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Generalizations of Determinants: Permanents, Immanants, etc.

  1. Dec 30, 2011 #1
    The determinant of a matrix is given by the well-known formula

    det(A) = sump parity(p) * producti = 1...n Ai,p(i)

    where the p's are all permutations of 1...n and A is a n*n matrix. Parity is +1 for an even permutation and -1 for an odd one.

    For a permanent, replace parity(p) with 1. For an immanant, replace parity(p) with X(q,p), a character of the symmetric group for irreducible representation q.

    Determinants show up in many places that matrices do. Permanents are the bosonic counterpart of the fermionic Slater determinant of multiplarticle wavefunctions. Immanants appear in the theory of Schur functions. But I've seen much less of permanents and immanants than determinants. Could that be due to not having certain nice properties that determinants have?

    In particular, determinants satisfy the multiplication law det(A.B) = det(A)*det(B) For permanents or immanants, however, one can find counterexamples. Is that law true for any generalization other than determinants themselves?

    Let's define a generalized determinant gdt(A) = sump X(p) * producti = 1...n Ai,p(i)

    and let's set gdt(A.B) = gdt(A)*gdt(B) for all possible A and B. What constraints can one find on the X(p)'s? Can one show that only X(p) = parity(p) will yield that multiplication law?

    So far, I've found that all p's where X(p) is nonzero must form a group, and that these nonzero X(p)'s with multiplication form an abelian quotient group of the group of p's. That includes determinants and permanents, and gdt's for various other groups of p's.
  2. jcsd
  3. Jan 7, 2012 #2

    Stephen Tashi

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    If you look only at nxn matrices A that represent permumutations of n things (so each row and each column is zero except for a single 1 in it somewhere) then gdt(A) (as I understand your definition) is a sum with only a single non-zero term in it that corresponds to the permutation represented by the matrix. So the rule gdt(AB) = gdt(A)gdt(B) implies, in the case of permutation matrices, that [itex] \chi(p_A p_B) = \chi(p_A) \chi(p_B) [/itex] where [itex] p_A [/itex] is the permutation represented by the matrix [itex] A [/itex] etc.

    I'd think that gdt( identity matrix) = 1, so [itex] \chi[/itex](identity permutation) = 1. Does this make the function [itex] \chi [/itex] itself a homomorphism from the symmetric group on n things to the real numbers? (Or am I just confusing myself by attempting to dabble in representation theory?)
  4. Jan 7, 2012 #3
    You've got it right. That's also true for matrices that are
    (permutation matrix) . (diagonal matrix from vector v)

    gdt(that matrix) = X(permutation) * product(v)

    If A and B have that form, then A.B also has that form, and from gdt(A.B) = gdt(A)*gdt(B), we find

    X(permutation 1)*X(permutation 2) = X(permutation 1 * permutation 2)

    For the identity permutation e, X(e)*X(e) = X(e*e) = X(e) meaning that X(e) is either 0 or 1.

    For all permutations p, X(p)*X(e) = X(p*e) = X(p). If X(e) = 0, then all X(p) is zero, a trivial solution. To get nontrivial ones, we must set X(e) = 1.

    Likewise, for all permutations p, X(p)*X(inverse(p)) = X(e) = 1
    This means that all X(p) must be nonzero, with |X(p)| = 1

    gdt(A) is summed over all permutations p with length n from A being a n*n matrix. These permutations form n-symbol symmetric group Sym(n), and we can now ask what homomorphisms to an abelian group that Sym(n) can possibly have. Homomorphisms are quotient groups of normal subgroups; the normal subgroup's elements map onto the homomorphism's identity.

    Every group G has two trivial normal subgroups, itself, G, and the identity group {e}. Their quotient groups are, respectively, {e} and G. This gives possible solution X(p) = 1.

    It can be shown that Sym(n) always has normal subgroup Alt(n), the alternating group for n symbols, the group of even permutations for n symbols. For n > 2, Alt(n) is a nontrivial normal subgroup of Sym(n), and all the Sym(n)'s have only one additional one:

    The Klein 4-group of all (2,2) permutations in Sym(4), isomorphic to Z(2)*Z(2).

    Its quotient group is, however, Sym(3), which is nonabelian.

    So Sym(n) has only two possible quotient groups, the identity group, and the quotient group of Alt(n), which is the group of permutation parities. Even = 1, odd = -1


    So I've shown that gdt(A.B) = gdt(A)*gdt(B) implies that X(p) = 1 (permanent) or X(p) = parity(p) (determinant)

    Since one can show that det(A.B) = det(A)*det(B) and find counterexamples to permament(A.B) = permanent(A)*permanent(B), one concludes that the only possible X(p) that makes that identity possible is the determinant one, X(p) = parity(p)
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