# Equivalent definitions of differential map

Let $f:M\rightarrow N$ be a smooth map between smooth manifolds, let $p\in M$ and $v\in T_{p}M$. Two different definitions of differential maps on tangent space: let $\gamma$ be a smooth curve on $M$ representing $v$ ($\gamma(0)=p,\gamma'(0)=v)$ and define $df_{p}(v)=(f\circ v)'(0)$.

Second definition: let $\mathcal{D}$ be a derivation at $p$, $g:N\rightarrow\mathbb{R}$ be a smooth function, then define $df_{p}(\mathcal{D})(g)=\mathcal{D}(g\circ f)$. We want to show that the two definitions of $df_{p}$ coincide.

Consider the matrix representation of the second definition of $df_{p}$ in local coordinates, this is: $$\left[ df_{p}=\frac{\partial\hat{f}^{j}}{\partial x^{i}}\right]$$

where $\hat{f}$ is the coordinate representation of $f$.

In the first definition, let $x_{1},\dots,x_{n}$ be local coordinates at $p$. Then, $\gamma(t)=(\gamma_{1}(t),\dots,\gamma_{n}(t))$ where $\gamma_{i}(t)=x_{i}(\gamma(t))$. The matrix representation of $df_{p}$ is:$$\left[ df_{p}=\sum_{i=1}^{n}\frac{\partial f}{\partial x^{i}}\right]$$

I might have gotten the matrix representation of $df_{p}$ in the first definition incorrectly, but of course if these matrices are the same with respect to these coordinates, this means that the two definitions are coincide. I am wondering are the steps I have done right.

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+1 for showing what you've done. But this problem can be solved without coordinates. Do you see how to associate a derivation to every curve? Or how to associate a tangent vector (of a curve) to every derivation? –  Jesse Madnick Sep 28 '11 at 4:31
Yes, I think that is just proving that the two different notions of a tangent vector: first, as an equivalence class of curves, and second as a derivation. But how to continue from there? –  776666 Sep 28 '11 at 4:44
Ah, sorry, I misread your question. But in the first definition, do you mean $df_p(v) = (f\circ \gamma)'(0)$? –  Jesse Madnick Sep 28 '11 at 4:57
Yes, that is my first definition. I think that my approach is correct? The matrices are the same: the first one is done in Einstein summation while the second one is not. –  776666 Sep 28 '11 at 5:04

Recall that if $\alpha$ is a curve in $M$ and $h\colon M \to \mathbb{R}$ is smooth, then the derivation associated to $\alpha'(0)$ is given by $$\alpha'(0)h = (h\circ \alpha)'(0).$$ We will use this fact twice.

As in your notation, we let $f\colon M \to N$ be a smooth map between manifolds, $p \in M$, $v \in T_pM$. Let $\gamma$ be a smooth curve on $M$ that represents $v$, and let $\mathcal{D}$ be the derivation at $p \in M$ that represents $v$. Let $g\colon N \to \mathbb{R}$ be a smooth function.

In the first definition, we have $df_p(v) = (f\circ \gamma)'(0)$, which is a tangent vector on $N$, and therefore may be regarded as a derivation at $f(p) \in N$. So, we can evaluate $df_p(v)$ at the function $g$:

\begin{align*} [df_p(v)](g) & = [(f\circ \gamma)'(0)](g) \\ & = (g \circ (f\circ\gamma))'(0) \\ & = ((g\circ f)\circ \gamma)'(0) \\ & = \gamma'(0)(g\circ f) \\ & = \mathcal{D}(g\circ f) \\ & = df_p(\mathcal{D})(g) \end{align*}

This shows that your two definitions coincide.

Edit: I realize that I have not addressed your actual question, which is whether or not your reasoning is correct. However, I think it is instructive nevertheless for you to see how to approach this problem without reference to coordinates and matrices.

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