[math-fun] Hardin 'nuff for you?
Hardin 'nuff for you? ********************* As a result of a serendipitous misunderstanding, I have recently derived much enjoyment from investigating this topic. In an effort to bring its puzzle aspects to a wider audience, convince everybody what a brilliant mathematician I am, and delay the looming necessity to write the results up properly, I have prepared the following sketch. Solutions to (some of) the numbered problems may be forthcoming should suspicion subsequently arise that anybody else gives a stuff. A `Hardin array' is defined to be a binary assignment to the nodes of a rectangular m x n grid, such that every 2 x 2 square has some diagonal given equal values; that is, no square is permitted to belong to the set C C C D D C D D D D C D D C C C [Circumventing proportional spacing propels the aliases C = 0 , D = 1 .] I am interested in evaluating T(m, n) = half the number of distinct m x n Hardin arrays, and in estimating its asymptotic behaviour for large m, n . It is not difficult to establish that there is a constant c such that as m, n -> oo , in a rough sense T(m, n) approximates c^(m n) -- rather more precisely, log T(m, n) ~ m n log c . Extrapolation via Aiken delta^2, Wynn rho, suggests (optimistically) that maybe c = 1.5396... . Easy combinatorial constructions establish the rigorous elementary bounds sqrt2 = 1.414213... < c < 1.618034... = tau . ((1)) For more refined bounds which approach the actual value, develop linear recurrences for small fixed m via classical Markov process analysis, then apply Fekete subadditivity (and stop dropping names). For example at m = 5 , we find that f(n) = T(m, n) satisfies f(n) - 16 f(n-1) + 65 f(n-2) - 92 f(n-3) + 48 f(n-4) - 8 f(n-5) = 0 , whence numerically f(n) = (1.39119...) (10.6829...)^n + (smaller terms ...) . Since any m x n Hardin array can be decomposed into roughly m/5 of these arrays, we deduce the upper bound c < 1.6060 = (10.6829)^(1/5) , improving on the earlier elementary bound. Applying similar methods at m = 20 yields tighter bounds 1.504113 < c < 1.555473 . ((2)) The polynomials associated with the recurrences apparently have unexpected properties, the least bizarre being irreducibility; all roots real; all roots positive (upper). ((3)) I'm not going to show you more of these polynomials here. At m = 17 say, they have degree 2123 , with some components over 1300 decimal digits! Establishing lower bounds involves instead a subset of Hardin arrays: those `bordered' with C along all four edges, and counted by S(m, n) . The border acts as cement, allowing m x n bordered arrays to be tiled together (borders overlapping), so building larger arrays of size (k m - k + 1) x (l n - l + 1) . A key stage in the recurrence construction involves characterising those distinct binary columns requiring consideration, leading to the following `accessibility' problem: Develop an algorithm to decide, in time and space linear in m , whether a given m-vector can occur as a column in a bordered m x n Hardin array (for sufficiently large n ). ((4)) For example with m = 15 columns CDDDDDDDDDDDDDC and CDDCCDDCDDCCDDC are accessible (via columns bordered by C...C) from an initial CCCCCCCCCCCCCCC ; however CDDCCCCCCCCCCCC is not. This puzzle has a wonderfully compact and elegant solution. Once you get it right, after the fourth or fifth fumble, that is. But having finally cracked it, count the number a(m) of accessible vectors, compared to the total number b(m) = 2^(m-2) of bordered vectors: m 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... a(m) 1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, ... b(m) 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, ... [I blush to confess that I went searching for a(m) in OEIS ; even then, I couldn't at first get my head around what it kept trying to tell me.] But amazingly, these seem to be simply the central binomial coefficients (Conjecture) a(m) = (m-1)_C_[(m-1)/2] ?! ((5)) though I have at present no idea why. Anyway on this assumption, Stirling's approximation yields a welcome reduction factor a(m)/b(m) ~ 1.595770 / sqrt(m) in the number of columns required -- a saving which squares up in space, a(m) representing the order of a rather hefty matrix. I mentioned previously the (presumed) conjecture -- mentioned [??] cryptically under OEIS A078099 -- to the effect that Hardin arrays correspond to grid-colourings of the same-size using three colours: here P, Q, R . Note that nodes sharing an edge are assigned distinct colours; and the top-left node of both grids remains constant, C and P resp. Specify a bijective mapping between m x n Hardin arrays and grid three- colourings. ((6)) While this is trickier than might be expected, there awaits again a solution both short and sweet. Appendix:: sample 9 x 9 Hardin array, together with corresponding grid 3-colouring: C D C C D C C C C P R P Q R Q P Q P D C C C C C D C D R P Q P Q P R P R C C D C C C C D D P Q R Q P Q P R Q C D D D C D C C D Q R P R Q R Q P R D D C D D C D C C R P Q P R Q R Q P ; D D D C D D D D C P R P Q P R P R Q C D D D D D D C C Q P R P R P R Q P D D D D C D C C C P R P R Q R Q P Q D D D D D C C D C R P R P R Q P R P and a bordered 8 x 8 Hardin array, illustrating some of 35 distinct accessible columns (or rows): C C C C C C C C C C D C D C C C C C C D C C C C C C D D D C D C C D D D C C C C C C D D D C C C C C C D C C C C C C C C C C C C Short tables -- best viewed sans pesky proportional spacing -- of T(m, n) , counting m x n Hardin arrays with C at top-left: m\n 1 2 3 4 5 6 7 1 1 2 4 8 16 32 64 2 2 6 18 54 162 486 1458 3 4 18 82 374 1706 7782 35498 4 8 54 374 2604 18150 126534 882180 5 16 162 1706 18150 193662 2068146 22091514 6 32 486 7782 126534 2068146 33865632 554916090 7 64 1458 35498 882180 22091514 554916090 13956665236 and S(m, n) , counting m x n Hardin arrays with C around border: m\n 1 2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 3 1 1 2 3 5 8 13 21 34 4 1 1 3 7 17 41 99 239 577 5 1 1 5 17 64 235 870 3211 11867 6 1 1 8 41 235 1322 7479 42267 238958 7 1 1 13 99 870 7479 64914 562213 4876632 8 1 1 21 239 3211 42267 562213 7474305 99480399 9 1 1 34 577 11867 238958 4876632 99480399 2033739170 Fred Lunnon Maynooth 22/02/14
Fred, very nice stuff. This reminds me o something that I worked on many years ago briefly discussed in A133130: Number of 0/1 colorings of an n X n square for which no 2 by 2 subsquare is monochromatic. I didn't mention it in the comments, but the matrices involved had a nice recursive structure which,upon proper scaling, converged to a fractal whose Hausdorff dimension was related to the rate of growth. Victor On Sat, Feb 22, 2014 at 8:10 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
mathematician I am, and delay the looming necessity to write the results
Victor, I'm unsure whether you meant this for the math-fun list? But anyway ... Your A133130 problem is indeed intimately similar to Hardin's, and I'm most interested to know more about how far you got with it: presumably you actually had to involve more general m x n rectangles? Do your recurrence polynomials have any striking properties? One instance: an extraordinary feature of the associated (monic) Hardin polynomial concerns its constant coefficient for m odd, which apparently equals (+/-) 2^k , where k equals the degree of the polynomial at m-1 ; while the constant coefficient of the bordered variant equals (+/-) 1 . [ m = 5 earlier has final coefficient -8 , while m = 4 has a cubic polynomial.] Please tell a little more about the intriguing relation you mentioned between the fractal dimension d of the adjacency matrix, and the rate of growth c^(m n) of the counting function. For the Hardin case, I find that d = Log(3)/Log(2) = 1.5849625 ... which does indeed look as if it might act as an upper bound on c ~ 1.54 ; but the mechanism behind such a connection remains a mystery to me. Fred Lunnon On 2/23/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, very nice stuff. This reminds me o something that I worked on many years ago briefly discussed in A133130: Number of 0/1 colorings of an n X n square for which no 2 by 2 subsquare is monochromatic. I didn't mention it in the comments, but the matrices involved had a nice recursive structure which,upon proper scaling, converged to a fractal whose Hausdorff dimension was related to the rate of growth.
Victor
On Sat, Feb 22, 2014 at 8:10 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fred, OEIS led me to the following recent paper of Kitaev and Remmel which looks relevant. I haven't had a chance to look at it yet. Victor http://arxiv.org/abs/1304.4286 PS. Sorry about the terse replies. I had an accident two weeks ago which damaged my shoulder and makes it hard for me to type very much. On Sun, Feb 23, 2014 at 12:05 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Victor,
I'm unsure whether you meant this for the math-fun list? But anyway ...
Your A133130 problem is indeed intimately similar to Hardin's, and I'm most interested to know more about how far you got with it: presumably you actually had to involve more general m x n rectangles?
Do your recurrence polynomials have any striking properties? One instance: an extraordinary feature of the associated (monic) Hardin polynomial concerns its constant coefficient for m odd, which apparently equals (+/-) 2^k , where k equals the degree of the polynomial at m-1 ; while the constant coefficient of the bordered variant equals (+/-) 1 . [ m = 5 earlier has final coefficient -8 , while m = 4 has a cubic polynomial.]
Please tell a little more about the intriguing relation you mentioned between the fractal dimension d of the adjacency matrix, and the rate of growth c^(m n) of the counting function. For the Hardin case, I find that
d = Log(3)/Log(2) = 1.5849625 ...
which does indeed look as if it might act as an upper bound on c ~ 1.54 ; but the mechanism behind such a connection remains a mystery to me.
Fred Lunnon
On 2/23/14, Victor Miller <victorsmiller@gmail.com> wrote:
Fred, very nice stuff. This reminds me o something that I worked on many years ago briefly discussed in A133130: Number of 0/1 colorings of an n X n square for which no 2 by 2 subsquare is monochromatic. I didn't mention it in the comments, but the matrices involved had a nice recursive structure which,upon proper scaling, converged to a fractal whose Hausdorff dimension was related to the rate of growth.
Victor
On Sat, Feb 22, 2014 at 8:10 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
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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 On Sunday, February 23, 2014, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Victor,
I'm unsure whether you meant this for the math-fun list? But anyway ...
Your A133130 problem is indeed intimately similar to Hardin's, and I'm most interested to know more about how far you got with it: presumably you actually had to involve more general m x n rectangles?
Do your recurrence polynomials have any striking properties? One instance: an extraordinary feature of the associated (monic) Hardin polynomial concerns its constant coefficient for m odd, which apparently equals (+/-) 2^k , where k equals the degree of the polynomial at m-1 ; while the constant coefficient of the bordered variant equals (+/-) 1 . [ m = 5 earlier has final coefficient -8 , while m = 4 has a cubic polynomial.]
Please tell a little more about the intriguing relation you mentioned between the fractal dimension d of the adjacency matrix, and the rate of growth c^(m n) of the counting function. For the Hardin case, I find that
d = Log(3)/Log(2) = 1.5849625 ...
which does indeed look as if it might act as an upper bound on c ~ 1.54 ; but the mechanism behind such a connection remains a mystery to me.
Fred Lunnon
On 2/23/14, Victor Miller <victorsmiller@gmail.com <javascript:;>> wrote:
Fred, very nice stuff. This reminds me o something that I worked on many years ago briefly discussed in A133130: Number of 0/1 colorings of an n X n square for which no 2 by 2 subsquare is monochromatic. I didn't mention it in the comments, but the matrices involved had a nice recursive structure which,upon proper scaling, converged to a fractal whose Hausdorff dimension was related to the rate of growth.
Victor
On Sat, Feb 22, 2014 at 8:10 PM, Fred Lunnon <fred.lunnon@gmail.com<javascript:;>> wrote:
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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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Mmm... I have essayed a cursory inspection of Lieb (1967). The first caveat is that it is upholstered with technical jargon that only another math. phys. specialist (which I surely have no claim to be) might be expected to follow. Nonetheless, I have come away with a strong impression of what has and has not been achieved. Bear in mind that a physicist's or engineer's concept of what constitutes "proof" is in general rather different from what a mathematician understands by the term --- not unreasonably, given that any intended application necessarily relies on intuitive judgement based on approximate experiment. Another difficulty is that the notation actually varies between sections: eg. "N" in section I corresponds to "N^2" in section II . More seriously, what is intended by the term "subspace" appears to have little connection with its customary meaning. And Jim Propp has already drawn attention to a lack of clarity concerning the invocation of the "Bethe ansatz". I want to try to establish the importance of such complaints. The proof depends on investigating the `transfer' matrix, that is the adjacency matrix of a graph, having 2^m vertices representing columns in an m x n Hardin array (grid 3-colouring, etc. etc.), with edges between columns which can validly lie adjacent. Under a suitable change of basis this matrix decomposes over the rationals into a number of smaller blocks: the corresponding subtotals, into which T(m, n) is partitioned by final column, form sequences as n varies which lie in a subspace associated with (remarkably) just one of these blocks. The first stage is to show that the block concerned is (some) one on which T(m, n) depends -- or to put it another way, that in the explicit formula as a sum of exponentials in n , the maximal eigenvalue gets a nonzero coefficient. Despite the plethora of wave-mechanical legerdemain that occupies sections II and III, I can find no evidence that the author has actually nailed this; indeed, at the start of section IV he seems to come close to admitting that it remains unsubstantiated. The second stage is to determine the maximum eigenvalue of the matrix, to which section IV devotes an impressive analytical manipulation which can quite possibly be made transparent, and in any case fairly certainly has managed to arrive at the correct answer: c = (4/3)^(3/2) . But since (presumably) we now have to hand the Mills-Zeilberger formula with asymptotic behaviour equivalent to T(n, n) , f(n) = 1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)! surely somebody whose calculus skills are less atrophied than my own can just roll up sleeves, take the logarithm, approximate the sums by integrals and deduce that f(n) ~ c^(n^2) ? Fred Lunnon On 2/24/14, Cris Moore <moore@santafe.edu> wrote:
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
I think f(n) ~ sqrt(27/16)^(n^2), which is not the Lieb constant. Charles Greathouse Analyst/Programmer Case Western Reserve University On Tue, Feb 25, 2014 at 11:04 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Mmm... I have essayed a cursory inspection of Lieb (1967).
The first caveat is that it is upholstered with technical jargon that only another math. phys. specialist (which I surely have no claim to be) might be expected to follow. Nonetheless, I have come away with a strong impression of what has and has not been achieved. Bear in mind that a physicist's or engineer's concept of what constitutes "proof" is in general rather different from what a mathematician understands by the term --- not unreasonably, given that any intended application necessarily relies on intuitive judgement based on approximate experiment.
Another difficulty is that the notation actually varies between sections: eg. "N" in section I corresponds to "N^2" in section II . More seriously, what is intended by the term "subspace" appears to have little connection with its customary meaning. And Jim Propp has already drawn attention to a lack of clarity concerning the invocation of the "Bethe ansatz". I want to try to establish the importance of such complaints.
The proof depends on investigating the `transfer' matrix, that is the adjacency matrix of a graph, having 2^m vertices representing columns in an m x n Hardin array (grid 3-colouring, etc. etc.), with edges between columns which can validly lie adjacent. Under a suitable change of basis this matrix decomposes over the rationals into a number of smaller blocks: the corresponding subtotals, into which T(m, n) is partitioned by final column, form sequences as n varies which lie in a subspace associated with (remarkably) just one of these blocks.
The first stage is to show that the block concerned is (some) one on which T(m, n) depends -- or to put it another way, that in the explicit formula as a sum of exponentials in n , the maximal eigenvalue gets a nonzero coefficient. Despite the plethora of wave-mechanical legerdemain that occupies sections II and III, I can find no evidence that the author has actually nailed this; indeed, at the start of section IV he seems to come close to admitting that it remains unsubstantiated.
The second stage is to determine the maximum eigenvalue of the matrix, to which section IV devotes an impressive analytical manipulation which can quite possibly be made transparent, and in any case fairly certainly has managed to arrive at the correct answer: c = (4/3)^(3/2) .
But since (presumably) we now have to hand the Mills-Zeilberger formula with asymptotic behaviour equivalent to T(n, n) ,
f(n) = 1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)!
surely somebody whose calculus skills are less atrophied than my own can just roll up sleeves, take the logarithm, approximate the sums by integrals and deduce that f(n) ~ c^(n^2) ?
Fred Lunnon
On 2/24/14, Cris Moore <moore@santafe.edu> wrote:
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
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Lieb's solution concerns the entropy of square ice with NO boundary conditions, or equivalently, with maximum entropy boundary conditions; I believe that maximum entropy boundary conditions in this case means that the arrows along each side of the square alternate in-out-in-out-... (This may even be a theorem: Scott Sheffield would know.) ASM's (alternating-sign matrices), as studied by Mills, Robbins, Rumsey, Zeilberger, Kuperberg, and many others, correspond to a very special subclass of states of the square ice model: those with "domain-wall boundary conditions", with all arrows pointing inward along the left and right sides of the square and outward along the top and bottom sides of the square. This subclass does not have full entropy; its growth rate is governed by a strictly smaller constant. Jim Propp On Tuesday, February 25, 2014, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Mmm... I have essayed a cursory inspection of Lieb (1967).
The first caveat is that it is upholstered with technical jargon that only another math. phys. specialist (which I surely have no claim to be) might be expected to follow. Nonetheless, I have come away with a strong impression of what has and has not been achieved. Bear in mind that a physicist's or engineer's concept of what constitutes "proof" is in general rather different from what a mathematician understands by the term --- not unreasonably, given that any intended application necessarily relies on intuitive judgement based on approximate experiment.
Another difficulty is that the notation actually varies between sections: eg. "N" in section I corresponds to "N^2" in section II . More seriously, what is intended by the term "subspace" appears to have little connection with its customary meaning. And Jim Propp has already drawn attention to a lack of clarity concerning the invocation of the "Bethe ansatz". I want to try to establish the importance of such complaints.
The proof depends on investigating the `transfer' matrix, that is the adjacency matrix of a graph, having 2^m vertices representing columns in an m x n Hardin array (grid 3-colouring, etc. etc.), with edges between columns which can validly lie adjacent. Under a suitable change of basis this matrix decomposes over the rationals into a number of smaller blocks: the corresponding subtotals, into which T(m, n) is partitioned by final column, form sequences as n varies which lie in a subspace associated with (remarkably) just one of these blocks.
The first stage is to show that the block concerned is (some) one on which T(m, n) depends -- or to put it another way, that in the explicit formula as a sum of exponentials in n , the maximal eigenvalue gets a nonzero coefficient. Despite the plethora of wave-mechanical legerdemain that occupies sections II and III, I can find no evidence that the author has actually nailed this; indeed, at the start of section IV he seems to come close to admitting that it remains unsubstantiated.
The second stage is to determine the maximum eigenvalue of the matrix, to which section IV devotes an impressive analytical manipulation which can quite possibly be made transparent, and in any case fairly certainly has managed to arrive at the correct answer: c = (4/3)^(3/2) .
But since (presumably) we now have to hand the Mills-Zeilberger formula with asymptotic behaviour equivalent to T(n, n) ,
f(n) = 1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)!
surely somebody whose calculus skills are less atrophied than my own can just roll up sleeves, take the logarithm, approximate the sums by integrals and deduce that f(n) ~ c^(n^2) ?
Fred Lunnon
On 2/24/14, Cris Moore <moore@santafe.edu <javascript:;>> wrote:
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<javascript:;>> 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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Exercise 0: Find a formula for the number of states of the n-by-n square ice model (on a generalized tix-tac-toe board with n horizontal lines and n vertical lines) with all arrows pointing inward along the left, right, top, and bottom sides of the square. Exercise 1: Find a formula for the number of states of the n-by-n square ice model with all arrows pointing inward along the left and bottom sides of the square and all arrows pointing outward along the right and top sides of the square. This should enhance your appreciation of the important role played by boundary conditions! Jim Propp On Tuesday, February 25, 2014, James Propp <jamespropp@gmail.com> wrote:
Lieb's solution concerns the entropy of square ice with NO boundary conditions, or equivalently, with maximum entropy boundary conditions; I believe that maximum entropy boundary conditions in this case means that the arrows along each side of the square alternate in-out-in-out-... (This may even be a theorem: Scott Sheffield would know.)
ASM's (alternating-sign matrices), as studied by Mills, Robbins, Rumsey, Zeilberger, Kuperberg, and many others, correspond to a very special subclass of states of the square ice model: those with "domain-wall boundary conditions", with all arrows pointing inward along the left and right sides of the square and outward along the top and bottom sides of the square. This subclass does not have full entropy; its growth rate is governed by a strictly smaller constant.
Jim Propp
On Tuesday, February 25, 2014, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Mmm... I have essayed a cursory inspection of Lieb (1967).
The first caveat is that it is upholstered with technical jargon that only another math. phys. specialist (which I surely have no claim to be) might be expected to follow. Nonetheless, I have come away with a strong impression of what has and has not been achieved. Bear in mind that a physicist's or engineer's concept of what constitutes "proof" is in general rather different from what a mathematician understands by the term --- not unreasonably, given that any intended application necessarily relies on intuitive judgement based on approximate experiment.
Another difficulty is that the notation actually varies between sections: eg. "N" in section I corresponds to "N^2" in section II . More seriously, what is intended by the term "subspace" appears to have little connection with its customary meaning. And Jim Propp has already drawn attention to a lack of clarity concerning the invocation of the "Bethe ansatz". I want to try to establish the importance of such complaints.
The proof depends on investigating the `transfer' matrix, that is the adjacency matrix of a graph, having 2^m vertices representing columns in an m x n Hardin array (grid 3-colouring, etc. etc.), with edges between columns which can validly lie adjacent. Under a suitable change of basis this matrix decomposes over the rationals into a number of smaller blocks: the corresponding subtotals, into which T(m, n) is partitioned by final column, form sequences as n varies which lie in a subspace associated with (remarkably) just one of these blocks.
The first stage is to show that the block concerned is (some) one on which T(m, n) depends -- or to put it another way, that in the explicit formula as a sum of exponentials in n , the maximal eigenvalue gets a nonzero coefficient. Despite the plethora of wave-mechanical legerdemain that occupies sections II and III, I can find no evidence that the author has actually nailed this; indeed, at the start of section IV he seems to come close to admitting that it remains unsubstantiated.
The second stage is to determine the maximum eigenvalue of the matrix, to which section IV devotes an impressive analytical manipulation which can quite possibly be made transparent, and in any case fairly certainly has managed to arrive at the correct answer: c = (4/3)^(3/2) .
But since (presumably) we now have to hand the Mills-Zeilberger formula with asymptotic behaviour equivalent to T(n, n) ,
f(n) = 1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)!
surely somebody whose calculus skills are less atrophied than my own can just roll up sleeves, take the logarithm, approximate the sums by integrals and deduce that f(n) ~ c^(n^2) ?
Fred Lunnon
On 2/24/14, Cris Moore <moore@santafe.edu> wrote:
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 pr
Is exercise 0 intended as a joke? If so, I get it. Heh-heh. On Tue, Feb 25, 2014 at 11:59 AM, James Propp <jamespropp@gmail.com> wrote:
Exercise 0: Find a formula for the number of states of the n-by-n square ice model (on a generalized tix-tac-toe board with n horizontal lines and n vertical lines) with all arrows pointing inward along the left, right, top, and bottom sides of the square.
Exercise 1: Find a formula for the number of states of the n-by-n square ice model with all arrows pointing inward along the left and bottom sides of the square and all arrows pointing outward along the right and top sides of the square.
This should enhance your appreciation of the important role played by boundary conditions!
Jim Propp
On Tuesday, February 25, 2014, James Propp <jamespropp@gmail.com> wrote:
Lieb's solution concerns the entropy of square ice with NO boundary conditions, or equivalently, with maximum entropy boundary conditions; I believe that maximum entropy boundary conditions in this case means that the arrows along each side of the square alternate in-out-in-out-... (This may even be a theorem: Scott Sheffield would know.)
ASM's (alternating-sign matrices), as studied by Mills, Robbins, Rumsey, Zeilberger, Kuperberg, and many others, correspond to a very special subclass of states of the square ice model: those with "domain-wall boundary conditions", with all arrows pointing inward along the left and right sides of the square and outward along the top and bottom sides of the square. This subclass does not have full entropy; its growth rate is governed by a strictly smaller constant.
Jim Propp
On Tuesday, February 25, 2014, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Mmm... I have essayed a cursory inspection of Lieb (1967).
The first caveat is that it is upholstered with technical jargon that only another math. phys. specialist (which I surely have no claim to be) might be expected to follow. Nonetheless, I have come away with a strong impression of what has and has not been achieved. Bear in mind that a physicist's or engineer's concept of what constitutes "proof" is in general rather different from what a mathematician understands by the term --- not unreasonably, given that any intended application necessarily relies on intuitive judgement based on approximate experiment.
Another difficulty is that the notation actually varies between sections: eg. "N" in section I corresponds to "N^2" in section II . More seriously, what is intended by the term "subspace" appears to have little connection with its customary meaning. And Jim Propp has already drawn attention to a lack of clarity concerning the invocation of the "Bethe ansatz". I want to try to establish the importance of such complaints.
The proof depends on investigating the `transfer' matrix, that is the adjacency matrix of a graph, having 2^m vertices representing columns in an m x n Hardin array (grid 3-colouring, etc. etc.), with edges between columns which can validly lie adjacent. Under a suitable change of basis this matrix decomposes over the rationals into a number of smaller blocks: the corresponding subtotals, into which T(m, n) is partitioned by final column, form sequences as n varies which lie in a subspace associated with (remarkably) just one of these blocks.
The first stage is to show that the block concerned is (some) one on which T(m, n) depends -- or to put it another way, that in the explicit formula as a sum of exponentials in n , the maximal eigenvalue gets a nonzero coefficient. Despite the plethora of wave-mechanical legerdemain that occupies sections II and III, I can find no evidence that the author has actually nailed this; indeed, at the start of section IV he seems to come close to admitting that it remains unsubstantiated.
The second stage is to determine the maximum eigenvalue of the matrix, to which section IV devotes an impressive analytical manipulation which can quite possibly be made transparent, and in any case fairly certainly has managed to arrive at the correct answer: c = (4/3)^(3/2) .
But since (presumably) we now have to hand the Mills-Zeilberger formula with asymptotic behaviour equivalent to T(n, n) ,
f(n) = 1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)!
surely somebody whose calculus skills are less atrophied than my own can just roll up sleeves, take the logarithm, approximate the sums by integrals and deduce that f(n) ~ c^(n^2) ?
Fred Lunnon
On 2/24/14, Cris Moore <moore@santafe.edu> wrote:
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 pr
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I had carelessly assumed that the boundary didn't affect the constant here --- thanks! WFL Charles Greathouse <charles.greathouse@case.edu> Tue, Feb 25, 2014 at 4:33 PM To: math-fun <math-fun@mailman.xmission.com> I think f(n) ~ sqrt(27/16)^(n^2), which is not the Lieb constant. Charles Greathouse James Propp <jamespropp@gmail.com> Tue, Feb 25, 2014 at 4:50 PM To: math-fun <math-fun@mailman.xmission.com> Lieb's solution concerns the entropy of square ice with NO boundary conditions, or equivalently, with maximum entropy boundary conditions; I believe that maximum entropy boundary conditions in this case means that the arrows along each side of the square alternate in-out-in-out-... (This may even be a theorem: Scott Sheffield would know.) ASM's (alternating-sign matrices), as studied by Mills, Robbins, Rumsey, Zeilberger, Kuperberg, and many others, correspond to a very special subclass of states of the square ice model: those with "domain-wall boundary conditions", with all arrows pointing inward along the left and right sides of the square and outward along the top and bottom sides of the square. This subclass does not have full entropy; its growth rate is governed by a strictly smaller constant. Exercise 0: Find a formula for the number of states of the n-by-n square ice model (on a generalized tix-tac-toe board with n horizontal lines and n vertical lines) with all arrows pointing inward along the left, right, top, and bottom sides of the square. Exercise 1: Find a formula for the number of states of the n-by-n square ice model with all arrows pointing inward along the left and bottom sides of the square and all arrows pointing outward along the right and top sides of the square. This should enhance your appreciation of the important role played by boundary conditions! Jim Propp On 2/25/14, Allan Wechsler <acwacw@gmail.com> wrote:
Is exercise 0 intended as a joke? If so, I get it. Heh-heh.
On Tue, Feb 25, 2014 at 11:59 AM, James Propp <jamespropp@gmail.com> wrote:
Exercise 0: Find a formula for the number of states of the n-by-n square ice model (on a generalized tix-tac-toe board with n horizontal lines and n vertical lines) with all arrows pointing inward along the left, right, top, and bottom sides of the square.
Exercise 1: Find a formula for the number of states of the n-by-n square ice model with all arrows pointing inward along the left and bottom sides of the square and all arrows pointing outward along the right and top sides of the square.
This should enhance your appreciation of the important role played by boundary conditions!
Jim Propp
On Tuesday, February 25, 2014, James Propp <jamespropp@gmail.com> wrote:
Lieb's solution concerns the entropy of square ice with NO boundary conditions, or equivalently, with maximum entropy boundary conditions; I believe that maximum entropy boundary conditions in this case means that the arrows along each side of the square alternate in-out-in-out-... (This may even be a theorem: Scott Sheffield would know.)
ASM's (alternating-sign matrices), as studied by Mills, Robbins, Rumsey, Zeilberger, Kuperberg, and many others, correspond to a very special subclass of states of the square ice model: those with "domain-wall boundary conditions", with all arrows pointing inward along the left and right sides of the square and outward along the top and bottom sides of the square. This subclass does not have full entropy; its growth rate is governed by a strictly smaller constant.
Jim Propp
On Tuesday, February 25, 2014, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Mmm... I have essayed a cursory inspection of Lieb (1967).
The first caveat is that it is upholstered with technical jargon that only another math. phys. specialist (which I surely have no claim to be) might be expected to follow. Nonetheless, I have come away with a strong impression of what has and has not been achieved. Bear in mind that a physicist's or engineer's concept of what constitutes "proof" is in general rather different from what a mathematician understands by the term --- not unreasonably, given that any intended application necessarily relies on intuitive judgement based on approximate experiment.
Another difficulty is that the notation actually varies between sections: eg. "N" in section I corresponds to "N^2" in section II . More seriously, what is intended by the term "subspace" appears to have little connection with its customary meaning. And Jim Propp has already drawn attention to a lack of clarity concerning the invocation of the "Bethe ansatz". I want to try to establish the importance of such complaints.
The proof depends on investigating the `transfer' matrix, that is the adjacency matrix of a graph, having 2^m vertices representing columns in an m x n Hardin array (grid 3-colouring, etc. etc.), with edges between columns which can validly lie adjacent. Under a suitable change of basis this matrix decomposes over the rationals into a number of smaller blocks: the corresponding subtotals, into which T(m, n) is partitioned by final column, form sequences as n varies which lie in a subspace associated with (remarkably) just one of these blocks.
The first stage is to show that the block concerned is (some) one on which T(m, n) depends -- or to put it another way, that in the explicit formula as a sum of exponentials in n , the maximal eigenvalue gets a nonzero coefficient. Despite the plethora of wave-mechanical legerdemain that occupies sections II and III, I can find no evidence that the author has actually nailed this; indeed, at the start of section IV he seems to come close to admitting that it remains unsubstantiated.
The second stage is to determine the maximum eigenvalue of the matrix, to which section IV devotes an impressive analytical manipulation which can quite possibly be made transparent, and in any case fairly certainly has managed to arrive at the correct answer: c = (4/3)^(3/2) .
But since (presumably) we now have to hand the Mills-Zeilberger formula with asymptotic behaviour equivalent to T(n, n) ,
f(n) = 1! 4! ... (3n-2)! / n! (n+1)! ... (2n-1)!
surely somebody whose calculus skills are less atrophied than my own can just roll up sleeves, take the logarithm, approximate the sums by integrals and deduce that f(n) ~ c^(n^2) ?
Fred Lunnon
On 2/24/14, Cris Moore <moore@santafe.edu> wrote:
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 pr
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (6)
-
Allan Wechsler -
Charles Greathouse -
Cris Moore -
Fred Lunnon -
James Propp -
Victor Miller