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Guillemin and Pollack asks:

Explicitly exhibit enough parameterizations to cover $S^1 \times S^1 \subset R^4$.

My solution is given by $8$ parameterizations (closely following their example of parameterizing a circle):

$$f_1(x,z) = (x, \sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_2(x,z) = (x, -\sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_3(x,z) = (x, \sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_4(x,z) = (x, -\sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_5(y,w) = (y, \sqrt{1 - y^2}, w, \sqrt{1 - w^2})$$ $$f_6(y,w) = (y, -\sqrt{1 - y^2}, w, \sqrt{1 - w^2})$$ $$f_7(y,w) = (y, \sqrt{1 - y^2}, w, -\sqrt{1 - w^2})$$ $$f_8(y,w) = (y, -\sqrt{1 - y^2}, w, -\sqrt{1 - w^2})$$

I believe these are all I would need In terms of parameterizations but I am unsure, can anyone confirm or deny this?

Dair
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  • You'd need to state the domains of your parameterizations. (It looks as if all of them are $(-1, 1) \times (-1, 1)$ or $[-1, 1] \times [-1, 1]$. ) In the first case, they're open sets, but don't cover; in the second case, they cover, but they're not open. If it's an open cover you need, then you're not done yet. – John Hughes Sep 26 '16 at 00:13
  • @JohnHughes: Yes the parameterizations are meant to cover $(-1, 1) \times (-1, 1)$. Hmm, if there are points not being covered by a particular parameterization, then I'm guessing I probably need another $8$ functions to consider $f_i(x,w)$ and $f_k(y,z)$... – Dair Sep 26 '16 at 00:19
  • Well...only another 4. Notice that your function $f_5$ is the same as your function $f_1$, just with different names on the arguments. – John Hughes Sep 26 '16 at 00:39
  • @JohnHughes: I guess that makes sense, just kind of weird how in the book they use four different parameterizations for $S^1$... I guess $2$ of them are redundant since $y$ could be replaced by $x$ in a couple cases. – Dair Sep 26 '16 at 00:44
  • Perhaps needless to say, all the plus signs under the radicals in the fourth coordinate should be minus signs. Separately, with the scheme of covering semicircles you do need eight charts; your $f_{1}$ through $f_{4}$, say, plus the four charts $$f_{5}(y, w) = (\sqrt{1 - y^{2}}, y, \sqrt{1 - w^{2}}, w)$$ (etc.) obtained by swapping in pairs the first two coordinates and the last two coordinates (and re-naming). Finally, if you use sine and cosine you can get away with four charts. – Andrew D. Hwang Sep 26 '16 at 01:05
  • @AndrewD.Hwang: Ok, I believe I have fixed those issues now. I messed up my reading of the book, when they use $y$ the switch them around like you have pointed out. (It also explains why the you use $y$) – Dair Sep 26 '16 at 01:18
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    @Dair: The maps $f_{1}$ through $f_{8}$ in your answer look fine. (They don't match $f_{1}$ through $f_{8}$ in your edited question, if it matters; the components still need to be swapped in pairs.) The other eight maps are unnecessary. :) – Andrew D. Hwang Sep 26 '16 at 10:30

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As pointed out by John Hughes and Andrew Hwang I have not considered every parameterization needed and made some mistakes in the existing parameterization. I have included an additional parameterizations and fixed the existing ones:

$$f_1(x,z) = (x, \sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_2(x,z) = (x, -\sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_3(x,z) = (x, \sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_4(x,z) = (x, -\sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_5(y,w) = (\sqrt{1 - y^2}, y, \sqrt{1 - w^2}, w)$$ $$f_6(y,w) = (-\sqrt{1 - y^2}, y, \sqrt{1 - w^2}, w)$$ $$f_7(y,w) = (\sqrt{1 - y^2}, y, -\sqrt{1 - w^2}, w)$$ $$f_8(y,w) = (-\sqrt{1 - y^2}, y, -\sqrt{1 - w^2}, w)$$

$$f_9(x,w) = (x, \sqrt{1 - x^2}, w, \sqrt{1 - w^2})$$ $$f_{10}(x,w) = (x, -\sqrt{1 - x^2}, w, \sqrt{1 - w^2})$$ $$f_{11}(x,w) = (x, \sqrt{1 - x^2}, w, -\sqrt{1 - w^2})$$ $$f_{12}(x,w) = (x, -\sqrt{1 - x^2}, w, -\sqrt{1 - w^2})$$ $$f_{13}(y,z) = (\sqrt{1 - y^2}, y, \sqrt{1 - z^2}, z)$$ $$f_{14}(y,z) = (-\sqrt{1 - y^2}, y, \sqrt{1 - z^2}, z)$$ $$f_{15}(y,z) = (\sqrt{1 - y^2}, y, -\sqrt{1 - z^2}, z)$$ $$f_{16}(y,z) = (-\sqrt{1 - y^2}, y, -\sqrt{1 - z^2}, z)$$

Note: $f_9$, $f_{10}$ ... are unnecessary, as an appropriate variable substitution can be made.

Dair
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  • All those plusses must be wrong, for if you look at the square of the third coord plus the square of the 4th coord for, say, $f_9$, you get $$ w^2 + 1 + w^2 = 1 + 2w^2,$$ which means that point is not on $S^1$ unless $w = 0$. – John Hughes Sep 26 '16 at 01:04
  • @JohnHughes: Whoops, didn't notice the $+$ sign. It should be $-$. Will change. – Dair Sep 26 '16 at 01:06