I’d like to take a break from the previous homotopy-theoretic series of posts and do something a bit more geometric here. I’ll describe the classical Atiyah-Bott fixed point formula for an elliptic complex and one of the applications in the paper. The ultimate goal is for me to understand some of the more recent rigidity results for genera.
1. The Atiyah-Bott fixed point formula
Let be a compact manifold, and suppose given an endomorphism with finitely many fixed points. The classical Lefschetz fixed-point formula counts the number of fixed points via the supertrace of the action of on cohomology . In other words, if is the fixed point set, we have
where is a sign related to the determinant of at .
Using the de Rham isomorphism, the groups are identified with the cohomology of a complex of sections of bundles
This is an example of an elliptic complex of differential operators: in other words, when one takes the symbol sequence at a nonzero cotangent vector, the induced map of vector spaces is exact. It is a consequence of this that the cohomology groups are finite-dimensional.
The Atiyah-Bott fixed point formula is a striking generalization of the previous fact. Consider an elliptic complex of differential operators on ,
where the are vector bundles over . The cohomology groups of this complex are finite-dimensional and provide a generalization (not much of a generalization, actually) of the index of an elliptic operator; they thus often hold significant geometric information about .
Given an endomorphism , we are interested in computing the trace of the action of on the cohomology . We don’t yet have enough information to do this: in the de Rham case, we could pull back a differential form. So we’re going to need to be able to pull back sections. We can do this if we have maps
given for each , and that they commute with the differential operators in the sequence above. This will often be the case if the vector bundles are constructed by a “natural” process (e.g., the de Rham or Dolbeaut complexes). In this case, pulling back by acts on the cohomology of the complex .
Then we have:
All the traces in the above expression are, of course, supertraces.
If and the were the identity, then this would be precisely the alternating sum of the dimensions of : in the case where the complex has length two, we would be computing precisely the index of an elliptic operator. This can, of course, be done via the Atiyah-Singer index theorem, but the present result is at the opposite end of the spectrum: we are assuming has isolated fixed points. Using an equivariant version of the index theorem, one can derive the Atiyah-Bott formula when is part of a compact subgroup of .
The first example of the fixed point formula occurs when is simply the de Rham complex of a manifold. In this case, the trace of on the cohomology is the Lefschetz number of the formula, while the local traces in (1) counts the number of fixed points with signs.
The generality of the formula, though, extends far beyond the case of the de Rham complex. Let’s consider an example which provides more information.
Let be an oriented Riemannian manifold even dimension, and an orientation-preserving isometry with finitely many fixed points. These are automatically nondegenerate: if is a fixed point and is a fixed tangent vector, then the geodesic in direction at will be fixed by .
The signature operator on is an elliptic operator
where denotes a decomposition (or -grading) of the Clifford bundle via a suitable parallel element in the Clifford bundle. This was discussed in a previous post; the reason for the name is that the index of is precisely the signature of if . Since is an isometry, yields an endomorphism of the signature operator (by pull-back of forms).
be the difference of the trace of on the kernel and cokernel of .
Suppose now . Then if is homotopic to the identity, this is the same as the signature. This follows by the same reasoning as in the previous analysis: the supertrace of is given by the difference of the trace of on the subspace where the cup square is positive definite and the trace of on .
By the Atiyah-Bott fixed-point formula, we can compute as
So, to compute this, one has to compute the action of on the fibers of and divide by the determinant of (on the tangent space).
3. An application
Here is an application of the preceding analysis. Let’s work out explicitly, in local coordinates, what the local terms in the fixed point formula above for look like. If is an isometry of a compact even-dimensional manifold with a fixed point at , we compute the local contribution .
First, we need to describe the construction
which sends an oriented, even-dimensional inner-product space to the -graded vector space . This is determined by the following two properties:
- If is the realification of a one-dimensional complex vector space , then
- The map is multiplicative from even-dimensional vector spaces to -graded vector spaces.
Let be an orientation-preserving isometry. We want to compute the supertrace
in terms of the eigenvalues of . We know that there is a decomposition of into a system of oriented 2-planes such that acts on each 2-plane by rotation by the complex number (i.e., rotation by the angle ). By multiplicativity and the one-dimensional case, we find
In particular, if is an isolated fixed point of the isometry (where ), and if are the angles of rotation of , we find that the local contribution to the fixed point formula is
In particular, we find:
An intriguing corollary to this is:
Corollary 4 (Atiyah-Bott; Conner-Floyd) Let be an automorphism of a compact, oriented manifold of positive dimension. Suppose the order of is for an odd prime . Then cannot have precisely one fixed point.
Proof: Assume has precisely one fixed point. We can choose a suitable Riemannian metric for which is an isometry, by an averaging procedure. Note that is even, because is an isometry without eigenvalue at any fixed point of .
In this case, we use the formula (2) where two things happen. First, has finite order, so that must be a sum of roots of unity and is in particular an algebraic integer. Second, the right-hand-side consists of simply a product
where the are the angles of rotation of ; these are nonzero multiples of . But this is neveran algebraic integer if there is more than one factor! In fact, the -adic valuation of the terms in the numerator is zero, while (since is odd) the -adic valuation of the terms in the denominator is positive.