# Prove that tangent spaces, modeled as equivalence classes of curves, are vector spaces

Starting with these definitions

• A curve on a manifold $$\mathcal M$$ is a smooth (i.e. $$C^{\infty}$$) map $$\sigma$$ from some open interval $$(-\epsilon,\epsilon)$$ of the real line into $$\mathcal M$$

• Two curves $$\sigma_1$$ and $$\sigma_2$$ are tangent at a point $$p$$ in $$\mathcal M$$ if (a) $$\sigma_1(0) = \sigma_2(0) = p$$ and (b) In some local coordinate system $$(x^1,x^2,\ldots,x^m)$$ around $$p$$, two curves are tangent in the usual sense as curves in $$\mathbb R^m$$, $$(x^i \circ \sigma_1)'(0) = (x^i \circ \sigma_2)'(0)$$ here, $$i=1,\ldots,m$$

• The tangent vector is defined as the equivalence class of curves in $$\mathcal M$$ where the equivalence relation between two curves is that they are tangent at point $$p$$.

• The tangent space is $$T_p\mathcal M$$ to $$\mathcal M$$ at point $$p$$ is the set of all tangent vectors at point $$p$$

I am trying to prove the tangent space at the point $$p$$ in a manifold $$\mathcal M$$ is a vector space.

I'm starting with $$v_1 \in T_p\mathcal M$$, and $$v_2 \in T_p\mathcal M$$, and I have the following definitions $$v_1 + v_2 := [\phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )] \\ r \ v_1 := [\phi^{-1}\circ \ (r \phi\ \circ \sigma_1)]\ \forall r \in \mathbb R$$

I want to show that $$v_1 + v_2 \in T_p \mathcal M$$ and $$r \ v_1 \in T_p \mathcal M$$

As $$v_1 ,v_2 \in T_p\mathcal M$$, then $$\sigma_1(0) = \sigma_2(0) = p$$

Now, for $$v_1 + v_2$$ be a vector at $$p$$ , $$\phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )(0) = p$$ $$\phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )(0) = \phi^{-1} \ (\phi\ ( \sigma_1(0)) + \phi\ (\sigma_2(0)) ) \\ = \phi^{-1}((\phi\ ( p) + \phi\ (p) )) \\ = \phi^{-1}( \ 2\phi\ ( p) ) \neq p$$

I can't prove the closure relations starting from the definitions, what I'm doing wrong?

## Edit:

The book I am following "Isham, Chris J. Modern differential geometry for physicists. Vol. 61. World Scientific, 1999.", takes a special chart $$(U,\phi)$$ such that $$\phi(p) = \mathbf 0 \in \mathcal M$$, using this choice

$$\phi^{-1}\circ \ (\phi\ \circ \sigma_1 + \phi\ \circ \sigma_2 )(0) = \phi^{-1}( \ 2\phi\ ( p) ) = \phi^{-1}(0) = p$$ So, the closure is proven under addition. But this chart is a special choice. But the definitions hold for any charts around $$p$$, so another choice of charts should give the same result.

• The definition of a manifold means it's always possible to find a chart that maps any point in the manifold to a defined point in the underlying vector space, so without loss of generality we can always choose a chart that maps $p$ to $0$. See Definition 2.1 in the book and figure 2.2 Aug 21, 2020 at 12:52
• @PeteBabe Now, I get it, the functions $\phi : U \to \mathbb R^m$ is defined as per my need. So I can choose $\phi(p) = \mathbf 0$ Aug 21, 2020 at 14:56

Tangent vectors a $$p \in M$$ are equivalence classes of smooth curves $$\sigma : (-\epsilon,\epsilon) \to M$$ such that $$\sigma(0) = p$$ ("smooth curves in $$M$$ through $$p$$"). Here $$\epsilon = \epsilon (\sigma)$$ is a parameter which may vary from curve to curve. The equivalence relation is given by $$\sigma_1 \sim \sigma_2$$ if $$(\phi \sigma_1)'(0) = (\phi \sigma_2)'(0)$$ for some chart $$\phi$$ around $$p$$. It is easy to verify that $$\sigma_1 \sim \sigma_2$$ iff $$(\phi \sigma_1)'(0) = (\phi \sigma_2)'(0)$$ for all charts $$\phi$$ around $$p$$.

Given a smooth curve $$\sigma : (-\epsilon,\epsilon) \to M$$ through $$p$$, you can of course define $$r \cdot \sigma : (-\epsilon/\lvert r \rvert,\epsilon/\lvert r \rvert) \to M, (r \cdot \sigma)(t) = \sigma (rt)$$. Unfortunately there is no similar definition of $$\sigma_1 + \sigma_2$$ for curves $$\sigma_i$$ in $$M$$ trough $$p$$. You try to add them via the definition $$\sigma_1 + \sigma_2 = \phi^{-1}(\phi\sigma_1 + \phi \sigma_2).$$ This exploits the fact that the chart $$\phi : U \to V \subset \mathbb R^n$$ take values in $$\mathbb R^n$$, but in general it does not work because you cannot be sure that $$\phi\sigma_1(t) + \phi \sigma_2(t) \in V$$ for $$\lvert t \rvert$$ sufficiently small. Not even $$\phi\sigma_1(0) + \phi \sigma_2(0) = \phi(p) + \phi(p) = 2\phi(p)$$ is in general contained in $$V$$.

The solution is to consider only charts such that $$\phi(p) = 0$$. This can always be achieved if we replace an arbitrary chart $$\phi$$ by $$T\phi$$ where $$T$$ is the translation by $$-\phi(p)$$. The same holds for your definition of $$r \cdot \sigma$$.

Doing so, you will see that you get in fact the structure of a vector space on $$T_p M$$. Formally I suggest to proceed as follows:

1. Show that $$\phi_* : T_pM \to T_0V, \phi_*([\sigma]) = [\phi\sigma]$$, is a bijection.

2. Show that $$T_0V$$ becomes a vector space via $$[\tau_1] + [\tau_2] = [\tau_1 + \tau_2]$$ and $$[r \cdot \tau] = [r \cdot \tau]$$, where $$(\tau_1 + \tau_2(t) = \tau_1(t)+ \tau_2(t)$$ and $$(r \cdot \tau)(t) = r \cdot \tau(t)$$. Note that there always exist a maximal interval on which $$\tau_1(t)+ \tau_2(t) \in V$$ and $$r \cdot \tau(t) \in V$$; we take these intervals as the domains of $$\tau_1 + \tau_2$$ and $$r \cdot \tau$$. It is then easy to see that the map $$\mathbb R^n \to T_0V, v \mapsto \tau_v$$ with $$\tau_v(t) = tv$$, gives an isomorphism of vector spaces whichs shows that $$\dim T_0V = n$$.

3. Observe that $$\phi_*$$ induces a unique structure of a vector space on $$T_pM$$ such that $$\phi_*$$ becomes an isomorphism of vector spaces.

4. At first glance it seems that the vector space structure on $$T_pM$$ depends on the choice of $$\phi$$. The final step will therefore be to prove that any two charts $$\phi_1, \phi_2$$ around $$p$$ with $$\phi_i(p) = 0$$ produce the same vector space structure on $$T_pM$$.

• wouldn't it be possible to, instead, define the sum of equivalence classes by sending $[\tau_1]+[\tau_2]=[t\mapsto \phi^{-1}(\phi(\tau_1(t))+\phi(\tau_2(t))-\phi(p))]$? In other words, using an affine transformation to sum the $\mathbb R^n$ vectors rather than just their sum. This would at least give a path mapping $0\mapsto p\in M$, as required
– glS
Dec 23, 2021 at 17:16
• @gls Yes, you can do that. It results in the same what I have done: You replace the chart $\phi$ by the translated chart $\psi = \phi - \phi(p)$ which satisfies $\psi(p) = 0$. Then you take the sum $\psi(\tau_1(t)) + \psi(\tau_2(t))$ and compute $\psi^{-1}(\psi(\tau_1(t)) + \psi(\tau_2(t)))$. Dec 24, 2021 at 8:04