24
$\begingroup$

This is migrated by math.stackexchange as I did not receive an answer. I do not know if it is too naive for this site.

I am taking a lectured class in Atiyah-Singer this semester. While the class is moving on really slowly (we just covered how to use Atiyah-Singer to prove Gauss-Bonnet, and introducing pseudodifferential operators), I am wondering how practical this theorem is. The following question is general in nature:

Suppose we have a PDE given by certain elliptic differential operator, how computable is the topological index of this differential operator? If we give certain boundary conditions on the domain (for example, the unit circle with a point removed, a triangle, a square, etc), can we extend the $K$-theory proof to this case? I know the $K$-theory rhetoric proof in literature, but to my knowledge this proof is highly abstract and does not seem to be directly computable. Now if we are interested in the analytical side of things, but cannot compute the analytical index directly because of analytical difficulties, how difficult is it to compute the topological index instead? It does not appear obvious to me how one may compute the Todd class or the chern character in practical cases.

The question is motiviated by the following observation: Given additional algebraic structure (for example, if $M$ is a homogeneous space, $E$ is a bundle with fibre isomorphic to $H$) we can show that Atiyah-Singer can be reduced to direct algebraic computations. However, what if the underlying manifold is really bad? What if it has boundaries of codimension 1 or higher?How computable is the index if we encounter an analytical/geometrical singularity?(which appears quite often in PDE).

On the other hand, suppose we have a manifold with corners and we know a certain operator's topological index. How much hope do we have in recovering the associated operator by recovering its principal symbol? Can we use this to put certain analytical limits on the manifolds(like how bad an operator on it could be if the index is given)?

$\endgroup$
0

3 Answers 3

10
$\begingroup$

As Johannes Ebert said, it's best if at first you stay away from boundary value problems. For some elliptic operators there may not even exist local boundary conditions satisfying the conditions guaranteeing Fredholmness; the Dolbeault operator is such an example. Therefore often one has to deal with pseudo-local boundary value problems such as the Atiyah-Patodi-Singer boundary condition.

A pseudo-diff operator on a closed manifold is Fredholm iff it is elliptic and the index is determined by the principla symbol, which is an element in the $K$-theory of a commutative algebra. For a boundary value problem Fredholmness is a much more subtle issue. It imposes restrictions on the type of boundary value conditions allowed (think Lopatinskii-Schapiro) and as Boutet de Monvel has shown almost four decades ago, the index is determined by the symbol of the problems which is an element in the $K$-theory of a certain non-commutative algebra; see e.g. this paper and the references therein.

The index of an operator on a closed manifold is eminently computable. In most geometric applications it can be reduced to the computation of the indices of a few classical operators: the spin and spin-c Dirac operators, the Hodge-de Rham operator (leading to the Gauss-Bonnet and the Hirzebruch signature operator), Dolbeault operator (leading to the Riemann-Roch-Hirzebruch formula).

The reduction to these cases requires good knowledge of representation theory, differential geometry and extensive familiarity with the theory of characteristic classes.

For manifolds with corners things are even more nebulous; same for most noncompact manifolds. In any case, to paraphrase one of my former professors, if you can describe a PDE problem explicitly, and you can prove its Fredholmness, then the index theorem will give you an answer as explicit as your question.

$\endgroup$
18
$\begingroup$

This isn't really going to be an answer, but it's too long to be a comment and I think it will be helpful.

First, as the other answerers pointed out it is a good idea to avoid boundary value problems and singularities at first. The usual index formula is not correct in this setting (even once you manage to formulate it correctly): there is a mysterious error term called the "eta invariant." The eta invariant is generally even harder to compute than the Todd class because it is non-local and it depends on a choice of Riemannian metric. It is defined in terms of the eigenvalues of your operator, and actually working these out requires quite a lot of symmetry so that you can do harmonic analysis.

Second, I don't think the real point of the index theorem is to numerically compute all of the integrals involved (though it's nice that you can in principle). It is often useful enough just to know that it is even possible to compute the index in terms of local data. For instance, this immediately implies that any invariant which can be expressed as the index of an operator is multiplicative under coverings, something which might not always be obvious. Also, if you understand the geometric structures which produced your operator then you can often refine the Atiyah-Singer integrand; for instance, it is useful to know that the Euler characteristic is the integral of curvature terms and that the signature of a 4-manifold is the integral of Pontryagin classes.

Third, in applications the goal is often to find geometric reasons why the index of an operator is $0$, and sometimes you can do this even if you don't know exactly what the integrand looks like. This sometimes happens in Seiberg-Witten theory, for instance, where the index of the Dirac operator tells you the dimension of a certain moduli space and thus you're pretty happy if it happens to be $0$.

Finally, I would argue that the actual cohomological formula for the index is sort of beside the point. The index theorem should really be viewed as a version of Poincare duality in K-theory, and all those messy characteristic classes are the price we pay for trying to stuff the theorem into ordinary cohomology. Working with index invariants at the level of K-theory is usually much easier and you can pick up on much more refined index invariants that don't necessarily have a local formula (such as the Clifford index in real K-theory).

$\endgroup$
5
  • $\begingroup$ @Paul Siegel: May I ask why we do not want to care about boundaries? After all pseduo-differential operators, etc is devised precisely to tackle the singularity we have in analysis. I feel if this is only we can do then I would have hugely disappointed. $\endgroup$ Apr 22, 2013 at 0:58
  • $\begingroup$ I certainly would not say that we don't want to care about boundaries - I'm personally very interested in the theory of eta invariants! It's just that index theory over manifolds with boundary and manifolds with singularities is much harder and more subtle, and it's better to get comfortable with closed manifolds first. In particular, there can't really be a K-theoretic formulation of the index theorem for manifolds with boundary because K-theory is homotopy invariant whereas the eta invariant is not. $\endgroup$ Apr 22, 2013 at 1:35
  • $\begingroup$ @Paul Siegel: I see. I guess I have to ask my professor about eta invariant in detail. I heard of it before but did not know it. $\endgroup$ Apr 22, 2013 at 3:29
  • $\begingroup$ If your main interest is in index theory on manifolds with boundary and manifolds with singularities, you should make sure to study the heat equation proof of the index theorem. The index formula arises by explicitly calculating coefficients in an asymptotic expansion of the heat kernel of the Dirac operator, and the same method can be made to work on manifolds with boundary (this is how the eta invariant most naturally appears). $\endgroup$ Apr 22, 2013 at 4:26
  • $\begingroup$ @Paul Siegel: I see. This is the track I am following in the class. We are computing the $t^{1/2}$ term in the heat kernel expansion. I cannot really follow the proof as the formula looks extremely complicated and unintuitive, which partly what motivates the question. $\endgroup$ Apr 22, 2013 at 7:27
10
$\begingroup$

''If we give certain boundary conditions on the domain (for example, the unit circle with a point removed, a triangle, a square, etc), can we extend the K-theory proof to this case?''

At least not in general. The first problem is to formulate the appropriate boundary value problem to have the Fredholm property, which is already a highly nontrivial task, especially if you wish to consider pseudodifferential operators. First, you would like to consider a local boundary condition, i.e. one of the form $P(s|_{\partial W})=0$, where $P$ is a vector bundle map. There is an early paper by Atiyah and Bott on an extension of the $K$-theoretic index formula for differential operators with such local boundary conditions. It turns out that there is a topological obstruction to the existence of a (local) boundary condition. For geometrically interesting operators as the signature and the Atiyah-Singer-Dirac operator, this obstruction is nonzero, and so a more sophisticated type of boundary conditions is required. The index theorem for operators with these boundary conditions was proven by Atiyah-Patodi-Singer in a sequence of three magnificient papers. The theorem is in terms of cohomology; and there is no formulation in terms of $K$-theory alone (at least not without Kasparov theory). I recommend you to ignore manifolds with boundary for the time being, and also all questions of singularities etc.

"It does not appear obvious to me how one may compute the Todd class or the Chern character in practical cases."

It is indeed not obvious at all how to compute these things. As a general rule, the analytical index is more difficult to compute than the topological one, because the amount of data is reduced. And both indices become harder to compute the less symmetric the manifold is. On a generic manifold, it is very difficult to compute characteristic classes (just as it is very difficult to compute anything in mathematics in a generic situation).

The strategy for calcluations is to start from a very symmetric situation, such as $CP^n$ or spheres, where the characteristic classes (and the analytic indices of the standard operators) are computable to other manifolds. Hirzebruch was a true master in computing these things. In his book ''Topological methods in algebraic geometry'' there is, for example, a formula how to calculate the characteristic numbers of complete intersection varieties in $CP^n$ and in a series of papers with Borel, he calculates characteristic classes of homogeneous spaces.

There is one exception, where the analytical index is easier to compute, namely if some reason (like the Weitzenböck formula) forces the operator to be invertible and so to have index zero.

$\endgroup$
1
  • $\begingroup$ May I ask if you meant the topological index is harder to compute instead? I feel the analytical index should be okay to compute (thanks for Getlzer's work). $\endgroup$ Dec 10, 2014 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.