# The unit tangent bundle for submanifold $M^{m}\subset \mathbb{R}^{n}$ is a (2m-1)-dim submanifold

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

Thanks

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

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.

• 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. Sep 5, 2020 at 22:53
• 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 $T_1 \Bbb R \cong \Bbb R$. Sep 10, 2020 at 3:10
• 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. Sep 10, 2020 at 3:13
• The previous comment should read, "... $g_p(X, X) \color{red}{\neq} 0$ ...". 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.