In this paper we give a new expansion, based on cyclicity of the trace, to study regularity properties of twisted expectations (X(s)) =Tr H (#U(%) X(s)). Here X(s)=X 0 e &s 0 Q 2 X 1 e &s 1 Q 2 } } } X k e &s k Q 2 is a product of operators X j , regularized by heat kernels e &s j Q 2 with s j >0. The twist groups # # Z 2 and U(%) # U(1) are commuting symmetries of Q 2 . The name ``holonomy expansion'' arises from picturing (X(s)) as a circular graph, with vertices in the graph representing the operators X j , in the order that they appear in the product, and the linesegment following X j representing the heat kernel e &s j Q 2
. The trace functional is cyclic, so the graph is circular. We generate our expansion by ``transporting'' a vertex X k around the circle, ending in its original position. We choose an X k that transforms under a one-dimensional representation of Z 2 _U(1). For % in the complement of the discrete set ( sing (where the group Z 2 _U(1) acts trivially on X k ) we obtain an identity between the original expectation and some new expectations. We study an example from supersymmetric quantum mechanics, with a Dirac operator Q(*) depending on a parameter * and with a U(1) group of symmetries U(%). We apply our expansion to invariants Z(*; %)=Z(Q(*); %) suggested by non-commutative geometry. These invariants are sums of expectations of the form above. We investigate this example as a first step toward developing an expansion to evaluate related invariants arising in supersymmetric quantum field theory. We establish differentiability of Z(*; , %) in * for * # (0, 1] and show Z(*; %) is independent of *. We wish to evaluate Z(*; %) at the endpoint *=0, but Z(0; %) is ill-defined. We regularize the endpoint, while preserving the U(%)-symmetry, by replacing Q(*) 2 with H(=, *)=Q(*) 2 += 2 |z| 2 . The regularized function Z(=, *; %) depends on all three variables =, *, %; for fixed %, it is differentiable in the unit (=, *) square, except at the origin. Using the holonomy expansion, we prove for fixed % Γ ( sing that Z(=, *; %) is also jointly continuous in (=, *), at the origin. As a consequence, if % Γ ( sing , then we can interchange limits and Z(*; %)=lim = Γ 0 Z(=, 0; %). We observe that the joint continuity of Z(=, *; %) in (=, *) is not uniform in %, and Z(=, *; %) is not jointly continuous for % # ( sing . But the limiting function Z(*; %) is continuous in %; so the =-limit also determines Z(*; %) for all %, including for % # ( sing . We use these facts to calculate Z(*; %). Our regularization destroys supersymmetry, but the holonomy expansion gives quantitative bounds on the error terms.