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NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY
AARON LANDESMAN
2 AARON LANDESMAN
4 AARON LANDESMAN
- INTRODUCTION Hiro Tanaka taught a course (Math 230a) on Differential Geometry at Harvard in Fall 2015. These are my “live-TEXed“ notes from the course. Conventions are as follows: Each lecture gets its own “chapter,” and appears in the table of contents with the date. Of course, these notes are not a faithful representation of the course, either in the mathematics itself or in the quotes, jokes, and philosophical musings; in particular, the errors are my fault. By the same token, any virtues in the notes are to be credited to the lecturer and not the scribe. 1 Please email corrections to [email protected].
(^1) This introduction has been adapted from Akhil Matthew’s introduction to his notes, with his permission.
NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY 5
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2.1. Logistics.
(1) Phil Tynan is the TF, who isn’t here (2) email: [email protected] (3) Hiro’s office is 341, office hours are Tuesday 1:30 - 2:30pm, and Wednesday 2-3pm. (4) Phil will have office hours 2-3pm on Thursdays, in office 536 and 532. (5) There will be homeworks, once a week, the first homework is due Sept 17. (6) When homework is graded, we will get a remark from Phil to see Hiro and Phil during office hours. You will not be numerically graded from week to week, but you have to come to them in person, so that we know what is going on. (7) There will be no midterm, but one take home final.
Remark 2.1. There are two words in the title of the course, Differential and Ge- ometry. This is not Riemannian geometry and we’ll discuss the difference later. “Differential” connotates calculus. You can ask how to do calculus on shapes likes triangles and cubes. To understand calculus, we will learn about manifolds, and calculus on manifolds. To understand geometry, we will think of a space together with some structure (possibly some type of metric).
Example 2.2. (1) Riemannian geometry (2) Symplectic geometry - use things like Hamiltonian to describe how vector spaces evolve. (3) Complex geometry - generalize complex analysis to shapes you can build with C n^ or CW complex. (4) Kahler geometry (5) Calabi-Yau geometry - study supersymmetric string theory
2.2. Lecture begins. Consider a curve γ : R → R n, t 7 → γ(t).
Definition 2.3. For γ a curve, we define the length of γ to be ∫
R
|γ′(t)| dt,
where
|γ′(t)| =
i
γ i′(t)^2 =
〈γ′, γ′〉
.
Remark 2.4. The inner product from Definition 2.3 should really be thought of as an inner product on Tγ(t) R n^ and not on R n. Even though these objects are isomorphic, they should not be thought of as “the same.”
Definition 2.5. Let U ⊂ R n^ be an open set. A function f : U → R m^ is called
- C^0 if it is continuous
- C^1 if it has partial derivatives (^) ∂x∂fi for i = 1 ,... , n which are all C^0
- Cr^ if it has all partial derivatives of order at most r which are all C^0
- C∞^ or smooth if f is Cr^ for all r.
NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY 7
Remark 2.16. Caution: Locally Euclidean does not imply Hausdorff. As a coun- terexample, consider the affine line with a doubled origin.
Definition 2.17. A topological space X is second countable if X admits a countable basis of open sets.
Definition 2.18. A basis for a topological space X is a collection of subsets Vα so that
(1) x = ∪αVα (2) For every α, β, one can cover Vα ∩ Vβ = ∪γVγ
Warning 2.19. The above definition 2.18 determines a topology, where the open sets are given by arbitrary unions of elements in the basis. However, if we are given a topology on X to start with, we will also need to require that every open set U ⊂ X can be written as a union of basis elements.
Example 2.20. Euclidean space ( R n) is second countable. To see this, take a count- able basis given by balls around all rational points with rational radii. Any sub- space of a second countable space is also second countable, by restricting the basis.
Remark 2.21. If X is a topological manifold, every connected component is of X will be a locally Euclidean, Hausdorff, second countable space. So, one can define a topological manifold to be something satisfying these three properties.
Definition 2.22. An open cover Uα is locally finite if for every x ∈ X, there exists W ⊂ X an open subset containing x such that W ∩ Uα 6 = 0 for only finitely many α.
Definition 2.23. A space X is paracompact if every open cover admits a locally finite refinement.
Definition 2.24. A topological manifold is a space X so that X is
(1) locally Euclidean (2) Hausdorff (3) Paracompact Paracompact allows you to turn local functions to global ones.
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Exercise 3.1. Let γ : R → R n^ be an immersion. Show there exists a diffeomor-
phism φ : R → R such that γ ◦ φ is parameterized by arc length, i.e., | d(γ dt◦φ )| = 1.
Remark 3.2. If you’re given a smooth curve in R n, we have an intuitive idea of what it means, but we can choose various parameterizations. We can choose a parameterization by arc length so that the amount of time traveled is the amount of distance traveled. This exercise looks a lot like a differential equation, which can be solved by the fundamental theorem of calculus.
Solution to exercise: take φ to be
∫s 0 |^
dγ dt
− 1 dt By the chain rule, d ds γ ◦ φ =
dγ dt
dφ ds
dγ dt
dγ dt
− 1 (3.1)
we use the fundamental theorem of calculus is employed to calculate the deriv- ative of φ.
8 AARON LANDESMAN
3.1. Curvature of curves.
Definition 3.3. Let γ : R → R n^ be an immersion. Define
~T : R → R nt 7 → γ˙ | γ˙|
The curvature vector at γ(t) is defined to be
~κ :=
d~T ds
d~T /dt ds/dt
Exercise 3.4. (1) Show ~κ ⊥ ~T. (2) If γ : R → R^2 has image a circle of radius R, show |~κ| = (^1) R. (3) If φ : R → R is a diffeomorphism, then the value of ~κ(γ(t)) = ~κ(γ ◦ φ(s)). Solution to exercise: (1) Consider the function t 7 → 〈~T (t), ~T (t)〉. This is a constant function. The de- rivative (^) dtd 〈~T (t), ~T (t)〉 = 〈 (^) dtd ~T (t), ~T (t)〉 + 〈~T (t), (^) dtd ~T (t)〉 = 2 〈 (^) dtd ~T (t), ~T (t)〉. (2) Choose γ : R → R^2 , t 7 → R · (cos t, sin t) So, ~T = (− sin t, cos t). Then |d~T /dt ds/dt
ds/dt
R
because we the circle is parameterized by t between 0 and 2π while the length of the circle is 2πR. (3) We use the chain rule. We write the circle in two ways.
Consider a hyperboloid in R^3. Say we want to know the curvature of the sur- face at x. We can define a normal vector to a tangent plane at a point. Given two vectors, a normal vector and a point in the plane, we can intersect the plane with a surface and obtain a curve. Given this curve, we know how to compute the curva- ture. Then, there are two principal vectors in the tangent space, one with minimal curvature and one with maximal curvature. The Gaussian curvature is then the product of the maximal and minimal curvature. This turns out to be independent of the embedding of the surface.
Remark 3.5. Curvature |κ(γ(t))| is the inverse radius of the best approximating circle at γ(t).
3.2. Manifolds. Recall the following definitions from the previous class:
Definition 3.6. An open cover Uα is locally finite if for every x ∈ X, there exists W ⊂ X an open subset containing x such that W ∩ Uα 6 = 0 for only finitely many α.
Example 3.7. Say Uα = Bα( 0 ), α ∈ Q , where Bα( 0 ) is a ball about the origin of radius α. Then, {Uα}α∈ Q is not a locally finite cover about 0. Similarly, if we only index over the integers, it is still not locally finite.
Definition 3.8. A space X is paracompact if every open cover admits a locally finite refinement, where a refinement is another cover so that each element of the new cover is contained in some element of the original cover.
Definition 3.9. A topological manifold is a space X so that X is
(1) locally Euclidean (2) Hausdorff
10 AARON LANDESMAN
(2) We have
φ∈Φ φ(x) =^1 , which makes sense as the sum is a finite sum, by the previous point. (3) For each φ ∈ Φ, there exists α so that. we have Supp(φ) ⊂ Uα.
Theorem 3.22. Given any set A ⊂ R n, and any open cover Uα, a partition of unity on A subordinate to Uα exists.
Proof. We prove this by breaking successively tackling more and more compli- cated types of sets A.
3.4. A is compact.
Lemma 3.23. For any open ball B(x, r) there exists a smaller open ball B(x, s) ⊂ B(x, r) and a smooth φ with φ|B(x,s) = 1 and φ| R n\B(x,r) = 0.
Proof. We can replace B(x, r) and B(x, s) by cubes S =
i(ai,^ bi)^ ⊂^ R^ =^
i(ci,^ di) by choosing s so that B(x, s) ⊂ S ⊂ R ⊂ B(x, r). So, it suffices to prove the theorem for cubes. Now, we have already shown this on problem set 5, problem 4c in the case n = 1. Let fi : R → R be a function which is 1 on on (ai, bi) and 0 outside of (ci, di). Then, f(x 1 ,... , xn) =
i fi(xi)^ is the desired function.^
In this case, for each x ∈ X, choose Bx to be an open ball so that there is some Uα with Bx ⊂ Uα, and choose Cx to be a smaller open ball so that x ∈ Cx ⊂ Bx, so that there exists a function which takes the value 1 in Cx and 0 outside of Bx. Then, take a finite cover of A by such balls Cx. call the associated functions ψi, with 1 ≤ i ≤ n. Define
φk =
ψk ∑n i= 1 φn
Observe that
∑^ k
i= 1
φi = 1.
This shows the φi sum to 1 everywhere. Additionally, each φi has support con- tained in the same Uα that ψi does.
3.5. A = ∪iAi with Ai compact and Ai ⊂ int(Ai+ 1 ). t Take our given open cover Uα of A. Construct Uiα an open cover of Bi = int(Ai+ 1 ) \ Ai− 2 , by defining Uiα = Uα ∩ Bi. Define Ci = Ai \
(Ai− 1 ). Then, Ci ⊂ Bi. Therefore, we can construct a partition of unity subordinate of Ci subordinate to Uiα, Let the partition of unity be denoted Φi. Define
σ(x) =
i∈ N ,φ∈φi
ψ(x).
Define φ(x) = ψ(x)/σ(x). Note that σ 6 = 0 on some open set containing A, since at each x ∈ A, some φ are strictly positive at x. Say x ∈ Ai, x ∈/ Ai− 1. Therefore, on the domain where σ 6 = 0 , we obtain there are only finitely many φ with φ(x) 6 = 0 , since we must have φ ∈ Φk for k ≤ i + 2 , and there are only finitely many such functions in each Φk. Additionally, the φi sum to 1 by construction, because we divided by their sum, σ.
NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY 11
3.6. A is open. Construct
Ai = {x ∈ A|d(x, ∂A) ≥
i
, |x| ≤ i}.
Observe this give a cover of A by sets as in the previous case.
3.7. A general. Say our open cover of A is Uα. Then, choose B = ∪αUα. Note that there is a partition of unity for B, which is also a partition of unity for A.
4.1. Partitions of Unity, Hiro’s version.
Exercise 4.1. (1) Consider j : R^2 → R^3 , (x, y) 7 → (x, cos y, sin y). Compute j∗gstd. (2) The arc length parameterization proof from last lecture is incorrect (some- thing about the chain rule being incorrectly applied) Why? Solution: (1) Recall j∗gstd : R^2 → M 2 × 2 ( R ). Note the image j( R^2 ) is a cylinder. To
compute the pullback of the inner product is given by dj(x,y) =
0 − sin y 0 cos y
Then, we compute g 11 = 1 , g 12 = 0 , g 22 = 1 , so it is the standard metric. We can see this also by computing j∗gstd = djt^ · dj (2) Look at the errata. To correctly parameterize curves, given γ : R → R n, consider the map ` : R → R , t 7 →
∫t 0 |^ γ˙|dt. Since^ γ^ is an immersion,^ γ^ has an inverse, so we see γ = γ ◦ `−^1 and we can find the derivative of γ.
Remark 4.2. From now on, we write X for a smooth manifold, but remember this also comes with the datum of an atlas A.
Remark 4.3. In the previous day, I added some notes I had written for a previ- ous class on partitions of unity. Here, we repeat the same thing, but with Hiro’s notation.
Definition 4.4. Let X be a smooth manifold. Fix an open cover U = {Uα}. A parti- tion of unity subordinate to U is a collection of smooth functions
fβ : X → R ≥ 0
so that
(1)
β∈B fβ(x) =^1 (2) For all β, Supp(fβ) =
x : fβ(x) 6 = 0
⊂ Uβ (3)
Supp(fβ) is locally finite
That is, for every x there is an open x ∈ W so that W ∩ Supp(fβ) 6 = ∅ for only finitely many β ∈ B.
Theorem 4.5. (Existence of partitions of unity) Let X be a C∞^ manifold. Then for all open covers U =
Uβ
, there exists a C∞^ partition of unity subordinate to U.
Remark 4.6. This is the way we’ll prove that any manifold admits a Riemannian metric, and many other foundational results. It will let us patch things on R n together.
NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY 13
Remark 4.10. It is possible Hiro came up with this proof but the inspiration came from Collins’ textbook which mentioned that Lemma 4.9 as crucial.
4.2. Submersions. We will treat the submersion theorem just inside R n. The prin- cipal for why we can do this is that anything you can do in R n, you can do for manifolds in general by piecing together open sets.
Definition 4.11. Let f : U → V be a smooth map. Let U ⊂ R n, V ⊂ R m^ open. Then, f is called a submersion at x ∈ U if dfx : TxU → Tf(x)V is a surjection.
Remark 4.12. (1) For f to be a submersion, n ≥ m. (2) if U → V is an inclusion of open sets with m = n, then f is a submersion. (3) f : (x 1 ,... , xn) 7 → (x 1 ,... , xm) is a submersion, because dfx is (Im×m 0 ).
Definition 4.13. f is a submersion if f is a submersion at all x ∈ U.
Theorem 4.14. (Submersion Theorem) Let f : U → V be a submersion. Then, for all y ∈ V, f−^1 (y) ⊂ U is a smooth submanifold.
Remark 4.15. This theorem will readily generalize to arbitrary manifolds, once we define the relevant terms.
The following definition was stated in class, but isn’t relevant to the submersion theorem
Definition 4.16. A continuous map f : X → Y between topological spaces is proper if for all K ⊂ Y compact, f−^1 (K) is compact.
Remark 4.17. The dimension of f−^1 (y) will be n − m if n = dim U, m = dim V.
Definition 4.18. A subset X ⊂ U ⊂ R n^ with U open is a smooth submanifold of U if for all x ∈ X there exists an open W ⊂ U and a smooth diffeomorphism φ : R n^ → W so that φ( R i) = X ∩ W with R i^ ⊂ R n^ some sub vector space.
Remark 4.19. A smooth submanifold of U is a smooth manifold.
Example 4.20. If f : R n^ → R , ~x 7 → |x|^2.
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5.1. Tangent Spaces.
Remark 5.1. A tangent vector gives me a way to take derivatives. Say we have U ⊂ R n, f : U → R. The derivative is a row vector with n entries. More ge- ometrically, we can discuss the derivative as follows: Given X ∈ TxU, we know how to compute the directional derivative of f at x in the direction of X, using X(f), Xx(f), X(x)(f), X(f)(x) when X is a vector field.
Question 5.2. What algebraic properties does Xx : C∞(U) → R satisfy?
Definition 5.3. Given a manifold X, we let C∞(X), C∞(X; R ) denote the set of smooth functions X → R.
What properties do tangent vectors satisfy?
14 AARON LANDESMAN
(1) Xx(af + g) = aXx(f) + Xx(g) for a ∈ R , f, g ∈ C∞(M) (2) Leibniz rule, Xx(fg) = Xx(f) · e(g) + e(f) · Xx(g).
Definition 5.4. Let A, B be commutative algebras over R. Fix an R algebra homo- morphism e : A → B. A derivation is a function D : A → B satisfying linearity and the Leibniz rule.
Example 5.5. (1) Take A = C∞(M), B = R , and e = evx : C∞(M) → R , f 7 → f(x). (2) A = C∞(M), B = A, e = id. (3) A = C∞(M), B = C∞(N), j : N → M, e : A → B, f 7 → f ◦ j.
Remark 5.6. In algebraic geometry, given a map of manifolds, we get a map of rings, and this operation similarly encodes the relative geometry of the rings.
Definition 5.7. Let M be a smooth manifold. Then, the tangent space of M at x ∈ M is denoted
(5.1) TxM := {D : C∞(M) → R derivations with respect to evx}.
We should verify things like (1) T 0 R n^ =∼ R n^ as vector spaces (2) Chain rule
Proposition 5.8. Let x ∈ U ⊂ M. Then, if f|U = gU with f, g ∈ C∞(M), then Xx ∈ TxM implies Xx(f) = Xx(g).
Proof. Choose some compact ball B with int(B) 3 x so that B ⊂ U. Fix h : M → R so that h|B = 1 and Supp h ⊂ U. Given a derivation Xx consider Xx(h · (f − g)) =
- By the Leibniz rule, we see
0 = Xx(h · (f − g)) = Xx(h) · (f − g)(x) + h(x) · Xx(f − g) = h(x) · Xx(f − g) = Xx(f) − Xx(g).
Proposition 5.9. Let j : N → M. Then, there exists a R linear map notated by any of dj|x, djx = dj(x) with x ∈ M TxN → Tj(x)M defined by
Xx 7 → (f 7 → Xx(f ◦ j))
Proof. Exercise
Proposition 5.10. Let N j −→ M h −→ L be C∞^ functions. Then the chain rule holds. That is,
d(h ◦ j)x = dhj(x) ◦ djx
Proof. Given f ∈ C∞(L), we have
d(h ◦ j)x(Xx)(f) = Xx(f ◦ (h ◦ j)) = Xx((f ◦ h) ◦ j) = djx(Xx)(f ◦ h) = dhj(x)djx(Xx)(f)
16 AARON LANDESMAN
Remark 5.17. For all y ∈ R n, there is a smooth diffeomorphism T y : R n^ → R n, x 7 → x + y. Then,
∂ ∂xi^
|y = d T y
∂xi^
Exercise 5.18. By the chain rule, and diffeomorphism j : M → N induces linear isomorphism dxj : TxM =∼ Tj(x)N.
5.2. Return to the submersion theorem. Recall:
Definition 5.19. Let F : M → N be smooth. A point y ∈ N is a regular value of f if for all x ∈ f−^1 (y), dfx is a surjection.
Example 5.20. f : R → R , t 7 → t^2 is regular whenever y ∈ R is nonzero.
Definition 5.21. A subset Z ⊂ M is called a smooth submanifold if for all z ∈ Z there is U ⊂ M open and Z ⊂ U and a smooth diffeomorphism h : V → U so that h( R m) = U ∩ Z, with R m^ ⊂ R n.
Theorem 5.22. Let M, N be smooth manifolds and f : M → N be smooth. Then, for all regular values y ∈ N, we have f−^1 (y) ⊂ M is a C∞^ submanifold.
Proof. Go to local charts M f −→ N ⊃ V 3 y. Then,
U V
φ(U) φ(V).
We now ask what f looks like in these coordinate charts. By definition of smooth- ness, ψ ◦ f ◦ φ−^1 : U → V is smooth. Since y is a regular value, d(ψ ◦ f ◦ φ−^1 )φ(x)
with x ∈ f−^1 (y) is a surjection. So, Tφ(x)U → Tφ(y)V is surjection. Without loss of generality, assume φ(x) = 0 ⊂ R m^ and φ(y) = 0 ∈ R n. By linear algebra, there is an invertible matrix A : R n^ → R n^ so that A ◦ d(ψ ◦ f ◦ φ−^1 = (In 0 ). So, the C∞^ function A ◦ ψ ◦ f ◦ φ−^1 : R m^ → R n^ has the derivative (In 0 ) at 0. Now, define a function and expand the function so that the derivative matrix is the identity matrix by the inverse function theorem, and make it the matrix mapping the coordinate matrix to a hyperplane.
6.1. Completing the submersion theorem. Hiro was up late last night, so he might be a little less active and a little more sarcastic or dismissive, but he said he’ll try not to be. The homework is due, emailed to Phil by 11:59pm tonight. Recall: Last time we defined TxM, tangent spaces, and started proving the sub- mersion theorem:
Theorem 6.1. If f : X → Y smooth and y ∈ Y is a regular value, then f−^1 (y) ⊂ X is a smooth submanifold.
NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY 17
Proof. As in Guilliman and Pollack, find coordinate charts U ⊂ X, B ⊂ Y so that y ∈ V and x ∈ f−^1 (y), x ∈ U so that
U V
φ(U) ψ(V)
with φ(U) ⊂ R n, φ(V) ⊂ R m. We can take the composite ψ ◦ f ◦ φ−^1 , where ψ ◦ f ◦ φ−^1 : (x 1 ,... , xn) 7 → (x 1 ,... , xm), where we are viewing R m^ ⊂ R n. Assuming without loss of generality ˜f = ψ ◦ f ◦ φ−^1 ( 0 ) = 0 , then ˜f−^1 ( 0 ) = ( 0 ,... , 0 .xm+ 1 ,... , xn). This finishes the proof because f−^1 (y) ∩ U = φ−^1 ( R n−m).
6.2. Lie Brackets.
Exercise 6.2. If f : R n^ → R , x 7 → |x|^2 , then f−^1 ( 1 ) = Sn−^1 is a smooth submanifold of R n, hence a C∞^ manifold.
Recall Xx : C∞(M) → R then Xx is a derivation with evx : C∞(M) → R. Let’s examine:
Definition 6.3. Define
Γ (TM) := { R linear derivations from C∞(M) to itself with respect to e = id}
= {X : C∞(M) → C∞(M) : X(af + g) = aX(f) + X(g), X(f · g) = X(f) · g + fX(g)}
Definition 6.4. An element X ∈ Γ (TM) is a vector field on M.
Remark 6.5. For every x ∈ M, we have a function Γ (TM) → TxM, X 7 → (Xx : C∞(M) → R , Xx := ev ◦ X. Then, Xx is a derivation because evx is a ring homo- morphism.
Remark 6.6. Geometrically, any vector field X in the of multivariable calculus gives a derivation C∞(M) → C∞(M) as follows: for all x ∈ M, consider the directional derivative of f in the direction of Xx. This gives me a new function X(f)(x) = Xx(f), the directional derivative.
Remark 6.7. Since any X : C∞(M) → C∞(M), we can try composing vector fields.
Proposition 6.8. Let X, Y ∈ Γ (TM) be vector fields. Define X ◦ Y − Y ◦ X := [X, Y]. Then,
(0) [•, • ] : Γ (TM) × Γ (TM) → Γ (TM). (1) [•, • ] is R bilinear. (2) [X, Y] = − [Y, X] (3) [•, • ] satisfies the Jacobi identity: [X, [Y, Z]] = [[X, Y] , Z] + [Y, [X, Z]]. That is, for every X ∈ Γ (TM), the operation Dx = [X, • ] is a derivation on [•, • ]. That is, Dx [Y, Z] = [DxY, Z] + [Y, DxZ].
Definition 6.9. Let V be a R vector space. Any bilinear map V × V → V, (x, y) 7 → [X, Y] is called a lie bracket if it satisfies (2) and (3) from Proposition 6.8. The pair (V, [•, • ]) is called a Lie algebra.
NOTES FOR MATH 230A, DIFFERENTIAL GEOMETRY 19
(3) and, by the way we define the smooth atlas, we will have local trivializa- tions. That is, we will have U ⊂ M open, x ∈ U and
U × R k^ TU TM
U M
Φ
where Φ is a diffeomorphism making
U × R k^ TU
U
pr π
commutes, where pr(y, v) = y, y ∈ U, v ∈ R k, and Φ(y, • ) : TyU → {y} R k is a linear isomorphism for all y ∈ U.
Definition 6.15. Let E be a smooth manifold together with a smooth map π : E → M and the structure of a R vector space on each π−^1 (x) so that for all x ∈ M there is a U with U open and Φ : E|U =∼ U × R k^ as in the above enumeration. Then, (E, π) is called a rank k vector bundle over M, and k can be any nonnegative integer.
Remark 6.16. E is like a bundle of vector spaces, one vector space for each x ∈ X. The condition of E and π being smooth means these vector spaces vary smoothly and piece together. Local triviality is mimicking the convenience of local charts. Now, we’ll topologize the tangent bundle.
Remark 6.17. Vector bundles are here to stay.
We’ll construct the tangent bundle as follows: (1) Take a sufficient open cover U = {Uα} (2) identify TUα =∼ Uα × R k, so that TUα inherits a C∞^ structure. (3) set an equivalence relation
α /^ ∼=:^ TM, where^ ∼^ says when^ V^ ∈^ TUα and v′^ ∈ TUβ come from the same tangent vector on M. Here is the construction:
Construction 6.18. Let A = {(U, φα)} be a smooth atlas for M. Consider the map U → φα(Uα) ⊂ R n, which is smooth by definition. So, for all x ∈ Uα I get a map TxUα → Tφα(x) R n. As sets, we obtain a map ∏
x∈Uα
TαUα →
Tφα(x) R n
For all x, this is an isomorphism of vector spaces. But, we know
Tφα(x) R n^ = span〈
∂x^1
|φα(x),... ,
∂xn^
|φα(x)〉
20 AARON LANDESMAN
So, we have an isomorphism
Tφα(x) R n^ =∼ {φα(x)} × R n, x 7 → (a 1 ,... , an)
where X =
ai (^) ∂x∂i |φα(x). So, we obtain a map ∐
x∈Uα
TxUα → φα(Uα) × R n
Let TUα =
x∈Uα TxUα^ be given the unique smooth structure making this a diffeomorphism. What is the equivalence relation? We have
∐ x∈Uα∩Uβ Tx
φαUα × R k^ φβ(Uβ × R k)
Uα × R k^ Uβ × R k
(Uα ∩ Uβ) × R k^ (Uβ ∩ Uα) × R k
(x, v) (x, d(φβ ◦ φ−α 1 )U)
Φα
Φβ
φ−α^1 ×id φ−β 1 ×id
that is, for all α, β we have a function γβα : Uα ∩ Uβ → GLk( R ), x 7 → d(φβ ◦ φ−α 1 ). You can check γαα(x) = id and γδβ ◦ γβα = γδα by the chain rule. The equivalence relation on TUα × R k^ ∼ (y, w) ∈ Uβ × R k^ which is equivalent to x = y and γβα(v) = w. Then, we can check that ∐
α
TUγ/ ∼=: TM
is a smooth manifold.
Recall, last time we defined
TM :=
x∈M
TxM
α∈A
Uα × R k
