Permanent vs. Determinant Frank Ban Introuction A major problem in theoretical computer science is the Permanent vs. Determinant problem. It asks: given an n by n matrix of ineterminates A = (a i,j ) an an m by m matrix B with entries that are affine linear functions of the entries of A, what is the smallest m such that the eterminant of B equals the permanent of A? In other wors, what is the complexity of writing the permanent in terms of the eterminant? At first, one might believe that there shoul not be much ifference because the two functions et A = n sgn(σ) σ S n an perma = σ S n i= a i,σ(i) n a i,σ(i) appear to be very similar. However it is conjecture that m > poly(n) asymptotically. It is strongly believe that this conjecture is true because its falsity woul imply P = NP. [] The best lower boun for the eterminantal complexity of perm n was obtaine by stuying the Hessian matrices of perm n an et m. In this paper, we will let T p (2) (x i,j ) i,j n enote the n 2 by n 2 matrix with the entry in row i, j an column k, l equal to i= 2 x i,j x k,l perm n (x i,j ) an we will let T (2) (x i,j) i,j n enote the corresponing matrix for et n.
2 Lower Boun Although the conjecture is that the permanent woul have super polynomial eterminantal complexity, the best lower boun attaine so far is merely quaratic. This boun is ue to Mignon an Ressayre. [2] Theorem 2. The eterminantal complexity of perm n is at least n 2 /2. The proof of their result epens on the existence of an n by n matrix C with perm n (C) equal to 0 an T p (2) (C) having full rank. We will assume that n 3 since the case of n = 2 is trivial. Lemma 2.2 Let n 3. Let C be an n by n matrix with C, = n + an C i,j = for (i, j) (, ). Then perm n (C) = 0 an rank(t (2) p (C)) = n 2. Proof By oing a Laplace expansion along the first row of C we get perm n (C) = (n )P n + (n )P n = 0, where P k is the permanent of the k by k matrix with all entries equal to. Note that P k is equal to k!. We claim that 2 x i,j x k,l perm n (C) = { (n 2)! if {i, j, k, l} 2(n 3)! otherwise Inee, ifferentiating perm n with respect to x i,j an x k,l is equivalent to eleting rows i an k an columns j an l an taking the eterminant of the remaining sub matrix. If {i, j, k, l}, then the remaining sub matrix is P n 2. Otherwise, the remaining sub matrix is the n 2 by n 2 matrix with upper left entry equal to n an all other entries equal to. Therefore, we can say 0 B B B B 0 C C T p (2) (C) = (n 3)! B C 0............ C B C C 0 where 0. 0... B = (n 2)....... 0 2
an 0 n 2 n 2 n 2 n 2 0 2 2. C = n 2 2 0........... 2 n 2 2 2 0 To show that T p (2) (C) has full rank, we show that it has a trivial kernel. Since B = (n 2)(J n I n ), then B only has eigenvalues 2 n an (n )(n 2) so it has full rank. Now let Cv = 0. By looking at the bottom two rows of C we have (n 2)v 2v 2 2v n 2 2v n = 0 an (n 2)v 2v 2 2v n = 0. This implies v n = v n. By a similar argument, v a = v b for all 2 a, b n. By consiering the first row of C, this implies that v 2 = v 3 = = v n = 0. It follows that v = 0an v = 0. Therefore, both B an C have full rank. x x 2 Now let x =. where each x i is a vector in R n. Arguing similarly x n as we i for the matrix C we can show that T p (2) (C) has full rank by first observing that C( x a x b ) = 0 for all 2 a, b n. Since C has full rank, then x a = x b for all 2 a, b n. Thus x 2 = x n. By consiering the first row of T p (2) (C) we see that x a = 0 for all 2 a n because B has full rank. It follows that x = 0 as well giving the esire result. We now present the proof of Mignon an Ressayre s lower boun using the matrix C from Lemma 2.2. Proof (of Theorem 2.) Let m be the eterminantal complexity of perm n. Then there exists a family of affine linear functions A k,l, for k, l m in the variables x i,j for i, j n, with perm n (x i,j ) = et m (A k,l (x i,j ) i,j n ). We can perform a translation on the coorinates x i,j. By this we mean, there exist homogeneous linear functions L k,l an a matrix of constants Y such that (A k,l (x i,j )) k,l = (L k,l (x i,j C i,j )) + Y. Thus perm n (x i,j ) = et m ((L k,l (x i,j C i,j )) + Y ) (2.) Wolog, we can [ apply ] a series of row an column operations to Y to put 0 0 it in the form. Since perm(c) is 0, then by equation (2.), Y has 0 I s 3
eterminant 0 an so oes not have full rank. Thus s < m. Encoe the row operations in a matrix P an the column operations in [ a matrix Q. Note ] etp 0 that if we left multiply the above eterminantal matrix by P [ ] 0 I m etq 0 an right multiply by Q then we o not change the value of 0 I m the eterminant. Since [ ] [ ] [ ] etp 0 etq 0 0 0 P Y Q = 0 I m 0 I m 0 I s [ ] 0 0 then wolog we can just let Y = in equation (2.). 0 I s By the multivariate chain rule, there exists an m 2 by n 2 matrix L such that T (2) p (x i,j ) = L T (T (2) (L k,l(x i,j C i,j ) + Y ) k,l )L The matrix L has its (k, l, i, j) entry given by x i,j L k,l (x i,j C i,j ). Therefore T p (2) (C) = L T T (2) (Y )L Thus, rank(t p (2) (C)) rank(t (2) (2) (Y )). By Lemma 2.2, rank(t p (C)) = n 2. Therefore it suffices to show that rank(t (2) (Y )) 2m. To see this, first consier the case when s = m. Note that 2 x i,j x k,l et m Y is non-zero if an only if one of the following hols:. (i, j) = (, ) an (k, l) = (t, t) for some t > 2. (i, j) = (, t) an (k, l) = (t, ) for some t > 3. (i, j) = (t, ) an (k, l) = (, t) for some t > 4. (i, j) = (t, t) an (k, l) = (, ) for some t > The above four conitions tell us that T (2) (Y ) has 3m 2 nonzero rows ( row for conition an m rows each for conitions 2, 3, an 4). Each of the m rows satisfying conition 4 are all copies of the same row with a in column (, ) an zeroes in every other column. 4
The 2m 2 rows satisfying conitions 2 an 3 have only a single nonzero entry (of value ) an each row contains a in a ifferent column. The row satisfying conition contains m nonzero entries. These nonzero entries are in columns that are not in the support of the rows satisfying conitions 2, 3, or 4. Thus, there are exactly 2m linearly inepenent rows in T (2) (Y ). Now consier the case when s = m 2. Note that 2 x i,j x k,l et m Y is non-zero if an only if i, j, k, l {, 2}. Then the number of non-zero entries in T (2) (Y ) is at most 4 which is certainly less than 2m as n grows. If s < m 2, then every entry of T (2) (Y ) is 0 so we are one. 3 Upper Bouns There is a combinatorial interpretation for perm n. Let G be a igraph on n vertices labelle {, 2,..., n} an let (x i,j ) i,j n be the weighte ajacency matrix for G. Then perm(x i,j ) is equal to the number of weighte irecte cycle covers of G. This is because perm(x i,j ) = x i,σ(i) σ S n i= an each cycle cover of G can be encoe as a permutation σ S n with σ(i) ientifying the irecte ege (i, σ(i)) in G. The eterminant can be interprete similarly, but each cycle cover woul be weighte by the sign of the permutation that it correspons to. Thus, if every cycle cover of a graph correspone to a permutation with even sign, then the permanent an the eterminant woul share this combinatorial interpretation. Using this iea, Grenet was able to provie an upper boun for the eterminantal complexity of the permanent. [3] Theorem 3. There exists a 2 n by 2 n matrix M with all entries of the form -, 0,, or x i,j such that et M = perm n (x i,j ). Proof Let G be a graph where the vertices are in bijection with subsets of {, 2,..., n} but the vertices corresponing to an the whole set {, 2,..., n} are ientifie as the same vertex v 0. It follows that G has 2 n vertices. 5 n
Suppose vertices v, w correspon to S, T {, 2,..., n} respectively. We will put a irecte ege with weight x i,j (with i, j n) between vertices v an w if an only if S = i +, T = i + 2, an T \ S = {j}. The vertex v 0 will have outgoing eges labelle x n,j an incoming eges labelle x i,n. Now put loops with weight on every vertex except v 0 to complete the ege set of G. Note that every non-loop cycle in G has the form x,σ(), x 2,σ(2),..., x n,σ(n) since each cycle can be seen as the number of ways to a elements to the empty set until you have the set {, 2,..., n}. Since every vertex except v 0 contains a loop, then all cycle covers of G consist of non-loop cycles containing v 0 an loops. Thus, every cycle cover of G correspons to an n-cycle. Note that the sign of an n-cycle is ( ) n. Let M be the ajacency matrix of G. It follows that the eterminant of M equals ( ) n σ S n x,σ() x 2,σ(2) x n,σ(n) which is equal to ±perm n (x i,j ). If et M = perm(x i,j ), then multiply the first row of M by - to get et M = perm(x i,j ). Figure shows the graph obtaine from the proof of Theorem 3. in the case when n = 3. It is unerstoo that the noes labelle an [3] = {, 2, 3} are ientifie together an S refers to the complement of S in [3]. 4 Glynn s Formula We can consier alternative methos of rewriting the permanent besies expressing it as a eterminant. Ryser was able to use inclusion-exclusion to write perm n as a sum of 2 n terms rather than a sum of n! terms. Glynn was able to provie another exponential expression for perm n, but he was able to write the polynomial as a sum of 2 n terms. [4] Theorem 4. We can write 2 n perm n = δ ( n ) n n δ i δ i x i,j k= j= i= where the outer sum is over all 2 n vectors δ {, } n with δ =. 6
x 0, x0,2 x 0,3 {} {2} {3} x,3 x,3 x,2 x,2 x, x, {} {2} {3} x 2, x 2,2 x 2,3 [3] Figure : n = 3 Proof Consier a monomial m in the x i,j variables on the RHS of our propose equality. Let λ i be the egree of m in the variables x i,j for fixe i an varying j. Let the coefficient of m be c. We have c = δ n i= δ λ i+ i = n i=2 δ i {,} (δ i ) λ i+ Thus c = 0 unless λ i is o for all i. Note that m has total egree n. Thus n = n i= λ i. Since λ =, then n = n i=2 λ i an if λ i is o for all i, then we must have λ i = for all i. Thus, the only such monomials m with non-zero coefficient woul have coefficient 2 n. These monomials woul have the form n i= x i,σ(i) where 7
σ S n. It follows that the RHS of our propose equality is equal to which is exactly 2 n perm n. σ S n 2 n n i= x i,σ(i) 5 Fano Schemes The Fano scheme of a space X, enote F k (X), parameterizes k-imensional planes that lie on X. Let D n an P n enote the space in P n2 cut out by the n by n eterminant an permanent respectively. Work by Chan an Ilten specifically stuies the Fano schemes of the n by n eterminant, F k (D n ), an the n by n permanent, F k (P n ). [5] It turns out the geometry of these schemes gives us information about the algebra of permanents. Lemma 5. It is impossible to write perm 3 as l q + l 2 q 2 where l, l 2 are linear forms in the variables x i,j an q, q 2 are quaratic forms in x i,j. Proof Suppose that perm 3 = l q + l 2 q 2. Then the space Y cut out by l an l 2 lies in the space cut out by perm 3. Since Y P 8 is cut out by two linear forms then it has coimension 2 which makes it a 6-imensional space. However, accoring to Table 2 in [5] the Fano scheme F k (P 3 ) is non-empty if an only if k 5. The result follows by contraiction. 6 The case of n = 3 a b c For this section we will let perm 3 be the permanent of e f. g h i Grenet s work gives a 7 by 7 matrix whose eterminant equals the 3 by 3 permanent. 8
0 a g 0 0 0 0 0 0 i f 0 a b c 0 0 0 0 c i perm e f = et 0 0 0 c 0 f g h i e 0 0 0 0 0 h 0 0 0 0 0 b 0 0 0 0 0 In conjunction with Theorem 2. an Theorem 3., we see that the eterminantal complexity of perm 3 is 5, 6, or 7. It remains an open question to etermine exactly which of the three numbers is the true eterminantal complexity. Another way of expressing of the permanent vs. eterminant problem is by saying that we want to fin matrices C, A,, A,2,..., A 3,3 M(m) such that M = C + x, A, + x,2 A,2 + + x 3,3 A 3,3 an perm n = et(m) with m minimal. We can state a technical result that puts restrictions on the eterminantal representation of perm 3. Lemma 6. Let n = 3. If m = 6, then the rank of C is 5. If m = 5, then the rank of C is 4. Proof We will prove only the case when m = 6. The case when m = 5 is similar. Note that by applying certain elementary row an column operations to M we o not change the value of its eterminant. Row operations can be encoe as left multiplication by a matrix an column operations can be encoe as right multiplication by a matrix. Suppose C has rank less than 3. Note that we can left multiply M by P an right multiply M by Q such that P CQ = Diag(,, 0, 0, 0, 0). Then et M = et P MQ. Since P MQ is a matrix with two entries of the form + l (where l is a linear form in the variables (x i,j )) an every other entry is a linear form in (x i,j ), then every term in the polynomial et P MQ has egree at least 4. Thus we cannot have perm 3 = et M. If C has rank exactly 3, then let P an Q be such that P CQ = Diag(,,, 0, 0, 0). It follows that the egree 3 part of et M = et P MQ is equal to et M where M is the lower-right 3x3 sub matrix of P MQ. This implies that 9
perm 3 can be written as the eterminant of a 3 by 3 matrix which is false because c(3) 5. Now suppose that C has rank equal to 4. Let P an Q be such that P CQ = Diag(,,,, 0, 0). The egree 2 part of et P MQ is equal to et M where M is the lower 2x2 sub matrix of P MQ. Thus et M = 0. [ We can ] apply further row operations to M to transform it into the form α β where α an β are linear forms in the variables a, b, c,, e, f, g, h, i. 0 0 We can exten these row operations to P MQ so that wolog + c c 2 c 3 e f c 4 + 2 c 5 c 6 e 2 f 2 P MQ = c 7 c 8 + 3 c 9 e 3 f 3 c 0 c c 2 + 4 e 4 f 4 g g 2 g 3 g 4 α β h h 2 h 3 h 4 0 0 where the inexe variables are linear forms in a, b, c,, e, f, g, h, i. The egree 3 part of et P MQ must be equal to perm 3. This gives us perm 3 = α(f h + f 2 h 2 + f 3 h 3 + f 4 h 4 ) + β(e h + e 2 h 2 + e 3 h 3 + e 4 h 4 ) which contraicts Lemma 5.. Finally suppose that C has full rank. Let P an Q be such that P CQ = I. Then et M = et P MQ woul contain a constant term so we coul not have et M = perm 3. The result follows. Glynn s formula (Theorem 4.) allows us to write perm 3 as (a + + g)(b + e + h)(c + f + i) (a + g)(b e + h)(c f + i) (a + g)(b + e h)(c + f i) + (a g)(b e h)(c f i) Another approach to etermining the eterminantal complexity of perm 3 might involve stuying Glynn s formula, since he is able to express the permanent as a sum of 4 proucts of 3 linear terms rather than a sum of 6 proucts of 3 linear terms. References [] M. Agrawal: Determinant versus permanent (2006) 0
[2] T. Mignon, N. Ressayre: A Quaratic Boun for the Determinant an Permanent Problem (2004) [3] B. Grenet: An Upper Boun for the Permanent versus Determinant Problem (202) [4] D. Glynn: The permanent of a square matrix (2009) [5] M. Chan, N. Ilten: Fano Schemes of Determinants an Permanents (203)