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Stream: learning: questions

Topic: Product of finitely presentable objects?

Martti Karvonen (Jun 04 2021 at 18:36):

If two groups $G$ and $H$ have finite presentations, $G\times H$ has a finite presentation as well, by taking the disjoint union of the two finite representations and adding relations saying that generators of $G$ commute with those of $H$. When can something like this be done in other algebraic theories?

More abstractly, under what conditions is a product of finitely presentable objects finitely presentable (where the abstract definition says that $X$ is f.p. if $hom(X,-)$ preserves filtered colimits)? In algebraic categories, any f.p. object can be written as a reflexive coequalizer of a diagram of free models, and as this is a shifted colimit, the product of two f.p. objects can be written as a reflexive coequalizer of a diagram of products of free models. So I guess the question reduces to: when is a product of free things finitely presentable?

I'm also interested in seeing counterexamples, i.e. two f.p. objects whose product is not.

Joe Moeller (Jun 04 2021 at 18:40):

It may be relevant that there is an intermediate step in what you described. If you just take the disjoint union of the presentations, then you get the free product $G \ast H$, which is the coproduct in the category of groups.

Martti Karvonen (Jun 04 2021 at 18:44):

I think that should be true very generally, so that a disjoint union of presentations gives the coproduct. The part that seems more specific to groups is that the product of two finitely generated things can be obtained from their coproduct by adding finitely many relations.

Reid Barton (Jun 04 2021 at 19:43):

Martti Karvonen said:

I'm also interested in seeing counterexamples, i.e. two f.p. objects whose product is not.

I described a kind of trivial example of this kind at https://mathoverflow.net/a/307344: take an algebraic theory given by infinitely many constant symbols, and two copies of the initial algebra.

Reid Barton (Jun 04 2021 at 19:45):

Another positive example is simplicial sets: the product of two simplices has a finite presentation by simplices.

Martti Karvonen (Jun 05 2021 at 02:42):

Thanks for the counterexample. Is there one when the algebraic theory can be given by finitely many operations and axioms?

Martti Karvonen (Jun 06 2021 at 03:35):

Semigroups give another easy counterexample: the free semigroup on one generator is $\mathbb{Z}^+$ but $\mathbb{Z}^+\times\mathbb{Z}^+$ is not finitely generated: for any finite $X\subset \mathbb{Z}^+\times\mathbb{Z}^+$ , consider the least $k$ such that $(1,k)$ is not in $X$.

Amar Hadzihasanovic (Jun 06 2021 at 04:42):

The example of semigroups seems to generalise like this: take a presentation of a theory and assign natural numbers $n_f > 0$ to all functional symbols and constants $f$ of the theory. Extend this to a “valuation” $v$ on the set of terms by

• $v(x) = 0$ for variables,
• $v(f(t_1, \ldots, t_n)) = n_f + v(t_1) + \ldots + v(t_n)$.

Suppose that $v$ is compatible with the equations of the theory, i.e. the two sides of each equation have the same valuation, and the theory has at least one symbol of arity > 0 (which implies that $v$ is unbounded). Then the product of two free algebras on finite nonempty sets is not finitely generated.

Amar Hadzihasanovic (Jun 06 2021 at 04:46):

In the case of semigroups, this comes from assigning, say, 1 to the multiplication (which is compatible with the associativity axiom).

Amar Hadzihasanovic (Jun 06 2021 at 04:46):

This covers e.g. all theories without equations and at least one (non-constant) operation...

Martti Karvonen (Jun 06 2021 at 12:53):

And similarly, having any kind of neutral element in the axiomatization results in an obstruction for such a valuation, as the corresponding constants should get assigned 0 in order to respect the equations.