ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 19 (2020) 351-362
https://doi.org/10.26493/1855-3974.2262.9b8
(Also available at http://amc-journal.eu)
On the Smith normal form of the Varchenko matrix*
*
Tommy Wuxing Cai
Department of Mathematics, University of Manitoba, Winnipeg, MB R3T 2N2 Canada
YueChen©
School of Sciences, South China University of Technology, Guangzhou 510640, PR China
Lili Mu t ©
School of Mathematics,Liaoning Normal University, Dalian 116029, PR China Received 24 February 2020, accepted 1 September 2020, published online 21 November 2020
Let A be a hyperplane arrangement in A isomorphic to Rn. Let Vq be the q-Varchenko matrix for the arrangement A with all hyperplane parameters equal to q. In this paper, we consider three interesting cases of q-Varchenko matrices associated to hyperplane arrangements. We show that they have a Smith normal form over Z[q].
Keywords: Hyperplane arrangement, Smith normal form, Varchenko matrix. Math. Subj. Class. (2020): 15A21, 52C35
1 Introduction
Let M be an n x n matrix over a commutative unital ring R. We say that M has a Smith normal form (SNF for short) over R if there are matrices P, Q G Rnxn such that det(P) and det(Q) are units in R and PMQ is a diagonal matrix diag(di, d2,..., dn) where d divides dj in R for all i < j.
* The first author and the third author would like to thank M.I.T. for their hospitality and the China Scholarship Council for their support. Part of the work was done when they were visiting the M.I.T. Department of Mathematics during the 2013-2014 academic year. They thank Richard Stanley for his comprehensive help on this work. The first author and the second author thank Naihuan Jing for his help. The authors thank the anonymous referee for his/her careful reading and helpful comments.
t Corresponding author. The author was supported by the National Natural Science Foundation of China (No. 11701249) and the Natural Science Foundation of Liaoning Province (Grant No. 2019-BS-152).
E-mail addresses: caiwx@scut.edu.cn (Tommy Wuxing Cai), 353043301@qq.com (Yue Chen), lly-mu@hotmail.com (Lili Mu)
Abstract
©® This work is licensed under https://creativecommons.org/licenses/by/4.0/
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Recently, there is an interest in SNF in combinatorics. A survey of this topic was given by Stanley in [11]. The SNF of a matrix of a differential operator was considered by Stanley and the first author in [2], where they proved a special case of a conjecture given by Miller and Reiner [7]. In [13], interesting results concerning the SNF of random integer matrix were found.
It is well known that M has an SNF if R is a principal ideal domain (PID), but not much is known for general rings. In this paper we are interested in the integer polynomial ring Z[q]. Some matrices in Z[q]nxn do not have an SNF over R. For example, it is not hard to show that [2 0] does not have an SNF over Z[q]. However, lots of matrices in Z[q]nxn do have sNf over Z[q]. For example, it is asked whether every matrix of the form A = (q°ij), where a^ are nonnegative integers, has an SNF over Z[q]. There is not a general solution to this question. But we could give a positive answer which arises from some special cases of geometrical structures. The matrices we are interested in are called Varchenko matrices (see [12]). These matrices are associated to a hyperplane arrangement (see Definition 1.2). The Varchenko matrix was studied in the papers of Varchenko [12], Schechtman and Varchenko [8], and Brylawski and Varchenko [1]. These matrices describe the analogue of Serre's relations for quantum Kac-Moody Lie algebras and are relevant to the study of hypergeometric functions and the representation theory of quantum groups [6]. Entries appearing in the diagonal of a Smith normal form of a matrix are called invariant factors. Applications of invariant factors of a q-matrix can be found in [3, 4, 9]. We are going to prove that Varchenko matrices associated to some hyperplane arrangements do have an SNF.
We use the notation and terminology on hyperplane arrangements in [10]. A finite (real) hyperplane arrangement A is a finite set of affine hyperplanes in some affine space A isomorphic to Rn.
For a hyperplane H in A, let
Ah = {H n H' : H' e A such that H' n H = 0 and H' = H}.
This is a hyperplane arrangement in the affine space H. We also write A - {H} for the arrangement from A with H removed.
Let A be a hyperplane arrangement in A. Then A is divided into some regions by these hyperplanes. Explicitly, a region is a connected component of A - |JHeA H. We let R(A) denote the set of regions of A.
Example 1.1. In the following picture, arrangement Ap is an example of the so-called peelable arrangement, which is treated in Section 2. Here we see straight lines a, b, c form a hyperplane arrangement in the plane R2. There are 7 regions of Ap which we denote by 1', 2', 3', 1, 2, 3,4. (We write it in this way for the example in Section 2.) The hyperplane arrangement A contains two affine hyperplanes A = b n a, B = b n c (two points in b).
Arrangement Ap is also an example of the regular n-gon arrangement Gn, which is treated in Section 4. It is a regular triangle arrangement. (Although in Figure 1 the central triangle is not so much like a equilateral triangle. This does not matter, because the Varchenko matrix that we are concerned with is a topological invariant.) As another example for the n-gon arrangement, a picture of the pentagon arrangement Gs is given in Section 4.
Arrangement C4 in Figure 1 is an example of arrangement Cn, which is treated in Section 3.
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d x c
(a) Arrangement Ap
(b) Arrangement C4
Figure 1: Arrangements Ap and C4.
Definition 1.2. Let A be a finite hyperplane arrangement and R( A) its set of regions, and let aH for H e A be indeterminates. The Varchenko matrix V = V (A) is indexed by R(A) with the entries given by
Vrr =	aH,
HeSepA(R,R')
(1.1)
where Sep^(R, R') is the set of hyperplanes in A which separate R and R'. We write Vq = Vq (A) for V(A) when we set each aH = q, an indeterminate, and call Vq the q-Varchenko matrix of A.
Thus (Vq)RR = q#Sep(R,R ). Also note that V(A) and Vq(A) are symmetric matrices with 1's on the main diagonal.
We are interested mostly in the q-Varchenko matrix Vq. We are going to prove that Vq(A) has an SNF over the ring Z[q] for the peelable arrangements (in Section 2), arrangement Cn (in Section 3) and regular n-gon arrangement Gn (in Section 4). (Since this ring is not a PID, an SNF does not a priori exist.) In Section 5, we compute the SNF of the Varchenko matrices for two arrangements which are not included in the previous sections.
2 Peelable hyperplane arrangements
Example 2.1. Let us look at the arrangement Ap in Example 1.1. Its Varchenko matrix
Vq = Vq (Ap) is
Vq
1	q	q2	q	q2	q3	q2 "
q	1	q	q2	q	q2	q3
q2	q	1	q3	q2	q	q2
q	q2	q3	1	q	q2	q
q2	q	q2	q	1	q	q2
q3	q2	q	q2	q	1	q
q2	q3	q2	q	q2	q	1
where the columns are indexed by the regions in the order 1', 2', 3', 1,2,3,4, and so are the rows. We will briefly show that this matrix has an SNF. We write Vq as a block matrix the
a
b
b
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way it is partitioned:
Vq
A1	B1	C1
A2	B2	C2
As	Bs	Cs
Notice that (Bi, Ci) = q(B2, C2) and A2 = qA^ (This is not a coincidence. We see that [B1, C1] is the submatrix indexed by 1', 2', 3' (rows) and 1, 2, 3,4 (columns), while [B2,C2] is the submatrix indexed by 1,2,3 (rows) and 1,2,3,4 (columns). There is one more line, line b, to separate regions i' and j' than regions i and j.) We can multiply by the following matrix on the left to cancel Bi :
P
Is -qls 0 0 Is 0 0 0 1
We have
As Vq is a symmetric matrix, so is PVPl. We thus have
	A1 - qA2	0	0
PVq =	A2	B2	C2
	As	Bs	Cs
	A1 - qA2	0	0
PVq P1 =	0	B2	C2
	0	Bs	Cs
(1 - q2)Ai 0
0
M1
where we write
M1 =
B2 Bs
C2 Cs
and we use that A2 = qA1. The matrix A1 is the q-Varchenko matrix of A£. (See Example 1.1 for the notation A£). The matrix M1 is the q-Varchenko matrix of Ap - {b}. We can use induction to transform PVqP1 into an SNF.
This example motivates us to define a peelable hyperplane in an arrangement.
Definition 2.2. Let A be a finite hyperplane arrangement and H be a hyperplane in A. We say that H is peelable (from A) if there is one side Hf of H such that if R is a region of A and R is in Hf, then R n H is the closure of a region of AH.
For example, the hyperplane b is peelable from Ap in Example 1.1. Let us see why this is. On the side above b there are three regions 1', 2' and 3'. For each one of these regions, the intersection of its closure with b is actually a closure of a region of A£. For instance, the closure of region 2' intersects b at a line section AB, and this line section is actually a closure of a region of A£. (In fact Ab has 3 regions: the part to the left of A, the part between A and B, and the part to the right of B.)
Theorem 2.3. Assume that H is peelable from A. Then there is a matrix P with entries in Z[q] such that det(P) = 1 and
PVq (A)P' :
(1 - q2)Vq (AH ) 0
0
Vq (A-{H })
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Remark 2.4. Under the same assumption, a similar result can be given for the Varchenko matrix V(A), and the proof is almost the same. Using this result, we can prove that the Varchenko matrix V(A) associated to a peelable hyperplane arrangement (as defined below) has a "diagonal form" in Z[aH : H € A], that is, we can find matrices P, Q whose determinants are units and PV (A)Q is a diagonal matrix. Let us mention that, subsequent to our work, Gao and Zhang [5] gave a necessary and sufficient condition on an arrangment A for V (A) to have a diagonal form.
The main idea of the proof of this theorem is in the previous example. We will give a rigorous proof in a while, in order to make sure there is no gap that might have occurred when we move from the more visualizable two-dimensional example.
Iteratively using this result, the Varchenko matrices of a special type of hyperplane arrangement can be shown to have an SNF.
Definition 2.5. Let A = {Hi, H2,..., Hm} be a finite hyperplane arrangement. We inductively define A to be peelable as follows.
1.	If m = 1 then A = {H1} is peelable.
2.	If there is one peelable hyperplane H in A such that both A - {H} and AH are peelable, then we say that A is peelable.
Now it is easy to see that we have the following result.
Corollary 2.6. The q-Varchenko matrix Vq (A) of a peelable hyperplane arrangement A has an SNF over Z[q]. Moreover, its SNF is of the form
diag((1 - q2)"1, (1 - q2)"2,..., (1 - q2)"^),
where 0 < n1 < n2 < ■ ■ ■ < nr is a sequence of nonnegative integers and r is the number of regions of A.
We will need the following two results, which are not hard to prove.
Lemma 2.7. Let H be a hyperplane in A. Assume that R is a region such that R n H contains a point which is an interior point of some region R1 in AH. Then R = R n H.
Lemma 2.8. Let H be a hyperplane in A. Assume that R is a region such that R n H is the closure of some region of AH. Then there is a unique region R' on the other side of H such that R' n H = R n H.
To simplify the wording of the proof of Theorem 2.3, we introduce a new notation.
Definition 2.9. Let A be a hyperplane arrangement. Let R1, R2 be two subsets of R(A). We denote by Vq (R1, R2) the submatrix of Vq (A) with rows indexed by R1 and column indexed by R2.
Now let us prove Theorem 2.3. Assume that H is peelable from A and Hf is a side of H with the properties as in Definition 2.2. Let R1,R2,... ,Rs be the set of the regions in Hf. Let H, A, Hf be as in the Definition 2.2. Let R = {1', 2',..., r'} denote the set of regions in Hf, and let R1 = {1,2,..., r} denote the corresponding regions on the other side of Hf as given by the previous lemma. Let R2 = {r +1,..., r + s} be the set of other
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regions. Let R' = {1, 2,..., r +1}, i.e., R' is the union of Ri and R2. It is not difficult to prove the following facts:
Vq (Ri, Ri) Vq (R', R') Vq (Ri, R') Vq (Ri, Ri)
Vq (AH ) Vq (A-{H }) qVq (Ri, R') qVq(Ri, Ri).
The q-Varchenko matrix V = V( A) has the following block matrix form: Vq (A)
Now an argument similar to Example 2.1 can be applied to prove the theorem.
Vq (Ri, Ri) Vq (Ri, Ri) Vq (R'i, R2) Vq (Ri, Ri) Vq (Ri, Ri) Vq (Ri, R2) Vq (R2, Ri) Vq (R2, Ri) Vq (R2, R2 )
3 The case that all lines go through the same point
From now on, we consider hyperplane arrangements in R2. Define Cn to be the arrangement consisting of n lines intersecting in a common point in R2. We prove that the q-Varchenko matrix V(n) associated to Cn has a Smith normal form (over Z[q], as usual). This matrix has the form
V(n)
1 q q2 q3
2
qq
2 3 4
q q2 q3 q4
q
„n-1
q
i-1
qn—i qn-2
q
q2
1
Remark 3.1. This matrix is an example of circulant matrices C(ci, c2,. defined by
C(ci, C2, . . . , Cn)
ci	c2	c3 . .	. cn—i	cn
cn	ci	c2 . .	. cn —2	cn—i
cn—i	cn	ci . .	. cn—3	cn—2
c2
c3 c4
Ci
.., cn) which is
(3.1)
We see that V(n) is circulant because the regions of Cn are in a circular mode. Similar but more complicated situations occur in the regular n-gon arrangement, which is considered in the next section.
Proposition 3.2. Let n be a positive integer. Then the Varchenko matrix V(n) has the following Smith normal form over Z[q]:
diag(1, 1 - q2, . . . , 1 - q2, (1 - q2)2, (1 - q2)(1 - q2n), . . . , (1 - q2)(1 - q2n)). (3.2)
n	n —2
Proof. First successively apply the row operations r - qri—i (i = n, n - 1,..., 2), rn+i - qrn+i+i (i = 1, 2,..., n - 1), r2n - qri. This transforms V(n) into the block
1
n
q
q
c
n
T. W. Cai, Y. Chen and L. Mu: On the Smith normal form of the Varchenko matrix	357
matrix
1 a O M
q O
0 ¡3 1 - q2
where M is a 2(n — 1) x 2(n — 1) matrix, a, 3 are row vectors, O is a zero column vector and ¡'s components are all multiples of 1 — q2. It's easy to see that we only need to find the Smith normal form of M. Factoring 1 — q2 out of M, one finds that
where
M = (1 — q2)
%-2
A = ^ qk Tk, B = J2
A B Bt A1
2
k=0
=q
k=0
■t-1-kfrrit\ k
(T l)A
and T = (tij) with tijj = 6i+i,j. Note that A is a unitriangular matrix; in particular, it is invertible in Z[q]. Multiplying M on the left by
we transform M into
P
(1 — q2)
I
—BlA
O
-i
A
B
OA1- B lA-iB
We see that we only need to find the Smith normal form of A1 — BlA iB, but it can be seen from the following lemma that its SNF is
diag(1 — q2,1 — q2n,..., 1 — q2n).
n-2
Now the SNF of V(n) follows.
Lemma 3.3. Let m x m matrix T = (tij) with ti,j = 6i+i,j. Let
(3.3)
□
i
i
A =$3 qkTk, B = ^ qm-k(T')k.
k=0
k=0
Then the matrix
C = (Im — qT' )(A' — B' A B)
is equal to a matrix with first row
(1 — q2, q3 — q2m+i, q4 — q2m, q5 — q2m-i, . . . , qm+i — qm+3)
the other diagonal entries all equal to 1 — q2m+2, and all other entries zero. Proof. First A-i = Im — qT, so
BlA-i = qmIm +y(q
m-i
m + ^ (qm-k — qm+2-k)Tk. k=i
I
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Then one computes BlA lB and finds it is equal to
M =
q""
qm+l

,2m+l
qm+l
qm+2
q5 - q
q2m 2m+l
„6
qm+2 qm+3
m+l
m+3
qm+2 _ qm+4 qm+3 - qm+5
q2m-1 _ q2m+1
2m
Now let N = (Im _ qTl)M. We find that the first row of N is the same as that of M, the other diagonal entries of N are all equal to q2m+2 and all other entries are zero. Now we see that C is as claimed in the lemma since
C = (Im _ qTl)Al _ (Im _ qTl)M = Im _ (Im _ qTl)M = Im _ N.
□
4 The case of regular n-gon arrangement Gn
Let Gn be the arrangement in R2 obtained by extending the sides of a regular n-gon. Let Vq(Gn) be the Varchenko matrix associated to Gn. We are going to prove the following
Theorem 4.1. Let Vq(Gn) be the Varchenko matrix associated to the regular n-gon arrangement Gn. A Smith normal form of Vq (Gn) over Z[q] is
diag(1,1 _ q2,..., 1 _ q2,(1 _ q2)2, . . . , (1 _ q2)2)
(4.1)
(p-l)n
where p is the integer part of (n + 1)/2.
The above result can be proved by using some results and tools in [3, 12]. But we want to prove it directly. First, it is easy to calculate the number of regions of the arrangement Gn. For instance, one uses the formula that the number of regions is one more than the sum of the number of the lines and the number of intersection points.
Lemma 4.2. The number of the regions associated to the regular n-gon arrangement Pn is np + 1, where p is the integer part of (n + 1) /2.
The main idea of the proof of Theorem 4.1 is to group the regions by their shapes. We then write the Varchenko matrix as a block matrix. The columns of each block are labeled by regions of a same shape and so are the rows of a block. For regions of the same shape, we order them clockwise. The key property of this treatment is that each block is a circulant matrix. Once we write the block matrix down, it will be relatively easy to do cancelations and turn it into an SNF, although it takes some space to write the process down. To show how to write the block matrix, we consider the example of G5. Then in the proof we write the block matrix for general n and then do the cancelation.
Example 4.3. We mark the regions of G5 (see Figure 2) as in the following.
They are regions A(j) (i = 1,2,..., 5; j = 1,2,3) together with a unmarked central region. Note that we mark the regions according to their shape. Precisely, for each j, the shape of the A(j) for i = 1, 2,..., 5 are the same. Let us call them the regions of type
2
3
4
q
q
q
3
q
4
q
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v(3) 4
v(3)
A.
(3) 2
(3)
Figure 2: Arrangement G5.
(j) A(j)
, A
(j)
We
j. For regions of the same type, we label them clockwise as AY', A call AY' the leading region of the type j regions. The union of the three leading regions A^1', Af, a13' is the region inside an exterior angle of the pentagon. (So is the union of three region Ai1', Ai2', Ai3'.) We obtain the Varchenko matrix
Vq (Gs) =
1	Qi	Q2	Qs'
Qi	Eii Ei2 Eis
Q2	E2i E22 E2S
Qs Esi Es2 Ess
where the first (block) column is indexed by the central non-marked region. For j = 2, 3,4, the jth block column is indexed by the type j regions. The block rows are indexed in the same way. For example, the rows of the matrix Ei2 are indexed by the type 1 regions and the columns of it are indexed by type 2 regions. Because regions of the same type are ordered in a circular mode, the blocks Ej should all be circulant matrices (see Remark 3.1). In fact, it can be checked that the blocks are as follows
Qk = (qk,qk,qk, qk, qk) Eii = C (1, q2, q2, q2, q2 ) E22 = C (1,q2,q4,q4,q2 )
for k = 1, 2, 3,
Ei2 = C(q, qS, qS, qS, q)
E23 = C(q, qS,q5,qS,q)
Eis = C (q2,q4,q4,q2,q2 ), Ess = C(1,q2,q4,q4, q2).
We then use Gaussian elimination (in blocks) to turn the matrix into an SNF. For instance, at the beginning, we subtract the q times of the third block row (Q2 E21 E22 E23) from the fourth block row (Q3 E31 E32 E33).
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Proof. We write the Varchenko matrix of Gn in the following form of block matrix:
Vq (Gn) =
1	Qi	Q2 . .	. Qp
Q1	En	E12 .	. Ei p
Q2	E21	E22 .	. E2 p
Qpp Epi EP2 .
Ep
where Ekl = E/k, Qk is the row vector
Qk = (qk ,qk ,...,q
(4.2)
and Ej (i < j) is a circulant matrix:
Ei.
cfqj-i, qj-i+2, qj-i+4, . . . , qj-i+2(i-i), qi+j, . . .
i+j
qj+i-2 q>+i-4
q_,
n+1-i-j
.,qj+i-2(i-l),qj-i,...,qj-i
v-v-'
Now we apply Gaussian elimination to Vq(Gn) and transform it into the desired diagonal form. We do this in blocks and we will use the multiplication of elementary block matrices to realize the elimination. We proceed in four steps.
Step 1: We first apply some row eliminations. Let
R
i=
"qInXl In
and Rk
-qIn In
for k > 2, where Inxl is a column of n 1's and Rk comes from the (block) identity matrix by adding the —q times of its (k - 1)th block row to it's kth block row. Now compute the matrix Mi = R1R2 • • • RpVq(Pn).
Step 2: We apply some column eliminations. Let
Si =
i -qllXn In
In-
and Sk
In -qIn
for k > 2, where Iixn is a row of n 1's and Sk comes from the (block) identity matrix by adding the —q times of its (k — 1)th block column to its kth block column. (So Sk = Tkl.) Now compute the matrix M2 = MiSp • • • S2Si.
k
i
I
i

I
I

I

i
I

I

I
T. W. Cai, Y. Chen and L. Mu: On the Smith normal form of the Varchenko matrix	361
Step 3: We apply some more row eliminations. Let
Tk
-qJ In
with J
0	1	0 .	.0	0
0	0	1 .	.0	0
0	0	0 .	.0	1
1	0	0 •	• 0	0
where Tk come from the (block) identify matrix by adding the —q J times the kth block row to the (k + 1)th block row. Now compute M3 = Ti • • • Tp-iM2. We find the Varchenko matrix is now transformed to
1
0	0	0 .	.0	0
D	N12	N13 .	. Nip-1	Nip
0	D'	N23 .	. N2p-i	N2p
0	0	0 .	. D'	Np-ip
0	0	0 .	.0	D'
M3 =
where D = (1 — q2)In, D' = (1 — q2)2In and all non-diagonal entries are the multiple of the diagonal entry on the same row. This ensures that we can do the following:
Step 4: We apply more column eliminations to cancel the non-diagonal entries. This does not change the diagonal entries of M3. We finish the proof as the diagonal of M3 is the same as that of (4.1).	□
i
I
I

5 Two more examples
We now simply say that a hyperplane arrangement A has SNF if its Varchenko matrix Vq (A) has an SNF over Z[q]. We can use Theorem 2.3 to give more examples of hyperplane arrangements who have SNF. For example, starting from an arrangement which has SNF, for instance Cn, we can keep adding straight lines to it. As long as every time the line added does not separate the set of intersection points of the previous arrangement, the new arrangement will have SNF. This helps us to construct lots of examples of hyperplane arrangements having SNF. We now give two examples which can not be constructed this way. We found that they both have SNF.
1.	The Shi arrangement S3 with hyperplanes x® — Xj =0,1 for 1 < i < j < 3. We write the multiplicity of a diagonal element in brackets following that entry. For instance, 1 — q2 [3] indicates that 1 — q2 occurs three times as a diagonal element of the SNF. The diagonal elements of the SNF of Vq(S3) are 1 [1], 1 — q2 [6], (1 — q2)2 [6], and (1 — q2)(1 — q6) [3].
2.	Define a hyperplane arrangement A in R3 by the equations x = 0, y = 0, z = 0,
x — y — z = 0. We verified that its q-Varchenko matrix has an SNF over Z[q], with
diagonal entries 1 [1], 1—q2 [4], (1 — q2)2 [6], (1 — q2)3 [2], and (1—q2)2(1 —q8) [1].
Based on the previous examples, it is natural to consider the following problem.
Problem 5.1. Do all hyperplane arrangements have SNF?
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ORCID iDs
Tommy Wuxing Cai © https://orcid.org/0000-0001-9816-4115
Yue Chen © https://orcid.org/0000-0002-8173-0894
Lili Mu © https://orcid.org/0000-0003-0911-519X
References
[1]	T. Brylawski and A. Varchenko, The determinant formula for a matroid bilinear form, Adv. Math. 129 (1997), 1-24, doi:10.1006/aima.1996.1530.
[2]	T. W. Cai and R. P. Stanley, The Smith normal form of a matrix associated with Young's lattice, Proc. Amer. Math. Soc. 143 (2015), 4695-4703, doi:10.1090/proc/12642.
[3]	G. Denham and P. Hanlon, On the Smith normal form of the Varchenko bilinear form of a hyperplane arrangement, Pacific J. Math. 181 (1997), 123-146, doi:10.2140/pjm.1997.181. 123.
[4]	G. Denham and P. Hanlon, Some algebraic properties of the Schechtman-Varchenko bilinear forms, in: L. J. Billera, A. Bjorner, C. Greene, R. E. Simion and R. P. Stanley (eds.), New Perspectives in Algebraic Combinatorics, Cambridge University Press, Cambridge, volume 38 of Mathematical Sciences Research Institute Publications, pp. 149-176, 1999, papers from the MSRI Program on Combinatorics held in Berkeley, CA, 1996 - 1997.
[5]	Y. Gao and Y. Zhang, Diagonal form of the Varchenko matrices, J. Algebraic Combin. 48 (2018), 351-368, doi:10.1007/s10801-018-0813-7.
[6]	P. Hanlon and R. P. Stanley, A q-deformation of a trivial symmetric group action, Trans. Amer. Math. Soc. 350 (1998), 4445-4459, doi:10.1090/s0002-9947-98-01880-7.
[7]	A. Miller and V. Reiner, Differential posets and Smith normal forms, Order 26 (2009), 197228, doi:10.1007/s11083-009-9114-z.
[8]	V. V. Schechtman and A. N. Varchenko, Arrangements of hyperplanes and Lie algebra homology, Invent. Math. 106 (1991), 139-194, doi:10.1007/bf01243909.
[9]	W. C. Shiu, Invariant factors of graphs associated with hyperplane arrangements, Discrete Math. 288 (2004), 135-148, doi:10.1016/j.disc.2004.07.009.
[10]	R. P. Stanley, An introduction to hyperplane arrangements, in: E. Miller, V. Reiner and B. Sturmfels (eds.), Geometric Combinatorics, American Mathematical Society, Providence, Rhode Island, volume 13 of IAS/Park City Mathematics Series, pp. 389-496, 2007, doi: 10.1090/pcms/013/08, lectures from the Graduate Summer School held in Park City, UT, 2004.
[11]	R. P. Stanley, Smith normal form in combinatorics, J. Comb. Theory Ser. A 144 (2016), 476495, doi:10.1016/j.jcta.2016.06.013.
[12]	A. Varchenko, Bilinear form of real configuration of hyperplanes, Adv. Math. 97 (1993), 110144, doi:10.1006/aima.1993.1003.
[13]	Y. Wang and R. P. Stanley, The Smith normal form distribution of a random integer matrix, SIAMJ. Discrete Math. 31 (2017), 2247-2268, doi:10.1137/16m1098140.