Transcendence of some Hilbert-Kunz multiplicities (modulo a conjecture)

TL;DR

This paper explores the potential transcendence of certain Hilbert-Kunz multiplicities in characteristic 2, contingent on a conjecture, linking these multiplicities to a known transcendental number.

Contribution

It demonstrates that assuming a specific conjecture, transcendental Hilbert-Kunz multiplicities can exist, connecting them to a known transcendental series.

Findings

Assuming the conjecture, transcendental multiplicities are possible.

A specific transcendental series is expressed as a Q-linear combination of Hilbert-Kunz multiplicities.

The work links algebraic geometry with transcendental number theory.

Abstract

Suppose that h in F[x,y,z], char F=2, defines a nodal cubic. In earlier papers we made a precise conjecture as to the Hilbert-Kunz functions attached to the powers of h. Assuming this conjecture we showed that a class of characteristic 2 hypersurfaces has algebraic but not necessarily rational Hilbert-Kunz multiplicities. We now show that if the conjecture holds, then transcendental multiplicities exist, and in particular that the number \sum\binom{2n}{n}^{2}/(65,536)^{n}, proved transcendental by Schneider, is a Q-linear combination of Hilbert-Kunz multiplicities of characteristic 2 hypersurfaces.

Full text

1 The power series $\theta_{g}$

We recall some notation and results from [3] that will be used throughout. $X$ is the vector space of functions $I\rightarrow Q$ where $I=[0,1]\cap Z[\frac{1}{2}]$. $F$ is a field of characteristic 2, while $q$ is always a power, $2^{n}$, of 2. For each $f\neq 0$ in the maximal ideal of $F[[x_{1},\ldots,x_{r}]]$ there is an element $\phi_{f}$ of $X$ whose value at $\frac{i}{q}$ is $q^{-r}e_{n}(f^{i})$ where $n\rightarrow e_{n}$ is the Hilbert-Kunz function. There is a symmetric bilinear product  # , $X\times X\rightarrow X$ with the following property. If $f$ and $g$ are in $F[[u_{1},\ldots,u_{r}]]$ and $F[[v_{1},\ldots,v_{s}]]$, and $h=f+g$, then \phi_{h}=\phi_{f}\mbox{,\#\,}\phi_{g}. $\Gamma$ is the Grothendieck group of isomorphism classes of finitely generated $F[T]$-modules annihilated by a power of $T$; see the material following Theorem 1.5 of [3] for the $Z$-basis $\lambda_{0},\lambda_{1},\ldots$ of $\Gamma$ and the multiplication on $\Gamma$. For $n\geq 0$ and $\alpha$ in $X$, $\mathcal{L}_{n}(\alpha)$ is the element $\sum_{0}^{q-1}\left(\alpha\left(\frac{i+1}{q}\right)-\alpha\left(\frac{i}{q}\right)\right)(-)^{i}\lambda_{i}$ of $\Gamma_{Q}=\Gamma\underset{Z}{\otimes}Q$. The product in $\Gamma_{Q}$ of $\mathcal{L}_{n}(\alpha)$ and $\mathcal{L}_{n}(\beta)$ is \mathcal{L}_{n}(\alpha\mbox{,\#\,}\beta).

Definition 1.1**.**

$\eta:\Gamma_{Q}\rightarrow Q$* is the linear map taking each $\lambda_{i}$ to $1$.*

Lemma 1.2**.**

If \gamma=\alpha\mbox{,\#\,}\beta then for each $n$, $\eta(\mathcal{L}_{n}(\gamma))=\eta(\mathcal{L}_{n}(\alpha))\cdot\eta(\mathcal{L}_{n}(\beta))$.

Proof.

If $k$ is the Nim-sum of $i$ and $j$ then $\lambda_{i}\cdot\lambda_{j}=\lambda_{k}$. It follows that $\eta(\lambda_{i}\lambda_{j})$ and $\eta(\lambda_{i})\eta(\lambda_{j})$ are both $1$ and that $\eta:\Gamma_{Q}\rightarrow Q$ is multiplicative. Now use the fact that $\mathcal{L}_{n}(\gamma)=\mathcal{L}_{n}(\alpha)\mathcal{L}_{n}(\beta)$. ∎∎

Lemma 1.3**.**

If $n\geq 1$, \eta(\mathcal{L}_{n}(\alpha))=2^{2n}(\phi_{uv}\mbox{,\#\,}\alpha)\left(\frac{1}{2^{n}}\right)-2^{2n-1}(\phi_{uv}\mbox{,\#\,}\alpha)\left(\frac{1}{2^{n-1}}\right).

Proof.

Write $\mathcal{L}_{n}(\alpha)$ as $\sum_{0}^{q-1}(-)^{i}a_{i}\lambda_{i}$ where $q=2^{n}$. Now 2^{2n}(\phi_{uv}\mbox{,\#\,}\alpha)\left(\frac{1}{2^{n}}\right) is the co-efficient of $\lambda_{0}$ in 2^{2n}\mathcal{L}_{n}(\phi_{uv}\mbox{,\#\,}\alpha)=2^{2n}\mathcal{L}_{n}(\phi_{uv})\mathcal{L}_{n}(\alpha). Since $e_{n}(u^{i}v^{i})=q^{2}-(q-i)^{2}$, $2^{2n}\mathcal{L}_{n}(\phi_{uv})=\sum_{0}^{q-1}(-)^{i}(2q-1-2i)\lambda_{i}$. Using the formula for the product in $\Gamma$ we find that 2^{2n}(\phi_{uv}\mbox{,\#\,}\alpha)\left(\frac{1}{2^{n}}\right) is $\sum_{0}^{q-1}(2q-1-2i)a_{i}$.

Rewrite the above as $\sum_{0}^{\frac{q}{2}-1}(2q-1-4i)(a_{2i}+a_{2i+1})-2(a_{1}+a_{3}+\cdots+a_{q-1})$. Now $\mathcal{L}_{n-1}(\alpha)$ is easily seen to be $\sum_{0}^{\frac{q}{2}-1}(-)^{i}(a_{2i}+a_{2i+1})\lambda_{i}$. The last paragraph, with $n$ replaced by $n-1$, shows that 2^{2n-2}(\phi_{uv}\mbox{,\#\,}\alpha)\left(\frac{1}{2^{n-1}}\right)=\sum_{0}^{\frac{q}{2}-1}(q-1-2i)(a_{2i}+a_{2i+1}). It follows that the right-hand side of Lemma 1.3 is $\sum_{0}^{q-1}a_{i}-2(a_{1}+a_{3}+\cdots+a_{q-1})=\eta(\mathcal{L}_{n}(\alpha))$. ∎∎

Corollary 1.4**.**

Let $g$ be a power series in $r$ variables over $F$, and $G=uv+g$, viewed as a power series in $r+2$ variables. Then for $n\geq 1$, $e_{n}(G)-2^{r+1}e_{n-1}(G)=2^{rn}\eta(\mathcal{L}_{n}(\phi_{g}))$.

Proof.

Set $\alpha=\phi_{g}$, and multiply the identity of Lemma 1.3 by $2^{rn}$. Then since \phi_{uv}\mbox{,\#\,}\alpha=\phi_{uv}\mbox{,\#\,}\phi_{g}=\phi_{G}, the right-hand side becomes $e_{n}(G)-\frac{1}{2}\cdot 2^{r+2}\cdot e_{n-1}(G)$, giving the corollary. ∎∎

Definition 1.5**.**

Let $r$, $g$ and $G$ be as in Corollary 1.4. Then $\theta_{g}$ in $Z[[w]]$ is $(1-2^{r+1}w)\sum{e_{n}}(G)w^{n}$.

Note that the power series $\theta_{g}$ converges in $|w|\leq\frac{1}{2^{r}}$, and that $\theta_{g}\left(\frac{1}{2^{r+1}}\right)$ is just the Hilbert-Kunz multiplicity of $G$. Furthermore the co-efficient of $w^{n}$ in $\theta_{g}$ is $e_{n}(G)-2^{r+1}e_{n-1}(G)$.

Suppose now that we have finitely many power series $g_{i}$ over $F$, that $g_{i}$ is in $r_{i}$ variables, and that the variables corresponding to distinct $g_{i}$ are pairwise disjoint.

Theorem 1.6**.**

$\theta_{\Sigma g_{i}}$* is the Hadamard product of the $\theta_{g_{i}}$.*

Proof.

Let $G_{i}=uv+g_{i}$ and $M=uv+\sum g_{i}$. Applying Corollary 1.4 to each $g_{i}$ and to $\sum g_{i}$ and using Lemma 1.2 repeatedly we find that $e_{n}(M)-2^{1+\Sigma r_{i}}e_{n-1}(M)$ is the product of the various $e_{n}(G_{i})-2^{1+r_{i}}e_{n-1}(G_{i})$. In other words the co-efficient of $w^{n}$ in $\theta_{\Sigma g_{i}}$ is the product of the co-efficients of $w^{n}$ in the various $\theta_{g_{i}}$. ∎∎

Corollary 1.7**.**

The Hilbert-Kunz multiplicity of $uv+\sum g_{i}$ is the value at $w=\frac{1}{2^{1+\Sigma r_{i}}}$ of the Hadamard product of the $\theta_{g_{i}}$.

Corollary 1.7 is the key to this note. It allows us to pass from the algebraic realm (the $\theta$ attached to $x^{3}+y^{3}+xyz$ is conjectured by us to be algebraic of degree 2 over $Q(w)$) to the transcendental realm by making use of the Hadamard product. We next give some easy results about $\theta$.

Theorem 1.8**.**

Let $g$ be an $r$-variable power series and $h=g^{2}$. Then $\theta_{h}=1+2^{r}w\cdot\theta_{g}$.

Proof.

If $i\leq q$, $\phi_{h}\left(\frac{i}{2q}\right)=\phi_{g}\left(\frac{2i}{2q}\right)$, while if $i\geq q$, $\phi_{h}\left(\frac{i}{2q}\right)=1$. It follows that $\mathcal{L}_{n+1}(\phi_{h})=\mathcal{L}_{n}(\phi_{g})$ for all $n$. Applying $\eta$ and multiplying by $2^{r(n+1)}$ we find from Corollary 1.4 that the left hand side of this equation becomes $e_{n+1}(H)-2^{r+1}e_{n}(H)$ where $H=uv+h$. And the right-hand side becomes $2^{r}(e_{n}(G)-2^{r+1}e_{n-1}(G))$ where $G=uv+g$. We have shown that the co-efficients of $w^{n+1}$ in $\theta_{h}$ and in $2^{r}w\cdot\theta_{g}$ are equal, giving the theorem. ∎∎

When $r=1$ and $g$ is a power of the variable it’s easy to calculate $\theta_{g}$. In particular we find:

Lemma 1.9**.**

When $r=1$,

1. (a)

$\theta_{x^{5}}-\theta_{x^{3}}=2\sum_{1}^{\infty}w^{2n}$** 2. (b)

$2\theta_{x^{3}}-\theta_{x^{5}}=1+2\sum_{0}^{\infty}w^{2n+1}$**

Now take $r=3$, fix $f$ in $F[[x,y,z]]$, and let $\theta_{f}=\sum a_{n}w^{n}$. We shall use Corollary 1.7 to show that certain infinite sums involving the $a_{n}$ are $Q$-linear combinations of Hilbert-Kunz multiplicities.

Theorem 1.10**.**

$\sum\frac{a_{2n}^{2}}{2^{16n}}$, $\sum\frac{a_{2n+1}^{2}}{2^{16n}}$, $\sum\frac{a_{2n}a_{2n+1}}{2^{16n}}$, $\sum\frac{a_{2n+1}a_{2n+2}}{2^{16n}}$ and $\sum\frac{a_{2n}a_{2n+2}}{2^{16n}}$ are all $Q$-linear combinations of Hilbert-Kunz multiplicities.

Proof.

Using Theorem 1.8 we find that the co-efficients of $w^{2n+2}$ in $\theta_{f}$ and $\theta_{f^{2}}$ are $a_{2n+2}$ and $8a_{2n+1}$. So the Hadamard product of $\theta_{f}$, $\theta_{f^{2}}$ and $\theta_{x^{5}}-\theta_{x^{3}}=2\sum_{0}^{\infty}w^{2n+2}$ is $16\sum_{0}^{\infty}a_{2n+1}a_{2n+2}w^{2n+2}$. Since $1+(3+3+1)=8$, Corollary 1.7 with $g_{1}=f$, $g_{2}=f(X,Y,Z)^{2}$ and $g_{3}=T^{3}$ or $T^{5}$ shows us that this Hadamard product, evaluated at $\frac{1}{2^{8}}$, is a difference of two Hilbert-Kunz multiplicities. Also, the co-efficients of $w^{2n+1}$ in $\theta_{f}$ and $\theta_{f^{2}}$ are $a_{2n+1}$ and $8a_{2n}$. So the Hadamard product of $\theta_{f}$, $\theta_{f^{2}}$ and $2\theta_{x^{3}}-\theta_{x^{5}}=1+2\sum w^{2n+1}$ is $1+16\sum_{0}^{\infty}a_{2n}a_{2n+1}w^{2n+1}$. As above we see that this Hadamard product, evaluated at $\frac{1}{2^{8}}$, is a $Z$-linear combination of two Hilbert-Kunz multiplicities. We have proved the third and fourth of the assertions of the theorem. To prove the first and the second we argue similarly with $g_{1}=f$ and $g_{2}=f(X,Y,Z)$. For the final assertion we take $g_{1}=f$ and $g_{2}=f(X,Y,Z)^{4}$, using Theorem 1.8 to calculate $\theta_{g_{2}}$. ∎∎

2 Transcendence results (modulo a conjecture)

Now let $r=3$, and $f=x^{3}+y^{3}+xyz$ be the defining equation of a nodal cubic. In Definitions 2.1 and 2.2 of [3] we constructed elements $\phi_{0},\phi_{1},\ldots$ of $X$; we further conjectured that $\phi_{f}=t+\phi_{0}$. (This is an alternative form of a conjecture we made earlier in [2].)

Lemma 2.1**.**

Let $A_{n}$ be the binomial co-efficient $\binom{2n}{n}$, so that $\sum A_{n}w^{2n}$ converges to $(1-4w^{2})^{-\frac{1}{2}}$ in the disc $|w|<\frac{1}{2}$. Then if the conjecture of [3] holds, $(1-6w+8w^{2})\theta_{f}=(1-2w)+(2w-8w^{2}-24w^{3})\sum A_{n}w^{2n}$.

Proof.

This is shown in the paragraph following Corollary 2.7 of [2]. Alternatively it is an easy consequence of Lemma 2.6 of [3], since in the notation of that lemma, $\sum e_{n}(uv+f)w^{n}$ is, granting the conjecture, equal to $\sum(2^{-n}+E_{1}(2^{-n}))(32w)^{n}$. ∎∎

Theorem 2.2**.**

If the conjecture of [3] holds, then the value of $\sum A_{n}^{2}\lambda^{2n}$ at $\lambda=\frac{1}{2^{8}}$ is a $Q$-linear combination of Hilbert-Kunz multiplicities.

Proof.

Suppose $\theta_{f}=\sum a_{n}w^{n}$. Comparing co-efficients of $w^{2n+2}$ in the identity of Lemma 2.1 we see that $a_{2n+2}-6a_{2n+1}+8a_{2n}=-8A_{n}$. So $64\sum A_{n}^{2}\cdot\left(\frac{1}{2^{16n}}\right)=\sum(a_{2n+2}-6a_{2n+1}+8a_{2n})^{2}\cdot\left(\frac{1}{2^{16n}}\right)$. Expanding and using Theorem 1.10 we get the result. ∎∎

It only remains to show that $\sum A_{n}^{2}\cdot\left(\frac{1}{2^{16n}}\right)$ is transcendental. Results of this sort about special values of hypergeometric functions were first shown by Schneider [4]; we’ll sketch a proof.

Lemma 2.3**.**

(Euler)* If $0\leq\lambda<\frac{1}{4}$, $\int_{-1}^{1}\frac{\mathrm{d}w}{\sqrt{(1-16\lambda^{2}w^{2})(1-w^{2})}}=\pi\sum A_{n}^{2}\lambda^{2n}$.*

Proof.

$(1-16\lambda^{2}w^{2})^{\frac{1}{2}}=\sum(4\lambda^{2})^{n}A_{n}w^{2n}$. So our integral is $\sum(4\lambda^{2})^{n}A_{n}\cdot\int_{-1}^{1}\frac{w^{2n}\mathrm{d}w}{\sqrt{1-w^{2}}}$. But as one learns in every introductory calculus course, $\int_{-1}^{1}\frac{w^{2n}\mathrm{d}w}{\sqrt{1-w^{2}}}=\frac{\pi A_{n}}{4^{n}}$. ∎∎

Now suppose that $\lambda$ in $(0,\frac{1}{4})$ is rational, and consider the affine curve $y^{2}=(1-16\lambda^{2}x^{2})(1-x^{2})$ defined over $Q$. The real locus of this curve has a single bounded component. This component is a simple closed curve containing $(-1,0)$ and $(1,0)$. $\frac{\mathrm{d}x}{y}$ is a $1$-form on our affine curve. When we integrate $\frac{\mathrm{d}x}{y}$ over the bounded component (with the clockwise orientation) the integrals over the pieces in the upper and lower half-planes are equal, and each is $\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{(1-16\lambda^{2}x^{2})(1-x^{2})}}$. So by Lemma 2.3 the integral of $\frac{\mathrm{d}x}{y}$ over the bounded component is $2\pi\sum A_{n}^{2}\lambda^{2n}$.

Lemma 2.4**.**

Suppose $\lambda$ in $(0,\frac{1}{4})$ is rational. Then there is an elliptic curve $E$ defined over $Q$, a holomorphic $1$-form on $E$ rational over $Q$ and a $1$-cycle on $E$, such that the integral of the form over the $1$-cycle is $2\pi\sum A_{n}^{2}\lambda^{2n}$.

Proof.

Projectify the affine curve $y^{2}=(1-16\lambda^{2}x^{2})(1-x^{2})$, and let $E$ be a desingularization over $Q$ of the resulting complex projective plane quartic. The $1$-form is the pull-back of $\frac{\mathrm{d}x}{y}$, and the $1$-cycle corresponds to the inverse image in $E$ of the bounded component of the real locus of the affine curve $y^{2}=(1-16\lambda^{2}x^{2})(1-x^{2})$. ∎∎

Theorem 2.5**.**

(Schneider)* Let $E$ be an elliptic curve defined over $Q$ and $\omega\neq 0$ be the integral over a $1$-cycle on $E$ of a holomorphic $Q$-rational $1$-form. The $\frac{\omega}{\pi}$ is transcendental.*

Proof.

On page 63 of [4], Schneider deduces this from his Theorem 13. Lang later put Theorem 13 into a more general setting. I’ll indicate how Theorem 2.5 above follows from this “Schneider-Lang theorem”.

We may assume that the curve is $y^{2}=4x^{3}-g_{2}x-g_{3}$ with $g_{2}$ and $g_{3}$ rational, and that the $1$-form is $\frac{\mathrm{d}x}{y}$. Let $\wp$ be the associated Weierstrass $\wp$-function. $\wp(z+\omega)=\wp(z)$ and $(\wp^{\prime})^{2}=4\wp^{3}-g_{2}\wp-g_{3}$. Choose $z_{0}$ with $\wp(z_{0})$ and consequently $\wp^{\prime}(z_{0})$ algebraic. Suppose that $\frac{\omega}{\pi}$ is algebraic. We define a subring $Y$ of the field of meromorphic functions on $\mathbb{C}$ — $Y$ is generated over $Q$ by $\wp(z+z_{0})$, $\wp^{\prime}(z+z_{0})$, $\mathrm{e}^{\frac{2\pi\mathrm{i}z}{\omega}}$, and the algebraic constants $\wp(z_{0})$, $\wp^{\prime}(z_{0})$ and $\frac{2\pi\mathrm{i}}{\omega}$. $Y$ is stable under $\frac{\mathrm{d}}{\mathrm{d}z}$. Furthermore the values of the 6 generators of $Y$ at the infinitely many points $\omega,2\omega,3\omega,\ldots$ all lie in a fixed number field. The Schneider-Lang theorem [1] then tells us that the transcendence degree of $Y$ over $Q$ is $\leq 1$ which is evidently false. ∎∎

Combining Lemma 2.4 with Theorem 2.5 we find:

Corollary 2.6**.**

If $\lambda$ in $(0,\frac{1}{4})$ is rational then $\sum A_{n}^{2}\lambda^{2n}$ is transcendental.

So if the conjecture of [3] holds, the transcendental number $\sum\frac{A_{n}^{2}}{2^{16n}}$ is a $Q$-linear combination of Hilbert-Kunz multiplicities, and transcendental Hilbert-Kunz multiplicities exist.