2
$\begingroup$

I have this problem and I don't know why I can't finish it:

Let $S=\{x\in \mathbb{R}^4\mid \vert\vert x \vert\vert=2\}$ the sphere of dimension $3$ and radius $2$. Let $T_+$ (resp. $T_-$) the set of the $x=(x_1,x_2,x_3,x_4)\in S$ such that $x_1^2+x_2^2\geq 2$ (resp. $x_1^2+x_2^2\leq 2$). Prove that $T_+$ and $T_-$ are manifolds with boundary, both diffeomorphic to $D^2\times S^1$ and that \begin{align*} S=T_+\cup T_-&& \text{and} && T_+\cap T_-=\partial T_-= \partial T_+ \end{align*}

I used the function $f\colon S\longrightarrow \mathbb{R}$ sending $f((x_1,x_2,x_3,x_4))=x_1^2+x_2^2-2$, and this has $df_x=(2x_1\ \ 2x_2\ \ 0\ \ 0)$, so it is only zero if $x_1=x_2=0$, and this value doesn't affect $0$ to be a regular value. With this function I have that $T_+$ and $T_-$ are manifolds and that their boundaries are the points where $x_1^2+x_2^2=2$ (this forces $x_3^2+x_4^2=2$, to keep the point in $S$), but I can't see why it is diffeomorphic to $D^2\times S^1$. Did I do something wrong?

$\endgroup$
2
$\begingroup$

Suggestion: Show that stereographic projection from $(0, 0, 0, 2)$ maps the rest of $S^{3}$ diffeomorphically to $\mathbf{R}^{3} \times \{0\}$. The common boundary, $$ \{(x_{1}, x_{2}, x_{3}, x_{4}) \in S^{3} : x_{1}^{2} + x_{2}^{2} = 2 = x_{3}^{2} + x_{4}^{2}\}, $$ maps to a torus; check that $T_{-}$ maps to the interior, a solid torus. (It's clear $T_{-}$ and $T_{+}$ are diffeomorphic by permuting coordinates.)

$\endgroup$
  • $\begingroup$ I know that, but I can't see why that helps me :( $\endgroup$ – MonsieurGalois Dec 2 '16 at 11:42
  • $\begingroup$ In fact what I see is that the boundary of $T_+$ ($T_-$) is $S^1\times S^1$, not $D^2\times S^1$ $\endgroup$ – MonsieurGalois Dec 2 '16 at 11:43
  • $\begingroup$ The boundary is $S^{1} \times S^{1}$; the solid tori $T_{\pm}$ are the components of the complement. :) $\endgroup$ – Andrew D. Hwang Dec 2 '16 at 11:49
  • 1
    $\begingroup$ OOOOOOOOOOOOOOOOOOOOOOOOOOH NOW I UNDERSTOOD. I though I needed to prove that $\partial T_{\pm}$ was diffeomorphic to $D^2\times S^1$. $\endgroup$ – MonsieurGalois Dec 2 '16 at 11:55
  • 1
    $\begingroup$ Thanks a lot! your last comment was just what I needed. $\endgroup$ – MonsieurGalois Dec 2 '16 at 11:55

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.