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Let $\phi$ an involution defined as follows:

$$\phi:\mathbb{N}^{2n}\longrightarrow \mathbb{N}^{2n}$$ $$(a_1,b_1,a_2,b_2,\dots,a_n,b_n)\longrightarrow (b_1,a_1,b_2,a_2,\dots,b_n,a_n).$$

Is there any polynomial injection between: $\mathbb{N}^{2n}/\phi$ and $\mathbb{N}$, where $\mathbb{N}^{2n}/\phi$ is the quotient space that identifies $x$ with $\phi(x)$?

Tio Miserias
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    Can you do the case $n=1$? I think that is where the difficulty lies. – GEdgar Aug 20 '22 at 14:02
  • Dear @GEdgar, the case $ n=1$ is this: https://math.stackexchange.com/questions/4504946/is-there-an-injective-polynomial-f-mathbbn2-s-2-longrightarrow-mathbb. However, this does not work, because, when $n$ is higher than $1$, the involution $\phi$ is not $S_2$. – Tio Miserias Aug 20 '22 at 14:10
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    Changing $\mathbb R$ to $\mathbb N$ makes the problem completely different. – GEdgar Aug 20 '22 at 14:12
  • Yes, you are absolutely right. I have "reduced" the problem. – Tio Miserias Aug 20 '22 at 14:14
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    Yes, please try not to completely change your question like this. I just had to delete a long draft of an answer which is now useless, and a waste of my time – David Sheard Aug 20 '22 at 14:15
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    @DavidSheard I think adjusting the problem within the first hour is generally OK. – GEdgar Aug 20 '22 at 14:18
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    Sorry, @DavidSheard. I thought that it would be easier to ask by $\mathbb{N}$, in place of $\mathbb{R}$. However, if you have some answer with $\mathbb{R}$, I would be grateful with you. – Tio Miserias Aug 20 '22 at 14:19
  • The question you linked is different from the $n = 1$ case here: an injection is not a bijection. – Anders Kaseorg Aug 24 '22 at 20:41
  • @AndersKaseorg Thank you Anders to pay your attention to this problem. I have changed the conditions. I admit also an injection. – Tio Miserias Aug 25 '22 at 15:13

1 Answers1

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In order to make the explanation simpler, let's rearrange the terms a little. Instead of alternating $a$'s and $b$'s, the involution $\phi$ will be defined as $\phi(\underline{a},\underline{b})=(\underline{b},\underline{a})$ where $\underline{a}$ is abbreviation for $a_1,\ldots,a_n$ (and likewise for other letters).

We'll use two ingredients:

  • A polynomial $f:\mathbb{N}^2\to \mathbb{N}$ which satisfies $f(x,y)=f(a,b)\iff (x,y)\in\{(a,b),(b,a)\}$.
  • An injective polynomial $g:\mathbb{N}^n\to \mathbb{N}$.

We can then define $$h(\underline{a},\underline{b}):=f(g(\underline{a}),g(\underline{b}))$$

Then:

  • $h$ is clearly polynomial.
  • $h(\underline{a},\underline{b})=h(\phi(\underline{b},\underline{a}))$ so $h$ is well-defined on $\mathbb{N}^{2n}/\phi$
  • $h(\underline{x},\underline{y})=h(\underline{a},\underline{b})$ implies, due to properties of $f$, that $$(g(\underline{x}),g(\underline{y})) \in \{(g(\underline{a}),g(\underline{b})), (g(\underline{b}),g(\underline{a}))\}$$ Injectivity of $g$ then implies $(\underline{x},\underline{y})\in \{(\underline{a},\underline{b}),(\underline{b},\underline{a})\}$ so $h$ is injective-up-to-$\phi$.

Thus, $h$ satisfies the required conditions.

A particular example of $h$ can be obtained by considering any injective polynomial $p:\mathbb{N}^2\to \mathbb{N}$ and setting:

  • $f(x,y)=p(x+y,x\cdot y)$
  • $g(x_1,\ldots,x_n)=p(x_1,p(\ldots,p(x_{n-1},x_n)))$
Peter Košinár
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