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 = ∑m_iχ_i.

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 m_i.

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 m_i, this matrix algebra occurring with multiplicity in the regular representation equal to the degree of the character χ_i. 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?