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I am confused with a little detail in Guillemin & Pollack's proof on Whitney embedding theorem. Please see page 54 in their book "Differential topology".

In the second paragraph of page 54, they said that

Recall that the map $\pi\circ F:X\rightarrow H$ is still an injective immersion for almost every $a\in S^{2k+1}$, so we may pick an $a$ that happens to be neither of the sphere's two poles.

I cannot figure this out. Here is my thinking: GP claimed "for almost every $a\in S^{2k+1}$" derived from Sard's theorem. According to the technique used in the proof of the early version of Whitney's theorem (see page 51), we should find a point $a\in \mathbb{R}^{2k+2}$ out of the image $h$ and $g$ to construct $H$ and collapse $\mathbb{R}^{2k+2}$ to $\mathbb{R}^{2k+1}$. In this case ($M=2k+2>2k+1$), every value in the image set is critical value and every value out of the image set is regular value. However, it is possible that the image sets (sets of critical value) of both $h$ and $g$ cover $S^{2k+1}/\{poles\}$, since $S^{2k+1}/\{poles\}$ is still of measure zero in $\mathbb{R}^{2k+2}$, which derives no contradiction. If this happens, we cannot claim the existence of such $a$ not being poles, depending on which the following discussion establishes.

BYW, at the point of what confusing me, we have one-to-one immersion $F:X\rightarrow \mathbb{R}^{2k+2}$. And $h:X\times X\times\mathbb{R}\rightarrow \mathbb{R}^{2k+2}$ is defined by $h(x,y,t)=t[F(x)-F(y)]$, and $g:T(X)\rightarrow \mathbb{R}^{2k+2}$ is defined by $g(x,v)=\mathrm{d}F_x(v)$.

Any details am I missing?

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  • $\begingroup$ Or is there any way to claim the existence of such point? $\endgroup$ – zenos Apr 5 '15 at 9:34
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    $\begingroup$ Can you also include the definition of $h$, $g$? $\endgroup$ – user99914 Apr 5 '15 at 9:47
  • $\begingroup$ @John I have added related definitions. Thanks! $\endgroup$ – zenos Apr 5 '15 at 9:55
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Scaling $a$ will not change $H$ or $\pi$; so it suffices to note that you can find some $a \in \mathbb{R}^{2k + 2} - \{0\}$ (perhaps not in $S^{2k + 1}$) that doesn't project down to either pole once you normalize it (and now it is true that the set of vectors which satisfy this have infinite measure) which satisfies your condition of $\pi \circ F$ being an injective immersion, and then normalize it.

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  • $\begingroup$ Get it! Thanks very much Pedro. This is a 'stupid' question and I was stuck with their 'misleading' words... $\endgroup$ – zenos Apr 5 '15 at 11:50

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