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.

In section 2, page 9 of Guillemin and Pollack's book $\textit{Differential Topology}$, he gave a proof that the dimension of the tangent space $T_x(X)$ is equal to the dimension of the manifold $X$. However I read the proof at least 5 times and had no idea what they were talking about.

"The dimension of the vector space $T_x(X)$ is, as you expect, the dimension $k$ of $X$. To prove this, we use the smoothness of he inverse $\phi^{-1}$. Choose an open set $W$ in $\mathbf{R}^N$ and a smooth map $\Phi': \mathbf{R}^N \rightarrow \mathbf{R}^k$ that extends $\phi^{-1}$. Then $\Phi'\circ\phi$ is the identity map of U, so the chain rule implies that the sequence of linear transformations

$\mathbf{R}^k\xrightarrow{d\phi_0}T_x(X)\xrightarrow{d\Phi_x'}\mathbf{R}^k$

is the identity map of $\mathbf{R}^k$. It follows that $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism, so $dim T_x(X)=k$."

Just for everyone's information, in GP's book, the convention is that $\phi$ is a diffeomorphism from $\mathbf{R}^k$ to a k-dimensional manifold $X\subset \mathbf{R}^N$ ($\phi :\mathbf{R}^k\rightarrow X$).

I have two questions here:

  1. why do we need to choose an open set $W$ in $\mathbf{R}^N$ and define $\Phi'$ which extends $\phi^{-1}$? Naively I would just take $\phi^{-1}$. Then $d\phi^{-1}$ can also map $T_x(X)$ to $\mathbf{R}^k$ without problems. The map $\Phi'$ seems a bit artificial and redundant here. In fact it seems really redundant to me as when you extend $\phi^{-1}$, you actually don't care the rest of the function except for the $\phi^{-1}$ part.

  2. The proof seems only to serve as a proof that dim $T_x(X)=k$. However we know the tangent space itself is a manifold(actually a hyperplane). Why don't we find a diffeomorphism between $T_x(X)$ and $\mathbf{R}^k$? I haven't proved but I think we can use $d\phi_0$ as the diffeomorphism.

Thanks a lot for everyone's help!

Regards, Evariste

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The strategy of the proof is to convert the statement about smooth maps between manifolds into a statement about linear maps between vector spaces. At this early point in the book, the authors have not yet presented a way to define $d(\phi^{-1})$, because $\phi^{-1}: V \rightarrow U$ has a domain $V$ that is an open subset of a manifold, whereas they have only defined differentials in terms of difference quotients, which require that the domain be a vector space, so that addition is defined; see the limit definition given on page 8 of your text.

Therefore, it is necessary to "straighten" $V$ by first extending $\phi^{-1}$ to a map between Euclidean spaces, and then defining the differential of the extension using difference quotients.

To address your second question, this proof shows precisely that $d\phi_0$ is a linear isomorphism between $\mathbb{R}^k$ and the tangent space $T_x(X)$. A linear map is infinitely differentiable, so it is also a diffeomorphism.

share|improve this answer
1  
To amplify on this, note that on pp. 1-2, they define smoothness of $f\colon: X\to\mathbb R^m$ for an arbitrary subset $X$ (manifold or not) by saying it may be locally extended to a smooth map on an open subset of $\mathbb R^n$ (when $X\subset\mathbb R^n$). By the way, there are a fair number of errors in Guillemin & Pollack. You can find my list of errata at math.uga.edu/~shifrin/MATH4220/Guillemin_Pollack_errata.pdf. –  Ted Shifrin May 21 '13 at 20:38
    
Thanks Chris and Prof. Shifrin for the help! It makes much more sense now. I'll read that section a few more times later. –  Evariste May 22 '13 at 3:59

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.