Introduction to manifolds M. L¨ubke April 17, 2018 1 Contents 1 Submanifolds of Euclidean space 4 1.1 Submanifolds of Rn ..................................... 4 1.2 Tangent spaces . 11 2 Differentiable manifolds 14 2.1 Manifolds . 14 2.2 Differentiable maps . 18 2.3 Tangent spaces . 23 2.4 The tangent map . 32 2.5 Submanifolds . 35 3 Differential forms 39 3.1 The exterior algebra of a manifold . 39 3.2 The exterior differential . 44 4 Integration on manifolds 47 4.1 Orientations on a manifold . 47 4.2 The integral . 51 4.3 Subsets with smooth boundary . 56 4.4 The Theorem of Stokes . 58 4.5 The Integral Theorem of Greene . 63 4.6 The Fixed Point Theorem of Brouwer . 64 5 Appendix: (Multi)Linear algebra 67 5.1 Duality . 67 5.2 Exterior powers . 70 5.3 Orientation of vector spaces . 74 5.4 Tensor products . 75 6 Vector bundles and connections 77 6.1 Vector bundles . 77 6.2 Connections in vector bundles . 85 2 In this reader, by differentiable we always mean ∞. C We will use the following notation. n m Let U R be open and f = (f1, . , fm): U R be a differentiable map. Then the Jacobian ∈ −→ (matrix) of f at x U is ∈ ∂f1 (x) ... ∂f1 (x) ∂f ∂x1 ∂xn i . .. Df(x) := (x) = . . ∂xj i=1,...,m j=1,...,n ∂fm (x) ... ∂fm (x) ∂x1 ∂xn We also recall the following results which should be well known from calculus. Theorem 0.0.1 (Implicit Function Theorem) n m n+m Consider R R = R with coordinates (x, y) = (x1, . , xn, y1, . , ym) . Given is an open n× m m subset U R R , a point p U, and a differentiable map f = (f1, . , fm): U R with ⊂ × ∈ −→ f(p) = 0 and ∂f det i (p) = 0 . ∂y 6 j i,j=1,...,m! n m Then there are open V R ,W R with p U 0 := V W U such that there is a unique ⊂ m ⊂ 1 ∈ × ⊂ differentiable map g : V R with U 0 f − (0) = (x, g(x)) x V . −→ ∩ { | ∈ } Theorem 0.0.2 (Inverse Function Theorem) n n Let U R be open, f = (f1, . , fn): U R a differentiable map, p U and ⊂ −→ ∈ det (Df(p)) = 0 . 6 n Then there is an open U 0 U , p U 0 , and an open V R such that f U : U 0 V is a ⊂ ∈ ⊂ | 0 −→ diffeomorphism. 3 1 Submanifolds of Euclidean space 1.1 Submanifolds of Rn Definition 1.1.1 A k-dimensional submanifold of Rn is a subset X Rn with the following pro- ⊂ perty: n For every p X there is an open Up R with p Up , and a differentiable map ∈ ⊂ ∈ n k f = (f1, . , fn k): Up R − − −→ such that f 1(0) = X U and − ∩ p rk (Df(p)) = n k . − Examples 1.1.2 1. We start by describing the dimensionally extreme cases. Claim: (a) The 0-dimensional submanifolds of Rn are precisely the discrete subsets. (b) The n-dimensional submanifolds of Rn are precisely the open subsets. Proof: (a) Recall that a subset X Rn is discrete if and only if for every p X there exists ⊂ ∈ an open subset U Rn such that p = X U. In this case, define ∈ { } ∩ n f : U R , f(x) := x p ; −→ − 1 then f is differentiable with X U = p = f (0) and rk(Df(p)) = rk(id n ) = n = n 0 . ∩ { } − R − This shows that X is a 0-dimensional submanifold. Conversely, let X Rn be a 0-dimensional submanifold and p X. Then there exists an open n ⊂ n 0 ∈ n 1 U R with p U and a differentiable map f : U R − = R such that X U = f − (0) ⊂ ∈ −→ ∩ and rkDf(p) = n . By the Inverse Function Theorem 0.0.2 we may assume (after making U smaller if necessary) that f is a diffeomorphism and in particular injective. This implies p = f 1(0) = X U. This shows that X is discrete. { } − ∩ (b) If X Rn is open, then for every p X take U = X and f 0 , the constant function ∈ 0 n n ∈ ≡ from U to 0 = R = R − ; the Jacobian of this map has rank 0, and therefore X is an { } n-dimensional submanifold. Conversely, let X Rn be an n-dimensional submanifold and p X. Then there exists an n ⊂ ∈n n 0 open U R with p U and a differentiable map f : U R − = R = 0 such that ⊂ 1 ∈ −→ { } n X U = f − (0) = U. This implies that for every p X there exists an open U R with ∩ ∈ ⊂ p U X, i.e. that X is an open subset of Rn. ∈ ⊂ 1 2 2 2 1 2. Let S = (x1, x2) R x1 + x2 = 1 be the unit circle. For every point p = (p1, p2) S { 2 ∈ | } 2 2 ∈ we choose U = R and f : U R , f(x1, x2) = x + x 1 . Then f is differentiable, it holds −→ 1 2 − S1 = S1 U = f 1(0) , the Jacobian of f is Df(x , x ) = (2x , 2x ) , and since (p , p ) = (0, 0) ∩ − 1 2 1 2 1 2 6 for all (p , p ) S1 , we have rkDf(p , p ) = 1 for all (p , p ) S1 . Therefore, S1 is a 1- 1 2 ∈ 1 2 1 2 ∈ dimensional submanifold of R2. 4 n n k 3. Suppose that X R is (locally) the zero set of a map f : U R − , but that f is not ⊂ −→ everywhere differentiable, or that for some p X U it holds rkDf(p) < n k . Does this ∈ ∩ − mean that X is not a submanifold of Rn? The answer to this question is no, as we see from the following example. 2 Consider the diagonal X := (x1, x2) R x1 = x2 ; intuitively one expects that this straight { ∈ | } line is a one-dimensional submanifold of R2. But X equals f 1(0) for the differentiable function f(x , x ) = (x x )2 = x2 2x x + x2 , − 1 2 1 − 2 1 − 1 2 2 whose Jacobian Df(x , x ) = (2x 2x , 2x + 2x ) = (0, 0) for all (x , x ) X, i.e. it 1 2 1 − 2 − 1 2 1 2 ∈ holds rkDf(x , x ) = 0 < 2 1 = 1 for all (x , x ) X. Nevertheless, X is indeed a 1- 1 2 − 1 2 ∈ dimensional submanifold, since it is also the zero set of the differentiable function x x whose 1 − 2 Jacobian (1, 1) has rank 1 everywhere. − The point we want to make here is that if an X is given as zero set of a map f which does not satisfy all of the conditions of the definition, it can nevertheless be a submanifold; the given f might just be the wrong choice. 4. We view elements of Rn as column vectors. n 1 A quadratic hypersurface in R is a subset X = fA,b,c− (0) where n n n t t fA,b,c : R R , fA,b,c(x) = x Ax + 2b x + c = aijxixj + 2 bixi + c , −→ i,jX=1 Xi=1 b1 . n with A = (aij)i,j=1,...,n a real symmetric (n n)-matrix, b = . R , and c R .X is × . ∈ ∈ b n called smooth or nondegenerate if A b det = 0 . bt c 6 Claim: A smooth quadratic hypersurface is an (n 1)-dimensional differentiable submanifold of − Rn. Proof: It suffices to show ∂f ∂f f (x) = 0 Df (x) := A,b,c ..., A,b,c (x) = 0 . A,b,c ⇒ A,b,c ∂x ∂x 6 1 n t It is easy to see that DfA,b,c(x) = (2Ax + 2b) . Assume that fA,b,c(x) = 0 and DfA,b,c(x) = 0 . The second condition means Ax + b = 0 or xtA = bt . In combination with the first condition it follows that btx + c = 0 . This implies − A b x Ax + b = = 0 , bt c 1 btx + c A b but this means that is not invertible; a contradiction. bt c 5 5. Let M(n) = M(n n, R be the vector space of real (n n)-matrices which is naturally identified 2 × × with Rn . Let S(n) = A M(n) A = At be the vector space of symmetric real (n n)-matrices; this { ∈ n(n+1) | } × is identified with R 2 by taking the coefficients on and above the diagonal as coordinates. Let I denote the n n unit matrix; then O(n) = A M(n) A At = I is the orthogonal n × { ∈ | ∙ n } group in dimension n, i.e. the group of orthogonal (n n)-matrices. Observe that an orthogonal 1 t × matrix A is invertible with inverse A− = A . n(n 1) Claim: O(n) is an 2− -dimensional submanifold of M(n). Proof: O(n) is the zero set of the differentiable map f : M(n) S(n) , f(A) := A At I . −→ ∙ − n For every A M(n) , the Jacobian ∈ Df(A): M(n) S(n) −→ is given by d d Df(A)(B) = (f(A + tB)) = A At + t(A Bt + B At) + t2B Bt I dt |t=0 dt ∙ ∙ ∙ ∙ − n |t=0 = A Bt + B At + 2tB Bt = A Bt + B At . ∙ ∙ ∙ |t=0 ∙ ∙ Take now A O(n) and C S(n) , and define B := 1 C A M(n) . Then Bt = 1 At C, ∈ ∈ 2 ∙ ∈ 2 ∙ and we get 1 Df(A)(B) = A At C + C A At = C. 2 ∙ ∙ ∙ ∙ This means that for every A O(n) the Jacobian Df(A) is surjective, i.e.