I am trying to understand the following definition of a tangent vector to a manifold which I came across in some lecture notes and how it relates to the three "standard definitions" as they can be found in for example Lee's book on smooth manifolds. The definition is
That $X:\mathcal{C}^{\infty}(M)\to\mathbb{R}$ is a linear map and it satisfies $Xf=(f\circ\gamma)'(0)$ made me guess that it probably also satisfies the Leibniz rule. Hence, we would have $X_{p}\in\{\text{Derivations at p}\}$.Would't we then have $T_{p}M\subset\{\text{Derivations at p}\}$. I would also like to have that or rather my guess is that $\{\text{Derivations at p}\}\subset T_{p}M$. I think this guess is due to the following proposition in Tu's book on smooth manifolds:
One of the things I don't get my head around too is that it also reminds me of the definition in terms of equivalence classes of curves, but apperently one can get away without them.
I would really appreciate some hints and tips. Many thanks in advance!
Your guess is correct.
First note that we have to interpret Definition 2.1 in the sense that there exists $\gamma$ such that for all $f$ one has $Xf = (f \circ \gamma)'(0)$, i.e. $\gamma$ does not depend on $f$.
Given $f,g \in C^\infty(M)$, we get $(f \cdot g) \circ \gamma = (f \circ \gamma) \cdot (g \circ \gamma)$ and therefore by the product rule of elementary calculus (applied to functions $(-\epsilon, \epsilon) \to \mathbb R$) $$X(f \cdot g) = ((f \cdot g) \circ \gamma)'(0) = (f \circ \gamma)'(0) \cdot (g \circ \gamma)(0) + (f \circ \gamma)(0) \cdot (g \circ \gamma)'(0)\\ = (Xf)\cdot g(p) + f(p) \cdot (Xg)$$ which means that $X$ is a derivation at $p$.
That each derivation $X$ at $p$ has the property in Definition 2.1 follows indeed from Tu's Proposition 8.17. There exists $c$ such that $c'(0) = X$, and by Tu's Proposition we have $$Xf = (f \circ c)'(0) .$$ Note that Tu's Proposition says that this is true for any such $c$.