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• CommentRowNumber1.
• CommentAuthorFinnLawler
• CommentTimeAug 16th 2010

New page: indecomposable object, following (what I think is) Johnstone's definition. I also found it in some online topos theory lecture notes by Ieke Moerdijk and Jaap van Oosten.

Lambek and Scott give a different definition in Intro. to Higher-order Cat. Log., p. 168. I'm not sure how it relates to Johnstone's.

I've also given a proof that indecomposable <=> connected in an extensive category. I'd be interested to know whether this hypothesis is the weakest possible, if anyone has any ideas (or just likely-looking references).

• CommentRowNumber2.
• CommentAuthorRodMcGuire
• CommentTimeAug 17th 2010

New page:indecomposable object

This definition (defined using coproduct) exactly corresponds in lattice theory to join irreducible (non existent page) which has the co-notion meet irreducible.

The time is ripe for some terminological cleanup so that ’indecomposable obect’ has a transparent co-notion. Various options (not necessarily disjoint) are:

1) Rename indecomposable to coindecomposable because it is defined with coproducts, leaving indecomposable to be defined with products.

2) Follow lattice theory and distinguish +indecomposable from ×indecomposable.

3) Rename indecomposable to irreducible because it as a notion is already in use at such places as irreducible closed subspace, irreducible topological space , and possibly irreducible representation.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeAug 17th 2010

Of the options listed, (2) is my favorite. Of course the two notions coincide where one has biproducts (and so it would be pointless to maintain the distinction in such contexts), as in classical representation theory.

I guess you yourself have some hesitancy about the last example listed in option (3), since you say “possibly”. As one knows from quiver representations, “irreducible” is a stronger notion than “indecomposable”.

Aesthetically, I don’t like the sound of “coindecomposable” in (1) for such a common notion, and I don’t think one needs to be so fussy here. Cf. the common usage “filtered colimit”, as opposed to “cofiltered colimit”.

• CommentRowNumber4.
• CommentAuthorFinnLawler
• CommentTimeAug 17th 2010

I agree with both of you – specifically, I think that ’indecomposable’ by itself isn’t very suggestive, but ’+-indecomposable’ would seem to mean what it says. I would have thought that ’indecomposable’ and ’irreducible’ were synonyms, but then I don’t know any representation theory. Meanwhile ’coindecomposable’ is even worse, I think, because it’s not at all obvious where the dualization is happening. So I vote for (2).

On an unrelated note, I’ve asked a question on MathOverflow about the Johnstone and Lambek–Scott definitions. I’ll transfer any interesting responses to nLab.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeAug 17th 2010
• (edited Aug 17th 2010)

I also like (2).

Cf. the common usage “filtered colimit”, as opposed to “cofiltered colimit”.

This is a technicality (indeed, the whole point of your example is that this is a technicality), but you’ve got it backwards; filtered colimits really are filtered, but filtered limits are really cofiltered.

[Edit: In fact, Todd did not have it backwards; I just read Todd backwards.]

The problem is that ‘co’ sometimes goes with products instead of with coproducts. Besides filters, another example is (co)induction and (co)recursion. In both of these cases, the sum-like notion was studied first, which is why the terminology came out that way.

Actually, if I were king of the world, I’d switch things around so that products were called ‘cosums’ and (co)limits followed suit. (The only big problem is that I don’t know any simple term to replace ‘coequaliser’.) But it is too late now!

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeAug 17th 2010

you’ve got it backwards

Hmm… I do? I took it that Rod was saying that the prefix “co” ought to be used consistently, so that “coindecomposable” would go with coproducts. I was merely pointing out that this is not observed elsewhere in category theory, using an example that is not my invention. Notice that I was not making a claim that “filtered colimit” was wrong, as you seem to interpret me as saying. In fact, I am in complete agreement with your penultimate paragraph.

• CommentRowNumber7.
• CommentAuthorRodMcGuire
• CommentTimeAug 17th 2010
• (edited Aug 17th 2010)

As an order chauvinist I would like the notions of indecomposible and inrreducible when they are the same to use the same term, irreducible (Though I have yet to look into those areas where they differ).

However both terms seem to have little meaning as English words and a more enlightening choice is available.

Consider ×indecomposible which basically means un-factorable. However we have simple word for this - “prime”. I think ×indecomposible directly carries over to numbers and defines the primes. +indecomposible would then become coprime and the only coprime number is 1. (this notion of coprime is not to be confused the notion of coprime as relatively prime.)

If I was king I would decree prime and coprime (or ×prime and +prime if one wants to avoid primacy). This approach is along the line of avoiding negative terms such as “non-empty” in favor of the positive “inhabited”.

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeAug 17th 2010

Hmm… I do?

• CommentRowNumber9.
• CommentAuthorTobyBartels
• CommentTimeAug 17th 2010

Consider ×indecomposible which basically means un-factorable. However we have simple word for this - “prime”.

There’s already a big mess here in ring theory.

We have the concepts of irreducible element (Wikipedia), meaning $\times$-irreducible (or equivalently $\times$-indecomposable) as you have said, and prime element, meaning generating a prime ideal. These are not the same, and what’s more, the usual concept of prime number corresponds to the former and not the latter!

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeAug 18th 2010

No problemo!

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeAug 18th 2010

I don’t think there is a serious enough problem here to merit changing established words. Words like “prime” and “irreducible” have heavy connotations in other contexts and I hesitate to try to reuse them to mean something fairly different here. In particular, already in lattice theory there are the notions of “prime filter” and “prime ideal” (one could argue that one of those should be “coprime” but I think they are both used prefixless), which are related but not quite the same as the meaning of “indecomposable” here.

I guess I’m saying that I agree with (2) – that we shouldn’t introduce new words but just clarify when necessary. But I also don’t have a problem with “indecomposable” referring to coproducts by default, if unqualified.

• CommentRowNumber12.
• CommentAuthorzskoda
• CommentTimeAug 18th 2010

Without a question, in algebra, and in representation theory proper, irreducible is stronger than indecomposable and the distinction is completely standard. The entry under the consideration here is about irreducible, not indecomposable. In many special categories the two notions happen accidentally to be equivalent.

• CommentRowNumber13.
• CommentAuthorTodd_Trimble
• CommentTimeAug 18th 2010
• (edited Aug 18th 2010)

in algebra, in representation theory proper, irreducible is stronger than indecomposable and the distinction is completely standard. The entry under the consideration here is about irreducible

The same conceptual distinction resides in lattice theory as well: “irreducible” in representation theory corresponds to ($\vee$)-“atomic” in lattice theory, whereas “indecomposable” means ($\vee$)-“irreducible” in lattice theory (as Rod is pointing out under his option (3)). The entry under consideration is actually about indecomposable object; it’s just that there is a terminological shift across fields where elsewhere it’s called “irreducible element”. But again, for special lattices such as Boolean algebras, the two notions happen to be equivalent.

A terminological mess, really, but that’s the way it is.

• CommentRowNumber14.
• CommentAuthorTobyBartels
• CommentTimeAug 18th 2010

I’ve written a discussion of terminology in the article. Is it helpful?

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeAug 18th 2010

I think it looks good; thanks Toby.

• CommentRowNumber16.
• CommentAuthorzskoda
• CommentTimeAug 18th 2010

Maybe we have to relate this to the discussion of reducibility and complete reducibility at some point.

• CommentRowNumber17.
• CommentAuthorMarc
• CommentTimeAug 18th 2010
In a topos, Lambek's definition of an indecomposable object is
equivalent to the requirement that the object in question is not
a union of proper subobjects (possibly taking care of the initial
object). This is actually an exercise in that section.
In fact, I think it is enough to have (Epi,StrongMono) factorizations
available to proof the equivalence. Anyway, in a distributive category
this condition (about subobjects) is equivalent to the usual
"indecomposable" because coproduct injections are monic.

As to the question about the requirements of "irreducible <=> connected",
I do not know for sure. One can consider the following three conditions
for an object c in a category K (I restrict to the binary case):

(I) c =/= 0 and for any coproduct diagram  a --> c <-- b    we have a=0 or b=0.(P) c =/= 0 and for any coproduct diagram  a --> c <-- b    we have  a --> c  or b --> c an isomorphism.(C) K(c,-): K --> Set  preserves coproducts

Then you have
(1) (I) ==> (P)  always holds.(2) (C) ==> (P)  holds whenever coproduct injections are monic.(3) (I) ==> (C)  holds whenever coproducts are stable under pullback and coproduct injections are monic.(4) (P) ==> (I)  holds whenever K has no proper preinitial objects    (i.e. every epi  0 --> p must be a isomorphism)

I hope I have not forgotten some obvious conditions. These days I do not
have internet access, so I cannot post many details. One the other hand,
the proofs are more or less straightforward.

So it seems that (I) <==> (C) holds e.g. in every distributive category
without proper preinitials. But are such categories already extensive?
I do not know.