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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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1  
What have you tried? –  lhf Dec 13 '11 at 12:03
2  
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

3 Answers 3

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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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It depends on how you define manifolds and what properties you are allowed to assume.

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 $d\psi = d (\varphi^{-1}) = (d\varphi)^{-1}$.)

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