0
$\begingroup$

This is a follow up to this question, in which the context is provided. I would like to see the equivalence of two notions of tangent spaces and differentials of smooth maps.

Guillemin and Pollack define the tangent space as follows. Let $\phi: U \subseteq \mathbb{R}^n \to M\subseteq \mathbb{R}^N$ be a coordinate map for the embedded submanifold $M$ s.t.\ $0 \in U$ and $\phi(0) = p$. Then $D\phi|_0$ is the derivative of $\phi$ at $0$. Then let $T_pM := D\phi|_0( \mathbb{R}^n)$ i.e. a certain $n$ dimensional vector subspace of $\mathbb{R}^N$. It is then proved that this definition is independent of $\phi$.

Tu defines the tangent space of the vector space of derivations of germs of $C^\infty$ functions at a point $p$.

How do these definitions correspond?

Under this correspondence, we should get a correspondence between notions of the differential of a smooth map. Let $M, N \subseteq \mathbb{R}^N$ be $m, n$ resp. dimensional embedded submanifolds. Let $f: M \to N$ be a smooth map between the two manifolds. From this we want to construct a vector space homomorphism from $T_pM \to T_{f(p)}N$ for any $p \in M$. Tu and G&P do this in two different ways.

G&P define $df_p$ as follows. Let $\phi: U \to M$ and $\psi: V \to N$, where $U,V$ are open subsets of $\mathbb{R}^m, \mathbb{R}^n$ resp., be coordinate maps centered at $p$ and $f(p)$ resp. Then let $h = \psi^{-1} \circ f \circ \phi: U \cap f^{-1}(\psi(V)) \to \mathbb{R}^n$. Then $h$ is smooth and we define $df_p = d\psi_0 \circ dh_0 \circ d\phi^{-1}_0$, where $d\psi_0$ is just the total derivative of $\psi$ at $0$. It is then verified this definition is in fact coordinate independent.

Tu defines the differential of $f$ by defining its action on $C^\infty$ real-valued functions (as this will uniquely determine a derivation): $f_*(v)(g) =v(g\circ f)$.

How do these definitions correspond?

$\endgroup$
2
  • $\begingroup$ The natural candidate for associating a derivation to a vector in G&P's definition of a tangent space is taking the directional derivative along that vector. This only makes sense though if smooth functions from the manifold to $\mathbb{R}$ can be extended to smooth functions on an neighborhood of any point. This is intrinsically possible in G&P's definition of smooth though it is not clearly possible in Tu's definition (this is part of the linked question). $\endgroup$ Jan 17 '20 at 0:46
  • $\begingroup$ Though for this to work, first we have to show we get a well defined derivation that is independent of choice of extension. Then we would have to verify it actually gives a vector space isomorphism. $\endgroup$ Jan 17 '20 at 0:49
0
$\begingroup$

I've written up a somewhat lengthy but very explicit demonstration of the correspondence between Tu's and G&P's definitions of smooth manifolds, smooth maps, tangent spaces, and differentials. I will leave a link to the pdf of the write up here for anyone who might want it:

https://drive.google.com/file/d/1WvfwaKvKLp13Swioqy0ElLb7zLxisi_D/view?usp=sharing

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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