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The graph of $xy = 1$ in $\Bbb C^2$ is connected. True or false?

I know that it is not connected in $\Bbb R^2$, but what is the case of $\Bbb C^2$?

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  • $\begingroup$ TIFR GS problem. $\endgroup$ – Dutta Nov 26 '13 at 1:39
  • $\begingroup$ Where are you studying poton? $\endgroup$ – Dutta Nov 26 '13 at 1:40
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The map $f:\Bbb C\setminus\{0\}\to\Bbb C^2$ given by $f(z)=(z,1/z)$ is continuous and maps $\Bbb C\setminus \{0\}$ onto the set $\{(z,w)\in\Bbb C^2:zw=1\}$. Since $\Bbb C\setminus \{0\}$ is connected, the image must also be connected.

Note that this argument fails in the real case, since $\Bbb R\setminus\{0\}$ is not connected.

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  • 2
    $\begingroup$ Great minds think alike ;) We posted at the same time. $\endgroup$ – Clive Newstead Aug 9 '12 at 11:20
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    $\begingroup$ Indeed, although in my case I think it was more of a Pavlovian response after too much exposure to Rudin's Principles. $\endgroup$ – Per Manne Aug 9 '12 at 11:22
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Every value of $x$ has a corresponding value of $y$ for which $xy=1$ except when $x=0$, since $\mathbb{R}$ and $\mathbb{C}$ are fields.

If $\Gamma$ is the function that takes a point $x \in \mathbb{C}$ to the corresponding point $(x, \frac{1}{x})$ on the graph $xy=1$, then $\Gamma$ must have domain $\mathbb{C}-\{0\}$.

But $\mathbb{C}-\{0\}$ is connected and $\Gamma : \mathbb{C}-\{0\} \to \mathbb{C}^2$ is continuous, so $$\Gamma(\mathbb{C} - \{0\}) = \{ (x,y) \in \mathbb{C}^2\, |\, xy=1 \}$$ is connected.

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Suppose that $w_1,w_2,z_1,z_2\in\Bbb C$ are such that $w_iz_i=1$ for $i=1,2$. There are positive $r_i,s_i\in\Bbb R$ and $\theta_i,\varphi_i\in[0,2\pi)$ such that $w_i=r_ie^{i\theta_i}$ and $z_i=s_ie^{i\varphi_i}$.

Show that the set in question contains a path from $\langle w_1,z_1\rangle$ to $\langle w_2,z_2\rangle$ consisting of a linear path from $\langle w_1,z_1\rangle$ to $\langle r_2e^{i\theta_1},s_2e^{i\varphi_1}\rangle$ followed by a path from $\langle r_2e^{i\theta_1},s_2e^{i\varphi_1}\rangle$ to $\langle w_2,z_2\rangle$ that takes each coordinate around a circular arc in its copy of $\Bbb C$.

Once you’ve done this, you’ll have shown that the set is path connected and a fortiori connected.

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Given any two points $z,w\in \Bbb C^\times$ consider an arbitrary path $\gamma:[0,1]\to\Bbb C$ with $\gamma(0)=z,\gamma(1)=w$ such that $\gamma(t)\ne0$ for all $t$ (convince yourself such a path exists necessarily). In particular, think about $(\gamma,1/\gamma):[0,1]\to\Bbb C^2$. Also note path-connectedness is stronger than connectedness.

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