# Lifting the Hasse invariant mod $2$

Katz defines in Section 2.0 $p$-adic properties of modular schemes and modular forms the Hasse invariant as a mod $p$ modular form $A$ of weight $p-1$. In other words, it is a section of $\omega^{\otimes p-1}$ on the compactified moduli stack of elliptic curves $\overline{\mathcal{M}}_{ell,\mathbb{F}_p}$.

We have a morphism from integral modular forms to mod $p$ modular forms, induced by the morphism $\omega^{\otimes p-1} \to \omega^{\otimes p-1}/p$ on $\overline{\mathcal{M}}_{ell,\mathbb{Z}}$. We say that the Hasse invariant $A$ is liftable to characteristic zero if it is in the image of this morphism. By the long exact sequence in cohomology $$H^0(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1})\to H^0(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1}/p) \xrightarrow{\partial} H^1(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1}) \xrightarrow{\cdot p}$$ the Hasse invariant is automatically liftable if $H^1(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1})$ has no $p$-torsion. This happens iff $p\geq 5$ and for $p=2,3$ the Hasse invariant is not liftable to an integral modular form (of level $1$).

If we consider $\Gamma_1(n)$-modular forms instead, the Hasse invariant becomes liftable to characteristic zero for $p=3$ and all levels $n\geq 2$, $(p,n)=1$, because then $H^1(\overline{\mathcal{M}}_1(n)_{\mathbb{Z}[\frac1n]}; \omega^{\otimes 2}) = 0$.

For what $n$ does the Hasse invariant lift to a characteristic zero $\Gamma_1(n)$-modular form, $(p,n)=1$, if $p=2$?

At the time of Katz's article the problem seems to have been unsolved as he just states (with a sketch of proof) that it is liftable if $n$ is divisible by $3,5,7$ or $11$, but he does not know it in other cases. In light of the long exact sequence above, the question is equivalent to $\eta = \partial(A)$ vanishing in (the $2$-torsion of) $H^1(\overline{\mathcal{M}}_1(n)_{\mathbb{Z}[\frac1n]}; \omega)$.

In particular, I would be interested in concrete examples (for $\Gamma_1(n)$ or $\Gamma(n)$) where the Hasse invariant is not liftable if such examples are known.

• What I would really like is to be thanked under the user name "Electric Penguin." – Electric Penguin Feb 2 '16 at 17:34
• The argument in your paper needs a minor modification. You write $L(0,\chi)(1 - \zeta) \equiv 1 + 1 \equiv 0 \mod 2$, but you probably want to write that $L(0,\chi)(1 - \zeta) \equiv 1 + 1 \equiv 2 \mod 2(1 - \zeta)$, since what you currently have written is not enough to deduce even that $L(0,\chi) \ne 0$. – Pound Sterling Feb 13 '17 at 21:44
• I think what I wrote was correct (as I had shown before that $L(0,\chi)$ is not congruent to $0$ mod $2$), but I agree that it was not well-written. I updated the proof and it should be on the arXiv tomorrow. – Lennart Meier Feb 20 '17 at 14:45


A lift always exists, even with coefficients in $\Z$. If suffices to consider the case when $p > 2$ is prime. (Added Note: there is a second elementary argument at the end of this post.)

Let $p$ be prime, and let $\chi$ be an odd character of $(\Z/p\Z)^{\times}$ of order $2^m$ for some $m$. This character is valued in the field $E = \Q(\zeta_{2^m})$, and is unique up to Galois conjugacy (this follows from the oddness assumption, which implies that $2^m$ is the largest power of $2$ dividing $p-1$). The fixed field of $\chi$ considered as a Galois character is the unique field $K \subset \Q(\zeta_p)$ which has degree $2^m$; it is totally complex.

Claim: The $L$ value $L(\chi,0)$ has $2$-adic valuation at most $(2^{m-1} - 1)/2^{m-1} < 1$.

In particular, $L(\chi,0)/2$ has negative valuation, and, as in the previous answer, this shows that the Eisenstein series $E_{1,\chi}$, suitably normalized, provides a lift of Hasse to $\Z[\zeta_{2^m}]$. Hence there is a multiple of $E_{1,\chi}$ with coefficients in $\Z[\zeta_{2^m}]$ which is congruent modulo $\pi = 1 - \zeta_{2^m}$ to $1$. Write this form as follows:

$$\sum_{k=1}^{2^{m-1} - 1} \zeta^k_{2^m} f_k \in \Z[\zeta_{2^m}][[q]],$$

where $f_k \in \Z[[q]]$. Since the Galois group preserves $M_1(\Gamma_1(p))$, all the forms $f_k$ lies in $_1(\Gamma_1(p))$, and thus so does the form

$$\sum_{k=0}^{2^{m-1} - 1} f_k \in \Z[[q]].$$

Yet this form is congruent modulo $\pi = 1 - \zeta_{2^m}$ to the multiple of $E_{1,\chi}$ above, which in turn is congruent modulo $1 - \zeta_{2^m}$ to $1$. Hence it yields an integral lift of the Hasse invariant.

Proof: It suffices to show that the norm of $L(\chi,0)$ to $\Q$ has valuation at most $2^{m-1} - 1$. We may identify this norm with the product

$$\prod_{\chi \text{ odd}} L(\chi,0)$$

Yet this is equal to the limit of $\zeta_K(s)/\zeta_{K^+}(s)$ at $s = 0$ where $K^{+}$ is the maximal totally real subfield of $K$. By the class number formula, this limit thus evaluates to:

$$\frac{w(K^{+})}{w(K)} \cdot \frac{h(K)}{h(K^+)} \cdot \frac{R(K)}{R(K^{+})}.$$

Now $w(K^{+}) = 2$, and $w(K) = 2$ or $2p$ depending on whether $p$ is a Fermat prime or not. Hence this factor can be ignored when computing the $2$-adic valuation. I claim that the $2$-parts of $h(K^{+})$ and $h(K)$ are both $1$. To see this, let $H$ be the $2$-part of the Hilbert class field of $K^{+}$ or $K$. Then $H/\Q$ is Galois and of $2$-power order. By class field theory (over $\Q$), $\Gal(H/\Q)^{\mathrm{ab}}$ is cyclic, because it is ramified only at $p$ and $p \ne 2$. Yet this implies that $\Gal(H/\Q)$ is cyclic, because it is a $2$-group with cyclic abelianization. But then $H$ is abelian, and so from the theory of cyclotomic fields we easily see that $H = K^{+}$ or $K$. Hence the factor from the class number is also prime to $2$. Finally, the ratio of regulators is equal to the index of the units $[U_K:U_{K^{+}}]$. Let $u \in U_K$. If $c \in \Gal(K/\Q)$ is complex conjugation, then a standard argument shows that $cu/u$ is a root of unity. Now $w(K) = |\mu_K|$ has either order $2$ or $2p$. In particular, after replacing $u$ by an odd power, we deduce that $cu = \pm u$. It follows that $u^2 \in K^{+}$, and that $V_K:=U_K/U_{K^{+}} \otimes \Z_2$ is an abelian group of exponent $2$. The power of $2$ dividing the index is $|V_K|$, so it suffices to show that the dimension of $V_K$ over $\F_2$ is at most $2^{m-1} - 1$. Since $-1 \in K^{+}$, there is a surjection:

$$U_K/(\pm 1,U^2_K) \rightarrow V_K.$$

By Dirichlet's unit theorem, we have

$$U_K \simeq \Z^{[K^{+}:\Q] - 1} \oplus \mu_K.$$

Since $\mu_K/{\pm 1} \otimes \Z_2$ is trivial, it follows that $\dim_{\F_2} V_K \le 2^{m-1} - 1$. This proves the claim. In fact, using circular units, one can show that $\dim_{\F_2} V_K = 2^{m-1} - 1$, so the upper bound is actually an equality.

In the special case when $p \equiv 3 \mod 4$, then $2^{m-1} - 1 = 0$ so $L(\chi,0)$ is a unit; this is the case considered by wrigley.

Remark: This argument is probably more complicated than necessary. There may well be an elementary way to prove directly that

$$\sum_{n=1}^{p-1} n \chi(n)$$

is not divisible by $2$ directly for the odd character $\chi$ of $2$-power order, but since the argument above was immediately apparent, I didn't try to look for such an argument.

Oh, actually, now that I make this comment, I see it is actually obvious. The character $\chi$ is valued in $E = \Q(\zeta_{2^{m}})$ which is a field of degree $2^{m-1}$, and the ring of integers has a basis $\zeta^i$ for $i = 0$ to $2^{m-1} - 1$. Moreover, $\chi$ exactly takes values in this basis. So it simply suffices to sum over the integers $n$ with $\chi(n) = 1$ or $\chi(n) = -1$ and show that this sum is not divisible by $2$, because then $L(\chi,0)$ will not be divisible by $2$. Yet these integers come in pairs $(n,p-n)$ whose difference is odd (so $n \chi(n) + (p-n) \chi(p-n) = \pm n - \pm (p-n) \equiv 1 \mod 2$), so it suffices to show that the total number of integers with $\chi(n) \in \{\pm 1\}$ is divisible by $2$ and not by $4$. Yet these integers are precisely the kernel of $\chi^2$, whose order has this property by definition. Oops! Well, I guess I may as well leave the class field theory argument here as well.

Added: I guess this last argument also shows that the coefficient of $\zeta^i$ is odd for any $i$, and hence

$$L(\chi,0) \equiv 1 + \zeta + \zeta^2 + \ldots + \zeta^{2^{m-1} -1} \mod 2.$$

Yet then

$$(1 - \zeta) L(\chi,0) \equiv 1 - \zeta^{2^{m-1}} = 1 + 1 \equiv 0 \mod 2(1 - \zeta),$$

and hence one sees directly that the valuation of $L(\chi,0)$ is $(2^{m-1} - 1)/2^{m-1}$.

• Very nice, thanks. Just one very small error: In the paragraph before the "Added:" you claim that the character takes values in the basis of $\zeta^i$ for $i=0, 2^{m-1}-1$, but it can also hit their negatives. – Lennart Meier Jan 18 '16 at 16:09
• I plan to use this result in a paper of mine. If you want to mentioned by real name in the referencing, please contact me. – Lennart Meier Jan 28 '16 at 12:32

This isn't a complete answer by any means but I've not got enough rep to comment. Using the well-known formula for weight 1 Eisenstein series one can certainly expand on the list that Katz gives. For example let $p$ be a prime congruent to $3$ mod $4$, so the quadratic character $\chi$ mod $p$ is odd, and let's consider the associated Eisenstein series (normalised as $c+q+(1+\chi(2))q^2+\cdots$). The constant term $c=L(\chi,0)/2=\frac{-1}{2p}\sum_{n=1}^{p-1}n\chi(n)$. Because $\chi(n)=+1$ or $-1$ we see that the sum is congruent to $\sum_{n=1}^{p-1}n=p(p-1)/2$ which is odd. In particular $c$ has 2-adic valuation $-1$. So twice this Eisenstein series lifts the mod 2 Hasse invariant, so one can add to Katz' list all positive odd integers which are multiples of a prime which is congruent to 3 mod 4.

• PS my formula for $c$ is valid for any odd character so one can finish Katz' proof by observing that for $p=5$ the two characters of conductor 5 and order 4 give $c=(3\pm\sqrt{-1})/10$ and the coeficients of $q^n$ for $n>0$ are all congruent mod the prime above 2 (because the characters are both trivial mod the prime above 2). So their sum has constant term 3/5 and all other coefficients even, so lifts Hasse mod 2. I don't immediately see how to generalise this to other primes 1 mod 4 though -- although I guess it could be possible. Look at characters of highest 2-power conductor? – wrigley Jan 15 '16 at 20:46
• PPS your link to Katz' paper is currently broken but the point about 3,5,7,9,11 is that every form, not just Hasse, lifts. For Hasse the problem is much easier in some sense because one can try and do it explicitly. – wrigley Jan 15 '16 at 21:12