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I have a question about the proof of Theorem 5 in Thurston's paper "A norm for the homology of 3-manifolds". It's the theorem that asserts that the fibered faces are, well, fibered. More precisely, fix a compact 3-manifold $M$, and let $B_x$ be the Thurston polytope. Then the theorem asserts that the set of elements of $H^1(M;\mathbb{R})$ which are representable by non-singular closed $1$-forms is a union of cones on open top-dimensional faces of $B_x$ (minus the origin).

Here's the part that I am having trouble with. Assume that $M$ is closed, and let $\alpha$ be any non-singular closed $1$-form. Thurston already proved that $\alpha$ lies in the cone on an open top-dimensional face of $B_x$. If $S$ is a surface in $M$, then denote by $[S] \in H^1(M;\mathbb{R})$ the Poincare dual cohomology class. Consider an incompressible surface $S$ in $M$ such that $[S]$ lies in the cone on the same open top-dimensional face of $B_x$ as $\alpha$. Most of the work of the proof goes into showing that linear combinations $t[S] + u \alpha$ with $t \geq 0$ and $u > 0$ are representable by nonsingular closed $1$-forms. That I have no problem with. However, Thurston then asserts that $[S]$ can be so represented, and the only justification he gives is that you can get it by "iterating this construction". I cannot figure out what he means here.

Can anyone help me? Thanks!

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Sorry for asking, but I have been trying to find this paper. Does anyone have a link for it please? –  user48617 Mar 23 at 15:14
    
@user48617: this is available (volume 59 Memoirs AMS, Number 339) at the AMS website. ams.org/books/memo/0339 –  Ian Agol Jun 2 at 17:41

1 Answer 1

up vote 7 down vote accepted

It's been a very long time since I've read this paper, and I haven't been able to find a copy online, so my apologies in advance if what I'm about to write is nonsense.

Think in terms of projective coordinates (i.e. I won't say "cone" any more). I think the lemma you describe also holds for surfaces $S$ on the boundary of the face in question. (If this is not the case, then never mind, I'll delete this answer.) That is, we can take $[S]$ to be a vertex of the face, and we know that $t[S] + u\alpha$ is representable by a nonsingular 1-form whenever $\alpha$ is, for $t \ge 0$ and $u>0$. In other words, starting at a representable point in the interior of the face, we can move a arbitrary distance toward any of the corners of the boundary of the face, so long as we stay in the interior. Since the interior of the face is the open convex hull of the corners, we can reach any point in the interior this way by letting $S$ run through the corners of the face.

(EDIT: As Agol points out in a comment below, instead of "corners" I should say rational points in the boundary of the face.)

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I think you're right, except that when you say "corners", I think you mean (rational) points on the boundary of the face. –  Ian Agol Jan 16 '12 at 17:18
    
Thanks Ian -- I'll make that correction. –  Kevin Walker Jan 16 '12 at 17:28
    
Thanks!!!!!!!!! –  Sasha Jan 17 '12 at 3:47
    
By the way, if you believe the lemma for $S$ in the interior of the face, you don't have to reprove it for $S$ in the boundary. Instead, you just have to continue the segment from $\alpha$ to $S$ a little further, to another rational point $S'$ still in the interior of the face, and apply the lemma to $S'$. Not that it will be much different to prove for points in the boundary. –  Lee Mosher Feb 29 '12 at 20:27

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