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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeFeb 24th 2010

am expanding the entry on smooth algebra (aka $C^\infty$-rings, see the query box discussion there): more examples, more properties, etc.

But have to interrupt now.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

Please somebody give me a sanity check on this. I ought to understand this, but I feel a little unsure that I am not overlooking something stupid.

I want to see if the forgetful functor $U : C^\infty Ring \to Alg_{\mathbb{R}}$ from $C^\infty$-rings aka smooth algebras to ordinary $\mathbb{R}$-algebras has a left adjoint.

Evidently the functor preserves all small limits, so I want to check if the adjoint functor theorem applies. Take the third of the three sufficient conditions listed there: $C^\infty Ring$ should be cototal and $Alg$ locally small. The latter is clearly true, so is $C^\infty Ring$ cototal?

It would seem to me that it clearly is, just as $Alg_{\mathbb{R}}$ is. But maybe I am making some dumb mistake.

let $X \in [C^\infty Ring, Set]$ be a functor, $A$ a $C^\infty$-ring and $Hom_{C^\infty Ring}(A,-)$ its image under the Yoneda embedding. Then

$Hom_{[C^\infty Ring,Set]}(X, Hom(A,-)) = \int_{B \in C^\infty Ring} Hom_{Set}(X(B), Hom(A,B) ) \,.$

In the integrand we have simply $|X(B)|$-many morphisms from $A$ to $B$. So this is

$\cdots \simeq \int_B Hom_{C^\infty Ring}(A, \prod_{X(B)} B) \simeq Hom_{C^\infty Ring}(A, \int_{B} \prod_{X(B)} B) = Hom_{C^\infty Ring^{op}}(\int_{B} \prod_{X(B)} B, A)$

This gives a left adjoint to the Yoneda embedding $C^\infty Ring^{op} \hookrightarrow [C^\infty Ring, Set]$ by $X \mapsto \int_B \prod_{X(B)} B$.

Is that right? Am I making a stupid mistake somewhere?

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeApr 26th 2010

Perhaps I’ll need a sanity check myself, but I think the left adjoint exists on very general grounds. I think I can simplify the argument as follows:

Suppose given any morphism of finitary monads $f: S \to T$. Here $S$ will be the free algebra monad on $Set$, $T$ will be the free smooth algebra monad, and

$f X: S X \to T X$

is the inclusion which takes an $X$-ary operation for the theory $S$ to the corresponding “smooth” $X$-ary operation for the theory $T$. The forgetful functor pulls back a $T$-algebra $\xi: T X \to X$ along $f X$ to obtain an $S$-algebra $\xi \circ f X: S X \to X$.

The left adjoint takes an algebra $\theta: S X \to X$ to the (reflexive) coequalizer of the pair

$T S X \stackrel{\overset{(\mu_T \circ T f)X}{\to}}{\underset{T\theta}{\to}} T X$

living in the category of $T$-algebras, where $\mu: T T \to T$ is the monad multiplication. This may be denoted $T \circ_S X$, by analogy with tensor products.

I expect a crucial observation that this construction works is that a finitary monad $T$ (meaning a monad which arises from a Lawvere theory) preserves reflexive coequalizers, more or less because finite power functors $x \mapsto x^n$ preserve reflexive coequalizers. Followed by a longish diagram chase which is more or less the same as for an analogous argument for modules over rings. I’ll try to check this more carefully in a bit.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 26th 2010

I think the left adjoint exists on very general grounds.

Okay, good, that’s what I expected. Thanks, Todd.

I would have taken this as evident had not somebody who ought to be expert on this made a remark that contradicted this in some informal discussion. I figured there must have been a misunderstanding, but now I got worried that I was overlooking something very simple.

All right, thanks again.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

I filled this discussion into the section The underlying ordinary algebra at smooth algebra.

Todd, please check to see if I reproduced your comment to your satisfaction. Otherwise please edit.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeApr 26th 2010

Todd,

over at smooth algebra you added a query box asking for my proposed proof that $C^\infty Alg$ is cototal. I stated that above in this thread here. Please have a look.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeApr 26th 2010

Right, I saw in # 1 that you asked whether $C^\infty Alg$ is cototal. Oh, I see now, you are sketching a proof; sorry. :-/ Let me have a closer look later; I want to see why this case would go through but doesn’t for the case of groups (according to Richard Wood).

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeApr 26th 2010

Okay, I took a look, and one problem I think I see with this formal argument is that the end you wrote down looks like a limit of a large diagram, and there’s no clear indication why it should exist in $C^\infty Alg$, which is small-complete. This is one of those annoying calculations where size may matter (annoying because it looks formally correct after all). The question remains whether $C^\infty Alg$ is nevertheless cototal after all; I’d have to think about it some more.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

Thanks, Tood. I appreciate that a lot.

right, size issues: but actually what I have in mind is the small site of Coo-rings, which is that of finitely generated Coo-rings only. That’s the one used in this business. I’ll add that qualifier to the Lab entry now. Wouldn’t that solve the problem? Ah, hm, then I have to argue that $\int_B \prod_{X(B)} B$ is still finitely generated. Hm.

I should admit that the argument here is inspired by what Bertrand Toen does around page 43 here. The adjunction he states there is precisely the one I am talking about, only that he considers k-algebras and not smooth algebras. But he has some careful universe-gymnastics earlier on. I should check that more carefully.

• CommentRowNumber10.
• CommentAuthorUrs
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

Hm, I realize that I don’t understand how the size issue is actually resolved in Toen’s article either. How embarrasing.

So he invokes a universe $\mathbb{U}$ and a bigger one $\mathbb{V}$ containing the former. Then $Alg_k$ for him is algebras in $\mathbb{V}$ (bottom of p. 37).

Then when he talks about that left adjoint we have been talking about, $Alg_k^{op} \leftarrow [Alg_k, Set_{\mathbb{V}}]$, given by $\mathcal{O}(F) = \int_{B \in Alg_k} \prod_{F(B)} B$ on p. 43, he remarks that $\mathcal{O}(F)$ is not generally in $\mathbb{U}$. But why is it in $\mathbb{V}$?? I think I am missing something here…

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeApr 26th 2010
• (edited Apr 26th 2010)

Oh, now I get it. We talk there about the adjunction

$Alg_k(\mathbb{V})^{op} \stackrel{\leftarrow}{\to} [Alg_k(\mathbb{U}), Set] \,.$

Okay, never mind me, I am just being stupid. But at least I understand it now. :-)

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeNov 15th 2016

I am moving the following ancient query box out of the entry smooth algebra to here:

+– {: .query} I don't see why this is a problem; it's not like our ‘$\infty$’ ever gets into a superscript. I find ‘$C^\infty$-ring’ much more descriptive than ’generalized smooth algebra’, in fact. —Toby

Urs Schreiber: I’ll see what to do about it here. Over at derived smooth manifolds and related entried before long we’ll have to be talking about “$\infty-C^\infty$-rings” or $C^\infty_\infty$-rings or the like, which is not good. Also, the term “$C^\infty$-ring” hides that it is necessarily an $\mathbb{R}$-algebras. Finally, the entire theory is really a special case of algebras over a Fermat theory and hence most other examples of a similar kind will by default be called algebras, not rings. For all these reasons I find “$C^\infty$-ring” an unfortunate term. But of course I am aware that it is entirely standard.

I am not sure about ring vs algebra, however, I am for using the word “smooth” instead of the symbol $C^\infty$. This is similar to the term “smooth map” to mean $C^\infty$ map. — Colin Tan

=–

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeNov 16th 2016

Now that this thread has been resurrected, let me return to my over-6-year-old promise in #8 to “think about it [whether $C^\infty Alg$ is cototal] some more”, in case it is of any interest.

Unless I’m missing something, it’s not cototal. We can adapt one of the standard proofs that the category of groups is not cototal, which I’ll recall now: if it were, then every continuous functor $G: Grp \to Set$ would have a left adjoint $F$ and therefore be representable (by $F 1$). Let $G_\alpha$ be a class of simple groups of cardinalities $\alpha$ without bound. Then consider $G = \prod_\alpha Grp(G_\alpha, -): Grp \to Set$, which indeed lands in $Set$ since for each group $H$ the factor $Grp(G_\alpha, H)$ is trivial as soon as $\alpha \gt card(H)$ and so doesn’t affect the overall product. The functor $G$ is continuous. But we cannot have $G \cong \hom(F, -)$ for any group $F$, else $\hom(G_\alpha, -)$ would be a retract of $\hom(F, -)$ for all $\alpha$, hence $G_\alpha$ would be a retract of $F$ for all $\alpha$, which is clearly impossible.

In the case of $C^\infty Alg$, we can create a large class of simple objects ($C^\infty$ fields): if $R$ is a $C^\infty$ integral domain (meaning $r \cdot -: R \to R$ is injective if $r \neq 0$), such as a free $C^\infty$ algebra $C^\infty(X)$ on a set $X$, then we can construct its field of fractions $\widebar{R}$ by taking $X \cong R \setminus \{0\}$, $X = \{x_r: r \neq 0\}$, and modding out $R[X] \coloneqq R \otimes_{C^\infty} C^\infty(X)$ by the $C^\infty$ ideal generated by elements $1 - r x_r$. Let $F_\alpha$ be a class of $C^\infty$ fields of cardinalities $\alpha$ without bound, and let $A_\alpha = \mathbb{R} \times F_\alpha$. The only non-trivial regular quotients of $A_\alpha$ are $F_\alpha$ and $\mathbb{R}$. If $A$ is any non-terminal object, then for all sufficiently large $\alpha$ there is exactly one map $A_\alpha \to A$, since as soon as $\alpha \gt card(A)$ the regular epi-mono factorization of such a map would have to take the form $A_\alpha \to \mathbb{R} \to A$ and $\mathbb{R}$ is initial.

Then as before, form $G = \prod_\alpha \hom(A_\alpha, -): C^\infty Alg \to Set$. By the same argument as before, this indeed lands in $Set$ and is continuous, but is not representable (we can argue that each $\hom(A_\alpha, -)$ is a retract of $G$, but for any object $B$ we cannot have $\hom(A_\alpha, -)$ a retract of $\hom(B, -)$ for all $\alpha$, by cardinality considerations).

• CommentRowNumber14.
• CommentAuthorTodd_Trimble
• CommentTimeNov 16th 2016
• (edited Nov 16th 2016)

(Hm, maybe make that “mod out by a maximal $C^\infty$ ideal containing the elements $1 - r x_r$”.)