Math Challenge - November 2021

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  • #26
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I'll take a go at 8b.

Let ##p(t) = T^3-2##. Notice that ##p## has not roots over ##\mathbb{F}_7## which can be seen by simply evaluating ##p## at each element of ##\mathbb{F}_7##.

We claim then ##p## is irreducible. If not then ##p## can be factored as ##p = f g## where ##f##, ##g## are irreducible and ##deg(f) + deg(g) = 3##. Since ##f## and ##g## are irreducible, ##deg(f), deg(g) > 0## and so either ##deg(f) = 1## or ##deg(g) = 1##. Suppose without loss of generality that ##deg(f) = 1##, then ##f## is linear and hence has a root of ##\mathbb{F}_7## which means that ##p## has a root over ##\mathbb{F}_7## which is a contradiction.

Since ##p## is irreducible it then ##(p)## must be maximal. Otherwise suppose there exists some ##q \in \mathbb{F}_7[T]## such that ##(p) \subset (q)##. Hence there exists an ##f \in \mathbb{F}_7[T]## such that ##p = qf##. Since ##p## is irreducible either ##q## is a unit (in which case ##(q) = \mathbb{F}_7[T]##) or ##f## is a unit (in which case ##(f) = (q)##). So indeed ##(f)## is maximal.

We now use the fact that given any ring ##R## with a maximal ideal ##M##, ##R/M## is a field. This follows from one of the standard isomorphism theorems which states that there is a one-to-one correspondence between the ideals of ##R/M## and the ideals in ##R## containing ##M##. So if ##M## is maximal, ##R/M## can not contain any proper ideals, and so must be a field.

To compute the number of elements in ##\mathbb{F}_7[T]/(T^3-2)## we note that since ##3## is prime each cosets can be represented by elements of the form ## c_1 + c_2 T + c_3 T^2## where ##c_i \in \mathbb{F}_7[T]##, and there are ##7^3 = 343## such elements.

To compute ##(T^2+2T+4)(2T^2+5)## in ##\mathbb{F}_7[T]/(T^3-2)## we first note that ##(T^2+2T+4)(2T^2+5) = 2T^4 + 2T^3 + 6T^2 + 3T + 6## in ##\mathbb{F}_7## then reducing ##mod (T^3-2)## gives ## 2(2)T + 2(2) + 6T^2 + 3T + 6## and so $$(T^2+2T+4)(2T^2+5) = 6T^2 + 3$$

To compute ##\frac{1}{T+1}## you can do a sort of "brute force" method. Let ##\frac{1}{T+1} = P(T) = a_0 + a_1T + a_2T^2## then ##(T+1)P(T) = (a_1+a_2)T^2 + (a_1 + a_0)T + (a_0 + 2a_2) = 1## which gives a system $$a_0 + a_2 = 1, a_1 + a_0 = 0, a_1 + a_2 = 0$$
which has a solution ##a_0 = 5, a_1 = 2, a+2 = 5##. So $$\frac{1}{T+1} =5 + 2T + 5T^2$$
 
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  • #27
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I'll take a go at 8b.

Let ##p(t) = T^3-2##. Notice that ##p## has not roots over ##\mathbb{F}_7## which can be seen by simply evaluating ##p## at each element of ##\mathbb{F}_7##.

We claim then ##p## is irreducible. If not then ##p## can be factored as ##p = f g## where ##f##, ##g## are irreducible and ##deg(f) + deg(g) = 3##. Since ##f## and ##g## are irreducible, ##deg(f), deg(g) > 0## and so either ##deg(f) = 1## or ##deg(g) = 1##. Suppose without loss of generality that ##deg(f) = 1##, then ##f## is linear and hence has a root of ##\mathbb{F}_7## which means that ##p## has a root over ##\mathbb{F}_7## which is a contradiction.

Since ##p## is irreducible it then ##(p)## must be maximal. Otherwise suppose there exists some ##q \in \mathbb{F}_7[T]## such that ##(p) \subset (q)##. Hence there exists an ##f \in \mathbb{F}_7[T]## such that ##p = qf##. Since ##p## is irreducible either ##q## is a unit (in which case ##(q) = \mathbb{F}_7[T]##) or ##f## is a unit (in which case ##(f) = (q)##). So indeed ##(f)## is maximal.

We now use the fact that given any ring ##R## with a maximal ideal ##M##, ##R/M## is a field. This follows from one of the standard isomorphism theorems which states that there is a one-to-one correspondence between the ideals of ##R/M## and the ideals in ##R## containing ##M##. So if ##M## is maximal, ##R/M## can not contain any proper ideals, and so must be a field.

To compute the number of elements in ##\mathbb{F}_7[T]/(T^3-2)## we note that since ##3## is prime each cosets can be represented by elements of the form ## c_1 + c_2 T + c_3 T^2## where ##c_i \in \mathbb{F}_7[T]##, and there are ##7^3 = 343## such elements.

To compute ##(T^2+2T+4)(2T^2+5)## in ##\mathbb{F}_7[T]/(T^3-2)## we first note that ##(T^2+2T+4)(2T^2+5) = 2T^4 + 2T^3 + 6T^2 + 3T + 6## in ##\mathbb{F}_7## then reducing ##mod (T^3-2)## gives ## 2(2)T + 2(2) + 6T^2 + 3T + 6## and so $$(T^2+2T+4)(2T^2+5) = 6T^2 + 3$$

To compute ##\frac{1}{T+1}## you can do a sort of "brute force" method. Let ##\frac{1}{T+1} = P(T) = a_0 + a_1T + a_2T^2## then ##(T+1)P(T) = (a_1+a_2)T^2 + (a_1 + a_0)T + (a_0 + 2a_2) = 1## which gives a system $$a_0 + a_2 = 1, a_1 + a_0 = 0, a_1 + a_2 = 0$$
which has a solution ##a_0 = 5, a_1 = 2, a+2 = 5##. So $$\frac{1}{T+1} =5 + 2T + 5T^2$$
Almost perfect! Only ##(T^2+2T+4)(2T^2+5)=6T^2.## The coefficient at ##T^3## is ##4,## not ##2.##
 
  • #28
mathwonk
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Nice work! Linear equations can always be used to find inverses as you do. Another trick is to find the "minimal polynomial" of an element, since if X^3 + aX^2 + bX + c = 0, then X^3 + aX^2 + bX = -c, and so X(X^2 + aX + b) = -c, and since we know the inverse of -c in the field of coefficients, we can divide by it. (In particular an element has an inverse iff its minimal poly has non zero constant term. This works in linear algebra too, where we recall the constant term of the characteristic polynomial is the determinant.)

This is brute force, but for linear polynomials like T+a, the force needed is minimal, since we can always substitute T = (T+a)-a into the given equation and expand. E.g. T+1 = S, gives T^3 = (S-1)^3 = S^3 - 3S^2 + 3S -1 = 2, so S^3 - 3S^2 + 3S = S(S^2-3S+3) = 3, so since 3.5 =1, we get S(5S^2 - S + 1) = 1, and thus 1/(T+1) = 1/S = 5S^2 -S + 1 = 5(T+1)^2 -(T+1) + 1 = 5T^2 + 2T +5.

Or just play with it, looking for a multiple of T+1 that is constant: E.g. here T^3 = 2, so T^3 + 1 = 3, so T^3 + 1 = (T+1)(T^2-T+1) = 3, and since 1/3 = 5, thus 1/(T+1) = 5(T^2-T+1) = 5T^2 -5T +5 = 5T^2 +2T +5.

I tried to use linear equations to show this ring is a field, by checking that we can always solve the system for an inverse but had trouble showing the determinant is non zero. For a general element a + bT + cT^2, I got determinant a^3 + 2b^3 + 4c^3 + abc, which apparently has no solutions mod 7, except a=b=c= 0. I could show this at least when abc = 0, i.e. when at least one coefficient is zero, (which includes the case of T+1), using the fact that mod 7 the only numbers having cube roots are 0,1,-1, and also when abc≠0 and all the cubes are equal, but did not pursue all other cases in general.

Your abstract method using maximal ideals of course works beautifully in general. You can also use abstract linear algebra by first showing that the product of two non zero polyomials of form a+bX+cX^2 cannot be divisible by T^3-2, (either directly as Artin does, following Gauss, or using unique factorization of polynomials), hence the map from our ring (a finite dimensional vector space) to itself defined by multiplication by a non zero polynomial a+bX+cX^2 is a linear injection, hence also surjective, hence multiplies some polynomial into 1.

Congratulations!

(What do you think of 8c? For 8d, I myself will need to recall Galois' construction of the field with 25 elements, a field I have never used. ... Oh yes, fresh_42 just showed us how to comnstruct the field of order 7^3 so we can do likewise for 5^2. I also need to know what the frobenius map of F25 is, .... ok it raises each element to the 5th power. Ah yes! it is additive, multiplicative, and fixes the subfield F5, hence is an F5 - linear map!...after much faulty calculation, I seem to have accidentally chosen a nice simple eigenbasis. By the way, in hindsight, what happens when you compose the frobenius with itself?)

Hint: For 8c:

If f.g = P, all with integer coefficients, what would be true mod 5? Is that possible?


By the way, as usual these are very nice problems. I think #8 in particular is wonderful. The technical difficulties are minimal, and the instructional value is high. It is not easy to come up with problems like this that are not routine, do not overwhelm, and teach you a lot.
 
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  • #29
mathwonk
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SPOILER, solutions of 8c, 8d:

8c: To be reducible it suffices to have a root, say X=a, since then by the root factor theorem, (X-a) is a factor. In each of the last three fields the polynomial has a root: in the reals, by the intermediate value theorem every odd degree polynomial takes both signs hence has a root; in the field with 2 elements X=1 is visibly a root since each of the 4 terms is congruent to 1, mod2; in the quotient ring, the meaning of the notation is that the polynomial in the bottom is set equal to zero, hence T itself is a root. Over the integers, the polynomial is irreducible, since it is congruent to a positive power of X, mod 5, but then both factors must also be congruent to a positive power of X, mod 5, which implies both have constant term divisible by 5, contradicting the fact that the constant term is not divisible by 25. By Gauss's lemma, an integer polynomial which factors with rational coefficients, has also integer coefficient factors, so it is also irreducible over the rationals.

8d: The non zero elements of a finite field form a cyclic group, so all 25 terms of the field F25 satisfy the polynomial X^25 = X, but not all satisfy X^5 = X. Since by definition the Frobenius map F takes an element a to F(a) = a^5, that means F^2 = Id, so F satisfies the minimal and characteristic polynomial X^2-1 = 0. Hence its eigenvalues are 1,-1, and it is diagonalizable as a 2x2 matrix, in a suitable basis, with those eigenvalues on the diagonal.
 
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  • #30
mathwonk
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SPOILER: solution for 9:

By definition of the tensor product, or by use of its basic property, a linear map out of VtensW is equivalent to a bilinear map out of VxW, which then is equivalent both to a linear map from V to W*, and to a linear map from W to V*. In particular, given f as in the problem, define a linear map V-->W* by sending v to the linear functional that sends w to f(vtensw). By what is given, no non zero v maps to the zero functional, i.e. this injects V linearly into W*. Hence dim(V) ≤ dim(W*) = dim(W). Similarly, W embeds in V*, so also dim(W) ≤ dim(V). Since V,W are finite dimensional of the same dimension, they are isomorphic, (but not naturally).
 
  • #31
mathwonk
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SPOILER: solution of #10:


By definition, V(Y^2-X^2) denotes the set of points in (X,Y) space whose coordinates satisfy the polynomial Y^2-X^2. Hence the real points consist of the points on the two lines X=Y and X+Y = 0, in the real plane. By definition Spec(R) is the set of prime ideals in the ring R, where an ideal J in R is prime if and only if the quotient ring R/J is a "domain", i.e. R/J has no non trivial divisors of zero. Equivalently, if and only if for elements a,b, in R, the product ab belongs to J if and only if at least one of a or b, or both, belongs to J.

If (a,b) is a solution of Y^2-X^2 = 0, the evaluation map C[X,Y]-->C, sending f(X,Y) to f(a,b), defines a surjective homomorphism C[X,Y]-->C with kernel the maximal ideal (X-a,Y-b). Y^2-X^2 belongs to this ideal as one sees by setting Y = (Y-b)+b and X=(X-a)+a and expanding Y^2-X^2. Hence (X-a,Y-b) defines an ideal J in R = C[X,Y]/(Y^2-X^2), such that R/J ≈ C, field. Thus J is a prime ideal in R. Moreover J determines the point (a,b) as the only common zero of all elements of J. Hence there is an injection from the infinite set of (real, hence also complex) points of V(Y^2-X^2) into Spec(R), that spectrum is infinite.

By definition, the Krull dimension of R is the length of the longest strict chain of prime ideals in R, ordered by containment, where the chain P0 < P1 <...< Pn has "length" = n. Since (X-Y) < (X-1,Y-1) is a chain of prime ideals in R of length 1, the Krull dimension is at least one. It already follows that R is not Artinian, since a ring is Artinian if and only if it is Noetherian, (which R is), and of Krull dimension zero, which R is not.

Now R actually has dimension one, since (X-Y) is a principal, hence minimal, prime in R, and when we mod out by it we get the principal ideal domain C[X,Y]/(X-Y) ≈ C[X]. In this ring an ideal is prime if and only if it is generated by an irreducible element, hence no non - zero prime ideal can be contained in another, since one of the irreducible generators would divide the other, contradicting the meaning of irreducible. Thus only one prime ideal, at most, can contain (X-Y). Similarly at most one can contain (X+Y). Thus any chain starting from either (X-Y) or (X+Y) has length at most one. But by definition of a prime ideal, any prime in R must contain zero, i.e. the product (X-Y)(X+Y), hence contains at last one of them.
 
  • #32
fresh_42
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SPOILER, solutions of 8c, 8d:

8c: To be reducible it suffices to have a root, say X=a, since then by the root factor theorem, (X-a) is a factor. In each of the last three fields the polynomial has a root: in the reals, by the intermediate value theorem every odd degree polynomial takes both signs hence has a root; in the field with 2 elements X=1 is visibly a root since each of the 4 terms is congruent to 1, mod2; in the quotient ring, the meaning of the notation is that the polynomial in the bottom is set equal to zero, hence T itself is a root.

Over the integers, the polynomial is irreducible, since it is congruent to a positive power of X, mod 5, but then both factors must also be congruent to a positive power of X, mod 5, which implies both have constant term divisible by 5, contradicting the fact that the constant term is not divisible by 25. By Gauss's lemma, an integer polynomial which factors with rational coefficients, has also integer coefficient factors, so it is also irreducible over the rationals.
Or short with Eisenstein and ##5##.
8d: The non zero elements of a finite field form a cyclic group, so all 25 terms of the field F25 satisfy the polynomial X^25 = X. Since by definition the Frobenius map F takes an element a to F(a) = a^5, that means F^2 = Id, so F satisfies the minimal and characteristic polynomial X^2-1 = 0. Hence its eigenvalues are 1,-1, and it is diagonalizable as a 2x2 matrix, in a suitable basis, with those eigenvalues on the diagonal.
... which results in ##
\begin{bmatrix}
1&0\\0&4
\end{bmatrix}
##
 
  • #33
mathwonk
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SPOILER: solution for 2:

Let the two disjoint lines be L and M, and p the point not on either of them. Then there is a unique plane A spanned by L and p, and a unique plane B spanned by p and M. Then A and B are distinct planes, since one of them contains L, the other contains M, and L and M cannot be in the same plane since they do not meet. Hence the planes A and B meet along a unique common line K. Since K consists of all points common to A and B, it contains p. Since K also lies in both planes A and B, it meets both L and M. If R were some other such line, R would contain p as well as a point of L, hence would meet A in 2 points, hence would lie in A. Similarly it would lie in B, so the only possible such line R is the unique line K of intersection of A and B.


That leaves 4b, 5 and 6. Are we stumped?!
 
  • #34
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SPOILER: solution for 2:

Let the two disjoint lines be L and M, and p the point not on either of them. Then there is a unique plane A spanned by L and p, and a unique plane B spanned by p and M. Then A and B are distinct planes, since one of them contains L, the other contains M, and L and M cannot be in the same plane since they do not meet. Hence the planes A and B meet along a unique common line K. Since K consists of all points common to A and B, it contains p. Since K also lies in both planes A and B, it meets both L and M. If R were some other such line, R would contain p as well as a point of L, hence would meet A in 2 points, hence would lie in A. Similarly it would lie in B, so the only possible such line R is the unique line K of intersection of A and B.


That leaves 4b, 5 and 6. Are we stumped?!
Doh! I was just working on sketching out a logical proof based on your axioms of P^3:
1) the two skew lines (g, h) do not meet, so they are in distict planes, A and B
1a) g is a line on A
1b) h is a line on B
2) A and B are distinct so they meet in a unique line, k != h or g
2a) h intersects A at exactly one point
2b) g intersects B at exactly one point
2c) k is a line in both planes
3) the point p is neither on g or h
3a) p and g lie in exactly one plane, C
3b) p and h lie in exacly one plane, D
4) C and D are distinct, so they meet in a unique line, l
5) I'm not sure here... but I think:
4a) C and A intersect along the line g (g is in both planes)
4b) D and B intersect along the line h (h is in both planes)
6) l intersects with both g and h
6a) since both l and g are on C, they meet at a unique point
6b) since both l and h are on D, they meet at a unique point
6c) The point p is on both C and D, so it must be also be on l

If that is more or less correct, I guess 2) becomes irrelevant, but let me know if I got anything wrong.
 
  • #35
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If that is more or less correct, I guess 2) becomes irrelevant, but let me know if I got anything wrong.
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.
 
  • #36
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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.
 
  • #37
martinbn
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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.
 
  • #38
mathwonk
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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?
 
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  • #39
martinbn
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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.
 
  • #40
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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.
 
  • #41
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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.)
 
  • #42
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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.
 
  • #43
martinbn
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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.
 

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