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The graph of $xy = 1$ is connected in $\mathbb{C}^2$.

The above statement is true. Why? Please show reason.

In $\mathbb{R}^2$ $xy = 1$ is not connected as it has two disjoint components in $1$-st and $3$-rd quadrant. I have no idea about this function in $\mathbb{C}^2$. I do not know multivariate complex analysis.

Thank you for your help.

Supriyo
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3 Answers3

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Given any point $(x,1/x)$ on the graph, the point $(xe^{i\theta}, (1/x)e^{-i\theta})$ is also on the graph for all $\theta$, and the set of all such points is connected. Now, choose $\theta$ so that $xe^{i\theta}$ is positive and real...

Micah
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  • I understand "Given any point $(x,1/x)$ on the graph, the point $(xe^{iθ},(1/x)e^{−iθ})$ is also on the graph for all $θ$".. but there after why is that collection is connected.. :O COuld you please eloborate this a bit.. –  Nov 23 '13 at 05:56
  • @Praphulla: Continuous images of connected sets are connected. – Micah Nov 23 '13 at 06:01
  • yes yes i know that continuous image of connected sets are connected but i could not relate that idea to the very next statement of yours.. do you mean as collection $\theta \in \mathbb{R}$ is connected so its image (for fixed $x$) $e^{i\theta}$is connected and so is the case of other coordinate ... :O –  Nov 23 '13 at 06:06
  • @Praphulla: Pretty much. Since each of its coordinates is continuous, the map $\theta \mapsto (xe^{i\theta}, (1/x)e^{-i \theta})$ is continuous for all nonzero $x$. – Micah Nov 23 '13 at 06:08
  • But then why does this idea does not work in $\mathbb{R}$? Just because i can not produce another point if i know one point on graph of $xy=1$? –  Nov 23 '13 at 06:10
  • @Praphulla:Because $e^{i\theta}$ is almost never real (and in particular the set of real values it assumes is certainly not connected!) – Micah Nov 23 '13 at 06:13
  • yes yes.. this does makes some sense to me... THank you :) –  Nov 23 '13 at 06:15
  • @Micah Thank you for your answer. It is clear now. – Supriyo Nov 23 '13 at 06:57
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Probably the easiest way to see that the statement is true is to show that the set $xy=1$ is path-connected.

Suppose that you have some pair $(x_1,y_1)$ for which $x_1y_1=1$, and likewise for $(x_2,y_2)$. As Micah pointed out, if $(x_1,y_1)$ is in the set, then $(x_1e^{i\theta},y_1e^{-i\theta})$ is as well for every $\theta\in [0,2\pi)$. So start the path by letting $\theta=0$ and let it vary until the modulus of both $x_1e^{i\theta}$ and $x_2$ are the same.

Now it's also true that if $(x_1,y_1)$ is in the set, then so is $(r\cdot x_1,\frac{1}{r}\cdot y_1)$ for every real $r>0$. Now that the modulus of $x_1e^{i\theta}$ and $x_2$ are the same, let $r=1$ and vary $r$ until $x_1\cdot re^{i\theta} = x_2$. The number $re^{i\theta}$ is, of course, just $x_2/x_1$.

The path traced out by $(x_1\cdot re^{i\theta},y_1\cdot \frac{1}{r}e^{-i\theta})$ as $\theta$ and $r$ vary will connect any $(x_1,y_1)$ to any $(x_2,y_2)$, and so the set is path-connected (and therefore connected).

Josephine Moeller
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A completely graphical way of thinking about this is to imagine the section of the graph in the first quadrant of the graph of $(\mathcal{Re}(x), \mathcal{Re}(y))$ as a rigid object. It's exactly the graph of the complex function $y = 1/x$ restricted to the positive real half-line in $x$ (or "the ray of $x$-values with argument $0$".)

Now let's sweep the ray around the origin in the complex $x$-plane. The graph of $y=1/x$ restricted to the $x$-values with argument $\theta$ looks exactly like the curve you started with, except that the values are rotated by an angle $-\theta$ about the origin in the complex $y$-plane.

After 90 degrees, $\pi/2$ radians, the ray picks out the positive imaginary $x$, and the reciprocals are negative imaginary $y$, for example. On the $(\mathcal{Im}(x), \mathcal{Im}(y))$-plane this picks out a curve in quadrant $4$, but it's otherwise exactly the same as the real curve in quadrant $1$ we started with

Thinking about the whole curve may make your eyes water, as it's a two-dimensional surface in four-dimensional space. But still it's very nicely behaved. In particular, once you've swept your $x$-ray through an angle of $\pi$ radians, so it lies back on the real $x$-graph, you've swept the $y$ values through $-\pi$ radians and they lie on the negative half of the real $y$-graph. This is the quadrant $3$ segment of the real graph.

And we can continue sweeping around, all the way back to the ray we started with; the $y$-values continue to sweep round their own axis in the opposite direction, and the graph closes up nicely into a connected single sheet.

HTFB
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