Talk:06-1350

From Drorbn
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Guys,

I thought it might be nice to establish a lively discussion board on which we can discuss, develop and clarify any, or all, aspects of material pertaining to the class! I'll try to get things started with something that I had on my mind.

From lectures, we've opted to get our teeth stuck into things from the outset, with the point being that a thorough discussion of the elementary concepts would eat up a substancial amount of time, from which the only outcome would be a rigorous confirmation of what we all already understood knot, and equivalence of knots, to mean.

Being eager, I thought I'd jot down a couple of sentences (having just finished the post, I can say that this is was a gross underestimate of length) indicating what I think the pedantic issues at stake to be, and was hoping that perhaps we could iron out a couple of things I find to be confusing.

As I see it, the first, and main, point that might warrent a discussion, is on the correspondence between smooth and peicewise linear embeddings of the circle in three space.

[Recall from the first lecture that we would like a knot to be an equivalence class of embeddings, where the embeddings can be taken to be peicewise linear or smooth and the equivalence relation is determined by ambient isotopy; namely f and g are said to be ambient isotopic embeddings if and only if we have a smooth/peicewise linear (depending on whether we're talking about smooth or peicewise linear embeddings) map H:S^3x[0,1]->S^3 subject to H(_,t) being a smooth/peicewise linear homeomorphism with H(_,0) the identity and H(f(S^1),1)=g(S^1).]

The first question is: what is the correspondence between the two theories? Clearly any peicewise linear knot may be smoothed at the corners, and any smooth embedding has a linear approximation. However, if we try to talk about isotopies of such knots we're talking about something living in S^3x[0,1]xS^3. I think smooth manifolds, of which the graph of H is one, can be triangulated, though I don't have a good reference for this. Further, if the peicewise linear isotopies could be "smoothed" then we could conclude that thinking of knots as being peicewise linear or smooth is really the same for our purposes. Does anyone know about this or have any good references?

The second question (which is really dependant on the first question) I had in mind was on the connection between representations of knots in the form of quadravalent planar diagrams up to Reidmeister moves, and knots up to ambient isotopy.

Firstly, well defined parallel projections of knots exist

[infact there is a lovely visual argument that show that almost all parallel projections are "nice" in the sence that the inverse image of points of the diagram have at most two preimages and such points are isolated... equate the projections with points on the sphere and observe that two linear segments of the knot project onto each other only on arcs of the sphere, and further that projections having preimage sets of cardinality greater or equal to two are isolated or again lie on an arc. Thus the measure of such a set is zero.]

Now ambient isotopies of peicewise linear isotopies of knots can be broken down into moves where you can either replace one segement of a knot with the other two edges of a triangle containing the fixed segment as an edge as long as the spanning triangle doesn't intersect the knot anywhere else and vica versa (a picture would be very handy here, but I don't really know how to use any such feature yet), and consequently we see the Reidemister moves of planar diagrams all correspond to isotopies. The other direction is Reidemeister's famous theorem (the equivalence that we assumed axiomatically) and is quite a complicated argument. The question that arises in my mind is that if the smooth and peicewise linear theories are not equivalent [see question one], then it would seem that Reidemeister's proof may not carry over!

I'd love to have this put straight in my mind, and hope that someone can help me via a reference or authorative confirmation.

Sorry this was long winded, confused and above all practically irrelevant to the course!

Fionntan