lSpace A convenient setting for differential geometry and global analysis (Rev #2, changes)

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Reference Details

cite key: MR764972
title: A convenient setting for differential geometry and global analysis
author: Michor, Peter
coden: CTGDBR
eprint: <>
fjournal: Cahiers de Topologie et G'eom'etrie Diff'erentielle
issn: 0008-0004
journal: Cahiers Topologie G'eom. Diff'erentielle
mrclass: 58C20 (18F15 58A05 58D15)
mrnumber: MR764972 (86g:58014a)
mrreviewer: Sadayuki Yamamuro
pages: 63ñ109
url: <>
volume: 25
year: 1984
refbase: 888

Commentary on the Papers

Section One: Kriegl’s Convenient Setting for Differential Calculus on Locally Convex Spaces

The first proper section in the article pertains to Kriel’s development of a linear differential calculus. This is essentially the same as that in [?] with the minor difference that the locally convex topological vector spaces are made bornological. As smoothness is a bornological concept, this makes no difference to their calculus. The result is the category of convenient vector spaces.

Section Two: Premanifolds and Pre-Vector Bundles

This section begins with the definition of a premanifold.

Definition

A premanifold $M$ consists of:

Two sets, $M$ and $T M$ , together with a map $\pi _{M} \colon T M \to M$ such that for each $x \in M$ , the fibre $T_{x} M \coloneqq \pi ^{-1}_{M}(x)$ is a convenient vector space.
A subset $\mathcal{S} (\mathbb{R},M)$ of $\Set(\mathbb{R},M)$ which is closed under smooth reparametrisation and contains the constant maps. Elements of this set are called smooth curves in $M$ .
For each $t \in \mathbb{R}$ , a map $\delta _{t} \colon \mathcal{S} (\mathbb{R},M) \to T M$ such that:
1. for $c \in \mathcal{S} (\mathbb{R},M)$ then $\pi _{M} \circ \delta _{t} (c) = c(t)$ ,
2. for $c \in \mathcal{S} (\mathbb{R},M)$ and $f \in Ci(\mathbb{R},\mathbb{R})$ then $\delta _{t}(c \circ f) = f'(t) \cdot \delta _{f(t)}(c)$
3. if $c$ is such that for all $t \in \mathbb{R}$ then $\delta _{t}(c) = 0_{c(t)}$ then $c$ is constant.
The point $\delta _{t}(c)$ is called the differential at $t$ of $c$ .
A map $\Pt\coloneqq \Pt^{T M} \colon \mathcal{S} (\mathbb{R},M) \times \mathbb{R} \to \mathcal{L} (T M, T M) \coloneqq \bigcup _{x,y \in M} \mathcal{L} (T_{x} M, T_{y} M)$ such that
1. for a smooth curve $c$ and $t \in \mathbb{R}$ then $\Pt(c,t) \in \mathcal{L} (T_{c(0)} M, T_{c(t)} M)$ ,
2. for a smooth curve $c$ then $\Pt(c,0) = \Id_{T_{c(0)} M}$ ,
3. For a smooth curve $c$ and $f \in Ci(\mathbb{R},\mathbb{R})$ then $\Pt(c,f(t)) = \Pt(c \circ f,t) \circ \Pt(c, f(0))$ .
The map $\Pt$ is called parallel transport.
For each $c \in \mathcal{S} (\mathbb{R},M)$ the map

$t \mapsto \Pt(c,t)^{-1} (\delta _{t} c) = \Pt(c( \cdot + t), -t)(\delta _{t} c) \colon \mathbb{R} \to T_{c(0)} M$

is a $Ci$ –curve in the bornological locally convex topological vector space $T_{c(0)} M$ .
A map $\Geo^{M} \colon T M \to \mathcal{S} (\mathbb{R},M)$ such that:
1. $\Geo^{M}(t,v_{x})(s) = \Geo^{M}(v_{x})(t \cdot s)$ ,
2. $\delta _{t} \Geo^{M}(v_{x}) = \Pt(\Geo^{M} (v_{x}), t)$ (this line doesn’t make sense: the left-hand side is a point in some tangent space, the right-hand side is an operator between tangent spaces),
3. $\delta _{t} \Geo^{M}(v_{x}) = \Pt(\Geo^{M}(v_{x}), t)(v_{x})$ ,
4. $\Geo^{M} (\delta _{t} \Geo^{M}(v_{x}))(s) = \Geo^{M}(v_{x})(s + t)$

A premanifold is topologised with the final topology with respect to the smooth curves (ie the family $\mathcal{S} (\mathbb{R},M)$ ).

Example

A finite dimensional manifold (in the usual sense) is a premanifold as follows:

The set $M$ is the underlying set of the manifold, $T M$ of its tangent bundle, and $\pi _{M} \colon T M \to M$ the projection. The fibres of $\pi _{M}$ are isomorphic to Euclidean spaces whence convenient vector spaces.
The smooth curves are the smooth curves $Ci(\mathbb{R},M)$ .
The differential is the obvious one: $\delta _{t} c = c'(t)$ .
To get the parallel transport and map $\Geo$ , pick a complete Riemannian metric on $M$ . Then $\Pt$ is the parallel transport operator and $\Geo$ is given by $\Geo(v_{x})(t) = \exp (t \cdot v_{x})$ .

Michor then defines pre-vector bundles.

Definition

Let $M$ be a premanifold. A pre-vector bundle over $M$ , written $(E,p,M)$ , consists of the following:

A set $E$ with a map $p \colon E \to M$ such that for $x \in M$ , the fibre, $E_{x} \coloneqq p^{-1}(x)$ , is a convenient vector space.
There is a parallel transport map $\Pt^{E} \colon \mathcal{S} (\mathbb{R},M) \times \mathbb{R} \to \mathcal{L} (E,E) \coloneqq \bigcup _{x,y \in M} \mathcal{L} (E_{x},E_{y})$ such that
1. $\Pt^{E}(c,t) \in \mathcal{L} (E_{c(0)}, E_{c(t)})$ ,
2. $\Pt^{E}(c,0) = \Id_{E_{c(0)}}$
3. $\Pt^{E}(c,f(t)) = \Pt^{E} (c \circ f, t) \circ \Pt^{E}(c,f(0))$

The primary example of a pre-vector bundle is the triple $(T M, \pi _{M}, M)$ of a premanifold. The main result on pre-vector bundles is the following.

Theorem

A pre-vector bundle can be naturally given the structure of a premanifold.

Let us note how the curves work, as this shows the importance of the transport mechanism built in to the structure. A curve $\alpha \colon \mathbb{R} \to E$ is smooth if the projection $\pi \circ \alpha \colon \mathbb{R} \to M$ is smooth and the transported curve $\Pt^{E}(\pi \alpha ,t)^{-1} (\alpha (t))$ in $E_{\pi \alpha (0)}$ is smooth.

This leads in the obvious way to an iterated family of tangent bundles.

Section Three: Smooth Mappings

Definition

Let $M,N$ be premanifolds. A map $f \colon M \to N$ is smooth if there is a sequence of maps $(T^{n} f)_{n \ge 0}$ with $T^{0} f = f$ and $T^{n} f \colon T^{n} M \to T^{n} N$ such that for each $n \in \mathbb{N}$ then

\delta _{0} (T^{n} f)_{*} = T^{n+1} f \delta _{0} \colon \mathcal{S} (\mathbb{R}, T^{n} M) \to T^{n+1} N.

The push-forward, $f_{*}$ , is the push-forward on smooth curves. Thus the above includes the statement that $f$ (and each $T^{n} f$ ) takes smooth curves to smooth curves. Hence $f$ (and each $T^{n} f$ ) is a morphism in our category $\mathcal{C}$ of generalised smooth spaces.

Lemma

Composition of smooth maps is smooth, and the identity is smooth.

The resulting category of premanifolds is denoted $\pMf$ . The set of smooth maps from $M$ to $N$ is written $\mathcal{S} (M,N)$ . It is a lemma in the article that $\mathcal{S} (\mathbb{R},M)$ is unambiguous.

Section Four: Smoothness of Certain Structure Mappings

This section is concerned with certain categorical constructions. It is shown that the product in $\pMf$ is obtained by taking the product in $\Set$ and putting on it a natural premanifold structure. The fibre product of pre-vector bundles over the same base is again a pre-vector bundle, as is the pullback of a pre-vector bundle by a smooth map.

The remainder of this section lays down some technical groundwork for proving that maps such as the projection map of a pre-vector bundle are smooth. This is highly technical.

Section Five: Pre-Vector Bundles in More Detail

This section provides more details on the structure of pre-vector bundles. It proves that certain structure maps are smooth. Then it goes on to show that various natural operations on pre-vector bundles produce additional pre-vector bundles (such as taking duals and mapping spaces). Also constructed are covariant differential operators.

Section Six: First Steps Towards Cartesian Closedness

Proposition

If $M$ and $N$ are premanifolds then $\mathcal{S} (M,N)$ satisfies the first three axioms for a premanifold in a natural way.

Interesting to note here is that $T \mathcal{S} (M,N)$ is defined as $\mathcal{S} (M, T N)$ .

Section Seven: Vector Bundles and Cartesian Closedness

Definition

A premanifold $M$ is called a manifold if its parallel transport and geodesic structure are smooth:

$\Pt^{T M} \colon \mathcal{S} (\mathbb{R},M) \times \mathbb{R} \to \mathcal{L} (T M, T M)$
$\Geo^{M} \colon T M \to \mathcal{S} (\mathbb{R}, M)$ , equivalently that either $\exp ^{M} \coloneqq \ev_{1} \circ \Geo^{M} \colon T M \to M$ or $(\Geo^{M})^{\wedge }\colon T M \times \mathbb{R} \to M$ is smooth.

Definition

A pre-vector bundle $(E,p,M)$ is a vector bundle if $M$ is a manifold and $\Pt^{E}$ is smooth.

The rest of this section is concerned with building up the theory to sufficient level that the main result can be proved.

Theorem

Let $M$ and $N$ be manifolds. Then $\mathcal{S} (M,N)$ is again a manifold.

Section Eight: A Miscellany

This contains a few additional results, including the characterisation of manifolds in the usual sense in the above setting.

Reworking in a Category of Generalised Smooth Spaces

Part of the data of a premanifold is a set, say $M$ , and a family of “declared” smooth curves in $M$ , the set $\mathcal{S} (\mathbb{R},M)$ . This set has to satisfy some conditions. The composition condition is enough to show that we can interpret the pair $(M, \mathcal{S} (\mathbb{R},M))$ in the context of [?]: namely that there is a category of smooth objects sitting underneath in which everything naturally lives.

Recall that to apply [?] we need a test category, and underlying category, and certain conditions. In this case, the test category is the one-object category corresponding to the monoid $Ci(\mathbb{R},\mathbb{R})$ . The underlying category is that of sets. The condition on the input functor (the curves) is simply that constant curves are smooth. The condition on the output functor (the functionals) is the saturation condition. This means that we can effectively ignore functionals when working in this category. Let us write this category as $\mathcal{C}$ .

Theorem

The category $\mathcal{C}$ is complete, cocomplete, and cartesian closed.

From [?,], the forgetful functor $\mathcal{C} \to \Set$ has both a left and right adjoint. It therefore preserves and creates both limits and colimits.

To show cartesian closedness, we need to show that the desired smooth structure on $\mathcal{C} (X,Y)$ is actually a smooth structure. That is, we want to declare a smooth curve $\mathbb{R} \to \mathcal{C} (X,Y)$ to be one such that the set-adjoint $\mathbb{R} \times X \to Y$ is smooth. We need to show that this satisfies the conditions. It is clearly functorial, and if $\alpha \colon \mathbb{R} \to \mathcal{C} (X,Y)$ is constant then its adjoint factors through the projection $\mathbb{R} \times X \to X$ and hence is smooth.

Now if $M$ is a premanifold then it is a smooth space in this category. The proof that a pre-vector bundle over $M$ is again a premanifold says that it is again a smooth space. The smooth curves $\alpha \colon \mathbb{R} \to E$ are those curves for which $\pi \alpha \colon \mathbb{R} \to M$ are smooth and $\Pt^{E}(\pi \alpha , t)^{-1} \alpha (t)$ is smooth in $E_{\pi \alpha (0)}$ . Thus, as noted, $T M$ is a premanifold and hence a smooth space.

Let us consider the structure maps. It is not possible to assume that these are smooth (in the category $\mathcal{C}$ ). The problem is that there is too little information about how these maps vary over $M$ . For the parallel transport map, everything is concentrated on a single path. The only condition relating to different paths is about reparametrisation. For the geodesic map, the conditions only relate to tangent vectors at a single point.

Let us, as an example, consider the parallel transport map. This works along curves, but we need to consider a family of curves which we can think of as a map $\alpha \colon \mathbb{R} ^{2} \to M$ . Then we have a source curve $\beta$ defined as $\beta (t) = \alpha (0,t)$ . Let $\eta \colon \mathbb{R} \to \mathbb{R}$ be a smooth map. Then we have a target curve $\gamma$ defined as $\gamma (t) = \alpha (\eta (t),t)$ . Actually, as one of the few things we do know about parallel transport is that it is invariant under reparametrisations, we can replace $\alpha$ by the map $(s,t) \mapsto \alpha (s \eta (t), t)$ and replace $\eta$ by the constant map at $1$ . We want to consider a smooth curve $\mathbb{R} \to T M$ above $\beta$ . Saying that it is smooth means that when transported back to $T_{\beta (0)} M$ we have a smooth curve there. Let $\nu \colon \mathbb{R} \to T_{\beta (0)} M$ be this smooth curve, so that the curve above $\beta$ is $t \mapsto \Pt(\beta ,t) \nu (t)$ .

We want to consider the result of parallel transporting this curve along $\alpha$ , in that $\Pt(\beta ,t) \nu (t)$ is transported along $\alpha (-,t)$ from $\alpha (0,t)$ to $\alpha (1,t)$ . To test the smoothness of the result, we parallel transport along $\gamma$ back to $\gamma (0) = \alpha (1,0)$ . Putting all of this together, we start with a smooth curve in $T_{\alpha (0,0)} M$ . We stretch this out along $\alpha (0,-)$ using parallel transport. Then transport each point along $\alpha (-,t)$ before collapsing back to $\alpha (1,0)$ . So the resulting curve is:

t \mapsto \Pt(\gamma ,t)^{-1} \Pt(\alpha _{t}, 1) \Pt(\beta ,t) \nu (t).

So for this to be smooth (whence $\Pt$ to be smooth), we need $\Pt$ to behave well as the underlying curve varies smoothly. This is not built in to the definition.

So let us build some smoothness into the definition of a premanifold. The central idea of a premanifold is that the space $T M$ behaves “like a tangent space”. There are a variety of notions of tangent space for a smooth space. Let us examine the definition of a premanifold to see if we can extract the essential conditions.

So we have a smooth spaces $M$ and $T M$ with a smooth map $T M \to M$ . The fibres of this map are convenient vector spaces (with the subspace structure). The first condition, which seems an eminently reasonable one, is the existence of the ~~“king~~ “taking ~~tangents’‘~~ tangents” map which we view as a smooth map $\delta \colon \mathcal{S} (\mathbb{R},M) \times \mathbb{R} \to T M$ (note that this combines the third and fifth conditions in the definition of a premanifold). We write this as $\delta \colon (\alpha ,t) \mapsto \delta _{t} \alpha$ . This should satisfy the following conditions:

$\pi \circ \delta _{t} \alpha = \alpha (t)$ ,
$\delta _{t} (\alpha \circ \beta ) = \beta '(t) \delta _{\beta (t)} \alpha$ ,
If $\delta _{t} \alpha = 0$ for all $t$ then $\alpha$ is constant.

The parallel transport mapping is somewhat like saying that $T M \to M$ is locally trivial, in that it allows us to compare distinct fibres. One could rename the concept of a pre-vector bundle something more intuitive like a transportable structure.

The geodesic map is then a partial inverse to the differential map. We view it as an $\mathbb{R}$ –equivariant smooth map $\Geo\colon T M \to \mathcal{S} (\mathbb{R},M)$ such that $\delta _{0} \Geo(v) = v$ . The other conditions relate to the parallel transport structure. This implies, amongst other things, that $\delta$ is surjective (and, indeed, a quotient map).

I wonder if it is possible to cook up a distinctly bizarre version of this. Let us take $M = \mathbb{R}$ (or $\mathbb{R} ^{n}$ ). For $T M$ , I want to take the largest possible space. So we look at $\mathcal{S} (\mathbb{R},M)$ , which is a vector space, and impose the conditions. The only serious condition is that $\delta _{t}(\alpha \circ \beta ) = \beta '(t) \delta _{\beta (t)} \alpha$ . Let us consider the fibre at $0$ . So then we consider curves $\mathbb{R} \to M$ taking $0$ to $0$ , and reparametrisations which preserve $0$ . Then for $\beta \in Ci_{0}(\mathbb{R},\mathbb{R})$ we consider the map $\mathcal{S} _{0}(\mathbb{R},M) \to \mathcal{S} _{0}(\mathbb{R},M)$ given by $\alpha \mapsto \alpha \circ \beta - \beta ' \cdot \alpha$ . This is linear, whence the quotient is a vector space. Obviously, we need to check to see if it is a convenient vector space.

We define parallel transport simply by translation. The geodesic map is perhaps a little trickier. We need a splitting map of this quotient. It is entirely possible that this does not exist.

In summary, there are two interesting themes from this definition.

The notion of a transportable structure; namely, one that admits a parallel transport operator.
The use of a splitting map as part of a characterisation of the notion of a tangent space, essentially the ability to exponentiate a tangent vector to a curve.

category: smootheology

Revision on June 20, 2013 at 02:23:07 by Andrew Stacey. See the history of this page for a list of all contributions to it.