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.

1 Introduction

Given a Fredholm map $f: M \to N$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}f: M \to N$ 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 $f : M \to N$$f : M \to N$ with the pullback of the determinant line bundle of the cokernel "bundle". Unless the dimension of $ker \, df_x$$ker \, df_x$ 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].

2 Definition

Let $f: M \to N$$f: M \to N$ be a Fredholm map. We equip $M$$M$ and $N$$N$ with Riemannian metrics. For each $x \in M$$x \in M$ we will construct an open neighborhood $U$$U$ and a line bundle $L$$L$ over $U$$U$ depending on a choice. For different choices one has explicit isomorphisms which are compatible with restrictions to open subsets of $U$$U$. These isomorphisms depend only on the metric and the differential of $f$$f$ 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 $U$$U$ of $x$$x$ and the desired line bundle over $U$$U$ we note that the arguments in [Jänich1965, Lemma 3] (he constructs for compact spaces a global object, but the first step is the local object, and that's all we need) implies that for each $x$$x$ in $M$$M$ there is an open neighborhood $U$$U$ and a finite dimensional subbundle $\xi$$\xi$ of $TU$$TU$ such that $ker \, df_y \subseteq \xi_y$$ker \, df_y \subseteq \xi_y$ for all $y \in U$$y \in U$. Let $\pi _{ \xi^\perp}$$\pi _{ \xi^\perp}$ be the orthogonal projection to the orthogonal complement of $\xi$$\xi$ in $TU$$TU$. Then $df \pi _{ \xi^\perp}$$df \pi _{ \xi^\perp}$ is a Fredholm bundle map and its kernel is the subbundle $\xi$$\xi$ [Jänich1965, p. 138]. Thus the cokernel bundle with fibre over $y \in U$$y \in U$ the cokernel of $df_y \pi _{ \xi_y^\perp}$$df_y \pi _{ \xi_y^\perp}$ is defined and denoted by $\eta (\xi)$$\eta (\xi)$. To avoid signs, when we define the product orientation and interchange the factors we always assume that $\eta(\xi)$$\eta(\xi)$ is even dimensional, which can be achieved by stabilizing $\xi$$\xi$ and by this $\eta (\xi)$$\eta (\xi)$, if necessary. We define a line bundle over $U$$U$ by $\det (\xi ) \otimes \det (\eta(\xi))$$\det (\xi ) \otimes \det (\eta(\xi))$, where $\det$$\det$ denotes the determinant line bundle, and denote it by $L (\xi)$$L (\xi)$. If $V \subseteq U$$V \subseteq U$ is open then $L (\xi)|_V = L(\xi|_V)$$L (\xi)|_V = L(\xi|_V)$.

If $\xi \subseteq \xi'$$\xi \subseteq \xi'$ is a subbundle, the differential of $f$$f$ and the Riemannian metric can be used to construct explicit isomorphisms $\varphi (\xi, \xi'): L(\xi) \to L(\xi')$$\varphi (\xi, \xi'): L(\xi) \to L(\xi')$. For this we note that - using the Riemannian metric - we have an embedding of $\eta$$\eta$ into $\eta'$$\eta'$ and that Jänich shows ([Jänich1965], p. 138) that the differential induces an isomorphism $\eta' /\eta \to \xi'/ \xi$$\eta' /\eta \to \xi'/ \xi$. Since for all $\xi$$\xi$ and $\xi'$$\xi'$ there is a common finite dimensional subbundle of $TU$$TU$ containing both one obtains isomorphisms $\varphi (\xi, \xi'): L(\xi) \to L(\xi')$$\varphi (\xi, \xi'): L(\xi) \to L(\xi')$ in general.

Now we glue all these line bundles over such open subsets $U$$U$ in $M$$M$ together using the isomorphisms $\varphi (\xi, \xi'): L(\xi) \to L(\xi')$$\varphi (\xi, \xi'): L(\xi) \to L(\xi')$ and the fact that everything is compatible with restrictions and the construction of $\varphi (\xi, \xi')$$\varphi (\xi, \xi')$ 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 $L(f)$$L(f)$ over $M$$M$.

Definition 2.1. An orientation of a Fredholm map $f: M \to N$$f: M \to N$ is an orientation of $L(f)$$L(f)$. If we replace the orientation on $L(f)$$L(f)$ by the opposite orientation we call the corresponding Fredholm map $-(f: M \to N)$$-(f: M \to N)$

Lemma 2.2. If $\ker df_x$$\ker df_x$ has constant dimension then

$\displaystyle L(f) = \det (\ker df )\otimes \det(coker \, df).$ $$\displaystyle L(f) = \det (\ker df )\otimes \det(coker \, df).$$

Proof. Take $\xi$$\xi$ to be $\ker df$$\ker df$ globally.

$\square$$\square$

3 Orientation of compositions

Given two Fredholm maps $f: M \to N$$f: M \to N$ and $g: N \to P$$g: N \to P$, it follows from the proof of [Jänich1965, Lemma 6] that there is an explicit isomorphism

$\displaystyle L(gf) \cong L (f) \otimes f^*(L(g)).$ $$\displaystyle L(gf) \cong L (f) \otimes f^*(L(g)).$$

Thus if both maps are oriented one obtains an induced orientation of the composition by using the isomorphism above to orient $L(gf)$$L(gf)$.

4 Product orientation

Given two oriented Fredholm maps $f: M \to N$$f: M \to N$ and $f': M' \to N'$$f': M' \to N'$, we would like to orient the product map $f\times f':M\times M' \to N\times N'$$f\times f':M\times M' \to N\times N'$. For this we consider the local construction of $L(f)|_U = det (\xi) \otimes det (\eta(\xi))$$L(f)|_U = det (\xi) \otimes det (\eta(\xi))$ and similarly $L(f')|_{U'} = det (\xi ') \otimes det (\eta(\xi'))$$L(f')|_{U'} = det (\xi ') \otimes det (\eta(\xi'))$ and obtain: $L(f \times f') |_{U \times U'} = det (\xi) \otimes det (\xi') \otimes det (\eta(\xi) )\otimes det (\eta (\xi '))$$L(f \times f') |_{U \times U'} = det (\xi) \otimes det (\xi') \otimes det (\eta(\xi) )\otimes det (\eta (\xi '))$. By interchanging the factors in the middle we identify this with $det (\xi) \otimes det (\eta(\xi)) \otimes det (\xi') \otimes det (\eta(\xi'))$$det (\xi) \otimes det (\eta(\xi)) \otimes det (\xi') \otimes det (\eta(\xi'))$. Recall that we always assume the $\eta(\xi)$$\eta(\xi)$ is even dimensional. Then we take the product orientation on $det (\xi) \otimes det (\eta(\xi)) \otimes det (\xi') \otimes det (\eta(\xi'))$$det (\xi) \otimes det (\eta(\xi)) \otimes det (\xi') \otimes det (\eta(\xi'))$ and equip $L(f \times f') |_{U \times U'}$$L(f \times f') |_{U \times U'}$ 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 $fp: M \times \mathbb R \to N$$fp: M \times \mathbb R \to N$, where $p$$p$ is the projection to $M$$M$. As explained before, this is an orientation of $\det (\ker dp) \otimes p^*(L(f))$$\det (\ker dp) \otimes p^*(L(f))$, and since $\ker dp = \mathbb R$$\ker dp = \mathbb R$, the trivial bundle, it is equivalent to an orientation of $\mathbb R$$\mathbb R$ and of $f$$f$. If we change the orientation on $\mathbb R$$\mathbb R$ 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 $\xi$$\xi$ as $TM$$TM$, then we get

$\displaystyle L(f) = \det (TM )\otimes f^*(\det (TN)).$ $$\displaystyle L(f) = \det (TM )\otimes f^*(\det (TN)).$$

If we denote by $H$$H$ the infinite dimensional separable Hilbert space, then for the product map $f\times id:M\times H\to N\times H$$f\times id:M\times H\to N\times H$ we have

$\displaystyle L(f \times \, id ) = p^*L(f),$ $$\displaystyle L(f \times \, id ) = p^*L(f),$$

where $p$$p$ is the projection from $M \times H$$M \times H$ to $M$$M$, since in this case we note that we can choose $\xi$$\xi$ to be $TM\times \{ 0\}$$TM\times \{ 0\}$.

Remark 6.1. The above implies that a map between oriented finite dimensional manifolds has a natural orientation in this sense.

7 References

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