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1. • CommentRowNumber101.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 25th 2021
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexAre we entering the world of [skew tableaux](https://en.wikipedia.org/wiki/Young_tableau#Skew_tableaux)?

   Are we entering the world of skew tableaux?

2. • CommentRowNumber102.
   • CommentAuthorUrs
   • CommentTimeMay 25th 2021
   • (edited May 25th 2021)
   • PermaLink
   Author: Urs
   Format: MarkdownItexAha, 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](https://ncatlab.org/nlab/show/Cayley%20distance%20kernel#CayleyStateOnTheGroupAlgebra) 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...

   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…

3. • CommentRowNumber103.
   • CommentAuthorUrs
   • CommentTimeMay 28th 2021
   • PermaLink
   Author: Urs
   Format: MarkdownItexFinally finding time to look into some of these counting formulas for the number $\left\vert sYT_n(N)\right\vert$ standard Young tableaux with $n$ boxes and at most $N$ rows. So coming back to around [#87](https://nforum.ncatlab.org/discussion/12920/cayley-mixture/?Focus=92750#Comment_92750): Looking now at the literature, I see that an asymptotic formula for $\left\vert sYT_n(N)\right\vert$ is the main result of [Regev 81](https://ncatlab.org/nlab/show/semistandard+Young+tableau#Regev81), whose theorem 2.10 says that: $$ \begin{aligned} \left\vert sYT_n(N) \right\vert & \; \overset{ n \to \infty }{\sim} \; \underset{ \gamma_N }{ \underbrace{ (2 \pi)^{ -(N-1)/2 } \cdot N^{ N^2 / 2 } } } \cdot n^{ - (N - 1)(N + 2)/4 } \cdot N^{ n } \cdot n^{ (N - 1)/2 } \cdot \underset{ \mathclap{ x_1 \geq \cdots \geq x_N } }{\int} \;\;\;\;\; \underset{ D(x_1, \cdots, x_N) }{ \underbrace{ \underset{i \lt j}{\prod} \big( x_i - x_j \big) } } e^{ - N \left\vert x\right\vert^2 /2 } d x_1 \cdots d x_{N} \\ & \;=\; (2 \pi)^{ -\tfrac{1}{2}( N - 1 ) } \cdot N^{ \tfrac{1}{2} N^2 + n} \cdot n^{ - \tfrac{1}{4}( N^2 + N ) } \cdot \underset{ \mathclap{ x_1 \geq \cdots \geq x_N } }{\int} \;\;\;\;\; \underset{i \lt j}{\prod} \big( x_i - x_j \big) e^{ - N \left\vert x\right\vert^2 /2 } d x_1 \cdots d x_{N} \end{aligned} \,. $$ 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 $N$ to the max-entropy of the Cayley state in the limit of large $n$ should be $$ S_0\big(p^{Cayley}\big) \;\;\underset{n \to \infty}{\sim}\;\; \frac{1}{2} N^2 ln(N) - \frac{1}{4}N^2 ln(n) + \mathcal{O}(N) $$

   Finally finding time to look into some of these counting formulas for the number standard Young tableaux with boxes and at most rows. So coming back to around #87:

   Looking now at the literature, I see that an asymptotic formula for is the main result of Regev 81, whose theorem 2.10 says that:

   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 to the max-entropy of the Cayley state in the limit of large should be

4. • CommentRowNumber104.
   • CommentAuthorUrs
   • CommentTimeMay 28th 2021
   • PermaLink
   Author: Urs
   Format: MarkdownItexAh, the last conclusion is not quite correct, as there is $N$-dependence also in the integration domain of the last factor. Since ${\vert x \vert}^2 \;=\; \sum_{i = 1}^N x_i^2 \;\sim\; \mathcal{O}(N)$, this last terms probably gives one more contribution at order $N^2$ and independent of $n$.

   Ah, the last conclusion is not quite correct, as there is -dependence also in the integration domain of the last factor. Since , this last terms probably gives one more contribution at order and independent of .

5. • CommentRowNumber105.
   • CommentAuthorUrs
   • CommentTimeMay 28th 2021
   • (edited May 29th 2021)
   • PermaLink
   Author: Urs
   Format: MarkdownItexWe 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: $$ \int_0^{\infty} R^{ N(N-1)/2 } e^{ - N R^2 /2 } \, d R \;\propto\; (N^{-1/2})^{ N(N - 1)/2 } \;=\; N^{ - N(N - 1)/4 } $$ Hope I have the factors right. Therefore, it looks like we get a max-entropy at large $n$ of the form $$ S_0(p^{Cayley}) \;\;\underset{n \to \infty}{\sim}\;\; a_0 + a_2(n) \cdot N + a_4(n) \cdot N^2 + a'_4(n) \cdot N^2 ln(N) \,, $$ Curiously, this would be the form of the entropy of Yang-Mills theory with $N$ colors (e.g. [(1.1) on p. 3](https://arxiv.org/pdf/2105.02101.pdf#page=3) in [arXiv:2105.02101](https://arxiv.org/abs/2105.02101)) -- the point being that besides plain powers of $N$, the leading contribution $\sim N^2 ln(N)$ is, in both cases, the second power of $N$ times the logarithm of $N$.

   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:

   Hope I have the factors right.

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

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

6. • CommentRowNumber106.
   • CommentAuthorUrs
   • CommentTimeMay 29th 2021
   • (edited May 29th 2021)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI'll stop doing incremental computations here in the comments, and instead put the computation into the entry ([here](https://ncatlab.org/nlab/show/semistandard+Young+tableau#AsymptoticFormulaForNumberOfSYTWIthBoundedHeight)). Following through along the above lines, I now get $$ \begin{aligned} \left\vert sYT_n(N) \right\vert & \; \overset{ n \to \infty }{\sim} \; const \cdot (2 \pi)^{ -\tfrac{1}{2}( N - 1 ) } \cdot N^{ \tfrac{1}{4} (N^2 + N) + \tfrac{1}{2} n} \cdot n^{ - \tfrac{1}{4}( N^2 - N ) } \end{aligned} $$

   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

7. • CommentRowNumber107.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 29th 2021
   • (edited May 29th 2021)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexAdding one more to the list > a(n) ~ 3 * 5^(n+5)/(8 * Pi *n^5) [A049401](https://oeis.org/A049401) > a(n) ~ 3/4 * 6^(n+15/2)/(Pi^(3/2)*n^(15/2)) [A007579](https://oeis.org/A007579) > a(n) ~ 45/32 * 7^(n+21/2)/(Pi^(3/2)*n^(21/2)) [A007578](https://oeis.org/A007578) > a(n) ~ 135/16 * 8^(n+14)/(Pi^2*n^14) [A007580](https://oeis.org/A007580) So you'd imagine it was > a(n) ~ $const \cdot N^{(n+\frac{N \cdot N-1}{4})}/(\pi^{c(N)}\cdot n^{\frac{N \cdot N-1}{4}})$ So $log(a(n)) ~ n log N + \frac{N \cdot N-1}{4} log N - \frac{N \cdot N-1}{4} log n$. Hmm, not quite tallying with yours.

   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) ~

   So .

   Hmm, not quite tallying with yours.

8. • CommentRowNumber108.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 29th 2021
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexThat $\tfrac{1}{2} n$ in your exponent of $N$ should just be $n$. Fixed that on the page. Getting closer. So why do you have $\tfrac{1}{4} (N^2 + N)$ where I have $\tfrac{1}{4} (N^2 - N)$? Is it that my 'const' is $N$-dependent? And my powers of $\pi$ are going up by $1/2$ every other step. So that will because there's a gamma function contributing a $\sqrt(\pi)$ for odd arguments.

   That in your exponent of should just be . Fixed that on the page.

   Getting closer.

   So why do you have where I have ?

   Is it that my ’const’ is -dependent?

   And my powers of are going up by every other step. So that will because there’s a gamma function contributing a for odd arguments.

9. • CommentRowNumber109.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 30th 2021
   • (edited May 30th 2021)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexThose estimates in #107 are due to Václav Kotěšovec who [conjectures](http://www.kotesovec.cz/math_articles/kotesovec_young_tableaux_conjecture.pdf): $$ a_N(n) \sim \frac{N^n}{\pi^{N/2}} \cdot \biggl( \frac{N}{n} \biggr)^{\frac{N(N-1)}{4}} \cdot \prod_{j = 1}^N \Gamma \biggl( \frac{j}{2} \biggr) . $$

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

10. • CommentRowNumber110.
    • CommentAuthorDavid_Corfield
    • CommentTimeMay 30th 2021
    • (edited May 30th 2021)
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexBut 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. $$ a_N(n) \sim \frac{N^n}{\pi^{N/2}} \cdot \biggl( \frac{N}{n} \biggr)^{\frac{N(N-1)}{4}} \cdot \prod_{j = 1}^N \Gamma \biggl(1+ \frac{j}{2} \biggr) \cdot 2^N \cdot \frac{1}{N!}. $$ Might be the same. (The shift in those two gamma terms would make for an extra $\frac{N!}{2^N}$.) In any case we'd need to know how the product of gamma functions varies with $N$. Perhaps the hint to use the Barnes G function helps. I got to some multiple of $N^2 ln N$ with it.

    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.

    Might be the same. (The shift in those two gamma terms would make for an extra .)

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

11. • CommentRowNumber111.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021
    • (edited Jun 1st 2021)
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexContinuing #109, Kotěšovec writes that $$ \prod_{j = 1}^N \Gamma \biggl( \frac{j}{2} \biggr) = \frac{G(N/2 + 1)G(N/2 + 1/2)}{G(1/2)}, $$ where $G$ is the [Barnes G-function](https://mathworld.wolfram.com/BarnesG-Function.html). (14) there suggests that the greatest term in $logG(1+z)$ is $\frac{1}{2} z^2 log z$.

    Continuing #109, Kotěšovec writes that

    where is the Barnes G-function.

    (14) there suggests that the greatest term in is .

12. • CommentRowNumber112.
    • CommentAuthorUrs
    • CommentTimeJun 1st 2021
    • PermaLink
    Author: Urs
    Format: MarkdownItexThanks 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 $N$ ($= l$) from Regev's F.2.10. But I don't see my mistake yet -- this here was my reasoning: There are $N(N-1)/2$ factors in $D = \underset{i \lt j}{\prod} (\cdots)$. With each factor proportional to $R$, these give a global factor of $R^{N (N-1)/2}$ in the integrand of a Gaussian integral over $R \in \mathbb{R}_{\geq 0}$ with standard deviation $\sigma = N^{-1/2}$, which hence yields "half" a Gaussian moment $\propto (N^{-1/2})^{N(N-1)/2} = N^{-\tfrac{1}{4}(N^2 - N)}$. Multiplied with the $N^{\tfrac{1}{2}N^2}$ hidden in Regev's "$\gamma$", this yields $N^{ \tfrac{1}{4}( N^2 + N ) }$. Or so it seems. (?)

    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 () from Regev’s F.2.10. But I don’t see my mistake yet – this here was my reasoning:

    There are factors in . With each factor proportional to , these give a global factor of in the integrand of a Gaussian integral over with standard deviation , which hence yields “half” a Gaussian moment . Multiplied with the hidden in Regev’s “”, this yields . Or so it seems. (?)

13. • CommentRowNumber113.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexWell aren't I seeing (#111) an extra $\sim \frac{1}{2}(N/2)^2 log(N/2) + \frac{1}{2}((N-1)/2)^2 log((N-1)/2) \sim \frac{1}{4} N^2 log N$? We would need it to be twice this.

    Well aren’t I seeing (#111) an extra ?

    We would need it to be twice this.

14. • CommentRowNumber114.
    • CommentAuthorUrs
    • CommentTimeJun 1st 2021
    • PermaLink
    Author: Urs
    Format: MarkdownItexOh, I see, right. (On the other hand, what "we need" would be a term $\tfrac{1}{2} N ln(N)$, not $\tfrac{1}{2} N^2 ln(N)$, 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](https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function) applies, which should be helpful.

    Oh, I see, right. (On the other hand, what “we need” would be a term , not , 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.

15. • CommentRowNumber115.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021
    • (edited Jun 1st 2021)
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexI was going on your > Multiplied with the $N^{\tfrac{1}{2}N^2}$ hidden in Regev's "$\gamma$"... so adding on $\tfrac{1}{2}N^2 ln(N)$. > 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 $\Gamma(1/2)$ and $\Gamma(1)$ and gain $\Gamma(1+ N/2)$ and $\Gamma(1 + (N-1)/2)$. The first two make $\sqrt{\pi}$. The latter two are this multiplied by a product of half-integers to $N/2$ which is $N!/2^N$.

    I was going on your

    Multiplied with the hidden in Regev’s ““…

    so adding on .

    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 and and gain and . The first two make . The latter two are this multiplied by a product of half-integers to which is .

16. • CommentRowNumber116.
    • CommentAuthorDavid_Corfield
    • CommentTimeJun 1st 2021
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexThe latter fact is just an instance of the Legendre relation you just mentioned, for $z = (N+1)/2$. This product of $\Gamma$ 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 $G$-function which is just an extended form of double factorial.

    The latter fact is just an instance of the Legendre relation you just mentioned, for .

    This product of 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 -function which is just an extended form of double factorial.

17. • CommentRowNumber117.
    • CommentAuthorUrs
    • CommentTimeJun 1st 2021
    • PermaLink
    Author: Urs
    Format: MarkdownItexI just meant that this "hidden" factor made the $N^2 ln(N)$-term come out as expected, but left the $N ln(N)$-term with the wrong sign. Anyways, as you observed, all this is besides the point, as there are more $N^k ln(N)$-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.

    I just meant that this “hidden” factor made the -term come out as expected, but left the -term with the wrong sign. Anyways, as you observed, all this is besides the point, as there are more -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.

18. • CommentRowNumber118.
    • CommentAuthorDavid_Corfield
    • CommentTimeAug 11th 2022
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexVague idea: I was wondering whether the 'Lorentzian' qualifier in [[Lorentzian polynomial]] should point us to something physics-related, when I recalled this thread from last year. Since we were looking at probability distributions over Young tableaux, it's interesting to see that Lorentzian polynomials (which are used to establish log-concavity of sequences, $a_k^2 \geq a_{k+1}a_{k-1}$, and so unimodality of sequences, as explained [here](https://mattbaker.blog/2019/08/30/lorentzian-polynomials/)) are cropping up in this area, as in * [[June Huh]], Jacob P. Matherne, Karola Mészáros, Avery St. Dizier, _Logarithmic concavity of Schur and related polynomials_ ([arXiv:1906.09633](https://arxiv.org/abs/1906.09633)) They're touching on some physics directly in * Petter Brändén, [[June Huh]], _Hodge-Riemann relations for Potts model partition functions_ ([arXiv:1811.01696](https://arxiv.org/abs/1811.01696))

    Vague idea: I was wondering whether the ’Lorentzian’ qualifier in Lorentzian polynomial should point us to something physics-related, when I recalled this thread from last year.

    Since we were looking at probability distributions over Young tableaux, it’s interesting to see that Lorentzian polynomials (which are used to establish log-concavity of sequences, , and so unimodality of sequences, as explained here) are cropping up in this area, as in

    • June Huh, Jacob P. Matherne, Karola Mészáros, Avery St. Dizier, Logarithmic concavity of Schur and related polynomials (arXiv:1906.09633)

    They’re touching on some physics directly in

    • Petter Brändén, June Huh, Hodge-Riemann relations for Potts model partition functions (arXiv:1811.01696)