9
$\begingroup$

Let $\phi: M \to N$ be an immersion from smooth manifold $M^m$ into $N^n$ ($\dim M = m$ and $\dim N = n$). Prove there exists smooth charts $(U,h)$ in $M$ with $p \in U$, $h(p) = 0$, and $(V,g)$ in $N$ with $\phi(p) \in V$, $g(\phi(p)) = 0$ such that the transition map $$g \circ \phi \circ h^{-1}(x_1, \dots, x_m) = (x_1, \dots, x_m, 0,\dots,0)$$ in a neighbourhood of $0$.

Consider the case for $\phi: W \to \Bbb R^n$ is smooth with $W $ be open in $\Bbb R^m$ containing $0$ and $\phi(0) = 0$ and the Jacobian of $\phi$ is $J(\phi)_{m \times m} \neq 0$. Then use Inverse Function Theorem on $F : W \times \Bbb R^{n - m} \to \Bbb R^n$ where $$F(x_1, \dots, x_n) = (\phi_1(x_1 ,\dots, x_m),\dots, \phi_m(x_1 ,\dots, x_m),x_{m+1},\dots,x_n).$$

So I showed that $J(F) = \begin{bmatrix} J(\phi) &0 \\ 0& I \end{bmatrix}_{n \times n}$ is of course invertible, then Inverse Function Theorem says I have open set $0 \in L \subset W \times \Bbb R^{n -m}$ and $F(0) \in F(L) \subset \Bbb R^n$ where $F$ is a local diffeomoprhism.

In this suggests I need to relate the given charts from the open sets I have shown here to finish off the transition map to show $$g \circ \phi \circ h^{-1}(x_1, \dots, x_m) = (x_1, \dots, x_m, 0,\dots,0)$$

If someone could help me glue everything together, it would be great.

$\endgroup$
13
  • $\begingroup$ Why the charts may not be smooth? $\endgroup$ – Martín-Blas Pérez Pinilla Nov 13 '14 at 18:08
  • 1
    $\begingroup$ Yes, before defining the differentiable structure. But after... $\endgroup$ – Martín-Blas Pérez Pinilla Nov 13 '14 at 18:20
  • $\begingroup$ @Martín-BlasPérezPinilla, I erase that comment. $\endgroup$ – Hawk Nov 13 '14 at 19:11
  • $\begingroup$ Actually, not $h\circ F$ but $F^{-1}\circ g$. And then you need another small variation, giving you a chart around $\phi(p)$ of the form $G\circ F^{-1}\circ g$. $\endgroup$ – Daniel Fischer Nov 14 '14 at 0:45
  • $\begingroup$ I didn't even look at the link. I should have used a different symbol. You start with two (smooth) charts $h$ mapping $p$ to $0$, and $g_1$ mapping $\phi(p)$ to $0$. That gives you $\phi_1 = g_1\circ \phi \circ h^{-1}$ as a function from an open neighbourhood $W$ of $0$ in $\mathbb{R}^m$ into $\mathbb{R}^n$. If necessary, permute the coordinates in $\mathbb{R}^n$ so that the first $m$ rows of $J(\phi_1)$ are linearly independent. The permutation is a diffeomorphism $P$ of $\mathbb{R}^n$. Let $g_2 = P\circ g_1$. Then you are in the "Consider the case" situation. Let $g_3 = F^{-1}\circ g_2$. ... $\endgroup$ – Daniel Fischer Nov 14 '14 at 1:02
8
+100
$\begingroup$

First, we use two charts, $(U,h)$ with $h(p) = 0 \in \mathbb{R}^m$, and $(V,g_1)$ with $g_1(\phi(p)) = 0 \in \mathbb{R}^n$ to transport the problem to Euclidean space. Shrinking $h$ if necessary to have $\phi(U) \subset V$, we obtain a smooth $\psi = g_1 \circ \phi \circ h^{-1} \colon \underbrace{h(U)}_{W} \to \mathbb{R}^n$ with $\psi(0) = 0$, and the rank of $J(\psi)(0)$ being $m$.

From this point on, we only consider local diffeomorphisms of $\mathbb{R}^n$ at $0$ - and possibly shrink $W$ if necessary - to achieve the desired representation.

If $n > m$, it need not be the case that the first $m$ rows of $J(\psi)(0)$ are linearly independent, so we need a permutation of the coordinates of $\mathbb{R}^n$ to achieve that. For $\pi \in S_n$, the map

$$P_\pi \colon (x_1,\dotsc,x_n) \mapsto (x_{\pi(1)},\dotsc,x_{\pi(n)})$$

is a diffeomorphism of $\mathbb{R}^n$ (with inverse $P_{\pi^{-1}}$), and for some $P = P_\pi$, the composition $\chi = P\circ \psi$ has the first $m$ rows of $J(\chi)(0) = J(P)(0)\cdot J(\psi)(0)$ linearly independent.

Now we are - except for notation - in the "consider the case" situation. We consider the map

$$F \colon W\times \mathbb{R}^{n-m} \to \mathbb{R}^n,\quad F(x_1,\dotsc,x_n) = \begin{pmatrix}\chi_1(x_1,\dotsc,x_m)\\ \vdots\\ \chi_m(x_1,\dotsc,x_m)\\x_{m+1}\\ \vdots\\ x_n\end{pmatrix}.$$

$F$ is smooth, and its Jacobi matrix at $0$ is

$$J(F)(0) = \begin{bmatrix} J(\chi^{(m)})(0) & 0 \\ 0 & I\end{bmatrix},$$

where $\chi^{(m)}$ denotes the first $m$ components of $\chi$. By construction, $J(\chi^{(m)})(0)$ is invertible, hence $J(F)(0)$ is invertible, and $F$ is a local diffeomorphism of $\mathbb{R}^n$ at $0$ by the inverse function theorem, say $F\lvert_{V_1}\colon V_1 \to V_2$ is a diffeomorphism with $V_1,V_2$ open neighbourhoods of $0$ in $\mathbb{R}^n$.

We shrink $W$ if necessary to have $\chi(W) \subset V_2$, and consider $\eta = (F\lvert_{V_1})^{-1} \circ \chi$. We have

$$\begin{pmatrix} \chi_1(x') \\ \vdots \\ \chi_m(x') \\ \chi_{m+1}(x')\\ \vdots \\ \chi_n(x')\end{pmatrix} = \chi(x') = (F\circ \eta)(x') = \begin{pmatrix} \chi_1(\eta^{(m)}(x'))\\ \vdots\\ \chi_m(\eta^{(m)}(x'))\\\eta_{m+1}(x')\\\vdots\\ \eta_n(x')\end{pmatrix},$$ where $\eta^{(m)}$ denotes the first $m$ components of $\eta$ and $x' \in W$. Since $\chi^{(m)}$ is a local diffeomorphism of $\mathbb{R}^m$ at $0$, it follows that $\eta^{(m)}(x') = x'$ in some neighbourhood $W'\subset \mathbb{R}^m$ of $0$. Shrink $W$ if necessary to assume $W = W'$. So we have

$$\eta(x_1,\dotsc,x_m) = (x_1,\dotsc, x_m, \eta_{m+1}(x_1,\dotsc,x_m),\dotsc, \eta_n(x_1,\dotsc,x_m)),$$

or, by slight abuse of notation, $\eta(x') = (x',\tilde{\eta}(x'))$.

Now consider the map

$$G\colon W\times \mathbb{R}^{n-m},\quad (x',x'') \mapsto (x', x'' - \tilde{\eta}(x')).$$

We have

$$J(G)(0) = \begin{bmatrix} I & 0 \\ J(\tilde{\eta})(0) & I\end{bmatrix},$$

which is evidently invertible, so the restriction of $G$ is a diffeomorphism between two open neighbourhoods $V_3$ and $V_4$ of $0$ in $\mathbb{R}^n$. If necessary, shrink $V_1$ and accordingly $V_2$, and consequently $W$ so that $V_1 \subset V_3$.

Then $\iota = G\circ \eta \colon x' \mapsto G(x',\tilde{\eta}(x')) = (x', \tilde{\eta}(x') - \tilde{\eta}(x')) = (x',0)$ is the representation we wanted, and unraveling the construction,

$$\iota = G\circ \eta = G\circ (F\lvert_{V_1})^{-1} \circ \chi = G\circ (F\lvert_{V_1})^{-1} \circ P \circ \psi = G\circ (F\lvert_{V_1})^{-1} \circ P \circ g_1 \circ \phi \circ h^{-1},$$

we see that the chart

$$g = G\circ (F\lvert_{V_1})^{-1} \circ P \circ g_1$$

gives the desired representation.

In particular, we see that for every chart $h$ around $p$ with $h(p) = 0$, there exists a chart $g$ around $\phi(p)$ such that $g\circ \phi \circ h^{-1}$ has the desired form on a neighbourhood of $0$ in $\mathbb{R}^m$.

$\endgroup$
2
  • $\begingroup$ May I ask where you get the map $G$ from? $\endgroup$ – Hawk Nov 18 '14 at 1:24
  • $\begingroup$ Looking at what $\eta$ does, $G$ is just the map we need to get the map into the form $(x_1,\dotsc,x_m) \mapsto (x_1,\dotsc,x_m,0,\dotsc,0)$. $\endgroup$ – Daniel Fischer Nov 18 '14 at 9:42

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.