Duality between K-theory and K-homology in the non-spin^c case.

Question

I posted this question on Math.SE (https://math.stackexchange.com/questions/409444/), but got no answer. So I repost it here.

Let M be a closed manifold. Then there is a cap product $K^\ast(M) \times K_\ast(M) \to K_\ast(M)$ between the K-theory of M and its K-homology. For a definition of it one could see my prior question on Math.SE about it: https://math.stackexchange.com/questions/402170/ - take there A = C(M).

Now if M is spin$^c$, it has a fundamental class $[M] \in K_\ast(M)$ and it is well-known that the cap product with $[M]$ induces the Poincare duality $K^\ast(M) \stackrel{\cong}\to K_\ast(M)$. (See also my other question about it: Duality between K-theory and K-homology in the non-compact, spin$^c$ case.)

In the book "Spin Geometry" by Lawson, Michelsohn it is shown (on page 257) that every class in $K_{cpt}(TM) \cong K_0(M)$ is the difference of two Atiyah-Singer operators with coefficients (if M is spin and even dimensional), i.e., this translates to saying that the Poincare duality map is surjective onto $K_0(M)$ in this case.

Then it is written: "For non-spin manifolds, one can argue similarly by using the signature operator with coefficients". This means that there is some class $[D] \in K_\ast(M)$ (the class of the signature operator) such that the cap product with [D] is onto on $K_0(M)$.

Now the question is, if this statement generalizes:

Is it crucial that it is the signature operator? Could we also take the Euler characteristic operator?

Is there always (i.e., in the non-spin$^c$ case and not only for even dimensional manifolds) a class $[D] \in K_\ast(M)$ such that the cap product with $[D]$ is onto on $K_\ast(M)$ (and not only on $K_0(M)$)?

Maybe even if M is not orientable? Or even when M is not a manifold?

Are there other sufficient conditions besides spin$^c$ such that such a map is injective?

If one of the statements above is true, it would be nice to have some references. Thanks!

Answer 1

This is nicely understood by duality (in the sense of dualizable objects in monoidal categories) in the monoidal category KK. The relevant results are collected and referenced on the nLab at Poincaré duality algebra:

For $X$ a closed manifold equipped with a twist $\chi$ (the class of a circle 2-bundle, realized as a $U(1)$bundle gerbe etc., let $C_\chi(X)$ be the corresponding twisted groupoid convolution C*-algebra, well defined up to Morita equivalence and hence KK-equivalence. This is the $C^\ast$-algebra such that

$$ KK_\bullet(\mathbb{C}, C_\chi(X)) \simeq K^{\bullet+\chi}(X) $$

is the $\chi$-twisted K-theory of $X$.

Then here is the statement: The dual object of $C_\chi(X)$ in $(KK, \otimes)$ is

$$ \left( C_\chi\left(X\right) \right)^\vee \simeq C_{-\chi - W_3(T X)}(X) \,, $$

where $W_3(T Q)$ is the third integral Stiefel-Whitney class of the tangent bundle of $X$.

This formula implies everything in this business.

For instance it implies that $C_\chi(X)$ is self-dual, and hence (since duality in KK is duality between K-homology and K-cohomology) that $X$ exhibits K-Poincaré duality, precisely if $\chi = 0$ and $W_3(T X) = 0$. The latter is precisely the condition of spin^c structure.

But you get much more by playing with the formula, and this is I think what you are after. For instance given a map of closed manifolds $i \colon Q \to X$ with $X$ carrying a twist $\chi$ and $Q$ carrying the twist $i^\ast \chi$, we get the corresponding pullback map of twisted convolution algebras

$$ i^\ast \colon C_\chi(X) \to C_{i^\ast \chi}(Q) \,. $$

Now since both objects have dualty, we can form the corresponding dual morphism by the general formula in monoidal categories:

$$ i_! := (i^\ast)^\vee : (C_{i^\ast \chi}(Q))^\vee \to (C_\chi(X))^\vee \,, $$

hence by the above

$$ i_! : C_{-i^\ast \chi - W_3(T Q)}(Q) \to C_{-\chi-W_3(T X)}(X) \,. $$

By just relabelling the original twist for transparency as $\chi \mapsto - \chi - W_3(T X)$ this is equivalently a morphism

$$ i_! : C_{i^\ast \chi + W_3(N_i Q)}(Q) \to C_{\chi}(X) \,. $$

(Here $N_i Q$ denotes the normal bundle of $Q$ in $X$ via $i$.)

By postcomposition in KK and using the general relation that $KK(\mathbb{C}, A) \simeq K(A)$ this yields a homomorphism

$$ i_! : K^{\bullet + i^\ast \chi + W_3(N_i Q)} \to K^{\bullet + \chi}(X) $$

from the $(i^\ast \chi + W_3(N_i Q))$-twisted K-theory of $Q$ to the $\chi$-twisted K-thory of $X$.

This is the general twisted Umkehr map in twisted K-theory. Notice again that it is simply the dual morphism of $i^\ast$ in $(KK,\otimes)$.

If now the normal bundle has $Spin^c$-structure then that term drops out, and so on. But in general it is like this.

If here we think of $X$ as the target spacetime for a type II superstring, of $\chi$ as the instanton sector of the B-field, and of $Q \hookrightarrow X$ as the worldvolume of a D-brane, then the $(i^\ast \chi + W_3(N_i Q))$-twisted K-cocycles on $Q$ are the (underlying instanton sectors of) the Chan-Paton gauge fields subject to the Freed-Witten-Kapustin anomaly cancellation condition and their image under the above twisted Umkehr map is the D_brane charge of the D-brane with that Chan-Paton bundle.

All this just means really: the dual map of $i^\ast$ picks up twists as above. See at Poincaré duality algebra for citations. All these results go through also for $G$-equivariant twisted K-theory.

Answer 2

For a non-Spin$^c$-manifold, Poincaré duality is best understood in the wrold of KK-theory. In G. Kasparov's paper on KK-theory and the Novikov conjecture, he shows that for any smooth compact manifold $M$, $C_0(M)$ is Poincaré dual in KK to $C_0(T^*M)$. In particular, the K-homology of $M$ is isomorphic to the K-theory of $T^*M$. The isomorphism is induced by the Dolbeault operator on $T^*M$, which is the Dirac operator for a particular Spin$^c$-structure. This isomorphism can also be identified with the inverse symbol map that constructs a K-homology class from a given symbol in the K-theory of $T^*M$; this is closely related to the Atiyah-Singer index theorem.

If $M$ is not Spin$^c$, then in general the K-homology and K-theory of $M$ are not isomorphic. Neither the signature operator nor the Euler characteristic operator are relevant here.

We may, however, replace $C_0(T^*M)$ by the Z/2-graded algebra of sections of the Clifford algebra bundle of $M$, Cliff($M$), which has the same K-theory. A Spin$^c$-structure on $M$ is equivalent to a Morita-equivalence between Cliff($M$) and $C_0(M)$ by an old article by Roger Plymen. This explains why the K-homology of $M$ is isomorphic to the K-theory of $M$ in the Spin$^c$-case but not in general.

Answer 3

I believe you will need to add some twisting to obtain Poincare duality. Specifically, if $w\in H^3(X,\mathbb{Z})$ represents the obstruction the your manifold possessing a $spin^c$ structure, then there is a Poincare duality isomorphism $K_*(M)=K^{d-\ast}(X)_{w}$ where $d=\dim M$ and $K^{d-\ast}(X)_w$ is $w$-twisted $K$-theory. This result is proved, for instance, as Proposition 9.2 of Ando-Blumberg-Gepner's paper http://arxiv.org/abs/1002.3004.

Answer 4

There is a paper by Jonathan Rosenberg with the title "The K-homology class of the Euler characteristic operator is trivial" see here, which proves that the only information contained in the class of the Euler characteristic operator is just the Euler characteristic, i.e. it lies is the image of $\mathbb{Z}= KO_0(pt) \to KO_0(M)$ induced by the inclusion of the basepoint.