How to show $UM$ is an embedded $(2m-1)$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$?

Question

Because $TM$ is an embedded $2m$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$,

we can define the smooth map $\varphi: TM\to \Bbb R$ by

$(x_1,\cdots,x_n,v_1,\cdots,v_n)\mapsto \sqrt{v_1^2+\cdots+v_n^2}$.

then we can get $UM=\varphi^{-1}(1)$.

If we can show $\forall (x,v)\in UM$, $d\varphi_{(x,v)}$ is surjective, then $UM=\varphi^{-1}(1)$ is a regular level set, then we can use the Corollary 5.14 to show $UM$ is an embedded $(2m-1)$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$.

But how to show $\forall (x,v)\in UM$, $d\varphi_{(x,v)}$ is surjective?

Answer 1

Let $F:M\to \Bbb R^n$ be the inclusion map and $dF: TM\to T\Bbb R^n$ the smooth map induced by $F$,

then we have $$dF:TM\to T\Bbb R^n,$$ $$(x,v)\mapsto (x,v).$$

$\forall x\in M$, we choose a smooth chart containing $x$ on $M$, then $dF$ has the following coordinate representation in terms of natural coordinates for $TM$ and $T\Bbb R^n$:

$dF(x^1,\cdots,x^m,v^1,\cdots,v^m)=(F^1(x),\cdots,F^n(x),\frac{\partial F^1}{\partial x^i}(x)v^i,\cdots,\frac{\partial F^n}{\partial x^i}(x)v^i).$

We composite $dF:TM\to T\Bbb R^n$ with $T\Bbb R^n\to \Bbb R$ defined by $(x,v)\mapsto |v|^2$, then we get $\Phi: TM\to \Bbb R$ defined by $(x,v)\mapsto |v|^2$. Correspondingly, $\Phi$ has the following coordinate representation:

$\Phi(x^1,\cdots,x^m,v^1,\cdots,v^m)=\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x)v^i)^2$.

Suppose $\Phi(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i)^2=1$.

Because we have

$\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n\frac{\partial F^k}{\partial x^1}(x_0)(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i),$

$\qquad \qquad \qquad \vdots$

$\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n\frac{\partial F^k}{\partial x^m}(x_0)(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i),$

then $v_0^1\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)+\cdots+v_0^m\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=2\sum\limits_{k=1}^n(\frac{\partial F^k}{\partial x^i}(x_0)v_0^i)^2=2$,

so at least one of $\frac{\partial\Phi}{\partial v^1}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m),\cdots,\frac{\partial\Phi}{\partial v^m}(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)$ is not equal to $0$, then $(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)$ is a regular point of $\Phi$ such that $\Phi(x_0^1,\cdots,x_0^m,v_0^1,\cdots,v_0^m)=1$, hence $\Phi^{-1}(1)$ is a regular level set.

By Corollary 5.14(Regular Level Set Theorem) of Introduction to Smooth Manifolds by Lee, $UM=\Phi^{-1}(1)$ is an embedded $(2m-1)$-dimensional submanifold of $TM$, thus an embedded $(2m-1)$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$.

Answer 2

Choose local coordinates $x_1,...,x_m$ for a small open set $V\subset M$. Then over $V$ we can assume that $TM$ is trivial. Thus if $\pi:TM\to M$ is the projection map then local coordinates for $\pi ^{-1}(V)=V\times\mathbb{R}^m$ are $x_1,...,x_m,v_1,...,v_m$ where $v_1,...,v_m$ are the standart coordinates for $\mathbb{R}^m$.

Moreover, by the Gramm-Schmiddt process, we assume that the trivialization is consistent with the scalar product on $TM$. This means that if $u,v\in T_xM$ are represented by $v_i,u_i$ then $<u,v>=\sum u_iv_i$, where the left hand side is just the scalar product in $\mathbb{R}^m$.

Now we look at $\psi:TM\to \mathbb{R}$ defined by $\psi(x,v)=||v||^2-1$. Then $UM=\psi^{-1}(0).$ To check that $0$ is indeed a regular value of $\psi$, it is enough to check it locally. Now locally in our coordinates this map is just: $\begin{equation} (x_1,...,x_m,v_1,...,v_m)\mapsto v_1^2+\dots +v_m^2 -1. \end{equation} $

And its gradiant is $(0,...,0,2v_1,...,2v_m)$, which is nonzero whenever $\psi=0$.

Remark: This could also be done with your choice of $\psi$.