TITLE: Generalization of the Sultan's dowry problem
QUESTION [10 upvotes]: We know the solution of the Sultan's dowry problem:
To reject the first $n/e$ candidates and then to select the first who exceeds the best of the sample.
How to find the best strategy if we want select the 2, or 3, or $k$ best instead of only one?

REPLY [0 votes]: For the general case, have a look at this article of Bruss and Paindaveine (2000) (Bruss, F., & Paindaveine, D. (2000). Selecting a sequence of last successes in independent trials. Journal of Applied Probability, 37(2), 389-399. doi:10.1239/jap/1014842544)


TITLE: A consequence of edge-criticality
QUESTION [0 upvotes]: Let $G$ be $\Delta$-edge-critical (that is, $G$ is $\Delta + 1$-edge-chromatic and removing
any edge of $G$ gives a subgraph which is at most $\Delta$-edge-chromatic, where $\Delta$ is the max degree of $G$). Let $d(x) + d(y) = \Delta + 2$, and let $\pi$ be a $\Delta$-coloring of $G-xy$. Let $S$ be the set of colors used in $\pi$
The article I'm reading goes on to claim without proof that $C_{\pi}^{\prime}(x)\cap C_{\pi}^{\prime}(y) = \emptyset$, and furthermore that $C_{\pi}^{\prime}(x)\cup C_{\pi}^{\prime}(y) = S$. (That is, there is no color absent at both $x, y$, but every color is absent at exactly one of $x,y$.)
The first statement seems to be proven by saying that if there were a color absent at both $x$ and $y$, removing the edge $xy$ from $G$ would not change the edge-chromaticity, but I don't trust my intuition. On the second statement I'm totally lost, because I can't even begin to see how applying the sum of the degrees helps.

REPLY [1 votes]: Your intuition on the first statement is correct, but why 'intuition'?
If a color is absent at both $x$ and $y$ you can use it to color $xy$ and get a $\Delta$-coloring of $G$.
This is logic.
Now note that in $G-xy$ the degreesum of $x$ and $y$ is exactly $\Delta$, i.e. the number of colors.
If no color is absent, every color is used exactly once.


TITLE: Is there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group?
QUESTION [19 upvotes]: Recall that for any space $X$, the cohomology $H^*X$ (always, in this post, with $\mathbb{Z}/2$-coefficients) has an action of the Steenrod algebra $\mathcal{A}$; that is, a natural morphism $\mathcal{A} \otimes H^*X \to H^*X$.  This is not a morphism of algebras, but $\mathcal{A}$ has a Hopf algebra structure such that the appropriate diagrams involving comultilication on $\mathcal{A}$ and $H^*X$ commute.
Consider now a space with finitely generated homology in each degree. Then the Steenrod algebra
acts on homology by duality, and dualizing this gives that the dual Steenrod algebra $\mathcal{A}^{\vee}$ coacts on the completed cohomology ring; that is, there is a map
$$ H^*X \to H^*X \hat{\otimes} \mathcal{A}^{\vee}$$
which is in fact a morphism of rings, and makes appropriate diagrams commute for products in $X$. It follows that when $X$ is a H space, and $H^*X$ is a Hopf algebra, then we have a coaction of $\mathcal{A}^{\vee}$ on the completed cohomology ring $H^{\star \star}X$ (or something like that). Anyway, the upshot of this is that $\mathrm{Spec} \mathcal{A}^{\vee}$ is a (noncommutative) group scheme because of the Hopf algebra structure, and what we really have is an action of this group scheme (in some sense, at least) on the formal scheme $\mathrm{Sppf} H^{\star \star}(X)$.
In the case where $X = \mathbb{RP}^\infty$, then the formal scheme just described is the formal additive group. Milnor's paper "The Steenrod algebra and its dual" shows that $\mathrm{Spec} \mathcal{A}^{\vee}$ is precisely the automorphism group scheme of this (i.e. a polynomial ring on variables in each power of $2$ minus one).
This is established by computation in Milnor's paper.
Q1: Is there a high-concept explanation of why this should the case?
Q2: What's the analog in characteristic $p \neq 2$?

REPLY [15 votes]: What I'd like to say is, to some degree, commentary and expansion on what some people have said in the comments above.
I feel that the answer to (Q1), about whether there is a "high-concept" explanation for the dual Steenrod algebra, is "no" at the current point in time.  Even further, I feel like the attempt to do so right now would be misleading.  We can assign some high-concept descriptions to phenomena in terms of formal group laws, but trying to give a high-concept reason why the Steenrod algebra takes the form it does would be misinterpreting the role of the Steenrod algebra.
We construct the stable homotopy category from the category of topological spaces.  Based on its construction, we can say a lot of things in older and more modern language.  It's a tensor triangulated category; it's the homotopy category of a symmetric monoidal stable model category; it's the homotopy category of a stable $\infty$-category; it's some kind of "universal" way to construct a stable category out of the category of topological spaces.  Based on this, we can describe a lot of the foundational properties of the stable homotopy category and the category of spectra.  However, there are a ton of categories that satisfy basically the same properties.  Without doing more work, we don't understand anything about specifics that distinguish the stable homotopy category from any other example.
These kinds of computations originate with the Hurewicz theorem, the Freudenthal suspension theorem, Serre's computation of the cohomology of Eilenberg-Mac Lane spaces and his method for computing homotopy groups iteratively, and Adams' method of stepping away from Postnikov towers and upgrading Serre's method into the Adams spectral sequence.  Before these, you would have no reason to necessarily believe that the stable homotopy category isn't equivalent to, say, the derived category of chain complexes over $\mathbb{Z}$ or some other weird differential graded algebra.  Before you have these, you don't have Milnor's computation of the homotopy groups of $MU$ or $MU \wedge MU$, and you don't have Quillen's interpretation in terms of the Lazard ring.
In short, understanding the Steenrod algebra and its dual are prerequisites for all the qualitative things we understand about the stable homotopy category to distinguish it from another example.  Right now there is not a door to the stable homotopy category that comes from formal-group data, as much as we would like one.  As a result, I guess I feel like trying to assign a high-concept description to the Steenrod algebra is backwards right now.
The fact that often computations come first and the conceptual interpretations second is something that gives the subject some of its flavor.  Are there high-concept explanations why vector bundles should have Stiefel-Whitney classes?  Why complex vector bundles should have Chern classes?  Why the "quadratic" power operations in mod-$2$ cohomology should generate all cohomology operations, and why all the relations are determined by those coming from composing two?
This is definitely not to say that such a description wouldn't be desirable.  It would be very desirable to have a more direct route from concepts to the stable homotopy category, because constructing objects that realize conceptual descriptions can be very difficult.


TITLE: Upper Bound on Surface Area of Algebraic Surfaces
QUESTION [3 upvotes]: I have a family of multi-variate polynomials, with bounded degree and a bound on the norm of the coefficients, say for example, all multi variate polynomials of the form:
$\{ a_1 xy + a_2x + a_3xyz: |a_1|,|a_2|,|a_3| \le 1\}$.
Each polynomial defines a surface in Euclidean space by considering its common roots i.e.
$A_p = \{(x,y,z): p(x,y,z)=0\}$. We assume that $p=0$ is not in our family, so all surfaces are proper hyper surfaces.
Given that the degree of the polynomials is uniformly bounded and the size of the coefficients is uniformly bounded, can I have a uniform upper bound on the surface area (Haussdorf measure) of the surfaces when intersected with a compact domain? i.e. bound on
$\sup_p \mathcal{H}(A_p\cap K)$
A theoretical bound will suffice.

REPLY [0 votes]: Here is a general answer: Let $S\subset R^n$ be a $k$-dimensional real-algebraic submanifold of degree $\le d$, meaning that every generic affine subspace of dimension $n-k$ intersects $S$ is at most $d$ points. Then for every round ball $B$ of radius $r$, we have
$$
Vol_{k}(S\cap B)\le C(n,k,d)r^n,
$$
where $C$ is some universal function independent of $r$ and $S$.
This is an application of Crofton's formula, see Theorem 4.3 in
A. Paiva, E. Fernandes, "Gelfand transforms and Crofton formulas",
http://www.math.poly.edu/~alvarez/pdfs/crofton.pdf (I am pretty sure one can find it elsewhere too, this is just a free source that I found).
Thus, in your setting, you need only to bound the degree of your surface and not the coefficients.


TITLE: Sturm-Liouville Problem
QUESTION [2 upvotes]: Possible Duplicate:
Sturm-Liouville Problem

How could one prove that there are at most countably many eigenvalues of the Sturm-Liouville problem $−Lu=ju$, $j$ = eigenvalue, and $u$ is in $C^2[a,b]$?
I have put some more thought and I am still confused about this problem.
Suppose I have $-Lu=\lambda u$ where $-Lu=(pu')'+qu.$  How can you justify that if $\left\{ c_{\alpha}\right\} _{\alpha\in\Gamma}$
is a set of non-zero real numbers such that $\mathbb{\sum}_{j=1}^{N}c_{\alpha_{j}}^{2}\leq1$
for every choice of $N\geq1$ and every choice of distinct $\alpha.$
Then $\Gamma$ is a countable set. How to proceed from there? Is there an easy solution? Thank you very much for help.

REPLY [1 votes]: $\sum_{j=1}^N c_{\alpha_j}^2\le 1$ implies $0< c_\alpha^2\le 1$ for each $\alpha\in\Gamma$.
Let $A_n$, $n\ge 1$, be the subset of $S=\{c_\alpha\}_{\alpha\in\Gamma}$ defined by
$$A_n = \left\{c_\alpha\in S : \frac {1}{n+1} < c_\alpha^2 \le \frac 1 n\right\}$$
$A_n$ must contain $n$ elements at most, otherwise
$$\sum_{c_\alpha\in A_n} c_\alpha^2 > \sum_{c_\alpha\in A_n} \frac{1}{n+1} = \frac{\text{Card}(A_n)}{n+1} \ge 1$$
Evidently $\cup_n A_n = S$; $S$ is a countable union of finite sets, so it is countable.


TITLE: Is n over itself, over itself ... recursively equal to one, or square-root of n?
QUESTION [0 upvotes]: Suppose that $x = \frac{n}{x} $ where n a constant real number and we wish to solve for x. Solving analytically, $x^2 = n \rightarrow x = \sqrt{n}$. But, suppose we plugged in the value of x recursively so that $x = \frac{n}{\frac{n}{\frac{n}{...}}}$. Does this mean that $x = \frac{n}{\frac{n}{\frac{n}{...}}} = \sqrt{n}$? Intuitivly, I would expect it to be 1.

REPLY [2 votes]: Usually we define infinite nesting as the $k\to\infty$ limit of the result with $k$ items. In this case the sequence is $x_0=n,\,x_{k+1}=n/x_k$, a sequence that alternates bewteen $n$ and $1$, and for $n\ne1$ doesn't converge.


TITLE: Why $\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\geq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|?$
QUESTION [2 upvotes]: Let $E$ be a complex Hilbert space.
For  $A\in\mathcal{L}(E)$, I want to show that
$$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|=\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.$$
Clearly, since
$$\{\|Ax\|;\;x\in E,\,\|x\|=1\}\subset \{\|Ax\|;\;x\in E,\,\|x\|\leq1\},$$
then
$$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\leq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|.$$

Why
  $$\sup_{\substack{x\in E,\\ \|x\|=1}}\|Ax\|\geq\sup_{\substack{x\in E,\\ \|x\|\leq1}}\|Ax\|?$$

REPLY [3 votes]: Hint: note that if $\|x\|\le1$, then $\|Au\|\ge\|Ax\|$ where $u=\frac{x}{\|x\|}$.


TITLE: Poisson process at $t=0$
QUESTION [0 upvotes]: In the next definition of a Poisson process:
A counting process $\left \{ N(t); t \geq 0 \right \}$ is said to be a Poisson process with rate $\lambda > 0$ if
1) $N(0)=0$
2) $N(t)$ has stationary and independent increments
3) $\mathbb{P}(N(h)=1) = \lambda h + o(h)$
4) $\mathbb{P}(N(h)\geq 2) =  o(h)$
Why can't we count an occurrence or event at the initial point $t=0$ in 1)?

REPLY [0 votes]: Nothing can happen while $t<0$ since we don't consider the negative time.
Now, $N(0) \ne 0$ means the event happens exactly at $t=0$.
What is the Probability that the event happens at $t=0$ exactly?
More generally, what is the probability, $Pr(X=c)$ when $X$ is a continuous random number and $c$ is a specific number?

Even though the process counts the number of occurrences, the process is the same as pointing occurrences on the timeline (as you mentioned). If the time is in between the time interval, the occurence is counted.
What is the probability that you point an occurrence on timeline, exactly at $t=0$?
 This question is the same as What is the probability of N(0) = 1.


text:
\begin{document}

\title[Topological invariants of groups and Koszul modules]{Topological invariants of groups and Koszul modules}

\author[M. Aprodu]{Marian Aprodu}
\address{Marian Aprodu: Simion Stoilow Institute of Mathematics
\hfill \newline\texttt{}
 \indent P.O. Box 1-764,
RO-014700 Bucharest, Romania, and \hfill  \newline\texttt{}
\indent  Faculty of Mathematics and Computer Science, University of Bucharest,  Romania}
\email{{\tt marian.aprodu@imar.ro}}

\author[G. Farkas]{Gavril Farkas}
\address{Gavril Farkas: Institut f\"ur Mathematik,   Humboldt-Universit\"at zu Berlin \hfill \newline\texttt{}
\indent Unter den Linden 6,
10099 Berlin, Germany}
\email{{\tt farkas@math.hu-berlin.de}}

\author[S. Papadima]{\c Stefan Papadima\textsuperscript{\textdagger} }
\address{\c Stefan Papadima: Simion Stoilow Institute of Mathematics \hfill \newline\texttt{}
\indent P.O. Box 1-764,
RO-014700 Bucharest, Romania}

\thanks{\c Stefan Papadima passed away on  January 10, 2018. }
\author[C. Raicu]{Claudiu Raicu}
\address{Claudiu Raicu: Department of Mathematics,
University of Notre Dame \hfill \newline\texttt{}
\indent 255 Hurley Notre Dame, IN 46556, USA, and \hfill\newline\texttt{}
\indent Simion Stoilow Institute of Mathematics, \hfill\newline\texttt{}
\indent  P.O. Box 1-764, RO-014700 Bucharest, Romania}
\email{{\tt craicu@nd.edu}}

\author[J. Weyman]{Jerzy Weyman}
\address{Jerzy Weyman: Department of Mathematics,
University of Connecticutt \hfill \newline\texttt{}
\indent Storrs, CT,
06269-3009, USA}
\email{{\tt  jerzy.weyman@uconn.edu}}

\begin{abstract}
We provide a uniform vanishing result for the graded components of the finite length Koszul module associated to a subspace $K\subseteq \bigwedge^2 V$, as well as a sharp upper bound for its Hilbert function. This purely algebraic statement has interesting applications to the study of a number of invariants associated to finitely generated groups, such as the Alexander invariants, the Chen ranks, or the degree of growth and nilpotency class. For instance, we explicitly bound the aforementioned invariants in terms of the first Betti number for the maximal metabelian quotients of (1) the Torelli group associated to the moduli space of curves; (2) nilpotent fundamental groups of compact K\"ahler manifolds; (3) the Torelli group of a free group.\end{abstract}

\maketitle

\section{Introduction}

A well-established way of studying the properties of a finitely generated group $G$ is by considering its associated graded Lie algebra
$$\mbox{gr}(G):=\bigoplus_{q\geq 1} \Gamma_q(G)/\Gamma_{q+1}(G),$$
defined in terms of its \emph{lower central series}
\[G=\Gamma_1(G)\geq \ldots \geq \Gamma_q(G)\geq \Gamma_{q+1}(G) \geq \ldots\mbox{ where }\Gamma_{q+1}(G)=[\Gamma_q(G),G].\]
The quotients $\mbox{gr}_q(G)=\Gamma_q(G)/\Gamma_{q+1}(G)$ are finitely generated abelian groups and their ranks $\phi_q(G):=\mbox{rk } \mbox{gr}_q(G)$ encode fundamental numerical information on $G$ and are in general extremely difficult to compute. For example, $\Gamma_2(G)=G'$ is the commutator subgroup of $G$ and $\mbox{gr}_1(G)$ is the abelianization of $G$, whose rank $b_1(G)$ is the first Betti number of $G$.

\vskip 4pt

Following Chen \cite{Chen}, Massey \cite{Massey}, Papadima-Suciu \cite{PS-imrn} and many others, a  more manageable task is to pass instead to
the \emph{metabelian quotient} $G/G''$, where $G''=[G',G']$ is the second commutator subgroup of $G$, and to consider instead its graded Lie algebra $\mbox{gr}(G/G'')$
and the corresponding \emph{Chen ranks}
$$\theta_q(G):=\phi_q(G/G'')=\mbox{rk } \mbox{gr}_q(G/G'').$$
One sample application of the results of our work is the following (see Theorem~\ref{thm:bound-chen}).

\begin{thm}\label{thm:Kahler}
 Let $G$ be the fundamental group of a compact K\"ahler manifold, and suppose its first Betti number is $n=b_1(G)\geq 3$. If $G/G''$ is nilpotent, then $\theta_q(G)=0$ for $q\geq n-1$ and
 \[\theta_q(G) \leq {n+q-3\choose n-1} \cdot \frac{(n-2)(n-1-q)}{q}, \quad\mbox{ for }q=2,\ldots,n-2.\]
\end{thm}

The first Chen rank always satisfies $\theta_1(G)=\phi_1(G)=b_1(G)$, and in general one has the inequality $\phi_q(G)\geq \theta_q(G)$. For basic properties of these invariants initially considered in knot theory we refer to \cite{PS-imrn}, whereas for applications of Chen ranks to hyperplane arrangements one can consult \cite{CS}, \cite{CSch} and the references therein.

\vskip 4pt

It turns out that that in many cases $\mbox{gr}(G/G'')$ is an object belonging essentially to commutative algebra, which can be studied to great effect with homological and algebro-geometric methods. The precise context where these methods are most fruitful is when $G$ is a \emph{$1$-formal group} in the sense of Sullivan \cite{Su77}. We recall that for instance fundamental groups of compact K\"ahler manifolds, hyperplane arrangement groups, or the Torelli group $T_g$ of the mapping class group are all known to be $1$-formal. For such groups we prove the following in Section~\ref{subsec:deg-growth}.

\begin{thm}\label{thm:growth-degree}
 Let $G$ be a finitely generated $1$-formal group, and let $M = G/G''$ denote its maximal metabelian quotient. If $\Gamma_q(M)$ is finite for $q\gg 0$ (in particular if $M$ is nilpotent) and if $n=b_1(G)\geq 3$, then
 \begin{enumerate}
  \item\label{it:conc-1} $\Gamma_{n-1}(M)$ is finite.
  \item\label{it:conc-2} $M$ contains a nilpotent finite index subgroup with nilpotency class at most $n-2$.
  \item\label{it:conc-3} The group $M$ has polynomial growth of degree $d(M)$ bounded above by
 \[d(M) \leq n + (n-2)\cdot{2n-3\choose n-4}.\]
 \end{enumerate}
\end{thm}

We recall that the \emph{nilpotency class} of a nilpotent group $H$ is the largest value of $c$ for which $\Gamma_c(H)\neq \{1\}$, and that a group is \emph{virtually nilpotent} if it contains a nilpotent finite index subgroup. We can then define the \defi{virtual nilpotency class (vnc)} of a group to be the minimal nilpotency class of a finite index subgroup, so that the second conclusion of Theorem~\ref{thm:growth-degree} states that the metabelian quotient $G/G''$ has $\vnc(G/G'')\leq n-2$. A celebrated theorem of Gromov \cite{gromov} asserts that virtually nilpotent groups are precisely the ones that exhibit polynomial growth, and our theorem gives a precise bound for the degree of growth of $M=G/G''$ solely in terms of the first Betti number of $G$. As we show in Section~\ref{subsec:deg-growth}, this bound is a consequence of the Bass--Guivarc'h formula and the estimate on $\theta_q(G)$ as in Theorem~\ref{thm:Kahler}. As explained in Example~\ref{ex=freenilp} and Remark~\ref{rem:freenilp}, in the absence of $1$-formality (or some appropriate replacement) there can be no such bound for either $\vnc(M)$ or for $d(M)$.

\vskip 5pt

An important special case of Theorem~\ref{thm:growth-degree} concerns the Torelli group $T_g$ which measures from a homotopical point of view the difference between the moduli space of curves of genus $g$ and that of principally polarized abelian varieties of dimension $g$. If $\mathrm{Mod}_g$ denotes the mapping class group, recall that the Torelli group is defined via the exact sequence
$$1\longrightarrow T_g\longrightarrow \mathrm{Mod}_g\longrightarrow \mbox{Sp}_{2g}(\mathbb Z)\longrightarrow 1.$$
Using Johnson's fundamental calculation $H^1(T_g,\mathbb Q)\cong \bigwedge ^3 H_{\mathbb Q}/H_{\mathbb Q}$, where $H$ is the first integral homology of a genus $g$ Riemann surface (and hence $\mbox{rk} (H)=2g$), together with the recent result by Ershov and He \cite[Corollary~1.5]{EH} asserting that $T_g/T_g''$ is nilpotent for $g\geq 12$, and the $1$-formality of $T_g$ proved in \cite{Hai}, our Theorem~\ref{thm:growth-degree} yields the following consequence (see Section~\ref{subsec:torelli}).

\begin{thm}\label{thmB}
If $g\geq 12$,  the metabelian quotient $T_g/T_g''$ has virtual nilpotency class at most $${2g\choose 3}-2g-2.$$
\end{thm}

\noindent We note that some of the results of \cite{EH} have been extended to $g\geq 4$ in \cite{CEP}, but the nilpotence of $T_g/T_g''$ is still open, preventing us from extending the range for $g$ in Theorem~\ref{thmB}.

We have similar applications  to the Torelli group $\OA_g$ of the free group $F_g$ on $g$ generators, discussed in Section~\ref{subsec:torelli}. We show in Theorem~\ref{thm=oan} that
\[\vnc(\OA_g/\OA_g'') \leq \frac{g(g+1)(g-2)}{2}-2\quad\mbox{ for all }g\geq 4,\]
which is an analogue of Theorem~\ref{thmB}. We note that in contrast with the Torelli groups~$T_g$, the groups $\OA_g$ are not known to be $1$-formal \cite[Question~10.6]{PS-johnson}.

\vskip 5pt

The proofs of Theorems~\ref{thm:Kahler} and~\ref{thm:growth-degree} are fundamentally algebraic. In order to make the transition to commutative algebra and algebraic geometry, for every finitely generated group $G$ we consider its \emph{Alexander invariant}
$$B(G):=H_1(G',\mathbb Z)=G'/G'',$$
viewed as a module over the group ring $\mathbb Z[G/G']$, with the action being given by conjugation. Using the \emph{augmentation ideal}
$I\subseteq \mathbb Z[G/G']$, one then constructs the graded module $\mbox{gr}\,B(G)$ over the ring $\mbox{gr } \mathbb Z[G/G']=\bigoplus_{q\geq 0} I^q/I^{q+1}$, by setting
$$\mbox{gr}_q B(G):=I^q\cdot B(G)/I^{q+1}\cdot B(G).$$
Massey proves in \cite{Massey} the existence of canonical isomorphisms $\Gamma_{q+2}(G/G'')\cong I^q\cdot B(G)$, which in particular yield
$$\theta_{q+2}(G)=\mbox{rk } \mbox{gr}_q B(G).$$
Following \cite{PS-imrn} and especially \cite{PS15}, it turns out that the graded ring $\mbox{gr}\,B(G)$ can be studied via Koszul cohomology using a purely algebraic construction which we describe next.

\vskip 5pt

\noindent {\bf Koszul modules.} Suppose that $V$ is an $n$-dimensional complex vector space and fix an $m$-dimensional subspace $K\subseteq \bigwedge^2 V$. We denote by $K^\perp=(\bigwedge^2V/K)^\vee\subseteq \bigwedge^2V^\vee$ the orthogonal of $K$.
We also denote by $S:=\mbox{Sym}(V)$ the polynomial algebra over $V$ and consider the Koszul complex
$$\cdots \longrightarrow \bigwedge^3 V\otimes S\stackrel{\delta_3}\longrightarrow \bigwedge^2 V\otimes S\stackrel{\delta_2}\longrightarrow V\otimes S\stackrel{\delta_1}\longrightarrow S\longrightarrow \mathbb C\longrightarrow 0.$$ According to \cite{PS15}, the  {\em Koszul module} associated to $(V,K)$ is the graded $S$-module

\[W(V,K):=\mathrm{Coker}\Bigl\{\bigwedge^3V\otimes S\longrightarrow \Bigl(\bigwedge^2V/K\Bigr)\otimes S\Bigr\},\]

where the map in question is the projection $\bw^2 V \oo S \to (\bigwedge^2V/K)\otimes S$ composed with the Koszul differential $\delta_3$. Our grading conventions are so that $W(V,K)$ is generated in degree $0$. Hence,
passing to individual graded pieces one has the identification $$W_q(V,K)\cong \bigwedge^2 V\otimes \mbox{Sym}^q(V)/\Bigl(\mbox{Ker}(\delta_{2,q})+K\otimes \mbox{Sym}^q(V)\Bigr).$$

\vskip 3pt

If $G$ is a finitely generated group, we define its \emph{Koszul module} $$W(G):=W\bigl(H_1(G,\mathbb C), K\bigr),$$ by setting
$K^{\perp}:=\mbox{Ker}\Bigl\{\cup_G:\bigwedge^2 H^1(G,\mathbb C)\rightarrow H^2(G,\mathbb C)\Bigr\}$, or equivalently by letting $K$ be the dual of $\Im(\cup_G)$, which embeds naturally into $\bigwedge^2 H_1(G,\mathbb C)$. It is shown in \cite{DHP} and \cite{PS-imrn} that the following inequality holds
\begin{equation}\label{eq:theta-less-W}
\theta_{q+2}(G)=\mbox{rk } \mbox{gr}_q B(G)\leq \mbox{dim } W_q(G),
\end{equation}
with equality if the group $G$ is \emph{1-formal}. We prove the following result about~$W(G)$.

\begin{thm}\label{thmA}
Let $G$ be a finitely generated group and assume that the cup product map $$\cup_G:\bigwedge^2 H^1(G,\mathbb C)\rightarrow H^2(G,\mathbb C)$$ does not vanish on any non-zero decomposable elements. If $n=b_1(G)\geq 3$, then
\begin{equation}\label{eq:vanishing-WG}
W_q(G)=0 \ \ \mbox{ for all } \ \ q\geq n-3.
\end{equation}
If moreover $M=G/G''$ satisfies the condition that $\Gamma_q(M)$ be finite for $q\gg 0$, then its virtual nilpotency class and degree of polynomial growth satisfy the inequalities in Theorem~\ref{thm:growth-degree}.
\end{thm}

\vskip 3pt

The vanishing (\ref{eq:vanishing-WG}) follows from Theorem~\ref{thm:WG-RG}, while the inequalities for $d(M)$ and $\vnc(M)$ are explained in the proof of Theorem~\ref{thm:growth-degree} (see Remark~\ref{rem:vanishing-resonance}). An important application of Theorem \ref{thmA} is when $X$ is a compact K\"ahler manifold and $G=\pi_1(X)$, in which case $H^1(G,\mathbb C)=H^1(X,\mathbb C)$. In this case, the kernel of the cup-product map $\cup_G$ equals the kernel of the map
$$\cup_X:\bigwedge^2 H^1(X,\mathbb C)\rightarrow H^2(X,\mathbb C).$$
By a well-known generalization of the Castelnuovo--de Franchis Theorem \cite{Cat}, the condition that $\cup_X$ does vanish on decomposable elements amounts to the existence of a fibration $X\rightarrow C$ over a smooth curve of genus $g\geq 2$. Thus, the first part of Theorem~\ref{thmA} applies to all compact K\"ahler manifolds $X$ with irregularity $q(X)\geq 3$ which are not fibred over curves.

\vskip 5pt

We obtain the vanishing in Theorem~\ref{thmA} as a special instance of a general vanishing result for Koszul modules, as explained next. We return to the general situation when $V$ is a vector space, $\iota: K\hookrightarrow \bigwedge^2 V$ is a fixed subspace, and $K^\perp:=\ker(\iota^{\vee})\subseteq \bigwedge^2 V^{\vee}$ is the space of skew-symmetric bilinear forms on $V$ which vanish identically on $K$. It is shown in \cite[Lemma 2.4]{PS15} that the support of the Koszul module $W(V,K)$ in the affine space $V^\vee$ coincides (if non-empty) with the \emph{resonance variety} $\mc{R}(V,K)$ defined as
$$\mc{R}(V,K):=\Bigl\{a\in V^\vee  \, | \, \mbox{ there exists }b\in V^\vee \mbox{ such that } a\wedge b\in K^\perp\setminus \{0\} \Bigr\}\cup \{0\}.$$
In particular, $W(V,K)$ has finite length, that is, $W_q(V,K)=0$ for $q\gg 0$ if and only $\mc{R}(V,K)=\{0\}$. This last condition is equivalent to the fact that the projective subspace $\bb{P}(K^{\perp})\subseteq\bb{P}(\bw^2 V^{\vee})$ is disjoint from the Grassmann variety $\Gr_2(V^{\vee})$ in its Pl\"ucker embedding. Our vanishing result applies uniformly to all subspaces $K$ satisfying this condition, as follows (see Theorem~\ref{thm=main}).

\begin{thm-main*}
\emph{Let $V$ be a complex $n$-dimensional vector space and let $K\subseteq\bw^2 V$ be a subspace such that $\mc{R}(V,K)=\{0\}$. We have that $W_{q}(V,K)=0$ for all $q\geq n-3$}.
\end{thm-main*}

In a forthcoming paper, we prove the equivalence between the Main Theorem  and Green's Vanishing Conjecture for syzygies of generic canonical curves. This leads to an alternate approach to Green's conjecture from the one of Voisin's \cite{V02}, \cite{V05}.

The vanishing result in the Main Theorem is responsible for the estimates of the virtual nilpotency classes that appear in Theorems~\ref{thm:growth-degree},~\ref{thmB} and~\ref{thmA}. In addition to the vanishing result, we also provide a sharp upper bound for the Hilbert function of a finite length Koszul module in Theorem~\ref{thm:HilbSeries}. In combination with (\ref{eq:theta-less-W}) this gives the bound on the Chen ranks in Theorem~\ref{thm:Kahler}, and in conjunction with the Bass--Guivarc'h formula (recalled as formula (\ref{thm:bass-guivarch})) it provides the estimate for the degree of polynomial growth in Theorem~\ref{thm:growth-degree}.

To emphasize the robust relationship between algebra and topology in the $1$-formal setting, we end our Introduction summarizing the connection between some of the invariants discussed in this setting.

\begin{thm}\label{thm:equivalences}
 Let $G$ be a finitely generated $1$-formal group, and assume that $n=b_1(G)\geq 3$. The following statements are equivalent:
 \begin{enumerate}
  \item[(a)] $\theta_q(G)=0$ for some (any) $q\gg 0$.
  \item[(b)]  $\theta_{n-1}(G)=0$.
  \item[(c)]  $\cup_G$ does not vanish on any non-zero decomposable elements.
  \item[(d)]  $W_q(G)=0$ for some (any) $q\gg 0$.
  \item[(e)]  $W_{n-3}(G)=0$.
 \end{enumerate}
\end{thm}

The equivalence of (c), (d), (e) comes from the Main Theorem, since all three conditions characterize finite dimensional Koszul modules. The equivalence of these statements with (a) and (b) follows from the fact that (\ref{eq:theta-less-W}) is an equality for $1$-formal groups.

\vskip 4pt

\begin{ack}
Above all, we acknowledge with thanks the contribution of Alex Suciu. This project started with the paper \cite{PS15}, and since then we benefited
from numerous discussions with him. We also thank A. Beauville, F. Campana, F. Catanese, D. Eisenbud, B. Farb, B. Klingler, P. Pirola, A. Putman and C. Voisin for interesting discussions related to this circle of ideas.

\vskip 3pt

{\small{Aprodu was partially supported by the Romanian Ministry of Research and
Innovation, CNCS - UEFISCDI, grant
PN-III-P4-ID-PCE-2016-0030, within PNCDI III. Farkas was supported by the DFG grant \emph{Syzygien und Moduli}. Raicu was supported by the Alfred P. Sloan Foundation and by the NSF Grant No.~1600765. Weyman was partially supported by the Sidney Professorial Fund and the NSF grant No.~1400740.
}}
\end{ack}

\section{Koszul modules}
\label{sec:kmod}

Throughout this article, we denote by $V$ a complex vector space of dimension $n\ge 2$, we write $V^{\vee}=\Hom_{\bb{C}}(V,\bb{C})$ for its dual, and $S=\Sym V$ for the symmetric algebra of~$V$. We consider the standard grading on $S$ where the elements in $V$ are of degree 1. We fix a subspace $K\subseteq \bigwedge^2V$ of dimension~$m$, we denote by $\iota:K\to \bw^2V$ the inclusion and let $K^\perp:=\ker(\iota^{\vee}) =\coker(\iota)^\vee$, where $\iota^{\vee}:\bw^2 V^{\vee} \lra K^{\vee}$ is the dual of $\iota$.

\subsection{Definitions and basic properties}
\label{subsec:defs}

We recall from \cite{PS15} the definition of Koszul modules. They are defined starting from the classical Koszul differentials
\[
\delta_p:\bigwedge^pV\otimes S\to \bigwedge^{p-1}V\otimes S,
\]
\[
\delta_p(v_1 \wedge \cdots \wedge v_p \oo f) = \sum_{j=1}^p (-1)^{j-1} v_1 \wedge \cdots \wedge \widehat{v_j} \wedge \cdots \wedge v_p \oo v_j f.
\]
Note that we have a decomposition $\delta_p=\bigoplus_q\delta_{p,q}$, where
\begin{equation}\label{eq:delta-pq}
\delta_{p,q}:\bw^p V \oo \Sym^q V \to \bw^{p-1}V \oo \Sym^{q+1}V,
\end{equation}
and that the differentials $\delta_p$ fit together into the Koszul complex which gives a minimal resolution by free graded $S$-modules of the residue field $\mathbb C$.

\medskip

\begin{defn}[{\cite[\S2]{PS15}}]
The {\em Koszul module} associated to the pair $(V,K)$ is a graded $S$-module denoted by $W(V,K)$,  and defined as
\begin{equation}\label{eq:def-Koszul-module}
W(V,K):=\mathrm{Coker}\Bigl\{\bigwedge^3V\otimes S\lra \Bigl(\bigwedge^2V/K\Bigr)\otimes S\Bigr\}
\end{equation}
where the map $\bigwedge^3V\otimes S\to (\bigwedge^2V/K)\otimes S$ is the composition of the quotient map $\bw^2 V \oo S \to (\bigwedge^2V/K)\otimes S$ with the Koszul differential $\delta_3$. The grading is inherited from the symmetric algebra if we make the convention that $\bigwedge^2V/K$ is placed in degree~$0$ and $\bw^3 V$ is in degree~$1$. The degree $q$ component of $W(V,K)$ is then given by
\begin{equation}
\label{eqnWqDef}
W_q(V,K)=\mathrm{Coker}\Bigl\{\bigwedge^3V\otimes \Sym^{q-1}V\lra \Bigl(\bigwedge^2V/K\Bigr)\otimes \Sym^qV\Bigr\}.
\end{equation}
Note that this grading convention guarantees that $W(V,K)$ is generated in degree $0$. In particular, if $W_q(V,K)=0$ for some $q\geq 0$, then $W_p(V,K)=0$ for all $p \ge q$.
\end{defn}

Since the Koszul complex is exact in homological degree one, we can realize $W(V,K)$ as the middle cohomology of the complex
\begin{equation}\label{eqn:W}
\xymatrixcolsep{5pc}
\xymatrix{
K \oo S \ar[r]^{\delta_2|_{K \oo S}} & V\oo S \ar[r]^{\delta_1} & S
}
\end{equation}
Using the fact that $\ker(\delta_{1,q+1}) = \im(\delta_{2,q})$, we obtain as in \cite[(2.1)]{PS15}
\begin{equation}
\label{eqn:Wq_2}
W_q(V,K)=\mathrm{Im}(\delta_{2,q})/\mathrm{Im}(\delta_{2,q}|_{K\oo S}).
\end{equation}

This construction is natural in the following sense: for $K \subseteq K'$, the identity map of $\bigwedge^2V\otimes S$
induces a surjective morphism of graded $S$-modules
\begin{equation}
\label{eq:wnat}
W(V,K) \twoheadrightarrow W(V,K').
\end{equation}

\begin{ex}\label{ex:extremal-Koszul}
In the extremal cases the Koszul modules are easy to compute:
\begin{itemize}
\item If $K=\bigwedge^2V$ then $W(V,K)=0$;
\item If $K=0$ then $W_q(V,K)=H^0\bigl(\mathbb{P}(V^\vee ),\Omega^1_{\mathbb P(V^{\vee})} (q+2)\bigr)\neq 0$ for all $q\geq 0$.
\end{itemize}
We will therefore assume when necessary that $1\leq m < {n\choose 2}$.
\end{ex}

\begin{ex}[{\cite[\S 6.2]{PS15}}]\label{ex:W(G)}
Any finitely generated group $G$ gives rise to a pair $(V,K)$ by letting $V:=H_1(G,\mathbb C)\cong  H^1(G,\mathbb C)^\vee$ and $K$ be the the dual of the image of the cup--product map $\cup_G:\bigwedge^2H^1(G,\mathbb C)\to H^2(G,\mathbb C)$. The corresponding Koszul modules are denoted $W(G)$. Conversely, it was noted in \cite[Proposition~6.2]{PS15} that every Koszul module $W(V,K)$ corresponding to a rational subspace $K\subseteq\bigwedge^2V$ has the form $W(G)$ for some finitely presented group $G$. A more thorough analysis of the Koszul modules associated to groups will be discussed in \S \ref{subsec:top}.
\end{ex}

\subsection{Resonance}
\label{subsec:resonance}

Inspired by work of Green and Lazarsfeld in~\cite{GL91}, Papadima and Suciu  defined in \cite{PS15} resonance varieties in a purely algebraic context as follows.

\begin{defn}\label{def:resonance}

The {\em resonance variety} associated  to the pair $(V,K)$ is the locus
\begin{equation}
\label{eq:defr}
\mathcal R(V,K):=\left\{a\in V^\vee  \, | \, \mbox{ there exists }b\in V^\vee \mbox{ such that } a\wedge b\in K^\perp\setminus \{0\} \right\}\cup \{0\}
\end{equation}
\end{defn}

The resonance variety $\mathcal{R}(V,K)$ is the union of $2$-dimensional subspaces of $V^{\vee}$ parametrized by the intersection $\mathbb{P}(K^\perp)\cap\mathrm{Gr}_2(V^\vee )$. More precisely, via the canonical diagram
\[
\xymatrixcolsep{5pc}
\xymatrix{
\left(\mathbb{P}V^\vee\times\mathbb{P}V^\vee\right)\setminus\mathrm{Diag}\ar[r]^{\pi} \ar[d]^{p_1}& \mathrm{Gr}_2(V^\vee )\\
\mathbb{P}V^\vee &
}
\]
$\mathcal{R}(V,K)$ is the affine cone over $p_1(\pi^{-1}(\mathbb P(K^\perp)\cap\mathrm{Gr}_2(V^\vee )))$. It was showed in \cite[Lemma~2.4]{PS15} that resonance and Koszul modules are strongly related: the support of the Koszul module in the affine space $V^\vee $ coincides with the resonance variety away from $0$, and hence $\mathcal R(V,K)=\{0\}$ if and only if $W(V,K)$ is of finite length. In particular
\begin{equation}
\label{eq:mainequiv}
\mathbb P(K^\perp)\cap\mathrm{Gr}_2(V^\vee )=\emptyset \Longleftrightarrow \mathcal R(V,K)= \{0\}
\Longleftrightarrow \dim_{\mathbb C}W(V,K) < \infty.
\end{equation}

It was also proved in \cite[Proposition~2.10]{PS15} that there exists a uniform bound $q(n,m)$ such that $W_q(V,K)=0$ for all $q\ge q(n,m)$
and for all $(V,K)$ satisfying  $\mathbb P(K^\perp)\cap\mathrm{Gr}_2(V^\vee )=\emptyset$.
A central goal of our paper is to determine this bound, which we do in Theorem~\ref{thm=main}. For dimension reasons, if $\mathbb P(K^\perp)\cap\mathrm{Gr}_2(V^\vee )=\emptyset$ then $m\ge 2n-3$. The
{\em borderline case} $m = 2n-3$ is therefore of maximal interest, as illustrated for instance in \S\ref{subsec:HilbSeries} below.

\

\section{Finite length Koszul modules}
\label{sec:vanish}

The goal of this section is to prove the main theorem of our paper, characterizing finite dimensional Koszul modules (those for which the equivalent conditions in (\ref{eq:mainequiv}) hold) in terms of the vanishing of a fixed graded component. More precisely, we prove:

\begin{thm}\label{thm=main}
Let $V$ be a vector space of dimension $n\geq 3$ and let $K\subseteq\bw^2 V$ be an arbitrary subspace. We have that
\begin{equation}\label{eq:thm-main-equiv}
\mc{R}(V,K) = \{0\} \Longleftrightarrow W_{n-3}(V,K) = 0.
\end{equation}
\end{thm}

The assumption that $n\geq 3$ in the above theorem is made in order to avoid trivialities: when $n=1$ we have $\bw^2 V=0$, while for $n=2$ we have that either $K=\bw^2 V$ or $K=0$, both of which have been discussed in Example~\ref{ex:extremal-Koszul}. In addition to the vanishing result above, in the borderline case $m=2n-3$ we give an exact description of the Hilbert series of the Koszul modules with vanishing resonance, which in turn provides an upper bound for such Hilbert series in the general case $m\geq 2n-3$, as follows.

\begin{thm}
\label{thm:HilbSeries}
If $\mathcal{R}(V,K) = \{0\}$, then for $q=0,\cdots,n-4$, we have that
\[
\dim\, W_q(V,K) \leq {n+q-1 \choose q} \frac{(n-2)(n-q-3)}{q+2},
\]
and equality holds for all $0\leq q\leq n-4$ when $\dim(K)=2n-3$.
\end{thm}

We prove Theorem~\ref{thm=main} using Bott's theorem and a hypercohomology spectral sequence that was featured in the work of Voisin on the Green conjecture \cite{V02}. We recall the special case of Bott's theorem needed for our argument in Section~\ref{subsec:bott}, and prove the vanishing theorem in Section~\ref{subsec:bordervanish}. The proof of Theorem~\ref{thm:HilbSeries} is presented in Section~\ref{subsec:HilbSeries}, and is summarized as follows. We reinterpret the characterization (\ref{eq:thm-main-equiv}) in terms of a transversality condition on $K$ with respect to the kernel of the Koszul differential $\delta_2$, which allows us to prove the conclusion of Theorem~\ref{thm:HilbSeries} in the borderline case. The general case of Theorem~\ref{thm:HilbSeries} reduces to the borderline case via a generic projection argument.

\subsection{Bott's Theorem for Grassmannians}\label{subsec:bott} Our main reference for this topic is \cite[Ch.~4]{weyman}.
Let $\GG=\Gr_2(V^{\vee})$ denote the Grassmannian of $2$--dimensional subspaces of $V^{\vee}$, which we think of by duality as parametrizing $2$-dimensional quotients of $V$. On $\GG$ we then have the tautological exact sequence
\begin{equation}\label{eq:tautological}
0\lra\cU \lra V\oo\mc{O}_{\GG}\lra\mc{Q}\lra 0,
\end{equation}
where $\cU$ (respectively $\mc{Q}$) denotes the universal rank $(n-2)$ sub--bundle (respectively rank $2$ quotient bundle) of the trivial bundle $V$. We write $\mc{O}_{\GG}(1)$ for $\bw^2\mc{Q}$, which is the Pl\"ucker line bundle giving an embedding of $\GG$ into $\bb{P}(\bw^2 V^{\vee})$ as the projectivization of the set of decomposable $2$-forms $a\wedge b$, with $a,b\in V^{\vee}$.

\vskip 4pt

We write $\bb{Z}^r_{\op{dom}}$ for the set of \emph{dominant weights} in $\bb{Z}^r$, that is, tuples $\nu=(\nu_1,\ldots,\nu_r)\in\bb{Z}^r$ with $\nu_1\geq\nu_2\geq\ldots\geq\nu_r$. We write $\bb{S}_{\nu}$ for the \emph{Schur functor} associated with $\nu$, and recall that if $\nu=(a,0,\ldots,0)$ then $\bb{S}_{\nu} = \Sym^a$, and if $\nu=(1,1,\ldots,1)$ then $\bb{S}_{\nu} = \bw^r$. A special case that will be of interest to us is when $\a=(\a_1,\a_2)\in\bb{Z}^2_{\op{dom}}$, in which case we have
\begin{equation}\label{eq:SS-alpha}
\bb{S}_{\a}\mc{Q} = \Sym^{\a_1-\a_2}\mc{Q} \,\oo \mc{O}_{\GG}(\a_2).
\end{equation}
The cohomology groups of sheaves of the form $\bb{S}_{\a}\mc{Q}\oo\bb{S}_{\b}\cU$ are completely described by Bott's Theorem \cite[Corollary~4.1.9]{weyman}. We will only need a special case of this theorem, which we record next.

\begin{thm}[Bott]\label{thm:bott}
 Let $\a\in\bb{Z}^2_{\op{dom}}$ and  $\b\in\bb{Z}^{n-2}_{\op{dom}}$, and let $\gamma=(\a|\b)\in\bb{Z}^n$ denote their concatenation. If $\gamma$ is dominant then
 \begin{equation}\label{eq:gamma-dominant}
 H^0(\GG,\bb{S}_{\a}\mc{Q}\oo\bb{S}_{\beta}\cU) = \bb{S}_{\gamma}V,\mbox{ and }H^j(\GG,\bb{S}_{\a}\mc{Q}\oo\bb{S}_{\beta}\cU)=0\mbox{ for all }j>0.
 \end{equation}
 If there exist $1\leq x<y\leq n$ with $\gamma_x-x=\gamma_y-y$, then
\begin{equation}\label{eq:Bott-vanishing}
H^j(\GG,\bb{S}_{\a}\mc{Q}\oo\bb{S}_{\beta}\cU)=0,\mbox{ for all }j\geq 0.
\end{equation}
In particular,
\begin{enumerate}
\item\label{it:1} If $\a_1\geq\a_2\geq 0$, then $H^0(\GG,\bb{S}_{\a}\mc{Q}) = \bb{S}_{\a}V$ and $H^j(\GG,\bb{S}_{\a}\mc{Q})=0$ for all $j>0$.
\item\label{it:2} For all $a,j\geq 0$ we have that $H^j(\GG,\Sym^a\mc{Q}\oo\cU) = 0$.
\item\label{it:3} If $-2\geq\a_1\geq 1-n$, or $-1\geq\a_2\geq 2-n$, then $H^j(\GG,\bb{S}_{\a}\mc{Q})=0$ for all $j\geq 0$.
\end{enumerate}
\end{thm}

To see how (\ref{it:1}) and (\ref{it:3}) follow from the general statement of the theorem we specialize to $\b=(0,0,\ldots,0)$, so that $\gamma=(\a_1,\a_2,0,\ldots,0)$. In case (\ref{it:1}) we have that $\gamma$ is dominant so we can apply (\ref{eq:gamma-dominant}), while in case (\ref{it:3}) we apply (\ref{eq:Bott-vanishing}) with $(x,y) = (1,1-\a_1)$ when $-2\geq\a_1\geq 1-n$, and with $(x,y) = (2,2-\a_2)$ when $-1\geq\a_2\geq 2-n$. To see how (\ref{it:2}) follows from (\ref{eq:Bott-vanishing}) we note that $\a=(a,0)$, $\b=(1,0,\ldots,0)$, and take $(x,y) = (2,3)$.

\subsection{Vanishing for Koszul modules}
\label{subsec:bordervanish}

We begin by giving a geometric construction of $\ker(\delta_1)$, where $\delta_1:V\oo S\lra S$ is the first Koszul differential. We let $\GG=\Gr_2(V^{\vee})$ as in the previous section, and consider
\[\mc{S} = \Sym_{\mc{O}_{\GG}}(\mc{Q}) = \mc{O}_{\GG} \oplus \mc{Q} \oplus \Sym^2\mc{Q} \oplus \cdots\]
as a sheaf of graded $\mc{O}_{\GG}$-algebras on $\GG$. Locally on $\GG$, the sheaf $\mc{S}$ can be identified with a polynomial ring in two variables, where $\mc{Q}$ is the space of linear forms. We can then define the Koszul differentials
\[\delta_1^{\mc{Q}} : \mc{Q} \oo_{\mc{O}_{\GG}} \mc{S} \lra \mc{S}\quad\mbox{ and }\quad\delta_2^{\mc{Q}} : \bw^2\mc{Q} \oo_{\mc{O}_{\GG}} \mc{S} \lra \mc{Q} \oo_{\mc{O}_{\GG}} \mc{S}\]
as in Section~\ref{subsec:defs}. Recalling that $\bw^2\mc{Q}=\mc{O}_{\GG}(1)$, we write $\bw^2\mc{Q} \oo_{\mc{O}_{\GG}} \mc{S}$ simply as $\mc{S}(1)$.

\begin{lem}\label{lem:H0-mcS} We have a commutative diagram of graded $S$--modules
\begin{equation}\label{eq:H0-mcS}
\begin{gathered}
\xymatrix{
& \bw^2 V \oo S \ar[rr]^{\delta_2} \ar[d]_{f_2} & & V \oo S \ar[rrr]^{\delta_1} \ar[d]_{f_1} & & & S \ar[d]_{f_0} \\
0\ar[r] & H^0(\GG,\mc{S}(1)) \ar[rr]^{H^0(\GG,\delta_2^{\mc{Q}})} & & H^0(\GG,\mc{Q} \oo \mc{S}) \ar[rrr]^{H^0(\GG,\delta_1^{\mc{Q}})} & & & H^0(\GG,\mc{S}) \\
}
\end{gathered}
\end{equation}
where the maps $f_1$ and $f_0$ are isomorphisms. Moreover, we have a natural isomorphism between $\ker(\delta_1)$ and $H^0(\GG,\mc{S}(1))$.
\end{lem}

 \begin{proof}
 If we identify $S\oo\mc{O}_{\GG}$ with $\Sym_{\mc{O}_{\GG}}(V\oo\mc{O}_{\GG})$ then the tautological quotient map $V\oo\mc{O}_{\GG}\onto\mc{Q}$ in (\ref{eq:tautological}) yields a natural surjection
 \[S\oo\mc{O}_{\GG} \onto \mc{S}\]
 which can be thought of locally on $\GG$ as realizing the polynomial ring $\mc{S}$ in two variables as a quotient of $S$ by the ideal generated by the linear forms in $\cU$. The naturality of the Koszul complex yields a commutative diagram
 \[
 \xymatrix{
& \bw^2 V \oo S \oo \mc{O}_{\GG} \ar[r] \ar[d] & V \oo S \oo \mc{O}_{\GG} \ar[r] \ar[d] & S \oo \mc{O}_{\GG} \ar[d] \\
0 \ar[r] & \bw^2 \mc{Q} \oo \mc{S} \ar[r] & \mc{Q} \oo \mc{S} \ar[r] & \mc{S} \\
 }
 \]
 where the bottom row is the full Koszul complex since $\mc{Q}$ has rank $2$. The diagram (\ref{eq:H0-mcS}) is then obtained by applying the global sections functor $H^0(\GG,\bullet)$. It is sufficient to prove that $f_0$ and $f_1$ are isomorphisms, since this implies that $f_1$ induces an isomorphism between $\ker(\delta_1)$ and $\ker(H^0(\GG,\delta_1^{\mc{Q}}))$, and the latter is in turn is isomorphic to $H^0(\GG,\mc{S}(1))$ via the map $H^0(\GG,\delta_2^{\mc{Q}})$.

\vskip 4pt

To prove that $f_0$ is an isomorphism it suffices to prove it is injective, since for each~$q$ we have that $H^0(\GG,\Sym^q\mc{Q}) = \Sym^q V$ by conclusion (\ref{it:1}) of Theorem~\ref{thm:bott}. If we let $Y = \ul{\rm{Spec}}_{\GG}(\mc{S})$ denote the total space of the bundle $\mc{Q}^{\vee}$ (see \cite[Exercise~II.5.18]{hartshorne}), then $Y$ is a subbundle of the trivial bundle $V^{\vee}$ over $\GG$, which can be described explicitly as
\[ Y = \Bigl\{ (c,[a\wedge b]) \in V^{\vee} \times \GG : c\in\mbox{Span}(a,b)\Bigr\}\]
If we write $\pi:Y\to V^{\vee}$ for the natural projection map, then $\pi$ is surjective and therefore the induced pull-back map $f_0:S\lra H^0(V^{\vee},\pi_*\mc{O}_Y)=H^0(\GG,\mc{S})$ is injective, as desired.

To prove that $f_1$ is an isomorphism, we observe that it factors as the composition of
\[V\oo S \overset{\op{id}_V\oo f_0}{\lra} V \oo H^0(\GG,\mc{S}) = H^0(\GG,V\oo\mc{S}),\]
which is an isomorphism, with the map $H^0(\GG,V\oo\mc{S})\lra H^0(\GG,\mc{Q}\oo\mc{S})$ induced by the tautological surjection $V\oo\mc{O}_{\GG}\onto\mc{Q}$. To prove that this map is an isomorphism, it is enough to check that
\[ H^0(\GG,\cU\oo\Sym^q\mc{Q}) = H^1(\GG,\cU \oo\Sym^q\mc{Q}) = 0\mbox{ for all }q\geq 0,\]
which follows from conclusion (\ref{it:2}) of Theorem~\ref{thm:bott}.
\end{proof}

We next recall that $W(V,K)$ can be thought of as the middle cohomology of (\ref{eqn:W}), or equivalently as
\[
\begin{aligned}
\xymatrixcolsep{5pc}
W(V,K) &= \coker\Bigl\{
\xymatrix{
K \oo S \ar[rr]^{\delta_2|_{K \oo S}} & & \ker(\delta_1)
}\Bigr\} \\
&\cong \coker\Bigl\{
\xymatrix{
K \oo S \ar[rr]^{f_2|_{K \oo S}} & & H^0(\GG,\mc{S}(1))
}\Bigr\} \\
\end{aligned}
\]
where the isomorphism above follows from Lemma~\ref{lem:H0-mcS}. Using again the isomorphism $f_0:S\lra H^0(\GG,\mc{S})$ and the fact that $K\oo H^0(\GG,\mc{S}) = H^0(\GG,K\oo\mc{S})$, we can rewrite
\begin{equation}\label{eq:WVK-from-cohom}
 W(V,K) = \coker\Bigl\{
\xymatrix{
H^0(\GG,K\oo\mc{S}) \ar[rr]^{H^0(\GG,\eta)} & & H^0(\GG,\mc{S}(1))
}\Bigr\}
\end{equation}
where
\begin{equation}\label{eq:def-eta}
\eta:K\oo\mc{S} \lra \mc{S}(1)\mbox{ is induced by }K\oo\mc{O}_{\GG}\hookrightarrow\bw^2 V \oo\mc{O}_{\GG} \onto \bw^2\mc{Q} = \mc{O}_{\GG}(1).
\end{equation}
Equipped with the description (\ref{eq:WVK-from-cohom}) of the Koszul modules, we can now give the proof of our main theorem.

\vskip 4pt

\begin{proof}[Proof of Theorem~\ref{thm=main}]
If $W_{n-3}(V,K)= 0$, then $\dim\, W(V,K)< \infty$, since the graded module $W(V,K)$ is generated in degree $0$.
It follows from \eqref{eq:mainequiv} that $\mathcal{R} (V,K)=\{ 0\}$.

\vskip 3pt

Conversely, suppose that $\mathcal{R} (V,K)=\{ 0\}$, which by (\ref{eq:mainequiv}) is equivalent to the fact that $\bb{P}(K^{\perp})$ is disjoint from the Grassmannian $\GG$, which in turn is equivalent to the fact that the linear system of divisors $(K,\mc{O}_{\GG}(1))$ is base point free, or to the fact that the map $K\oo\mc{O}_{\GG}\lra\mc{O}_{\GG}(1)$ is surjective. The Koszul complex associated to this map is therefore exact, and takes the form
\[\mc{K}^{\bullet}: 0\lra \bw^m K\oo \mc{O}_{\GG}(1-m) \lra \cdots \lra \bw^2 K\oo \mc{O}_{\GG}(-1) \lra K \oo \mc{O}_{\GG} \lra \mc{O}_{\GG}(1) \lra 0\]
where we use a cohomological grading with $\mc{K}^{-i} = \bw^i K \oo \mc{O}_{\GG}(1-i)$. Tensoring with $\Sym^{n-3}\mc{Q}$ preserves exactness, so the hypercohomology of $\mc{K}^{\bullet}\oo\Sym^{n-3}\mc{Q}$ vanishes identically. The hypercohomology spectral sequence associated with $\mc{K}^{\bullet}\oo\Sym^{n-3}\mc{Q}$ then yields that
 \[ E_1^{-i,j} = H^j\Bigl(\GG,\bw^i K\oo \mc{O}_{\GG}(1-i) \oo \Sym^{n-3}\mc{Q}\Bigr)  \Longrightarrow 0.\]
 Since $\Sym^{n-3}\mc{Q}\oo\mc{O}_{\GG}(1-i) = \bb{S}_{(n-2-i,1-i)}\mc{Q}$ by (\ref{eq:SS-alpha}), we can rewrite the terms in the spectral sequence as
 \begin{equation}\label{eq:Eij}
  E_1^{-i,j} = H^j\Bigl(\GG,\bw^i K \oo \bb{S}_{(n-2-i,1-i)}\mc{Q}\Bigr) = \bw^i K \oo H^j\bigl(\GG,\bb{S}_{(n-2-i,1-i)}\mc{Q}\bigr).
 \end{equation}
 We have using (\ref{eq:WVK-from-cohom}) the following identification
 \[W_{n-3}(V,K) = \coker\bigr\{E^{-1,0}_1 \lra E^{0,0}_1\bigr\}\]
 and we suppose by contradiction that this map is not surjective. Since $E_{\infty}^{0,0}=0$, there must be some \emph{non-zero} differential
 \[E^{-r,r-1}_r \lra E_r^{0,0}\mbox{ for }r\geq 2,\mbox{ or }E_r^{0,0} \lra E_r^{r,1-r}\mbox{ for }r\geq 1.\]
 Since $E_1^{r,1-r}=0$ for all $r\geq 1$ it follows that $E_r^{r,1-r}=0$ as well, so the latter case does not occur. To prove that the former case does not occur either and obtain a contradiction, it suffices to check that $E^{-r,r-1}_1=0$ for all $r\geq 2$, which follows from (\ref{eq:Eij}) if we can prove that
 \begin{equation}\label{eq:vanishing-Hr-1}
 H^{r-1}\bigl(\GG,\bb{S}_{(n-2-r,1-r)}\mc{Q}\bigr) = 0\mbox{ for }r\geq 2.
 \end{equation}
 If we write $\a=(n-2-r,1-r)$ then for $2\leq r\leq n-1$ we have that $-1\geq\a_2\geq 2-n$, and for $n\leq r\leq 2n-3$ we have that $-2\geq\a_1\geq 1-n$, so (\ref{eq:vanishing-Hr-1}) follows from conclusion (\ref{it:3}) of Theorem~\ref{thm:bott}. When $r>2n-3$ we get that $r-1>2n-4=\dim(\GG)$, so (\ref{eq:vanishing-Hr-1}) is a consequence of Grothendieck's vanishing theorem \cite[Theorem~III.2.7]{hartshorne}.
\end{proof}

\subsection{The Hilbert series of Koszul modules}\label{subsec:HilbSeries}
The goal of this section is to prove Theorem~\ref{thm:HilbSeries}. Recall from equation \eqref{eqn:Wq_2} that $W_q(V,K)$ is obtained as the cokernel of the map
\begin{equation}\label{eq:presentation-Wq}
\delta_{2,q}:K\otimes \Sym^q V \lra \Im(\delta_{2,q})
\end{equation}
and in particular we have that $W_q(V,K) \neq 0$ when $\dim(\Im(\delta_{2,q}))>\dim(K\otimes \Sym^qV)$. We say that the {\em divisorial case} occurs when the map (\ref{eq:presentation-Wq}) is given by a square matrix, i.e.
\[
\mathrm{dim}\, (K\otimes \Sym^qV)=\mathrm{dim}\left(\mathrm{Im}(\delta_{2,q})\right).
\]

\begin{prop}
\label{prop:nonzero}
Let $K\subseteq \bigwedge^2V$ be an arbitrary subspace of dimension $m=2n-3$. For any $0\le q\le n-4$ we have $W_q(V,K)\ne 0$, and $q=n-3$ is a divisorial case.
\end{prop}

\proof
Using the fact that $\dim\Sym^k V= {n+k-1 \choose k}$ and $\dim(K)=2n-3$ we obtain
\[
\mathrm{dim}\, (K\otimes \Sym^qV)=(2n-3){n+q-1 \choose q}.
\]
The exactness of the Koszul complex implies that $\delta_{1,q+1}$ is surjective and that we have the equality $\Im(\delta_{2,q}) = \ker(\delta_{1,q+1})$, so
\[
\mathrm{dim}(\Im(\delta_{2,q}))= \mathrm{dim}\, (V\otimes \Sym^{q+1}V) - \mathrm{dim}\, (\Sym^{q+2}V) = n\cdot{n+q\choose q+1} - {n+q+1\choose q+2},
\]
A direct computation leads for $q\ge 0$ to the formula

\begin{equation}
\label{eqn:dimW}
\mathrm{dim}(\Im(\delta_{2,q})) - \mathrm{dim}\, (K\otimes \Sym^qV)= {n+q-1 \choose q} \frac{(n-2)(n-q-3)}{q+2},

\end{equation}
which is positive for $0\leq q\leq n-4$ and vanishes for $q=n-3$, proving that $W_q(V,K) \neq 0$ for $0\leq q\leq n-4$, and that the case $q=n-3$ is divisorial.
\endproof

We next analyze in more detail the divisorial case, where the vanishing of $W_{n-3}(V,K)$ can be rephrased into a transversality condition as follows.

\begin{lem}
\label{lem:Schubert1}
Let $K\subseteq \bigwedge^2V$ be a subspace of dimension $2n-3$. The following are equivalent:
\begin{enumerate}
\item[(a)] $W_{n-3}(V,K) = 0$.
\item[(b)] $(K\otimes \Sym^{n-3}V)\cap \ker(\delta_{2,n-3})=0$.
\item[(c)] $(K\otimes \Sym^{q}V)\cap \ker(\delta_{2,q})=0$ for all $q\le n-3$.
\end{enumerate}
\end{lem}

\proof
It is clear that (c) implies (b), while for the reverse implication it suffices to observe that multiplication by a linear form in $V$ yields for each $q\geq 0$ an injective map
from $(K\otimes \Sym^{q}V)\cap \ker(\delta_{2,q})$ to $(K\otimes \Sym^{q+1}V)\cap \ker(\delta_{2,q+1})$.

Using the isomorphism $\im(\delta_{2,n-3}) = (\bw^2 V \oo \Sym^{n-3}V) / \ker(\delta_{2,n-3})$ and (\ref{eqn:Wq_2}) we get
\begin{equation}\label{eq:Wn-3=quot}
W_{n-3}(V,K)=\bigl(\bigwedge^2V\otimes \Sym^{n-3}V\bigr)/\bigl(K\otimes \Sym^{n-3}V+\ker(\delta_{2,n-3})\bigr).
\end{equation}
Since we are in the divisorial case we have $\dim(\im(\delta_{2,n-3}))=\dim(K\otimes \Sym^{n-3}V)$, thus
\begin{equation}\label{eq:dimensions-match}
\mathrm{dim}(\bigwedge^2V\otimes \Sym^{n-3}V)=\mathrm{dim}(K\otimes \Sym^{n-3}V)+\mathrm{dim}(\ker(\delta_{2,n-3})).
\end{equation}
Combining (\ref{eq:Wn-3=quot}) with (\ref{eq:dimensions-match}) we conclude that $W_{n-3}(V,K)=0$ if and only if the sum $K\otimes \Sym^{n-3}V+\ker(\delta_{2,n-3})$ is direct, proving the equivalence of (a) and (b).
\endproof

\begin{cor}\label{cor:maximal-rank}
If $K\subseteq\bigwedge^2V$ has dimension $2n-3$ and $\mathcal{R}(V,K) = \{0\}$, then the map (\ref{eq:presentation-Wq}) is of maximal rank for all $q$.
\end{cor}

\proof
Using (\ref{eqn:dimW}), it suffices to prove that (\ref{eq:presentation-Wq}) is injective for $q\leq n-3$ and surjective for $q\geq n-3$. Since $\mathcal{R}(V,K) = \{0\}$, we know from Theorem~\ref{thm=main} that $W_{n-3}(V,K)=0$, hence the assertion in Lemma~\ref{lem:Schubert1}(c) holds, proving the injectivity of (\ref{eq:presentation-Wq}) in the range $q\leq n-3$. The surjectivity of (\ref{eq:presentation-Wq}) is equivalent to the vanishing of $W_q(V,K)$, which is true for $q\geq n-3$ because $W_{n-3}(V,K)=0$ and $W(V,K)$ is generated in degree $0$.
\endproof

We now have all the tools necessary to prove Theorem~\ref{thm:HilbSeries}.

\begin{proof}[Proof of Theorem~\ref{thm:HilbSeries}]
We write $m=\dim(K)$ as usual, and suppose first that $m=2n-3$. By Corollary~\ref{cor:maximal-rank} we know that the maps (\ref{eq:presentation-Wq}) have maximal rank for all $q$. Since their cokernels are the graded components of $W(V,K)$, it follows from (\ref{eqn:dimW}) that
\[\dim\, W_q(V,K) = {n+q-1 \choose q} \frac{(n-2)(n-q-3)}{q+2}\mbox{ for all }q=0,\cdots,n-4.\]

We next consider the case $m>2n-3$. If we set $\GG :=\mathrm{Gr}_2(V^\vee )\subseteq \mathbb P(\bigwedge^2V^\vee )$ then~(\ref{eq:mainequiv}) implies that $\bb{P}(K^{\perp}) \cap \GG = \emptyset$. We consider the linear projection away from $\bb{P}(K^{\perp})$
\[ \pi : \mathbb P(\bigwedge^2V^\vee ) \dashrightarrow \bb{P}^{m-1}\]
Since $\dim(\GG)=2n-4$ it follows that $\pi(\GG)$ is a subvariety of $\bb{P}^{m-1}$ of codimension at least $m-1 - (2n-4) = m-2n+3$, hence a general linear subspace $\bb{P}^{m-2n+2} \subset \bb{P}^{m-1}$ is disjoint from $\pi(\GG)$, i.e. it satisfies $\pi^{-1}(\bb{P}^{m-2n+2}) \cap \GG = \emptyset$. Since $\pi^{-1}(\bb{P}^{m-2n+2})$ is a linear subspace of $\bb{P}(\bigwedge^2V^\vee )$ of codimension $2n-3$ containing $\bb{P}(K^{\perp})$, we can write $\pi^{-1}(\bb{P}^{m-2n+2})=\bb{P}(K'^\perp)$ for some $K'\subseteq K$ with $\dim(K')=2n-3$. It follows from \eqref{eq:wnat} (with the roles of $K$ and $K'$ exchanged) that $\dim\, W_q(V,K)\leq\dim\, W_q(V,K')$ for all $q$, which combined with the case $m=2n-3$ proves the desired inequality and completes our proof.
\end{proof}

\begin{rmk}
\label{rem:bigu}
A special case of Theorem~\ref{thm:HilbSeries} is obtained by taking $V = \Sym^{n-1}\bb{C}^2$ and $K=\Sym^{2n-4}\bb{C}^2$, in which case $W(V,K)$ is the \emph{Weyman module} $W(n-1)$ (see \cite[Section~5]{PS15} and \cite[Section~3.I.B]{E91}). It follows that Theorem~\ref{thm:HilbSeries} computes the Hilbert series of the said module, answering a problem left open in \cite[Section~1.2]{PS15}. For $n\le 8$, Alex Suciu used a computer search to construct subspaces $K_n \subseteq \bigwedge^2 \mathbb{C}^n$ with $m=2n-3$ and $\mathcal{R}(\mathbb{C}^n, K_n)= \{0\}$. He also computed the Hilbert series of the Koszul modules $W(\mathbb{C}^n, K_n)$. He conjectured the existence of a family of subspaces $K_n \subseteq \bigwedge^2 \mathbb{C}^n$ with $m=2n-3$ and $\mathcal{R}(\mathbb{C}^n, K_n)= \{0\}$, for all values of $n$, with Hilbert series given by the formula from Theorem~\ref{thm:HilbSeries}. Our results prove that the expectations were correct.
\end{rmk}

\section{Topological invariants of groups}
\label{subsec:top}

In this section we explain the topological motivation for investigating Koszul modules, and derive a number of consequences of our results to the study of invariants of fundamental groups. Some of our main applications concern groups $G$ that are $1$-formal (in the sense of Sullivan \cite{Su77}), where a direct relationship between certain invariants of $G$ and those of the corresponding Koszul modules can be derived. The class of $1$-formal groups contains important examples such as the fundamental groups of compact K\"ahler manifolds, hyperplane arrangement groups, or the Torelli group of the mapping class group of a genus $g$ surface.

\subsection{The Koszul module and the resonance variety of a group G}
\label{subsec:WG-RG}

We recall from Example~\ref{ex:W(G)} the construction of the \emph{Koszul module $W(G)$} associated to a finitely generated group $G$. The module $W(G):=W(V,K)$ is obtained by setting $V:= H_1(G, \mathbb{C})$, and letting $K^{\perp} \subseteq \bigwedge^2 V^\vee $ be the kernel of the cup-product map
\[\cup_{G} : \bigwedge^2 H^1(G, \mathbb{C}) \to H^2(G, \mathbb{C}).\]
To the Koszul module $W(G)$ we can associate its \emph{resonance variety $\mc{R}(G)$} as in Definition~\ref{def:resonance}. Alternatively, we can define $\mc{R}(G)$ without reference to Koszul modules as follows. For an element $a\in H^1(G,\mathbb C)$, we consider the complex of abelian groups
$$(G,a):\qquad 0 \longrightarrow H^0(G,\mathbb C)\stackrel{\cup a}\longrightarrow H^1(G,\mathbb C)\stackrel{\cup a}\longrightarrow H^2(G,\mathbb C)\stackrel{\cup a} \longrightarrow \cdots.$$
The resonance variety $\mc{R}(G)$ is then obtained as
$$\cR(G)=\bigl\{a\in H^1(G,\mathbb C): H^1(G,a)\neq 0 \bigr\}.$$

In the $1$-formal context, $\mc{R}(G)$ has yet another interpretation given as follows.

\begin{rmk}\label{rem:jump-loci} For a character $\rho\in \widehat{G}:=\mbox{Hom}(G,\mathbb C^*)$, we denote by $\mathbb C_{\rho}$ the $\mathbb C[G]$-module $\mathbb C$ with multiplication given by
$g\cdot z:=\rho(g)z$. The resonance variety $\cR(G)$ turns out to be an infinitesimal analogue of the much studied \emph{characteristic variety}
$$\mathcal{V}(G):=\bigl\{\rho\in \widehat{G}: H^1(G,\mathbb C_{\rho})\neq 0\bigr\},$$
which already appeared in the work of Green--Lazarsfeld \cite{GL87}, Beauville \cite{Be}, or Libgober \cite{Li}. It is shown in \cite[Theorems A and C]{DPS} that when $G$ is $1$-formal, $\cR(G)$ is isomorphic to the tangent cone at $1$ of the characteristic variety~$\mathcal{V}(G)$.
\end{rmk}

Reinterpreted in the context of groups, our Theorems~\ref{thm=main} and~\ref{thm:HilbSeries} imply the following. We write $b_1(G) := \dim\, H_1(G,\bb{C})$ for the \emph{first Betti number} of the group $G$.

\begin{thm}\label{thm:WG-RG}
 Let $G$ be a finitely generated group and let $n=b_1(G)$. If $n\geq 3$ then we have the equivalence
 \[\mc{R}(G)=\{0\} \Longleftrightarrow W_{n-3}(G)=0.\]
 Moreover, if the equivalent statements above hold then
\[ \dim\, W_q(G) \leq {n+q-1 \choose q} \frac{(n-2)(n-q-3)}{q+2}\quad\mbox{ for }0\leq q\leq n-4.\]
\end{thm}

\vskip 3pt

This theorem can be used to obtain upper bounds on the Chen ranks of a group $G$ with vanishing resonance, once we establish a relationship between the said ranks and the Hilbert function of $W(G)$. This is achieved via the study of the Alexander invariant of $G$ in the next section.

\subsection{Alexander invariants.}
\label{subsec:alexander}
Alexander-type invariants have a long history in topology, starting with the definition of the Alexander polynomial of knots and links \cite{Alex}.
For a connected CW--complex $X$ with fundamental group $G= \pi_1(X)$, each surjective group homomorphism $\alpha :G \to H$ gives rise to a sequence of
Alexander-type invariants of $X$, namely the twisted homology groups $H_i(X, L)$, where $L$ is the local system $\mathbb{C}[H]$.
By construction, these invariants are $\mathbb{C}[G]$--modules. When $i=1$ the construction depends solely
on $G$, is purely algebraic and can be carried out for any group. The classical Alexander invariant of a knot $Y\subseteq S^3$ corresponds to the case
when $X:=S^3\setminus Y$ is the knot complement, $i=1$ and  $\alpha:G \to \mathbb{Z}$ is the abelianization map.

\vskip 3pt

\begin{defn}\label{alexinvarians}
If $G$ is a finitely generated group, its \emph{Alexander invariant} is defined as $B(G):=H_1(G',\mathbb Z)=G'/G''$, viewed as a module over the group ring
$\mathbb Z[G_{\mathrm{ab}}]$, where $G_{\mathrm{ab}}:=G/G'$, the action being given by conjugation. We write $B(G)_{\mathbb C}:=B(G)\otimes_{\mathbb Z} \mathbb C$.
\end{defn}

Geometrically, if $X$ is a CW--complex with $\pi_1(X)\cong G$ and $X_{\mathrm{ab}} \rightarrow X$ is the maximal abelian cover corresponding to the group $H_1(X,\mathbb Z)$,
then $B(G)$ can be identified with $H^1(X',\mathbb Z)$, viewed as a module over $G_{\mathrm{ab}}\cong H_1(X,\mathbb Z)$.

\vskip 3pt

The Alexander invariant is equipped with the $I$--adic filtration, where
$I:=\mbox{Ker}(\epsilon)$ is the kernel of the augmentation map $\epsilon:\mathbb Z[G/G'] \rightarrow \mathbb Z$.
A key result of Massey \cite{Massey} asserts that for each $q\geq 0$, one has an isomorphism
\begin{equation}\label{eq:Massey}
\mathrm{gr}_q B(G):=\mathrm{gr}_q^{I} B(G)\cong \mathrm{gr}_{q+2} (G/G'').
\end{equation}
In particular, the generating series of the Chen ranks of $G$ is given as a Hilbert series $$\sum_{q=0}^{\infty} \theta_{q+2}(G)t^q=\mbox{Hilb}\bigl(\mbox{gr } B(G)_{\mathbb C},t\bigr).$$

\vskip 4pt

\begin{rmk}
The Chen ranks  are usually more easily computable than the largely inaccessible lower central ranks of a group $G$. An instance of this is the case of hyperplane
arrangement groups associated to an arrangement $\mathcal{A}=\{H_1, \ldots, H_{n}\}$ of hyperplanes in $\mathbb C^m$. Whereas the fundamental group $G(\mathcal{A}):=\pi_1\bigl(\mathbb C^m-\cup_{H\in \cA} H\bigr)$ of the arrangement is \emph{not} determined by the intersection lattice of $\mathcal{A}$, the lower central series and the corresponding ranks $\phi_q(G(\mathcal{A}))$ and $\theta_q\bigl(G(\mathcal{A}))$  are, and in particular they are combinatorial objects. An explicit formula for $\phi_q(G(\cA))$ remains elusive, but the Chen ranks are subject to a conjecture of Suciu, proven in many cases in \cite{ScSu} and in full generality in \cite{CSch}. It asserts that for $q\gg 0$, one has
the formula
$$\theta_q\bigl(G(\cA)\bigr)=\sum_{r\geq 1} h_r \theta_q(F_{r+1}),$$
where $h_r$ is the number of $r$-dimensional components of the resonance variety $\mathcal{R}(G(\cA))$ and the Chen ranks of the free group are given by (see \cite{Chen})
\[\theta_1(F_{r+1})=r+1\quad\mbox{ and }\quad\theta_q(F_{r+1})=(q-1){q+r-1\choose q}\mbox{ for }q\geq 2.\]
\end{rmk}

\vskip 4pt

The Alexander invariants of $G$ exhibit the same type of behaviour as their cousins, the topological jump loci $\mathcal{V}(X):=\mathcal{V}(\pi_1(X))$ from Remark~\ref{rem:jump-loci}. They are complicated objects encoding useful information, with complexity bounded by invariants depending on the cohomology ring. Work from \cite{DHP, Massey, PS-imrn, PS-johnson, Sulli} implies the existence of a natural surjection

\[
W(G) \twoheadrightarrow \mathrm{gr}\, B(G)_{\mathbb C}.
\]
In particular, for all $q\geq 0$,

\begin{equation}
\label{eq=bw}
\mbox{dim } \mathrm{gr}_q B(G)_{\mathbb C} \leq \mbox{dim } W_q(G),
\end{equation}
with equality when the group $G$ is $1$--formal \cite[Corollary~6.1]{PS15}.

\vskip 4pt

One should regard the relation between $\mathrm{gr}\, B(G)_{\mathbb C}$ and $W(G)$ as a commutative analogue of the relationship between the Lie algebra $\mbox{gr}(G)_{\mathbb C}$ and the holonomy Lie algebra of $G$, whose definition we recall next. Denoting by $\partial_G:\bigwedge^2 H_1(G,\mathbb C)\rightarrow H_2(G,\mathbb C)$ the dual of the cup product map $\cup_G$, one defines following Chen \cite{Chen} the \emph{holonomy Lie algebra} $$\mathfrak{H}(G):=\mathbb L\bigl(H_1(G,\mathbb C)\bigr)/\mbox{Ideal}\bigl(\mbox{Im}(\partial_G)\bigr),$$
where $\mathbb L\bigl(H_1(G,\mathbb C)\bigr)$ is the free complex Lie algebra generated by $H_1(G,\mathbb C)$.
Work of Sullivan \cite{Su77} implies that for a $1$-formal group $G$ one has an isomorphism of Lie algebras
$$\mbox{gr}(G)_{\mathbb C}\cong \mathfrak{H}(G),$$
that is, the entire lower central series of $G$ is determined by the first cup-product map.

\vskip 4pt

In the following applications we relate the $I$--adic filtrations of the Alexander invariant $B(G)$  of $G$
to the cohomology ring of $G$ in low degrees in a more precise way.

\begin{thm}
\label{thm=maingroups}
Let $G$ be a finitely generated group with $b_1(G)\geq 3$. If $\mathcal{R} (G)=\{0\}$, then
\[\bigl(I^q \cdot B(G)\bigr)_{\bb{C}}=\bigl(I^{q+1} \cdot B(G)\bigr)_{\bb{C}}\]
for all $q\ge b_1(G)-3$.
\end{thm}

\proof
Combine Theorem \ref{thm:WG-RG} with \eqref{eq=bw}.
\endproof

We recall that a module $M$ over a group ring is nilpotent if $I^q \cdot M=0$ for some $q$, where $I$ is the augmentation ideal.

\begin{cor}
\label{cor=nilpmod}
If $\mathcal{R} (G)=\{0\}$ and $B(G)$ is nilpotent, then $\bigl(I^{b_1(G)-3} \cdot B(G)\bigr)_{\bb{C}}=0$. If moreover $G$ is $1$--formal, then
$\dim \, B(G)_{\mathbb C}= \dim \, W(G)< \infty$.
\end{cor}

In general, the Alexander invariant of a group may not be nilpotent as a module, but this is the case when the group itself is nilpotent. Recall that a group $G$ is said to be nilpotent if there exists $c\geq 1$ with $\Gamma_c(G)=\{1\}$. The largest $c$ such that $\Gamma_c(G)\neq \{1\}$ is called the \emph{nilpotency class} of $G$, and is denoted $\nc(G)$. A group $G$ is \emph{virtually nilpotent} if there exists a finite index nilpotent subgroup $H\leq G$. We define the \emph{virtual nilpotency class} $\vnc(G)$ of $G$ to be the following quantity
\begin{equation}\label{eq:def-vnc}
\vnc(G) := \min\bigl\{\nc(H):H\leq G\mbox{ is a nilpotent finite index subgroup}\bigr\}.
\end{equation}
Since every virtually nilpotent group contains a torsion free finite index subgroup, and since the nilpotency class of a torsion free nilpotent group equals that of any of its finite index subgroups, it follows that for a virtually nilpotent group $G$ we have
\begin{equation}\label{eq:vncG=ncH}
\vnc(G) = \nc(H)\mbox{ for any torsion free nilpotent subgroup }H\leq G\mbox{ of finite index}.
\end{equation}
One useful bound for the virtual nilpotency class of $G$ is obtained as follows.

\begin{lemma}\label{lem:vnc}
If $G$ contains a torsion-free finite index subgroup $H$ and $\Gamma_c(G)$ is finite, then $\vnc(G)=\nc(H)\leq c-1$.
\end{lemma}

\begin{proof}
Since $\Gamma_c(H)\leq\Gamma_c(G)$ and $\Gamma_c(H)$ is torsion-free, it follows that $\Gamma_c(H)=\{1\}$ and therefore $\nc(H)\leq c-1$. The equality $\vnc(G)=\nc(H)$ follows from (\ref{eq:vncG=ncH}).
\end{proof}

\begin{cor}
\label{cor=nilpgr}
If $G$ is nilpotent and $\mathcal{R} (G)=\{0\}$, then $\bigl(I^{b_1(G)-3} \cdot B(G)\bigr)_{\bb{C}}=0$. It follows that $\vnc(G/G'')\leq b_1(G)-2$.
\end{cor}

\begin{proof}
 Since $G$ is nilpotent, the Alexander invariant is also nilpotent so the conclusion $\bigl(I^{b_1(G)-3} \cdot B(G)\bigr)_{\bb{C}}=0$ follows from Corollary~\ref{cor=nilpmod}. It follows moreover from (\ref{eq:Massey}) and \eqref{eq=bw} that $\theta_q(G)=0$ for $q\geq b_1(G)-1$. Since $G/G''$ is nilpotent, we must have that $\theta_q(G)$ is finite for $q\geq b_1(G)-1$, so $\vnc(G/G'')\leq b_1(G)-2$ by Lemma~\ref{lem:vnc}.
\end{proof}

The conclusion of Corollary~\ref{cor=nilpgr} \emph{does} require the vanishing resonance assumption.

\begin{ex}
\label{ex=freenilp}
Indeed, let $F_n$ be the free group on $n\ge 2$ generators, and consider the nilpotent group $G:= F_n/\Gamma_k(F_n)$, where $k\geq 3$.
As explained for instance in \cite[Remark~2.4]{MP}, the resonance variety of a group depends only on its third nilpotent quotient.
Since $G/\Gamma_3(G)\cong F_n/\Gamma_3(F_n)$, it follows that $\mathcal{R} (G)\cong \mathcal{R} (F_n)\cong \mathbb{C}^n$. Note also that $b_1(G)=n$.
On the other hand, as already pointed out, the Hilbert series
$\sum_{q\ge 1} \theta_q(F_n) t^q$ was computed by Chen \cite{Chen} and {\em all} of its coefficients are strictly positive. Since $\theta_q(F_n)=\theta_q(G)$ for $q<k$, it follows from (\ref{eq:Massey}) that $I^q\cdot B(G)\neq 0$ for $q<k-2$, thus the nilpotency class cannot be bounded solely in terms of $n=b_1(G)$.
\end{ex}

In fact the resonance variety of a nilpotent group of nilpotency class $2$ can be arbitrarily complicated, cf. \cite[Remark 2.4]{MP}. However, this phenomenon does not occur if we assume $1$-formality. The resonance variety of a finitely generated, nilpotent, $1$-formal group $G$ must vanish,  see \cite[Lemma 2.4]{CT} and the proof of Theorem~\ref{thm:bound-chen} below. Theorem~\ref{thm:Kahler} in the Introduction is the special case when $G$ is a K\"ahler group of the following.

\begin{thm}\label{thm:bound-chen}
 Let $G$ be a finitely generated $1$-formal group, and suppose its first Betti number is $n=b_1(G)\geq 3$. If $G/G''$ is nilpotent, then $\theta_q(G)=0$ for $q\geq n-1$ and
 \[\theta_q(G) \leq {n+q-3\choose n-1} \cdot \frac{(n-2)(n-1-q)}{q}\quad\mbox{ for }q=2,\ldots,n-2.\]
\end{thm}

\begin{proof} The condition that $G/G''$ be nilpotent implies that $\theta_q(G)=\phi_q(G/G'')=0$ for $q\gg 0$. Since $G$ is $1$-formal, we have using (\ref{eq:Massey}) and \eqref{eq=bw} that $\dim\,W_q(G)=\theta_{q+2}(G)$ for all $q\geq 0$. In particular $W(G)$ is a finite dimensional module, hence $\mc{R}(G)=\{0\}$ and Theorem~\ref{thm:WG-RG} applies to give the desired conclusions about $\theta_q(G)=\dim\,W_{q-2}(G)$.
\end{proof}

We proved in Theorem \ref{thm=main} that $W_{n-3}(V,K)=0$ when $\mathcal{R} (V,K)=\{ 0\}$, and therefore $W_q(V,K)=0$ for all $q\ge n-3$.
In the borderline case $m=2n-3$, we infer from Proposition \ref{prop:nonzero} that this is the best possible vanishing result
for Koszul modules. In the general case, the following examples coming from nilpotent groups show that the bound $q(n,m)= n-3$ may be improved, at least in some cases.

\begin{ex}
\label{ex=heis}
For $k\ge 1$, we denote by $H_k$ the fundamental group of the {\em Heisenberg nilmanifold} of dimension $2k+1$.
It is well-known that $H_k$ is finitely generated and $\Gamma_3(H_k)= \{ 1\}$. We set
$V:= H_1(H_k, \mathbb{C})$ and denote by $K^{\perp}\subseteq \bigwedge^2 V^{\vee}$ the kernel of the cup-product map as before. If $k\ge 2$,
we know from \cite{Mac} that $H_k$ is $1$--formal and $\mathcal{R}(H_k)=\{ 0\}$. In this case, $b_1(H_k)=2k$ and $K\subseteq \bigwedge^2 V$
has codimension $1$. The fact that $\Gamma_3(H_k)= \{ 1\}$ and \eqref{eq=bw} together imply that $W_1(H_k)=0$, but note that $1<b_1(H_k)-3$ as soon as $k>2$.
\end{ex}

\subsection{A bound on the degree of growth of a group.}\label{subsec:deg-growth}

The goal of this section is to provide a proof of Theorem~\ref{thm:growth-degree}. Suppose $G$ is a finitely generated group and let $S\subseteq G$ be a finite set of generators. For an integer $m\geq 0$, we denote by $\ell_G(m)=\ell_{G,S}(m)$ the number of elements $g\in G$ expressible as a product of $m$ elements from $S\cup S^{-1}$. Bass \cite{bass} and Guivarc'h \cite{guivarch} showed that a finitely generated nilpotent group $G$ has polynomial growth. More precisely, the inequality $\ell_G(m)\leq C m^{d(G)}$
holds for $m\gg 0$, where $d(G)$ is the \emph{degree of polynomial growth} of the group, which in the case of nilpotent groups is given by the  formula

\begin{equation}\label{thm:bass-guivarch}
d(N) = \sum_{q\geq 1} q\cdot\phi_q(N).
\end{equation}

We also recall a celebrated theorem of Gromov \cite{gromov} which asserts that conversely, the virtually nilpotent groups are precisely the ones that have polynomial growth. If $G$ is merely virtually nilpotent, the Bass-Guivarc'h formula (\ref{thm:bass-guivarch}) fails as seen for instance by taking $G$ to be the infinite dihedral group, where $d(G)=1$, whereas $\phi_q(G)=0$ for all $q$. For a quantitative version of Gromov's result, we refer to \cite[Theorem 4]{M}.

\vskip 3pt

Before stating the following result, we recall that the \emph{Hirsch index} of a finitely generated nilpotent (or more generally polycyclic) group $G$ is defined to be the quantity
$h(G):=\sum_{q\geq 1} \phi_q(G)$, see \cite[Chapter~1]{segal}. It is well-known that for a normal subgroup $N\trianglelefteq G$ one has the additivity property $h(G)=h(N)+h(G/N)$.

\begin{lem}\label{lem:ineq-sumphi}
 Let $G$ be a finitely generated group, and $H$ be a finite index subgroup of~$G$. For every $k\geq 1$ we have that
 \begin{equation}\label{eq:sum-phi-ineq}
 \sum_{q=1}^k \phi_q(H) \geq \sum_{q=1}^k \phi_q(G).
 \end{equation}
 Moreover, if $[H\cap\Gamma_{k+1}(G):\Gamma_{k+1}(H)]<\infty$, then the above inequality is in fact an equality.
\end{lem}

\begin{proof} The quotient $\ol{H}:=H/\Gamma_{k+1}(H)$ is nilpotent and its Hirsch number is given by
\[h(\ol{H}) = \sum_{q=1}^k \phi_q(H).\]
A similar formula holds for $\ol{G}:=G/\Gamma_{k+1}(G)$. Since $H\leq G$, we must also have that $\Gamma_{k+1}(H)\leq\Gamma_{k+1}(G)$, and therefore we obtain an induced group homomorphism
\[ \psi : \ol{H} \lra \ol{G}.\]
Since $[G:H]<\infty$,  the image of $\psi$ has finite index in $\ol{G}$, therefore $h(\op{Im}\psi) = h(\ol{G})$. Since $\op{Im}\psi$ is a quotient of $\ol{H}$, it follows that $h(\ol{H})=h(\ker\psi) + h(\op{Im}\psi)\geq h(\op{Im}\psi)$, which proves that $h(\ol{H})\geq h(\ol{G})$. If we assume that $\Gamma_{k+1}(H)$ has finite index in $H\cap\Gamma_{k+1}(G)$ then $\ker\psi$ is a finite group and therefore $h(\ker\psi)=0$, which implies $h(\ol{H})=h(\ol{G})$.
\end{proof}

In the next proof, recall that $M:=G/G''$ denotes the maximal metabelian quotient of a $1$-formal finitely generated group $G$.

\begin{proof}[Proof of Theorem~\ref{thm:growth-degree}]
Since $\Gamma_q(M)$ is finite for $q\gg 0$, it follows that $\theta_q(G)=\phi_q(M)=0$ for $q\gg 0$. Using (\ref{eq:Massey}) and the fact that (\ref{eq=bw}) is an equality when $G$ is $1$-formal, we obtain
\[\dim\, W_q(G) = \phi_{q+2}(M) = 0 \mbox{ for }q\gg 0.\]
It follows that $W(G)$ is finite dimensional, which by (\ref{eq:mainequiv}) and Theorem~\ref{thm=main} implies that $W_q(G)=0$ for $q\geq n-3$ and thus $\phi_{q}(M)=0$ for $q\geq n-1$. Combining this with the fact that $\Gamma_q(M)$ is finite for $q\gg 0$, we conclude that in fact $\Gamma_{n-1}(M)$ is finite proving conclusion (\ref{it:conc-1}) of Theorem~\ref{thm:growth-degree}.

 We note that if $N$ is a finite normal subgroup of $M=G/G''$ then the groups $M$ and $M/N$ have the same growth rate. Taking $N=\Gamma_q(M)$ where $q$ is such that $\Gamma_q(M)$ is finite, we get that $M$ has the same growth rate as the nilpotent group $M/N$, and in particular $M$ is virtually nilpotent by Gromov's theorem. We can then apply Lemma~\ref{lem:vnc} to deduce part (\ref{it:conc-2}) of Theorem~\ref{thm:growth-degree}.

Consider now $H$ to be a nilpotent finite index subgroup of $M$. Using the last part of Lemma~\ref{lem:ineq-sumphi} and the fact that $\Gamma_{n-1}(M)$ is finite we get that
\begin{equation}\label{eq:sum-phi-equal}
\sum_{q=1}^{n-2} \phi_q(H) = \sum_{q=1}^{n-2} \phi_q(M).
\end{equation}
Applying the Bass--Guivarc'h Formula \ref{thm:growth-degree} to $H$ yields
\[d(M) = d(H) = \sum_{q\geq 1} q\cdot \phi_q(H) = (n-2)\cdot\left(\sum_{q=1}^{n-2} \phi_q(H)\right) - \sum_{k=1}^{n-3}\left(\sum_{q=1}^k \phi_q(H)\right)\]
\[\overset{(\ref{eq:sum-phi-ineq}),(\ref{eq:sum-phi-equal})}{\leq} (n-2)\cdot\left(\sum_{q=1}^{n-2} \phi_q(M)\right) - \sum_{k=1}^{n-3}\left(\sum_{q=1}^k \phi_q(M)\right) = \sum_{q = 1}^{n-2} q\cdot \phi_q(M)\]
\[=\phi_1(M) + \sum_{q = 0}^{n-4} (q+2)\cdot \dim\, W_q(G).\]
Using the fact that $\phi_1(M)=b_1(G) = n$ and the inequality in Theorem~\ref{thm:HilbSeries}, it follows that
\[ d(M) \leq n + \sum_{q=0}^{n-4} {n+q-1\choose q} \cdot (n-2)\cdot(n-q-3) = n + (n-2)\cdot{2n-3\choose n-4}.\qedhere\]
\end{proof}

\begin{rmk}\label{rem:freenilp}
Just as in the case of the nilpotency class, in the absence of $1$-formality there can be no uniform bound in terms of $b_1(G)$ for the degree of polynomial growth of $M=G/G''$. Indeed, if we take $G= F_n/\Gamma_k(F_n)$ as in Example~\ref{ex=freenilp}, then $M$ is nilpotent. Combining~(\ref{thm:bass-guivarch}) with the fact that $\phi_q(M) = \theta_q(G) = \theta_q(F_n)$ is strictly positive for $q<k$, we obtain that $d(M)$ can be made arbitrarily large if we let $k\to\infty$.
\end{rmk}

\begin{rmk}\label{rem:vanishing-resonance}
 The only place in the proof of Theorem~\ref{thm:growth-degree} that required $1$-formality is the assertion that the vanishing of the Chen ranks implies the vanishing of $W_q(G)$. If we replace the $1$-formality assumption with the hypothesis that $\mc{R}(G)=\{0\}$ then the vanishing of $W_q(G)$ follows from Theorem~\ref{thm:WG-RG}, and the proof of Theorem~\ref{thm:growth-degree} carries through without any further modifications.
\end{rmk}

\subsection{K\"ahler groups.}\label{kaehlercsoport} Our results can be applied to analyze the nilpotency class of fundamental groups of compact K\"ahler manifolds.
Suppose $X$ is a compact K\"ahler manifold and $G:=\pi_1(X)$, hence $H_1(G,\mathbb Z)\cong H_1(X,\mathbb Z)$. The first thing we observe is that
$\cR(G)\neq 0$ if and only if the $(2,0)$-part of the cup product map
$$\psi_X=\cup_X^{2,0}:\bigwedge^2 H^0(X,\Omega_X^1)\rightarrow H^0(X,\Omega_X^2)$$
is zero on a degenerate element $0\neq \omega_1\wedge \omega_2\in \bigwedge^2 H^0(X,\Omega_X^1)$. This follows because
$\cup_X^{0,2}$ is the conjugate of $\cup_X^{2,0}$, whereas the $(1,1)$-part of the cup-product in the Hodge decomposition
$\cup_X^{1,1}:H^{1,0}(X)\otimes H^{0,1}(X)\rightarrow H^{1,1}(X)$ cannot vanish on decomposable tensors. Following Catanese's
generalization of the Castelnuovo-de Franchis Theorem~\cite{Cat}, it follows that $\cR(X)\neq 0$ if and only if $X$ is fibred onto
a curve $C$ of genus at least~$2$. In this case, we have a surjection $\pi_1(X)\twoheadrightarrow \pi_1(C)$, thus $\pi_1(X)$ is not nilpotent.
We say that in this case $X$ is \emph{fibred}. Since $\pi_1(X)$ is $1$-formal, we obtain as a special case of Theorem~\ref{thm:growth-degree} the following.

\begin{thm}\label{kaehler-csoport}
Suppose that $X$ is a non-fibred compact K\"ahler manifold. If $\pi_1(X)/\pi_1(X)''$ is nilpotent, then its virtual nilpotency class is at most $2q(X)-2$.
\end{thm}

Constructing explicit K\"ahler manifolds with nilpotent fundamental group has proven to be a challenging task. Examples of K\"ahler groups with a non-trivial nilpotent filtration (in fact, of nilpotency class $2$) have been produced first by Campana \cite{Cam}. On the other hand, a K\"ahler group
is either fibred or each of its solvable quotients is virtually nilpotent \cite{Del}. In particular, solvable K\"ahler groups are virtually nilpotent.

\vskip 3pt

A recent example is provided by the Schoen surfaces discovered in \cite{Schoe} and studied further in \cite{CLMR}. They are minimal algebraic surfaces $X$ of general type having
invariants $$q(X)=4, \ p_g(X)=5, \ K_X^2=16, \ h^{1,1}(X)=12.$$
 Remarkably, $p_g(X)=2q(X)-3$, that is, we are in  the divisorial case when $\mbox{Ker}(\psi_X)$ may not contain decomposable elements. That is the case, for Schoen surfaces are not fibred. In fact $\mbox{Ker}(\psi_X)$ is of dimension one, implying $\mbox{dim } \mbox{Ker}(\cup_X)\in \{6,7\}$. It is unknown whether the fundamental group of such an $X$ is nilpotent. Applying Theorem \ref{kaehler-csoport}, we obtain that the virtual nilpotency class of $\pi_1(X)/\pi_1(X)''$ is at most $6=b_1(X)-2$.

\subsection{The Torelli group of the mapping class group.}\label{subsec:torelli}
A particularly interesting class of applications of our results  is provided by the {\em Torelli groups} $T_g$ of genus  $g\ge 2$.
We fix  a closed oriented surface $\Sigma_g$ of genus $g$ and denote by $\pi_g:=\pi_1(\Sigma_g)$ its fundamental group and by $H:=H_1(\Sigma_g,\mathbb Z)$ its first homology. Let $\mathrm{Mod}_g$ be the mapping class group  of isotopy classes of orientation-preserving homeomorphisms of $\Sigma_g$.  If $\mathcal{X}_g$ denotes the Teichm\" uller space of genus $g$, then  $\mathrm{Mod}_g$ is the orbifold
fundamental group of the moduli space $\cM_g$ of smooth curves  of genus $g$,  which can be realized as the quotient $\cM_g\cong \mathcal{X}_g/\mathrm{Mod}_g$.  Nielsen theory offers an alternative description
$\mathrm{Mod}_g\cong \mathrm{Out}^+(\pi_g)$ of the mapping class group as the group of orientation preserving outer isomorphisms of~$\pi_g$. The Torelli group consists of those outer automorphisms of $\pi_g$ that act trivially on~$H$. The \emph{Torelli space}
$$\mathcal{T}_g:=\mathcal{X}_g/T_g$$
can then be thought of as the moduli space of pairs $[C,\alpha_1, \ldots, \alpha_g, \beta_1, \ldots,  \beta_g]$, consisting of a smooth curve $C$ of genus $g$ and a symplectic basis of $H_1(C,\mathbb Z)$.

Johnson \cite{J} defined the surjective homomorphism
$$\tau:T_g\rightarrow \bigwedge^3 H/H,$$
where the $\mathrm{Sp}_{2g}(\mathbb Z)$-equivariant injective map $H\hookrightarrow \bigwedge^3H$ is given by $z\mapsto z\wedge \omega$, where $\omega:=\alpha_1\wedge \beta_1+\cdots +\alpha_g\wedge \beta_g\in \bigwedge^2 H$, with $(\alpha_1, \ldots, \alpha_g,\beta_1, \ldots, \beta_g)$ being a symplectic basis of $H$. Then he proved that $T_g$ is finitely generated and that the \emph{Johnson kernel}
defined by the exact sequence
$$1\longrightarrow K_g\longrightarrow T_g\longrightarrow \bigwedge^3 H/H\longrightarrow 1,$$
is precisely the subgroup of $\mathrm{Mod}_g$ generated by Dehn twists along separated simple closed curves in $\Sigma_g$. Finally, Johnson \cite{J2}  showed that the abelianization of $T_g$ is given (up to $2$-torsion) by the map $\tau$, that is one has an isomorphism
$$\tau_*:H_1(T_g,\mathbb Q)\cong \bigwedge^3 H_{\mathbb Q}/H_{\mathbb Q},$$
which in particular implies that
$$b_1(T_g) = {2g\choose 3}-2g.$$

For  $g\ge 4$, the Torelli group is $1$-formal  \cite{Hai, Hai2} and  $\mathcal{R} (T_g)=\{ 0\}$, see \cite[Theorem~4.4]{DP}. Furthermore, Dimca, Hain and Papadima \cite[Corollary~3.5]{DHP} proved that $B(T_g)_{\mathbb C}=H_1(K_g,\mathbb C)$ is a nilpotent module over $\mathbb C[T_g/T_g']$. We deduce from Corollary~\ref{cor=nilpmod} the following new estimate of nilpotency of the Alexander invariant of $T_g$.

\begin{thm}\label{thm:Alexander-Tg}
 If $g\geq 4$ and $I \subseteq\bb{C}[T_g/T_g']$ denotes the augmentation ideal, then
 \[I^q\cdot B(T_g)_{\mathbb C}=0 \mbox{ for } q\geq {2g\choose 3}-2g-3.\]
\end{thm}

More recently, it was shown in \cite[Corollary 1.5]{EH} that the metabelian quotient $T_g/T_g''$ is nilpotent for $g\geq 12$, and in particular it follows that Theorem~\ref{thm:growth-degree} applies for the Torelli groups. It follows that $\vnc(T_g/T_g'')\leq b_1(T_g)-2$, which is the statement of Theorem~\ref{thmB} in the Introduction. Of course, the groups $T_g$ and $\mathrm{Mod}_g$ are very far from being nilpotent. In fact, if $G\leq \mathrm{Mod}_g$ is a subgroup, either $G$ is virtually abelian, or it contains a non-abelian free subgroup, see \cite{McC}.

\vskip 3pt

\subsection{The Torelli group of a free group.} More generally, for  an arbitrary fintely generated group $G$, we  define its \emph{Torelli group} $T(G)\subseteq \mathrm{Out}(G)$ to be the subgroup of
outer automorphisms of $G$ consisting of automorphisms which induce the identity on $H_1(G, \mathbb{Z})$. The class of right-angled Artin groups, associated to
finite simple graphs, is of considerable interest in geometric group theory. The groups $G$ in this class interpolate between free abelian groups
(corresponding to full graphs) and free groups (corresponding to discrete graphs). Of particular interest is the case when $G=F_g$ is a free group on $g$ generators $x_1, \ldots, x_g$.
The corresponding Torelli group $T(F_g)$ denoted by $\OA_g$, is thus defined by the following exact sequence
$$1\longrightarrow \OA_g\longrightarrow \mbox{Out}(F_g)\longrightarrow GL_g(\mathbb Z)\longrightarrow 1.$$

The Culler-Vogtmann \emph{outer space} $X_g$ of marked metric graphs of rank $g$ can be thought of as the classifying space for $\OA_g$, see \cite{CV}. The moduli space of graphs $X_g/\OA_g$ is (essentially) the moduli space $\cM_g^{\mathrm{tr}}$ of tropical curves of genus $g$, see \cite{CMV} and references therein.
In analogy with $\OA_g$, one defines the subgroup $\IA_g$ of the group $\mathrm{Aut}(F_g)$ of all automorphisms of the free group which induce the identity on $H_1(F_g, \mathbb{Z})$. Both $\IA_g$ and $\OA_g$ are finitely generated for $g\geq 3$ by \cite{Mag}, and there exist natural quotient maps $\IA_g\onto\OA_g$.

\vskip 3pt

It is known that for $g\ge 4$ one has $\mathcal{R} (\OA_g)=\{ 0\}$, see \cite[Theorem~9.7]{PS-johnson}. Moreover, the first Betti number of $\OA_g$ has been computed by Kawazumi \cite{K}, see also \cite[Theorem~9.1]{PS-johnson} and the references therein and is given by
$$ b_1(\OA_g) = \frac{g(g+1)(g-2)}{2}.$$
Note \cite{Mag} that $\IA_g$ is generated by the following set of \emph{Magnus generators} of $\mbox{Aut}(F_g)$
$$\alpha_{ij}: x_i\mapsto x_jx_ix_j^{-1}, \ \ x_{\ell}\mapsto x_{\ell} \ \mbox{for } \ell\neq i, \mbox{ and }$$
$$\alpha_{ijk}: x_i\mapsto x_i\cdot [x_j,x_k], \ \ x_{\ell}\mapsto x_{\ell} \mbox{ for } \ell\neq i,$$
where the indices are subject to the conditions $1\leq i\neq j\leq g$ in the first case, respectively $1\leq j<k\leq g$ and $i\neq j,k$, in the second case. One has
$\frac{g^2(g-1)}{2}=b_1(\IA_g)$ generators, see again \cite{K}. When passing to $\OA_g$, one notices the relation $\alpha_{1,\ell} \cdots  \alpha_{g,\ell}=1$ for $\ell=1, \ldots, g$ and one is left with $b_1(\IA_g)-g=\frac{g(g+1)(g-2)}{2}$ generators, which explains the formula for $b_1(\OA_g)$.

\vskip 3pt

It is shown in \cite[Corollary 1.5]{EH} that the metabelian quotients of $\IA_g$ are nilpotent for $g\geq 4$, so the same is true for the metabelian quotients of $\OA_g$. Applying Theorem~\ref{thmA} with $G=\OA_g$ we obtain the following result.
\begin{thm}\label{iag}
\label{thm=oan}
If $g\geq 4$ then the metabelian quotient $\OA_g/\OA_g''$ has virtual nilpotency class at most $\frac{g(g+1)(g-2)}{2}-2$.
\end{thm}

Using Corollary~\ref{cor=nilpmod} we also obtain an analogue of Theorem~\ref{thm:Alexander-Tg} for the Alexander invariant of $\OA_g$.

config: arxiv


TITLE: Probability of a machine with two switches failing
QUESTION [1 upvotes]: A certain type of switch has a probability p of working when it is turned on. Machine A, which has two of these switches, needs both such switches to be switched on to work, and machine B (also with two of the same switch) only needs at least one of them switched on to work.
How can I calculate the probability that each machine will fail to work?

REPLY [2 votes]: Assuming that the switches fail or do not fail independently the probability of both failing (for machine B) is the product of the probabilities: $p^2$.  The probability that either switch fails (for machine A) is the complement of the probability that neither fails, or $1 - (1 - p)^2 = p(2 - p)$.
So the probability of A failing is $p(2 - p)$ (either switch fails) and the probability of B failing is $p^2$ (both switches fail).  You can check that when $0 < p < 1$ the probability that A fails is greater than the probability that B fails, which is intuitively correct since the condition that makes A fail is "easier" than the condition that makes B fail.

REPLY [1 votes]: Let $ q=1-p $ be the probability of a switch failing and let $ P(n) $ be the probability that, for a given machine, n switches succeed. For machine A to fail, you need to find $ P_a( \leq 1)=P_a(1)+P_a(0) $ because as long as one switches fails, the whole thing does. For machine B to fail, you need $ P_b(0) $ as both switches must fail. Assuming that the success of one switch does not affect the probability for the other, you can invoke the product rule; that is, the probability of an event for one switch and an even for another coinciding is the product of their probabilities. From there, hopefully you can figure out the rest.

REPLY [1 votes]: Prob of both switches on A working: $p^2$
Prob of one of switches on B working = prob NOT both of them fail: $1-(1-p)^2$
so $p^2(1-(1-p)^2)$ is the probability both is working


TITLE: divisors of $p^4+1$ of the form $kp+1$
QUESTION [23 upvotes]: In group theory the number of Sylow $p$-subgroups of a finite group $G$, is of the form $kp+1$.
So it is interesting to discuss about the divisors of  this form. As I checked it seems that for an odd prime $p$, there is not any divisor $a$ of $p^4+1$, where $1<a<p^4+1$ and $a=kp+1$, for some $k>0$.
Could you help me about this question? If it is true how we can prove it?
Also when I checked the same fact for $p^8+1$, we get many counterexamples. What is the difference between $4$ and $8$?

REPLY [37 votes]: There's none indeed.
Lemma: if $1<m<n$ are coprime integers then $mn+1$ does not divide $n^4+1$.
First observe that for any $m,n$, of $mn+1$ divides $n^4+1$, then it divides $n^4m^4+m^4=((nm)^4-1)+1+m^4$, and since $mn+1$ clearly divides $((nm)^4-1)$, we deduce that it also divides $1+m^4$.
To prove the lemma, assume the contrary. As we have just seen, $mn+1$ divides $m^4+1$ as well. Write $(m^4+1)/(mn+1)=k$, and $k=(\ell m+r)$ with $0\le r\le m-1$. Then $m^4+1=(\ell m+r)((mn+1)=mN'+r$, so $r=1$. So $(\ell m+1)$ divides $m^4+1$. Then $m$ and $\ell$ are coprime: indeed, we have $(m\ell +1)(mn+1)=m^4+1$, so $\ell(mn+1)-m^3=-n$. If a prime $p$ were dividing both $m$ and $\ell$ then it would also divide $n$, contradicting that $m$ and $n$ are coprime, so $m$ and $\ell$ are indeed coprime. We have $\ell<n$ because otherwise
$$m^4+1=(mn+1)(m\ell+1)\ge (mn+1)^2> (m^2+1)^2>m^4+1.$$ So we found a new pair $(m,\ell)$ with $\max(m,\ell)<n$, with $m\ell+1$ dividing both $m^4+1$ and $\ell^4+1$. So, assuming that $n$ is minimal, we're done unless $\ell=1$. This happens if $m+1$ divides $m^4+1$, and since $m+1$ divides $m^4-1$ as well, if this occurs then $m+1=2$, contradicting $m>1$.
(Note: without the coprime assumption the conclusion fails, as $(m,n)=(m,m^3)$ for $m\ge 2$, e.g. $(m,n)=(2,8)$, satisfies $mn+1|n^4+1$.)
Proposition If $p$ is prime then $p^4+1$ has no divisor of the form $kp+1$ except $1$ and $p^4+1$.
Proof: write $p^4+1=(kp+1)(k'p+\ell)$ with $0\le\ell\le p-1$; then $\ell=1$. So exchanging $k$ and $k'$ if necessary we can suppose $k\le k'$.
If by contradiction $k$ is divisible by $p$ then $k'\ge k\ge p$, so $p^4+1\ge (p^2+1)^2>p^4+1$, contradiction. So $k$ and $p$ are coprime, and the lemma yields a contradiction.

REPLY [2 votes]: Let $p$ be prime. Assume that $q \mid p^4+1$, where $q = ap+1$ for some
$a \in \mathbb{N}$.
Then $(p^4+1)/q \equiv 1 \!\! \mod p$, hence
$p^4+1 = (ap+1)(bp+1)$ for some $b \in \mathbb{N}$.
Now we have $p^4 = abp^2+(a+b)p$, respectively, $p^3-abp-a-b = 0$. Therefore, effectively
we are looking for solutions to the diophantine equation $x^3-xyz-y-z=0$ in positive
$x$, $y$, $z$, and prime $x$.
When allowing negative (or zero) $x, y, z$, we would get a lot of solutions,
and for positive $x, y, z \leq 1000$, we still find (up to switching of $y$ and $z$)
$$
  (x,y,z) \in \{(8,30,2),(27,240,3),(30,112,8),(112,418,30)\}.
$$
I don't know whether there is a solution with prime $x$ as required --
however we have a diophantine equation of total degree $3$ in $3$ variables,
and such are (at least potentially) hard.


TITLE: Partial Derivative Stickler.
QUESTION [0 upvotes]: I am having trouble with a question with partial derivatives.
Here is the question:

Let $\rho = \sqrt{x^2 + y^2 + z^2}$
Show that $$\frac{\partial ^2\rho}{\partial x^2} + \frac{\partial ^2\rho}{\partial y^2}+\frac{\partial ^2\rho}{\partial z^2} = \frac{2}{\rho}$$

All I am getting is that the LHS is equal to O.
A square root is something to the power of 0.5
$$
\begin{eqnarray}
\rho &=& \left(x^2\right)^{1/2} + \left(y^2\right)^{1/2} + \left(z^2\right)^{1/2}\\
&=&x^1 + y^1 + z^1\\
&=& x + y + z.
\end{eqnarray}
$$
Therefore,
$$
\frac{\partial \rho}{\partial x} = \frac{\partial \rho}{\partial y} = \frac{\partial \rho}{\partial z} = 1
$$
Furthermore,
$$
\frac{\partial^2 \rho}{\partial x^2} = \frac{\partial^2 \rho}{\partial y^2} = \frac{\partial^2 \rho}{\partial z^2} = 0
$$
But this isn't equal to $2/ \rho$.
Any help here is appreciated. Apologies for the bad formatting. I don't know how to make it any nicer.

REPLY [2 votes]: $$
\frac{\partial \rho}{\partial x} = \frac{1}{2}\frac{2x}{\sqrt{x^2+y^2+z^2}} = \frac{x}{\rho} \tag{*}
$$
similarly for the other derivatives
$$
\begin{eqnarray}
\frac{\partial \rho}{\partial y} &=& \frac{y}{\sqrt{x^2+y^2+z^2}} = \frac{y}{\rho},\\
\frac{\partial \rho}{\partial z} &=& \frac{z}{\sqrt{x^2+y^2+z^2}} = \frac{z}{\rho}.
\end{eqnarray}
$$
then taking the second derivative for the x. We use the previous result from Eq. (*) as follows
$$
\frac{\partial^2\rho}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{x}{\rho}\right)
$$
this leads to
$$
\frac{\partial^2\rho}{\partial x^2} = \frac{1}{\rho} - \frac{x}{\rho^2}\frac{x}{\rho} = \frac{1}{\rho} - \frac{x^2}{\rho^3}.
$$
similary we have
$$
\begin{eqnarray}
\frac{\partial^2 \rho}{\partial y^2} &=& \frac{1}{\rho} - \frac{y^2}{\rho^3},\\
\frac{\partial^2 \rho}{\partial z^2} &=& \frac{1}{\rho} - \frac{z^2}{\rho^3}.
\end{eqnarray}
$$
all together
$$
\begin{eqnarray}
\frac{\partial^2\rho}{\partial x^2} + \frac{\partial^2\rho}{\partial y^2} + \frac{\partial^2\rho}{\partial z^2} &=& \frac{3}{\rho} - \frac{x^2+y^2+z^2}{\rho^3}\\
&=& \frac{3}{\rho}-\frac{\rho^2}{\rho^3}\\
&=& \frac{3}{\rho}-\frac{1}{\rho} = \frac{2}{\rho}.
\end{eqnarray}
$$


TITLE: What is the physical basis behind the independence assumption in Bell's theorem?
QUESTION [1 upvotes]: Bell's theorem rests on two main assumptions: locality and independence. Locality has its basis in relativity and it seem to be a property of all known physical theories accepted today (General relativity and the quantum field theory).
The independence assumption (also known as "freedom", "free-will", "no-conspiracy") implies that the hidden variables and the measurement settings are independent variables. This seems intuitively appealing, but is there any reason to accept this assumption?
The modern theories I mentioned above are field theories. In such theories the state of a subsystem is not independent of the state of the entire system. Stars in a galaxy do not move independently of other stars, charged particles do not move independently of other charges even if the distance separating them is large. But the source of the entangled particles and the detectors are nothing but such subsystems of the whole experimental setup.  They are nothing more than large groups of charged particles (electrons and nuclei/quarks). So, it seems to me, that the independence assumption is in fact contradicted, not supported by the mainstream physical theories.

REPLY [3 votes]: Experimental science is based on the idea that what you measure is independent from your measuring device. Or at the very least that it is not a part of the measuring device.
What can you make of any experimental result if each experiment is taken to be a specific expression of the whole (aka the universe), without any sensible division between an observed system and a measuring system?
Without that independence assumption, there is no science. And since researchers over the ages actually did find more or less valid laws of physics, through uncountable experiments, such independence somehow holds. Rejecting it would immediately raise the question of why do experiments actually yield results from which one can infer generic laws and principle governing the behavior of (sub)systems?
It is for good reasons that explaining away the violation of Bell's inequalities by giving up this principle is seen as a "conspiracy theory": it would mean that while the universe appears to follow specific patterns implying the separation of systems, everything is actually connected together in the background so tightly that all that happens is strictly predetermined (one then talks of superdeterminism).
Now perhaps the true nature of things is beyond these distinctions, as in this zen quote:

The blue mountain is the father of the white cloud. The white cloud is
  the son of the blue mountain. All day long they depend on each other,
  without becoming dependent on each other. The white cloud is always
  the white cloud. The blue mountain is always the blue mountain.

But is this what physics is about?