Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know that what taking square roots for reals, we can choose the standard square root in such a way that the square root function is continuous, with respect to the metric.

Why is that not the case over $\mathbb{C}$, with respect the the $\mathbb{R}^2$ metric? I suppose what I'm trying to ask is why is there not continuous function $f$ on $\mathbb{C}$ such that $f(z)^2=z$ for all $z$?

This is what I was reading, but didn't get:

Suppose there exists some $f$, and restrict attention to $S^1$. Given $t\in[0,2\pi)$, we can write $$ f(\cos t+i\sin t)=\cos(\psi (t))+i\sin(\psi (t)) $$ for unique $\psi(t)\in\{t/2,t/2+\pi\}$. (I don't understand this assertion of why the displayed equality works, and why $\psi$ only takes those two possible values.) If $f$ is continuous, then $\psi:[0,2\pi)\to[0,2\pi)$ is continuous. Then $t\mapsto \psi(t)-t/2$ is continuous, and takes values in $\{0,\pi\}$ and is thus constant. This constant must equal $\psi(0)$, so $\psi(t)=\psi(0)+t/2$. Thus $\lim_{t\to 2\pi}\psi(t)=\psi(0)+\pi$.

Then $$ \lim_{t\to 2\pi} f(\cos t+i\sin t)=-f(1). $$ (How is $-f(1)$ found on the RHS?) Since $f$ is continuous, $f(1)=-f(1)$, impossible since $f(1)\neq 0$.

I hope someone can clear up the two problems I have understanding the proof. Thanks.

share|improve this question
    
See also math.stackexchange.com/questions/89883/… –  Gerry Myerson Feb 27 '12 at 5:48
    
Regarding the formula for $f$, you are familiar with how to compute square roots of complex numbers in polar form right? And if not, you could at least verify that the two stated possibilities are square roots. How many square roots can a complex number have? –  Hurkyl Feb 27 '12 at 6:25
    
@Hurkyl I see now, for $z$ on the unit circle, $z=e^{it}$. So the square roots are $e^{it/2}$ and $e^{i(t/2+\pi)}$. Thanks. Do you see how the $-f(1)$ is determined? –  Botts Feb 27 '12 at 6:32
    
@Botts: Yes. That's what the previous paragraph is all about: using the formula for $f$ to compute that equality. –  Hurkyl Feb 27 '12 at 6:33
    
OK, it all makes sense now. Thanks. –  Botts Feb 27 '12 at 6:44
add comment

3 Answers 3

Here is a proof for those who know a little complex function theory.

Suppose $(f(z))^2=z$ for some continuous $f$.
By the implicit function theorem, $f(z)$ is complex differentiable (=holomorphic) for all $z\neq0$ in $\mathbb C$.
However since $f$ is continuous at $0$, it is also differentiable there thanks to Riemann's extension theorem.
Differentiating $z=f(z)^2$ at $z=0$ leads to $1=2f(0)f'(0)=2\cdot0\cdot f'(0)=0 \;$. Contradiction.

share|improve this answer
add comment

Start at $z=1$, compute $\sqrt z$ as you move around the origin on a circle of radius 1, and look at what happens when you get back to $z=1$.

share|improve this answer
1  
He's describing the very proof you're looking at, but in more geometric terms. –  Hurkyl Feb 27 '12 at 6:23
add comment

Here is a proof for those who know a little topology.

Suppose $(f(z))^2=z$ for some continuous $f$.
The covering space $p:\mathbb C^*\to \mathbb C^*: z\mapsto z^2$ then has $f$ as a section: $p\circ f=Id_{\:\mathbb C^*}$.
You get for the induced morphisms on fundamental groups (based at $1$, say)
$p_*=2\cdot1_{\mathbb Z}:\pi_1(\mathbb C^*)=\mathbb Z \to \pi_1(\mathbb C^*)=\mathbb Z:n\mapsto 2n\quad$ and
$f_*=a\cdot1_{\mathbb Z}:\pi_1(\mathbb C^*)=\mathbb Z\to \pi_1(\mathbb C^*)=\mathbb Z:n\mapsto an\quad$ (for some unknown $a\in \mathbb Z$).

It follows from functoriality that $p_*\circ f_*=2a\cdot1_{\mathbb Z}=(p\circ f)_*=Id_{\pi_1(\mathbb C^*)}=1\cdot1_{\mathbb Z}:\mathbb Z\to \mathbb Z$.
Hence $2a=1\in \mathbb Z. $ Contradiction.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.