# Identification of each tangent space $T_pV$ with $V$ itself?

I found this statement from my text very confusing:

What does it mean by identification of each tangent space $T_pV$ with $V$ itself? - what does "identification" really mean here?

If it means isomorphic, then it conflicts with my understanding that each tangent space $T_pV$ is locally isomorphic to $V$.

What is $Xf$? I don't understand the expression. I don't know what is the function $f$.

Thank you very much for your help.

• "Identification" means "pretend things are one and the same," which we're perfectly within our rights to do here because they are all isomorphic. Why do you think isomorphism "conflicts" with local isomorphism? – anon Jun 27 '13 at 0:42
• But we're not considering $S^1$, the text says quite clearly we're specifically considering the situation where we have a vector space as a manifold, in which case yes there is a canonical isomorphism between $V$ and all of its points' tangent spaces. If a text for instance were to say "polynomials have derivatives as follows ..." the reaction "but it isn't possible to take derivatives of functions in general because they may not be differentiable" is out of place! – anon Jun 27 '13 at 0:47
• As the definition states, it is the directional derivative of $f$ with respect to the vector $X$. – anon Jun 27 '13 at 0:51
• $f$ is any differentiable function we care to speak of. – anon Jun 27 '13 at 1:06
• Have you studied multivariable calculus before? In that subject, we talk about tangent vectors as elements of the space itself, and consider directional derivatives accordingly! – anon Jun 27 '13 at 1:12

What is implicit in the author's comment is the view of a tangent vector as a linear functional on the space of smooth functions defined near $p$.
The author's point is that when the manifold happens to be a vector space $V$, there is a natural choice for such a functional, namely the one given by the formula $Xf = \frac{d}{dt}\big\vert_{t=0}$, etc. In this case the roundabout approach is clearly redundant, but it needs to be shown that it is consistent with what we expect it to be intuitively, namely the directional derivative.
One can think of the tangent spaces as the "directions one can move" (which include both speed and direction, i.e. they are vectors) on a manifold at a point $p$. It's not hard to convince yourself that if you look at a point on a vector space (which you should think of as flat in some sense), the directions you can move correspond to all the vectors in the space. If I start at a point of $\mathbb{R}^n$, I can move on $\mathbb{R}^n$ in any direction from $\mathbb{R}^n$. Not so if I start at a point on the two-sphere in $\mathbb{R}^3$, where I can move only in directions tangent to the two-sphere.