Concrete description of the tensor product of suplattices? I know the universal property of the tensor product of suplattices: sup-preserving maps $A \otimes B \to C$ represent bi-sup-preserving maps $A \times B \to C$.
But I want to know how to build $A \otimes B$ concretely. For example, if $A$ and $B$ are finite powersets what is $A \otimes B$?
Thank you.
 A: Well, here's a slick way to compute the tensor product of two power sets.  First, $A\otimes \{0,1\}\cong A$: bi-sup-preserving maps $f:A\times \{0,1\}\to C$ are in natural bijection with sup-preserving maps $g:A\to C$ by sending $g$ to the map $f$ defined by $f(a,0)=0$ and $f(a,1)=g(a)$.  Second, tensor products distribute over coproducts (which are the same as products): bi-sup-preserving maps $g:A\times \prod B_i\to C$ are in natural bijection with families of bi-sup-preserving maps $f_i:A\times B_i\to C$, by defining $g(a,(b_i))=\bigvee_i f_i(a,b_i)$.
Putting these together, a power set is just a product of copies of $\{0,1\}$, so $$P(X)\otimes P(Y)\cong\{0,1\}^X\otimes \{0,1\}^Y\cong (\{0,1\}\otimes \{0,1\})^{X\times Y}\cong \{0,1\}^{X\times Y}\cong P(X\times Y).$$  Concretely, the universal bi-sup-preserving map $P(X)\times P(Y)\to P(X\times Y)$ is just $(a,b)\mapsto a\times b$.

Here is a general explicit construction of $A\otimes B$, though I imagine in practice it is often not so easy to use.  If $S\subseteq A\times B$ and $(a,b)\in A\times B$, say that $b\in B$ is the $S$-partner of $a$ if for all $b'\in B$, $(a,b')\in S$ iff $b'\leq b$.  Similarly, say that $a$ is the $S$-partner of $b$ if for all $a'\in A$, $(a',b)\in S$ iff $a'\leq a$.  Say that $S$ is good if every element of $A$ and $B$ has an $S$-partner.  (Equivalently, $S$ is good if for each $a\in A$, $\{b\in B:(a,b)\in S\}$ is a complete ideal in $B$, and for each $b\in B$, $\{a\in A:(a,b)\in S\}$ is a complete ideal in $A$.)   Let $G$ be the set of all good subsets of $A\times B$, ordered by inclusion; this poset will be our tensor product of $A$ and $B$.  The intuition here is that a good subset $S$ represents the formal join of the elements $a\otimes b$ over all $(a,b)\in S$.
I claim first that $G$ is a suplattice.  Joins in $G$ are complicated to describe; it is easier to instead prove the equivalent statement that $G$ has all meets.  Indeed, if $(S_i)$ is a family of good sets, then $\bigcap S_i$ is a good set: if $b_i$ is the $S_i$-partner of $a$ for each $i$, and then $\bigwedge b_i$ is the $\bigcap S_i$-partner of $a$.
Next, I claim there is a bi-sup-preserving map $\mu:A\times B\to G$ defined by $$\mu(a,b)=\{(a',b'):a'\leq a,b'\leq b\}\cup A\times\{0\}\cup\{0\}\times B.$$  First, this set $\mu(a,b)$ is always good: if $a'\leq a$, then the $\mu(a,b)$-partner of $a'$ is $b$, and otherwise it is $0$, and similarly for elements of $B$.  To see that $\mu$ preserves sups in the first variable, let $(a_i)$ be a family of elements of $A$ and $b\in B$.  It is clear that $\mu(\bigvee a_i,b)\supseteq \mu(a_i,b)$ for each $i$.  To prove that $\mu(\bigvee a_i,b)=\bigvee\mu(a_i,b)$, suppose $S\in G$ contains $\mu(a_i,b)$ for each $i$; we must show that $\mu(\bigvee a_i,b)\subseteq S$.  Then for any $b'\leq b$, $(a_i,b')\in S$ for each $i$.  If $a$ is the $S$-partner of $b'$, this means that $a\geq a_i$ for all $i$ so $a\geq\bigvee a_i$.  Thus $(a',b')\in S$ for all $a'\leq \bigvee a_i$, and so $\mu(\bigvee a_i,b)\subseteq S$.  Thus $\mu$ preserves sups in the first variable, and by symmetry also preserves sups in the second variable.
Finally, I claim this $\mu$ is the universal bi-sup-preserving map out of $A\times B$.  To prove this, let $f:A\times B\to C$ be any bi-sup-preserving map.  Note that if $S\in G$ and $(a,b)\in S$, then $\mu(a,b)\subseteq S$.  Thus every element of $G$ is a union of sets of the form $\mu(a,b)$, so the image of $\mu$ generates $G$ as a suplattice.  So it suffices to show that $f$ factors through $\mu$ (uniqueness is automatic).  To do so, define $g:G\to C$ by $$g(S)=\bigvee_{(a,b)\in S}f(a,b).$$  It is easy to see that $g(\mu(a,b))=f(a,b)$, so it suffices to show that $g$ is sup-preserving.  So, suppose $(S_i)$ is a family of elements of $G$; it is clear that $g$ is order-preserving so $g(\bigvee S_i)\geq \bigvee g(S_i)$.  To prove the reverse inequality, suppose $c\in C$ is such that $c\geq g(S_i)$ for all $i$; we must show that $c\geq g(\bigvee S_i)$.  Let $$S=\{(a,b)\in A\times B:f(a,b)\leq c\}.$$  Since $f$ is bi-sup-preserving, $S$ is good: if $a\in A$, the $S$-partner of $a$ is just the join of all $b$ such that $f(a,b)\leq c$, and similarly for elements of $B$.  So $S\in G$, and $S\supseteq S_i$ for each $i$ since $c\geq g(S_i)$ for each $i$.  Thus $S\supseteq \bigvee S_i$, and $g(S)\geq g(\bigvee S_i)$.  But we also have $c\geq g(S)$ by definition of $S$, and hence $c\geq g(\bigvee S_i)$, as desired.
To connect this construction to the power set example above, if $A=P(X)$ and $B=P(Y)$, then the canonical isomorphism $P(X\times Y)\to G$ sends a subset $Z$ of $X\times Y$ to the set of all $(a,b)$ such that $Z$ contains the rectangle $a\times b$.  The good subsets of $P(X)\times P(Y)$ are exactly the collections of rectangles that are "saturated", in that every rectangle contained in their union is in the collection.  In general, you can think of the tensor $a\otimes b\in A\otimes B$ as representing the "rectangle $a\times b$", and a general element of the tensor product is a formal union of such rectangles, with two such formal unions being equal if they generate the same good subset of $A\times B$.

Finally, here is another neat way to think about this construction.  We can identify a good set $S\subseteq A\times B$ with the function $f:A\to B$ sending an element of $A$ to its $S$-partner.  Conversely, given a function $f:A\to B$, we can define $S=\{(a,b)\in A\times B:b\leq f(a)\}$.  For this set to be good, we additionally need every element of $B$ to have an $S$-partner, and this will be true iff $f$ turns joins into meets, i.e. iff $f$ is a sup-preserving map $A\to B^{op}$.  The inclusion order on $S$ corresponds to the pointwise order on such functions $f$, or the opposite pointwise order when we think of the codomain as $B^{op}$.  So, $A\otimes B$ can be identified with $\operatorname{Hom}(A,B^{op})^{op}$.
You can also derive this description via some abstract nonsense.  Recall that the category of suplattices is self-dual, by the contravariant functor taking a suplattice to its opposite and a sup-preserving map to its right adjoint (which is inf-preserving), and this gives not just a bijection but an order-isomorphism $\operatorname{Hom}(A,B)\cong \operatorname{Hom}(B^{op},A^{op})$.  Now, morphisms $A\otimes B\to C$ are by definition in natural bijection with morphisms $A\to \operatorname{Hom}(B,C)\cong \operatorname{Hom}(C^{op},B^{op})$.  These are then in natural bijection with morphisms $A\otimes C^{op}\to B^{op}$, which are also in natural bijection with morphisms $C^{op}\to\operatorname{Hom}(A,B^{op})$.  Finally, by the self-duality, these are in natural bijection with morphisms $\operatorname{Hom}(A,B^{op})^{op}\to C$.  So, by Yoneda, this identifies $A\otimes B$ with $\operatorname{Hom}(A,B^{op})^{op}$.
