Question on Smooth Functions

February 11, 2008

Posted by Urs Schreiber

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There must be some standard textbook reference for the following statement:

The functor from smooth manifolds to algebras which sends each smooth manifold to the algebra of smooth functions on it, and which sends morphisms of smooth manifolds to the algebra homomorphisms obtained by the corresponding pullbacks of functions $(X \stackrel{\phi}{\to} Y) \mapsto ( C^\infty(X) \stackrel{\phi^*}{\leftarrow}C^\infty(Y))$ is full: every algebra homomorphism between algebras of smooth functions comes from a pullback along a smooth map of the underlying smooth manifolds.

This is of course closely akin to Gel’fand’s equivalence (for instace recalled as theorem 1 in Spaceoids), although a little different.

A proof should consist of two steps:

a) every homomorphism of function algebras comes from pullback along a map of the underlying sets.

b) only pullback along smooth maps will take all smooth functions to smooth functions.

What’s a good reference for this?

Posted at February 11, 2008 8:53 AM UTC

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Re: Question on Smooth Functions

Natural Operations in Differential Geometry has a simple proof of this statement. It is Corollary 35.10. The preceding material in Section 35 is overkill for this result; 35.9 has an elementary proof in the “short story” in the introduction to Chapter VIII.

Posted by: Evan Jenkins on February 11, 2008 1:52 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Thanks!!

Posted by: Urs Schreiber on February 11, 2008 2:13 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Another reference is

Ieke Moerdijk and Gonzalo Reyes. Models for Smooth Infinitesimal Analysis, Springer-Verlag, New York (1991).

On p. 30 appears Theorem 2.8: Let $\mathbf{M}$ be the category of manifolds and smooth maps. Then the contravariant functor

$\mathbf{M} \to C^{\infty}-rings, M \mapsto C^{\infty}(M)$

is full and faithful, maps into finitely presented $C^{\infty}$ -rings, and sends transversal pullbacks to pushouts. (See also proposition 1.5, p. 21.)

This result is crucial to their later development of topos-theoretic models for smooth analysis.

Posted by: Todd Trimble on February 11, 2008 2:26 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Thanks!!

Posted by: Urs Schreiber on February 11, 2008 2:35 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Thanks again to Evan Jenkins and Todd Trimble for the references. I have now included a pointer to these at the end of section 5.1 “Differential forms on smooth spaces”.

This entire section is crying for being extended to a more comprehensive discussion in its own right, but for the moment it is what it is.

But here is a related question, concerning, once again, the internal hom (or lack thereof) for differential graded commutative algebras (DGCAs):

for any two DGCAs $A$ and $B$ we are calling (definition 4) $maps(B,A)$ the sheaf on smooth test domains (manifolds, say) which sends each smooth test domain $U$ to the set $U \mapsto Hom_{DGCAs}(B, A \otimes \Omega^\bullet(U)) \,.$

So let’s look at what this does in the case that $A$ and $B$ are of the form $A = \Omega^\bullet(X)$ and $B = \Omega^\bullet(Y)$ . Then

$U \mapsto Hom_{DGCAs}(\Omega^\bullet(Y), \Omega^\bullet(X) \otimes \Omega^\bullet(U)) \hookrightarrow Hom_{DGCAs}(\Omega^\bullet(Y), \Omega^\bullet(X \times U)) = Hom_{manifolds}(X \times U , Y) \,.$

The inclusion is due to the fact that not every smooth function on $X \times U$ is a tensor product of smooth functions on $X$ with smooth functions on $U$ .

On the other hand, the last term on the right

$U \mapsto Hom_{manifolds}(X \times U,Y)$

would be the inner hom in sheaves on test domains (still for the case that $X$ and $Y$ are test domains themselves, i.e. representable sheaves).

So it looks as if the failure of the inclusion $C^\infty(U) \otimes C^\infty(V) \hookrightarrow C^\infty(U \times V)$ to be a bijection is one crucial indicator for the failure of $(B,A) \mapsto \mathrm{maps}(B,A)$ to be an inner hom.

On the other hand, we could consider forming a completed tensor product, some “fusion product” which remedies this.

Such a completion would strongly remind me of the discussion of saturating sheaves which we are having over here.

While it looks as if this situation wants some conclusion to be drawn from it, I cannot say that I am seeing this conclusion at the moment…

Posted by: Urs Schreiber on February 11, 2008 3:20 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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So let’s look at what this does in the case that $A$ and $B$ are of the form $A = \Omega^\bullet(X)$ and $B = \Omega^\bullet(Y)$ .

Oohhh!!! Now I get it! I’m very glad you did that little calculation. For some reason I had a mental block about this $maps(B, A)$ business, and hadn’t realized there was this calculation sitting in the background. That’s very helpful motivation.

Hmm… okay, let me think. There is a tensor product which does what you want it to; it’s the tensor product for $C^{\infty}$ -rings, discussed in the Moerdijk-Reyes book. This is discussed in chapter I, which lays the foundation for the rest of the book.

It’s possible then that you don’t want to think of these $\Omega^\bullet(U)$ just as DGCA’s, because they in fact admit a lot more structure: that of DG $C^{\infty}$ -ring.

What’s a $C^{\infty}$ -ring? Think of an ordinary (commutative) ring as a structure in which you can interpret polynomials. Then a $C^{\infty}$ -ring is a structure in which you can interpret not just polynomials, but smooth functions generally.

This is made precise through Lawvere algebraic theories. For example, let $\mathbf{A}^{op}$ be the category opposite to the category of free finitely generated $\mathbb{R}$ -algebras (i.e., polynomial algebras in finitely many variables). Then $\mathbf{A}^{op}$ is equivalent to a Lawvere theory, a category with finite products in which every object is a (chosen) $n$ -fold product $1^n$ of a basic object $1$ . An $\mathbb{R}$ -algebra $E$ is equivalent to a product-preserving functor

$F: \mathbf{A}^{op} \to Set$

where $F(A)$ is of the form $\mathbb{R}-Alg(A, E)$ . For example, the operation of multiplication on $E$ is given by the map

$E \times E \cong \mathbb{R}-Alg(\mathbb{R}[x, y], E) \to \mathbb{R}-Alg(\mathbb{R}[z], E) \cong E$

induced by the unique algebra map $\mathbb{R}[z] \to \mathbb{R}[x, y]$ that sends $z$ to $x y$ .

For a $C^{\infty}$ -ring, we expand the Lawvere algebraic theory to include more operations (smooth operations). Thus, consider the category $S$ whose objects are (chosen) products $\mathbb{R}^n$ and smooth maps between them. This is a category of (smooth) spectra which plays the role of $\mathbf{A}^{op}$ above. Hence, this category is itself a Lawvere theory, and a $C^{\infty}$ -ring is by definition a product-preserving functor

$S \to Set.$

This is exactly what Lawvere was talking about! So we have this conjugation adjoint pair

$Set^{S^{op}} \stackrel{\leftarrow}{\to} Set^S$

which takes a sheaf $X$ to a cosheaf, but Lawvere is saying that this cosheaf is nothing but the associated $C^{\infty}$ -algebra: conjugation as a contravariant functor factors through

$Set^{S^{op}} \to Prod(S, Set) = C^{\infty}-Ring$

where $Prod$ indicates product-preserving functors. On the other hand, the conjugation functor running the other way factors through, according to the Lawvere quote, the topos of canonical sheaves on $S$ .

Aha!!!

I do believe some things are coming together, Urs.

Posted by: Todd Trimble on February 11, 2008 5:04 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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hadn’t realized there was this calculation sitting in the background

Oh, I see. I am glad I said what I just said, then.

An $\mathbb{R}$ -algebra $E$ is equivalent to a product-preserving functor $F : \mathbf{A}^{op} \to Set$

While not directly relevant to what we are discussing here, let me just highlight that this (kind of) statement seems to be very important for the general discussion we had in A topos for algebraic quantum theory:

There David Corfield asked, paraphrased, what general motivation one might have to consider non-commutative algebras as presheaves over commutative algebras in a way that is a simple variation of the theme of the remark you just made.

Maybe we are even approaching a good general answer to that question.

but Lawvere is saying that this cosheaf is nothing but the associated $C^\infty$ -algebra

Let me see if I am following. The point is that the image of any presheaf $X$ under this “conjugation” operation

$Set^{S^{op}} \to Set^S$ $X \mapsto \bar X := Hom_{Set^{S^{op}}}(X,-)$

always has the property that it is not just any functor

$S \to Set$

but a product-preserving one, in that $\bar X : U \times V \mapsto \bar X(U \times V) = \bar X(U) \times \bar X(V)$

Wait, this is probably supposed to be obvious. But I am not sure I see why products are preserved here necessarily. (Possibly after I hit “post” it will become clear ;-)

Posted by: Urs Schreiber on February 11, 2008 5:34 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Let me see if I am following. The point is that the image of any presheaf X under this “conjugation” operation is not just any functor but a product-preserving one?

Yep! That’s what I was saying. (And yes, it should be clear! ;-) It’s in the vein of the Yoneda lemma.)

By the way, I quoted a result from Moerdijk and Reyes on $C^{\infty}$ -rings whereas you were interested in algebras or $DG$ -algebras. Happily, though, the result you asked for is given in their corollary 3.7. But I reckon it was a fortuitous misconstrual on my part! :-)

Posted by: Todd Trimble on February 11, 2008 6:01 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Yep! That’s what I was saying.

All right.

Next, I gather, you were suggesting that I regard DGCAs as “DG $C^\infty$ -rings” instead of as just DG algebras.

Right now I am not entirely sure what that definition should look like. Problem is that I know what a function with values in $U \times V$ is, but not what a “differential form with values in $U \times V$ ” should be.

I could try to approach it more formally, less heuristically: maybe I need to consider productr-preserving co-pre-sheaves with values in $Vect$ ?

Sorry for being dense. I’d be grateful for hints.

Posted by: Urs Schreiber on February 11, 2008 7:15 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I gather, you were suggesting that I regard DGCAs as “DG $C^{\infty}$ -rings” instead of as just DG algebras.

I was just suggesting that DGCA’s of the form $\Omega^\bullet(U)$ can be naturally regarded as DG $C^{\infty}$ -rings. I was also suggesting that the contravariant functor

$\Omega^\bullet(-): S \to C^{\infty}-rings$

takes products to coproducts; the coproduct of $C^{\infty}$ -rings is what I was calling the $C^{\infty}$ -tensor product (just as the coproduct of ordinary commutative algebras is the ordinary tensor product). Hence we have (for somewhat formal reasons)

$\Omega^\bullet(U \times V) \cong \Omega^\bullet(U) \otimes_{C^{\infty}} \Omega^\bullet(V)$

as you seemed to want for the “fusion product”.

There’s also a way of beefing up an ordinary DGCA to a DG $C^{\infty}$ -ring, by a kind of tensoring construction which is left adjoint to a forgetful functor

$DG C^{\infty}-Ring \to DGCA$

and this left adjoint would of course preserve coproducts.

I’d like to think about this idea of internal homs of algebras that you’ve been discussing in a possibly more general context relating to conjugation. I have a memory from long ago of a talk by Lawvere in which he may have been trying out a similar thing, of enriching the algebras in the topos, or something like that! [I was a young and very green graduate student at the time, so my memory of that event may not be too accurate.]

Posted by: Todd Trimble on February 11, 2008 9:11 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Todd,

hope you don’t mind me asking again but:

what I am not seeing is how that functor $\Omega^\bullet(-) : S \to C^\infty-rings$ is supposed to be defined.

Posted by: Urs Schreiber on February 11, 2008 9:20 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Sorry – I can see that what I wrote might have been confusing!

$C^{\infty}Alg$ will denote the category of $C^{\infty}$ -algebras, and $S$ the category whose objects are $\mathbb{R}^n$ , one for each $n$ , and smooth maps between them. There are actually two functors which I’m considering. One is this functor

$C^{\infty}(-): S \to C^{\infty}Alg$

with which we more or less started this thread (except that I’ve now beefed up the structure, from commutative algebras to $C^{\infty}$ -algebras). The other is a slightly fancier construction

$\Omega^\bullet(-): S \to DGC^{\infty}Alg.$

I’d like to talk about the first functor first, as the formal development is just a little simpler in that case. The second develops along similar lines, I think.

A $C^{\infty}$ -algebra is, by definition, a product-preserving functor

$F: S \to Set.$

This is supposed to be a smooth analogue of the following definition of an ordinary commutative algebra (à la Lawvere algebraic theories): a product-preserving functor

$F: T \to Set$

where $T$ is the category whose objects are $\mathbb{R}^n$ , one for each $n$ , and whose maps are polynomial maps

$\langle p_1(x_1, \ldots, x_m), \ldots, p_n(x_1, \ldots, x_m) \rangle: \mathbb{R}^m \to \mathbb{R}^n$

between them.

Next, there is a contravariant functor

$C^{\infty}(-): S \to C^{\infty}Alg$

which takes an object $\mathbb{R}^m$ to the hom-functor $C^{\infty}(\mathbb{R}^m, -): S \to Set$ . This representable functor preserves finite products, and hence is a $C^{\infty}$ -algebra, by definition.

(Remark: as I’m sure you realize, representable functors $hom(A, -): C \to Set$ preserve all limits that happen to exist in $C$ . In fact, the very way one defines limits in a general category is makes this true; for example, for finite products, we have

$hom(A, X \times Y) \cong hom(A, X) \times hom(A, Y),$

essentially as a restatement of the universal property, that maps into a product correspond to pairs of maps.)

Each $C^{\infty}$ -algebra $F: S \to Set$ has an underlying set, defined to be $U(F) = F(\mathbb{R})$ . As usual, we sometimes abuse language and name a $C^{\infty}$ -algebra by its underlying set. By the Yoneda lemma, we have

$hom(C^{\infty}(\mathbb{R}, -), F) \cong F(\mathbb{R}) \cong Set(1, U(F))$

and from this it follows that the representable functor $C^{\infty}(\mathbb{R}, -)$ is the free $C^{\infty}$ -algebra on one generator. Similarly,

$hom(C^{\infty}(\mathbb{R}^m, -), F) \cong F(\mathbb{R}^m) \cong F(\mathbb{R})^m \cong Set(m, U(F))$

[where the middle isomorphism obtains because $F$ is product-preserving]. This says that $C^{\infty}(\mathbb{R}^m, -)$ is the free $C^{\infty}$ -algebra on $m$ generators. It is the analogue of a polynomial algebra.

Here is another point of view on this. If $A$ , $B$ are two $C^{\infty}$ -algebras, define their $C^{\infty}$ -tensor product to be their coproduct in the category of $C^{\infty}$ -algebras (these coproducts always exist, a fact we shall assume for now). Let $U, V$ be two objects of $S$ , which induce free $C^{\infty}$ -algebras $C^{\infty}(U), C^{\infty}(V): S \to Set$ . Then

$C^{\infty}(U \times V) \cong C^{\infty}(U) \otimes_{C^{\infty}} C^{\infty}(V).$

This follows from the following line of reasoning:

$\array{ hom(C^{\infty}(U) \otimes_{C^{\infty}} C^{\infty}(V), F) & \cong & hom(C^{\infty}(U), F) \times hom(C^{\infty}(V), F) & [universal property of coproduct] \\ & \cong & F(U) \times F(V) & [by the Yoneda lemma] \\ & \cong & F(U \times V) & [since F preserves products] \\ & \cong & hom(C^{\infty}(U \times V), F) & [by the Yoneda lemma] }$

and since this holds true for all $F$ , we conclude $C^{\infty}(U) \otimes_{C^{\infty}} C^{\infty}(V) \cong C^{\infty}(U \times V)$ , by one more application of the Yoneda lemma.

Now, assuming the standard fact that categories of algebras admit all limits, we can extend (uniquely up to isomorphism) to a cocontinuous $C^{\infty}$ -algebra functor defined on all presheaves:

$C^{\infty}(-): Set^{S^{op}} \to (C^{\infty}Alg)^{op}.$

The fact that the restriction

$S \to (C^{\infty}Alg)^{op}$

preserves products implies that the extension also preserves products. This is actually the cartesian case of a general fact about Day convolution, which we’ve discussed before. But it means that for general presheaves $X, Y$ , we have

$C^{\infty}(X \times Y) \cong C^{\infty}(X) \otimes_{C^{\infty}} C^{\infty}(Y).$

Now, all of the above is supposed to lift to an analogous development for the fancier construction

$\Omega^\bullet(-): Set^{S^{op}} \to DGC^{\infty}Alg$

of which $C^{\infty}(-)$ is the degree 0 part. I’m tempted to let you figure this out yourself in your preferred idiom! But I also want to think about it further myself, to see if I can concoct a slick way of saying it (some buzzwords in my head: acyclic models, left Kan from DGCA to DGC $^{\infty}Alg$ , Koszul complexes).

Sorry if this is an unsatisfactory response! Let me think a little further about what I want to say about $\Omega^\bullet$ .

Posted by: Todd Trimble on February 12, 2008 3:16 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Thanks for the long and detailed explanation. Very nice.

The cause of my puzzlement was the definition of $\Omega^\bullet(-) : S \to somewhere$ which I had gotten the – wrong – impression you were already taking for granted in your previous comments.

So let me make a suggestion for how to proceed:

to recap, originally we started from the fact that differential forms on an arbitrary presheaf had a very nice definition in that we could simple write

$\Omega^\bullet(X) := Hom_{Set^{S^{op}}} (X, \Omega^\bullet(-))$

where $\Omega^\bullet(-) : U \to \Omega^\bullet(U)$

sends each test domain to the set of differential forms on it.

Now the idea is that, while nice, this is slightly too simple minded: why do we consider just differential forms here, as opposed to, say, forms with values in some vector space $V$ ?

Since we should either consider just the canonical vector space $\mathbb{R}$ or else all of them, we should hence consider actually the operation which sends two vector spaces $U$ and $V$ to the collection of $V$ -valued forms on $U$ .

So let me take our category of test objects now to be

$S := \{objects are finite dimensional vector spaces, morphisms are smooth maps between these\}$

and consider

$\Omega^\bullet : S^{op} \times S \mapsto Set$ $(U,V) \mapsto \Omega^\bullet(U,V) \,.$

I should be specific about what I mean by the right hand side here:

$\Omega^0(U,V)$ is just the collection of smooth maps from $U$ to $V$

$\Omega^{n \geq 1}(U,V)$ is the collection of smooth maps $\wedge^n T U \to V$ which are linear on each fiber.

Let $f : V_1 \to V_2$ be a morphism in $S$ , hence a smooth map between manifolds which happen to be vector spaces, then take

$f_* : \Omega^0(U,V_1) \to \Omega^0(U,V_2)$

to be the ordinary push-forward of functions and

$f_* \Omega^{n \geq 1}(U,V_1) \to \Omega^{n \geq 1}(U,V_2)$

the ordinary push-forward of forms, as usual.

Accordingly the “janusian” deRham thing is now

$\Omega^\bullet : S^{op} \to Set^S$ $\Omega^\bullet : U \mapsto (V \mapsto \Omega^\bullet(U,V))) \,.$

And in fact in the image of this functor we find just product preserving co-presheaves, so that I can consider

$\Omega^\bullet : S^{op} \to C^\infty Alg \,.$

But maybe for the moment it is helpful not to add that additional notational layer.

Possibly I am on a different track now than you had in mind, but I feel like keep going.

Now I’d define the collection of $V$ -valued differential forms on an arbitrary $Set$ -valued presheaf $X \in Set^{S^{op}}$

$\Omega^\bullet(X,V) := Hom_{Set^{S^{op}}}( X, \Omega^\bullet(-,V) ) \,.$

That seems to make sense.

Then $\Omega^\bullet(X,-) \in Set^S$

and I guess

$\Omega^\bullet(X,-) \in C^\infty Alg \,.$

Posted by: Urs Schreiber on February 12, 2008 8:25 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I like this very much. It seems to hang together very nicely.

And it’s a nice instance of the general formalism that David and I were discussing here: starting with a 2-relation (a functor $C \times D \to Set$ , in this case your $\Omega^\bullet(-, -): S^{op} \times S \to Set$ ), we get a contravariant adjunction [categorified Galois connection] between $Set^C$ and $Set^D$ .

In one direction, the contravariant functor

$Set^{S^{op}} \to Set^S$

is what you wrote down: $X \mapsto \Omega^\bullet(X, -)$ . And this indeed factors through $C^{\infty}Alg$ as you guessed.

In fact, implementing Lawvere’s suggestion, what we should get is a janusian/ambimorphic contravariant adjunction

$Sh(S) \stackrel{\leftarrow}{\to} C^{\infty}Alg$

where on the left is the category of canonical sheaves on $S$ . So it’s all sticking together very well.

Fresh. Word.

Posted by: Todd Trimble on February 12, 2008 10:30 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Okay, good. Thanks for your reply.

Next I need to think about how to incorporate what you said about the completed “fusion” product of DGCAs in this context.

But tomorrow.

Posted by: Urs Schreiber on February 12, 2008 10:44 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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For a $C^\infty$ -algebra $A$ , it is the “underlying set”

$A(\mathbb{R})$

which carries the structure of an algebra.

So, it seems the following should make sense:

A $C^\infty$ -DGCA $A$ should be a $C^\infty$ -algebra with the property that its underlying set $A(\mathbb{R})$ carries the structure of a DGCA.

Then a morphisms of $C^\infty$ -DGCAs should be a morphisms of $C^\infty$ -algebras respecting that DGCA structure on the “underlying sets”.

Thus we get a category

$C^\infty DGCA \,.$

It seems there is a canonical inclusion functor

$DGCAs \hookrightarrow C^\infty DGCAs$ obtained by $A \mapsto \Omega^\bullet(X_A)$

where $X_A : U \mapsto Hom_{DGCAs}(A,\Omega^\bullet(U))$ is the sheaf defined by $A$ and where

$\Omega^\bullet : Set^{S^{op}} \to C^\infty DGCAs$

is the construction I described in my previous comment.

So:

Is $C^\infty DGCAs$ closed?

I’d be tempted to define (as I did before, without the $C^\infty$ -tensor product)

for $A$ and $B$ $C^\infty DGCAs$

$(B,A) \mapsto \Omega^\bullet( maps(B,A) )$

where

$maps(B,A) : U \mapsto Hom_{C^\infty DGCAs}(B, A \otimes_{C^\infty} \Omega^\bullet(U)) \,.$

I’d need to check if

$Hom_{C^\infty DGCAs}( C, A \otimes_{C^\infty} B ) \simeq Hom_{C^\infty DGCAs}( \Omega^\bullet(maps(C,A)), B ) \,.$

Posted by: Urs Schreiber on February 13, 2008 8:24 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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A $C^\infty$ -DGCA $A$ should be a $C^\infty$ -algebra with the property that its underlying set $A(\mathbb{R})$ carries the structure of a DGCA.

I have to admit that I was getting a little annoyed with myself yesterday, trying to make proper sense of $DGC^\infty$ -algebras. I thought it would be something very easy and straightforward, but then found myself getting confused.

One sticking point is that commutativity in the graded context is a skewed commutativity, so it won’t do to start with a $C^\infty$ -algebra (which is commutative in the ordinary sense) and then slap a grading on it.

A slightly less superficial point is that $C^\infty$ -algebras are defined in terms of (cartesian) Lawvere theories, whereas it would be nice to have something more ‘portable’ or operadic, which would carry over straightforwardly to the monoidal category $Vect$ .

Maybe the solution is obvious and I’m not seeing it.

Posted by: Todd Trimble on February 13, 2008 9:03 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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sticking point

Ah, right. My fault.

Actually, I don’t like my last proposal anymore for other reasons, too: if I am going to embed $DGCAs \hookrightarrow somewhere$ to use nice properties in $somewhere$ I can just as well stick with my previous approach, which used $DGCAs \hookrightarrow Set^{S^{op}}$ and then the internal hom in $Set^{S^{op}}$ .

Posted by: Urs Schreiber on February 13, 2008 10:01 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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[Recall, above and a few weeks back, we (Todd Trimble and I, mostly) were trying to find the right generalization of “ $C^\infty$ -algebras” from Moerdijk-Reyes’ book, which are product-preserving co-presheaves on smooth test domains to “ $C^\infty DGCAs$ ”: $C^\infty$ -differential graded commutative algebras. ]

I am currently thinking about this approach:

a $C^\infty$ DGCA should be at least a DGCA whose degree 0 part is a $C^\infty$ -algebra. The idea is: this characterization is already a complete characterization.

So:

We have a forgetful functor

$ev_{\mathbb{R}} : C^\infty-Algebras \to CommAlg$

which sends each $C^\infty$ -algebra $A$ to its “underlying algebra”

$A \mapsto A(\mathbb{R})$

and the functor

$(\cdot)^0 : DGCAs \to CommAlg$

which sends each DGCA $A$ to its degree 0 part

$A \mapsto A^0 \,.$

proposed definition: The category $C^\infty DGCAs$ of $C^\infty DGCAs$ is the pullback $\array{ C^\infty DGCAs &\to& DGCAs \\ \downarrow && \downarrow^{(\cdot)^0} \\ C^\infty-Algebras &\stackrel{\mathrm{ev}_\mathbb{R}}{\to}& CommAlg }$

So a $C^\infty DGCA$ is a pair, consisting of an ordinary DGCA $\mathbf{A}$ and a $C^\infty$ -algebra $A(\cdot)$ , such that there is an isomorphism of commutative algebras $\mathbf{A}^0 \simeq A(\mathbb{R}) \,.$

Notice that, in particular, ordinary algebras of smooth differential forms are $C^\infty-DGCAs$ in the obvious way:

$\Omega^\bullet(\mathbb{R}^n) \in C^\infty DGCAs \,.$

Since underlying any morphism of $DGCAs$ is always a morphism of the corresponding commutative algebras in degree 0, we are bound to inherit much of the nice structure on $C^\infty-Algebras$ this way.

In particular, it is easy to see that the coproduct

$\otimes_\infty$

on $C^\infty-DGCAs$ is automatically the “completed tensor product” in that

$\Omega^\bullet(\mathbb{R}^n) \otimes_\infty \Omega^\bullet(\mathbb{R}^m) \simeq \Omega^\bullet(\mathbb{R}^n \times \mathbb{R}^m) \,.$

This follows from the corresponding fact for the underlying degree 0 part

$\Omega^0(\mathbb{R}^n) = C^\infty(\mathbb{R}^n)$

and the obervation on p. 22 of Moerdijk-Reyes (which in turn is a consequence of their proposition 1.1 that $C^\infty(\mathbb{R}^n)$ is the free $C^\infty$ -algebra on $n$ -generators and the universal property of the coproduct):

$\Omega^0(\mathbb{R}^n) \otimes_\infty \Omega^0(\mathbb{R}^m) \simeq \Omega^0(\mathbb{R}^n \times \mathbb{R}^m) \,.$

From this the statement follows using the fact that a DGCA morphism

$\Omega^\bullet(X) \to \Omega^\bullet(Y)$

is entirely fixed by its action on the degree 0-part, because the rest follows from the fact that differential forms can locally be written as

$\omega = \omega_0 \wedge d \omega_1 \wedge d\omega_2 \wedge \cdots \wedge d \omega_n$

with $\omega_i \in C^\infty(X)$

and the fact that a DGCA morphism has to commute with differentials:

$\phi (\omega) = \phi(\omega_0) \wedge \phi(d \omega_1) \wedge \phi(d\omega_2) \wedge \cdots \wedge \phi(d \omega_n)$ $= \phi(\omega_0) \wedge d \phi(\omega_1) \wedge d\phi(\omega_2) \wedge \cdots \wedge d \phi(\omega_n) \,.$

So that’s a first important consistency check that the definition of $C^\infty DGCAs$ above makes good sense.

Posted by: Urs Schreiber on March 5, 2008 9:47 PM | Permalink | Reply to this

Re: Question on Smooth Functions

janusian/ambimorphic

We really ought to come to a decision about this.

Posted by: David Corfield on February 13, 2008 8:27 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I suppose you’re right that it’s a little awkward of me to keep going on like that, although I’m not sure the need to decide is pressing for any other reason.

(In case anyone is wondering what the heck we’re talking about, it all started back here.)

If ‘janusian’ is what other people are using and I’m the only one who ever uses ‘ambimorphic’, then I guess it makes sense for me to step down. However, I would like to argue once more for ‘ambimorphic’: I think ‘janusian’ just sounds a little too cute or preciously literary or something, whereas ‘ambimorphic’ sounds sort of neutral and scientific to my ear, while still being faithful to the intended meaning. One could counter-argue that ‘janusian’ is also faithful to the meaning, and somewhat more vivid than ‘ambimorphic’, but sorry: for me the cutesiness outweighs the vividness.

When I was in grad school, our topology professor (Julius Shaneson) was talking to us about surgery (on manifolds), and mentioned ‘surge’ being used as a verb. At the time it was invented, it was funny and people liked it (I’m imagining quips like, “let’s surge ahead!”); thirty years down the road, the joke wasn’t funny any more, and people were stuck with this silly word. I’m not saying that would necessarily happen to ‘janusian’, but I do think ‘ambimorphic’ is a little safer.

Looking back at the thread which started this, there were other suggestions too. Tom seemed ambivalent about ‘janusian’; David and Toby Bartels [who has apparently stopped commenting here] seemed to like ‘janusian’; John Armstrong and I didn’t like it so much. John Baez said too bad ‘ambidextrous’ was already taken. I don’t think Urs said anything. No one but me has commented on ‘ambimorphic’.

Then again, maybe we can propose something better than any of these!

Posted by: Todd Trimble on February 13, 2008 10:53 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Todd, I’m fine with ambimorphic. I prefer it to janusian, about which I share your reservations. Actually, I don’t think I’d ever heard the word ambimorphic before you suggested it, so it’s completely neutral for me — except on odd occasions when it makes me think of frogs and newts.

Part of the reason why I’ve been quiet about this is that I’m aware of not being very good at choosing terminology. I only know which word I don’t like.

Posted by: Tom Leinster on February 14, 2008 1:57 AM | Permalink | Reply to this

Re: Question on Smooth Functions

OK, I’m with you.

Even overlooking the use of the word in the spoof paper A Simple Proof of the Riemann Hypothesis by Leon Oiler, John Grauss, Joe Lagrunge, John Dirishlay, and Joe Fouray:

Sasaki constructed several ambimorphic solutions,

and the mention of ‘Journal of Ambimorphic Information’ in the bibliography.

Someone evidently finds the name amusing.

Posted by: David Corfield on February 14, 2008 8:57 AM | Permalink | Reply to this

Re: Question on Smooth Functions

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Well, as I say I don’t think we have to decide right away. What I’ll do is try ‘ambimorphic’ on for size for myself, and if other people like and start using it too, we can (for the purposes of this blog) stick with it. If at this point someone prefers ‘janusian’, he or she can either declare that, or [more quietly and surreptitiously] slip in the usage in a comment somewhere when the opportunity presents itself. Or suggest something else.

But (sad to say), my prediction is that none of this will matter anyway: people will continue using ‘schizophrenic’, despite Tom’s valid objections. Established terminology in mathematics has a terrible inertia: it would take a lot of momentum to get enough people to be bothered to say, “ambimorphic, the-concept-formerly-known-as schizophrenic”. (What do people call Prince these days, anyway?)

Posted by: Todd Trimble on February 14, 2008 2:05 PM | Permalink | Reply to this

Re: Question on Smooth Functions

The Artist Formerly Known As The Artist Formerly Known As Prince?

Posted by: Tom Leinster on February 14, 2008 4:26 PM | Permalink | Reply to this

Re: Question on Smooth Functions

People call him “Prince”, because that’s his name.

Look, I know this isn’t the place to go into all the details, but Prince really did have a point to make. In fact, if he did the same thing today he’d be positively mainstream. He was just a decade too early, as usual.

If you want the full story, Todd, I’d be glad to fill you in over email.

Posted by: John Armstrong on February 14, 2008 5:04 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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After I posted, I googled “Prince” and read the Wikipedia article, and indeed Prince is part of his birth name, which I hadn’t known. And I read some of the history behind the various name changes, which I found interesting. (And you’re right – he apparently was making a well-taken point, which got lost in the hay that people were making over what could easily pass for affectation.)

I wasn’t trying to make fun of Prince, by the way – it was an honest question. I honestly didn’t know.

I did find it funny, however, that as a child he was known as “Skipper”.

Posted by: Todd Trimble on February 14, 2008 7:19 PM | Permalink | Reply to this

Re: Question on Smooth Functions

Oh, I didn’t think you were making fun. I just find that nobody knows the real story behind his name changes, and I was thinking our hosts would rather the explanation take place somewhere other than their comment threads.

Posted by: John Armstrong on February 14, 2008 9:15 PM | Permalink | Reply to this

Re: Question on Smooth Functions

Toby Bartels [who has apparently stopped commenting here]

I’ve even stopped regularly reading, but every once in a while I search for my name here.

Anyway, I just want to say that I like the term “surgery” too!

Posted by: Toby Bartels on February 27, 2008 1:50 AM | Permalink | Reply to this

Re: Question on Smooth Functions

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Sorry to spoil the fun but: could somebody just recall the precise definition of that concept whose best name we are trying to find?

I take it that it’s an object which we want to hom into. But what else?

Posted by: Urs Schreiber on February 14, 2008 2:32 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I’ll be just shy of absolutely precise. The way I’ve generally seen it is that an ambimorphic structure consists of two structures on the same underlying object $S$ [let’s say a set for now] which “commute” with one another, say a $C$ -structure and a $D$ -structure, such that the lift

$C^{op} \to D$

of $hom_C(-, S): C^{op} \to Set$ , and the lift

$D^{op} \to C$

of $hom_D(-, S): D^{op} \to Set$ , are adjoint in the appropriate sense. (Most of the time where I’ve seen the concept applied, this adjunction is actually an equivalence between $C^{op}$ and $D$ .)

A classical example is Stone duality. The set $\mathbf{2}$ carries both a Boolean algebra structure and a structure of Stone space (a compact Hausdorff totally disconnected space), in such a way that it induces a contravariant equivalence between the category of Boolean algebras $C$ and the category of Stone spaces $D$ .

Posted by: Todd Trimble on February 14, 2008 2:51 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Thanks.

[let’s say a set for now]

Okay, and now let’s say it more generally.

I think you told me that the fact that we have this adjunction between presheaves on smooth test domains and DGCAs

$A \mapsto X_A := Hom_{DGCA}(A, \Omega^\bullet(-))$ $X \mapsto \Omega^\bullet(X) := Hom_{Set^{S^{op}}}(X,\Omega^\bullet)$

is an example for an ambimorphic object $\Omega^\bullet$ .

But, apart from the very suggestive notation, I am having trouble thinking in a precise way of “ $\Omega^\bullet$ ” as one abstract thing realized in two ways in these two maps.

In the second case it is clear: here $\Omega^\bullet$ is regarded as a set-valued presheaf.

In the first case, do I have to regard it as a DGCA-valued presheaf to make its ambimorphicity manifest?

Posted by: Urs Schreiber on February 14, 2008 5:01 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I’d say in both cases, you have to regard it as a DGCA-valued presheaf to make its ambimorphicity manifest.

In the second case, you want to use the available structure on $\Omega^\bullet$ in order to put a DGCA structure on $Hom_{Set^{S^{op}}}(X, \Omega^\bullet)$ , and that’s to see $\Omega^\bullet$ as a DGCA-valued presheaf. Same for the first case.

Posted by: Todd Trimble on February 14, 2008 7:30 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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What’s the right reference to give when I tell somebody about how $\Omega^\bullet$ is schizophrenic (I guess I have to at least mention that word for purposes of reference).

Is it Johnstone’s Stone Spaces, or is it Lawvere-Rosebrugh’s Sets for mathematics ?

Posted by: Urs Schreiber on February 14, 2008 7:43 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I don’t know. I think the word (in this sense) has been around since at least the 60’s.

Anyone?

Posted by: Todd Trimble on February 14, 2008 7:53 PM | Permalink | Reply to this

Re: Question on Smooth Functions

I think the term “schizophrenic object” was coined by Harold Simmons. I don’t know in which paper, but I can ask him.

Ah, Google turns up this posting of Peter Freyd, which quotes a book review by Peter Johnstone, in which he says:

Again, the 1982 paper in which Harold Simmons coined the term “schizophrenic object” [Topology Appl. 13 (1982), no. 2, 201–223; MR 83f:18006], for a set with two commuting algebraic structures, is not there, although Clark and Davey freely use this term in their text.

Posted by: Robin on February 14, 2008 11:26 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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Thanks, Robin. 1982! That seems like such a late date for such a basic concept. (Think of all the antecedents: Stone duality, Pontryagin duality, duality between finite posets and finite distributive lattices, etc., etc.)

My memory is playing tricks on me: I thought I had seen it in a Midwest Category Seminar somewhere. Signs of my impending dotage…

Posted by: Todd Trimble on February 15, 2008 12:16 AM | Permalink | Reply to this

Re: Question on Smooth Functions

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Another question:

one might ask: why this ambimorphic object and not another.

So in the example I like to talk about (which happens to be on-topic for this particular thread, so I don’t feel bad about going on about it), one could ask:

why not instead consider just $C^\infty$ , the algebra valued sheaf of functions.

Of course everybody can consider whatever he or she likes, but there is of course a deeper reason here for emphasizing DGCAs so much, and I am wondering if that deeper reason can be phrased somehow nicely in terms of general nonsense about duality and ambimorphicity:

so the point is this: DGCAs are $\infty$ -like, while mere algebras are not.

And it’s that “ $\infty$ -structure” which should correspond to the “ $\infty$ -structure” on smooth spaces.

Hm, can we make this more precise? Are there model category structures around here? On sheaves over smooth test domains?

Is there anything to be said about ambimorphicity-induced duality of model categories?

Posted by: Urs Schreiber on February 14, 2008 8:04 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I am currently writing this up in a hopefully eventually coherent set of notes.

And I realize that I don’t quite follow the last step:

could you say again how exactly it follows from the fact that

$C^\infty : S \to C^\infty Algebras$

preserves products (your above proof of that I follow without problem) that also

$C^\infty : Set^{S^{op}} \to C^\infty Algebras$

preserves products?

(I’ll try to go back to what you said about Day convolution and see if I can figure it out myself…)

Posted by: Urs Schreiber on March 6, 2008 7:11 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I might be able to save you a little time. Here is a slightly informal way of putting it.

Let $F$ , $G$ be two objects of $Set^{S^{op}}$ , and write each as a colimit of representables, say $colim_i hom(-, s_i)$ and $colim_j hom(-, t_j)$ . Then, because the cartesian product $F \times G$ preserves colimits in each of its separate arguments $F$ and $G$ , we have

$F \times G \cong colim_{i, j} hom(-, s_i) \times hom(-, t_j) \cong colim_{i, j} hom(-, s_i \times t_j).$

Now the extension of $C^{\infty}(-): S \to C^{\infty}Alg^{op}$ to $C^{\infty}(-): Set^{S^{op}} \to C^{\infty}Alg^{op}$ is (essentially by definition) cocontinuous. It follows that

$C^{\infty}(F \times G) \cong C^{\infty}(colim_{i, j} hom(-, s_i \times t_j)) \cong colim_{i, j} C^{\infty}(hom(-, s_i \times t_j)),$

and since $C^{\infty}(hom(-, s)) = C^{\infty}(s)$ by definition, we may continue this string of isomorphisms:

$colim_{i, j} C^{\infty}(hom(-, s_i \times t_j)) \cong colim_{i, j} C^{\infty}(s_i \times t_j).$

Finally, since $C^{\infty}: S \to C^{\infty}Alg^{op}$ preserves products, we may continue this as:

$colim_{i, j} C^{\infty}(s_i) \times C^{\infty}(t_j) \cong (colim_i C^{\infty}(s_i)) \times (colim_j C^{\infty}(t_j))$

and eventually unwind the calculation we did above to conclude this is isomorphic to $C^{\infty}(F) \times C^{\infty}(G)$ . (As I say, this proof is a little bit informal, but rest assured it’s essentially correct.)

Posted by: Todd Trimble on March 6, 2008 8:38 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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There is this other question here which I haven’t managed to satisfactorily answer:

Let $A$ and $B$ be two DGCAs, of which $A$ is assumed to be a qDGCA, meaning that it is free as a graded-commutative algebra (not necessarily as a differential algebra), and assume it is generated as such from a finite-dimensional positively graded vector space.

Then there is that presheaf:

$X : U \mapsto Hom_{DGCAs}(A, B \otimes \Omega^\bullet(U))$

(ordinary tensor product of DGCAs here again) and the claim is that one gets a $(k-p)$ -form on this presheaf for each pair consisting of

- a homogeneous element $a$ of $A$ of degree $k$

- a current $c$ on $B$ of degree $p$

(a current is a smooth map $B \to \mathbb{R}$ , a generalization of the notion of distributions to DGCAs, and it’s of homogeneous degree $k$ if it vanishes on all homogeneous elements of $B$ not of degree k).

This works by sending each $f \in Hom_{DGCAs}(A, B \otimes \Omega^\bullet(U))$ to

$c(f(a)) \in \Omega^\bullet(U)$

(as described in (what is now) prop. 1 on p. 19 and applied for instance in 9.3.1, p. 86).

What I couldn’t satisfactorily answer so far:

Question: do all forms on that presheaf of maps arise from sums of wedge products of forms coming from such pairs?

Posted by: Urs Schreiber on February 13, 2008 11:08 PM | Permalink | Reply to this

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Re: Question on Smooth Functions

$MathML-enabled post (click for more details).$

I’m climbing down the comment tree to respond to Urs’s comment back here.

This looks like a very interesting resolution to the problem, Urs, which I’d like to test out a little. So you’re defining the category of $C^{\infty}DGCA$ ’s as a weak pullback (an iso comma object) of the forgetful functor $U: C^{\infty}Alg \to CommAlg$ along the degree 0 functor $DGCA \to CommAlg$ . So a $C^{\infty}$ DGCA is defined to be a triple $(A, C, U A \stackrel{\sim}{\to} C^0)$ .

I’m pretty sure that limits in the category $C^{\infty}DGCA$ are computed in the obvious way (by taking limits in the first and second components $A$ and $C$ ). This relies on the fact that the functors $U$ and $(-)^0$ preserve limits. If $U$ and $(-)^0$ also preserved colimits, then a similar statement could be made about colimits in $C^{\infty}DGCA$ . However, $U$ does not preserve colimits ( $(-)^0$ does), so I don’t think I understand how colimits work in $C^{\infty}DGCA$ . One thing I’d like to understand is how coproducts work, in detail.

(Put slightly differently, I hear a voice from afar crying, ¡Arriba! ¡Arriba! ¡Andale! ¡Andale! ¡Yii-haa!, and I’m saying, Wait! Wait for me! Slow down!)

Posted by: Todd Trimble on March 7, 2008 4:09 PM | Permalink | Reply to this

Re: Question on Smooth Functions

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I don’t think I understand how colimits work in $C^\infty DGCA$

Then it’s more subtle than I thought it is.

Maybe before we try to understand those colimits in full generality, let’s make a quick check for the important special case which I mentioned:

Let $CE(g)$ be a DGCA whose underlying GCA is freely generated on a positively graded finite dimensional vector space. So it naturally extends to the structure of a $C^\infty DGCA$ , the degree 0 part being functions on the point.

Let $A$ be any $C^\infty DGCA$ .

Does the coproduct of $CE(g)$ with $A$ exist in $C^\infty DGCAs$ ?

I thought it exists and happens to coincide with the ordinary tensor product $CE(g)\otimes A$ .

Wait for me! Slow down!

Right, I might well have overtaken myself here, as you suggest. Hereby I have come back and halted. (Have to catch a train in a minute anyway and then there is someone waiting for me who doesn’t allow me to do math anyway… ;-)

As you might recall, a while back I suggested somewhere that a $C^\infty DGCA$ should be a $C^\infty$ algebra in degree 0, and a “ $C^\infty$ algebra module” in degree $k \gt 0$ , whatever that might mean. Maybe we need to say it that way after all.

Do you have an idea what a good notion of $C^\infty$ algebra module would be? That sounds like an interesting question even apart from our application here.

(Hm, now that I think of it, Moerdijk & Reyes must surely have thought of this. But they don’t seem to have the book here at the train station…)

Posted by: Urs Schreiber on March 7, 2008 7:22 PM | Permalink | Reply to this

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