## More questions involving characteristic 2 theta series identities

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In my answer to my earlier question, "Existence of certain identities involving characteristic 2 thetas", I established some curious identities when the thetas have prime "level" congruent to 1 mod 4 or to 3 mod 8. This question concerns the case when the level is 7 mod 8.

I reprise notation from earlier questions. l is an odd prime and [j] is the sum of the x^(n^2), where n runs over the integers congruent to j mod l; we view the "theta series" [j] as elements of Z/2[[x]]. F is the power series x+x^9+x^25+x^49+x^81..., G=F(x^l) and H=G(x^l). My identities involve G,H and the various [j].

There is evidently a unique C in Z/2[[x]], having constant term 0, with C^2+C=G+H. I showed that when l is 1 mod 4 or 3 mod 8 (or when l=7), then C can be written explicitly as a polynomial in the [j]. Here is what the computer suggests when l=7 mod 8 and is < 50. First some notation. If (r,s,t) is a triple of integers, we define C(r,s,t) to be the sum of the power series [rj][sj][tj] where j runs from 1 to (l-1)/2. Define C(r,s,t,u) similarly. (When l is 3 mod 8, I showed that C is C(1,1,t) where t^2 is congruent to -2 mod l).

(1) When l=7, I can show that C=C(1,1,1,2)+C(1,2,3)

(2) When l=23 I think that C=C(3,3,1,2)+C(1,3,6)

(3) When l=31 I think that C=C(3,3,2,3)+C(2,3,7) (In my original post I wrote C(2,5,8), but C(2,3,7)=C(2,5,8))

(4) When l=47 I think that C=C(3,3,2,5)+C(2,3,9)

(Note that the sum of the squares of 3,3,2 and 5 is 47, etc.)

QUESTION 1: Can one establish the truth of (2),(3) and (4)? Kevin Buzzard explained to me that it's enough to show that the power series expansions agree up to a certain exponent, but I'm not sure what that exponent is, and I doubt that I have the computer power.

QUESTION 2: Are there identities like those above for l>50? And if so, what are these identities explicitly?

EDIT: Let V be the space spanned by the C(r1,r2,r3,r4) with r1=r2 and l dividing the sum of the squares of r1,r2,r3 and r4, together with the C(s1,s2,s3) with l dividing the sum of the squares of s1,s2 and s3. When l=7 mod 16 I can use Jacobi's 4-square theorem to show that C is in V. It's then possible to prove identities like those of (2) above by exploiting the geometry of of Spec R where R is the subring of Z/2[[x]] generated by the theta series [j].

-----One can show that an element of V has at most l(l-1)(l+1)/6 poles, counted with multiplicity, on the obvious projective completion of this curve. So if it has a zero of large enough order at the origin, it vanishes. I applied this technique for various l congruent to 7 mod 16; the results boggled my mind. It's only necessary to use 2 terms in the power series expansion of each theta series. When l=23, I got (2) above.

When l=71, I found that C=C(3,3,2,7)+C(5,6,9)

When l=103, I got 5 different expressions for C! Explicitly:

a----C(3,3,6,7)+C(2,9,11)

b----C(7,7,1,2)+C(5,9,10)

c----C(5,5,2,7)+C(1,3,14)

d----C(3,3,2,9)+C(6,7,11)

e----C(1,1,1,10)+C(1,6,13)

It seems possible to me that in general, for l=7 mod 8, one gets h/4 formulae of this sort where h is the class-number of Q(Root(-2l)). I've discussed the case l=31 in the comment to ARupinski. When l=47, I can show that C(3,3,2,5)+C(2,3,9)=C(1,1,3,6)+C(3,6,7). So if (4) above holds, there's a second formula for C in this case, just as in the case l=31. But I can't prove that C is in V when l=15 mod 16.

UPDATE__ Suppose l is 7 mod 8; consider the vectors W in Z^3 with(W,W)=2l. There is a group of order 48 operating on the set of such W by permutation and sign change of co-ordinates; the group operates without fixed points. So if there are 12h such W there are h/4 orbits under the group action.

----Ira Gessel's calculations, carried out for l<1500, indicate that there is an involution, O-->O' on the set of orbits, which has the following property. Let O be any of the (h/4) orbits and (r1,r2,r3) be a representative of O with r1 even (so that r2 and r3 are odd). Then if (s1,s2,s3) is a representative of O', we have the explicit identity C((r1)/2,(r1)/2,(r2+r3)/2,(r2-r3)/2)+C(s1,s2,s3)=C.

----But to know what these conjectured(but true beyond possibility of doubt) equations are for l>1500, we need to describe the involution. Franz Lemmermeyer suggested that the involution comes from an involution on a set of equivalence classes of quadratic forms of discriminant -8l. This is surely the case; I'll explain what the involution on the forms is, and how to transfer it to the orbits.

----Consider positive quadratic forms rx^2+2sxy+ty^2 with s^2-rt=-2l. Gauss showed that these fall in exactly h equivalence classes under the action of SL_2(Z), where 12h is the number of W with (W,W)=2l; we'll be interested in GL_2 equivalence however. Since rt=2l+s^2, we find that mod 16, rt is 2,7,14 or 15. This can be used to show that one of the following possibilities must occur:

a.--- Every non-zero n represented by the form is the product of an integer that is 1 or 7 mod 8 by a power of 2.

b.---Every non-zero n represented by the form is the product of an integer that is 3 or 5 mod 8 by a power of 2.

----In the first case we say that the form is in the principal genus, while in the second that it is in the non-principal genus. There are (h/4) GL_2 classes in the non-principal genus. Furthermore there is an involution on this set of classes taking the class of rx^2+2sxy+2ty^2 to the class of 2rx^2+2sxy+ty^2. I'll call this involution "composition with 2x^2+ly^2".

----I now describe a map from the set of (h/4) orbits to the set of (h/4) classes. The map can be shown to be onto, and so is bijective. When we transfer composition with 2x^2+ly^2 to the set of orbits, we get our desired involution; one which is in complete accord with Gessel's calculations. Suppose (W,W)=2l. Let W# consist of all elements of Z^3 orthogonal to W. We attach to W the class of the form (xU+yV,xU+yV), where U and V are a basis of W#. This class is evidently independent of the choice of basis; one can show that it consists of forms of discriminant -8l and lies in the non-principal genus. This gives the desired map from orbits to classes of forms; as I've indicated it is bijective.

EXAMPLE____Take l=1567, and W=(3,25,50) so that (W,W)=2l. Let O be the orbit of W. I'll calculate O', and write down the conjectured equations coming from O and O'. A basis for W# consists of U=(0,2,-1) and V=(25,1,-2). Then (U,U)=5, (U,V)=4, (V,V)=630, and a form attached to O is 5x^2+8xy+630y^2. Composition with 2x^2+1567y^2 takes this to 10x^2+8xy+315y^2. So we seek U' and V' with (U',U')=10, (U',V')=4, and (V',V')=315. Take U'=(3,1,0). A little experimenting, writing 315 as a sum of 3 squares, shows that we should take V'=(5,-11,13). Then W' which is orthogonal to U' and V' can be taken to be their vector product (13,-39,-38). So O' is the orbit of (13,38,39). And one of our predicted expressions for C is C(25,25,11,14)+C(13,38,39), while another is C(19,19,13,26)+C(3,25,50).

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Extending your comment that $3^2+3^2+2^2+5^2 = 47$ etc., is there any reason you can see as to why the sum of the squares of the arguments of the second function is always twice the I-value except in case (3) where it is triple the I-value? – ARupinski Sep 5 2011 at 5:06
In that case, it also could be twice the l-value; multiplication by 12 takes(2,5,8) to(-7,-2,3). So C(2,5,8)=C(2,3,7) when l=31. But I have no idea what happens in general. When l is 7 mod 16, I think I see  how to write C as a sum of various C(r1,r2,r3) and various C(s1,s2,s3,s4). But my argument fails when l is 15 mod 16. – paul Monsky Sep 5 2011 at 12:18
@ARupinski---Quadratic forms arguments(Jacobi's 4 square theorem is the key!) will give (2) above. When l=31, similar arguments show that C(3,3,2,3)+C(2,3,7) is equal to C(1,1,2,5)+C(1,5,6). But I can't use these arguments to prove (3); the trouble is that 31=15 mod 16. – paul Monsky Sep 6 2011 at 23:54

## 1 Answer

I can now, with less computer calculation than I'd feared, answer Question 1. (I'll say more about Question 2 later).

Lemma:__ Let V be the vector space over Z/2 spanned by the C(r1,r2,r3) and the C(s1,s2,s3,s4). If an element of V has its power series expansion divisible by x^(l^2), it is 0.

To see this, let K be an algebraic closure of Z/2, S' be the subring of K[[x]] generated over K by the [j] and L be the field of fractions of S'. I can show that L/K is the function field of a curve, that there are exactly l(l-1)(l+1)/24 valuation rings in L/K that don't contain S', and that each of [1],...,[l-1] has a simple pole at each of them. So an element of V has at most l(l-1)(l+1)/6 poles in L/K, counted with multiplicity.

Also, the localization of S' at the maximal ideal generated by [1],...,[l-1] is dominated by exactly (l-1)/2 valuation rings in L/K. I can show that for each r prime to l there is an automorphism of L/K taking [j] to [rj] for each j, and that these automorphisms act transitively on this set of valuation rings. Now the elements of V are fixed by these automorphisms. So an element of V whose power series expansion is divisible by x^(l^2) has zeros of order at least l^2 at each of these valuation rings. Since (l^2)(l-1)/2 > l(l-1)(l+1)/6, such an element must vanish.

Suppose now that l=23. Since l=7 mod 16, my answer to my question "Existence of certain identities..." shows that C is in V. The lemma then shows that to prove 2. it's enough to show that C(3,3,1,2)+C(1,3,6)+C is divisible by x^529 in Z/2[[x]]. Now C(3,3,1,2)+C(1,3,6) is a sum of monomials in the [j]. Replacing each [j] by the sum of the first 2 terms in its power series expansion only modifies the sum by something divisible by x^529, and we're reduced to an easy computer calculation.

Establishing 3. and 4. is harder since we don't know in advance that C is in V. I'll use:

Theorem____Suppose there is an element R of V such that R+C is divisible by x^(d+1) where d=l(l+1)(l+1). Then R=C.

To see this, recall that F=x+x^9+x^25+..., that G=F(x^l), that H=G(x^l) and that C^2+C=G+H. Now there is a symmetric degree l+1 2-variable polynomial P over Z/2 with P(F,G)=0; furthermore P(z,H) is monic in z of degree l+1. This is discussed in my question "What's known about the mod 2 reduction...". Suppose a^l=1. Replacing x by ax^l in the identity P(F,G)=0, we get a root of P(z,H)=0 of the form ax^l+... This gives us l distinct roots, with G among them. By symmetry P(H,G)=0. So H(x^l) is still another root, and P(z,H) factors into linear factors over K[[x]].

Now R^2+R+G+H=(R+C)^2+ (R+C), and so is divisible by x^(d+1). Since P(G,H)=0, P(R^2+R+H,H) is divisible by x^(d+1). Also R^2+R has poles of order at most 8 at each valuation ring in L/K that doesn't contain S', while H has poles of order at most 12. It follows that P(R^2+R+H,H) has at most (l+1)l(l-1)(l+1)/2 =d(l-1)/2 poles, counted with multiplicity, in L/K. Arguing as in the proof of the lemma we see that it has more than d(l-1)/2 zeros, counted with multiplicity. So it vanishes, and R^2+R+H is a root of P(z,H)=0. Examining the roots of this equation we see that R^2+R+H can only be G. So R^2+R=G+H, and R=C.

Suppose now that l=31. To prove 3. we now see that it suffices to show that C(3,3,2,3)+C(2,3,7)+C is divisible by x^(186^2), since 186^2 >(31)(32)(32). This is carried out as in the case l=23, but now we have to use the first 12 terms in the power series expansion of each [j] rather than just the first 2. The treatment of 4. is similar, but now we show divisibility by x^(329^2) using the first 14 terms in the power series expansion of each [j].

EDIT___(UPDATE ON QUESTION 2)

Ira Gessel has now carried out computer calculations for all l<1500; here's a summary of his remarkable results. Consider the triples (r1,r2,r3) where the squares of r1,r1,r2 and r3 sum to l. To avoid duplicates, normalize each such triple so that r1,r2 and r3 are positive with r2>r3. Ira finds that for each such triple there is a unique second (normalized) triple (s1,s2,s3) such that the power series C(r1,r1,r2,r3)+C(2s1,s2+s3,s2-s3)+C is divisible by x^(l^2). Furthermore (r1,r2,r3)-->(s1,s2,s3) is an involution on the set of normalized triples with at most 1 fixed point.

When l=7 mod 16, the argument I gave above when l=23 then shows that for each normalized triple (r1,r2,r3) and the corresponding (s1,s2,s3) we have the identity C=C(r1,r1,r2,r3)+C(2s1,s2+s3,s2-s3); note that the squares of 2s1, s2+s3, and s2-s3 sum to 2l. So, for example, when l=1447 we get 17 distinct formulae for C.

When l=15 mod 16, it seems certain that once again C(r1,r1,r2,r3) and C(s1,s2,s3) sum to C. This could be proved by extending Ira's calculations to prove divisibility by x^(d+1) where d=l(l+1)(l+1), as in my treatment of l=31 and l=47. But I think this extension is unnecessary, and that one may deduce the identities simply from the divisibility by x^(l^2) established by Ira. If my idea for demonstrating this works out I'll post it as a comment.

But great mysteries remain. Why should this all be true? And can one describe the mysterious involution (r1,r2,r3)-->(s1,s2,s3) explicitly? By the way, it's known that the number of normalized triples is odd or even according as n is 7 or 15 mod 16. The proof of this goes back to Hasse; one shows that the number of triples is h/4 where h is the class number of Q(Root(-2l)) and uses results of Gauss on representations by sums of 3 squares, together with some genus theory for binary quadratic forms.

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In the theorem I state above I can now show that d can be replaced by l(l+1). As a result, to verify the analogues of (3) and (4) for any l that is 15 mod 16, it's enough to use 2 terms in the power series expansion of each [j]. And so Ira's calculations which led to h/4 identities for each l<1500 and congruent to 7 mod 16, extended a minute bit further, will do the same for l that are 15 mod 16. Here's a proof of my assertion. If x^(d+1) divides R+C, it divides (R+C)(R+C+1)=R^2+R+G+H. Now my (self-accepted) answer to "Existence of certain identities.." shows that (to be continued) – paul Monsky Sep 28 2011 at 14:58
R^2+R+G+H lies in S', is stabilized by the automorphisms [j]-->[rj] of S', and has poles of order no more than 12     at each of          the l(l-1)(l+1)/24 valuation rings in L/K that do not contain S'. If it's divisible by x^(d+1) it has at least (d+1)(l-1)/2 zeros counted with multiplicity. So it has more zeros than poles, and must vanish. It follows that R=C. – paul Monsky Sep 28 2011 at 15:06
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