# Vector on a manifold

I can't fully understand the step that is made in this diff. geometry textbook. It says:

For $c:(a,b)$ $\rightarrow$ $M$ and function $f: M$ $\rightarrow$ $\mathbb{R}$ where $(a,b)$ is an open interval containing $t=0$. We define the tangent vector at $c(0)$ as a direction derivative of a function $f(c(t))$ along the curve $c(t)$ at $t=0$. The rate of change of $f(c(t))$ at $t=0$ along the curve is

$$\left. \frac{df(c(t))}{dt} \right|_{t=0}.$$

In terms of the local coordinate, this becomes

$$\left. \frac{\partial f}{\partial x^{\mu}}\frac{dx^{\mu}(c(t))}{dt} \right|_{t=0}$$

where $M$ is of course a manifold with chart $(U,\phi)$ and so $f \circ \phi^{-1} : \mathbb{R}^{m} \rightarrow \mathbb{R}$. My problem is understanding where the $x^{\mu}$ are found, what is means by $x^{\mu}(c(t))$ and also how $f$ can be differentiated by a point $x^{\mu}$?

Hopefully I've defined everything needed to answer this.

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Which differential geometry text-book? – B11b Dec 1 '13 at 0:46
I only have an extract of it so I'm actually not sure on the name nor author – Phibert Dec 1 '13 at 1:00
do Carmo's book "Differential forms and applications" (40 page) may help you – HK Lee Dec 1 '13 at 1:07

$x^{\mu}$ is abstract index notation to indicate the coordinates given by $\phi$. That is, $\phi = (x^1, x^2, ..., x^n)$ where $n$ is the dimension of the manifold. Conflating $\mu$ as the integer index on the functions with the generic abstract index notation, then $x^{\mu}$ is a function $U \to \Bbb{R}$. Now, if we let $f_{\phi} := f \circ \phi^{-1} : \phi(U) \subset \Bbb{R}^n \to \Bbb{R}$ then $$\frac{\partial f}{\partial x^{\mu}} = \partial_{\mu} f_{\phi}$$ where $\partial_{\mu}$ is the old familiar $\mu$th partial derivative you learned in multivariable calculus. Finally, for $c$, notice that $f \circ c : \Bbb{R} \to \Bbb{R}$, so $\frac{d}{dt} \big(f \circ c\big)$ makes sense from calculus. Now, we can write $$f \circ c = (f \circ \phi^{-1}) \circ (\phi \circ c) = f_{\phi}\big( \phi \circ c \big)$$ Let's finally realize that $\phi \circ c = (x^1 \circ c, x^2 \circ c, ..., x^n \circ c) : (a,b) \to \Bbb{R}^n$ is a curve, and again from multivariable calculus you learn that the chain rule gives you $$\frac{d}{dt}\big(f_{\phi} ( \phi \circ c) \big) = \operatorname{grad}(f_{\phi}) \cdot \dfrac{d (\phi \circ c)}{dt} = \sum_{\mu=1}^n \partial_{\mu}f_{\phi} \dfrac{d (x^{\mu}\circ c)(t)}{dt}$$ The $\cdot$ after the first equality was the dot product. Now, using that $\partial_{\mu} f_{\phi} = \frac{\partial f}{\partial x^{\mu}}$ and using the Einstein summation convention you get that $$\frac{d}{dt} \big(f_{\phi} ( \phi \circ c) \big)= \sum_{\mu=1}^n \partial_{\mu}f_{\phi} \dfrac{d (x^{\mu}\circ c)(t)}{dt} = \frac{\partial f}{\partial x^{\mu}} \frac{d x^{\mu}(c(t))}{dt}$$