Understanding the completed stalk of the dualizing sheaf of a family of nodal curves at a node

Question

Let $f : C\rightarrow S$ be a proper flat morphism whose geometric fibers are connected nodal curves.

Let $P\in C$ be a closed point with image $s\in S$, and suppose $P$ is a 'node' - that is, if $\widehat{\mathcal{O}}_s$ is the completed etale local ring of $S$ at $s$, then we have: $$\widehat{\mathcal{O}}_P = \widehat{\mathcal{O}}_s[[x,y]]/(xy-a)$$ where $a$ is in the maximal ideal $m_s$ of $\widehat{\mathcal{O}}_s$.

Let $\omega_C = \omega_{C/S}$ be the dualizing sheaf of $C/S$.

Question 1 - How can we describe the completed stalk $\widehat{\omega_{C,P}}$?

If $S = \text{Spec }k$ for an algebraically closed field $k$, then if $\nu : C'\rightarrow C$ is the normalization map, then it's well known that

$$\omega_{C}(U) = \{w\in \Omega^1_{C'/S}(Q_1+Q_2)(\nu^{-1}(U)) : \text{res}_{Q_1}w + \text{res}_{Q_2}w = 0\}$$ where $Q_1,Q_2$ run over all distinct pairs of points in $C'$ which map to the same point $P$ in $C$.

This seems to imply that $\widehat{\omega_{C,P}}$ is the $\big(\widehat{\mathcal{O}_{P}} = k[[x,y]]/(xy)\big)$-submodule of $$\widehat{\Omega^1_{C'/S,Q_1}}(Q_1)\oplus\widehat{\Omega^1_{C'/S,Q_2}}(Q_2) = k[[x]]\frac{dx}{x}\oplus k[[y]]\frac{dy}{y}$$ generated by $w(x,y) := \left(\frac{dx}{x},-\frac{dy}{y}\right)$.

Does something like this still hold in the general case when $S$ isn't Spec $k$? What exactly would the description be?

Part of the reason for wanting to understand this is the following: In general there is always a canonical map $\theta : \Omega_{C/S}\rightarrow\omega_{C/S}$. If $e$ is an integer coprime invertible on $S$, then at the level of completed stalks, a paper I'm reading makes the claim: $$\theta(d(x^e+y^e)) = e(x^e-y^e)\tilde{w}(x,y)$$ where $\tilde{w}(x,y)$ is a generator of $\widehat{\omega_{C,P}}$ which reduces to $w(x,y)$ in the fiber above $s$ (actually the paper seems to call these completed stalks the 'etale local picture', so I guess they also want to assume that the residue field of $\widehat{\mathcal{O}_s}$ is separably closed, to parallel 'strict henselization')

Certainly, we have $$d(x^e+y^e) = e(x^{e-1}dx +y^{e-1}dy)$$ Thus, if we could write $\tilde{w}(x,y) = \text{"}\frac{dx}{x} - \frac{dy}{y}\text{"}$, in a way such that $x\frac{dy}{y} = y\frac{dx}{x} = 0$, then the claim would follow. This makes sense at the image of $P$ in the geometric fiber of $s$, but is it possible to make sense of this "computation" over $\widehat{\mathcal{O}}_P$?

Answer 1

So as it turns out the key fact is the beautiful description of dualizing sheaves in terms of 'generalized determinants' given in Knudsen-Mumford's "The Projectivity of the Moduli Space of Stable Curves" (I and II).

There, in general given a scheme $X$ and a perfect complex of $\mathcal{O}_X$-modules $F^\bullet$, which on an open cover $\{U\}$ of $X$ is quasi-isomorphic to a bounded complex of finite-free $\mathcal{O}_X$-modules $G^\bullet_U$, one can define the generalized determinant bundle $\det(F^\bullet)$, which is locally given by the alternating tensor product of determinants of the free sheaves $G^i$. $$\det(F^\bullet)|_U = \bigotimes_{i\ge 0} (\det(G^i))^{(-1)^i}$$ This generalized $\det$ commutes with arbitrary base change and is invariant under quasi-isomorphisms, so in practice to compute it on a coherent sheaf $F$, it suffices to let $G^\bullet$ be a bounded finite locally free resolution of $F$ and apply the above formula.

In our case, we want to take $F^\bullet$ to be the complex $\Omega^1_{C/S}$ concentrated in degree zero, and we'd like to understand the etale local picture of $\det(\Omega^1_{C/S})$. Since the fibers of $C$ have at worst ordinary double points, we may find an etale covering $\{u : U\rightarrow C\}$ such that each $U$ admits a regular immersion $j : U\hookrightarrow M$ with $M$ smooth of relative dimension 2 over $S$ (needs justification). In fact one should be able to take $M = \mathbb{A}^2_S$.

Letting $I$ be the sheaf of ideals of $U\hookrightarrow M$, we then have an exact sequence (needs justification) $$0\rightarrow I/I^2\rightarrow j^*\Omega^1_{M/S}\rightarrow\Omega^1_{U/S}\rightarrow 0$$ where the first two terms are locally free. Since $\det$ commutes with base change, and $u$ is etale, we have $$u^*\omega_{C/S} = u^*\det(\Omega^1_{C/S}) = \det(u^*\Omega^1_{C/S}) = \det(\Omega^1_{U/S}) = (I/I^2)^\vee\otimes \big(\wedge^2 j^*\Omega^1_{M/S}\big)$$ Fixing a point $P\in C$, and passing to the limit over all etale neighborhoods of $P$, the exact sequence above becomes: $$0\rightarrow \widehat{I}_P/\widehat{I}_P^2\rightarrow \widehat{j^*\Omega^1_{M/S,P}}\rightarrow \widehat{\Omega^1_{C/S,P}}\rightarrow 0$$ or more concretely, $$0\rightarrow \frac{(xy-a)}{(xy-a)^2}\rightarrow \left(\frac{\widehat{\mathcal{O}}_s[[x,y]]}{(xy-a)}\right)(dx\oplus dy)\rightarrow\frac{\widehat{\mathcal{O}}_s[[x,y]]}{(xy-a)}((dx\oplus dy)/(ydx+xdy)\rightarrow 0$$ where now the first two terms are free, and the map between them is given by $$xy-a\mapsto d(xy-a) = ydx + xdy$$ Assuming $\det$ commutes with ind-etale localization (needs justification?), we get $$\widehat{\omega_{C/S,P}} = (\widehat{I}_P/\widehat{I}_P^2)^\vee\otimes\big(\wedge^2 \widehat{j^*\Omega^1_{M/S,P}}\big)$$ Let $z\in \widehat{I/I^2}_P$ be a basis, with dual basis $z^\vee$. Then the canonical map $\theta : \widehat{\Omega^1_{C/S,P}}\rightarrow\widehat{\omega_{C/S,P}}$ is given by sending $w\in\widehat{\Omega^1_{C/S,P}}$ to $$\theta(w) = z^\vee\otimes (dz\wedge \tilde{w})$$ where $\tilde{w}$ is any lift of $w$ to $j^*\Omega^1_{M/S}$, and one may check that this map does not depend on the choice of the basis $z$ or on the lift $\tilde{w}$. In particular, we may choose $z = xy-a$, and we would get: $$\theta(dx) = (xy-a)^\vee\otimes ((ydx+xdy)\wedge dx = (xy-a)^\vee\otimes x(dy\wedge dx)$$ $$\theta(dy) = (xy-a)^\vee\otimes ((ydx+xdy)\wedge dy = (xy-a)^\vee\otimes y(dx\wedge dy)$$ The choice of coordinates $x,y$ in the presentation of $\widehat{\mathcal{O}_{P}}$ determines a basis of $\widehat{\omega_{C/S,P}}$ given by $$w(x,y) := (xy-a)^\vee\otimes(dy\wedge dx)$$ where we have $$x\cdot w(x,y) = \theta(dx)\quad\text{and}\quad y\cdot w(x,y) = -\theta(dy)$$ Thus, $w(x,y)$ is basically "$dx/x$" (ie, $\frac{1}{x}\theta(dx)$), or equivalently "$-dy/y$". Thus, we may compute: $$(x^e-y^e)w(x,y) = x^ew(x,y) - y^ew(x,y) = x^{e-1}\theta(dx) + y^{e-1}\theta(dy)$$ as desired.