nLab
Frölicher spaces and Isbell envelopes

Frölicher Spaces and Isbell Envelopes

Frölicher spaces are examples of generalised smooth spaces. The category of Frölicher spaces is also closely related to the concept of the Isbell envelope of a category.

Definition

Let \mathcal{R} denote the category with one object and morphism set C (,)C^\infty(\mathbb{R},\mathbb{R}).

There is a close relationship between Frölicher spaces and the subcategory of E()E(\mathcal{R}), the Isbell envelope of \mathcal{R}, of those objects satisfying Isbell duality.

Proposition

An object of E()E(\mathcal{R}) that satisfies Isbell duality is a Frölicher space.

Proof

Let X=(C,F)X = (C,F) be a generalised \mathcal{R}-object satisfying Isbell duality. From the page about the Isbell envelope, XX is concrete. Let X|X| denote the set of constant elements of CC. By concreteness, CC injects into Set(,X)Set(\mathbb{R},|X|) and FF injects into Set(X,)Set(|X|,\mathbb{R}). For clarity, we shall distinguish between an element of CC and its image in Set(,X)Set(\mathbb{R},|X|) writing α\alpha for the former and α|\alpha| for the latter. We shall do similarly for elements of FF.

Suppose that β:X\beta : \mathbb{R} \to |X| is such that ϕβC (,)|\phi| \circ \beta \in C^\infty(\mathbb{R}, \mathbb{R}) for all ϕF\phi \in F. Then the map ϕϕβ\phi \mapsto |\phi|\circ \beta is a natural transformation FC (,)F \to C^\infty(\mathbb{R},\mathbb{R}) (it obvious commutes with the left C (,)C^\infty(\mathbb{R},\mathbb{R})-actions). Hence it is an element of CC. That is, there is an element α\alpha of CC such that α(ϕ)=ϕβ\alpha(\phi) = |\phi| \circ \beta for all ϕF\phi \in F.

Let us compare α|\alpha| with β\beta. Let us put δ t:\delta_t : \mathbb{R} \to \mathbb{R} as the constant function at tt then for all ϕF\phi \in F,

(1)ϕ(α(t))=ϕαδ t=α(ϕ)δ t=ϕβδ t=ϕ(β(t))=ϕ(β(t)) \phi \circ (|\alpha|(t)) = \phi \circ \alpha \circ \delta_t = \alpha(\phi) \circ \delta_t = |\phi| \circ \beta \circ \delta_t = |\phi|(\beta(t)) = \phi \circ (\beta(t))

But if α|\alpha| and β\beta differ, then they differ at some tt \in \mathbb{R}. Now for xyXx \ne y \in |X|, as XX satisfies Isbell duality, there must be some ϕF\phi \in F such that ϕxϕy\phi \circ x \ne \phi \circ y. Hence α=β|\alpha| = \beta and so β\beta is in the image of CC in Set(,X)Set(\mathbb{R}, |X|).

For the other part, let θ:X\theta : |X| \to \mathbb{R} be such that θαC (,)\theta \circ |\alpha| \in C^\infty(\mathbb{R},\mathbb{R}) for all αC\alpha \in C. Then, exactly as above, we define a natural transformation CC (,)C \to C^\infty(\mathbb{R},\mathbb{R}) by αθα\alpha \mapsto \theta \circ |\alpha|. This corresponds to some ψF\psi \in F and it remains to compare ψ|\psi| with θ\theta. This is simpler since xXx \in |X| is an element of CC and so ψ(x)=ψ(x)=θx=θ(x)|\psi|(x) = \psi(x) = \theta \circ |x| = \theta(x).

However, not all Frölicher spaces can be obtained in this manner. The simplest example is the following:

(2)({0,1}, {0,1},) (\{0,1\},\mathbb{R}^{\{0,1\}},\mathbb{R})

where this is taken to mean that all curves are smooth and only the constant functionals are smooth. The problem here is that there are far more curves than the functionals warrant. Put another way, the functionals cannot distinguish between the points of the set.

All Frölicher spaces satisfy half of the requirements for Isbell duality: the functions are always the natural transformations of the curves.

Proposition

The object of E()E(\mathcal{R}) corresponding to a Frölicher space is FF-saturated.

Proof

Let (X,C,F)(X,C,F) be a Frölicher space. Recall from the page about the Isbell envelope that FF-saturated means that FF is precisely the set of C (,)C^\infty(\mathbb{R}, \mathbb{R})-homs CC (,)C \to C^\infty(\mathbb{R},\mathbb{R}).

We know that every element of FF gives a map CC (,)C \to C^\infty(\mathbb{R}, \mathbb{R}) which commutes with the right actions. As CC contains the constant functions at the points of XX, this assignment is injective.

Let ϕ:CC (,)\phi : C \to C^\infty(\mathbb{R},\mathbb{R}) commute with the right actions. Define ϕ:X|\phi| : X \to \mathbb{R} by ϕ(x)=ϕ(δ x)|\phi|(x) = \phi(\delta_x) where δ xC\delta_x \in C is the constant function at xx. Then for αC\alpha \in C, ϕα(t)=ϕ(α(t))=ϕ(αδ t)=ϕ(αδ t)=ϕ(α)δ t=ϕ(α)(t)|\phi| \circ \alpha (t) = |\phi|(\alpha(t)) = |\phi|(\alpha \circ \delta_t) = \phi(\alpha \circ \delta_t) = \phi(\alpha) \circ \delta_t = \phi(\alpha)(t). Hence ϕαC (,)|\phi| \circ \alpha \in C^\infty(\mathbb{R}, \mathbb{R}) for all αC\alpha \in C and so ϕ|\phi| is in FF.

Hence (X,C,F)(X,C,F) is FF-saturated.

The other half is more complicated. Using concreteness one can see that the essence depends on comparing the set XX with the set of constant natural transformations FC (,)F \to C^\infty(\mathbb{R}, \mathbb{R}). That is, those natural transformations γ\gamma with the property that γ(ϕ)\gamma(\phi) is a constant function in C (,)C^\infty(\mathbb{R}, \mathbb{R}) for all ϕF\phi \in F.

Lemma

A constant natural transformation FC (,)F \to C^\infty(\mathbb{R}, \mathbb{R}) is a function α:F\alpha : F \to \mathbb{R} with the property that α(ϕψ)=ϕ(α(ψ))\alpha(\phi \circ \psi) = \phi(\alpha(\psi)).

Let us write F|F| for this set.

Now every point of XX defines a constant natural transformation via evaluation but the map XFX \to |F| need be neither injective nor surjective. However, we can determine conditions on when it is injective or surjective. Injectivity is related to a fairly simple condition (as indicated by the example earlier).

Definition

A Frölicher space is said to be Hausdorff if the smooth functions separate points.

Proposition

A Frölicher space (X,C,F)(X,C,F) is Hausdorff if and only if the map XFX \to |F| is injective.

Proof

The key point here is that an element of F|F| is completely determined by its effect on functions in FF. Thus x,yXx, y \in X are such that they define the same element of F|F| if and only if ϕ(x)=ϕ(y)\phi(x) = \phi(y) for all ϕF\phi \in F. This means that the smooth functions do not separate xx and yy. Hence (X,C,F)(X,C,F) is Hausdorff if and only if XFX \to |F| is injective.

It is simple to construct non-Hausdorff Frölicher spaces. Indeed, the example earlier was one.

Surjectivity is more complicated. As currently stated, not even very simple Frölicher spaces satisfy the surjectivity condition.

Lemma

The Frölicher space defined by the usual structure on 2\mathbb{R}^2 does not satisfy the surjectivity condition.

Proof

We need to construct a map α:FC (,)\alpha : F \to C^\infty(\mathbb{R}, \mathbb{R}) such that α(ψϕ)=ψ(α(ϕ))\alpha(\psi \circ \phi) = \psi(\alpha(\phi)) but which does not correspond to a point in 2\mathbb{R}^2.

A non-zero point p 2p \in \mathbb{R}^2 defines an element π p\pi_p of FF by composing orthogonal projection 2p\mathbb{R}^2 \to \langle p \rangle with the map p\langle p \rangle \to \mathbb{R} defined by p1p \mapsto 1. If p,q 2p, q \in \mathbb{R}^2 are not collinear then the only situation in which ψπ p=ϕπ q\psi \circ \pi_p = \phi \circ \pi_q is if ψ\psi and ϕ\phi are constant functions. To see this, let pp' be orthogonal to pp and qq' orthogonal to qq be such that the orthogonal projection of pp' to the line spanned by qq is qq, and similarly qq' maps to pp. Then pp' and qq' span 2\mathbb{R}^2 and

(3)ψπ p(λp+μq) =ψπ p(μq) =ψ(μ) ϕπ q(λp+μq) =ϕπ q(λp) =ϕ(λ) \begin{aligned} \psi \circ \pi_p(\lambda p' + \mu q') &= \psi \circ \pi_p(\mu q') &&= \psi(\mu) \\ \phi \circ \pi_q(\lambda p' + \mu q') &= \phi \circ \pi_q(\lambda p') &&= \phi(\lambda) \end{aligned}

As this holds for all λ,μ\lambda, \mu we see that ψ\psi and ϕ\phi are constant functions.

Moreover, the functions π p\pi_p are “initial” in FF in the sense that if π p=ψϕ\pi_p = \psi \circ \phi for some ϕF\phi \in F then ϕ\phi is of the form θπ p\theta \circ \pi_p. We thus conclude that in defining a map α:F\alpha : F \to \mathbb{R} such that α(ψϕ)=ψ(α(ϕ))\alpha(\psi \circ \phi) = \psi(\alpha(\phi)) we have free choice on the values α(π p)\alpha(\pi_p). However, the maps FC (,)F \to C^\infty(\mathbb{R}, \mathbb{R}) which come from evaluation do not have this free choice: their value on π p\pi_p is completely determined by the values on π x\pi_x and π y\pi_y.

However, all is not lost. The set of functions in a Frölicher space has much more structure than simply composition by functions from C (,)C^\infty(\mathbb{R}, \mathbb{R}).

Lemma

The set FF of functions in a Frölicher space is a commutative \mathbb{R}-algebra.

Proof

Let (X,C,F)(X,C,F) be a Frölicher space. Let ϕ,ψ,θF\phi, \psi, \theta \in F. Then ϕα\phi \circ \alpha, ψα\psi \circ \alpha, and θα\theta \circ \alpha lie in C (,)C^\infty(\mathbb{R}, \mathbb{R}). Thus as C (,)C^\infty(\mathbb{R}, \mathbb{R}) is a ring,

(4)ϕα+(θα)(ψα)=(ϕ+θψ)α \phi \circ \alpha + (\theta \circ \alpha) \cdot (\psi \circ \alpha) = (\phi + \theta \cdot \psi) \circ \alpha

is in C (,)C^\infty(\mathbb{R}, \mathbb{R}). As this holds for all αC\alpha \in C, ϕ+θψF\phi + \theta \cdot \psi \in F. It is commutative because C (,)C^\infty(\mathbb{R}, \mathbb{R}) is commutative. Finally we note that there is an obvious ring homomorphism F\mathbb{R} \to F sending λ\lambda to the function xλx \mapsto \lambda.

This suggests that we should consider a Frölicher space not as a pair of functors , opSet\mathcal{R}, \mathcal{R}^{op} \to Set but as a pair of functors Set\mathcal{R} \to Set and opAlg\mathcal{R}^{op} \to Alg.

There is yet more structure on FF. Not only can we compose element of FF with elements of C (,)C^\infty(\mathbb{R}, \mathbb{R}) but if ϕF\phi \in F is a particular element then we can compose ϕ\phi with an element of C (imϕ,)C^\infty(\im \phi, \mathbb{R}). This suggests that we ought to enlarge the category \mathcal{R} so that its objects are the power set 𝒫()\mathcal{P}(\mathbb{R}).

(An alternative to this extension is to insist that the natural transformations be continuous with respect to a topology on FF compatible with the compact-open topology on C (,)C^\infty(\mathbb{R}, \mathbb{R}).)

With these two augmentations, a constant natural transformation from F(A)C (,A)F(A) \to C^\infty(-,A), with AA \subseteq \mathbb{R}, defines an algebra homomorphism F()F(\mathbb{R}) \to \mathbb{R} with the property that α(ψ)imψ\alpha(\psi) \in \im \psi for all ψF()\psi \in F(\mathbb{R}).

Lemma

Such a natural transformation, α\alpha, has the property that for any pair ϕ,ψF()\phi, \psi \in F(\mathbb{R}) there is a point xXx \in X such that α(ψ)=ψ(x)\alpha(\psi) = \psi(x) and α(ϕ)=ϕ(x)\alpha(\phi) = \phi(x).

Proof

Consider the function

(5)θ=(ψα(ψ)) 2+(ϕα(ϕ)) 2. \theta = (\psi - \alpha(\psi))^2 + (\phi - \alpha(\phi))^2.

As α\alpha is an algebra homomorphism, α(θ)=0\alpha(\theta) = 0. Since α(θ)imθ\alpha(\theta) \in \im \theta, there is thus some xXx \in X such that θ(x)=0\theta(x) = 0. For this xx we therefore have that ψ(x)=α(ψ)\psi(x) = \alpha(\psi) and ϕ(x)=α(ϕ)\phi(x) = \alpha(\phi).

This clearly extends to any finite family. Indeed, from the proof of this result we deduce that the family

(6)𝒜:={{x:ϕ(x)=α(ϕ)}:ϕF()} \mathcal{A} := \big\{\{x : \phi(x) = \alpha(\phi)\} : \phi \in F(\mathbb{R})\big\}

is directed and thus defines a filter on XX. If our natural transformation α\alpha is not represented by a point on XX then this filter will have empty intersection.

Now consider the subfamily of F()F(\mathbb{R}) consisting of those functions ϕ\phi with the property that imϕ[0,1]\im \phi \subseteq [0,1] and α(ϕ)=1\alpha(\phi) = 1. It is clear that if we restrict to this family then we still get all of 𝒜\mathcal{A}. We can order this family; there are two equivalent (but not isomorphic) orderings:

  1. ϕψ\phi \leq \psi if ϕ(x)ψ(x)\phi(x) \leq \psi(x) for all xXx \in X, or
  2. ϕψ\phi \leq \psi if ϕ 1((0,1])ψ 1((0,1])\phi^{-1}((0,1]) \subseteq \psi^{-1}((0,1]).

This family is directed (downwards) since ϕψϕ\phi \psi \leq \phi (and ψ\psi). In any reasonable topology on FF then this net converges to the zero function: on any compact subset of XX then we have (ϕ)0(\phi) \to 0 uniformly. Therefore if we simply add the condition that our natural transformation α\alpha be continuous (something we might have been ready to do anyway) we see that it must be represented by an element of XX since otherwise we have α(0)=limα(ϕ)=1im0\alpha(0) = \lim \alpha(\phi) = 1 \notin \im 0.

Thus XX is the set of continuous algebra homomorphisms FF \to \mathbb{R} and we finally see the relationship between Hausdorff Frölicher spaces as objects in the Isbell envelope of R\mathbf{R} satisfying Isbell duality.

Revised on September 12, 2010 21:06:49 by Toby Bartels (98.19.57.199)