How Do I show this Map is a Surjective Submersion?

Question

Question

Let $S^2\subseteq \mathbb{R}^3$ be the 2-sphere, and define $TS^2 = \{(p,v)\in \mathbb{R}^3\times \mathbb{R}^3|p\in S^2, v\in T_pS^2\}$. It may be regarded as a level set of the function $F:\mathbb{R}^6\rightarrow \mathbb{R}^2$, $F(p,v) = (p.p, p.v)$. Let $M = \{(p,v)\in TS^2|v.v = 1\}$. Show that each of the two maps $H_i:M\rightarrow S^2$ given by $H_1(p,v)=p$, $H_2(p,v)=v$ is a surjective submersion $M\rightarrow S^2$, with fibers diffeomorphic to $S^1$.

Note: There are some sub-questions that precede this. So, I have already shown that $TS^2$ is a 4-dimensional sub-manifold of $\mathbb{R}^6$. I have also shown that $M$ is a 3-dimensional sub-manifold of $\mathbb{R}^6$. For this question, I have shown that $H_1$ is a surjective submersion with fibers diffeormorphic to $S^1$. However, it seems $H_2$ is not surjective. And I can’t show that the fibers are diffeomorphic to $S^1$. Your help would be greatly appreciated.

Answer 1

To prove that $M$ is a 3-dimensional submanifold of $\mathbb R^6$ you certainly considered the smooth map $$G : \mathbb R^6 \to \mathbb R^3, G(p,v) = (p\cdot p, p \cdot v, v \cdot v)$$ and showed that $(1,0,1)$ is a regular value of $G$. This implies indeed that $M = G^{-1}(1,0,1)$ a 3-dimensional submanifold of $\mathbb R^6$. Note that $$M = \{ (p,v) \in \mathbb R^6 \mid p, v \in S^2, p \cdot v = 0 \}. \tag{1}$$

You also proved that $H_1 : M \to S^2$ is a surjective submersion with fibers diffeomorphic to $S^1$.

Now consider $\phi : \mathbb R^6 \to \mathbb R^6, \phi(p,v) = (v,p)$. This is a diffeomorphism. Clearly $\phi(M) = M$ because by (1) for $(p,v) \in \mathbb R^6$ we have $(p,v) \in M$ iff $\phi(p,v) = (v,p) \in M$. Hence $\phi$ restricts to a diffeomorphism $\psi : M \to M$.

Clearly $H_2 = H_1 \circ \psi$, thus $H_2$ a surjective submersion with fibers diffeomorphic to $S^1$.

Update.

Let us prove that $H_1$ is a surjective submersion. By the above considerations the same is true for $H_2$ because $M$ is symmetric in the variables $p, v$. Alternatively one can modify the $H_1$-proof in the obvious way.

1. $H_1$ is surjective with fibers diffeomorphic to $S^1$.

Let us first rewrite $(1)$ in a convenient form. We have $T_p S^2 = \{ v \in \mathbb R^3 \mid p \cdot v = 0 \} = p^\bot = $ orthogonal complement of the $1$-dimensional subspace spanned by $p$. Let $S^1_p$ denote the unit circle in $T_pS^2$. Then $$M = \bigcup_{p \in S^2}\{ p \} \times S^1_p .$$

This shows that $H_2 : M \to S^2$ is surjective with fibers $H_2^{-1}(p) = \{ p \} \times S^1_p$ which are diffeomorphic to $S^1$.

2. $H_1$ is submersion.

The tangent space $T_{(p,v)}M$ at $(p,v) \in M$ is given as $\ker D_{(p,v)} G$. See Gross and Meinrenken Chapter 5.2.3 "Tangent Spaces of Submanifolds", p. 117. To compute it, we use the Jacobian matrix of $G$ which is the $(3 \times 6)$-matrix

$$J_{(p,v)} G = \begin{pmatrix} 2p & 0 \\ v & p \\ 0 &2v \end{pmatrix}$$ Note that both of the above "vectorial" columns represent three "scalar" columns.

Let $a, b \in \mathbb R^3$ and $a^t, b^t$ the corresponding column vectors. Then $(a,b) \in \ker D_{(p,v)} G$ iff $$J_{(p,v)} G \cdot \begin{pmatrix} a^t \\ b^t \end{pmatrix} = 0 .$$ This gives the three equations

• $2p \cdot a = 0$
• $v \cdot a + p \cdot b = 0$
• $2v \cdot b = 0$

In other words $$T_{(p,v)} M = \{ (a,b) \in \mathbb R^6 \mid p \cdot a = 0, v \cdot b = 0, v \cdot a + p \cdot b = 0 \}.$$

We have to analyze $$T_{(p,v)}H_1 : T_{(p,v)}M \to T_{H_1(p,v))}S^2 = T_p S^2 .$$

Let $\pi_1 : \mathbb R^6 \to \mathbb R^3$ be the projection onto the first $3$ coordinates. Then $D_{(p,v)}\pi_1 : T_{(p,v)}\mathbb R^6 = \mathbb R^6 \to \mathbb R^3 = T_p\mathbb R^3$ is nothing else than $\pi_1$ (this is true for all linear maps).

$T_{(p,v)}H_1$ is the restriction of $D_{(p,v)}\pi_1 = \pi_1$. Its kernel is therefore $$\ker T_{(p,v)}H_1 = \{ (a,b) \in T_{(p,v)} M \mid a = 0\} = \{ (0,b) \in \mathbb R^6 \mid v \cdot b = p \cdot b = 0 \} \\= \{0\} \times \operatorname{Span}(p, v)^\bot .$$

Here $\operatorname{Span}(p, v)$ denotes the linear subspace spanned by $p$ and $v$. Since $p \cdot v = 0$, this subspace is $2$-dimensional and its orthogonal complement in $\mathbb R^3$ is $1$-dimensional. The image of $T_{(p,v)}H_1$ is isomorphic to $T_{(p,v)}M / \ker T_{(p,v)}H_1$, thus it has dimension $2$ which means that $T_{(p,v)}H_1$ is surjective.

Here is an alternative proof which shows that $H_1$ is a submersion.

Given $(p,v) \in M$, the idea is to find an open neighborhood $U$ of $p$ in $S^2$ and a smooth map $\phi : U \to M$ such that $\phi(p) = (p,v)$ and $H_1 \circ \phi = i : U \hookrightarrow S^2$. Then $T_{(p,v)}H_1 \circ T_p\phi = T_p i = id$ which shows that $T_{(p,v)}H_1$ must be surjective.

This requires to find a smooth map $g : U \to \mathbb R^3$ such that $g(p) = v$ and $g(q) \in S^1_q \subset T_qS^2$.

Since $p \cdot v = 0$, i.e. $p,v$ are linearly independent, we try the Ansatz $g(q)= a p + b v$ with $a, b \in \mathbb R \setminus \{0\}$. Let us begin by taking $b = 1$ and weakening the requirement to $g(q) \in T_qS^2$. This means $$q \cdot g(q) = a (q \cdot p) + q \cdot v = 0 $$ which is equivalent to $$a = - \frac{q \cdot v}{q \cdot p}. $$ $a$ is well-defined on $V = \{ q \in \mathbb R^3 \mid q \cdot p \ne 0\}$. This is an open neigborhood of $p$ in $\mathbb R^3$. Then indeed $$\gamma : V \to \mathbb R^3, \gamma(q) = - \frac{q \cdot v}{q \cdot p}p + v$$ is a smooth map with $\gamma(q) \in T_qS^2$ for $q \in U = V \cap S^2$. Since $p$ and $v$ are linearly independent, we see that $\gamma(q) \ne 0$. Hence also

$$g : V \to \mathbb R^3, g(q) = \frac{\gamma(q)}{\lVert \gamma(q) \rVert}$$ is smooth and has the property $g(q) \in S^1_q$ for $q \in U$. Now we can take $\phi(q) = (q, g(q))$.