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Suppose I have two ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$, each of type $A \times B$

I want to combine them into a single pair of type $A^2 \times B^2$:

$(a_1,b_1)$ OP $(a_2,b_2)$ = $( (a_1,a_2), (b_1,b_2) )$

Is there a name for this operation, or a standard operator? Someone suggested $\triangle$, but I've only ever seen that used for the Laplacian. It's called "zip" in functional programming.

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I don't think there's any standard terminology, though it looks similar to a tensor product, so you could use $\otimes$. In any case, you can use whatever symbol you want as long as you define it. – Yuval Filmus May 23 '11 at 16:53
Thanks. That's kind of what I thought, but I didn't want to write it down in a paper and then have someone say "Why didn't you use the standard 'foo' notation?!" I'll use $\otimes$ – ljp May 23 '11 at 17:00
Does it actually matter to your application that you don't use $((a_1, b_1), (a_2, b_2))$? (If so I'll delete my answer.) – Qiaochu Yuan May 23 '11 at 18:27
My application uses functions on pairs of $A \times B$ pairs. I'm trying to avoid writing them all out as $(a_i,b_i)$ just to make things cleaner. – ljp May 23 '11 at 18:56

I have never before seen a standard and well established notation for that.

As I usually say, in the absence of a certain notation, one can always choose a known one - or invent one, and be detailed and consistent about using it.

In this case, if you have chosen $\otimes$ just make sure this notation is not used for anything else standard in your field, and be sure to define it in your work.

Usually it works just fine.

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It seems to me more natural to write $(a_1, b_1) \times (a_2, b_2) \in (A \times B) \times (A \times B)$ than anything else. You can use this for arbitrary products, writing $a \times b \in A \times B$ by analogy with the tensor product. The reason I suggest you do this is that for sets $\times$ is obviously commutative but there are situations where we want to use something that behaves like $\times$ but is not obviously commutative, so there's no reason to needlessly switch around factors if you don't need to.

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Qiaochu, this depends on the context. In set theory, where ordered pairs are sets as well this might turn up to be a very bad choice of operator. – Asaf Karagila May 23 '11 at 18:08
@Asaf: well, then you could just write $((a_1, b_1), (a_2, b_2))$. – Qiaochu Yuan May 23 '11 at 18:18
@Qiaochu: I do not follow. $(a,b)\times (c,d)\neq ((a,b),(c,d))$ in general, or in fact - at all. The $\times$ operator is slightly overused and has too many meanings - most of those that I know are coherent if you consider the categorical view (i.e. they end up acting like the Cartesian product of sets), but since the OP did not specify any context it is hard to say whether or not $\times$ fits well, my hunch is not positive regarding this though. – Asaf Karagila May 23 '11 at 18:22
@Asaf: I'm not a set theorist. When I refer to either $x \times y$ or $(x, y)$ I mean the canonical map $1 \to X \times Y$ coming from the inclusions $1 \to X, 1 \to Y$ corresponding to $x$ and $y$. I guess I assumed that the order was not particularly important, mainly because I cannot imagine an application where one actually has to switch the middle two factors. But perhaps the OP has something very specific in mind. – Qiaochu Yuan May 23 '11 at 18:26
@ljp: okay, and probably you use $\otimes$ both for the tensor product of two vectors and for the tensor product of two vector spaces. It's a completely sensible overload. – Qiaochu Yuan May 23 '11 at 18:51

You are trying to transform an element of $(A\times B )\times (A\times B)$ into an element of $(A\times A )\times (B\times B)$.

If you view an ordered pair as a vector with two coordinates, and an ordered pair of ordered pairs as a $2\times2$ matrix, then this could be viewed as a kind of


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