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The eta invariant was introduced by Atiyah, Patodi, and Singer. It roughly measures the asymmetry of the spectrum of a self-adjoint elliptic operator with respect to the origin. In the paper "Exotic structures on 4-manifolds detected by spectral invariants" Stolz showed that the eta invariant $\eta(M,g,\phi)$ of the twisted Dirac operator on a smooth closed 4 manifolds with Riemannian metric $g$ and $\text{Pin}^{+}$ structure $\phi$ is a $pin^{+}$ bordism invariant.

As we know, for an oriented 4 manifold $M$, the Hirzebruch Signature Theorem implies that $\text{index}(D)=\text{sign}(M)$, where the left hand side is the index of the signature operator of M (the analytic signature), and the right hand side is the topological signature (the signature of a quadratic form on $H^{2k}(M)$ defined by the cup product). Moreover, we have $\text{sign}(M) = \int_M L(p_1,\dots,p_n)$, where $L$ is the Hirzebruch $L$-Polynomial, and $p_i$ the Pontryagin numbers of $M$.

According to my understanding, the twisted Dirac operator is the unoriented generalizaton of the signature operator, and the eta invariant is the unoriented generalization of the signature of a manifold.

So I was wondering if one can have a similar topological description of the eta invariants $\eta(M,g,\phi)$ of a $\text{Pin}^{+}$ manifold in terms of topological data like the Stiefel Whitney numbers or Pontryagin numbers of $M$?.

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First, in general, the eta invariant for a self-adjoint elliptic operator on closed manifold is not a topological invariant. It depends on the geometric structure of the manifold. For example in spin case, it depends on the spin structure. So we cannot expect that it can be writed only by characteristic numbers.

Next, the eta invariant is not local. In the second page of the original paper "Spectral Asymmetry and Riemannian Geometry I" by Atiyah-Patodi-Singer, they gave an nonlocal example about a suitable len space. So we cannot expect that it can be writed by an integral of some local terms.

So the result of Stolz is a surprising but not a general result.

As I know, if you want to write the eta invariant as a local expression or the spin structure, there are two point of views.

The first one followed by Atiyah-Patodi-Singer that if the manifold $M$ is the boundary of some manifold $W$, then the reduced eta invariant can be written as $$\eta(M)=\mathrm{APS-index}(D^W)-\int_W \text{some characteristic class of}\ W.$$ So after mod $\mathbb{Z}$, we can regard the eta invariant of $M$ as an integral of local terms of $W$.

The second point of view is for the spin case. In this case, the eta invariant has a geometric expression that $$\eta(M)=\frac{1}{\sqrt{\pi}}\int_0^{\infty}t^{1/2}\mathrm{Tr}[D\exp(-tD^2)]dt.$$ Here $D$ is the Dirac operator. So we can study the eta invariant using the heat kernel.

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    $\begingroup$ Thanks. Can you also give a detailed reference on the second view point? $\endgroup$
    – Zitao Wang
    Commented Apr 23, 2014 at 23:29
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    $\begingroup$ Maybe the first time of this point of view is (2.22) in maths.ed.ac.uk/~aar/papers/bismfree2.pdf $\endgroup$
    – Bo Liu
    Commented Apr 24, 2014 at 11:38
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    $\begingroup$ But the well defined property of this integral is not easy. For this, you can see the small paper cim.nankai.edu.cn/members/user/weiping/40.pdf $\endgroup$
    – Bo Liu
    Commented Apr 24, 2014 at 11:45
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First, some comments. "A Dirac operator" is an operator naturally associated to a bundle which is a module over the Clifford algebra of the tangent bundle. "A twisted Dirac operator" (in the sense of the paper you liked to) is a Dirac operator in which the $Cl(\mathbb{R}^n)$ module at each fiber extends to a $Cl(\mathbb{R}^{n+1})$ module. "The twisted Dirac operator" corresponds to a specific choice of Clifford module outlined in the paper. Now, any Dirac operator (twisted or not) has an associated eta invariant. The signature operator is an example of a Dirac operator, so in particular there is an eta invariant associated to the signature operator. So your analogy is flawed: the index of twisted Dirac is to the signature as the eta invariant of twisted Dirac is to the eta invariant of the signature operator.

The best geometric interpretation of the eta invariant is that it is the "correction term" for the index theorem on manifolds with boundary. If $M$ is a manifold with boundary whose metric is a product near the boundary then you might expect from thinking about the Gauss-Bonnet theorem that the signature of $M$ is just the integral in the usual signature theorem, but this is not correct. The difference is essentially the eta invariant of the signature operator. The eta invariant for other Dirac operators has a similar interpretation.

Now an attempt to answer your question. I haven't done any calculations with the twisted Dirac operator used in Stolz's paper, but in general the eta invariant can't be calculated using standard characteristic numbers. One way to see this is to calculate examples: the eta invariant associated to the standard Dirac operator is nontrivial for lens spaces but it is $0$ for the sphere. If it were simply some sort of characteristic number then it would be the integral of local data and hence it would be multiplicative for coverings.

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  • $\begingroup$ Thanks for the clarification. By Corollary 5.2 of Stolz's, in the case when M is orientable, the signature of the manifold is related to the eta invariant of the Dirac operator in a simple way. Namely, $\eta(M,g,\phi)=1/16\text{sign}(M) \text{mod}2\mathbb{Z}$. So the fact that the signature is computable in terms of the Pontryagin numbers imply that the eta invariant of the Dirac operator be computable in terms of Pontryagin numbers. So it would be nice to see analogous things to hold for $pin^{+}$ manifolds. $\endgroup$
    – Zitao Wang
    Commented Apr 15, 2014 at 21:03
  • $\begingroup$ the eta invariant may depend on the (s)pin structure. So it is still possible to write it as a local expression in terms of the characteristic numbers and the (s)pin structure. $\endgroup$
    – Zitao Wang
    Commented Apr 15, 2014 at 23:26
  • $\begingroup$ It is not immediately clear to me how you may work with eta invariant if there is no spin structure or other geometric structure available. You can work with eta invariants for a generalized Dirac operator, not necessarily a classical Dirac operator. But how this is related to Pin structure is quite involved and unclear to me. $\endgroup$ Commented Feb 26, 2015 at 7:20
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The eta invariant is typically defined for Dirac operators on odd-dimensional Riemann manifolds. They depend on the various geometric data needed to define them (metrics, connections etc) and for this reason they are notoriously difficult to compute. If the underlying geometric structure has a bit of symmetry (read Lie group action) then the computation may be possible. For one such instance see this paper.

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