# Show that $(f\circ c_{1})'(0)=(f\circ c_{2})'(0)$

I am trying to solve a problem in Jeffrey Lee's book "Manifolds and Differential Geometry" Ex 2.5:

Let $c_{1}$ and $c_{2}$ be smooth curves mapping into a smooth manifold $M$, each with open interval domains containing $0\in \mathbb{R}$ and with $c_{1}(0)=c_{2}(0)$. Show that

$(f\circ c_{1})'(0)=(f\circ c_{2})'(0)$

for all smooth $f$ iff the curves $\textbf{x}\circ c_{1}$ and $\textbf{x}\circ c_{2}$ have the same velocity vector in $\mathbb{R}^{n}$ for some and hence any chart $(U,\textbf{x})$.

Comments and attempt: What I understand is that $c_{1}$ and $c_{2}$ are mappings from $(-\epsilon, \epsilon)$ into $M$. I don't think it makes sense to show that both $\textbf{x}\circ c_{1}$ and $\textbf{x}\circ c_{2}$ to have the same velocity vector everywhere on $(-\epsilon, \epsilon)$ because two curves passing through the same point $0$ may have the same velocity vector just at the point $t=0$ but elsewhere their paths might diverge and attain different velocity vectors.

And I am not even sure where does $f$ maps $M$ to. I assume that $f$ maps $M$ to $\mathbb{R}$ (or probably even $\mathbb{R^{m}}$)

So I am modest here and try to prove the statement for just $t=0$. I have two requests: 1) to check if my workings are correct under the new assumption and 2) to tell me if it is possible to show how both of the curves have the same velocity vector on the whole of interval.

Here I go!

For =>, we may just take $f$ to be the coordinate function $\mathbb{x}$. For <=, we have

$\begin{eqnarray} (f\circ c_{1})'(0)&=&\frac{d}{dt}f\circ c_{1}(t)|_{t=0}=(D_{\textbf{x}\circ c_{1}(0)}f\circ \textbf{x}^{-1})\frac{d}{dt}\textbf{x}\circ c_{1}(t)|_{t=0}\\ &=& (D_{\textbf{x}\circ c_{2}(0)}f\circ \textbf{x}^{-1})\frac{d}{dt}\textbf{x}\circ c_{2}(t)|_{t=0}=(f\circ c_{2})'(0) \end{eqnarray}$

Am I doing this correctly?

Now how do I extend the result to $(-\epsilon,\epsilon)$?

Many thanks!

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It does not appear to me that you are answering this correctly, but my confusion might be a result of not knowing the notation. I do not have access to Lee's book, but this is what I imagine that the author has in mind:

For a smooth $n$-dimensional manifold $M$, there are a variety of ways to represent or define a tangent vector $V \in T_{p}M$, where $p$ is a point in $M$. One way is to define how the vector acts on smooth functions (i.e. define vectors at a point as derivations of real-valued functions), and another is through an equivalence class of curves passing through the point $p$. It appears that the author is trying to set up the relationship between the two.

Now, as for the questions at hand, i.) it does not appear that you are going about this correctly (this might be notational), and ii.) you are not able extend the definition to the interval $(-\epsilon, \epsilon)$.

Regarding ii.): Two curves $c_{i} : (-\epsilon, \epsilon) \to M$, $i = 1, 2$, satisfying $c_{1}(0) = c_{2}(0)$ determine the same tangent vector at the point $c_{1}(0)$ if and only if $c^\prime_{1}(0) = c^\prime_{2}(0)$ in some (and hence any) coordinate chart. This is a condition of first order contact at a point, and, thus, one cannot extend the result to an entire interval. An easy way to see way to see that this can't be extended is the following. Consider the point $\mathbf{x}(p) \in \mathbf{x}(U) \subset \mathbb{R}^{n}$ and let $v \in T_{\mathbf{x}(p)}$ be a geometric tangent vector (in $\mathbb{R}^{n}$) that corresponds to $V \in T_{p}(M)$. Then you can find any number of curves passing through the point $\mathbf{x}(p)$ at $t = 0$ with tangent vector $v$ and they will all correspond to curves in $M$ with tangent vector $V \in T_{p}M$ via $\mathbf{x}^{-1}$, so there is no reason to believe that you can extend the equality beyond $t = 0.$

Now, towards a proof of the original statement.

$(\rightarrow):$ Let $(U, \mathbf{x})$ be any coordinate chart, where $\mathbf{x} = (x^1, \ldots , x^{n})$. Now, apply the hypothesis to each of the coordinate functions $x^{i}$. Denoting the component functions of $c_{i}$ in the given coordinate system by $c_{i}^{j}$, ($i =1, 2$, $j = 1..n$), then you obtain

$$\frac{dc_{i}^{j}}{dt} = \left(x^{j} \circ c_{i}\right)^\prime,$$ and evaluating at $t = 0$ and applying the hypothesis gives the desired result.

$(\leftarrow):$ Assume that there exists a coordinate chart $(U, \mathbf{x})$ such that $\mathbf{x} \circ c_{1}$ and $\mathbf{x} \circ c_{2}$ have the same tangent vector at $t = 0$. Using the notation above for the component functions of $c_{i}$, then the tangent vectors at $t = 0$ in the given coordinates are

$$\left( \frac{dc_{1}^{1}}{dt}, \ldots ,\frac{dc_{1}^{n}}{dt}\right) = \left( \frac{dc_{2}^{1}}{dt}, \ldots ,\frac{dc_{2}^{n}}{dt}\right),$$ where evaluation at $t = 0$ has been suppressed.

But the action of the velocity vector arising from a curve $c : (-\epsilon, \epsilon) \to M$ as a derivation on smooth functions $f$ is defined by $$c_{*}(\frac{d}{dt}\Big\vert_{t= 0})f = \left(f \circ c\right)^\prime(0).$$ In the the given coordinate system we have $$c_{*}(\frac{d}{dt} \Big\vert_{t = 0}) = \frac{dc^{j}}{dt}\Big\vert_{t = 0} \frac{\partial}{\partial x^{j}}\Big\vert_{c(0)},$$ where there is summation over $j$ in the above. The result should now follow immediately by replacing $c$ with $c_{i}$ ($i = 1, 2$) and making use of the hypothesis.

This ended up being a little longer than anticipated and I am not sure how much it help. It will mostly depend on the order that the material is presented in Lee's text, but I am happy to try and clear up any confusion created by my (long-winded) response.

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Thanks! I have this feeling that my answer is similar to mine... but it clears things up a lot! – enoughsaid05 Jan 29 '13 at 19:48