# Verify ten axioms of a tangent space

So I know that tangent space as a set of linear approximation of all tangent vectors. And consequently, a tangent vector can be defined at a point in a vector space, as a order of $n$-tuples $v_p=\{a_1…,a_n\}p$ in which exist a parameterized curve $c:I→\mathbb{R}^n$ which derivative at 0 have the property $c(0)=p$ and $c'(0)=v_p=\{a_1, …,a_n\}p$

Because in the world of tangent spaces we are working with the properties of vector spaces, tangent vectors in the tangent space are defined by two operations: i) Vector addition. ii) Scalar multiplication and must satisfy 10 axioms.

My question is: How can I prove the ten axioms?

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What have you tried? –  lhf Dec 13 '11 at 12:03
Axioms are not meant to be proven. –  user13838 Dec 13 '11 at 13:12
List the vector space axioms you wish to prove and tell us which ones you have trouble with. –  Jeff Dec 13 '11 at 14:08
For example I need to check that if i take a tangent vector v in R^3 and a tangent vector w in R^3, their sum v + w is also a tangent vector in R^3 –  anilorap Dec 13 '11 at 14:34

if your definition is in terms of $\gamma'(0)$ for some curve $\gamma:(-1,1)\to M$, then given a curves with $\gamma_v'(0)=v, \gamma_w(0)=w$, use these to construct curves $\gamma_{av}, \gamma_{v+w}, \gamma_{-v}$ etc.

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You don't need to prove that the set of tangent vectors of a variety $M$ at a certain point $p$ satisfies all the axioms of vector space in order to see that it is a vector space.

To be specific, in the case that the variety $M \subset \mathbb{R}^3$ is a regular parametrized surface and

$$\varphi : U \longrightarrow M$$

is a parametrization of $M$ ($U$ and open set of $\mathbb{R}^2$), if $p = \varphi(q)$, you just need to prove that the tangent space of $M$ at the point $p$, $T_pM$, is the image of a vector space by a linear map. Namely,

$$T_pM = \mathrm{d}\varphi_q (\mathbb{R}^2) \ ,$$

where

$$\mathrm{d}\varphi_q : \mathbb{R}^2 \longrightarrow \mathbb{R}^3$$

is the differential of $\varphi$ at $q$ (which is a linear map by definition).

You can find a proof of this result, for instance, in do Carmo's "Differential geometry of curves and surfaces"(section 2.4, proposition 1, page 83).

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My favorite way is to use things like the constant rank theorem which allows to say that a manifold M embedded in euclidean space $R^n$ is locally the set of zeroes of some functions $f_1,\ldots,f_n$ and that the tangent space at a point $p$ is given by the subspace orthogonal to the derivatives of the $f_j$. (The set of vectors orthogonal to some set is always vector space).
On the other hand, combining the approaches given in the other two answers, if you have a parameterisation of your manifold $\varphi: R^d \to R^n$ around a point $p$ then the axioms of a vector space can easily be checked. First you have to define scalar multiplication and addition (and then associativity and distribution law follow from properties of derivatives).
Given a smooth curve $c: I \to M$ (where I is an open interval around zero), with $c(0)=p$ and $c'(0) = v$, you can define $c'(t) = c(\lambda t)$, which defines scalar multiplication. (In fact, $dc' = \lambda dc$, by linearity of the differential.)
Given two curves $c_1,c_2: I \to M$, consider $\psi \circ c_1, \psi \circ c_2$, where $\psi = \varphi^{-1}$ is the inverse of the parameterisation. Finally consider the curve $c = \varphi \circ (\psi \circ c_1 + \psi \circ c_2)$, this defines addition. (In fact $$dc = d\varphi d\psi dc_1 + d\varphi d\psi dc_2 = dc_1 + dc_2 ,$$ by additivity of the differential and the fact that $d\psi = d (\varphi^{-1}) = (d\varphi)^{-1}$.)