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Let $K$ be a finite extension of $\mathbb{Q}_p$. If $A_K$ is a semistable Abelian variety over $K$, then we have a Frobenius endomorphism on $H_{dR}^1(A_K)$, whose definition depends on a choice of a branch of the $p$-adic logarithm on $K^\times$. The unit root subspace $W$ of $H_{dR}^1(A_K)$ is defined as the slope zero subspace for the action of Frobenius. This subspace is independent of the choice of Frobenius. If $A_K$ is ordinary, then the dimension of $W$ is equal to the dimension of $A_K$.

Let us denote by $A^{\prime}_K$ the dual of $A_K$. Suppose $A_K$ is a split semistable abelian variety. Then under the cup product $$\cup: H_{dR}^1(A_K) \times H_{dR}^1(A^{\prime}_K) \rightarrow K$$ the unit root subspaces are orthogonal. This result implies that the unit root subspace of a curve is isotropic.


Now let $$C: y^2 = x^5+b_1x^4+b_2x^3+b_3x^2+b_4x+b_5\in\mathbb{Z}[x]$$ be a hyperelliptic curve that has semistable reduction at a prime $p$. The forms $\omega_0,\omega_1,\omega_2,\omega_3$ are a basis for $H_{dR}^1(C)$ where $\omega_i = x^i\frac{dx}{2y}$. There are effectively computable constants $\alpha,\beta,\gamma,\delta\in\mathbb{Q}_p$ such that the subspace of $H_{dR}^1(C)$ generated by \begin{align*} \eta_0 &= \big(3x^3+3b_1x^2-\alpha x - (3 b_1 b_2-b_1 \alpha+3 b_3+3\delta)\big)\omega_0 \\ \eta_1 &= \big(x^2-\beta x-(b_2-b_1\beta+3\gamma)\big)\omega_0 \end{align*} is isotropic with respect to the cup product. Clearly, it is $2$-dimensional. Based on some other work, I suspect that this subspace is nothing but the unit root subspace, at least when $\text{Jac}(C)$ is ordinary. What I need is a method to check my claim.


Let $R$ denote the ring of integers of $K$ and let $F$ be an $n$-dimensonal commutative formal group over $R$. The cohomology groups $H_{dR}^i(F/R)$ are the $R$-modules obtained by taking the cohomology of the formal de Rham complex $\Omega^{\cdot}_{F/R}$. We define the Dieudonne module of $F$, $D(F/R)$, to be the $R$-submodule of $H_{dR}^1(F/R)$ consisting of the primitive elements, $$D(F/R) = \{a\in H_{dR}^1(F/R) \mid m^*(a) = pr^*_1(a) + pr^*_2(a)\}$$ where $m,pr_1,pr_2: F\times F\to F$ be the group law and the two projections. Set $D(F/K):= D(F/R)\otimes_R K$.

Let $T = (T_1,T_2,\dots,T_n)$ be a system of coordinates of $F$. Then the affine algebra $A(F)$ of $F$ is isomorphic to $R[[T]]$ and the exterior differentiation $d$ induces an isomorphism between $D(F/K)$ and $$\small \frac{\{f\in K[[T]]: f(0_{F}) = 0, df\in d(R[[X]])_K, f(X+_{F}Y)) - f(X) - f(Y) \in R[[X,Y]]_K\}}{\{f\in R[[T]]_K: f(0_{F}) = 0\}}.$$


Now let $A$ be the Neron model of $A_K$. Then $H_{dR}^1(A)_K := H_{dR}^1(A)\otimes_R K\cong H_{dR}^1(A_K).$ We formally complete $A$ along the identity of its special fiber, $A^f$. This is a commutative formal group over $R$ and the restriction map $$\text{res}_{A^f}^A : H_{dR}^1(A)\to H_{dR}^1(A^f)$$ takes values in $D(A^f/R)$. We denote by $\text{res}_A$ the composition: $$H_{dR}^1(A_K) \cong H_{dR}^1(A)_K\to D(A^f/K).$$

If $A_K$ is split, then we have an exact sequence of $K$-vector spaces $$0 \rightarrow W \rightarrow H_{dR}^1(A_K) \xrightarrow{\text{res}_A} D(A^f/K) \rightarrow 0.$$ In particular, the unit root subspace $W$ is the kernel of the restriction map $\text{res}_A$.


Set $J_{\mathbb{Q}_p}=\text{Jac}(C)$. We denote by $\text{res}_C$ the composition: $$H_{dR}^1(C) \cong H_{dR}^1(J_{\mathbb{Q}_p})\xrightarrow{\text{res}_J} D(J^f/\mathbb{Q}_p).$$ By above, every $\omega_i=x^i\frac{dx}{2y}$ corresponds to some $f_i\in \mathbb{Q}_p[[T_1,T_2]]$.

QUESTION: How can I determine $f_i$'s explicitly?

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