# 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.