Ribbon category editionITT: discuss mathematicsFormerly: >>16777973Photo credit: https://indico.in2p3.fr/event/32579/contributions/143313/attachments/86648/130492/Vlaar.pdf
this is an ill-formed question, but when combining concepts in advanced physics, people seem to tensor product lots of spaces together; why not direct sum?
>>16803028Roughly:- a basis for [math] V \oplus W [/math] is of a union of a basis for V with a basis for W .- a basis for [math] V \otimes W [/math] consists of pairs (a,b) where a,b respectively are bases for V,W.In quantum mechanics, V would be the Hilbert space of quantum states of one physical system, and same for W for another physical system. Technical note: for an orthonormal basis of V generally we take an eigenbasis [math] B \subset V [/math] of some Hermitian operator corresponding to some physical observable (position, momentum, energy, etc) of the physical system of V. In stat mech they may often call such B the "set of quantum states" of V.Usually a state of the combined physical system of V and W corresponds to an ordered pair (a,b) where a,b respectively are states of V,W. (Instead of a choice of *either* a state of V or a state of W.) This is why we take the ombined physical system to have Hilbert space [math] V \otimes W [/math] instead of [math] V \oplus W [/math].
>>16803049i get all this, but i don't think this really answers the question in my mind.another way of thinking about it is: why, when combining certain spaces in physics, should one expect the dimension of the resulting space to be the product of the dimensions of the constituent spaces (which would indicate a tensor product) instead of their sum (which would indicate a direct sum)?how does one know when to use either tensor product or direct sum?for instance, in statistics, the joint probability of independent variables is the product of their probabilities, while the combined probability of mutually exclusive events is the sum of their probabilities. understanding this, one can determine when to multiply and when to add based on what events you are interested in knowing the probability of.if a particle can described by e.g. a tuple (spin state, position state), why should the resulting space have a dimension equal to the product of the constituent states?once again, i'm probably stating my question in a poor manner because i don't quite know what i don't know.
>>16803076When you put two systems together, they cease being independent, and you have to model them jointly; the state of system B can "twist" system A into having an incomparable state space relative to the state space of A but with a different state of B. Direct sums don't let you do this, anything you can do with a direct sum is too decomposable into "here's the A part, there's the B part." For example, the only interactions you can model with direct sums would be transitions from one system to another (like, say, creation/annihilation in Fock space); you can't model, say, when the internal evolution of one system depends directly on the state of the other.
>>16803028Tensor product allows to talk about pairs that aren't really interacting with each other. If I want to talk about a man and his dog I saw walking down the street, the analogy would be that the object I saw is within the set Men [math] \otimes [/math] Dog. If I want to talk about types of people, I can talk about a man, I can talk about women, and if I want to talk about a person named Sam who is either an man or a woman, then the analogy is that they're within Men [math] \oplus [/math] Women.In physics, we can talk about the angular momentum of particle 1, the angular momentum of a different particle 2. If we talk about the overall object of particle 1 and particle 2 together and something about it's angular momentum, do we use the sum or product?
>>16803223>Sam is either a man or a womenOr some linear combination
>>>/b/940574809Abstract algebra.
Failed my second semester bachelor analysis exam. Gonna write the retake exam. But my real problem is that I don't really know why I even failed the first one. And to look into my exam I only have an appointment one day before the retake exam. I thought that I got it. Idk why I became this retarded. Taylor polynome, multi dimensional integration, derivatives. It isn't that hard. It's so fucking over.
>>16803028Direct sum is a coproduct. Tensor product is a monoidal product. Two different things. Products of representations admit decompositions into direct sums of irreps, which is a basic result of representation theory.
>>16803825>It's so fucking over.And>It isn't that hard.Don't really go together.
>>16803832>Direct sum is a coproduct. Tensor product is a monoidal product.Both are functors [math] \mathcal{C} \times \mathcal{C} \rightarrow \mathcal{C} [/math], however. They just have different properties.>Two different things.I'm sure the person you're replying to wasn't implying they're the same thing.
>>16803841Maybe they mean it shouldn't have been that hard, and the fact that it was means it's "over" as they said
>>16803857>>16803841It's over became a standard phrase for me. But idk if I am able to study it. I was skipping a lot of lectures and was spending a good time in my fraternity with drinking and smoking. Anyways you got any tips or resources for me? Was reading a bit from the Königsberger Analysis 2.
>>16798772>the problem where a bug starts walking from one vertex of the box and it finds the shortest possible route to a point that you chose on its surface... what is the maximum distance that it would be possible to force the bug to walk>>16798877 if the final position P is coplanar to V, then the shortest distance is trivial. Else, position the vertex V at the origin and the box the 1st quadrant with P on the top face. Call the length of the side touching the 2nd quadrant W (x axis) and the length of the side touching the 4th quadrant D (y axis), and the height of the box H. If [math] w < d [/math], simply swap the the x and y coord. of P, then revert back afterwards if needed. Wlog, let [math] d \leq w [/math]. The coordinates of P will be determinedIf [math] h \leq w [/math], the longest of the minimized paths is simply the one where P is furthest away from V in the corner at coordinates (w,d,h), getting to the top from S1, and the distance is [math] \sqrt{w^2 + (h+d)^2} [/math]For any other case, the following must be done. For brevity, let [math] \tfrac{d}{h} = \bar{d},\ \tfrac{w}{h} = \bar{w} [/math]. If [math] \bar{d}\bar{w} (\bar{d} + \bar{w}) + 2\bar{w}^2 + 4\bar{d}\bar{w} + \bar{d}^2 + \bar{d} - 2 \geq 0 [/math], the coordinate of P is the same as above, but with shortest path entering the top via S2 with distance [math] \sqrt{(w+d)^2 + h^2} [/math]If instead [math] \bar{d}\bar{w} (\bar{d} + \bar{w}) + 2\bar{w}^2 + 4\bar{d}\bar{w} + \bar{d}^2 + \bar{d} - 2 \leq 0 [/math], then the coordinate of P is at [math] (w-a, d-a, h) [/math], where [math] \displaystyle a = \frac{ bd + h - w }{ c-b },\ b = \tfrac{h-w}{h+w},\ c = \tfrac{ 2h+d }{ d } [/math]and the distance of the longest of the minimized paths is [math] \sqrt{(w+d - a)^2 + (h+a)^2} [/math]The partition into 4 sections is two triangles on opp. sides and a line connecting their closest vertices to make 2 trapezoids. Takes only 5 straight lines to construct. No need for AI bs
>>16804365>enter through S1, or S2Well, sometimes you can enter through something else like S3. All the boundaries of the partition are shared so there are a couple of ways to get to that same spot with equal distance.
I'm making an anki deck to memorize a bunch of math constants and formulas, what should I put in there?I'm doing this mostly for fun, or for the party trick effect. Currently I have the useless trig formulas and identities (csc, sec, cot). Inclusion orders for fields to rings. I wanna put more probability stuff and category theory - maybe some things could actually be useful?
>>16803856>both are functorsNot very informative. Anything's a functor if you think hard enough. Monoidal products require additional structure in the form of unitor and associator nat transformations. Coproducts are very special monoidal products satisfying a universal property. Tensor products satisfy their own universal property that "enlarges" a category to the smallest cartesian-closed category containing it.In case of representation theory, vector spaces come with a natural coproduct monoidal structure, but no product monoidal structure. Hence the tensor product.>I'm sure the person you're replying to wasn't implying they're the same thing.The anon's wondering why not use one instead of the other. In reality, both are used. The Clebsch-Gordan decomposition is the relationship between tensor products and direct sums.
is this >>>/g/106777016>a single qubit's state is given by a|0> + b|1> where a and b are complex numbers such that aa* + bb* = 1.>however, we don't give a shit about the global phase, so the state of a single qubit can be modeled using the bloch sphere. (mathematicians call this the hopf fibration)true?
>>16804921Yes, but using the term Hopf fibration in this context is only explanatory for people familiar with QFT or pure mathematical jargon.Its describing rotations as [math] U(1) [/math] elements and then using this "Hopf fibration" you get that [eqn]S^3/U(1) \cong S^2 [/eqn], which is the Block sphere.Explaining Hopf fibrations requires a lot of differential geometry, but you can watch this video for an idea: https://www.youtube.com/watch?v=dkyvZo68IoM
Anyone here has a PDF of Herstein's Topics in Algebra? I just got the book from the library (the 2nd edition) and compared it to the PDF that I have (also the 2nd) and to my surprise the PDF version is missing the last two sections of chapter 2 (Direct Products and Finite Abelian Groups). Can anyone check their PDFs so I can see if this is a problem with the version I have or if it's a general issue with all PDF versions of the book? I find this super strange because generally if an author decides to cut or add chapters they call it a new edition, but the physical book says it's still the 2nd edition.
>>16804921i wrote this. i'll add some commentary on the quantum physics side. the reason we don't care about global phase is that it will never affect the magnitude of any state, so it will never affect the probability distribution of any measurement we make. thus, |S> = c |S> for any state S and any complex number c with magnitude 1.
I just invited an interesting factIf I construct a polynomial where it hits everything rational number on a function in an interval, because polynomials are continuous it means that for any continuous function it is an infinite polynomial.Here’s why, if you subtract by the function and one value is the same it means that the root exists, which is always true. So a sufficient polynomial exists for the rationals which are countable.I just invented this mathematical law. I’m going to call it the bad ass polynomial law.
>>16805676>infinite polynomialThere is no such thing.
>>16805676Learn to write coherently first
>>16805867>INFINITY ISNT REALOk the error limits to zero as the order increasesAt some point you need grow up and realize enumeration exists.>>16806144Not my problem.
>>16803023If I split 10 square feet (each side 10 feet long and 10 feet wide) into 4 parts, each of them are 5 square feet, yes)So, if I split 1 square mile (each side1 mile long and 1 mile wide) into 4 parts, is the result 4 1320 square feet or 4 2640 square feet?
>>16803023How many 5 x 7 x 11 binary tensors have no all-zero or all-one rows in any of the three directions?
>>16806376Each of the former are 25, each of the latter are 6969600.
>>16806381Sorry for being obtuse, but do you mean that both 1320 and 2640 are 6969600 or that 1320 is, like 5, 25?
>>16806387Each "is," I'm tired lol. 2640^2 = that
>>16803825Do all the problems in the book. Find other books and do the problems in them.
>>16805676I can't tell what you are trying to say, but I think you should look into the Stone–Weierstrass theorem.
I've never been good with combinatorics and this has me stumped.For fixed numbers [math] a_{1},...,a_{k},b_{1},...,b_{k} [/math], what is the number of odd permutations [math] \sigma \in S_{n} [/math] such that for all [math] i [/math], [math] \sigma(a_{i})=b_{i} [/math] expressed in terms of [math] k [/math] nd [math] n [/math] ? For a given permutation satisfying the conditions there are obviously [math] (n-k)! [/math] possible permutations total and from a few examples I would assume there are [math] (n-k)!/2 [/math] odd permutations, however I have no clue how to prove that. Any hints?
>>16805209>S^3/U(1)I don't want to be the nitpicky autist itt, but S^3 is a topological manifold, while U(1) is a Lie group. Two different categories :)
>>16807652it suffices to assume [math] a_i=b_i [/math] for all [math] i [/math]
>>16807720>S^3 is a topological manifold, while U(1) is a Lie groupAnon............................................
>>16807792I'm assuming the argument for that is that there exists some even permutation [math] \tau [/math] such that for our given permutation [math] \sigma [/math] satisfying the conditions, [math] \tau(b_{i})=a_{i} [/math] for all [math] i[/math], and then [math] c=\tau\sigma[/math] permutes only those elements not equal to the [math] a_{i} [/math] and gives us my assumed number of odd permutations and so the original permutations are also odd since [math] \sigma =\tau^{-1}c [/math].
>>16807875basically correct, but>so the original permutations are also oddnot necessarily
>>16807875>>16807889oh, you declared at the start [math] \tau [/math] to be evenin that case yes, but it's maybe not completely obvious that such an even permutation exists
>>16807898That is indeed the current problem.
>>16807720You can still quotient S^3 by the U(1) action on it you retard. You get a manifold but not a Lie group.
>>16807898I guess I define [math] \tau [/math] such that [math] \tau^{2}(b_{i})=a_{i} [/math] for all [math] i[/math]. That feels like cheating, can I just do that?
[math]f,g:X\to Y[/math] are continuous and [math]Y[/math] has the order topology. How can I prove that [math]\{x\in X\,|\,g(x)<f(y)\}[/math] is open?
>>16808019not without justification, most permutations aren't squaresbut ultimately for what you're trying to prove it doesn't matter if [math] \tau [/math] is even or odd, you still get the same result
>>16808081Yeah that made it click for me, cheers.
>>16808058This solves it if I'm not mistaken.
>>16808188Slight typo, forgot to mention that [math] x \in V \cap W [/math] as well (which also should be obvious) in both cases
I just want to make sure I am not fucking up anything here, I think I got how they are getting: [math]\frac{g'(s)}{g(s)}= 1+\frac{1}{2}\ln(\pi)+\frac{1}{2} \gamma[/math][math]g(s)=(s-1)\pi^{-\frac{s}{2}} \Gamma (\frac{s}{2}+1)=(s-1)\pi^{-\frac{s}{2}} \Pi(\frac{s}{2})[/math][math]g'(s)=\pi^{-\frac{s}{2}}\Pi(\frac{s}{2})+(s-1) \cdot -\frac{1}{2}\ln(\pi) \pi^{-\frac{s}{2}} \Pi(\frac{s}{2})+(s-1)\pi^{-\frac{s}{2}}\cdot -\frac{1}{2}\Pi'(\frac{s}{2})[/math][math]\dfrac{g'(x)}{g(x)} =\dfrac{\pi^{-\frac{s}{2}}\Pi(\frac{s}{2})-\frac{1}{2} (s-1) \ln(\pi) \pi^{-\frac{s}{2}} \Pi(\frac{s}{2})-\frac{1}{2}(s-1)\pi^{-\frac{s}{2}}\Pi'(\frac{s}{2})}{(s-1)\pi^{-\frac{s}{2}} \Pi(\frac{s}{2})}[/math][math]=\frac{1}{s-1}-\frac{1}{2}\ln(\pi)-\frac{1}{2}\frac{\Pi'(\frac{s}{2})}{\Pi(\frac{s}{2})}[/math]let s=0[math]-1-\frac{1}{2}\ln(\pi)-\frac{1}{2} \gamma[/math]if I am wrong somewhere just let me know
>>16807827A Lie group is a SMOOTH manifold AND a group. There’s no canonical choice of group structure (or even a smooth atlas for >3 dimensions) for top manifolds.>>16807949Quotients are only well-defined within a single category. You can think of group action as a functor, but then quotients turn out to be ill-defined. It’s better to think of group objects within a particular category and group actions as image factorizations. So strictly speaking your approach isn’t rigorous.
math btfoRe(z) = 1/2 + AI
>Statistical Methods & Data Analysis>Real Analysis>Probability & Stochastic Processes I>Probability & Stochastic Processes II>Monte Carlo Methods>Bayesian Statistics>Probabilistic Graphical Models>Theory of Machine Learning>Stochastic Optimization & Control>Causal InferenceMath pros..... I already have a BS in math. This is my outline for an MS in Math since I have to work for a living and can't do a PhD, and don't want to do a Data Science, Statistics, or Computer Science degree.When you look at this list, do you see "Mathematician" but also "Yeah, he can have a real job in tech/defense sector."?
>>16808538What you said is flawed in multiple ways:1. S^3 does have a canonical Lie group structure as SU(2), with it containing U(1) as a (non-normal) Lie subgroup. (The action of U(1) on S^3 via this construction agrees with action of U(1) on S^3 viewing S^3 as the unit sphere in [math] \mathbb{C}^2 [/math].)2. You can quotient any topological space X by a group action on X, or more generally by any equivalence relation on X. You don't even need categories to do this.Seriously, pick up a book sometime.
>>16808674>can't do a PhDOut of curiosity, why?
>>16808813>S^3 does have a canonical Lie group structure as SU(2)It depends on your definition of "canonical." It would be most accurate to say that [math] S^3 [/math] can be assigned a Lie group structure which is unique up to Lie group isomorphism but which isn't "canonical" in the functorial sense because, say, [math] \text{Diff}(S^3) [/math] acts transitively on [math] S^3 [/math] and so can't distinguish an identity point.
>>16808815Because I have to work for a living. I have children to feed.
>>16809091Euler had 13 children, no PhD, was blind in 1 eye, invented new math while Napoleon's army storming the his city. You have no excuse if you really want this.
> order a book from a local university on the Riemann Zeta function> try to understand the Hadamard Product related to the Riemann Zeta function> Hadamard Product references back to Riemann's original work... in German> The book I have does not elaborate any further "The two paragraphs follow the formula for ξ(s) are the most difficult portion of Riemann's paper"Proof by "this is really complicated and in German" has to beat out "proof is trivial and left as an exercise for the reader."
>>16809039there is no canonical definition of canonical in math
>>16809212One of the better comments on this thread
>>16805285https://marinazahara22.wordpress.com/wp-content/uploads/2013/10/i-n-herstein-topics-in-algebra-2nd-edition-1975-wiley-international-editions-john-wiley-and-sons-wie-1975.pdf
Why does the term magma even exist? All theorems about magmas are just theorems about laws of composition, we don't need to introduce a useless term like magma that is then immediately discarded.
>>16809133>no PhDSo basically you're saying he was also too busy to get a PhD.
>>16808674With the current job market, there's zero reason to do anything probability and statistics. It's all overtaken by ML slop. Either do pure ML courses, or tailor your profile to something non-stochastic: computational geometry, FEA, etc.
>>16809600Lol what is the point of posting this, nobody needs this 50 year old book
A tensor by definition is "diagonalizable".Yeah?It's a 2x2+ grid of values that are independent but have some idea of a product given a permutation of a row and column value.I haven't looked into it very hard but "a tensor is something that transforms like a tensor" seems to how far most people understand what this really is. They all "get it" and they know it's invariant but they don't really know why.
