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the order 2 group generated by $-1$. Note that if

If you take a submonoid that uses only one prime, it is quasi-isometric to a quotient of the hyperbolic plane $\mathbb H^2$ minus a horoball, since the affine group by an action of $\mathbb R$ Z$, which is multiplication by$p$in the upper half-space model. To see this, place a subgroup of dots labeled by integer$n$at position$(n*p^k, p^k)$in the isometry group upper half plane, for every pair of integers$\mathbb H^2$that acts simply transitively, (n,k)$, and any left-invariant metric on it is hyperbolicconnect them by horizontal line segments and by vertical line segments whenever points are in vertical alignment. The missing horoball is due to the absence quotient of integers $< 1$.

If you use only $n$ primes, upper half plane by the quasi-isometry type hyperbolic isometry $(x,y) \rightarrow (p*x, p*y)$ has a copy of the resulting Cayley graph for this monoidcan be thought. This is also quasi-isometric to the 1-point union of as a skewed metric on two copies of the product with a simplexhyperbolic plane, $\mathbb H^2 \times \Delta^{n-1}$, where one for negative integers, one for positive integes. It's a given fun exercise,using say $y$-coordinate, p = 2$. Start from 0, and recursively build the graph by connecting$\Delta^{n-1}$direction is the set of ways that n$ to $y$ is a product of the given set of primes raised n+1$by one color arrow, and$n$to non-negative fractional exponents$2*n$by another color arrow. If we assign each you arrange positive integers in a spiral, you can make a neat drawing of this graph (or the infinite set of generators corresponding graph for a lengthdifferent prime.) The negative integers look just the same, perhaps length 1but with the successor arrow reversed. If you use several primes, then we the picture gets more complicated. In any case, one can form an take rescaled limits of these graphs, based at sequence of points, and get asymptotic cone cones for the monoid. The graph is not homogeneous, so there is not just as you do for one limit. Another point of view is to take limits of$\mathbb Z$without rescaling, butwith a group$k$-tuple of constants$(n_1, \dots , n_k)$. The asymptotic cone for set of possible identities among polynomials in$\mathbb H^2$cross k$ variables is compact, so there is a crazily-branching treecompact space of limit rings for $\mathbb Z$ with $k$ constants. I think Perhaps this is begging the simplex directionjust turns into a Euclidean factor, but I haven't thought it through carefullyquestion: the identitites that define the limits correspond to diophantine equations that have infinitely many solutions.Rescaling may eliminate some of this complexity.

$\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on the some asymptotic cone for the affine monoid for $\mathbb Z$.

With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest.

2 removed horoball from hyperbolic plane.

Interesting question.

Here's a thought:

You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms $x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times the order 2 group generated by $-1$. Note that if you take a submonoid that uses only one prime, it is quasi-isometric to the hyperbolic plane $\mathbb H^2$ minus a horoball, since the affine group of $\mathbb R$ is a subgroup of the isometry group of $\mathbb H^2$ that acts simply transitively, and any left-invariant metric on it is hyperbolic. The missing horoball is due to the absence of integers $< 1$.

If you use only $n$ primes, the quasi-isometry type of the resulting monoid can be thought of as a skewed metric on the product with a simplex, $\mathbb H^2 \times \Delta^{n-1}$, where for a given $y$-coordinate, the $\Delta^{n-1}$ direction is the set of ways that $y$ is a product of the given set of primes raised to non-negative fractional exponents.

If we assign each of the infinite set of generators a length, perhaps length 1, then we can form an asymptotic cone for the monoid, just as you do for a group. The asymptotic cone for $\mathbb H^2$ cross is a crazily-branching tree. I think the simplex direction just turns into a Euclidean factor, but I haven't thought it through carefully.

A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to $\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on the asymptotic cone for the affine monoid for $\mathbb Z$.

With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest

1

Interesting question.

Here's a thought:

You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms $x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times the order 2 group generated by $-1$. Note that if you take a submonoid that uses only one prime, it is quasi-isometric to the hyperbolic plane $\mathbb H^2$, since the affine group of $\mathbb R$ is a subgroup of the isometry group of $\mathbb H^2$ that acts simply transitively, and any left-invariant metric on it is hyperbolic.

If you use only $n$ primes, the quasi-isometry type of the resulting monoid can be thought of as a skewed metric on the product with a simplex, $\mathbb H^2 \times \Delta^{n-1}$, where for a given $y$-coordinate, the $\Delta^{n-1}$ direction is the set of ways that $y$ is a product of the given set of primes raised to non-negative fractional exponents.

If we assign each of the infinite set of generators a length, perhaps length 1, then we can form an asymptotic cone for the monoid, just as you do for a group. The asymptotic cone for $\mathbb H^2$ cross is a crazily-branching tree. I think the simplex direction just turns into a Euclidean factor, but I haven't thought it through carefully.

A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to $\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on the asymptotic cone for the affine monoid for $\mathbb Z$.

With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest