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Take $S^3$ to be the three-sphere, that is, $S^3=\lbrace (x_1,x_2,x_3,x_4):x_1^2+x_2^2+x_3^2+x_4^4=1\rbrace$. Using the stereographic projection, $S^3=\mathbb{R}^3\cup \lbrace \infty \rbrace.$ Can someone explain how the complement of the solid torus (centered at the origin) $S^1\times D^2$, where $D^2$ is a 2-disk, is also a torus? I am reading Milnor's paper "On Manifolds Homeomorphic to the 7-Sphere," and this is a prerequisite to understand how Milnor glues the surfaces of two tori of the form $S^3 \times D^4$ in $S^6$ to create an exotic $7$-sphere.

The Substitute
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2 Answers2

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One solid torus is $x_1^2 + x_2^2 \leq \frac{1}{2}.$ The other solid torus is $x_3^2 + x_4^2 \leq \frac{1}{2}.$ The intersection, a torus, is...

Will Jagy
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  • Ah, definitely easier to see this in $S^4$ rather than $\mathbb R^3\cup {\infty}$. Although not entirely obvious why $x_1^2+x_2^2\leq \frac{1}{2}$ is a solid torus. (I see it, almost, but having a little trouble.) – Thomas Andrews Mar 19 '13 at 19:44
  • Why is $x_1^2 + x_2^2 \le \frac{1}{2}$ a torus? – The Substitute Mar 19 '13 at 19:46
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    Let $(x,y,\cos \theta,\sin\theta)\in D^2\times S^1$. Then send $$(x,y,\cos \theta,\sin\theta) \to (x/\sqrt 2,y/\sqrt 2, \sqrt{1-(x^2-y^2)/2}\cos\theta,\sqrt{1-(x^2-y^2)/2}\sin \theta)$$

    Show that this is $1-1$ into $S^4$ and that it is onto the set $x_1^2+x_2^2\leq \frac{1}{2}$

     – Thomas Andrews Mar 19 '13 at 19:53
  • @ThomasAndrews, yes. – Will Jagy Mar 19 '13 at 19:59
  • @ThomasAndrews, what is that map doing? – The Substitute Mar 20 '13 at 00:12
  • Other than showing that the set $x_1^2+x_2^2\leq 1/2$ in $S^4$ is the homeomorphic image of the solid torus? The point is that if you know that $x_1^2+x_2^2\leq 1/2$ then $(x_3,x_4)\neq 0$ so $(x_3/\sqrt{x_3^2+x_4^2},x_4/\sqrt{x_3^2+x_4^2})$ is on the unit circle. But $x_3^2+x_4^2=1-x_1^2+x_2^2$. @TheSubstitute – Thomas Andrews Mar 20 '13 at 00:23
  • @WillJagy, how is the intersection a torus? The intersection of those two sets is $S^3$? I was expecting the intersection of my tori to be empty. – The Substitute Mar 20 '13 at 15:04
  • @ThomasAndrews, don't you mean $S^3$? Thanks for the input guys. – The Substitute Mar 20 '13 at 15:50
  • @TheSubstitute, the union of the two sets is $\mathbb S^3.$ The intersection is $x_2^2 + x_2^2 = x_3^2 + x_4^2 = 1/2. $ If you look at this with the mapping of Thomas you get $\mathbb S^1 \times \mathbb S^1$ – Will Jagy Mar 20 '13 at 18:19
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The boundary of $x_1^2+x_2^2 < \frac{1}{2}$ (in the $x_1x_2$-plane) is a circle, with equation $x_1^2+x_2^2 = \frac{1}{2}$. Likewise for the boundary of $x_3^2+x_4^2 < \frac{1}{2}$. The boundary of either region is thus the cartesian product of two circles, a torus.

frabala
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WJM3
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