The Collatz Conjecture function should induce a collection of Grothendieck groups, one for each $n \in \Bbb{Z}$ or $\Bbb{N}$. Their properties?

Question

This question is about the Collatz conjecture.

Let $\Bbb{N}$ include $0$. The Collatz conjecture function is given by: $$ f: \Bbb{N} \to \Bbb{N}, \\ f(n) = \begin{cases} \dfrac{n}{2}, \text{ if } n = 0 \pmod 2,\\ \dfrac{3n + 1}{2}, \text{ if } n = 1\pmod 2 \text{ and } n \gt 1\\ \end{cases} $$

Now, break up the conjecture into an infinite number of cases, one for each $n \in \Bbb{N}$ as is what usually happens when mathematicians collectively attack problems. The result is a proven special cases section on the conjecture's Wikipedia page.

We can also extend $f$ to $\Bbb{Z}$ as is and get several all-negaive value loops.

Anyway, let the domain and range space be $\Bbb{N}$ for now. We know that $\Bbb{N}$ forms a commutative monoid under addition. Define the equivalence relation:

Fix $n \in \Bbb{N}$. Define the equivalence relation on $\Bbb{N}$:

$$ i\sim j \iff f^i(n) = f^j(n) $$

Then $i\sim j, i' \sim j' \implies$ $i + i' \sim j + j'$ so that the equivalence relation respects $+$ on $\Bbb{N}$.

Then you can form a congruence monoid $N = \Bbb{N} / \sim$. For example for $n = 1$ this is $f^0(1) = 1$, $f^1(1) = 2$, $f^2(1) = 1, \dots$ so that the equivalence classes are $2 \Bbb{N}$ and $2\Bbb{N} + 1$.

Therefore we say that the Collatz conjecture induces the group $\Bbb{Z}/2 = G(M/\sim) = $ the Grothendieck group of $\Bbb{N}/\sim$.

That makes perfect sense! There's a loop of "length 2" there namely $(1,2,1,2,1,\dots)$ where entries are iterate values $f^i(1), i \geq 0$ (so the entries are $0$-based indexed).

Therefore, I reckon that the Grothendieck groups $G(\Bbb{N}/\sim_{n})$ where $\sim_n$ means $n$ is the input to $f^i(n)$ in the definition of $\sim$ must have some property that flags whether or not $n$ gives an example to the conjecture, is there?

I don't think it's finiteness. For example $n$ terminating at $1$ gives a finite group as well as $n$ going off and then self-looping somewhere else would give a finite group.

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Further Attempt.

The $f^i(n)$ sequence can either shoot off to infinity, self-loop somewhere (here at $i = 0$ or further at $i = x \gt 0$), or it can drop back down to a value $m \lt n$. In the first case we have that $G_n := G(\Bbb{N}/\sim_n)$ is trivial. In the second case we have:

$$ n, n_1, n_2, \dots, n_j, n_{j+1}, \dots, n_{k-1}, n_k = n_j, n_{j+1}, \dots $$

for some $n_k \gt n_{j} \geq n$. And in the third case we have:

$n, n_1, n_2, \dots, n_{k-1}, n_{k} = 2, 1, 2, 1, \dots$ for some $k \gt 0$, since by induction on $n$ we've handled all cases $m \lt n$ and they go to the 1-2 loop.

The Grothendieck group in the second case is that of the monoid:

$$ \{[0], [1], [2], \dots, [j-1], [j],[j+1], \dots, [k-1] \} $$

What is the Grothendieck group isomorphic to in this case?

Answer 1

First of all we might want to note that there is nothing special about the Collatz function used for the construction here. In fact, we can even look at any function $f: X \to X$ for any set $X$ and construct an equivalence relation $\sim$ on $\mathbb{N}$ for any $x \in X$ just the way you did. You are effectively describing the equivalence relation defined by the fibres of $\varphi_{f,x}: \mathbb{N} \to X, i \mapsto f^{i}(x)$.

There are essentially two possibilities, either there exist some distinct $i$ and $j$ such that $f^i(x) = f^j(x)$ or there don't (i.e. $\varphi_{f,x}$ is injective). In the latter case $\sim$ is trivial and $G(\mathbb{N}/\!\sim) \:\cong G(\mathbb{N}) \cong \mathbb{Z}$. You wrote that this case would give a trivial Grothendieck group but that is actually not the case, a trivial Grothendieck group corresponds to one of the $f^i(x)$ being a fixed point (this follows from what is following).

The other option is to have $f^{i}(x) = f^{j}(x)$ for some distinct $i$ and $j$. Let us assume $i < j$ and that $j$ is minimal, so that $f^{0}(x),f^{1}(x),\dots, f^{j-1}(x)$ are the distinct values of $\varphi_{f,x}$.

The Grothendieck group $G(\mathbb{N}/\sim)$ is thus finite and moreover cyclic since $\mathbb{N}/\sim$ is generated by the equivalence class of $1$.

Since $i \sim j$ we have $j - i = 0$ in $G(\mathbb{N}/\sim)$. It turns out that $G(\mathbb{N}/\sim)$ is cyclic of order $j - i$. In fact, if we had $m = 0$ in $G(\mathbb{N}/\sim)$ for some $0 < m < j -i$, then $m + k = k$ in $\mathbb{N}/\sim$ for some $k$, in other words $f^{m + k}(x) = f^{k}(x)$. After using $f^{i}(x) = f^{j}(x)$ multiple times, this gives $f^{r}(x) = f^{s}(x)$ for some $r < s < j$ which contradicts the minimality of $j$.

Finally, for the Collatz Conjecture, note that the orbit $1,2,1,\dots$ of length two is actually the only orbit of length two occuring for the Collatz function $f$. Thus $n$ satisfying the Collatz conjecture is equivalent to $G(\mathbb{N}/\sim_{n})$ being cyclic of order two.