The tangent bundle to an infinite-dimensional manifold

Suppose that $A,B$ are smooth ($\mathrm C^\infty$) manifolds, and denote by $\hom(A,B)$ the set of $\mathrm C^\infty$-maps $A \to B$. It is a perfectly well-defined set, but often one wants more. For example, I regularly would like to talk about (small, or maybe infinitesimal) variations of a smooth map $A \to B$, and I regularly want to talk about smooth functions on the space of maps. Which is to say, there are many good reasons to want to have some sort of "infinite-dimensional manifold" which I will call $\newcommand\Maps{\underline{\operatorname{Maps}}} \Maps(A,B)$, whose set of points is $\hom(A,B)$.

A standard way of defining $\Maps(A,B)$ is to choose some regularity ($\mathrm C^\infty$ ends up not being a good choice, as the construction will not yield something with an inverse function theorem) and to try to first topologize and then assign a "smooth structure" to the set of functions $A\to B$ with the prescribed regularity. This works well for certain analytic applications, and allows for explicit constructions. But it has the disadvantage of making all facts demand on some real analysis, and the only thing I'm worse at than real analysis is complex analysis.

A different approach is to say that what "possibly-infinite-dimensional manifold" means is presheaf on the category of manifolds, or maybe restrict just to some well-behaved class of presheaves (e.g. sheaves for your favorite Grothendieck topology). Then you could define $\Maps(A,B)$ to be the sheaf

$$\Maps(A,B) = \hom( - \times A, B)$$

This definition answers some questions immediately. For example, there is a sheaf

$$\mathbb R = \hom (-, \mathbb R)$$

and a "smooth $\mathbb R$-valued function on $\Maps(A,B)$" is precisely a natural transformation of sheaves $\Maps(A,B) \to \mathbb R$. Note that the set of smooth $\mathbb R$-valued functions is actually an $\mathbb R$-algebra, because there is a product of sheaves

$$\Maps(A,B)^{\times 2} = \hom(- \times A,B)^{\times 2}$$

(right hand side is product of sets), and also the addition and multiplication maps $\mathbb R^{\times 2} \to \mathbb R$ are smooth. So we get an algebra $\mathrm C^\infty(\Maps(A,B))$ (which is even itself an infinite-dimensional smooth manifold in the same sense).

OK, so I've proposed some definition of "the smooth manifold $\Maps(A,B)$". I have "families of smooth maps", because I can map into $\Maps(A,B)$ from, say, curves, or spaces. What I really want, in addition to infinitesimal variations, are flows. So I'd like a "tangent bundle $\mathrm T (\Maps(A,B))$".

Here are at least three definitions I could think of:

1. There is a sheaf on manifolds called "$\operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2)$". Actually, it's easiest to define as a cosheaf; whether by definition or theorem, it satisfies that, for any finite-dimensional manifold $M$, $$\Maps( \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2) \to M ) = \text{the total space of } \mathrm T M$$ The bundle structure on $\mathrm T M$ corresponds to the unique inclusion $\lbrace \text{pt.}\rbrace \to \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2)$; the vector bundle structure from the endomorphisms $\Maps(\operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2) \to \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2)) = \mathbb R$. So one could define: $$\mathrm T (\Maps(A,B)) = \Maps( \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2), \Maps(A,B)).$$

2. One could reasonably say that a "vector bundle on $\Maps(A,B)$" should be determined by (indeed, the same as) its pulled-back vector bundle along any map $S \to \Maps(A,B)$. Given $f: S \to \Maps(A,B)$, which is to say $f\in \hom(S\times A, B)$, one can define an infinite-dimensional vector bundle $V \to S$ by saying that for any open set $U \subseteq S$, $$\Gamma_U(V) = \Gamma_{U \times A}(f^\ast \mathrm T B)$$ This is the way I usually think about the tangent bundle to $\Maps(A,B)$. More precisely, I usually define $\mathrm T(\Maps(A,B))$ by saying that over each point $f\in \hom(A,B)$, the fiber is $\mathrm T_f\Maps(A,B) = \Gamma_A(f^\ast \mathrm T B)$, and hoping that my readers don't ask me to explain in what sense these fibers glue together into a smooth bundle.

3. I have above defined an algebra $\mathrm C^\infty( \Maps(A,B))$. A tangent vector could be defined as a "point derivation": a map $v : \mathrm C^\infty(\Maps(A,B)) \to \mathbb R$ that satisfies that $v(\alpha\beta) = v(\alpha) \, \beta(f) + \alpha(f)\, v(\beta)$ for some $f\in \operatorname{spec} \mathrm C^\infty( \Maps(A,B))$ (or maybe just for those $f \in \hom(A,B)$). More generally, you could give this a smooth structure by explaining the notion of "derivation" internal to the world of sheaves. I think equivalently, for each $f : S \to \Maps(A,B)$, there is a restriction map $f^\ast: \mathrm C^\infty(\Maps(A,B)) \to \mathrm C^\infty(S)$, and you can consider the space of derivations over this map. It would be the space of sections of $f^\ast \mathrm T (\Maps(A,B))$, and this data should glue together into some form of "vector bundle".

My question is:

Are definitions 1,2,3 naturally equivalent? If not, which is best, and why? Or should I abandon all three in favor of some other definition?

• Maybe you already know about this, but you might want to look at the 1974 'Stable Mappings and their Singularities', by M. Golubitsky and V. Guillemin. That book contains an extensive discussion (and motivation) of the kinds of properties that you'd want for the `manifold' of smooth maps between two finite dimensional manifolds, and it contains a very nice treatment of Mather's fundamental work in this area. May 25 '11 at 17:23
• I was thinking about this issue of defining a tangent bundle for infinite dimensional manifolds when I stumbled upon this question. Did you arrive at some interesting conclusion since then? So far my thoughts are that one can break the problem into 3 steps: 1) Finding a good notion of smooth infinite dimensional vector spaces and maps between them (in particular it better be cartesian closed) 2) Finding the correct fibration over this category whose fibers are categories of vector bundles 3) Finding the section of this fibration corresponding to the tangent bundle. Jun 26 '18 at 13:28

I don't know about the abstract nonsense, but I do know that in the concrete world of infinite dimensional manifolds then there are different, and inequivalent, notions of "tangent bundle". In particular, if one defines the kinematic tangent bundle as via equivalence classes of curves and the operational tangent bundle via derivations of functions then the operational one contains the kinematic one, but the two are not, in general, isomorphic. The difference depends on the linear structure and so depends very much on the choice of model space.

For the details, you should read the relevant section in Kriegl and Michor's book.

• If you use the Whitney topology on $C^\infty(A,B)$, the modelling topological vector spaces are the spaces $\Gamma^\infty_c(f^*TB)$ of smooth sections of $f^*TB$ with compact support, $f\in C^\infty(A,B)$. In order to get a smooth manifold structure, one needs to use the so-called $c^\infty$ topology on $\Gamma^\infty_c(f^*TB)$, which is the final topology induced by all smooth curves therein and therefore finer than its standard topology. Nonetheless, for this particular modelling vector space, kinematical and operational tangent vectors do coincide. Dec 23 '16 at 18:22

This is not a complete answer, since I will only discuss the abstract nonsense in ideas 1 and 2.

Regarding your idea 1, the Grothendieck $\underline{\operatorname{Hom}}$ presheaf that you call $\underline{\operatorname{Maps}}$ admits a tautological adjunction, so you have three equivalent ways to define the tangent space of the presheaf $\underline{\operatorname{Maps}}(X,Y)$:

1. Your way: $\underline{\operatorname{Maps}}(\operatorname{Spec} \mathbb{R}[\epsilon]/(\epsilon^2), \underline{\operatorname{Maps}}(X, Y))$
2. The usual "restriction of scalars" way: $\underline{\operatorname{Maps}}(X \times \operatorname{Spec} \mathbb{R}[\epsilon]/(\epsilon^2), Y)$.
3. $\underline{\operatorname{Maps}}(X, TY)$.

The third definition makes it pretty clear that your construction in point 2 is equivalent to the construction in point 1.

I am having difficulty extracting a single precise definition from the discussion in point 3, so I don't know how to make a comparison.

• Ah, great, that was the argument I was missing. What I wanted to say in 3 was simply that any time I have a commutative algebra, I can talk about derivations thereof, and these should fit together into a "tangent bundle". May 25 '11 at 14:30

Perhaps I am misunderstanding your question, but to put a Banach manifold structure on mapping spaces between finite dimensional manifolds is not that hard. Start by choosing $f \in C^\infty$. Using that the exponential map is a local diffeomorphism, we can find a neighbourhood of the zero section in $C^0 \Gamma f^* (TB)$ (i.e. continuous sections of the pullback of the tangent bundle), called $V_0^f$, say, such that the map $\phi_0^f \colon V_0^f \to C^0 (A,B)$ given by $\phi_0^f (\mathcal{S})(a):= \mathrm{exp}_{f(a)}(\pi^* (f) \mathcal{S}(a))$ is locally invertible. Then choose these inverse maps to be your charts around $f$. By restricting the domains of these charts appropriately, you get a Banach manifold structure on any space $C^r (A,B)$ and indeed a Hilbert manifold structure on $H^s (A,B)$ which is rather nice.

• Oh, and then it is not hard to show that we can identify the tangent spaces by [T_f C^r (A,B)= \{v \colon A \to TB | v(a) \in T_{f(a)}B\}]. So in fact $T_{id}C^r(A,A)$ is precisely the space of $C^r$ vector fields on $A$. May 25 '11 at 8:39
• Oh, I know that it's not that hard. I'm just not very good at real analysis, and so try to avoid it whenever I can mumble some categorical incantations. Note that there's no way to decide which of the $C^r$ or $H^s$ spaces is "the right" one, and for most applications it doesn't matter which you pick. But note that any such choice does not recover the Grothendieck-style definition, because there are more "smooth maps from point" to the analytic choices than to the sheaf --- and if you want the "points" to be just the smooth maps, you're stuck: $C^\infty(A,B)$ is Frechet but not Banach. May 25 '11 at 14:39
• This is a clear answer, can anyone recommend a book on the subject, please? Thank you Dec 23 '16 at 17:21
• @TheoJohnson-Freyd the compact-open $C^\infty$ topology is the one which turns $C^\infty(A,B)$ into a metrizable topological space (provided $A$ is $\sigma$-compact and $B$ is metrizable, to be more precise), but this topology is not a manifold topology if $A$ is not compact and $B$ is not contractible, since in this case this topology is not locally arcwise connected. The Whitney $C^\infty$ topology described in Fly by Night's answer above solves this problem partially, yielding a topological but not smooth manifold structure. Dec 23 '16 at 17:49
• The reason is that the modelling topological vector spaces are spaces of smooth sections with compact support for a certain family of pullback vector bundles over $A$, and for these the topology which yields a smooth manifold topology on $C^\infty(A,B)$ with the atlas written in Michael Coffey's answer is actually finer than the standard inductive limit topology. This is well described in Sections 41 and 42 of the book of Kriegl and Michor. In either case the resulting manifold topology is metrizable if and only if $A$ is compact. Dec 23 '16 at 18:01

You said that you'd often like to talk about (small, or maybe infinitesimal) variations of a smooth map. There is a well known (in some circles at least) topology on the space of smooth maps.

It's called the Whitney $C^{\infty}$-topology.

First you define the Whitney $C^k$-topology using the natural projection $\pi : C^{\infty}(A,B) \twoheadrightarrow J^k(A,B),$ where the base space is the k-jet space; which is identified with $\mathbb{R}^p$, for some $p$ depending on $\dim(A)$ and $\dim(B),$ and of course on $k.$ As a basis for the $C^k$-topology on $C^{\infty}(A,B)$ we take the preimages of the open sets in $J^k(A,B)$ under the metric topology induced from $\mathbb{R}^p.$

Informally, the metric on the jet space measures the difference between derivatives. For example, let $f : \mathbb{R} \to \mathbb{R}$ and, for some $x \in \mathbb{R},$ consider $j_x^kf \in J^k(\mathbb{R},\mathbb{R}) \cong \mathbb{R}^{k+2}.$ We have

$j_x^kf \mapsto (x,f(x),f'(x),\ldots,f^{(k)}(x)) \in \mathbb{R}^{k+2}.$

If $W^k$ denotes the set of open sets of $C^{\infty}(A,B)$ under the $C^k$-topology, then the Whitney $C^{\infty}$-topology on $C^{\infty}(A,B)$ is defined to be the topology whose basis is $W$, where

$W := \bigcup_{k=0}^{\infty} W^k.$

The Whitney $C^{\infty}$-topology makes $C^{\infty}(A,B)$ into a Baire space.

See pages 42 - 50 of Golubitsky & Guillemin, "Stable Mappings and their Singularities", (1974).