harmonic functions and the chromatic polynomial R. Kenyon (Brown) based on joint work with A. Abrams, W. Lam
The chromatic polynomial with n colors. G(n) of a graph G is the number of proper colorings (adjacent vertices have di erent colors) (n) =n(n 1)(n 2) satisfies a contraction-deletion rule: G(n) = G e (n) G/e(n) but is #P -hard to compute in general.
A graph G =(V,E) The Dirichlet problem c : E! R >0 the edge conductances 1 B V boundary vertices u : B! R boundary values 4 3 2 Find f : V! R harmonic on V \ B and f B = u. 6 5 0= f(x) = X y x c e (f(x) f(y)) 0 f is the unique function with f B = u minimizing the Dirichlet energy E(f) = X e=xy c e (f(x) f(y)) 2 { edge energy
A harmonic function induces a compatible orientation: an acyclic orientation with no internal sources or sinks, and no oriented paths from lower boundary values to higher boundary values. current flows downhill 3 2 1 9 6 8 7 4 5 10 12 11 We let be the set of compatible orientations How many are there?
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Fixed energy problem:
Can we adjust edge conductances so that all bulbs burn with the same brightness?
0<latexit sha1_base64="ezpm10dbcmtonytiar5aneq3koq=">aaab53icbzc7sgnbfibpxlumt6ilzwaqrmkukdrgazawcbgmkcxhdni2gtm7u8zmcihkcwwsvgx9ft/azrdxcik08yebj/8/hznnhkng2rjut5nbwv1b38hvfra2d3b3ivsh9zrjfeofjsjrzzbqffyib7gr2ewv0jgu2aghn5o88yhk80temwgkqux7kkecuwotutspltyyoxvzbm8opevpsqukalvo8avdtvgwozrmuk1bnpuayesv4uzgundonkaudwgpwxyljveho+mgy3jins6jemwfngtq/u4y0vjryrzaypiavl7mjuz/wssz0uuw4jlndeo2+yjkbdejmwxnulwhm2jogtlf7aye9amiznjbfowrvmwvl8e/k1+wvbpbql7bthk4gmm4bq/ooqq3uamfgca8wqu8og/os/pmvm9kc8685xd+ypn4asfnjig=</latexit> sha1_base64="zgsmmecjd8rqitsuq0dd70lehqy=">aaab43icbzc/tsmwemyv5v8j/wori0wfxfqlmaalvgjhlbkhldqoctxra+o4ke0gvvgfgiubwhky3ocnt8fno0dlj1n66fvu5lulusg18bxvp7syura+ud50t7z3dvcq7v6dtjlfmgcjsfqrohoflxgybgs2uou0jgq2o9hnng8+odi8kfdmngiy04hkfc6osdad161uvzpxicydp4fq9edzoua38txpjsyluromqnzt30tnmfnlobm4ctuzxpsyer1g26kkmeowlwadkgpr9eg/ufzjqwr3d0doy63hcwqry2qgejgbmv9l7cz0l8kcyzqzknnso34mienidgvs4wqzewmllcluzyvssbvlxt7gtufwf1dehuc0dlnzq/urmkkmh3aej+ddodthfhoqaaoez3ifn+frexhez4ulz95xah/kfpwaasqnjw==</latexit> <latexit sha1_base64="zgsmmecjd8rqitsuq0dd70lehqy=">aaab43icbzc/tsmwemyv5v8j/wori0wfxfqlmaalvgjhlbkhldqoctxra+o4ke0gvvgfgiubwhky3ocnt8fno0dlj1n66fvu5lulusg18bxvp7syura+ud50t7z3dvcq7v6dtjlfmgcjsfqrohoflxgybgs2uou0jgq2o9hnng8+odi8kfdmngiy04hkfc6osdad161uvzpxicydp4fq9edzoua38txpjsyluromqnzt30tnmfnlobm4ctuzxpsyer1g26kkmeowlwadkgpr9eg/ufzjqwr3d0doy63hcwqry2qgejgbmv9l7cz0l8kcyzqzknnso34mienidgvs4wqzewmllcluzyvssbvlxt7gtufwf1dehuc0dlnzq/urmkkmh3aej+ddodthfhoqaaoez3ifn+frexhez4ulz95xah/kfpwaasqnjw==</latexit> sha1_base64="kzkc+j1bzcesn8pjn8nngu772bg=">aaab43icbvbns8naej3urxq/qlcvi0xwvbiv2osuvhisygyhdwwznbrrn5uwuxfk6c/w4kg9+p+8+w/ctjlo64obx3szzmylmsg18bxvp7kxubw9u9119/ypdo9q7vgjtnpfmgcpsfu3ohoflxgybgr2m4u0iqr2osnt3o88o9i8lq9mmmgy0jhkmwfuwoneg9tqxsnbgkwtvyr1knee1l76w5tlcurdbnw653uzcquqdgccz24/15hrnqej7fkqayi6lbahzsi5vyyktputachc/t1r0etrarlzzosasv715uj/xi838xvycjnlbivblopzquxk5l+tivfijjhaqpni9lbcxlrrzmw2rg3bx315nqsxjwbdr7duyiyqcapncae+xeel7qanatbaeie3eheenffny9lyccqje/gd5/mhfrwlww==</latexit> 1<latexit sha1_base64="nmmvvjje9oqjlvtems385xabkrk=">aaab53icbzc7sgnbfibpxlumt6ilzwaqrmkukdrgazawcbgmkcxhdni2gtm7u8zmcihkcwwsvgx9ft/azrdxcik08yebj/8/hznnhkng2rjut5nbwv1b38hvfra2d3b3ivsh9zrjfeofjsjrzzbqffyib7gr2ewv0jgu2aghn5o88yhk80temwgkqux7kkecuwotutcpltyyoxvzbm8opevpsqukalvo8avdtvgwozrmuk1bnpuayesv4uzgundonkaudwgpwxyljveho+mgy3jins6jemwfngtq/u4y0vjryrzaypiavl7mjuz/wssz0uuw4jlndeo2+yjkbdejmwxnulwhm2jogtlf7aye9amiznjbfowrvmwvl8e/k1+wvbpbql7bthk4gmm4bq/ooqq3uamfgca8wqu8og/os/pmvm9kc8685xd+ypn4aslqjik=</latexit> sha1_base64="agxkianpjqdec6rrdpwur13gjuw=">aaab53icbzdpsgmxemzn/vvrv6phl8eiecq7elavwvdisqxxftqlznpznjabxzksueqfwishfa8+i2/gzbcx3fagrr8efnzfdjmzmbvcg9f9dpawv1bx1gsbxc2t7z3d0t7+vu4yxdbniuhum6qabzfog24enlofna4fnslbzsrvpklspjf3zphienoe5bfn1fir7nvkzbfi5ikl4m2gfp15lqvwkx21uwnlypsgcap1y3nte4yompwjhbfbmcausghtycuipdhqyjqpoibh1umskfh2suny93fhimzad+pqvsbu9pv8njh/y1qzis6cezdpzlcy6udrjohjygrr0uukmrfdc5qpbmclre8vzcbepmip4m2vvaj+aewy4txdcvukpirairzbcxhwdlw4hrr4wadhcv7g1xlwnp03531auutmeg7gj5yph91mjru=</latexit> <latexit sha1_base64="agxkianpjqdec6rrdpwur13gjuw=">aaab53icbzdpsgmxemzn/vvrv6phl8eiecq7elavwvdisqxxftqlznpznjabxzksueqfwishfa8+i2/gzbcx3fagrr8efnzfdjmzmbvcg9f9dpawv1bx1gsbxc2t7z3d0t7+vu4yxdbniuhum6qabzfog24enlofna4fnslbzsrvpklspjf3zphienoe5bfn1fir7nvkzbfi5ikl4m2gfp15lqvwkx21uwnlypsgcap1y3nte4yompwjhbfbmcausghtycuipdhqyjqpoibh1umskfh2suny93fhimzad+pqvsbu9pv8njh/y1qzis6cezdpzlcy6udrjohjygrr0uukmrfdc5qpbmclre8vzcbepmip4m2vvaj+aewy4txdcvukpirairzbcxhwdlw4hrr4wadhcv7g1xlwnp03531auutmeg7gj5yph91mjru=</latexit> sha1_base64="wfkrejigk4eg0alix3fhbcbxw4w=">aaab53icbvbns8naej3ur1q/qh69lbbbu0m8qbcpephygrgfnptndtku3wzc7kyoob/aiwcvr/4lb/4bt20o2vpg4pheddpzwlrwbvz32ymtrw9sbpw3kzu7e/sh1cojb51kiqhpepgotkg1ci7rn9wi7kqkarwkbifj25nffkkleslvzstfikzdyspoqlfsy+txa27dnyoseq8gnsjq7fe/eooeztfkwwtvuuu5qqlyqgxnaqevxqyxpwxmh9i1vniydzdpd52sm6smsjqow9kqufp7iqex1pm4tj0xnso97m3e/7xuzqkriocyzqxktlguzykyhmy+jgoukbkxsyqyxe2thi2ooszybco2bg/55vxix9sv617lrtvuijtkcaknca4exeid7qajpjbaeizxehmenrfn3flytjacyuyy/sd5/ahmx4yb</latexit> Let :(0, 1) E! [0, 1) E be the map from conductances to energies. Example a d c b e (a, b, c, d, e) = ( a(bd + cd + ce + de) 2 (ab + ac + ae + bc + bd + cd + ce + de) 2, b(ae + cd + ce + de) 2 (ab + ac + ae + bc + bd + cd + ce + de) 2, c(bd ae) 2 (ab + ac + ae + bc + bd + cd + ce + de) 2, d(ab + ac + ae + bc) 2 (ab + ac + ae + bc + bd + cd + ce + de) 2, e(ab + ac + bc + bd) 2 (ab + ac + ae + bc + bd + cd + ce + de) 2 )
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[0, 1) E be the map from conductances to energies. Theorem 1: For any 2 and {E e > 0} there is a unique choice of conductances {c e } for which the associated harmonic function realizes this data. Thm 2: The harmonic functions h with energies {E e } are the solutions to the system of equations 8x 2 V \ B, 0 = X y x h(x) E e h(y) exactly <latexit sha1_base64="yauk7eho0fjhq+asejs7yaqahko=">aaab+3icbvdlssnafj3uv62vajdubovgqitdqbspuhfzwdhcg8pkctmontyymygh1f9x40lfrt/izr9xkmahrqcgdufcx9zjjzxjzvnfrm1tfwnzq77d2nnd2z8wd4/uzzwkcg6neswghphawqsoyordibfaqo9d35tdf37/ayrkcxsnsgtckewifjbkljbgznnuzsgf+hmmj4qqno3nltw2subvylekhsr0xubxyi9pgkkkkcdsdm0ruw5ohgkuw7wxsiukhm7ibiaariqe6ebl4jk+1yqpg1jofylcqr87chjkmywergyjmsplrxd/84apci7cnevjqicii0vbyrgkczee9pma4lxncbvm/xxtkre6ap1xq4dgl5+8spxo+7jt33za3asqjto6rifodnnohhxrdeohb1guowf0it6mj+pfedc+fqu1o+ppoj8wpn8aujgu/w==</latexit> sha1_base64="yauk7eho0fjhq+asejs7yaqahko=">aaab+3icbvdlssnafj3uv62vajdubovgqitdqbspuhfzwdhcg8pkctmontyymygh1f9x40lfrt/izr9xkmahrqcgdufcx9zjjzxjzvnfrm1tfwnzq77d2nnd2z8wd4/uzzwkcg6neswghphawqsoyordibfaqo9d35tdf37/ayrkcxsnsgtckewifjbkljbgznnuzsgf+hmmj4qqno3nltw2subvylekhsr0xubxyi9pgkkkkcdsdm0ruw5ohgkuw7wxsiukhm7ibiaariqe6ebl4jk+1yqpg1jofylcqr87chjkmywergyjmsplrxd/84apci7cnevjqicii0vbyrgkczee9pma4lxncbvm/xxtkre6ap1xq4dgl5+8spxo+7jt33za3asqjto6rifodnnohhxrdeohb1guowf0it6mj+pfedc+fqu1o+ppoj8wpn8aujgu/w==</latexit> the enharmonic equation energy harmonic Thm 3: The number of solutions N = satisfies the contraction-deletion rule N G = N G e + N G/e.
Cor: Let G k be obtained from G by adding k vertices as boundary, attached to all vertices of G. Then for any choice of k distinct boundary values, the number of compatible orientations is G(1 k). p 2 e G
Lemma: The Jacobian determinant of has the form det J = Y (h(x) h(y)) 2. e=xy Corollary: <latexit sha1_base64="0zn0fht8e/pb18lvalc8clrnlnm=">aaab+hicbvbns8naej3ur1q/oh69lbbbu0l68emghv48vjc20jay2w7apzvdslsphnb/4swdild/ijf/jds2b219mpb4b4azewhcmtae9+2unja3tnfku5w9/ypdi/f45enlvbeaemml6oryu84edqwznhysrxecctooj825355spzkujyzlad/gi8eirrcx0sb1814yoazuknosstvzwk16nw8bte78glshqgvgfvwgkqqxfyzwrhxx9xltz7eyjha6q/rstrnmjnheu5ykhfpdzxexz9cfvyyoksqwmgih/p7icax1foe2m8zmrfe9ufif101ndn3pmuhsqwvzlopsjoxe8xjqkclkdm8swuqxeysiy6wwmtasig3bx315nqt12k3nf6hxg3dfgmu4g3o4bb+uoah30iiacezhgv7hzcmdf+fd+vi2lpxi5ht+wpn8acmkk0y=</latexit> <latexit sha1_base64="0zn0fht8e/pb18lvalc8clrnlnm=">aaab+hicbvbns8naej3ur1q/oh69lbbbu0l68emghv48vjc20jay2w7apzvdslsphnb/4swdild/ijf/jds2b219mpb4b4azewhcmtae9+2unja3tnfku5w9/ypdi/f45enlvbeaemml6oryu84edqwznhysrxecctooj825355spzkujyzlad/gi8eirrcx0sb1814yoazuknosstvzwk16nw8bte78glshqgvgfvwgkqqxfyzwrhxx9xltz7eyjha6q/rstrnmjnheu5ykhfpdzxexz9cfvyyoksqwmgih/p7icax1foe2m8zmrfe9ufif101ndn3pmuhsqwvzlopsjoxe8xjqkclkdm8swuqxeysiy6wwmtasig3bx315nqt12k3nf6hxg3dfgmu4g3o4bb+uoah30iiacezhgv7hzcmdf+fd+vi2lpxi5ht+wpn8acmkk0y=</latexit> <latexit sha1_base64="0zn0fht8e/pb18lvalc8clrnlnm=">aaab+hicbvbns8naej3ur1q/oh69lbbbu0l68emghv48vjc20jay2w7apzvdslsphnb/4swdild/ijf/jds2b219mpb4b4azewhcmtae9+2unja3tnfku5w9/ypdi/f45enlvbeaemml6oryu84edqwznhysrxecctooj825355spzkujyzlad/gi8eirrcx0sb1814yoazuknosstvzwk16nw8bte78glshqgvgfvwgkqqxfyzwrhxx9xltz7eyjha6q/rstrnmjnheu5ykhfpdzxexz9cfvyyoksqwmgih/p7icax1foe2m8zmrfe9ufif101ndn3pmuhsqwvzlopsjoxe8xjqkclkdm8swuqxeysiy6wwmtasig3bx315nqt12k3nf6hxg3dfgmu4g3o4bb+uoah30iiacezhgv7hzcmdf+fd+vi2lpxi5ht+wpn8acmkk0y=</latexit> <latexit sha1_base64="0zn0fht8e/pb18lvalc8clrnlnm=">aaab+hicbvbns8naej3ur1q/oh69lbbbu0l68emghv48vjc20jay2w7apzvdslsphnb/4swdild/ijf/jds2b219mpb4b4azewhcmtae9+2unja3tnfku5w9/ypdi/f45enlvbeaemml6oryu84edqwznhysrxecctooj825355spzkujyzlad/gi8eirrcx0sb1814yoazuknosstvzwk16nw8bte78glshqgvgfvwgkqqxfyzwrhxx9xltz7eyjha6q/rstrnmjnheu5ykhfpdzxexz9cfvyyoksqwmgih/p7icax1foe2m8zmrfe9ufif101ndn3pmuhsqwvzlopsjoxe8xjqkclkdm8swuqxeysiy6wwmtasig3bx315nqt12k3nf6hxg3dfgmu4g3o4bb+uoah30iiacezhgv7hzcmdf+fd+vi2lpxi5ht+wpn8acmkk0y=</latexit> <latexit sha1_base64="0zn0fht8e/pb18lvalc8clrnlnm=">aaab+hicbvbns8naej3ur1q/oh69lbbbu0l68emghv48vjc20jay2w7apzvdslsphnb/4swdild/ijf/jds2b219mpb4b4azewhcmtae9+2unja3tnfku5w9/ypdi/f45enlvbeaemml6oryu84edqwznhysrxecctooj825355spzkujyzlad/gi8eirrcx0sb1814yoazuknosstvzwk16nw8bte78glshqgvgfvwgkqqxfyzwrhxx9xltz7eyjha6q/rstrnmjnheu5ykhfpdzxexz9cfvyyoksqwmgih/p7icax1foe2m8zmrfe9ufif101ndn3pmuhsqwvzlopsjoxe8xjqkclkdm8swuqxeysiy6wwmtasig3bx315nqt12k3nf6hxg3dfgmu4g3o4bb+uoah30iiacezhgv7hzcmdf+fd+vi2lpxi5ht+wpn8acmkk0y=</latexit> (1 k) = 1 m Z m Y e=xy (h(x) h(y)) 2 Z dvol where the integral is over the m-simplex m of conductances summing to 1, and Z = P e=xy c xy(h(x) h(y)) 2.
<latexit sha1_base64="uggo4plvsfj3qrbyh33c+ge3dyk=">aaacgnicbvbntwixeo36ififevtssezwqna5qbdd4swbmiiqachdmkbdt920sxpc+b9e/ctepkjxzrz4b+wcbwvf0ut1vzlp54wxfbz9/9tbwl5zxvvpbgq3t7z3dnn7+3dwj4zdlwuptt1kfqrquewbeuqxarafemrh4dl1a/dgrndqfocxtclwu6iroemntxolwrucqga6ljjpnbv9/ucxz9bdjbvajmmhmw1qn1gwt9q5vf/0j6cljjirpjmh0s59njuajxeo5jjz2wj8gfsjzlbwcensm7eqmz5gpwg4qlgetjwa7damx07p0k427iike/v3x4hf1g6j0fvgdpt23kvf/7xggt2z1kiooefqfppqn5eunu2doh1hgkmcosk4ee6vlpezyrxdnfkxqjc/8ikplornxecmlc9fznlikenyraokikektk5ihvqjj4/kmbysn+/je/hevy9p6zi36zkgf+b9/qdek6bd</latexit> sha1_base64="uggo4plvsfj3qrbyh33c+ge3dyk=">aaacgnicbvbntwixeo36ififevtssezwqna5qbdd4swbmiiqachdmkbdt920sxpc+b9e/ctepkjxzrz4b+wcbwvf0ut1vzlp54wxfbz9/9tbwl5zxvvpbgq3t7z3dnn7+3dwj4zdlwuptt1kfqrquewbeuqxarafemrh4dl1a/dgrndqfocxtclwu6iroemntxolwrucqga6ljjpnbv9/ucxz9bdjbvajmmhmw1qn1gwt9q5vf/0j6cljjirpjmh0s59njuajxeo5jjz2wj8gfsjzlbwcensm7eqmz5gpwg4qlgetjwa7damx07p0k427iike/v3x4hf1g6j0fvgdpt23kvf/7xggt2z1kiooefqfppqn5eunu2doh1hgkmcosk4ee6vlpezyrxdnfkxqjc/8ikplornxecmlc9fznlikenyraokikektk5ihvqjj4/kmbysn+/je/hevy9p6zi36zkgf+b9/qdek6bd</latexit>.<latexit sha1_base64="kpncj9u+vvgthv6zgsmj1b2xrwo=">aaab53icbvbns8naej34wetx1aoxxsj4ckkv6kukxjy2ygyhdwwznbrrn5uwuxfk6s/w4khfq3/jm//gbzudtj4yelw3w8y8kbncg8/7dtbwnza3tks75d29/ypdythxg05zxtbgquhvo6iabzcygg4etjofniketqlr7cxvpahspjx3zpxhmncb5dfn1fip6fyqvc/15icrxc9ifqo0epwvbj9leylsmeg17vhezsijvyyzgdnyn9eyutaia+xykmmcopzmd52sc6v0szwqw9kqufp7ykitrcdjzdstaoz62zuj/3md3mrx4ytlldco2wjrnatiujl7mvs5qmbe2blkfle3ejakijjjsynbepzll1djuhovxb9zq9zvijrkcapncae+xeid7qabatbaeizxehmenrfn3flytk45xcwj/ihz+qpi3oya</latexit> <latexit sha1_base64="kpncj9u+vvgthv6zgsmj1b2xrwo=">aaab53icbvbns8naej34wetx1aoxxsj4ckkv6kukxjy2ygyhdwwznbrrn5uwuxfk6s/w4khfq3/jm//gbzudtj4yelw3w8y8kbncg8/7dtbwnza3tks75d29/ypdythxg05zxtbgquhvo6iabzcygg4etjofniketqlr7cxvpahspjx3zpxhmncb5dfn1fip6fyqvc/15icrxc9ifqo0epwvbj9leylsmeg17vhezsijvyyzgdnyn9eyutaia+xykmmcopzmd52sc6v0szwqw9kqufp7ykitrcdjzdstaoz62zuj/3md3mrx4ytlldco2wjrnatiujl7mvs5qmbe2blkfle3ejakijjjsynbepzll1djuhovxb9zq9zvijrkcapncae+xeid7qabatbaeizxehmenrfn3flytk45xcwj/ihz+qpi3oya</latexit> <latexit sha1_base64="kpncj9u+vvgthv6zgsmj1b2xrwo=">aaab53icbvbns8naej34wetx1aoxxsj4ckkv6kukxjy2ygyhdwwznbrrn5uwuxfk6s/w4khfq3/jm//gbzudtj4yelw3w8y8kbncg8/7dtbwnza3tks75d29/ypdythxg05zxtbgquhvo6iabzcygg4etjofniketqlr7cxvpahspjx3zpxhmncb5dfn1fip6fyqvc/15icrxc9ifqo0epwvbj9leylsmeg17vhezsijvyyzgdnyn9eyutaia+xykmmcopzmd52sc6v0szwqw9kqufp7ykitrcdjzdstaoz62zuj/3md3mrx4ytlldco2wjrnatiujl7mvs5qmbe2blkfle3ejakijjjsynbepzll1djuhovxb9zq9zvijrkcapncae+xeid7qabatbaeizxehmenrfn3flytk45xcwj/ihz+qpi3oya</latexit> <latexit sha1_base64="kpncj9u+vvgthv6zgsmj1b2xrwo=">aaab53icbvbns8naej34wetx1aoxxsj4ckkv6kukxjy2ygyhdwwznbrrn5uwuxfk6s/w4khfq3/jm//gbzudtj4yelw3w8y8kbncg8/7dtbwnza3tks75d29/ypdythxg05zxtbgquhvo6iabzcygg4etjofniketqlr7cxvpahspjx3zpxhmncb5dfn1fip6fyqvc/15icrxc9ifqo0epwvbj9leylsmeg17vhezsijvyyzgdnyn9eyutaia+xykmmcopzmd52sc6v0szwqw9kqufp7ykitrcdjzdstaoz62zuj/3md3mrx4ytlldco2wjrnatiujl7mvs5qmbe2blkfle3ejakijjjsynbepzll1djuhovxb9zq9zvijrkcapncae+xeid7qabatbaeizxehmenrfn3flytk45xcwj/ihz+qpi3oya</latexit> <latexit sha1_base64="kpncj9u+vvgthv6zgsmj1b2xrwo=">aaab53icbvbns8naej34wetx1aoxxsj4ckkv6kukxjy2ygyhdwwznbrrn5uwuxfk6s/w4khfq3/jm//gbzudtj4yelw3w8y8kbncg8/7dtbwnza3tks75d29/ypdythxg05zxtbgquhvo6iabzcygg4etjofniketqlr7cxvpahspjx3zpxhmncb5dfn1fip6fyqvc/15icrxc9ifqo0epwvbj9leylsmeg17vhezsijvyyzgdnyn9eyutaia+xykmmcopzmd52sc6v0szwqw9kqufp7ykitrcdjzdstaoz62zuj/3md3mrx4ytlldco2wjrnatiujl7mvs5qmbe2blkfle3ejakijjjsynbepzll1djuhovxb9zq9zvijrkcapncae+xeid7qabatbaeizxehmenrfn3flytk45xcwj/ihz+qpi3oya</latexit> Proof of Theorems 1 and 2: 0= h(x) = X y x c e (h(x) h(y)) recall E xy = c xy (h(x) h(y)) 2 = X y x h(x) E e h(y) (One needs also show that all solutions are real.) Solutions of the enharmonic equation are critical points of the functional <latexit sha1_base64="fujnjokcrfkbgg9go7tcrszrgpw=">aaacmnicbvbnswmxfmzwr1q/qh69bivgqez2ol5e8ckeklor1fky6vsbzczr8iku4n/y4h/xiighfa/+clntea0opbhm5vf4e2dswazd56awnt0zo1ecly0sli2vlffxzq12hkoda6nnrcwsskgggqilxgqgwbplambxh7nfvavjhvzn2m+gnbirjrlbgxqpuz4+1dll1fkduowbbdvjjtvkcao3bhijzadlrqbfkzttquf3phgk5xkmo+vkwa2hoh9jncyvmka9u3687gruuldijbo2fyuztgfm+dss7kqxzklg+dw7gpaniqvg24phz3d0yytdmmjjryedqj83biy1tp/gppky7nljlxf/81ook932qkjmisg+opq4svhtvedafqy4yr4nbnqj5b4xxthxxpilrjmv/ywnwnwvgp3ukgf74zakzinskm0skr1yqi5intqij/fkibyst+aheaneg49rtbcmd9bjlwsfxzpaq3y=</latexit> <latexit sha1_base64="fujnjokcrfkbgg9go7tcrszrgpw=">aaacmnicbvbnswmxfmzwr1q/qh69bivgqez2ol5e8ckeklor1fky6vsbzczr8iku4n/y4h/xiighfa/+clntea0opbhm5vf4e2dswazd56awnt0zo1ecly0sli2vlffxzq12hkoda6nnrcwsskgggqilxgqgwbplambxh7nfvavjhvzn2m+gnbirjrlbgxqpuz4+1dll1fkduowbbdvjjtvkcao3bhijzadlrqbfkzttquf3phgk5xkmo+vkwa2hoh9jncyvmka9u3687gruuldijbo2fyuztgfm+dss7kqxzklg+dw7gpaniqvg24phz3d0yytdmmjjryedqj83biy1tp/gppky7nljlxf/81ook932qkjmisg+opq4svhtvedafqy4yr4nbnqj5b4xxthxxpilrjmv/ywnwnwvgp3ukgf74zakzinskm0skr1yqi5intqij/fkibyst+aheaneg49rtbcmd9bjlwsfxzpaq3y=</latexit> <latexit sha1_base64="fujnjokcrfkbgg9go7tcrszrgpw=">aaacmnicbvbnswmxfmzwr1q/qh69bivgqez2ol5e8ckeklor1fky6vsbzczr8iku4n/y4h/xiighfa/+clntea0opbhm5vf4e2dswazd56awnt0zo1ecly0sli2vlffxzq12hkoda6nnrcwsskgggqilxgqgwbplambxh7nfvavjhvzn2m+gnbirjrlbgxqpuz4+1dll1fkduowbbdvjjtvkcao3bhijzadlrqbfkzttquf3phgk5xkmo+vkwa2hoh9jncyvmka9u3687gruuldijbo2fyuztgfm+dss7kqxzklg+dw7gpaniqvg24phz3d0yytdmmjjryedqj83biy1tp/gppky7nljlxf/81ook932qkjmisg+opq4svhtvedafqy4yr4nbnqj5b4xxthxxpilrjmv/ywnwnwvgp3ukgf74zakzinskm0skr1yqi5intqij/fkibyst+aheaneg49rtbcmd9bjlwsfxzpaq3y=</latexit> <latexit sha1_base64="fujnjokcrfkbgg9go7tcrszrgpw=">aaacmnicbvbnswmxfmzwr1q/qh69bivgqez2ol5e8ckeklor1fky6vsbzczr8iku4n/y4h/xiighfa/+clntea0opbhm5vf4e2dswazd56awnt0zo1ecly0sli2vlffxzq12hkoda6nnrcwsskgggqilxgqgwbplambxh7nfvavjhvzn2m+gnbirjrlbgxqpuz4+1dll1fkduowbbdvjjtvkcao3bhijzadlrqbfkzttquf3phgk5xkmo+vkwa2hoh9jncyvmka9u3687gruuldijbo2fyuztgfm+dss7kqxzklg+dw7gpaniqvg24phz3d0yytdmmjjryedqj83biy1tp/gppky7nljlxf/81ook932qkjmisg+opq4svhtvedafqy4yr4nbnqj5b4xxthxxpilrjmv/ywnwnwvgp3ukgf74zakzinskm0skr1yqi5intqij/fkibyst+aheaneg49rtbcmd9bjlwsfxzpaq3y=</latexit> <latexit sha1_base64="fujnjokcrfkbgg9go7tcrszrgpw=">aaacmnicbvbnswmxfmzwr1q/qh69bivgqez2ol5e8ckeklor1fky6vsbzczr8iku4n/y4h/xiighfa/+clntea0opbhm5vf4e2dswazd56awnt0zo1ecly0sli2vlffxzq12hkoda6nnrcwsskgggqilxgqgwbplambxh7nfvavjhvzn2m+gnbirjrlbgxqpuz4+1dll1fkduowbbdvjjtvkcao3bhijzadlrqbfkzttquf3phgk5xkmo+vkwa2hoh9jncyvmka9u3687gruuldijbo2fyuztgfm+dss7kqxzklg+dw7gpaniqvg24phz3d0yytdmmjjryedqj83biy1tp/gppky7nljlxf/81ook932qkjmisg+opq4svhtvedafqy4yr4nbnqj5b4xxthxxpilrjmv/ywnwnwvgp3ukgf74zakzinskm0skr1yqi5intqij/fkibyst+aheaneg49rtbcmd9bjlwsfxzpaq3y=</latexit> M(h) = Y e h(x) h(y) E e. Note log M(h) is strictly concave on each polytope F = {h sign(dh) = }.
Proof of Theorem 3: Recall that E e = c e (h(x) h(y)) 2 = I e V e where I e = c e (h(x) h(y)) and V e = h(x) h(y). When E e! 0, either I e! 0 or V e! 0 (or both). In the first case, delete the edge; in the second contract the edge. Conversely, the operation of contracting or deleting is reversible by adding in an edge of small energy.
applications
Smith diagram of a planar network (with a harmonic function) v 1 3 2 1 6 4 3. 4. 9 8 7 5 6. 7. 10 12 11 9. 10. 1 1 v 0 voltage = y-coordinate edge = rectangle current = width conductance = aspect ratio energy = area
This graph has 12 acyclic orientations with source at 1 and sink at 0. v 1 ( = 1) 3 2 1 6 4 4. 9 8 7 5 3. 6. 7. 10 12 11 9. 10. 1 v 0 1 width(1) is the root of a polynomial: 2315250000z 12 107438625000z 11 +2230924692500z 10 27361273241750z 9 + 220350695004825z 8 1225394593409700z 7 +4817113876088640z 6 13468300499707200z 5 + 26554002301384704z 4 35985219877131264z 3 +31817913970765824z 2 16489700865736704z+ 3791571715620864 = 0
3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 3. 4. 6. 7. 9. 10. 1 1 1 2 3 4 5 6 7 8 9 10 11 12
One of the 10 599 area-1 rectangulations based on the 40 40 grid 0 (64/27) n2 (1+o(1)) 0.8 1 0.6 0 Z 2,directedS&W 0.4 0.2 500 1000 1500 2000 2500 3000
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Bipolar orientations on Z 2 and square ice (with Miller, She eld, Wilson)
add a row of edges around the boundary oriented N and W
From each black vertex, the two outgoing green arrows separate the incoming and outgoing black arrows From each face center, outgoing green arrows point to the face max and min.
Bend outgoing edges right if from vertices, left if from faces.
The associated peano curve, colored according to distance traveled
(Conjectural) SLE 12 scaling limit
Thank you