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.

  • 4
    $\begingroup$ 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... $\endgroup$ – GEdgar Nov 18 '14 at 19:46
  • 4
    $\begingroup$ I hate tetration. $\endgroup$ – Pedro Tamaroff Nov 18 '14 at 19:58
  • 5
    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Nov 18 '14 at 20:05
  • 2
    $\begingroup$ @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. $\endgroup$ – user180918 Nov 18 '14 at 20:16
  • 2
    $\begingroup$ I see someone asked a related question 15 years ago here. Write him a mail and ask if something came of it. $\endgroup$ – Nikolaj-K Nov 18 '14 at 21:29

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 $a,b$.

  • $\begingroup$ 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. $\endgroup$ – user180918 Nov 19 '14 at 0:44
  • $\begingroup$ 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 $\dim(b)$! $\endgroup$ – Martin Brandenburg Nov 19 '14 at 8:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Nov 19 '14 at 8:25

Your Answer

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