Recent content by Infrared

@mathwonk I think you have it all there! An isotropic subspace can have dimension at most half the dimension of the total space and since the first cohomology of ##M_g## over a field has dimension ##2g##, the result is immediate.

Some hints for the remaining problems :) 7)In general, for a finite group ##G##, the number of commuting pairs ##(g,h)\in G\times G## is ##|G| \cdot \left(\text{number of conjugacy classes of G}\right).## To prove this, find a formula for the number of elements which commute with a given ##g\in...

This is correct but this problem was intended for much less advanced students than you :)

A bit of googling finds a fun argument giving examples: Let ##f## be an irreducible monic degree 4 polynomial in ##\mathbb{Z}[x]## satisfying: 1) ##f## has exactly two real roots. 2) The coefficient of ##x^i## is the same as the coefficient of ##x^{4-i}.## An example of such a polynomial is ##...

This is not the case. For example, ##\frac{3}{5}+\frac{4}{5}i## is on the unit circle, but the angle ##\text{arcsin}(4/5)## is not a rational multiple of pi (the only time ##x/\pi## and ##\sin(x)## can both be rational is when ##\sin(x)=0,\pm 1/2,\pm 1##).

The solutions by @PeroK and @mfb for the first problem look correct! I'll also share how I counted. There are ##\frac{8!}{2!2!2!}=7!## ways to arrange the 8 pieces. If we choose a setup uniformly at random, there is a 4/7 probability that the two bishops have opposite colors, and then...

Very nice @julian! Your identity ##\sum_{n=1}^\infty\frac{\sin(nx)}{n}=\frac{\pi-x}{2}## is much easier to get with Fourier series though have you to work backwards a little bit to realize that this is the function you're taking a Fourier series of. @anuttarasammyak This looks very slick, but a...

The Math challenge threads have returned! Rules: 1. You may use google to look for anything except the actual problems themselves (or very close relatives). 2. Do not cite theorems that trivialize the problem you're solving. 3. Do not solve problems that are way below your level. Some problems...

@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...

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...

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.

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

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.##

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))##.

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.