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(Note: This is a related question, but my specific doubt isn't addressed in that, so I'll ask this new question)

Ref. 'Core Principles of Special and General Relativity' by Luscombe

Suppose we have a smooth curve $\gamma:I\to M$ from an open interval $I\subset\mathbb{R}$ to a manifold $M$, and let $f:M\to\mathbb{R}$ be smooth. We denote the generic argument of $\gamma$ by $u$, so $\gamma(u)\in M$ for any $u\in I$. The set of all smooth functions from $M\to\mathbb{R}$ at $p$ is denoted by $\mathcal{F}_p$. Quoting the rest from the book:

The tangent space at $p\in M$, the set of all directional derivatives at $p$, is denoted by $T_p(M)$. For fixed $\gamma$, the directional derivative $\mathbf{t}_{\gamma}(f)\equiv \text{d}(f\circ\gamma)/\text{d}u\ |_p$ is an operator that maps functions $f\in\mathcal{F}_p$ onto numbers, $\mathbf{t}_{\gamma}:\mathcal{F}_p\to\mathbb{R}$. To cement the relation between directional derivatives and vectors, we require that $\mathbf{t}$ possess the properties customarily associated with derivatives, that for fixed $\gamma$ and for $f,g\in\mathcal{F}_p$ and $\alpha,\beta\in\mathbb{R}$:

  • $\mathbf{t}$ is linear: $\mathbf{t}(\alpha f+\beta g)=\alpha\mathbf{t}(f)+\beta\mathbf{t}(g)$;
  • $\mathbf{t}$ satisfies the derivation property (product rule of calculus): $\mathbf{t}(fg)=f(p)\mathbf{t}(g)+g(p)\mathbf{t}(f)$.

Both points are easily checked. The linearity property establishes $T_p(M)$ as a vector space.

The thing distinguishing the vectors in $T_p(M)$ is $\gamma$ - i.e. each vector is tangent w.r.t. some curve, and that curve is identified with $\gamma$. So to establish $T_p(M)$ as a vector space, how's proving the linearity property in point 1 going to help? Shouldn't we be proving that if $\mathbf{t}_{\gamma}$ and $\mathbf{t}_{\eta}$ lie in $T_p(M)$, then so does $\alpha\mathbf{t}_{\gamma}+\beta\mathbf{t}_{\eta}$, and so on?

Is it true that $\mathcal{F}_p$ itself is a vector space? I'm guessing if that's the case, I can use the fact that the set of linear maps from one vector space to another is itself a vector space.

Shirish Kulhari
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    You haven't told us what $F_p$ denotes, so it's hard for us to say whether it's a vector space. If it's "germs of functions at $p$", then the answer is "yes, it's a vector space." Indeed, the set of germs is defined exactly so that the vector-space axioms will hold. – John Hughes Jun 21 '20 at 17:53
  • @JohnHughes: Sorry! I'll edit the question. $\mathcal{F}_p$ is the set of all smooth functions $M\to\mathbb{R}$ at $p$. – Shirish Kulhari Jun 21 '20 at 17:55
  • Yes, you are correct about what should be proved in order to establish that $T_p(M)$ is a vector space. It's unclear to me what exactly the author had in mind there – Ben Grossmann Jun 21 '20 at 17:58
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    While it is true that the set of linear maps from one vector space to another is itself a vector space, we are looking at the set of linear maps that satisfy the derivation property. It is not generally true that a collection of vectors satisfying some property forms a subspace; we should show that if $\mathbf t_1,\mathbf t_2$ are linear maps with the derivation property, then so is $\alpha \mathbf t_1 + \beta \mathbf t_2$. – Ben Grossmann Jun 21 '20 at 18:01
  • @Omnomnomnom: Ah yes you're right! So merely proving the linearity of $\mathbf{r}$ in the arguments doesn't mean much. – Shirish Kulhari Jun 21 '20 at 18:05
  • It could perhaps be noted that the set of maps satisfying the derivation property is linear because it is the intersection of the kernels of the linear maps $\phi_{f,g}:L(\mathcal F_p,\Bbb R) \to \mathcal F_p$ defined by $$ \phi_{f,g}(\mathbf t) = \mathbf t(f)g + f \mathbf t(g) - \mathbf t(fg), $$ but I doubt that this is what is being alluded to – Ben Grossmann Jun 21 '20 at 18:06
  • Ah, well...if $F_p$ is just the set of "smooth functions at $p$", then it isn't a vector space, so that line of proof isn't really going anywhere. Omnonnonnonm's recent comment describing what's needed to show it's a vector space is spot-on; I think that the author was just a little too glib when writing that property 1 establishes that it's a vector space. – John Hughes Jun 21 '20 at 18:08
  • @Omnomnomnom: Thanks! One last thing - Is there a way to separate $f$ out of the expression $d(f\circ\gamma)/du$? I'm guessing I can't write it as $\frac{\partial f}{\partial\gamma}\frac{d\gamma}{du}$ because in that case $d\gamma/du$ wouldn't make much sense? – Shirish Kulhari Jun 21 '20 at 18:22
  • I would say that $\frac{\partial f}{\partial\gamma}$ doesn't make much sense either. Ultimately, the point of defining the tangent space is so that we can make sense of either of these derivatives – Ben Grossmann Jun 21 '20 at 18:26
  • No problem.${}{}$ – Ben Grossmann Jun 21 '20 at 18:29

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