Why does directional derivative kernel contain only zero vectors?

[CMoore]

Hello,

I am trying to understand how the geometric tangent space $$\mathbb{R}^n_a$$ given by

$$\begin{displaymath}<br /> \mathbb{R}^n_a = \{(a,v) | v \in \mathb {R}^n\}<br /> \end{displaymath}$$

is isomorphic to the space of all derivations of $$C^{\infty}(\mathbb{R}^n)$$ at $$a$$, denoted by $$T_a(\mathbb{R}^n)$$.

According to the book "Introduction to Smooth Manifolds" by John M. Lee, an isomorphism between these spaces is given by a map that sends each $$v_a$$ in $$\mathbb{R}^n_a$$ to the operator that represents the directional derivative evaluated at the point $$a$$ in the direction of $$v$$. If $$\phi$$ denotes the proposed isomorphism, we can write

$$\begin{displaymath}<br /> \phi (v_a)(f) = \widetilde{v}_a(f) = v^i \dfrac{\partial f}{\partial x^i}(a) <br /> \end{displaymath}$$

for any $$f \in C^{\infty}(\mathbb{R}^n)$$.

To show that the (clearly linear) map $$\phi$$ is an isomorphism we must show that it is a bijection. Now, to prove that $$\phi$$ is 1-1, from linear algebra we know that it is sufficient to show that the kernel of $$\phi$$ contains only the $$0$$ vector in $$\mathbb{R}^n_a$$, denoted by $$0_a$$. This means that we must show that the only element of $$\mathbb{R}^n_a$$ that satisfies $$\phi (v_a)(f) = 0$$ is $$0_a$$.
So, suppose
$$\begin{displaymath}<br /> v^i \dfrac{\partial f}{\partial x^i}(a) = 0<br /> \end{displaymath}$$
If we can show that each component $$v^i = 0$$ injectivity will follow - and this is where I am stuck. Any ideas?

Thanks

 

[matt grime]

why not choose to evaluate the directional derivatives on nice choices of f, eg the function that picks out components.

 

[CMoore]

Indeed; that is what the author does. He uses the standard coordinate functions for this purpose. But how can such a selection be justified? The relationship should hold not just for particular choices of f but for any f. Obviously, I'm being dense here, but I just can't see it...

 

[matt grime]

Eh? You want to show that the resulting object (thing with $\partial_x$) is not zero, it suffices to demonstrate so by finding a function, f, on which it doesn't vanish. Since v_i is not zero for some i, it follows that evaluated on the function that gives the i'th coordinate (the derivative is just v_i) is not zero.