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For $P$ a plane in $\mathbb{R^3}$, let $\pi_P:\mathbb{R}^3 \to P$ denote the orthogonal projection onto $P$. Suppose that $g: S^1 \to \mathbb{R}^3$ is a smooth embedding. Show that there exists a plane $P$ for which $\pi_{P}\circ g$ is an immersion.

Let $\psi: S^1 \to\mathbb{RP}^2$ be a map defined by $p \to [g_{*}(\frac{d}{dt}|_p)]$. I claim that $\psi$ is not onto. Since $\dim(S^1) \lt \dim (\mathbb{RP}^2)$, every point of $S^1$ is a critical point and by Sard's Theorem, $\psi(S^1)$ has measure zero in $\mathbb{RP}^2$. Hence $\psi(S^1)$ can't be whole of $\mathbb{RP}^2$. Let $L \in \mathbb{RP}^2 \setminus \psi(S^1)$. Now consider the projection $\pi_L$. I think that $\pi_L \circ g$ is an immersion. I am kind of stuck here. How do I show this??

Thanks for the help!!

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You're right there.

Note, however, that you've defined $\pi_P$ for planes $P$, not lines $L$. So, having chosen $L$, you want to write $\pi_P$ where $P$ is the plane orthogonal to $L$.

What's the derivative of $\pi_P\circ g$? The chain rule says it's $(\pi_P\circ g)'(t)=d\pi_P(g'(t))= \pi_P(g'(t))$. Why is this everywhere nonzero? The kernel of $\pi_P$ is, as you said, the set of vectors orthogonal to $P$.

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  • $\begingroup$ $\pi_L(g'(t))=0$ iff $g'(t)$ is orthogonal to $L$ right? $\endgroup$ – tattwamasi amrutam Aug 12 '17 at 22:56
  • $\begingroup$ Ah, I see the confusion. You defined $\pi_P$ for planes $P$ and $L$ is a line. So you need not $\pi_L$, but what? $\endgroup$ – Ted Shifrin Aug 12 '17 at 23:09
  • $\begingroup$ I am a little confused here. $\endgroup$ – tattwamasi amrutam Aug 12 '17 at 23:11
  • $\begingroup$ See my edit. It's a good lesson, by the way, to make sure you're being consistent with your own notation :) $\endgroup$ – Ted Shifrin Aug 12 '17 at 23:15
  • $\begingroup$ I see now. Orthogonal to $P$ would mean that I am parallel to $L$ but by the choice of $L$ , I am not. Is this right? $\endgroup$ – tattwamasi amrutam Aug 12 '17 at 23:22

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