Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've a doubt about the tangent space to a manifold. Let $M$ be a $n$-manifold and let $p\in M$, I've heard that the tangent space $T_pM$ at $p$ is the first order approximation of $M$ near $p$ in the same way that the tangent hyperplane to the graph of a function $f : \mathbb{R}^n \to \mathbb{R}$ is the first order approximation to the graph of $f$.

This is really intuitive, but how do I show that ? I mean, I'm using the definition of tangent space with derivations, how do I show that that abstract set associated with each point of the manifold gives the first order approximation to the manifold ? Is this fact already built in into the definition somehow or we should prove it ? If we should prove it, can someone give a hint ? I don't want the full proof, just a hint to begin the proof.

Thanks in advance for your aid, and sorry if this question is too trivial.

share|improve this question
    
What is your definition of first order approximation to a manifold? –  Andrew Oct 9 '12 at 20:54
    
@Andrew that is really the point. It's something intuitive that I want to make precise. For instance, If I have a surface, it's tangent plane gives a first order approximation to the surface near the tangency point. In particular, I mean that it gives a linear approximation to the surface. It's like the example I gave: if $f : \mathbb{R}^n \to \mathbb{R}$ is differentiable, then Taylor's first order formula allows us to approximate the graph of $f$ by it's tangent plane. Intuition tells that the same relationship exists between manifolds and tangent space, but I want to make it precise. –  user1620696 Oct 9 '12 at 20:59
add comment

2 Answers

up vote 1 down vote accepted

In the case of a function $f:\Bbb R^n\to\Bbb R,$ the first order approximation to $f$ at a point $x$ is a linear function, say $Df,$ which agrees with $f$ at $x,$ and closely approximates nearby values of $f.$ Geometrically, it turns out that the graph of $Df$ is a hyperplane locally tangent to the graph of $f.$

In the case of manifolds, a priori we may not have a local embedding, as for graphs, and more importantly, the manifold itself need not be defined as the graph of a particular function. So strictly speaking, there is no first order approximation of some function lurking around. But, it still makes sense to consider a vector space which locally best approximates the manifold. The way this is done is via derivatives and derivations, which ultimately carry much more structure than a simple vector space would, letting us do calculus locally on manifolds, but intuitively one can just think of the tangent space as a local approximation of the manifold.

share|improve this answer
1  
Andrew, so it's really built in into the definition the fact that the tangent space approximates the manifold ? I think I understood the point. Thanks for your answer. –  user1620696 Oct 10 '12 at 22:22
    
That's exactly it. You're welcome! –  Andrew Oct 10 '12 at 22:37
add comment

In principle the tangent space T_pM is an abstract thing. But any manifold can be embedded in some R^n (Whitney theorem), and there T_pM is a true linear subspace (up to a translation) and the tangent vector to a curve in M is a true vector and so on... so you can do the same reasoning that you do in analysis

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.