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    • CommentRowNumber101.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 25th 2021

    Are we entering the world of skew tableaux?

    • CommentRowNumber102.
    • CommentAuthorUrs
    • CommentTimeMay 25th 2021
    • (edited May 25th 2021)

    Aha, that’s a plausible possibility!

    Checking this is essentially straightforward, though it requires some work: Using our discussion at Cayley state on the group algebra we have its “density matrix” realization and “just” need to work out a tractable expression for the partial trace of that over the span of a Sym-subgroup inclusion.

    Maybe next week…

    • CommentRowNumber103.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2021

    Finally finding time to look into some of these counting formulas for the number |sYT n(N)|\left\vert sYT_n(N)\right\vert standard Young tableaux with nn boxes and at most NN rows. So coming back to around #87:

    Looking now at the literature, I see that an asymptotic formula for |sYT n(N)|\left\vert sYT_n(N)\right\vert is the main result of Regev 81, whose theorem 2.10 says that:

    |sYT n(N)| n(2π) (N1)/2N N 2/2γ Nn (N1)(N+2)/4N nn (N1)/2x 1x Ni<j(x ix j)D(x 1,,x N)e N|x| 2/2dx 1dx N =(2π) 12(N1)N 12N 2+nn 14(N 2+N)x 1x Ni<j(x ix j)e N|x| 2/2dx 1dx N. \begin{aligned} \left\vert sYT_n(N) \right\vert & \; \overset{ n \to \infty }{\sim} \; \underset{ \gamma_N }{ \underbrace{ (2 \pi)^{ -(N-1)/2 } \cdot N^{ N^2 / 2 } } } \cdot n^{ - (N - 1)(N + 2)/4 } \cdot N^{ n } \cdot n^{ (N - 1)/2 } \cdot \underset{ \mathclap{ x_1 \geq \cdots \geq x_N } }{\int} \;\;\;\;\; \underset{ D(x_1, \cdots, x_N) }{ \underbrace{ \underset{i \lt j}{\prod} \big( x_i - x_j \big) } } e^{ - N \left\vert x\right\vert^2 /2 } d x_1 \cdots d x_{N} \\ & \;=\; (2 \pi)^{ -\tfrac{1}{2}( N - 1 ) } \cdot N^{ \tfrac{1}{2} N^2 + n} \cdot n^{ - \tfrac{1}{4}( N^2 + N ) } \cdot \underset{ \mathclap{ x_1 \geq \cdots \geq x_N } }{\int} \;\;\;\;\; \underset{i \lt j}{\prod} \big( x_i - x_j \big) e^{ - N \left\vert x\right\vert^2 /2 } d x_1 \cdots d x_{N} \end{aligned} \,.

    Here under the braces in the first row I am indicating Regev’s notation which I have expanded out (using the definitions given on the bottom of his p. 3), and in the second line I have collected exponents.

    This would mean that the leading contributions in NN to the max-entropy of the Cayley state in the limit of large nn should be

    S 0(p Cayley)n12N 2ln(N)14N 2ln(n)+𝒪(N) S_0\big(p^{Cayley}\big) \;\;\underset{n \to \infty}{\sim}\;\; \frac{1}{2} N^2 ln(N) - \frac{1}{4}N^2 ln(n) + \mathcal{O}(N)
    • CommentRowNumber104.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2021

    Ah, the last conclusion is not quite correct, as there is NN-dependence also in the integration domain of the last factor. Since |x| 2= i=1 Nx i 2𝒪(N){\vert x \vert}^2 \;=\; \sum_{i = 1}^N x_i^2 \;\sim\; \mathcal{O}(N), this last terms probably gives one more contribution at order N 2N^2 and independent of nn.

    • CommentRowNumber105.
    • CommentAuthorUrs
    • CommentTimeMay 28th 2021
    • (edited May 29th 2021)

    We may compute the order of the integral term in #103 by decomposing it into an integral over a sector of a unit sphere (which gives a constant) times a Gaussian moment:

    0 R N(N1)/2e NR 2/2dR(N 1/2) N(N1)/2=N N(N1)/4 \int_0^{\infty} R^{ N(N-1)/2 } e^{ - N R^2 /2 } \, d R \;\propto\; (N^{-1/2})^{ N(N - 1)/2 } \;=\; N^{ - N(N - 1)/4 }

    Hope I have the factors right.

    Therefore, it looks like we get a max-entropy at large nn of the form

    S 0(p Cayley)na 0+a 2(n)N+a 4(n)N 2+a 4(n)N 2ln(N), S_0(p^{Cayley}) \;\;\underset{n \to \infty}{\sim}\;\; a_0 + a_2(n) \cdot N + a_4(n) \cdot N^2 + a'_4(n) \cdot N^2 ln(N) \,,

    Curiously, this would be the form of the entropy of Yang-Mills theory with NN colors (e.g. (1.1) on p. 3 in arXiv:2105.02101) – the point being that besides plain powers of NN, the leading contribution N 2ln(N)\sim N^2 ln(N) is, in both cases, the second power of NN times the logarithm of NN.

    • CommentRowNumber106.
    • CommentAuthorUrs
    • CommentTimeMay 29th 2021
    • (edited May 29th 2021)

    I’ll stop doing incremental computations here in the comments, and instead put the computation into the entry (here).

    Following through along the above lines, I now get

    |sYT n(N)| nconst(2π) 12(N1)N 14(N 2+N)+12nn 14(N 2N) \begin{aligned} \left\vert sYT_n(N) \right\vert & \; \overset{ n \to \infty }{\sim} \; const \cdot (2 \pi)^{ -\tfrac{1}{2}( N - 1 ) } \cdot N^{ \tfrac{1}{4} (N^2 + N) + \tfrac{1}{2} n} \cdot n^{ - \tfrac{1}{4}( N^2 - N ) } \end{aligned}
    • CommentRowNumber107.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 29th 2021
    • (edited May 29th 2021)

    Adding one more to the list

    a(n) ~ 3 * 5^(n+5)/(8 * Pi *n^5) A049401

    a(n) ~ 3/4 * 6^(n+15/2)/(Pi^(3/2)*n^(15/2)) A007579

    a(n) ~ 45/32 * 7^(n+21/2)/(Pi^(3/2)*n^(21/2)) A007578

    a(n) ~ 135/16 * 8^(n+14)/(Pi^2*n^14) A007580

    So you’d imagine it was

    a(n) ~ constN (n+NN14)/(π c(N)n NN14)const \cdot N^{(n+\frac{N \cdot N-1}{4})}/(\pi^{c(N)}\cdot n^{\frac{N \cdot N-1}{4}})

    So log(a(n))~nlogN+NN14logNNN14lognlog(a(n)) ~ n log N + \frac{N \cdot N-1}{4} log N - \frac{N \cdot N-1}{4} log n.

    Hmm, not quite tallying with yours.

    • CommentRowNumber108.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 29th 2021

    That 12n\tfrac{1}{2} n in your exponent of NN should just be nn. Fixed that on the page.

    Getting closer.

    So why do you have 14(N 2+N)\tfrac{1}{4} (N^2 + N) where I have 14(N 2N)\tfrac{1}{4} (N^2 - N)?

    Is it that my ’const’ is NN-dependent?

    And my powers of π\pi are going up by 1/21/2 every other step. So that will because there’s a gamma function contributing a (π)\sqrt(\pi) for odd arguments.

    • CommentRowNumber109.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 30th 2021
    • (edited May 30th 2021)

    Those estimates in #107 are due to Václav Kotěšovec who conjectures:

    a N(n)N nπ N/2(Nn) N(N1)4 j=1 NΓ(j2). a_N(n) \sim \frac{N^n}{\pi^{N/2}} \cdot \biggl( \frac{N}{n} \biggr)^{\frac{N(N-1)}{4}} \cdot \prod_{j = 1}^N \Gamma \biggl( \frac{j}{2} \biggr) .
    • CommentRowNumber110.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 30th 2021
    • (edited May 30th 2021)

    But F.4.5.1 of the paper by Regev you mention in #105 concerns this value and looks slightly different. He also has a product of gamma function values at half-integers.

    a N(n)N nπ N/2(Nn) N(N1)4 j=1 NΓ(1+j2)2 N1N!. a_N(n) \sim \frac{N^n}{\pi^{N/2}} \cdot \biggl( \frac{N}{n} \biggr)^{\frac{N(N-1)}{4}} \cdot \prod_{j = 1}^N \Gamma \biggl(1+ \frac{j}{2} \biggr) \cdot 2^N \cdot \frac{1}{N!}.

    Might be the same. (The shift in those two gamma terms would make for an extra N!2 N\frac{N!}{2^N}.)

    In any case we’d need to know how the product of gamma functions varies with NN. Perhaps the hint to use the Barnes G function helps. I got to some multiple of N 2lnNN^2 ln N with it.

    • CommentRowNumber111.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021
    • (edited Jun 1st 2021)

    Continuing #109, Kotěšovec writes that

    j=1 NΓ(j2)=G(N/2+1)G(N/2+1/2)G(1/2), \prod_{j = 1}^N \Gamma \biggl( \frac{j}{2} \biggr) = \frac{G(N/2 + 1)G(N/2 + 1/2)}{G(1/2)},

    where GG is the Barnes G-function.

    (14) there suggests that the greatest term in logG(1+z)logG(1+z) is 12z 2logz\frac{1}{2} z^2 log z.

    • CommentRowNumber112.
    • CommentAuthorUrs
    • CommentTimeJun 1st 2021

    Thanks for the replies! And for spotting F.4.5.1 in Regev, I hadn’t seen that.

    So I must have been making a mistake in extracting the powers of NN (=l= l) from Regev’s F.2.10. But I don’t see my mistake yet – this here was my reasoning:

    There are N(N1)/2N(N-1)/2 factors in D=i<j()D = \underset{i \lt j}{\prod} (\cdots). With each factor proportional to RR, these give a global factor of R N(N1)/2R^{N (N-1)/2} in the integrand of a Gaussian integral over R 0R \in \mathbb{R}_{\geq 0} with standard deviation σ=N 1/2\sigma = N^{-1/2}, which hence yields “half” a Gaussian moment (N 1/2) N(N1)/2=N 14(N 2N)\propto (N^{-1/2})^{N(N-1)/2} = N^{-\tfrac{1}{4}(N^2 - N)}. Multiplied with the N 12N 2N^{\tfrac{1}{2}N^2} hidden in Regev’s “γ\gamma”, this yields N 14(N 2+N)N^{ \tfrac{1}{4}( N^2 + N ) }. Or so it seems. (?)

    • CommentRowNumber113.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021

    Well aren’t I seeing (#111) an extra 12(N/2) 2log(N/2)+12((N1)/2) 2log((N1)/2)14N 2logN\sim \frac{1}{2}(N/2)^2 log(N/2) + \frac{1}{2}((N-1)/2)^2 log((N-1)/2) \sim \frac{1}{4} N^2 log N?

    We would need it to be twice this.

    • CommentRowNumber114.
    • CommentAuthorUrs
    • CommentTimeJun 1st 2021

    Oh, I see, right. (On the other hand, what “we need” would be a term 12Nln(N)\tfrac{1}{2} N ln(N), not 12N 2ln(N)\tfrac{1}{2} N^2 ln(N), no?)

    But maybe better to go with Regev’s theorem than with Kotěšovec’s conjecture: To Regev’s product of Gamma functions the Legendre relation applies, which should be helpful.

    • CommentRowNumber115.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021
    • (edited Jun 1st 2021)

    I was going on your

    Multiplied with the N 12N 2N^{\tfrac{1}{2}N^2} hidden in Regev’s “γ\gamma“…

    so adding on 12N 2ln(N)\tfrac{1}{2}N^2 ln(N).

    But maybe better to go with Regev’s theorem than with Kotěšovec’s conjecture

    Regev’s F 4.5.1 is equal to Kotěšovec’s conjecture, right? The former’s product is shifted so we lose a Γ(1/2)\Gamma(1/2) and Γ(1)\Gamma(1) and gain Γ(1+N/2)\Gamma(1+ N/2) and Γ(1+(N1)/2)\Gamma(1 + (N-1)/2). The first two make π\sqrt{\pi}. The latter two are this multiplied by a product of half-integers to N/2N/2 which is N!/2 NN!/2^N.

    • CommentRowNumber116.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021

    The latter fact is just an instance of the Legendre relation you just mentioned, for z=(N+1)/2z = (N+1)/2.

    This product of Γ\Gamma functions at half-integers is a product of these products taken at odd and then even half-integers. That’s why Kotěšovec has them in terms of the Barnes GG-function which is just an extended form of double factorial.

    • CommentRowNumber117.
    • CommentAuthorUrs
    • CommentTimeJun 1st 2021

    I just meant that this “hidden” factor made the N 2ln(N)N^2 ln(N)-term come out as expected, but left the Nln(N)N ln(N)-term with the wrong sign. Anyways, as you observed, all this is besides the point, as there are more N kln(N)N^k ln(N)-term hidden in the “combinatorial prefactors”, which I hadn’t appreciated.

    Regev’s F 4.5.1 is equal to Kotěšovec’s conjecture, right?

    Oh, okay, I hadn’t seen this yet.

    Maybe in Regev’s form the Gauss multiplication formula lends itself more naturally than Barne’s G-function to get all the terms, but I haven’t worked it out yet. Need to go offline now for a bit.