Inverse of parametric equation.

Question

I'm going over a script on Tangent and Normal spaces and reading the definition for Tangent space of $T_p^{koord}M = d\phi(h(p))$ where $M \subset \mathbb{R}^n$ is a d-dimensional submanifold, $H: U \to U'$ an exterior map, $h$ an interior map and $\phi = h^{-1}$ a parametrization.

Now I'm wondering about $d\phi(h(p))$ because if I imagine $\phi(t)$ as a normal parameterization, concatenating it with its inverse would just result in $dt$, right? And how can that be the Tangent space? Am I understanding something wrong?

Edit: an exterior map is a function that maps any part of a manifold onto an open subset $U' \subset \mathbb{R}^n$ to make it flat, and the interior map is (as far as I understand) just the exterior map on a particular part of the manifold so that we could write: $h = H |_{M \cap U}: M \cap U \to (\mathbb{R}^d \times 0^{n-d})\cap U'$. By the way, if anybody could also help me understand what $0^{n-d}$ is supposed to denote I'd be very thankful since I can't find an explanation anywhere.

You should edit your question to explain what an exterior / interior map is. — Paul Frost, Feb 06 '20 at 19:52
$0^{n-d}$ denotes the point $0 \in \mathbb R^{n-d}$. Thus $\mathbb{R}^d \times 0^{n-d}$ is the $d$-dimensional subspace of $\mathbb R^n$ whose elements have the form $(x_1,\ldots,x_d,0,\ldots,0)$. — Paul Frost, Feb 06 '20 at 20:16
@PaulFrost thanks. And how can I think of the inverse of a parametrization? Let's say $\phi (t) = (t^2, t,1)$, how would I get to an inverse $\phi^{-1}$? — psyph, Feb 06 '20 at 20:20
Identifying the open subset $(\mathbb R^d \times 0^{n-d}) \cap U$ of $\mathbb R^d \times 0^{n-d}$ with the open subset $V \subset \mathbb R^d$ obtained by removong the last $n-d$ coordinates, we get a chart $h : M \cap U \to V$. Its inverse $\phi$ is a local parametrization of $M$. In your comment you alreday have a parametrization of a $1$-dimensional submanifold of $\mathbb R^3$. Why do you want to compute $h$? — Paul Frost, Feb 06 '20 at 20:31
Because I'm trying to understand the definition of the Tangent space, I just can't wrap my head around the $d \phi(h(p))$... — psyph, Feb 06 '20 at 20:36

score 1 · Accepted Answer · answered Feb 06 '20 at 20:48

It is difficult to answer your question without knowing the context. Howewer, it seems that you only consider submanifolds $M$ of Euclidean spaces.

Identifying the open subset $(\mathbb R^d \times 0^{n-d}) \cap U$ of $\mathbb R^d \times 0^{n-d}$ with the open subset $V \subset \mathbb R^d$ obtained by removong the last $n-d$ coordinates, we get a chart $h : M \cap U \to V$. Its inverse $\phi : V \to M \cap U$ is a local parametrization of $M$. Regarding it as a (smooth!) map $\phi : V \to \mathbb R^n$, we can form its derivative $d\phi(x) : \mathbb R^d \to \mathbb R^n$ at $x \in V$. This is a linear map having rank $d$. Now define for $p \in M \cap U$ $$T_pM = \text{im}(d\phi(h(p)). $$ This is a $d$-dimensional linear subspace of $\mathbb R^n$. You can accept this as a definition, but if you want motivation, have a look at

The motivation for a tangent space

Equivalent definition of a tangent space?

What is my mistake in the following proof?.

Inverse of parametric equation.

1 Answers1