This paper aims to study the Betti homology and de Rham cohomology of twisted symmetric powers of the Kloosterman connection of rank two on the torus.
We compute the period pairing and, with respect to certain bases, interpret these associated period numbers in terms of the Bessel moments.
Via the rational structures on Betti homology and de Rham cohomology, we prove the $\mathbb{Q}$-linear and quadratic relations
among these Bessel moments.

§ INTRODUCTION

Let ${\mathrm{Kl}}_{2}$ be the Kloosterman connection (of rank two) on the torus $\mathbb{G}_{\mathrm{m},z}=\Spec\left(\mathbb{Q}[z,z^{-1}]\right)$ corresponding to the differential operator $(z\partial_{z})^{2}-z$. (For details, see section <ref>.) In [8], in order to study the symmetric powers $\Sym^{k}{\mathrm{Kl}}_{2}$, Fresán, Sabbah, and Yu consider the following settings. Let $\left[2\right]:\mathbb{G}_{m,t}\rightarrow \mathbb{G}_{m,z}$ be the double cover induced by $z\mapsto t^{2}$. One obtains the pullback connection

\widetilde{{\mathrm{Kl}}}_{2}=\left[2\right]^{+}{\mathrm{Kl}}_{2}.

The structure of $\widetilde{{\mathrm{Kl}}}_{2}$ is much simpler since it is the restriction to $\mathbb{G}_{m}$ of the Fourier transform of a regular holonomic module on the affine line. In addition, the symmetric power $\Sym^{k}{\mathrm{Kl}}_{2}$ appears in the pushforward $\left[2\right]_{+}\Sym^{k}{\widetilde{\mathrm{Kl}}}_{2}$ naturally by the decomposition

\left[2\right]_{+}\Sym^{k}{\widetilde{\mathrm{Kl}}}_{2}\cong \Sym^{k}{\mathrm{Kl}}_{2}\oplus \sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2},

where $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}=\left(\mathcal{O}_{\mathbb{G}_{m}},{\mathrm{d}}+\frac{{\mathrm{d}}z}{2z}\right)\otimes \Sym^{k}{\mathrm{Kl}}_{2}$. In [9], they compute the de Rham cohomology and Betti homology for $\Sym^{k}{\mathrm{Kl}}_{2}$. In this paper, we study the analogues for $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$.

§.§ Historical results and our results

Let $I_{0}(t)$ and $K_{0}(t)$ be modified Bessel functions. Define the Bessel moments

\begin{align}\label{Bessel_moments}
\end{align}

provided that $0\leq a\leq k$ are non-negative integers, $b\in\mathbb{Z}$, and the convergence of this integral. The particular Bessel moments of the form
appear in two-dimensional quantum field theory as Feynman integrals [5, 1, 3]. In mathematical viewpoints, these moments are realized as period integrals of $\Sym^k\mathrm{Kl}_{2}$. For the details, we refer to [9]. In that paper, they developed the Hodge theory on
symmetric powers of the generalized Kloosterman connection $\mathrm{Kl}_{n+1}$ of rank $(n+1)$.

§.§.§ Sum rule identities
In <cit.>, the authors provide the following conjecture on the $\mathbb{Q}\left(\pi\right)$ linear relation of Bessel moments which is called “sum rule” in their paper.

For each pair of integers $\left(n,k\right)$ with $n\geq 2k\geq 2$, the following combination of Bessel moments vanish

\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^{m}\binom{n}{2m}\pi^{n-2m}{\mathrm{IKM}}_{2n}(n-2m,n-2k)=0.

Later in <cit.>, Zhou uses the Hilbert transformation to prove this conjecture. Moreover, he also proves a “sum rule”:

For each pair of integers $\left(n,k\right)$ with $n-1\geq 2k\geq 2$, the following conbination of Bessel moments vanish

\sum_{m=1}^{\lfloor (n+1)/2\rfloor}(-1)^{m}\binom{n}{2m-1}\pi^{n-2m+1}{\mathrm{IKM}}_{2n}(n-2m+1,n-2k-1)=0.

When the involved exponents of $t$ in (<ref>) are odd, these two identities are both reproved by Fresán, Sabbah, and Yu [9] using the algebraic geometry method. In this paper, we provide proof of these two identities involving even powers of $t$ using a similar approach in section <ref>. For example, we have the following result:

[Corollary <ref>]
For $k=4r+4$ and $k^{\prime}=\lfloor\frac{k-1}{2}\rfloor$,

\[
\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2i)
= \begin{cases}
(-1)^r\pi^{2r+2}{\mathrm{IKM}}_{k}(2r+2,2i) & \text{if $0\leq i\leq r$}, \\
& \text{if $r+1\leq i\leq k^{\prime}$}.
\end{cases}
\]

The notation $\mathrm{IKM}^{\mathrm{reg}}(2r+2,2i)$ above is the regularized Bessel moments (see Lemma <ref>). Roughly speaking, the regularized Bessel moments are the divergence Bessel moments but modifying their singularities. Therefore, our sum rule generalizes the sum rule in [17, 1].

§.§.§ $\mathbb{Q}$-dimension of Bessel moments
In [18], Zhou consider the $\mathbb{Q}$-vector subspace spanned by the Bessel moments in $\mathbb{C}$. This vector subspace is finite-dimensional due to the sum rule. Similarly, we have the following upper bound of the dimension.

For any $k$ and any $0\leq a\leq k^{\prime}=\lfloor\frac{k-1}{2}\rfloor$, the dimension of the $\mathbb{Q}$-vector space generated by the Bessel moments has upper bound:

\dim {\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j\in\left\{0\right\}\cup \mathbb{N}\right\}\leq k^{\prime}+1.

For $k$ even, the dimension of the $\mathbb{Q}$-vector space generated by the regularized Bessel moments has an upper bound:

\dim {\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}\left(k/2,2j\right)\mid j\in\left\{0\right\}\cup \mathbb{N}\right\}\leq k^{\prime}+1.

Note that our statement involves the regularized Bessel moments. This conclusion is a more general result than the one given by Zhou.

§.§.§ Quadratic relations of Bessel moments

In [7], the authors prove a general result of quadratic relations between periods given by a self-dual connection. We apply this result and obtain the quadratic relation between the Bessel moments. Using the basis of cohomologies, let $B$ be the topological pairing on Betti homology, $D$ be the Poincaré pairing on de Rham cohomology, and $P$  be the period pairing between these two homology and cohomology. Then, we write down the quadratic relations on these periods:

\begin{align}\label{quadratic_relation_formula}
\end{align}

Note that $P$ and $P_{\mathrm{c}}$ consist of $\mathbb{Q}$-linear combinations of Bessel moments and regularized Bessel moments. Moreover, due to the rational structure of Betti homology and de Rham cohomology, the corresponding pairing matrices $D, B$ consist of rational numbers.

§.§.§ Determinants of Bessel moment matrix
Another interesting result is to compute the determinants of certain matrices consisting of Bessel moments. In <cit.>, Broadhurst conjectures two formulae of the determinants of the following two $k\times k$ matrices $\mathbf{M}_{k}$ and $\mathbf{N}_{k}$ involving the Bessel moments:

\begin{align*}
\mathbf{M}_{k} &=({\mathrm{IKM}}_{2k+1}(a,2b-1))_{1\leq a,b\leq k},\\
\mathbf{N}_{k} &=({\mathrm{IKM}}_{2k+2}(a,2b-1)_{1\leq a,b\leq k}.
\end{align*}

Later, in [16], Zhou uses an analytic method to prove these two determinant formulae. Using the similar method as Zhou, in Corollary <ref>, we give explicit determinant formulae for the following two $r\times r$ matrices:

\begin{align*}
M_{r}&=({\mathrm{IKM}}_{2r-1}(i-1,2j-2))_{1\leq i,j\leq r},\\
N_{r}&=({\mathrm{IKM}}_{2r}(i-1,2j-2))_{1\leq i,j\leq r}.
\end{align*}

For $r\geq 1$, we have

\begin{align*}
\det M_{r}&=\sqrt{\pi}^{r(r+1)}\sqrt{2}^{r(r-3)}\prod_{a=1}^{r-1}\dfrac{a^{r-a}}{\sqrt{2a+1}^{2a+1}},\\
\det N_{r}&=\dfrac{\sqrt{\pi}^{(r+1)^{2}}}{\Gamma\left(\frac{r+1}{2}\right)}\dfrac{1}{\sqrt{2}^{r(r+3)}}\prod_{a=1}^{r-1}\dfrac{(2a+1)^{r-a}}{(a+1)^{a+1}}.
\end{align*}

§.§ Approach

In [2], Bloch and Esnault study the irregular connections on curves and provide the associated homology theory. Due to their results, we study the de Rham cohomology and Betti homology of $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ on $\mathbb{G}_{\mathrm{m}}$
and provide explicit bases in order to find the periods.

In section <ref>, we introduce the twisted symmetric power of Kloosterman connection $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$, which is the main object in this paper. We discuss the rational structures on the de Rham cohomology and Betti homology of the connection. Moreover, since the connection is self-dual, we introduce its algebraic and topological self-pairings. These pairings will play an important role in our computations.

In section <ref>, we study the de Rham cohomology and the de Rham cohomology with compact support on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ and write down some elements in these two cohomologies. Next, we introduce the Poincaré pairing between them and compute the pairing with respect to the elements we have written down. Using the dimension result of de Rham cohomology, along with the non-vanishing determinant of the Poincaré pairing, in Corollary <ref>, we conclude that the elements we write down in de Rham cohomology are bases.

We study parallelly the Betti homology on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ in section <ref>. Since our ambient space is a non-compact space $\mathbb{C}^{\times}$, we need to modify our Betti homology theory by allowing the chain to go to $0$ or $\infty$. By controlling the growth behaviors of the horizontal sections, we study the moderate decay Betti homology and rapid decay Betti homology on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$. Similarly, We first write down some elements in the moderate decay homology and rapid decay homology and compute their topological pairing explicitly. Moreover, by the duality of de Rham cohomology and Betti homology, the dimension of Betti homology is the same as the de Rham cohomology. Together with the topological pairing, we conclude that they are bases in Corollary <ref>.

Finally, in section <ref>, we compute the period pairing between the de Rham cohomologies and the Betti homologies and interpret them in terms of the Bessel moments. Note that our variety $\mathbb{G}_{m}=\Spec\mathbb{Q}[z,z^{-1}]$ and the connection $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ are defined over $\mathbb{Q}$ and therefore, the de Rham cohomology and Betti homology are naturally endowed with a $\mathbb{Q}$-vector space structure. From the dimension constraint of homologies, after computing the periods pairing, we obtain the $\mathbb{Q}$-linear relation of Bessel moments (Formula <ref>) and an upper bound of $\mathbb{Q}$-dimension of space spanned by the Bessel moments (Theorem <ref>). In addition, the self duality of $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ gives quadratic relations between these Bessel moments (<ref>).

In appendix <ref>, we provide an accurate analysis of the symmetric powers of the modified Bessel differential operator. The first usage belongs to section <ref>, which enables us to determine the dimension of the de Rham cohomology $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$. The second usage belongs to appendix <ref>, which allows us to analyze the leading term of the Vanhove operator. This helps us to obtain the determinant formula (Formula <ref>).

§ THE KLOOSTERMAN CONNECTION AND ITS TWISTED SYMMETRIC POWERS

§.§ Self-duality and pairing on $\mathrm{Kl}_{2}$

The connection ${\mathrm{Kl}}_{2}=(\mathcal{O}_{\mathbb{G}_{m}}v_{0}\oplus\mathcal{O}_{\mathbb{G}_{m}}v_{1},\nabla)$
consists of a free sheaf of rank $2$ on $\mathbb{G}_{m}=\Spec\mathbb{Q}[z,z^{-1}]$ with basis of sections $v_{0}$ and $v_{1}$. The connection $\nabla$ on ${\mathrm{Kl}}_{2}$ is given by

\begin{align*}
z\nabla\left(v_{0},v_{1}\right) = \left(v_{0},v_{1}\right)\begin{pmatrix}
0 & z\\
1 & 0
\end{pmatrix}{\mathrm{d}}z.
\end{align*}

We refer to <cit.><cit.> for more details.
Note that $\mathrm{Kl}_{2}$ is self-dual in the sense that there exists an algebraic horizontal pairing $\langle\ ,\, \rangle_{\text{alg}}$ on $\mathrm{Kl}_{2}$:

\left(\left\langle v_{i},v_{j}\right\rangle_{\mathrm{alg}}\right)_{0\leq i,j\leq 1}=
\begin{pmatrix}
0 & 1\\
-1 &0
\end{pmatrix}

such that $\lambda :\mathrm{Kl}_{2}\rightarrow\mathrm{Kl}_{2}^{\vee}$ by $\left(v_{0},v_{1}\right)\mapsto \left(v_{1}^{\vee},-v_{0}^{\vee}\right)$ makes the following diagram commutes.

\begin{tikzcd}[ampersand replacement=\&]
{\mathrm{Kl}}_{2}\times {\mathrm{Kl}}_{2} \arrow[rr, "{\langle\ ,\, \rangle_{\text{alg}}}"] \arrow[d, "\lambda\times 1"'] \& \& (\mathcal{O}_{\mathbb{G}_{m}},\mathrm{d}) \\
{\mathrm{Kl}}_{2}^{\vee}\times {\mathrm{Kl}}_{2} \arrow[rru, "{\text{natural}}"']                                                  \&       \&
\end{tikzcd}

Recall that the modified Bessel function $I_{0}(t)$ and $K_{0}(t)$ satisfy the differential equation $(t^{2}\partial_{t}^{2}+t\partial_{t}-t^{2})y=0$ and the Wronskian relation

\begin{equation}\label{eq:Wronskian_IK}
I_0(t)K_0'(t) - I_0'(t)K_0(t) = \frac{-1}{t}.
\end{equation}
After rescaling the modified Bessel function by

\begin{align*}
B_{0}(z)=2K_{0}(2\sqrt{z}), &\qquad
\end{align*}

the horizontal sections of $\nabla$ on ${\mathrm{Kl}}_{2}$ have basis

\begin{align}\label{horizontal_sections_Kl2}
e_{0}=\frac{1}{2}(A_{0}v_{1}-A_{1}v_{0}),\qquad e_{1}=\frac{1}{2\pi\sqrt{-1}}(B_{0}v_{1}-B_{1}v_{0}).
\end{align}

To see this, note that $A_{0}$ and $B_{0}$ are annihilated by the operator $\left(z\partial_{z}\right)^{2}-z$ and real-valued on the principal branch $\mathbb{R}_{>0}$. This gives

\[ \partial_{z}A_{1}(z)=A_{0}(z),\qquad\partial_{z}B_{1}(z)=B_{0}(z). \]

Together with the Wronskian relation
$A_{0}B_{1} - A_{1}B_{0} = 2$, the two sections $e_{0}$ and $e_{1}$ are indeed the horizontal sections. Let $\mathrm{Kl}_{2}^{\nabla}$ be the local system of $\mathbb{Q}$-vector space generated by $e_{0},e_{1}$ and there exists a topological pairing $\langle\ ,\, \rangle_{\mathrm{top}}=2\pi\sqrt{-1}\langle\ ,\, \rangle_{\mathrm{alg}}$:

\left(\left\langle e_{i},e_{j}\right\rangle_{\mathrm{top}}\right)_{0\leq i,j\leq 1}=
\begin{pmatrix}
0 & 1\\
-1 &0
\end{pmatrix}.

§.§ Rational structures and pairings on $(\mathcal{O}_{\mathbb{G}_{m}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$.

Consider the cyclic Galois cover $\left[2\right]:\mathbb{G}_{m,t}\rightarrow\mathbb{G}_{m,z}$ induced by $z\mapsto t^{2}$, where $\mathbb{G}_{m}=\Spec\mathbb{Q}[x,x^{-1}]$. Let $T=(\mathcal{O}_{\mathbb{G}_{m,t}},{\mathrm{d}})$ be the trivial connection on $\mathbb{G}_{m,t}$.
View $\mathcal{O}_{\mathbb{G}_{m,z}}$ as the $\mathbb{Q}[t,t^{-1}]$-module $\mathbb{Q}[t^2,t^{-2}]$.
The pushforward connection $\left[2\right]_{+}T$ decompose into

(\mathcal{O}_{\mathbb{G}_{m,z}},{\mathrm{d}})\oplus \left(t\cdot \mathcal{O}_{\mathbb{G}_{m,z}},{\mathrm{d}}\right).

The second component $\left(t\cdot \mathcal{O}_{\mathbb{G}_{m,z}},{\mathrm{d}}\right)$ is isomorphic to $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ via the following diagram

\begin{equation}\label{iso_of_connection_sqrt(z)}
\begin{tikzcd}
{\mathcal{O}_{\mathbb{G}_{m},z}} \arrow[rr, "\mathrm{d}+\frac{\mathrm{d}t}{t}=\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z}"] \arrow[d, "t"', "\wr"] && {\mathcal{O}_{\mathbb{G}_{m},z}\otimes \Omega_{\mathbb{G}_{m},z}^{1}} \arrow[d, "t", "\wr"'] \\
{t\cdot \mathcal{O}_{\mathbb{G}_{m},z}} \arrow[rr, "\mathrm{d}"]                                       && {t\cdot \mathcal{O}_{\mathbb{G}_{m},z}\otimes \Omega_{\mathbb{G}_{m},z}^{1}}
\end{tikzcd}
\end{equation}

The dual connection of $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ is given by $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}-\frac{1}{2}\frac{\mathrm{d}z}{z})$,
and the two are isomorphic via multiplication by $z$, that is, the following diagram commutes.

\begin{tikzcd}
{\mathcal{O}_{\mathbb{G}_{m},z}} \arrow[r, " \mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z}"] \arrow[d, "z"', "\wr"] & {\mathcal{O}_{\mathbb{G}_{m},z}\otimes \Omega_{\mathbb{G}_{m},z}^{1}} \arrow[d, "z", "\wr"'] \\
{\mathcal{O}_{\mathbb{G}_{m},z}} \arrow[r, "\mathrm{d}-\frac{1}{2}\frac{\mathrm{d}z}{z}"]                                       & {\mathcal{O}_{\mathbb{G}_{m},z}\otimes \Omega_{\mathbb{G}_{m},z}^{1}}
\end{tikzcd}

This induces an algebraic horizontal pairing $\langle\ ,\, \rangle_{\mathrm{alg}}$ on $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ given by

\langle 1,1\rangle_{\mathrm{alg}} = z.

On the other hand,
the rational structure of the local system of horizontal sections of $(\mathcal{O}_{\mathbb{G}_{m,t}},\mathrm{d})$ is generated by $1$. Under the isomorphism (<ref>), the rational structure of local system of horizontal sections of $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$ is generated by
$\frac{1}{t}=\frac{1}{\sqrt{z}}$. Its dual connection $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}-\frac{1}{2}\frac{\mathrm{d}z}{z})$ has local system of horizontal sections generated by $\sqrt{z}$. This induces a rational topological pairing $\langle\ ,\, \rangle_{\mathrm{top}}$ on $(\mathcal{O}_{\mathbb{G}_{m,z}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z})$

\langle\frac{1}{\sqrt{z}},\frac{1}{\sqrt{z}}\rangle_{\mathrm{top}} = 1.

§.§ Algebraic and topological pairings on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$

The $k$-th symmetric product of $\mathrm{Kl}_{2}$, $\Sym^{k}\mathrm{Kl}_{2}$, is a rank $k+1$ free sheaf on $\mathcal{O}_{\mathbb{G}_{m}}$ with basis of sections

v_{0}^{a}v_{1}^{k-a} = \frac{1}{|\mathfrak{S}_{k}|}\sum_{\sigma\in\mathfrak{S}_{k}}\sigma\left(v_{0}^{\otimes a}\otimes v_{1}^{\otimes k-a}\right)\qquad a=0,1,\ldots,k,

where $\mathfrak{S}_{k}$ is the symmetric group on $k$ elements. It is endowed with the induced connection from $\left(\mathrm{Kl}_{2},\nabla\right)$. After twisting with the connection $\left(\mathcal{O}_{\mathbb{G}_{m}},\mathrm{d}+\frac{1}{2}\frac{{\mathrm{d}z}}{z}\right)$, we define

\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}=\left(\mathcal{O}_{\mathbb{G}_{m}},{\mathrm{d}}+\frac{1}{2}\frac{{\mathrm{d}}z}{z}\right)\otimes \Sym^{k}{\mathrm{Kl}}_{2}.

The induced connection $\nabla$ on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ is given by

\begin{align}\label{connection_on_twisted_symmetric_Kl2}
\nabla v_{0}^{a}v_{1}^{k-a}&=(k-a)v_{0}^{a+1}v_{1}^{k-a-1}{\mathrm{d}}z+\dfrac{a}{z}v_{0}^{a-1}v_{1}^{k-a+1}{\mathrm{d}}z+\dfrac{1}{2z}v_{0}^{a}v_{1}^{k-a}{\mathrm{d}}z.
\end{align}

Note that $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$ is the same sheaf as $\Sym^{k}{\mathrm{Kl}}_{2}$ but endowed with a different connection.

Via the self-duality on ${\mathrm{Kl}}_{2}$ and on $\left(\mathcal{O}_{\mathbb{G}_{m}},\mathrm{d}+\frac{1}{2}\frac{\mathrm{d}z}{z}\right)$, we have the perfect algebraic pairing $\left\langle\ ,\ \right\rangle_{\mathrm{alg}}$ on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$:

\begin{tikzcd}
\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\times \sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2} \arrow[rr, "{\left\langle\ ,\ \right\rangle_{\mathrm{alg}}}"] &  & {\left(\mathcal{O}_{\mathbb{G}_{m}},{\mathrm{d}}\right)}
\end{tikzcd}

given by

\left\langle v_{0}^{k-a}v_{1}^{a},v_{0}^{k-b}v_{1}^{b}\right\rangle_{\mathrm{alg}}=z\delta_{k,a+b}(-1)^{a}\dfrac{a!b!}{k!}=(2\pi\sqrt{-1})^{k}\left\langle e_{0}^{k-a}e_{1}^{a},e_{0}^{k-b}e_{1}^{b}\right\rangle_{\mathrm{alg}}.

The local system $(\sqrt{z}\Sym^{k}\mathrm{Kl}_{2})^{\nabla}$ is a $\mathbb{Q}$-vector space generated by the horizontal sections

\begin{equation}\label{eq:horizontal_basis}
\dfrac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a} = \dfrac{1}{\sqrt{z}}\sum_{\sigma\in\mathfrak{S}_{k}}\sigma\left(e_{0}^{\otimes a}\otimes e_{1}^{\otimes k-a}\right),\quad a=0,1,\cdots,k,
\end{equation}

which are the product of the horizontal sections to the connection $\left(\mathcal{O}_{\mathbb{G}_{m}},\mathrm{d}+\frac{1}{2}\frac{{\mathrm{d}}z}{z}\right)$ and $\Sym^{k}\mathrm{Kl}_{2}$. The topological pairing $\langle\ ,\, \rangle$ on $\mathrm{Kl}_{2}^{\nabla}$ induces a topological pairing on $(\Sym^{k}\mathrm{Kl}_{2})^{\nabla}$ and thus on $(\sqrt{z}\Sym^{k}\mathrm{Kl}_{2})^{\nabla}$:

\begin{tikzcd}
(\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2})^{\nabla}\times (\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2})^{\nabla} \arrow[rr, "{\left\langle\ ,\ \right\rangle_{\mathrm{top}}}"] &  & \mathbb{Q}.
\end{tikzcd}

This pairing reads

\left<\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a},\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}\right>_{\mathrm{top}}=\delta_{k,a+b}(-1)^{k-a}\dfrac{a!b!}{k!}.

§ THE DE RHAM COHOMOLOGY

§.§ Dimension of $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$

For the connections $\Sym^{k}{\mathrm{Kl}}_{2}$ and $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$, we have

\dim H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\Sym^{k}{\mathrm{Kl}}_{2}\right)=\dim H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)=\left\lfloor\dfrac{k+1}{2}\right\rfloor =k^{\prime}+1.

In <cit.>, we have the following result for the Euler characteristic of a connection.

On $\mathbb{G}_{m}$ with parameter $z$, write $\theta_{z}=z\frac{\mathrm{d}}{{\mathrm{d}}z}$, and consider a non-zero operator $L=\sum z^{i}P_{i}(\theta_{z})$. Define integers $a,b$ by

a:=\max\left\{i\mid P_{i}\neq 0\right\};\ \ \ \ b:=\min\left\{i\mid P_{i}\neq 0\right\}.

Then $\chi\left(\mathbb{G}_{m},\mathcal{D}/\mathcal{D}L\right)=-\left(a-b\right)$.

Now, the differential operator on $\mathbb{G}_{m}$ associated with ${\mathrm{Kl}}_{2}$ is given by $L= \theta_{z}^{2}-z$.[We may view this connection as the changing variable of modified Bessel operator $L=\theta^{2}-t^{2}$ by $z=t^{2}/4$, that is, by $\partial_{t}=\sqrt{z}\partial_{z}$.] Then the differential operator for $\Sym^{k}{\mathrm{Kl}}_{2}$ is given by $k$-th symmetric power $L_{k+1}$ of $L$. Proposition <ref> gives $a=\left\lfloor\frac{k+1}{2}\right\rfloor$ and $b=0$. Hence, we conclude that

\chi\left(\mathbb{G}_{m},\Sym^{k}{\mathrm{Kl}}_{2}\right)=-\left\lfloor\frac{k+1}{2}\right\rfloor.

Moreover, we know that the solution to the differential equation $L_{k+1}Y=0$ is given by $A_{0}^{i}B_{0}^{k-i}$ for $i=0,1,2,\cdots k$. Note that $A_{0}$ is holomorphic at $0$ and has exponential growth near infinity, and $B_{0}$ has a log pole at $0$. These imply all of the solutions $A_{0}^{i}B_{0}^{k-i}$ are not algebraic solutions, and thus $H^{0}_{\mathrm{dR}}\left(\mathbb{G}_{m},\Sym^{k}{\mathrm{Kl}}_{2}\right)=0$. Hence, combining the fact that $H^{2}_{\mathrm{dR}}\left(\mathbb{G}_{m},\Sym^{k}{\mathrm{Kl}}_{2}\right)=0$ by Artin vanishing theorem, we conclude that $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\Sym^{k}{\mathrm{Kl}}_{2}\right)$ has dimension $\left\lfloor\frac{k+1}{2}\right\rfloor$.

For $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$, note that the corresponding differential operator becomes $\frac{1}{\sqrt{z}}L_{k+1}\sqrt{z}$. Now, using the fact that $\frac{1}{\sqrt{z}}\theta_{z}\sqrt{z}=\theta_{z}+\frac{1}{2}$, we have $\frac{1}{\sqrt{z}}L\sqrt{z}=\sum z^{i}P_{i}\left(\theta_{z}+\frac{1}{2}\right)$ whenever $L=\sum z^{i}P_{i}(\theta_{z})$. Therefore, we again obtain that $a=\lfloor\frac{k+1}{2}\rfloor$, $b=0$ and thus

\chi\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)=-\left\lfloor\frac{k+1}{2}\right\rfloor.

Moreover, the solutions to the differential operator $\frac{1}{\sqrt{z}}L_{k+1}\sqrt{z}$ are $\frac{1}{\sqrt{z}}A_{0}^{i}B_{0}^{k-i}$ for $i=0,1,\cdots,k$ and similarly, $H^{0}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)=0$. Hence, we conclude that $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ has dimension $\left\lfloor\frac{k+1}{2}\right\rfloor$ by Artin vanishing theorem.

In <cit.>, Katz provides the proof only on the base field $\mathbb{C}$. However, we need to determine the dimension of de Rham cohomology as a $\mathbb{Q}$-vector space. Thanks to <cit.>, we know that the change of de Rham cohomology on a characteristic zero field extension is just changing the base field:

and the dimension is preserved.

§.§ Compactly supported de Rham cohomology

Consider the $k^{\prime}+1$ elements $\left\{v_{0}^{k}z^{i}\frac{{\mathrm{d}z}}{z}\right\}_{i=0}^{k^{\prime}}$ in $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$. We will prove these elements form a $\mathbb{Q}$-basis through computation of the Poincaré pairing. (See Corollary <ref>.) The Poincaré dual of the de Rham cohomology is the de Rham cohomology with compact support.
Note that an element in $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{\mathrm{m}}\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ is represented by a triple $\left(\xi,\eta,\omega\right)$, where $\omega\in H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{\mathrm{m}},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ and $\xi,\eta$ are formal solutions to $\nabla\xi = \nabla\eta =\omega$ at $0$ and $\infty$ respectively. The solutions are provided by the following lemma.

Suppose that $k\equiv 0,1,3\pmod{4}$. For $0\leq i\leq k^{\prime}$, there exists $\left(\xi_{i},\eta_{i}\right)\in\left(\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)_{\widehat{0}}\oplus \left(\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)_{\widehat{\infty}}$ such that $\nabla\xi_{i}=\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z}$.

On the other hand, if $k\equiv 2\pmod{4}$, say $k=4r+2$. For $0\leq i\leq k^{\prime}$ with $i\neq r$, there exists $\left(\xi_{i},\eta_{i}\right)\in\left(\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)_{\widehat{0}}\oplus \left(\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)_{\widehat{\infty}}$ such that

\nabla\xi_{i}=\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{{\mathrm{d}}z}{z},

where $\gamma_{k,n}\in\mathbb{Q}$ are the coefficients in the asymptotic expansion of $(-A_{0}\left(z\right)B_{0}\left(z\right))^{k/2}$
given by (<ref>) below.

Near $0$, we want to find

\[ \xi_{i}=\sum_{a=0}^{k}\xi_{i,a}(z)v_{0}^{a}v_{1}^{k-a}\in\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a} \]

such that $\nabla\xi_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}$.
Using the connection formula (<ref>) on $\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}$, we need to solve:

\begin{align*}
\frac{{\mathrm{d}}}{\mathrm{d}z}\xi_{i,k}(z)+(k-1)\xi_{i,k-1}(z)+\frac{1}{2z}\xi_{i,k}(z)&=z^{i-1},\\
\frac{{\mathrm{d}}}{\mathrm{d}z}\xi_{i,a}(z)+(k-a+1)\xi_{i,a-1}(z)+\frac{a+1}{z}\xi_{i,a+1}(z)+\frac{1}{2z}\xi_{i,a}(z)&=0 {\text{ for }}a=1,2,\cdots,k-1,\\
\frac{{\mathrm{d}}}{\mathrm{d}z}\xi_{i,0}(z)+\frac{1}{z}\xi_{i,1}(z)+\frac{1}{2z}\xi_{i,0}(z)&=0.
\end{align*}

Write $\xi_{i,a}=\sum_{n=0}^{\infty}\xi_{i,a,n}z^{n}$. We solve $\xi_{i,a,n}$ recursively on $n$. Suppose that we have solved $\xi_{i,a,j}$ for $j<n$. Compare the coefficient of $z^{n-1}$ of the above system of equations and get

\left(
\begin{array}{ccccc}
&&& &n+\frac{1}{2}\\
\end{array}\right)
\left(\begin{array}{c}\xi_{i,0,n}\\ \xi_{i,1,n}\\\xi_{i,2,n}\\\vdots\\\xi_{i,k,n}\end{array}\right)=
{\text{lower order combinations}}.

Since the first square matrix is invertible, $\xi_{i,a,n}$ is determined uniquely. Thus, we find $\xi_{i}\in\displaystyle\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}$ such that $\nabla\xi_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}$. In $k\equiv 2\pmod{4}$ case, we only need to replace $\xi_{i}$ by $\xi_{i}-\gamma_{k,i-r}\xi_{r}$.

Next, we turn to investigate the formal solutions at $\infty$ using horizontal frames. Note that the modified Bessel functions have the asymptotic expansions at $\frac{1}{t}$

\begin{align}
&\sim e^t\frac{1}{\sqrt{2\pi t}}
\sum_{n=0}^\infty \frac{((2n-1)!!)^2}{2^{3n} n!} \frac{1}{t^n},
&& |\arg t| < \frac{1}{2}\pi \label{expansion_I_0_at_inf}\\
&\sim e^{-t} \sqrt{\frac{\pi}{2t}}
\sum_{n=0}^\infty (-1)^n\frac{((2n-1)!!)^2}{2^{3n} n!} \frac{1}{t^n},
&& |\arg t| < \frac{3}{2}\pi \label{expansion_K_0_at_inf}\\
&\sim \frac{1}{2t}
\sum_{n=0}^\infty \frac{((2n-1)!!)^3}{2^{3n}n!}\frac{1}{t^{2n}}.\label{expansion_I_0K_0_at_inf}
\end{align}

Let $w=\frac{1}{z}$ be the local coordinate at $z=\infty$. For $k$ even, by the last asymptotic expansion, we have

\begin{align}\label{expansion_of_(-AB)^half}
(-A_{0}(z)B_{0}(z))^{k/2}\sim w^{1/4}\sum_{n=0}^{\infty}\gamma_{k,n}w^{n},
\end{align}

where $\gamma_{k,0}=1$ and $\gamma_{k,n}>0$ for all $n>0$. For convenience, we set $\gamma_{k,j}=0$ for all $j<0$.

Following the notation in (<ref>), let us set $\overline{e}_{1} = \pi\sqrt{-1}e_{1}$. Then $\frac{e_{0}^{a}\overline{e_{1}}^{k-a}}{\sqrt{z}}$ are horizontal sections. Using the Wronskian relation $A_{0}B_{1}-A_{1}B_{0}=2$, we have $v_{0}=B_{0}e_{0}-A_{0}\overline{e}_{1}$. Let $w=\frac{1}{z}$ and we obtain

\begin{align*}
\end{align*}

By the expression (<ref>), since there is a factor $w^{1/4}$, we first solve $\eta_{i,a}$ in $\mathbb{Q}((w^{1/4}))$ such that

\begin{align*}
\mathrm{d}\eta_{i,a}=\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}.
\end{align*}

Finally, we will prove that the formal solution lies in $\mathbb{Q}((w))$ by showing that it is invariant by the Galois action $\sigma\in\operatorname{Gal}(\mathbb{Q}((w^{1/4}))/\mathbb{Q}((w)))$.

We have the asymptotic expansion

I_0(t) = \sum_{k=0}^{\infty}\frac{\left(\frac{1}{4}z^{2}\right)^{k}}{\left(k!\right)^{2}}.

K_0(t) =  -\left(\log\frac{z}{2}+\gamma\right)I_0(t)+\sum_{k=1}^{\infty}H_{k}\frac{\left(\frac{1}{4}z^{2}\right)^{k}}{\left(k!\right)^{2}}.

where $H_{k}=\sum_{j=1}^{k}\frac{1}{j}$ is the harmonic constant.

A_{0}(z)\sim -2\sum_{k=0}^{\infty}\frac{z^{k}}{\left(k!\right)^{2}}=-2\cdot F_{1}(z).

B_{0}(z)\sim 4\left(\log\sqrt{z}+\gamma\right)\sum_{k=0}^{\infty}\frac{z^{k}}{\left(k!\right)^{2}}+2\sum_{k=1}^{\infty}H_{k}\frac{z^{k}}{\left(k!\right)^{2}}=2\log z\cdot F_{2}(z)+4\gamma \cdot F_{3}(z).

where $F_{1},F_{2},F_{3}\in 1+z\mathbb{R}\llbracketz\rrbracket$.
This gives

(-A_{0})^{a}B_{0}^{k-a}z^{i-1/2}=z^{i-1/2}\left(\sum_{m=0}^{k-a}C_{m}\left(\log z\right)^{m}G_{m}(z)\right).

for some constant $C_{m}\in\mathbb{R}$ and $G_{m}(z)\in 1+z\mathbb{R}\llbracketz\rrbracket$.
Then we may take $\xi_{i,a}$ with the expansion

\xi_{i,a}\sim\sum_{m=0}^{k-a}D_{m}z^{i+1/2}\left(\log z\right)^{m}H_{m}(z)

for some constant $D_{m}\in\mathbb{R}$ and $H_{m}(z)\in 1+z\mathbb{R}\llbracketz\rrbracket$.

Near $\infty$, we have the expansion

\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}=
\begin{cases}
\sqrt{\pi}^{k-2a}e^{-2(k-2a)/\sqrt{w}}w^{k/4-i-3/2}\cdot F_{i,a}, & a\neq\frac{k}{2}\\
w^{k/4-i-3/2}\left(\sum_{n=0}^{\infty}\frac{\left(\left(2n-1\right)!!\right)^{3}}{2^{5n}n!}w^{n}\right)^{k/2}, & a=\frac{k}{2}
\end{cases}

where $F_{i,a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$.

When $a\neq \frac{k}{2}$, we can find an antiderivative $\eta_{i,a}$ of $\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}$ with the expansion

\begin{align}\label{expansion_eta_i_a}
\eta_{i,a}=\frac{\sqrt{\pi}^{k-2a}}{k-2a}e^{-2(k-2a)/\sqrt{w}}w^{k/4-i}\cdot G_{i,a}
\end{align}

for some $G_{i,a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$. We analyze $\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}$. Write $e_{0}^{k-a}\overline{e}_{1}^{a}$ back to the expression in basis $v_{0}^{b}v_{1}^{k-b}$:

\begin{align*}
&=2^{-k}e^{2(k-a)\sqrt{z}}\frac{1}{\sqrt{\pi}^{k-a}}(F_{1}v_{0}-F_{2}v_{1})^{k-a}\cdot e^{-2a\sqrt{z}}\sqrt{\pi}^{a}(G_{1}v_{1}-G_{2}v_{0})^{a}\\
\end{align*}

where $F_{1},F_{2},G_{1},G_{2}\in z^{1/4}\mathbb{Q}\llbracket z^{-1/4}\rrbracket$. Thus,

\begin{align*}
\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}&= \frac{2^{-k}}{k-2a}w^{k/4-i+1/2}G_{i,a}(F_{1}v_{0}-F_{2}v_{1})^{k-a}(G_{1}v_{1}-G_{2}v_{0})^{a},
\end{align*}

where $F_{1},F_{2},G_{1},G_{2}\in z^{1/4}\mathbb{Q}\llbracket z^{-1/4}\rrbracket$. Then, we conclude that $\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}$ has no exponential factor as a combination of monomials $v_{0}^{k-b}v_{1}^{b}$.

Let $\sigma :w^{1/4}\mapsto \sqrt{-1}w^{1/4}$ be the generator of the Galois group of the extension $\mathbb{C}\left(w^{1/4}\right)$ of $\mathbb{C}\left(w\right)$. The $\sigma$ action on $A_{i},B_{i}$ is given by

\begin{align*}
\sigma\left(A_{j},B_{j}\right)&=\left(\frac{1}{\pi\sqrt{-1}}B_{j},-\pi\sqrt{-1}A_{j}\right) {\text{ for }}j=0,1,
\end{align*}

and thus on $e_{0},\overline{e}_{1}$ by

\begin{align*}
\sigma\left(e_{0},\overline{e}_{1}\right)=\left(\frac{1}{\pi\sqrt{-1}} \overline{e}_{1},-\pi\sqrt{-1}e_{0}\right);&\quad
\sigma\left(e_{0}^{k-a}\overline{e}_{1}^{a}\right)=(\sqrt{-1})^{-k}\pi^{2a-k}e_{0}^{a}\overline{e}_{1}^{k-a}.
\end{align*}

Moreover, we have

\sigma\left(\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}\right)=\eta_{i,k-a}\frac{e_{0}^{a}\overline{e}_{1}^{k-a}}{\sqrt{z}}.

Hence, when $k\equiv 1,3\pmod{4}$, we take an element

\displaystyle\eta_{i}=-\sum_{a=0}^{k}\binom{k}{a}\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}\in\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}.

This gives $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}.$

When $k=4r+4$ and $a=2r+2$, the exponents of $w$ of expansion of $\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}$ are in $\frac{1}{2}+\mathbb{Z}$ and one takes

where $G_{i}\in 1+w\mathbb{Q}\llbracket w\rrbracket$. More precisely, we have

G_{i}= 1+\sum_{n=1}^{\infty}\frac{r-i+1/2}{r-i+1/2+n}\gamma_{k,n}w^{n}.

Moreover, $\eta_{i,2r+2}\frac{(e_{0}\overline{e}_{1})^{2r+2}}{\sqrt{z}}$ has no exponential factor as a combination of monomials $v_{0}^{k-b}v_{1}^{b}$ and is invariant under $\sigma$. Hence, when $k\equiv 4\pmod{4}$, we take an element

\eta_{i}=-\sum_{a=0}^{k}\binom{k}{a}\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}\in\displaystyle\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}

This gives $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}$.

Now, suppose that $k=4r+2$, a positive integer congruent to $2$ modulo $4$, and $a=2r+1$.
Using the expansion (<ref>), we have the residue:

\Res_{w}\frac{(-A_{0})^{a}B_{0}^{k-a}}{w^{i+3/2}}=\gamma_{k,i-r},

which vanishes if and only if $i\leq r-1$. Therefore, for $i\geq r$, there exists

\eta_{i,2r+1}\sim \frac{1}{r-i} w^{r-i}\cdot H_{i}

such that

\mathrm{d}\eta_{i,2r+1}=\left(w^{-i-3/2}-\gamma_{k,i-r}w^{-r-3/2}\right)(-A_{0}B_{0})^{2r+1}\mathrm{d}w

where $H_{i}\in 1+w\mathbb{Q}\llbracket w\rrbracket$. Also, $\eta_{i,2r+1}\frac{(e_{0}\overline{e}_{1}))^{2r+1}}{\sqrt{z}}$ is invariant under $\sigma$. Moreover, $\eta_{i,2r+1}\frac{(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}}$ has no exponential factor as a combination of monomials $v_{0}^{k-b}v_{1}^{b}$. Thus, we have

\begin{align*}
v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{\mathrm{d}z}{z}&=\nabla\left(-\sum_{\substack{a=0\\a\neq k/2}}^{k}\binom{k}{a}(\eta_{i,a}-\gamma_{k,i-r}\eta_{r,a})\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}-\binom{k}{k/2}\eta_{i,2r+1}\frac{e_{0}^{2r+1}\overline{e}_{1}^{2r+1}}{\sqrt{z}}\right)
\end{align*}

and hence we find an element $\eta_{i}$ in $\displaystyle\bigoplus_{a=0}^{k}\mathbb{Q}\llbracket z\rrbracket v_{0}^{a}v_{1}^{k-a}$ such that $\nabla\eta_{i}=v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{{\mathrm{d}}z}{z}$.

Now, we define some elements in the de Rham cohomology and the de Rham cohomology with compact support.

In the de Rham cohomology $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$,
the classes $\omega_i$ are given as follows.

* When $k\equiv 0,1,3\bmod{4}$, define the $k^{\prime}+1$ elements:

\omega_{k,i}=v_{0}^{k}z^{i}\frac{\mathrm{d}z}{z} {\text{ for }}i=0,1,2,\cdots,k^{\prime}.

* When $k\equiv 2\bmod{4}$, define the $k^{\prime}$ elements:

\omega_{k,i}=
\begin{cases}
v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z} & k=4r+2 {\text{ and }}0\leq i< r;\\
v_{0}^{k}z^{i}\frac{{\mathrm{d}}z}{z}-\gamma_{k,i-r}v_{0}^{k}z^{r}\frac{{\mathrm{d}}z}{z} & k=4r+2 {\text{ and }}r+1\leq i\leq 2r.
\end{cases}

From the above lemma, we define the elements in the compactly supported de Rham cohomology.

In the compactly supported de Rham cohomology $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$,

* when $k\equiv 0,1,3\pmod{4}$, for $0\leq i\leq k^{\prime}$, we define $k^{\prime}+1$ elements

\widetilde{\omega}_{k,i}=(\xi_{i},\eta_{i},\omega_{k,i}) {\text{ for }}0\leq i\leq k^{\prime},

where $\nabla \xi_{i}=\nabla\eta_{i}=\omega_{k,i}$.

* When $k\equiv 2\pmod{4}$, we define $k^{\prime}$ elements

\widetilde{\omega}_{k,i}=(\xi_{i},\eta_{i},\omega_{k,i}) {\text{ for }} 0\leq i\leq r-1 {\text{ and }} r+1\leq i\leq k^{\prime},

where $\nabla \xi_{i}=\nabla\eta_{i}=\omega_{k,i}$.

* In the case that $k\equiv 2\pmod{4}$, write $k=4r+2$, we define

\widehat{m}_{2r+1}=\left(0,\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},0\right)\in H^{1}_{\mathrm{dR},c}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right).

Note that $\widehat{m}_{2r+1}$ is defined only in the case when $k\equiv 2\pmod{4}$. In the next subsection, we will prove these elements really form a basis of $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$. (See Corollary <ref>.) Further, we define the middle part de Rham cohomology, $H^{1}_{\mathrm{mid}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$, to be the image of $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ in $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$. We therefore have

\begin{align*}
\omega_{k,i}\in H^{1}_{\mathrm{mid}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right) & {\text{ for }} 0\leq i\leq k^{\prime} {\text{ when }} k\equiv 0,1,3\pmod{4};\\
\omega_{k,i}\in H^{1}_{\mathrm{mid}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right) & {\text{ for }} 0\leq i\leq k^{\prime},\, i\neq r {\text{ when }} k=4r+2.
\end{align*}

We may regard $H^{1}_{\mathrm{mid}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ as a quotient of $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ containing the class of elements $\widetilde{\omega}_{k,i}$. In the next subsection, we will prove these elements form a basis. (See Corollary <ref>.)

§.§ Poincaré pairing
We have the following Poincaré pairing between the de Rham cohomology and the compactly supported de Rham cohomology. Note that the algebraic pairing $\left\langle\ ,\ \right\rangle_{\mathrm{alg}}$ is introduced in section <ref>.

\begin{tikzcd}[row sep=tiny]
{H^{1}_{\text{dR},c}\left(\mathbb{G}_{m},\sqrt{z}{\text{Sym}}^{k}{\text{Kl}}_{2}\right)\otimes H^{1}_{\text{dR}}\left(\mathbb{G}_{m},\sqrt{z}{\text{Sym}}^{k}{\text{Kl}}_{2}\right)} \arrow[rr, "{\left\langle\ ,\ \right\rangle_{\text{Poin}}}"] &  & \mathbb{Q}\left(-k-1\right)                                                                                \\
{(\widehat{m},\omega)\otimes\eta} \arrow[rr]                                                                                &  & {\Res_{z}\left\langle\widehat{m}_{0},\eta\right\rangle_{{\mathrm{alg}}}+\Res_{w}\left\langle\widehat{m}_{\infty},\eta\right\rangle_{{\mathrm{alg}}}}
\end{tikzcd}

Under the notation as in Definition <ref>, for $j\geq 0$, we have the Poincaré pairing

\begin{align*}
\left\langle \widetilde{\omega}_{k,i}, v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z} \right\rangle_{\mathrm{Poin}}&=
\begin{cases}
0 & {\text{if }}i+j\leq k^{\prime}-1,\ k:{\text{arbitrary,}}\\
(-2)^{k^{\prime}}\frac{k^{\prime}!}{k!!}& {\text{if }}i+j= k^{\prime},\ k\equiv 1,3\bmod{4},\\
\frac{\binom{k}{k/2}}{2^{k}(r-i+1/2)}& {\text{if }}i+j= k^{\prime},\ k\equiv 0\bmod{4},\ k=4r+4,\\
-\frac{\binom{k}{k/2}}{2^{k}(r-i)}& {\text{if }}i+j= k^{\prime},\ k\equiv 2\bmod{4},\ k=4r+2,\\
* &{\text{if }}i+j> k^{\prime}.
\end{cases}
\end{align*}

Moreover, if $k=4r+2$, we have

\left\langle\widehat{m}_{2r+1},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{Poin}}=\begin{cases}
0 & {\text{if }}j<r,\\
\gamma_{k,j-r} & {\text{if }}j\geq r.
\end{cases}

In particular, the Poincaré pairing between these elements are non-degenerate.

We first discuss the residue at $z=0$. For any $k$ and $0\leq i,j\leq k^{\prime}$, we compute

\begin{align*}
\Res_{z}\left\langle \xi_{i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}&=\sum_{a=0}^{k}\Res_{z}\left\langle \xi_{i,a}v_{0}^{a}v_{1}^{k-a},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}\\
&=\Res_{z}\left\langle \xi_{i,0}v_{1}^{k},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}\\
\end{align*}

where $\xi_{i,0}\in\mathbb{Q}\llbracket z\rrbracket$.

Next, we discuss the residue at $z=\infty$. When $k\equiv 1,3\pmod{4}$ and for any $0\leq i,j\leq k^{\prime}$, we compute

\begin{align*}
\Res_{w}\left\langle\eta_{i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}} &=-\sum_{a=0}^{k}\binom{k}{a}\Res_{w}\left\langle\eta_{i,a}\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}\\
\begin{cases}
0 & {\text{ if }}i+j\leq k^{\prime}-1,\\
(-2)^{k^{\prime}}\frac{k^{\prime}!}{k!!} & {\text{ if }} i+j=k^{\prime},\\
* & {\text{ if }}i+j\geq k^{\prime}+1.
\end{cases}
\end{align*}

where $G_{i,a},F_{j,k-a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$ and the last equality follows from <cit.>.

When $k\equiv 0\pmod{4}$, write $k=4r+4$. For any $0\leq i,j\leq k^{\prime}$, we compute

\begin{align*}
\Res_{w}\left\langle\eta_{i},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}&=\frac{1}{2^{k}}\sum_{\substack{a=0\\a\neq\frac{k}{2}}}^{k}\frac{(-1)^{a}\binom{k}{a}}{k-2a}\Res_{w}\left(w^{(k-1)/2-i-j-1}G_{i,a}F_{j,k-a}{\mathrm{d}}w\right)\\
&\qquad+\frac{\binom{k}{k/2}}{2^{k}(r-i+1/2)}\Res_{w}\left(w^{k^{\prime}-i-j-1}\cdot G_{i}F_{2r+2}{\mathrm{d}}w\right)\\
\begin{cases}
0 & {\text{ if }}i+j\leq k^{\prime}-1,\\
\frac{\binom{k}{k/2}}{2^{k}(r-i+1/2)} & {\text{ if }}i+j=k^{\prime},\\
* & {\text{ if }}i+j\geq k^{\prime}+1.
\end{cases}
\end{align*}

where $G_{i,a},F_{j,k-a}\in 1+\sqrt{w}\mathbb{Q}\llbracket\sqrt{w}\rrbracket$, $G_{i}\in 1+w\mathbb{Q}\llbracket w\rrbracket$ and $F_{2r+2}=\left(\sum_{n=0}^{\infty}\frac{\left(\left(2n-1\right)!!\right)^{3}}{2^{5n}n!}w^{n}\right)^{2r+2}$.

When $k\equiv 2\pmod{4}$, the computation is similar to the case $k\equiv 0\pmod{4}$.

Finally, we compute

\begin{align*}
\Res_{w}\left\langle\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\rangle_{\mathrm{alg}}&=-\sum_{b=0}^{k}\binom{k}{b}\Res_{w}\left\langle\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},\frac{(-A_{0})^{b}B_{0}^{k-b}}{w^{j+3/2}}\frac{e_{0}^{k-b}\overline{e}_{1}^{b}}{\sqrt{z}}{\mathrm{d}}w\right\rangle_{\mathrm{alg}}\\
0 & {\text{if }}j< r,\\
\gamma_{k,j-r} & {\text{if }}j\geq r.
\end{cases}
\end{align*}

Combining these residues, we obtain this proposition.

Putting the dimension result in Proposition <ref> and the non-vanishing determinant of Poincaré pairing (Proposition <ref>) together, we have the following corollary.

Let $k$ be a positive integer.

* $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ has basis

* $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ has basis

\[
\begin{cases}
\left\{\widetilde{\omega}_{k,j}\right\}_{j=0}^{k^{\prime}} & {\text{ if }}k\equiv 0,1,3\pmod{4},\\
\left\{\widetilde{\omega}_{k,j}\right\}_{j=0}^{r-1}\cup \left\{\widetilde{\omega}_{k,j}\right\}_{j=r+1}^{k^{\prime}}\cup \left\{\widehat{m}_{2r+1}\right\} & {\text{ if }}k\equiv 2\pmod{4} {\text{ with }} k=4r+2.
\end{cases}
\]

* $H_{{\mathrm{dR,mid}}}^{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ has basis

\[
\begin{cases}
\left\{\omega_{k,i}\right\}_{i=0}^{k^{\prime}} & {\text{ if }}k\equiv 0,1,3\pmod{4},\\
\left\{\omega_{k,i}\right\}_{i=0}^{r-1}\cup \left\{\omega_{k,i}\right\}_{i=r+1}^{k^{\prime}} & {\text{ if }} k\equiv 2\pmod{4} {\text{ with }} k=4r+2.
\end{cases}
\]

§ THE LOCAL SYSTEM AND THE ASSOCIATED HOMOLOGY

Recall the basis $e_0,e_1$ of the local system ${\mathrm{Kl}}_{2}^{\nabla}$
from (<ref>).
Note that the modified Bessel function $I_0(t)$ is entire. On the other hand, $K_0(s)$ extends analytically to a multivalued function on $\mathbb{C}^\times$ satisfying the monodromy $K_0(e^{2\pi\sqrt{-1}}t) = K_0(t) - 2\pi\sqrt{-1} I_0(t)$. This implies $e_{0},e_{1}$ undergo the monodromy action $T:\left(e_{0},e_{1}\right)\mapsto\left(e_{0},e_{1}+e_{0}\right)$ near $0$.
Then the basis in (<ref>) of the local system $(\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2})^{\nabla}$
satisfies $T:\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}\mapsto \frac{-1}{\sqrt{z}}e_{0}^{a}(e_{1}+e_{0})^{k-a}$ near $0$.

§.§ Rapid decay cycles

Write $k^{\prime}=\left\lfloor\frac{k-1}{2}\right\rfloor$. Denote the chains on $\mathbb{C}^{\times}$:

\begin{align*}
\sigma_{0}&={\text{the unit circle}}, {\text{ starting at }}1 {\text{ and oriented counterclockwise}};\\
\sigma_{+}&=\text{the interval $\left[1,\infty\right)$}, {\text{ starting at }}1 {\text{ toward }}+\infty.
\end{align*}

By the asymptotic expansion (<ref>), (<ref>), the horizontal sections $\frac{e_{0}^{a}e_{1}^{k-a}}{\sqrt{z}}$ decay exponentially along $\sigma_{+}$ for $a=0,1,\ldots,k^{\prime}$. Then, we have the following lemma describing the elements in the rapid decay homology.

The rapid decay cycles $\left\{\delta_{b}\right\}_{b=0}^{k^{\prime}}$ in $H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ are defined by

\begin{align}\label{rapid_decay_cycle_delta_b}
\delta_{b}&=\sigma_{+}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\dfrac{1}{2}\sigma_{0}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}+\sum_{n=1}^{k-b}d_{k-b}(n)\sigma_{0}^{2n}\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b},
\end{align}

where $d_{n}(i)$ satisfies

\begin{align*}
\sum_{i=1}^{n}d_{n}(i)\left(2i\right)^{m}&=-\dfrac{1}{2}, {\text{ for }}m=1,2,\cdots,n.
\end{align*}

In fact, by Cramer's rule, one solves $d_{n}(i)=\frac{(-1)^{i}}{n!2^{n+1}}\binom{n}{i}\frac{\left(2n-1\right)!!}{2i-1}$ uniquely.

It remains to prove that $d_{n}(i)$ makes $\delta_{b}$ into a cycle, that is, $\partial\delta_{b}=0$.

\begin{align*}
\partial\delta_{b}&= \frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}-\dfrac{1}{2}\left(\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}+\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+e_{0})^{k-b}\right)\\
\qedhere
\end{align*}

From this lemma, we have $k^{\prime}+1$ elements in the rapid decay homology $H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{\mathrm{m}},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$. At the end of this section, we will prove these elements form a basis (see Corollary <ref>).

§.§ Moderate decay cycles

Define one more chain

\begin{align*}
\mathbb{R}_{+}&= \text{the half line $\left[0,\infty\right)$, starting at $0$ toward $+\infty$}.
\end{align*}

Note that the modified Bessel function $K_{0}\left(t\right)$ has log pole at $0$, so the horizontal sections $\frac{e_{0}^{a}e_{1}^{k-a}}{\sqrt{z}}$ decay moderately along $\mathbb{R}_{+}$ near $0$ for $a=0,1,\ldots,\lfloor\frac{k}{2}\rfloor$. Moreover, by the expression (<ref>), $(I_{0}K_{0})^{a}$ decay polynomially along $\mathbb{R}_{+}$ near $\infty$. Then, we define the moderate decay cycles in $H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$

\begin{align}\label{moderate_decay_cycle_gamma_a}
\gamma_{a}=\mathbb{R}_{+}\otimes \frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}, {\text{ for }}a=0,1,2,\cdots,\lfloor\dfrac{k}{2}\rfloor.
\end{align}

They are indeed a cycle. The proof is the same as the above lemma by taking the limit as the radius of $\sigma_{0}$ tends to $0$. This gives a natural map

\begin{tikzcd}[row sep=tiny]
{H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}^{\nabla}_{2}\right)} \arrow[rr] \arrow[rr] &  & {H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{m},\sqrt{z}{\Sym^{k}}{\mathrm{Kl}}^{\nabla}_{2}\right)} \\
\delta_{b} \arrow[rr, maps to]                                                                               &  & \gamma_{b}                                                                             \\
\end{tikzcd}

for $b=0,1,\cdots,k^{\prime}$.[In fact, this is an isomorphism when $k\equiv 0,1,3\pmod{4}$ and has one-dimensional kernel when $k\equiv 2\pmod{4}$. See Corollary <ref>.]

The following lemma shows when $k\equiv 2\pmod{4}$, $\displaystyle\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\delta_{2j}$ belongs to the kernel of this map.

Let $\rho :\left\{\left(x,y\right)\in\mathbb{R}^{2}\mid 0<x,y,x+y<1\right\}\rightarrow\mathbb{C}$ be the open simplicial $2$-chain

\rho\left(x,y\right)=\tan\dfrac{\pi\left(x+y\right)}{2}\exp\left(4\sqrt{-1}\tan^{-1}\dfrac{y}{x}\right)

that covers $\mathbb{C}$ once. If $k$ is even, the singular chain

\Delta=\rho\otimes\left(\frac{1}{\sqrt{z}}(e_{1}-e_{0})^{k/2}\otimes e_{1}^{k/2}\right)

has moderate growth. In $H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$, one has

\begin{align*}
\sum_{j=0}^{k/4}\binom{k/2}{2j}\mathbb{R}_{+}\otimes \frac{1}{\sqrt{z}}e_{0}^{2j}e_{1}^{k-2j}&=\sum_{j=0}^{k/4}\binom{k/2}{2j}\gamma_{2j}=0 {\text{ if }}k\equiv 0\bmod{4};\\
\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\mathbb{R}_{+}\otimes \frac{1}{\sqrt{z}}e_{0}^{2j}e_{1}^{k-2j}&=\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\gamma_{2j}=0 {\text{ if }}k\equiv 2\bmod{4}.
\end{align*}

One computes $\partial\Delta$:

\begin{align*}
\partial\Delta &=\mathbb{R}_{+}\otimes\left(\frac{1}{\sqrt{z}}(e_{1}-e_{0})^{k/2}\otimes e_{1}^{k/2}\right)+\mathbb{R}_{+}\otimes\left(\frac{1}{\sqrt{z}}e_{1}^{k/2}\otimes(e_{0}+e_{1})^{k/2}\right)\\
&=\sum_{i=0}^{k/2}(-1)^{i}\binom{k/2}{i}\mathbb{R}_{+}\otimes \frac{1}{\sqrt{z}}e_{0}^{i}e_{1}^{k-i}+\sum_{i=0}^{k/2}\binom{k/2}{i}\mathbb{R}_{+}\otimes\frac{1}{\sqrt{z}}e_{0}^{i}e_{1}^{k-i}.
\end{align*}

Thus, the result follows.

Here, we have written down the $1+\lfloor\frac{k}{2}\rfloor$ elements $\left\{\gamma_{a}\right\}_{a=0}^{\lfloor k/2\rfloor}$ in the moderate decay homology $H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{\mathrm{m}},\sqrt{z}\Sym{\mathrm{Kl}}_{2}^{\nabla}\right)$. At the end of this section, we will prove that these elements form a basis modulo the linear relation given in the above lemma (see Corollary <ref>).

Similar to the middle part de Rham cohomology in the previous section, we define the middle part Betti homology $H_{1}^{{\mathrm{mid}}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ to be the image of $H_{1}^{\mathrm{rd}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ in $H_{1}^{\mathrm{mod}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$. More precisely, we have

\begin{align*}
\gamma_{i}\in H_{1}^{{\mathrm{mid}}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right) & {\text{ for }}0\leq i\leq k^{\prime} {\text{ when }}k\equiv 0,1,3\pmod{4};\\
\gamma_{i}\in H_{1}^{{\mathrm{mid}}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right) & {\text{ for }}1\leq i\leq k^{\prime} {\text{ when }}k\equiv 2\pmod{4}.
\end{align*}

Also, we may regard $H_{1}^{{\mathrm{mid}}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ as the quotient of $H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ containing the class of elements $\delta_{b}$. At the end of this section, we will prove these elements form a basis (see Corollary <ref>).

§.§ Betti intersection pairing

We use the topological pairing $\left\langle\ ,\ \right\rangle_{\mathrm{top}}$ introduced in section <ref> to define the Betti intersection pairing

\begin{tikzcd}[row sep=tiny]
{H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{m},\sqrt{z}{\Sym^{k}{\mathrm{Kl}}^{\nabla}_{2}}\right)\times H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}^{\nabla}_{2}\right)} \arrow[rr, "{\left\langle\ ,\ \right\rangle_{\mathrm{Betti}}}"] &  & \mathbb{Q}                                                                                \\
{\left(\delta=\sigma\otimes s_{\sigma},\gamma =\tau\otimes s_{\tau}\right)} \arrow[rr,maps to]                                                                                &  & {\displaystyle\sum_{\sigma\cap\tau}\left\langle s_{\sigma},s_{\tau}\right\rangle_{\mathrm{top}}}
\end{tikzcd}

where $\sigma\cap\tau$ is the topological intersection. At each topological intersection, the corresponding topological pairing is the pairing of sections at that point.

To compute the topological pairing with respect to the elements we had written down, we need to introduce the Euler numbers and Euler polynomials. The Euler polynomials $E_{n}(x)$ are given by the following power series, and we define the numbers $E_{n}$ for $n\geq 0$ as in [13],

\sum_{n=0}^{\infty}E_{n}(x)\frac{z^{n}}{n!}=\frac{2}{e^{z}+1}e^{xz},\ \ E_{n}(0)=E_{n}.

The first few $E_{n}$ are

\begin{tabular}{c|c|c|c|c|c|c}
$E_{0}$ & $E_{1}$ & $E_{2}$ & $E_{3}$ & $E_{4}$ & $E_{5}$ & $E_{6}$ \\ \hline
$1$     & $-1/2$  & $0$     & $1/4$   & $0$     & $-1/2$  & $0$
\end{tabular}

Note that we have the inversion formula for Euler polynomials,

Taking value at $x=0$, we get

\begin{align}\label{Euler_number_relation}
\sum_{k=0}^{n-1}\binom{n}{k}E_{k}=-2E_{n}.
\end{align}

We have the Betti intersection pairing

\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}=(-1)^{a}\frac{\binom{k-b}{a}}{\binom{k}{a}}
\frac{-1}{2}E_{k-b-a}=\frac{(-1)^{a+1}}{2}\frac{(k-a)!(k-b)!}{k!}\frac{E_{k-a-b}}{(k-a-b)!}

for $b=0,\cdots,k^{\prime}$ and $a=0,\cdots,\left\lfloor\frac{k}{2}\right\rfloor$.

Fix some $-\pi<\theta_{0}<0$ and let $x_{0}=\exp(\sqrt{-1}\theta_{0})$. To compute the pairing $\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}$, we move the ray $\sigma_{+}$ by adding the scalar $(x_{0}-1)$ and let the circle $\sigma_{0}$ start at $x_{0}$. Then the component $\sigma_{0}^{j}\otimes \frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}$ in the deformed $\delta_{b}$ meets $\gamma_{a}$ topologically $j$ times at the same point $+1\in\mathbb{C}^{\times}$. At the $i$-th intersection, the factor $\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}$ becomes $(-1)^{i-1}\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+(i-1)e_{0})^{k-b}$ and we have

\left\langle(-1)^{i-1}\frac{1}{\sqrt{z}}e_{0}^{b}(e_{1}+(i-1)e_{0})^{k-b},\frac{1}{\sqrt{z}}e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}=(-1)^{i-1}(i-1)^{k-b-a}(-1)^{a}\dfrac{\binom{k-b}{a}}{\binom{k}{a}}.

By adding these contributions, we obtain

\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}=\dfrac{\binom{k-b}{a}}{\binom{k}{a}}(-1)^{a}\sum_{n=1}^{k-b}d_{k-b}(n)T_{k-b-a}(2n),

T_{n}(k)=\sum_{\ell =0}^{k-1}(-1)^{\ell}\ell^{n}=-1+2^{n}-\cdots +(-1)^{k-1}(k-1)^{n}.

Kim [13] gave the following relation for $T_{n}(k)$:

T_{n}(k)=\dfrac{(-1)^{k+1}}{2}\sum_{\ell =0}^{n-1}\binom{n}{\ell}E_{\ell}k^{n-\ell}+\dfrac{E_{n}}{2}\left(1+(-1)^{k+1}\right).

Now, we have the following computation

\begin{align*}
\sum_{n=1}^{k-b}d_{k-b}(n)T_{k-b-a}(2n)&=\sum_{n=1}^{k-b}d_{k-b}(n)\left[\dfrac{-1}{2}\sum_{\ell =0}^{k-a-b-1}\binom{k-b-a}{\ell}E_{\ell}(2n)^{k-a-b-\ell}\right]\\
&=\dfrac{-1}{2}\sum_{\ell =0}^{k-a-b-1}\binom{k-a-b}{\ell}E_{\ell}\sum_{n=1}^{k-b}d_{k-b}(n)(2n)^{k-a-b-\ell}\\
&=\dfrac{1}{4}\displaystyle\sum_{\ell =0}^{k-b-a-1}\binom{k-b-a}{\ell}E_{\ell}
\end{align*}

and thus by (<ref>),

\[
\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}=\dfrac{-1}{2}E_{k-a-b}.
\qedhere
\]

Consider the $(k^{\prime}+1)\times (k^{\prime}+1)$ pairing matrix

\begin{cases}
\left(\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}\right)_{\ 0\leq b\leq k^{\prime},\ 0\leq a\leq\lfloor \frac{k}{2}\rfloor} & {\text{ if }}k {\text{ is odd,}}\\
\left(\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}\right)_{0\leq b\leq k^{\prime},\ 1\leq a\leq\frac{k}{2}} & {\text{ if }}k {\text{ is even.}}
\end{cases}

Using the pairing formula

\left<\delta_{b},\gamma_{a}\right>_{\mathrm{Betti}}=\frac{(-1)^{a+1}}{2}\frac{(k-a)!(k-b)!}{k!}\frac{E_{k-a-b}}{(k-a-b)!},

when $k$ is even, we have

\left(\begin{array}{ccc}
\frac{(-1)^{2}}{2}\frac{(k-1)!k!}{k!}\frac{E_{k-1}}{(k-1)!} & \cdots & \frac{(-1)^{k/2+1}}{2}\frac{(k/2)!k!}{k!}\frac{E_{k/2}}{(k/2)!}\\
\vdots & \ddots &\vdots \\
\frac{(-1)^{2}}{2}\frac{(k-1)!(k/2+1)!}{k!}\frac{E_{k/2}}{(k/2)!} & \cdots &\frac{(-1)^{k/2+1}}{2}\frac{(k/2)!(k/2+1)!}{k!}\frac{E_{1}}{\left(1\right)!}
\end{array}\right)

and that

\left(\begin{array}{ccc}
\frac{(-1)}{2}\frac{(k-1)!(k-1)!}{(k-1)!}\frac{E_{k-1}}{(k-1)!} & \cdots & \frac{(-1)^{k/2}}{2}\frac{(k/2)!(k-1)!}{(k-1)!}\frac{E_{k/2}}{(k/2)!}\\
\vdots & \ddots &\vdots \\
\frac{(-1)}{2}\frac{(k-1)!(k/2)!}{(k-1)!}\frac{E_{k/2}}{(k/2)!} & \cdots &\frac{(-1)^{k/2}}{2}\frac{(k/2)!(k/2)!}{(k-1)!}\frac{E_{1}}{\left(1\right)!}
\end{array}\right).

Then we obtain the relation

\begin{equation}\label{Betti_relation}
B_{k}=-\frac{1}{k} \diag(k,k-1,\cdots,k/2+1)\cdot B_{k-1}.
\end{equation}

Thus, $B_{k}$ and $B_{k-1}$ have the same rank whenever $k$ is even. Moreover, we may compute the determinant of $B_{k}$ explicitly given in the following proposition.

The determinant of $B_{k}$ is given by the following.

* When $k$ is odd, we have

\begin{align*}
\det B_{k}&=2^{-k-1}\prod_{a=1}^{k^{\prime}}a^{k^{\prime}+1-2a}(2a+1)^{k^{\prime}-2a}.
\end{align*}

* When $k$ is even, we have

\begin{align*}
\det B_{k}&=(-1)^{(k^{\prime}+1)(k^{\prime}+3)}2^{-k}\prod_{a=1}^{k^{\prime}}(a+1)^{k^{\prime}-2a-1}(2a+1)^{k^{\prime}+1-2a}.
\end{align*}

In particular, they are all non-vanishing.

Set $\mathcal{E}_{2n-1}=(-1)^{n}2^{2n-1}E_{2n-1}$. Apply the result <cit.> in the following computations.

When $k=2k'+1$ is odd, we have

\begin{align*}
\det B_{k} &=\frac{(-1)^{(k'+1)(k'+2)/2}}{(2\cdot k!)^{k'+1}}\left[\prod_{i=k'+1}^{k}i!\right]^{2}\det\left(\begin{array}{ccc}
\frac{E_{k}}{k!} & \cdots &\frac{E_{k'+1}}{(k'+1)!}\\
\vdots & \ddots &\vdots \\
\frac{E_{k'+1}}{(k'+1)!} & \cdots &\frac{E_{1}}{\left(1\right)!}
\end{array}\right)\\
&=\frac{1}{(2\cdot k!)^{k'+1}}\left[\prod_{i=k'+1}^{k}i!\right]^{2}\frac{1}{2^{(k'+1)^{2}}}\det\left(\begin{array}{ccc}
\frac{\mathcal{E}_{k}}{k!} & \cdots &\frac{\mathcal{E}_{k'+1}}{(k'+1)!}\\
\vdots & \ddots &\vdots \\
\frac{\mathcal{E}_{k'+1}}{(k'+1)!} & \cdots &\frac{\mathcal{E}_{1}}{\left(1\right)!}
\end{array}\right)\\
&=\frac{1}{2^{(k'+1)(k'+2)}(k!)^{k'+1}}\left[\prod_{i=k'+1}^{k}i!\right]^{2} 2^{k'^{2}}\frac{k'!}{k!}\prod_{j=1}^{k'}\frac{(j-1)!^{2}}{(2j-1)!^{2}}\\
\end{align*}

When $k=2k'+2$ is even, we have

\begin{align*}
\det B_{k}&=\frac{(-1)^{(k'+1)(k'+4)/2}}{(2\cdot k!)^{k'+1}}\left[\prod_{i=k'+1}^{k-1}i!\left(i+1\right)!\right] \det\left(\begin{array}{ccc}
\frac{E_{k-1}}{(k-1)!} & \cdots &\frac{E_{k'+1}}{(k'+1)!}\\
\vdots & \ddots &\vdots \\
\frac{E_{k'+1}}{(k'+1)!} & \cdots &\frac{E_{1}}{\left(1\right)!}
\end{array}\right)\\
&=\frac{(-1)^{(k'+1)(k'+4)/2}}{(2\cdot k!)^{k'+1}}\left[\prod_{i=k'+1}^{k-1}i!\left(i+1\right)!\right] \\
\frac{\mathcal{E}_{k-1}}{(k-1)!} & \cdots &\frac{\mathcal{E}_{k'+1}}{(k'+1)!}\\
\vdots & \ddots &\vdots \\
\frac{\mathcal{E}_{k'+1}}{(k'+1)!} & \cdots &\frac{\mathcal{E}_{1}}{\left(1\right)!}
\end{array}\right)\\
&=\frac{(-1)^{(k'+1)(k'+3)}}{2^{(k'+1)(k'+2)}(k!)^{k'+1}}\left[\prod_{i=k'+1}^{k-1}i!\left(i+1\right)!\right] 2^{k'^{2}}\frac{k'!}{(k-1)!}\prod_{j=1}^{k'}\frac{(j-1)!^{2}}{(2j-1)!^{2}}\\
\qedhere
\end{align*}

Finally, we conclude the basis of Betti homologies.

The natural map

\begin{tikzcd}
{H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{m},\sqrt{z}{\Sym^{k}{\mathrm{Kl}}^{\nabla}_{2}}\right)} \arrow[rr] \arrow[rr] &  & {H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{m},\sqrt{z}{\Sym^{k}{\mathrm{Kl}}^{\nabla}_{2}}\right)}
\end{tikzcd}

sending $\delta_{b}$ to $\gamma_{b}$ is an isomorphism when $k\equiv 0,1,3\pmod 4$ and has a one-dimensional kernel when $k\equiv 2\pmod 4$. Moreover, we find the following.

* $H^{\mathrm{rd}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ has basis $\left\{\delta_{b}\right\}_{b=0}^{k^{\prime}}$.

* $H^{\mathrm{mod}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ has basis

\[
\begin{cases}
\left\{\gamma_{a}\right\}_{a=0}^{k^{\prime}} & {\text{ if }} k {\text{ is odd.}}\\
\left\{\gamma_{a}\right\}_{a=1}^{k/2} & {\text{ if }} k {\text{ is even.}}
\end{cases}
\]

* $H^{\mathrm{mid}}_{1}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)$ has basis

\[
\begin{cases}
\left\{\gamma_{a}\right\}_{a=0}^{k^{\prime}} & {\text{ if }}k\equiv 0,1,3\pmod{4}\\
\left\{\gamma_{a}\right\}_{a=1}^{k^{\prime}} & {\text{ if }}k\equiv 2\pmod{4}\\
\end{cases}
\]

From the duality of Betti homology and de Rham cohomology, the dimension of rapid decay homology and moderate decay homology are both $k^{\prime}+1$ by Proposition <ref>. The above non-vanishing determinant of $B_{k}$ shows that the natural map is an isomorphism when $k\equiv 1,3\pmod 4$, and thus the elements form a basis. When $k\equiv 2\pmod{4}$, Lemma <ref> describes the one-dimensional kernel of the natural map. Moreover, $B_{k}$ has full rank $k/2$ when $k$ is even by the relation (<ref>). Hence, we conclude that the natural map is an isomorphism when $k\equiv 0\pmod{4}$.

§ TWISTED MOMENTS AS PERIODS

In this section, we compute the periods pairing of the basis of de Rham cohomology and Betti homology in Corollary <ref> and Corollary <ref>. Also, we interpret these periods as the Bessel moments and regularized Bessel moments.

§.§ Bessel moments and regularized Bessel moments

The Bessel moments are defined by

\begin{align*}
\end{align*}

provided the convergence of the integral, that is, for non-negative integers $k,a,b$ satisfying $a\leq k^{\prime}$, $b\geq 0$ or $a=\frac{k}{2}$, $0\leq b<k^{\prime}$. The justification is given in the following lemma. Moreover, if $a=\frac{k}{2}$ and $b\geq k^{\prime}$, by analyzing the singular integral, we could define the regularized Bessel moments $\mathrm{IKM}_{k}^{\mathrm{reg}}\left(\frac{k}{2},b\right)$ by subtracting the singular part of the integral. The precise definition is also given in the following lemma.

The integration of the Bessel moments

converge for non-negative integers $k,a,b$ satisfying $a\leq k^{\prime}$, $b\geq 0$ or $a=\frac{k}{2}$, $0\leq b<k^{\prime}$. Moreover, in the case that $a=\frac{k}{2}$ and $b\geq k^{\prime}$ with $b$ even, we may modify the singular integral. In other words, the following two limits exist for $2j\geq k^{\prime}$:

\begin{align*}
{\mathrm{IKM}}^{\mathrm{reg}}_{k}\left(\frac{k}{2},2j\right)&:=\lim_{t\rightarrow\infty}\left(\int_{0}^{t}(I_{0}K_{0})^{2r+2}s^{2j}{\mathrm{d}}s-\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}t^{2m+1}}{2^{k-2j+2m}(2m+1)}\right)\ \ {\text{if }}k=4r+4,\\
{\mathrm{IKM}}^{\mathrm{reg}}_{k}\left(\frac{k}{2},2j\right)&:=\lim_{t\rightarrow\infty}\left(\int_{0}^{t}(I_{0}K_{0})^{2r+2}s^{2j}{\mathrm{d}}s-\frac{2\gamma_{k,j-r}}{2^{k-2j}}\int_{0}^{t}\frac{\mathrm{d}s}{s}-\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}t^{2m+2}}{2^{k-2j+2m+1}(2m+2)}\right)\ \ {\text{if }}k=4r+2.
\end{align*}

Near $0$, we have the asymptotic

\begin{align}
K_0(t)&=-\left(\gamma +\log\frac{t}{2}\right)+O\left(t^{2}\log t\right),\label{expansion_of_K_0_at_0}
\end{align}

where $\gamma$ is the Euler constant. Then, the integration $\int_{0}^{1}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t$ converge for all $0\leq a\leq\frac{k}{2}$ and any $b\geq 0$.

Near $\infty$, from (<ref>) and (<ref>), when $0\leq a\leq k^{\prime}$, $I_{0}^{a}(t)K_{0}^{k-a}(t)$ is exponential decay and hence the integral $\int_{1}^{\infty}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t$ converges.

When $a=\frac{k}{2}$, near $\infty$, by (<ref>), we have the asymptotic expansion

\left(I_{0}K_{0}\right)^{k/2}t^{2j}=\frac{1}{2^{k/2}}\sum_{n=0}^{\infty}\gamma_{k,n}4^{n}t^{-2n-k/2+2j}.

Using the fact that $\int_{1}^{\infty}t^{\alpha}{\mathrm{d}}t$ converges if and only if $\alpha <-1$, the singular part of the integral
\int_{0}^{t}\left(I_{0}K_{0}\right)^{k/2}s^{2j}{\mathrm{d}}s

\begin{align}
\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}}{2^{k-2j+2m}}\frac{t^{2m+1}}{\left(2m+1\right)} & {\text{ if }}k=4r+4\label{Bessel_moments_singular_term_4r+4}\\
\frac{2\gamma_{k,j-r}}{2^{k-2j}}\int_{0}^{t}\frac{\mathrm{d}s}{s}+\sum_{m=0}^{j-r-1}\frac{\gamma_{k,j-r-1-m}}{2^{k-2j+2m+1}}\frac{t^{2m+2}}{(2m+2)}& {\text{ if }}k=4r+2\label{Bessel_moments_singular_term_4r+2}
\end{align}

Hence, after subtracting the singular part of the integral, we obtained the regularized Bessel moments.

For $a=\frac{k}{2}$ and $b\geq k^{\prime}$ with odd $b$, the integral $\int_{0}^{\infty}I_{0}^{a}(t)K_{0}^{k-a}(t)t^{b}{\mathrm{d}}t$ also diverges. We may similarly define the regularized Bessel moments in this case. However, we will only use the modified Bessel moments with $b$ even, so we only write the case when $b$ is even for convenience.

§.§ Period pairing and compactly supported period pairing

Define the period pairing using topological pairing introduced in section <ref>

% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12GYAzHACgASAfWABGAL4A9YJwC2dHAAsATnOAqoEiZ259+8xUoBGx4AHEJouRNKc4ARxU5gALwmz2C5WuABlAE8baWAAaw9DH3UAaQZtUQAmEM4wOmMGOh12FSwAcyUcAEpOHPyi0UiTMwBhbU48OXgAAl1eAUEZSQqvI18oACU6rjaDHuVTCytgGzt2KB4VOgBjcQj2R2c3bU9vVXVA4Jlwnd6YuKmkkryC4uzr8pPxmu0QW3RMXHxCFATyKlpGCw2JUJrVXv8YFBcvAiKAFhA5EgyCAcBAkGJqPRmKxEBxhnxdHQwLluJwmrMWncyiUiSSYN1dr40DAVC8JBQJEA
\begin{tikzcd}
{H_{1}^{\mathrm{rd}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)_{\mathbb{C}}\times H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)_{\mathbb{C}}} \arrow[rr, "{\left\langle\ ,\ \right\rangle_{\mathrm{per}}}"] &  & \mathbb{C}
\end{tikzcd}

\left\langle\sigma\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b},\omega\right\rangle_{\mathrm{per}}=\int_{\sigma}\frac{1}{\sqrt{z}}\left\langle e_{0}^{b}e_{1}^{k-b},\omega\right\rangle_{\mathrm{top}}.

Also, we define the compactly supported period pairing

% https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAB12GYAzHACgASAPWABGAL4B9YJwC2dHAAsATnOBQASqQDGEiZ259+8xUoBG54AHFpwORNKcoPFXR3iJs9nACOKnGAAL31vBWU1YABlAE8HCVEAay9TCPUAaQZ9GQAmA3YVLABzJRwASk5CkvKZVIsrAGF9Tjw5eENeAUEZSVE6yLloZq5Ok3Zw+ps7Bycff0CQsLNI2PiklPHljKy7PL72MDpzBjp8qtKKguKL2s3lS2AmiRBHdExcfEIUHPIqWkYWGw6g8mi8-jAoEV4ERQK4IHIkGQQDgIEgxNR6MxWIgOCM+IY6GAitxOAACWakyrXHCVQnEmC3CaRNAwFS6fRgiRAA
\begin{tikzcd}
{H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)_{\mathbb{C}}\times H_{1}^{\mathrm{mod}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}^{\nabla}\right)_{\mathbb{C}}} \arrow[rr, "{\left\langle\ ,\ \right\rangle_{\mathrm{per,c}}}"] &  & \mathbb{C}
\end{tikzcd}

\begin{align*}
\left\langle(\xi,\eta,\omega),\sigma\otimes\frac{1}{\sqrt{z}}e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{per,c}}&=\int_{\sigma}\frac{1}{\sqrt{z}}\left\langle\omega , e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{top}}
-\frac{1}{\sqrt{z}}\left\langle\eta, e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{top}}+\frac{1}{\sqrt{z}}\left\langle\xi,e_{0}^{b}e_{1}^{k-b}\right\rangle_{\mathrm{top}}.
\end{align*}

Note that the order of homology and cohomology in these two pairing are different. This is because we want to write down the matrix expression of quadratic relation (<ref>) preventing the transpose notation.

The period pairing of the rapid decay cycle $\delta_{b}$ in (<ref>) and the de Rham cohomology $\omega_{k,j}$ in Definition <ref> is given by

\left\langle \delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}=(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}{\mathrm{IKM}}_{k}(b,2j)

for $0\leq b\leq k^{\prime}$ and $0\leq j\leq k^{\prime}$.

Denote $\varepsilon \sigma_{0}$ to be the scaling of the chain $\sigma_{0}$, that is, $\varepsilon\sigma_{0}$ is a chain of a circle of radius $\varepsilon$. Then, by the scaling on the chain, we have

\begin{align*}
\left\langle \delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}&=\left\langle \varepsilon\delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}\\ &=\int_{\varepsilon\left(\sigma_{+}-\frac{1}{2}\sigma_{0}+\sum_{n=1}^{k-b}d_{k-b}(n)\sigma_{0}^{2n}\right)}\frac{1}{\sqrt{z}}\left\langle e_{0}^{b}e_{1}^{k-b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{top}}\\
\end{align*}

Changing of coordinate by $z=\frac{t^{2}}{4}$, the first term becomes

When $\varepsilon\rightarrow 0^{+}$, this term tends to $(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}{\mathrm{IKM}}_{k}(b,2j)$.

For the other two terms, note that $\int_{\varepsilon\sigma_{0}^{p}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z$ tends to zero as $\varepsilon\rightarrow 0^{+}$ for the following reason. As $s\rightarrow 0^{+}$, we have the asymptotic expansion (<ref>) and (<ref>). Then, as $\varepsilon\rightarrow 0^{+}$ for all $j\geq 0$, we have the estimate

\begin{align*}
\left|\int_{\varepsilon\sigma_{0}^{p}}z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\mathrm{d}z\right|&\leq \int_{\varepsilon\sigma_{0}^{p}}\left|z^{j-1}\sqrt{z}I_{0}(2\sqrt{z})^{b}K_{0}(2\sqrt{z})^{k-b}\right|\left|\mathrm{d}z\right|\\
&\leq \int_{0}^{2\pi p}\varepsilon^{j-1}\sqrt{\varepsilon}\left|I_{0}\left(2\sqrt{\varepsilon e^{i\theta}}\right)^{b}K_{0}\left(2\sqrt{\varepsilon e^{i\theta}}\right)^{k-b}\right|\varepsilon{\mathrm{d}}\theta\\
&\leq \varepsilon^{j}\sqrt{\varepsilon}\int_{0}^{2\pi p}\left|\gamma + \log\sqrt{\varepsilon e^{i\theta}}\right|^{k-b}{\mathrm{d}}\theta\\
&=\varepsilon^{j}\sqrt{\varepsilon}\int_{0}^{2\pi p}\left|\gamma + \log\sqrt{\varepsilon }+\frac{1}{2}i\theta\right|^{k-b}{\mathrm{d}}\theta\rightarrow 0.
\qedhere
\end{align*}

The compactly supported period pairing of the compactly supported de Rham cohomology $\widetilde{\omega}_{k,j}$ in Definition <ref> and moderate decay cycle $\gamma_{a}$ in (<ref>) is given by

\left\langle\widetilde{\omega}_{k,j},\gamma_{a}\right\rangle_{\mathrm{per,c}}=2^{k-2j}(-1)^{k-a}(\pi\sqrt{-1})^{a}\cdot{\mathrm{IKM}},

where $0\leq a\leq \lfloor k/2\rfloor, 0\leq j\leq k'$
with $j\neq r$ if $k\equiv 2\pmod{4}$,

\[ {\mathrm{IKM}}=\begin{cases}
& \text{if $4\mid k, a=k/2, r+1\leq j\leq  k^{\prime}$}, \\
& \text{if $4\mid (k+2), 0\leq a\leq k', r+1\leq j\leq k'$}, \\
& \text{if $4\mid (k+2), a=k'+1, r+1\leq j\leq k'$}, \\
& \text{otherwise}.
\end{cases}\]

Moreover, when $k=4r+2$, we have

\left\langle\widehat{m}_{2r+1},\gamma_{a}\right\rangle_{\mathrm{per,c}}=\delta_{a,2r+1}(\pi\sqrt{-1})^{a}2^{k}\frac{1}{\binom{k}{k/2}}.

When $k\equiv 1,3\pmod{4}$. We compute the compactly supported period pairing

\begin{align*}
\left\langle\left(\xi_{j},\eta_{j},\omega_{k,j}\right), \gamma_{a}\right\rangle_{\mathrm{per,c}}&=\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}} z^{j}\frac{{\mathrm{d}}z}{z}-\frac{1}{\sqrt{z}}\left\langle -\sum_{c=0}^{k}\binom{k}{c}\eta_{j,c}\frac{e_{0}^{k-c}\overline{e}_{1}^{c}}{\sqrt{z}} ,e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}\\
&\hspace{20pt}+\frac{1}{\sqrt{z}}\left\langle \sum_{c=0}^{k}\xi_{j,c}v_{0}^{c}v_{1}^{k-c} ,e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}\\
\end{align*}

where the last equality follows by the change of variable $z=\frac{s^{2}}{4}$. The first term converges by Lemma <ref>. Since $k>2a$, by (<ref>), the second term tends to zero as $s\rightarrow\infty$. The third term tends to zero as $s\rightarrow 0$ since all $\xi_{j,c}\in\mathbb{Q}\llbracket\frac{s^{2}}{4}\rrbracket$ and the topological pairing gives a factor $\frac{s^{2}}{4}$.

When $k\equiv 0\pmod{4}$, write $k=4r+4$. We compute the compactly supported period pairing

\begin{align*}
\left\langle\left(\xi_{j},\eta_{j},\omega_{k,j}\right), \gamma_{a}\right\rangle_{\mathrm{per,c}}&=\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}} z^{j}\frac{{\mathrm{d}}z}{z}-\frac{1}{\sqrt{z}}\left\langle -\sum_{c=0}^{k}\binom{k}{c}\eta_{j,c}\frac{e_{0}^{k-c}\overline{e}_{1}^{c}}{\sqrt{z}},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}\\
\end{align*}

where the last equality is the change of variable $z=\frac{s^{2}}{4}$. The third term tends to zero as $s\rightarrow 0$ since all $\xi_{j,c}\in\mathbb{Q}\llbracket\frac{s^{2}}{4}\rrbracket$ and the topological pairing gives a factor $\frac{s^{2}}{4}$. By the same argument above, when $a=0,1,\cdots,\frac{k-2}{2}=k^{\prime}$, that is, $k>2a$, we have that the first term converges and the second term tends to zero as $s\rightarrow\infty$.

Now, we turn to analyze the case that $a=\frac{k}{2}$. The pairing becomes

\left\langle\left(\xi_{j},\eta_{j},\omega_{k,j}\right), \gamma_{a}\right\rangle_{\mathrm{per,c}}=2^{k-2j}(-\pi\sqrt{-1})^{a}\int_{0}^{s}I_0(t)^{a}K_0(t)^{k-a}s^{2j}{\mathrm{d}}s+(-\pi\sqrt{-1})^{a}\eta_{j,2r+2}

This term converges as $s\rightarrow\infty$ for the following reason:

The singular part of the integral $\left(I_{0}K_{0}\right)^{2r+2}s^{2j}$ is given by (<ref>) and $\eta_{j,2r+2}$ has expansion

\eta_{j,2r+2}\sim\frac{2^{2r-2j+1}}{r-j+1/2}s^{2j-2r-1}\cdot G_{i}\sim 2^{2r-2j+1}\sum_{n=0}^{\infty}\frac{2^{2n}\gamma_{k,n}}{r-j+1/2+n}s^{2j-2r-1-2n}

Thus, both of the singular terms cancel.

When $k\equiv 2\pmod{4}$, write $k=4r+2$. Recall in Definition <ref>, we have $k^{\prime}+1$ elements in $H^{1}_{{\mathrm{dR,c}}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$:

\widetilde{\omega}_{k,j}=\left(\xi_{j},\eta_{j},\omega_{k,j}\right),\ \ j=0,1,\cdots,\widehat{r},\cdots,k^{\prime} {\text{ and }}\widehat{m}_{2r+1}=\left(0,\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},0\right).

If we use the convention that $\gamma_{k,p}=0$ whenever $p<0$, we rewrite

\begin{align*}
\widetilde{\omega}_{i,k}&=\left(\xi_{i},\eta_{i},\omega_{i,k}\right)\\
&\hspace{20pt}-\sum_{\substack{a=0\\a\neq k/2}}^{k}\binom{k}{a}(\eta_{i,a}-\gamma_{k,i-r}\eta_{r,a})\frac{e_{0}^{k-a}\overline{e}_{1}^{a}}{\sqrt{z}}-\binom{k}{k/2}\eta_{i,2r+1}\frac{(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},\\
\end{align*}

In the pairing $\left\langle\widetilde{\omega}_{j,k},\gamma_{a}\right\rangle_{\mathrm{per,c}}$, the third term

tends to zero as $s\rightarrow 0$ since all $\xi_{j,c}\in\mathbb{C}\llbracket\frac{s^{2}}{4}\rrbracket$ and the topological pairing gives a factor $\frac{s^{2}}{4}$.
The other two terms are equal to

\begin{align*}
&\left(\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}} z^{j}\frac{{\mathrm{d}}z}{z}-\gamma_{k,j-r}\int_{\mathbb{R}_{+}}\frac{1}{\sqrt{z}}\left\langle v_{0}^{k},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}} z^{r}\frac{{\mathrm{d}}z}{z}\right)\\
&-\frac{1}{\sqrt{z}}\left\langle -\sum_{\substack{b=0\\b\neq k/2}}^{k}\binom{k}{b}\left(\eta_{j,b}-\gamma_{k,j-r}\eta_{r,b}\right)\frac{e_{0}^{k-b}\overline{e}_{1}^{b}}{\sqrt{z}}-\binom{k}{k/2}\eta_{j,2r+1}\frac{(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}}, e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}\\
&\hspace{20pt}+\frac{1}{z}\left\langle\sum_{\substack{b=0\\b\neq k/2}}^{k}\binom{k}{b}\left(\eta_{j,b}-\gamma_{k,j-r}\eta_{r,b}\right)e_{0}^{k-b}\overline{e}_{1}^{b}+\binom{k}{k/2}\eta_{j,2r+1}(e_{0}\overline{e}_{1})^{2r+1},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}
\end{align*}

We analyze the convergence of these terms. When $2a<k$ or $j<r$, the integral $\int_{0}^{t}I_0(s)^{a}K_0(s)^{k-a}s^{2j}{\mathrm{d}}s$ converges as $t\rightarrow\infty$ by Lemma <ref>. The second term is equal to

\begin{align*}
\end{align*}

By the expansion of $\eta_{i,a}$:

\eta_{i,a}\sim\frac{\sqrt{\pi}^{k-2a}}{k-2a}e^{-(k-2a)s}\left(\frac{4}{s^{2}}\right)^{k/4-i}\cdot G_{i,a},

where $G_{i,a}\in 1+\frac{2}{s}\mathbb{Q}\llbracket\frac{2}{s}\rrbracket$, this term tends to $0$ as $s\rightarrow\infty$.

When $a=\frac{k}{2}$ and $j\geq r$, the integral $\int_{0}^{t}I_0(s)^{a}K_0(s)^{k-a}s^{2j}{\mathrm{d}}s$ has the singular part (<ref>). The second term is equal to

\[ (-1)^{2r+1}(\pi\sqrt{-1})^{2r+1}\eta_{j,2r+1}
=(-1)^{2r+1}(\pi\sqrt{-1})^{2r+1}\frac{1}{r-j}\left(\frac{4}{s^{2}}\right)^{r-j}\cdot H_{i} \]

where $H_{i}\in 1+\frac{4}{s^{2}}\mathbb{Q}\llbracket\frac{4}{s^{2}}\rrbracket$. Thus, the singular part of this term is

In consequence, the singular parts cancel.

Finally, for $a=0,1,\cdots,\frac{k}{2}$, we have

\begin{align*}
\left\langle\left(0,\frac{2^{k}(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},0\right),\gamma_{a}\right\rangle_{\mathrm{per,c}}&=-\frac{2^{k}}{\sqrt{z}}\left\langle\frac{(e_{0}\overline{e}_{1})^{2r+1}}{\sqrt{z}},e_{0}^{a}e_{1}^{k-a}\right\rangle_{\mathrm{top}}\\
\qedhere
\end{align*}

Form the period matrix $P$ using period pairing under the basis $\left\{\delta_b\right\}_{b=0}^{k^{\prime}}$ of $H_{1}^{\mathrm{rd}}$ and $\left\{\omega_{k,j}\right\}_{j=0}^{k^{\prime}}$ of $H^{1}_{\mathrm{dR}}$.

P=(P_{bj}) = \left\langle \delta_{b},v_{0}^{k}z^{j}\frac{\mathrm{d}z}{z}\right\rangle_{\mathrm{per}}=(\pi\sqrt{-1})^{b}(-1)^{k-b}2^{k-2j}{\mathrm{IKM}}_{k}(b,2j).

for $0\leq b\leq k^{\prime}$ and $0\leq j\leq k^{\prime}$. Then, $P$ is invertible.

In fact,

\begin{align*}
\det P&=(\pi\sqrt{-1})^{k^{\prime}(k^{\prime}+1)/2}(-1)^{(2k-k^{\prime})(k^{\prime}+1)/2}2^{(k-k^{\prime})(k^{\prime}+1)}\det\left({\mathrm{IKM}}_{k}(b,2j)\right),
\end{align*}

\begin{align*}
\det\left({\mathrm{IKM}}_{k}(b,2j)\right)=
\begin{cases}
\det\left(M_{k^{\prime}+1}\right) & {\text{if }}k{\text{ is odd;}}\\
\det\left(N_{k^{\prime}+1}\right) & {\text{if }}k{\text{ is even.}}\\
\end{cases}
\end{align*}

The definition of $M_{r}$ and $N_{r}$ are given in the appendix <ref> and its determinant is given in Corollary <ref> explicitly.

§.§ -linear and quadratic relations on Bessel moments

We have now developed all the tools and computations to see the wonderful results in $\mathbb{Q}$-linear and quadratic relations on Bessel moments.

For $k=4r+4$,

\begin{align*}
\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2i)
= \begin{cases}
(-1)^r\pi^{2r+2}{\mathrm{IKM}}_{k}(2r+2,2i) & \text{if $0\leq i\leq r$}, \\
& \text{if $r+1\leq i\leq k^{\prime}$}.
\end{cases}
\end{align*}

For $k=4r+2$,

\begin{align}\label{4r+2_algebraic_relation}
\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2i)
= \begin{cases}
0 & \text{if $0\leq i< r$}, \\
\gamma_{k,i-r}2^{2i-2r}
\displaystyle\sum_{j=0}^{r}\binom{k/2}{2j}(-1)^{j}\pi^{2j}{\mathrm{IKM}}_{k}(2j,2r)
& \text{if $r< i\leq 2r$}.
\end{cases}
\end{align}

By Lemma <ref>, we know that

\begin{align*}
\sum_{j=0}^{k/4}\binom{k/2}{2j}\gamma_{2j}&=0 {\text{ if }}k\equiv 0\pmod{4};\\
\sum_{j=0}^{\left(k-2\right)/4}\binom{k/2}{2j}\gamma_{2j}&=0 {\text{ if }}k\equiv 2\pmod{4}.
\end{align*}

Then take pairing with $\widetilde{\omega}_{k,i}$ in the compactly supported de Rham cohomology. Combining the result in Proposition <ref>, we obtain the desired algebraic relation.

The above linear algebraic relations
for $i$ in the range $0\leq i\leq r$, under the name sum rule identities, are previously proved by analytic method in [17] (see <cit.> for $k\equiv 2\pmod{4}$ and <cit.> for $k\equiv 4\pmod{4}$).

For any $k$ and any $0\leq a\leq k^{\prime}$, the dimension of the $\mathbb{Q}$-vector space generated by the Bessel moments has an upper bound:

\dim {\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j\in\left\{0\right\}\cup \mathbb{N}\right\}\leq k^{\prime}+1.

For $k$ is even, the dimension of the $\mathbb{Q}$-vector space generated by the regularized Bessel moments has an upper bound:

\dim {\mathrm{span}}_{\mathbb{Q}}\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}(k/2,2j)\mid j\in\left\{0\right\}\cup \mathbb{N}\right\}\leq k^{\prime}+1.

In the even case, note that when $0\leq j\leq \left\lfloor\frac{k-1}{4}\right\rfloor =r$, we do not need to regularize the Bessel moments, that is, ${\mathrm{IKM}}^{\mathrm{reg}}_{k}(k/2,2j)={\mathrm{IKM}}_{k}(k/2,2j)$ (see Lemma <ref>).

We know that the dimension of $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ and $H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ are $k^{\prime}+1$.

Now, since $\left\{v_{0}^{k}z^{j}\frac{{\mathrm{d}}z}{z}\right\}_{j=0,\cdots,k^{\prime}}$ form a basis of $H^{1}_{\mathrm{dR}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$, we may write $v_{0}^{k}z^{s}\frac{{\mathrm{d}}z}{z}$ into the $\mathbb{Q}$-linear combination of the basis. Then after we take the period pairing with the rapid decay cycle $\delta_{a}$ (see Proposition <ref>), the $\mathbb{Q}$-linear relation becomes a $\mathbb{Q}$-linear relation to the Bessel moments

\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j=0,\cdots,k^{\prime}\right\}\cup \left\{{\mathrm{IKM}}_{k}(a,2s)\right\}.

For $k$ is even, similarly, when $s>k^{\prime}$, write $\widetilde{\omega}_{k,s}\in H^{1}_{\mathrm{dR,c}}\left(\mathbb{G}_{m},\sqrt{z}\Sym^{k}{\mathrm{Kl}}_{2}\right)$ into the $\mathbb{Q}$-linear combination of basis. Then taking the compactly supported period paring (see Proposition <ref>), the $\mathbb{Q}$-linear equivalence become the $\mathbb{Q}$-linear relation to the regularized Bessel moments

\[
\left\{{\mathrm{IKM}}_{k}(a,2j)\mid j=0,\cdots,r-1\right\}\cup\left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}(a,2j)\mid j=r,\cdots,k^{\prime}\right\}\cup \left\{{\mathrm{IKM}}^{\mathrm{reg}}_{k}(a,2s)\right\}.
\qedhere
\]

In order to write down the explicit $\mathbb{Q}$-linear combination in the above corollary, we need to trace back to the proof of Lemma <ref> in <cit.>. In [6], they provide a recurrence to find out the $\mathbb{Q}$-linear combination for Bessel moments by analyzing the symmetric power of the modified Bessel differential operator. Moreover, there is a similar result in [18] for the $\mathbb{Q}$-linear dependence for Bessel moments ${\mathrm{IKM}}_{k}(a,2j-1)$. Our result is parallel to his result.

Using the basis of $H^{{\mathrm{rd}}}_{1}$, $H^{{\mathrm{mod}}}_{1}$, $H_{\mathrm{dR}}^{1}$, and $H_{\mathrm{dR,c}}^{1}$ described in Corollaries <ref>, <ref>, we form the pairing matrices that we have defined in previous texts:

* $B$, the Betti intersection pairing between $H^{{\mathrm{rd}}}_{1}$ and $H^{{\mathrm{mod}}}_{1}$. (Proposition <ref>.)

* $D$, the Poincaré pairing between $H_{\mathrm{dR,c}}^{1}$ and $H_{\mathrm{dR}}^{1}$. (Proposition <ref>.)

* $P$, the period pairing between $H^{{\mathrm{rd}}}_{1}$ and $H_{\mathrm{dR}}^{1}$. (Proposition <ref>.)

* $P_{\mathrm{c}}$, the period pairing between $H_{\mathrm{dR,c}}^{1}$ and $H^{{\mathrm{mod}}}_{1}$. (Proposition <ref>.)

* $B_{\mathrm{mid}}$, the Betti pairing on $H^{\mathrm{mid}}_{1}$.

* $D_{\mathrm{mid}}$, the Poincaré pairing on $H_{\mathrm{mid}}^{1}$.

* $P_{\mathrm{mid}}$, the period pairing between $H^{\mathrm{mid}}_{1}$ and $H_{\mathrm{mid}}^{1}$.[Note that $B,D,P,P_{\mathrm{c}}$ are square matrices of size $k^{\prime}+1$ and that $B_{\mathrm{mid}},D_{\mathrm{mid}},P_{\mathrm{mid}}$ are of size $k^{\prime}+1-\delta_{4\mathbb{Z}+2,k}$. When $k\equiv 0,1,3\pmod{4}$, we have $B=B_{\mathrm{mid}}$, $D=D_{\mathrm{mid}}$, and $P_{\mathrm{mid}}=P=P_{\mathrm{c}}^{t}$.]

Then we have the algebraic quadratic relations

\begin{align}
\end{align}

This quadratic relation is a general phenomenon on periods of meromorphic flat connection on complex manifolds. We refer to [7] for more details.

From this proposition, when $k\equiv 0,1,3\pmod{4}$, we see the Bessel moments have quadratic relation given by (<ref>). On the other hand, when $k\equiv 2\pmod{4}$, the relation involves some combination of Bessel moments and regularized Bessel moments in the matrix $P_{\mathrm{c}}$. In the following discussion, we provide another expression of this relation, and we will see the pure quadratic relation involving only Bessel moments.

When $k\equiv 2\pmod{4}$, write $k=4r+2$ and define two $(k^{\prime}+1)\times k^{\prime}$ matrices with rational coefficients:

\begin{align*}
\left(\begin{array}{cccc}
I_{r} & & 0 &\\
0 & -\gamma_{k,1} & \cdots & -\gamma_{k,k^{\prime}-r}\\
0 & & I_{r} &
\end{array}\right)\\
0 & -\binom{k/2}{2} & 0 & -\binom{k/2}{4} & \cdots & 0 & -\binom{k/2}{k^{\prime}}\\
& & & I_{k^{\prime}} &
\end{array}\right)
\end{align*}

By the linear relations (<ref>) in Corollary <ref>, we have

Also, $P_{\mathrm{mid}}$ is obtained by deleting the first row of $L_{k}P_{\mathrm{mid}}$. Set $\widetilde{B}=L_{k}B_{\mathrm{mid}}L_{k}^{t}$ and $\widetilde{D}=R_{k}D^{-1}_{\mathrm{mid}}R_{k}^{t}$ which are square matrices of size $k^{\prime}+1$ with rational coefficients. Then $B_{\mathrm{mid}}$ is obtained by deleting the first row and column from $\widetilde{B}$. Therefore, the quadratic relation (<ref>) (involving linear combinations of Bessel moments) now becomes

(involving pure Bessel moments).

The matrices $\widetilde{B}$ and $\widetilde{D}$ in the above expression are singular because of the linear relations (<ref>) in Corollary <ref>. Note that this expression is equivalent to the middle part quadratic relation (<ref>) together with linear relations (<ref>).

When $k=4r+2$, the middle part period matrix is a $k^{\prime}\times k^{\prime}$ matrix given by

P_{\mathrm{mid}}=\left(\left\langle\delta_{b},\omega_{k,i}\right\rangle_{\mathrm{per}}\right)_{b=1,\cdots,k^{\prime},\ i=0,\cdots,\hat{r},\cdots,k^{\prime}}.

The determinant of this matrix $P_{\mathrm{mid}}$ is given by

\det P_{\mathrm{mid}}=\pi^{r(k+1)}\sqrt{-1}^{r(k^{\prime}-1)}\frac{2^{r(2r+1)}}{r!}\prod_{a=1}^{k^{\prime}}\frac{(2a+1)^{k^{\prime}+1-a}}{(a+1)^{a+1}}.

Note that $P_{\mathrm{mid}}$ appears in the upper left of the compactly supported period pairing matrix $P_{\mathrm{c}}$. Now, just take determinant on (<ref>) and then use the results of Propositions <ref>, <ref>, and Remark <ref>.

§ THE BESSEL OPERATOR AND DETERMINANTS OF BESSEL MOMENTS

§.§ Symmetric power of the modified Bessel differential operator

Consider the Weyl algebra $\mathbb{Q}\langle t,\partial_{t}\rangle$ consisting of ordinary differential operators. Write $\theta=t\partial_{t}$. The modified Bessel differential operator is an element in the subalgebra
$\mathbb{Q}\langle t^{2},\theta\rangle$
given by $L_{2}=\theta^{2}-t^{2}$.
The corresponding solutions are the modified Bessel functions $I_{0}(t)$ and $K_{0}(t)$.
The $n$-th symmetric power $L_{n+1}\in\mathbb{Q}\left\langle\theta,t^{2}\right\rangle$
of $L_{2}$ has order $n+1$ and the corresponding solutions are $I_{0}^{a}(t)K_{0}^{n-a}(t)$ for $0\leq a\leq n$.
By [4, 6],
the operator $L_{n+1}=L_{n+1,n}$ can be obtained by the recurrence relation as follows:

\begin{align}\label{symmetric_of_bessel_recurrence}
\begin{split}
L_{0,n}&=1, \\
L_{1,n}&=\theta, \\
L_{k+1,n}&=\theta L_{k,n}-t^{2}k\left(n+1-k\right)L_{k-1,n},\  1\leq k\leq n .
\end{split}
\end{align}

Here we provide two more concrete results about the operator $L_{n+1}$.

Put the degree on $\mathbb{Q}\langle t,\theta\rangle$ as $\deg t=\deg \theta=1$.
The associated graded ring $\operatorname{gr}\mathbb{Q}\langle t,\theta\rangle = \mathbb{Q}[\overline{t},\overline{\theta}]$
is a polynomial ring
where $\overline{t}$ and $\overline{\theta}$ are the images of $t$ and $\theta$, respectively.

The image of $L_{n+1}$ in $\mathbb{Q}[\overline{t},\overline{\theta}]$ is the polynomial

\begin{equation}\label{result}
\overline{L}_{n+1}(\overline{t},\overline{\theta}) = \begin{cases}
\prod_{i=1}^r \left(\overline{\theta}^2 - (2i-1)^2\overline{t}^2\right)
& \text{if $n+1 = 2r$ is even}, \\
\overline{\theta}\prod_{i=1}^r \left(\overline{\theta}^2 - (2i)^2\overline{t}^2\right)
& \text{if $n+1 = 2r+1$ is odd}.
\end{cases}
\end{equation}

Taking the images in $\mathbb{Q}[\overline{t},\overline{\theta}]$
of the relation (<ref>),
we obtain $\overline{L}_{n+1}=\overline{L}_{n+1,n}$

\begin{align*}
\overline{L}_{0,n}&=1, \\
\overline{L}_{1,n}&=\overline{\theta}, \\
\overline{L}_{k+1,n}&=\overline{\theta}\,\overline{L}_{k,n}-t^{2}k\left(n+1-k\right)\overline{L}_{k-1,n},\  1\leq k\leq n .
\end{align*}

The formula (<ref>) is then a consequence of the following combinatorics lemma.

For any $m\in\mathbb{N}$,
set the recurrence for $\lambda_{n,m}(x), n\in\mathbb{N}\cup\{0\}$,

\begin{align*}
\lambda_{0,m}&=1, \\
\lambda_{1,m}&=x, \\
\lambda_{k+1,m}&=x\lambda_{k,m}-k\left(m+1-k\right)\lambda_{k-1,m},\quad k\geq 1.
\end{align*}

Then we have

\begin{align}\label{modresult}
\lambda_{m+1,m}(x)=\begin{cases}\prod_{i=1}^{r}\left(x^{2}-(2i-1)^{2}\right), & m+1=2r,\\ x\prod_{i=1}^{r}\left(x^{2}-(2i)^{2}\right), & m+1=2r+1.\end{cases}
\end{align}

Notice that $\lambda_{i,m}$ is a monic integral polynomial of degree $i$ for any $m$.
Consider the formal generating function[This generating function satisfies the differential equation $-y^{4}f^{\prime\prime}(y)-(2y^{3}-my^{3})f^{\prime}(y)+(1-xy+my^{2})f(y)=1.$]:

An induction on $i$ immediately yields the relation
for any $m$ and $x$.[Equality also holds when viewed as the solution of the corresponding differential equations.] In other words,
for all $i$.
Therefore we obtain

\[ \lambda_{m+1,m}(x-1)+\lambda_{m+1,m}(x+1)=2\lambda_{m+1,m-1}(x)=2x\lambda_{m,m-1}(x) \]

by the recurrence.
Thus, since $\lambda_{m+1,m}(x)$ is a monic polynomial of degree $m+1$, it is uniquely determined by the above functional equation when the polynomial $\lambda_{m,m-1}(x)$ is given. Hence, by the induction, it suffices to show that

\begin{align*}
\prod_{i=1}^{r}\left((x-1)^{2}-(2i-1)^{2}\right)+\prod_{i=1}^{r}\left((x+1)^{2}-(2i-1)^{2}\right)&=2x^{2}\prod_{i=1}^{r-1}\left(x^{2}-(2i)^{2}\right);\\
\end{align*}

which are straightforward to verify.

Write $L_{n+1}$ into the form $\sum_{i} t^{i}P_{i}\left(\theta\right)$, where $P_{i}(x)\in\mathbb{Q}\left[x\right]$. Define the integers $a,b$ by

a=\max\left\{i\mid P_{i}\neq 0\right\};\quad b=\min\left\{i\mid P_{i}\neq 0\right\}.

Then we have $a=2\left\lfloor\frac{n+1}{2}\right\rfloor$ and $b=0$.

By the recurrence (<ref>), if we set $\deg t=1$ and $\deg \theta=0$, we easily see that $L_{j}$ has degree $2\left\lfloor\frac{j}{2}\right\rfloor$ by the interchanging relation $\theta t=t+t\theta$. Thus, we have $a=2\left\lfloor\frac{n+1}{2}\right\rfloor$. On the other hand, if we set $\deg t=0$ and $\deg \theta=1$, we see that the leading term of $L_{k+1}$ is given by $\theta^{k+1}$. Therefore, we conclude that $b=0$ by Proposition <ref>.

§.§ Two-scale Bessel moments

Throughout this paper, we take for granted the properties of modified Bessel functions $I_0(t), K_0(t)$ in the treatise [15].

Recall the Bessel moments ${\mathrm{IKM}}_{k}(a,b)$ are defined in section <ref>. For $r = 1,2,\cdots$,
define the two $r\times r$ matrices

\begin{align*}
M_r &= \begin{pmatrix}
\end{pmatrix}_{1\leq i,j\leq r}, \\
N_r &= \begin{pmatrix}
\end{pmatrix}_{1\leq i,j\leq r}.
\end{align*}

We aim to determine the two scalars $\det M_r$, $\det N_r$
adapting the inductive methods explored by Zhou [16].

For the initial values,
we have <cit.>

\begin{equation}\label{eq:M_1}
M_1 = \int_0^\infty K_0(t)\,{\mathrm{d}} t = \frac{\pi}{2},
\end{equation}

and, by <cit.>,
one has

\begin{align*}
N_1 = \int_0^\infty K_0^2(t)\,{\mathrm{d}} t
&= \frac{1}{2}\int_0^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty
e^{-2t\cosh x\cosh y} \,{\mathrm{d}} x{\mathrm{d}} y{\mathrm{d}} t \\
&= \frac{1}{4}\int_{-\infty}^\infty\int_{-\infty}^\infty
\frac{{\mathrm{d}} x{\mathrm{d}} y}{\cosh x\cosh y} \\
&= \frac{\pi^2}{4}.
\end{align*}

For $r = 0,1,\cdots$,
let $\omega_{2r+1}(x)$ be the Wronskian
of the $(2r+1)$ functions $f_i(x)$

\begin{equation}\label{eq:f_i}
f_i(x) = \begin{cases}
\int_0^\infty I_0(xt)I^{i-1}_0(t)K_0^{2r-i+1}(t) \,{\mathrm{d}} t, & 1\leq i\leq r, \\
\int_0^\infty K_0(xt)I^{i-r-1}_0(t)K_0^{3r-i+1}(t) \,{\mathrm{d}} t, & r < i \leq 2r+1.
\end{cases}
\end{equation}

The functions $f_i$ are well-defined analytic on the interval $(0,2)$
and hence so is $\omega_{2r+1}$.
In particular,
$\omega_1(x) = \int_0^\infty K_0(xt) \,{\mathrm{d}} t = \frac{\pi}{2x}$
by (<ref>).

For $r = 1,2,\cdots$,
let $\omega_{2r}(x)$ be the Wronskian of the $2r$ functions $g_i(x)$

\[ g_i(x) = \begin{cases}
\int_0^\infty I_0(xt)I^{i-1}_0(t)K_0^{2r-i}(t) \,{\mathrm{d}} t, & 1\leq i\leq r, \\
\int_0^\infty K_0(xt)I^{i-r-1}_0(t)K_0^{3r-i}(t) \,{\mathrm{d}} t, & r < i \leq 2r.
\end{cases}\]

All entries in the Wronskian matrix
are well-defined analytic functions on the interval $(0,1)$
and so is $\omega_{2r}(x)$.

The determinant $\omega_{k}(x)$ and its evaluation at $x=1$ are given by the following formulae:

* For $r = 1,2,\cdots$,

\begin{align*}
\omega_{2r+1}(x) &= \frac{(-1)^{\frac{r(r+1)}{2}}}{2}
\left[ \frac{1}{x^2}
\prod_{i=1}^r \frac{(2i)^2}{\left((2i)^2-x^2 \right)}\right]^{\frac{2r+1}{2}}
\Gamma\Big(\frac{r+1}{2}\Big)^2 \big(\det N_r\big)^2, \\
\omega_{2r+1}(1) &= (-1)^{r(r+1)/2} \det M_r \cdot \det M_{r+1}.
\end{align*}

* For $r = 2,3,\cdots$,

\begin{align*}
\omega_{2r}(x) &= (-1)^{\frac{r(r+1)}{2}}
\left[ \frac{1}{x}
\prod_{i=1}^r \frac{(2i-1)^2}{\left((2i-1)^2-x^2 \right)}\right]^r
\big(\det M_r\big)^2, \\
\lim_{x\to 1^-} 2^r(1-x)^r\omega_{2r}(x) &=
(-1)^{r(r+1)/2} (r-1)! \det N_{r-1} \cdot \det N_r.
\end{align*}

The above proposition leads to the recursive formulae

\begin{align}
\nonumber
\det M_r \cdot \det M_{r+1}
&= \frac{1}{2}
\left[\frac{2^rr!\sqrt{2r+1}}{(2r+1)!!}\right]^{2r+1}
\Gamma\Big(\frac{r+1}{2}\Big)^2 \big(\det N_r\big)^2
& (r\geq 1), \\
\label{eq:N-M}
\det N_{r-1} \cdot \det N_r
&= \frac{2^r}{(r-1)!}
\left[\frac{(2r-1)!!\sqrt{2r}}{2^rr!}\right]^{2r}\big(\det M_r\big)^2
& (r\geq 2).
\end{align}

With the initial data $M_1 = \frac{\pi}{2}, N_1 = \frac{\pi^2}{4}$ and the relation

\Gamma\left(\frac{r}{2}\right)\Gamma\left(\frac{r+1}{2}\right)=\frac{\left(r-1\right)!}{2^{r-1}}\sqrt{\pi},

one immediately obtains the following results by induction.

For positive integers $r$, we have

\begin{align*}
\det M_{r}&=\sqrt{\pi}^{r(r+1)}\sqrt{2}^{r(r-3)}\prod_{a=1}^{r-1}\frac{a^{r-a}}{\sqrt{2a+1}^{2a+1}}, \\
\det N_{r}&=\frac{1}{\Gamma\left(\frac{r+1}{2}\right)}\frac{\sqrt{\pi}^{(r+1)^{2}}}{\sqrt{2}^{r(r+3)}}\prod_{a=1}^{r-1}\frac{(2a+1)^{r-a}}{(a+1)^{a+1}}.
\end{align*}

In particular,
the two scalars $\sqrt{(2r-1)!!}\pi^{-m_r}\det M_r$
and $\pi^{-n_r}\det N_r$
are positive rational numbers, where $m_{r}=\frac{r(r+1)}{2}$ and $n_{r}=\left\lfloor\frac{(r+1)^{2}}{2}\right\rfloor$.

§.§ The Vanhove operators

The adjoint $L_{n+1}^*$ of $L_{n+1}$ is derived under the convolution
$(t,\partial_t) \mapsto (t,-\partial_t)$
(so $\theta \mapsto -(\theta+1)$)
and hence the leading term
of the signed adjoint $\Lambda_{n+1} = (-1)^{n+1}L_{n+1}^*$
equals $\overline{L}_{n+1}(\overline{\theta},\overline{t})$ by Proposition <ref>.
For $F(xt) = I_0(xt), K_0(xt)$
and $G(t) = I_0^a(t)K_0^{n-a}(t)$,
we have, by integration by parts,

\[ \int_0^\infty (\Lambda_{n+1}F(xt)) G(t) \,{\mathrm{d}} t
= (-1)^{n+1}\int_0^\infty F(xt) (L_{n+1}G(t)) \,{\mathrm{d}} t = 0. \]

The Vanhove operator
$V_{n+1} \in \mathbb{Q}\left\langle \partial_x,x^{\pm 1} \right\rangle$
is of order $(n+1)$ such that $V_{n+1}F(xt) = \Lambda_{n+1}F(xt)$.
So one has $V_{n+1}f_i = 0$ for $f_i(x)$ in (<ref>)
and consequently $\omega_{n+1}(x)$
satisfies a first order linear differential equation (See (<ref>) below).

Let $\lambda_{n+1}(x) = \overline{L}_{n+1}(1,x^{-1}) \in \mathbb{Q}[x^{-1}]$
of order $2\left\lfloor\frac{n+1}{2}\right\rfloor$ with respect to $x^{-1}$.
Let $\theta_x = x\partial_x$.
One has

\begin{align*}
&= \lambda_{n+1}(x)\theta_x^{n+1}
+ (n+1)\left[ \lambda_{n+1}(x) + \frac{x\lambda_{n+1}'(x)}{2} \right]\theta_x^n
+ \delta_1 \\
&= x^{n+1}\lambda_{n+1}(x)\partial_x^{n+1}
+ \frac{n+1}{2}x^n\left[ (n+2)\lambda_{n+1}(x) + x\lambda_{n+1}'(x) \right]\partial_x^n
+ \delta_2
\end{align*}

where $\delta_1,\delta_2$ are of order at most $(n-1)$
with respect to $\partial_x$ in $\mathbb{Q}\langle\partial_x,x^{\pm 1}\rangle$.

By Vanhove [14],
there exists $\widetilde{L}_{n-1} \in \mathbb{Q}\langle\partial_x,x^{\pm 1}\rangle$ of order $(n-1)$
such that

\[ t\widetilde{L}_{n-1}F(xt) = \Lambda_{n+1}\frac{F(xt)}{t}. \]

The operator $\widetilde{L}_{n-1}$ is of the form

\[ \widetilde{L}_{n-1} = x^2\lambda(x)\theta_x^{n-1}
+ x^2\left[ 2(n-1)\lambda(x) + \frac{n-1}{2}x\lambda'(x) \right]\theta_x^{n-2}
+ \widetilde\delta \]

where $\widetilde\delta$ is of order at most $(n-3)$
with respect to $\partial_x$ in $\mathbb{Q}\langle\partial_x,x^{\pm 1}\rangle$[Comparing $\widetilde{L}_{n-1}(\theta_x)$
with Zhou's Vanhove operator $\widetilde{L}_{n-1}(\theta_u)$,
we set his variable $u = x^2$
and multiply $\widetilde{L}_{n-1}(\theta_u)$ by $2^{n-1}$.].

\[ \Delta_n(\theta_t)
= \Lambda_{n+1}(\theta_t) - \Lambda_{n+1}(\theta_t -1). \]

$\theta_t \frac{1}{t} = \frac{1}{t}(\theta_t -1)$ in $\mathbb{Q}\left\langle\partial_t,t\right\rangle$,
we have

\begin{align*}
\Lambda_{n+1}(\theta_t)F(xt)
&= t\Lambda_{n+1}(\theta_t)\frac{F(xt)}{t} + \Delta_n(\theta_t)F(xt) \\
&= \big[ t^2\widetilde{L}_{n-1}(\theta_x) + \Delta_n(\theta_t) \big]F(xt).
\end{align*}

$t^2F(xt) = \frac{1}{x^2}\theta_x^2F(xt)$,
we have

\begin{align*}
&= \widetilde{L}_{n-1}(\theta_x)\frac{1}{x^2}\theta_x^2F(xt) \\
&= \frac{1}{x^2}\widetilde{L}_{n-1}(\theta_x-2)\theta_x^2F(xt)
\end{align*}

and the differential operator reads

\[ \lambda(x)\theta_x^{n+1} + \frac{n-1}{2}x\lambda'(x)\theta_x^n + \delta_3 \]

where $\delta_3$ is of order at most $(n-1)$
with respect to $\partial_x$ in $\mathbb{Q}\langle\partial_x,x^{\pm 1}\rangle$.

On the other hand, since
$\theta_tF(xt) = \theta_xF(xt)$ and by Proposition <ref>,
we have

\[ \Delta_n(\theta_t)F(xt) =\left[\Lambda_{n+1}\left(\theta\right)-\Lambda_{n+1}\left(\theta-1\right)\right]F\left(xt\right)=
\left[ \left( (n+1)\lambda(x) + x\lambda'(x)\right)\theta_x^n
+ \delta_4 \right] F(xt) \]

where $\delta_4$ is of order at most $(n-1)$
with respect to $\partial_x$ in $\mathbb{Q}\langle\partial_x,x^{-1}\rangle$.
Therefore the leading two terms of $V_{n+1}$ are determined.

§.§.§ Rationality of $\omega_{n+1}(x)$

Lemma <ref> yields

\begin{equation}\label{odeequ}
\omega_{n+1}'(x) = -\frac{n+1}{2x}
\left[ (n+2) + \frac{x\lambda_{n+1}'(x)}{\lambda_{n+1}(x)} \right] \omega_{n+1}(x).
\end{equation}

Since $\omega_{n+1}(x)$ takes real values on $(0,1)$,
one obtains

\[ \omega_{n+1}(x) = C_{n+1}
\left[ (-1)^{\left\lfloor\frac{n+1}{2}\right\rfloor}x^{n+2}\lambda_{n+1}(x) \right]^{-\frac{n+1}{2}} \]

for some real constant $C_{n+1}$
for each $n = 0,1,\cdots$.
We shall determine $C_{n+1}$
by investigating the limiting behavior of $\omega_{n+1}(x)$
as $x \to 0^+$.

§.§ Singularities of $\omega_{n+1}(x)$

For $F(xt) = I_0(xt)$ or $K_0(xt)$,
we have

\[ \partial_xF(xt) = tF'(xt),
\quad
\partial_x^2F(xt) = -\frac{t}{x} F'(xt) + t^2F(xt). \]

So $\omega_{n+1}(x)$ coincides with the determinant of the matrix
$\Omega_{n+1}(x)$ of size $(n+1)$ whose $(i,j)$-entry is

\[\begin{cases}
\int_0^\infty I_0(xt) I_0^{j-1}(t) K_0^{n-j+1}(t) t^{i-1} \,{\mathrm{d}} t,
& 1\leq j\leq \left\lfloor\frac{n+1}{2}\right\rfloor, i =1,3,\cdots, 2\left\lfloor\frac{n}{2}\right\rfloor +1, \\
\int_0^\infty tI_0'(xt) I_0^{j-1}(t) K_0^{n-j+1}(t) t^{i-2} \,{\mathrm{d}} t,
& 1\leq j\leq \left\lfloor\frac{n+1}{2}\right\rfloor, i =2,4,\cdots, 2\left\lfloor\frac{n+1}{2}\right\rfloor, \\
\int_0^\infty K_0(xt) I_0^{j-r-1}(t) K_0^{n-j+r+1}(t) t^{i-1} \,{\mathrm{d}} t,
& \left\lfloor\frac{n+1}{2}\right\rfloor< j\leq n+1, i =1,3,\cdots, 2\left\lfloor\frac{n}{2}\right\rfloor+1, \\
\int_0^\infty tK_0'(xt) I_0^{j-r-1}(t) K_0^{n-j+r+1}(t) t^{i-2} \,{\mathrm{d}} t,
& \left\lfloor\frac{n+1}{2}\right\rfloor< j\leq n+1, i =2,4,\cdots, 2\left\lfloor\frac{n+1}{2}\right\rfloor.
\end{cases} \]

§.§.§ Properties of $I_0(t)$ and $K_0(t)$

We collect some properties of the modified Bessel functions
$I_0(t)$ and $K_0(t)$
in order to obtain information of $\omega_{n+1}(x)$
as $x\to 0^+, 1^-$.

The function $I_0(t)$ is entire and even;
it is real and increasing on the half line $[1,\infty)$.
The function $K_0(t)$ has a logarithmic pole at $x=0$;
it is real and decreasing on $(0,\infty)$.
On the half plane $\Re(t)>0$,
we have the asymptotic approximations

\[ I_0(t) = \frac{e^t}{\sqrt{2\pi t}}\left( 1+O\Big(\frac{1}{t}\Big) \right),
\quad
K_0(t) = \sqrt{\frac{\pi}{2t}}e^{-t}\left( 1+O\Big(\frac{1}{t}\Big) \right) \]

as $t\to\infty$.
In particular,
for a positive integer $a$,

\[ [I_0(t)K_0(t)]^a - \frac{1}{(2t)^a} = O\Big(\frac{1}{t^{a+1}}\Big) \]

as $t \to \infty$ along the real line.
One has the boundedness

\[ \sup_{t>0}\frac{|tK'_0(t) + 1|}{t(1+|\log t|)} < \infty. \]

Let $c=0,1,\cdots$.
One has the evaluation <cit.>

\[ \int_0^\infty K_0(t)t^c\,{\mathrm{d}} t
= 2^{c-1}\Gamma\Big(\frac{c+1}{2}\Big)^2. \]

§.§.§ Integrations

With the data collected above,
we list some consequences
for the integrals appeared in $\Omega_{n+1}(x)$.

For $0\leq a<b$ and $c\geq 0$,
one obtains

\begin{align}
\label{eq:simple_Kxt}
\int_0^\infty K_0(xt)I_0^a(t)K_0^b(t)t^c \,{\mathrm{d}} t
&= O(\log x), \\
\label{eq:simple_I'xt}
\int_0^\infty I_0'(xt)I_0^a(t)K_0^b(t)t^c \,{\mathrm{d}} t
&= O(x)
\end{align}

\begin{align}
\nonumber
\int_0^\infty tK'_0(xt)I_0^a(t)K_0^b(t)t^c \,{\mathrm{d}} t
&= \frac{-1}{x}\left[ \int_0^\infty I_0^a(t)K_0^b(t)t^c \,{\mathrm{d}} t
-\int_0^\infty \big(xtK_0'(xt)+1\big) I_0^a(t)K_0^b(t)t^c \,{\mathrm{d}} t \right] \\
\label{eq:simple_K'xt}
&= \frac{-1}{x}\int_0^\infty I_0^a(t)K_0^b(t)t^c \,{\mathrm{d}} t + O(\log x)
\end{align}

as $x \to 0^+$.
For $0\leq c\leq a$ and as $x\to 0^+$, we thus have

\begin{align}
\label{eq:KxtIK_1}
\int_0^\infty K_0(xt)I_0^a(t)K_0^a(t)t^c \,{\mathrm{d}} t
&= O\Big(\int_0^\infty K_0(xt) \,{\mathrm{d}} t\Big)
= O\Big(\frac{1}{x}\Big), \\
\label{eq:K'xtIK_1}
\int_0^\infty tK_0'(xt)I_0^a(t)K_0^a(t)t^c \,{\mathrm{d}} t
&= O\Big(\int_0^\infty tK_0'(xt) \,{\mathrm{d}} t\Big)
= O\Big(\frac{1}{x^2}\Big).
\end{align}

If $0\leq a <c$ and as $x\to 0^+$, then

\begin{align}
\nonumber
\int_0^\infty K_0(xt)I_0^a(t)K_0^a(t)t^c {\mathrm{d}} t
&= \int_0^\infty \frac{K_0(xt)t^{c-a}}{2^a} {\mathrm{d}} t
+ \int_0^\infty K_0(xt)\Big[ I_0^a(t)K_0^a(t) - \frac{1}{(2t)^a}\Big]t^c {\mathrm{d}} t \\
\label{eq:KxtIK_2}
&= \frac{2^{c-2a-1}}{x^{c-a+1}} \Gamma\Big(\frac{c-a+1}{2}\Big)^2
+ O\Big(\frac{1}{x^{c-a}}\Big), \\
\nonumber
\int_0^\infty tK_0'(xt)I_0^a(t)K_0^a(t)t^c {\mathrm{d}} t
&= \int_0^\infty \frac{tK_0'(xt)t^{c-a}}{2^a} {\mathrm{d}} t
+ \int_0^\infty tK_0'(xt)\Big[ I_0^a(t)K_0^a(t) - \frac{1}{(2t)^a}\Big]t^c {\mathrm{d}} t \\
\nonumber
&= \frac{c-a+1}{2^ax} \int_0^\infty K_0(xt)t^{c-a} \,{\mathrm{d}} t
+ O\Big(\frac{1}{x^2}\Big) \\
\label{eq:K'xtIK_2}
&= O\Big(\frac{1}{x^{c-a+2}}\Big).
\end{align}

On the real line,
we have <cit.>

\[ \lim_{x\to 1^-} \frac{\int_0^\infty I_0(xt)K_0(t) \,{\mathrm{d}} t}{-\log(1-x)}
= \frac{1}{2} \]

and for $c = 0,1,\cdots$,

\[ \lim_{x\to 1^-}(1-x)^{c+1}\int_0^\infty I_0(xt)K_0(t)t^{c+1} \,{\mathrm{d}} t
= \frac{c!}{2}
= \lim_{x\to 1^-}(1-x)^{c+1}\int_0^\infty tI_0'(xt)K_0(t)t^c \,{\mathrm{d}} t. \]

Therefore for $a\geq 1, a> c$ and $x\to 1^-$,
one has

\begin{align}
\nonumber
\int_0^\infty I_0(xt)I_0^{a-1}(t)K_0^a(t)t^c \,{\mathrm{d}} t
&= \int_0^\infty I_0(xt)K_0(t)\big[I_0^{a-1}(t)K_0^{a-1}(t)t^c\big] \,{\mathrm{d}} t \\
\nonumber
&= O\Big(\int_0^\infty I_0(xt)K_0(t) \,{\mathrm{d}} t \Big) \\
\label{eq:IxtIK_1}
&= O\big(\log (1-x)\big), \\
\nonumber
\int_0^\infty tI_0'(xt)I_0^{a-1}(t)K_0^a(t)t^c \,{\mathrm{d}} t
&= \int_0^\infty tI_0'(xt)K_0(t)\big[I_0^{a-1}(t)K_0^{a-1}(t)t^c\big] \,{\mathrm{d}} t \\
\nonumber
&= O\Big(\int_0^\infty tI'_0(xt)K_0(t) \,{\mathrm{d}} t \Big) \\
\label{eq:I'xtIK_1}
&= O\Big(\frac{1}{1-x}\Big).
\end{align}

If $c\geq a\geq 1$ and $x\to 1^-$, then

\begin{align}
\nonumber
\int_0^\infty I_0(xt)I_0^{a-1}(t)K_0^a(t)t^c \,{\mathrm{d}} t
= &\int_0^\infty \frac{I_0(xt)K_0(t)t^{c-a+1}}{2^{a-1}} \,{\mathrm{d}} t \\
&\hspace{20pt}+ \int_0^\infty I_0(xt)K_0(t)
\Big[ I_0^{a-1}(t)K_0^{a-1}(t) - \frac{1}{(2t)^{a-1}}\Big]t^c \,{\mathrm{d}} t
\end{aligned} \\
\label{eq:IxtIK_2}
&= \frac{(c-a)!}{2^a(1-x)^{c-a+1}}
+ o\Big(\frac{1}{(1-x)^{c-a+1}}\Big), \\
\nonumber
\int_0^\infty tI'_0(xt)I_0^{a-1}(t)K_0^a(t)t^c \,{\mathrm{d}} t
= &\int_0^\infty \frac{tI'_0(xt)K_0(t)t^{c-a+1}}{2^{a-1}} \,{\mathrm{d}} t \\
&\hspace{20pt}+ \int_0^\infty tI'_0(xt)K_0(t)
\Big[ I_0^{a-1}(t)K_0^{a-1}(t) - \frac{1}{(2t)^{a-1}}\Big]t^c \,{\mathrm{d}} t
\end{aligned} \\
\label{eq:I'xtIK_2}
&= \frac{(c-a+1)!}{2^a(1-x)^{c-a+2}}
+ o\Big(\frac{1}{(1-x)^{c-a+2}}\Big).
\end{align}

§.§.§ Evaluation of $\omega_{2r+1}(x)$ at $x=1$

All entries of $\Omega_{2r+1}(x)$ can be evaluated at $x=1$.
We move $2i$-th row to row $i$ in $\Omega_{2r+1}(1)$
for $1\leq i\leq r$
and then subtract $j$-th column by $(r+j+1)$-st column
for $1\leq j\leq r$.
By (<ref>) on the upper-left block,
we obtain

\[ \omega_{2r+1}(1) = (-1)^{\frac{r(r+1)}{2}}
\det \begin{pmatrix}
M_r & * \\
0 & M_{r+1}
\end{pmatrix}. \]

§.§.§ Behavior of $\omega_{2r+1}(x)$ as $x \to 0^+$

Fix $r\geq 1$.
We move row $(2i-1)$ of $\Omega_{2r+1}(x)$
to row $i$ for $1\leq i \leq r$,
which creates a sign $(-1)^{r(r-1)/2}$
to the determinant $\omega_{2r+1}(x)$.
As $x\to 0^+$,
the resulting matrix decomposes into
$(r,r,1)\times (r,r,1)$ blocks of the form

\[\begin{pmatrix}
N_r + o(1) & O(\log x) & O\big(\frac{1}{x^r}\big) \\
O(x) & \frac{-1}{x}N_r + O(\log x) & O\big(\frac{1}{x^r}\big) \\
O(1) & O(\log x) &
\frac{1}{2x^{r+1}}\Gamma\big(\frac{r+1}{2}\big)^2
+ O\big(\frac{1}{x^r}\big)
\end{pmatrix}\]

by direct evaluation and (<ref>)
in the left three blocks,
(<ref>) and (<ref>)
in the middle,
(<ref>), (<ref>),
(<ref>) and (<ref>)
in the last column.
The leading term of $\omega_{2r+1}(x)$,
which is of order $x^{-(2r+1)}$,
comes from the diagonal blocks
and one gets

\[ \lim_{x\to 0^+} x^{2r+1}\omega_{2r+1}(x)
= (-1)^{\frac{r(r+1)}{2}} \frac{1}{2}
\Gamma\Big(\frac{r+1}{2}\Big)^2 {\det}^2 N_r. \]

§.§.§ Behavior of $\omega_{2r}(x)$ as $x \to 1^-$

Fix $r\geq 2$.
We move $2i$-th row of $\Omega_{2r}(x)$
to row $i$ for $i=1,2,\cdots, (r-1)$
and $r$-th column to the last,
which adds a sign $(-1)^{r(r+1)/2}$
to the determinant $\omega_{2r}(x)$.
We subtract $j$-th column by $(r+j)$-th
for $j = 1,2,\cdots, (r-1)$.
As $x\to 1^-$,
the resulting matrix decomposes into
$(r-1,r,1)\times (r-1,r,1)$ blocks of the form

\[\begin{pmatrix}
N_{r-1} + o(1) & O(1) & O\big(\frac{1}{(1-x)^{r-1}}\big) \\
0 & N_r + o(1) & O\big(\frac{1}{(1-x)^{r-1}}\big) \\
O(1) & O(1) & \frac{(r-1)!}{2^r(1-x)^r} + o\big(\frac{1}{(1-x)^r}\big)
\end{pmatrix}\]

by (<ref>) and direct evaluation
in the left three blocks,
direct evaluation
in the middle,
(<ref>), (<ref>),
(<ref>) and (<ref>)
in the last column.
The leading term of $\omega_{2r}(x)$,
which is of order $(1-x)^{-r}$,
comes from the diagonal blocks.
It yields

\[ \lim_{x\to 1^-} (1-x)^r\omega_{2r}(x)
= (-1)^{\frac{r(r+1)}{2}} \frac{(r-1)!}{2^r} \det N_{r-1} \det N_r. \]

§.§.§ Behavior of $\omega_{2r}(x)$ as $x \to 0^+$

Fix $r\geq 2$.
We move row $(2i-1)$ of $\Omega_{2r}(x)$
to row $i$ for $1\leq i \leq r$,
which adds a sign $(-1)^{r(r-1)/2}$
to the determinant $\omega_{2r}(x)$.
As $x\to 0^+$,
the resulting matrix decomposes into
four blocks of equal size of the form

\[\begin{pmatrix}
M_r + o(1) & O(\log x) \\
O(x) & \frac{-1}{x}M_r + O(\log x)
\end{pmatrix}\]

by direct evaluation and (<ref>)
in the left two blocks
and (<ref>) and (<ref>)
in the right.
This leads to

\[ \lim_{x\to 0^+} x^r\omega_{2r}(x)
= (-1)^{\frac{r(r+1)}{2}} {\det}^2 M_r. \]

Proposition <ref> indeed
holds for $\omega_2(x)$ by the same analysis
if we set $\det N_0=1$;
it is consistent with the relation (<ref>) for $r=1$.

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