3 ms·
So is it just a fancy way of saying "every transformation has an axis or origin"?
by mjfisher 3y ago
So is it just a fancy way of saying "every transformation has an axis or origin"?
- Elyra 3y agoThis is also how I interpreted it, which I take it is probably incorrect and would like to know why.
- nerdponx 3y agoOne of the pre-conditions of the theorem is that the domain and codomain are the same. So that's one way to satisfy the theorem. But it's not really intuitive. If you gently swirl a bottle of water, somewhere in the bottle, some water molecule did not move. It's not necessarily in the center of anything. The fixed point could be anywhere. The theorem is about existence only, not construction. (I am not a professional mathematician, I might also be wrong.)
- tomrod 3y agoNo, you are on point.
- l__l 3y agoThat's not the best way to view it. You can prove it by showing that for an n-dimensional disc, there is no contraction to its boundary; which I think is a bit more illustrative of what this FPT is doing
- tsimionescu 3y agoWell, some continuous functions are simple and it's highly intuitive why this applies to them. Others are very weird and it's not intuitive at all that they should have such a point. And of course, some functions/transformations are not continuous, and those may not have such an "axis of origin" at all.
- r0uv3n 3y agoI don't quite know what you mean by that, but consider that this is not true for e.g. a torus or a sphere (take the function mapping every point to its antipodal point). The fact that the underlying space is contractible is very important here.