K & N Theorems / Results Section 1.1

Proposition 1. Let M be a manifold, p\in M, and x(t) a curve of class C^1, a\leq x\leq b, such that x(t_0)=p. Let X:\mathbf{F}(p)\to \mathbb{R} be defined by Xf=(df(x(t))/dt)_{t_0}, which satisfies the properties

  1. X is a linear mapping of \mathbf{F}(p) into \mathbb{R}, and
  2. X(fg)=(Xf)g(p)+f(p)(Xg) for all f,g\in\mathbf{F}(p).

Then the set of mappings X satisfying these conditions is a real vector space and: (i) The set of vectors Xf at p is a vector subspace of dimension n, and (ii) the collection

(\partial/\partial u_1)_p,\ldots,(\partial/\partial u_n)_p

is the basis for the set of vectors at p.

Proof. Given two such mappings X and Y along with constants \alpha,\beta\in\mathbb{R}, proving that the above space is a real vector space is straightforward as it suffices to show that X+Y and that \alpha X,\beta Y satisfy the above conditions. Indeed, linearity in both cases is immediate upon the assumption that X,Y are linear as in condition (1). Also, (X+Y)(fg)=X(fg)+Y(fg) and (\alpha X)(fg)=\alpha(X(fg)), whereby the claims will be immediate. It suffices to show that the vectors Xf satisfy condition (i) and that the collection of partial differentials above form a basis as claimed in (ii). This will (essentially) be done in one step.

To that end, first let p\in M with neighborhood U\ni p be arbitrary and choose a coordinate basis u_1,\ldots,u_n for U. Let x(t) be an arbitrary curve for which p=x(t_0) for some t_0\in[a,b], and let u_j=x_j(t), j=1,\ldots, n, be the equations for x in terms of u_1,\ldots,u_n. Let f\in\mathbf{F}(p) be arbitrary. Computing the derivative at t_0 of f(x(t)) with respect to t, it follows that

(df(x(t))/dt)_{t_0}=\sum_{j}(\partial f / \partial u_j)_p\cdot (dx_j(t)/dt)_{t_0}.

Therefore, any tangent vector at p is immediately a linear combination of partial derivative operators (\partial/\partial u_i)_p, i=1,\ldots, n, with components \xi_j= (dx_j(t)/dt)_{t_0}. Moreover, given a linear combination \sum_j \xi_j(\partial/\partial u_j), one can define a curve parametrically by u_j=u_j(p)+\xi_j t for j=1,\ldots,n. Doing so and considering for arbitrary f\in\mathbf{F}(p) the quantity Xf for the curve u_j, j=1,\ldots, n, at t=0 yields that

Xf = (df(u(t))/dt)=\sum_j (\partial f/\partial u_j)_p\cdot (du_j/dt)_{t=0}.

The first term in the above product is because t=0, and the second term reduces to \xi_j since d(u_j(p)+\xi_j t)/dt=\xi_j for all j. Hence, the vector tangent to the curve u_j, j=1,\ldots,n, at t=0 is precisely \sum_j \xi_j(\partial/\partial u_j)_p. The claim will thus be proven if it can be shown that the collection (\partial/\partial_j)_p, j=1,\ldots,n, is linearly independent.

Suppose, then, that \sum_j \xi_j(\partial f/\partial u_j)_p=0 for all f\in\mathbf{F}(p). In particular, considering the k^{th} projection function, it would have to follow that

\sum_j \xi_j(\partial u_k/\partial u_j)_p=0 for k=1,\ldots,n,

and because (\partial u_k/\partial u_j)=\delta_{kj}, it follows that the sum above is equal to \xi_k for k=1,\ldots, n. Hence, the original sum \sum_j \xi_j(\partial/\partial_j)_p equals zero precisely when the coefficients \xi_j equal zero for all j. Thus, linear independence is proved and so the collection of vectors at p is a vector space of dimension n with the basis claimed above.    \square

Proposition 2. If X,Y\in\mathbf{X}(M), then the bracket [X,Y]:R_M\to R_M is a vector field, where R_M is the ring of functions on M and where [X,Y]f = X(Yf)-Y(Xf) for all f\in R_M.

Proof. Let X,Y\in\mathbf{X}(M). To show that [X,Y] is a vector field, it suffices to show that [X,Y] assigns for all f\in R_M a vector X_p to each point p\in M. To begin, define a local coordinate system u_1,\ldots u_n and write X=\sum_j \xi_j(\partial/\partial u_j) and Y=\sum_j \eta_j(\partial/\partial u_j). Then per the definition of the bracket,

[X,Y]f = \sum_j \xi_j D_j(\sum_k \eta_k (\partial f/\partial u_k))-\sum_j \eta_j D_j(\sum_k \xi_k(\partial f/\partial u_k))

where D_j denotes the partial differential operator with respect to u_j. Combining the above expression, it follows that $

[X,Y]f=\sum_{j,k}\left\{\xi_k(\partial \eta_k/\partial u_k)-\eta_k(\partial\xi_k/\partial u_k)\right\}(\partial f/\partial u_j),

which is a vector whose components are given by \beta_k =\sum_k \xi_k(\partial \eta_k/\partial u_k)-\eta_k(\partial\xi_k/\partial u_k). Hence, [X,Y] is a vector field.    \square

Proposition 3. Let f be a mapping of M into M' and p a point of M.

  1. If (f_*)_p is injectie, there exist a local coordinate system u_1,\ldots,u_n in a neighborhood U of p and a local coordinate system v_1,\ldots,v_{n'} in a neighborhood of f(p) such that v_i(f(q))=u_i(q) for q\in U and i=1,\ldots,n. In particular, f is a homeomorphism of U onto f(U).
  2. If (f_*)_p is surjective, there exist a local coordinate system u_1,\ldots,u_n in a neighborhood U of p and a local coordinate system v_1,\ldots,v_{n'} of f(p) such that v_i(f(q))=u_i(q) for q\in U and i=1,\ldots,n'. In particular, the mapping f:U\to M' is open.
  3. If (f_*)_p is a linear isomorphism of T_p(M) onto T_{f(p)}(M'), the f defines a homeomorphism of a neighborhood U of p onto a neighborhood V of f(p) and its inverse f^{-1}:V\to U is also differentiable.

Proof.

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