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


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


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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Ted Shifrin May 21 '13 at 20:38
  • $\begingroup$ Thanks Chris and Prof. Shifrin for the help! It makes much more sense now. I'll read that section a few more times later. $\endgroup$ – Evariste May 22 '13 at 3:59

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