- #1

- 165

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Let ##\gamma:I\to M## be a smooth curve from an open interval ##I\subset\mathbb{R}## to a manifold ##M##, and let ##f:M\to\mathbb{R}## be smooth. We denote the generic argument of ##\gamma## by ##u##, so ##\gamma(u)\in M## for any ##u\in I##. The set of all smooth functions from ##M\to\mathbb{R}## at ##p## is denoted by ##\mathcal{F}_p## (so that ##f\in\mathcal{F}_p##). The definition of tangent space elements is given as:

The directional derivative on ##M## was defined indirectly via a curve ##\gamma## as follows:The tangent space at ##p\in M##, the set of all directional derivatives at ##p##, is denoted by ##T_p(M)##. For fixed ##\gamma##, the directional derivative ##\mathbf{t}_{\gamma}(f)\equiv d(f\circ\gamma)/du\ |_p## is an operator that maps functions ##f\in\mathcal{F}_p## onto numbers, ##\mathbf{t}_{\gamma}:\mathcal{F}_p\to\mathbb{R}##.

[/QUOTE]Let ##p\in M## be such that ##\gamma(c)=p## for ##c\in I\subset\mathbb{R}##. For a smooth function ##f## on ##M##, the composite map ##f\circ\gamma:\mathbb{R}\to\mathbb{R}## is a function on ##\mathbb{R}##. The directional derivative on ##M##,$$\frac{d(f\circ\gamma)}{du}\bigg|_c\equiv\lim_{s\to 0}\frac{1}{s}[(f\circ\gamma)(c+s)-(f\circ\gamma)(c)]$$

[/QUOTE]

Now to define the basis of ##T_p(M)##, we consider coordinate curves ##\gamma^i## passing through ##p\in M##. Let ##p##, with chart ##(U,\phi)##, have the coordinates ##(x^1(p),\ldots,x^n(p))##. The coordinate curves are specified by ##\phi^{-1}:\mathbb{R}\to M## as follows:$$\gamma^i(u)\equiv \phi^{-1}(x^1(p),\ldots,x^{i-1}(p),x^i(p)+u,x^{i+1}(p),\ldots,x^n(p))$$

Then it's written:

And this last part is pretty confusing. I evaluated the LHS as follows:$$\lim_{s\to 0}\frac{1}{s}[(f\circ\phi^{-1})(x^1(p),\ldots,x^i(p)+s,\ldots,x^n(p))-(f\circ\phi^{-1})(x^1(p),\ldots,x^i(p),\ldots,x^n(p))]$$ $$=\frac{\partial (f\circ\phi^{-1})}{\partial x^i}\ \bigg|_{\phi(p)}$$The directional derivative associated with ##\gamma^i## at ##p## defines the partial derivative of ##f## with respect to ##x^i##, $$\frac{d(f\circ\gamma^i)}{du}\ \bigg|_{u=0}=\frac{\partial f}{\partial x^i}\ \bigg|_p$$

which is not the same as the RHS in the quoted equation, so I don't know where that came from.

Another doubt is how do I reconcile the basis element of ##T_p(M)## with the differential operator ##\partial/\partial x^i##? As per the definition ##\mathbf{t}_{\gamma}(f)\equiv d(f\circ\gamma)/du\ |_p##, which means $$\mathbf{t}_{\gamma^i}(f)= d(f\circ\gamma^i)/du\ |_p = d(f\circ\gamma^i)/du\ |_{u=0}$$

I showed above that ##d(f\circ\gamma^i)/du\ |_{u=0}## is equal to ##\partial (f\circ\phi^{-1})/\partial x^i\ |_{\phi(p)}##. So

**finally**$$\mathbf{t}_{\gamma^i}(f)=\partial (f\circ\phi^{-1})/\partial x^i\ |_{\phi(p)}$$

I'm pretty sure (at least intuitively) that ##\{\mathbf{t}_{\gamma^i}\}## (the subscript reads ##\gamma^i##) forms the basis of ##T_p(M)##. I can only claim that ##\mathbf{t}_{\gamma^i}=\partial/\partial x^i## when ##\mathbf{t}_{\gamma^i}(f)=\partial f/\partial x^i## for all ##f\in\mathcal{F}_p##. From the above equation, I have no way to isolate ##f## on the RHS, so with what sorcery can I show that ##\mathbf{t}_{\gamma^i}(f)=\partial f/\partial x^i## for all ##f\in\mathcal{F}_p##?