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GRAPHENE NANOCONES and PASCAL MATRICES
– some determinantal conjectures –

Luca Guido Molinari L. G. Molinari, Physics Department “Aldo Pontremoli”, Università degli Studi di Milano and I.N.F.N. sez. di Milano, Via Celoria 16, 20133 Milano, Italy. [email protected]
Abstract.

I conjecture three identities for the determinant of adjacency matrices of graphene triangles and trapezia with Bloch (and more general) boundary conditions. For triangles, the parametric determinant is equal to the characteristic polynomial of the symmetric Pascal matrix. For trapezia it is equal to the determinant of a sub-matrix. Finally, the determinant of the tight binding matrix equals its permanent. The conjectures are supported by analytic evaluations and Mathematica, for moderate sizes. They establish connections with counting problems of partitions, lozenge tilings of hexagons, dense loops on a cylinder.

Key words and phrases:
graphene, adjacency matrix, Pascal matrix, plane partition, lozenge tiling.
2010 Mathematics Subject Classification:
05B20 (Primary) 15A15, 82D80, 05B45 (Secondary)

1. Introduction

Physics often entails intriguing mathematics, and Carbon Nanocones are no exception. They are conical honeycomb lattices of Carbon atoms, with a cap of Carbon rings causing different cone apertures [5]. The ones with smallest angle, near 1919^{\circ}, were sinthetized by Ge and Sattler in 1994, in the hot vapour phase of carbon; the others were found by accident, in pyrolysis of heavy oil [17].

Refer to caption
Figure 1. Scanning electron microscope images of carbon nanocones (size order 1μm1\mu m) produced at the Kvaerner Carbon Black & Hydrogen Process [16].

The simplest description for such structures is through the adjacency matrix. Chemists give weight β-\beta to lattice links among atoms, and name it Hückel matrix. Because of the CnC_{n} symmetry of nanocones, the Hückel matrix can be restricted to a single honeycomb triangle, with Bloch boundary conditions e±i2π/ne^{\pm i2\pi/n} on two sides. Properties of the spectrum, based on graph theory and symmetries, were studied in [15]. Recently, based on numerics, new rules were established for the filling of levels, that differ from the ordinary Hückel rules [4].
This paper originates from a question by an author in [4]: what is the determinant of the Huckel matrix of a nanocone? The answer here conjectured opens a nice connection with Pascal matrices and combinatorics.

Refer to caption
Figure 2. Six lines cut graphene into triangles with zig-zag rows of 1, 3, 5, … atoms. Cones are obtained by re-joining 3, 4, 5 triangles, with a cap-ring of 3, 4, 5 atoms.

It is convenient to replace the Bloch factors by free parameters xx and yy, or even by parameters xix_{i} and yiy_{i} for each row.
The size of a Hückel matrix with nn rows is (n+1)2(n+1)^{2}, which is the number of atoms in the triangle.
012 H2=[S001001y101110100x1100100001y2110101011101x210]\displaystyle\small{H_{2}=\left[\begin{array}[]{c|ccc|ccccc}S_{0}&0&1&&&&&&\\ \hline\cr 0&0&1&y_{1}&0&1&&&\\ 1&1&0&1&0&&0&&\\ &x_{1}&1&0&0&&&1&\\ \hline\cr&0&0&0&0&1&&&y_{2}\\ &1&&&1&0&1&&\\ &&0&&&1&0&1&\\ &&&1&&&1&0&1\\ &&&&x_{2}&&&1&0\end{array}\right]}


Example. A triangle with one hexagon and its Hückel matrix. In this representation, the diagonal blocks are the rows of the graph, with 1,3,5 atoms. Edges are 1s in the matrix. Dots connect row extrema with weights xi=yi=1x_{i}=y_{i}=1 (S0=x0+y0S_{0}=x_{0}+y_{0}).

The determinants of small Hückel matrices are evaluated:

detH0=\displaystyle\det H_{0}= x+y\displaystyle x+y
detH1=\displaystyle\det H_{1}= x2+3xy+y2\displaystyle x^{2}+3xy+y^{2}
detH2=\displaystyle\det H_{2}= x3+9x2y+9xy2+y3\displaystyle x^{3}+9x^{2}y+9xy^{2}+y^{3}
detH3=\displaystyle\det H_{3}= x4+29x3y+72x2y2+29xy3+y4\displaystyle x^{4}+29x^{3}y+72x^{2}y^{2}+29xy^{3}+y^{4}
detH4=\displaystyle\det H_{4}= x5+99x4y+626x3y2+626x2y3+99xy4+y5\displaystyle x^{5}+99x^{4}y+626x^{3}y^{2}+626x^{2}y^{3}+99xy^{4}+y^{5}
detH5=\displaystyle\det H_{5}= x6+351x5y+6084x4y2+13869x3y3+6084x2y4+351xy5+y6\displaystyle x^{6}+351x^{5}y+6084x^{4}y^{2}+13869x^{3}y^{3}+6084x^{2}y^{4}+351xy^{5}+y^{6}
detH6=\displaystyle\det H_{6}= x7+1275x6y+64974x5y2+347020x4y3+347020x3y4\displaystyle x^{7}+1275x^{6}y+64974x^{5}y^{2}+347020x^{4}y^{3}+347020x^{3}y^{4}
+64974x2y5+1275xy6+y7\displaystyle\quad+64974x^{2}y^{5}+1275xy^{6}+y^{7}

\bullet The determinants are homogeneous palindromic polynomials with positive coefficients.
\bullet If the parameters are x1,x2,,ynx_{1},x_{2},...,y_{n} are kept different, the determinants are homogeneous polynomial of degree n+1n+1 in the variables, with coefficients that are perfect squares of integers (this feature was noticed to me by Andrea Sportiello).

Their evaluation for small size nn with Schur’s formula for block matrices showed that the determinant of a Hückel matrix of size (n+1)2(n+1)^{2} eventually reduces to the determinant of a matrix of binomials of size (n+1)(n+1) (a step is illustrated in section 8). Binomials occur in many counting problems of benzenoids [6, 9].
After discovering the beautiful Pascal matrices in Lunnon’s paper [19], I found with Mathematica that the polynomial coefficients in detHn\det H_{n} for accessible values of nn, match the sequences A045912 in OEIS [22] for the characteristic polynomials of Pascal matrices. This led me to the first conjecture; the others then came out.

Conjectures 1 and 2 imply symmetries of the eigenvalues of Hückel matrices of nanocones and the larger family of graphannulenes, that support the generalized Hückel rules for the electronic structure proposed by Evangelisti et al. in ref.[4].

Before stating the conjectures, let’s introduce Pascal matrices.

2. Pascal matrices

There are various forms of Pascal matrices, and much literature [19, 7, 24, 10]. The lower triangular Pascal matrix PnP_{n} has size n+1n+1 and elements that are the binomial coefficients of the Pascal triangle. The inverse matrix also has binomial coefficients, but with alternating signs. For example:

P4=[111121133114641],P41=[111121133114641]\displaystyle P_{4}=\left[\begin{array}[]{ccccc}1&&&&\\ 1&1&&&\\ 1&2&1&&\\ 1&3&3&1&\\ 1&4&6&4&1\end{array}\right],\quad P_{4}^{-1}=\left[\begin{array}[]{ccccc}1&&&&\\ -1&1&&&\\ 1&-2&1&&\\ -1&3&-3&1&\\ 1&-4&6&-4&1\end{array}\right]

The product Qn=PnPnTQ_{n}=P_{n}P_{n}^{T} is the symmetric positive Pascal matrix. It has unit determinant and matrix elements (Qn)ij=(i+jj)(Q_{n})_{ij}=\binom{i+j}{j} i,j=0,,ni,j=0,...,n:

Q4=[111111234513610151410203515153570]\displaystyle Q_{4}=\left[\begin{array}[]{ccccc}1&1&1&1&1\\ 1&2&3&4&5\\ 1&3&6&10&15\\ 1&4&10&20&35\\ 1&5&15&35&70\end{array}\right]

Since the eigenvalues are strictly positive, the coefficients of the polynomial det(z+Qn)\det(z+Q_{n}) are positive. Moreover the eigenvalues come in pairs: qn+1j=1/qjq_{n+1-j}=1/q_{j}; if the matrix size is odd (nn even), an eigenvalue of QnQ_{n} is 1.
Note that the entry (i,j)(i,j) (independent of nn) counts the number of paths with i+ji+j steps \rightarrow or \downarrow that connect the corner (0,00,0) with (i,ji,j). For example, (3,43,4) can be reached in 35=(73)35=\binom{7}{3} different ways from (0,00,0).

3. The first conjecture (honeycomb triangles)

A graphene triangle and its Hückel matrix are shown below. For later convenience the atoms are marked as red and blue (two triangular sub-lattices of hexagonal graphene).
43210 H5(𝐱,𝐲)=[T0R1TR1R4TR4T4]\displaystyle H_{5}({\bf x},{\bf y})=\left[\begin{array}[]{cccc}T_{0}&R_{1}^{T}&&\\ R_{1}&\ddots&\ddots&\\ &\ddots&\ddots&R_{4}^{T}\\ &&R_{4}&T_{4}\end{array}\right]
A block Tm(xm,ym)T_{m}(x_{m},y_{m}) is a square matrix of size 2m+12m+1 that describes a row of 2m+12m+1 atoms with boundary parameters xm,ymx_{m},y_{m}: T0(x0,y0)=x0+y0T_{0}(x_{0},y_{0})=x_{0}+y_{0},

,T1(x1,y1)=[01y1101x110],T2(x2,y2)=[0100y2101000101000101x20010],\displaystyle{\small,\quad T_{1}(x_{1},y_{1})=\left[\begin{array}[]{ccc}0&1&y_{1}\\ 1&0&1\\ x_{1}&1&0\end{array}\right],\quad T_{2}(x_{2},y_{2})=\left[\begin{array}[]{ccccc}0&1&0&0&y_{2}\\ 1&0&1&0&0\\ 0&1&0&1&0\\ 0&0&1&0&1\\ x_{2}&0&0&1&0\end{array}\right],\ldots}

The size of HnH_{n} is 1+3++(2n+1)=(n+1)21+3+...+(2n+1)=(n+1)^{2}. The blocks RmR_{m} are (2m+1)×(2m1)(2m+1)\times(2m-1), with unit elements for the vertical edges in the graph, joining atoms in rows m1m-1 and mm:

R1=[010],R2=[000100000001000],R3=[00000100000000000100000000000100000],\displaystyle{\small R_{1}=\left[\begin{array}[]{c}0\\ 1\\ 0\end{array}\right],\quad R_{2}=\left[\begin{array}[]{ccc}0&0&0\\ 1&0&0\\ 0&0&0\\ 0&0&1\\ 0&0&0\end{array}\right],\quad R_{3}=\left[\begin{array}[]{ccccc}0&0&0&0&0\\ 1&0&0&0&0\\ 0&0&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{array}\right],\quad\ldots}

Conjecture 1: The determinant of Hn(𝐱,𝐲)H_{n}({\bf x},{\bf y}) is equal to the determinant of the following matrix of size n+1n+1:

(7) [xn+yn(n1)yn(n2)yn(n3)yn±(nn)yn(n1)xnxn1+yn1(n11)yn1(n12)yn1(n1n1)yn1(n2)xn(n11)xn1xn2+yn2(n21)yn2x1+y1(11)y1(nn)xn(n1n1)xn1(n2n2)xn2(11)x1x0+y0].\displaystyle\left[\begin{array}[]{cccccc}x_{n}+y_{n}&-\binom{n}{1}y_{n}&\binom{n}{2}y_{n}&-\binom{n}{3}y_{n}&...&\pm\binom{n}{n}y_{n}\\ \binom{n}{1}x_{n}&x_{n-1}+y_{n-1}&-\binom{n-1}{1}y_{n-1}&\binom{n-1}{2}y_{n-1}&...&\mp\binom{n-1}{n-1}y_{n-1}\\ \binom{n}{2}x_{n}&\binom{n-1}{1}x_{n-1}&x_{n-2}+y_{n-2}&-\binom{n-2}{1}y_{n-2}&...&\vdots\\ \vdots&\vdots&\vdots&...&...&\vdots\\ \vdots&\vdots&\vdots&...&x_{1}+y_{1}&-\binom{1}{1}y_{1}\\ \binom{n}{n}x_{n}&\binom{n-1}{n-1}x_{n-1}&\binom{n-2}{n-2}x_{n-2}&...&\binom{1}{1}x_{1}&x_{0}+y_{0}\\ \end{array}\right].

Example:

detH3=|x3+y33y33y3y33x3x2+y22y2y23x32x2x1+y1y1x3x2x1x0+y0|\displaystyle\det H_{3}=\left|\begin{array}[]{cccc}x_{3}+y_{3}&-3y_{3}&3y_{3}&-y_{3}\\ 3x_{3}&x_{2}+y_{2}&-2y_{2}&y_{2}\\ 3x_{3}&2x_{2}&x_{1}+y_{1}&-y_{1}\\ x_{3}&x_{2}&x_{1}&x_{0}+y_{0}\end{array}\right|
=(x0+y0)[x1x2x3+x2x3y1+x1x2y3+x1y2y3+x1x3y2+x3y1y2+x2y1y3+y1y2y3\displaystyle=(x_{0}+y_{0})[x_{1}x_{2}x_{3}+x_{2}x_{3}y_{1}+x_{1}x_{2}y_{3}+x_{1}y_{2}y_{3}+x_{1}x_{3}y_{2}+x_{3}y_{1}y_{2}+x_{2}y_{1}y_{3}+y_{1}y_{2}y_{3}
+4x2y2(x3+y3)+9x3y3(x1+y1+x2+y2)]+x1x2x3(y1+y2+y3)+(x1+x2)x3y1y2\displaystyle+4x_{2}y_{2}(x_{3}+y_{3})+9x_{3}y_{3}(x_{1}+y_{1}+x_{2}+y_{2})]+x_{1}x_{2}x_{3}(y_{1}+y_{2}+y_{3})+(x_{1}+x_{2})x_{3}y_{1}y_{2}
+x1x2(y1+y2)y3+x2x3y2y3+(x1+x2+x3)y1y2y3+4x3y3(x1y2+x2y1)+9x1x3y1y3\displaystyle+x_{1}x_{2}(y_{1}+y_{2})y_{3}+x_{2}x_{3}y_{2}y_{3}+(x_{1}+x_{2}+x_{3})y_{1}y_{2}y_{3}+4x_{3}y_{3}(x_{1}y_{2}+x_{2}y_{1})+9x_{1}x_{3}y_{1}y_{3}

It is a sum of monomials of degree 44 with coefficients that are perfect integer squares.

With all xk=xx_{k}=x and yk=yy_{k}=y, we obtain the interesting identity:

Proposition 3.1.
(8) detHn(x,y)=xn+1det(Qn+yxIn+1)\displaystyle\boxed{\det H_{n}(x,y)=x^{n+1}\det\left(Q_{n}+\frac{y}{x}I_{n+1}\right)}

where QnQ_{n} is the symmetric Pascal matrix.

Proof.

The (n+1)×(n+1)(n+1)\times(n+1) matrix in (7) is the sum of lower and upper triangular matrices related to the Pascal matrix: x(JnPnTJn)+y(JnPn1Jn)x(J_{n}P_{n}^{T}J_{n})+y(J_{n}P_{n}^{-1}J_{n}), where

Jn=(11)J_{n}=\left(\begin{smallmatrix}&&1\\ &\reflectbox{$\ddots$}&\\ 1&&\end{smallmatrix}\right)

is the inversion matrix. Then:

detHn(x,y)\displaystyle\det H_{n}(x,y) =det(xPnT+yPn1)\displaystyle=\det(xP_{n}^{T}+yP_{n}^{-1})
=det(xPn1)det(PnPnT+y/x)\displaystyle=\det(xP_{n}^{-1})\det(P_{n}P_{n}^{T}+y/x)
=xn+1det(Qn+y/x)\displaystyle=x^{n+1}\det(Q_{n}+y/x)

More generally: detHn(𝐱,𝐲)=det(JnPnTJnX+YJnPn1Jn)=det(PnJnXJn+JnYJnPn1)\det H_{n}({\bf x},{\bf y})=\det(J_{n}P_{n}^{T}J_{n}X+YJ_{n}P_{n}^{-1}J_{n})=\det(P_{n}J_{n}XJ_{n}+J_{n}YJ_{n}P_{n}^{-1}) where X=diag[xi]X={\rm diag}[x_{i}] and Y=diag[yi]Y={\rm diag}[y_{i}]. If YY is the identity matrix, then detHn(𝐱,1)=(x0xn)det(Qn+diag(1/x0,,1/xn))\det H_{n}({\bf x},1)=(x_{0}...x_{n})\det(Q_{n}+{\rm diag}(1/x_{0},...,1/x_{n})). For example: the size of H3H_{3} is N=16N=16, and

detH3(𝐱,1)=(x0x3)det[1+1/x011112+1/x134136+1/x210141020+1/x3].\det H_{3}({\bf x},1)=(x_{0}...x_{3})\det\left[\begin{array}[]{cccc}1+1/x_{0}&1&1&1\\ 1&2+1/x_{1}&3&4\\ 1&3&6+1/x_{2}&10\\ 1&4&10&20+1/x_{3}\end{array}\right].

In the next section I report the very few known analytic values of detHn\det H_{n}, given by eq.(8) for special values of the ratio ω=y/x=e2iθ\omega=y/x=e^{2i\theta}. They result from amazing combinatorial problems, that are here reported for the delight of the reader.

4. Pascal matrix, partitions and lozenge tilings

The characteristic polynomial of the symmetric Pascal matrix appears in diverse combinatorial problems with the following generalization:

(9) det[(m+j+kk)+ωδjk]0j,knm=0,1,2,\displaystyle\det\left[\binom{m+j+k}{k}+\omega\delta_{jk}\right]_{0\leq j,k\leq n}\quad m=0,1,2,\ldots

George Andrews [2] in 1979, in proving the weak Macdonald conjecture for certain plane partitions, obtained the closed expression of (9) with ω=1\omega=1. He exploited identities for hypergeometric series. For the Pascal matrix (m=0m=0) it greatly simplifies:

(10) det(Qn+In+1)=k=0n13k+1k!(3k+2)!(2k)!(2k+1)!\displaystyle\det(Q_{n}+I_{n+1})=\prod_{k=0}^{n}\frac{1}{3k+1}\frac{k!(3k+2)!}{(2k)!(2k+1)!}

This number is also the determinant of the periodic Hückel matrix, detHn(1,1)\det H_{n}(1,1).

A plane partition π\pi of NN of shape (a,b,c)(a,b,c) is an array a×ba\times b of numbers cπij0c\geq\pi_{ij}\geq 0 (i=1,..,ai=1,..,a, j=1,..,bj=1,..,b) with the property that all rows and columns are weakly decreasing, and the sum of all πij\pi_{ij} is NN.
These are two plane partitions of 38 in (3,5,7)(3,5,7):

553225521042110766615321010000\displaystyle\begin{array}[]{ccccc}5&5&3&2&2\\ 5&5&2&1&0\\ 4&2&1&1&0\end{array}\qquad\begin{array}[]{ccccc}7&6&6&6&1\\ 5&3&2&1&0\\ 1&0&0&0&0\end{array}

A plane partition may be represented by stacking πij\pi_{ij} cubes on each square cell (i,j)(i,j) in the rectangle a×ba\times b. A stacking of cubes defines ascending staircases that correspond to sets of non-intersecting walks in a lattice. The beautiful Gessel-Viennot theorem [11] counts them as determinants of certain binomial matrices111see also the nice presentation by Aigner in [1].

It was later discovered [13] that the projection of the bounding box a×b×ca\times b\times c on a oblique plane normal to (1,1,1) is an hexagon H(a,b,c)H(a,b,c) with side-lengths a,b,c,a,b,ca,b,c,a,b,c (in cyclic order) and angles π/3\pi/3 in a triangular lattice. Each stacking is one-to-one with a lozenge tiling of the hexagon. A lozenge is the union of two triangular unit cells of the foreground triangular lattice, and has angles π/3\pi/3 and 2π/32\pi/3.

Refer to caption
Figure 3. A lozenge tile of H(6,6,6)H(6,6,6) in Class 10. The cube stacking represents the partition with shape 6×6×66\times 6\times 6 with rows: 6,6,6,5,4,3; 6,6,5,3,3,2; 6,5,5,3,3,1; 5,3,3,1,1,0; 4,3,3,1,1,0; 3,2,1,0,0,0 (https://mathematica.stackexchange.com/questions/ 228422/lozenge-tilings)

Plane partitions were classified by Stanley and others into 10 symmetry classes and enumerated ([18] is a nice review by Christian Krattenthaler).
Class 1 are the unrestricted partitions. They were enumerated by Percy MacMahon (1896): the number of all plane partitions in (a,b,c)(a,b,c), i.e. lozenge tilings of H(a,b,c)H(a,b,c), is

H(a)H(b)H(c)H(a+b+c)H(a+c)H(a+c)H(b+c)\frac{H(a)H(b)H(c)H(a+b+c)}{H(a+c)H(a+c)H(b+c)}

where H(k)=1!2!(k1)!H(k)=1!2!...(k-1)!.
The other extremum is Class 10, with totally symmetric self-complementary plane partition in (2n,2n,2n)(2n,2n,2n). Viewed as a stack of cubes in the cubic box of side 2n2n, a TSSC partition has the property of being equal to its complement (the void in the box, see fig.3) and, viewed as a tiling, it is invariant under rotations by π/3\pi/3 and reflections in the main diagonals of the hexagon. The enumeration was finally obtained by Andrews [3]:

A(n)=k=0n1(3k+1)!(n+k)!A(n)=\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}

A(n)A(n) is also the number of n×nn\times n matrices {0,±1}\{0,\pm 1\} whose row-sums and column-sums are all 1, and in every row and column the non-zero entries alternate in sign [26].

With the aid of the identity

(11) 1=k=0nk!(n+k+1)!(2k)!(2k+1)!\displaystyle 1=\prod_{k=0}^{n}\frac{k!(n+k+1)!}{(2k)!(2k+1)!}

the numbers (10) can be rewritten as follows:

(12) det(Qn+In+1)=A(n+1)k=0n3k+23k+1\displaystyle\det(Q_{n}+I_{n+1})=A(n+1)\prod_{k=0}^{n}\frac{3k+2}{3k+1}

Ciucu, Eisenköbl, Krattenthaler and Zare evaluated weighted sums over the tilings of the hexagon H(a,b+m,c,a+m,b,c+m)H(a,b+m,c,a+m,b,c+m) with a central triangular hole of sides m,m,mm,m,m. With a=b=c=n+1a=b=c=n+1 the counts are given by the determinant (9) with weights ω=±1,ei2π/3,eiπ/3\omega=\pm 1,e^{i2\pi/3},e^{i\pi/3} (equations 3.3, 3.4, 3.5 in [8]).
With m=0m=0 they simplify to eqs.(13), (14) and (15) given below. They are the weighted counts πω#π\sum_{\pi}\omega^{\#\pi} (#π\#\pi is the number of cubes on the main diagonal) over the cyclically symmetric partitions π\pi with shape (n+1)3(n+1)^{3} (Corollary 8 of [8]). In other words, the determinants give the weighted sums over the tilings of H(n+1,n+1,n+1)H(n+1,n+1,n+1) that are invariant under rotations by π/3\pi/3 around the center of the hexagon (Class 3). The different weights ω\omega give:

(13) det(Q2n+1I2n+2)=(1)n+1[k=0nk!(3k+1)!(2k)!(2k+1)!]4=(1)n+1A(n+1)4\displaystyle\det(Q_{2n+1}-I_{2n+2})=(-1)^{n+1}\left[\prod_{k=0}^{n}\frac{k!(3k+1)!}{(2k)!(2k+1)!}\right]^{4}=(-1)^{n+1}A(n+1)^{4}

Up to sign, it coincides with detH2n+1(1,1)\det H_{2n+1}(1,-1). For an odd size det(Q2nI2n+1)=0\det(Q_{2n}-I_{2n+1})=0. Some numbers are listed by Stembridge (1998), who first studied this case222page 6 in http://www.math.lsa.umich.edu/ jrs/papers/strange.pdf.
Eqs. 3.4 and 3.5 with m=0m=0 have simpler expressions given in Table 1 of [21]:

(14) det(Qn+ei2π3In+1)=eiπ3(n+1)A(n+1)\displaystyle\det(Q_{n}+e^{i\tfrac{2\pi}{3}}I_{n+1})=e^{i\frac{\pi}{3}(n+1)}A(n+1)
(15) det(Qn+eiπ3In+1)=eiπ6(n+1)AHT(n+1)2×{1n odd3n even.\displaystyle\det(Q_{n}+e^{i\tfrac{\pi}{3}}I_{n+1})=e^{i\frac{\pi}{6}(n+1)}A_{HT}(n+1)^{2}\times\begin{cases}1&\text{$n$ odd}\\ \sqrt{3}&\text{$n$ even}\end{cases}.

The numbers AHT(n)A_{HT}(n) enumerate the half-turn symmetric alternating sign n×nn\times n matrices333It is the subclass of alternating sign n×nn\times n matrices such that Aij=An+1i,n+1jA_{ij}=A_{n+1-i,n+1-j}. They are related to the ice model of statistical mechanics. [23]:

(16) AHT(2n)=k=0n1(3k)!(3k+2)![(n+k)!]2AHT(2n+1)=n!(3n)![(2n)!]2AHT(2n)\displaystyle A_{HT}(2n)=\prod_{k=0}^{n-1}\frac{(3k)!(3k+2)!}{[(n+k)!]^{2}}\qquad A_{HT}(2n+1)=\frac{n!(3n)!}{[(2n)!]^{2}}A_{HT}(2n)

The generalized Pascal matrices also occur in the works by Mitra and Nienhuis [20, 21]. On an infinite cylindric square lattice of even circumference LL, they consider coverings of closed paths that at each vertex make a left or right turn with equal probability. Two paths can have a vertex in common, but do not intersect.
The probability P(L,m)P(L,m) that a point (not a vertex) of the cylinder is surrounded by mm loops is guessed as Q(L,m)/AHT(L)2Q(L,m)/A_{HT}(L)^{2}, where Q(L,m)Q(L,m) is expressed with the polynomial coefficients of the Pascal matrix. At the same time, P(L,m)P(L,m) is estimated by Coulomb gas techniques. The following asymptotics is obtained:

(17) det(QL1+iIL)\displaystyle\det(Q_{L-1}+iI_{L}) =iL/2L5/48AHT(L)2×\displaystyle=i^{L/2}L^{-5/48}A_{HT}(L)^{2}\times
[0.810997530.028861L3/2+0.021012L2+a7L7/2+]\displaystyle\left[0.81099753-\frac{0.028861}{L^{3/2}}+\frac{0.021012}{L^{2}}+\frac{a_{7}}{L^{7/2}}+...\right]

The large nn (i.e. L+1L+1) expansion of (8) for any value |ω|=1|\omega|=1 is discussed by Mitra [21].

The Hückel matrices Hn(θ)=Hn(eiθ,eiθ)H_{n}(\theta)=H_{n}(e^{-i\theta},e^{i\theta}) are Hermitian and their determinants can be written as polynomials of cosθ\cos\theta. The known results for the Pascal matrix give:

detHn(0)=A(n+1)k=0n3k+23k+1\displaystyle\det H_{n}(0)=A(n+1)\prod_{k=0}^{n}\frac{3k+2}{3k+1}
detH2n+1(π2)=[A(n+1)]4,detH2n(π2)=0\displaystyle\det H_{2n+1}(\tfrac{\pi}{2})=[A(n+1)]^{4},\qquad\det H_{2n}(\tfrac{\pi}{2})=0
detHn(π3)=A(n+1)\displaystyle\det H_{n}(\tfrac{\pi}{3})=A(n+1)
detH2n+1(π6)=[AHT(2n+2)]2,detH2n(π6)=[AHT(2n+1)]23\displaystyle\det H_{2n+1}(\tfrac{\pi}{6})=[A_{HT}(2n+2)]^{2},\qquad\det H_{2n}(\tfrac{\pi}{6})=[A_{HT}(2n+1)]^{2}\sqrt{3}

The following table is a check. It shows the first values of A(n)A(n), AHT(n)A_{HT}(n) and detHn(θ)\det H_{n}(\theta), that coincide with the values of the determinants listed in introduction. The value for θ=π/4\theta=\pi/4 is compared with the expansion (17) by Mitra; the values are very close.

nAAHTθ=0π/6π/3π/2π/4Mitra22220323708237313210242247069.9964421014522523429052625429252674114027436741316713166.706743614082654058823218348028077227218348588\displaystyle\begin{array}[]{c|c|c||c|c|c|c|c||c}n&A&A_{HT}&\theta=0&\pi/6&\pi/3&\pi/2&\pi/4&{\rm Mitra}\\ \hline\cr 2&2&2&20&3^{2}\sqrt{3}&7&0&8\sqrt{2}&\\ 3&7&3&132&10^{2}&42&2^{4}&70&69.996\\ 4&42&10&1452&25^{2}\sqrt{3}&429&0&526\sqrt{2}&\\ 5&429&25&26741&140^{2}&7436&7^{4}&13167&13166.70\\ 6&7436&140&826540&588^{2}\sqrt{3}&218348&0&280772\sqrt{2}&\\ 7&218348&588&&&&&&\\ \hline\cr\end{array}

5. The second conjecture (honeycomb trapezia)

If rows 0,1,,k10,1,...,k-1 are removed from a honeycomb triangle, a trapezium results, with rows of lengths 2k+1,,2n+12k+1,...,2n+1. The corresponding Hückel matrix Hk,n(𝐱,𝐲)H_{k,n}({\bf x},{\bf y}) is obtained by deleting the first k2k^{2} rows and columns of Hn(𝐱,𝐲)H_{n}({\bf x},{\bf y}). Its size is (n+1)2k2=(2k+1)++(2n+1)(n+1)^{2}-k^{2}=(2k+1)+...+(2n+1).

For example, the graph and the matrix H6,7H_{6,7} (size 2828) are:
115113 [T6R7TR7T7]\displaystyle\left[\begin{array}[]{cc}T_{6}&R_{7}^{T}\\ R_{7}&T_{7}\end{array}\right]

Conjecture 2: The determinant of the Hückel matrix Hk,n(𝐱,𝐲)H_{k,n}({\bf x},{\bf y}) of a honeycomb trapezium with rows with 2k+12k+1, 2k+32k+3, … , 2n+12n+1 sites, size (n+1)2k2(n+1)^{2}-k^{2}, is equal to the determinant of the following matrix of size n+1kn+1-k:

detHk,n(𝐱,𝐲)=|xn+yn(n1)yn(n2)yn±(nnk)yn(n1)xnxn1+yn1(n11)yn1(n1nk1)yn1(nnk)xn(n1nk1)xn1xk+yk|\displaystyle\det H_{k,n}({\bf x},{\bf y})=\left|\begin{array}[]{ccccc}x_{n}+y_{n}&-\binom{n}{1}y_{n}&\binom{n}{2}y_{n}&...&\pm\binom{n}{n-k}y_{n}\\ \binom{n}{1}x_{n}&x_{n-1}+y_{n-1}&-\binom{n-1}{1}y_{n-1}&...&\mp\binom{n-1}{n-k-1}y_{n-1}\\ \vdots&\vdots&\vdots&...&\vdots\\ \binom{n}{n-k}x_{n}&\binom{n-1}{n-k-1}x_{n-1}&...&...&x_{k}+y_{k}\\ \end{array}\right|

In these examples the determinants of Hückel matrices of size 2828, 5151 and 6464 are evaluated as matrices of size 22, 33 and 44 (Sn=xn+ynS_{n}=x_{n}+y_{n}):

detH6,7=\displaystyle\det H_{6,7}= |S77y77x7S6|=S6S7+72x7y7\displaystyle\left|\begin{array}[]{cc}S_{7}&-7y_{7}\\ 7x_{7}&S_{6}\end{array}\right|=S_{6}S_{7}+7^{2}x_{7}y_{7}
detH7,9=\displaystyle\det H_{7,9}= |S99y936y99x9S88y836x98x8S7|=S9S8S7+82x8y8S9+92x9y9S7+362x9y9S8\displaystyle\left|\begin{array}[]{ccc}S_{9}&-9y_{9}&36y_{9}\\ 9x_{9}&S_{8}&-8y_{8}\\ 36x_{9}&8x_{8}&S_{7}\\ \end{array}\right|=S_{9}S_{8}S_{7}+8^{2}x_{8}y_{8}S_{9}+9^{2}x_{9}y_{9}S_{7}+36^{2}x_{9}y_{9}S_{8}
detH6,9=\displaystyle\det H_{6,9}= |S99y936y984y99x9S88y828y836x98x8S77y784x928x87x7S6|=S9[S6S7S8+72S8x7y7+82S6x8y8\displaystyle\left|\begin{array}[]{cccc}S_{9}&-9y_{9}&36y_{9}&-84y_{9}\\ 9x_{9}&S_{8}&-8y_{8}&28y_{8}\\ 36x_{9}&8x_{8}&S_{7}&-7y_{7}\\ 84x_{9}&28x_{8}&7x_{7}&S_{6}\end{array}\right|=\,S_{9}[S_{6}S_{7}S_{8}+7^{2}S_{8}x_{7}y_{7}+8^{2}S_{6}x_{8}y_{8}
+(28)2S7x8y8]+x9y9[92S6S7+(36)2S6S8]+(63)2x7y7x9y9\displaystyle+(28)^{2}S_{7}x_{8}y_{8}]+x_{9}y_{9}[9^{2}S_{6}S_{7}+(36)^{2}S_{6}S_{8}]+(63)^{2}x_{7}y_{7}x_{9}y_{9}
+(84)2(x7x8+y7y8+4y7x8+4x7y8+16x8y8)x9y9\displaystyle+(84)^{2}(x_{7}x_{8}+y_{7}y_{8}+4y_{7}x_{8}+4x_{7}y_{8}+16x_{8}y_{8})x_{9}y_{9}

It appears that all coefficients are perfect squares.

6. The third conjecture (permanents)

The equality of the permanent and the determinant of a matrix is a rare circumstance [12, 25]. Numerical checks on Hückel matrices for triangles and trapezia of small sizes, support the conjecture:

Conjecture 3. The permanent and the determinant of Hn(𝐱,𝐲)H_{n}({\bf x},{\bf y}) coincide. The same is true for Hk,n(𝐱,𝐲)H_{k,n}({\bf x},{\bf y}).

Accordingly, detHn=σH1,σ1H2,σ2HN,σN\det H_{n}={\sum}_{\sigma}H_{1,\sigma_{1}}H_{2,\sigma_{2}}...H_{N,\sigma_{N}} where σ\sigma are even permutations of 1,,N1,...,N, with N=(n+1)2N=(n+1)^{2}. Each nonzero term contains n(n+1)n(n+1) unit factors and n+1n+1 factors xjx_{j} or yky_{k} and is visualized as NN arrows 1σ11\to\sigma_{1}NσNN\to\sigma_{N} along the edges of the graph. All vertices of the graph participate with one outgoing and one incoming arrow that make closed oriented self-avoiding loops and dimers (two opposite arrows on the same edge). Such a configuration is a “loop covering” GG of the graph.

Theorem (Harari [14]): Let GG be a covering of the graph with dimers and oriented loops. If ϵG\epsilon_{G} is the number of dimers and oriented loops of even length in GG, and pGp_{G} is the product of the values of the edges in GG, then:

(18) detHn=G(1)ϵGpG\displaystyle\det H_{n}=\sum_{G}(-1)^{\epsilon_{G}}p_{G}

For honeycomb triangles and trapezia only even permutations contribute to the determinant, ϵG=1\epsilon_{G}=1.

7. Some Proofs

Some facts about determinants of Hückel matrices of graphene triangles and trapezia are now proven.

Proposition 7.1.

detHn(x,y)\det H_{n}(x,y) is a homogeneous palindromic polynomial of degree n+1n+1 in xx and yy:

detHn(x,y)=(xn+1+yn+1)+c1(xny+xyn)+c2(xn1y2+x2yn1)+\displaystyle\det H_{n}(x,y)=(x^{n+1}+y^{n+1})+c_{1}(x^{n}y+xy^{n})+c_{2}(x^{n-1}y^{2}+x^{2}y^{n-1})+\dots
Proof.

Consider the diagonal matrix Pn=diag[(1)(1,1t,1)(1,1t,1,1t,1)]P_{n}={\rm diag}[(1)(1,\frac{1}{t},1)(1,\frac{1}{t},1,\frac{1}{t},1)...], of size 1+3++(2n+1)=(n+1)21+3+...+(2n+1)=(n+1)^{2}. It is detPn=tn(n+1)/2\det P_{n}=t^{-n(n+1)/2} and

(19) tPnHn(x,y)Pn=Hn(tx,ty)\displaystyle tP_{n}H_{n}(x,y)P_{n}=H_{n}(tx,ty)

In particular: detHn(detPn)2t(n+1)2=detHn(tx,ty)\det H_{n}(\det P_{n})^{2}t^{(n+1)^{2}}=\det H_{n}(tx,ty), i.e.

(20) tn+1detHn(x,y)=detHn(tx,ty)\displaystyle t^{n+1}\det H_{n}(x,y)=\det H_{n}(tx,ty)

Therefore, the nonzero terms in the expansion of detHn(x,y)\det H_{n}(x,y) are the products H1i1HN,iNH_{1i_{1}}...H_{N,i_{N}} that contain monomials xkyn+1kx^{k}y^{n+1-k}.
This property and the symmetry detHn(x,y)=detHn(y,x)\det H_{n}(x,y)=\det H_{n}(y,x) imply the statement. The first coefficient is 1 because detHn(x,0)=xn+1\det H_{n}(x,0)=x^{n+1}. ∎

As the result (20) only depends on the positions of the non-zero matrix elements and not on their values, every single product H1,σ1HN,σNH_{1,\sigma_{1}}...H_{N,\sigma_{N}} that does not contain a monomial of degree n+1n+1 is identically zero.

The graph of the triangle of graphene with n+1n+1 rows has b=2n+1b=2n+1 blue dots at the end of the rows, b=12n(n1)b^{\prime}=\tfrac{1}{2}n(n-1) blue dots in bulk, r=12n(n+1)r=\tfrac{1}{2}n(n+1) red dots. The total number of vertices is b+b+r=(n+1)2b+b^{\prime}+r=(n+1)^{2}.
Depending on the numbering of vertices, one has different matrix representations of the graph.

Proposition 7.2.

The coefficients of the expansion of detHn(𝐱,𝐲)\det H_{n}({\bf x},{\bf y}) in monomials of degree n+1n+1, are perfect squares of integers.

Proof.

Let us number the blue dots at the row ends from 11 to 2n+12n+1, the remaining bb^{\prime} blue dots from 2n+22n+2 to b+bb+b^{\prime} and the red dots from b+b+1b+b^{\prime}+1 to (n+1)2(n+1)^{2}. The result is the block representation of the adjacency matrix:

(24) H~n=[B0C0OρCTρTOr]\displaystyle\tilde{H}_{n}=\left[\begin{array}[]{c|cc}B&0&C\\ \hline\cr 0&O^{\prime}&\rho\\ C^{T}&\rho^{T}&O_{r}\end{array}\right]

- BB has size (2n+1)×(2n+1)(2n+1)\times(2n+1) and elements in {0,xi,yi}\{0,x_{i},y_{i}\},
- CC is the matrix (2n+1)×r(2n+1)\times r of connections (0,1)(0,1) of row-ends with red dots,
- ρ\rho has elements 0,10,1 that are the connections of inner blue with red dots,
- the bb^{\prime} blue vertices are not inter-connected: this is the zero b×bb^{\prime}\times b^{\prime} matrix OO^{\prime} ,
- the rr red vertices are not inter-connected: this is the zero r×rr\times r matrix OrO_{r}.
For example (S0=x0+y0S_{0}=x_{0}+y_{0}):
0123456780S0110y1112x101130y214x2015011611100071110008111000\displaystyle\begin{array}[]{c||ccccc|c|ccc}&0&1&2&3&4&5&6&7&8\\ \hline\cr\hline\cr 0&S_{0}&&&&&&1&&\\ 1&&0&y_{1}&&&&1&1&\\ 2&&x_{1}&0&&&&1&&1\\ 3&&&&0&y_{2}&&&1&\\ 4&&&&x_{2}&0&&&&1\\ \hline\cr 5&&&&&&0&&1&1\\ \hline\cr 6&1&1&1&&&&0&0&0\\ 7&&1&&1&&1&0&0&0\\ 8&&&1&&1&1&0&0&0\\ \end{array} 026148573

The permutations of rows (1,2), (3,4) and (7,8) bring the matrix to a block form that is diagonal in the parameters, associated to a graph that topologically differs only in the end dots, that are now self-linked. The rest of the graph is unchanged by the axial symmetry. As a consequence the new matrix H^n\hat{H}_{n} is symmetric:
0123456780S011x1112y1113x214y215011611100071110008111000\displaystyle\begin{array}[]{c||ccccc|c|ccc}&0&1&2&3&4&5&6&7&8\\ \hline\cr\hline\cr 0&S_{0}&&&&&&1&&\\ 1&&x_{1}&&&&&1&&1\\ 2&&&y_{1}&&&&1&1&\\ 3&&&&x_{2}&&&&&1\\ 4&&&&&y_{2}&&&1&\\ \hline\cr 5&&&&&&0&&1&1\\ \hline\cr 6&1&1&1&&&&0&0&0\\ 7&&&1&&1&1&0&0&0\\ 8&&1&&1&&1&0&0&0\\ \end{array} 016238574

According to Prop.7.1, the determinant is a sum of homogeneous terms of degree n+1n+1. An expansion by diagonal of the matrix gives:

detHn(𝐱,𝐲)=sign×𝝉S0τ0x1τ1ynτ2nC𝝉\det H_{n}({\bf x},{\bf y})={\rm sign}\times{\sum}_{\boldsymbol{\tau}}S_{0}^{\tau_{0}}x_{1}^{\tau_{1}}...y_{n}^{\tau_{2n}}C_{\boldsymbol{\tau}}

where τi=0,1\tau_{i}=0,1 and i=02nτi=n+1\sum_{i=0}^{2n}\tau_{i}=n+1. The coefficient C𝝉C_{\boldsymbol{\tau}} is a principal minor of H^n\hat{H}_{n} of size (n+1)2n(n+1)^{2}-n, obtained by deleting the nn rows and nn columns that contain parameters with power 1, the other parameters being put equal 0. The principal minor times sign is a squared determinant.
In the example, the sign is (1)3(-1)^{3} and the coefficient (including sign) of S0x1y1S_{0}x_{1}y_{1} is

C1,1,1,0,0=|010101100011110000011000101000|=\displaystyle-C_{1,1,1,0,0}=-\left|\begin{array}[]{ccc|ccc}0&&&1&0&1\\ &0&&1&1&0\\ &&0&0&1&1\\ \hline\cr 1&1&0&0&0&0\\ 0&1&1&0&0&0\\ 1&0&1&0&0&0\end{array}\right|= |101110011|2=22\displaystyle\left|\begin{array}[]{ccc}1&0&1\\ 1&1&0\\ 0&1&1\\ \end{array}\right|^{2}=2^{2}

Remark 7.3.

The signed principal minor has the meaning of matching number of the subgraph (Kasteleyn theorem). In the example the subgraph (in the left) admits only two coverings with dimers (i.e. matchings) that are shown.

162857162857162857

The same tricks used for honeycomb triangles apply to trapezia with simple modifications:

Proposition 7.4.

1) detHk,n(x,y)\det H_{k,n}(x,y) is a homogeneous palindromic polynomial of degree n+1kn+1-k.
2) detHk,n(𝐱,𝐲)=(xn+yn)detHk,n1(𝐱,𝐲)xnyndetHk,n(𝐱,𝐲)\det H_{k,n}({\bf x},{\bf y})=(x_{n}+y_{n})\det H_{k,n-1}({\bf x^{\prime}},{\bf y^{\prime}})-x_{n}y_{n}\det H^{\prime}_{k,n}({\bf x^{\prime}},{\bf y^{\prime}}), where Hk,nH^{\prime}_{k,n} is the matrix Hk,nH_{k,n} with the two rows and two columns of xnx_{n},yny_{n} deleted.
3) If the parameters are distinct, the coefficients of the expansion of the determinant in monomials are perfect squares of integers.

8. Analytic reduction for small matrices

I now describe an analytic approach to the evaluation of the determinants. The key facts are the peculiar structure of the inverse of the diagonal blocks of HnH_{n} and Schur’s formula for determinants of block matrices.
Recall that TnT_{n} has size 2n+12n+1.

Lemma 8.1.

RnTTn1(xn,yn)RnR_{n}^{T}T_{n}^{-1}(x_{n},y_{n})R_{n} is a rank-1 matrix of size 2n12n-1. For example:

R3TT31(x3,y3)R3=x3y3x3+y3𝐮3𝐮3T,𝐮3T=[+1,0,1,0,+1]\displaystyle R_{3}^{T}T_{3}^{-1}(x_{3},y_{3})R_{3}=-\frac{x_{3}y_{3}}{x_{3}+y_{3}}{\bf u}_{3}{\bf u}_{3}^{T},\quad{\bf u}_{3}^{T}=\left[\begin{array}[]{cccccc}+1,&0,&-1,&0,&+1\end{array}\right]
R4TT41(xn,yn)R4=x4y4x4+y4𝐮4𝐮4T,𝐮4T=[+1,0,1,0,+1,0,1,]\displaystyle R_{4}^{T}T_{4}^{-1}(x_{n},y_{n})R_{4}=\frac{x_{4}y_{4}}{x_{4}+y_{4}}{\bf u}_{4}{\bf u}_{4}^{T},\quad{\bf u}_{4}^{T}=\left[\begin{array}[]{ccccccc}+1,&0,&-1,&0,&+1,&0,&-1,\end{array}\right]
Proof.

The matrix (x+y)Tn1(x,y)(x+y)T_{n}^{-1}(x,y) is the following:
\bullet if n=2k+1n=2k+1 the matrix has size 4k+34k+3 and it is

[(1y1y)(1y1y)(1y1xxyyxyyxyy1yxyy1x1y1y1y1y1xxyxxyyxyyxyyy1x1x1y1y11xxyx1x1x1x1]\displaystyle\left[\begin{array}[]{cccccccccccc}(-1&y&1&-y)&(-1&y&1&-y)&\ldots&(-1&y&1\\ x&-xy&y&xy&-y&-xy&y&1&-y&&-xy&y\\ 1&x&-1&y&1&-y&1&y&1&\ldots&-y&-1\\ -x&xy&x&-xy&y&xy&-y&-xy&y&\ldots&&-y\\ -1&-x&1&x&-1&y&1&-y&1&\ldots&&1\\ x&-xy&-x&&&&&&&&&\ldots\\ ...&&&&&&&&&&&\ldots\\ 1&x&-1&-x&&&&&&1&x&-1\end{array}\right]

\bullet if n=2kn=2k the matrix has size 4k+14k+1 and it is

[(1y1y)(1y1y)1y)1xxyyxyyxyyxyyxyy1x1y1y1y1y1xxyxxyyxyyxyyxyy1x1x1y1y1y11xxyx1x1xx1]\displaystyle\left[\begin{array}[]{ccccccccccccc}(1&y&-1&-y)&(1&y&-1&-y)&\ldots&&-1&-y)&1\\ x&xy&y&-xy&-y&xy&y&-xy&-y&\ldots&&-xy&-y\\ -1&x&1&y&-1&-y&1&y&-1&-y&\ldots&&-1\\ -x&-xy&x&xy&y&-xy&-y&xy&y&-xy&\ldots&&y\\ 1&-x&-1&x&1&y&-1&-y&1&y&-1&&1\\ x&xy&-x&&&&&&&&&&\ldots\\ ...&&&&&&&&&&&&\ldots\\ 1&-x&-1&x&&&&&&&&x&1\end{array}\right]

(brackets are introduced to ease the pattern). If xy=1xy=1 the matrix is Toeplitz.
Multiplication of T1T^{-1} with the rectangular matrices deletes the first and last rows and columns, and replaces even rows and columns with zeros.
For example: R3TT31(x,y)R3R_{3}^{T}T_{3}^{-1}(x,y)R_{3} and R4TT41(x,y)R4R_{4}^{T}T_{4}^{-1}(x,y)R_{4} are:

xyx+y[10+10100000+1010+10000010+101],xyx+y[+1010+101000000010+1010+10000000+1010+101000000010+1010+1]\displaystyle\frac{xy}{x+y}\left[\begin{array}[]{ccccc}-1&0&+1&0&-1\\ 0&0&0&0&0\\ +1&0&-1&0&+1\\ 0&0&0&0&0\\ -1&0&+1&0&-1\end{array}\right],\quad\frac{xy}{x+y}\left[\begin{array}[]{ccccccc}+1&0&-1&0&+1&0&-1\\ 0&0&0&0&0&0&0\\ -1&0&+1&0&-1&0&+1\\ 0&0&0&0&0&0&0\\ +1&0&-1&0&+1&0&-1\\ 0&0&0&0&0&0&0\\ -1&0&+1&0&-1&0&+1\end{array}\right]

The matrices are then written in terms of vectors. ∎

Proposition 8.2.

detHn\det H_{n} coincides with the determinant of the bordered matrix of size n2+1n^{2}+1:

(27) detHn=det[xn+yn()nxnUTynUHn1(𝐱,𝐲)]\displaystyle\det H_{n}=\det\left[\begin{array}[]{c|c}x_{n}+y_{n}&(-)^{n}x_{n}U^{T}\\ \hline\cr y_{n}U&H_{n-1}({\bf x^{\prime},y^{\prime}})\end{array}\right]

where UnT=[0,0,𝐮n]U_{n}^{T}=[0,\ldots 0,{\bf u}_{n}] begins with (n1)2(n-1)^{2} zeros, followed by 𝐮n{\bf u}_{n} that has 2n12n-1 alternating components 1,0,1,0,1,1,0,-1,0,1,....

Proof.

Let us evaluate detHn\det H_{n} with Schur’s formula :

det[ABCD]=detDdet(ABD1C).\displaystyle\det\left[\begin{array}[]{cc}A&B\\ C&D\end{array}\right]=\det D\det(A-BD^{-1}C).

In this case, the full matrix is HnH_{n}, D=Tn(xn,yn)D=T_{n}(x_{n},y_{n}). With detTn=xn+yn\det T_{n}=x_{n}+y_{n}, the formula and prop.8.1 give:

detHn=(xn+yn)det[Hn1(1)nxnynxn+ynUnUnT]\displaystyle\det H_{n}=(x_{n}+y_{n})\det\left[H_{n-1}-(-1)^{n}\frac{x_{n}y_{n}}{x_{n}+y_{n}}U_{n}U_{n}^{T}\right]

If a0a\neq 0 is a number, Schur’s formula states (Sherman-Morrison-Woodbury formula):

det[a𝐛T𝐜M]=adet(M1a𝐜𝐛T)\displaystyle\det\left[\begin{array}[]{cc}a&{\bf b}^{T}\\ {\bf c}&M\end{array}\right]=a\det(M-\frac{1}{a}{\bf c}{\bf b}^{T})

With a=xn+yna=x_{n}+y_{n}, M=Hn1M=H_{n-1}, 𝐛T=(1)nxnUnT{\bf b}^{T}=(-1)^{n}x_{n}U^{T}_{n} and 𝐜=ynUn{\bf c}=y_{n}U_{n} the result follows. ∎

These are the first bordered matrices:

detH1=|x1+y1x1y1x0+y0|,detH2=|x2+y20x20x20x0+y0010y2001y101101y20x110|\displaystyle\det H_{1}=\left|\begin{array}[]{c|c}x_{1}+y_{1}&-x_{1}\\ \hline\cr y_{1}&x_{0}+y_{0}\end{array}\right|,\quad\det H_{2}=\left|\begin{array}[]{c|cccc}x_{2}+y_{2}&0&x_{2}&0&-x_{2}\\ \hline\cr 0&x_{0}+y_{0}&0&1&0\\ y_{2}&0&0&1&y_{1}\\ 0&1&1&0&1\\ -y_{2}&0&x_{1}&1&0\end{array}\right|
detH3=|x3+y30000x30x30x30x0+y001000001y101011010000x11001y300000100y20110100y30010100100101y3x20010|\displaystyle\det H_{3}=\left|\begin{array}[]{c|ccccccccc}x_{3}+y_{3}&0&0&0&0&-x_{3}&0&x_{3}&0&-x_{3}\\ \hline\cr 0&x_{0}+y_{0}&0&1&0&0&&&&\\ 0&0&0&1&y_{1}&0&1&&&\\ 0&1&1&0&1&0&&0&&\\ 0&0&x_{1}&1&0&0&&&1&\\ y_{3}&0&0&0&0&0&1&0&0&y_{2}\\ 0&&1&&&1&0&1&0&0\\ -y_{3}&&&0&&0&1&0&1&0\\ 0&&&&1&0&0&1&0&1\\ y_{3}&&&&&x_{2}&0&0&1&0\end{array}\right|

The bordering vectors and the off diagonal matrices RkR_{k} have non-zero elements in different rows and columns. This makes it possible to iterate the procedure with Schur’s formula, that eliminates a block but adds a border. In the end, the determinant of the matrix of size (n+1)2(n+1)^{2} condensates to a bordered matrix of size n+1n+1. This was done by hand for nn up to 5 to guess the binomial pattern of the reduced matrix, and thus the connection with the Pascal matrix.

Aknowledgements

I thank prof. Stefano Evangelisti for addressing me to the study of determinants of Hückel matrices. This, unexpectedly, turned me to the beautiful mathematics of Pascal matrices and combinatorics (that are in The Book [1]).
I thank prof. Andrea Sportiello for his advices and insight for a possible proof.

References

  • [1] Martin Aigner and Günter M. Ziegler, Proofs from the Book, 5th Edition, Springer 2014.
  • [2] G. E. Andrews, Plane Partitions (III): The Weak Macdonald Conjecture, lnventiones Math. 53 (1979) 193–225. https://doi.org/10.1007/BF01389763
  • [3] G. E. Andrews, Plane Partitions, V: The T.S.S.C.P.P. conjecture, J. Combin. Theory Ser. A 66 (1994), 28–39. https://doi.org/10.1016/0097-3165(94)90048-5
  • [4] Y. B. Apriliyanto, S. Battaglia, S. Evangelisti, N. Faginas-Lago, T. Leininger, and A. Lombardi, Toward a generalized Hückel rule: the electronic structure of Carbon Nanocones, Journal of Physical Chemistry A 125 (2021) 9819–9825.
    https://doi.org/10.1021/acs.jpca.1c06402
  • [5] A. T. Balaban, D.J. Klein and X. Liu, Graphitic cones, Carbon 32 (2) (1994) 357–359.
    https://doi.org/10.1016/0008-6223(94)90203-8
  • [6] O. Bodroza, I. Gutman, S. I. Cyvin and R. Tosić, Number of Kekulé structures of hexagon-shaped benzenoids, J. Math. Chem. 2 (1988) 287–298. https://doi.org/10.1007/BF01167208
  • [7] R. Brawer and M. Pirovino, The linear algebra of the Pascal matrix, LAA 174 (1992). 13–23.
    https://doi.org/10.1016/0024-3795(92)90038-C
  • [8] M. Ciucu, T. Eisenkölbl, C. Krattenthaler, D. Zare, Enumeration of lozenge tilings of hexagons with a central triangular hole, J. Combin. Theory Ser. A 95 (2001) 251–334.
    https://doi.org/10.1006/jcta.2000.3165
  • [9] T. Doslic, Perfect matchings, Catalan numbers and Pascal’s triangle, Mathematics Magazine 80 (3) (2007) 219–226. https://www.jstor.org/stable/27643031
  • [10] A. Edelman and G. Strang, Pascal matrices, The American Mathematical Monthly 111 n.3 (2004) 189–197. https://doi.org/10.2307/4145127
  • [11] I. Gessel and G. Viennot, Binomial determinants, paths and hook length formulae, Adv. Math. 58 (1985) 300–321. https://doi.org/10.1016/0001-8708(85)90121-5
  • [12] P. M. Gibson, Conversion of the permanent into the determinant, Proc. AMS 27 (3) (1971) 471–476. https://www.ams.org/journals/proc/1971-027-03/S0002-9939-1971-0279110-X/S0002-9939-1971-0279110-X.pdf
  • [13] D. Guy and C. Tomei, The problem of calissons, Amer. Math. Monthly 96 (5) (1989) 429–431. https://doi.org/10.2307/2325150
  • [14] F. Harari, The Determinant of the Adjacency Matrix of a Graph, SIAM Review 4 (1962) 202–210. https://doi.org/10.1137/1004057
  • [15] H. Heiberg–Andersen and A T. Skjeltorp, Spectra of conic carbon radicals, Journal of Mathematical Chemistry 42 (4) (2007) 707–727. https://doi.org/10.1007/s10910-006-9103-z
  • [16] Picture by D. Knudsen Kenneth, CC BY-SA 3.0,
    https://commons.wikimedia.org/w/index.php?curid=10522963
  • [17] H. Heiberg-Andersen, A. T. Skjeltorp and K. Sattler, Carbon nanocones: A variety of non-crystalline graphite, Journal of Non-Crystalline Solids 354 47-51 (2008) 5247–5249.
    https://doi.org/10.1016/j.jnoncrysol.2008.06.120
  • [18] C.  Krattenthaler, Plane partitions in the work by Richard Stanley and his school, in ’The Mathematical Legacy of Richard P. Stanley’ P. Hersh, T. Lam, P. Pylyavskyy and V. Reiner (eds.), Amer. Math. Soc., R.I., 2016, pp. 246-277. https://arxiv.org/abs/1503.05934.
  • [19] W. F. Lunnon, The Pascal matrix, The Fibonacci Quaterly 15 (1977) 201–204.
    https://www.fq.math.ca/Scanned/15-3/lunnon.pdf
  • [20] S. Mitra and B. Nienhuis, Exact conjectured expressions for correlations in the dense O(1) loop model on cylinders, J. Stat. Mech. (2004) P10006.
    https://doi.org/10.1088/1742-5468/2004/10/P10006
  • [21] S. Mitra, Exact asymptotics of the characteristic polynomial of the symmetric Pascal matrix J. Combinatorial Theory Series A 116 (2009) 30–43.
    https://doi.org/10.1016/j.jcta.2008.04.004
  • [22] On-Line Encyclopedia of Integer Sequences. https://oeis.org/A045912
  • [23] A. V. Razumov and Yu. G. Stroganov, Enumerations of half-turn symmetric alternating-sign matrices of odd order, Theor. Math. Phys. 148 (2006) 1174–1198.
    https://doi.org/10.1007/s11232-006-0111-8
  • [24] L. Yates, Linear algebra of Pascal matrices (2014)
    https://www.gcsu.edu/sites/files/page-assets/node-808/attachments/yates.pdf
  • [25] W. McCuaig, Polya’s permanent problem, Journal of Combinatorics 11 (2004) R79 (83 pp).
    https://doi.org/10.37236/1832
  • [26] D. Zeilberger, Proof of the alternating sign matrix conjecture, Electron J. Combinatorics 3 (1996) 1213 (84pp). https://doi.org/10.37236/1271