The simplicity of Lieb's constant (4/3)^(3/2) has long made me dream of an elementary combinatorial proof, say at the same level of technicality as the entropy of domino tilings of the square lattce (Jim has a lovely review paper about this, which I cribbed from for my book). But the 6-vertex ice model and the Bethe ansatz seems totally different from the permanent-determinant trick we can use for planar perfect matchings (dominos, rhombi, etc). To put it differently, there seem to be at least two reasons why a stat mech model can be "exactly solvable", and they seem incomparable to each other. Cris On Feb 24, 2014, at 1:25 PM, James Propp <jamespropp@gmail.com> wrote:
I should say that I've never read Lieb's paper. I did once try to read Rodney Baxter's proof of Lieb's result, but there was one place (an appeal to a broad principle called the "Bethe ansatz" that in some contexts is a theorem and in others is merely a universally-believed conjecture) where I couldn't supply the missing details.
That was twenty years ago.
At some point I intend to try to find out (probably via MathOverflow) whether anyone's created a write-up that fills in all the details. My gloomy guess is that the answer is "no": all the people who'd be qualified to write such an article have other projects that they're more excited about.
I'd be happy to be wrong about this!
Jim Propp
On Mon, Feb 24, 2014 at 1:58 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 2/24/14, James Propp <jamespropp@gmail.com> wrote:
The constant c is exactly 4/3 to the power of 3/2, or 1.5396007... See the Wikipedia article en.wikipedia.org/wiki/Lieb's_square_ice_constant
To see what square ice has to do with 3-colorings, see http://jamespropp.org/faces.pdf
Jim Propp
c = (4/3)^(3/2) -- gobsmacked!
Amost as astonishing is that my extrapolation turned out to be accurate to all four places of decimals given: some compensation for having my lovingly tended borders so brutally uprooted ...
Notice that the `proper' grid 3-colourings described in Jim's excellently readable survey article do not correspond to bordered Hardin arrays; what recurrences do their generalisation to m x n rectangles satisfy?
My assumption that it was necessary to consider rectangles seems to have been somewhat off the mark; though the Elliot Lieb paper does utilise the same matrices, there called `transfer' matrices. Apparently proper colourings of a n x n square array are enumerated by the explicit formula
1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)! ;
what is striking about this is that it has a form totally different from the explicit exponential sums for T(m, n) etc., for fixed m and variable n .
So how does all this stuff relate to Victor Miller's variant, avoiding monochrome 2 x 2 square factors? For fixed m the same sort of recurrence analysis is applicable, yielding polynomials with even more outrageous degrees, still mysteriously irreducible with real roots.
The counting function, V(m, n) say, has subadditive logarithm, yielding progressive upper bounds; and we can border with alternating boundary to get lower bounds -- but this time the interior turns out to be unrestricted, so there is no need to consider a second function.
For fixed m and varying n we have constants a_m, b_m such that
V(m, n) = a_m (b_m)^n + (smaller terms) ;
and progressive two-sided bounds for (the new) c
(b_m)^( 1/(m+1) ) < c < (b_m)^( 1/m ) ,
yielding at m = 18
1.754240 < c < 1.809880 .
Extrapolating from these bounds suggest c ~ 1.8020 .
Is this also a simple algebraic number? And is there also a nice factorial formula for square Miller arrays?
Short table of V(m, n) , counting m x n Miller arrays (unrestricted):
m\n 1 2 3 4 5 6 7
1 2 4 8 16 32 64 128 2 4 14 50 178 634 2258 8042 3 8 50 322 2066 13262 85126 546410 4 16 178 2066 23858 275690 3185462 36806846 5 32 634 13262 275690 5735478 119310334 2481942354 6 64 2258 85126 3185462 119310334 4468252414 167341334542 7 128 8042 546410 36806846 2481942354 167341334542 11282914491066
(Maybe these ought to be halved, by fixing the top left assignment?)
Fred Lunnon
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