Math(s)! Currently: Induction Failure

Discussion in 'General Chatter' started by Exohedron, Mar 14, 2015.

  1. Exohedron

    Exohedron Doesn't like words

    What does that one look like? Like, can you list particular basis elements?
  2. syntheme

    syntheme Active Member

    It's not easy on mobile but i^2=-1, a^2=-3, ai=-ia gets you the rational basis at least
  3. syntheme

    syntheme Active Member

    You can check that under the 3-adics it's not the matrix ring, distinguishing it from the Hurwitz quaternions
  4. Exohedron

    Exohedron Doesn't like words

    Okay, so you take the Lipschitz quaternions and scale two of the basis elements by √3. I'm not seeing the S3 right off the bat, but it is definitely an order, as are all similar scalings.
  5. Exohedron

    Exohedron Doesn't like words

    Actually, I already mentioned scaling issues as one of the things I was planning to ignore. Oh well. I guess I was also implicitly using the quaternions as in the ones built on the reals, rather than a more abstract quaternion algebra, so scaling is basically a free action once you've dealt with your compatibility issues.
    Last edited: Oct 6, 2017
  6. syntheme

    syntheme Active Member

    (-1+a)/2 is a cube root of unity, taken to its inverse by conjugation with i. That's sufficient to get 2S_3, though usually you'd describe it as a dihedral group.
  7. Exohedron

    Exohedron Doesn't like words

    Ah, so it's not actually what I initially thought, which is a lattice with a basis of unit-length vectors that form a double cover of S3. Rather, you have a bunch of cube roots of -8. Cool.
  8. Exohedron

    Exohedron Doesn't like words

    Okay, I think I've talked about well-orderings before, but now I can't remember, and I'm somehow too lazy to go read through my own posts. Anyway:

    A linear order on a set is a relation generally called "less than or equal to" which we'll write as ≤, and it obeys a few simple rules:
    For all a in the set, a ≤ a.
    For all a and b in the set, either a ≤ b or b ≤ a or both.
    For all a and b in the set, if a ≤ b and b ≤ a, then a = b.
    For all a, b and c in the set, if a ≤ b and b ≤ c, then a ≤ c.
    Fairly reasonable, I think.
    A set with a linear order is often called an ordered set.
    If we have an ordered set, then we say that it has a least element if there is an element a in the set such that for any b in the set, a ≤ b.
    Given an ordered set, any subset of that set inherits an ordering in the natural manner: for a and b in the subset, a ≤ b in the subset if a ≤ b in the original set.
    An ordered set is well-ordered if every subset of the set has a least element.
    A classic example is the natural numbers: the natural numbers are ordered in the usual manner, and every subset of the natural numbers has a least element. In contrast, the integers with the usual order are not well-ordered, because you can take the entire set of integers as your subset, and then there's no least integer.
    A linear order on a set that yields a well-ordered set is called a well-ordering.
    The well-ordering principle says that for any set, you can find a well-ordering on that set. In ZF, the well-ordering principle is equivalent to the axiom of choice.

    Given two ordered sets S and T, we can consider a map f from S to T that preserves the order relation: if a ≤ b for a and b in S, then we'd want f(a) ≤ f(b) in T. Given two well-ordered sets, we can compare them using this idea: we say that S ≤ T if there is a map f from S to T that is injective and preserves the order relation. Basically, we're saying that S fits inside T. This gives an ordering on ordinals, and indeed a well-ordering, although the ordinals do not form a set.

    Claim: The well-ordering principle is equivalent to the statement that cardinalities are linearly ordered.


    Suppose that the well-ordering principle is true. Then given two sets of the given cardinalities we can stick a well-ordering on each, and compare them, and that will give us our order. Note that it doesn't matter which well-ordering we stick on our two cardinals, because if one ordinal has bigger cardinality than another, then the first ordinal cannot be less than or equal to the second ordinal.

    Now suppose that the cardinalities are linearly ordered. Then for any set S, there is a well-ordered set T (this is the hard part, proven by Hartog) such that the cardinality of S is less than the cardinality of T, and so we can inject S into T. Then the resulting set inherits a well-ordering, so S also gets a well-ordering.

    So the upshot is that without the axiom of choice, i.e. the well-ordering principle, cardinality doesn't work as well as we want it to. For instance, you can find sets X that aren't bijective to {1,...,n} for any finite n, but where you can't find an injection from N into X.
  9. Exohedron

    Exohedron Doesn't like words

    Quantum Computation

    Suppose you have a bit, in the computer sense. It can either take the value 0, or the value 1. If you have two bits, then together they can take the values 00, 01, 10, or 11. And so on.

    Now suppose you have a quantum bit, or qubit, can take a bunch of different values. In particular, there is a basis state that we call "o", and a basis state that we call "1", and a qubit then exists in a superposition of those two states. I.e. for a given qubit q we have |q> = a|0> + b|1> where a and b are both complex numbers.
    Suppose we have two qubits. Now we have four basis states, |00>, |01>, |10>, and |11>, and our pair of qubits generally exists in a superposition of those four states:
    |q1q2> = a|00> + b|01> + c|10> + d|11>.
    And you can see that as we add in qubits, the number of coefficients, a, b, c, d etc, grows exponentially.
    Now suppose we had a function f that takes in bits and returns bits. Then we can consider what this function does to qubits.
    We would generally think that f(|00>) should yield |f(00)>, and similarly with f(|01>) and so on. So we then have that
    f(|q1q2>) = a|f(00)> + b|f(01)> + c|f(10)> + d|f(11)>.
    But this only gives us a means of shuffling basis states around. We can also do more complicated things that change the coefficients. For instance, we might consider a function g that sends |00> and |01> to themselves, but sends |10> and |11> to -|10> and -|11> respectively. So then
    g(|q1q2>) = a|00> + b|01> - c|10> - d|11>.
    Or maybe h sends |00> to |00>+|10> and |10> to |00> - |10>. Then
    h(|q1q2>) = (a+c)|00> + b|01> + (a-c)|10> + d|11>.
    And more complicated stuff.
    So we have a general setup now: you take a bunch of qubits, which are in some superposition, and you apply a function to them. Because of the rules of quantum mechanics, this function should be linear: in other words, you can figure out what it does to the superposition by seeing what it does to each basis state and then adding those results together with the appropriate coefficients.
    And that's what a quantum computation is, for the most part.

    Now suppose you had a bunch of n-bit strings, i.e. strings of 0s and 1s, and only one of them fit some criterion, but you really have no way of knowing which one until you actually check that correct one.
    So classically all you could do is just check all of them, one by one.
    Quantumly, what you can do is take a superposition of all the n-qubit states, and run that superposition through quantum operations until you end up with another superposition of n-qubit states, where the coefficient of the state corresponding to the correct string is a lot larger (in the usual complex absolute value sense) than the coefficients of the other states, so that when you observe all n qubits, you probably get the correct string out.
    This is how a quantum computer "tries all the possibilities at once", using a superposition so that you're only doing one computation at any given time, it's just a very different type of computation than a classical computer would be doing.

    Now, there are some limits to what quantum operations you can actually do. There's observation, where you can observe some number of your qubits, and each observed qubit will return either a 0 or a 1. You can then do classical computations on these bits.
    You can introduce new qubits, although you need to be prepared to either have no idea whatsoever what their values are, or do a bunch of work to wrangle them into known states. Initializing a qubit to a particular value is hard.
    Or you can apply a unitary transformation. It's the complex version of a rotation, in that if we view a quantum state as a vector with complex coefficients, then a unitary transformation doesn't change the complex norm of our vector. More explicitly, if our basis states are |00>, |01>, |10> and |11>, we define our norm as
    N(|q1q2>) = |a|2+|b|2+|c|2+|d|2
    where | | is the complex absolute value. Then a unitary transformation U is one that doesn't change the norm:
    N(U|q1q2>) = N(|q1q2>)

    These operations are actually kind of restrictive. For instance, we can't duplicate quantum states. The No-Cloning theorem says that we can't take two qubits whose states we don't know, and somehow make the state of the second qubit the same as the first one. At least, we can't do so with any sort of reliability; if it happens at all, we didn't cause it. So this means there' s no copying a qubit, whereas in regular programming you can always just set a new variable to the value of an already-declared variable whenever you want.

    As a result of the restrictiveness of quantum programming, we only really have two quantum algorithms that people know to work and have applications for. Notably, both of these algorithms have application to breaking modern cryptography.

    Grover's algorithm is good for searching through lists of things where finding the wrong thing doesn't tell you much about how to find the right thing. It gives a quadratic speedup, i.e. if you have x2 things to search through, Grover promises to take O(x) time to find what you're searching for. This allows for speedups in, say, guessing passwords or trying to break hashes.

    Shor's algorithm is good for finding subgroups in an abelian group, if you have a classical function that takes the same value on all elements of that subgroup and different values elsewhere. This ends up being polynomial in the logarithm of the size of the group, i.e. in (quantum) polynomial time. The main speedup here is that using quantum computation, you can perform a discrete Fourier transform on n samples in O(log(n)2) time.
    This can be leveraged to solve certain supposedly NP problems like the discrete logarithm problem or to factor integers, which in turn breaks our public key cryptography.
    Last edited: Oct 22, 2017
  10. Exohedron

    Exohedron Doesn't like words

    Note to self about 2-group actions and 2-torsors*

    Putting this here because I keep forgetting it and I also keep forgetting that I have an actual math blog. Also that math blog is attempting to be somewhat organized, whereas here I am under no compunction to keep anything in order.


    A group, viewed as a one-object groupoid, acts on a set S via a functor to the automorphism group Aut(S). In particular, the 1-morphisms of the group map to automorphisms of the set. If you're looking at (right) torsors of a group G, you want to consider a set T with an object for each 1-morphism in G, and don't make it monoidal. We might as well label the objects of T by the 1-morphisms of G.
    A 1-morphism g in G acts on T by right-multiplication by g, which sends an object s in T to the object sg.
    A G-torsor is now a set S with an isomorphism to T, via this isomorphism, we can create a G-action on S.

    A 2-group, viewed as a one-object 2-groupoid, acts on a category C via a weak 2-functor to the automorphism 2-group Aut(C). In particular, the 1-morphisms of the 2-group map to automorphisms of the category, and the 2-morphisms map to auto-(natural isomorphisms), which I'm not calling natural automorphisms because its feels like that name implies that the source and target of a natural automorphism are the same map, whereas I want a natural isomorphism between automorphisms.
    If you're looking at (right) 2-torsors of a 2-group G, you want to consider a category T with an object for each 1-morphism in G, and a 1-morphism for each 2-morphism in G, and don't make it monoidal. We might as well label the objects and morphisms of T by the 1-morphisms and 2-morphisms of G.
    A 1-morphism g in G to the automorphism rmg, i.e. right-multiplication by g, which sends an object s in T to the object sg, and sends the morphism r:s->t to the morphism rmg(r) = rmg◦r◦rmg-1
    Note that the 1-morphism in G acts on both the objects and morphisms of T.
    Now the 2-morphisms of G act by f:g=>h in G getting mapped to the natural isomorphism that sends rmg to rmh. On the level of objects, f sends sg to sh by lm(f) = lms◦f◦lms-1, where lm is left-multiplication.
    A G-2-torsor is now a category C with an equivalence to T; via this equivalence, we can create a G-2-action on C. Note that since equivalences aren't necessarily bijections on the level of objects, you need to specify both the map from C to T and the map from T to C.

    * Here I'm using the notion of a torsor as "thing which is isomorphic to the thing acting on it", rather than a sheaf of such things over a space. This is similar to how I will talk about a tensor as an element of a tensor product of vector spaces, rather than a section of a tensor bundle over a space.
    Last edited: Oct 25, 2017
  11. Exohedron

    Exohedron Doesn't like words

    The Hopf Fibration

    Consider a surface, S, like a torus or a 2-sphere, or the plane with the origin taken out. We can draw loops on S, and sometimes we can fill in the loops to make disks. For instance, any loop on the 2-sphere can be filled in, but a loop that goes around the hole of a torus can't be filled in without leaving the torus. So this is one way to detect holes in your surface.
    Perhaps a better way to talk about these loops is as continuous functions. Consider the circle, S1, and consider a function f that takes points on S1 to points in S, and does so continuously. When we fill in the image of f, one way to think about that is to consider the disk D2, i.e. what you get by filling in the circle, and ask if there is a continuous function F from D2 to S such that the boundary points of D2 get mapped by F to the same points that the points of S1 get mapped to by f. Then the interior points of D2 get mapped by F to the "filling in" of the original loop.
    So now we can see pretty clearly that certain loops on the torus can't be filled in, because we can consider continuous functions from the disk to the torus and consider their images, like pepperoni on a donut.*
    In contrast, loops on the 2-sphere can always be filled in.
    We say that a loop that can be filled in is nullhomotopic. This is a fancy word made up of two words, meaning "homotopic to a null map", where "null" means maps everything to a single point, and "homotopic" is a complicated thing that I'm not going to get into here.

    Of course, we can also consider maps from S1 to a circle. Can we fill these in? Well, sometimes. Consider the loop that starts at a point on the circle, starts to go out along the circle, then changes its mind and goes back to the starting point. That's a loop. And you can fill it in pretty easily, since it doesn't really have any area to fill in.
    In contrast, the identity map from S1 to the circle can't be filled in. So, as we expected, the circle has a hole in the middle.

    If for a geometric thing X, all of the loops in X can be filled in, i.e. all loops in X are nullhomotopic, then we say that X is simply connected. You might remember this from the post on the Poincare Conjecture/Perelman Theorem.

    We can also talk about maps from the 2-sphere S2 to things. For instance, we can map S2 to the 2-sphere pretty easily via the identity map. Can we fill it in? That is, can we make a function from the solid ball (or 3-disk) to the 2-sphere that matches the identity map on the surface of the ball?
    It turns out that we can't. That emptiness in the middle of the 2-sphere is really there, so we can't fill it in.

    What about maps from the 2-sphere to the circle? Well, in that case it turns out there are no continuous maps from the 2-sphere to the circle that can't be filled in, i.e. all maps from S2 to the circle are nullhomotopic. The circle has a hole in it, but it's a different kind of hole than the emptiness in the middle of the 2-sphere.

    Now for the fun bit. Consider functions from the 3-sphere S3 to things. For instance, to the circle. Are there maps from S3 to the circle that aren't nullhomotopic?
    Nope. Indeed, for every n > 1, every map from Sn to the circle is nullhomotopic. There are no "higher-dimensional" holes in the circle. And I suppose this makes intuitive sense.

    But what about the 2-sphere? For the 2-sphere, it turns out, there are maps from S3 to the 2-sphere that aren't nullhomotopic. Here's one:

    Consider the 3-sphere as the set of points (a, b) in C2 where |a|2+|b|2 = 1; if you write out the complex absolute value in terms of pairs of real numbers, you will see that this description works.
    Now consider the pair [a:b] as a projective point, i.e. all you care about is the ratio between a and b. If you think back to my posts on projective spaces, you'll see that this gets us a point in CP1, which is the 2-sphere. And indeed, you get all of the points on the 2-sphere. Also, this map is continuous. So we have a continuous map from S3 to the 2-sphere.
    And this map is not nullhomotopic!
    So...does the 2-sphere have a 4-dimensional hole in it? Somehow?

    Let's take a closer look at this map. Consider the function that takes a pair of complex numbers (a,b) to [a:b]. Suppose that a/b is not 0 or infinity, so some nonzero, finite complex number q. Well, if a/b = q, and c/d = q, then what can we say about a, b, c and d?
    Since we're looking at points (a,b) and (c,d) on the 3-sphere, and we've decided that a/b and c/d are neither 0 nor infinity, we get that all four of a, b, c and d are nonzero.
    Also, if a/b = c/d, then we get that a/c = b/d. In other words, there is some complex number p such that a = cp and b = dp. And if both (a,b) and (c,d) are on the 3-sphere, then we actually get that p must be a unit complex number, i.e. |p| = 1.
    The unit complex numbers form a circle, so the set of points (x,y) on S3 that map to a particular point q on the 2-sphere has the shape of a circle!
    This also works for the points (a, 0) and (0, b): since |a|2+|b|2 = 1, in the case of (a,0), a has to be a unit complex number, and indeed all points (a, 0) get mapped to the same point [1:0]; similarly for (0,b), b has to be a unit complex number, and all points (0, b) get mapped to [0:1].
    So in other words, if you take the 2-sphere and replace each point with a circle in the correct manner, you get a 3-sphere.

    This map is called the Hopf Fibration, and was the first indication that mathematicians had that this notion of "higher-dimensional holes" was actually quite complicated.

    There are also higher-dimensional versions of the Hopf Fibration. Consider the 7-sphere, viewed as pairs of quaternsions (a, b) such that |a|2 + |b|2 = 1. Then consider the map to the projective space HP1, which is the 4-sphere, via sending (a,b) to [a:b]. We thus get a map from S7 to S4, and the set points on S7 that map to a particular point q on the 4-sphere has the shape of a 3-sphere.
    Similarly, for the 15-sphere S15, considered as pairs of octonions (a,b) such that |a|2+|b|2 = 1, we get a map to OP1, which is an 8-sphere, and the set of points on S15 that map to a particular point q on the 8-sphere has the shape of a 7-sphere.

    There's also a lower-dimensional version. Consider the 1-sphere, i.e. the circle S1, as a pair of real numbers (a, b) such that |a|2+|b|2 = 1. Then consider the map to the projective space RP1, which is the circle, that sends (a,b) to [a:b]. Two points (a, b) and (c, d) get mapped to the same poitn [a:b] if a/b = c/d, i.e. if a = c and b = d or if a = -c and b = -d. So the set of points on S1 that map to a particular point on the circle has the shape of 2 points, i.e. S0.

    And all of these maps are not nullhomotopic, i.e. they can't be filled in without leaving whatever object you're trying to map to.

    In general, for n > 1, the (n+1)-sphere Sn+1 actually has a non-nullhomotopic map to the n-sphere Sn. So in general, only the circle S1 is actually as simple as it seems in terms of holes.

    The question of what kind of holes an object has has sprawled a very large branch of mathematics called homotopy theory. As you can see, simple geometric ideas like "dimension" don't tell us as much as we might want them to; the 2-sphere has a "4-dimensional" hole in it!

    *Don't knock 'til you've tried it.**
    **I've never tried it.
  12. Exohedron

    Exohedron Doesn't like words

    I'm realizing that not having the first post be an introductory post with a table of contents is making it hard for me to remember what I've actually written about so far.
  13. Exohedron

    Exohedron Doesn't like words

    It's too bad most symmetric-key crypto systems these days are so terribly inelegant. I mean, sure, that same inelegance helps make them secure, but still. I'd love to make a post about them, but most are just "do this thing a bunch of times until all the bits look random."
  14. Exohedron

    Exohedron Doesn't like words

    Is there someone who is good with PDEs who can explain why the solutions to the wave equation with an even number of space dimensions are so different from the solutions with an odd number of space dimensions? I mean, I know that to get solutions with an even number of space dimensions we start with the odd case and look for solutions that are independent of one of the space dimensions, but I have no understanding of why we do this rather than directly going for the solutions using the same method that we used for the odd-dimensional case.
  15. seebs' mom

    seebs' mom Yes, really!

  16. Exohedron

    Exohedron Doesn't like words

    In an attempt to figure out what the connection the Quanta article mentions is more explicitly, I came across this paper on Arithmetic Chern-Simons theory and a talk he gave at a conference on Arithmetic Topological Quantum Field Theory. I think these describe what the Quanta article is talking about, although I haven't had the time and probably don't have the background to really look into either of those.
  17. Exohedron

    Exohedron Doesn't like words

    Vertex Curvature

    We have a notion of "curvature" for surfaces, where a dome or a cup or a saddle is somehow fundamentally different than a flat plane in a geometric sense, and that a rolled-up piece of paper is still "flat" from the point of view of something living on it (assuming they can only see a short distance).
    One issue with this notion of curvature is that it requires some amount of smoothness to the thing we're looking at. But what about if our shape has corners or edges?

    Suppose we have a cube (well, the surface at least). The sides of the cube are flat, and the edges are straight. We are usually pretty okay with these facts. But there is a nice theorem called the Gauss-Bonnet theorem that relates the curvature of a surface to its topology, and a cube is topologically a sphere. And spheres are curved. So where is the curvature of a cube?

    A while back I mentioned various types of "curvature" that can be measured by someone living on a surface. The one I want to use here is measuring the size of a "circle". In particular, to find the curvature at a point on the surface, we take all of the points that are a fixed distance d away from the point; these points form a loop, and we measure the length l(d) of the loop. For a flat surface, the length of the loop should be 2πd. For a cup or a dome, you'll find that you get a length that's somewhat less than 2πd, and for a saddle you'll end up with slightly more. We can call the value 2πd-l(d) the "defect" of the loop. There's a formula for relating the defect to the curvature, but for smooth surfaces it's kind of complicated.
    Fortunately for our case, it gets a lot simpler.

    So let's look at our cube once again. The curvature certainly isn't in the faces. Those are flat. And it's probably not on the edges; you can fold an edge into a flat piece of paper, and that edge will look just like the edge on the cube.
    So it has to be hiding in the vertex.
    And indeed, if you take just a vertex and the edges and faces immediately touching it, and ignore the rest of the cube, it looks kind of dome-like, right? So it should have positive curvature somewhere.
    So let's take a corner and draw a loop around it.
    If we consider all the points on the cube that are a constant distance d from the vertex along the surface of the cube, we end up drawing on the three faces that are touching the corner. And on each face we draw a quarter circle of radius d, and hence contributes an arc length of (1/2)πd. So our formula for the length of the loop is (3/2)πd. So the defect is (1/2)πd, which would correspond to a fourth square face touching the vertex if we were looking at a flat surface rather than a cube.
    Note that the defect is just a constant times d, so really the interesting value is just (1/2)π. We call that the "vertex curvature" of our vertex. So the vertices of a cube each have vertex curvature (1/2)π

    Is this the right notion of vertex curvature? Well, let's compare to a sphere.
    For a sphere of radius r, each point on the sphere has curvature 1/r2 according to the usual smooth-surface definition of curvature. If we integrate that over the surface of the sphere, that's just multiplication by 4πr2 since we're integrating a constant value, so the total curvature of the sphere is 4π. Note that the total curvature doesn't depend on the size of the sphere. The Gauss-Bonnet theorem states that for smooth surfaces that are topologically equivalent to spheres, the total curvature is going to be 4π (in general, for smooth surfaces without boundary the total curvature will be 2π times the Euler characteristic).
    Now if we take our cube, with all of its curvature in the eight corners, and "integrate" where we integrate the face curvatures over each face (giving 0), and integrate the edge curvatures (which I haven't defined, but here it's 0) along each edge (giving 0), and "integrate" (i.e. sum) the vertex curvatures for each vertex, we end up with 8 * (1/2)π = 4π, same as the sphere.
    So we have that, if we use this notion of vertex curvature, then the cube has the same total curvature as the sphere, it's just concentrated at the vertices!
    And you can check that this is true for the other regular polyhedra: the tetrahedron has four vertices, each with vertex curvature π, the octahedron has six vertices, each with vertex curvature (2/3)π, the dodecahedron has twenty vertices, each with vertex curvature (1/5)π, and the icosahedron has twelve vertices, each with vertex curvature (1/3)π.

    So this seems like it gives the correct notion of curvature for polyhedra, in that it gives the same total curvature as for a sphere and hence, as promised, the total curvature is the same for surfaces that are topologically the same, despite being different as geometric objects. This is the Gauss-Bonnet theorem extended to polyhedra.
    Note that there are of course surfaces that aren't smooth and aren't polyhedra, with curved edges and faces. The Gauss-Bonnet theorem still holds for them, but then I would need to actually talk about curvature in general and I'd rather not get too far into the weeds.

    So one last thing. This was inspired by someone trying to tile a sphere with equilateral triangles and noting the various degrees of the vertices in the tiling. Some vertices ended up with seven triangles around them, and some of the vertices had five and the rest of the vertices had six. The vertices with six triangles around them were flat, and the vertices with five looked like cups or domes, and the vertices with seven around them looked like saddles.
    Now, a computation of the Euler characteristic will tell you the relation between the number of vertices with five triangles around them compared to the number of vertices with seven. A completely equivalent computation will tell you that the vertex curvature of the vertices with five triangles is (1/3)π, and the vertex curvature of the vertices with seven triangles is -(1/3)π, so to get a total curvature of 4π we need to have exactly twelve more vertices with five triangles than with seven.
    For instance, an icosahedron fits this exactly, having twelve vertices with five triangles and zero vertices with seven.

    Fun fact: this notion of "vertex curvature", extended upwards a dimension, gives us the gravitational curvature that you see from black holes, cosmic strings, and massive point particles; they all have some "angle defect" from what is expect in flat space.
    Last edited: Dec 8, 2017
    • Informative x 1
  18. Exohedron

    Exohedron Doesn't like words

    Apparently there's some drama in the ABC fandom. I wish I could comment meaningfully on it, but I'm not enough of an expert on either the material or the history to really say much more than that it really does seem to indicate a breakdown in the ways the mathematical community is used to for establishing trust.
  19. Exohedron

    Exohedron Doesn't like words

    Quantum Dodecahedra

    Over here on G+, I had a nice conversation with two people about what we call a "quantum dodecahedron". Here by "quantum dodecahedron" we mean a waveform for a stationary boson with the rotational symmetries of a dodecahedron centered at the center of the boson. In particular, we were looking at wavefunctions with well-defined total spin.

    Let's unpack that a bit.

    We start with the wavefunction. There are a lot of ways to describe a wavefunction for a stationary boson. One way is via what are called "spherical harmonics", which are like trig functions but instead of being circle-based, they're based on the 2-sphere.
    Now suppose that we have a dodecahedron. We can rotate the dodecahedron by various angles to get an identical dodecahedron. Call the group of such rotations SH3, for reasons I'll get into. We say that a function f is SH3-invariant if, for all rotations r in SH3 and all points s on the sphere, f(s) = f(r(s)). In other words, rotating everything doesn't change the value that f takes.
    So any SH3-invariant spherical harmonic f gives us a quantum dodecahedron.

    The first question for any definition is: does such a thing exist?
    The answer in this case is yes. The obvious one is the case where f = 0 everywhere. That's boring, but it works. A slightly more interesting, but still boring, example is f = 1 everywhere. Bleh. Are there more interesting ones?

    A quantum particle can have what is called "spin". This is like angular momentum, only it applies to quantum particles that don't appear to be moving. So for instance, a stationary electron can still have a magnetic field due to its quantum mechanical spin even though it isn't moving in the classical sense.
    What does that mean for wavefunctions? All spherical harmonics can be written in terms of basis spherical harmonics that have spins associated with them, specifically two numbers: total spin, and spin around a particular axis. These basis spherical harmonics are often denoted by Ylm where l is the total spin and m is the spin around, say, the z-axis.
    So now we can ask about quantum dodecahedra with a well-defined spin.
    Note that we can only ask about total spin, because if we specify the spin around the z-axis, we lose the ability to specify the spin around any other axes (due to the Uncertainty Principle), but being SH3-invariant would force us to specify those other spins. So no quantum dodecahedron can have a well-defined spin around the z-axis, only a well-defined total spin.

    Spherical harmonics are kind of awkward to work with. Instead, we can extend a spherical harmonic, which is really only defined on the sphere, to a function of the entirety of 3-space. For any given spherical harmonic there are multiple ways to do this. The best way is to use solid harmonics, which most of the time are just called harmonic functions.
    There is an interesting fact that every spherical harmonic corresponds to a single harmonic function, but harmonic functions are less awkward.
    For instance, if we start with a spherical harmonic with a well-defined total spin l, its corresponding harmonic function will be a homogeneous polynomial in three variables with degree l.
    Note that for a quantum dodecahedron, the corresponding harmonic function will also be SH3-invariant. We might as well call this harmonic function a solid quantum dodecahedron.
    Now a linear combination of polynomials of the same degree doesn't change the degree, and if they're homogeneous or harmonic or SH3-invariant then so is the linear combination. Hence a linear combination of solid quantum dodecahedra of a given total spin l are still solid quantum dodecahedra with well-defined total spin l, and so we should really be asking about linearly independent solid quantum dodecahedra of a given total spin l.

    So we have the final version of the question: given a number l, how many linearly independent homogeneous SH3 harmonic polynomials of degree l are there? How many truly distinct quantum dodecahedra of total spin l are there?

    The answer ends up being kind of interesting: It's the number of ways you can write l as a sum of 6s, 10s, and at most one 15. So for l = 16 there is 1 linearly independent quantum dodecahedron, for 30 there are 2, for 60 there are 3.
    Here's first few non-boring cases with values indicated by colors, rotating for some reason:
    Pictures by Greg Egan.

    The reason for this answer is kind of involved. Look here if you want to see me drop a bunch of probably-unnecessarily-powerful representation theory on (a generalization of) this problem.
    Last edited: Jan 5, 2018
  20. syntheme

    syntheme Active Member

    Honestly this problem seems to me to be pretty much Chevalley-Todd-Shephard so I don't think the representation theory you're using is unnecessarily powerful. In three dimensions I think you can also derive your solution from the McKay correspondence and deformations of C[x,y]#G but if anything that's more powerful representation theory.
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