proof that tangent space is a vector space

As I know so far (correct me if I am wrong): The tangent space $T_p(M)$ at a point $p$ of a manifold is defined as the set of all tangent vectors at $p$ where the tangent vector $\dot{\gamma}_p$ to a curve $\gamma(t)$ is defined as the map which assigns to each real valued function $f$ of M the derivative "along" $\gamma(t)$: $$\dot{\gamma}_p: C^{\infty}(M,\mathbb{R})\rightarrow \mathbb{R},\ f\mapsto \frac{\mathrm{d}}{\mathrm{d}t}(f\circ\gamma)\bigg|_p$$ (I assumed $C^{\infty}$)

How can I now show that $T_p(M)$ is a vector space? I don't even know how addition in this space is defined...

• Wouldn't Mathematics be better for that? – user36790 Dec 9 '16 at 15:31
• Well, you can add two maps and multiply them by a scalar; that's all you need for a vector space to be such. – gented Dec 9 '16 at 16:14
• How would a map $\dot{\gamma}_1+\dot{\gamma}_2$ act on a function f? I don't know how this addition is defined. – user126452 Dec 9 '16 at 16:21
• @user126452 as $f\mapsto\dot{\gamma}_1(f)+\dot{\gamma}_2(f)$ – coconut Dec 9 '16 at 16:42

Since $\dot{\gamma}_p$ is a $C^\infty(M)\rightarrow\mathbb{R}$ map, if $\dot{\gamma}_p$ and $\dot{\eta}_p$ are both such maps, then you can easily define their sum as $$(\dot{\gamma}_p+\dot{\eta}_p)(f)=\dot{\gamma}_p(f)+\dot{\eta}_p(f).$$ However, if you defined tangent vectors not as point-derivations of $C^\infty(M)$ (or derivations of the germs of such functions at $p$), then you need to show that this sum is also the tangent vector of some curve.
This is not trivial, because the set of (smooth) curves of $M$ do not form a vector space, unlike say, curves in $\mathbb{R}^n$, where you can easily add two curves together. The solution is to go to $\mathbb{R}^n$!
Let $(U,\psi)$ be a local chart of $M$ that contains $p$, and let $\psi(p)=(x^1(p),...,x^n(p))$. Let $\gamma$ and $\eta$ be two curves that both pass through $p$ at $t_0$. Then $\psi\circ\gamma$ and $\psi\circ\eta$ are both curves in $\mathbb{R}^n$.
Let us define the sum of $\gamma$ and $\eta$ as $\psi^{-1}\circ(\psi\circ\gamma+\psi\circ\eta)$. It should be clear that this is the sum of the representations of the two curves (representations in $\mathbb{R}^n$ of course), which are mapped back to the manifold $M$ by the chart map.
This, in general, will not be independent of the chart, but you can show (do it!), that the "sums" of the curves $\gamma$ and $\eta$ are in the same equivalence class for all charts containing $p$. Ergo, $$\left.\frac{d}{dt}\psi^{-1}(\psi\circ\gamma+\psi\circ\eta)\right|_{t=t_0}$$ is actually independent of the chart map $\psi$.
• This helped a lot, thank you! However I am not able to show the equivalence because one cannot apply the inverse function theorem (which is only defined for functions $\mathbb{R}^n\rightarrow \mathbb{R}^m$) and I have no idea how to proceed without it... I guess I first have to read on. – user126452 Dec 10 '16 at 22:02