# Local index formula for >ungraded< elliptic operators

Let $P\colon E \to F$ be an elliptic pseudodifferential operator over $M$. Assuming that $P$ defines a finitely summable Fredholm module, we may apply the Chern-Connes character to it to get a cyclic cocycle and this may then be mapped into the homology of $M$, i.e., we have $ch_\ast(P) \in H_\ast(M)$.

On the other hand, we may do the clutching construction using the symbol $\sigma(P)$ of $P$ to get an element $[\sigma_P] \in K_{cpt}^0(TM)$, apply the usual cohomological Chern character, integrate over the fiber, and finally multiply with the Todd genus of $M$, to get the index class $ind(P) = \pi_! ch^\ast [\sigma_P] \wedge Td(M) \in H^\ast(M)$. (I think I forgot here some sign $(-1)^?$.)

If $P$ is a graded operator (so that $E$ and $F$ are the positive and negative parts, respectively, of the vector bundle on which $P$ acts) we have $ch_\ast(P) \in H_{ev}(M)$ and Connes and Moscovici proved in Cyclic Cohomology, The Novikov Conjecture And Hyperbolic Groups that this is the Poincare dual of $ind(P)$ (up to some constants that only depend on the degree $q$ in which we compare these classes).

But if $P$ is an ungraded operator (in this case we would have $E=F$) we get $ch_\ast(P) \in H_{odd}(M)$ since now $P$ itself defines an odd $K$-homology class. But the Poincare dual of $ind(P)$ still lives in $H_{ev}(M)$, i.e., we can not compare them.

What is the correct formulation of the local index formula in the case of ungraded operators? Is it in the literature somewhere?

I expect somehow a small fix to the construction of the index class $ind(P)$ such that it lives then in the correct degrees (maybe something like taking a suitable product with $S^1$ similarly as the trick used by Connes and Moscovici in their above cited paper where they reduced from the case of odd-dimensional manifolds to even-dimensional ones).

• There are a number of index theorems for operators whose K-homology classes live in the odd degree groups for whatever reason, but typically you need the operator to have some extra structure. Do you have a specific operator in mind? – Paul Siegel Dec 2 '14 at 16:25
• Typically, you would replace the operator $P$ by $P \oplus P^{\ast}: E \oplus F \to E \oplus F$. This is graded, and the index (in the graded sense) is the usual index. An ungraded operator has an index in $K^1$, but only if it is self-adjoint. – Johannes Ebert Dec 2 '14 at 21:45
• Yes, I assume that $P$ is essentially self-adjoint so that it indeed defines an element of $K_1(M)$. I also assume the manifold to be compact (and without boundary). Applying the homological Chern character (constructed by using Connes' cyclic cohomology), we get an element of $H_\ast(M)$ and I want to identify the Poincare dual of it. If $P$ is graded the Poincare dual is, up to some universal constants, $ind(P)$ which is computed / constructed using the clutching construction. This is a generalization of Atiyah-Singer. But what is this Poincare dual now in the ungraded case? – AlexE Dec 3 '14 at 8:27

If $P\colon E \to E$ is elliptic and self-adjoint, then its symbol $\sigma(P) \colon \pi^\ast E \to \pi^\ast E$ is a self-adjoint automorphism, where $\pi\colon SM \to M$ denotes the unit sphere bundle. So we have a splitting $\pi^\ast E = E^+ \oplus E^-$, where $E^+$ and $E^-$ are fiber wise spanned by the positive, resp. negative, eigenvalues of $\sigma(P)$, and we define the symbol class $\sigma_P$ of $P$ as $\sigma_P := [E^+] \in K^0(SM)$. To get the index class, we do now the same as in the graded case: $ind(P) := (-1)^{n(n+1)/2} \pi_! ch^\ast [\sigma_P] \wedge Td(M) \in H^\ast(M)$.