freecounterpoint: Why CDs beat Rot's inequalities

Thursday, December 10, 2009

Why CDs beat Rot's inequalities

10 comments:

Arelcao Akleos said...: No, she has a thing for peasants who carried arms for Lord Marlborough. Lucky me.; 12/10/2009 3:23 AM
Mr roT said...: Your chain of logic eludes me.; 12/10/2009 3:26 AM
Tecumseh said...: AA needs some codfish to fire up his logic.; 12/10/2009 7:52 AM
Tecumseh said...: As for Herr Rott, he flunked the computation. It's actually $\pi_1(X) = \mathbb Z \times (\mathbb Z \ast\mathbb Z)$. Tsk, tsk, tsk. Go back to the corner.; 12/10/2009 7:55 AM
Tecumseh said...: Exactly. The circle in the middle links the two outside, so that its meridian commutes with the two nearby ones. If the circles were completely apart, you'd get what you wrote. But they aren't, so things partially commute. Hah! Where's my Perrier Jouet?; 12/10/2009 9:19 AM
Mr roT said...: It's in your dreams! The goddam $S^1$s are linked, not glued!

Powww! There's no basepoint!

You owe me, you schmuck!!!; 12/10/2009 9:34 AM
Tecumseh said...: Of course, I'm talking about the complement of those circles -- that's where that beautiful girl extends her curves, after all. The circles themselves are not interesting -- they are not even connected. So in that case, \pi_1 can only be computed for each component individually, in which case you get a stupid \Z for each one -- still not waht you said, by a mile.

I suggest, stop diggin', Herr Rott, and poney up, by adding 3 VCPs to the tab. The tab, you know.; 12/10/2009 10:26 AM
Mr roT said...: You should've written $X^c$. Pay up.; 12/10/2009 10:34 AM
Tecumseh said...: X always denote the complement in such a situation. Trust me.; 12/10/2009 5:52 PM
Mr roT said...: Trust me. I'm a doctor.; 12/10/2009 6:52 PM

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