Let M be a noncompact manifold with two ends.Then we can construct a line(geodesic for infinite time) on M.The Busemann function for the line $\beta $ is regular on M.By morse theory,M is homeomorphic to $R \times {\beta ^{ - 1}}\left( 0 \right)$.Then the metric on M can be written as ${g_M} = d{t^2} + {g_t} = d{t^2} + {\varphi ^2}\left( t \right){g_0}$.Where ${g_0}$ is the metric for ${\beta ^{ - 1}}\left( 0 \right)$.Can some one point it out where I am wrong and give a counterexample?
-
3$\begingroup$ Removing two points from a 2-dimensional torus gives a manifold with two ends, but it is not the product of $\mathbb R$ and another manifold because that other manifold would have to be a circle for dimension reasons. The claim that the Busemann function of a line has no critical points is false. $\endgroup$– Igor BelegradekApr 5, 2013 at 15:19
-
$\begingroup$ Although your assumption that the Busemann function is regular is false in general, if $M$ has non-positive curvature, then it follows from the Toponogov splitting theorem that $M \cong \mathbb{R}\times Y$. See 10.5.1: math.psu.edu/petrunin/papers/alexandrov/bbi.pdf $\endgroup$– Ian AgolApr 5, 2013 at 17:49
-
2$\begingroup$ @Agol: I believe you mean non-negative curvature. $\endgroup$– Mohan RamachandranApr 5, 2013 at 18:51
-
$\begingroup$ Yes, of course Mohan (at first I wrote positive, then changed it incorrectly), and also complete. $\endgroup$– Ian AgolApr 5, 2013 at 21:20
1 Answer
Take the infinite ladder surface, a 2-manifold with 2 ends and infinitely generated $H_1$. One description, an embedding in $\mathbb{R}^3$, is the boundary of an $\epsilon$-neighborhood of the ladder graph $$(\{0,1\} \times 0 \times \mathbb{R}) \cup ([0,1] \times 0 \times \mathbb{Z}) $$ In this description, the geodesic $\beta$ will just be a vertical line, and the Busemann function is just projection to the $z$-axis.
Another description is to start with the closed surface of genus 2 whose fundamental group has the presentation $\langle a,b,c,d \,\, | \,\, [a,b][c,d]=1 \rangle$ and then take the infinite cyclic regular covering space associated to the homomorphism to $\mathbb{Z}$ defined by $a \mapsto 1$, $b,c,d, \mapsto 0$. In this description, the geodesic $\beta$ will be the lift of a closed geodesically embedded circle representing the conjugacy class of $a$ in the fundamental group.
Assuming that by a "regular function" you mean one for which every value is a regular value, the error in your proof is simply the statement that the Busemann function is regular. For any complete $\mathbb{Z}$-invariant Riemannian metric on the infinite ladder surface, even if the Busemann function is Morse there will be infinitely many singular points of index $1$.
-
$\begingroup$ Lee: OP's definition of regular might require "no singular values". $\endgroup$– MishaApr 5, 2013 at 14:23
-
$\begingroup$ That was also my understanding of the OP's intention; I edited my answer to that effect. But the OP's statement that the Busemann function is regular is incorrect. $\endgroup$ Apr 5, 2013 at 14:32