are speaking about! Bookmarked. Please also discuss with my web site =).

We will have a link exchange contract among us

]]>Hmm, the Wolf Prize… how likely is it that a mathematician of Arab descent would accept a prize that is presented by the President of the state of Israel?

]]>You’re both quite welcome, and thanks for the praise! A small correction: I exaggerated the difficulty of proving that every diffeomorphism of a circle extends to a disk. If you consider radius-preserving diffeomorphisms, it suffices to show that every diffeomorphism of the circle is isotopic to an isometry. (In a high enough dimension, not every extension to the ball can be realized by an isotopy leading to a radius-preserving diffeomorphism.) Yes, you could use the heat equation, but a better idea is to convert to a diffeomorphism of an interval and then simply gradually average it with the identity. Still, even though it is a very easy case of the result, you can already see a shadow of some of the difficulties later on.

]]>Thanks Greg! This is really excellent.

]]>Oops, I wrote something stupid there. The equation is a codimension-three condition rather than a codimension-one condition of course (if we restrict attention to the 3-sphere, that is) so it’s not obvious on dimension grounds. So I presume it’s related to the fact that there are many half turns in SO(2), double covers, etc. I will do the exercise just to lodge this properly where it ought to have been in my brain.

]]>Thanks for those remarks. I have to confess that I had never really thought about the question of how many square roots -1 had in the quaternions (though a similar problem arose in the Princeton Companion, where what I wrote suggested that it had three and someone objected and said what about -i, -j, -k — this is what underlies what I wrote in this article), though as soon as one *does* think about it it is sort of obvious that a single polynomial equation like that should have infinitely many roots in a three-dimensional space. (At some point I’ll actually do the exercise of calculating what they are.)

As it happens, the version available on their website was already supposed to have been updated to one where I improved the wording in a few places (e.g. not writing “quite exceptional” twice in the first paragraph of Section 2). In the light of your remarks, I have updated it further and asked them to put the new version on the website instead. I hope that will happen soon. (If I can’t get them to do it, then I’ll simply put it up here.)

Edit: they’ve put a corrected version there. In this one I don’t imply so strongly that there are only finitely many square roots of -1 (though the innocent reader might assume this because I don’t explicitly point out that there are more than just i, j and k), and I’ve got my 3 and 4 the right way round …

]]>Thank you very much for this extremely nice explanation — just the kind of thing I was hoping for.

]]>– on p.14, the number of non-cancellable words of length n is of course 4.3^(n-1) and not as written.

– A subtler problem (subtle to the point that it’s arguable if it’s a problem at all) is the comment about 6 square roots of -1 in the ring of quaternions. One introduces these roots to define the quaternions, but once this is done we find that -1 actually has infinitely many distinct square roots, not just those 6. It’s possible to misinterpret this paragraph as saying that those 6 are the only square roots of -1.

]]>Imagine the task of smoothing a piecewise linear manifold. (Topological manifolds are a decidedly harder starting point.) What do you really do? A PL manifold, since it comes with a triangulation, is already smooth on an open dense set, but at other points it is locally cone-like, like the tip of a sharpened pencil or crinkled aluminum. So the task at hand is to smooth the cone-like points. And to do that, you should:

(1) Work your way down from codimension 1, the facets, to codimension 0, the vertices. After all, most of a k-dimensional face is far away from the (k-1)-skeleton of the manifold, so whatever you do on the lower-dimensional kinks won’t affect what you must do there; on the other hand, what you do to the edges CAN effect what you do to the vertices, etc.

(2) Smooth a conical structure, at a vertex say, by removing a neighborhood of the apex and gluing in a ball smoothly. Or if it is not at a vertex, it is locally the Cartesian product of such an apex and a vector space. This is like smoothing the tip of a pencil; what was a point before becomes a disk that has some small but non-zero area.

(3) Now suppose that everything works as you expected it to up to dimension 6. (The suppose is true, by the way.) And suppose that you have a 7-manifold and you do smooth every part of the triangulation except at the end for the vertices. To smooth a vertex, remove an open neighborhood to leave a 6-sphere boundary; you want to glue in a 7-ball along that 6-sphere. Two different gluings will give the same answer if their composition is a diffeomorphism that extends to the 7-ball. So okay, Diff(S^6) is a group, and it has a subgroup Diff_0(S^6) of diffeomorphisms that extend to the 7-ball. It turns out that they are not the same and the quotient has 28 elements. They could in principle give you 28 different smooth structures on the same PL 7-manifold, and in the case of the 7-sphere, they actually do.

Is it so hard to believe that not every diffeomorphism of S^6 extends to B^7? That which is hard to prove in one case, might well be outright false in another case. Can you prove that every diffeomorphism of a circle extends to a disk? Well, there is a proof that relaxes a diffeomorphism of a circle gradually using the heat equation, but that’s hardly a trivial proof. Can you prove that every diffeomorphism of a 2-sphere extends to a 3-ball? That’s also true, but it’s not easy to prove either. (It’s a theorem of Smale and Munkres.) You have to first accept that this non-trivial work is the real work that you have to do smooth a PL manifold. And second, it is yet harder in higher dimensions — it is really a formidable theorem of Cerf and Hatcher that every diffeomorphism of a 3-sphere extends to a 4-ball — and eventually it’s false.

The non-existence of a smooth structure is a similar story. Imagine Munkres procedure again in 8 dimensions. Without any claims to uniqueness, you can smooth the triangulation somehow down to the vertices. But, wrench in the works, the link of a vertex could be one of Milnor’s exotic 7-spheres. So that is NOT the boundary of a smooth 8-ball (which only has one smooth structure, by the way) and there does not exist a way to blunt that vertex. You might wonder if there is some way to compensate by altering the earlier gluings, but it is sometimes an unavoidable obstruction.

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