13
$\begingroup$

Let $E$ be an elliptic curve defined over $\mathbf{Q}$. Let $p \geq 5$ be a prime of additive reduction for $E$.

Let $f$ be the newform associated to $E$, and let $\pi$ be the irreducible admissible representation of $G=\mathrm{GL}_2(\mathbf{Q}_p)$ associated to $f$ (the so-called local component of $f$ at $p$). Then $\pi$ has conductor $p^2$. Assume that $\pi$ is supercuspidal. By the classification of supercuspidal representations, there exists an irreducible representation $\xi : \mathrm{GL}_2(\mathbf{F}_p) \to \mathrm{GL}(V)$ such that $\pi \cong \mathrm{Ind}_K^G \xi$ where $K$ is the maximal compact-mod-center subgroup of $G$ given by $K=\mathbf{Q}_p^\times \cdot \mathrm{GL}_2(\mathbf{Z}_p)$. The representation $\xi$ has dimension $p-1$.

The classification of irreducible representations of $\mathrm{GL}_2(\mathbf{F}_p)$ is well-known, and in our case $\xi$ arises from a character $\phi : \mathbf{F}_{p^2}^\times \to \mathbf{C}^\times$. More precisely we have the relation $\operatorname{Tr}(\xi(g)) = - (\phi(g)+\phi(g^p))$ for every element $g$ in $\mathbf{F}_{p^2}$ not in $\mathbf{F}_p$.

Now since $f$ arises from an elliptic curve, the representation $\xi$ has trivial central character so that $\phi |_{\mathbf{F}_p^\times}=1$. This implies $\phi$ has order dividing $p+1$ and $\operatorname{Tr}(\xi(g)) = - (\phi(g)+\overline{\phi}(g))$. Moreover the representation $\xi$ can be realized over $\mathbf{Q}$, which implies that $\phi+\overline{\phi}$ takes values in $\mathbf{Q}$. Therefore the possibilities are very restricted: $\phi$ has order 3, 4 or 6. Note that this implies $p \equiv -1 \mod{} 3, 4 \textrm{ or } 6$ respectively.

Now comes my question: is there a simple way to tell whether $\phi$ has order 3, 4 or 6 in terms of $E$? By the local Langlands correspondence, I would expect a condition depending only on the local Galois representation associated to $E$. Moreover, when $p \equiv -1 \mod{12}$, does every possible order for $\phi$ occur?

$\endgroup$
9
  • 2
    $\begingroup$ Doesn't this come from the $p$-valuation of the discriminant of $E$? Namely, the order is $12/\gcd(v_p(\Delta),12)$. $\endgroup$ – GuestPoster May 9 '15 at 7:04
  • $\begingroup$ @GuestPoster I've written a Magma code and for $p=11$ it seems to be the case that $\phi$ has order 6, 4, 3, 3 according to whether $v_p(\Delta)$ is 2, 3, 4, 8, so your guess seems right. For $p=5$ I have examples where $(v_p(\Delta),\textrm{ord}(\phi))=(2,3),(2,6),(4,3),(4,6),(8,3),(8,6)$ so $v_p(\Delta)$ seems not sufficient to determine $\phi$. For $p=17$ I have a $(2,3)$-example. If you could elaborate on your comment, this would make a nice answer! $\endgroup$ – François Brunault May 9 '15 at 9:52
  • $\begingroup$ I've never actually seen the specific question of the order of $\phi$ being asked, but it must be equivalent to something that is known in this genre (for instance, that the order of the inertia group is same as the order of $\phi$). Unfortunately, for $p\ge 5$ everyone seems to assume this type of knowledge, or at best cites Serre's 1972 Inventiones paper or Tate's algorithm. For instance page 3-4 of Kraus's work (on field extensions to prescribe good reduction, as in Will Sawin's answer) eudml.org/doc/155566 where a bit more of a proof is given in Section 2. $\endgroup$ – GuestPoster May 9 '15 at 12:05
  • 1
    $\begingroup$ Another possibly useful paper is "Euler factors determine local Weil representations" by the Dokchitsers. By my understanding, the local Weil representation is thus determined by the Euler factor of $E$ over the field of good reduction, and this field follows for $p\ge 5$ from the discriminant valuation. Similarly, the minimal model of $E$ over this field and the subsequent Euler factor from point-counting are also immediate (as in Will Sawin's answer). $\endgroup$ – GuestPoster May 9 '15 at 12:09
  • $\begingroup$ I have added a rough explanation of the relation between the determinant and the inertia to my answer. $\endgroup$ – Will Sawin May 9 '15 at 13:22
10
$\begingroup$

Because $p \geq 5$, the ramification of the Galois representation is tame, hence the action of the inertia group on that Galois representation factors through a cyclic group. For the exact same defined-over-$\mathbb Q$ reasons, the image of the inertia group has order $1$, $2$, $3$, $4$, or $6$. If it's $1$ or $2$, the representation is a $1$-dimensional character of the inertia group tensor a two-dimensional unramified representation. Because all irreducible unramified representations over a local field are one-dimensional, the Galois representation is not irreducible, so does not correspond to a supercuspidal representation.

So the order possibilities are $3$, $4$, and $6$.

You might guess that the order of the inertia group corresponds exactly to the order of that character. As far as I know, this is correct, and is true more generally for tamely ramified Galois representations / automorphic representations that arise from induction of representation $GL_n(\mathbb F_p)$ in the manner you describe, but I don't actually know anything about the local Langlands correspondence.

Assuming it's correct, we can see that every possible order occurs. The curve $y^2=x^3-p$ has inertia of order $6$ and is supercuspidal when $p \equiv -1$ mod $3$, $y^2=x^3-px$ is similar for order $4$, and $y^2=x^3- p^2$ for order $3$.

To prove this, first check that $y^2=x^3-1$ and $y^2=x^3-x$ are unramified outside $2$ and $3$ by computing the discriminants of the polynomials: $27$ and $4$. The curves I wrote down are all twists of those two that are trivialized over $\mathbb Q(p^{1/6})$, $\mathbb Q(p^{1/4})$, and $\mathbb Q(p^{1/3})$ respectively. These are Galois extensions whose inertia group at $p$ has order $n=6$, $4$, or $3$. The isomorphism is by multiplying $x$ by the square of the $n$th root of $p$ and multiplying $y$ by the third power. Using this we can see how the inertia group acts: It acts on the $n$th root by multiplying by an $n$th root of unity, so it acts on the curve by multiplying $x$ by the second power of that $n$th root of unity and $y$ by a third power of the $n$th root of unity. This is a CM automorphism of the curve of order $n$ and acts faithfully on the Tate module, so $n$ is the order of the inertia group acting on the Tate module.

To tell whether the representation is irreducible or reducible you look at the Frobenius action by conjugation on the inertia group. If it's trivial, then the Tate module splits into two distinct characters of the inertia group. If it's nontrivial, then the two characters are Galois conjugate to each other and cannot be separated. The conjugation action is exactly raising to the power of $p$ mod $n$, so is nontrivial when $p \equiv -1$ mod $n$.

Another way to get these answers would be to apply Tate's algorithm to compute the Neron model type (II, III, IV respectively) and then using the formula that for $p>5$ determines the Galois representation from the Neron model type. This would let you construct many more examples.

So indeed all occur when $p \equiv -1$ mod $12$, assuming my claim about the local Langlands correspondence is correct.


Here's how to relate the order of inertia to the $p$-adic valuaton of the discriminant, when a curve has potentially good reduction. Observe that for a curve with semistable, the discriminant is naturally a section of the $12$th power of the relative canonical bundle - in other words, its a modular form of weight $12$. For a curve with good reduction, the discriminant is nonvanish.

So for an elliptic curve with potentially good reduction, if the discriminant has $p$-adic valuation $v$, then over a field extension with good reduction, the relative canonical bundle is shifted from the relative canonical bundle of the original curve by $p^{v/12}$. I mean the natural map from one to the other is multiplication by $p^{v/12}$. This gives the Galois action on the relative canonical bundle of a smooth model - it's multiplication by an $n$th root of unit where $n$ is the denominator of $v/12$. Then the Galois action on the relative de Rham cohomology is the sum of that Galois character and its dual. By comparing the relative de Rham cohomology to the Tate module, we get that the order of the inertia group is also $n$.

One way to check for potentially good reduction is to check that the $p$-adic valuation of the $j$ invariant is nonnegative.

$\endgroup$
3
  • $\begingroup$ Thanks Will for your answer. Do you have a reference for determining the Galois representation from the Kodaira type for $p>5$? $\endgroup$ – François Brunault May 8 '15 at 16:52
  • $\begingroup$ @FrançoisBrunault No, I don't remember where I heard it. Let me flesh out my other argument instead, which I realized is simpler. $\endgroup$ – Will Sawin May 8 '15 at 19:19
  • 1
    $\begingroup$ @FrançoisBrunault: This is e.g. in Serre's paper "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques", p.312 (and he refers to Néron, I think) $\endgroup$ – Tim Dokchitser May 9 '15 at 17:34
3
$\begingroup$

Jared Weinstein and I worked out an algorithm which explicitly computes the character $\phi: \mathbf{F}_{p^2}^\times \to \mathbf{C}^\times$ (more precisely, a conjugate pair of characters) using modular symbols. You can read about it in our paper. It's implemented in both Sage and Magma.

Here's the computation in Sage for the elliptic curve with Cremona label 121a:

sage: f = Newform('121a')
sage: Pi = LocalComponent(f, 11)
sage: Pi.species()
'Supercuspidal'
sage: Pi.characters()
[
Character of unramified extension Q_11(s)* (s^2 + 7*s + 2 = 0), of level 1, mapping s |--> d, 11 |--> 1,
Character of unramified extension Q_11(s)* (s^2 + 7*s + 2 = 0), of level 1, mapping s |--> -d + 1, 11 |--> 1
]
sage: xi1, xi2 = _
sage: xi1.base_ring()
Number Field in d with defining polynomial x^2 - x + 1
sage: xi1.multiplicative_order()
6
$\endgroup$
1
  • $\begingroup$ Thanks David, I'm indeed using your algorithm to compute the examples in my comment to @GuestPoster above. I will see if I can find more patterns. $\endgroup$ – François Brunault May 9 '15 at 10:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.