The n-Category Café

I have restructured a bit and added a section (number 2 at the moment) on the general abstract nonsense of space and quantity. I followed Lawvere’s Taking categories seriously, but varied slightly. Please check.

(I haven’t said anything about the dotted arrows that Lawvere has on p. 17. I guess these are supposed to allude to the kind of “saturation” or “completion” operation which we keep talking about?)

I have added a statement, right now proposition 5, which says that for all $C^\infty DGCAs$ $A$ and $g$ any $L_\infty$-algebra we have

$$CE(g)\otimes_\infty A \simeq CE(g) \otimes A \,.$$

I think this is essentially obvious, using the fact that the GCA underlying $CE(g)$ is freely generated in positive degree. But please check.

space.pdf

I have tuned and expanded a bit the intro on the general abstract nonsense about conjugation and ambi-duality.

Have discarded the notion of $C^\infty DGCA$ alltogether, since it turned out to be less obvious than I thought it would – and since I realize that I don’t actually need it!

Added two quick examples for the integration theory of $L_\infty$-algebras.

Stated the definition of the integral $$\int_\Sigma A$$

of a general $L_\infty$-algebra valued form $A$ on the smooth space $X$ over smooth space $\Sigma$.

(Compare the notes on Integration without integration #. More to come later.)

I was asked by private email how to define vector fields and curves integrating them on smooth spaces conceived in terms of sheaves.

I think this way:

using the internal hom, both the space of forms as well as the space of functions on our smooth space $X$ is again a smooth space.

Then I say that a vector field on X is a smooth $C^\infty(X)$-linear map

$$v : \Omega^1(X) \to C^\infty(X) \,.$$

A curve

$$\gamma : \mathbb{R} \to X$$

is parallel to this vector field if the diagram

$$\array{ \Omega^1(X) &\stackrel{v}{\to}& C^\infty(X) \\ \downarrow^{\gamma^*} && \downarrow^{\gamma^*} \\ \Omega^1(\mathbb{R}) &\stackrel{\partial_t}{\to}& C^\infty(\mathbb{R}) }$$

commutes, where $\partial_t$ denotes the canonical vector field on the real line.

Abject apologies for being a bit late to this party; had a bit of trouble getting a pumpkin (can’t be a Norwegian myth).

All of the suggested starting categories have a separator and this has serious consequences for Isbell conjugation. (To throw a link to the discussion over there, after working some of this out on paper I tried a few MathSciNet searches to see if anyone else had come up with it; several of the searches I tried used MSC codes to try to restrict the results to something useful. For some reason, the words “conjugation” and “dual” are fairly common in mathematics.)

So let $\mathcal{C}$ be a category with a separator, say $S$. Let $X$ be a presheaf on $\mathcal{C}$ (oh how I’d love to be able to do macros in iTeX); that is, as I’ve learnt, a contravariant functor from $\mathcal{C}$ to $Set$.

For objects $A, B$ in $\mathcal{C}$ we therefore get a set map

$$\mathcal{C}(A,B) \to Set(X(B), X(A))$$

functorial in both $A$ and $B$. We can turn this around slightly to get a set map

$$X(B) \to Set(\mathcal{C}(A,B), X(A))$$

still functorial in both $A$ and $B$. In terms of elements (gosh, did I really use that word?), this is

$$x \mapsto \big( f \mapsto X(f)x \big)$$

In particular, for a $\mathcal{C}$-morphism $g : B \to C$ we get a commutative diagram

$$\begin{aligned} X(C) & \overset{\quad\quad}{\to} && Set(\mathcal{C}(S,C), X(S)) \\ X(g) \downarrow & && \downarrow \\ X(B) & \overset{\quad\quad}{\to} && Set(\mathcal{C}(S,B), X(S)) \end{aligned}$$

where the right-hand vertical arrow is the obvious one. Hence we can define a new functor $Y : \mathcal{C}^{op} \to Set$ by

$$Y(B) = Im\big( X(B) \to Set(\mathcal{C}(S,B), X(B)) \big)$$

There is an obvious natural transformation $X \to Y$ which consists of surjections.

Let us denote Isbell conjugation by primes. From the natural transformation $X \to Y$ we obtain a natural transformation $Y' \to X'$. I claim that this is a natural isomorphism.

Let us start with injectivity.

Let $B$ be an object of $\mathcal{C}$ and let $f \in X'(B)$. This is a natural transformation $X \to B^*$ so for an object $A$ of $\mathcal{C}$ we have a map of sets

$$f_A : X(A) \to B^*(A) = \mathcal{C}(A,B).$$ In particular, we get $f_S : X(S) \to \mathcal{C}(S,B)$.

Suppose that $f, g \in X'(B)$ are such that $f \ne g$. Then there is some object $A$ of $\mathcal{C}$ such that $f_A \ne g_A : X(A) \to \mathcal{C}(A,B)$. As these are set maps, there is some $x \in X(A)$ such that $f_A(x) \ne g_A(x) \in \mathcal{C}(A,B)$. These are now morphisms in $\mathcal{C}$. As $S$ is a separator, there is some morphism $a : S \to A$ such that $f_A(x)a \ne g_A(x)a$. Now as $f$ and $g$ are natural transformations,

$$f_A(x)a = \mathcal{C}(a, B) f_A(x) = f_S (X(a)x)$$

and similarly for $g$. Hence the element $X(a)x \in X(S)$ distinguishes $f_S$ and $g_S$. Thus the map

$$X'(B) \to Set(X(S), \mathcal{C}(S,B)), \quad f \mapsto f_S$$

is injective.

This holds with $Y$ in place of $X$ and we have a commuting diagram.

$$\begin{aligned} X'(B) & \overset{\quad\quad}{\to} && Set(X(S), \mathcal{C}(S,B)) \\ \uparrow &&& \uparrow \\ Y'(B) & \overset{\quad\quad}{\to} && Set(Y(S), \mathcal{C}(S,B)) \end{aligned}$$

with both horizontal maps being inclusions.

Now

$$Y(S) = Im\big(X(S) \to Set(\mathcal{C}(S,S), X(S))\big)$$

where the map is $x \mapsto (f \mapsto X(f)x)$. There is also a map

$$Set(\mathcal{C}(S,S), X(S)) \to X(S)$$

given by evaluation on the identity. The composition is the identity on $X(S)$ and thus $Y(S) \cong X(S)$. Moreover, this isomorphism is that appearing in the natural transformation $X \to Y$. Thus in the above diagram, the right-hand vertical map is an isomorphism. The left-hand map is thus injective.

Now let us consider surjectivity.

Let $f \in X'(B)$. This is a natural transformation $X \to B^*$. We need to show that it factors through $Y$. Let $A$ be an object in $\mathcal{C}$ and suppose that $x, y \in X(A)$ are such that $f_A(x) \ne f_A(y)$ in $\mathcal{C}(A,B)$. Then there is a morphism $a : S \to A$ such that $f_A(x)a \ne f_A(y)a$. Using again the fact that $f$ is a natural transformation, $f_A(x)a = f_S(X(a)x)$ and similarly for $y$. Thus $X(a)x \ne X(a)y$ in $X(S)$. Hence the maps $\mathcal{C}(S,A) \to X(S)$ defined by $x$ and $y$ differ. That is, the images of $x$ and $y$ in $Y(A)$ are distinct.

This shows that $f_A$ factors through $Y(A)$ and hence, as $X \to Y$ consists of surjections, we can build a natural transformation $f^Y : Y \to B^*$, such that under $Y' \to X'$, $f^Y$ maps to $f$. Thus $Y'(B) \to X'(B)$ is surjective.

We therefore conclude that the natural transformation $Y' \to X'$ is a natural isomorphism.

We have

$$\begin{aligned} Y(A) &\subseteq Set(\mathcal{C}(S,A), X(S)) \\ Y'(A) &\subseteq Set(X(S), \mathcal{C}(S,B)) \end{aligned}$$

we also have a pairing compatible with these inclusions; that is (on objects)

$$Y(A) \times Y'(B) \to \mathcal{C}(A,B) \subseteq Set\big(\mathcal{C}(S,A), \mathcal{C}(S,B)\big)$$

(inclusion as $S$ is a separator).

From this it is clear that $Y''$ is given by

$$Y''(A) = \big\{ c : \mathcal{C}(S,A) \to X(A) \mid f c \in \mathcal{C}(A,B) \forall B \in \mathcal{C}, f \in Y'(B)\big\}$$

Lo and behold! A Frölicher space! Or rather, A Frölicher $\mathcal{C}$-object.

Thus stability under Isbell conjugation is a very strong condition indeed; it forces quasi-representability and also the saturation condition. This suggests to me that there is a qualitative difference between quasi-representability and non-quasi-representable. As I conjectured, there is a functor from all presheaves to quasi-representable ones, and of course then one can (may I say “should”?) saturate with respect to duality. So every presheaf has an underlying Frölicher object, but this assignment is by no means injective. In fact, contrary to what I thought, the fibres of this functor do have some interesting stuff in them.

For example, if we take the simplified version of Urs’ favourite functor; namely 1-forms, then the associated Frölicher space is the single point (since $\Omega^1(pt) = \{0\}$).

It turns out, therefore, that Urs and I are looking at orthogonal concepts!

One possible area for further study is to see what happens if you vary the separator. It doesn’t have to be a singleton point in our categories of interest. For example, if we enlarge our category of “known smooth stuff” to include $\mathbb{R}^{\infty}$, the direct limit of the $\mathbb{R}^n$s, and $X = \Omega^1$ to be 1-forms, then by taking $\mathbb{R}^{\infty}$ as the separator we find that

$$\Omega^1(M) \to Set\big(C^\infty(\mathbb{R}^{\infty},M), \Omega^1(\mathbb{R}^\infty)\big)$$

is injective, and so $\Omega^1$ is quasi-representable. Of course, it is no longer quasi-representable in the strictest sense because we do not have an injection

$$\Omega^1(M) \subseteq Set(\vert M \vert, X)$$

for any set $X$. We only get this type of quasi-representability if our separator is a singleton point.

So perhaps we should define quasi-$S$-representable where $S$ is a separator in the category $\mathcal{C}$, by which we mean that the natural transformation which on objects is

$$X(B) \to Set\big(\mathcal{C}(S,B), X(S)\big)$$

consists of injections.

However, varying the singleton won’t change the fact that Isbell conjugation (now thinking of one of our smoothish categories) results in strictly quasi-representable presheaves since the argument above made no assumptions on the separator $S$ so it certainly does work for $S$ the singleton point.

Right, it’s lunchtime so I’m stopping here.

Andrew

Just a thought on a few definitions. In a sense, the discussion on smooth spaces is about finding the right home for things like $\Omega^\bullet(M)$. The problem is that by having a specific example in mind, the temptation is always to find the one in which the example sits in a “nice” fashion.

The principle I would like to apply is: only trim things down for a reason.

So at first sight, $\Omega^\bullet(M)$ is a presheaf on the category of all smooth manifolds (maybe even manifolds with corners). So we start with that category. In that, we can define the “nice spaces” and “very nice spaces” as you do; namely, as sheafs and as presheaves stable under Isbell conjugation.

Then we look at our subcategories. First, sheaves. As these are sheaves, they are completely determined by their effect on open subsets. As we are dealing with manifolds, these are open subsets of Euclidean spaces and in fact we can restrict to simply Euclidean spaces themselves (though if we have corners we need quadrants, I suppose).

Then we look at “very nice spaces”. By my previous comment, these are Frölicher spaces. These, as we know, are completely specified by their restriction to the real line.

So in each case we do end up with being able to define the category of “generalised smooth objects” on a smaller category than that of all smooth manifolds. But just because we can do it with the smaller categories doesn’t mean that we should do it with the largest category. After all, if we took the line that we should use the smallest throughout then we would use the full subcategory with one object which is the real line.

I guess that this is more style than substance, though.

Andrew