Math Challenge - July 2023

[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.

 

[AndreasC]

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

Will it? I think it should be just a bunch of closed loops connected by a single lines in 2d, then sphere-like things connected by lines in 3d, etc.

 

[Infrared]

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

It's still true in that case! Just to fill in the details in @mathwonk's argument: If $f$ has no fixed points, then $\mathbb{R}^n\to\mathbb{R}^n/(x\sim f(x))=:X$ is a double cover, and hence $\pi_1(X)=\mathbb{Z}/2.$ Also $\pi_k(X)=\pi_k(\mathbb{R}^n)=0$ for $k>1.$ So $X$ is a $K(\mathbb{Z}/2,1)$ (meaning that its only nontrivial homotopy group is $\mathbb{Z}/2$ in degree 1). Being a $K(G,n)$ uniquely defines a cell complex up to homotopy type (though this is not exactly Whitehead- in general, the homotopy groups of a cell complex do not determine its homotopy type). So, $X$ is homotopy equivalent to $\mathbb{R}P^\infty,$ which has nonzero homology in arbitrarily high degrees, which cannot be true for $X$, which is an $n$-dimensional manifold.

To deal with the case $f^k=Id,$ all that changes is that the group $\mathbb{Z}/2$ is replaced by $\mathbb{Z}/k$ and $\mathbb{R}P^\infty$ is replaced by a Lens space.

I'm a bit busy at the moment but I'll take a look at the other ideas for this problem as soon as I can.

 

[mathwonk]

for #9, I use the cell by cell construction of P^infinity to define a map, cell by cell, from P^infinity to any other K(Z/2Z,1) space, then whitehead gives the homotopy equivalence.

Hints/Spoilers: for 5,7:
for #5: (assuming the matrices have entries from a field, as suggested by the term "vector space" for the set of solutions X); sending X to AXB changes X linearly from an arbitrary linear map k^p-->k^m, into an arbitrary linear map k^s-->k^r, so this linear operation has kernel of dimension mp-rs.

for #7, one can use certain covering spaces, in particular, spaces X,Y, with X a subspace of Y, and Y a cover of X. one can even take them to be 1 dimensional simplicial complexes. The point is that a covering space map induces injection on fundamental groups. (I do not know if one can find an abelian solution.)

 

[mathwonk]

In #9, if f has no fix point, then if gives a free action by the group Z/2Z on R^n. Then as Infrared points out, we get an equivalence relation by setting x ≈ f(x) and the quotient map to the set of equivalence classes, R^n-->X = R^n/{x≈f(x)} is actually a covering space map with fibers of degree 2. Thus X is a manifold of dimension n, with fundamental group Z/2Z, and contractible universal covering R^n. Since maps of simply connected spaces lift to covering spaces, and homotopies push down, it follows that all homotopy groups of X in degree ≥ 2 are trivial, since this holds for R^n.

Now this reminds us of real projective space P. I.e. P^2 has fundamental group Z/2Z, but has higher homotopy groups, since it is covered by S^2 which has them. but P^3 also has π1 = Z/2Z, and no π2, since S^3 has none. So if we crank up to P^infinity, we kill all πk, for k ≥2, while keeping π1 = Z/2Z. Then we ask whether this would force our X above to be homotopy equivalent to P^infinity. Indeed it does, as we can deduce from Whitehead's theorem, provided we find a map from P^infinity to our X that induces these isomorphisms on homotopy groups.

'This follows from the construction of P^infinity by adding cells one at a time to P^2. First we map S^1 to our X as a representative of the generator of π1. Then since twice that map is homotopic to zero, we can extend twice that map to the disc, but since P^2 is constructed by gluing a disc to S^1 by twice the identity, this means we can extend our map of S^1 to X, to a map of P^2 to X. Now if we map S^2 onto P^2 by the antipodal map, we get a map of S^2 into X, identifying antipodal points, and since X has no π2, the map extends to a map of the solid ball into X, also identifying antipodal points of the boundary sphere, hence defining a map of P^3 into X. Continuing to infinity, I hope we get a map of P^infinity to our X, inducing isomorphisms of all homotopy groups, hence by Whitehead, defining a homotopy equivalence of P^infinity with our X. But this is a problem, since P^infinity is infinite dimensional and our Xn is a finite quotient of an n manifold. So can we make this a contradiction?
One way is to compute the homology groups of these spaces. An n manifold has them only up to at most dimension n, and we claim that P^infinity has them in infinitely many degrees, but why?

I guess I don't know this homology theory, so need to learn it before answering further. but people usually compute homology by a maier vietoris sequence.

ok there is an explanation in Greenberg chapter 19, of how to compute the change in homology from adding an n cell to a space, and it gives projective space as an example, and in the case of real projective space with Z/2Z coefficients, every time you add a new n-cell you get a new copy of Z/2Z in n dimensional homology . Thus for P^infinity you have Z/2Z as homology in every dimension. In particular it is not homotopy equivalent to any finite dimensional (connected) manifold. qed. X does not exist.

 

[mathwonk]

now for #7: by the Seifert-VanKampen theorem, the fundamental group of a wedge of n circles, is the free group on n generators, and a covering map induces injections of fundamental groups. Now you may be able to see how to construct a 2:1 covering map from a wedge of 3 circles to a wedge of 2 circles; just put the end circle over the end circle, then wrap the middle circle twice around the other circle, and then map the other end circle down to the first circle. This implies the free group on 3 generators is isomorphic to a subgroup of the free group on 2 generators. Now a free Group on n generators is not isomorphic to the free group on m generators unless n=m, since modding out by the commutator subgroup would imply the isomorphism of free abelian groups G,H on n ≠ m generators. That fails by modding out further by the subgroups 2G, 2H, forcing isomorphism of vector spaces G/2G, H/2H, over the field Z/2Z, but of different dimensions.

challenge: does an example exist with abelian groups????

 

[Infrared]

@mathwonk's answer for 7 is correct, but just to make his comments a little more explicit: if you pick a double cover of $S^1\vee S^1$ (which will be homeomorphic to a wedge of three circles, though not with only one shared wedge point) and then pick three elements which freely generate its fundamental group, they will descend to elements of $\pi_1(S^1\vee S^1)$ which freely generate a free subgroup of rank 3. For the double cover as I drew it, this means that $a,b^2,bab^{-1}$ generate a free subgroup of rank 3 inside of the free group generated by $a$ and $b.$ I'm also pretty sure you don't need any topology and can just argue directly that there are no relations between $a,b^2,bab^{-1}$ or whatever the elements you found were.

challenge: does an example exist with abelian groups????

Apparently yes. Here is a construction of an abelian group $A$ which is isomorphic to $A\oplus\mathbb{Z}^2$ but not isomorphic to $A\oplus\mathbb{Z}.$

 

[mathwonk]

interesting. that abelian example seems to have taken several months for some well known mathematicians to come up with.

and thank you, I guess my example is only homotopic to a true wedge of 3 circles. I.e. I added the third circle on at a smooth point of the wedge of the first two.

 

[AndreasC]

now for #7: by the Seifert-VanKampen theorem, the fundamental group of a wedge of n circles, is the free group on n generators, and a covering map induces injections of fundamental groups. Now you may be able to see how to construct a 2:1 covering map from a wedge of 3 circles to a wedge of 2 circles; just put the end circle over the end circle, then wrap the middle circle twice around the other circle, and then map the other end circle down to the first circle. This implies the free group on 3 generators is isomorphic to a subgroup of the free group on 2 generators. Now a free Group on n generators is not isomorphic to the free group on m generators unless n=m, since modding out by the commutator subgroup would imply the isomorphism of free abelian groups G,H on n ≠ m generators. That fails by modding out further by the subgroups 2G, 2H, forcing isomorphism of vector spaces G/2G, H/2H, over the field Z/2Z, but of different dimensions.

challenge: does an example exist with abelian groups????

Woah I did think free groups but it never occured to me the free group on 3 generators is isomorphic to a subgroup of the free group on 2. It's very counter intuitive.

 

[Greg Bernhardt]

Last day for the July challenges! Let's go!

 

[mathwonk]

I think I gave the idea and answer to #5, 3 weeks back in post #63, but here goes again: I will do it two different ways.

1) To satisfy AXB = 0, X only has to map the s dimensional image of B into the (m-r) dimensional kernel of A, and such linear maps have dimension s(m-r). X can do anything to the complement of the image of B, i.e. can send that (p-s) dimensional complement into anything in the m dimensional domain space of A. Such complementary maps have dimension m(p-s). So in sum, the dimensions of the space of such X equals s(m-r) + m(p-s) = mp-rs.

2) I.e. sending X to AXB, changes X from a map from p space to m space, into a map from the image of B to the image of A, i.e. from s space to r space. Thus sending X to AXB is a surjective (as is "easily shown") linear map from a space of dimension mp to a space of dimension rs, hence the kernel, i.e. those X with AXB = 0, has dimension mp-rs.