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In category theory there are definitions for $A\oplus B$, $A\times B$ and $A^B$ via universal properties. I wonder if it is possible to isolate a particular universal property to represent the tetration of $A,B$ which we denote it by $A\uparrow B$. Intuitively $A\uparrow B$ is $\underbrace {A^{A^{A^{.^{.^{.}}}}}}_{B-times}$.

Question: What is the category theoretic definition of $A\uparrow B$ object?

Remark: Regarding the comments on finding some examples of tetration of two mathematical objects, I think this is exactly the difficulty of the problem. It seems there is no intuition about tetration and other hyperoperators out of number theory. But I think there is an "implicit" way to describe such an object in category theory via the notions of "exponentiation" and "limit" objects. In fact I hope one may give me a purely abstract way of defining tetration of two objects via categorical constructions that could be used as a base of definition for tetration of two objects in different contexts.

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    First you need some examples. Tetration of groups. Tetration of vector spaces. And so on. In particular, say what you mean by $B$-times where $B$ is a group... – GEdgar Nov 18 '14 at 19:46
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    I hate tetration. – Pedro Nov 18 '14 at 19:58
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    I'm not aware of any reasonably general definition of $A \uparrow B$ where $B$ is anything other than a non-negative integer; categorically this corresponds to taking iterated exponential objects. I think in general tetration is a pretty unnatural operation to look at, and the lack of an obvious answer to this categorical question is one of the more compelling pieces of evidence in favor of that. – Qiaochu Yuan Nov 18 '14 at 20:05
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    Half a year ago I also asked a question about the number-analogy for categories and in particular the exponentiation operation here. The exponential object isn't a limit, so I don't think there'll be a good limit which end up as $A^{A^{...}}$. I'd ask myself the question what properties the number $a^{a^{a^{...}}}$ has, which can be expressed more polynomial. Maybe you find some analogous object then. What's a combinatorial problem with the answer being the tetration? – Nikolaj-K Nov 18 '14 at 20:13
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    @QiaochuYuan About unnaturality of tetration, I was agree with you but as well as I learned more about it from recursion theory in connection with Ackermann function and Chaitin's incompleteness theorem, I found out that it is a really natural continuation of exponentiation operator and at least as useful as successor, addition, multiplication and exponentiation with a really deep and completely unknown theory. It seems tetration is a forgotten operator in our maths. –  Nov 18 '14 at 20:16
  • @AliSadeghDaghighi: at least as useful as addition!! – Nikolaj-K Nov 18 '14 at 20:17
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    @NikolajK Consider our mathematical world without exponentiation. Many deep theorems of number theory, differentiatl equations, analysis, calculus, and other parts of maths have a neat formulation in terms of exponentiation. In many aspects "exponentiation" and "tetration" are similar and very unnatural with respect to addition and multiplication. For example note that both are non-commutative but addition and multiplication commute. It seems concluding the unntaturality of tetration just because we have no intuition about it, is as same as discussing on unnaturality of 4-dimensional spaces. –  Nov 18 '14 at 20:24
  • @NikolajK Regarding your subtle observation that "exponentiation is not a limit of multiplication" and so "we should not expect that tetration be a limit of exponentiation", I am completely agree with you and this makes the problem of "finding a true interpretation for tetration" even more complicated. It seems your suggested approach on searching for a combinatorial property which have a solution in terms of tetration is the correct and only possible approach. –  Nov 18 '14 at 20:35
  • @NikolajK See my question here and also Graham's number. –  Nov 18 '14 at 20:38
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    I see someone asked a related question 15 years ago here. Write him a mail and ask if something came of it. – Nikolaj-K Nov 18 '14 at 21:29
  • @NikolajK Interesting! Thanks. –  Nov 18 '14 at 22:07

1 Answers1

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Let $a,b$ be objects of a closed symmetric monoidal category. Then $a^b$ may be written for the internal hom $\underline{\hom}(b,a)$. In fact, then we have the usual laws such as $a^{b+c}=a^b \times a^c$ and $(a^b)^c = a^{b \times c}$.

Now let us iterate this. $a^a = \underline{\hom}(a,a)$, $a^{a^a} = \underline{\hom}(\underline{\hom}(a,a),a)$, etc. We can define $^{n} a=a^{a^{a^{a^\dotsc}}}$ for every $n < \omega$. Assume that every object is dualizable (for example, consider the category of finite-dimensional vector spaces over some field), so that $a^b = b^* \otimes a$. Then one shows by induction that $$^n a = \left\{\begin{array}{c}(a^*)^{\otimes \frac{n}{2}} \otimes a^{\otimes \frac{n}{2}} & n \text{ even} \\ (a^*)^{\otimes \frac{n-1}{2}} \otimes a^{\otimes \frac{n+1}{2}} & n \text{ odd}\end{array}\right.$$ This case distinction indicates that it is impossible to give a natural definition of $^b a$ for objects $a,b$.

Martin Brandenburg
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  • Thanks for your interesting answer. Would you please explain more on your final claim that "this case distinction indicates that it is impossible to give a natural definition of $a\uparrow b$ for objects $a,b$"? However I think tetration of $a$, $b$ should not be considered as limit of exponentiations. –  Nov 19 '14 at 00:44
  • I think that every sensible definition of $^{b} a$ should satisfy $^{n \cdot 1} b = {}^{n} a$ for $n < \omega$, where $n \cdot 1$ is a direct sum of $n$ copies of $1$. In the case of f.d. vector spaces this means that $^{b} a$ should be isomorphic to $^{n} a$ when $n = \dim(b)$. But it is weird that this depends on the parity of $\dim(b)$! – Martin Brandenburg Nov 19 '14 at 08:23
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    Here is a related problem: Is there a natural (in particular, functorial) definition of $b \cdot a$ which satisfies $(n \cdot 1) \cdot a = a^{\otimes n}$? I don't think so, because $n \times m$-matrices don't induce linear maps $a^{\otimes m} \to a^{\otimes n}$. In the case of tetration, it is even worse. – Martin Brandenburg Nov 19 '14 at 08:25
  • What about defining a functor $\mathfrak T:\mathcal C\to\mathcal C$ a tetration functor for that closed symmetric monoidal category iff there exists an object $a\in\mathcal C$ such that $$\underline{\rm hom}(\mathfrak T(b),a)\simeq \mathfrak T(b+1)$$maybe using a coproduct or the terminal object? – MphLee Jun 16 '21 at 15:16