## The Abelian House is not closed

Today I will talk about $\displaystyle{\frac{97 + 26 \sqrt{13}}{27} = 7.064604\ldots}$

For an algebraic integer $\alpha$, the house $\overline{|\alpha|}$ is the absolute value of the largest conjugate of $\alpha.$ Kronecker proved the following:

1. If $\overline{|\alpha|} \le 1$, then either $\alpha = 0$ or a root of unity.
2. If $\overline{|\alpha|} \le 2$ and $\alpha$ lives in a CM field, then $\overline{|\alpha|} = 2 \cos \pi/N$.

The first claim is well known. The second claim follows from the first: the CM condition implies that the conjugates of the squares of the absolute value are the squares of the absolute values of the conjugates. Hence, if $\zeta^2 + \zeta^{-2} = \overline{|\alpha|}^2 - 2$, then $\zeta$ must be a root of unity by part one. On the other hand, beyond these two results, the respective values of $\overline{|\alpha|}$ are dense in $[1,\infty)$ (general case) and $[2,\infty)$ (CM case). There are a number of ways to modify this problem. One way is to replace the largest conjugate $\overline{|\alpha|}$ by the $d$-power mean of the absolute values:

$\displaystyle{M_n(\alpha) = \left(\frac{1}{[{\mathbf{Q}}(\alpha):{\mathbf{Q}}]} \sum |\sigma \alpha|^n \right)^{1/n}}.$

For such a construction, it makes the most sense to assume either that $\alpha$ is totally real or lives in a CM field, so that $|\sigma \alpha| = \sigma |\alpha|$. For example, if one lets

$\mathfrak{M}_n = \{x \in (1,\infty) | x = M_n(\alpha), \ \sigma c \alpha = c \sigma \alpha\},$

then Chris Smyth shows (MR0736460) that, for all $n > 0$, the smallest elements of ${\mathfrak{M}}_n$ are isolated, whereas ${\mathfrak{M}}_n$ is dense for sufficiently large $x$. In this post, we shall be interested in what happens when one restricts to the class of cyclotomic integers. Namely, let

${\mathfrak{M}}^{{\mathrm{ab}}}_n = \{x \in (1,\infty) | x = M_n(\alpha) \ \alpha \in \mathbf{Q}^{\mathrm{ab}}\}.$

In particular, when $n = \infty$, we obtain the set ${\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}$ consisting of the values $\overline{|\alpha|}$ for cyclotomic integers $\alpha$. We call ${\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}$ the Abelian House. As already noted, the values of ${\mathfrak{M}}^{{\mathrm{ab}}}_{\infty} \cap [1,2]$ consist of elements of the form $2 \cos(\pi/N)$, which includes $2$ as a limit point. However, the spectrum of ${\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}$ for a short while beyond $2$ is once again discrete. For example, the main theorem of MR3119783 (previously discussed here) completely computes ${\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}$ in the interval $[0,(5.04)^{1/2}]$ — it has a second limit point at $\sqrt{5} = 2.2360679\ldots$ and is once again discrete beyond this point. The case $n =2$ was studied in Cassels (MR0246852) and in MR2786219. In particular, Theorem 9.1.1 of MR2786219 is eqivalent to:

Proposition: The set ${\mathfrak{M}}^{{\mathrm{ab}}}_2 = \overline{{\mathfrak{M}}^{{\mathrm{ab}}}_2} \subset {\mathbf{R}}$ is closed.

Note that $M_2(\alpha)^2 =: \mathcal{M}(\alpha) \in {\mathbf{Q}}$ (the notation $\mathcal{M}$ being used in ibid, so this closed subset is countable and is thus very far from being dense. Moreover $M_{2n}(\alpha)^n = M_2(\alpha^n)$, so the theorem above implies that the closure $\overline{{\mathfrak{M}}^{{\mathrm{ab}}}_{2n}}$ is also countable and lives inside ${\mathbf{Q}}^{1/n} \cap {\mathbf{R}} \subset {\overline{\mathbf{Q}}} \cap {\mathbf{R}}$. The main goal of the current post is to generalize this result to the abelian house.

Theorem: The closure of the abelian house $\overline{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$ is a subset of ${\overline{\mathbf{Q}}} \cap {\mathbf{R}}$. If $S \subset {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$ is bounded, then $\liminf S \in {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$. However, $\limsup S$ is not necessarily in ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$, that is, the abelian house itself is not closed.

One application of this is to the possible index of subfactors (see here and here for an overview of the problem):

Corollary A: Let $\alpha \in {\mathbf{R}} \setminus {\mathbf{R}} \cap {\overline{\mathbf{Q}}}$ be a real transcendental number. Then there does not exist a finite depth subfactor $A < B$ of index in the range $(\alpha - \epsilon, \alpha + \epsilon)$ for some $\epsilon > 0$.

Corollary B: Let $\alpha \in {\mathbf{R}} \cap {\overline{\mathbf{Q}}}$ be an algebraic number. Then there does not exist a finite depth subfactor $A < B$ of index in the range $(\alpha, \alpha + \epsilon)$ for some $\epsilon > 0$.

Corollary C: The set of indices of finite depth subfactors is a well-ordered subset of $\mathbf{R}$ of ordinal type $\omega^{\omega}$. (Random aside: Just like volumes of $3$-manifolds according to Thurston and Jørgensen.) I’m assuming here that it is easy enough to construct subfactors of index

$\prod_{i=1}^{n} 4 \cos^2(\pi/p_i)$

for distinct odd prime numbers $p_i$.

Since the main context here is that such indices arise as the spectral eigenvalue of graphs, it might be helpful (for contrast) to note that this latter spectrum is dense in $[\sqrt{2+\sqrt{5}},\infty)$ (MR986863).

This theorem came from my bag of thesis problems. I actually expected it to be the case that ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$ was closed, but this turns out to be completely false. On the other hand, the argument I had in mind to prove this theorem was roughly correct. On the third hand, it turns out that the solution to this problem was almost entirely included in a paper of A.Jones from the ’70s (MR0437466) (using the method I roughly had in mind).

Rational Linear subspace of $({\mathbf{R}}/{\mathbf{Z}})^k$ : Consider the standard torus ${\mathbf{T}}:=({\mathbf{R}}/{\mathbf{Z}})^k$ with coordinates $(x_1,\ldots,x_k)$. We define a rational linear subspace $V$ of ${\mathbf{T}}$ to be the subspace cut out by any number of equations of the form:

$\sum a_{i,j} x_i = c_j$

for integers $a_{i,j}$ and elements $c_j \in {\mathbf{Q}}/{\mathbf{Z}}$. Topologically, $V$ is finite disjoint union of tori. Any connected component of $V$ is also a rational linear subspace. If all the $c_j = 0$, then we call $V$ a rational linear subgroup. Call a point ${\underline{x}} \in V$ rational if ${\underline{x}} = (x_1,\ldots,x_k)$ where $x_i \in {\mathbf{Q}}/{\mathbf{Z}}$. Given $V$, the rational numbers $c_j$ have a common denominator; let $M$ denote some integer divisible by this common denominator. The map $[m]: {\mathbf{T}} \rightarrow {\mathbf{T}}$ given by multiplication by $m$ preserves $V$ whenever $m \equiv 1 \mod M$.

Definition: For any rational point ${\underline{x}}$ on $V$ and an admissible integer $M$, let $L({\underline{x}}) = L_M({\underline{x}})$ denote the (finite) set of rational points of the form $[m] {\underline{x}} \in V$ for all $m$ satisfying the following two conditions:

1. $m \equiv 1 \mod M$,
2. $m$ is prime to $N$, where ${\underline{x}} = (x_1,\ldots,x_k)$ are elements of ${\mathbf{Z}}[1/N]/{\mathbf{Z}}$.

Of course, this definition comes from looking at the exponent of the conjugates of root of unity which fix an $M$th root of unity. We call $L({\underline{x}})$ the line through ${\underline{x}}$. The notion of line depends on a choice of integer $M$, although replacing $M$ by a multiple only (at worst) decreases the size of $L({\underline{x}})$. Our main technical lemma is the following:

Lemma A: Let $S \subset V$ be any set of rational points, and let $M$ be admissible for $V$. Then the closure of $W = \bigcup L({\underline{x}})$ of all lines $L({\underline{x}}) = L_M({\underline{x}})$ for ${\underline{x}}\in S$ is a union of connected rational subspaces $W^0$ of $V$.

We shall apply this theorem to $V = {\mathbf{T}}$ with $M =1$. However, in order to prove the result (by induction), it is easier to prove this more general statement.

Example: Suppose that $n = 2$, $V = {\mathbf{T}}$, and $M = 2$. If ${\underline{x}} = (1/2,1/q)$, then the line $L({\underline{x}})$ consists of points of the form $(1/2,p/q)$ with $p$ odd and prime to $q$. The closure of all such points is the rational subspace $x_1 = 1/2$.

We proceed by induction on the dimension of $V$. We may first claim that we can assume $V = V^0$ is connected. The connected components of a rational subspace are obtained by replacing the linear equations by their saturation. However, this requires introducing numerators into the constants $c_j$, and so for this step (as well as several others) we must allow the auxiliary integer $M$ to increase. If $V$ is connected, it suffices to show that if the closure is not dense, then the points all lie on a (finite union of) co-dimension $\ge 1$ rational subspaces $W$ of $V$, and then apply the inductive hypothesis.

Choose a rational base point ${\underline{v}} \in V$. After increasing $M$ again if necessary, we may assume that $M {\underline{v}} = 0$. Under this assumption, translation by ${\underline{v}}$ preserves lines and sends $V$ to a connected rational linear subgroup of ${\mathbf{T}}$. After an integral change of basis, any connected rational linear subgroup is linearly equivalent to one of the form $a_i x_i = 0$ for $a_i$ either zero or one, and thus, again without loss of generality, we may assume that $V = {\mathbf{T}}$. Now suppose that ${\underline{v}} \in V$ is a point which is not in the closure of the set of lines. Because the complement of the closure is open, we may assume that ${\underline{v}}$ is rational. Hence, once more translating by ${\underline{v}}$ and increasing $M$ if necessary, it suffices to show that either $0$ is in the closure of the set of lines, or the points are all contained in a subvariety defined by a linear equation. Let ${\underline{x}} = (x_1,\ldots,x_k) \in S$, where one may think of the $x_i$ as being lifted to ${\mathbf{Q}}$. The problem is to construct an integer $n$ with $n \equiv 1 \mod M$ and $(n,N) = 1$ such that if $\|x\|$ denotes the nearest integer to $x$, then $\|n x_i\| < \epsilon$ for all $i$, or to show that all the $x_i$ satisfy some linear relation in ${\mathbf{Q}}/{\mathbf{Z}}$. Without the congruence condition on $M$, this is exactly a lemma proved by Davenport and Schinzel in MR0205926. Their proof does not obviously extend to this case, however. I had an idea to replace this analytic argument by using an idea of Cassels using the geometry of numbers. Write $x_i = a_i/N$ where $N$ is the smallest common denominator (so the greatest common divisor of the $a_i$ is one). Let $\Lambda \subset {\mathbf{Z}}^k$ denote the lattice

$\Lambda := \{\lambda \equiv m (a_1,a_2, \ldots, a_k) \mod N, \quad m \in {\mathbf{Z}}\}.$

The basic idea is to break up the problem into two steps: first, find an element of $\Lambda$ of small length. If this element reduces under the natural map to $\mathbf{Z}/N\mathbf{Z}$ to a multiple $m$ of $(a_1,\ldots,a_k)$ which is prime to $N$ and $1 \mod M$, then one wins. If not, deform the element both other small vectors, and use the fact that (in an arithmetic progression) one doesn’t have to go very far to find elements prime to $N$ (by Iwaniec, $\log(N)$ or so will suffice). In the end, it turns out that this improved version is essentially proved by Jones in MR0437466. (For me, it is easiest to modify his proof of Theorem 1 than read the notation in some of the latter theorems, but all of the required content is here.) In fact, the application Jones had in mind was almost identical to the topic of this post, namely, to study the higher derived sets of ${\mathfrak{M}}^{{\mathrm{ab}}}_n$. For some reason, however, he did not seem to notice the implication that $\overline{{\mathfrak{M}}^{{\mathrm{ab}}}_n}$ was a subset of ${\overline{\mathbf{Q}}}$, possibly because his formalism was less algebraic than what we consider below.

Consider an infinite set $S$ of roots of $k$-tuples of roots of unity $(x_1, \ldots, x_k)$ which is closed under the action of $\mathrm{Gal}({\overline{\mathbf{Q}}}/{\mathbf{Q}})$, and view it as a subset of ${\mathbf{G}}^k_m$. Say that a set of $k$-tuples of units are constrained by a $k$-tuple of integers $h = (h_1,\ldots,h_k)$ if, for all such tuples,

$(x_1)^{h_1} (x_2)^{h_2} \ldots (x_k)^{h_k} \in \zeta^{{\mathbf{Z}}},$

for some fixed root of unity $\zeta$. Since this property is preserved under taking $d$th roots for any fixed $d \in {\mathbf{Z}}$, we also insist that each constraining $k$-tuple consists of co-prime integers. A constraint cuts out a subvariety $Z_h$ of ${\mathbf{G}}^{k}_{m}$, which in general is not geometrically connected. The intersection of any finite number of subvarieties $Z_{h_i}$ is determined by the saturation of the subgroup of ${\mathbf{Z}}^k$ generated by the $h_i$. In particular, there exists a maximal finite set of $h_i$ such that $Z:= \cap Z_{h_i}$.

Theorem: The supremum of the elements

$|y_1 + y_2 + \ldots + y_k|^2$

in $S$ is equal to the supremum of the quantity

$|z_1 + z_2 + \ldots + z_k|^2,$

over $(z_1,\ldots,z_k) \in (S^1)^k \cap Z^0$, where $Z^0 \subset Z$ is some geometrically connected component of $Z^0$, and $Z$ is a variety cut out by constraints for finitely many $k$-tuples. The infimum of the houses

$\overline{|y_1+ y_2 + \ldots + y_k|}$

is realized by an element of $S$.

Pulling back under the isomorphism $\exp: {\mathbf{T}} \rightarrow (S^1)^k \subset {\mathbf{G}}^k_m$, the pre-image of any geometrically connected component $Z^0$ is a connected rational linear subspace of ${\mathbf{T}}$, and conversely any connected rational linear subspace gives rise to such a $Z^0$. Write the pre-image of ${\underline{y}} \in S$ as ${\underline{x}} = (x_1, \ldots, x_k)$. Suppose that ${\underline{y}} = (y_1,\ldots,y_k)$ where each $y_i$ is a roots of unity in ${\mathbf{Q}}(\zeta_N)$ (with $N$ divisible by $M$), so that the denominators of the $x_i$ divide $N$. The action of $G:=\mathrm{Gal}({\mathbf{Q}}(\zeta_N)/{\mathbf{Q}}(\zeta_M))$ on ${\underline{x}}$ via $\exp^*$ sends ${\underline{x}}$ to $m x$ for some $m$ with $(n,m) = 1$. In particular, the conjugates on $V$ precisely cut out the line $L_M({\underline{x}})$ of ${\mathbf{T}}$ (with $M = 1$). It follows from Lemma A that the closure of $\exp^*(S)$ consists of a finite union of connected rational subspaces $W = \coprod W^0$, and hence the closure of $S$ is the finite union of the sets $(S^1)^k \cap Z^0$ for a finite number of geometrically connected $Z^0$. This proves the claim concerning the supremum. For the infimum, we argue as follows. There exists a component $Z^0$ such that the infimum of the largest conjugate of $y_1 + \ldots y_k$ on $Z^0$ is equal to the infimum of the houses of elements of $S$. Let the supremum of $|z_1 + z_2 + \ldots + z_k|^2$ on this space be $\beta$. If the desired infimum is equal to $\beta$, then all elements must the same house, and the result follows immediately. If not, there exists a subset of $S$ whose largest conjugates are bounded by $\beta - \epsilon$. But such a set can no longer be dense in $(S^1)^k \cap Z^0$. Hence, replacing $S$ by this smallest set, we may reduce the dimension of $Z^0$. Continuing this process, we reduce to the case when either $Z^0$ is a point or all the houses of elements are the same, and in either case the result follows. Note that the supremum of an algebraic function on $(S^1)^k \cap Z$ will automatically be algebraic — essentially by a rigidity argument. Alternatively, one can write down the equations required for a point to be a local minima, and observe that they are algebraic. To finish the proof of the main claim (except for the claim that ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$ is not itself closed) it suffices to note, following a result of Loxton, that any cyclotomic integer of absolute value at most $B$ can be written as a sum of (at most) $L(B)$ roots of unity, so, when dealing with the closure of ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$, it suffices to consider sums of $k$ roots of unity for a fixed $k$.

Returning to what Jones does, his main result is to consider the sets ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}(k)$ of cyclotomic integers which are the sum of $k$ roots of unity, and then prove that the $k-1$st derived set consists precisely of the element $\{k\}$. In our context, it may seem as though the $n$-th derived set should consist precisely of the maxima of the natural function on the sets $(S^1)^k \cap Z$ where $Z$ has codimension $n$. However, there is an extra degeneracy coming from the fact that multiplication by a root of unity doesn’t change the house — so we may insist from the start that $1$ is always one of the roots of unity of $S$, imposing the condition $x_1 = 1$.

The abelian house is not closed

We now prove that $\overline{{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}} \ne {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$ by constructing an explicit element of $\overline{{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}}$ not in ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$. Indeed, the corresponding element will neither be cyclotomic nor an algebraic integer (although it will be algebraic). Consider the set of cyclotomic algebraic integers:

$\beta = \zeta^2 + \zeta - \zeta^{-1}$

$\gamma = \zeta^2 + \zeta + \omega \zeta^{-1}$

where $\omega$ is a cube root of unity, and $\zeta$ is (say) and $p$th root of unity for prime $p$. For large $p$, the Galois conjugates of $\zeta$ become dense in the unit circle. It follows that the supremum of $\overline{|\beta|}^2$ is the square of the maximum of the quantity

$|X^2 + X - X^{-1}|$

over $|X|=1$, and similarly the supremum of $\overline{|\gamma|}^2$ is the maximum of the two quantities

$|X^2 + X + \omega X^{-1}|, \quad |X^2 + X - \omega^{-1} X^{-1}|,$

over the same region. One can compute this maximum, and it turns out, perhaps surprisingly, that it is equal to the value

$\displaystyle{\frac{97 + 26 \sqrt{13}}{27} = 7.064604\ldots}$

in the first case, which is not an algebraic integer and so not in ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$, and is equal in the second case to

$\displaystyle{\frac{1}{27} \cdot \theta = 8.096242\ldots}$

where

$\theta^5 - 446 \theta^4 + 62377 \theta^3 - 3023244 \theta^2 + 57168180\theta - 351065988 = 0,$

and $K = {\mathbf{Q}}(\theta)$ has discriminant $2^2 \cdot 3^5 \cdot 15619$ and Galois closure $S_5$. These are, perhaps, surprisingly ugly numbers for fairly simple looking maximization problems. It is clear, of course, that neither of these numbers lies in ${{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$, so this proves $\overline{{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}} \ne {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$. Moreover, I think it quite likely (and quite provable, perhaps with a certain amount of computational effort) that $\displaystyle{\frac{97 + 26 \sqrt{13}}{27} = 7.064604\ldots}$ is the smallest number in $\overline{{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}} \setminus {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}$.

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