The Abelian House is not closed

Posted on August 3, 2014 by Persiflage

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 \(\overline{{{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}} \setminus {{\mathfrak{M}}^{{\mathrm{ab}}}_{\infty}}\).

Posted in Mathematics | Tagged , , , , , , , , , , , | 2 Comments

The distribution of Hecke eigenvalues, part II

Posted on August 1, 2014 by Persiflage

Here are some numbers from KB promised in my last post.

“For the first 61595 newforms of squarefree level coprime to 15 here’s
the field extension of Z/3Z generated by the \(a_5\) field extensions:”

\([\mathbf{F}_3(a_5):\mathbf{F}_3]\) Total Number Number of Galois conjugacy classes Density of forms Density of conjugacy classes
Totals: 61595 10740 1 1
\(1\) 4623 4623 0.07505 0.4304
\(2\) 2492 1246 0.04046 0.1160
\(3\) 2397 799 0.03892 0.07439
\(4\) 2476 619 0.04020 0.05764
\(5\) 2600 520 0.04221 0.04842
\(6\) 2142 357 0.03478 0.03324
\(7\) 2289 327 0.03716 0.03045
\(8\) 2008 251 0.03260 0.02337
\(9\) 1962 218 0.03185 0.02030
\(10\) 1530 153 0.02484 0.01425
\(11\) 1837 167 0.02982 0.01555
\(12\) 1656 138 0.02689 0.01285
\(13\) 1612 124 0.02617 0.01155
\(14\) 1638 117 0.02659 0.01089
\(15\) 1455 97 0.02362 0.009032
\(16\) 1088 68 0.01766 0.006331
\(17\) 1292 76 0.02098 0.007076
\(18\) 1008 56 0.01636 0.005214
\(19\) 1159 61 0.01882 0.005680
\(20\) 1120 56 0.01818 0.005214
\(21\) 987 47 0.01602 0.004376
\(22\) 990 45 0.01607 0.004190
\(23\) 966 42 0.01568 0.003911
\(24\) 1056 44 0.01714 0.004097
\(25\) 1100 44 0.01786 0.004097
\(26\) 650 25 0.01055 0.002328
\(27\) 783 29 0.01271 0.002700
\(28\) 868 31 0.01409 0.002886
\(29\) 551 19 0.008946 0.001769
\(30\) 420 14 0.006819 0.001304
\(31\) 775 25 0.01258 0.002328
\(32\) 800 25 0.01299 0.002328
\(33\) 759 23 0.01232 0.002142
\(34\) 374 11 0.006072 0.001024
\(35\) 490 14 0.007955 0.001304
\(36\) 576 16 0.009351 0.001490
\(37\) 592 16 0.009611 0.001490
\(38\) 380 10 0.006169 0.0009311
\(39\) 429 11 0.006965 0.001024
\(40\) 680 17 0.01104 0.001583
\(41\) 492 12 0.007988 0.001117
\(42\) 294 7 0.004773 0.0006518
\(43\) 258 6 0.004189 0.0005587
\(44\) 308 7 0.005000 0.0006518
\(45\) 180 4 0.002922 0.0003724
\(46\) 322 7 0.005228 0.0006518
\(47\) 282 6 0.004578 0.0005587
\(48\) 144 3 0.002338 0.0002793
\(49\) 147 3 0.002387 0.0002793
\(50\) 350 7 0.005682 0.0006518
\(51\) 561 11 0.009108 0.001024
\(52\) 260 5 0.004221 0.0004655
\(53\) 106 2 0.001721 0.0001862
\(54\) 378 7 0.006137 0.0006518
\(55\) 0 0 0 0
\(56\) 112 2 0.001818 0.0001862
\(57\) 171 3 0.002776 0.0002793
\(58\) 406 7 0.006591 0.0006518
\(59\) 236 4 0.003831 0.0003724
\(60\) 120 2 0.001948 0.0001862
\(61\) 183 3 0.002971 0.0002793
\(62\) 62 1 0.001007 0.00009311
\(63\) 378 6 0.006137 0.0005587
\(64\) 320 5 0.005195 0.0004655
\(65\) 130 2 0.002111 0.0001862
\(66\) 132 2 0.002143 0.0001862
\(67\) 201 3 0.003263 0.0002793
\(68\) 68 1 0.001104 0.00009311
\(69\) 276 4 0.004481 0.0003724
\(70\) 140 2 0.002273 0.0001862
\(71\) 284 4 0.004611 0.0003724
\(72\) 144 2 0.002338 0.0001862
\(73\) 292 4 0.004741 0.0003724
\(74\) 74 1 0.001201 0.00009311
\(75\) 0 0 0 0
\(76\) 152 2 0.002468 0.0001862
\(77\) 0 0 0 0
\(78\) 78 1 0.001266 0.00009311
\(79\) 79 1 0.001283 0.00009311
\(80\) 160 2 0.002598 0.0001862
\(81\) 81 1 0.001315 0.00009311
\(82\) 0 0 0 0
\(83\) 83 1 0.001348 0.00009311
\(84\) 168 2 0.002727 0.0001862
\(85\) 85 1 0.001380 0.00009311
\(86\) 0 0 0 0
\(87\) 0 0 0 0
\(88\) 0 0 0 0
\(89\) 89 1 0.001445 0.00009311
\(90\) 0 0 0 0
\(91\) 0 0 0 0
\(92\) 0 0 0 0
\(93\) 0 0 0 0
\(94\) 0 0 0 0
\(95\) 95 1 0.001542 0.00009311
\(96\) 0 0 0 0
\(97\) 0 0 0 0
\(98\) 0 0 0 0
\(99\) 0 0 0 0
\(100\) 0 0 0 0
\(101\) 0 0 0 0
\(102\) 0 0 0 0
\(103\) 0 0 0 0
\(104\) 104 1 0.001688 0.00009311

I’ve presented the numbers KB send me in various ways. The first column simply counts the field generated by \(a_5\). The second column normalizes by the order of the field. This is a little like counting two representations which differ by an automorphism of the coefficient field as being “the same.” The final two comments are then the proportion of the first two columns overall.

I’m really not quite sure what to make of this data. It does suggest that A is false, which is perhaps not surprising. It’s not terribly overwhelming evidence for B, but then, law of smaller numbers and all.

AV’s suggestion in the comments that the constants \(C_q\) should be independent of \(q\) must refer to the constants in the second last column, I believe. Of course, it might be the case that \(\mathbf{F}_3(a_5)\) is smaller than \(\mathbf{F}_3(a_2,a_5,a_7,a_{11},\ldots)\), so these numbers aren’t exactly the same as the fields generated by the mod-\(p\) reductions of the eigenforms. If you squint, the numbers in this column do look somewhat constant for \(n < 10\) or so. One can even argue that \(n = 1\) might be artificially inflated exactly because the phenomenon of "slipping into a subfield" mentioned above. So I'm giving the points to AV. (Yes, that's right, there were points available and you missed out because you didn’t play the game.)

Posted in Mathematics | Tagged , , , , , | Leave a comment

The distribution of Hecke eigenvalues, Part I

Posted on July 29, 2014 by Persiflage

Here is a question I raised at the Puerto-Rico conference during one of the “problem sessions.” Toby Gee seems to remember that I had some half-baked heuristics that predicted both A and B below, but perhaps one of my readers has a more sophisticated suggestion, or even a similarly wild guess (or even a similarly contradictory collection of guesses).

Fix a pair of distinct odd primes \(p\) and \(l\). Now consider a random normalized Hecke eigenform \(f = \sum a_n q^n \in \overline{\mathbf{Z}}\) of weight two and level \(\Gamma_0(N)\), where \(N\) is squarefree and prime to both \(p\) and \(l\). Now take the Hecke eigenvalue \(a_{l}\) and reduce it modulo a random prime \(\mathfrak{p}\) above \(p\).

Question: As one ranges over all newforms of conductor \(< X\), what is the resulting distribution — if it even exists — of \(a_l \in \overline{\mathbf{F}}_p\)?

Let me not be too precise about what random means — for example, there's a question about whether one wants to normalize in some way for Galois conjugates of eigenforms, but none of this will really matter for the very weak questions I have in mind. For example, consider the following two possibilities:

  1. A: The element \(a_l\) lies in \(\mathbf{F}_p\) at least 100% of the time.
  2. B: The element \(a_l\) lies in \(\overline{\mathbf{F}}_p \setminus \mathbf{F}_p\) at least 100% of the time.

Here by 100% I mean as a proportion of all forms as \(X \rightarrow \infty\), although I confess that I can’t even rule out the extreme version of A where 100% really means every single form.

The specifications on the level are designed to rule out some “trivial” examples. At level \(\Gamma_0(N)\) with \(N\) squarefree there will be no CM-forms, which is one cheap way to generate large coefficient fields. The level also prevents twisting by characters (an even cheaper trick). Finally, in general, it is possible to generate large coefficient fields for Galois representations by imposing a local condition at some auxiliary prime \(q\). For example, one can impose some supercuspidal condition so that the *local* residual representation at \(q\) does not land in \(\mathrm{GL}_2(\mathbf{F}_{p^m})\) for any \(m\) not divisible by an arbitrary fixed integer chosen in advance. However, this too is not possible if \(\pi_q\) is forced to be either unramified or special (up to unramified quadratic twist).

Note that there do exist infinitely many semi-stable modular elliptic curves, so \(a_l\) will lie in \(\mathbf{F}_p\) at least infinitely often. This disproves the “extreme” version of B, but doesn’t go very far towards disproving the asymptotic version of B. As for A, every single time you write down a normalized eigenform with coefficients in some field \(E \ne \mathbf{Q}\), you disprove the extreme version of A for a positive density of pairs \((p,l)\). But no finite collection of such forms can disprove A even for a single \(l\) and varying \(p\), because there will always be (many) primes which split completely in any finite collection of number fields.

Here are three questions:

  1. Can you disprove the extreme version of A for all \(p\) and \(l\)?
  2. Can you disprove the super-extreme version of A, namely, show that for all primes \(p\), there exists a newform of squarefree level \(N\) prime to \(p\) such that the residual representation is not defined over \(\mathbf{F}_p\)? (equivalently, replace \(a_l\) by the collection of all \(a_l\) with \(l\) prime to \(Np\).)
  3. Can you give any heuristic that suggests that either A or B (in the weaker form) is either strong or true?
  4. Do you have any guesses as to the distribution of the \(a_l\)?

Right now, as you read this, KB’s computer is churning away in sage generating some data, which will be the topic of Part II. But until then, I would like to hear your opinions/guesses. For me, I think that A is probably false, but I honestly have no feeling for B.

Posted in Mathematics | Tagged , , | Leave a comment

En Passant IV

Posted on July 19, 2014 by Persiflage

My students Richard Moy and Joel Specter have uploaded their paper on partial weight one Hilbert modular forms, previously discussed here, to the ArXiv.

Germany now leads the world in both soccer and perfectoid spaces.

This is a recipe for a moist tasty almond cake. Also: quick way to make dessert with stale mint oreo cookies: add brandy and mash together, then add coconut, and, voilà, instant rum balls.

SiriusXM is truly, truly disappointing. They don’t seem to understand that when presenting classical music one should always indicate who the performer is. On top of that, SiriusXM just cancelled one third of its classical stations (that would be one out of three).

Here is Existential Comics, the best comic strip I’ve found since I learnt about Dinosaur Comics from BZ.

I still go the the Bourgeois Pig to chat with Matt, but I don’t buy anything; instead, I stop off at Intelligentsia first and get a Cortado (which they call a Gibraltar) instead.

What were the two topics which dominated the political conversation in Australia when I visited in 2011? Asylum policy and the Carbon Tax (Caahbin Tax). What are the topics today? The very same. Is it possible that Abbott could make Little Johnny H seem enlightened? Oh, talking about Australia, perhaps you would like to play the white Australia game.

Justice
This is where I like to sit when I talk to Toby about math. My math comes with the imprimatur of justice!

Posted in Waffle | Tagged , , , , , , , , , , , , , , , , , , | Leave a comment

A public service announcement concerning Fontaine-Mazur for GL(1)

Posted on July 12, 2014 by Persiflage

There’s a rumour going around that results from transcendence theory are required to prove the Fontaine-Mazur conjecture for \(\mathrm{GL}(1)\). This is not correct. In Serre’s book on \(\ell\)-adic representations, he defines a \(p\)-adic representation \(V\) of a global Galois group \(G_F\) to be rational if it is unramified outside finitely many primes and if the characteristic polynomials of \(\mathrm{Frob}_{\lambda}\) actually all lie in some fixed number field \(E\) rather than over \(\mathbf{Q}_p\). Certainly being rational is a consequence of occurring inside the etale cohomology of a smooth proper scheme \(X\), and one might be motivated to make a conjecture in the converse direction assuming that \(V\) is absolutely irreducible. But being “rational” is just a rubbish definition (sorry Serre), a mere proxy for the correct notion of being potentially semistable at all primes dividing \(p\) (“geometric,” given the other assumptions on \(V\)). And the implication

A character \(\chi: G_F \rightarrow \overline{\mathbf{Q}}_p\) is Hodge-Tate \(\Rightarrow \chi\) is automorphic

doesn’t require any transcendence results at all. One can’t really blame Serre for not coming up with the Fontaine-Mazur conjecture in 1968. The reason for this confusion seems to be the proof of Theorem stated on III-20 of Serre’s book on abelian \(\ell\)-adic representations (with the modifications noted in the updated version of Serre’s book), namely:

Theorem (Serre-Waldschmidt): If \(V\) is an abelian representation of \(G_F\) which is rational, then \(V\) is locally algebraic.

This argument (even for the case when \(F\) is a composite of quadratic fields, the case considered by Serre) requires some transcendence theory. But the implication \(V\) is abelian and Hodge-Tate \(\Rightarrow V\) is locally algebraic (also proved in Serre) only uses Tate era p-adic Hodge theory. The other ingredients for Fontaine-Mazur are as follows: First, there is the classification of algebraic Hecke characters (due to Weil, I think). A key point here is that the algebraicity forces the unit group to be annihilated by some element in the integral group ring. However, the representation \(V\) occurs in \(\mathcal{O}^{\times}_F \otimes \mathbf{C}\) with dimension \(\dim(V|c = 1)\) if \(V\) is non-trivial, so this forces the existence of representations \(V\) of \(G\) on which \(c = – 1\), corresponding to CM subfields. The final step is the theory of CM abelian varieties. So although the result is non-trivial, you can be rest assured, gentle reader, that you are not secretly invoking subtle transcendence results every time you twist an automorphic Galois representation by a Hodge-Tate character and claim that the result is still automorphic.

Posted in Mathematics | Tagged , , , | Leave a comment

Report from Luminy

Posted on July 8, 2014 by Persiflage

For how long has Luminy been infested with bloodthirsty mosquitoes? The combination of mosquitoes in my room with the fact that my bed was 6′ long with a completely unnecessary headboard (which meant that I had to sleep on an angle with my ankles exposed) did not end well.

As for the math, there were plenty of interesting talks, most of which I will not discuss here. Jan Nekovar gave a nice talk explaining how one could prove that the cohomology of compact Shimura varieties of \(\mathrm{GL}(2)\)-type were semi-simple. For concreteness, imagine that \(X\) is a Hilbert modular surface associated to a real quadratic field \(F\). Suppose that \(\rho\) is the Galois representation associated to a cuspidal Hilbert modular form of parallel weight two. Then the Langlands-Kottwitz method shows that the semi-simplification of \(\rho^{\otimes 2}\) should occur inside \(H^2\). On the other hand, this argument only ever deals with the trace of Hecke operators and so cannot say anything about semi-simplifications. Nekovar’s argument is to use the Eichler-Shimura relation applied to partial Hecke operators for primes which split completely in both \(F\) and the corresponding reflex field. The point is that these operators satisfy a quadratic relation (with distinct eigenvalues for generic elements of the Galois group), and so act semi-simply on \(H^2\) (imagine everything is compact here). Then, by pure group theory, if the image of \(\rho\) is large enough, the sheer number of such elements is enough to force semi-simplicity. It is perhaps useful to note that if \(V\) is a representation such that \(V^{ss} = (\rho^{\otimes 2})^{ss}\) and \(V\) is geometric and pure, then \(V\) should automatically be semi-simple. This follows from any number of combinations of bits of the standard conjectures, but one way to see it is that if \(W\) is geometric and of weight zero, then (by Bloch-Kato) one should have \(H^1_f(F,W) = 0\). The relevant \(W\) in the example above is \(\mathrm{ad}^0(\rho)\). So in fact one can give an alternate proof of the theorem using the full power of modularity lifting theorems, providing one is willing to omit finitely many primes \(p\). This is really an explanation of why Jan’s result is nice! For example, as soon as one replaces \(\rho^{\otimes 2}\) by \(\rho^{\otimes n}\), one has to start dealing with \(H^1_f(F,\mathrm{Sym}^{2n}(\rho)(-n))\), which gets a little tricky.

Boars
These fellows turned up for the Bouillabaisse.

Ana Caraiani talked about a very nice result concerning the sign of Galois representations associated to torsion classes for \(\mathrm{GL}(n)/F^{+}\) for totally real fields \(F^{+}\) (this was joint work with Bao Le Hung). Namely, the trace of any complex conjugation lies in \(\{-1,0,1\}\) (in fact, the result identified the exact characteristic polynomial, which is more general in small characteristic). The basic strategy is to follow Scholze’s construction and “reduce” the problem to the case of essentially self-dual forms, where one has previous results by Taylor, Bellaïche-Chenevier, and Taïbi. However, there is a problem, which is that the regular self-dual automorphic forms one finds congruences with need not be globally irreducible, and perhaps not even cuspidal. Suppose one can show that they decompose into an isobaric sum \(\pi = \boxplus \pi_i\) where the \(\pi_i\) are self-dual. One runs into problems if too many of the \(\pi_i\) are of dimension \(n_i\) with \(n_i\) odd. However, by considering the weights, only one of the \(\pi_i\) can be odd, because otherwise the Hodge-Tate weight zero would occur with multiplicity which would violate the fact that \(\pi\) is regular. There is still something to check for the even \(n_i\) also, because previous results required some sign assumption on the character \(\eta\) such that \(\pi = \pi^{\vee} \eta\). I believe that even getting to this point required a further assumption on the torsion class not coming from the boundary. In the boundary case, there was also a reduction/induction case which also required careful handling of “odd” dimensional pieces, and some computation of a restriction of Hecke operators from the relevant Parabolic/Levi which required a sign to come out correctly. One clever technical step was working with the cohomology of adelic quotients \(G(F)\backslash G(\mathbf{A})/UK\) where \(K\) is a maximal compact of \(G(\mathbf{R})\) rather than the connected component \(K^0\). The advantage of this is that, in the odd dimensional case, this pins down the trace of complex conjugation to be \(+1\) rather than \(\pm 1\). This is clear when \(n = 1\), and that one should expect it to be true follows for \(n\) odd by taking determinants.

Peter Scholze gave a talk on his new functor. The basic elements in the construction of this functor are as follows. The Gross-Hopkins period map allows one to view the (infinite level) Lubin-Tate tower as a \(\mathrm{GL}_n(F)\)-torsor over the (\(D^{\times}\) Severi-Brauer variety) \(\mathbf{P}^{n-1}_{\mathbf{C}_p}\). So, given an admissible representation \(\pi\), one can form the “local system” \(\mathscr{F}_{\pi}\) on the base, and then take its cohomology. The key technical point of this construction is to show that the result is admissible for \(D^{\times}\), which amounts to proving finiteness of \(K\)-invariants for suitable compact open \(K\) of \(D^{\times}\). The first step is to pull back to the (lowest level) part of the Lubin-Tate tower, which one can do because the GH map splits. Now the map from infinite level to the base of the Lubin-Tate tower is really a \(\mathrm{GL}_n(\mathcal{O}_F)\)-torsor, so one only has to consider the restriction of \(\pi\) to \(\mathrm{GL}_n(\mathcal{O}_F)\). But then using the admissibility of \(\pi\), one can look instead at the regular representation of \(\mathrm{GL}_n(\mathcal{O}_F)\). Now, by some sort of Shapiro’s Lemma, one can pull everything back up to infinite level. At infinite level, however, one can replace the Lubin-Tate space by the corresponding Drinfeld tower. Now taking \(K\)-invariants is something that is “easy” to do, because there is an action of \(K\) on the space, and the quotient by \(K\) is some sufficiently nice object for which one has (again by Peter) some nice finiteness theorems for cohomology. I should probably have mentioned that at some point we are working with coefficients in \(\mathcal{O}^+/p\), i.e. in the almost world. The main application in the talk was to show that when one patches completed cohomology (a la CEGGPS), then one can recover the Galois representation from the result. This essentially amounts to showing that when one patches together suitable admissible \(\pi_i\), one can also patch the functor. This requires more than admissibility of the functor, but some sort of “uniform” admissibility (which is always required for patching). I think the key point here is that if \(\pi_i\) is something patched with a group of diamond operators \(\Delta\), then \(\pi_i\) has a filtration by \(|\Delta|\) copies of the original \(\pi\), and so \(\mathscr{F}_{\pi_i}\) has a corresponding uniformly bounded filtration by \(\mathscr{F}_{\pi}\), and so \(H^{n-1}(\mathbf{P}^{n-1}_{\mathbf{C}_p},\mathscr{F}_{\pi})^K\) has length at most \(|\Delta|\) times the corresponding length for the (fixed for all time) version for \(\pi\). On the other hand, Peter instead pulled out a new piece of kit by patching using ultra-filters. My own feeling about logic is that it is never really necessary to prove anything, and I think PS agreed that it wasn’t strictly required for this particular application. Now I understand that my prejudice may not be justified (for example, it is probably hard to prove various identities concerning orbital integrals in small characteristic directly), but I think it applies in this case. Plus, as a purely expositional remark, if you are going to whip out ultrafilters during a number theory talk then everyone is just going to talk about ultrafilters rather than the beautiful construction!

Posted in Mathematics, Uncategorized | Tagged , , , , , , , | Leave a comment

An Obvious Claim

Posted on July 6, 2014 by Persiflage

It’s been a while since I saw Serre’s “how to write mathematics badly” lecture, but I’m pretty sure there would have been something about the dangers of using the word “obvious.” After all, if something really is obvious, then it shouldn’t be too difficult to explain why. It is especially embarrassing when someone asks you to clarify a remark/claim in one of your papers which you claim is “obvious” and you find yourself having no idea what the implicit argument was supposed to be. Such a thing happened recently to me, when Toby asked me to explain why the following was true:

Claim: Let \(N \equiv 3 \mod 4\) be prime, and let \(\epsilon\) be the fundamental unit of \(K = \mathbf{Q}(\sqrt{N})\). Then \(\epsilon = a + b \sqrt{N}\) where \(a\) is even and \(b\) is odd.

Proof of Claim: Between Toby, Kevin, and myself, we managed to come up with the argument below, following a suggestion of Toby Rebecca Bellovin: It’s easy enough to see (obvious) that \(a\) and \(b\) are integers and \(N(\epsilon) = 1\). Hence, it suffices to rule out the case that \(b\) even and \(a\) odd. Write \(a^2 – N b^2 = 1\). It follows that \(a^2 \equiv 1 \mod N\), and since \(N\) is prime, that \(a \equiv \pm 1 \mod N\). Assuming that \(a\) is odd, write \(a = 2NA \pm 1\), and \(b = 2B\). Then the equation above becomes

\(A(NA \pm 1) = B^2.\)

Without loss of generality, assume that \(A\) is positive. Then this equation implies that \(A\) and \(NA \pm 1\) are squares, say \(A = d^2\) and \(NA \pm 1 = c^2\). But then

\(c^2 – N d^2 = (NA \pm 1) – N A = \pm 1,\)

and hence \(\eta = c + N \sqrt{d}\) is a (smaller) unit (in fact, \(\eta^2 = \pm \epsilon\)), contradicting the assumption that \(\epsilon\) was a fundamental unit. \(\quad \square\)

This argument is really a 2-descent on the unit group. As Kevin remarked: “So this is a descent argument in a completely elementary situation which I don’t think I’d ever seen before and which proves something that I don’t think I knew … What’s ridiculous is that if the equation had been a cubic and we were after rational solutions then I would have instantly leapt on descent as one of my main tools for attacking it :-/ We live and learn!”

So what was I thinking when I wrote the paper? The actual claim in the paper is this: “If \(H’\) is the (2 part of the) strict ray class group of \(K\) of conductor \((2)\), then \(H = H’\), where \(H\) is the (2 part of the) class group. The “argument” is as follows:

The proof of [the above] is even more straightforward: it follows immediately from a consideration of the units in \(\mathcal{O}^{\times}_K\) and the exact sequence

\(\mathcal{O}^{\times}_K \rightarrow (\mathcal{O}_K/2 \mathcal{O}_K)^{\times} \rightarrow H’ \rightarrow H \rightarrow 0.\)

Well, at least the word obvious was only implicit here. I could try to place the blame on my co-author Matt here, but honestly the phrasing of the claim does sound a little like something I would write.

Next up: a report from Luminy!

Posted in Mathematics | Tagged , , , , | Leave a comment

Huuuuuge piles of cash

Posted on June 23, 2014 by Persiflage

As widely reported today, the first of the “Breakthrough” prizes in mathematics have been announced. Leaving aside the question as to whether such awards are sensible (Persiflage is more sympathetic to capitalist principles than your average pinko marxist mathematician), I think we should all be happy that the inaugural awardees are beyond reproach, something which could not have been taken for granted. So well done to whoever was on the prize committee.

Of the awardees that I know personally, my impression is that, at least in part, they would feel mildly embarrassed by the large amount of cash involved, even if they (presumably) feel gratified by the deserved acknowledgement of their contributions to mathematics. I do hear, however, that a bottle of Chateau d’Yquem 1967 does wonders to wash away any last remaining vestiges of embarrassment, and I am more than willing to help out with the consumption of said beverage if required.

Posted in Mathematics, Politics | Tagged , | Leave a comment

There are non-liftable weight one forms modulo p for any p

Posted on June 10, 2014 by Persiflage

Let \(p\) be any prime. In this post, we show that there is an integer \(N\) prime to \(p\) such that \(H^1(X_1(N),\omega_{\mathbf{Z}})\) has a torsion class of order \(p\). Almost equivalently, there exists a Katz modular form of level \(N\) and weight one over \(\mathbf{F}_p\) which does not lift to characteristic zero. We shall give two different arguments. The first argument will have the virtue that the torsion class is non-trivial after localization at a maximal ideal \(\mathfrak{m}\) which is new of level \(N\). The second argument, in contrast, will produce torsion classes at fairly explicit levels. Neither proof, unfortunately, implies the existence of interesting Galois representations unramified at \(p\) with image containing \(\mathrm{SL}_2(\mathbf{F}_p)\). Rather, the classes will come from deformations of characteristic zero classes. (This post is an elaboration of my comment here.)

The first argument: Let \(K/\mathbf{Q}\) be an imaginary cubic extension unramified outside \(p\) with Galois closure \(L/\mathbf{Q}\) with Galois group \(S_3\). There is a corresponding Galois representation:

$latex \rho: G_{\mathbf{Q}} \rightarrow \mathrm{Gal}(L/\mathbf{Q}) = S_3
\rightarrow \mathrm{GL}_2(\mathbf{Q}_p).$

This representation is modular. Suppose for convenience that \(p > 3\). Associated to \(\rho\) is an absolutely irreducible residual representation \(\overline{\rho}\). Let \(R\) denote the corresponding universal unramified deformation. The only characteristic zero deformations are dihedral. Let \(R^{\mathrm{dh}}\) denote the corresponding universal unramified dihedral deformation ring. It’s easy to identify this ring explicitly; it is

\( R^{\mathrm{dh}} = \mathbf{Z}_p[C_E \otimes \mathbf{Z}_p],\)

where \(C_E\) is the class group of the imaginary quadratic subfield \(E\) of \(L\). The ring \(R\) will fail to be \(\mathbf{Z}_p\)-flat exactly when \(R \ne R^{\mathrm{dh}}\). Fortunately, this can be determined purely from the reduced tangent space of \(R\). Note that

\(\mathrm{ad}^0(\rho) \simeq \rho \oplus \eta,\)

where \(\eta\) is the quadratic character of \(E/\mathbf{Q}\). The reduced tangent space of \( R^{\mathrm{dh}}\) is the Bloch–Kato Selmer group \(H^1_f(\mathbf{Q},\overline{\eta})\), where \(H^1_f\) denotes the subring of cohomology classes which are unramified everywhere. So it all comes down to finding \(K\) so that \(H^1_f(\mathbf{Q},\overline{\rho})\) is non-zero. However, an elementary argument using inflation-restriction shows that this is equivalent to showing that the class number \(h_K\) of \(K\) is divisible by \(p\). So we are done provided that we can find a suitable \(K\) with class number divisible by \(p\). (I should mention, of course, that we are using the theorem that \(R = \mathbf{T}_{\mathfrak{m}}\) which was proved by David Geraghty and me.) The last step follows from the lemma below; the argument is essentially taken from this paper of Bilu–Luca.

Lemma: Fix a prime \(p > 3\). There exists an imaginary cubic field \(K/\mathbf{Q}\) of
discriminant prime to \(p\) and class number divisible by \(p\).

Proof: Consider the field \(K = \mathbf{Q}(\theta)\), where

\((\theta^2 + 1)(\theta – t^p + 1) – 1 = 0\),

and \(t\) is an element of \(\mathbf{Q}\) to be chosen later. Note that \(\theta^2 + 1\) is manifestly a unit in \(K\). We may compute that

\((\theta^2 + \theta + 1) \theta = (1 + \theta^2) t^p.\)

Since \((\theta,\theta^2 + \theta + 1) = (1)\) is trivial, it follows that \((\theta) = \mathfrak{a}^p\) for some ideal \(\mathfrak{a}\). We shall show that, for a suitably chosen \(t\), the element \(\mathfrak{a}\) is non-trivial in the class group. If \(\mathfrak{a}\) is trivial, then, up to a unit, \(\theta\) is a \(p\)th power. On the other hand, the rank of the unit group of \(K\) is one, and \(\theta^2 + 1\) is a unit. Hence, it suffices to choose a \(t\) such that:

  1. \(\theta^2 + 1\) generates a subgroup of \(\mathcal{O}^{\times}_K\) of index prime to \(p\). Equivalently, \(\theta^2 + 1\) is not a perfect \(p\)th power in \(K\).
  2. None of the elements \(\theta (\theta^2 + 1)^i\) for \(i = 0,\ldots,p-1\) is a perfect \(p\)th power in \(K\).
  3. The polynomial defining \(K\) is irreducible.
  4. The discriminant of \(K\) is not a square.
  5. The discriminant of \(K\) is prime to \(p\).

By working over the function field \(\mathbf{Q}(t)\) instead of \(\mathbf{Q}\), one finds that the first four conditions hold for all \(t \in \mathbf{Q}\) outside a thin set. (The discriminant \(\Delta\) is always negative, so the signature of the field is always \((1,1)\).) On the other hand, the discriminant of the defining polynomial is \(-3 \mod t\), so if one (for example) takes \(t\) to be an integer divisible by \(p\) then the discriminant will be prime to \(p\). Note that the set of integers divisible by \(p\) will contain elements not in any thin set, because the number of integral points of height at most \(H\) in a thin set is \(o(H)\).

Second Argument: Let \(E = \mathbf{Q}(\sqrt{-23})\), and let \(L = E[\theta]/(\theta^3 – \theta + 1)\) be the Hilbert class field of \(E\). There is a weight one modular form of level \(\Gamma_1(23)\) and quadratic character corresponding to the Galois representation:

\(\rho: \mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{Q}_p).\)

Lemma: Let \(p > 3\). Let \(q = x^2 + 23 y^2\) be a prime such that \(q \equiv 1 \mod p\). Equivalently, let \(q\) be a prime which splits completely in \(L(\zeta_p)\). Then

\(\# H^1(X(\Gamma_1(23) \cap \Gamma_0(q)),\omega)^{\mathrm{tors}}\)

is divisible by \(p\). More generally, for any prime \(q\), the quantity above is divisible by any prime divisor of \(a^2_q – (1+q)^2\), and \(a_{\ell} \in \{2,0,-1\}\) for a prime \(\ell\) is the coefficient of \(q^{\ell}\) in \(q \prod (1-q^n) (1 – q^{23 n})\).

Proof: This follows from “level–raising” in characteristic \(p\) for weight one forms. Under the hypothesis that \(a^2_q – (1+q)^2\), we find that there is more cohomology (over \(\mathbf{Z}_p\)) in level \(\Gamma_1(23) \cap \Gamma_0(q)\) than is accounted for by oldforms. Assuming that there is no torsion, this is inconsistent with the fact that there are no newforms in characteristic zero, because weight one forms cannot be Steinberg at any place. (The easiest way to see this is that the eigenvalue of \(U_q\) would have to be non-integral — it also follows on the Galois side from local-global compatibility, but this is overkill.) Note that level–raising in this context does not follow from classical level–raising — for the details I refer to you my fifth lecture in Barbados on non-minimal modularity lifting theorems in weight one.

Posted in Mathematics | Tagged , , , | Leave a comment

Math and Genius

Posted on June 6, 2014 by Persiflage

Jordan Ellenberg makes a compelling case, as usual, on the pernicious cultural notion of “genius.” Jordan’s article also brought to mind a thought provoking piece on genius by Moon Duchin here (full disclosure: the link on Duchin’s website has the folowing caveat: note: juvenilia! Not that I hate it).

I should first clarify that I am essentially in agreement with many of Jordan’s points, as well as his implicit and explicit recommendations. That said, there is a secondary argument which portrays mathematics as a grand collective endeavour to which we can all contribute. I think that this is a little unrealistic. In my perspective, the actual number of people who are advancing mathematics in any genuine sense is very low. This is not to say that there aren’t quite a number of people doing *interesting* mathematics. But it’s not so clear the extent to which the discovery of conceptual breakthroughs is contingent on others first making incremental progress. This may sound like a depressing view of mathematics, but I don’t find it so. Merely to be an observer in the progress of number theory is enough for me — I know how to prove Fermat’s Last Theorem, how exciting is that*?

Having said this, there are two further points on which I do agree with Jordan. The first is that it is a terrible idea to *prejudge* who will be the ones to make progress, at least any more than necessary (judgements of various levels are completely built in to the academic world in the context of hiring, grants, etc.). This is consistent with Jordan’s point that one should view accomplishments rather than people as genius. The second is that the mere notion of genius has detrimental psychological effects on mathematicians. The depressing effect on enthusiasm for mathematics amongst students has already been mentioned, but I also wanted to note a contrasting effect. There’s a tendency amongst a certain group of students — for concreteness imagine a certain flavour of Harvard male undergraduate — who buys into the notion of genius with themselves as an example. This doesn’t do any favours for anyone; obviously not fellow students who are put off, but also the students themselves who are tempted to learn fancy machinery —- that’s what a genius does! — rather than to really understand the basics. Two clarifying points: first, this in no way applies to all Harvard undergraduates — some of the smartest I have met have been very humble and level headed. Second, when I single out Harvard undergraduates, I am implying a correlation rather than a causation: this is not at all the fault of the Harvard faculty who are generally no nonsense. Then again, Harvard does have Math 55, which is somewhat of a petri dish for this sort of attitude.

Having written this, I now see there is a practical problem in the classroom which I don’t know how to solve. What do you do when you are teaching and one student sticks out like a sore thumb because their understanding of the material (seemingly) far exceeds the rest of the class?

(*) OK, I have not read all the details of the proof of cyclic base change for \(\mathrm{GL}(2)\).

Posted in Mathematics, Politics, Waffle | Tagged , , , | Leave a comment