arXiv subjects
Publications and source records attributed to M. Walze.
Let (A,H,F) be a p-summable Fredholm module where the algebra A= C Γis generated by a discrete group of unitaries in L(H) which is of polynomial growth r. Then we construct a spectral triple (A,H,D) with F= sign D which is q-summable for each q > p+r+1. In case (A,H,F) is (p,\infty)-summable we obtain (q,\infty)-summability of (A,H,D) for each q > p+r+1.
The purpose of this article is to apply the concept of the spectral triple, the starting point for the analysis of noncommutative spaces in the sense of A.~Connes, to the case where the algebra $\cA$ contains both bosonic and fermionic degrees of freedom. The operator $\cD$ of the spectral triple under consideration is the square root of the Dirac operator und thus the forms of the generalized differential algebra constructed out of the spectral triple are in a representation of the Lorentz group with integer spin if the form degree is even and they are in a representation with half-integer spin if the form degree is odd. However, we find that the 2-forms, obtained by squaring the connection, contains exactly the components of the vector multiplet representation of the supersymmetry algebra. This allows to construct an action for supersymmetric Yang-Mills theory in the framework of noncommutative geometry.
We derive an action for gravity in the framework of non-commutative geometry by using the Wodzicki residue. We prove that for a Dirac operator $D$ on an $n$ dimensional compact Riemannian manifold with $n\geq 4$, $n$ even, the Wodzicki residue Res$(D^{-n+2})$ is the integral of the second coefficient of the heat kernel expansion of $D^{2}$. We use this result to derive a gravity action for commutative geometry which is the usual Einstein Hilbert action and we also apply our results to a non-commutative extension which, is given by the tensor product of the algebra of smooth functions on a manifold and a finite dimensional matrix algebra. In this case we obtain gravity with a cosmological constant.