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I am trying to understand how the geometric tangent space [tex]\mathbb{R}^n_a[/tex] given by

[tex]

\begin{displaymath}

\mathbb{R}^n_a = \{(a,v) | v \in \mathb {R}^n\}

\end{displaymath}

[/tex]

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

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 [tex]v_a[/tex] in [tex]\mathbb{R}^n_a[/tex] to the operator that represents the directional derivative evaluated at the point [tex]a[/tex] in the direction of [tex]v[/tex]. If [tex]\phi$[/tex] denotes the proposed isomorphism, we can write

[tex]

\begin{displaymath}

\phi (v_a)(f) = \widetilde{v}_a(f) = v^i \dfrac{\partial f}{\partial x^i}(a)

\end{displaymath}

[/tex]

for any [tex]f \in C^{\infty}(\mathbb{R}^n)[/tex].

To show that the (clearly linear) map [tex]\phi[/tex] is an isomorphism we must show that it is a bijection. Now, to prove that [tex]\phi[/tex] is 1-1, from linear algebra we know that it is sufficient to show that the kernel of [tex]\phi[/tex] contains only the [tex]0[/tex] vector in [tex]\mathbb{R}^n_a[/tex], denoted by [tex]0_a[/tex]. This means that we must show that the only element of [tex]\mathbb{R}^n_a[/tex] that satisfies [tex]\phi (v_a)(f) = 0[/tex] is [tex]0_a[/tex].

So, suppose

[tex]

\begin{displaymath}

v^i \dfrac{\partial f}{\partial x^i}(a) = 0

\end{displaymath}

[/tex]

If we can show that each component [tex]v^i = 0[/tex] injectivity will follow - and this is where I am stuck. Any ideas?

Thanks