Definition of Tangentspace

Question

Given a smooth manifold $M \ni p$ of dimension $m$ and smooth curves $\gamma,\delta:(-\epsilon,\epsilon) \to M$ with $\gamma(0) = \delta(0) = p$, I got a relation which identifies $\gamma$ and $\delta$ iff $D_0(\kappa \gamma) = D_0(\kappa \delta)$ where $\kappa: U \to \kappa (U)$ is some chart with $\kappa (p) = 0 \in \mathbb R^m$.

Now I want to see why this relation does not depend on $\kappa$. I think if $\kappa'$ is another chart at $p$ we just can write $$ D_0(\kappa'\gamma) = D_0(\kappa' \kappa^{-1}\kappa \gamma) = D_{\kappa(0)}(\kappa'\kappa^{-1})D_0(\kappa \delta) = D_0(\kappa'\delta). $$ Is it as simple as that ? (I am new to the subject so I just want to get everything right.)

The usual derivative, i.e. $D_0(\kappa \delta)$ will be a $m\times 1$ matrix. — , Mar 19 '14 at 13:05
Charts are homeomorphisms $\kappa: U \to \kappa(U)$ where $U \subset M$ open en $\kappa(U) \subset \mathbb R^m$. — , Mar 19 '14 at 13:06
So, changing the charts is just a reparametrization of the curve — Semsem, Mar 19 '14 at 13:08

score 0 · Accepted Answer · answered Mar 19 '14 at 13:19

0

That looks exactly right to me. It's just the chain rule. The one slightly tricky point is that for this to work, your other chart $\kappa'$ generally has to have $\kappa' \circ \kappa^{-1}$ be smooth, so that you can differentiate it.

Fortunately, that's one of the conditions for the charts in an atlas: the transition function between any two charts must be smooth. (And now you see why that's a condition: it lets definitions like this one make sense!)

answered Mar 19 '14 at 13:19

John Hughes

93,729

Great. That was the answer I was looking for. Indeed, I did not check wether the maps are differentiable or not, so now I am aware of that. – Mar 19 '14 at 13:22

Definition of Tangentspace

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