# index of a family of Dirac operators in $K^1$

Suppose I have a family of Dirac operators over a compact base space B. From the paper of Atiyah and Singer about skew adjoint Fredholm operators we know that it has an index in $K^1(B)$.

Suppose furthermore I know "a lot" about these Dirac operators (like their spectrum, eigenspaces etc.) and $B$ is a simple space like e.g. a torus where I know everything about $K^1$ and the cohomology.

What methods are there to give an explicit description of the index in $K^1(B)$ or its image in the odd-dimensional cohomology?

Any suggestions or references welcome.

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There's a lovely example of such a calculation, applying the relevant A-S index theorem to calculate the (odd) Chern character of the index, in Kronheimer-Mrowka's book "Monopoles and 3-manifolds", section 35. –  Tim Perutz Oct 22 '10 at 15:48

Whether the following is useful might depend on your concrete example. Because you are mentioning $K^1$ instead of $K^0$, I assume that your Dirac operator is ungraded (if it is graded, the index should be in $K^0$). The graded case is the ordinary Atiyah-Singer family index theorem. There is a version for the ungraded case, with index in $K^1$, which is expressed by the same formula, except that the symbol class is in $K^1$ and its definition is slightly more complicated than in the $K^0$-case. This is explained in Atiyah-Patodi-Singer, "Spectral asymmetry and Riemann geometry III" (you do not have to read parts I and II, but only one section of part III, nevertheless, these are beautiful papers). There is a cohomological formula that can be derived from the K-theory formula, much in the same way as in the ordinary $K^0$-case. In other words, the family index in $K^1$ is as computable as the ordinary one. By the way: if you take a graded operator, you get - forget the grading - an ungraded operator. Its $K^1$-index is always zero.

If your base space is a circle and if you really know the spectrum of the operator at any point of $B$, there is a nice spectral-theoretic way for computing the index, by the so-called "spectral flow". The spectral flow of a map $f: S^1 \to Fred_{sa}$ is the geometric intersection number of $f$ with the sub"manifold" of noninvertible operators in the space of selfadjoint Fredholms. Sounds complicated, but what you have to do is to count how often an eigenvalue crosses zero. A nice explanation of the spectral flow is in Booss-Woichechowski's book "Boundary value problems for Dirac-operators". If $B$ is a $2$-torus, then $K^1 (B)$ should be $Z^2$, the factor being detected on the two circles, and this reduces the problem to the case $B=S^1$.

A last comment: if some miracle happens and the dimension of the kernel is constant, then your computation is VERY easy. The family index in $K^1 (B)$ will be zero in that case. I wrote down a simple proof of this fact in http://arxiv.org/PS_cache/arxiv/pdf/0902/0902.4719v3.pdf, Theorem 4.2.1.

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Thank you very much so far! An additional question: If B happens to be a 3-torus, what might I try to calculate the "$H^3$-part" of $K^1(B)$ ? –  J. Fabian Meier Nov 22 '10 at 9:59
Can't you do the following? By the Künneth formula (proven by Atiyah, of course), $K^* (T^3) = K^* (S^1)\otimes K^* (S^1) \otimes K^* (S^1)$. The K-theory of the circle is $K^0 (S^1)=Z$ and $K^1(S^1)=Z$. –  Johannes Ebert Nov 22 '10 at 10:13
Sorry, I should have explained my question in more detail: I mean, you explained above what to do in the case of a 2-torus. But in the case of the 3-torus, the image of an index element in $K^1(B)$ splits into a part in $H^1(B)$ and a part in $H^3(B)$. If I understood the situation correctly I can find out about the $H^1(B)$-part by calculating three appropriate spectral flows. But what about the image of the index in $H^3(B)$? –  J. Fabian Meier Nov 22 '10 at 11:57
I see; on the fundamental group, you can only detect a $Z^3$-factor of $K^1 (T^3)$. For the last one, you need some information on $H^3$ indeed, which is not captured by the spectral flow. Did you try to compute the Chern character of the index, using the index formula for selfadjoint operators? –  Johannes Ebert Nov 22 '10 at 13:10

Two years ago my student Daniel Cibotaru wrote a dissertation entitled

Localization formulae in odd K-theory

in which he answers precisely this question in great generality.

More precisely, given a smooth family $(T_b)_{b\in B}$ of complex, Fredholm selfadjoint operators parametrized by a compact, connected, oriented smooth manifold $B$ he describes explicitely a (stratified) cycle in $B$ that is Poincare dual to the odd Chern character of this family. This cycle is non-homogeneous, i.e., it is a sum of cycles of various codimensions. The codimension $1$-part is the so called Maslov class or spectral flow class and it is more or less known. Daniel has a very nice description of the codimension $3$-part. For example if $B$ is a $3$-manifold, then the family defines a cohomology class in $H^3(B,\mathbb{Z})\cong \mathbb{Z}$ and Daniel explains how to compute this as a signed count of points in $B$.

More precisely, for a generic family $(T_b)_{b\in B}$, the locus of points $b$ where $\ker T_b\neq 0$ is an oriented surface $S\subset B$ and the family of vector spaces

$$S\ni s\mapsto \ker T_s$$

is a complex line bundle over $S$.The above integer is none other than the degree of this complex line bundle.

The higher codimension parts have explicit but rather complicated descriptions.

The philosophy of his dissertation can be easily described: he constructs a smooth model for the classifying space of $K^1$. This is an infinite dimensional Grassmanian-like object $\mathcal{X}$ equipped with a Schubert-like stratification by strata of finite codimensions. He then shows that the closures of the Schubert strata determine cohomology classes forming an integral basis of the cohomology of $\mathcal{X}$. The whole thing has a strong symplectic flavor.

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