0: I came across this on Wikipedia (I think while clicking on fixed point theorems):
First, note that if we have an involution (meaning self inverse) on a finite set, then the number of fixed points has the same parity as the size of the set.
Next, we’ll prove Fermat’s sum of 2 squares theorem. Suppose we have a prime p>2. If it’s the sum of two squares, then exactly one of them must be even (because p is odd), and thus must be a multiple of 4. This proof is due to Zagier, with the visual interpretation by Spivak, supposedly math overflow
Therefore, a prime is a sum of two squares when x^2 + 4 y^2 = p
Now let’s look at the set {(x,y,z) : x^2 + 4 y z = p}. Clearly if we have a fixed point of the obvious involution that swaps y and z, we’ll have proven that p is a sum of two squares.
Now, we’ll come up with another involution. Or maybe you, dear reader, can? The first step is to come up with a geometric interpretation of a point (x,y,z) in the set.
Make a square of side x, and put 4 y by z rectangles around it like a ‘windmill’ (precise arrangement details are in the earlier link if you are confused).
Now our involution:
There’s often a different way to cut the shape into a central square surrounded by rectangles.
The only way to have a fixed point is for the area to be a big square with four squares removed. This will leave a square behind, along with four long rectangles. If x=y, then p = x(x+4z), but then since p is prime, x must be 1 and p must be x+4z = 1 + 4z. So for p = 4k + 1, we can make the fixed point (1,1,k), which is unique. Thus there is at least one fixed point of the swapping map, so we can write p as a sum of two squares.
1: Generalizing:
The lemma that is not Burnside’s (which, incidentally, was what initially enticed me into learning group theory) states that if you have a finite group G acting on a finite set X, then the number of orbits equals the average number of fix points of g (averaged over G).
That is, |X/G| = 1/|G| \sum_g |X^g| (if you know representation theory: this is also the number of copies of the trivial representation in the permutation representation, because the projection to the trivial representation is 1/|G| \sum_g g, and this gives a vector parallel to \sum_{x in O} e_x for every orbit O, and these are linearly independent for to different orbits).
Thus, |G| |X/G| = \sum_g |X^g|. If we take this mod n for t = |G|, we’ll get 0 = \sum_g |X^g| mod n, or in other words the sum of the number of fixed points for each g must divide n.
The fixed points of the identity are everything, so \sum_{g ≠ e} |X^g| = -|X| mod n
If we have an involution, then we have a Z/2Z action, and that’s why the parity must be the same.
2: A trick: Say we have a finite degree field extension E of K, and a group of field automorphisms of E that fix K. Then not only do we have a represention on E thought of as a vector space/module over K, but we also have a representation on the abelian group of units E^× under multiplication thought of as a Z-module.
So we can do things like map to the copies of an irreducible representation τ via p = \sum_g χ_τ(g^{-1}) g, and then also project to the trivial representation via the product \prod_g g. The copies of the trivial representation will form the fixed field of the group; so if the fixed field is K then we know we are in K.
Important special case: if the group is abelian then the irreducible characters are homomorphisms (because the irreducible representations are one dimensional and so the trace is just literally basically the same as the matrix), and so h p α = χ_τ(h)^{-1} p α. So for example if the galois group is cyclic of order n, we can take τ as the representation that scales by ζ where ζ is some primitive nth root of unity that we assume the base field has; then this reads g^k pβ = ζ^k pβ for generator of the field extension β. This then means that if pβ ≠ 0 then it isn’t fixed by any subgroup. Since getting 0 would mean it’s zero on any element, we’d then have a linear dependenc of n-1 basis vectors, which is impossible. Now \prod_k g^k pβ is fixed and so is in K, and specifically must equal (\prod_{k=0}^{n-1} ζ^k) (pβ)^n; which is just p β because the roots of unity multiply to 1 (since you have 1 + n-1 = n, 2 + n −2 = n, etc. around the circle). So the point is that (pβ)^n must be equal to some element in K, yet the earlier powers can’t have been because then we’d have a smaller degree field extension and thus smaller degree galois subgroup; thus we have a radical extension of degree n, with pβ generating the field E over K.
This is basically Hilbert’s Theorem 90, and I’ve seen it before. The interesting bit is the intuition that now will hopefully let me remember it easily: first we project to a good representation that isn’t fixed by any subgroup (and thus isn’t any smaller field), then we use the fact that we have field automorphisms to project to the trivial representation as a product.
I’m sure people are aware of that intuition, but unfortunately I hadn’t seen anyone else actually point out that we have a representation on the abelian group aka Z-module of units of the field.
3: Idle question, from the representation theory stuff and the way I did Sperner’s lemma: given a finite group G, what’s the product of all the elements?
So far: split the group into the subgroups generated by some gs. Choose an exhaustive set without repeats. Now, odd order elements will end up contributing nothing from their subgroup, for g^k will pair with g^{n-k} to make the identity. For odd order elements, we’ll have an extra g.
Example: the cyclic group of order n gives the identity for odd n, and n/2 for even n.
Direct products will also give products of values. Now, I wonder what having a quotient would tell us.
April 22
0: I came across this on Wikipedia (I think while clicking on fixed point theorems):
First, note that if we have an involution (meaning self inverse) on a finite set, then the number of fixed points has the same parity as the size of the set.
Next, we’ll prove Fermat’s sum of 2 squares theorem. Suppose we have a prime p>2. If it’s the sum of two squares, then exactly one of them must be even (because p is odd), and thus must be a multiple of 4. This proof is due to Zagier, with the visual interpretation by Spivak, supposedly math overflow
Therefore, a prime is a sum of two squares when x^2 + 4 y^2 = p
Now let’s look at the set {(x,y,z) : x^2 + 4 y z = p}. Clearly if we have a fixed point of the obvious involution that swaps y and z, we’ll have proven that p is a sum of two squares.
Now, we’ll come up with another involution. Or maybe you, dear reader, can? The first step is to come up with a geometric interpretation of a point (x,y,z) in the set.
Make a square of side x, and put 4 y by z rectangles around it like a ‘windmill’ (precise arrangement details are in the earlier link if you are confused).
Now our involution:
There’s often a different way to cut the shape into a central square surrounded by rectangles.
The only way to have a fixed point is for the area to be a big square with four squares removed. This will leave a square behind, along with four long rectangles. If x=y, then p = x(x+4z), but then since p is prime, x must be 1 and p must be x+4z = 1 + 4z. So for p = 4k + 1, we can make the fixed point (1,1,k), which is unique. Thus there is at least one fixed point of the swapping map, so we can write p as a sum of two squares.
1: Generalizing:
The lemma that is not Burnside’s (which, incidentally, was what initially enticed me into learning group theory) states that if you have a finite group G acting on a finite set X, then the number of orbits equals the average number of fix points of g (averaged over G).
That is, |X/G| = 1/|G| \sum_g |X^g| (if you know representation theory: this is also the number of copies of the trivial representation in the permutation representation, because the projection to the trivial representation is 1/|G| \sum_g g, and this gives a vector parallel to \sum_{x in O} e_x for every orbit O, and these are linearly independent for to different orbits).
Thus, |G| |X/G| = \sum_g |X^g|. If we take this mod n for t = |G|, we’ll get 0 = \sum_g |X^g| mod n, or in other words the sum of the number of fixed points for each g must divide n.
The fixed points of the identity are everything, so \sum_{g ≠ e} |X^g| = -|X| mod n
If we have an involution, then we have a Z/2Z action, and that’s why the parity must be the same.
2: A trick: Say we have a finite degree field extension E of K, and a group of field automorphisms of E that fix K. Then not only do we have a represention on E thought of as a vector space/module over K, but we also have a representation on the abelian group of units E^× under multiplication thought of as a Z-module.
So we can do things like map to the copies of an irreducible representation τ via p = \sum_g χ_τ(g^{-1}) g, and then also project to the trivial representation via the product \prod_g g. The copies of the trivial representation will form the fixed field of the group; so if the fixed field is K then we know we are in K.
Important special case: if the group is abelian then the irreducible characters are homomorphisms (because the irreducible representations are one dimensional and so the trace is just literally basically the same as the matrix), and so h p α = χ_τ(h)^{-1} p α. So for example if the galois group is cyclic of order n, we can take τ as the representation that scales by ζ where ζ is some primitive nth root of unity that we assume the base field has; then this reads g^k pβ = ζ^k pβ for generator of the field extension β. This then means that if pβ ≠ 0 then it isn’t fixed by any subgroup. Since getting 0 would mean it’s zero on any element, we’d then have a linear dependenc of n-1 basis vectors, which is impossible. Now \prod_k g^k pβ is fixed and so is in K, and specifically must equal (\prod_{k=0}^{n-1} ζ^k) (pβ)^n; which is just p β because the roots of unity multiply to 1 (since you have 1 + n-1 = n, 2 + n −2 = n, etc. around the circle). So the point is that (pβ)^n must be equal to some element in K, yet the earlier powers can’t have been because then we’d have a smaller degree field extension and thus smaller degree galois subgroup; thus we have a radical extension of degree n, with pβ generating the field E over K.
This is basically Hilbert’s Theorem 90, and I’ve seen it before. The interesting bit is the intuition that now will hopefully let me remember it easily: first we project to a good representation that isn’t fixed by any subgroup (and thus isn’t any smaller field), then we use the fact that we have field automorphisms to project to the trivial representation as a product.
I’m sure people are aware of that intuition, but unfortunately I hadn’t seen anyone else actually point out that we have a representation on the abelian group aka Z-module of units of the field.
3: Idle question, from the representation theory stuff and the way I did Sperner’s lemma: given a finite group G, what’s the product of all the elements?
So far: split the group into the subgroups generated by some gs. Choose an exhaustive set without repeats. Now, odd order elements will end up contributing nothing from their subgroup, for g^k will pair with g^{n-k} to make the identity. For odd order elements, we’ll have an extra g.
Example: the cyclic group of order n gives the identity for odd n, and n/2 for even n.
Direct products will also give products of values. Now, I wonder what having a quotient would tell us.