A problelm

Given a finite permutation group G on a set X, the permutation character π of G is the function on G mapping an element g to its number of fixed points in X. This is a character of G, the function giving the trace of a matrix representation of G (in this case, the representation by permutation matrices). So it can be decomposed into irreducible characters of G; say,

π = ∑miχi.

Now the number of orbits of G is equal to the multiplicy m0 of the trivial character χ0 in this expression (by the Orbit-Counting Lemma), while the rank of G (the number of orbits on X2) is equal to the sum of squares of the multiplicities mi.

Thus, the permutation character is the sum of the trivial character and one non-trivial irreducible if and only if G is doubly transitive. An old result asserts that the permutation character cannot have the form χ0+mχ for a non-trivial irreducible χ if m > 1.

The proof goes like this. By Jordan’s Theorem, since G is transitive, it contains an element g with no fixed points. Now the expression for π shows that χ(g) = −1/m. But any character value must be an algebraic integer, and so an integer if it is also rational, as this value is.

Now I have described before the notion of coherent configuration, a combinatorial gadget which describes (among other things) the orbits of a permutation group G on the set of ordered pairs. The relation matrices for the group orbits span an algebra (over the complex numbers) which is a direct sum of complete matrix algebras of dimensions equal to the multiplicities mi, this matrix algebra occurring with multiplicity in the regular representation equal to the degree of the character χi. For a general coherent configuration, we have a similar algebraic theory, but we do not have the group to give us these numbers.

Problem Is there a coherent configuration which “looks like” one coming from a group whose permutation character has this forbidden form, that is, the sum of a 1-dimensional algebra and a complete matrix algebra of degree greater than i?

We no longer have Jordan’s theorem to help us. I suspect that there is a simple algebraic argument which substitutes, but I cannot at the moment see how it would go. Any ideas?

About Peter Cameron

I count all the things that need to be counted.
This entry was posted in open problems and tagged , . Bookmark the permalink.

2 Responses to A problelm

  1. Misha says:

    Hi Peter,

    I think the answer is “no”. If the adjacency algebra A of a CC (X,S) is isomorphic to C\oplus M_d(C), d > 1, then (X,S) is homogeneous. A has two irreducible characters \chi_0 and \chi_1 with \chi_0(A_0)=1,\chi_1(A_0)=d (A_0 is the identity matrix). The character \chi of the standard module CX has a decomposition: \chi = chi_0 + m\chi_1 for some positive integer m. Then $\chi(A_0)=|X|$ implies $|X|=1+md$. If A_i, i >0 is a basic matrix, then $\chi(A_i)=0$ implying $\chi_1(A_i)m+k_i=0$ (k_i is the valency of A_i). Since $\chi_1(A_i)$ is an algebraic integer, the valency $k_i$ is divisible by $m$. Therefore, $k_i \geq m$ for each $i > 0$. Thus $|X|\geq 1 + m(|S|-1) = 1 + md^2$. Together with $|X|=1+md$ we obtain $d=1$. I hope this argument is mistake-free.

    More general statements are proven in [1](Theorem 1) and [2] (Theorem 6).
    If you assume that A \cong M_a(C)\oplus M_d(C) with a > 1 than it becomes possible. Inhomogeneous coherent algebra could be a direct sum of only two full matrix algebras: M_a(C)\oplus M_a(C). It happens if the CC comes from a system of linked block designs with the same parameters.

    All the best, Misha

    [1] H. Blau, Association schemes, fusion rings, C-algebras, and reality-based algebras where all nontrivial multiplicities are equal”, J Algebr Comb (2010) 31: 491–499.
    [2] Herman, A.; Muzychuk, M.; Xu, B. The recognition problem for table algebras and reality-based algebras. J. Algebra 479 (2017), 173–191.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.