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We can see the vectors of the tangent space $T_pM$ to a smooth manifold as velocities of curves. This is elaborated here.

Each velocity $\gamma'(0)$ corresponds to a derivation $D_{\gamma}(f) = (f \circ \gamma)'(0)$ as seen in the Wikipedia article.

But why is every derivation also a velocity (vector)?

Jean Valjean
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1 Answers1

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Once you choose local coordinates, a basis for the derivations is $\left\{\dfrac{\partial}{\partial x^1}\Big|_p,\dots, \dfrac{\partial}{\partial x^n}\Big|_p\right\}$. Then there's an obvious curve with tangent vector $\sum\limits_{i=1}^n a_i \dfrac{\partial}{\partial x^i}\Big|_p$.

Ted Shifrin
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  • Oh, I think I see this. Let $(x,U)$ be a chart near $p \in U$. Let $\gamma(t) = x(p) + t(a_1,\dots,a_n)$ be a curve in $\mathbb{R}^n$. Then, $x^{-1} \circ \gamma$ is a curve in the manifold such that $(x^{-1} \circ \gamma)(0) = p$, and it has the required tangent vector. – Jean Valjean Jun 26 '14 at 23:12
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    Parfaitement. Très bien. :) – Ted Shifrin Jun 26 '14 at 23:15
  • Haha. Thanks a lot for your help. +1 and accepted. – Jean Valjean Jun 26 '14 at 23:16