1
$\begingroup$

This material is from a class I am taking so some definition might be different from normal sense. So let me define some necessary concepts first and ask question.

Definition Let $F:M\rightarrow N$ be smooth map between two smooth manifolds.

Then $F$ is called a smooth embedding if it is an immersion(i.e, $F_* : T_pM\rightarrow T_{F(p)}M$ is injective $\forall p\in M$) and $F(M)\subseteq N$ is given with the subspace topology and $F:M\rightarrow F(M)$ is homeomorphism.

--

$S^{n-1}\subseteq \mathbb{R}^n$ is a sphere.

--

Setting : For any $v\in S^{n-1}$, there is a smooth map $\pi_v:\mathbb{R}^n\rightarrow \mathbb{R}^{n-1}$ such that $$ (\pi_v)_*:T_x\mathbb{R}^n\rightarrow T_{\pi_{v}(x)}\mathbb{R}^{n-1} \hspace{3mm}\forall x\in \mathbb{R}^n.$$

with

$$ker(\pi_v)_*=span\{ v \}.$$

--

My Question

There were two statement the professor did not prove.

(1) $\pi_v|_M$ is an immersion iff $$ v\neq \frac{w}{|w|} \hspace{4mm}\forall w\in TM\setminus \{0\}.$$

(2) $\pi_v|M$ is $1$ to $1$ iff $$v\neq \frac{p-q}{|p-q|}\hspace{4mm}\forall p,q\in M\hspace{2mm}\text{ with } p\neq q.$$

I am not sure why the two state above are true. I understood that we are projecting our manifold in the direction of $v$. But I think there would be no problem even if $TM$ contains just one of tangent vector from $span\{v\}$. I will be thanking to any comments.

$\endgroup$
2
$\begingroup$

For $v$ a unit vector, the map $\pi_v$ is $$ u \mapsto u - (u \cdot v) v $$ whose kernel is exactly $H = span{v}$. If $T_xM$, the tangent space to $M$ at $x$, intersects $H$ only in the zero-vector, then the restriction of $\pi_v$ to $T_xM$ has maximal rank, for it sends nothing but the zero-vector to zero.

That means that $(\pi_v)_*$, on each tangent space, $T_xM$, is an isomorphism, which meets your definition for "immersion."

The second claim seems clear. Suppose that $\pi_v(P) = \pi_v(Q)$ for points $P, Q \in M \subset \Bbb R^n$. Then

$$ P - (P \cdot v) v = Q - (Q \cdot v)v $$ Doing a little algebra, we get $$ (P - Q) = ((P-Q) \cdot v)v. $$ So $P-Q$ must be a multiple of $v$. And that means that $p-q$, when normalized, must actually be $\pm v$. If it's $-v$, then swapping the roles of $P$ and $Q$ gives you $+v$.

Similarly, if $Q = P + v$, then it's clear, by direct computations, that $\pi_v(Q) = \pi_v(P)$. So that covers both "if" and "only if".

$\endgroup$
  • $\begingroup$ Thank you! and sorry for confusion! I misdefined the immersion so I corrected. But your answer is still valid! $\endgroup$ – Lev Ban Oct 26 '18 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.