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Near the end of this MAA piece about elliptic curves, the author explains why the complex domain of the cosine function is a sphere: since it's periodic, its domain can be taken as a cylinder, wrapping up the real axis. And because cosine of $\theta\pm i\infty$ is $\infty$, the two ends of the cylinder can be identified with a single point $\infty$. Ok, great, but this sounds to me like a pinched torus. Can I have a clearer explanation why this is a sphere?

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Ted Shifrin
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ziggurism
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  • I think this answer by Robert Israel probably explains it, but can it be made clearer? http://math.stackexchange.com/a/231930/16490 – ziggurism Jul 28 '16 at 06:48
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    Is it really true that $\cos(\theta+i\infty)=\infty$? Isn't $\infty$ an essential singularity of $\cos$? – ziggurism Jul 28 '16 at 15:38
  • The answer by Robert Israel suggests that in fact the cylinder, not the sphere, is the correct answer. – ziggurism Jul 28 '16 at 15:39

2 Answers2

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Let me start by saying you are right in a sense, and I think the article is at best being unclear, but the larger point that the equation of ellipse cuts out a sphere is also right. There are a few things going on, hence a long answer below.

What one is trying to describe is the topology of a set defined by a quadratic equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Of course it is crucially important to explain what values $x$ and $y$ are allowed to take. In the context to the article, that means whether one allows only real values or complex ones as well (this is the question of choosing the field of coefficients), and whether one allows infinite values and in what sense (this is a choice of compactification). Both of these choices affect the topology of the solution set. If only finite real values are allowed the result is the familiar ellipse in the plane (which has topology of a circle); now one has various options about how to treat adding infinite values of $x$ and $y$. However the (real) ellipse does not go to infinity in any way (it stays in finite part of the plane; or, in other words, it is already compact so needs no compactification; algebraically, no real vector with large modulus can solve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$). Thus for this equation over real numbers the second choice is irrelevant. However (spoiler!) it is not irrelevant over complex numbers.

To better see what's going on let's stick to the real variables for now, but consider the equation $uw=1$ instead. The solutions to this in $\mathbb{R}^2$ define a hyperbola and this has topology of 2 lines. Now if we compactify $\mathbb{R}^2$ to the sphere - treating all points at infinity as one - we get one extra point, and the topology of solution set is a wedge of two circles. If we treat $u$-infinities and $v$-infinities as different (compactify each $\mathbb{R}$ to $S^1$, so $\mathbb{R}^2$ is compactified to $S^1\times S^1$), then we get 2 extra points ("$(\infty, 0)$" and "$(0, \infty)$") and a topology of a single circle. We can also compactify $\mathbb{R}^2$ to $\mathbb{R}P^2$ (and still get a circle), or to $\mathbb{D}^2$ (and get two closed segments) or any other number of things, some more natural than others. To finish with the discussion over $\mathbb{R}$, note that if we consider $(u-v)(u+v)=1$ instead of $uv=1$, the two curves in $\mathbb{R}^2$ are isomorphic, but in the compactification $S^1\times S^1$ we get now a wedge of two circles instead of a single circle. This is because the coordinate change of $\mathbb{R}^2$ given by $u_n=u-v$, $v_n=u+v$ taking curve $(u-v)(u+v)=1$ to $u_n v_n=1$ does not extend to these compactifications. This never happens for $\mathbb{R}P^2$, which is one big reason why this is usually the prefered choice of compactification.

Now over complex numbers, the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ has a solution set in $\mathbb{C}^2$ which is a cylinder (aka tangent bundle of $S^1$). Then we can compactify by adding one point for "all infinity" and get a "pinched torus" with both infinities of the cylinder filled in by that one extra point; or by adding a point at infinity to each coordinate separately - this is what is done in the article - and still get a "pinched torus", since $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ forces $x$ to have infinite modulus as soon as $y$ does and vice versa, so the only added point is $(\infty, \infty)$. However, if we compactify to $\mathbb{C}P^2$ we get two extra points ($[a:ib:0]$ and $[a:-ib:0]$), and topology of a sphere. This is in a sense the "right" compactification of the image, sitting inside the "right compactification" $\mathbb{C}P^2$.

Finally, we should note that this is not so much about the domain of $\cos$ but more about the image of $z \to (a\cos z, b\sin z)$ in $\mathbb{C}^2$ and its compactifications. Whether we can extend $\cos z$ or $\sin z$ or $(\cos z, \sin z)$ to a map of some compactification of $\mathbb{C}$ to somewhere depends on where (which extension/compactification of $\mathbb{C}$ or $\mathbb{C}^2$) we are mapping to, and does not really have so much intrinsic meaning.

Max
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  • I think I see it now. The top of the cylinder has "slope" $+ib/a$ and the bottom has slope $-ib/a$, hence they attach to different points in the $\mathbb{C}P^1$ at infinity. Almost exactly the same way that the two asymptotes of the real hyperbola have slopes $\pm b/a$ and so attach to different points in $\mathbb{R}P^1$. – ziggurism Jul 29 '16 at 16:44
  • So are we saying that the map $z \mapsto (\cos z,\sin z)$ can take its domain as all of $\mathbb{C}P^1$? What about the essential singularity at infinity in the trig functions? So perhaps we're only allowed to speak of the compactification/closure of the image of the map $\mathbb{C}\to \mathbb{C}^2\hookrightarrow\mathbb{C}P^2$? – ziggurism Jul 29 '16 at 16:51
  • There are again several things to say. One is that we are not trying to extent "at infinity on $\mathbb{C}P^1$, only on the infinity of the strip, which is a different thing; on the "infinity on the strip" imaginary part goes to infinity, so $\cos z$ goes to infinity as well (not true for other "real infinity directions''). So we can define $\cos z$ as a continuous map from the strip with sides identified and the two points added (which is topologically a sphere) to $\mathbb{C}P^1$. This map is 2 to 1 except for values $\pm 1$ ($\cos -z=\cos z$, but $z=0=-z$ and $z=\pi=-z=-\pi$ on the strip). – Max Jul 29 '16 at 22:12
  • Another thing to note is that this extension is an extension as a map to $\mathbb{C}P^1$. If we wanted to extend as map to something else (say to a map to $\mathbb{R}P^2$, a bit unorthodox but why not...) we might not be able to do that (as $\cos z$ can approach infinity through any real direction: fix a real part $r$ between $-\pi$ and $\pi$; now increase imaginary part to plus infinity; you will approach infinity from angle $-r$). – Max Jul 29 '16 at 22:34
  • Similar considerations apply to extending the map $z \to (\cos z, \sin z)$. It can be continuously extended to the domain (strip with boundary identified, with 2 points added) as a map to $\mathbb{C}P^2$ (in which case it's a homeomorphism onto its image), or as a map to $\mathbb{C}P^1\times \mathbb{C}P^1$ (in which case it glues the two infinity points together). – Max Jul 29 '16 at 22:39
  • OK so you're saying cosine isn't continuous on the whole Riemann sphere, but it can be extended from its domain cylinder to $\pm i\infty$. In this context is it holomorphic at $i\infty$? – ziggurism Jul 30 '16 at 14:28
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Trigonometric functions are circular functions. Complex cosine is: cos z = (e^(iz) + e^(-iz))/2 where z = x + iy where x and y are real numbers and z is a complex number. Suppose we restrict z so modulus z = 1 = modulus (x + iy). Hence, the image of z is the unit circle in the complex plane. What if we were to look at the map X = (x, y)? Let X = (0, 0) correspond to an origin. Then, apparently, the image cos(z) = cos(0) = 1. The unit sphere has radius = 1. Consider any rotation of the plane we choose to call the complex plane passing through any point we have chosen to call the origin, namely X = (0, 0) with the property modulus z = 1. Again, this is the unit circle on that particular complex plane. When modulus z = r for any non-negative r, we have an entire complex plane which may be mapped onto the unit sphere. So much for the prelims. Perhaps making more sense write z = (e^x)(e^(iy) which has modulus r = e^x. Then realize cos(z) has modulus 1 and is entire as (x,y) range over all reals. Recalling the pre-image of cos(z) was z = x + iy, the domain of cos(z) may be mapped onto the unit sphere by starting at z, drawing a secant to the North Pole of the unit sphere centered at (0, 0, 1) and calling the point P on the unit sphere where the secant first pierces the unit sphere starting from z in the complex plane and repeating this for every z in the complex plane. The result is the points P trace out the unit sphere, so that by reverse stereographic projection, we have the unit sphere of all such points P as our domain for cos(z). For finite r = e^x, the North Pole of the unit sphere mapping gets omitted. Now let x approach infinity, so r = e^x approaches infinity. Identify the North Pole of the unit sphere with the point at infinity, and I believe the stereographic mapping for z = (e^x)(e^iy) becomes onto for all real x and y. I hope this helps.

Richard Anthony Baum
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