Pretty swirly things

[simont]

I've just put up a web page of pretty swirly fractals, one of which is animated and thus doubly swirly. Anyone who's into that sort of thing, have a look at http://www.chiark.greenend.org.uk/~sgtatham/newton/. Gets a bit mathematical in places; read the maths if you care, skip down to the pretty pictures if you don't :-)

(Yes, as it happens, I have been spending a lot of today waiting for long test runs to finish. How can you tell?)

Comments

[ewx.livejournal.com]

I did the Newton-Raphson fractals years ago, but it never occurred to me to animate them. Cute.

[kaet.livejournal.com]

(re: unfactorized polynomials on that other guy's page)

You could use the Newton Raphson technique to find the zeros! :)

[simont]

Don't think it didn't occur to me! :-) But since all of them are degree-4 or lower, that would feel sort of dishonourable. Cubics and quartics are feasible to solve exactly, after all :-)

[kaet.livejournal.com]

I accuse you, Mr Tatham, of being a Mathematician. Dishonourable indeed, :)! I suppose you're going to say that taking the result and looking it up in a large database of common constants and then checking if it works anlytically with the nearest entry is dishonorouble true? :)

As a mathematician, though, can I ask you a topology question? I'm just starting to learn topology, and aren't sure I've got the axioms sorted in my head.

I've heard a rumour that given an Euler number and whether something is orientable or not, then you precisely determine a topological solid. Furthermore, upon closer questioning, I've hear that there are only two non-orientable solids, the projective plane and the Klein bottle, modulo extra handles and, more importantly only one kind of orientable solid, the sphere (modulo extra handles). So if you take your average sphere and dig an 'ole through it, you've clearly got a torus. But what if you tie a simple knot in the 'ole? Clearly, because of the eminent folks, this is still a torus (it'll have the same Euler number and is still orientable). But how do you squidge it about? I can think of two possible answers. The first is that my mind isn't yet warped enough and there is a perfectly plausible way of deforming a sphere with a knotted hole into a torus in three dimensions. The second is that if the going gets tough you're allowed to invent extra dimensions to help you undo your knots.

Which, in your expert opinion, is it to be?

[simont]

I think it's both, in fact.

I'm fairly sure it's possible to continuously deform a torus-with-knotted-hole into a torus-with-unknotted-hole using only the three dimensions you're given and nothing up your sleeve. I've got a Martin Gardner book which shows the transformation if you're interested, but essentially the trick involves moving one end of the hole around to the other end and then further into the hole itself.

However, that doesn't deal with the case of a torus where the substance, rather than the hole in the middle, is tied in a knot. I don't think you can continuously deform that one into an ordinary unknotted torus; so in that situation you'd have to invent an extra dimension in which to untie it. So although as it happens you don't need the extra dimensions for the knotted-hole case, you would have been allowed to use them if you'd needed them.

My understanding of topology, though, is that all this continuous-deformation stuff is actually a red herring. The true definition of topological equivalence is that there exists a bijection from the points of one surface to the points of the other, such that neighbourhoods are preserved (points close together on one surface end up close together on the other). Now clearly a continuous deformation of one surface into the other (using any number of additional dimensions) implies the existence of such a bijection, but it doesn't work the other way round - two surfaces can be shown to be topologically equivalent without the need to exhibit such a continuous deformation.

[ex-lark-asc.livejournal.com]

Ooooh, pretty animation!

(Just thought I'd put in a word for the maths-free ferret-shock sufferers out there ;)

[kaet.livejournal.com]

Thanks!

My main concern was that I thought it was obvious with the extra dimensions and if there's no way to do it in three, I shouldn't waste time trying to warp my brain around it.

I'd come across this bijection definition, but I thought that the two things were equivalent (given everyone 'knows' that's what topological deformation is). I shall endeavour to abandon it, :). It makes more sense when you're dealing with immersions anyway, I think, to think of them as 1-to-1 mappings of neighbourhoods rather than the screwy one about intersections in n-2 dimensions, or whatever it is, so the deformation thing probably gives up even the advantage of being more intuitive when you get to more involved stuff, anyway.

I had also suspected that it was that my brain wasn't warped enough to untie the knot in three, :). Some of those shapes are just, just not natural, :).

[simont]

It may be the case that any two surfaces equivalent by the bijection definition are also deformable into each other given sufficiently many dimensions. Certainly I can't immediately think of a counterexample...

I was wrong about the internally knotted torus, incidentally. The book I mentioned does contain diagrams for untying a knot in the hole of a torus, but only in the case where the torus has two holes and the unknotted one goes through the middle of the knot in the other. An internally knotted one-hole torus can't be untied (in three dimensions) any more than an externally knotted one; as a simple proof, consider a line drawn along the surface such that it goes through the knotted hole, loops round the outside of the torus and joins back up with itself. This closed curve is tied in a simple overhand knot, and hence no continuous deformation of the torus can transform it into an unknotted curve. (I feel silly for not having spotted that to start with.)