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.

Is the circle compact in $\mathbb{P}_{2}(\mathbb{C})$?

Here what I did: I considered the circle in $\mathbb{C}^2$ is $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$. The projective closure in $\mathbb{P}_{2}(\mathbb{C})$ is $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}$. The points at infinity are $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}\cap\{{x}_{0}=0\}=\{[0,1,i],[0,1,-i]\}$. So for every open cover ${\{{A}_{i}\}}_{i\in I}$ there will be a $j$ and a $k$ in $I$ fow which $[0,1,i]\in{A}_{j}$ and $[0,1,-i]\in{A}_{k}$. Therefore ${\{{A}_{i}\}}_{i\in I-\{j,k\}}$ is an open cover of the affine part $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}$ and (if what i wrote is correct) the circle in $\mathbb{P}_{2}(\mathbb{C})$ is compact iff the circle $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$ is compact in $\mathbb{C}^2$. A subspace is compact in $\mathbb{C}^2$ iff it's close and limitated. Considering the continuous function $f:\mathbb{C}^2\to\mathbb{C}$ defined by $(x,y)\mapsto x^2+y^2$, the circle is ${f}^{-1}(1)$ so it's closed ($\mathbb{C}$ is T1, the points are closed). But it's limitated? Seems to me it is not.

Can anyone help me. Is what i wrote correct? Is $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$ limitated in $\mathbb{C}^2$?

share|improve this question
2  
$\{(x,y)\in\mathbb{C}^2\ | \ x^2+y^2=1\}$ is not compact (or bounded="limitated"), that's why there are points at infinity in the projective closure. for instance, given any $x$, you have solutions $(x,\pm\sqrt{1-x^2})$ so take some sequence of $x$'s going off to infinity to see it is unbounded (or not "limitated") –  yoyo Jan 27 '12 at 19:05
1  
Here is the first error I found: If $A_j$ and $A_k$ contain your two points at infinity, it does NOT follow that $\{A_i\}_{i\in I\setminus\{j,k\}}$ is an open cover for the affine part. For example, you could take the cover containing only the trivial open set. You can repair this problem by removing the points at infinity from each of the open sets in your cover to get an open cover of the affine part. –  Aaron Jan 27 '12 at 19:10
    
Thank you very much... really usefull. So my reasoning is correct (with the correction Aaron made). Thanks again for your time and disponibility. P.S : And yoyo sorry for the word limitated, I've wrote the question quickly and I've wrongly translated the word from Italian. –  Lorenzo Rossi Jan 27 '12 at 20:49
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.