Partial derivative is tangent vector of a manifold

Question

I know that this has been discussed a lot but I couldnt find something that really helps me out. I am new to differential geometry and I try to understand the following.

Definition: Let $M$ be a smooth manifold with chart $(U,x)$ around $p$ \begin{align} \left.\dfrac{\partial}{\partial x^i}\right|_p :C^\infty_p(M,\mathbb{R}) \ni f \rightarrow \left.\dfrac{\partial(f\circ x^{-1})}{\partial x^i}\right|_{x(p)}\in\mathbb{R} \end{align} are real tangent vectors for $i=1,\ldots,\text{dim}(M)$

First of all: Why do we define the partial derivative evaluated at $p$ in this strange way? How can one check if this acutally satisfies the Leibnitz-Rule? In the skript I found it just says that this is easy to see. Sorry for this beginner question, but I am not really experienced as you might see.

Answer 1

The partial derivative is defined that way because it is easier. To see this, imagine if you were the first one to define partial differentiation on a manifold. You know a lot about partial derivatives in ${R}^n$ and are used to it. You have also a map from your manifold to ${R}^n$ thanks to charts (i.e. your coordinates). Since you already have a connection with your manifold and ${R}^n$, you realise you can just take the derivative of your function on ${R}^n$ using your chart! The formula basically says, if you want to take derivative of a function on a manifold, just evaluate it at ${R}^n$ using your coordinates and take that function's derivative where you know already a lot about differentiation. This definition essentially removes the necessity to create a new notion of derivative and use your earlier results for flat spaces!

Seen this way, it is apparent why it satisfies Leibniz rule. You know it is obeyed on ${R}^n$ and since you take the derivative on ${R}^n$ (specifically on your chart) and not on your manifold, you can reuse your results for ${R}^n$!

Answer 2

Why do we define the partial derivative evaluated at $p$ in this strange way?

Let $V\subseteq\mathbb R^n$ be an open set and $f:V\rightarrow \mathbb R$ a real valued function on $V$. Then for each $i=1,2,\dots,n$ the partial derivative (if exists) is defined by $$ (D_if)(r^1,\dots,r^n):=\lim_{h\rightarrow 0}\frac{f(r^1,\dots,r^i+h,\dots,r^n)-f(r^1,\dots,r^i,\dots,r^n)}{h}. $$In particular, this essentially requires the function to be defined on some open set of $\mathbb R^n$.

Let $f\in C^\infty_p(M)$ be a smooth function germ near $p\in M$ of the $n$-manifold $M$. This is to say, $f$ is represented by a smooth function defined near $p$ and two smooth functions defined near $p$ represent the same germ $f$ if and only if they agree on some neighborhood of $p$.

We cannot define the partial derivative of $f$ (or any representative) directly, because $M$ is not an open set of $\mathbb R^n$. However suppose that $(U,x)$ is a local coordinate system near $p$. Then $x:U\rightarrow V\subseteq\mathbb R^n$ is a homeomorphism between $U\subseteq M$ and $V\subseteq \mathbb R^n$. We may suppose without loss of generality that the germ $f$ has a representative defined on $U$ and let us also write $f$ for this representative. Then the function$$ \hat f:=f\circ x^{-1} $$ is a smooth function defined on $V\subseteq\mathbb R^n$ which is an open set of $\mathbb R^n$. Explicitly, we have$$ \hat f(x^1(p),\dots,x^n(p))=f(p) $$ for each $p\in U$. We can now take the partial derivatives of this function, so we define $$ \frac{\partial f}{\partial x^i}(p):=(D_i\hat f)(x(p)). $$ We have to do this because we cannot take partial derivatives on $M$ directly as $f$ is not a multivariable function. But coordinate maps can be used to represent $f$ as the multivariable function $\hat f$, which is then partially differentiable. Hence the "strange" procedure.

How can one check if this acutally satisfies the Leibnitz-Rule?

Using the previous notation, we have for any pair of germs $f,g\in C^\infty_p(M)$ (identified with their representatives defined on the coordinate domain $U\subseteq M$)$$ \frac{\partial(fg)}{\partial x^i}(p)=D_i(\widehat{fg})(x(p)) \overset{(1)}{=} D_i(\hat f\hat g)(x(p)) \overset{(2)}{=} (D_i\hat f)(x(p))\hat g(x(p))+\hat f(x(p))(D_i\hat g)(x(p)) \\ =\frac{\partial f}{\partial x^i}(p)g(p)+f(p)\frac{\partial g}{\partial x^i}(p), $$where at (1) we have used $$ \widehat{fg}=(fg)\circ x^{-1}=(f\circ x^{-1})(g\circ x^{-1})=\hat f\hat g, $$i.e. that composition distributes with respect to multiplication and at (2) we have used that the ordinary partial derivative $D_i$ in $\mathbb R^n$ obeys the Leibniz rule.

Answer 3

However you define what an abstract tangent vector is, you can always do the following: Given a tangent vector $v$ at a point $p$ (i.e, $p\in T_pM$), there exists a curve $c(t)$ such that $c(0)=p$ and $c’(0)=v$. The directional derivative of a function $f$ at $p$ in the direction $v$ is defined to be $$D_vf(p)=\left.\frac{d}{dt}\right|_{t=0}f(c(t)).$$ It can be shown that the value of this remains the same, no matter what curve is used, as long as the velocity of the curve at $p$ is $v$. Also, it is easy to check that $D_v$ obeys the Leibniz rule (i.e., it is a derivation).

Now suppose there is a coordinate chart $$\Phi: U \rightarrow M$$ such that $\Phi(0)=p$. Then for the $i$th coordinate there is a curve $$c_i(t)=\Phi(0,\dots,t,\dots,0),$$ where all of the coordinates are held constant equal to $0$ except the $i$th which is set equal to $t$. One usually denotes the velocity vector of $c_i$ at $p$ by $$\partial_i = c_i’(0)\in T_pM.$$ The partial derivative of $f$ at $p$ is denoted and defined to be $$\partial_if(p) = D_{\partial_i}f = \left.\frac{d}{dt}\right|_{t=0}f(c_i(t)) = \partial_i(f\circ\Phi)(0).$$ Note that in this formula $\partial_i$ has, strictly speaking, two different but closely related definitions.