polynomial map φ : R 8 R k that identifies M = U(2) as a (real) hypersurface of dimension d = 8 r in R 8 = M(2, C) if rk(dφ) A r (constant) at all points of M. In the next exercise you are asked to find valid choices of d coordinates x i1,...,x id from x 1,..., x 8 that parametrize M near p. As in the IFT, given a base point g M you must find a partition of coordinate indices [1, 8] = J J such that the restriction (π J M ) : M R J is an admissible coordinate chart near g Exercise. Consider the identity element g = e = I 2 2 in G = U(2). 1. Produce a partition of [ 1, 8] = J J such that the projection π J : R 8 R J restricts to SU(2) to give a standard chart defined near the identity I 2 2. This will show that dim R (M) = J = Determine all partitions [1, 8] = J J that produce admissible chart maps near the identity in SU(2). 3. Produce an explicit partition [ 1, 8] = J J with J = 3 that does not yield valid coordinates describing SU(2) near the identity element I Exercise. Perform the calculations outlined in the previous exercise for the group G = SO(3), identifying the matrix A = [a ij ] with the vector x = (x 1,, x 9 ) = (a 11,, a 13 ; ; a 31,, a 33 ) R Exercise. We have already shown that G = SL(n, R) is a hypersurface in M(n, R) with dim R (G) = n 2 1. Taking n = 3 and identifying matrices A M(3, R) with vectors x = (a 11, a 12, a 13 ; ; a 31, a 32, a 33 ) R 9, 1. Exhibit a partition [1, 9] = J J such that the projection (π J G ) : G R J = R 8 yields a standard chart on SL(3, R) near the identity element I. 2. Does every choice J = {1 i 1 < < i 8 9} of eight matrix entries (out of 9) yield a standard coordinate chart about the identity element I 3 3 in G? If not, exhibit a choice for which this fails. ( ) Repeat (1.) taking A = as the base point in SL(3, R). The need for detailed matrix calculations of this sort can sometimes be avoided by appeal to a theorem of Elie Cartan Theorem (E. Cartan). Every closed subgroup G M(n, R) can be covered with 91

18 C -related charts (x α, U α ) with values in some R d that make it into a real Lie group. No hypothesis is made here that G is a level set for some C map φ : R m R n ; and even if G happens to be a level set of this kind, there is no need to investigate its rank on G. The subgroup G could even be a closed discrete subgroup (a zero-dimensional Lie group consisting of isolated points in matrix space) such as G = Z R, or SL(n, Z) = {A M(n, R) : det(a) = +1 and all entries a ij are integers}, for which differentiability considerations are irrelevant. Translations and Automorphisms on Matrix Lie Groups. There are some obvious remarks to be made about any matrix Lie group G. If g G we can define left and right translation operators λ x, ρ x : G G, letting λ x (g) = x g and ρ x (g) = g x. These are continuous and invertible maps with the properties λ e = id G λ x y = λ x λ y λ x 1 = (λ x ) 1, and likewise for right translations, except that ρ x y = ρ y ρ x They are invertible bicontinuous maps when G is given the obvious metric it inherits from the surrounding space M(n, F). Furthermore, all these maps are diffeomorphisms: when λ x or ρ x are described in terms of chart coordinates they (and their inverses) become C maps of R d R d, where d = dim R (G). The action of an element g G by conjugation yields an inner automorphism of the group, i g (x) = gxg 1 for all x G. This is a diffeomorphisms of G G because i g = λ g ρ 1 g is a composite of C maps. (Note that λ x, ρ y commute for all x, y G.) Its inverse is the inner automorphism (i g ) 1 = i g 1, and the correspondence g i g is a homomorphism from G into the full group Aut(G) of automorphisms of G, because i e = id G and i g1g 2 = i g1 i g2 If G GL(n, R) is a matrix Lie group its C structure is obtained by identifying G M(n, R) = R n2 and then covering G with standard charts (x α, U α ) constructed via the IFT, where U α is an open subset in G and x α is a bicontinuous map from U α to the open set V α = x α (U α ) in coordinate space R d. As in the IFT, the chart maps x α are just restrictions to G of various projection maps π J : R n2 R J corresponding to some splitting R n2 = R J R J with J = d. Because of the way standard charts are defined, it is not hard to verify that a C map ψ : R k M(n, R) whose range happens to lie within the set G M(n, R), is automatically a C map ψ : R k G when G is given its standard structure as a real C manifold Exercise. Left translations λ x are obviously bijective maps on G. Prove that they are C maps for the standard C structure on G. Thus if if (x α, U α ), (y β, U β ) are standard charts about a point p G and its image λ x (g) = x p, explain why λ x becomes a C map F d F d when described in local chart coordinates i.e. why is the coordinate map y = Φ(x) = y β λ x x 1 α (x) 92

19 a C map F d F d wherever it is well-defined? Note: Since (λ x ) 1 = λ x 1 is also a translation, λ x is a C diffeomorphism of G. Likewise for right translations ρ x Lemma. Let G be a matrix Lie group in M(n, R) and f : R k M(n, R) = R n2 a C map. If G is given the standard C structure as a smooth real hypersurface and if range(ψ) G, then f : R k G is automatically a C map i.e. if x 0 R k, p 0 = ψ(x 0 ) and if (x α, U α ) is a standard chart on G about p 0, then is a C map between Euclidean spaces. x α ψ : R k R d X.3. The Exponential Map Exp : M(n, R) GL(n, R). The matrix groups above were all described as smooth d-dimensional hypersurfaces M embedded in some Euclidean space R m with m d. At each base point p M there is a well-defined d-dimensional tangent space TM p, a translate TM p = p + E p of some d-dimensional vector subspace E p R m (see Figure 10.7). It is an almost universal convention to indicate tangent vectors in TM p by capital roman letters X, Y,..., or by X p, Y p,... when it is necessary to indicate the base point. The tangent hyperplane TM p itself is not itself a vector subspace in matrix space (for one thing, it does not contain the zero element), but we may nevertheless define vector space operations on tangent vectors attached to the same base point p, and think of TM p = p + E p as a vector space, by referring everything back to E p by translation. Vector Operations in TM p. Given tangent vectors X 1, X 2 in TM p form their sum X 1 + X 2 in TM p by (50) (i) Subtracting p from each to get actual vectors v k = X k p in E p, then add these to get v 1 + v 2 in E p where (+) makes sense. (ii) Then move the result back into the tangent space to get X 1 + X 2 = p + (v 1 + v 2 ) in TM p. Scalar multiples λ X = p+(λ v) in TM p are defined similarly. Note carefully that there is no natural way to define the sum of two tangent vectors attached to different base points p q in M. Equipped with these operations the tangent hyperplane TM p becomes a vector space isomorphic to E p under the bijection v v + p, v E p. To carry out calculations with tangent vectors we shall exploit the close connection between derivatives of smooth curves in R m passing through p and tangent vectors in TM p. A parametric curve in E = R m is a continuous map γ : [a, b] R m, and if {e j } is the standard basis in R m we may describe γ by writing γ(t) = m x j (t)e j. This is a C curve if the scalar components x j (t) are smooth functions. Note: When E = C m a parametric curve takes the form γ(t) = j=1 m z j (t)e j with complex coefficients z j (t) = x j (t) + 1y j (t), j=1 93

20 Figure The tangent hyperplane TM p to a smooth d-dimensional hypersurface in Euclidean space R m is made up of vectors p + v attached to the base point p, where v lies in a certain vector subspace E p passing through the origin in R m. Vector space operations are introduced into the tangent space by parallel transport between vectors X = p+v TM p = p+e p and v E p, which actually is a vector subspace of R m. and γ(t) is a C curve if the real and imaginary parts x j, y j : [a, b] R are real-valued C functions. In this situation the vector derivative takes the form γ (t) = m j=1 dz j dt e j = m j=1 ( dx j dt + 1 dy j dt )e j 3.1. Definition. Let M be any smooth d-dimensional manifold embedded in Euclidean space R m and consider any C curve γ(t) m j=1 x j(t)e j in R m that remains in M for all t. By Lemma 2.12, γ is a C map from R into M equipped with its standard C structure, so if (x α, U α ) is a standard chart on M the map x α γ(t) is differentiable from R R d. Any C curve γ(t) with values in R m is differentiable, with vector derivative γ (t 0 ) = dγ dt (t 0) = lim t 0 γ(t 0 + t) γ(t 0 ) t = dx 1 dt e dx m dt e m (limit computed in R m ) for all a < t 0 < b. Moreover, if γ(t) remains in M for all t and passes through p when t = t 0, then the vector derivative γ (t 0 ) must lie in the hyperplane tangent to M at p. Therefore γ (t 0 ) is an element of the tangent space TM p, and in fact every tangent vector at p arises this way, as the vector derivative of a smooth curve in M passing through p. In discussing tangent vectors to smooth curves we may as well assume t 0 = 0 because we can always reparametrize γ via γ(t) γ(t) = γ(t t 0 ) to make γ(0) = p without changing the tangent vector at p. Lie Algebra of a Matrix Lie Group. If M is a smooth hypersurface in Euclidean space R m the tangent space TM p at p M acquires a vector space structure as explained in (50). But if the hypersurface is a matrix Lie group G, the tangent space TG e at the identity element acquires additional algebraic structure, induced by the algebraic operations in G itself, and becomes a Lie algebra Definition. An abstract Lie algebra over F is a vector space V over F equipped with a bracket operation B(X, Y ) = [X, Y ] that is: 94