Let $M$ be a(n abstract) smooth manifold and $\alpha(t):\mathbb{R}\to \phi(U)\subseteq{\mathbb{R}^n}$ be defined by $\alpha(t)=(t+b_1,b_2,...,b_n)$ for suitable constants $b_i$. I would like to know whether this object $$\dot{\gamma}(t) = \frac{\partial}{\partial x^1}\Bigg|_{\gamma(t)}$$

is something different from the inverse image $\phi^{-1}\circ\alpha$ near $t$ of the representation of $\gamma$ which is just $\alpha$ by the inverse of the chart $(U,\phi)$.

Or whether it is literally the same?

  • 1
    $\begingroup$ This seems a bit muddled. So $\gamma(t) = \phi^{-1}\circ\alpha(t)$ is the curve in $M$. $\alpha = \phi\circ\gamma$ is its representation in the chart $\phi$ (not inverse). But, yes, the convention is that if we write $\phi = (x^1,\dots,x^n)$, then $\partial/\partial x^1$ is the tangent vector to the $x^1$-curve in $M$ passing through your point. That's what your curve $\gamma$ is. $\endgroup$ – Ted Shifrin Nov 4 '17 at 18:56
  • $\begingroup$ Could you put "the inverse image... near $t$ of the representation of $ \gamma$..." in functional notation? It is hard for me to tell what you mean. $\endgroup$ – Andrew Tindall Nov 4 '17 at 18:57
  • $\begingroup$ I would like to make clear what is really meant by a tangent vector when we deal by an abstract manifold, i.e. not the one embedded in $\mathbb{R}^N$.Is a tangent vector literally the same as the curve $\gamma$ itself? $\endgroup$ – user122424 Nov 4 '17 at 19:05

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