Orientation of Fredholm maps between Hilbert manifolds
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Though one can define a notion of an oriention of a manifold in the setting of Banach manifolds (namely, for Fredholm manifolds) as was done in [Elworty&Tromba1970], it seems that the more natural concept is that of an orientation of a map between such manifolds. In this note we define a notion of an orientation of a Fredholm map between Hilbert manifolds. We don't claim any originality. Related notions were given in [Morava1968], [Fitzpatrick&Pejsachowicz&Rabier1994] and [Benevieri&Furi1998] for non-linear maps between Banach spaces. The definition of an orientation is given in terms of a line bundle, the determinant line bundle. Our construction of the determinant line bundle agrees with that of [Wang2005] but he defines orientation only for index 0 Fredholm maps (since he is interested in the degree of proper Fredholm maps). We also add some useful information like orientation of a composition of two maps and product orientations.
Given a Fredholm map we want to construct the determinant line bundle. The rough idea is to consider the tensor product of the determinant line bundle of the kernel "bundle" of the differential of with the pullback of the determinant line bundle of the cokernel "bundle". Unless the dimension of is constant, these are not bundles and thus one has to make sense of these constructions. We will do this by following Jänich's construction of an index bundle for a Fredholm map. This is similar to the construction of the determinant line bundle in [Quillen1985].
Let be a Fredholm map. We equip and with Riemannian metrics. For each we will construct an open neighborhood and a line bundle over depending on a choice. For different choices one has explicit isomorphisms which are compatible with restrictions to open subsets of . These isomorphisms depend only on the metric and the differential of and so one can glue these bundles together to obtain the desired global line bundle. Since the space of metrics is convex the resulting bundle is also independent of the metric.
To find such an open neighborhood of and the desired line bundle over we note that the arguments in [Jänich1965, Lemma 3] (he constructs for compact spaces a global object, but the first step is the local object, and that's all we need) implies that for each in there is an open neighborhood and a finite dimensional subbundle of such that for all . Let be the orthogonal projection to the orthogonal complement of in . Then is a Fredholm bundle map and its kernel is the subbundle [Jänich1965, p. 138]. Thus the cokernel bundle with fibre over the cokernel of is defined and denoted by . To avoid signs, when we define the product orientation and interchange the factors we always assume that is even dimensional, which can be achieved by stabilizing and by this , if necessary. We define a line bundle over by , where denotes the determinant line bundle, and denote it by . If is open then .
If is a subbundle, the differential of and the Riemannian metric can be used to construct explicit isomorphisms . For this we note that - using the Riemannian metric - we have an embedding of into and that Jänich shows ([Jänich1965], p. 138) that the differential induces an isomorphism . Since for all and there is a common finite dimensional subbundle of containing both one obtains isomorphisms in general.
Now we glue all these line bundles over such open subsets in together using the isomorphisms and the fact that everything is compatible with restrictions and the construction of based on the differential (and the global Riemannian metric) fulfills the cocycle conditions, since they are induced by differentials. This is our determinant line bundle over .
Definition 2.1. An orientation of a Fredholm map is an orientation of . If we replace the orientation on by the opposite orientation we call the corresponding Fredholm map
Lemma 2.2. If has constant dimension then
Proof. Take to be globally.
3 Orientation of compositions
Given two Fredholm maps and , it follows from the proof of [Jänich1965, Lemma 6] that there is an explicit isomorphism
Thus if both maps are oriented one obtains an induced orientation of the composition by using the isomorphism above to orient .
4 Product orientation
Given two oriented Fredholm maps and , we would like to orient the product map . For this we consider the local construction of and similarly and obtain: . By interchanging the factors in the middle we identify this with . Recall that we always assume the is even dimensional. Then we take the product orientation on and equip with the induced orientation given by the isomorphism interchanging the middle factors.
5 Orientation of the cylinder and of the boundary of a Fredholm map
A useful thing for coboridsm is to orient the cylinder , where is the projection to . As explained before, this is an orientation of , and since , the trivial bundle, it is equivalent to an orientation of and of . If we change the orientation on we obtain the opposite orientation. Using this and a collar we orient the boundary of a Fredholm map.
6 Orientation of maps in finite dimensions
This construction also works for maps between finite dimensional manifolds. In this case we can choose globally as , then we get
If we denote by the infinite dimensional separable Hilbert space, then for the product map we have
where is the projection from to , since in this case we note that we can choose to be .
Remark 6.1. The above implies that a map between oriented finite dimensional manifolds has a natural orientation in this sense.
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