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Let X be an n-dimensional differentiable manifold and $p \in X$ . Let $(U, h, V )$ for X around p with coordinates $(x_1 , . . . , x_n )$ in V , and let $v_i , i = 1, . . . , n$ , be the basis of $T_{p}^{alg}X$ associated to this chart.

Let $f : X \rightarrow R$ be a differentiable function. One could try to define the gradient $\bigtriangledown f(p) \in T_{p}^{alg}X$ as:

$$\bigtriangledown f(p) \equiv \displaystyle\sum\limits_{i=1}^n \frac{\partial(f \circ h^{-1})}{\partial x_{i}}(h(p)\cdot v_{i})$$

The question is: Is this a good definition, i.e. independent of the chosen chart?

Now it seems like it should be, as charts are one-to-one. I however have found the question posed in a different problem set which says to show that this is not a good definition. I don't see why.

If I did something such as define $$\bigtriangledown g(p) \equiv \displaystyle\sum\limits_{i=1}^n \frac{\partial(g \circ h'^{-1})}{\partial x_{i}}(h'(p)\cdot v_{i})$$ Where I am looking at the chart $(W, h', Z)$ and both charts $h,h'$ map to $\mathbb{R^n}$ hence they map to eachother. Thus somehow I can take $h'$ and turn it in to some function of $h$. This would be one way to show that the gradient is independent of the chart. I have no idea how to show this though. I have a posted question about the algebraic tangent spaces here:

Algebraic Tangent Space and Vector - an intuitive understanding?

Thus can see a definition of the $\bigtriangledown f(p) \in T_{p}^{alg}X$.

Thank you for your help,

Brian

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  • $\begingroup$ @Relativeo No problem. Comment removed. $\endgroup$ – Git Gud Jun 7 '13 at 13:46

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