If the fundamental group of the complement of a knot in $\mathbb R^3$ has $n$ generators, must its total curvature be atleast $(2π)n$ ? ( Milnor,in his 1950 paper "On the total curvature of knots" http://jstor.org/pss/1969467, has shown how to assign an integer termed as crookedness to a knot and that crookedness is related to the total curvature of the knot. I am not sure about the relevence, though.)
$\begingroup$
$\endgroup$
7
-
1$\begingroup$ See mathoverflow.net/questions/32245/… I believe your question is essentially answered there although you need to relate crossing number to minimal number of generators. $\endgroup$– Ryan BudneyCommented Aug 18, 2016 at 5:59
-
1$\begingroup$ The crookedness as defined by Milnor is the expected number of the restriction to the knot of a random linear function on $\mathbb{R}^3$, where the randomness is uniform in terms of their unit normal defining a linear function. $\endgroup$– Liviu NicolaescuCommented Aug 18, 2016 at 10:33
-
$\begingroup$ @RyanBudney : I also believe that crookedness in fact equals n, the number of generators. $\endgroup$– ChaitanyaCommented Aug 19, 2016 at 4:47
-
$\begingroup$ @LiviuNicolaescu : Could you provide a reference for this fact ? It seems very interesting and reminds me of the Cauchy-Crofton formula. Thanks for the response. $\endgroup$– ChaitanyaCommented Aug 19, 2016 at 4:54
-
$\begingroup$ www3.nd.edu/~lnicolae/knot-curv.pdf $\endgroup$– Liviu NicolaescuCommented Aug 19, 2016 at 8:33
|
Show 2 more comments