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.

  • $\begingroup$ 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

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 $\mathbb{R}^m$.

  • $\begingroup$ 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

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