CHAPTER 5 TANGENT VECTORS

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1 CHAPTER 5 TANGENT VECTORS In R n tangent vectors can be viewed from two ersectives (1) they cature the infinitesimal movement along a ath, the direction, and () they oerate on functions by directional derivatives. The first viewoint is more familiar as a concetual viewoint from Calculus. If a oint moves so that its osition at time t is ρ(t), then its velocity vector at ρ(0) is ρ (0), a tangent vector. Because of the concetual familiarity, we will begin with the first viewoint, although there are technical difficulties to overcome. The second interretation will be derived as a theorem. The second viewoint is easier to generalize to a manifold. For instance, oerators already form a vector sace. It is the second viewoint that ultimately lays the more imortant role. Suose M is an n-manifold. If m M, then we define a tangent vector at m as an equivalence class of aths through m. Equivalent aths will have the same derivative vector at m and so reresent a tangent vector. The set of all tangent vectors at m forms the tangent sace. The descrition and notation of tangent vectors in R n from the advanced Calculus setting and in the resent setting is discussed in Remark 5.9***. Definition 5.1***. Suose M is a manifold. A ath is a smooth ma ρ : ( ɛ, ɛ) M, where ɛ > 0. As was mentioned, if M = R n, then ρ (0) is the velocity vector at ρ(0). We also recall, from advanced Calculus, the relationshi between the derivative ma and the directional derivative, (1) Dρ(0)(1) = D 1 ρ(0) = ρ (0) Definition 5.***. Suose M is a manifold and m M A tangent vector at m is an equivalence class of aths α with α(0) = m. Let (U, φ) be a coordinate chart centered at dφ α(t) m, two aths α and β are equivalent if dφ β(t) =. Denote the equivalence class of a ath α by [α]. We can icture [α] as the velocity vector at α(0). We next observe that the equivalence class doesn t deend on the secific choice of a coordinate chart. If (W, ψ) is another coordinate neighborhood centered at m, then ψ α = ψ φ 1 φ α, and, we use formula (1), coyright c 00 dψ α(t) = D(ψ φ 1 )(φ(m)) D(φ α)(0)(1). Tyeset by AMS-TEX 1

2 CHAPTER 5 TANGENT VECTORS The diffeomorhisms ψ φ 1 and φ α are mas between neighborhoods in real vector saces, so dφ α(t) = dφ β(t) if and only if dψ α(t) = dψ β(t). Therefore the notion of tangent vector is indeendent of the coordinate neighborhood. If ρ : ( ɛ, ɛ) M is a ath in M with ρ(0) = m, then [ρ] is a tangent vector to M at m and is reresented by the ath ρ. Consistent with the notation for R n, we can denote [ρ] by ρ (0). Let T M m denote the set of tangent vectors to M at m. Other common notations are M m and T m M. Theorem 5.3***. Suose M, N, and R are manifolds. (1) If φ : M N is a smooth ma between manifolds and m M then there is an induced ma φ m : T M m T N φ(m). () If ψ : N R is another smooth ma between manifolds then (ψ φ) m = ψ φ(m) φ m. This formula is called the chain rule. (3) If φ : M M is the identity then φ m : T M m T M m is the identity. If φ : M N is a diffeomorhism and m M then φ m is 1-1 and onto. (4) T M m is a vector sace of dimension n, the dimension of M, and the induced mas are linear. The induced ma φ m is defined by φ m ([α]) = [φ α]. Notice that if M = R m, N = R n, then we have a natural way to identify the tangent sace and the ma φ. We have coordinates on the tangent sace so that [φ α] = dφ α(t) and φ m ([α]) = Dφ(m)(α (0)). The induced ma φ m is also commonly denoted T φ or dφ. These results follow for neighborhoods in manifolds since these are manifolds too. Also note that if there is a neighborhood U of m M and φ U is a diffeomorhism onto a neighborhood of φ(m) then φ m is an isomorhism. Proof. (1) If φ : M N is a smooth ma and m M then there is an induced ma φ m : T M m T N φ(m) defined by φ m ([α]) = [φ α]. We need to show this ma is

3 CHAPTER 5 TANGENT VECTORS 3 well-defined. Take charts (U, θ) on N centered on φ(m) and (W, ψ) on M centered on m. If [α] = [β], then dψ α(t) dψ β(t) = (θ φ ψ 1 dψ α(t) ) ( ) = (θ φ ψ 1 dψ β(t) ) ( ) d (θ φ ψ 1 ψ α)(t) = d (θ φ ψ 1 ψ β)(t) d (θ φ α)(t) = d (θ φ β)(t) so φ m is well defined on equivalence classes. () If φ : M N and ψ : N R are smooth mas, then (ψ φ) m ([α]) = [ψ φ α] = ψ φ(m) ([φ α]) = ψ φ(m) φ m ([α]). (3) I M m ([α]) = [I M α] = [α]. If φ φ 1 = I M then φ (φ 1 ) = I M = I T Mm. Also, if φ 1 φ = I M, then (φ 1 ) φ = I T Mm. Therefore φ is a bijection and (φ ) 1 = (φ 1 ). (4) Let (U, φ) be a coordinate neighborhood centered at m. We first show that T R n 0 is an n-dimensional vector sace. Since R n requires no coordinate neighborhood (i.e., it is itself), [α] is equivalent to [β] if and only if α (0) = β (0): two aths are equivalent if they have the same derivative vector in R n. Every vector v is realized by a ath α v, α v (t) = tv. This identification gives T R n 0 the vector sace structure. We show that the linear structure is well defined on T M m. The linear structure on T M m is induced by the structure on T R n 0 (where [α]+k[β] = [α +kβ] and induced mas are linear) via the coordinate mas. If (V, ψ) is another chart centered at m, then the structure defined by ψ and φ agree since (φ ψ 1 ) is an isomorhism and (φ ψ 1 ) ψ = φ. We can give exlicit reresentatives for linear combinations of aths in the tangent sace T M m. In the notation of the roof of Theorem 5.3*** art 4, k[α] + c[β] = [φ 1 (kφ α + cφ β)] Note that the coordinate chart serves to move the aths into R n where addition and multilication makes sense. Before we turn to the second interretation of a tangent vector as a directional derivative, we ause for a hilosohical comment. We first learn of functions in our grade school education. We learn to seak of the function as a whole or its value at articular oints. Nevertheless, the derivative at a oint does not deend on the whole function nor is it determined by the value at a single oint. The derivative requires some oen set about a oint but any oen set will do. If M is a manifold and m M, then let G m be the set of functions defined on some oen neighborhood of m.

4 4 CHAPTER 5 TANGENT VECTORS Definition 5.6***. A function l : G m R is called a derivation if for every f, g G m and a, b R, (1) l(af + bg) = al(f) + bl(g) and () l(fg) = l(f)g(m) + f(m)l(g) Denote the sace of derivations by D. The roduct rule which occurs in the definition is called the Leibniz rule, just as it is in Calculus. Proosition 5.7***. Elements of T M m oerate as derivations on G m. In fact there is a linear ma l : T M m D given by v l v. The theorem is straightforward if the manifold is R n. If v T R n x, then the derivation l v is the directional derivative in the direction v, i.e., l v (f) = Df(x)(v). On a manifold the argument is really the same, but more technical as the directions are more difficult to reresent. We will see in Theorem 5.8*** that the derivations are exactly the directional derivatives. df α(t) Proof. If α : (( ɛ, ɛ), {0}) (M, {m}) reresents v then define l v (f) =. The fact that l v is a linear functional and the Leibniz rule follow from these roerties of the derivative. To show that l is a linear ma requires calculation. Suose (U, φ) is a coordinate chart centered at m. If [α] and [β] are equivalence classes that reresent tangent vectors in T M m and c, k R, then φ 1 ((kφα(t) + cφβ(t))) reresents k[α] + c[β]. Hence, l k[α]+c[β] (f) = df(φ 1 ((kφα(t) + cφβ(t)))) ( ) d(kφα(t) + cφβ(t)) = f φ 1 ( = f φ 1 k d(φα(t) + c d(φβ(t)) ) ( ) ( ) dφα(t) = kf φ 1 dφβ(t) + cf φ 1 = k df(φ 1 (φ(α(t)))) + c df(φ 1 (φ(β(t)))) = k df((α(t))) + c df((β(t))) = kl [α] (f) + cl [β] (f) Lines 3 and 4 resectively follow from the linearity of the derivative and the total derivative ma. Therefore l is linear. The second interretation of tangent vectors is given in the following Theorem.

5 CHAPTER 5 TANGENT VECTORS 5 Theorem 5.8***. The linear ma l : T M m D given by v l v is an isomorhism. The elements of T M m are the derivations on G m. Proof. We first note two roerties on derivations. (1) If f(m) = g(m) = 0, then l(fg) = 0 Since l(fg) = f(m)l(g) + g(m)l(f) = () If k is a constant, then l(k) = 0 Since l(k) = kl(1) = k(l(1) + l(1)) = kl(1), l(k) = l(k) and l(k) = 0. We now observe that l is one-to-one. Let (U, φ) be a coordinate chart centered at m. Suose v 0 is a tangent vector. We will show that l v 0. Let φ (v) = w 1 R n. Note that w 1 0. Then [φ 1 (tw 1 )] = v where t is the real variable. Let w 1,, w n be a basis for R n and π( n i=1 a iw i ) = a 1. Then l v (π φ) = l [φ 1 (tw 1 )](π φ) = dπ(φ(φ 1 (tw 1 ))) = w 1 = w 1. Next we argue that l is onto. Let (U, φ) be a coordinate chart centered at m and let e i for i = 1,, n be the standard basis for R n. We consider the ath t φ 1 (te i ) and comute some useful values of l, i.e., the artial derivatives. l [φ 1 (te i )](f) = dfφ 1 (te i ) = fφ 1 Let x i (a 1,, a n ) = a i. Suose d is any derivation. We will need to name certain values. Let d(x i φ) = a i. These are just fixed numbers. Suose f is C on a neighborhood of m. Taylor s Theorem says that for in a neighborhood of 0 R n, n f φ 1 () = f φ 1 f φ 1 n ( 0) + x i=1 i x i () + R ij ()x i ()x j () 0 i,j=1 where R ij () = 1 0 (t 1) f φ 1 t x j are C functions. So, n f φ 1 n f = f(m) + x i φ + (R ij φ) (x i φ) (x j φ). 0 i=1 We now aly d. By (), d(f(m)) = 0. Since x j φ(m) = 0, the terms d((r ij φ) (x i φ) i,j=1 (x j φ)) = 0 by (1). Also, d( f φ 1 x i φ) = a i l [φ 1 (te i )](f). Hence, d = l P n 0 and l is onto. 0 i=1 a i[φ 1 (te i )],

6 6 CHAPTER 5 TANGENT VECTORS Remark 5.9***. Tangent vectors to oints in R n. The usual coordinates on R n give rise to standard coordinates on T R n. Let e i = (0,, 0, 1, 0,, 0) with the only nonzero entry in the i-th sot. The ath in R n defined by α i (t) = te i + is a ath with α i (0) =. Its equivalence class [α i ] is a vector in T R n and we denote it. In Advanced Calculus, the ordered basis x 1,, x n is the usual basis in which the Jacobian matrix is usually written and sets u a natural isomorhism T R n = R n. The reader should notice that the isomorhism is only natural because R n has a natual basis and is not just an abstract n-dimensional vector sace. If ρ is a ath in R n, then ρ (0) T ρ(0) R n via this isomorhism. This notation is also consistant with the oerator notation (the second interretation) since, In the first line, the tangent vector function f. (f) = [f α i ] = d f(te i + ) = f R n = T R n x= oerates via the second interretation on the Examle 5.10***. T M x for M an n-dimensional submanifold of R k. Suose M R k is a submanifold and i : M R k is the inclusion. Take (U x, φ) a slice coordinate neighborhood system for R k centered at x as secified in the definition of a submanifold, Definition 3.***, φ : U x U 1 U. Under the natural coordinates of T R k x = R k, T M x = φ(u 1 {0}) R k and i x has rank n. To see these facts, note that φ i (φ Ux M) 1 : U 1 {0} U 1 U is the inclusion. So, rank(i ) = rank((φ i φ Ux M) ) = n. Under the identification T R k x = R k, φ x (R n {0}) = Dφ(x)(R n {0}) R k. This is the usual icture of the tangent sace as a subsace of R k (i.e., shifted to the origin) that is taught in advanced Calculus. Examle 5.11***. T S n x for Sn R n+1, the n-shere. n+1 This is a secial case of Examle 5.10***. Suose (x 1,, x n+1 ) S n, i.e., x i = 1. One of the x i must be nonzero, we assume that x n+1 > 0. The other cases are analogous. The inclusion from the Imlicit Function Theorem is φ R n(x 1,, x n ) = (x 1,, x n, 1 n i=1 x i ) so Dφ R n(x 1,, x n )(v 1,, v n ) = (v 1,, v n, n i=1 x iv i 1 n i=1 x i )). i=1

7 Since x n+1 > 0, x n+1 = 1 n i=1 x i CHAPTER 5 TANGENT VECTORS 7 T (x1,,x n+1 )S n = {(v 1,, v n, ) and the tangent sace is n i=1 x iv i x n+1 ) v i R} n+1 = {(w 1,, w n+1 ) w i x i = 0} Examle 5.1***. Recall that O(n) Mat n n = R n is a submanifold of dimension n(n 1) which was shown in Examle 3.7***. Then, we claim, if and only if XA 1 is skew. i=1 X T A O(n) Mat n n This comutation is a continuation of Examle 3.7***. Suose A O(n). Since O(n) = f 1 (I), T A O(n) Ker(Df(A)). The dimension of the kernel and the dimension of T A O(n) are both n(n 1). Therefore T A O(n) = Ker(Df(A)). It is enough to show that Ker(Df(A)) {X XA 1 is skew} since the dimension of {X XA 1 is skew} is the dimension of Skew n n = n(n 1) (from Examle.8d***). So it is enough to show that XA 1 is skew. Again, from Examle 3.7***, Df(A)(X) = AX T + XA T. AX T = XA T. Since A O(n), A 1 = A T. So, Therefore XA 1 is skew. (XA 1 ) T = (XA T ) T = AX T = XA T = XA 1 If Df(A)(X) = 0, then Examle 5.13***. Recall that S(n, R) Mat n n = R n is a submanifold of dimension n(n+1) which was shown in Examle 3.9***. Then, we claim, if and only if JXA 1 is symmetric. X T A S(n, R) Mat n n This comutation is a continuation of Examle 3.9***. Suose A S(n, R). Since S(n, R) = f 1 (J), T A S(n, R) Ker(Df(A)). The dimension of the kernel and the dimension of T A S(n, R) are both n(n+1). Therefore T A S(n, R) = Ker(Df(A)). It is enough to show that Ker(Df(A)) {X JXA 1 is symmetric} since the dimension of {X JXA 1 is symmetric } is the dimension of Sym n n = n(n+1) (from Examle.8c***). So it is enough to show that JXA 1 is symmetric. Again, from Examle 3.9***, Df(A)(X) = AJX T + XJA T. If Df(A)(X) = 0, then AJX T = XJA T. Since A S(n, R), A 1 = JA T J T. So, (JXA 1 ) T = (JXJA T J T ) T = ( JAJX T J T ) T = JXJ T A T J T Therefore XA 1 is symmetric. = JXJA T J T as J T = J by Lemma 3.8*** = JXA 1

8 8 CHAPTER 5 TANGENT VECTORS Remark 5.14***. Notation for Tangent vectors The sace R n comes equied with a canonical basis e 1,, e n which allows us to ick a canonical basis for T R n x. For an n-manifold M, T M doesn t have a natural basis. We can give coordinates on T M in terms of a chart. Suose that (U, φ) is a chart for a neighborhood of U M. Write φ = (φ 1,, φ n ) in terms of the coordinates on R n. Hence, φ i = x i φ. We can imort the coordinates T R n φ(). Let φ i = φ 1 ( ) φ() As a ath φ i is the equivalence class of φ 1 (te i + φ()). As an oerator, φ i (f) = f φ 1. φ()

9 CHAPTER 5 TANGENT VECTORS 9 Exercise 1***. Suose F : R 4 R by Exercises F ((w, x, y, z)) = (wxyz, x y ). Comute F and be exlicit in exhibiting the bases in the notation used in Remark 5.9***. Where is F singular? The reader may wish to review Examle.10** and Exercise 4*** from chater 3 for the following exercise. Exercise ***. Let g((x, y)) = x + y and h((x, y)) = x 3 + y. Denote by G g and G h the grahs of g and h which are submanifolds of R 3. Let F : G g G h by F : ((x, y, z)) = (x 3, xyz, x 9 + x y). The reader may wish to review Examle.10** and Exercise *** from chater 3. a. Exlicitly comute the derivative F and be clear with your notation and bases. b. Find the oints of G g where F is singular. What is the rank of F for the various singular oints G g. Exercise 3***. Let F : R 3 S 3 be defined by F ((θ, φ, η)) = (sin η sin φ cos θ, sin η sin φ sin θ, sin η cos φ, cos η). Use the charts from stereograhic rojection to comute F in terms of the bases discussed in Remark 5.9*** and Remark 5.14***.