This is homework so no answers please

Here is the problem: Show that $UM:=\{(x,v)\in T\mathbb{R}^{n}:x\in M^{m}, v\in T_{x}M^{m},|v|=1\}$ is a (2m-1)-dim submanifold of $T\mathbb{R}^{n}$.

My attempt is ridiculously long, so I was wondering if I can get hints for a shorter one.

Also, is there a rigorous way to shrink open sets? At location (***), I need to shrink an open set $\pi^{-1}(U)$, to fit its image into another open set V. The problem is that both sets $\pi^{-1}(U)$, V are arbitrary and so I am not sure how to do rigorous shrinking.

Here is my attempt:

The goal is to show $UM$ satisfies the (2m-1) local slice criterion in $T\mathbb{R}^{n}$ for chart $(W,f)$. We will build f as a composition of charts.

$\blacktriangleright$ Since $M^{m}$ is an embedded submanifold, for each $p\in M$ there exists chart $(U,\phi)$ in $\mathbb{R}^{n}$ containing it s.t. $\phi(U\cap M)=\{(x_{1},...,x_{n})\in U: x_{m+1}=...=x_{n}=0\}$. The associated map for the tangent bundle is:

$\widetilde{\phi}:\pi^{-1}(U)\to \mathbb{R}^{2n}$ defined as $\widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=(x_{1},...,x_{n},v_{1},...,v_{n})$, where $\pi:T\mathbb{R}^{n}\to \mathbb{R}^{n}$.

Therefore, for $p\in M$ we have $\widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m},0,...,0)$.

$\blacktriangleright$ The idea is to compose $\widetilde{\phi}$ by diffeomorphic map $(v_{1},...,v_{m})\mapsto (v_{1},...,v_{m-1},0)$, where $|(v_{1},...,v_{m})|=1\Leftrightarrow (v_{1},...,v_{m})\in \mathbb{S}^{m-1}$, and so we get a chart on $\pi^{-1}(U)\cap UM$.

The $\mathbb{S}^{m-1}$ is an embedded (m-1)-dim submanifold of $\mathbb{R}^{m}$ and so we have some $(V,\psi)$ s.t. $\psi(V\cap \mathbb{S}^{m-1})=\{(x_{1},...,x_{m})\in V: x_{m}=0\}$. Then define $g:\mathbb{R}^{n}\times V\times \mathbb{R}^{n}\to \mathbb{R}^{n}$ as: $$g(x_{1},...,x_{n},v_{1},...,v_{n})=(x_{1},...,x_{n},\psi(v_{1},...,v_{m}),v_{m+1},...,v_{n})$$. The map g is a diffeomorphism as it is the identity in $\mathbb{R}^{n}\times \mathbb{R}^{n}$ and the diffeomorphic chart $\psi$ on V.

$\blacktriangleright$ We will show that the desired map is $f:=g\circ \widetilde{\phi}$ and $W:=\pi^{-1}(U)\cap UM$.\ Well-defined: g's domain is $\mathbb{R}^{n}\times V\times \mathbb{R}^{n}$, so we need to make sure $\pi_{(n+1,n+m)}(\widetilde{\phi}(\pi^{-1}(U) ))\subset V$, where $\pi_{(n+1,n+m)}:\mathbb{R}^{2n}\to \mathbb{R}^{m}$ be the projection onto $n+1,...,n+m$ coordinates. (***)

The map $g\circ \widetilde{\phi}:\pi^{-1}(U)\to \mathbb{R}^{2n}$ is a composition of diffeomorphisms and thus a coordinate chart on $\pi^{-1}(U)$. Now we need to show where it sents $UM\cap \pi^{-1}(U)$.

Derivations in $UM\cap \pi^{-1}(U)$ are tangent to M and so $\widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m},0,...,0)$. Since $(v_{1},...,v_{m})\in \mathbb{S}^{m-1}$, we get $\psi(v_{1},...,v_{m})=(v_{1},...,v_{m-1},0)$ and so $$g\circ \widetilde{\phi}(v_{i}\frac{\partial }{\partial x_{i}}|_{p})=g(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m},0,...,0)=$$ $$=(x_{1},...,x_{m},0,...,0,v_{1},...,v_{m-1},0,0,...,0)$$.

Thus, $g\circ \widetilde{\phi}(UM\cap \pi^{-1}(U))=\{(x_{1},...,x_{n},v_{1},...,v_{n})\in g\circ \widetilde{\phi}(\pi^{-1}(U)): x_{m+1}=...=x_{n}=v_{m}=...=v_{n}=0\}$.


  • 2
    $\begingroup$ Think of $UM$ as a submanifold of $TM$ with one restriction. That is, consider $UM$ as the preimage of $1$ under $\phi:TM\to \mathbb{R}, (x,v)\to |v|^2.$ $\endgroup$
    – mfl
    Oct 28, 2014 at 0:00
  • 1
    $\begingroup$ i knew there was a level set trick. Thanks $\endgroup$
    – TKM
    Oct 28, 2014 at 4:06

4 Answers 4


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 $T\Bbb R^n\approx \Bbb R^n\times \Bbb R^n$.


It's much easier to avoid the use of coordinate charts altogether and using the properties of level sets of smooth functions.

Hint Denote by $g$ the metric induced on $M$ by pulling back the Euclidean metric on $\Bbb R^n$ via the inclusion map $M \hookrightarrow \Bbb R^n$. We can identify $g$ with the smooth map $$\hat g : TM \to \Bbb R , \qquad (p, X) \mapsto g_p(X, X)$$ that maps a vector $X$ to the square of its norm.

Additional hint By definition, $UM$ is the level set $\hat g^{-1}(1)$. So, if $1$ is a regular value of $\hat g$, that is, that $\hat g$ has constant rank $1$ on $UM$, then $UM$ is an embedded submanifold of $TM$ of codimension $1$.

Remark Notice that we only used the embedding to identify the metric on $M$. So, the embedding is irrelevant in the sense that the argument applies just as well to an abstract Riemannian manifold.

  • 1
    $\begingroup$ About your additional hint: is there an easy way of seeing that 1 is a regular value of $\hat{g}$? I tried to compute derivatives for $\hat{g}$ in local coordinates, but this just got me a lot of calculations that lead me nowhere, I was wondering if there is an easier way to show that. $\endgroup$
    – Andrew V
    Sep 5, 2020 at 22:53
  • 1
    $\begingroup$ It suffices to show that $\hat g$ has rank $1$ everywhere on the unit sphere bundle $UM \subset TM$, and in particular it suffices to show that $\hat g_p$ has rank $1$ on the unit sphere $U_p M \subset T_p M$. The tangent map to $\hat g_p$ at $X \in T_p M$ is a map $T_X \hat g: T_X T_p M \to T_1 \Bbb R$. But both $T_p M$ and $T_1 \Bbb R$ are vector spaces, so there are canonical identifications $T_X T_p M \cong T_p M$ and $T_1 \Bbb R \cong \Bbb R$. $\endgroup$ Sep 10, 2020 at 3:10
  • 2
    $\begingroup$ Thus, we can identify the tangent map $T_X \hat g_p$ with a map $T_p M \cong \Bbb R$, and unwinding definitions gives that $$T_X \hat g_p \cdot X = 2 g_p(X, X) = 2 \hat g_p(X) ,$$ so the tangent map $\hat g_p$ has rank $1$---and thus so does $T_X \hat g$---everywhere that $g_p(X, X) = 0$, that is, for all nonzero vectors $X$, including in particular those of unit length. $\endgroup$ Sep 10, 2020 at 3:13
  • 1
    $\begingroup$ The previous comment should read, "... $g_p(X, X) \color{red}{\neq} 0$ ...". $\endgroup$ Feb 2, 2022 at 21:04

So, actually $(x,v)\in TM^m $ a manifold of dimension $2m $. Define a function $f:TM^m \rightarrow \mathbf {R} $ as $f (x,v)=|v|$; assuming everything is embedded in some big euclidean space. Check if pre-image theorem works.


The solution is wrong.

$v_1^2+\dots+v_m^2=1$ is not equivalent to $|v|=1$, because the choice of the chart is random.


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