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A vanishing group-ring coefficient sum pairs off opposite-signed equal labels
Statement
Let be a group and let and satisfy in , or for a single element . If , then there are indices with and . In particular, in the single-monomial case with the multiset of signed labels contains an opposite-signed pair of equal labels.
Facts & Assumptions
Given: A group , a finite list of elements and signs , together with the equality , or the equality for a single element .
The classes , , form a -basis of : the group ring is the free left -module on the set , and every element of has a unique expression with finite and , so two such expressions are equal if and only if they have the same coefficient at every label; in particular the expression of has every coefficient , and the expression of a single basis vector has coefficient at and coefficient at every other label (The group ring of finitely supported formal -linear combinations of group elements, The group ring is a unital -algebra with basis , and each is a unit of ).
Proof
Group the terms of the sum by label: for each set , a finite sum that is nonzero only for the finitely many occurring labels, so that is the expansion of the left-hand side in the basis of [F1]. By the uniqueness of that expansion, the equality holds exactly when for every , and the equality holds exactly when and for every .
Assume no two indices carry equal labels with opposite signs, so that for every occurring label all terms with share one sign and is the number of occurrences of , hence a nonzero integer. If , then step 1.1 forces to vanish for every label, contradicting the nonzero coefficient of each occurring label; therefore in the zero-sum case some pair of indices has equal labels and opposite signs.
Assume no two indices carry equal labels with opposite signs and consider the single-monomial case with . By step 1.1 every label must have , and under the assumption every occurring label has a nonzero coefficient, so no label other than occurs and all terms carry the label . Then , where counts the indices with , and the equation with gives an odd and ; hence some index has sign and some index has sign , both with label , so an opposite-signed pair of equal labels exists in the single-monomial case as well.
Steps 2.1 and 2.2 settle the zero-sum and the single-monomial case respectively, so under the stated hypotheses and there are always indices with and . The argument used only the basis expansion of , no property of beyond it and no choice principle.
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Sources
- Wolfgang Lück, A Basic Introduction to Surgery Theory (ICTP lecture notes, 27 October 2004; complete author text) (standard reference, not scraped)
- Andrew Ranicki, Algebraic and Geometric Surgery (Oxford Mathematical Monographs, electronic edition) (standard reference, not scraped)