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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 $\left\vert sYT_n(N)\right\vert$ standard Young tableaux with $n$ boxes and at most $N$ rows. So coming back to around #87:

   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, 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

   $$S_0\big(p^{Cayley}\big) \;\;\underset{n \to \infty}{\sim}\;\; \frac{1}{2} N^2 ln(N) - \frac{1}{4}N^2 ln(n) + \mathcal{O}(N)$$
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 $N$-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 $n$.

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:

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

   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 in arXiv: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 $N$.

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

   $$\begin{aligned} \left\vert sYT_n(N) \right\vert & \; \overset{ n \to \infty }{\sim} \; const \cdot (2 \pi)^{ -\tfrac{1}{2}( N - 1 ) } \cdot N^{ \tfrac{1}{4} (N^2 + N) + \tfrac{1}{2} n} \cdot n^{ - \tfrac{1}{4}( N^2 - N ) } \end{aligned}$$
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) ~ $const \cdot N^{(n+\frac{N \cdot N-1}{4})}/(\pi^{c(N)}\cdot n^{\frac{N \cdot N-1}{4}})$

   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.

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 $\tfrac{1}{2} n$ 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.

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:

   $$a_N(n) \sim \frac{N^n}{\pi^{N/2}} \cdot \biggl( \frac{N}{n} \biggr)^{\frac{N(N-1)}{4}} \cdot \prod_{j = 1}^N \Gamma \biggl( \frac{j}{2} \biggr) .$$
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.

    $$a_N(n) \sim \frac{N^n}{\pi^{N/2}} \cdot \biggl( \frac{N}{n} \biggr)^{\frac{N(N-1)}{4}} \cdot \prod_{j = 1}^N \Gamma \biggl(1+ \frac{j}{2} \biggr) \cdot 2^N \cdot \frac{1}{N!}.$$

    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.

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

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

    where $G$ is the Barnes G-function.

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

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 $N$ ($= l$) 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. (?)

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 $\sim \frac{1}{2}(N/2)^2 log(N/2) + \frac{1}{2}((N-1)/2)^2 log((N-1)/2) \sim \frac{1}{4} N^2 log N$?

    We would need it to be twice this.

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 $\tfrac{1}{2} N ln(N)$, 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 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 $N^{\tfrac{1}{2}N^2}$ 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 $N!/2^N$.

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 $z = (N+1)/2$.

    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.

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 $N^2 ln(N)$-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.

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, $a_k^2 \geq a_{k+1}a_{k-1}$, 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)