Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
+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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.