MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
0

In Gromov's paper regarding this proof, he says that by taking the best possible $H$ , we can get the following result: Let $k>0 , x$ be such that $ 1-x \leq K \int_0^{\infty} J_{k,H} (t) dt $ and $ x \leq K \int_0^{\infty} J_{k,-H} (t) dt $ for some constant $K$ . He then says that this implies that $ K \geq \phi \circ \Phi^{-1}(x) $ , where:

$\phi(t) = \frac{sin^{n-1} ( \sqrt{k} t) }{\int_0^{\pi/\sqrt{k}}sin^{n-1} (\sqrt{k} s ) ds}$

$ \Phi(t)=\int_0 ^t \phi(s)ds $ , $J_{k,H} (t) = ( sLk ' (t) + \frac{H}{n-1} s_k(t) ) _+ ^{n-1} $ $ s_k(t) = \frac{sin(\sqrt{k} t ) }{\sqrt{k} } $ .

does someone have an idea about this conclusion? How did he deduce this fact?

thanks !

[http://www.ihes.fr/~gromov/PDF/11[33].pdf ]

flag
2 
@Jeremy, do you think it might be possible to understand your question? (Better say what is the place in the paper your are trying to understand.) – Anton Petrunin Jul 24 at 12:26
2 
Or at least provide more context for these constants, functions, and formulas. – Deane Yang Jul 24 at 12:38
I really cannot find these formulas in the paper you quoted. What is the precise page number? – YangMills Jul 28 at 16:00

Your Answer

Get an OpenID
or

Browse other questions tagged or ask your own question.