#S_2

[loving-n0t-heyting]

a much tamer realisation i had abt ultrafilters the same day is that u can think of the ultrafilter lemma as a weakening of countable choice wherein a "pseudochoice" function on a denumerable set of sets selects not an element (aka a principal ultrafilter) of each set in the set but a possibly free ultrafilter (a "fake" element) on each

the trick is to take the family S_n of sets indexed by the naturals (S_0, S_1, S_2...) on which the pseudochoice function is to be defined, then consider the sequence T [S_0, (S_0 x S_1), (S_0 x S_1 x S_2), ...]; then the filter {U{T_m | m ≥ n} | n is natural} on the elements of elements of T has an ultrafilter F extending it. for every n, F contains the set of all sequences in UT of length greater than n; ie, F "represents" a "fake" omega-sequence. and for any n and X a subset of S_n, F will contain either the set of sequences s in UT of length ≥n with s_n in X or the set of sequences s in UT of length ≥n with s_n in (S_n \ X), and the Xes into which this fake nth entry of the fake omega-sequence falls will be finitarily consistent bc F is a filter, meaning it defines an ultrafilter on each S_n.

there is smth neat to me abt this, the way the mere guarantee of an ultrafilter can get so close to simulating a sequence in Π_n(S_n) without that product even being inhabited (if we let AC_omega, but not the ultrafilter lemma, fail). its familiar thinking of an ultrafilter as a fictitious or simulated element of an infinite set, but here an ultrafilter apes (up to finitary specifications) being an element in good standing of not just a finite set but an empty one!!

there might be a way to extend this trick to arbitrarily large families but i couldnt trivially come up with one; tried and failed to exploit the UL's consequence of the linear orderability of all sets (another cutesy way it stands on a spectrum w Ac ie the well ordering theorem!). if not i dont grok why countability is so important

this struck me while reading thru an exposition of godels proof of the completeness theorem—mostly very dry and bookkeep-y in a typically logic-y way but at the final moment, where the magic happens, it implicitly invokes a choice function on a denumerable set of nested sets of truth-value assignments, which i realised could be replaced with an ultrafilter extending the sequence considered as a filter of its 1st entry (whose restriction to each subsequent entry in the sequence is in turn an ultrafilter on it) w/o detriment to the functions use in the proof—naturally, bc the completeness theorem itself is equivalent to the ultrafilter lemma! much easier to see in the usual henkin construction but it was cool to see it peeking thru even in the original proof

[xeter-group]

Exceptional Automorphisms of S_6

The symmetric group S_6 has a special property that S_n does not have for ANY n ≠6. Really? 6, of all numbers?? How odd.

For any group G, and every g in G, the conjugation map f_g: h ↦ g h g^-1 is an automorphism of G. That is, it is an isomorphism from G -> G. The group of these particular automorphisms (under composition) is called the group of inner automorphisms of a group. Did you know that for every symmetric group S_n EXCEPT n=6, this is the entire automorphism group?

For some reason, S_6 has basically ONE (and only one) weird automorphism. What I mean by this is if we denote the inner automorphisms of S_6 by Inn(S_6), then S_6/Inn(S_6) ~= Z_2. (In other words, you can pick a single automorphism sigma such that the automorphisms of S_6 that are not inner automorphisms can be written as a composition of inner automorphisms and this one sigma.)

So we have this weird situation where |Aut(S_2)|=|Aut(C_2)| = 1, which is kinda trivial, |Aut(S_6)| = 2n!, and |Aut(S_n)| = n! otherwise. Kinda weird, huh?

What is also interesting is the proofs that there are no outer automorphisms for n≠6. Basically you can eliminate all but finitely many n, then you can pick off cases, and you are left with an n=6 shaped hole in your cases that CANNOT be filled, which feels so weird to me. If I was proving this myself I'd be going crazy having proven it for every case except 6, and having to resort to some proof which for some reason doesn't work for finitely many values.

Construction

There are so many different ways to construct the weird automorphism of S_6 - I have some links at the bottom. I particularly like the graph theory/geometric ones. Something about using factorisations of K_6 or a dodecahedron just makes the 6 feel more unique. I will admit I don't think I understand what is fundamentally special about 6 enough yet, though, on a philosophical level.

Practically, though, the constructions all basically boil down to the fact you can put a copy of S_5 inside S_6 in a way that isn't the obvious way, which you can only do for n=6. That is the special part.

After that, you have these 6 cosets of S_5 inside S_6 that S_6 acts upon. In other words, each element of S_6 permutes the 6 cosets of S_5 living inside it. But the group of permutations of 6 cosets is S_6. So we have a mapping from elements of S_6 to S_6 - an automorphism! Is this automorphism an inner automorphism? Each construction shows why they are outer differently, but a common theme is to show that the mapping S_6 to S_6 does not take transpositions to transpositions, but inner automorphisms preserve the cycle type.

Some proofs and examples

The wikipedia article. I like the construction about graph partitions. It does not, however, have much detail sadly.

Fortunately, the graph partitions thing from wikipedia is explained here. It's very short and to the point, also quite nice to look at:

Fairly elementary explanations, followed by more intense ones:

Requires only basic group theory to understand the first few explanations they provide (although it isn't trivial if you just learned group theory). Bonus points for "MyStIc PeNtAgOnS" (capitalisation mine):

Allegedly useful, but I can't grab a copy: (I think its on mathscinet under MR1240362 as per David Leep's personal website's publications section)

Combinatorial Structure of the automorphism group of S6 by T.Y. Lam and David B. Leep, Expositiones Mathematicae11 (1993), no. 4, pp. 289-903.

The comment by Matthew Towers here is also interesting:

[co-xeter-group]

what's the siney graph in your header?

I encourage anyone reading this to read as much or as little as they want. I hope that anyone can understand the detailed explanation, and that most math undergrads can understand most of the observations I make. The generalisations might not be so accessible all the time.

To put it tersely, its the projection of the barycentric subdivision of a tetrahedron onto the 2-sphere, visualised on a rectangle via the mercator projection.

This is closely related to Coxeter groups, the classification of polytopes, the classification of straight line Coxeter groups, the classification of regular tilings of surfaces of constant curvature, and Schl\"afli symbols.

A more detailed explanation:

Imagine taking a tetrahedron, putting a dot on the middle of each vertex, edge, and face, and connecting them all up with straight lines along its surface. This is the so called "barycentric subdivision". Then consider the origin to be in the middle of the tetrahedron, and then project the tetrahedron (and the lines we drew on it) onto a sphere. We use the mercator projection to view it like a map, but we still think of it as lying on the sphere (see below mp4 of said sphere with the barycentric subdivision drawn on it). Each face of the tetrahedron could be imagined to be coloured a certain colour, so v_2 in the diagram on my header is the vertex at the centre of the yellow face, v_1 is the vertex in the centre of one of the yellow faces edges, and similarly for v_0. These appear to be connected by curved lines, but these are straight lines on the surface of the sphere.

Assume the tetrahedron and sphere are embedded in R^3 and share a common centre at the origin.

Some interesting observations:

each of these lines we drew now gets turned into a great circle, which corresponds to a plane going through the origin

reflections in these great circles preserves the lines we drew, and correspond to automorphisms of the tetrahedron

each triangle in the subdivision has angles pi/3 radians, pi/3 radians, and pi/2 radians.

the symmetry group of the tetrahedron is S_4, where the adjacent transpositions correspond to permutations of the faces (or if you like, vertices)

the symmetric group S_4 has a presentation , where the s_i are adjacent transpositions of 4-tuples

The exponents of the (s_is_j) terms above exactly match the denominators of the angles of the triangle mentioned above

we can pick one triangle on the sphere and consider the reflections in (the faces corresponding to) its edges, denoted s_0, s_1, and s_2. These reflections permute the coloured faces of the tetrahedron, or if you like, its vertices.

we can repeatedly apply these reflections to flip across an edge or vertex of the yellow face, rotate about the centre of the yellow face (e.g. s_0 s_1), and transpose the yellow face with any other face

In this manner we can represent every symmetry of the barycentric subdivision, and by extension, the tetrahedron, in terms of these three reflections. If you don't see this, consider the effect of conjugation.

The sphere is a surface of constant curvature

This tiling generated by the tetrahedron is a regular tiling of the sphere

In short, the tetrahedron has a symmetry group S_4 (often called A_4 in analogy with the Dynkin Diagram) that has a presentation in terms of three reflections, which act transitively on this barycentric subdivision. The angles of the barycentric subdivision correspond to the relations of the presentation. One can generalise this observation and use it to classify polyhedra.

Some theoretical results.

A Coxeter group is a group W accompanied by a set of generators S = {s_1, s_2, ..., s_n} \subseteq W, such that W = <s_i | (s_is_j)^m(i,j) = 1>, where m(i,j) is an integer at least 1, m(i,i) = 1, and m(i,j) > 1 if i != j. These relations turn out to exactly correspond to the relations necessary to define a finite system of reflections in (n+1)-dimensional space.

By polytope, I mean a bounded convex polytope.

The regular tilings of the sphere correspond to regular polytopes, which correspond to the finite irreducible Coxeter groups whose Dynkin diagrams have straight lines

We can define a polytope to be regular if the automorphism group of the polytope acts transitively on the regions of the barycentric subdivision (or equivalently, its "flags"), which corresponds to chains of i-faces of the polytope ordered by inclusion

To go from a polytope to its Coxeter group, you take its automorphism group to get the group structure, and do a similar thing to above to find the generators, you arrange some hyperplanes so that their reflections satisfy the relations of the Coxeter group, generate a system of hyperplanes closed under reflection, and intersect this with an (n-1)-sphere to get the barycentric subdivision, from which you can recover a polytope and its dual polytope, which have isomorphic Coxeter groups

The regular tilings of the plane correspond to the affine irreducible Coxeter groups with straight line Dynkin diagrams.

One can study the regular tilings of hyperbolic space and classify those Coxeter groups too.

The E_8 lattice, which gives solution to 8-dimensional sphere packing has a a load of other interesting properties, corresponds to the Coxeter group E_8, via a certain semiregular polytope which is the convex hull of some lattice points.

The classification of regular (n-dimensional) polytopes and regular tilings of R^n is via the classification of Coxeter groups (and by extension Dynkin diagrams with certain properties)

There is an elegant classification according to Bourbaki that resembles the typical intuitive classification of regular polyhedra and regular tilings of R^2

Fun extensions

There are a lot of ways to represent a symmetry of a polytope/element of a Coxeter group in terms of the reflections/generators s_i. Is there an easy way to determine whether your representation of the symmetry/group element is the shortest? Yes! In fact, you can construct a DFA on the generators in the finite case.

The Cayley graph of a Coxeter group is Hamiltonian

My pfp shows the duality between an octahedron and a cube. If you draw a vertex at the centre of each face of the cube, and take the convex hull of the vertices, you get an octahedron. Note that vertices of the octahedron correspond to faces of the cube, edges of the octahedron correspond to edges of the cube, and faces of the octahedron correspond to vertices of the cube. Two vertices of the octahedron are incident with each other when the corresponding faces of the cube share an edge, and so on. To put it formally, the poset of i-faces of the octahedron and the poset of i-faces of the cube, both under inclusion, have an anti-isomorphism between them. This causes them to have isomorphic symmetry groups.

The cube has its Coxeter group with relations (s_0s_1)^4 = (s_1s_2)^3 = (s_0s_2)^2 = s_i^2 = 1. Note that here the 4 and 3 are different numbers, and the cube has a dual of an octahedron. In the case of the tetrahedron, the exponents are the same, and the tetrahedron is self dual. In general, finite irreducible Coxeter groups with straight line Dynkin diagrams correspond to self dual polytopes exactly when their Dynkin diagrams are "reversible".

This is heavily related to how the cube and octahedron have reversed Schl\"afli symbols.

[tanydi]

[scottbird]

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[questioning-my-identity-element]

Some basic general topology

I want to do a little bit of topology later, but in order to do that, I gotta get the basics down for anyone who's gonna read it. Anyway, I will talk about the definition of a topological space, open sets, and closed sets.

So, we define a topological space X like this. You have a collection of subsets of X called T (usually tau, but idk if everyone has support for that character). The collection satisfies the following properties

X and the empty set are in T

T is closed under arbitrary unions (you can union uncountably many sets and they stay in T always)

T is closed under finite intersections (you can intersect finitely many sets, but not necessarily infinitely many)

The elements of T are called open sets, T is called a topology, and (X, T) is called a topological space. If S is a subset of X such that S complement is in T, then S is called a closed set.

Now, that definition is really abstract. It's hard to see what motivation we have for defining it like this. The intuition comes from a certain type of topological space called metric spaces, which are defined as follows: X is a metric space with metric d that maps to real numbers, if for any elements x, y, z in X

d(x, y) > 0 if x≠y,

d(x, x) = 0,

d(x, y) = d(y, x),

d(x, y) ≤ d(x, z) + d(z, y).

Okay, so that's still abstract, right? Well, think of d as distance; the first one says that distance is positive, the second says that the distance between a point and itself is 0, the third says that the distance between one point and another is equal to the distance between the other point and this one, and the final one says that a direct route from one point to another is shorter than an indirect route, stopping at another point along the way.

Okay, so how is this related to topological spaces. Well, in order to clarify that, we need to know what an open ball is: define an open ball centered at y (of radius r>0) to be {x | d(x,y)<r}. Then, a set S is open if for every element x of S, there is an open ball centered at x such that the open ball is contained completely within S. A set S is closed if, for any point y in X (not necessarily S) such that every open ball centered at y contains an element of S, y is in S.

Here's some pictures that I stole off the internet to illustrate the concepts:

Notice how in the first one, every point has a disk (an open ball) centered at it contained in the set. For the second one, notice how once you get small enough, there will no longer be any points in Y contained in a ball centered at the black point, so the black point doesn't need to be in Y.

So, does this fit our definition of a topological space? Well, in order for it to fit our definition, we need X and the empty set to be open, we need arbitrary unions of our open sets to be open, we need finite intersections of our open sets to be open, and we need complements of our closed sets to be open. That's a lot, but we're gonna do it, and then I'll make another topology post (that isn't covered in baby Rudin) later.

First, the empty set is an open set because it is vacuously true that every point..., the empty set has no points. The whole metric space X is also an open set because all of the open balls are necessarily subsets of X, so every point has an open ball contained in X.

Next, unions, suppose you have a collection of open sets S_i, where i can be spanned by any index set (not necessarily S_1, S_2, ...). Let x be in the union of the S_i, then x is in at least one of the S_i, call it S_k. By definition of an open set, there is a ball centered at x that is contained completely in S_k, but S_k is a subset of the union, so the ball is also contained in the union, meaning the union is an open set.

Next, let S_1, S_2, ..., S_n be open sets and consider their intersection. Let x be a point in the intersection and let r_1, r_2, ..., r_n be the radii of the balls centered at x that are contained in S_1, S_2, ..., S_n. Then, let r be the minimum of those r_i. Then, the ball of radius r centered at x is contained within the intersection because it is contained in all of the other balls, meaning it is a subset of each of the S_i. Thus, the intersection is an open set.

Finally, let S be a closed set. Then, let x be in the complement. Suppose, for contradiction, that there is no radius r such that a ball of radius r is contained completely within the complement. Then, for every radius r, there is a point y not in the complement, i.e. in S, such that y is contained in the ball of radius r centered at x. Therefore, S can't be a closed set by definition, a contradiction. Thus, S complement is an open set. (I technically could have just used contrapositive, but I am unsure whether that will be more or less clear.)

Alright, that was a lot. But I have two more posts to get to my favorite proof in all of math (maybe later today because I'm bored).

[numb3rth30ry]

Georg Cantor’s mind-bending Diagonal argument.

The argument itself is remarkably succinct, despite the fact that it constructively establishes the existence of “large infinities.”

We begin by considering the set of all infinite sequences of binary digits, called T.

Cantor’s argument shows that, for any list of these sequences (e.g. above, in black and red), there is always an s (in blue) that does not appear in the list.

To find s, we begin with the elements along the diagonal. We take the sequence of their complements (ones become zeros, zeros become ones) to be the digits of s.

Observe that s cannot appear anywhere on the list, since the nth digit of each s_n differs from that of s (as they are complementary). The first digit of s is different from the first digit of s_1, the second digit of s is different from that of s_2, etc.

No matter how we arrange a list, and no matter how large we make it, we can always find an element of T not on the list.

Even if we let our list contain (countably) infinite[ly] many terms, T contains all the elements of our list and s, so it is larger.

Thus, very loosely, there are infinite sets larger than other ones. 

A major consequence of Cantor’s results is that set of real numbers is “denser” than the integers.

Set theory is otherworldly. Mathematics is beautiful. <3

[trashbrigade]

There’s an old Roger Miller song that has the following lyrics: Ya can't roller skate in a buffalo herd, But you can be happy if you've a mind to, Ya can't take a shower in a parakeet cage, But you can be happy if you've a mind to, Well, ya can't go a-swimmin' in a baseball pool, But you can be happy if you've a mind to, Ya can't drive around with a tiger in your car, But you can be happy if you've a mind to, Ya can't go fishin' in a watermelon patch, But you can be happy if you've a mind to… Now you’ll notice that the song has the upbeat refrain: You can be happy, if you’ve a mind to, but I think the overarching message is quite negative. He’s telling people over and over what they CAN’T do, and nothing drives me crazier than that kind of pessimistic messaging. Let’s prove Roger Miller wrong and do one of the things his song tells us we can’t (be careful, please, especially with the buffalo and tigers.). Set your video to music. The music, of course, should be the corresponding section of the Roger Miller song we are discussing.

[kiiamn]

:(

But its a quick fix i think, (THIS IS NOT QUICK AT ALL, I CAN FEEL A MORE ELEGANT SOLUTION IN MY BONES BUT CANT REACH IT) (I figured it out within like, an hour, but solving a problem and communicating it are two different things, and apparently I am much less motivated at the second)

Its still n-3 for odd n, as those cant have perpendicular diagonals.

Why no ⟂ diagonals for odd n:

you can get from any diagonal to another diagonal by shifting the points by one-side increments, which rotates the diagonal by 180/n, which means that the angle between two diagonals is always an integer multiple of 180/n.

If we want the angle to be 90, then 90/(180/n) must be an integer, 90/(180/n) is n/2, so n must be even.

playing around with n=4, 6, and 8, it seems that n-2 is the answer. You can connect all diagonals coming from one point, and draw a diagonal between the two points adjacent to the point that the middle diagonal connects to.

Link

this sets a minimum at least, which is nice.

How I'm gonna think about the diagonals:

All the diagonals are either intersecting at right angles, or not intersecting other diagonals at all.

If you imagine a reg. polygon with only the perpendiculars, it divides the area into a bunch of sectors, where the rest of the lines must fit into, and can't influence each other.

All the sectors that have points on them (from the original reg. polygon) are composed of an arc with equally spaced points (adj. points connected) on some shape that only has right angles on it. (only right triangle or rectangle I think, but that doesn't really matter)

These are the only possible shapes, by induction, assuming there is only one "chunk" of perpendicular lines, which is always true. (you figure out why, its not too hard)

from now on, the number of perpendiculars will be p, and the amount of sectors with usable points on them will be s.

Non Intersecting Diagonals in 1 Sector:

There will be some amount of non perpendicular diagonals here. The diagonal connecting the furthest two points (AD in diagram above) can always exist, as there cannot be a line intersecting it, so it will exist in the optimal configuration.

If we connect that line, we just get a problem equivalent to the one I solved in the first part of my first rb, so there is a maximum of n-3 more diagonals added, or n-2 total, including the AD line.

Non Intersecting Diagonals Total:

Splitting with perpendiculars is pretty much identical to splitting with non-intersecting diagonals in the first fart of the first rb, in that the total amount of points over all the sectors increases by 2 for each perpendicular. But each corner is a point used on a non usable sector, so we need to subtract those.

Adding it all up, the sum of all the points across all s sectors is n+2s-k (k=# of corners).

or:

n+2p-k = S_1 + S_2 + ... + S_s

Where S_i is the number of points in the ith sector

Since the amount of diagonals in the ith sector is S_i - 2, the sum will then be S_1 + S_2 + ... + S_s with a -2 for every usable sector.

So the total amount of non intersecting diagonals are:

n+2p-k-2s

or the total amount of diagonals, after adding the p perpendicular lines:

n+3p-k-2s

Getting rid of "s"

Each Sector needs at least 2 sides (outside of the ones in the arc) that need to come from the end of a perpendicular. Each perpendicular diagonal can participate in 4 sides (2 sectors worth), because they have two ends that can pull double duty if two sectors share the side.

But if there's a corner, both sides in the corner lose the ability to double duty for a total loss of 2 sides (1 sector's worth)

so s=2p-k, thus the maximum of diagonals is:

n+3p-k-2s

n+3p-k-2(2p-k)

n-p+k

we want to maximize n-p+k, which means as many corners as we can get out of as few perpendiculars as we can get.

Corners and Perpendiculars

How many corners per perpendicular can you eek out?

First off, the max is one corner per diagonal, as each corner requires two ends, and there are two ends per diagonal that cannot be reused.

For more detail, lets look at one "bit" where a bit is a section of one chunk where all the diagonals are connected by corners, clearly, the only possible bits are a single diagonal, an L shape, a ㄷ shape, and a rectangle.

The maximizing case of n=p is only possible with all rectangles, as all other shapes have exposed edges. Rectanges can all be categorized by their height, from 0 to the diameter:

now, all the sides of every rectangle must be intersected. For vertical lines, to be intersected, the intersecting rectangle must be shorter than the intersected.

But there must always be a shortest rectangle, and so there must be a rectangle with their vertical lines not intersected, which is a contradiction.

(and with the same logic one for one rectangle's horizontal lines)

having one less corner doesn't work either:

One lost corner means two exposed edges, and all the other types of bits have exactly two exposed edges, so there must be one and only one non rectangle bit.

It cant be the L or ㄷ, as if that were possible, you could complete the square and create a p=k situation, possible because if one side of a rectangle is intersected, the opposite side is too.

The line doesn't work either, as if the line is removed, there is a set of only rectangles, which always have a pair of horizontal lines and a pair of vertical lines intersecting, and a single line can't intersect both.

k=p-2, is in fact possible, by example, only two perpendicular diagonals, which have two diagonals and no corners.

So putting that into the formula above:

n-p+k

n-p+(p-2)

n-2

Is the maximum amount of diagonals one can have given the constraints.

A nice problem:

Let n > 3 be a positive integer. Find the maximum number of diagonals of a regular n-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other.

(IMO Shortlist 2016 C5)

[max1461]

Oh I misinterpreted this. Thought you took his direct product with a multiplicative subset S and then modded him out by the equivalence relation (r_1, s_1) ~ (r_2, s_2) iff there exists t in S such that t(s_1r_2 - s_2r_1) = 0.

Sorry, we localized your boyfriend. Yeah. A lot of his character got lost in translation. Had to change his name to one that would make sense to a western audience too. My deepest apologies.

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[unofficialchronicle]

Link to a First Nations Writing Journal, winter Issue. 

Their website: https://fourwindslitmag.wordpress.com/

[etemysore]

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[arxt1]

On Detecting Equatorial Symmetry Breaking with LISA. (arXiv:2201.03569v1 [gr-qc])

The equatorial symmetry of the Kerr black hole is generically broken in models of quantum gravity. Nevertheless, most phenomenological models start from the assumption of equatorial symmetry, and little attention has been given to the observability of this smoking gun signature of beyond-GR physics. Extreme mass-ratio inspirals (EMRIs), in particular, are known to sensitively probe supermassive black holes near their horizon; yet estimates for constraints on deviations from Kerr in space-based gravitational wave observations (e.g. with LISA) of such systems are currently based on equatorially symmetric models. We use modified "analytic kludge" waveforms to estimate how accurately LISA will be able to measure or constrain equatorial symmetry breaking, in the form of the lowest-lying odd parity multipole moments $S_2, M_3$. We find that the dimensionless multipole ratios such as $S_2/M^3$ will typically be detectable for LISA EMRIs with a measurement accuracy of $\Delta(S_2/M^3) \sim 1\%$; this will set a strong constraint on the breaking of equatorial symmetry.

from gr-qc updates on arXiv.org https://ift.tt/3njcg02

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