Math Challenge - July 2020

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  • #36
Adesh
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A more serious flaw is that not every continuous function has a (pointwise convergent) Fourier series so your argument does not work.
It is given that the function is bounded and periodic therefore it does have a convergent Fourier series.
 
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  • #37
Adesh
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take a function ##f(x)=2+\cos(2\pi x)## and try to prove this equality
Yes I got ##a_0 \gt \frac{c_0}{c’_0}##.
 
  • #38
It is given that the function is bounded and periodic therefore it does have a convergent Fourier series.

This is false. As a corollary of the uniform boundedness principle you can prove that there are Fourier series that do not converge pointwise to the function itself. See Papa Rudin for a proof.

You always have ##L_2##-convergence but that's not enough for what you did.
 
  • #39
lavinia
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I make a rule not to answer Math Challenges but if anyone wants I can throw out some thoughts for problem 8.
 
  • #40
mathwonk
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I am interested in your thoughts on prob. #8. The only solution I know is via non trivial properties of Eilenberg Maclane spaces, i.e. if a K(G,1) has a finite dimensional CW structure, then G is torsion free. Might be enough to know the cohomology of (infinite) lens space.
 
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  • #41
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@lavinia I don't mind you giving your thoughts (unless you've seen an equivalent problem before). I proposed it knowing that it might be less accessible than most of the other problems.

@mathwonk There is a solution only using standard tools in a first algebraic topology class, but it might be a little tricky to find (hint: by Whitehead, an equivalent problem is to show that the universal cover of ##X## is not contractible)
 
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  • #42
mathwonk
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Nice hint. (That seems to do it assuming your compact manifolds have no boundary.) But I confess that theorem of Whitehead was not in my alg top course (although postnikov towers were). To be fair, his presumed parent theorem that a map of CW complexes induces a homotopy equivalence if it induces isoms on homotopy groups was. Of course Whitehead's 1978 book, in which I have seen the theorem cited, came out after my graduate school experience. And I am probably confusing JHC with GW.
 
  • #43
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To be fair, his presumed parent theorem that a map of CW complexes induces a homotopy equivalence if it induces isoms on homotopy groups was.
This was the theorem I was referring to. Did you have something else in mind?
 
  • #44
mathwonk
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No but I was wondering where the map comes from. In case all groups vanish, do you just map in a one point space, and deduce that the space is homotopy equivalent to a point? That's a nice example of a case where just knowing the groups does all the work! ( I have always internalized the warning about that theorem that just knowing two spaces have the same homotopy groups does not (usually) mean they have the same homotopy type, unless those group isomorphisms are induced by a map of the spaces.)
 
  • #45
zinq
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8. Let be a compact manifold M such that π1(M) is finite and nontrivial. Show that πn(M) is also non-trivial for some n ≥ 2.

Let G = π1(M). Suppose that for all n ≥ 2, πn(M) is trivial. Then M is a K(G,1) (Eilenberg-MacLane) space. Let g ∈ G be any nontrivial element and let H denote the finite subgroup of G generated by g. H is a nontrivial finite cyclic group, say H = Zp. Let M' → M be the covering space of M corresponding to the subgroup H of G = π1(M). That is, π1(M') = Zp.

Then by standard covering space theory, πn(M') = πn(M) is trivial by assumption. Hence M' is a K(Zp, 1).

Since the group H determines the homotopy type of any K(H, 1), this implies that each K(Zp, 1) space is homotopy equivalent to the lens space Lp = S / Zp, where S is the infinite ascending union of S1 ⊂ S3 ⊂ ... ⊂ S2n-1 ⊂ ..., where

S2n-1 = {(z1, ..., zn) ∈ ℂn | |z1|2 + ... + |zn|2 = 1},

and Zp acts on S via multiplication by exp(2πi/p) on each component.

However, we know (Hatcher, Algebraic Topology (2001), Example 3.41, p. 251) that each even-dimensional cohomology group H2k(Lp; Zp) is nonzero.

This shows that any K(Zp, 1) — and in particular the covering space M' — cannot be a finite cell complex. But since every compact manifold is a finite cell complex, and every finite covering space of a finite cell complex is also a finite cell complex, this is contradicts our assumption that M is a K(G,1). Hence πn(M) is also non-trivial for some n ≥ 2, QED.
 
  • #46
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@mathwonk Right, this is the one case where it is possible! Of course the other pitfall is that you have to verify your space is in fact a cell complex.

@zinq Yes, that's right, nice work. I think this was the argument @mathwonk had in mind in post 40.

I'll just point out a short/elementary but tricky solution: Suppose for contradiction the universal cover ##\tilde{X}## were contractible. The Lefschetz fixed point theorem applies, and so in particular any covering transformation of ##\tilde{X}\to X## has a fixed point and is thus the identity. This contradicts ##\pi_1(X)\neq 1##. Hence ##\tilde{X}## has a nontrivial homotopy group, necessarily in degree at least ##2##, and so does ##X##.
 
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  • #47
Adesh
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I tried something few things more, I worked on finding the minimum sufficient codition for ##\int_{0}^{1} \frac{f(x+a)}{f(x)}\geq 1## but couldn’t anything better than ##\frac{f(x+a)}{f(x)} \geq##.

If ##a## is an integer then ##\frac{f(x+a)}{f(x)}## will be 1, if ##a## is not an integer then let ##\{a\}## denote the fractional part of ##a##, so we have
$$
\int_{0}^{1}\frac{f(x+\{a\})}{f(x)} \geq 1$$
But again not much useful.

Can I get a little hint about this problem? (If a little hint can solve the whole problem then, no issues, I shall wait for the solution).

Thanks @wrobel for this question.
 
  • #48
nuuskur
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I understand we can assume without loss [itex]|a|<1[/itex]. I tried using Lagrange's MVT (integrand is continuous), that means the integral is equal to [itex]\frac{f(u+a)}{f(u)}[/itex] for some [itex]u\in (0,1)[/itex]. No brilliant ideas, though.
 
  • #49
zinq
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I'll just point out a short/elementary but tricky solution: Suppose for contradiction the universal cover were contractible. The Lefschetz fixed point theorem applies, and so in particular any covering transformation of has a fixed point and is thus the identity. This contradicts . Hence has a nontrivial homotopy group, necessarily in degree at least , and so does .

Wow, that is really elegant!

P.S. Before posting, I had not seen any prior posts on Problem 8. except the statement of the problem.
 
  • #50
lavinia
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A couple of ideas: Maybe all wrong

Assume that the manifold is aspherical in the sense that all of its homotopy groups of dimension 2 through n are zero where n is its dimension.

Proof 1:By the exact homotopy sequence of the fibration the universal covering manifold is also aspherical and since it is also simply connected Hurewicz's Theorem says that its integer homology is zero in dimensions 1 to n. So it has Euler characteristic 1. The covering projection divides the Euler characteristic by the order, p, of the fundamental group of the base manifold so the base manifold has Euler characteristic 1/p. This can't happen because Euler characteristics are integers. ( This Euler characteristic argument I think uses that the base manifold is a finite CW complex so I am not sure if all compact manifolds are accounted for. )

Proof 2: If a finite p-group acts on a compact topological space and that space's homology mod p is acyclic, then the action has a fixed point. In this case the universal cover's homology is mod p acyclic for all p. So any p subgroup of the group of covering transformation must have a fixed point.

This theorem of p group actions is due to P. A. Smith.
 
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  • #51
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@lavinia Manifolds are CW complexes by Morse theory, and a CW complex must be finite to be compact (otherwise picking an interior point from each cell gives an infinite discrete subspace), so I think that part of the argument is fine. The asphericity assumption looks like the main issue to me.

Your second argument has the same idea as my argument in post 46. You can take a look there to see one way to sharpen it.
 
  • #52
lavinia
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@lavinia Manifolds are CW complexes by Morse theory, and a CW complex must be finite to be compact (otherwise picking an interior point from each cell gives an infinite discrete subspace), so I think that part of the argument is fine. The asphericity assumption looks like the main issue to me.

Your second argument has the same idea as my argument in post 46. You can take a look there to see one way to sharpen it.
The assumption of aspherical is weaker than the problem demanded., It is used for getting a contradiction. The argument is correct.

Morse theory I think works for smooth manifolds but I don't know what works for other manifolds.
 
  • #53
Infrared
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I see, I didn't read carefully. It looks fine then.
 
  • #54
zinq
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"Morse theory I think works for smooth manifolds but I don't know what works for other manifolds."

Yup. PL manifolds are immediately CW complexes. I'm not sure about those non-PL and not even triangulable manifolds they now know exist in all dimensions ≥ 4. (I just read somewhere that an argument of Milnor shows that all topological manifolds have the homotopy type of a CW-complex.)
 
  • #55
Summary:: Functional Analysis, Topology, Differential Geometry, Analysis, Physics
Authors: Math_QED (MQ), Infrared (IR), Wrobel (WR), fresh_42 (FR).

(solved by @Isaac0427 with complex numbers, completely real solution still possible)
I'll leave that for someone else to answer-- I thought of the complex solution first for some reason, but I see that there's a much easier way to go about it.
 
  • #56
fresh_42
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I'll leave that for someone else to answer-- I thought of the complex solution first for some reason, but I see that there's a much easier way to go about it.
The solution I have in mind is basically only a real version of your solution, but it uses a function which helps here. That's why I wrote that remark. Something left to learn.
 
  • #57
The solution I have in mind is basically only a real version of your solution, but it uses a function which helps here. That's why I wrote that remark. Something left to learn.
The other one I had in mind is an inductive proof.
 
  • #58
fresh_42
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The other one I had in mind is an inductive proof.
I think formally it is always an induction. Even your pairing is formally an induction since commutativity is only defined for two factors. My solution works with matrices, but of course it is only another formalism than yours.
 
  • #59
Adesh
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Another method for Question 15.

By Stirling’s approximation we have ##n! =\sqrt{2\pi n} \left( \frac{n}{e}\right)^n ##.
Therefore,
$$
(n!)^2= 2\pi n \left(\frac{n}{e}\right)^{2n}
$$
And
$$
(2n)!= 2 \sqrt{\pi n} 4^n\left( \frac{n}{e}\right)^{2n}$$
$$\frac{(2n)!}{(n!)^2}= \frac{4^n}{\sqrt{\pi n}}
$$
So, we got to compare ##\frac{4^n}{\sqrt{\pi n}}## and ##\frac{4^n}{n+1}##.
My claim: ##\sqrt{\pi n} \lt n+1##,
$$
\pi n \lt n^2 + 1 + 2n
$$ above statement is true for n=2 and n=3 by trial and error method. For ##n \gt 3## we have
$$
\pi n \lt n^2 \\
\therefore
\pi n \lt n^2 +1 + 2n
$$
Hence, ##\sqrt{\pi n} \lt (n+1)##, therefore
$$
\frac{4^n}{\sqrt {\pi n}} \gt \frac{4^n}{n+1}
$$
 
  • #60
fresh_42
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Another method for Question 15.

By Stirling’s approximation we have ##n! =\sqrt{2\pi n} \left( \frac{n}{e}\right)^n ##.
Therefore,
$$
(n!)^2= 2\pi n \left(\frac{n}{e}\right)^{2n}
$$
And
$$
(2n)!= 2 \sqrt{\pi n} 4^n\left( \frac{n}{e}\right)^{2n}$$
$$\frac{(2n)!}{(n!)^2}= \frac{4^n}{\sqrt{\pi n}}
$$
So, we got to compare ##\frac{4^n}{\sqrt{\pi n}}## and ##\frac{4^n}{n+1}##.
My claim: ##\sqrt{\pi n} \lt n+1##,
$$
\pi n \lt n^2 + 1 + 2n
$$ above statement is true for n=2 and n=3 by trial and error method. For ##n \gt 3## we have
$$
\pi n \lt n^2 \\
\therefore
\pi n \lt n^2 +1 + 2n
$$
Hence, ##\sqrt{\pi n} \lt (n+1)##, therefore
$$
\frac{4^n}{\sqrt {\pi n}} \gt \frac{4^n}{n+1}
$$
Nice idea! The problem with Stirling is, that your equalities are only approximations. So in order for proof to count, you have to take the error margins into account. Especially the division needs watching, since a small error could theoretically turn into something large if you divide it. So you need to use expressions with upper and lower bounds.
 
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  • #61
lavinia
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I am interested in your thoughts on prob. #8. The only solution I know is via non trivial properties of Eilenberg Maclane spaces, i.e. if a K(G,1) has a finite dimensional CW structure, then G is torsion free. Might be enough to know the cohomology of (infinite) lens space.
Another way at your argument is to invoke the theorem that the manifold ##M## together with the covering ##M^{*}→M## must be the universal classifying space for principle discrete ##π_1(M)## bundles. ##M## then has the cohomology of its fundamental group and since ##π_1(M)## is finite and non-trivial it has non-zero cohomology in unbounded dimensions. One can get away with a cyclic subgroup of prime order. For all of this I think you need ##M^{*}## to be weakly contractible which is a stronger result than is necessary to answer the question. This is the same as @zinq 's argument I think.

BTW: If one argues by contradiction that the manifold has no homotopy groups in dimension 1 through n then the manifold can not be closed for then Hurewicz's Theorem would say that the nth homotopy group of the universal covering manifold is isomorphic to ##Z##. You can also argue that the boundary can have only 1 connected component from the exact homology sequence of the pair. I tried taking this idea further but didn't succeed.

One thought was that a contractible compact manifold with boundary reminds one of Brouwer's fixed point theorem and one can ask when such manifolds can be made convex with respect to some geometry.
 
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  • #62
Adesh
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Nice idea! The problem with Stirling is, that your equalities are only approximations. So in order for proof to count, you have to take the error margins into account. Especially the division needs watching, since a small error could theoretically turn into something large if you divide it. So you need to use expressions with upper and lower bounds.
Oh thank you for pointing that out! Yes for small ##n## we must take care of errors.
 
  • #63
Adesh
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Nice idea! The problem with Stirling is, that your equalities are only approximations. So in order for proof to count, you have to take the error margins into account. Especially the division needs watching, since a small error could theoretically turn into something large if you divide it. So you need to use expressions with upper and lower bounds.
Okay, in this article the error bounds for ##n!## are given as
$$
\sqrt{2\pi n} \left( \frac{n}{e} \right)^n \leq n! \leq \sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}$$
So, working with error bounds we have
$$
\frac{1}{
{2\pi n} \left( \frac{n}{e} \right)^{2n} ~e^{\frac{1}{6n}}} \leq \frac{1}{(n!)^2} \leq \cdots $$
$$2\sqrt{\pi n} ~4^n ~\left(\frac{n}{e}\right)^{2n} \leq (2n)! \leq \cdots $$

$$\frac{
4^n}
{\sqrt{\pi n} e^{\frac{1}{6n} }} \leq \frac{(2n)!}{(n!)^2} \leq \cdots$$
Now, it is left to show that ##\sqrt{\pi n} e^{\frac{1}{6n}} \lt n+1##. Which we can easily prove by noting that we have to compare

$$\pi ~n~e^{\frac{1}{3n}} ~and~n^2 +1 +2n$$

if ##n\gt 3~~ (n\in \mathbb N)## then we have

$$\pi e^{ \frac{1}{3n}} n \lt n^2 $$
$$\implies ~~\pi e^{ \frac{1}{3n}} n \lt n^2 +1 + 2n $$
For ##n=2 , 3## we can do trial and error.

Hopefully, we are done. Please point if there are some errors, or you have more elegant way for doing this.
 
  • #64
fresh_42
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Hopefully, we are done. Please point if there are some errors, or you have more elegant way for doing this.
No, that's ok. Maybe you should have mentioned that you can only do this, because the numbers are positive. And one doesn't say "trial and error" here. Better say: "we check for ..." but you could have simply said that it is true for ##n=2##: ##\pi \cdot 2 \cdot e^{\frac{1}{6}} \approx 7.423 < 8 < 9 = 2^2+1+2\cdot 2## and then observed that the RHS grows faster than the LHS.
 
  • #65
mathwonk
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In reference to post #51, does a CW complex which is homotopy equivalent to a compact manifold have to also be compact? I.e. does one know that a compact topological n manifold has a structure of finite CW complex? (The paper of Milnor I have seen referenced apparently proved only that there is a countable such CW complex.)
 
  • #66
Infrared
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@mathwonk It certainly doesn't have to be compact (e.g. ##\mathbb{R}## is a CW complex homotopy equivalent to a point) but I did find a reference that you can always find a homotopy equivalent finite CW complex to a compact manifold (boundary allowed): http://people.math.harvard.edu/~lurie/281notes/Lecture34-Part3.pdf I haven't tried to read it. There's also a sketch here: https://math.stackexchange.com/questions/1648250/spaces-homotopy-equivalent-to-finite-cw-complexes (again I haven't tried to read it)
 
  • #67
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Thanks for the link to the nice, if brief, notes. They do not actually prove the desired result, referring to a "non trivial" theorem of Chapman. I have no hope of getting through the works of chapman, but the (first) link you gave states a very nice result, apparently having the homotopy type of a finite CW complex is equivalent to having the homotopy type of a compact manifold (possibly with boundary)!

(I had also seen the second link but found it somewhat unclear.)
 
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  • #68
zinq
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Please forgive my primitive typography. (In what follows, Int denotes the integral from 0 to 1.)

Let f : R —> R be continuous such that for all x, f(x+1) = x and f(x) > 0.

Let a = p/q be a rational number in lowest terms.

Then

Int ( (f(x + p/q) / f(x)) * (f(x + 2*p/q) / f(x + p/q)) * ... * (f(x + q*p/q) / f(x + (q-1)*p/q)) )^(1/q) dx

= Int 1 dx = 1

because by periodicity the integrand is the constant function = 1^(1/q) = 1 for all x.

Since any (arithmetic mean) ≥ (the corresponding geometric mean), we get

(1/q) * Int (f(x + p/q) / f(x)) + ... + (f(x + q*p/q) / f(x + (q-1)*p/q)) dx ≥ 1

But the integrals from 0 to 1 of all the summands are equal by periodicity. Hence

Int( f(x+p/q) / f(x) dx ≥ 1

But

G(a) = Int f(x+a)/f(x) dx

is a continuous function of a, and rationals are dense in R. Hence by taking a sequence of rationals p/q converging to an arbitrary a, we get that

G(a) ≥ 1

for all a.
 
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  • #69
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3 has been solved by @zinq
another solution is by Jensen's inequality:
$$\ln\int_0^1\frac{f(x+a)}{f(x)}dx\ge\int_0^1\ln f(x+a)-\ln f(x) dx=0$$
 
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  • #70
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I don't know if you're allowed to answer/give hints, but for problem 14, I had this thought:

We have two equations and two unknowns. This is a good start. Although, I don't see any way we can solve for ##x## in terms of ##y## (I've really tried), which makes it hard to make progress. Although I recognize that ##\sin^n{(x+\frac{\pi}{2})} = \cos^n{(x)}##. Does that mean for in order for there to be any solutions at all that $$\sin^n{(x+\frac{\pi}{2})} - \cos^n{(x)} = y^4+\left(x + \frac{\pi}{2} \right)y^2-4y^2+4 - x^4+x^2y^2-4x^2+1 = 0$$

I might be completely wrong in my reasoning, but I was just wondering.

Disclaimer: I've never done a "find all solutions x,y for a function" type problem.
 

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