Challenge Math Challenge - July 2023

[AndreasC]

It is kind of insane to me that #7 could possibly have a solution... Thus far all I have figured out is that it is definitely an infinite group, and obviously the subgroups have to be proper... But that is simple, I can't think of what it might be however.

Similarly, #9 is bothering me a lot. I tried to use Brouwer like I used for the n=1 case, but while I can easily come up with a compact set that is mapped to itself, coming up with a convex one is hard.

 

[wrobel]

the set in Brouwer theo. can only be homeomorphic or even it can be a retraction to (of) the convex one but it does not help me. I do not see any suitable invariant compact sets

 

[AndreasC]

the set in Brouwer theo. can only be homeomorphic or even it can be a retraction to (of) the convex one but it does not help me

Right. I have an idea for n=2 as well. But I find it kinda hard to generalize it.

 

[AndreasC]

I can prove #9 if I could prove several other hunches I have that probably have to do with algebraic topology (which I don't really know much at all about).

I will lay out my idea here, and maybe someone can do something with it. My goal is to come up with a set homeomorphic to a compact convex set that is mapped to itself. If that is done, then Brouwer settles it.

The idea is, I take a point, say 0. Assume it is not fixed. Then f(0) is some other point. I include them both in one compact, convex set, call it A. Then f(A) has a non-zero intersection with A, and the union of these sets is also compact, and mapped to itself.

This union in general has a hole, similar to the hole in a torus, but it is simply connected. That is, we can contract the subspace into a closed loop-like object, or something higher dimensional.

We can close the holes using other compact sets. We thus get another closed set that is homeomorphic to a convex one, call it A'. Take its union with f(A'). This time, their union again may not be contractible to a point (so homeomorphic to a convex set, at least that's my assertion), if it has a hole then it is a higher dimensional hole, that is, it contracts to something like a typical 2-sphere, or something higher dimensional. Not sure how to prove it, but I'm guessing it has something to do with the fact that the interection of the two (contractible to a point) sets is now simply connected. If we do this process again then the intersection will not just be simply connected, but also contractible to a loop at worst, etc. That's my hunch.

We can continue doing this, until we have no dimensions left in the space. Every time the hole goes up at least one dimension, if one exists. So if its dimension would have to be greater than n, then we can conclude that there is no hole any more.

In the case where n=1, the set we are looking for is the union of A and f(A). In the case where n=2, we need f(A') and A', etc.

 

[mathwonk]

@AndreasC: What about this version of your idea? choose any point p in the plane, and join p to f(p) by a line segment L. Then consider the simple closed curve f(L). If L meets f(L) only at the endpoints, then their union bounds a compact set K homeomorphic to a disc. Then K must be mapped to a compact connected set with the same boundary curve, hence to itself. Apply Brouwer to K.

If L does meet f(L), choose a point q on L such that f(q) is also on L, and the distance from q to f(q) is minimal. Let M be the sub line segment joining q to f(q). Then no interior point of M can map to a point of M, i.e. M and f(M) meet only at endpoints, so apply the earlier argument to the set bounded by M union f(M).

 

[wrobel]

a hypothesis on #9

 

[Throwaway_for_June]

a hypothesis on #9

Spoiler: #9

I was thinking something similar-ish. Except that my $\psi: \mathbb{R}^n \rightarrow \mathbb{R}^n, \psi \left( x \right) =\frac{ x + f \left( x \right) } {2}$ so that any fixed point of $\psi$ would also be a fixed point of $f$ but I couldn't see a way to make it work. Using the norm is clever.

 

[AndreasC]

@AndreasC: What about this version of your idea? choose any point p in the plane, and join p to f(p) by a line segment L. Then consider the simple closed curve f(L). If L meets f(L) only at the endpoints, then their union bounds a compact set K homeomorphic to a disc. Then K must be mapped to a compact connected set with the same boundary curve, hence to itself. Apply Brouwer to K.

If L does meet f(L), choose a point q on L such that f(q) is also on L, and the distance from q to f(q) is minimal. Let M be the sub line segment joining q to f(q). Then no interior point of M can map to a point of M, i.e. M and f(M) meet only at endpoints, so apply the earlier argument to the set bounded by M union f(M).

I think this runs into two problems: first of all, it works on 2 dimensions, but in 3 dimensions, there are many possible sets bounded by the curve, and they are not mapped to themselves. This is essentially the same reason I ran into the problem of holes. I like that you used a line segment though, it's easier to think about that way.

The other problem is that it's not immediately clear to me why you can find a minimal distance. However, I think starting from a line the way you described, we can use my construction of taking the union of L and f(L), closing the holes to get L', then taking the union f(L') and L' etc. This time it is a bit easier to think about because what we are doing is basically stitching together simply connected closed sets at their boundary. There has to be a theorem that tells us if we stitch together 2 n-dimensional sets like that and we get a hole, then that hole is at least (n+1)-dimensional. If that really is true then we are done!

 

[AndreasC]

I think that's something like the Jordan curve theorem actually. Some googling revealed the Jordan-Brouwer separation theorem, which I think is essentially what we need to complete it.

 

[wrobel]

The solution of 9 under slightly stronger conditions.

Assume in addition that $f\in C^1(\mathbb{R}^n,\mathbb{R}^n)$ and all the critical points of a function
$$\psi(x):=|x|^2+|f(x)|^2$$ are contained in a compact set.

Observe that
$\psi(x)\to \infty$ as $|x|\to\infty.$

Thus $\psi$ attains its minimum in $\mathbb{R}^n$ say at $x_0$.

From all these observations it follows that a set $K_c=\{\psi(x)\le \psi(x_0)+c\}$ is homeomorphic to the closed ball if only $c>0$ is large enough.

Note that $\psi$ is an invariant function: $\psi\circ f=\psi$.
So that $K_c$ is an invariant set: $f(K_c)\subset K_c$.

The Brouwer fixed point theorem concludes the proof.

 

[mathwonk]

@AndreasC: As to a minimum distance existing, the function "distance between x and f(x)" is continuous on the compact set L, so has a minimum. and yes the Jordan curve theorem is used, and yes it apparently only works well in the plane.

Come to think of it, using the Brouwer theorem would not be a direct proof anyway, since the proof of Brouwer is to assume no fix point and derive a contradiction. So what if one assumes f has no fix point. Then one seems to get a rather interesting double covering map: R^n-->R^n/{x≈f(x)} = Y.

 

[Infrared]

@mathwonk @AndreasC This argument looks correct for the case $n=2$, nice job.

I'm not sure if it can be done for $n>2$ with just the Brouwer theorem though.

 

[mathwonk]

yes I have added a comment on higher dimensions.

 

[Infrared]

I agree that considering the double cover $\mathbb{R}^n\to\mathbb{R}^n/(x\sim f(x))$ that arises if $f$ has no fixed points is a great way to continue!

 

[mathwonk]

From there, I need Whitehead and the homology of P^infinity, to finish. Maybe there is a more elementary way?

 

[Infrared]

From there, I need Whitehead and the homology of P^infinity, to finish. Maybe there is a more elementary way?

Feel free to look for one, but I think I'm allowed to ask one question using more advanced techniques per thread :)

This was a HW question back when I took alg top in undergrad and that was the intended solution.

 

[mathwonk]

well I enjoyed reviewing this material from alg top, but for me it was 2nd yr grad school! I've been holding this solution for a couple days since so many nice ideas are generated, as long as it is open.

By the way, I may be missing something, but this argument does not seem to use fof. = id, just that f is a homeomorphism. Thus it seems to show every homeomorphism of a nice contractible space has a fixed point?

Oh yes, and I had another idea, of extending f to a homeomorphism F of the n-sphere, fixing infinity, and concluding via some lefschetz type result that since F has at least one fix point, but should have zero or 2 (?), then f also has one. This would require computing the index of the fix point of F at infinity, to be ± 1.

 

[Infrared]

By the way, I may be missing something, but this argument does not seem to use fof. = id, just that f is a homeomorphism. Thus it seems to show every homeomorphism of a nice contractible space has a fixed point?

Translation on $\mathbb{R}^n$ would be a counterexample to that! You need the fact that $f\circ f=1$ to describe the homology of $\mathbb{R}^n/(x\sim f(x))$.

 

[mathwonk]

yes I see that too. trying to see now where I used f^2 = id. oho! thank you!
I also added another idea/speculation above, on extending to the sphere.

it seems the extension of a translation to the n - sphere would have a fixed point of index 2 (or 0?) at infinity. (I am thinking mainly in R^2.). Obviously this is idea is not mature.

 

[Infrared]

yes I see that too. trying to see now where I used f^2 = id. oho! thank you!
I also added another idea/speculation above, on extending to the sphere.

it seems the extension of a translation to the n - sphere would have a fixed point of index 2 (or 0?) at infinity. (I am thinking mainly in R^2.)

I had that idea too (and then apply Lefschetz of course), but I confused myself quite a bit in trying to compute the degree of the fixed point at infinity using only $f\circ f=Id.$

 

[mathwonk]

well what about for n=2? it seems one would get a double cover of a surface by the 2-sphere with only one branch point, contradicting riemann-hurwitz.

so maybe one version of the index at infinity is the degree of the cover minus one, i.e. 2-1 = 1 (if orientation preserving).

 

[AndreasC]

@mathwonk @AndreasC This argument looks correct for the case $n=2$, nice job.

I'm not sure if it can be done for $n>2$ with just the Brouwer theorem though.

My argument is about dimensions higher than the plane. Basically you do the same thing you did for the plane, but once more for 3 dimensions, and then again for 4, etc. In 3 dimensions, the set A' is now essentially a 2-disk in 3d space, but f(A') is not the same as A'. However it also is homeomorphic to a 2-disk, and they share a boundary. If they are not the same, their union is essentially a 2-sphere, that is mapped to itself. The interior (A'') is a 3d ball, and because its border is mapped to itself, then A'' is mapped to itself. We do the same for 4 dimensions once more.

 

[mathwonk]

for me the hard part is getting f(A') disjoint from A'

 

[Infrared]

well what about for n=2? it seems one would get a double cover of a surface by the 2-sphere with only one branch point, contradicting riemann-hurwitz.

so maybe one version of the index at infinity is the degree of the cover minus one, i.e. 2-1 = 1 (if orientation preserving).

How are you getting a (branched) double cover of a sphere? $f$ is a homeomorphism so a degree 1 map. Maybe I misunderstand?

 

[mathwonk]

Z/2Z acts on the sphere by a group of homeomorphisms {F,id}, where F fixes only infinity, hence there is again an identification space surface. maybe this is ok only in dimension 2.

 

[Infrared]

Sorry I guess I use the terminology a little bit differently. To me, a "double cover of X" means a 2:1 map with target X.

 

[mathwonk]

Yes that is what I meant to say. If f is a homeomorphism of R^2 with no fix point and fof = id, extend f to a homeomorphism F of S^2 fixing only infinity, then π:S^2-->S^2/{x≈F(x)} = Y is the double cover of Y I meant. Y should be a surface, and π may violate Riemann-Hurwitz ?. Sorry for the imprecise stuff. This could still be wrong, and is more unclear for n > 2, to me.

 

[wrobel]

What about the case $f^k=\mathrm{id},\quad k=3,4,\ldots$? I mean continuous $f$; my trick remains valid here.

 

[AndreasC]

for me the hard part is getting f(A') disjoint from A'

But why does it have to be disjoint? The whole point is that it is not disjoint, they share a boundary. Do you mean their interiors being disjoint? Well, I think they don't have to be disjoint. I think Brouwer should work even if the union is homeomorphic to a bunch of disks connected by line segments or whatever.

 

[mathwonk]

AndreasC: yes, disjoint except for their boundaries, otherwise the intersection might be quite wild.

Infrared: as to lefschetz fix point index, it seems that the graph of F, since F = F^-1, in the product space, is symmetric about the diagonal, so it should be "transverse" to the diagonal, hence have fix point index one? of course F is not smooth.

In fact since in Dold, the fix point index is the degree of a map on local homology groups, is it possible that since F takes an open nbhd of infinity homeomorphically to another open nbhd of infinity, that this forces the index to be one? I have not understood Dold's definitions very well though....Oops, there are obvious counterexamples. I am confusing the degree of g with the degree of id-g.