Challenge Math Challenge - July 2023

[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.