# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 29th 2021

Page created, but author did not leave any comments.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeMay 9th 2021

added under “Properties” (here) brief statement of and pointer to the hook-content formula.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMay 15th 2021
• (edited May 15th 2021)

I have expanded out the statement of the definition, making everything fully explicit; also defining standard Young diagrams.

Also added a minimum of an Idea-section.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeMay 15th 2021
• (edited May 15th 2021)

In the discussion of the relation to Schur polynomials I have made more explicit (here) what the formula reduces to for Schur polynomials of a finite number of variables

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeMay 19th 2021

• CommentRowNumber6.
• CommentAuthorDavid_Corfield
• CommentTimeMay 29th 2021

Removed a $1/2$ which shouldn’t be there.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJun 1st 2021
• (edited Jun 1st 2021)

Now I have put it all together and summed up (here) the full asymptotic expansion of the number of height-bounded standard Young tableaux. Currently I get this expression:

\begin{aligned} & ln \big( \left\vert sYT_n(N) \right\vert \big) \\ & \; \underset{ { n \to \infty } \atop { N \to \infty } }{\sim}\; n ln(N) - \tfrac{1}{4}N^2 ln(n) + \tfrac{1}{4}N ln(n) \\ & \phantom{ \underset{ { n \to \infty } \atop { N \to \infty } }{\sim}\; } + \tfrac{1}{2} N^2 ln(N) - \left( \tfrac{3}{8} + \tfrac{ln(2)}{4} \right) N^2 - \tfrac{1}{4} N ln(N) + \tfrac{ln(2)}{2} N - \tfrac{1}{6} ln(N) \\ & \; \phantom{ \underset{ { n \to \infty } \atop { N \to \infty } }{\sim}\; } + \big( 2 \zeta^'(-1) - \tfrac{ln(2)}{6} - ln G(1/2) \big) \end{aligned}

I have checked factors a few times, but it won’t hurt to check again… (and thanks for catching a wrong factor in #6!)

• CommentRowNumber8.
• CommentAuthorDavid_Corfield
• CommentTimeJun 1st 2021

Are there issues about taking limits in an order?

Next would be to get an approximation for min-entropy.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeJun 1st 2021
• (edited Jun 1st 2021)

Are there issues about taking limits in an order?

I was wondering about that, too. Maybe this needs more thinking.

Incidentally, I was just turning again to Mkrtchyan 14. Its main Theorem 1.1. (p. 2) says something about the limit $n , N \to \infty$ taken with fixed proportionality, namely such that $n \propto N^2$.

In words, the statement of the theorem is roughly that in this limit the log of the Schur-Weyl probability density becomes “almost certainly approximately constant” , and that this constant scales as $\sqrt{n}$, hence as $N$ – which should mean that in this limit the Shannon entropy of the Schur-Weyl measure is approximately unity times this constant $\sim \sqrt{n} \cdot H_c$.

But the Shannon entropy of the Schur-Weyl measure is a lower bound for our entropy.

Not sure yet if this is good for anything, just saying.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeJun 2nd 2021

In particular, this result (mentioned in #9) seems to imply that in the limit $n \propto N^2 \to \infty$ our Cayley-state probability distribution is

$(\lambda, i_\lambda) \;\mapsto\; \frac{ e^{\sqrt{n} \cdot H_c} }{dim\big(S^{(\lambda)}\big)} \;=\; \frac{ e^{\sqrt{n} \cdot H_c} }{ \left\vert sYTableaux_\lambda \right\vert } \,.$

This would reduce us to computing asymptotic numbers of standard Young tableaux without bound on their height, which is a question with some classical results to it…

• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeJun 3rd 2021

That’s interesting. There’s a sign out, no?

$(\lambda, i_\lambda) \;\mapsto\; \frac{ e^{-\sqrt{n} \cdot H_c} }{ \left\vert sYTableaux_\lambda \right\vert } \,.$

So we have the hook content formula for $\left\vert sYTableaux_\lambda \right\vert$.

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeJun 3rd 2021
• (edited Jun 3rd 2021)

Moreover, if we assume that the Schur-Weyl- and Plancherel-measures are asymptotically indistinguishable (as suggested on p. 3 here, though the way in which the double scaling limit is taken may matter) then we could use in the above formula the asymptotic estimate

$ln \big( \left\vert sYTableaux_\lambda \right\vert \big) \;\sim\; \frac{c}{2}\sqrt{n} - \tfrac{1}{2}ln(n!)$

(from here). Naively plugging in these asymptotic estimates into the formula for the Shannon entropy would then yield the following asymptotic expression for the entropy of the Cayley state:

\begin{aligned} - \underset{ \mathclap{ {\lambda \in Part(n)} \atop { 1 \leq i_\lambda \leq \left\vert sYTableaux_\lambda \right\vert } } }{\sum} p^{Cay}(\lambda,i_\lambda) \cdot ln p^{Cay}(\lambda,i_\lambda) & = - \underset{ \mathclap{ {\lambda \in Part(n)} } }{\sum} p^{SW}(\lambda) \cdot ln \frac{e^{- H\cdot \sqrt{n}}}{ \left\vert sYoungTableaux_\lambda\right\vert } \\ & \underset{n = c N^2 \to \infty}{\sim} \underset{\lambda \in Part(n)}{\sum} e^{- \sqrt{n}\cdot H} \big( (- H - \tfrac{c}{2}) \cdot \sqrt{n} + \tfrac{1}{2} ln(n!) \big) \\ & \;\sim\; e^{\sqrt{n}\cdot (- H + \pi \sqrt{2/3} )} \big( (- H - \tfrac{c}{2}) \cdot \sqrt{n} + \tfrac{1}{2} ln(n!) \big) \end{aligned}

In the last step I used the Hardy-Ramanujan asymptotics for $\left\vert Part(n)\right\vert \sim \pi \sqrt{2n/3 }$ (taken from MO:103070 for the moment).

Not claiming that this is correct, as there is handwaving involved in putting these different asymptotics together this way. But maybe it helps to think along these lines.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeJun 3rd 2021

Oh, I only now see #11. Yes, let me check the sign.

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeJun 3rd 2021

By the way, a joint scaling limit of just this form

$n = c N^2 \to \infty$

$N = \lambda \left(\tfrac{1}{g}\right)^2 \to \infty$

with $\lambda$ the ’t Hooft coupling.

• CommentRowNumber15.
• CommentAuthorDavid_Corfield
• CommentTimeJun 3rd 2021

Unfortunately I’ve got to spend today composing a reply to reviewers of a grant bid, with pretty poor prospects of success. This looks much more interesting.

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeJun 3rd 2021

Sure, no rush. And I need to get back to equivariant bundles.

BTW, I see now that the real asymptotic expansion of the partition function has a $1/n$ factor (now here).

With that included, the expansion of the Cayley entropy in #12 is of the form

$e^{ - a \cdot \sqrt{n} } \big( b + c \cdot n^{-1/2} + d \cdot n^{- 1/2} ln(n) \big) \,.$

If here we use #14 to identify $n \propto g^{-2}$ and think of $g$ as a coupling constant, then this has the form of a non-perturbative contribution to a trans-series expansion of a Feynman series (here):

$e^{- \alpha /g} \big( \beta + \gamma \cdot g + \delta \cdot g ln(g) \big)$
• CommentRowNumber17.
• CommentAuthorDavid_Corfield
• CommentTimeJun 3rd 2021

if we assume that the Schur-Weyl- and Plancherel-measures are asymptotically indistinguishable

I added a couple of papers to Pierre-Loïc Méliot which are relevant. Will need more time to digest.

• CommentRowNumber18.
• CommentAuthorDavid_Corfield
• CommentTimeJun 4th 2021

At the bottom of page 2 of

there is discussion of the relationship between Schur-Weyl- and Plancherel-measures.

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeJun 4th 2021

Could you check if you are pointing to the intended paper? On relation between the two measures this one seems to just mention that both have the same limit shape of Young diagrams for $c = 0$. (Incidentally, from p. 5 in Meliot 11 I take it that they don’t quite have the same limit shape for $c \neq 0$, but the statement there remains vague.)

I had been going by p. 3 of Mkrtchyan 11, which claims that at fixed $n$ the Schur-Weyl measure converges pointwise to the Plancherel measure as $N \to \infty$, with pointer, “for example”, to Olshanski 09 – I had dug that out but still can’t read it, since it is in Russian. But it would be necessary to see the details, since we would need such a convergence to hold not for fixed $n$, but for $n \propto N^2 \to \infty$.

On the other hand, for the argument in #12 to work we’d need less than the two measures to converge – It would be sufficient that they have the same null sets in the limit.

• CommentRowNumber20.
• CommentAuthorDavid_Corfield
• CommentTimeJun 4th 2021

Sorry I got the wrong one. It’s bottom of p. 2 of

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeJun 4th 2021

Thanks. So that’s indeed the kind of discussion as on that p. 5 in Meliot 11 that I had seen.

I suppose the fact that the limit shapes for $c \gt 0$ are different (though close, apparently) for the two measures means that the measures themselves can’t quite converge in the double scaling limit $\sqrt{n} = c N \to \infty$. Maybe they could still be close enough to have the same asymptotic null sets?

By the way, I was vaguely wondering if those limit shapes are the answer to our quest for the maximum of the Schur-Weyl distribution: We had seen that the naive idea that the maximum is at $\lambda = (n)$ receives some correction. But those limit shapes could possibly be thought of as a deformation of $\lambda = (n)$ by adding a small(er) amount of boxes in further rows. This intuition is supported by the graphics on the very top of p. 4 of the article you point to, for what it’s worth.

So I am wondering: Does the Schur-Weyl distribution in the limit $\sqrt{n} = c N \to \infty$ take its maximum on that “deformed LSKV” limit shape? Is that maybe implied by those “central limit theorems” that these authors discuss, saying that the Schur-Weyl measure is a Gaussian fluctuation around that limit shape, of sorts (if I am understanding that right)?