Math Challenge - November 2021

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In summary, In summary, we discussed various mathematical topics including analysis, projective geometry, ##C^*##-algebras, group theory, Markov processes, manifolds, topology, Galois theory, linear algebra, and commutative algebra. We also solved various problems, including the continuity of inverse functions, determining the existence of a straight line intersecting two skew lines, positive functionals in ##C^*##-algebras, and irreducibility and aperiodicity of Markov chains. We also proved properties of normable topological vector spaces, orientability of manifolds, and isomorphisms between vector spaces. Lastly, we solved equations and inequalities involving real numbers, sequences, and functions
fresh_42 said:
This is too vague and looks too Euclidean. How do you construct ##A## and ##B## with those properties? Anyway, the correct solution has been given in post #33.
Yeah, I wasn't finished when mathwonk posted the answer. I was working from the axioms in post #14. After review, it looks like statements 3, 4, and 6 are pretty close.

6
mathwonk said:
That leaves 4b, 5 and 6. Are we stumped?!
What is the difference between 4.a and 4.b? These are the matrices of the corresponding Moebius' transformations ##\alpha## and ##\beta##, and a composition of transformations has the product of the matrices.

Nice! I was thinking they should be equivalent, but was trying to do it via 4b acting on the real plane, the complement of infinity. Your idea seems to use the realization of the riemann sphere as the projective complex line. Maybe you could give a little more detail.

\\SPOILER: #6:
as for 6, some hints are: an n form determines a b-orientation, and then to get an a-orientation, choose any atlas whose domain sets are connected. Then each chart will determine a pullback of the standard n form which is everywhere in that chart either a positive or negative multiple of the given one. If negative, compose the chart with an oreintation reversing linear isomorpohism of n space. The resulting family of charts, all of which pull the standard form back to the given one on M, is an a -orientation.

To go the other way, use an oriented atlas of charts to pull back copies of the standard form, then paste them together with a partition of unity, and use the property that the atlas is oriented, to check that at each point, the constructed form is a sum of positive multiples of one of them, hence non zero.

an a -orientation gives an "orientation" by using the standard vector fields defined by the charts in an oriented atlas. To go the other way, choose an atlas made up of charts subordinate to the domains of the vector fields verifying the orientation property. Then maybe wedge them together to give a non zero field of n -vectors, hence by duality an n form in that chart, and glue them into an n form with a partition of unity?

martinbn and fresh_42
mathwonk said:
Nice! I was thinking they should be equivalent, but was trying to do it via 4b acting on the real plane, the complement of infinity. Your idea seems to use the realization of the riemann sphere as the projective complex line. Maybe you could give a little more detail.
What I meant is that the two groups are isomorphic. For fractional linear transformations composition is given by the multiplication of the corresponding matrices. Similarly for takinf inverse. So maming the two generators of 4b to those of 4a and extending it multiplicatively will be well defined and an isomorphism.

martinbn said:
What I meant is that the two groups are isomorphic. For fractional linear transformations composition is given by the multiplication of the corresponding matrices. Similarly for takinf inverse. So maming the two generators of 4b to those of 4a and extending it multiplicatively will be well defined and an isomorphism.
There is a reason why I put them into one problem and not two. Problems of the same number should be solved as a whole. Unfortunately, members torpedoed this plan - pretty much from the beginning.

martinbn
martinbn, I understood you meant they were isomorphic, but I think you still have to prove it. One way to prove it is to give a bijection between the riemann sphere and the projective line under which the two actions correspond. Perhaps you were taking all that for granted. (As a naive person like me might attempt, of course one cannot so easily define a homomorphism just from the matrix group in 4b to the group in 4a, by saying what it does to the generators, unless we know already the matrix group they generate is free.)

I seem to be on a slightly different page from others, or just further behind, so let me explain my viewpoint. I apologize if this is more algebro - geometric than desired, and/or too elementary to be useful.

I was trying to relate problems 4a and 4b by letting a matrix act on the riemann sphere, by having the matrix with rows [1 0], [2 1], send x+iy to x + (2x+y)i, and the matrix with rows [1 2], [0 1], send x+iy to (x+2y) + iy, both of which of course send infinity to itself. I.e. I was letting x+iy be the point of real 2 space with coordinates (x,y) and then letting the (real) matrix act on this vector. So, obviously this didn’t work, since the given fractional transformations do not both fix infinity.

Now martinbn gave the correct action, which is the natural linear action of a complex 2x2 matrix matrix on the projective complex line. I.e. consider non zero vectors [u,v] and [z,w] in complex 2 space as equivalent if they are proportional, i.e. [u,v], with u,v, both complex, is equivalent to any vector of form [tu,tv] with t≠0. Then a vector [u,v] with v≠0 is equivalent to a unique vector of form [z,1], namely [u.v] ≈ [u/v,1], so take z = u/v. All vectors [u,v] with v=0 are equivalent to each other, in particular [u,0] ≈ [1,0].

The resulting identification space, the complex projective line, is in natural bijective correspondence with the riemann sphere by letting [u,v] correspond to [u/v,1] <—-> z = u/v, if v ≠ 0, and letting [1, 0] <—-> infinity. In the other direction, a finite complex number z corresponds to [z,1], and infinity corresponds to [1,0] in the projective line.

Now a complex matrix with rows [a b], [c d], sends the vector [u, v] to [au+bv, cu+dv], and it sends the equivalent vector [tu, tv] to an equivalent vector, namely [t(au+bv), t(cu+dv)]. Hence a complex 2x2 matrix acts on the projective line, and hence also on the riemann sphere.

If z is a finite point of the riemann sphere, corresponding to the vector [z,1], then [a b], [c d], sends it the vector [az+b, cz+d], which corresponds to the point (az+b)/(cz+d) of the riemann sphere. I.e. this is a finite point iff cz+d is not zero. If we want to act on the point at infinity, we take the corresponding vector [1,0], and it then goes to the vector [a, c], which corresponds to the finite point a/c if c≠0, and to infinity if c=0. Oh yes, since we want the action on the riemann sphere to be a bijection, we assume that ad-bc ≠ 0.

So this shows how a complex matrix acts on the riemann sphere in a way that exactly corresponds to the action of a fractional transformation. Moreover, matrix multiplication corresponds to composition of mappings of complex 2 space, hence also of the projective line, and of the riemann sphere. Note however that the matrix [a b], [c d], acts in the same way as the matrix [ta tb], [tc td], for t≠0. So the matrix group is not isomorphic to the group of fractional transformations, since any diagonal matrix with equal diagonal entries corresponds to the identity fractional transformation, i.e. two matrices define the same transformation iff their entries are proportional.

We can however restrict to the subgroup of matrices with determinant one, and then the map from matrices to transformations is still a surjective homomorphism, and has kernel of order two, consisting of the identity matrix and minus the identity matrix. Now restrict this map to the subgroup generated by the given matrices in problem 4b. It follows that this is a surjective homomorphism from the subgroup defined in 4b, to the subgroup of fractional transformations defined in 4a. We must still show this is injective, and hence an isomorphism, since if minus the identity matrix were in the subgroup generated by the matrices in 4b, the map would be 2 to 1 instead of an isomorphism. But since 4a has been solved, that subgroup is free, and hence there is a map back, which is inverse to this map on generators, hence everywhere.

This may all be standard, but for me it is complex enough to deserve some details. At any rate, this is just an elaboration of martinbn's cogent observation.

fresh_42
mathwonk said:
martinbn, I understood you meant they were isomorphic, but I think you still have to prove it. One way to prove it is to give a bijection between the riemann sphere and the projective line under which the two actions correspond. Perhaps you were taking all that for granted. (As a naive person like me might attempt, of course one cannot so easily define a homomorphism just from the matrix group in 4b to the group in 4a, by saying what it does to the generators, unless we know already the matrix group they generate is free.)
We don't need to know that it is free. My point was that multiplication and inverse of the matrices corresponds to composition and inverse of the linear fractional transformations. For example the inverse of ##f(x)=\frac{ax+b}{cx+d}## is ##f^{-1}(x)=\frac{a'x+b'}{c'x+d'}##, where ##\begin{bmatrix}a'&b'\\c'&d' \end{bmatrix}## is the inverse of ##\begin{bmatrix}a&b\\c&d \end{bmatrix}##

So, if we have a relation in group 4.b, say ##A^{n_1}B^{m_1}\dots A^{n_k}B^{m_k}=I ##, then the element ##A^{n_1}B^{m_1}\dots A^{n_k}B^{m_k} ## is mapped to ##\alpha^{n_1}\beta^{m_1}\dots \alpha^{n_k}\beta^{m_k}##, which is going to be a linear fractional transformation with matrix the matrix ##A^{n_1}B^{m_1}\dots A^{n_k}B^{m_k} ## , so it will be the identity. So the map is well defined.

fresh_42