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[math-fun] Room temperature superconductor
by Eugene Salamin 15 Oct '20

15 Oct '20

Superconductor at 15 C, but at 267 GPa pressure. https://news.yahoo.com/super-material-raises-hope-energy-094906169.html https://www.nature.com/articles/s41586-020-2801-z -- Gene

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[math-fun] Homology intersection and cup product on cohomology in a manifold
by Dan Asimov 15 Oct '20

15 Oct '20

I wrote this up for a friend and thought I'd share it with math-fun. It assumes you have some familiarity with * the idea of a manifold (which basically means a topological space that locally looks like Euclidean space R^n for some fixed n). For instance Euclidean space R^n, or any open subset of it, is an n-dimensional manifold. Any surface is a 2-dimensional manifold, and the spatial universe (or at least a snapshot of it at a fixed time) is a 3-dimensional manifold. and * the idea of a k-dimensional homology class (which basically describes a k-dimensional "hole" in a topological space X by something like a k-dimensional manifold S that conforms to the border of the hole). Think for instance of X = the plane R^2 with all the points at a distance of less than 1 from the origin removed, and then S could be the unit circle, which is a 1-dimensional homology class in X. Then H_i(X) is the *group* of i-dimensional homology classes. It's always a finitely generated abelian group if X is a reasonably nice space like a compact manifold. -------------------------------------------------------------------------- Given two *homology* classes, say a ∊ H_p(M), b ∊ H_q(M) where M is an n-dimensional compact connected manifold, imagine that each is represented by a submanifold of M (not always true, but useful for heuristics). Say a = [A] and b = [B] for compact submanifolds A, B of M of dimensions p, q respectively. [X] denotes the homology class of X. Then by a small perturbation we can assume A ⫚ B (transversality). If p + q < n then A ∩ B will be empty. So Assume p + q ≥ n: Assume p + q ≥ n: Then A ∩ B will be of dimension (p+q) - n. This means [A∩B] ∊ H_r(M) where r = (p+q) - n. This defines a *product* on the Poincaré dual* cohomology classes [A]*, [B]*, [A∩B]*: [A]* ∪ [B]* = [A∩B]*, i.e., a product from H^(n-p) ⊗ H^(n-q) —> H^(n-((p+q) - n), i.e. But letting P = n-p, Q = n-q, we see that this is a bilinear product H^P(M) × H^Q(M) —> H^(P+Q)(M). This nice dimension-summing property makes the total cohomology group H*(M) = ⊕ H^j(M) (summed over 0 ≤ j ≤ n) into a graded ring. This beautiful structure encapsulates a great deal of the topology of M. For instance each orientable surface M_g of genus g has H*(M_g) = H^0 ⊕ H^1 ⊕ H^2 = Z ⊕ Z^(2g) ⊕ Z where H^1(M_g) has the 2g generators g_j, h_j, 1 ≤ j ≤ g with g_j ∪ g_k = 0 = h_j ∪ h_k and { 1 if j = k g_j ∪ h_k = -h_k ∪ g_j = { { 0 if j ≠ k, where 1 generates Z = H^2(M_g). —Dan ————— * Think of the Poincaré dual (cohomology class) of a homology class as describing how that class intersects with all the other homology classes (of any dimension). This is extremely analogous to thinking of a vector v (in a vector space with inner product) as how it dot-products (w |—> <v,w>) with all the other vectors.

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[math-fun] Do "Hawking Points" Actually Exist?
by Brad Klee 15 Oct '20

15 Oct '20

I was wondering about this question recently, so I was happy to get a dissenting opinion via Forbes: https://www.forbes.com/sites/startswithabang/2020/10/08/no-roger-penrose-we… But the tone of the Forbes article is too businesslike for my taste, and does it mention that Penrose himself describes the idea as "crazy"? See also: https://www.youtube.com/watch?v=YTttUigXulk When you watch this video, how can you help but wonder about the incredible mystery of what exactly is going on in the background out there? Maybe universal reincarnation is a plausible cosmogony? We don't know. Fifty or one hundred years from now we could have better gravity detection, and a better idea of what is going on behind (or before) the CMB. Until then... Anyways, congratulations Penrose! More and more cheers and congratulations for unexcelled creativity! --Brad

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[math-fun] Disks in Euclidean approximation metrics ...
by Colin Wright 15 Oct '20

15 Oct '20

I've been asked a question to which I don't know the answer. Although this is not unusual, I'm wondering if there is an answer at all. If there is one, the audience here is likely to know of it. Let's start with the well-known to give a context. Embed Z^2 in R^2, and define the distance (0,0) to (x,y) to be |x|+|y|. Disks around a point are square diamonds. We now depart into vaguely familiar territory, but I'm finding it hard to pin down exactly what the question is, so I'm going to ask for some leniency, and for some creativity in finding the right question that's close to what I might be asking. When we use the Taxi-Cab metric above we are saying that the distance from (0,0) to (1,1) is 2. So let's short-circuit that and say that the distance from (0,0) to (1,1) is sqrt(2), and that we can only get somewhere by taking diagonal hops followed by taxi-cab journeys, and the *distance* is the minimum taken over all possible journeys. Disks around (0,0) are now octogons. I think. Are they regular? I suspect not. And then add that we are permitted to make a knight's move of distance sqrt(5). So journeys now consist of diagonal steps, knight's move, and taxi-cab journeys, and the distance from (a,b) to (c,d) is the minimum over all such journeys. Disks around (0,0) are now 16-gons, but are not regular. Obviously we can continue, but the question is: * Have these specific n-gons been studied? * Do these specific n-gons have names? * Do they have any interesting properties? Thanks for reading, I look forward to your thoughts, which I'll pass on to my interlocutor. Colin -- The power of accurate observations is commonly called cynicism by those who haven't got it. -- George Bernard Shaw

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[math-fun] Quaternion eigenvalues & eigenvectors
by Henry Baker 14 Oct '20

14 Oct '20

Eigenvalues and eigenvectors are typically associated with matrices, not quaternions. But there is a standard injection of quaternions into 2x2 complex matrices: The quaternion Q=A+Bj, for a pair A,B of complex numbers, can be represented by the 2x2 complex matrix: [ A B ] Q = [-B' A'] If A=ar+i*ai and B=br+i*bi, then the matrix for our quaternion Q is: [ ar+i*ai br+i*bi] Q = [-br+i*bi ar-i*ai] Q's eigenvalues are lambda = ar +- i*sqrt(ai^2+br^2+bi^2) = sQ +- i*|VQ|, where sQ=ar, VQ=(ai,br,bi). We seek an invertible matrix X, s.t. [sQ+i*|VQ| ] (1/X)QX = [ 0 sQ-i*|VQ|] = L Here, the matrix X is the matrix of eigenvectors; i.e., the columns of X are eigenvectors for the matrix Q. But *because the eigenvalues are complex conjugates*, the diagonal matrix of eigenvalues L can also represent the quaternion L =(sQ+|VQ|*i)+0j. We have thus converted our *matrix* eigenvalue problem into a *quaternion* problem: We seek a quaternion X, s.t. (1/X)QX = L = (sQ+i*|VQ|) + 0j But we already know the solution to the equation (1/X)QX = L, due to a previous posting! If (1/X)QX = L => QX = XL then the following X is a solution: X = 1-(VQ/|VQ|)(VL/|VL|) = 1-(VQ/|VQ|)(VL/|VQ|) = 1-(VQ/|VQ|)(i*|VQ|/|VQ|) = 1-(VQ/|VQ|)*i = 1-(VQ*i)/|VQ| Now we can clear fractions for X: X = |VQ| - VQ*i QX = XL Q(|VQ| - VQ*i) = (|VQ| - VQ*i) (sQ + |VQ|*i) (sQ+VQ) (|VQ| - VQ*i) = (|VQ| - VQ*i) (sQ+|VQ|*i) sQ*|VQ| - sQ*VQ*i +VQ*|VQ|- VQ*VQ*i = (|VQ| - VQ*i) (sQ+|VQ|*i) sQ*|VQ| - sQ*VQ*i + |VQ|*VQ + |VQ|^2*i = (|VQ| - VQ*i) (sQ+|VQ|*i) = |VQ|*sQ + |VQ||VQ|*i - (VQ*i)*sQ - (VQ i)*|VQ|*i = sQ*|VQ| + |VQ|^2*i - sQ*VQ*i - |VQ|*VQ*i*i = sQ*|VQ| + |VQ|^2*i - sQ*VQ*i + |VQ|*VQ = sQ*|VQ| - sQ*VQ*i + |VQ|*VQ + |VQ|^2*i QED Note that X = |VQ| - VQ*i is a full quaternion whose scalar and vector parts are: sX = |VQ|+(VQ.i) VX = - (VQ x i) = i x VQ where VQ.i is the classical vector dot product, and i x VQ is the classical vector cross product for the 'i' vector and the 'VQ' vector. We thus have an elegant one-step solution for the eigenvalue/eigenvector problem in quaternions: L contains both eigenvalues and X contains both eigenvectors. No polynomials, no determinants, no Gaussian eliminations. We do have one scalar square root needed to calculate |VQ|, but no other square roots are necessary unless we wish to normalize X. A normalized X/|X| provides the special unitary matrix that diagonalizes Q, should that be necessary.

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Re: [math-fun] Factoring/normalizing quaternions
by Dan Asimov 13 Oct '20

13 Oct '20

----- By 'complex number' A, I sometimes mean the quaternion A+0j, constructed from the pair of complex numbers (A,0). This subgroup of the quaternions is isomorphic to the standard complex numbers. Yes, I'm abusing the notation, but mathematicians often abuse notation. :-) ----- That's not abuse of notation. It's a mistake. —Dan

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Re: [math-fun] Factoring/normalizing quaternions
by Dan Asimov 13 Oct '20

13 Oct '20

I skipped over sections of this post by Henry Baker to try to find the points that it included. One point seems to be this: ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. —Dan

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Re: [math-fun] Factoring/normalizing quaternions
by Henry Baker 13 Oct '20

13 Oct '20

By 'complex number' A, I sometimes mean the quaternion A+0j, constructed from the pair of complex numbers (A,0). This subgroup of the quaternions is isomorphic to the standard complex numbers. Yes, I'm abusing the notation, but mathematicians often abuse notation. :-) At 05:33 PM 10/13/2020, Dan Asimov wrote: >Henry, Despite whatever else you say, you also said: >----- >We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. >----- >As you have defined then, a "complex number" in the quaternions is a quaternion of form a + bi. > >So maybe that's not what you mean, or maybe you don't mean that you have factored a quaternion into two complex numbers. > >But those are things you said, and that was my point. > >ÂDan > >>Hi Dan: >> >>There's a lot of subtlety in my post that a quick scan might miss. >> >>There's a *huge* difference between 'i' and 'j', so the eye might not >>pick up on this subtlety. In the quaternions, 'i' is perpendicular to >>'j', and 'i' doesn't commute with 'j'. BTW, 'k' is there: it is >>simply represented by 'ij' rather than by a separate symbol. >> >>My post is basically a rehash of the standard theorem about factoring >>SU(2) matrices into (diagonal complex)(real cos/sin)(diagonal complex) >>SU(2) matrices, but it does it completely within the quaternion >>framework. >> >>At 02:16 PM 10/13/2020, Dan Asimov wrote: >>>I skipped over sections of this post by Henry Baker to try to find the points that it included. >>> >>>One point seems to be this: >>> ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- >>>But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. >>> >>> ÂDan

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Re: [math-fun] Factoring/normalizing quaternions
by Dan Asimov 13 Oct '20

13 Oct '20

Henry, Despite whatever else you say, you also said: ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- As you have defined then, a "complex number" in the quaternions is a quaternion of form a + bi. So maybe that's not what you mean, or maybe you don't mean that you have factored a quaternion into two complex numbers. But those are things you said, and that was my point. —Dan >Hi Dan: > >There's a lot of subtlety in my post that a quick scan might miss. > >There's a *huge* difference between 'i' and 'j', so the eye might not >pick up on this subtlety. In the quaternions, 'i' is perpendicular to >'j', and 'i' doesn't commute with 'j'. BTW, 'k' is there: it is >simply represented by 'ij' rather than by a separate symbol. > >My post is basically a rehash of the standard theorem about factoring >SU(2) matrices into (diagonal complex)(real cos/sin)(diagonal complex) >SU(2) matrices, but it does it completely within the quaternion >framework. > >At 02:16 PM 10/13/2020, Dan Asimov wrote: >>I skipped over sections of this post by Henry Baker to try to find the points that it included. >> >>One point seems to be this: >> ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- >>But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. >> >>Dan

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Re: [math-fun] Factoring/normalizing quaternions
by Henry Baker 13 Oct '20

13 Oct '20

Hi Dan: There's a lot of subtlety in my post that a quick scan might miss. There's a *huge* difference between 'i' and 'j', so the eye might not pick up on this subtlety. In the quaternions, 'i' is perpendicular to 'j', and 'i' doesn't commute with 'j'. BTW, 'k' is there: it is simply represented by 'ij' rather than by a separate symbol. My post is basically a rehash of the standard theorem about factoring SU(2) matrices into (diagonal complex)(real cos/sin)(diagonal complex) SU(2) matrices, but it does it completely within the quaternion framework. At 02:16 PM 10/13/2020, Dan Asimov wrote: >I skipped over sections of this post by Henry Baker to try to find the points that it included. > >One point seems to be this: > >----- >We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. >----- >But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. > >ÂDan

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