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• CommentRowNumber1.
• CommentAuthorAndrew Stacey
• CommentTimeFeb 23rd 2010

Added a diagram at commutative algebraic theory using the totally awesome SVG Editor.

it now does itex!

This was a ridiculously simple diagram to do.

• CommentRowNumber2.
• CommentAuthorHarry Gindi
• CommentTimeFeb 23rd 2010
• (edited Feb 23rd 2010)

The matrices don't show up in opera.

• CommentRowNumber3.
• CommentAuthorAndrew Stacey
• CommentTimeFeb 23rd 2010

Gosh! There's no satisfying some people, is there?

Seriously, exactly which bit doesn't show up? Better still, email me a screenshot.

• CommentRowNumber4.
• CommentAuthorEric
• CommentTimeFeb 23rd 2010

Awesome :)

• CommentRowNumber5.
• CommentAuthorTim_Porter
• CommentTimeFeb 23rd 2010

@Andrew. Lovely but are there some righthand ) s that are either missing or swallowed by abigger bracket?

• CommentRowNumber6.
• CommentAuthorAndrew Stacey
• CommentTimeFeb 23rd 2010

I've enlarged the boxes that contain the MathML bits. Does it look better now? What browser (and version) are you using?

I just tried Opera and none of the MathML comes out, but then I think that Opera's rendering of MathML is broken as a whole (at least, it was; not sure what the current state of affairs is). Jacques' blog is a must-read on things like that.

The SVG-editor should now be considered fully usable. There will, no doubt, be bugs and "features" but Jacques says that he thinks the main issues have been sorted out and would like lots of people to try it out. I'll put up more detailed instructions at the SVG Editor HowTo when I get a moment.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeFeb 29th 2012

I have added a brief Idea-sentence to commutative algebraic theory together with a brief mentioning of monoidal mondads.

Since there is also a good list of references, the following query box, which used to be there, should be obsolete, and so I am hereby moving it from there to here:

[old query box:]

+– {: .query} Can we have some please! This is something I want to use soon so it’d be nice to know where the details were originally worked out. —Andrew

Mike: We should really have a separate page on monoidal monads since that notion is more general than its restriction to the case of algebraic theories. =–

[end old query box]

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 29th 2012

I made some changes to the Idea-sentence at commutative algebraic theory. Notice that commutativity (with respect to permuting arguments) is much stronger than what is implied by the notion of commutative algebraic theory; for example, the theory generated by a single binary operation $\ast$ and subject to the equation $(a \ast b) \ast (c \ast d) = (a \ast c) \ast (b \ast d)$ is a commutative theory, but you get a different 4-ary operation if you permute $a$ and $b$.

This entry deserves further expansion, which I will try to get to later.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeFeb 29th 2012

I still think we ought to have a separate page on monoidal monads. But I guess I don’t think that strongly enough yet to take the time to write one myself.

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 29th 2012

Mike, I agree with you, and I was also hoping to get to that myself as well.

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeMar 1st 2012

I have added more material to commutative algebraic theory, saying something about the relation to monoidal monad, for instance.

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeMar 2nd 2012

Nice, thank you! I’m a bit confused about the tensor product of Lawvere theories; when you say every theory is a monoid for $\otimes$ do you mean in a unique or canonical way? That seems to say that every $T$-algebra is automatically a $T$-algebra internal to $T$-algebras; isn’t that more of a property of commutative theories?

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2012

Mike, thanks for having a look – I probably goofed up here; I was in a hurry. I’ll need to think about what I wanted to say. (I thought I eventually wanted to say that the Kronecker product $\otimes$ is a coproduct for commutative theories, and I must have had an analogy with rings vaguely in mind when I pulled the goof.)

• CommentRowNumber14.
• CommentAuthorZhen Lin
• CommentTimeMar 2nd 2012

I’m slightly confused. At some point in the article we seem to go from considering general monads to strong monads…?

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2012

Zhen, the monads considered in the article commutative algebraic theory are monads on $Set$, where they all automatically carry canonical strengths. But there is an interesting circle of ideas surrounding commutative monads and monoidal monads (as explored in Kock’s article) when one moves to more general monads on monoidal $V$.

• CommentRowNumber16.
• CommentAuthorZhen Lin
• CommentTimeMar 2nd 2012

Ah. I didn’t know about that. Is the strength supposed to be obvious? I’m not seeing a way to get a canonical map $A \times T B \to T(A \times B)$ for an arbitrary monad. Or is this a special fact about commutative monads?

• CommentRowNumber17.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2012

It’s only obvious after you think of a strength as more or less the same as an enrichment of a functor (plus the fact that all functors are automatically $Set$-enriched). If $T$ is enriched (and we write internal homs as exponentials), then a strength $A \times T B \to T(A \times B)$ can be defined as the mate of the composite

$A \to (A \times B)^B \to T(A \times B)^{T B}$

where the first arrow is the unit of the $\times-hom$ adjunction and the second is an instance of enrichment. Conversely, if you have a strength, you get an enrichment by taking the mate of the composite

$A^B \times T(B) \stackrel{strength}{\to} T(A^B \times B) \stackrel{T(eval)}{\to} T(A)$
• CommentRowNumber18.
• CommentAuthorZhen Lin
• CommentTimeMar 2nd 2012

Very clearly explained, thanks! Perhaps it should be added to the strong monad article.

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeMar 2nd 2012
• (edited Mar 3rd 2012)

Just for amusement value, a more concrete set-based way to say it is: a point $a\in A$ defines a map “pair with $a$$B\to A\times B$. Apply $T$ to that map to get $T B \to T(A\times B)$. Since there’s one such map for each $a\in A$, we have $A\times T B \to T(A\times B)$. If you interpret this in the internal logic of any ccc, with $T$ an enriched functor, you get the proof that Todd gave. (And I suppose you could interpret it in the “internal linear logic” of an arbitrary closed monoidal category too.)

• CommentRowNumber20.
• CommentAuthorPaoloPerrone
• CommentTimeJan 14th 2020