I know Law has a tensor product, is closed with respect to that tensor product, and it has coproducts. Does it have products?

My best guess at the cartesian product of Lawvere theories is the "intersection" of the theories: say $Th_1$ has a sort $X,$ function symbols $f_i\colon X^{n_i} \to X$ and a set of equations $R$; say $Th_2$ has a sort $Y,$ function symbols $g_j\colon Y^{m_j} \to Y$ and a set of equations $S$.

Then the product will have a sort $X\times Y$, function symbols given by all pairs $(f_i, g_j)\colon (X\times Y)^{n_i = m_j} \to (X\times Y)$ with the same signature, and the "intersection" $R \cap S$ of the equations.

For instance, suppose $Th_1$ is the theory of abelian groups with sort $A$ and $Th_2$ is the theory of monoids with sort $M$; then the result would have one sort $A\times M$, function symbols $(+, \cdot)\colon (A\times M) \times (A\times M) \to (A\times M)$ and $(0, e)\colon 1 \to (A\times M)$ subject to associativity and unit laws, but not commutativity. That is, the product is just the theory of monoids.

  • 1
    $\begingroup$ You can't really mean the intersection of the equations? Some of the equations deal with one sort, and some of them deal with the other sort... I can kind of guess what you mean, but your example is a little degenerate... $\endgroup$ – David Roberts Jan 26 '17 at 5:38
  • $\begingroup$ You're right; I put "intersection" in quotes because I'm not sure exactly how it goes. $\endgroup$ – Mike Stay Jan 26 '17 at 14:51

According to

Fajtlowicz, S. Birkhoff's theorem in the category of non-indexed algebras. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 17 1969 273-275.

a product of algebras and varieties was introduced by W. Narkiewicz. The terminology ``nonindexed product'' was used. It is stated in the above paper that the nonindexed product is the category-theoretic product in the category of nonindexed algebras.

This is basically the thing that you are describing: the product of the clones of the varieties within the category of clones, or the product of their algebraic theories within the category of algebraic theories. Given varieties $\mathcal U$ and $\mathcal V$, the models of ${\mathcal U}\times {\mathcal V}$ are those algebras isomorphic to a set-theoretical product of some algebra $A\in \mathcal U$ with some algebra $B\in \mathcal V$, whose $n$-ary operations are pairs $(f,g)$ where $f$ is an $n$-ary operation of $A$ and $g$ is an $n$-ary operation of $B$. It is clear how these operations should act on $A\times B$, namely $$ (f,g)((a_1,b_1),\ldots,(a_n,b_n)) = (f(a_1,\ldots,a_n),g(b_1,\ldots,b_n)). $$

The conjecture that the product of the variety $\mathcal A$ of abelian groups with the variety $\mathcal M$ of monoids is the variety of monoids is not correct. There are two isotypes of monoids of size $2$, but there are $3$ isotypes of algebras in $\mathcal A\times \mathcal M$ that have size $2$.

Let me edit this to respond to comments: "isotype" = "isomorphism type". To expand on the preceding paragraph, let $A_2$ be the $2$-element group. Let $MA_2$ be the $2$-element monoid that is a group, and let $MS_2$ be the $2$-element monoid that is a semilattice. Let $*$ denote a $1$-element algebra of any type. Then the $3$ isotypes of $2$-element algebras in ${\mathcal A}\times {\mathcal M}$ are $A_2\times *$, $*\times MA_2$ and $*\times MS_2$. Observe that $A_2\times *$ and $*\times MA_2$ are not isomorphic, since there is a binary operation of ${\mathcal A}\times {\mathcal M}$ of the form $(f,g)(x,y)=(x+y,x)$. This agrees with the group operation on $A_2\times *$ but not on $*\times MA_2$.

Let me edit again to respond (slightly) to a question about identities satisfied by a product ${\mathcal U}\times {\mathcal V}$. Given varieties $\mathcal U$ and $\mathcal V$ and their identities, the identities of ${\mathcal U}\times {\mathcal V}$ have been worked out. I learned about a number of sources from Walter Taylor:

Taylor, Walter The fine spectrum of a variety. Algebra Universalis 5 (1975), no. 2, 263-303. [See Proposition 0.9]

McKenzie, Ralph On spectra, and the negative solution of the decision problem for identities having a finite nontrivial model. J. Symbolic Logic 40 (1975), 186–196.

García, O. C.; Taylor, W. The lattice of interpretability types of varieties. Mem. Amer. Math. Soc. 50 (1984), no. 305, v+125 pp. [See Definition preceding Proposition 3.]

I won't list the details here, but will note that this issue (identities defining the product) has a complicated history. Taylor mentioned the names: Newman, Foster, Pixley, Knoebel, Gratzer, Lakser, Plonka, Hu, Kelenson, Bernardi, Draskovicova, Chang, Jonsson, Tarski, Fajtlowicz, Lawvere, McKenzie, Garcia, Taylor.

  • $\begingroup$ Thank you! I'm not familiar with isotypes; can you list the five isotypes you mentioned explicitly? $\endgroup$ – Mike Stay Jan 26 '17 at 16:32
  • $\begingroup$ Also, your answer seems to be the only place online with the phrase "isotype of algebras"; is there some other term that's more commonly used? $\endgroup$ – Mike Stay Jan 26 '17 at 17:09
  • 1
    $\begingroup$ your answer seems to be the only place online with the phrase "isotype of algebras": I use this term in lectures. I find it easier to say or write on the board "there are exactly two isotypes which have property $X$" instead of "there are exactly two structures up to isomorphism which have property $X$". $\endgroup$ – Keith Kearnes Jan 26 '17 at 22:44
  • $\begingroup$ Ah, so there is something in the equations enforcing commutativity of the abelian group part. Do the equations from M and A hold on the projections of the new morphisms independently? E.g. $(+, \cdot)$ is a binary operation on the sort $\mathcal{A}\times \mathcal{M}$ subject to the equation $$\pi_\mathcal{A} \circ (+,\cdot)((u,v),(x,y)) = \pi_\mathcal{A} \circ (+,\cdot)((x,y),(u,v))$$ (and all the other equations, of course)? $\endgroup$ – Mike Stay Jan 27 '17 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.