text:
\begin{document}
\title{On the de Rham homology of affine varieties in characteristic 0}
\author{Nicole Bridgland}
\thanks{NSF support through grants DMS-1500264 and DMS-1800355 is gratefully acknowledged}
\address{Department of Mathematics, University of Minnesota, Minneapolis, MN 55455}
\email{bridg130@umn.edu}
\maketitle

\begin{abstract}
Let $k$ be a field of characteristic 0, let $S$ be a complete local ring with coefficient field $k$, let $k[[x_1,\dots,x_n]]$ be the ring of formal power series in variables $x_1,\dots, x_n$ with coefficients from $k$, let $k[[x_1,\dots,x_n]]\to S$ be a $k$-algebra surjection and let $E_\bullet^{\bullet,\bullet} $ be the associated Hodge - de Rham spectral sequence for the computation of the de Rham homology of $S$. Nicholas Switala \cite{switala} proved that this spectral sequence is independent of the surjection beginning with the $E_2$ page, and the groups $E^{p,q}_2$ are all finite-dimensional over $k$.

In this paper we extend this result to affine varieties. Namely, let $Y$ be an affine variety over $k$, let $X$ be a non-singular affine variety over $k$, let $Y\subset X$ be an embedding over $k$ and let $E_\bullet^{\bullet,\bullet} $ be the associated Hodge - de Rham spectral sequence for the computation of the de Rham homology of $Y$. Then this spectral sequence is independent of the embedding beginning with the $E_2$ page, and the groups $E^{p,q}_2$ are all finite-dimensional over $k$.
\end{abstract}

\section{Introduction}
In \cite{hartshorne}, Hartshorne developed algebraic de Rham homology and cohomology theories using embeddings of schemes over a field $k$ of characteristic zero.  The algebraic de Rham homology and cohomology are known to be independent of the embedding, and both are known to be finite-dimensional over k.

Hartshorne defined the de Rham homology of a closed subscheme $Y$ of a smooth scheme $X$ over $k$ as the hypercohomology of the de Rham complex on $X$ with supports in $Y$, which he establishes to be independent of $X$ and the embedding.  Since this is computed from the hypercohomology of a complex, we may also compute it via a spectral sequence, known in this case as the Hodge-de-Rham spectral sequence. The first page of this spectral sequence is given by $E_1^{n-p,n-q}=H_Y^{n-q}(X, \Omega_X^{n-p})$, where $\Omega^{\bullet}_X$ denotes the de Rham complex on $X$, and the abutment of this spectral sequence is $H^{dR}_{p+q}(Y)$.  While the abutment is known to be independent of the embedding $Y \rightarrow X$, the spectral sequence is not.

In \cite{switala}, Switala showed the embedding-independence of the spectral sequence, starting with the $E_2$ page and up to degree shift, in the case where $X$ is the spectrum of a complete local ring.  Additionally, he showed that the $k$-spaces appearing on the $E_2$ page are finite dimensional.  Since these dimensions are independent of the embedding, these are invariants associated to the variety $Y$.  In this paper we show the corresponding results for embeddings of affine varieties into arbitrary smooth affine varieties. Our theorem is the following:

\begin{theorem}
Let $Y$ be an affine variety over a field $k$ of characteristic 0, and let $Y \rightarrow X$ be an embedding of $Y$ into a smooth affine variety $X$ over $k$.  Starting with the $E_2$ page, the Hodge-de-Rham spectral sequence associated to this embedding is independent of the choice of $X$ and embedding, up to a bidegree shift.  Also, the modules appearing on the $E_2$ page, and hence on every $E_r$ page for $r\geq 2$, are finite-dimensional $k$-spaces.
\end{theorem}

The precise statement of embedding-independence is as follows:  given two embeddings $Y \rightarrow X$ and $Y \rightarrow X'$ of an affine variety $Y$ into smooth affine varieties $X$ and $X'$ with $\dim X = s$ and $\dim X' = s'$, there are two associated spectral sequences $E^{\bullet, \bullet}_{\bullet}$ and ${E'}_{\bullet}^{\bullet, \bullet}$.  The theorem asserts that there is a morphism $E_{\bullet}^{\bullet, \bullet} \rightarrow {E'}^{\bullet, \bullet}_{\bullet}$ with bidegree shift $(s'-s, s'-s)$, which is an isomorphism on the $E_2$-page and later.  Precise definitions of the notions of bidegree shifts and morphisms of spectral sequences are provided in section 2 on the preliminaries.

The results in this paper are from the author's thesis, supervised by Gennady Lyubeznik.  We would like to thank Professor Lyubeznik both for suggesting the problem and for his guidance and support.
\section{Preliminaries}

\subsection{$\mathcal{D}$-modules}
Let $k$ be a field of characteristic $0$, and let $T=k[x_1,\dots ,x_n]$.  We write $\mathcal{D}(T,k)$ for the subring of End$_k$($T$) generated by $T$ and the differential operators $\delta_1,\dots ,\delta_n$, where $\delta_i=\frac{\partial}{\partial x_i}$, the usual partial differentiation with respect to $x_i$.  The ring $\mathcal{D}(T,k)$ is noncommutative.  We may consider a filtration on $\mathcal{D}(T,k)$ by degree, that is $\mathcal{F}_0 = k$ and $\mathcal{F}_v=\{\text{$k$-span of monomials in the variables } x_1, x_2, \dots, x_n, \delta_1, \delta_2, \dots \delta_n \text{ of degree} \leq v\}$ for $v > 0$.  The graded ring associated to this filtration, gr $\mathcal{D}(T,k)$, is commutative.  A filtration on a $\mathcal{D}(T,k)$-module is called {\it good} if gr $M$ is finitely generated as a gr $\mathcal{D}(T,k)$-module. $M$ is a finitely generated left $\mathcal{D}(T,k)$-module, if and only if there is a good filtration of $M$ (Propositions 1.2.6 and 1.2.7 in \cite{bjork}).

Given a finitely generated $\mathcal{D}(T,k)$-module $M$, we may define the dimension of $M$ as the dimension of gr $M$ as a gr $\mathcal{D}(T,k)$-module.  Let $\Gamma_v$ be a good filtration of $M$, and consider the Hilbert function $h(v) = \dim_k(\Gamma_v)$.  Hilbert proved that for large enough $v$, this function agrees with a polynomial in $v$.  The dimension of $M$ is defined to be the degree of this polynomial.  (More details can be found in section 1.9 of \cite{eisenbud}, where this statement appears as Theorem 1.11, for example.)

Bernstein's theorem states that for any finitely generated $\mathcal{D}(T,k)$-module $M$, the dimension of $M$ is bounded between $n$ and $2n$.  Those finitely generated modules of dimension $n$ form the Bernstein class of $D(T,k)$-modules and are called holonomic.  Holonomic modules form a full subcategory of $D(T,k)$-modules, closed under formation of submodules, quotient modules, and extensions. (Prop 1.5.2 in \cite{bjork})

If $M$ is a holonomic $D$-module, and $f$ is an element of $T$, then $M_f$ is a holonomic $D$-module  (Theorem 3.4.1 in \cite{bjork}). It follows by considering the Koszul complex that local cohomology modules of a holonomic module are holonomic \cite{lyubeznik}.

\subsection{K\"ahler Differentials}
Here we summarize fundamentals from \cite{eisenbud}.  Let $R$ be a ring, let $S$ be an $R$-algebra and consider the set of $R$-linear derivations of $S$: $R$-linear maps $d$ from $S$ to some $S$-module $M$ satisfying $d(ab) = ad(b) + bd(a)$.  The module of K\"{a}hler Differentials of $S$ over $R$, written $\Omega_{S/R}$, is the dual object to the set of $k$-linear derivations of $S$.  It is the $S$-module generated by $\{ds | s \in S\}$ with relations $d(ss')=sds' + s'ds$ and $d(as + a's')=ads+a'ds'$ for any $s$ and $s'$ in $S$, and any $a, a'$ in $R$.  Observe that these two relations together require that $da=0$ for any element $a \in R$.  We can consider $d$ to be a map from $S$ to $\Omega_{S/R}$, and we call $d$ the universal $R$-linear derivation.  It has the following universal property: given any $S$-module $M$ and a derviation $d': S \rightarrow M$, there exists a unique $S$-linear homomorphism $e: \Omega_{S/R} \rightarrow M$ with $e \circ d = d'$.

For a simple example, consider the case in which $R=k$, and $S=k[x_1,\dots,x_n]$.  In this situation, $\Omega_{S/R}$ is the $S$-module generated by $dx_1, \dots dx_n$.

We need two additional properties of the module $\Omega_{S/R}$ for our results.  First, we need the conormal sequence, which gives us a relationship between the modules of differentials for a given surjection of algebras. The following is proposition 16.3 in \cite{eisenbud}.

\begin{prop}\label{conormal}  If $\pi:T \rightarrow S$ is an surjection of $R$-algebras and $I$ is the kernel of $\pi$, then there is an exact sequence of $S$-modules $$I/I^2 \rightarrow S\otimes_T\Omega_{T/R} \rightarrow \Omega_{S/R} \rightarrow 0.$$
where the left-most map is given by the restriction of $d$ to $I$, and the middle map is given by $s\otimes dt \mapsto sd\pi(t)$
\end{prop}

We also need to know the relationship between the formation of the module of differentials and localizations, namely that they commute.  This is proposition 16.9 in \cite{eisenbud}.

\begin{prop}\label{localdiff}
Let $S$ be a $R$-algebra and let $U$ be a multiplicative subset of $S$.  Then $$\Omega_{S[U^{-1}]/R} \cong S[U^{-1}] \otimes_S \Omega_{S/R}$$
\end{prop}

\subsection{Preliminaries on de Rham cohomology}
Now let $S$ be a $k$-algebra, and let $\Omega^i(S) = \wedge^i\Omega_{S/k}$.  Then there is a complex $$0 \rightarrow S \rightarrow \Omega^1(S) \rightarrow \Omega^2(S) \rightarrow \dots$$ where the map from $\Omega^p(S)$ to $\Omega^{p+1}(S)$ is $k$-linear and sends $sds_1\wedge\dots\wedge ds_p$ to $ds\wedge ds_1\wedge\dots\wedge ds_p$.  This is well-defined because of properties of derivations, and is a complex because the usual wedge rules apply: $dx_i\wedge dx_i = 0$ for all $i$, and $dx_i \wedge dx_j = - dx_j\wedge dx_i$ for all $i$ and $j$.

Let $M$ be a $\mathcal{D}(T,k)$-module, where $T=k[x_1,\dots, x_n]$ as before. The de Rham complex of $M$, denoted $\Omega^\bullet(M)$, is the complex $$0 \rightarrow M \rightarrow \Omega^1(M) \rightarrow \Omega^2(M) \rightarrow \dots$$ where $\Omega^p(M) = M\otimes_T\Omega^p(T)=\bigoplus_{i_1<i_2< \cdots< i_p} M dx_{i_1} \wedge \dots \wedge dx_{i_p}$ and the map from $\Omega^p(M)$ to $\Omega^{p+1}(M)$ sends $\oplus m_{i_1,\dots, i_p} dx_{i_1} \wedge \dots \wedge dx_{i_p}$ to $\sum_{j=1}^n \oplus \delta_j(m_{i_1,\dots, i_p})dx_j\wedge dx_{i_1}\wedge \dots \wedge dx_{i_p}$, where the direct sum on the right hand side is taken over all $i_1 < i_2 \dots < i_p$ and $j$.  This sum is an element of $\Omega^{p+1}$ because the usual wedge rules apply.

The $p$-th de Rham cohomology of a $\mathcal D(T,k)$-module $M$, denoted $H^p(\Omega^\bullet(M))$, is the $p$-th cohomology group of $\Omega^\bullet(M)$.

We will need the following theorem, which is Theorem 6.1 in chapter 1 of \cite{bjork}, later.
\begin{theorem}\label{findim}
If $M$ is a holonomic $\mathcal{D}(T,k)$-module, where $T=k[x_1,...,x_n]$ is the ring of polynomials in variables $x_1,\dots,x_n$ over $k$, then $H^p(\Omega^{\bullet}(M))$ are finite dimensional vector spaces over $k$.
\end{theorem}

\subsection{Preliminaries on Spectral Sequences}
The following is a synopsis of necessary definitions and results from Weibel's introduction to the subject \cite{weib}.
\begin{definition}
A cohomology spectral sequence in an abelian category $\mathcal A$ is a family of objects $\{E^{p,q}_{r}\}$ defined for all integers $p,q$, and for all integers $r>a$ for some $a\geq 0$, with differential maps
$d_r^{p,q}:E^{p,q}_{r}\rightarrow E^{p+r,q-r+1}_r$, with the property that $E_{r+1}^{p,q}$ is isomorphic to the cohomology module $\ker (d_r^{p,q}:E^{p,q}_{r}\rightarrow E^{p+r,q-r+1}_r)/\im (d_r^{p-r,q+r-1}:E^{p-r,q+r-1}_{r}\rightarrow E^{p,q}_r)$.
\end{definition}

The collection of objects in the spectral sequence of the form $E^{p,q}_{r}$ for fixed $r$, together with their associated differentials are referred to as the $E_r$-{\it page} of the spectral sequence.

The spectral sequence in this paper is an example of a {\it first-quadrant spectral sequence}, so named because the only nonzero terms are those in the first quadrant.  That is, $E^{p,q}_{r}=0$ for $p<0$ or $q<0$.  Since for any index in the spectral sequence, one can choose $r$ large enough that the differentials mapping to and from $E^{p,q}_{r}$ are maps to or from positions not in the first quadrant, such spectral sequences have the property that for any fixed $p, q$, there is some $r$ such that $E^{p,q}_{k}=E^{p,q}_{r}$ for $k>r$.  We write $E^{p,q}_{\infty}$ for the entry at that $r$.

A spectral sequence is said to converge to its abutment $H^*$ if $E^{p,q}_{\infty}$ give the quotients of a finite filtration of $H^{*}$.

Double complexes are a common source of spectral sequences. Namely, let $C^{\bullet, \bullet}$ be a first quadrant double complex (i.e. $C^{p,q}=0$ for $p<0$, or $q<0$) with vertical differentials $d_{ver}^{p,q}:C^{p,q}\to C^{p,q+1}$ and horizontal differentials $d_{hor}^p:C^{p,q}\to C^{p+1,q}$. This yields a spectral sequence with $E_0^{p,q}=C^{p,q}$ and $d_0^{p,q}=d_{ver}^{p,q}$ where $d_{ver}^{p,q}:C^{p,q}\to C^{p, q+1}$ is the vertical differential of the double complex, while on the $E_1$ page, $E_1^{p,q}=H^q(C^{p,\bullet})$, the $q$-th cohomology of the $p$-th column, and $d_1:E_1^{p,q}\to E_1^{p+1,q}$ is induced by the horizontal differential in the double complex (this is known as the column-filtered spectral sequence of $C^{\bullet,\bullet}$; there also is a row-filtered spectral sequence which we will not need). This spectral sequence abuts to $H^\bullet({\rm Tot}(C^{\bullet,\bullet}))$, the cohomology of the total complex of $C^{\bullet, \bullet}$, in the following sense. Let $C^{\bullet,\bullet} _{\geq t}$ be $C^{\bullet,\bullet}$ truncated at column $t$, i.e. the double subcomplex of $C^{\bullet,\bullet}$ defined by $C^{p,q}_{\geq t}=C^{p,q}$ if $p\geq t$ and $C^{p,q}_{\geq t}=0$ if $p<t$. Then $C^{\bullet,\bullet}_{\geq t+1}$ ia a double subcomplex of $C^{\bullet,\bullet}_{\geq t}$ and there is a filtration $\dots\supset F^{t}H^\bullet\supset F^{t+1}H^\bullet\supset\dots$ on $H^\bullet({\rm Tot}(C^{\bullet,\bullet}))$, namely, $F^tH^\bullet$ is the image of $H^\bullet({\rm Tot}(C^{\bullet,\bullet}_{\geq t}))$ under the natural map induced by the natural inclusion Tot$(C^{\bullet,\bullet}_{\geq t})\subset {\rm Tot}(C^{\bullet,\bullet})$ and $F^{t}H^n/F^{t+1}H^n\cong E_\infty^{t,n-t}$.

Let $F:\mathcal A\to \mathcal B$ be a left-exact additive functor from an abelian category $\mathcal A$ to an abelian category $\mathcal B$ and let $K^\bullet$ be a complex in $\mathcal A$ concentrated in degrees $p\geq 0$. The hyperderived functors $R^qF(K^\bullet)$ are defined as $H^q({\rm Tot} (F(L^{\bullet,\bullet})))$, the $q$-th cohomology of the total complex of $F(L^{\bullet,\bullet})$ where $L^{\bullet,\bullet}$ is a Cartan-Eilenberg resolution of $K^\bullet$; these hyperderived functors are independent of the particular Cartan-Eilenberg resolution. The spectral sequence of the double complex $F(L^{\bullet,\bullet})$ described in the preceding paragraph also is independent of the particular Cartan-Eilenberg resolution (starting from the $E_1$ page) and it abuts to $R^\bullet F(K^\bullet)$. The $E_1^{p,q}$ entry is $E_1^{p,q}=H^q(F(L^{p,\bullet}))=R^qF(K^{p})$, the $q$-th derived functor of $F$ evaluated at $K^p$.

A morphism of spectral sequences $f:\{E^{\bullet,\bullet}_\bullet\} \rightarrow \{{E'}^{\bullet,\bullet}_\bullet\}$ consists of homomorphisms $f^r:E^{\bullet, \bullet}_r \rightarrow E'^{\bullet,\bullet}_r$, that commute with the differentials in the spectral sequence.

\begin{lemma}[Lemma 5.2.4 in \cite{weib}]\label{mapping} Let $f:\{E^{p,q}_r\} \rightarrow \{E'^{p,q}_r\}$ be a morphism of spectral sequences such that $f^{r_0}:E^{p,q}_{r_0} \rightarrow E'^{p,q}_{r_0}$ is an isomorphism for some $r_0$, then $f^r$ is an isomorphism for all $r>r_0$.  In particular $f^{\infty}: E^{p,q}_{\infty} \rightarrow E'^{p,q}_{\infty}$ is an isomorphism.
\end{lemma}

We will also need the following Lemma, which is Lemma 2.13 in \cite{switala}:

\begin{lemma}\label{canusesequence}
Let $\mathcal A$ be an Abelian category with enough injectives, $\mathcal B$ another Abelian category, and $F:\mathcal A \rightarrow \mathcal B$ a left-exact additive functor.  Let $K^\bullet$ be a complex in $\mathcal A$ that is concentrated in degrees $p \geq 0$ and let $L^{\bullet, \bullet}$ be a double complex in $\mathcal A$ whose objects $L^{p,q}$ are $F$-acyclic.  Suppose additionally that for each $p$, $L^{p, \bullet}$ is a resolution of $K^p$.  Then the spectral sequence for the hyperderived functors of $F$ applied to $K^\bullet$, which starts $E^{p,q}_1 = R^qF(K^p)$ and abuts to ${\bf R}^{p+q}F(K^{\bullet})$ is isomorphic to the column-filtered spectral sequences of the double complex $F(L^{\bullet, \bullet})$
\end{lemma}

\begin{lemma}[see page 30 in \cite{grothendieck}]\label{dbtoss}
A morphism of double complexes induces a morphism of corresponding spectral sequences.
\end{lemma}

\begin{lemma}\label{cptoss}
A morphism of complexes induces a morphism of corresponding spectral sequences for a hyperderived functor.
\end{lemma}

\begin{proof}
Let $f:C^{\bullet} \rightarrow C'^{\bullet}$ be a morphism of complexes.  Let $\mathcal J^{\bullet,\bullet}$ and $\mathcal J'^{\bullet, \bullet}$ be Cartan-Eilenberg resolutions of $C^{\bullet}$ and $C'^{\bullet}$, respectively.  Then $f$ induces a map $\mathcal J^{\bullet,\bullet} \rightarrow \mathcal J'^{\bullet, \bullet}$ that is unique up to homotopy (see page 33 in \cite{grothendieck}), which induces a map $F(\mathcal J^{\bullet,\bullet}) \rightarrow F(\mathcal J'^{\bullet, \bullet})$ that is unique up to homotopy.  Thus from \autoref{dbtoss} this induces a morphism of spectral sequences for the hyperderived functors of a functor $F$ applied to $C^{\bullet}$ and $C'^{\bullet}$.  Chain homotopic maps induce the same morphisms on cohomology, so this is well-defined.
\end{proof}

We use \autoref{canusesequence} to construct the Hodge-de-Rham spectral sequences for our embeddings.  Given a de Rham complex, we take injective resolutions to obtain a double complex, and use its column-filtered spectral sequence.

Finally we recall the notion of degree-shifted morphisms of spectral sequences.

\begin{definition}
[see Definition 2.1 in \cite{switala}] Let $E$ and $E'$ be two spectral sequences with respective differentials $d$ and $d'$. If $a,b\in \mathbb Z$, a morphism $u:E\to E'$ of bidegree $(a,b)$ is a family of morphisms $u^{p.q}_r:E^{p,q}_r\to E'^{p+a,q+b}_r$ such that the $u_r^{p,q}$ are compatible with the differentials and $u^{p,q}_{r+1}$ is induced on cohomology by $u_r^{p,q}$.
\end{definition}

As is pointed out in \cite{switala} right after Definition 2.1, a degree-shifted analogue of Lemma 2.1 also holds. It is not hard to see that degree-shifted versions of Lemmas 2.3 and 2.4 hold as well (that is, a morphism of double complexes of bidegree $(a,b)$ induces a morphism of corresponding spectral sequences of the same degree, and a morphism of complexes $f:C^\bullet\to \mathcal C^\bullet$ of degree $n$, i.e. $f$ sends $C^t$ to $\mathcal C^{t+n}$ for every $t$,  induces a morphism of corresponding spectral sequences of bidegree $(0,n)$).

\section{A morphism of spectral sequences}
The main result of this section is the following.
\begin{prop}\label{morphism}
Let $Y$ be an affine variety, and let $Y \rightarrow X$ be an embedding of $Y$ into a smooth affine variety $X$.  Let $X \rightarrow \mathbb{A}^n$ be an embedding of $X$ into affine space.
There is a morphism of spectral sequences between the Hodge-de-Rham local hyperhomology spectral sequence associated to the embedding $Y\rightarrow X$, and the Hodge-de-Rham local hyperhomology spectral sequence associated to the embedding $Y \rightarrow \mathbb{A}^n$ given by the composition $Y \rightarrow X \rightarrow \mathbb {A}^n$.  This morphism is of bidegree shift $(n-s, n-s)$, where $s=\dim X$.
\end{prop}

The proof results from a series of lemmas.  Throughout this section, let $T$ be the ring $k[x_1,...,x_n]$, the coordinate ring of $\mathbb A^n$.  Let $Y=V(I)$ be an affine variety with an embedding to a smooth affine variety $X=V(J)$ of dimension $s<n$, itself embedded into $\Spec T$, where $J \subset I \subset T$. We write $R$ for $T/J$ and $S$ for $T/I$.

\begin{lemma}
Let $\hat{T}$ be the $J$-adic competion of $T$.  The natural map $T \rightarrow \hat{T}$ induces a map of de Rham complexes $\Omega^{\bullet}(T) \rightarrow \hat{T} \otimes_T \Omega^{\bullet}(T)$, given by $a \mapsto 1 \otimes a$.  According to \autoref{cptoss}, this map of complexes induces a morphism of spectral sequences corresponding to local hypercohomology of the de Rham complex of $T$ with support in $I$ and the local hypercohomology of the complex $\hat{T}\otimes_T \Omega^{\bullet}(T)$ with support in $I$.  This morphism of spectral sequences is an isomorphism starting on the $E_1$ page and is degree-preserving, i.e it is of bidegree shift (0,0).
\end{lemma}

\begin{proof}
Write $E_\bullet^{\bullet,\bullet}$ for the spectral sequence corresponding to local hypercohomology of the de
 Rham complex of $T$, and ${\bf E}_\bullet^{\bullet,\bullet}$ for the spectral sequence corresponding to local hypercohomology of the complex $\hat{T}\otimes_T \Omega^{\bullet}(T)$. The $E_1$ pages of these spectral
sequences consist of the modules $E_1^{p,q} = H_I^{q}(T \otimes_T \Omega^p(T)) = H^q_I(T) \otimes_T \Omega^p(T)$ and ${\bf E}_1^{p,q}= H^q_I(\hat{T} \otimes_T \Omega^p(T)) = H^q_I(\hat{T}) \otimes_T  \Omega^p(T)$.  The map $E_1^{p,q} \rightarrow {\bf E}_1^{p,q}$, is induced by the map $H_I^q(T) \rightarrow H^q_I(\hat{T})$, which in turn is induced by the natural map $T \rightarrow \hat{T}$.  This map $H_I^q(T) \rightarrow H^q_I(\hat{T})=H^q_I(T)\otimes_T\hat{T}$ is an isomorphism since $\hat{T}$ is $J$-adically complete.  Thus $E_1^{p,q} \rightarrow {\bf E}_1^{p,q}$ is an
  isomorphism, so from \autoref{mapping} the map $E_\bullet^{\bullet,\bullet} \rightarrow {\bf E}_\bullet^{\bullet,\bullet}$ is an isomorphism, starting on the $E_1$ page.
\end{proof}

Since $\hat{T}$ is naturally a $D(T,k)$-module, the complex $\hat{T}\otimes_T\Omega^{\bullet}(T)$ is the de Rham complex of $\hat{T}$.

\begin{lemma}
There exists an morphism of spectral sequences $E_\bullet^{\bullet, \bullet} \rightarrow \tilde{\bf E}_\bullet^{\bullet, \bullet}$ of bidegree shift $(n-s,0)$, where $E_\bullet^{\bullet, \bullet}$ is the local hyperhomology spectral sequence with support in $I$ associated to de Rham complex of $\hat{T}$, and $\tilde {\bf E}_\bullet^{\bullet, \bullet}$ is the local hyperhomology spectral sequence with support in $I$ associated to the de Rham complex of $H^{n-s}_J(\hat{T})$.
\end{lemma}

\begin{proof}
Let $\mathcal J^{\bullet}$ be a minimal injective resolution of $\hat{T}$.  Since $\hat{T}$ is Gorenstein, this is the Cousin complex $C^\bullet$, which is defined by $C^{-2}(\hat{T})=0$, $C^{-1}(\hat{T})=\hat{T}$, and when $i\geq 0$, $C^i(\hat{T})=\oplus(\coker d^{i-2})_{\p}$, where the direct sum extends over all $\p$ of height $i$, and the differentials are the canonical localization maps.  These modules are $D$-modules, and the maps are $D$-linear maps (see Lemma 2.25 in \cite{switala}).

Consider the double complex $\mathcal J^{\bullet, \bullet}$ where $\mathcal J^{p,q}=\mathcal J^q\otimes_{\hat{T}} \Omega^p(\hat{T})$.  The columns of this double complex are the minimal injective resolutions of $\Omega^t$ and the rows are the de Rham complexes of $\mathcal J^t$. Now in the complex $\Gamma_J(\mathcal J)^{\bullet, \bullet}$ given by applying the functor $\Gamma_J(-)$ to each entry, the $p$th row, where $p=n-s$, is the first nonzero row, since ht$(J)$ is $n-s$.  The vertical cohomology in degree $n-s$ of this complex is $H^{n-s}_J(\Omega^{\bullet}(\hat{T}))$ and the vertical cohomology in degrees $>n-s$ is zero, since $J$ is locally a complete intersection of height $n-s$. So by performing a row shift of $s-n$ on the double complex $\Gamma_J(\mathcal J)^{\bullet, \bullet}$, we obtain the double complex for the de Rham complex of $H^{n-s}_J(\hat{T})$, i.e. the columns are injective resolutions of $H^{n-s}_J(\hat{T})\otimes_T\Omega^t$ and the rows are the de Rham complexes of $\mathcal J^t$.  Write $\mathcal J_0^{\bullet, \bullet}$ for that complex.

Now, since $\Gamma_I(\Gamma_J(\mathcal J^{p,q}))=\Gamma_I(\mathcal J^{p,q})$, by applying $\Gamma_I$ to each complex we may identify $\Gamma_I(\mathcal J^{\bullet, \bullet})$ with $\Gamma_I(\mathcal J_0^{\bullet,\bullet})$, by a row shift of $n-s$ on $\Gamma_I(\mathcal J^{\bullet, \bullet})$.  This isomorphism induces a morphism on spectral sequences by \autoref{cptoss}
\end{proof}

We now construct a morphism between the de Rham complex of $X$ and the de Rham complex of $H_J^{n-s}(\hat{T})$. The morphism will be shifted in degrees by $n-s$, i.e.~the degree $i$ piece of the de Rham complex of $X$ will map to the degree $i+n-s$ piece of the de Rham complex of  $H_J^{n-s}(\hat{T})$. This will produce the other $n-s$ half of the bidegree shift in the isomorphism of spectral sequences. For this we will need a few more lemmas.

Recall that a $k$-algebra $S$ is {\it formally smooth} over $k$ if given any $k$-algebra $A$ and ideal $\mathfrak A\subset A$ of square zero, every $k$-algebra map $S\to A/\mathfrak A$ lifts to a $k$-algebra map $S\to A$ (See Definition 2.1 of chapter 17 of \cite{cring}). The basic example of a formally smooth $k$-algebra is the polynomial ring $k[x_1,\dots,x_n]$ (Example 2.2 in chapter 17 of \cite{cring}). We quote the following result of Grothendieck.

\begin{theorem}
[\cite{cring} Ch.~17, 2.4] Suppose $T$ is a formally smooth $k$-algebra and let $J\subset T$ be an ideal. Then $R=T/J$ is a formally smooth $k$-algebra if and only if the conormal sequence $$J/J^2\to \Omega_{T/k} \otimes_TR\to\Omega_{R/k}\to 0$$ is split exact.
\end{theorem}

We use this result in the proof of the following lemma.

\begin{lemma}
There exists a $k$-algebra homomorphism $\phi:R=\hat T/J\to \hat T$ such that the composition $\psi\circ\phi:R\to R$ is the identity, where the homomorphism $\psi:\hat T\to \hat T/J\cong R$ is the natural surjection.
\end{lemma}

\emph{Proof.} Since in our case the polynomial ring $T=k[x_1,\dots,x_n]$ is indeed formally smooth over $k$ and the conormal sequence is indeed split exact (since $R=T/J$ is locally a complete intersection, see Ex 17.12 in \cite{eisenbud}), we conclude that $R$ is formally smooth over $k$. Thus there exists a lifting $R\to \hat T$ (Remark on pages 199-200 in \cite{matsumura}). \qed

From now on we regard $\hat T$ as an $R$-algebra via $\phi$.  We will denote the module of K\"ahler differentials of $T$ as an $R$-algebra as $\Omega_{T/R}$.  Next we quote the following result.

\begin{lemma}
[Lemma 10.130.11 in \cite{cring}] Let $R\to S$ be a ring map. Let $I\subset S$ be an ideal. Let $n\geq 1$ be an integer. Set $S'=S/I^{n+1}$. The map $\Omega_{S/R}\to \Omega_{S'/R}$ induces an isomorphism $$\Omega_{S/R}\otimes_SS/I^n\to\Omega_{S'/R}\otimes_{S'}S/I^n.$$
\end{lemma}

We use this result in the proof of the following lemma.

\begin{lemma}
The $R$-algebra structure on $\hat T$ induces a natural injective $\hat T$-module map $$\hat T\otimes_R\Omega_{R/k}\to \hat T\otimes_T\Omega_{T/k}$$ and a direct sum decomposition $$\hat T\otimes_T\Omega_{T/k}=(\hat T\otimes_R\Omega_{R/k})\oplus P$$ where $P$ is a projective $\hat T$-module such that $P/JP\cong J/J^2$.
\end{lemma}

\emph{Proof.} Let $R\subset T/J^{n+1}$ be the image of $R=\phi(R)\subset \hat T$ under the natural surjection $\hat T\to T/J^{n+1}$. The ring maps $k\to R\to T/J^{n+1}$ induce the standard exact sequence of $T/J^{n+1}$-modules $$T/J^{n+1}\otimes_R\Omega_{R/k}\to \Omega_{(T/J^{n+1})/k}\to \Omega_{(T/J^{n+1})/R}\to 0.$$
Tensoring this exact sequence with $T/J^n\otimes_{T/J^{n+1}}$ and considering that the tensor product is right exact we get the exact sequence $$T/J^{n}\otimes_R\Omega_{R/k}\to T/J^n\otimes_{T/J^{n+1}}\Omega_{(T/J^{n+1})/k}\to T/J^n\otimes_{T/J^{n+1}}\Omega_{(T/J^{n+1})/R}\to 0.$$ According to the preceding lemma, the term in the middle is isomorphic to $T/J^n\otimes_T\Omega_{T/k}$ while the term on the right is isomorphic to $T/J^n\otimes_T\Omega_{\hat{T}/R}$ (considering that $\hat T/J^{n+1}\cong T/J^{n+1}).$ Hence the exact sequence becomes $$T/J^{n}\otimes_R\Omega_{R/k}\to T/J^n\otimes_T\Omega_{T/k}\to  T/J^n\otimes_T\Omega_{\hat{T}/R}\to 0.$$

We claim that the term on the right is a projective, i.e.~locally free, $T/J^n$-module of rank $n-s$, where $s$ is the dimension of $R$. It is sufficient to prove this locally, so let $f\in R\subset \hat T$ be an element such that $J_f$ is a complete intersection, generated by $n-s$ elements $z_1,\dots, z_{n-s}$. In this case $(T/J^{n+1})_f=R_f[z_1,\dots,z_{n-s}]/(z_1,\dots,z_{n-s})^{n+1}$ where $R_f[z_1,\dots,z_{n-s}]$ is the polynomial ring in $n-s$ variables $z_1,\dots,z_{n-s}$ over $R_f$. Since the module $\Omega_{(R_f[z_1,\dots,z_{n-s}])/R_f}$  is the free $R_f[z_1,\dots,z_{n-s}]$-module of rank $n-s$, we conclude from the preceding lemma, that $(T/J^n\otimes_T\Omega_{(T/J^{n+1})/R})_f=(T/J^n)_f\otimes_T\Omega_{(R_f[z_1,\dots,z_{n-s}])/R_f}$ is the free $(T/J^n)_f$-module of rank $n-s$. This proves the claim.

Since $\Omega_{R/k}$ is a projective, i.e.~locally free $R$-module of rank $s$, the module $(T/J^n)\otimes_R\Omega_{R/k}$ is a locally free, i.e.~projective $T/J^n$-module of rank $s$. Since the module in the middle is a free module of rank $n$ and the module on the right is a projective $T/J^n$-module of rank $n-s$, i.e.~the rank in the middle is the sum of the two ranks at the ends, we conclude that the map on the left is injective. Since this is a short exact sequence of projective $T/J^n$-modules, it splits, i.e.~we get a direct sum decomposition $$T/J^n\otimes_T\Omega_{T/k}=(T/J^n\otimes_R\Omega_{R/k})\oplus (T/J^n\otimes_T\Omega_{\hat T/R}).$$

As $n$ runs through all positive integers these direct sum decompositions form an inverse systems in which all maps are surjective. Taking the inverse limits we get the direct sum decomposition of $J$-adic completions $$\hat \Omega_{T/k}=(\hat T\otimes_R\Omega_{R/k})\oplus\hat\Omega_{\hat T/R}.$$ Since $\Omega_{T/k}$ is a finitely generated $T$-module, $\hat \Omega_{T/k}=\hat T\otimes_T\Omega_{T/k}$. The position of $\hat T\otimes_R\Omega_{R/k}$ as a submodule in $\hat T\otimes_T\Omega_{T/k}$ is induced by the maps on the left hand sides in the above exact sequences which are in turn induced by the $R$-algebra structure of $\hat T$. This proves the first statement of the lemma.

$\Omega_{\hat T/R}$ is not a finitely generated $\hat T$-module; yet its $J$-adic completion $\hat \Omega_{\hat T/R}$ is the inverse limit of projective $T/J^n$-modules $T/J^n\otimes_T\Omega_{\hat T/R}$ of ranks $n-s$ and all maps in this inverse system are surjective, hence $\hat \Omega_{\hat T/R}$ is a projective $\hat T$-module of rank $n-s$. Setting $P=\hat \Omega_{\hat{T}/R}$, this proves the direct sum decomposition statement of the lemma.

Note that $P/JP=\hat T/J\otimes_{\hat{T}} \hat \Omega_{\hat T/R}=T/J\otimes_T\Omega_{\hat T/R}$, since all completions are the $J$-adic completions.  The conormal sequence $$J/J^2 \rightarrow T/J\otimes\Omega_{\hat T/R} \rightarrow \Omega_{R/R} \rightarrow 0$$ is split exact, from 10.130.10 in \cite{stacks}, since the surjection $\hat{T} \rightarrow T/J$ has a right inverse by Lemma 3.3. Since $\Omega_{R/R}=0$, this implies that $J/J^2 \rightarrow  T/J\otimes \Omega_{\hat T/R}$ is an isomorphism, and thus $P/JP \cong J/J^2$. \qed

Setting $\hat T\otimes_R\Omega_{R/k}=Q$ we get a direct sum decomposition  $\hat T\otimes_T\Omega_{T/k}=Q\oplus P$. Thus if $i\geq n-s$, then $\wedge^{n-s}P\otimes_T\wedge^{i-(n-s)}Q$ is a submodule of $\wedge^i(\hat T\otimes_T\Omega_{T/k})=\hat T\otimes_T\Omega^i(T)$ (in fact, $\wedge^{n-s}P\otimes_T\wedge^{i-(n-s)}Q$ is a direct summand of $\wedge^i(\hat T\otimes_T\Omega_{T/k})=\hat T\otimes_T\Omega^i(T)$).

\begin{lemma}
The differential in the de Rham complex $\hat T\otimes_T\Omega_{T/k}$ sends the submodule $\wedge^{n-s}P\otimes_T\wedge^{i-(n-s)}Q$ of $\hat T\otimes_T\Omega^i(T)$ to the submodule $\wedge^{n-s}P\otimes_T\wedge^{i-(n-s)+1}Q$ of $\hat T\otimes_T\Omega^{i+1}(T)$. Thus there is a subcomplex $$0\to \wedge^{n-s}P\to \wedge^{n-s}P\otimes_T Q\to \wedge^{n-s}P\otimes_T\wedge^2Q\to\dots\to \wedge^{n-s}P\otimes_T\wedge^sQ\to 0$$ of the de Rham complex $\hat T\otimes_T\Omega_{R/k}$ shifted in degrees by $n-s$ (i.e.~the degree $i$ piece of the subcomplex sits in the degree $i+(n-s)$ piece of the de Rham complex).
\end{lemma}

\emph{Proof.} Since the de Rham complex commutes with localization, it is enough to prove this statement after localization at every maximal ideal $\mathfrak m$ of $\hat T$. Since completion of a local ring at the maximal ideal is faithfully flat, for every submodule $L\subset N$ of a $\hat T$-module $N$, we have that $\tilde L\cap N=L$, where $\tilde L$ is the $\mathfrak m$-adic completion of $L$. Hence it is enough to prove the statement after completion with respect to $\mathfrak m$.

Let $x_1,\dots, x_s\in R$ generate the maximal ideal $\mathfrak m\cap R$ of $R$ and let $y_1,\dots, y_{n-s}\in J$ generate $J_{\mathfrak m}$. The $\mathfrak m\cap R$-adic completion of $R$ is the formal power series ring $R'=k[[x_1,\dots, x_s]]$ and the $\mathfrak m$-adic completion of $\hat T$ is the formal power series ring $S=R'[[y_1,\dots, y_{n-s}]]$.  The $\mathfrak m$-adic completion of $\Omega_{T/k}$, i.e.~$S\otimes_T\Omega_{T/k}$, is the free $S$-module on $dx_1,\dots, dx_s,dy_1,\dots, dy_{n-s}$ and the submodule $S\otimes_R\Omega_{R/k}$ of $S\otimes_T\Omega_{T/k}$, whose existence is claimed in Lemma 3.5, is the free $S$-submodule generated by $dx_1,\dots,dx_s$.

Set $\tilde y=dy_1\wedge\dots\wedge dy_{n-s}$. The module $S\otimes_T(\wedge^{n-s}P\otimes_T\wedge^{i-(n-s)}Q)$ is the free $S$-module on all elements $dx_{j_1}\wedge\dots \wedge dx_{j_{i-(n-s)}}\wedge\tilde y$ as $\{j_1,\dots,j_{i-(n-s)}\}$ run through all $(i-(n-s))$-element subsets of $\{1,\dots s\}$, for every $i$. The de Rham differential clearly sends every $dx_{j_1}\wedge\dots \wedge dx_{j_{i-(n-s)}}\wedge\tilde y$ to the $S$-submodule of $S\otimes_T(\wedge^{n-s}P\otimes_T\wedge^{i-(n-s)+1}Q)$ spanned by all elements $dx_{j_1}\wedge\dots \wedge dx_{j_{i-(n-s)+1}}\wedge\tilde y$. \qed

\begin{lemma}
Let $M$ be the submodule of $H^{n-s}_J(\hat T)\otimes_T\wedge^{n-s}P$ annihilated by $J$. Then $M\cong T/J=R$.
\end{lemma}

\begin{proof}\label{annihilator} The composition of functors ${\rm Hom}_{\hat T}(\hat T/J,-)={\rm Hom}_{\hat T}(\hat T/J,\Gamma_J(-))$ leads to the spectral sequence $$E_2^{p,q} = {\rm Ext}^p_{\hat T}(\hat T/J,H^q_J(\hat T))\Longrightarrow {\rm Ext}^{p+q}_{\hat T}(\hat T/J,\hat T).$$ Since $\hat T/J$ is regular, $H^q_J(\hat T)\ne 0$ if and only if $n-s=q$. Thus the spectral sequence yields an isomorphism $${\rm Hom}_{\hat T}(\hat T/J,H^{n-s}_J(\hat T))\cong {\rm Ext}^{n-s}_{\hat T}(\hat T/J,\hat T).$$

Recall that $P/JP\cong J/J^2$. Thus there is a surjection $P\to J/J^2$. Since $P$ is projective, this surjection lifts to a $\hat T$-linear
map $\phi:P\to J$. Let $J'\subset J$ be the image of $\phi$. By Nakayama's lemma, $J'_\mathfrak m=J_\mathfrak m$ for every maximal ideal $\mathfrak m$ that contains $J$. We claim that every maximal ideal of $\hat T$ contains $J$. Indeed, assume $\mathfrak m$ is a maximal ideal that does not contain $J$. Then $\mathfrak m+J=\hat T$, i.e.~there exist elements $y\in J$ and $x\in \mathfrak m$ with $x+y=1$. But then $x=1-y$ is invertible in $\hat T$, for $(1-y)^{-1}=1+y+y^2+y^3+\dots\in \hat T$. Therefore $x$ cannot be contained in any maximal ideal. This contradiction proves the claim. Thus $J'_\mathfrak m=J_\mathfrak m$ for every maximal ideal $\mathfrak m$ of $\hat T$, i.e.~$J'=J$. In other words, the map $\phi:P\to J$ is surjective.

The map $\phi:P\to J$ can be completed to a Koszul resolution $\wedge^\bullet P$ of $\hat T/J$ as follows:
$$0\to \wedge^{n-s}P\to \wedge^{n-s-1}P\to \dots\to\wedge ^2P\to \wedge P(=P)\to \wedge^0P(=\hat T)\to0$$
where the $\hat T$-linear differential $d_t:\wedge^tP\to\wedge^{t-1}P$ is given by $$d_t(x_1\wedge\dots\wedge x_t)=\Sigma_j(-1)^j\phi(x_j)(x_1\wedge\dots\wedge x_{j-1}\wedge x_{j+1}\dots\wedge x_t).$$
This complex is a resolution of $\hat T/J$ in the sense that its 0-th homology is $\hat T/J$ while all other homology groups vanish.

The module Ext$^{n-s}_{\hat T}(\hat T/J,\hat T)$ is the $(n-s)$th homology of the complex Hom$_{\hat T}(\wedge^\bullet P,\hat T)$, i.e.~it is the cokernel of the map $${\rm Hom}_{\hat T}(d_{n-s},\hat T):{\rm Hom}_{\hat T}(\wedge^{n-s-1}P, \hat T)\to {\rm Hom}_{\hat T}(\wedge^{n-s}P,\hat T).$$
The image of this map is $J{\rm Hom}_{\hat T}(\wedge^{n-s}P,\hat T)$, i.e.~$${\rm Ext}^{n-s}_{\hat T}(\hat T/J,\hat T)\cong {\rm Hom}_{\hat T}(\wedge^{n-s}P,\hat T)/J{\rm Hom}_{\hat T}(\wedge^{n-s}P,\hat T)\cong {\rm Hom}_{T/J}(\wedge^{n-s}P/J\wedge^{n-s}P, T/J).$$
Since $\wedge^{n-s}P/(J\wedge^{n-s}P)\cong \wedge^{n-s}(P/JP)=\wedge^{n-s}(J/J^2)$, we conclude that $${\rm Ext}^{n-s}_{\hat T}(\hat T/J,\hat T)\cong {\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J).$$
Thus the module ${\rm Hom}_{\hat T}(\hat T/J,H^{n-s}_J(\hat T))$, i.e.~the submodule of $H^{n-s}_J(\hat T)$ annihilated by $J$, is isomorphic to ${\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J).$

Let $M={\rm Hom}_{\hat T}(\hat T/J, H^{n-s}_J(\hat T)\otimes_T\wedge^{n-s}P)$, i.e.~$M$ is the submodule of $H^{n-s}_J(\hat T)\otimes_T\wedge^{n-s}P$ annihilated by $J$. Since the module $\wedge^{n-s}P$ is flat over $T$, we conclude that $M\cong {\rm Hom}_{\hat T}(\hat T/J, H^{n-s}_J(\hat T))\otimes_T\wedge^{n-s}P$. This, as has just been proven, is isomorphic to ${\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J)\otimes_T\wedge^{n-s}P$. Since the module ${\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J)$ is annihilated by $J$, this tensor product is isomorphic to ${\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J)\otimes_{T/J}(\wedge^{n-s}P/J\wedge^{n-s}P)$ which (since $\wedge^{n-s}P/J\wedge^{n-s}P=\wedge^{n-s}(P/JP=\wedge^{n-s}(J/J^2)$ is isomorphic to ${\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J)\otimes_{T/J}\otimes\wedge^{n-s}(J/J^2)$. This last module is isomorphic to $T/J$ since $\wedge^{n-s}(J/J^2)$ is an invertible $T/J$-module and ${\rm Hom}_{T/J}(\wedge^{n-s}(J/J^2),T/J)$ is the inverse of this module in the Picard group of $T/J$.
\end{proof}

At last we are ready to prove Proposition 3.1.

\emph{Proof of Proposition 3.1.} It is enough to prove that the de Rham complex of $R$ is a subcomplex of the de Rham complex of $H^{n-s}_J(\hat T)$, shifted in degrees by $n-s$ (i.e.~the degree $i$ piece of the de Rham complex of $R$ sits in the degree $i+n-s$ piece of the de Rham complex of $H^{n-s}_J(\hat T)$).

Denote by $\wedge^{n-s}P\otimes_T\wedge^\bullet Q$ the subcomplex of $\hat T\otimes_T\Omega^\bullet(T)$ whose existence was proved in Lemma 3.6. Tensoring with $H^{n-s}_J(T)$ we see that $H^{n-s}_J(T)\otimes_T\wedge^{n-s}P\otimes_T\wedge^\bullet Q$ is a subcomplex of $H^{n-s}_J(T) \otimes_T \Omega^\bullet(T)$ shifted in degrees by $n-s$ (tensoring this subcomplex results in a subcomplex because in every degree $i$ the submodule $\wedge^{n-s}P\otimes_T\wedge^iQ$ is a direct summand of $\hat T\otimes_T\Omega^{i+n-s}(T)$).

We claim that the differentials in the complex $H^{n-s}_J(T) \otimes_T \wedge^{n-s}P\otimes_T\wedge^\bullet Q$ send elements annihilated by $J$ to elements annihilated by $J$ (this is certainly not true for the differentials in the ambient complex $H^{n-s}_J(T) \otimes_T \Omega^\bullet(T))$, i.e.~${\rm Hom}_T(T, H^{n-s}_J(T) \otimes_T \wedge^{n-s}P\otimes_T\wedge^\bullet Q)$ is a complex. It is sufficient to show this after completion at every maximal ideal of $\hat T$, in which case (using notation from the proof of Lemma 3.6) $H^{n-s}_J(\hat T)$ is the free $R'$-module on the monomials $y_1^{i_1}\cdots y_{{n-s}}^{i_{n-s}}$ where $i_1,\dots, i_{n-s}$ run through all negative integers $\leq -1$ and the annihilator of $J$ in $H^{n-s}_J(\hat T)$ is the $R'$-submodule spanned by $(y_1\cdots y_{n-s})^{-1}$. The differentials in the complex $H^{n-s}_J(T) \otimes_T \wedge^{n-s}P\otimes_T\wedge^\bullet Q$ involve differentiations only with respect to $x_1,\dots, x_s$ and leave the $y_1,\dots, y_{n-s}$ alone. Thus they send the submodule spanned by $(y_1\cdots y_{n-s})^{-1}$ to the submodule spanned by $(y_1\cdots y_{n-s})^{-1}$. This proves the claim. Thus ${\rm Hom}_T(T, H^{n-s}_J(T) \otimes_T \wedge^{n-s}P\otimes_T\wedge^\bullet Q)$ is a complex, in fact a subcomplex of $H^{n-s}_J(T) \otimes_T \wedge^{n-s}P\otimes_T\wedge^\bullet Q$.

Lemma 3.7 shows that the degree 0 piece of this complex is isomorphic to $R=T/J$, while the formula $\hat{T}/J \otimes_{\hat{T}} Q = (\hat{T}/J) \otimes_{\hat{T}} \hat{T} \otimes_R \Omega_{R/k}= \hat{T}/J \otimes_R \Omega_{R/k} = \Omega_{R/k}$ shows that the degree 1 piece is isomorphic to $\Omega_{R/k}$. Thus one sees (again by going to the completion at every maximal ideal and considering the explicit formulas for the differentials in the subcomplex) that the differentials in this subcomplex are the differentials in $\Omega^\bullet(R)$, the de Rham complex of $R$. This completes the proof of Proposition 3.1. \qed

\section{The isomorphism on the $E_2$ page}

\subsection{The isomorphism}
We now show the morphism of spectral sequences found in the previous section is an isomorphism starting at the $E_2$ page. That is, we show the following:

\begin{prop} The map induced on the $E_2$ page by the morphism of complexes in Proposition 3.1 is an isomorphism.
\end{prop}

The general approach is to localize, demonstrate an isomorphism on the localizations by inducting on the difference in dimension, and then glue the local isomorphisms together.  Again, let $Y \rightarrow X$ be an embedding of $Y$ into an arbitrary smooth affine scheme $X$, and let $X \rightarrow \mathbb A^n$ be an embedding of $X$ as a closed subscheme of $\mathbb{A}^n$.  Let $S$, $R$ and $T$ be the coordinate rings associated with these varieties.  That is, $Y= \Spec S,\ X=\Spec R$, and $T=k[x_1,...,x_n]$.  Let $I$ be the kernel of the surjection $T \rightarrow S$, and let $J$ be the kernel of the surjection $T \rightarrow R$.

Since $X$ is a smooth scheme, $\Omega_{R/k}$ is locally free of rank $s=\dim X$.  Thus we may find $f_1, \dots ,f_m \in R$ so that the $\Spec R_{f_i}$ form an open cover of $\Spec R$, and so that for each $f_i$ there exist $y_{i1},\dots,y_{is} \in R$ so $dy_{i1},\dots,dy_{is}$ generate $\Omega_{R_{f_i}/k}$ as a $T_{f_i}$-module.

Write $Y_{f_i}$ for $\Spec S_{f_i}$ and $X_{f_i}$ for $\Spec R_{f_i}$.  We show our claim first for the local case $Y_{f_i} \rightarrow X_{f_i}$
and then show how to extend this to the situation $Y \rightarrow X$.

Fix some $f_i$ and say $\Omega_{R_{f_i}/k}$ is generated by $dy_1,...,dy_s$. We know there is an exact sequence for differentials over k from \autoref{conormal}:

$$J_{f_i}/J_{f_i}^2 \rightarrow R_{f_i} \otimes_{T_{f_i}} \Omega_{T_{f_i}/k} \rightarrow \Omega_{R_{f_i}/k}\rightarrow 0,$$
which shows that we can extend the set of generators of $\Omega_{R_{f_i}/k}$ to a generating set of $\Omega_{T_{f_i}/k}$ by picking $n-s$ elements $dz_1,...,dz_{n-s}$ with $J_{f_i}=(z_1\ ...\ z_{n-s})$.

 \begin{lemma} In this situation, we may view the $J_{f_i}$-adic completion of $T_{f_i}$, which we will denote $\hat{T}_{f_i}$, as a power series ring $\hat{T}_{f_i} \cong (T_{f_i}/J)[[z_1,\dots, z_{n-s}]]=R_{f_i}[[z_1,\dots,z_{n-s}]]$.
 \end{lemma}

 \begin{proof}
 Since $R_{f_i}$ is formally smooth over $k$, there exists a ring homomorphism $\phi:R_{f_i}\to \hat T_{f_i}$ such that the composition of $\phi$ with the natural surjection $\hat T_{f_i}\to \hat T_{f_i}/J$ is the identity map $R_{f_i}\to \hat T_{f_i}/J=R_{f_i}$. Denote $\phi(R_{f_i})$ by $R_{f_i}$. Since $\hat T_{f_i}$ is complete and $J$ is an ideal of height $n-s$ generated by $z_1,\dots,z_{n-s}$, it follows that   $T_{f_i}=R_{f_i}[[z_1,\dots,z_{n-s}]]$.
 \end{proof}

\begin{lemma}\label{completion} The modules in the complex $H^{\bullet}_I(T_{f_i})$ have a natural structure as $\hat{T}_{f_i}$-modules.
 \end{lemma}
\begin{proof}   Since $I \subset J$, every element of those modules is killed by a power of $J$, so each inherits a $\hat{T}_{f_i}$-module structure from its $T_{f_i}$ structure.  It's well-known that local cohomology commutes with completions, and it follows that each module is isomorphic to the local cohomology of the completion as a $\hat{T}_{f_i}$-module.
\end{proof}

At this point we need a definition of the plus operation from \cite{emilyluis}:

\begin{definition}
Let $A$ be a noetherian ring, and let $B=A[[z]]$.  The plus operation, is the functor from the category of $A$-modules to the category of $B$-modules given by $M_+ = M\otimes_A(B_z/B)=M\cdot\frac{1}{z}\oplus M\cdot\frac{1}{z^2}\oplus\dots\oplus M\cdot \frac{1}{z^i}\oplus\dots.$
\end{definition}

Multiplication by $z$ snds $\frac{m}{z^i}$ to $\frac{m}{z^{i-1}}$. Thus the annihilator of $z$ in $M_+$ is the $A$-module $M\cdot\frac{1}{z}$. If $A$ is a regular finitely generated $k$-algebra and $M$ is a $D(A,k)$-module, then $M_+$ is a $D(B,k)$-module (namely, $\partial_z$ sends $\frac{m}{z^i}$ to $(-i)\frac{m}{z^{i+1}}$).

Lemma 3.5 says that $\Omega_{B/k}=(B\otimes_A\Omega_{A/k})\oplus P$ and in this particular case $P=Bdz$, the free module of rank one generated by $dz$.
Hence there is a chain map of de Rham complexes $\phi^\bullet:\Omega^\bullet (M)\to \Omega^\bullet (M_+)$, where $\Omega^\bullet(M)$ is the de Rham complex of $M$ in the category of $A$-modules while $\Omega^\bullet(M_+)$ is the de Rham complex of $M_+$ in the category of $B$-modules and the chain map sends $m\otimes\omega\in M\otimes_A\Omega^i(A)$ to $(m\otimes\omega)\cdot\frac{1}{z}dz\in M_+\otimes_B\Omega^{i+1}(B)$. Thus the chain map sends $\Omega^\bullet(M)$ to the annihilator of the ideal $J=(z)$ in the complex $P\otimes_B\wedge^\bullet(B\otimes_A\Omega_{A/k})$, exactly as in the proof of Proposition 3.1.

\begin{prop}
This chain map $\phi^\bullet:\Omega^\bullet(M)\to \Omega^\bullet(M_+)$ induces an isomorphism on homology, i.e.~$\phi^i_*:h^p(\Omega^\bullet(M))\to h^{p+1}(\Omega^\bullet(M_+))$ is an isomorphism for all $p$.
\end{prop}

\emph{Proof.} Consider the short exact sequence of modules
\begin{equation}\label{ses} 0 \rightarrow M _+\otimes_A \Omega^{p-1}(A) \rightarrow M_+ \otimes_B \Omega^p (B) \rightarrow M_+\otimes_A \Omega^p (A) \rightarrow 0,
\end{equation}
 where the first map is the wedge map with $dz$, and the second map acts by sending $dz$ to $0$.  This induces a short exact sequence on complexes, and so induces a long exact sequence in de Rham cohomology:
$$\cdots \rightarrow h^{p-1}(M_+ \otimes_A \Omega^{\bullet}(A)) \rightarrow h^p(M_+ \otimes_B\Omega^{\bullet}(B)) \rightarrow h^p (M_+ \otimes_{A} \Omega^{\bullet}(A)) \rightarrow h^p (M_+ \otimes_A \Omega^{\bullet}(A)) \rightarrow \cdots$$ where the last map is the connecting homomorphism.  Switala showed that the connecting homomorphism map is the same as $\partial_z$. (Lemma 2.22 in \cite{switala})

Hence the last map above is $\partial z: h^p (M_+ \otimes_{A} \Omega^{\bullet}(A)) \rightarrow h^p (M_+ \otimes_A \Omega^{\bullet}(A))$.  Since the maps in the complex $\Omega^\bullet(A)$ are unaffected by $z$ or $dz$, it follows that $$h^p (M_+ \otimes_{A} \Omega^{\bullet}(A))=h^p(M\otimes_A\Omega^\bullet(A))\cdot\frac{1}{z}\oplus h^p(M\otimes_A\Omega^\bullet(A))\cdot\frac{1}{z^2}\oplus\dots.$$ Therefore $$\ker(\partial z: h^p (M_+ \otimes_{A} \Omega^{\bullet}(A)) \rightarrow h^p (M_+ \otimes_A \Omega^{\bullet}(A)))= 0$$ and $$\coker(\partial z: h^p (M_+ \otimes_{A} \Omega^{\bullet}(A)) \rightarrow h^p (M_+ \otimes_A \Omega^{\bullet}(A)))= h^p(M\otimes_A\Omega^\bullet(A))\cdot\frac{1}{z}.$$

Therefore $h^{p+1}(\Omega^\bullet(M_+))=h^p(M\otimes_A\Omega^\bullet(A))$ and the isomorphism is induced by the chain map $M\otimes_A\Omega^\bullet(A)\to M_+\otimes_B\Omega^\bullet(B)$ that sends $M\otimes_A\Omega^i(A)$ to $M\otimes_A\Omega^i(A)\cdot\frac{1}{z}dz\in \Omega^{p+1}(B)$, exactly as claimed. \qed

Nu\'{n}\~{e}z-Betancourt and Witt showed the following:

\begin{lemma}[Lemma 3.9 in \cite{emilyluis}]\label{plusop}
Let $A$ be a noetherian ring, $M$ an $A$-module, and $B=A[[x]]$.  For every ideal $I \subseteq A$ any $i \in \mathbb N$, $$(H^i_I(M))_+ \cong H^{i+1}_{(I,x)}(M\otimes_A B)$$
\end{lemma}

\begin{prop}
Let $f_i\in T$ be such that the ideal $J_{f_i}$ is a complete intersection, i.e.~$J_{f_i}=(z_1,\dots,z_{n-s})$. Let $\psi^\bullet:\Omega^\bullet(R_{f_i})\to \Omega^\bullet(H^{n-s}_J(T_{f_i}))$ be the morphism of complexes from the proof of Proposition 3.1 and let $H^q_I(\psi^\bullet):\Omega^\bullet(H^q_I(R_{f_i}))\to \Omega^\bullet(H^q_I(H^{n-s}_J(T_{f_i})))$ be the induced morphism on the de Rham complexes of local cohomology modules. Considering that $H^q_I(H^{n-s}_J(T_{f_i}))\cong H^{q+n-s}_I(T_{f_i})$, the morphism $H^{n-s}_I(\psi^\bullet)$ induces an isomorphism on homology, i.e.~ the map $\psi^p_*:h^p(\Omega^\bullet(H^q_I(R_{f_i})))\to h^{p+n-s}(\Omega^\bullet(H^{q+n-s}_{I}(T_{f_i})))$ is an isomorphism.
\end{prop}

\emph{Proof.} According to \autoref{completion}, after renaming $R_{f_i}$ as $A$ we may replace $T_{f_i}$ by the power series ring $B=A[[z_1,\dots,z_{n-s}]]$.
Now let $B_t=A[[z_1,\dots,z_t]]$ for $1\leq t\leq k$. Then $B_{t}=B_{t-1}[[z_{t}]]$ and therefore iterating the plus construction $t$ times we get by induction on $t$ that $(\dots(H^i(A)_+)_+\dots)_+=H^{i+t}_{(I, z_1,\dots,z_t)}(B_t)$.

The morphism of complexes $H^{n-s}_I(\psi^\bullet):\Omega^\bullet(H^{n-s}_I(A))\to \Omega^\bullet(H^{n-s}_{(I,z_1,\dots,z_{n-s})}(B))$ is the composition $\psi_1^\bullet\circ\psi_2^\bullet\circ\dots\circ\psi^\bullet_{n-s}$, where $\psi_i^\bullet:\Omega^\bullet(H^{i-1}_{(I,z_1,\dots,z_{i-1})}(B_{i-1}))\to \Omega^\bullet (H^i_{(I,z_1,\dots,z_i)}(B_i))$ is the morphism of Lemma 4.2. Since by Lemma 4.2 each morphism in the composition induces an isomorphism on homology, so does the composition. \qed

We can now prove Proposition 4.1:

\emph{Proof.} Let $\psi^\bullet:\Omega^\bullet(R)\to \Omega^\bullet(H^{n-s}_J(T))$ be the morphism of complexes in Proposition 3.1. It induces a morphism $\psi^\bullet_{f_i}:\Omega^\bullet(R_{f_i})\to \Omega^\bullet(H^{n-s}_J(T_{f_i}))$ for every $i$. By proposition 4.2 this map induces an isomorphism $(\psi_{f_i}^p)_*:h^p(\Omega^\bullet(H^q_I(R_{f_i})))\to h^p(\Omega^\bullet(H^{q+n-s}_I(T_{f_i})))$ on homology, for all $p$.

For any $R$-module or $T$-module $M$, the Cech complex: $$0 \rightarrow M \rightarrow \oplus M_{f_i} \rightarrow \dots \rightarrow M_{f_1\dots f_m} \rightarrow 0$$ is exact. Hence the induced morphism of complexes $H^q_I(\psi^\bullet):\Omega^\bullet(H^q_I(R))\to \Omega^\bullet(H^q_I(H^{n-s}_J(T)))$ induces (considering that $H^q_I(H^{n-s}_J(T))\cong H^{q+n-s}_I(T)$, a commutative diagram of complexes with exact rows:
\begin{scriptsize}
$$\begin{CD}
0@>>>\Omega^\bullet(H^q_I(R))@>>>\oplus\Omega^\bullet(H^q_I(R_{f_i}))@>>>\dots@>>>\Omega^\bullet(H^q_I(R_{f_1\cdots f_m}))@>>>0\\
@.@VV\psi^\bullet V@VV\oplus\psi^\bullet_{f_i}V@.@VV\psi^\bullet_{f_1\cdots f_m}V\\
0@>>>\Omega^\bullet(H^{q+n-s}_I(T))@>>>\oplus\Omega^\bullet(H^{q+n-s}_I(T)_{f_i})@>>>\dots@>>>\Omega^\bullet(H^{q+n-s}_I(T)_{f_1\cdots f_m})@>>>0\\
\end{CD}$$
\end{scriptsize}
where the vertical maps $\oplus\psi_{f_{i_1}\cdots f_{i_j}}^\bullet$ are direct sums of localizations of the vertical map $\psi^\bullet$ on the very left of the diagram. We have shown that each $\psi^\bullet_{{f_1}\cdots f_{i_j}}$ induces an isomorphism on homology. Hence so does every vertical map $\oplus\psi_{f_{i_1}\cdots f_{i_j}}^\bullet$.

Let $K^\bullet_{R,j}$ be the kernel of the chain map $\oplus\Omega^\bullet(H^q_I(R)_{f_{i_1}\cdots f_{i_j}}))\to \oplus\Omega^\bullet(H^q_I(R_{f_{i_1}\cdots f_{i_j}f_{i_{j+1}}}))$ in the top row of the diagram and let $K^\bullet_{T,j}$ be the kernel of the corresponding chain map $\oplus\Omega^\bullet(H^{q+n-s}_I(T_{f_{i_1}\cdots f_{i_j}}))\to \oplus\Omega^\bullet(H^{q+n-s}_I(T_{f_{i_1}\cdots f_{i_j}f_{i_{j+1}}}))$ in the bottom row. Both $K^\bullet_{R,j}$ and $K^\bullet_{T,j}$ are complexes and the vertical maps in the diagram induce a morphism of complexes $\psi^\bullet_{K,j}:K^\bullet(R,j)\to K^\bullet(T,j)$.

We claim that the induced maps on homology $(\psi^p_{K,j})_*:h^p(K^\bullet(R,j))\to h^p(K^\bullet(T,j))$ are isomorphisms for all $p$. To prove this claim we use descending induction on $j$. To begin the induction, let $j=m-1$. The above diagram induces the following commutative diagram with short exact rows.

$$\minCDarrowwidth10pt\begin{CD}
0@>>>K^\bullet_{R,m-1}@>>>\oplus_j\Omega^\bullet(H^q_I(R_{{f_1}\cdots f_{i_{j-1}}f_{i_{j+1}}\cdots f_m}))@>>> \Omega^\bullet(H^q_I(R_{{f_1\cdots f_m}}))@>>> 0\\
@. @VV\psi^\bullet_{K,m-1}V @VV\oplus_j\psi^\bullet_{1,\dots,j-1,j+1,\dots,m}V@VV\psi^\bullet_{{f_1}\cdots f_m}V\\
0@>>>K^\bullet_{T,m-1}@>>>\oplus_j\Omega^\bullet(H^{q+n-s}_I(T_{{f_1}\cdots f_{i_{j-1}}f_{i_{j+1}}\cdots f_m}))@>>> \Omega^\bullet(H^{q+n-s}_I(T_{{f_1\cdots f_m}}))@>>> 0\\
\end{CD}$$

Let us write, for the sake of legibility, $A_R^{\bullet} = \oplus_j\Omega^\bullet(H^q_I(R_{{f_1}\cdots f_{i_{j-1}}f_{i_{j+1}}\cdots f_m}))$, $A_T^{\bullet}$ = $\oplus_j\Omega^\bullet(H^{q+n-s}_I(T_{{f_1}\cdots f_{i_{j-1}}f_{i_{j+1}}\cdots f_m}))$, and $p' = p+n-s$. Then this commutative diagram induces the following commutative diagram where the rows are the long exact sequences produced by the short exact sequences in the rows of the preceding diagram.

\begin{footnotesize}
$$\minCDarrowwidth10pt\begin{CD}
h^{p-1}(A_R^{\bullet}) @>>> h^{p-1}(\Omega^\bullet(R_{{f_1\cdots f_m}})) @>>> h^p(K^\bullet_{R,m-1})@>>> h^{p}(A_R^{\bullet})@>>> h^{p}(\Omega^\bullet(R_{{f_1\cdots f_m}}))\\
@VV\cong V @VV\cong V @VVV @VV \cong V @VV\cong V\\
h^{p'-1}(A_T^{\bullet}) @>>> h^{p'-1}(\Omega^\bullet(H^{n-s}_J(T_{{f_1\cdots f_m}})))@>>> h^{p'}(K^\bullet_{T,m-1}) @>>> h^{p'}(A_T^{\bullet})@>>> h^{p'}(\Omega^\bullet(H^{n-s}_J(T_{{f_1\cdots f_m}}))) \\
\end{CD}$$
\end{footnotesize}

The two vertical maps on the left and the two vertical maps on the right have been shown to be isomorphisms. By the 5-lemma the map in the middle is an isomorphism as well. This finishes the proof of the claim in the case $j=m-1$.

Now let $j<m-1$ and assume the claim proven for $j+1$. Since the rows in the very first commutative diagram in this proof are exact, we get the following commutative diagram with exact rows, where the direct sum in the middle is over all $j$-element subsets of the set $\{1,\dots,m\}$.

$$\minCDarrowwidth10pt\begin{CD}
0@>>>K^\bullet_{R,j}@>>>\oplus\Omega^\bullet(R_{{f_{i_1}}\cdots f_{i_j}})@>>> K^\bullet_{R,j+1}@>>> 0\\
@. @VV\psi^\bullet_{K,j}V @VV\oplus\psi^\bullet_{f_{i_1}\cdots f_{i_j}}V@VV\psi^\bullet_{K,j+1}V\\
0@>>>K^\bullet_{T,j}@>>>\oplus\Omega^\bullet(H^{n-s}_J(T_{f_{i_1}\cdots f_{i_j}}))@>>> K^\bullet_{T,j+1}@>>> 0\\
\end{CD}$$

This commutative diagram induces the following commutative diagram where the rows are the long exact sequences produced by the short exact sequences in the rows of the preceding diagram.  Again, we write $p'$ for $p+n-s$.

\begin{footnotesize}
$$\minCDarrowwidth10pt\begin{CD}
h^{p-1}(\oplus\Omega^\bullet(R_{{f_{i_1}}\cdots f_{i_j}}) @>>> h^{p-1}(K^\bullet_{R,j+1}) @>>> h^p(K^\bullet_{R,j})@>>> h^{p}(\Omega^\bullet(R_{{f_{i_1}}\cdots f_{i_j}})@>>> h^{p}(K^\bullet_{R,j+1})\\
@VV\cong V @VV\cong V @VVV @VV \cong V @VV\cong V\\
h^{p'-1}( \oplus\Omega^\bullet(H^{n-s}_J(T_{f_{i_1}\cdots f_{i_j}})) @>>> h^{p'-1}(K^\bullet_{T,j+1})@>>> h^{p'}(K^\bullet_{T,j}) @>>> h^{p'}(\oplus\Omega^\bullet(H^{n-s}_J(T_{f_{i_1}\cdots f_{i_j}}))@>>> h^{p'}(K^\bullet_{T,j+1}) \\
\end{CD}$$
\end{footnotesize}

The first and the third vertical maps in this diagram have been shown to be isomorphisms, while the second and the foruth vertical maps are isomorphisms by the induction hypothesis. The 5-lemma implies that the vertical map in the middle is an isomorphism as well. This completes the proof of the claim.

In particular, the map $\psi^\bullet_{K,1}:K^\bullet(R,1)\to K^\bullet(T,1)$ induces for all $p$ isomorphisms $(\psi^p_{K,1})_*:h^p(K^\bullet(R,1))\to h^p(K^\bullet(T,1))$ on homology.

The very first commutative diagram in this proof induces the following commutative diagram with short exact rows.

$$\minCDarrowwidth10pt\begin{CD}
0@>>>\Omega^\bullet(R)@>>>\oplus\Omega^\bullet(R_{f_i})@>>>K^\bullet_{R,1}@>>>0\\
@.@VV\psi^\bullet V@VV\oplus\psi^\bullet_{f_i}V@VV\psi^\bullet_{K,1}V\\
0@>>>\Omega^\bullet(H^{n-s}_J(T))@>>>\oplus\Omega^\bullet(H^{n-s}_J(T)_{f_i})@>>>K^\bullet_{T,1}@>>>0\\
\end{CD}$$

This diagram similarly produces a commutative diagram with long exact sequences sequences in the rows and the 5-lemma then implies that the chain map $\psi^\bullet$ induces isomorphisms on homology.\qed

\subsection{Proof of finite-dimensionality of the modules appearing on each page}
\begin{prop}
Consider an embedding $Y\rightarrow X$ of an affine scheme $Y$ into a smooth affine scheme $X$.  The modules appearing on the $E_2$ page of the spectral sequence associated to $Y\rightarrow X$ are finite-dimensional. These modules are finite-dimensional, and thus the modules appearing in either sequence on any page $E_r$, $r \geq 2$ are finite-dimensional.
\end{prop}

\begin{proof}
Due to the embedding-independence shown before, we may consider the case $X=\Spec k[x_1,...,x_n]$, and write $T$ for $k[x_1,...,x_n]$.  Write $S$ for the ring with $Y = \Spec S$, and let $I$ be the kernel of the surjection $T \rightarrow S$.
The $E_1$ page of the embedding is given by $E_1^{p,q} = H_{Y}^q(X,\Omega^p_{X})$, or equivalently, $E_1^{p,q}=H^q_{I}(T\otimes\Omega^p(T))$.  $T$ is the ring of polynomials in $n$ variables, so $\Omega^p(T)$ is free of rank $n$.  Local cohomology commutes with direct sums, so this is the same as
$H_{I}(T)\otimes_{T}\Omega^p(T)$.  Now as before, the maps in any row are the de Rham differentials of $H^q_{I}(T)\otimes_{T}\Omega^{\bullet}$(T), so the modules on the $E_2$ page - the cohomology objects of these rows - are $h^p(H^q_{I}(T)\otimes_{T}\Omega^{\bullet}(T))$.  This shows the first part of our proposition.

The second part of the statement follows from \autoref{findim}, because $H^q_{I}(T)$ is a holonomic $\mathcal{D}(T,k)$-module.
 \end{proof}

config: arxiv


TITLE: Ricatti Differential equation and initial value problem
QUESTION [1 upvotes]: I have to find a polynomial that solves the following Riccati-equation.
$y'=\frac {-2}{x}-2xy+\frac {2y^2}{x}$
How to solve it?
Furthermore I need to solve the initial value problem for arbitrary $(x_{0}, y_{0})$ out of $\mathbb {R}\setminus{0}$
Thanks for your help

REPLY [2 votes]: From the form of the ODE we observe that the solutions $y(x)$ are even functions. So, the change of variable $X=x^2$ allows a first simplifixation. Then, the usual change of function for Riccati's ODE leads to a linear second order ODE.


text:
\begin{document}
\maketitle
\begin{abstract} We present a family of sharpness examples for Falconer-type single dot product results. In particular, for $d\geq 2,$ for any $s<\frac{d+1}{2},$ we construct a Borel probability measure $\mu$ satisfying the energy estimate $I_s(\mu)<\infty,$ yet the estimate
\begin{equation} \label{monalisa} (\mu \times \mu)\{(x,y):1\leq x\cdot y \leq 1+\epsilon\} \leq C\epsilon \end{equation} does not hold with constants independent of $\epsilon$. It is known (\cite{EIT11}) that such an estimate always holds with $C$ independent of $\epsilon$ if $I_{\frac{d+1}{2}}(\mu)<\infty$. Thus our estimate proves the sharpness of the dimensional threshold in this result and generalizes similar results (\cite{Mat95}, \cite{IS16}) established in the case when the dot product $x \cdot y$ is replaced by the Euclidean distance function $|x-y|$, or, more generally, ${||x-y||}_K$, the distance that comes from the norm induced by a symmetric convex body $K$ with a smooth boundary and non-vanishing curvature. Our constructions are partially based on ideas that come from discrete incidence theory.

\end{abstract}

\section{Introduction}
In \cite{Fal86}, Falconer conjectured that any subset of $\mathbb R^d$ with Hausdorff dimension greater than $\frac{d}{2}$ would determine a set of distances with positive Lebesgue measure. In the same paper, he proved that a weaker version of the conjecture holds for sets whose Hausdorff dimension is more than $\frac{d+1}{2}.$ While this threshold has been lowered over the years (see \cite{Mat87, Mat95, W99, Erd05, GIOW}), the full conjecture is still open. The key to Falconer's original result was based on the following lemma, which states that if a certain energy estimate holds, then we are guaranteed a bound on the product measure of pairs of points separated by approximately any fixed distance.
\begin{lemma}\label{key}
Given a compactly supported Borel measure $\mu$, there exists a constant $C>0$ such that
\begin{equation}\label{muIncidence} (\mu\times\mu)\{(x,y): 1 \leq |x-y| \leq 1+\epsilon\} \leq C I_{\frac{d+1}{2}}(\mu) \epsilon,\end{equation} where
\begin{equation}\label{muEnergy} I_s(\mu) \equiv \int\int |x-y|^{-s}d\mu(x)d\mu(y). \end{equation} \end{lemma}

In particular, this lemma says that if $\mu$ is supported on a set of Hausdorff dimension $>\frac{d+1}{2}$, then the estimate (\ref{muIncidence}) holds with uniform constants. It is also not difficult to see that the proof easily extends to the case when the Euclidean norm $|\cdot|$ is replaced by ${|| \cdot ||}_K$, the norm induced by a symmetric convex body $K$ with a smooth boundary and non-vanishing Gaussian curvature.

In \cite{Mat95}, Mattila showed that Falconer's lemma is sharp in the sense that if $d=2$, then for any $s<\frac{3}{2},$ there exists a measure obeying $I_s(\mu)<\infty$, and yet
$$ \limsup_{\epsilon \to 0} \epsilon^{-1} (\mu\times\mu)\{(x,y): 1 \leq |x-y| \leq 1+\epsilon\}=\infty.$$

This result was extended to three dimensions by the authors of this paper (\cite{IS16}). In higher dimensions, the sharpness of Falconer's lemma is still open in the case of the Euclidean distance, but in the same paper, the authors of this showed that for any $s<\frac{d+1}{2},$ there exists a compactly supported Borel measure $\mu$ with $I_s(\mu)<\infty$, $s<\frac{d+1}{2}$,  and yet
$$ \limsup_{\epsilon \to 0} \epsilon^{-1} (\mu\times\mu)\{(x,y): 1 \leq {||x-y||}_K \leq 1+\epsilon\}=\infty,$$ where ${||\cdot||}_K$ is the norm induced by a symmetric convex body $K$ obtained by gluing the paraboloid in such a way that results in a smooth symmetric convex body with a smooth boundary and non-vanishing Gaussian curvature.

In recent decades, the Falconer distance problem has been generalized in a variety of directions, each having its own geometric, analytic, and combinatorial nuances. For example, given a compact set $E$ in ${\Bbb R}^d$, $d \ge 2$, we can ask whether the dot product set
$$\Pi(E)=\{x \cdot y: x,y \in E\}$$ has positive Lebesgue measure. A step in this direction was taken by Suresh Eswarathasan, the first listed author of this paper, and Krystal Taylor (\cite{EIT11}) who proved that the Lebesgue measure of $\Pi(E)$ is indeed positive if the Hausdorff dimension of $E$ is greater than $\frac{d+1}{2}$. They proved this by showing that the analog of Lemma \ref{key} holds if $|x-y|$ is replaced by $x \cdot y$, i.e.
\begin{equation} \label{dpfalconer} (\mu \times \mu)\{(x,y): 1 \leq x \cdot y \leq 1+\epsilon \} \leq C I_{\frac{d+1}{2}}(\mu) \epsilon. \end{equation}

Indeed, they proved a much more general result where the conclusion of the Falconer lemma holds if $|x-y|$ is replaced by any $\phi(x,y)$ smooth away from the diagonal and having a non-zero Monge-Ampere determinant. In this paper we prove that (\ref{dpfalconer}) is sharp. To state this precisely, we introduce some notation and state our main result. Here and throughout, $X \lesssim Y$ means that there exists a uniform constant $C>0$ such that $X \leq CY$. Also, we write $X\approx Y$ to mean that both $X\lesssim Y$ and $Y\lesssim X.$

\begin{theorem}\label{main}
For any dimension $d\geq 3,$ and any $s\in\left(\frac{d}{2},\frac{d+1}{2}\right),$ there exists a Borel measure $\mu$ on $\mathbb R^d$ such that $I_s(\mu)\approx 1$ and for any $\epsilon >0,$
$$(\mu \times \mu) \left\{(x,y) : 1 \leq x \cdot y \leq 1+ \epsilon \right\} \approx \epsilon^{\frac{2s}{d+1}}.$$

In particular,
$$ \limsup_{\epsilon \to 0} \epsilon^{-1} (\mu \times \mu) \left\{(x,y) : 1 \leq x \cdot y \leq 1+ \epsilon \right\}=\infty.$$
\end{theorem}

\vskip.125in

\begin{remark} The analogous statement of Theorem \ref{main} for two dimensions was shown by Suresh Eswarathasan, the first listed author of this paper, and Krystal Taylor (\cite{EIT11}) using a simpler construction. The construction used in this paper can also work when $d=2,$ but the estimates become more delicate, and the analysis becomes much more cumbersome than for it is for $d\geq 3.$ Because a simpler proof for the two dimensional case already exists, we only prove the result for higher dimensions here.
\end{remark}

\begin{remark} The authors believe that the estimate
\begin{equation} \label{generalMAest} (\mu \times \mu) \{(x,y): t \leq \phi(x,y) \leq t+\epsilon \} \lesssim I_{\frac{d+1}{2}}(\mu) \epsilon, \end{equation} which is shown in (\cite{EIT11}) to hold for all functions $\phi$ satisfying the non-vanishing Monge-Ampere determinant condition, is sharp in the sense that for any $s<\frac{d+1}{2}$ there exists a compactly supported Borel measure $\mu$ with $I_s(\mu)<\infty$ for which (\ref{generalMAest}) fails. We hope to address this issue in the sequel. \end{remark}

The authors would like to thank Adam Sheffer for helpful conversations about the discrete incidence construction in this paper, and Thang Pham for pointing out an error in an earlier version.

\vskip.25in

\section{Proof of Theorem \ref{main}}

\vskip.125in

\subsection{Preliminaries}
We begin with the celebrated Szemer\'edi-Trotter Theorem, from \cite{ST83}.
\begin{theorem}\label{ST}[Szemer\'edi-Trotter]
Given a set of $n$ points and $m$ lines in $\mathbb R^2$, the number of incidences is bounded above by
$$I \lesssim (mn)^\frac{2}{3}+m+n.$$
\end{theorem}
We now record some illustrative discrete constructions and indicate why they cannot be extended to the continuous setting as in Theorem \ref{main}.
\begin{proposition}\label{silly}
In $\mathbb R^2,$ for any large finite $n\in \mathbb N$, there exists a set $E$ of $n$ points with $\approx n^2$ occurrences of the dot product zero. In $\mathbb R^3,$ for any large finite $n\in \mathbb N$, there exists a set $E$ of $n$ points with $\approx n^2$ occurrences of any dot product.
\end{proposition}
\begin{proof} In $\mathbb R^2,$ for we arrange $n/2$ points along the $x$-axis and $n/2$ points along the $y$-axis. Note that any point on the $x$-axis is orthogonal to any point on the $y$-axis, and there are $n^2/4$ such point pairs. In $\mathbb R^3,$ for any $\alpha \in \mathbb R,$ we arrange $n/2$ points along the line $\{(1,y,0):y\in \mathbb R\}$ and arrange $n/2$ points along the line $\{(\alpha,0, z):z\in \mathbb R\}$. Similar to the previous case, the dot product of any point from the first line with any point from the second line will be $\alpha$, and we will again have $n^2/4$ such point pairs.
\end{proof}
In a rough sense, both of the constructions in Proposition \ref{silly} are too ``low-dimensional" to be used to construct a measure as in Theorem \ref{main}. To quantify this, if we follow the procedure detailed below for either of these constructions, we will not get the corresponding energy bound, which would lead to an unbounded $I_s(\mu)$ for $s>1$. It should be noted that both of these examples were inspired by the celebrated Lenz construction in $\mathbb R^4$. It consists of $n/2$ points on a unit circle in the first two dimensions and $n/2$ points on a circle in the second two dimensions, and has $n^2/4$ occurrences of the distance $\sqrt 2$, measured between points from the different circles.

\subsection{Constructing the measure}
First, for any large, finite $n\in \mathbb N$, we will construct a set of $\approx n$ points in $[0,2]^d$ that has $\approx n^\frac{2d}{d+1}$ pairs of points whose dot product is 1. This construction is motivated by a well-known sharpness example for Theorem \ref{ST}. Next, we will quantify some properties of the point set. Finally, we will use an infinite sequence of such sets to generate a measure $\mu$ satisfying the properties of Theorem \ref{main}.
\subsubsection{Discrete construction}\label{discreteSet}
Fix a large, finite, $q\in \mathbb N,$ such that $q^{d+1}\approx n$. Define
$$A:=\left\{\frac{q+i}{2q}:i=1,\dots, q\right\},$$
and
$$B:=\left\{\frac{q^2+i}{2q^2}:i=1,\dots,q^2\right\}.$$
Let $A^k$ denotes a Cartesian product of $k$ copies of $A$, and define $$E:=A^{d-1}\times B \subset [0,1]^2.$$ We can see that $|E|=q^{d+1}.$ Next, we turn our attention to another set of points defined by $A$ and $B$. Let $(-A)$ denote the additive complements of $A$, namely $(-A):=\{-a:a\in A\}.$
$$F:=\left\{\left(\frac{-m_1}{b}, \frac{-m_2}{b}, \dots, \frac{-m_{d-1}}{b}, \frac{1}{b}\right):m_{j}\in(-A), b\in B \right\}.$$
Finally, given a $d$-tuple, $(m_1, m_2, \dots, m_{d-1},b),$ define the hyperplane
$$H(m_1, m_2, \dots, m_{d-1},b) := \left\{ (x_1, x_2, \dots, x_d)\in \mathbb R^d : x_d = \left(\sum_{j=1}^{d-1}m_jx_j\right) + b \right\}.$$
The family of these hyperplanes we will consider is
$$\mathcal H :=\left\{H(m_1, m_2, \dots, m_{d-1},b):m_j\in(-A), b\in B \right\}.$$
We now show that these hyperplanes are level sets for points in $F.$
\begin{lemma}\label{hyp}
For every $f\in F$, there is a unique $H_f\in\mathcal H$ so that for any $x\in H_f,$ we have $f\cdot x =1.$
\end{lemma}
\begin{proof}
To see this, fix a $d$-tuple, $(m_1, m_2, \dots, m_{d-1},b)\in(-A)^{d-1}\times B,$ and compute the dot product of the element of $F$ and any point on the hyperplane associated to the same $d$-tuple. So the associated point $f\in F$ will be
$$f= \left(\frac{-m_1}{b}, \frac{-m_2}{b}, \dots, \frac{-m_{d-1}}{b}, \frac{1}{b}\right),$$
and we will consider an arbitrary point $x\in H(m_1, m_2, \dots, m_{d-1},b).$ Their dot product will be
\begin{align*}
f\cdot x &= \left(\frac{-m_1}{b}, \frac{-m_2}{b}, \dots, \frac{-m_{d-1}}{b}, \frac{1}{b}\right) \cdot \left(x_1, x_2, \dots, x_{d-1}, \left(\sum_{j=1}^{d-1}m_jx_j \right)+b \right)\\
&=\left(\sum_{j=1}^{d-1}\frac{-m_jx_j}{b}\right) + \left(\sum_{j=1}^{d-1}\frac{m_jx_j}{b}\right) +\frac{b}{b}=1.
\end{align*}
\end{proof}
We now pull this all together to quantify how many point pairs in $E\cup F$ determine the dot product 1.
\begin{proposition}\label{discreteDP}
There are $\approx q^{2d}$ point pairs determining the dot product 1 in $E\cup F.$
\end{proposition}
\begin{proof}
Fix an arbitrary point $f\in F$. Notice that the associated hyperplane, $H_f\in \mathcal H,$ consists of points that have dot product 1 with $f.$ However, each $H_f$ will contain $\approx q^{d-1}$ points from $E$. Since there are $q^{d+1}$ choices for $f,$ we have a total of $\approx q^{d+1}q^{d-1}=q^{2d}$ point pairs whose dot product is 1, as claimed.
\end{proof}

\subsubsection{Separation}

In order to construct our measure from the discrete point sets $E$ and $F$, we will need to show that these sets are not too ``low-dimensional" as were the point sets from Proposition \ref{silly}. To be sure we will need to show that the points are separated, and that they will not concentrate mass too much. We make this precise below.

By the definitions of $A$ and $B$, we know that the minimum difference between distinct coordinates of points in $E$ is $q^{-2}.$ This gives us
\begin{equation}\label{Esep}
\min_{\substack{p, p'\in E\\ p\neq p'}}{|p-p'|}\gtrsim \frac{1}{q^2}.
\end{equation}
However, $F$ will take a little more work.

\begin{lemma}\label{Sep}
$$\min_{\substack{p, p'\in F\\ p\neq p'}}{|p-p'|}=\frac{1}{q^2}.$$
\end{lemma}
\begin{proof}
Consider two arbitrary distinct points, $p,p'\in F.$
$$p= \left(\frac{-m_1}{b}, \frac{-m_2}{b}, \dots, \frac{-m_{d-1}}{b}, \frac{1}{b}\right),$$
and
$$p'= \left(\frac{-m_1'}{b'}, \frac{-m_2'}{b'}, \dots, \frac{-m_{d-1}'}{b'}, \frac{1}{b'}\right),$$
where the $m_j$ and $m_j'$ come from $A$ and $b,b'\in B.$ We now split into two cases: the case where $b\neq b',$ and the case where $b=b'.$

If $b\neq b',$ we have that, for appropriate choices of $1 \leq i_d,i_d'\leq q^2,$
$$|p-p'|\geq \left|\frac{1}{b}-\frac{1}{b'}\right| = \left|\frac{b'-b}{bb'}\right|=\left|\frac{\frac{q^2+i_d'}{2q^2}-\frac{q^2+i_d}{2q^2}}{\left(\frac{q^2+i_d}{2q^2}\right)\left(\frac{q^2+i_d'}{2q^2}\right)}\right|$$
$$=2q^2\left| \frac{(q^2+i_d')-(q^2+i_d)}{(q^2+i_d)(q^2+i_d')}\right|=2q^2\left|\frac{i_d'-i_d}{q^4+q^2(i_d+i_d')+i_di_d'}\right|,$$
which, because $i_d,i_d'\leq q^2,$ is bounded below by
$$\geq 2q^2\left|\frac{i_d'-i_d}{q^4+q^2(q^2+q^2)+q^2q^2} \right|\geq 2q^2\left|\frac{1}{q^4+q^2(q^2+q^2)+q^2q^2} \right|\gtrsim q^{-2},$$
where we used the fact that $i_d\neq i_d'$ in the last line. Putting this together, we obtain that for $p$ and $p'$ whose last coordinates are different,
\begin{equation}\label{dthDiff}
|p-p'|\gtrsim q^{-2}.
\end{equation}

If $b=b',$ then for $p$ and $p'$ to be distinct points, they must differ in some other coordinate. Suppose that $p$ has the value $\frac{-m}{b}$ in that coordinate, and $p'$ has the value $\frac{-m'}{b}$ in the same coordinate, for distinct $m,m'\in (-A)$. Then we can be assured that for appropriate $1 \leq i,i'\leq q$ and $1\leq i_d\leq q^2,$ we have
$$|p-p'|\geq\left|\frac{-m}{b}-\frac{-m'}{b}\right|=\left|\frac{-\left(\frac{-(q+i)}{2q}\right)}{\frac{q^2+i_d}{2q^2}}-\frac{-\left(\frac{-(q+i')}{2q}\right)}{\frac{q^2+i_d}{2q^2}}\right|$$
$$=q\left|\frac{(q+i)-(q+i')}{q^2+i_d}\right|\geq q\left|\frac{i-i'}{q^2+i_d} \right|,$$
which, because $i_d \leq q^2,$ is bounded below by
$$\geq q\left| \frac{i-i'}{2q^2}\right|\geq q\frac{1}{2q^2}\gtrsim q^{-1},$$
where we used the fact that $i\neq i',$ as they correspond to the distinct values of $m$ and $m'$ in the coordinate where $p$ and $p'$ differ. Putting these together, we get for distinct $p$ and $p'$ with the same last coordinate,
\begin{equation}\label{ithDiff}
|p-p'|\gtrsim q^{-1}.
\end{equation}
Combining \eqref{dthDiff} and \eqref{ithDiff} yields the desired result.
\end{proof}
Notice that by \eqref{Esep} and Lemma \ref{Sep}, we get the following.
\begin{proposition}\label{sep}
There are no more than a constant number of points of $E \cup F$ in any cube of side-length $q^{-2}.$
\end{proposition}
\subsubsection{Discrete energy}
Now that we know our discrete points are sufficiently separated, we turn our attention to showing that they are also not clumped up too much. We quantify this by calculating the discrete versions of the associated energy integrals, namely
$$I_s'(X,Y)=\frac{1}{{n \choose 2}}\sum_{\substack{p \in X,\\ p'\in Y,\\ p\neq p'}}|p-p'|^{-s},$$
with the convention that we write $I_s'(X)$ in place of $I_s'(X,X).$ Also notice that $I_s'(X,Y)=I_s'(Y,X),$ by definition.
We will break the total discrete energy of our point set into three parts, each treated separately:
$$I_s'(E \cup F)= I_s'(E) + 2I_s'(E,F) + I_s'(F).$$
The discrete energy calculation for $E$ is given as Lemma 2.1 in \cite{IS16}. This gives
\begin{equation}\label{enE}
I_s'(E)\lesssim 1.
\end{equation}
Now we compute the discrete energy for $F$, recalling that $n\approx q^{d+1}.$
$$
I_s'(F)=\frac{1}{{n \choose 2}}\sum_{\substack{p,p' \in E,\\ p\neq p'}}|p-p'|^{-s}\approx q^{-2d-2}\sum_{\substack{p,p' \in F,\\ p\neq p'}}|p-p'|^{-s}= I + II,
$$
where $I$ is the sum over $p$ and $p'$ whose first $(d-1)$ coordinates are the same, and $II$ is its complement.
\subsubsection{Bounding $I$}
To estimate $I$, we mirror the derivation of \eqref{dthDiff}, and use similar notation for coordinates of points. Define $\mathcal F$ to be the set of point pairs of $F$ that agree in the first $(d-1)$ coordinates. That is,
$$\mathcal F :=\{(p, p')\in F\times F :p_d\neq p_d', p_j=p_j', j=1,\dots,d-1\}.$$
 Because the sum is over pairs of points that agree on the first $(d-1)$ coordinates, we only need to focus on the difference in the final coordinate. Recall that coordinates $p_j$ are written in terms of their respective indices $i_j$ as before.
\begin{align*}
I &\approx n^{-2}\sum_{(p, p')\in \mathcal F}|p_d-p_d'|^{-s}\lesssim q^{-2d-2}\sum_{p\in F}\sum_{\substack{i_d'\in[1..q^2],\\ i_d'\neq i_d}}\left|\frac{2q^2}{q^2+i_d}-\frac{2q^2}{q^2+i_d'}\right|^{-s}\\
&\leq q^{-2d-2}\sum_{i_1=1}^q\cdots\sum_{i_{d-1}=1}^q\sum_{i_d=1}^{q^2}\sum_{\substack{i_d'\in[1..q^2],\\ i_d'\neq i_d}}(2q^2)^{-s}\left|\frac{1}{q^2+i_d}-\frac{1}{q^2+i_d'}\right|^{-s}\\
&\lesssim q^{-2s-2d-2}q^{d-1}\sum_{i_d=1}^{q^2}\sum_{\substack{i_d'\in[1..q^2],\\ i_d'\neq i_d}}\left|\frac{1}{q^2+i_d}-\frac{1}{q^2+i_d'}\right|^{-s}.
\end{align*}
The partial sum with $i_d'<i_d$ is the same as its complement, so we rewrite it as twice the sum with $i_d'>i_d$. We will also use the fact that $1\leq i_d,i_d'\leq q^2$ to bound the denominator below.
\begin{align*}
I &\lesssim q^{-2s-d-3}\cdot 2 \sum_{i_d=1}^{q^2-1}\sum_{i_d' =i_d+1}^{q^2}\left|\frac{(q^2+i_d')-(q^2+i_d)}{q^4+q^2(i_d+i_d')+i_di_d'}\right|^{-s}\\
&\lesssim q^{-2s-d-3}\sum_{i_d=1}^{q^2-1}\sum_{i_d' = i_d+1}^{q^2}\left|\frac{i_d'-i_d}{4q^4}\right|^{-s}\lesssim q^{-2s-d-3} q^{4s}\sum_{i_d=1}^{q^2-1}\sum_{i_d' = i_d+1}^{q^2}(i_d'-i_d)^{-s}.
\end{align*}
We reparameterize the sum by setting $j:=i_d'-i_d$. We then approximate the sum by an integral.
\begin{align*}
I &\lesssim q^{2s-d-3}\sum_{i_d=1}^{q^2-1}\sum_{i_d' = i_d+1}^{q^2}(i_d'-i_d)^{-s}=q^{2s-d-3}\sum_{j=1}^{q^2-1}(q^2-j)j^{-s}\\
&\approx q^{2s-d-3}\int_1^{q^2}(q^2-x)x^{-s}dx\\
&=q^{2s-d-3}\left(\left[(q^2-x)-\left(\frac{x^{1-s}}{1-s}\right)\right]_1^{q^2}-\left[ \frac{x^{2-s}}{(1-s)(2-s)}\right]_1^{q^2} \right)\\
&\approx q^{2s-d-3}\left(q^2-\frac{q^{4-2s}}{(1-s)(2-s)} \right).
\end{align*}
Recall that by assumption, we have that $s>(d/2)>1,$ and notice that if $1<s<2,$ the second term in the parentheses is still bounded by a constant times $q^2.$ Putting this all together with the assumption that $s<(d+1)/2,$ we get
$$I\lesssim q^{2s-d-1}\lesssim 1.$$
\subsubsection{Bounding $II$}
The estimate of $II$ is similar, though significantly more involved.
\begin{align*}
II &= q^{-2d-2}\sum_{\substack{p\neq p' \in F,\\ (p, p')\notin \mathcal F}}|p-p'|^{-s}.
\end{align*}
We now estimate $II$ by replacing the $\ell^2$ distance by $\ell^1$ distance, losing at most a constant in the process.
\begin{align*}
II&\leq q^{-2d-2}\sum_{\substack{p\neq p' \in F,\\ (p, p')\notin \mathcal F}}\left(\sum_{j=1}^d |p_j-p_j'|^2\right)^{-\frac{s}{2}}\lesssim q^{-2d-2}\sum_{\substack{p\neq p' \in F,\\ (p, p')\notin \mathcal F}}\left(\sum_{j=1}^d |p_j-p_j'|\right)^{-s}.
\end{align*}
We then rearrange the terms of the sum so that the largest difference between coordinate indices from the first $(d-1)$ is recorded in the first coordinate, with $i_1>i_1'.$ So we can continue bounding the above sum by
\begin{align*}
&\leq q^{-2d-2}2(d-1)\sum_{i_1'=1}^{q-1}\sum_{i_1=i_1'+1}^q\sum_{\substack{1\leq i_j,i_j'\leq q\\j=2, \dots, (d-1)}}\sum_{1\leq i_d,i_d' \leq q^2}\left(\sum_{j=1}^d |p_j-p_j'|\right)^{-s}.\\
\end{align*}
If we separate out the contribution of the difference between the $dth$ coordinates to the innermost sum, and absorb the multiplicative constants, we get
\begin{align*}
&\lesssim q^{-2d-2}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q\sum_{\substack{1\leq i_j,i_j'\leq q\\j=2, \dots, (d-1)}}\sum_{1\leq i_d,i_d' \leq q^2}\\
&\qquad\qquad\left(\left|\frac{1}{\frac{q^2+i_d}{2q^2}}-\frac{1}{\frac{q^2+i_d'}{2q^2}}\right|+\sum_{j=1}^{d-1} \left|\frac{\frac{q+i_j}{2q}}{\frac{q^2+i_d}{2q^2}}-\frac{\frac{q+i_j'}{2q}}{\frac{q^2+i_d'}{2q^2}}\right|\right)^{-s}.\\
\end{align*}
Recalling that each of the differences in the first $(d-1)$ coordinates are dominated by the difference in the first coordinate, as per our reordering, we get that this is bounded above by
\begin{align*}
&\lesssim q^{-2d-2}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q\sum_{\substack{1\leq i_j,i_j'\leq q\\j=2, \dots, (d-1)}}\sum_{1\leq i_d,i_d' \leq q^2}\\
&\qquad\qquad\left(\left|\frac{2q^2}{q^2+i_d}-\frac{2q^2}{q^2+i_d'}\right|+(d-1) \left|\frac{q^2+qi_1}{q^2+i_d}-\frac{q^2+qi_1'}{q^2+i_d'}\right|\right)^{-s}.\\
\end{align*}
Since the summands now no longer depend on the middle $(d-2)$ pairs of indices, we will remove them from consideration. Because the sum was reordered so that $|i_j-i_j'|\leq(i_1-i_1')$ for all $j\in [2.. (d-1)],$ we have control on the number of pairs of indices in each such dimension. More precisely, we have that for any $j\in[2..(d-1)],$ we have that the number of pairs $(i_j,i_j')$ contributing to this sum is no more than $2q(i_1-i_1')$, because we have $q$ choices for $i_j$, then no more than $2(i_1-i_1')$ choices for $i_j'$ within range. So the sum above can be bounded by
\begin{align*}
&\lesssim q^{-2d-2}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q \left(q(i_1-i_1')\right)^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}\\
&\qquad\qquad\left(\left|\frac{2q^2}{q^2+i_d}-\frac{2q^2}{q^2+i_d'}\right|+(d-1) \left|\frac{q^2+qi_1}{q^2+i_d}-\frac{q^2+qi_1'}{q^2+i_d'}\right|\right)^{-s}\\
&\lesssim q^{-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}\left(\left|\frac{q^2(i'_d-i_d)}{q^4+q^2(i_d+i_d')+i_di_d'}\right|+\right.\\
&\qquad\qquad\left. (d-1) \left|\frac{q^3(i_1-i_1')+q^2(i_d'-i_d)+q(i_1i_d'-i_1'i_d)}{q^4+q^2(i_d+i_d')+i_di_d'}\right|\right)^{-s},\\
\end{align*}
which, by recalling that $1\leq i_d,i_d'\leq q^2$ is bounded above by
\begin{align*}
&\lesssim q^{-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}\left(\left|\frac{q^2(i'_d-i_d)}{4q^4}\right|+\right.\\
&\qquad\qquad\left. (d-1) \left|\frac{q^3(i_1-i_1')+q^2(i_d'-i_d)+q(i_1i_d'-i_1'i_d)}{4q^4}\right|\right)^{-s}\\
&\lesssim q^{-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}q^{4s}\\
&\qquad\qquad\left(\left|q^2(i'_d-i_d)\right|+ (d-1) \left|q^3(i_1-i_1')+q^2(i_d'-i_d)+q(i_1i_d'-i_1'i_d)\right|\right)^{-s}.\\
\end{align*}
By adding and subtracting $i_1i_d$ within the final set of parentheses, this expression is
\begin{align*}
&= q^{4s-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}\left(\left|q^2(i'_d-i_d)\right|\right.+\\
&\qquad\qquad\left. (d-1) \left|q^3(i_1-i_1')+q^2(i_d'-i_d)+q(i_1i_d'-i_1i_d+i_1i_d-i_1'i_d)\right|\right)^{-s}\\
&= q^{4s-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}\left(\left|q^2(i'_d-i_d)\right|\right.+\\
&\qquad\qquad\left. (d-1) \left|q^3(i_1-i_1')+q^2(i_d'-i_d)+qi_1(i_d'-i_d)+qi_d(i_1-i_1')\right|\right)^{-s}\\
&= q^{4s-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}\left(\left|q^2(i'_d-i_d)\right|\right.+\\
&\qquad\qquad\left. (d-1) \left|(q^3+qi_d)(i_1-i_1')+(q^2+qi_1)(i_d'-i_d)\right|\right)^{-s}\\
&= q^{4s-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}(g(i_1, i_1', i_d, i_d'))^{-s},\\
\end{align*}
where we define $$g(i_1, i_1', i_d, i_d'):=|q^2(i_d'-i_d)|+ (d-1) |(q^3+qi_d)(i_1-i_1')+(q^2+qi_1)(i_d'-i_d)|.$$
We now need to look at these terms closely. If the following inequality holds for some positive constant $c$,
\begin{equation}\label{goodMiddle}
(q^3+qi_d)(i_1-i_1')+(q^2+qi_1)(i_d'-i_d)\geq cq^3(i_1-i_1'),
\end{equation}
then we can safely ignore the $q^2(i_d'-i_d)$ part of the term, and we get that
$$g(i_1, i_1', i_d, i_d')\gtrsim q^3(i_1-i_1').$$
However, it's possible that the factor $(i_d'-i_d)$ is so negative, that it could make \eqref{goodMiddle} fail to hold. In this case, we then have that for any constant $c>0$,
$$(q^2+qi_1)(i_d-i_d')>(q^3+qi_d-cq^3)(i_1-i_1').$$
Recalling that $i_1<q,$ this tells us that for any $c>0,$
$$2q^2(i_d-i_d')>(q^3+qi_d-cq^3)(i_1-i_1'),$$
which tells us that again
\begin{align*}
g(i_1, i_1', i_d, i_d')&= |q^2(i_d'-i_d)|+ (d-1) |(q^3+qi_d)(i_1-i_1')+(q^2+qi_1)(i_d'-i_d)|\\
&\gtrsim \frac{1}{2}(q^3+qi_d-cq^3)(i_1-i_1')\gtrsim q^3(i_1-i_1')\\
\end{align*}
So we can continue our analysis of $II$ by bounding each of these terms by some positive constant multiple of $q^3(i_1-i_1')$. Continuing, we now have that
\begin{align*}
II&\lesssim q^{4s-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}\sum_{1\leq i_d,i_d' \leq q^2}(q^3(i_1-i_1'))^{-s}\\
&=q^{4s-d-4}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_1')^{d-2}q^4q^{-3s}(i_1-i_i')^{-s}\\
&\lesssim q^{s-d}\sum_{i_1=1}^{q-1}\sum_{i_1=i_1'+1}^q(i_1-i_i')^{d-s-2}\\
&=q^{s-d}\sum_{j=1}^{q-1} (q-j)j^{d-s-2},
\end{align*}
where in the last step, we reparameterized the sum with $j:=i_1-i_1'.$ We now estimate this sum by an integral to get that the previous expression is bounded above by
\begin{align*}
&\leq q^{s-d} \int_1^q (q-x)x^{d-s-2}dx\\
&= q^{s-d}\left(\left[(q-x)\left(\frac{x^{d-s-1}}{d-s-1}\right)\right]_1^q-\int_1^q\frac{x^{d-s-1}}{d-s-1}(-1)dx\right)\\
&=q^{s-d}\left(\frac{-(q-1)}{(d-s-1)}+\frac{q^{d-s}-1}{(d-s-1)(d-s)}\right)\lesssim 1,
\end{align*}
where the last step follows by our assumption that $\frac{d}{2}<s<\frac{d+1}{2}.$

To conclude the energy estimate on $F$, we combine the above bounds on $I$ and $II$ to get
\begin{equation}\label{enF}
I_s'(F)\lesssim 1.
\end{equation}

Finally, we compute the energy between the sets $E$ and $F$. We break up the sum coordinatewise into

\begin{align*}
I_s'(E,F)&=\frac{1}{{n \choose 2}}\sum_{\substack{p \in E,\\ p'\in F}}|p-p'|^{-s}\lesssim n^{-2}\sum_{\substack{p\in E,\\ p'\in F}}\sum_{j=1}^d|p_j-p_j'|^{-s}\\
&= n^{-2}\sum_{\substack{i_j,i'_j\in A\\j=1,\dots, (d-1)}}\sum_{b,b'\in B}\left(\left[\sum_{j=1}^{d-1}\left|\frac{q+i_j}{2q}-\frac{\frac{q+i_j'}{2q}}{b'}\right|\right]+\left|\frac{q^2+b}{2q^2}-\frac{1}{b'}\right|\right)^{-s}\\
&\lesssim q^{-2d-2}\sum_{\substack{i_j,i'_j\in A\\j=1,\dots, (d-1)}}\sum_{b,b'\in B}\left|\frac{q^2+b}{2q^2}-\frac{1}{b'}\right|^{-s}\\
&= q^{-2d-2}(2q^2)^s\sum_{b,b'\in B}(b')^s\sum_{\substack{i_j,i'_j\in A\\j=1,\dots, (d-1)}}\left|b'q^2+bb'-2q^2\right|^{-s}\\
&\lesssim q^{2s-2d-2}\sum_{b,b'\in B}\sum_{\substack{i_j,i'_j\in A\\j=1,\dots, (d-1)}}\left|b'q^2+bb'-2q^2\right|^{-s},\\
\end{align*}
Where we have used the fact that any value of $b'$ is in the interval $\left(\frac{1}{2},1\right]$ in the last step. Using this fact again, we see that $\left|b'q^2+bb'-2q^2\right|\geq \frac{3}{2}q^2-1\geq q^2,$ so we can continue
\begin{align*}
I_s'(E,F)&\lesssim q^{2s-2d-2}\sum_{b,b'\in B}\sum_{\substack{i_j,i'_j\in A\\j=1,\dots, (d-1)}}\left|b'q^2+bb'-2q^2\right|^{-s}\\
&\lesssim q^{2s-2d-2}\sum_{b,b'\in B}\sum_{\substack{i_j,i'_j\in A\\j=1,\dots, (d-1)}}|q^2|^{-s}\\
&\lesssim q^{2s-2d-2}\cdot q^4\cdot q^{2(d-1)}\cdot q^{-2s}\lesssim 1.
\end{align*}
This gives us that
\begin{equation}\label{enEF}
I_s'(E,F)\lesssim 1.
\end{equation}

\subsubsection{Continuous construction}
Given an $s<\frac{d+1}{2},$ and an $\epsilon >0,$ fix a scale $q\approx \epsilon^{-\frac{s}{d+1}}$ and decompose $\mathbb R^d$ into a lattice of half-open cubes of side-length $\epsilon$, that is, translates of $[0,q^{-2})^d.$ By construction, we know that $E\cup F\subset [0,2]^d.$ Select any cube with a point from either $E$ or $F$ in it. Proposition \ref{sep} tells us that no square of side-length $q^{-2}\geq \epsilon$ can have more than a constant number of points in it, so we should have selected about $n$ cubes. Call this set of cubes $R.$ Now, we define $\mu$ to be supported on the union of these cubes, and so that $\int d\mu = 1.$ So we will have that the $\mu$ measure of a cube of side-length $\epsilon$ will be $\epsilon^s.$ More precisely,
\begin{equation}\label{muDef}
d\mu(x) = \epsilon^{s-d}\sum_{r\in R}\chi_r(x)dx.
\end{equation}
Now, we compute the energy of $\mu,$ which by \eqref{enE}, \eqref{enF}, and \eqref{enEF}, is
$$I_s(\mu)\approx I_s'(E\cup F)= I_s'(E)+2I_s'(E,F)+I_s'(F)\lesssim1.$$

Finally we compute the dot product version of the Falconer estimate with $\epsilon = q^{-\frac{d+1}{s}}$ and using Proposition \ref{discreteDP}.
$$(\mu\times\mu)\left\{(x,y)\in[0,2]^d:1 \leq x\cdot y \leq 1+\epsilon \right\}\approx q^{2d}\cdot \epsilon^{2s}\approx \epsilon^{\frac{2s}{d+1}},$$
as claimed.


TITLE: How many cuts are required for a weighted-proportional cake-cutting?
QUESTION [4 upvotes]: In proportional cake-cutting, there are $n$ agents with equal entitlements to a "cake" (an interval). Each agent $i$ has a nonatomic value measure $V_i$ over the cake, and it is required to create a partition of the cake, $X_1,\dots,X_n$, such that:
$$\forall i: {V_i(X_i)\over V_i(Cake)} \geq {1 \over n}$$
It is known that a proportional partition can always be done with $n-1$ cuts, giving each agent a connected piece, e.g. using the last diminisher procedure.
In weighted-proportional cake-cutting, the agents have different entitlements. With every agent $i$ is associated an integer weight $p_i$. Let $Q = p_1+\cdots+p_n$; it is required to partition the cake such that:
$$\forall i: {V_i(X_i)\over V_i(Cake)} \geq {p_i \over Q}$$
There is a simple but inefficient way to find a weighted-proportional division:

Create, for each agent $i$, $p_i$ duplicates.
Run a classic cake-cutting algorithm on the $Q$ duplicates.
Give each original agent the pieces allocated to all its dopplegangers.

This is inefficient since it makes $Q-1$ cuts on the cake, giving each agent a large number of "crumbs" (when $Q$ is large).
When there are $n=2$ agents, there is a solution that requires only 2 cuts, regardless of $Q$. To simplify the presentation, assume that the entire cake is worth exactly $Q$ to both agents. Also, assume that the two endpoints of the cake are identified, so that it is topologically a circle.

Agent #1 makes $Q$ marks on the circle, such that the piece between each two consecutive marks is worth for him exactly $1$.
Agent #2 evaluates, for each mark, the sequence of $p_2$ consecutive pieces starting that mark and going clockwise, and selects the sequence most valuable in his eyes. By the pigeonhole principle, the value of this sequence for him is at least $p_2$ (since the sum of values of the $Q$ sequences is $p_2 Q$).
Agent #1 receives the remaining sequence of $p_1$ pieces. The value of this sequence for him is exactly $p_1$.

How many cuts are required when there are three or more agents?
The following example shows that at least $2n-2$ cuts might be required. The valuations are:
Agent 1:  1 0 1 0 1 0 1 ... 1 0 1
Agent 2:  0 1 0 0 0 0 0 ... 0 0 0
Agent 3:  0 0 0 1 0 0 0 ... 0 0 0
Agent 4:  0 0 0 0 0 1 0 ... 0 0 0
...
Agent n:  0 0 0 0 0 0 0 ... 0 1 0

and the weights are:

$p_1 = n^2$,
$p_i=1$ for every agent $i\geq 2$.

Agent 1 must receive more than $(n-1)/n$ of the cake, so he must receive something from each of his $n$ desired slices. Each of the other $n-1$ agents  must still receive a positive slice. Hence, we must use two cuts for each agent $i\geq 2$, and the total number of required cuts is $2(n-1)$.
We saw above that this lower bound is tight for $n=2$, since there always exists a weighted-proportional division with 2 cuts. So the following question is interesting for $n\geq 3$ agents:
Does there always exist a weighted-proportional partition with $2n-2$ cuts?

REPLY [1 votes]: We can do an inductive construction for $n(n-1)$ cuts. For two agents, you gave the construction. For $n>2$, divide the cake proportionally among the first $n-1$ (into ratios of $\frac{p_i}{p_1+\cdots+p_{n-1}}$). Then the $n$th agent comes along and wants to take a fraction of $\frac{p_n}{Q}$ from each of the others (then he'll have $\frac{p_n}{Q}$ of the cake). Each of the other agents wants to make sure that he retains a fraction of $1-\frac{p_n}{Q}$, since then he'll have a total of $\left(1-\frac{p_n}{Q}\right)\left(\frac{p_i}{p_1+\cdots+p_{n-1}}\right)=\frac{p_i}{Q}$. They can achieve this using 2-agent division, at the cost of 2 cuts (even though the piece they're dividing may be disconnected; just pretend it's a circle and ignore the rest of the cake).
That means that the $n$th agent coming in makes $2(n-1)$ cuts, so the total number of cuts is
$$
\sum_{k=1}^n 2(k-1)=n(n-1)
$$


text:
\begin{document}

\maketitle

\begin{abstract}
    A key problem in multiobjective linear programming is to find the
    set of all efficient extreme points in objective space.
    In this paper we introduce oriented projective geometry
    as an efficient and effective framework for solving this problem.
    The key advantage of oriented projective geometry is that we can
    work with an ``optimally simple'' but unbounded efficiency-equivalent
    polyhedron, yet apply to it the familiar theory and algorithms that
    are traditionally restricted to bounded polytopes.
    We apply these techniques to Benson's outer approximation algorithm,
    using oriented projective geometry to remove an exponentially large
    complexity from the algorithm and thereby remove a significant burden
    from the running time.

    \medskip
    \noindent \textbf{AMS Classification}\quad
        Primary
        90C29,

        90C05;

        Secondary
        51A05,

        52B05

    \medskip
    \noindent \textbf{Keywords}\quad Multiobjective linear programming,
        efficient outcome set, oriented projective geometry,
        outer approximation, polytope complexity
\end{abstract}

\input{intro.tex}
\input{prelim.tex}
\input{proj.tex}
\input{eff.tex}
\input{outer.tex}
\input{conc.tex}

\section*{Acknowledgements}

The first author is supported by the Australian Research Council
under the Discovery Projects funding scheme (project DP1094516).

\small
\bibliographystyle{amsplain}
\bibliography{pure}

\bigskip
\noindent
Benjamin A.~Burton \\
School of Mathematics and Physics, The University of Queensland \\
Brisbane QLD 4072, Australia \\
(bab@debian.org)

\bigskip
\noindent
Melih Ozlen \\
School of Mathematical and Geospatial Sciences, RMIT University \\
GPO Box 2476V, Melbourne VIC 3001, Australia \\
(melih.ozlen@rmit.edu.au)

\end{document}


TITLE: When are dual modules free?
QUESTION [15 upvotes]: Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^*$ be the set of $y$ in the dual vector space, $V^*$, such that $\langle x, y \rangle \in A$ for every $x \in T$.
Under what hypotheses on $A$ can I conclude that $T^*$ is a free $A$-module? My current conjecture is that this holds whenever $A$ is a UFD. (Of course, it trivially holds if $A$ is a PID.)
Here are a few ideas of mine. Define the rank of $T$ to be $\dim_K T \otimes_A K$. I can show that, if $A$ is a UFD, then $T^*$ is free for $T$ of rank $1$. For any $T$, we can make a short exact sequence
$$0 \to S \to T \to U \to 0$$
where $S$ is rank $1$ and $U$ is torsion free with rank one less than $T$. So we have
$$0 \to U^* \to T^* \to S^* \to \mathrm{Ext}^1(U,A) \to \cdots$$.
This looks like a good start, but I don't know how to control that Ext group. I suspect that one of you does!
This is motivated by Kevin Buzzard's question about matrix rings.

REPLY [0 votes]: Generalizing your counter-example a bit:
Over a noetherian normal domain, being a reflexive is the same as being a second syzygy (by Auslander). Take any second syzygy of a finite length module, it will have depth 2 - not enough to be depth of the ring in general.


TITLE: Line Integral to Find Work on Slope (Without Explicit Use of Vector Calculus Format)
QUESTION [1 upvotes]: Ok, so my math class has introduced the Line Integral as a way to find the work done on a two-dimensional slope by gravity on an object traversing any distance on the slope. This is all supposed to be useful for Maxwell's equations later, but I know nothing of those right now. (P.S. We use the original texts a lot of times, with some supplementary material.)
The equation we're given, (which is actually straight out of Maxwell's "Treatise On Vectors and Vector Operations") is this: $L = \int_0^S R \space cos \space\epsilon \space \mathrm{d}x$
S is the distance on the slope. R is here the force of gravity. Epsilon is the angle between the force of gravity and the tangent of the slope. Now I understand how to do this sort of thing with a straight line. It would just be the equation without the integral. $R \space cos \space\epsilon$ is just the projection of the gravity vector onto the slope, so its the "living" force that's acting in the direction down the slope. The rest of R is countered by the normal force. dx would be an actual non-infinitesimal distance. And so we'd have force x   distance which gives us work
I don't understand how this integral is working though. Why would we try to find the area under the graph of the work done if we're trying to find the work done? I also don't understand how the change of epsilon is determined in this equation.
The book is here: http://books.google.com/books?hl=fr&lr=&id=WtExAQAAMAAJ&oi=fnd&pg=PR1&dq=The+operation+of+integration+of+the+resolved+part+of+a+vector+quantity&ots=jD6WFkOpwj&sig=GJWbq0KEb5iHrvSnvubKuMTWvMg#v=onepage&q=The%20operation%20of%20integration%20of%20the%20resolved%20part%20of%20a%20vector%20quantity&f=false
You can use the tab at the right to navigate to page 13, where this extract is from.
I'm sorry if this doesn't make much sense. But thanks for any help!

REPLY [0 votes]: Note that the integral in the quote has $s$ as parameter and integration variable, not $x$. Written in really long form this would be
$$\int_0^S R(\vec P(s))\cos\epsilon(s)\,ds$$
where $\vec P(s)=(x(s),y(s),z(s))$ and $R(\vec P(s))=\|\vec R(\vec P(s))\|$ and
$$\cos\epsilon(s)=\frac{\langle \vec R(\vec P(s)), \dot{\vec P}(s)\rangle}{\|\vec R(\vec P(s))\|\,\|\dot{\vec P}(s)\|}$$
and since it is assumed that $s$ corresponds to the path length, $\|\dot{\vec P}(s)\|=1$.
Thus another form to write the line integral is
$$\int_\Gamma \langle \vec R(\vec P), d\vec P\rangle$$
So you see that there is no area directly involved, it is all more complicated.
Think about the path $\Gamma$ consisting of many small straight segments $[\vec P_{k-1},\vec P_k]$, $k=1,...,N$, and that the force $\vec R$ is nearly constant along each segment. Then you can compute the work along each small segment without integration, and summing up all small segments gives an approximation of the integral formula,
$$\sum_{k=1}^N \langle \vec R(\vec P_k), \vec P_k-\vec P_{k-1}\rangle.$$


TITLE: Measuring background radiation
QUESTION [3 upvotes]: We tried to measure background radiation using a geiger counter for a experiment at school. The meter showed  $0.12$-$0.21$ microSv/h during the day averaging at about $0.14$ mcSv/h.
As we tried to see ways how to shield incoming radiotion nothing seemed to work. Taking a cue from nuclear power plant we put the meter into a watertigt glasscontainer and sank it in the gym pool so that the container had a good three feet of water all around. The measurement did not budge at all. We tried all settings fast/slow/auto. The geiger meter should be okay as it had recently been calibrated.
Can someone explain.

REPLY [0 votes]: Thans for all answers and links.
If cosmic radiation is about 0.03 mcSv/h we should still see a drop.
There is no power plant in the area.
The pool was an indoor pool and it is in the basement which has a solid concrete ceiling.
How would I set this experiment to show that the counter goes down preferably to zero.