I hope that this is a foolish question and that someone can make quick meat out of it. If [itex] \gamma: S^1 \to M [/itex] is a loop on an arbitrary manifold M, our goal is to analyze the tangent vectors to [itex] \gamma [/itex] when the loop is traversed in the opposite direction.(adsbygoogle = window.adsbygoogle || []).push({});

Let [itex] \iota: S^1 \to S^1 [/itex] be the map which reverses your favourite parameterization of [itex] S^1[/itex]. That is, if [itex] S^1 = \mathbb R/2\pi\mathbb Z [/itex] then [itex] \iota(\theta) = -\theta[/itex]; if [itex] S^1 = \{ z \in \mathbb C: |z|=1 \} [/itex] then [itex] \iota(z) = \bar z [/itex], etc. I think it is clear that these are identical maps given appropriate composition with parameterization isomorphism.

We want to compute [itex] (\gamma \circ \iota)'(\theta) [/itex]. We may equivalently write this as any of the following equivalent expressions

[tex] \begin{align*}

(\gamma \circ \iota)'(\theta_0) &= d(\gamma\circ\iota)\left.\frac{d}{d\theta}\right|_{\theta_0} = (d\gamma_{\iota(\theta_0)} \circ d\iota_{\theta_0}) \left.\frac{d}{d\theta}\right|_{\theta_0} \\

&= d\gamma_{\iota(\theta_0)} (\iota'(\theta_0))

\end{align*}

[/tex]

where in the latter equation, we view [itex]\iota [/itex] as a loop on [itex] S^1[/itex].

We now determine the action of [itex] d\iota[/itex]. For the sake of example, let us fix [itex] S^1 = \mathbb R/2\pi \mathbb Z [/itex], [itex] \theta_0 \in S_1 [/itex]. If [itex] X \in T_{\theta_0} S^1 [/itex] then set [itex] \beta: \mathbb R\to S^1 [/itex] such that [itex] \beta(0) = \theta_0, \beta'(0) = X [/itex] so that

[tex] \begin{align*}

d\iota_{\theta_0} \left.\frac{d}{d\theta}\right|_{\theta_0} &= \left.\frac{d}{dt}\right|_{t=0} \iota(\beta(t)) \\

&= -\left.\frac d{dt} \right|_{t=0} \beta(t) \\

&= - X

\end{align*}[/tex]

Hence we find that

[tex] \begin{align*}

(\gamma\circ\iota)'(\theta_0) &= d\gamma_{\iota(\theta_0)} d\iota_{\theta_0} \left. \frac d{d\theta}\right|_{\theta_0}\\

&= -d\gamma_{\iota(\theta_0)} X \\

&= -\gamma'(X)

\end{align*}

[/tex]

This is of course non-sense in general, since [itex] d\iota_{\theta}(X) = -X [/itex] implies that [itex] X [/itex] lives in the wrong space; and one would suspect that [itex] (\gamma(-\theta))' = -\gamma'(-\theta) [/itex], but this extra negative sign in the argument seems to disappear. However, everything works out in the end since in fact, [itex] \iota [/itex] is easily seen to be a diffeomorphism hence [itex] d\iota [/itex] is an isomorphism on the tangent spaces, and we get

[tex] d\iota(X) = -d\iota(X), \qquad (\gamma(-\theta))' = -\gamma(-\theta). [/tex]

So ultimately, the reason why it looks wrong in the first place is that there is an implicit identification of [itex] -X [/itex] with [itex] -d\iota(X) [/itex].

My question is as follows: In my experience such identifications are made explicitly by the mathematician, yet in this case the isomorphism "hid" itself. If one had naively carried through with the computation, one would have actually gotten the wrong answer ([itex] -\gamma'(\theta) \neq -\gamma'(-\theta) [/itex] in general!). Where did the isomorphism go? Was there some assumption above that made the identification implicit?

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# Derivative of a curve traversed in the opposite direction

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