text:
\begin{document}
\title{Periodic String Complexes over String Algebras}

 \author{Andr\'es Franco*, Hern\'an Giraldo*, and Pedro Rizzo*.}

\date{\today}

\begin{abstract}
In this paper we develop combinatorial techniques for the case of string algebras with the aim to give a characterization of string complexes with infinite minimal projective resolution. These complexes will be called \textit{periodic string complexes}. As a consequence of this characterization, we give two important applications. The first one, is a sufficient condition for a string algebra to have infinite global dimension. In the second one, we exhibit a class of indecomposable objects in the derived category for a special case of string algebras. Every construction, concept and consequence in this paper is followed by some illustrative examples.
\end{abstract}
\maketitle

Keywords: String algebras, String Complexes, Derived Categories, Projective resolutions, Infinite global dimension.

\vskip5pt
\noindent

\vskip5pt
\noindent
 * Instituto de Matem\'aticas, Universidad de Antioquia, Medell\'in-Colombia

 e-mail: andres.francol@udea.edu.co, \ hernan.giraldo@udea.edu.co, \ pedro.hernandez@udea.edu.co.

\section{Introduction}
Projective resolutions have played an important role in the development of algebra with many applications in representation theory, ring theory, homological theory and algebraic geometry (\cite{GSZ} and references therein). Many theoretical, combinatorial and computational techniques  have been introduced in order to obtain these resolutions (\cite{FGKK}, \cite{GHZ}). In this paper we develop combinatorial techniques for the case of string algebras with the aim to give a characterization of string complexes with infinite minimal projective resolution. These complexes will be called \textit{periodic string complexes}.

As a consequence of this characterization, we give two important applications. The first one, is a sufficient condition for a string algebra to have infinite global dimension. In the second one, we use the functor in \cite{Be-Me} to prove that string complexes are indecomposable objects in the derived category of a special case of string algebras. In this special case we say that the string algebra satisfies the \textit{unique maximal path property}. Every construction, concept and consequence in this paper is followed by some illustrative examples.

Actually, this work has deep connections with the ideas and consequences in \cite{FGR}. In fact, some results of this paper were announced without proof in \cite{FGR}. However, we consider these results could have independent interest, hence detailed proofs are presented here.

In general, it is not easy to describe the indecomposable objects in the derived category of an algebra. In \cite{Be-Me}, the authors solved this problem when the algebra is Gentle. A recent work in \cite{FGR}, using an adaptation of the constructions in \cite{Be-Me}, the authors prove that string and band complexes are indecomposable objects in the bounded derived category of a new class of algebras: the String Almost Gentle algebras.

On the other hand, P. Bergh, Y. Han and D. Madsen in \cite{Be-Han-Ma} (see also \cite{Ha-Za}) gave a sufficient condition for a monomial algebra to have infinite global dimension. In this paper we will give an analogous sufficient condition for the case of a string algebra using different combinatorial techniques.

This paper is organized as follows. In section \ref{Sec-Pre}, we fix some notations and present some background material about well-known results in derived categories, string algebras and string complexes.

Then, in section \ref{Sec:PSC} we state and prove the main Theorem. We introduce a class of complexes characterized by having infinite minimal projective resolution. These complexes will be called \textit{periodic string complexes}. The main technique is the generalization of the combinatorial sets $Q_c$, $GSt_c$, $GSt^c$ introduced in  \cite{Be-Me} and the proof follows some ideas in \cite{FGR}.

Finally, in section \ref{Sec:App} we introduce and prove two important applications of our main result, as we mentioned above.

\section{Preliminaries}\label{Sec-Pre}

\subsection{About derived categories of modules over finite dimensional algebras}
Let $A$ be a finite-dimensional algebra of the form $kQ/I=k(Q, I)$ over an algebraically closed field $k$, where $I$ is an admissible ideal of $kQ$ and $Q$ is a finite quiver. As usual, we denote by $Q_0$ (resp. by $Q_1$) the set of vertices (resp. the set of arrows) of $Q$. Let $A-\text{mod}$ be the category of finitely generated left $A$--modules.

We denote by $e_i$ the trivial path at vertex $i\in Q_0$ and by $P_i=Ae_i$ the corresponding indecomposable projective $A$--module.  We will denote by $\textbf{Pa}$ the set of all paths of $k(Q,I)$, that is, the set of all paths of $Q$ that are outside $I$ and by $\textbf{Pa}_{\geq l}$ (resp. by $\textbf{Pa}_{> l}$) the set of all paths in $\textbf{Pa}$ of length greater than or equal to a fixed non-negative integer $l$ (resp. greater than $l$). Every element of $A=k(Q,I)$ can be represented uniquely by a linear combination of elements in $\textbf{Pa}$, and hence we can assume that $\textbf{Pa}$ is a basis for $A$.

The set $\textbf{M}$ of \textit{maximal paths} in $\textbf{Pa}$ has a very important role throughout this theory. A path $w=w_1\cdots w_n$ in  $k(Q,I)$ is \textit{maximal} in $k(Q,I)$ if for all arrows $a,b \in Q_1$, we have that  $aw$ and $wb$ are not paths in $k(Q,I)$. Furthermore, a nontrivial path $w$ of $Q$ is in $\textbf{Pa}$ if and only if it is a sub-path of a maximal path  $\widetilde{w}$ that is not in $I$ (that is, of an element of $\textbf{M}$). This maximal path has the form $\widetilde{w}=\hat{w}w\bar{w}$ with $\hat{w}, \bar{w}\in \textbf{Pa}$. For future reference, the path $\hat{w}\in\textbf{Pa}$ is called a \emph{left completion of $w$}.

We denote by $D(A)$ (resp. $D^-(A)$ or $D^b(A)$) the derived category of $A-\text{mod}$ (resp. the derived category of right bounded complexes of $A-\text{mod}$ or the derived category of bounded complexes of $A-\text{mod}$). We denote by $C^b(\text{pro} \ A)$ (resp. $C^-(\text{pro} \ A)$ or $C^{-,b}(\text{pro} \ A)$) the category of bounded projective complexes (resp. of right bounded projective complexes or right bounded projective complexes with bounded cohomology).

Also, we denote by $K^b(\text{pro} \ A)$ (resp. $K^-(\text{pro} \ A)$ or $K^{-,b}(\text{pro} \ A)$) the corresponding homotopy categories to $C^b(\text{pro} \ A)$ (resp. $C^-(\text{pro} \ A)$ or $C^{-,b}(\text{pro} \ A)$).

By $\mathfrak{p}(A)$ we denote the full subcategory of $C^b(\text{pro} \ A)$ defined by the projective complexes such that the image of every differential map is contained in the radical of the corresponding projective module. It is well known that $\mathfrak{p}(A)$ is a Krull-Schmidt category.

Any projective complex is the sum of two complexes: one complex in $\mathfrak{p}(A)$ and another one isomorphic to the zero object in the derived category (because all differential maps are 0's or isomorphisms). Hence, we can assume that all the complexes we deal with are in $\mathfrak{p}(A)$.

Now, since every projective module is a finite direct sum of indecomposable projective modules, each morphism (differential) in the complex $P^{\bullet}$ is given by a block matrix of size $\sum_id_{i,j}\times\sum_id_{i,j+1}$ if the differential goes from the place $j$ to the place $j+1$. Here $d_{i,j}$ represents how many times the indecomposable projective module $P_i$ appears as a direct summand in each $P^j$ in the complex $P^{\bullet}$ and we will use the notation $P_i^{d_{i,j}}$. Thus, each block gives the component of the morphism corresponding to each pair of indecomposables. That is, each block corresponds to a morphism $P_r^{d_{r,j}}\longrightarrow P_s^{d_{s,j+1}}$.

It is well known that the paths $w\in\textbf{Pa}$ such that $s(w)=r$ and $t(w)=s$ form a basis for $\text{Hom}(P_r,P_s)$. In the particular case of the category $\mathfrak{p}(A)$ we can assume that only the paths $w\in\textbf{Pa}_{\geq 1}$ are involved, because trivial paths give isomorphisms.

If $w\in\textbf{Pa}_{\geq 1}$ is one of these paths, it defines the morphism $p(w): P_r\longrightarrow P_s$, given by the multiplication times $w$ on the right: $u\mapsto v=uw$. Thus, any homomorphism from $P_r$ to $P_s$ is associated to a linear combination of paths like  $w$.

An important and useful result, which we will use later, is that $D^b(A)$ is equivalent to $K^{-,b}(\text{pro} \ A)$ (see, \cite{KoZi}, Prop 6.3.1, p.113). In fact, this result allows us to classify the indecomposable objects in $D^b(A)$. Precisely,

\begin{proposition}\label{prop:1}
There exist spectroids $\text{ind} \ D^b(A)$ and $\text{ind} \ \mathfrak{p}(A)$  of $D^b(A)$ and $\mathfrak{p}(A)$, respectively, such that the set of objects of the spectroid satisfies
$$
\text{ind}_0 \ D^b(A)= \text{ind}_0 \ \mathfrak{p}(A) \cup \{\beta(M^\bullet)^\bullet \ | \ M^\bullet \in \mathcal{X}(A) \}.
$$
\end{proposition}
Here, $\mathcal{X}(A)$ is a fixed set of representatives of the quotient of the set
$$
\overline{\mathcal{X}(A)}:=\{M^\bullet\in \text{ind}_0 \ \mathfrak{p}(A) \ | \ P_{\beta(M^\bullet)^\bullet}^\bullet \notin K^b(\text{pro} \ A) \}
$$
by the equivalence relation $\cong_{\mathcal{X}}$. This equivalence relation is defined on $\overline{\mathcal{X}(A)}$ in $K^{-,b}(\text{pro} \ A)$, by
$$
M^\bullet\cong_{\mathcal{X}}N^\bullet \quad \text{if and only if} \quad
P_{\beta(M^\bullet)^\bullet}^\bullet\cong P_{\beta(N^\bullet)^\bullet}^\bullet
$$
where $P_{\beta(M^\bullet)^\bullet}^\bullet$ is the projective resolution of the good truncation $\beta(M^\bullet)^\bullet$ of the complex $M^{\bullet}$. Precisely, if $P^{\bullet}\in C^{b}(\text{pro} \ A)$, $P^{\bullet}\neq 0^{\bullet}$ and if $t$ is the greatest integer such that $P^i=0$ for $i<t$, then $\beta(P^{\bullet})^{\bullet}$, \emph{the good truncation of $P^{\bullet}$ below $t$}, is the complex given by
$$
\beta(P^{\bullet})^i=\left\{ \begin{array}{lcc}
                    P^i,       &   \text{if}  & i\geq t, \\
\ker \partial_{P^{\bullet}}^t, & \text{if}    & i=t-1,\\
                    0,         &              & \text{otherwise.}
\end{array}
\right.
$$
$$
\partial_{\beta(P^{\bullet})^{\bullet}}^i=\left\{ \begin{array}{lcc}
      \partial_{P^{\bullet}}^i,       &   \text{if}  & i\geq t, \\
\iota_{\ker \partial_{P^{\bullet}}^t} & \text{if}    & i=t-1 \\
                           0,         &              & \text{otherwise,}
\end{array}
\right.
$$
where $\iota_{\ker \partial_{P^{\bullet}}^t}$ is the inclusion map. More details in \cite{KoZi}, \cite{Be-Me}.

\begin{remark}\label{Rem:remark1}
If $A$ has finite global dimension, then $\mathcal{X}(A)=\emptyset$ and $\text{ind}_0 \ D^b(A)= \text{ind}_0 \ \mathfrak{p}(A)$.
\end{remark}

\subsection{About string algebras}
For the benefit of the reader we begin this section by recalling the definition of a special biserial algebra, and, in particular, of a string algebra.

\begin{definition}
 The algebra $A=kQ/I$ is called \emph{special biserial} if it satisfies the following conditions:
	\begin{enumerate}
		\item Any vertex of $Q$ is the starting point of at most two arrows. Any vertex of $Q$ is the ending point of at most two arrows.
		\item Given an arrow $a\in Q_1$, there is at most one arrow $b\in Q_1$ with $s(b)=t(a)$ and $ab\notin I$.
		\item Given an arrow $a\in Q_1$, there is at most one arrow $c\in Q_1$ with $t(c)=s(a)$ and $ca\notin I$.
	\end{enumerate}
\end{definition}

It is a well known fact that, for special biserial algebras, $I$  can be generated by zero relations and by commutativity relations. More details in \cite{SW}.

\begin{definition}
 A special biserial algebra $A=kQ/I$ is called a \emph{string algebra} if, additionally, $I$ is generated by zero relations, i.e., by paths of length greater than or equal to 2.
\end{definition}

The first preliminary result of our interest on string algebras, establishes the uniqueness of the left completion of an arrow $a$, denoted by $\hat{a}$.

\begin{lemma}\label{lem-left-completion}
 Let $A=kQ/I$ be a string algebra and let $a\in Q_1$. Then $\hat{a}$ is unique.
\end{lemma}

\begin{proof}
	Suppose there are two different left completions $\hat{a}$ and $\hat{b}$ of $a$.

	$$
	\xymatrix@R=5mm{\ar@{~>}[rrd]^{\hat{a}}&&&\\
	                           &&\ar[r]^a&\\
                    \ar@{~>}[rru]_{\hat{b}}&&           &}
	$$
	Since $\hat{a}a$ and $\hat{b}a$ are elements in $\textbf{Pa}_{>0}$, then we have two compositions that not lie in $I$, which is a contradiction with the definition of a string algebra.
\end{proof}

Our next goal is to study the structure of the kernel of a morphism $p(w)$ for some $w\in\textbf{Pa}_{>0}$. From the definition of a string algebra, we know that there are at most two arrows $a$ and $b$ such that $t(a)=t(b)=s(w)$. We know also that it cannot happen that $aw\neq 0$ and $bw\neq 0$. Thus, at least one of $aw$ or $bw$ must be zero. Suppose, without loss of generality, that $bw=0$. For string algebras it can occur that $aw\neq 0$ but there could be a path $p$ of the form $p=p_1\cdots p_r a$, where $p_1,\dots, p_r\in Q_1$, such that $pw=0$.  Let us denote by $a^*$ the smallest path of this form (in case it exists). Then, $a^*$ is the smallest subpath of $\hat{a}a$ with the property $a^*w=0$.

According to the above notation, we state the next Lemma, which will allow us to study the structure of $\ker p(w)$ for some $w\in\textbf{Pa}_{>0}$, and, consequently, the structure of the minimal projective resolutions of complexes. This result is a generalization of Lemma 5 in \cite{Be-Me} (also of Lemma 28 in \cite{FGR}).

\begin{lemma}[Lemma 43 \cite{FGR}]\label{Lem-St}
 Let $A=kQ/I$ be a string algebra and let $w\in\emph{\textbf{Pa}}_{>0}$. Then the general structure of the kernel of $p(w)$ is $\ker p(w)=Aa^*\oplus Ab$.
\end{lemma}

\begin{proof}[Proof]
Here we consider the general case in which $a^*w=0$ and $bw=0$ (we do not exclude the possibility that $a^*=a$). It is clear that $Aa^*\oplus Ab \subseteq \ker p(w)$. For the other inclusion, let $u$ be an element of $\ker p(w)$. Then, since $A$ is a string algebra $u$ must have either $a$ or $b$ as its last arrow. In the latter case $u\in Ab$. In the former case, by the minimality of $a^*$ we have that $a^*$ is a subpath of $u$ and hence $u\in Aa^*$ (here we are using the fact that both $u$ and $a^*$ are subpaths of $\hat{a}a$, which is unique as a consequence of Lemma \ref{lem-left-completion}).
\end{proof}

\begin{remark}
Two important examples of string algebras are gentle algebras \cite{Be-Me} and string almost gentle (briefly, SAG) algebras \cite{FGR}. Is it important to notice that there exist fundamental differences with these latter cases. The first one, is related with the smallest paths in Lemma \ref{Lem-St}, which are arrows themselves. Precisely, for Gentle algebras, if there are two arrows $a$ and $b$ ending at the vertex $s(w)$, then, necessarily, $bw=0$, $aw\neq 0$ and, so $b^*=b$. For SAG algebras it could happen that $aw=0$ and $bw=0$, and hence $a^*=a$ and $b^*=b$. The second one, is that for string algebras it could happen that different maximal paths have common arrows. However, as the following example (\ref{Ex1:2}) shows, there are string algebras, which are neither gentle nor SAG algebras, that satisfy the unique maximal path property. Recall that this property is automatically satisfied by all gentle and SAG algebras (see \cite{Be-Me} and \cite{FGR}, respectively)
\begin{example}\label{Ex:Str:1}
\begin{enumerate}
\item\label{Ex1:1} Consider the bound quiver $(Q, I)$ where
$$
Q: \xymatrix{1\ar[r]^a&2\ar[r]^b&3\ar[r]^c&4}
$$
and $I=\langle abc\rangle$. Then $A=kQ/I$ is a string algebra which is neither, a gentle algebra nor a SAG algebra. In this case the set of maximal paths is $\emph{\textbf{M}}=\{ab, bc\}$. Thus, the arrow $b$ belongs to two different maximal paths, and so maximal paths are not unique for a given arrow.
\item\label{Ex1:2} Consider the bound quiver $(Q, I)$
$$
\xymatrix@C=4.5mm{ & 3\ar[dl]_c&                &&&6\ar[ld]_{c'}\\
		                              1\ar[rr]_a&& 2\ar[rr]_d\ar[ul]_b&&4\ar[rr]_{a'}&&5\ar[ul]_{b'}}
$$
where $I=\langle abc, ad, da', a'b'c'\rangle$. Then $A=kQ/I$ is a string algebra over $k$ that is neither, gentle nor SAG. The set of maximal paths is $\emph{\textbf{M}}=\{bcab, d, b'c'a'b'\}$. Thus, every arrow belongs to a unique maximal path.
\end{enumerate}
\end{example}
\end{remark}

\subsection{String complexes}
 The {\it string complexes} were introduced by Bekkert and Merklen in \cite{Be-Me}, which are associated to {\it generalized strings}. As we will see, some of these complexes in $D^b(A)$ constitute a class having infinite minimal projective resolution, when $A$ is a string algebra.

Let $A=kQ/I$ be a string algebra. Recall that, for each arrow $a\in Q_1$, we denote by $a^{-1}$  its \textit{formal inverse}, which verifies  $s(a^{-1})=t(a)$, $t(a^{-1})=s(a)$ and $(a^{-1})^{-1}=a$. Similarly, if $w=a_1\cdots a_n$ is a path in $Q$, the \textit{inverse path} of $w$ is given by $w^{-1}=a_n^{-1}\cdots a_1^{-1}$.  Then $s(a_i^{-1})=t(a_i)=s(a_{i+1})=t(a_{i+1}^{-1})$ and it is clear that $s(w^{-1})=t(w)$ and $t(w^{-1})=s(w)$.

Now, a \textit{walk} $\omega$ (resp. a \textit{generalized walk}) of length $n>0$ is a sequence $w_1\cdots w_n$, where each $w_i$ is either of the form $w$ or $w^{-1}$, $w$ being an arrow (resp. a path of positive length) and such that $t(w_i)=s(w_{i+1})$ for $i=1,\dots, n-1$. It is clear that $s(\omega)=s(w_1)$ and $t(\omega)=t(w_n)$. The notion of the \textit{inverse} of a walk (resp. of a generalized walk) is defined analogously as that for a path. Thus, the passage to inverses is an involutory transformation.

A \textit{closed walk} (resp. a \textit{closed generalized walk}) is a walk $\omega$ (resp. a generalized walk) such that $t(\omega)=s(\omega)$. In this case we consider its \textit{rotations} (or \textit{cyclic permutations}), denoted by $\omega[j]$, which are given by $\omega[j]=w_{j+1}\cdots w_nw_1\cdots w_j$, for $j=1,\dots,n-1$.

The \textit{product} or \textit{concatenation} of two walks (resp. of two generalized walks) $\omega=w_1\cdots w_n$ and $\omega'=w_1'\cdots w_n'$ is defined as the walk (resp. generalized walk) $\omega\omega'=w_1\cdots w_nw_1'\cdots w_n'$, whenever $t(w_n)=s(w_1')$.

We consider the following equivalence relation on the set of generalized walks, denoted by $\cong_s$: If $\upsilon$ and $\omega$ are two generalized walks, then
$$
\upsilon\cong_s \omega\quad \text{if and only if}\quad \upsilon=\omega \ \text{or} \ \upsilon=\omega^{-1}.
$$

\begin{definition}
A \emph{string} is a walk $\omega=w_1\cdots w_n$ such that $w_{i+1}\neq w_i^{-1}$ for $1\leq i <n$ and such that no sub-word of $\omega$ or $\omega^{-1}$ is in $I$. The set of all strings in $(Q, I)$ will be denoted by $St$.
\end{definition}

\begin{example}
Consider the quiver $$Q: \ \xymatrix{1 \ar@<0.7ex>[r]^{a} \ar@<-0.5ex>[r]_c&2\ar@<0.7ex>[r]^{b} \ar@<-0.5ex>[r]_d &3}$$ with $I=\langle ad, cb\rangle$. We have that $\omega=abd^{-1}c^{-1}$ is a string but $\omega'=abb^{-1}c^{-1}$ and $\upsilon=abd^{-1}c^{-1}ad$ are not.
\end{example}

Now, we denote by $\overline{GSt}$ the set of all generalized walks $\omega=w_1\cdot w_2\cdots w_n$ satisfying

\begin{itemize}
	\item If $w_i, w_{i+1} \in \textbf{Pa}_{>0}$, then $w_iw_{i+1}\in I$.
	\item If $w_i^{-1}, w_{i+1}^{-1}\in \textbf{Pa}_{>0}$, then $w_{i+1}^{-1}w_i^{-1}\in I$.
	\item If $w_i, w_{i+1}^{-1} \in \textbf{Pa}_{>0}$ or $w_i^{-1}, w_{i+1} \in \textbf{Pa}_{>0}$, then $w_iw_{i+1}\in St$.
\end{itemize}

We use the notation $GSt$ for a fixed set of representatives of the quotient of $\overline{GSt}$ over the equivalence relation $\cong_s$ together with all trivial paths. The elements of $GSt$ are called \textit{generalized strings}.

\begin{example}\label{Ex:Gst}
Consider the string algebra $A=kQ/I$ given by the quiver
$$
Q: \ \xymatrix{1 \ar@<0.7ex>[r]^{a} \ar@<-0.5ex>[r]_c&2\ar@<0.7ex>[r]^{b} \ar@<-0.5ex>[r]_d &3}
$$
with $I=\langle ab, cd\rangle$. Then $\omega=a^{-1}\cdot c\cdot d\cdot (cb)^{-1}$ is a generalized string.
\end{example}

\begin{remark}
Notice that in example \ref{Ex:Gst} we use a different notation in order to distinguish  generalized strings from strings. Thus, we use a dot $\cdot$ between $w_i$ and $w_{i+1}$ in a generalized string. For instance, in the quiver  $Q: \ \xymatrix{2 \ar[r]^{a} & 1 & 3\ar[l]_{b}}$ we have that $ab^{-1}$ is a string and $a\cdot b^{-1}$ is a generalized string.
\end{remark}

Now, for a generalized walk $\omega=w_1\cdot w_2\cdot\cdots\cdot w_n$ we introduce the function $\mu_{\omega}: \{0,1,\ldots, n\}\longrightarrow\mathbb{Z}$, defined by
$\mu_{\omega}(0):=0$ and
$$\mu_{\omega}(i):=\left\{ \begin{array}{ll}
\mu_{\omega}(i-1)+1 ,  &  \text{if} \ \ w_i\in\textbf{Pa}_{>0}, \\
\mu_{\omega}(i-1)-1 ,  &  \text{if} \ \ w_i^{-1}\in\textbf{Pa}_{>0}.
\end{array}
\right.$$

Thus, we set $\mu(\omega):=\min_{1\leq i\leq n}\{\mu_{\omega}(i)\}$. In example \ref{Ex:Gst}, for $\omega=a^{-1}\cdot c\cdot d\cdot (cb)^{-1}$, we have $\mu_{\omega}(0)=0, \mu_{\omega}(1)=-1, \mu_{\omega}(2)=0, \mu_{\omega}(3)=1, \mu_{\omega}(4)=0$, and hence $\mu(\omega)=-1$.

\begin{remark}
	In practice we will assume in general that $\mu_{\omega}(0)\leq \mu_{\omega}(n)$. This is possible because if $\mu_{\omega}(0)\geq\mu_{\omega}(n)$, then $\mu_{\omega^{-1}}(0)\leq\mu_{\omega^{-1}}(n)$ and $\omega^{-1}\cong_s \omega$. For instance, in the quiver of the previous example, for $\omega=b^{-1}\cdot a^{-1}$ we have that $\mu_{\omega}(0)=, \mu_{\omega}(1)=-1, \mu_{\omega}(2)=-2$ and for $\omega^{-1}=a\cdot b$, $\mu_{\omega^{-1}}(0)=0, \mu_{\omega^{-1}}(1)=1, \mu_{\omega^{-1}}(2)=2$.
\end{remark}

Now, for every generalized string, we will associate a finite projective complex called a \textit{string complex}.

\begin{definition}\label{def-st-comp}
Let  $\omega=w_1\cdot w_2\cdots w_n$ be a generalized string. Then  $P_{\omega}^\bullet$ is the projective complex $\xymatrix{\cdots P_{\omega}^i\ar[r]^{\partial_w^i}&P_{\omega}^{i+1}\cdots}$ defined as follows. The modules are given by
\begin{equation}\label{def-st-comp-mod}
P_{\omega}^i=\displaystyle\bigoplus_{j=0}^n\delta(\mu_{\omega}(j),i)P_{c(j)}
\end{equation}
where $c(0)=s(w_1)$, $c(j)=t(w_j)$ for $j>0$ and $\delta$ is the Kronecker delta.

	The differential maps are given by $\partial_{\omega}^i=\left(\partial_{jk}^i\right)_{1\leq j,k\leq n}$, where
	\begin{equation}\label{def-string-complex-differentials}
	\partial_{jk}^i:=\left\{ \begin{array}{ll}
	p(w_{j+1}) ,  &  \text{if} \ \ w_{j+1}\in\emph{\textbf{Pa}}_{>0}, \mu_{\omega}(j)=i \ \text{and} \ k=j+1, \\
	p(w_j^{-1}),  &  \text{if} \ \ w_j^{-1}\in\emph{\textbf{Pa}}_{>0} , \mu_{\omega}(j)=i \ \text{and} \ k=j-1, \\
	0,                &  \text{otherwise}.
	\end{array}
	\right.
	\end{equation}
	Also, for each trivial generalized string $e_i^{\pm 1}$, let us denote by $P_{e_i^{\pm} 1}^\bullet$, the following projective complex
	$$\xymatrix{\cdots\ar[r]&0\ar[r]&P^0=P_i\ar[r]^(0.7){\partial^0}&0\ar[r]&\cdots}$$
\end{definition}
\begin{example}
	Let $A=kQ/I$ be the string algebra given by the bound quiver $$Q: \ \xymatrix{1 \ar@<0.7ex>[r]^{a} \ar@<-0.5ex>[r]_c&2\ar@<0.7ex>[r]^{b} \ar@<-0.5ex>[r]_d &3}$$ with $I=\langle ab, cd\rangle$ and let $\omega$ be the generalized string $\omega=a^{-1}\cdot c\cdot d \cdot (cb)^{-1}=w_1\cdot w_2\cdot w_3\cdot w_4$. Then, as we have seen above, $\mu_{\omega}(0)=0, \mu_{\omega}(1)=-1, \mu_{\omega}(2)=0, \mu_{\omega}(3)=1, \mu_{\omega}(4)=0$. Thus, from \eqref{def-st-comp-mod} we have
	\begin{itemize}
		\item For $i=-1$, we have $P_{\omega}^{-1}=\displaystyle\bigoplus_{j=0}^4\delta(\mu_{\omega}(j),-1)P_{c(j)}=P_{c(1)}=P_{t(w_1)}=P_1$.
		\item For $i=0$, we have $P_{\omega}^{0}=\displaystyle\bigoplus_{j=0}^4\delta(\mu_{\omega}(j),0)P_{c(j)}=P_{c(0)}\oplus P_{c(2)}\oplus P_{c(4)}=P_2\oplus P_2\oplus P_1$.
		\item For $i=1$, we have $P_{\omega}^{1}=\displaystyle\bigoplus_{j=0}^4\delta(\mu_{\omega}(j),1)P_{c(j)}=P_{c(3)}=P_{t(w_3)}=P_3.$
	\end{itemize}
Now, let us calculate de differential maps. According to \eqref{def-string-complex-differentials},
$$\partial_{\omega}^{-1}=\left(\partial_{jk}^{-1}\right)_{0\leq j,k\leq 4}=\begin{pmatrix}
0                  & 0 & 0                 & 0  & 0\\
\boxed{p(a)}& 0 & \boxed{p(c)}& 0  & \boxed{0}\\
0                  & 0 & 0                & 0  & 0\\
0                  & 0 & 0                & 0  & 0\\
0                  & 0 & 0                & 0  & 0
\end{pmatrix}\longleftrightarrow\begin{pmatrix}
p(a) & p(c) & 0
\end{pmatrix}$$
and
$$\partial_{\omega}^{0}=\left(\partial_{jk}^{0}\right)_{0\leq j,k\leq 4}=\begin{pmatrix}
0      & 0 & 0    & \boxed{0}       & 0\\
0      & 0 & 0    & 0                    & 0\\
0      & 0 & 0    & \boxed{p(d) } & 0\\
0      & 0 & 0    & 0                    & 0\\
0      & 0 & 0    & \boxed{p(cb)} & 0
\end{pmatrix}\longleftrightarrow\begin{pmatrix}
0 \\ p(d) \\ p(cb)
\end{pmatrix}$$

Hence, the string complex associated to $w$ is
$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r] & 0\ar[r] & P_1\ar[r]^(0.3){\partial_{\omega}^{-1}}&P_2\oplus P_2\oplus P_1\ar[r]^(0.7){\partial_{\omega}^0}&P_3\ar[r]&0\ar[r]&\cdots}$$

We observe that there is a ``graphic way'' to construct this complex, as follows. We can represent the generalized string $\omega=a^{-1}\cdot c\cdot d \cdot (cb)^{-1}$ as
$$
\xymatrix{1\ar[rrd]_c&&2\ar[ll]_{a^{-1}}&&\\
                               &&2\ar[rr]^d        &&3\ar[dll]^{(cb)^{-1}}\\
                               &&1                     &&             }
$$
In the figure above, we put the vertices $s(w_1)=2$, $t(w_1)=1$, $t(w_2)=2$, $t(w_3)=3$ and $t(w_4)=1$ sequentially, such that we go to the left if $w_i$ is an inverse path and we move to the right if $w_i$ is a direct path. By replacing the vertex $i$ by the projective $P_i$ and writing the corresponding morphisms induced between these projective modules ($p(w_i): P_{s(w_i)}\longrightarrow P_{t(w_i)}$ if $w_i\in\textbf{Pa}_{\geq 1}$ or $p(w_i^{-1}): P_{t(w_i)}\longrightarrow P_{s(w_i)}$ if $w_i^{-1}\in\textbf{Pa}_{\geq 1}$), we obtain

$$
\xymatrix{P_1\ar[rr]^{p(a)}\ar[rrd]_{p(c)}&&P_2&&\\
	&&P_2\ar[rr]^{p(d)}        &&P_3\\
	&&P_1\ar[urr]_{p(cb)}                     &&             }
$$

This diagram represents the complex (we form direct sums by columns)

$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r] & 0\ar[r] & P_1\ar[r]^(0.3){\partial_{\omega}^{-1}}&P_2\oplus P_2\oplus P_1\ar[r]^(0.7){\partial_{\omega}^0}&P_3\ar[r]&0\ar[r]&\cdots}$$
\end{example}

\section{Periodic string complexes}\label{Sec:PSC}
In this section we will give a necessary and sufficient condition for a string complex to have infinite minimal projective resolution when the algebra is string. These complexes are called \textit{periodic string complexes.} We give a generalization of the cyclic sets $Q_c$, $\overline{GSt}_c$, $\overline{GSt}^c$ in \cite{Be-Me}, in order to characterize string complexes with infinite minimal projective resolution.

Let us begin by defining a condition of minimality on paths in the kernel of a morphism $p(w)$. Notice that this condition is automatic for the cases of gentle and SAG algebras, in which the minimal generators of kernels are arrows.

\begin{definition}
	Let $w, u\in\emph{\textbf{Pa}}_{>0}$ with $t(w)=s(u)$ and such that $wu=0$. We say that $w$ is \textit{minimal for $u$}  if no proper subpath $w'$ of $w$ with $t(w')=s(u)$ verifies $w'u=0$.
\end{definition}
Thus, $a^*$ in the notation of the Lemma \ref{Lem-St} satisfies this minimality condition, that is, $a^*$ is minimal for $w$.

Let us define a new set, which generalizes the sets of cyclic arrows $Q_c$ in \cite{Be-Me} and $Q_c^*$ in \cite{FGR}. Let $\textbf{Pa}_c$ be the set of paths $w\in\textbf{Pa}_{>0}$ for which there exist paths $w_m, w_{m-1}, \dots, w_1\in \textbf{Pa}_{>0}$ such that $t(w_m)=s(w)$, $t(w_i)=s(w_{i+1})$ for $i=1,\dots, m-1$, $s(w_1)=t(w_j)$ for some $1\leq j\leq m+1$, where $w_{m+1}=w$, with $w_iw_{i+1}=w_jw_1=0$. In addition we require that $w_i$ is minimal for $w_{i+1}$ and $w_j$ is minimal for $w_1$.

It is clear that $Q_c\subseteq Q_c^*\subseteq \textbf{Pa}_c$, and, in the case of gentle algebras (resp. SAG algebras), we have $Q_c=\textbf{Pa}_c$ (resp. $Q_c^*=\textbf{Pa}_c$). The elements of $\textbf{Pa}_c$ are called \textit{cyclic paths}.

\begin{example}\label{Ex:Str:2}
Let $(Q, I)$ be the bound quiver
	$$ \xymatrix@C=4.5mm{              & 3\ar[dl]_c&                &\\
		1\ar[rr]_a&                & 2\ar[ul]_b&}
	$$
	with $I=\langle ca, abc\rangle$. Then $A=kQ/I$ is a string algebra over $k$, which is neither gentle nor SAG algebra. In this case we have that $Q_c$ and $Q_c^*$ are empty sets but $\emph{\textbf{Pa}}_c=\{a,bc,c,ab\}$.

	Now, consider the generalized string $\omega=w_1=bc$. Then, $l(\omega)=1>0$, $\mu(\omega)=0$ and $\exists a\in\emph{\textbf{Pa}}_c$ such that $a\cdot \omega=a\cdot bc\in GSt$. The complex $P_{\omega}^\bullet$ is
	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_2\ar[r]^{p(bc)}&P_1\ar[r]&0\ar[r]&\cdots}$$

	Since $abc=0$ and $ca=0$, then $\ker p(bc)=Aa$, $\ker p(a)=Ac$, $\ker p(c)=Aab$ and $\ker p(ab)=Ac$. Thus, the (minimal) projective resolution $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet$ is

	$$
	\xymatrix{\cdots\ar[r]&P_1\ar[r]^{p(ab)}&P_3\ar[r]^{p(c)}&P_1\ar[r]^{p(ab)}&P_3\ar[r]^{p(c)}&P_1\ar[r]^{p(a)}&P_2\ar[r]^{p(bc)}&P_1\ar[r]&0}
	$$

	Hence $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$.

   $$
   \xymatrix{  &            & \ar@{~>}[rr]^{bc} &  &\\
                & \ar[ru]^a \ar@{~>}@/^{4mm}/[rd]^a  &                         &   & \\
                 \ar@[]+<0.4cm,0.2cm>; [ur]+<-0.01cm,-0.3mm>^c&    & \ar@{~>}@/^{5mm}/[]+<0.05mm, 0.2mm>;[ll] +<0.4cm,0.1cm>^b     &   &  \\
                            &    &                           &   &}
   $$
\end{example}
In this example we observe that $a$ is minimal for $bc$ and when we calculate the projective resolution of $P_{\omega}^\bullet$ we get the cycle $\{c, ab\}$, which is a subset of $\textbf{Pa}_c$. The consequence of this is that $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$. This motivates one of the conditions in the following generalization of the special sets.

We denote by $\overline{GSt}_{cp}$ the set of generalized strings $\omega=w_1\cdot w_2\cdots w_n$ of positive length such that $\mu(\omega)=0$, there exists $w\in\textbf{Pa}_c$  with $w\cdot\omega\in GSt$  or $w\in\ker p(w_{l}^{-1})\cap \ker p(w_{l+1})$ for some even index $l> 0$ with $\mu_{\omega}(l)=0$,  where $w$ is minimal for $w_1$ or minimal for $w_{l}^{-1}$ and $w_{l+1}$, respectively.

Also, we denote by $\overline{GSt}^{cp}$ the set of generalized strings $\omega=w_1\cdot w_2\cdots w_n$ of positive length such that $\mu(\omega)=\mu_{\omega}(n)$ and there exists $w\in \textbf{Pa}_c$ such that $\omega\cdot w^{-1}\in GSt$.

Thus, we have that these sets reduce to the corresponding one for the case of string almost gentle algebras in \cite{FGR} (resp. gentle algebras \cite{Be-Me}).

Using this new special sets we can generalize the characterization of periodic string complexes for SAG algebras given in Theorem 31, part 2 in \cite{FGR}. We will divide this characterization into several lemmas here below.

\begin{lemma}\label{Lemma7-n=1-converse}
	Let $\omega=w_1$ be a generalized string. If $\omega\in \overline{GSt}_{cp}$ or $\omega\in \overline{GSt}^{cp}$, then $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$.
\end{lemma}
\begin{proof} We will consider only the case $\omega\in \overline{GSt}_{cp}$. The other case is similar.

	If $\omega=w_1\in \overline{GSt}_{cp}$, then there exists $w\in\textbf{Pa}_c$ such that $w\cdot\omega\in GSt$ with $w$ minimal for $w_1$ and $\mu(\omega)=0$.

	Since $\mu(\omega)=0$, then $w_1\in\textbf{Pa}_{>0}$ and the condition $w\cdot \omega\in GSt$ implies $ww_1=0$. Now, since $A=k(Q,I)$ is a string algebra, there are at most two arrows $a$ and $b$ ending at vertex $s(w_1)$.

	$$\xymatrix{            &\ar[rd]^{b}&                             &&& \\
		                           &                &\ar@{~>}[rrr]^{w_1}&&& \\
	                               &\ar[ru]_a   &                                      \\
   \ar@{~}[rruu]^(0.4)w&                &                             &&& }$$

   We know that it cannot occur that $aw_1\neq 0$ and $bw_1\neq 0$. Suppose, without loss of generality, that $bw_1=0$ and $w$ is a subpath of $\hat{a}a$. (Recall that $\hat{a}$ is unique for string algebras as a consequence of Lemma \ref{lem-left-completion}, and so $a^*=w$. Also, it is possible that $w=a$). Therefore, by Lemma \ref{Lem-St}, it follows that $\ker p(w_1)=Aw \oplus Ab$. (If $bw_1\neq 0$, then $aw_1=0$, $w=a$ and $\ker p(w_1)=Aa$, as in the gentle case. However, we consider here the more general case $\ker p(w_1)=Aw\oplus Ab$).

   We have $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$. Then the complex $P_{\omega}^\bullet$ is

   $$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{s(w_1)}\ar[r]^{p(w_1)}&P_{t(w_1)}\ar[r]&0\ar[r]&\cdots}$$

   In order to determine $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet$, the projective resolution of $P_{\omega}^\bullet$, we first note that, since $$p(w): P_{s(w)}\longrightarrow Aw \quad \text{and} \quad p(b): P_{s(b)}\longrightarrow Ab$$
   are projective covers, then $\ker p(w_1)=Aw\oplus Ab$ can be covered with $P^{-1}=P_{s(w)}\oplus P_{s(b)}$ and the epimorphism
   $$
   \begin{array}{rcl}
   \partial^{-1}: P_{s(w)}\oplus P_{s(b)} & \longrightarrow & Aw\oplus Ab\\
                                        (u,v)           & \mapsto             & (uw,vb)
   \end{array}
   $$
   where $\partial^{-1}=\begin{pmatrix}
   p(w)  & 0 \\ 0  &  p(b)
   \end{pmatrix}$. Besides, $\ker \partial^{-1}=\ker p(w)\oplus \ker p(b)$.
   Since $w\in\textbf{Pa}_c$, by Lemma \ref{Lem-St} we have $\ker p(w)=Au_m\oplus Ac$, where $c$ is an arrow such that $t(c)=s(w)$ and $u_m$ is a cyclic path minimal for $w$. Therefore $\ker\partial^{-1}=Au_m\oplus Ac\oplus\ker p(b)$. Furthermore, $\ker p(b)$ is at most two-dimensional, i.e., $\ker p(b)=Ab_1^*\oplus Ab_2$, for some arrows $b_1, b_2$ ending at $s(b)$ and $b_1^*$ is the smallest subpath of $\widehat{b_1}b_1$ such that $b_1^*b=0$ and $b_2b=0$ (recall that this is the general case in Lemma \ref{Lem-St}).

   $$\xymatrix{&       \ar[rd]^{b_2}     &               &                                                            &&&\\
  &                                                 	&\ar[rd]^{b}&                                                           &&& \\
   &                      	\ar[ru]_{b_1}     &                &\ar@{~>}[rrr]^{w_1}                              &&& \\
  &                                               	&\ar[ru]_a   &                                                                     \\
   &	\ar@{~}[rruu]^(0.4)w                  &                &                                                             &&& \\
                        \ar[ru]^c	&               & & \ar@{~>}[llu]^{u_m}                              &\ar@{.}[l]&&}$$
   Thus, $\ker p(b)$ can be covered by $P_{s(b_1^*)}\oplus P_{s(b_2)}$ and the corresponding epimorphisms. Hence, we can cover $\ker\partial^{-1}=Au_m\oplus Ac\oplus Ab_1^*\oplus Ab_2$ with $P^{-2}=P_{s(u_m)}\oplus P_{s(c)}\oplus P_{s(b_1^*)}\oplus P_{s(b_2)}$ and the epimorphism
   $$\partial^{-2}=\begin{pmatrix}
   p(u_m) & 0      & 0            & 0 \\
   0          & p(c) & 0            & 0 \\
   0          & 0     & p(b_1^*) & 0 \\
   0          & 0     & 0            & p(b_2)
   \end{pmatrix}$$
   where $\ker \partial^{-2}=\ker p(u_m)\oplus\ker p(c)\oplus\ker p(b_1^*)\oplus\ker p(b_2)$. Again, since $w\in\textbf{Pa}_c$,  there is a cyclic path $u_{m-1}$, minimal for $u_m$, such that $u_{m-1}u_m=0$ and thus $Au_{m-1}$ is a direct summand of $\ker \partial^{-2}$. Hence $\ker \partial^{-2}$ can be covered by some projective $P^{-3}$ which has $P_{s(u_{m-1})}$ as a direct summand.

   If we continue in this fashion, by the definition of the cyclic set of paths $\textbf{Pa}_c$, in finitely many steps we will obtain again $P_{s(u_j)}$, for some $1\leq j\leq m+1$, as a direct summand in the projective resolution $P_{\beta(P_{\omega  }^\bullet)^\bullet}^\bullet$ and, since $Q_1$ is finite, eventually the other summands will disappear.  We can repeat the process over the subset $\{u_1, u_2, \dots, u_j\}$ of the cyclic set $\textbf{Pa}_c$. This shows that $$P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$$
\end{proof}

\begin{lemma}\label{Lemma7-n=2-converse}
Let $\omega=w_1\cdot w_2$ be a generalized string.	If $\omega\in \overline{GSt}_{cp}$ or $\omega\in \overline{GSt}^{cp}$, then $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$
\end{lemma}

\begin{proof}
	As before, we only consider the case $\omega\in \overline{GSt}_{cp}$.
	If $\omega=w_1\cdot w_2\in \overline{GSt}_{cp}$, then $\mu(\omega)=0$ and there exists $w\in\textbf{Pa}_c$ such that $w\cdot\omega=w\cdot w_1\cdot w_2\in GSt$ where $w$ is minimal for $w_1$.

Now, for $w_2$ we have two possibilities: $w_2\in\textbf{Pa}_{>0}$ or $w_{2}^{-1}\in\textbf{Pa}_{>0}$.\\

	\textit{Case 1} : if $w_2\in\textbf{Pa}_{>0}$, then $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$, $\mu_{\omega}(2)=2$ and $w_1w_2\in I$. The complex $P_{\omega}^\bullet$ is

	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{s(w_1)}\ar[r]^(0.5){p(w_1)}&P_{t(w_1)}\ar[r]^{p(w_2)}&P_{t(w_2)}\ar[r]&0\ar[r]&\cdots}$$
	As in Lemma \ref{Lemma7-n=1-converse}, $\ker p(w_1)=Aw\oplus Ab$ has an infinite (minimal) projective resolution (because $w\in \textbf{Pa}_c$) and hence
	$$P_{\beta(P_{\omega }^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$$

	\textit{Case 2}: if $w_{2}^{-1}\in\textbf{Pa}_{>0}$, then $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$, $\mu_{\omega}(2)=0$ and $w_1w_2\in St$. The complex $P_{\omega}^\bullet$ is

	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{s(w_1)}\oplus P_{t(w_2)}\ar[r]^(0.6){\partial_{\omega}^0}&P_{t(w_1)}\ar[r]&0\ar[r]&\cdots}$$ where $\partial_{\omega}^0=\begin{pmatrix}
	p(w_1) \\ p(w_2^{-1})
	\end{pmatrix}$.

	We first determine $\ker \partial_{\omega}^0$. For this, $\begin{pmatrix}
	u & v
	\end{pmatrix}\in\ker \partial_{\omega}^0$ if and only if $uw_1+vw_{2}^{-1}=0$ with $u\in P_{s(w_1)}$ and $v\in P_{t(w_2)}$.

	Since $A=k(Q,I)$ is a string algebra, there are at most two arrows $a$ and $b$ such that $t(a)=t(b)=s(w_1)$ and at most two arrows $c$ and $d$ such that $t(c)=t(d)=t(w_2)=s(w_{2}^{-1})$.
$$
\xymatrix@R=4.5mm{ \ar[rd]^{b}               &                                &&&\\
                                                &\ar@{~>}[rrr]^{w_1} &&&\ar@{~>}[ddlll]^{w_{2}} \\
                 \ar[ru]_a\ar[rd]^{d} &                                &&&\\
                                                &                                &&&\\
                  \ar[ru]_c                &                                &&&}
$$
Notice that it is not possible that $aw_1\neq 0$ and $bw_1\neq 0$ (similarly for $c$ and $d$). Suppose, without loss of generality, that $bw_1=0$, $cw_{2}^{-1}=0$ and $w$ is a subpath of $\hat{a}a$. Since $\omega\in \overline{GSt}_{cp}$, then $ww_1=0$ and therefore $w=a^*$. Thus $\ker p(w_1)=Aw\oplus Ab$. (If $bw_1\neq 0$, then $aw_1=0$, $w=a=a^*$ and $\ker p(w_1)=Aa$). Similarly, in the more general case we have that $\ker p(w_{2}^{-1})$ is of the form $Ac\oplus Ad^*$, where it is possible that $d^*=d$.

Let $\{e_{s(w_1)}, u_1,\dots, u_s\}$ and $\{e_{t(w_2)}, v_1,\dots, v_t\}$ be bases of $P_{s(w_1)}$ and $P_{t(w_2)}$, respectively. Then
\begin{eqnarray*}
	u & = & \alpha_0e_{s(w_1)}+\alpha_1u_1+\cdots+\alpha_su_s\\
	v & = & \beta_0e_{t(w_2)}+\beta_1v_1+\cdots+\beta_tv_t
\end{eqnarray*}
for some scalars $\alpha_0, \alpha_1, \dots, \alpha_s, \beta_0, \beta_1, \dots, \beta_t$. Now, suppose that $w$ is a subpath of $u_1, \dots, u_k$, that $u_{k+1}, \dots, u_{k+r}$ have $a$ as their last arrow but are smaller paths than $w$ and $u_{k+r+1}, \dots, u_s$ have $b$ as their last arrow (reordering the basis if necessary).

Analogously, suppose that $d^*$ is a subpath of $v_1, \dots, v_q$, that $v_{q+1}, \dots, v_{q+l}$ have $d$ as their last arrow but are smaller paths that $d^*$, and $v_{q+l+1}, \dots, v_t$ have $c$ as their last arrow.

Then, from the condition $uw_1+vw_{2}^{-1}=0$  and the minimality of $w$ and $d^*$ it follows that
\begin{eqnarray*}
0 & = & (\alpha_0e_{s(w_1)}+\alpha_1u_1+\cdots+\alpha_su_s)w_1+(\beta_0e_{t(w_2)}+\beta_1v_1+\cdots+\beta_tv_t)w_{2}^{-1} \\
   & = & \alpha_0w_1+\alpha_{k+1}u_{k+1}w_1+\cdots \alpha_{k+r}u_{k+r}w_1 \\
   &    & + \beta_0w_{2}^{-1}+\beta_{q+1}v_{q+1}w_{2}^{-1}+\cdots+\beta_{q+l}v_{q+l}w_{2}^{-1}
\end{eqnarray*}

Since $w_1w_2\in St$, then $w_1, u_{k+1}w_1,\dots, u_{k+r}w_1, w_{2}^{-1}, v_{q+1}w_{2}^{-1},\dots, v_{q+l}w_{2}^{-1}$ are linearly independent, and hence

$$\alpha_0=\alpha_{k+1},\cdots=\alpha_{k+r}=\beta_0=\beta_{q+1}=\cdots=\beta_{q+l}=0$$
Thus, $$u=\alpha_1u_1+\cdots+\alpha_ku_k+\alpha_{k+r+1}u_{k+r+1}+\cdots+\alpha_su_s\in Aw\oplus Ab$$ and
$$v=\beta_1v_1+\cdots+\beta_qv_q+\beta_{q+l+1}v_{q+l+1}+\cdots+\beta_tv_t\in Ad^*\oplus Ac$$
that is, $$\ker\partial_{\omega}^0=Aw\oplus Ab\oplus Ad^*\oplus Ac$$
We can cover this kernel with $P^{-1}=P_{s(w)}\oplus P_{s(b)}\oplus P_{s(d^*)}\oplus P_{s(c)}$ and the epimorphism $\partial^{-1}:P^{-1}\longrightarrow \ker\partial_{\omega}^0$, where
$$\partial^{-1}=\begin{pmatrix}
p(w) & 0     & 0         & 0\\
0     & p(b) & 0         & 0\\
0     & 0     & p(d^*)  & 0\\
0     & 0     & 0         & p(c)
\end{pmatrix}$$
Notice that $\ker\partial^{-1}=\ker p(w)\oplus\ker p(b)\oplus \ker p(d^*)\oplus \ker p(c)$. Now, since $w\in\textbf{Pa}_c$, then, as in Lemma \ref{Lemma7-n=1-converse} we have that $\ker p(w)=Au_m\oplus Af$, where $u_m$ is a cyclic path minimal for $w$ and $f$ is an arrow such that $t(f)=s(w)$. In addition, each of $\ker p(b)$, $\ker p(d^*)$ and $\ker p(c)$ is at most two-dimensional as a $k$-vector space and each one can be covered by some projectives $P_{r_b}\oplus P_{l_b}$, $P_{r_{d^*}}\oplus P_{l_{d^*}}$ and $P_{r_c}\oplus P_{l_c}$, respectively. Thus, $\ker \partial^{-1}$ can be covered by $P^{-2}=P_{s(u_m)}\oplus P_{s(f)}\oplus P_{r_b}\oplus P_{l_b}\oplus P_{r_{d^*}}\oplus P_{l_{d^*}}\oplus P_{r_c}\oplus P_{l_c}$ and the epimorphism
$$\partial^{-2}=\begin{pmatrix}
p(u_m) & 0        & \cdots & 0 \\
0          & *        & \cdots & 0 \\
 \vdots & \vdots &\ddots &  \vdots  \\
 0         &  0    &\cdots     & *
\end{pmatrix}$$
where $\ker \partial^{-2}=\ker p(u_m)\oplus M$, and $M$ is an $A$-module. Now, since $w\in \textbf{Pa}_c$, we have that $\ker p(u_m)=Au_{m-1}\oplus Ag$, where $u_{m-1}$ is a cyclic path minimal for $u_m$ and $g$ is an arrow ending at $s(u_m)$. Hence, we can cover $\ker\partial^{-2}$ by some projective $P^{-3}$ which has $P_{s(u_{m-1})}$ as a direct summand. If we continue the reasoning in the same way, by the definition of $\textbf{Pa}_c$, in finitely many steps we will get again $P_{s(u_j)}$, for some $1\leq j\leq m+1$, as a direct summand in the projective resolution $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet$ and the other summands will eventually vanish because $Q_1$ is finite. We can repeat indefinitely the same process over the subset $\{u_1, u_2, \dots, u_j\}$ of the cyclic set of paths $\textbf{Pa}_c$. The conclusion is that $$P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$$
\end{proof}

\begin{lemma}\label{lemma7-n-converse}
	Let $\omega=w_1\cdot w_2\cdots w_n$ be a generalized string. If $\omega\in \overline{GSt}_{cp}$ or $\omega\in \overline{GSt}^{cp}$, then $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$
\end{lemma}
\begin{proof}
	Let $\omega=w_1\cdot w_2\cdots w_n$ be a generalized string. Here we will consider the case $\omega\in \overline{GSt}_{cp}$. The case $\omega\in \overline{GSt}^{cp}$ is similar.

	If $\omega\in \overline{GSt}_{cp}$, then $\mu(\omega)=0$ and there exists $w\in\textbf{Pa}_c$ such that $w\cdot \omega=w\cdot w_1\cdot w_2\cdots w_n\in GSt$ or such that $w\in\ker p(w_{l}^{-1})\cap\ker p(w_{l+1})$ for some even index $l> 0$ for which $\mu_{\omega}(l)=0$, where $w$ is minimal for $w_1$ or $w$ is minimal for $w_{l}^{-1}$ and $w_{l+1}$, respectively

	Also, we will consider the case where $w\cdot \omega\in GSt$. The case in which $w\in\ker p(w_{l}^{-1})\cap\ker p(w_{l+1})$ is completely analogous. Since $\mu(\omega)=0$, it follows that $w_1\in\textbf{Pa}_{>0}$ and hence $ww_1=0$.

	Since $A=k(Q, I)$ is a string algebra, there are at most two arrows $a$ and $b$ such that $t(a)=t(b)=s(w_1)$. As before, we consider the general situation in which $bw_1=0$ and $w=a^*$. Thus $\ker p(w_1)=Aw\oplus Ab$.

	We must consider two cases:\\

	\textit{Case 1:} $w_2\in\textbf{Pa}_{>0}$. In this case we have $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$, $\mu_{\omega}(2)=2$ and $w_1w_2\in I$. We represent this situation as follows:

	$$
	\xymatrix{\cdot\ar@{~>}[rr]^{w_1}&&\cdot\ar@{~>}[rr]^{w_2}&&\cdot\ar@{.}[dl]\ar@{.}[r]&\\
		&&                                    &&}
	$$
	The possibilities for some of the next values of $\mu_{\omega}$ are:

	$$\mu_{\omega}(3)=\left\{\begin{array}{l}
	1 \\
	3
	\end{array}\right. ,\quad \mu_{\omega}(4)=\left\{\begin{array}{l} 0\\2\\4\end{array}\right.,\quad \mu_{\omega}(5)=\left\{\begin{array}{l} 1\\3\\5\end{array}\right.$$
	Thus, in the first two places of the complex $P_{\omega}^\bullet$ we have
	$$P_{\omega}^0=P_{s(w_1)}\oplus0\oplus0\oplus0\oplus \delta(\mu_{\omega}(4),0)P_{t(w_4)}\oplus\cdots$$
	and
	$$P_{\omega}^1=0\oplus P_{t(w_1)}\oplus0\oplus \delta(\mu_{\omega}(3),1)P_{t(w_3)}\oplus 0 \oplus \delta(\mu_{\omega}(5),1)P_{t(w_5)}\oplus\cdots$$
	The complex $P_{\omega}^\bullet$ is of the form
	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{\omega}^0\ar[r]^{\partial_{\omega}^0}&P_{\omega}^1\ar[r]^{\partial_{\omega}^1}&P_{\omega}^2\ar[r]&\cdots}$$
	where $$\partial_{\omega}^0 = \begin{pmatrix}
	0         & p(w_1)   &   0       &  0         & 0 &\cdots  & 0          \\
	0         & 0           &   0       &  0         & 0 &\cdots  & 0          \\
	0         & 0           &   0       &  0         & 0 &\cdots  & 0          \\
	0         & 0           &   0       &  0         & 0 &\cdots  & 0          \\
	0         & 0           &   0       &  \ast     & 0 &\cdots & 0          \\
	\vdots & \vdots   & \vdots & \vdots   & 0 &\ddots & \vdots  \\
	0         & 0          &  0        & 0          & 0 & \cdots & 0
	\end{pmatrix}
	$$
	and the morphism $\ast: \delta(\mu_{\omega}(4), 0)P_{t(w_4)}\longrightarrow \delta(\mu_{\omega}(3),1)$ is $p(w_4^{-1})$ or zero. Also, it is possible that other entries in the matrix are nonzero, but the important fact is that they are not in the second column.

	Now, $(u_0, u_1, \dots , u_n)\in \ker \partial_{\omega}^0$ if and only if $u_0w_1=0$ together with other equations. This shows that $u_0\in\ker p(w_1)=Aw\oplus Ab$ and hence $Aw\oplus Ab$ is a direct summand of $\ker \partial_{\omega}^0$. As in Lemma \ref{Lemma7-n=1-converse}, since $w\in\textbf{Pa}_c$, it follows that $Aw$ has an infinite (minimal) projective resolution, whence the same is true for $\ker\partial_{\omega}^0$. We conclude that  $$P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$$

	\textit{Case 2:} $w_{2}^{-1}\in\textbf{Pa}_{>0}$. In this case we have $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$, $\mu_{\omega}(2)=0$ and $w_1w_2\in St$. Additionally, since $\mu(\omega)=0$, then $w_3\in\textbf{Pa}_{>0}$, that is $\mu_{\omega}(3)=1$. The following diagram represents the situation.

	$$
	\xymatrix{\cdot\ar@{~>}[rrr]^{w_1}&&&\cdot\ar@{~>}[dlll]^{w_2}&&\\
		\cdot\ar@{~>}[rrr]_{w_3}&&& \cdot\ar@{.}[dl]\ar@{.}[r]                  &&\\
		&&&                                                 &&}
	$$
	The possibilities for some of the next values of $\mu_\omega$ are
	$$\mu_{\omega}(4)=\left\{\begin{array}{l}
	0 \\
	2
	\end{array}\right. ,\quad \mu_{\omega}(5)=\left\{\begin{array}{l} 1\\3\end{array}\right.$$
	Therefore, for $P_{\omega}^0$ and $P_{\omega}^1$ we have
	$$P_{\omega}^0=P_{s(w_1)}\oplus0\oplus P_{t(w_2)}\oplus0\oplus \delta(\mu_{\omega}(4),0)P_{t(w_4)}\oplus0\oplus\cdots$$
	and
	$$P_{\omega}^1=0\oplus P_{t(w_1)}\oplus0\oplus P_{t(w_3)}\oplus 0 \oplus \delta(\mu_{\omega}(5),1)P_{t(w_5)}\oplus 0\oplus\cdots$$
	The complex $P_{\omega}^\bullet$ is of the form
	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{\omega}^0\ar[r]^{\partial_{\omega}^0}&P_{\omega}^1\ar[r]^{\partial_{\omega}^1}&P_{\omega}^2\ar[r]&\cdots}$$
	where $$\partial_{\omega}^0 = \begin{pmatrix}
	0         & p(w_1)           &   0       &  0         &\cdots  & 0          \\
	0         & 0                   &   0       &  0         &\cdots  & 0          \\
	0         & p(w_2^{-1})   &   0       & p(w_3)   &\cdots  & 0          \\
	0         & 0                   &   0       &  0         &\cdots  & 0          \\
	\vdots & \vdots           & \vdots & \vdots   & \ddots & \vdots  \\
	0         & 0                   &  0        & 0          &  \cdots & 0
	\end{pmatrix}
	$$
	and possibly other entries of the matrix are nonzero but they are not in the second column.
	Now, $(u_0, u_1, u_2, \dots, u_n)\in \ker\partial_{\omega}^0$ if and only if $u_0w_1+u_2w_{2}^{-1}=0$, $u_2w_3+\delta(\mu_{\omega}(4),0)u_4w_{4}^{-1}=0$ and other equations. This implies, by the similar arguments as in Lemma \ref{Lemma7-n=2-converse}, case 2, that $u_0\in Aw\oplus Ab$ and $u_2\in \ker p(w_2^{-1})\cap\ker p(w_3)=\left(Ad^*\oplus Ac\right)\cap \ker p(w_3)$, where $b, c, d\in Q_1$ are such that $t(b)=s(w_1)$ and $t(c)=t(d)=s(w_{2}^{-1})=t(w_2)$. The important facts are that $Aw$ is a direct summand of $\ker\partial_{\omega}^0$ and $w\in\textbf{Pa}_c$. Thus, as in Lemmas \ref{Lemma7-n=1-converse} and \ref{Lemma7-n=2-converse} we conclude that $$P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A).$$
\end{proof}
Now, we state and prove the converse of the previous results
\begin{lemma}\label{Lemma7-n=1-direct}
	Let $\omega=w_1$ be a generalized string. If $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$, then either $\omega\in \overline{GSt}_{cp}$ or $\omega\in \overline{GSt}^{cp}$.
\end{lemma}
\begin{proof}
	Let $\omega=w_1\in GSt$. Suppose first that $w_1\in\textbf{Pa}_{>0}$, so $\mu_{\omega}(0)=0$ and $\mu_{\omega}(1)=1$. Then $\mu(\omega)=0$ and the complex $P_{\omega}^\bullet$ is
	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{s(w_1)}\ar[r]^{p(w_1)}&P_{t(w_1)}\ar[r]&0\ar[r]&\cdots}$$
	The hypothesis $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$ implies that $\ker p(w_1)\neq 0$. Besides, since $A$ is a string algebra, the possibilities for this kernel are: $\ker p(w_1)=Aa^*$ or $\ker p(w_1)=Aa^*\oplus Ab$, where $a, b$ are arrows ending at the vertex $s(w_1)$.

	$$\xymatrix{            &\ar[rd]^{b}&                             &&& \\
		&                &\ar@{~>}[rrr]^{w_1}&&& \\
		&\ar[ru]_a   &                                      }
	$$

	We will consider the general case $\ker p(w_1)=Aa^*\oplus Ab$ (it is possible that $a^*=a$). By the hypothesis, $Aa^*\oplus Ab$ has infinite minimal projective resolution, which means that the same is true for either $Aa^*$ or $Ab$. Suppose, without loss of generality, that the minimal projective resolution of $Aa^*$ is not bounded. The projective cover of $Aa^*$ is given by $p(a^*): P_{s(a^*)}\longrightarrow Aa^*$. Again, the hypothesis allows us to choose a path $a_m^*$ such that $a_m$ is an arrow with $t(a_m)=s(a^*)$, $a_m^*a^*=0$ and $\ker p(a_m)$ has infinite minimal projective resolution.

	Analogously, there is a path $a_{m-1}^*$ where $a_{m-1}$ is an arrow with $t(a_{m-1})=s(a_m^*)$, $a_{m-1}^*a_m^*=0$ and $\ker p(a_{m-1}^*)$ has infinite minimal projective resolution. If we continue the process in the same way, since $\textbf{Pa}_{>0}$ is finite, we see that there exist paths $w=a^*, w_m=a_m^*, w_{m-1}=a_{m-1}^* \dots, w_1=a_1^*$ such that $t(w_m)=s(w)$, $t(w_i)=s(w_{i+1})$ for $i=1,\dots, m-1$, $s(w_1)=t(w_j)$ for some $1\leq j\leq m+1$, where $w_{m+1}=w$, and  $w_iw_{i+1}=w_jw_1=0$. By our construction, it is clear that $w_i$ is minimal for $w_{i+1}$ and $w_j$ is minimal for $w_1$.  Since $ww_1=0$, we have shown that there exists $w\in\textbf{Pa}_c$ such that $w\cdot \omega\in GSt$ and $w$ is minimal for $w_1$. Therefore $\omega\in \overline{GSt}_{cp}$.

	Finally, if $w_{1}^{-1}\in\textbf{Pa}_{>0}$, by dual arguments it can be shown that $\omega\in\overline{GSt}^{cp}$.
\end{proof}

\begin{lemma}\label{Lemma7-n=2-direct}
	Let $\omega=w_1\cdot w_2\in GSt$. If $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$, then either $\omega\in\overline{GSt}_{cp}$ or $\omega\in\overline{GSt}^{cp}$.
\end{lemma}
\begin{proof}
	Let $\omega=w_1\cdot w_2$ be a generalized string. Suppose that $w_1\in\textbf{Pa}_{>0}$. We will show that $\omega\in\overline{GSt}_{cp}$. If $w_{1}^{-1}\in\textbf{Pa}_{>0}$, by dual arguments it can be shown that $\omega\in\overline{GSt}^{cp}$. Now, we have two cases to consider: $w_2\in\textbf{Pa}_{>0}$ or $w_{2}^{-1}\in\textbf{Pa}_{>0}$.\\

	\textit{Case 1}: If $w_2\in\textbf{Pa}_{>0}$, $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$, $\mu_{\omega}(2)=2$ and $w_1w_2\in I$. It follows that $\mu(\omega)=0$ and the complex $P_{\omega}^\bullet$ is

	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{s(w_1)}\ar[r]^(0.5){p(w_1)}&P_{t(w_1)}\ar[r]^{p(w_2)}&P_{t(w_2)}\ar[r]&0\ar[r]&\cdots}$$

	We know that $\ker p(w_1)=Aa^*\oplus Ab$ for some arrows $a, b$ such that $t(a)=t(b)=s(w_1)$. The hypothesis $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$ implies that this kernel has an infinite minimal projective resolution. Thus, we are in the same position as in Lemma \ref{Lemma7-n=1-direct}. Therefore, using the same arguments we have that there exist $w\in \textbf{Pa}_c$ such that $w\cdot \omega\in GSt$, where $w$ is minimal for $w_1$. Hence $\omega\in\overline{GSt}_{cp}$. \\

	\textit{Case 2}: If $w_{2}^{-1}\in\textbf{Pa}_{>0}$, $\mu_{\omega}(0)=0$, $\mu_{\omega}(1)=1$, $\mu_{\omega}(2)=0$ and $w_1w_2\in St$. Thus, $\mu(\omega)=0$ and the complex $P_{\omega}^\bullet$ is

	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{s(w_1)}\oplus P_{t(w_2)}\ar[r]^(0.6){\partial_{\omega}^0}&P_{t(w_1)}\ar[r]&0\ar[r]&\cdots}$$ where $\partial_{\omega}^0=\begin{pmatrix}
	p(w_1) \\ p(w_2^{-1})
	\end{pmatrix}$.
	Since the hypothesis implies that $\ker\partial_{\omega}^0\neq 0$ and $A$ is a string algebra, the general form of this kernel is $\ker\partial_{\omega}^0=\ker p(w_1)\oplus \ker p(w_{2}^{-1})=Aa^*\oplus Ab\oplus Ad^*\oplus Ac$ for some arrows $a, b, c, d$ such that $t(a)=t(b)=s(w_1)$, $t(c)=t(d)=s(w_{2}^{-1})=t(w_2)$ (see Lemma \ref{Lemma7-n=2-converse}), and at least one direct summand is different from zero.
	$$
	\xymatrix@R=4.5mm{ \ar[rd]^{b}               &                                &&&\\
		&\ar@{~>}[rrr]^{w_1} &&&\ar@{~>}[ddlll]^{w_{2}} \\
		\ar[ru]_a\ar[rd]^{d} &                                &&&\\
		&                                &&&\\
		\ar[ru]_c                &                                &&&}
	$$

	Since $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$, then either $\ker p(w_1)$ or $\ker p(w_{2}^{-1})$ has an infinite minimal projective resolution.\\

	If $\ker p(w_1)=Aa^*\oplus Ab$ has infinite minimal projective resolution, then the same is true for either $Aa^*$ or $Ab$ and we proceed as in Lemma \ref{Lemma7-n=1-direct} to show that there exists $w\in\textbf{Pa}_c$ such that $w\cdot \omega\in GSt$ with $w$ minimal for $w_1$. Therefore $w\in \overline{GSt}_{cp}$.\\

	If $\ker p(w_{2}^{-1})=Ad^*\oplus Ac$ has infinite minimal projective resolution, the same holds for either $Ad^*$ or $Ac$. In this case we consider $\omega^{-1}=w_{2}^{-1}\cdot w_{1}^{-1}$ instead of $\omega$ and use the same arguments as in Lemma \ref{Lemma7-n=1-direct} to show that there is a path $w\in\textbf{Pa}_c$ such that $ww_{2}^{-1}=0$ (that is, $w\cdot \omega^{-1}\in GSt$), where $w$ is minimal for $w_{2}^{-1}$. This means that $\omega^{-1}\in\overline{GSt}_{cp}$, that is $\omega\in\overline{GSt}_{cp}$. (Here, we are using the fact that $\omega\cong_s \omega^{-1}$ and $P_{\omega}^{\bullet}\cong P_{\omega^{-1}}^{\bullet}$ in $D^b(A)$).
\end{proof}

Now, we use Lemmas \ref{Lemma7-n=1-direct} and \ref{Lemma7-n=2-direct} to prove the same result for the general case.

\begin{lemma}\label{Lemma7-n-direct}
	Let $\omega=w_1\cdot w_2\cdots w_n$ be a generalized string. If $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$, then either $\omega\in\overline{GSt}_{cp}$ or $\omega\in\overline{GSt}^{cp}$.
\end{lemma}
\begin{proof}
	As in the previous lemma, we will show that if $w_1\in\textbf{Pa}_{>0}$, then $\omega\in\overline{GSt}_{cp}$. By dual arguments it can be shown that if $w_{1}^{-1}\in\textbf{Pa}_{>0}$, then $\omega\in\overline{GSt}^{cp}$.

	Let us begin by assuming that $\mu(\omega)=0$, that is, we suppose that $\omega$ has as number of direct paths greater than or equal to the number of inverse paths.  We are allowed to do so since, if $\omega$ has more inverse than direct paths, we consider $\omega^{-1}$ instead of $\omega$.
 	If $w_1\in \textbf{Pa}_{>0}$, as $\mu(\omega)=0$, then $\mu_{\omega}(1)=1$ and the possibilities for some of the next values of $\mu_{\omega}$ are

 	$$\mu_{\omega}(2)=\left\{\begin{array}{l}
 	0 \\
 	2
 	\end{array}\right. ,\quad \mu_{\omega}(3)=\left\{\begin{array}{l} 1\\3\end{array}\right. ,\quad \mu_{\omega}(4)=\left\{\begin{array}{l} 0\\2\\4 \end{array}\right., \quad \mu_{\omega}(5)=\left\{\begin{array}{l} 1\\3\\5 \end{array}\right.
 	$$
 	Thus, if $\mu_{\omega}(l)=0$ with $l>0$, then $l$ is an even number greater than or equal to two and less that $n$. Also, it is possible that $\mu_{\omega}(n)=0$ when $n$ is even. It follows that the complex $P_{\omega}^\bullet$ is
 	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_{\omega}^0\ar[r]^{\partial_{\omega}^0}&P_{\omega}^1\ar[r]^{\partial_{\omega}^1}&P_{\omega}^2\ar[r]&\cdots}$$
 	where the general form of $\ker\partial_{\omega}^0$ is either one of the following
 	\begin{enumerate}
 		\item\label{ker:1} $\ker\partial_{\omega}^0\cong \ker p(w_1)\oplus\cdots\oplus \ker p(w_{l}^{-1})\cap \ker p(w_{l+1})\oplus\cdots$, whenever $n$ is odd and $\mu_{\omega}(l)=0$ for some $2\leq l\leq n-1$.
 		\item\label{ker:2} $\ker\partial_{\omega}^0\cong \ker p(w_1)\oplus\cdots\oplus \ker p(w_{l}^{-1})\cap \ker p(w_{l+1})\oplus\cdots\oplus \ker p(w_{n}^{-1})$, whenever $n$ is even, there is an even index $2\leq l\leq n-2$ with $\mu_{\omega}(l)=0$ and $\mu_{\omega}(n)=0$.
 		\item\label{ker:3} $\ker\partial_{\omega}^0\cong \ker p(w_1)\oplus\cdots\oplus \ker p(w_{l}^{-1})\cap \ker p(w_{l+1})\oplus\cdots$, whenever $n$ is even, there is an even index $2\leq l\leq n-2$ with $\mu_{\omega}(l)=0$ and $\mu_{\omega}(n)>0$.

 	\end{enumerate}
 Now, since $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$, then $\ker\partial_{\omega}^0$ has an infinite minimal projective resolution. If $\ker\partial_{\omega}^0$ is as in (\ref{ker:1}), then at least one direct summand also has an infinite minimal projective resolution. If it is the case for $\ker p(w_1)$, we proceed as in Lemma \ref{Lemma7-n=1-direct}, to show that there is a path $w\in\textbf{Pa}_c$ such that $w\cdot \omega\in GSt$, where $w$ is minimal for $w_1$, and hence $\omega\in \overline{GSt}_{cp}$.

 If the minimal projective resolution of $\ker p(w_{l}^{-1})\cap\ker p(w_{l+1})$ is not bounded for some $2\leq l\leq n-1$, the same arguments allow us to show that there exist $w\in\textbf{Pa}_c$ such that $w\in\ker p(w_{l}^{-1})\cap\ker p(w_{l+1})$, where $w$ is minimal for both $w_{l}^{-1}$ and $w_{l+1}$. Therefore, the conclusion follows.

 If $\ker\partial_{\omega}^0$ is as in (\ref{ker:3}), the reasoning is the same. Finally, if $\ker\partial_{\omega}^0$ is as in (\ref{ker:2}) and $\ker p(w_{n}^{-1})$ has infinite minimal projective resolution, we consider ${\omega}^{-1}$ instead of $\omega$ and, as in the proof of Lemma \ref{Lemma7-n=2-direct}, case 2, it can be shown that $\omega\cong_s {\omega}^{-1}\in\overline{GSt}_{cp}$.
\end{proof}

Now, if we put Lemmas \ref{Lemma7-n=1-converse} to \ref{Lemma7-n-direct} together, we get the following theorem, which provides a characterization for a string complex to be periodic in the case of a string algebra.

\begin{theorem}\label{Lemma7-String}
	Let $A=k(Q, I)$ be a string algebra. If $\omega=w_1\cdot w_2\cdots w_n$ is a generalized string, then $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$ if and only if either $\omega\in\overline{GSt}_{cp}$ or $\omega\in\overline{GSt}^{cp}$.
\end{theorem}

\section{Applications}\label{Sec:App}
Two important applications of the Theorem \ref{Lemma7-String}, which we detail below, are related to the global dimension and the indecomposables of the derived category of a string algebra.

\subsection{About of the global dimension of a string algebra}
 Let $A$ be a finite dimensional $k$-algebra. Recall that given an $A$-module $M$, the \textit{projective dimension} of $M$, denoted by $pd \ M$ is the smallest integer $d$ such that there exists a projective resolution of the form
$$\xymatrix{0\ar[r]&P_d\ar[r]&P_{d-1}\ar[r]&\cdots\ar[r]&P_1\ar[r]&P_0\ar[r]&M}.$$
If no resolution exists, then we say that $M$ has infinite projective dimension. The \textit{global dimension of $A$}, denoted by $\text{gl.dim} A$ is defined as the supremum of the projective dimensions of all $A$-modules, that is, $$\text{gl.dim} A:=\text{sup}\{pd \ M \mid M\in A-\text{mod}\}.$$

Theorem \ref{Lemma7-String} is very important because it deals with the structure of a string algebra. That is, it establishes when a string algebra has infinite global dimension. We state this in the following result.

\begin{theorem}\label{Th:gl.dim-St}
	Let $A=k(Q, I)$ be a string algebra. If $\overline{GSt}_{cp}\neq\emptyset$ or $\overline{GSt}^{cp}\neq\emptyset$, then $\text{gl.dim} A=\infty$.
\end{theorem}

\begin{proof}
	Suppose that $\overline{GSt}_{cp}\neq\emptyset$. The case $\overline{GSt}^{cp}\neq\emptyset$ is similar.
If $\omega=w_1\cdot w_2\cdots w_n\in \overline{GSt}_{cp}$, then $\mu(\omega)=0$ and there exists $w\in \textbf{Pa}_c$ such that $w\cdot \omega=w\cdot w_1\cdot w_2\cdots w_n\in GSt$ or  $w\in\ker p(w_{l}^{-1})\cap\ker p(w_{l+1})$ for some even index $l> 0$ with $\mu_{\omega}(l)=0$.

	Again, we will consider the case where $w\cdot \omega\in GSt$. The case in which $w\in\ker p(w_{l}^{-1})\cap\ker p(w_{l+1})$ is completely analogous. If $w\cdot \omega\in GSt$ and $\mu(\omega)=0$, necessarily $w_1\in\textbf{Pa}_{>0}$ and hence $ww_1=0$. According to Lemma \ref{Lem-St}, the general structure of $\ker p(w_1)$ is $\ker p(w_1)=Aw\oplus Ab$, where $b$ is an arrow such that $t(b)=s(w_1)$.
	Following the proof of Lemma \ref{lemma7-n-converse}, since $w\in\textbf{Pa}_c$, we get that $Aw$ has infinite minimal projective resolution and this implies that $\text{gl.dim} A=\infty$.
\end{proof}

\begin{example}\label{Example-Qc*} Let $Q$ be the bound quiver

	$$ \xymatrix{&1\ar[r]^a&2\\
		5\ar@(ul,dl)[]_{x} \ar[r]_d &3\ar[r]_c\ar[ur]^b&4}
	$$
with $I=\langle db, dc, x^2, xd\rangle$. Then $A=kQ/I$ is a string algebra over $k$. In this case we have that $\emph{\textbf{Pa}}_c=\{x,d,c,b\}$.

	Now, consider the generalized string $\omega=w_1\cdot w_2\cdot w_3=a\cdot b^{-1}\cdot c$. Then, $l(\omega)=3>0$, $\mu(\omega)=0$ and $\exists d\in \emph{\textbf{Pa}}_c$ such that $d\in \ker p(w_{2}^{-1})\cap\ker p(w_3)=\ker p(b)\cap \ker p(c)$, with $\mu_{\omega}(2)=0$. That is, $\omega\in \overline{GSt}_{cp}$.

	According to Definition \ref{def-st-comp}, the complex $P_{\omega}^\bullet$ is
	$$P_{\omega}^\bullet: \ \xymatrix{\cdots\ar[r]&0\ar[r]&P_1\oplus P_3\ar[r]^{\partial_{\omega}^0}&P_2\oplus P_4\ar[r]&0\ar[r]&\cdots}$$

	where $\partial_w^0=\begin{pmatrix}
	p(a)  & 0\\
	p(b)  & p(c)
	\end{pmatrix}$

	Now, $\begin{pmatrix}u & v\end{pmatrix}\in\ker \partial_{\omega}^0$ if and only if $ua+vb=0$ and $vc=0$. Since $u\in P_1$ and $v\in P_3$, then $u=\alpha_0e_1$ and $v=\beta_0e_3+\beta_1d$ for some scalars $\alpha_0, \beta_0, \beta_1\in k$.

	From $ua+vb=0$, it follows that $\alpha_0a+\beta_0b=0$, and hence $\alpha_0=\beta_0=0$. Notice that the condition $vc=0$ is already satisfied. Therefore, we have that $u=0$ and $v=\beta_1d\in Ad=\ker p(b)\cap\ker p(c)$ and thus $$\ker \partial_{\omega}^0\cong \ker p(b)\cap\ker p(c)=Ad$$ (in general, we have for this case that $\ker\partial_{\omega}^0=\ker p(w_1)\oplus\ker p(w_2^{-1})\cap\ker p(w_3)$).

	The projective cover of this kernel is $p(d): P_5\longrightarrow Ad$ and $\ker p(d)=Ax$. Since $x^2=0$, it is clear that the minimal projective resolution $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet$ of $P_{\omega}^\bullet$ is

	$$\xymatrix@C=4mm@R=4.7mm{\cdots\ar[rr]&&P_5\ar[rr]^{p(x)}\ar@{->>}[dr]&                              &P_5\ar[rr]^{p(d)}\ar@{->>}[dr]&                           &P_{\omega}^0\ar[rr]^{\partial_{\omega}^0}&&P_{\omega}^1\ar[rr]&&0\ar[r]&\cdots\\
		&&                                                             &Ax\ar@{^(->}[ru]&                                                    & Ad\ar@{^(->}[ru]&                                       &&                       && }
	$$
	that is, $P_{\beta(P_{\omega}^\bullet)^\bullet}^\bullet\notin K^b(\text{pro} \ A)$.

	Notice also that, since the generalized string $\omega=w_1\cdot w_2\cdot w_3=a\cdot b^{-1}\cdot c \in\overline{GSt}_{cp}$, then, according to Theorem \ref{Th:gl.dim-St}, we have that $\text{gl.dim} A=\infty$.
\end{example}

\subsection{About the indecomposables in $D^b(A)$}
 For string algebras with the property that every arrow belongs to a unique maximal path, as in example \ref{Ex:Str:1}(2), the functor $F: \mathfrak{p}(A)\longrightarrow s(\mathcal{Y}, k)$ in \cite{FGR} assigns to every  generalized strings $\omega$ an indecomposable object in $D^b(A)$. In fact, this is a consequence of Theorem \ref{Lemma7-String} and Proposition \ref{prop:1}. Precisely,

\begin{theorem}
	Let $A=k(Q,I)$ be a string algebra with the property that every arrow belongs to a unique maximal path. Then
	\begin{eqnarray*}
		\text{ind}_0  D^b(A) &\supseteq&\{T^i(P_w^\bullet) \mid w\in GSt, i\in\mathbb{Z}\}\dot{\cup}\\
		&  &\{T^i(\beta(P_{\omega}^\bullet)^\bullet) \mid \omega\in \overline{GSt}_{cp}, i\in\mathbb{Z}\}\dot{\cup}\\
		&  & \{T^i(\beta(P_{\omega}^\bullet)^\bullet) \mid \omega\in \overline{GSt}^{cp}\setminus \overline{GSt}_{cp}, i\in\mathbb{Z}\}.
	\end{eqnarray*}
	where $T$ is the translation functor.
\end{theorem}
\begin{proof}
The proof follows the same reasoning as the Main Theorem in \cite{FGR} (Theorem 37), because the important fact is the \textit{unique maximal path property}. In fact, Lemmas 24 and 35 in \cite{FGR} are still valid in this case, which are essential in the proof and are based on the uniqueness of maximal paths.
\end{proof}

\subsection*{Acknowledgments}
This paper was prepared for the proceedings of the meeting {\it Geometry in Algebra and Algebra in Geometry}, GAAG-V, held in Medell\'{i}n in 2019. The first author would like to thank the organizers for the invitation to speak and to Colciencias (Beca Doctorado Nacional Colciencias, Convocatoria 647 de 2014). The authors are also grateful to CODI (Universidad de Antioquia, U de A). The GAAG-V and the third author, were partially supported by CODI, University of Antioquia, project 2017-15756 Stable Limit Linear Series on Curves.

config: arxiv


TITLE: Are there any zero divisors in this ring?
QUESTION [1 upvotes]: Definition: Zero-Divisors.
A nonzero element $a$ in a commutative ring $R$ is called a zero divisor
  if there is a non zero element $b\in R$ such that $ab=0$.

Consider the set $\mathbb Z$ with the operations $\oplus$ and $\otimes$ defined for
$a, b \in \mathbb Z$ by $a\oplus b=a+b−1$ and $a \otimes b = ab − (a + b) + 2$.
My question

Are there any zero divisors in this ring?
Do we need to check zero divisor in case of non-commutative ring.

The solution I have for zero divisor part is

Solution: No, there are no zero divisors in this ring. It is crucial to remember that
      the “zero element” in this ring is the additive identity, namely, 1. So, we need to check
      whether there are elements a = 1 and b = 1 such that a ⊗ b = 1, i.e.
      ab − (a + b) + 2 = 1
      (a − 1)(b − 1) = 0
      The only way for this equation to hold is that either a = 1 or b = 1. Thus there can be
      no zero divisors in this ring.

Why we are checking a ⊗ b = 1 ,we should check a ⊗ b = 0 where $a,b \neq 0$

REPLY [2 votes]: As suggested by anon in his comment to OP, note that the map $\varphi: \Bbb{Z} \to \Bbb{Z}$ given by $\varphi(x) = x - 1$ is an isomorphism between the two rings
$$
\varphi: (\Bbb{Z}, \oplus, \otimes) \to (\Bbb{Z}, +, \cdot).
$$
In fact $$\varphi(x) + \varphi( y) = x - 1 + y - 1 = \varphi(x \oplus y),$$ and $$\varphi(x) \cdot \varphi( y) = (x - 1) \cdot (y - 1) = xy - (x+y) + 1 = \varphi(x \otimes y).$$
Everything becomes easy then, because it's the usual ring structure on the integers we are talking about, only in disguise.

REPLY [0 votes]: $ab-(a+b)+2=1\Leftrightarrow (a-1)(b-1)=0\Leftrightarrow a=1$ or $b=1$, which are the additive identity.


TITLE: Does Newton's third law remain totally unchanged even in Einstein's theory?
QUESTION [1 upvotes]: Why Newton's third law remain unchanged still now in relativity theory (as for example that is why we feel weight due to equal but opposite reaction of the Earth's surface)?

REPLY [5 votes]: Newton't third law is just the conservation of momentum, so the question is whether conservation of momentum applies in general relativity. This turns out to be a rather complicated issue, and for the details I suggest you look at General relativity and the conservation of momentum.
However you specifically ask why we feel weight due to equal but opposite reaction of the Earth's surface. For an observer spacetime locally appears to be flat i.e. locally special relativity applies rather than general relativity, and four-momentum is conserved in special relativity. Since you and the Earth are stationary with respect to each other we can ignore relativistic effects and Newton's third law applies. Things would be a bit different if you and the Earth were colliding at a significant fraction of the speed of light!


TITLE: Bound for Analytic Function on Unit Disk
QUESTION [5 upvotes]: The following is an old qualifying exam problem that I can't seem to piece together:
Suppose we have an analytic function $f$ on the unit disk $\mathbb{D}$ s.t. $|f|  \leq 1$. Show $$ \frac{|f(0)|-|z|}{1-|f(0)||z|} \leq |f(z)| \leq \frac{|f(0)|+|z|}{1+|f(0)||z|}  $$.
I've tried two things. First, decomposing $f$ into its real and imaginary parts and applying the Harnack inequality to each (after adding 1 so that it is nonnegative) and then piecing them together so they say something about $f$. I couldn't get that to look close to the inequality.  Second, I define $h(z) = \frac{f(z)-f(0)}{1+|f(0)|}$ and apply the Schwartz lemma. This comes close but I couldn't get it to work.

REPLY [0 votes]: Let $f(0) = a$ and consider the function $\phi_a(z)  = \dfrac{a-z}{1-\bar{a}z}$.
Apply Schwarz lemma on $(\phi_a \circ f)(z)$ to use the result that $|(\phi_a \circ f)(z)| \leq |z|$. From this, using triangle and reverse triangle inequality you can arrive at
$\dfrac{|f(0)|-|z|}{1-|f(0)||z|} \leq |f(z)|$.
Similarly apply Schwarz lemma on $(\tilde{\phi}_a \circ f)(z)$ where $\tilde{\phi}_a(z) = \dfrac{z-a}{1-\bar{a}z}$.
This will lead you to
$ |f(z)| \leq \dfrac{|f(0)|+|z|}{1+|f(0)||z|}$


TITLE: How many arrangements? $(a, b, c, d, x, y, z, w)$
QUESTION [2 upvotes]: I would appreciate if somebody could help me with the following problem:
Q: How many arrangements?

My Study
I solved by mathematica11.0
f[n_] := {a, b, c, d, x, y, z, w} /.
  Solve[0 < a < b < c < d < n && 0 < x < y < z < w < n && a < x &&
    b < y && c < z && d < w, {a, b, c, d, x, y, z, w}, Integers]
f[9]=490

REPLY [1 votes]: This is only a partial solution.
The problem is somehow related to Young tableaux.  If we were required to use all the numbers from $1$ to $8$, we could compute the number of tableau of this shape with the hook-length formula which in this instance gives $$\frac{8!}{5\cdot4\cdot3\cdot2\cdot4\cdot3\cdot2\cdot1}=14$$
Here are the hook lengths:
$$5\ 4\ 3\ 2\\
  4\ 3\ 2\ 1$$
It is apparent that we can have any or all of the equalities $b=x,\ c=y, d=z$. For example, if
$$abcd\\
xyzw$$  is a standard tableau, we could replace it by either $$abcd\\byzw$$ or $$axcd\\xyzw$$  For each of the $3$ diagonals, we have $3$ choices: leave it unchanged, or make it constant in one of two ways, which brings us up to $27\cdot14=378$ solutions.
So far all our solutions have $a=1,\ w=8$, but there are others, for example $$2345\\3456$$  I haven't been able to think of a simple way to count these.
I've Googled "Young tableaux with repeated elements," but I didn't find anything bearing directly on this problem.  Admittedly, I've just spent a few minutes on this.  This paper discusses a number of variants, but a brief perusal didn't turn up anything directly relevant.


TITLE: Determine if $f(x)=\int_0^x\frac{1}{t}\ln\left(\frac{1}{1-t}\right)$ is n-times differentiable
QUESTION [1 upvotes]: The function $f$ is defined as
$$f(x)=\int _{ 0 }^{ x }{ \frac { 1 }{ t }  } \ln\left(\frac { 1 }{ 1-t }\right)\, dt$$
By using the second fundamental theorem of calculus, we get the first derivative:
$$f^{ (1) }( x ) ={ \frac { 1 }{ x } } \ln\left(\frac { 1 }{ 1-x } \right)$$
My question is: Is there a quick way to determine if $f(x)$ is $n$-times differentiable on $(-1,1)$ without taking any further derivatives but still somehow get the $n$th derivative term? I want to do a Taylor expansion around $x=0$.

REPLY [3 votes]: We have the series
$$
\log\left(\frac1{1-x}\right)=\sum_{k=1}^\infty\frac{x^k}{k}
$$
Thus, the function in question is a Polylogarithm
$$
\operatorname{Li}_2(x)=\sum_{k=1}^\infty\frac{x^k}{k^2}
$$
which is infinitely differentiable on $(-1,1)$.


TITLE: How Quantum Information can be important for fundamental physics?
QUESTION [6 upvotes]: It seems that Quantum Information has been seriously considered as something very important for fundamental physics, and in particular, to quantum gravity.
For example, this is the topic of this year's prospects in theoretical physics. So we see that it seems some quite big names on the community, which are mainly focused on purely theoretical and fundamental problems, like Witten and Susskind, are taking Quantum Information quite seriously on this.
The program description of this event says:

PiTP 2018 is titled "From Qubits to Spacetime," and will cover topics
  ranging from the connections between quantum information and the
  strucuture of spacetime, to how information is manipulated by the
  dynamics and how quantum effects can be included in black hole
  thermodynamics.

Now, on the other hand, every time I begin taking a look on it, it appears to be the opposite. It seems the deal with Quantum Information is all about how to produce a quantum computer. So it seems in the end, that the deal with Quantum Information is not exactly on "how does nature behave fundamentaly", but rather, "how can we harness nature's quantum behavior to create technology".
This can been for instance in quantum algorithms, quantum error correction, quantum criptography, and so forth. So, as in computer science, this seems to have the clear intention not on understanding nature very deeply, but rather on harnessing it to create technology. In particular here one is trying to use the properties of quantum systems to create better computers.
This is all nice in itself, but I can't seem how all of this can be connected to fundamental physics to the point of being related to quantum gravity, string theory, and so forth. I mean, what nature cares about algorithms, error correction and cryptography? These are things that are of interest for us, humans, which want better technology. I'm failing to imagine why a fundamental force of nature as gravity, in its most fundamental form, would even be related to these things.
So the question here is: if on the one hand, quantum information seems extremely tied to technology development in computer science, how can it be so important as it seem to be, in the understanding of fundamental physics and in particular, quantum gravity and string theory?
Edit: Perhaps the OP wasn't clear. I'm not asking, what is the intent of quantum information. I'm saying the following: if we search on the web about QI, we mostly find applied stuff aimed at quantum computers. Algorithms, codes and so forth. Nature doesn't care whether or not we build computers. So how can these things (algorithms, codes, etc) have any relation whatsoever with nature's fundamental workings, and for example the nature of spacetime (see citation above)? That is the objective question. I'm not saying a field can't have an applied and a theoretical part. Obviously it can (as far as I know, GR has been used in the development of the GPS). What I'm failing to see is where the theoretical part is on QI. Is how a field that seems to be pure computer science have anything to do with fundamental physics.

REPLY [3 votes]: (My background: I'm not a fundamental physicist nor a quantum information theorist, although I work with a few of the latter.)
Information theory is a framework for studying and characterising randomness. In particular, information theory is good at answering questions about random variables such as "how random" or how strongly correlated random variables are. It turns out that such properties are often characterised by non-linear functionals of the probability distribution, such as the entropy, mutual information, etc.
Quantum information theory is a framework for studying and characterising randomness in quantum mechanics. It provides tools for answering questions such as how strongly correlated two parts of a quantum system are. Usually such properties are well characterised by non-linear functions of the density operator, such as the (entanglement) entropy, the mutual information etc. This is rather different from the usual agenda of physicists (fundamental or otherwise) who are typically interested in observables/correlation functions, which are linear functions of the density operator.
In my (necessarily limited) experience, quantum information theorists are often interested in questions of a kinematical nature, e.g. how to classify quantum states according to their entanglement properties, or what kinds of quantum states can be transformed into each other under very general constraints like unitarity, locality, energy conservation etc., without worrying about the detailed dynamics of such a transformation. On the other hand, many physicists are deeply concerned with writing down the actual Lagrangian describing the fundamental dynamics of the Universe. In one (debatable) sense, quantum information theory is "even more fundamental" than this noble endeavour, since many of its theorems will continue to be applicable even if one day we discover a much better Lagrangian than the Standard Model. However, I suspect that the apparent interest in quantum information theory shown by certain people working on fundamental dynamics is more related to questions such as the black-hole thermodynamics, entanglement entropy of quantum-field vacua, holographic tensor network states etc.


TITLE: Finite extension with abelian Galois group
QUESTION [2 upvotes]: I'm facing this question:
Let $E/F$ be a finite field extension such that its Galois group $\text{Gal}(E/F)$ is abelian. Is it necessary that $E/F$ is a Galois extension?
Attempt:
My first guess is no, but apparently I can't come up with any counter example. Maybe choosing $F$ not to be a perfect field would help, but I'm not sure in what direction to move to get an abelian Galois group.
My attempt to prove the statement was also unfruitful; I tried showing that the field of fixed elements of $E$ through $\text{Gal}(E/F)$ is precisely $F$, namely $E^{\text{Gal}(E/F)}=F$, which I know is equivalent to $E/F$ being a Galois extension. The inclusion $F\subset E^{\text{Gal}(E/F)}$ is always true; for the other inclusion: $[E:F]=[E:E^{\text{Gal}(E/F)}]\cdot[E^{\text{Gal}(E/F)}:F]$, but since $E/F$ is finite, $[E:E^{\text{Gal}(E/F)}]=|\text{Gal}(E/F)|$. I suppose that, if this is the right way, the fact that the Galois group is abelian comes in here. But I can't prove that $[E:F]=|\text{Gal}(E/F)|$ with only this assumption.
Any counterexample or hint is greatly appreciated.

REPLY [4 votes]: Hint: The automorphism group of a nontrivial non-Galois field extension can be trivial! (hence abelian)

Take $\mathbb Q(\sqrt[3]2)/ \mathbb Q$: any automorphism fixes $\sqrt[3]2$, hence $\operatorname{Gal}(\mathbb Q(\sqrt[3]2)/ \mathbb Q) = 1$!


TITLE: Why is mean free time inversely proportional to the electric field strength?
QUESTION [1 upvotes]: I was looking at the mean free time formula in the Drude model, and realized that it is proportional to 1/E. I am having trouble grasping this idea. I am aware that in practical cases the drift velocity is very small compared to the effective speed of the electrons. But if the force applied by the electric field increased appreciably, then a greater than average fraction of the electrons would only have a component only in that direction. If the electric field were to be increased until the effective speed became negligible, it is clear to see that the mean free time would increase.

REPLY [3 votes]: Keep in mind that that there is more going on, and that thinking about electrons as a gas is going to have limitations.
As you increase the electric field and the electrons accelerate, if the mean free path between scattering is the same, the mean velocity will be higher and the mean free time between collisions will be shorter.  So that checks out.
However the material also has a band structure. In different directions depending on the type of crystal the electrons are moving through.
So the resultant mobility or conductivity would be according to the orientation of the field. So if you applied fields in different directions you could get different nobilities or conductivities, and if you wanted to you can write he conductivity as a tensor. Sometimes people working with the Hall effect do that.
But if you start turning up the electric field, you can’t do that forever. At some point you will reach a saturation velocity. Looking at it from a band structure point of view, that would depend on the band structure. Or if you have a lot of scattering for some reason perhaps limited by the mean free path.
So the Drude model is really useful conceptually but doesn’t necessarily capture all the physics.


TITLE: Second order derivation of Quadratic form
QUESTION [4 upvotes]: I would like to find the second order derivative of a Quadratic form. Assume we have a random complex column vector $x$ and a real constant value $C$. I am interested in computing the following:
$$ \frac{\partial^2}{\partial x^H \partial x}(x^H C x) = ~? $$
The differentiation of the quadratic form is:
$$ d(x^H C x) = 2 \mathcal{R}( x^H C dx) $$
where $\mathcal{R}(\cdot)$ denotes the real part.
Are the following expressions correct?
$$d^2(x^H C x) = 2\mathcal{R}(C dx^H dx)$$
since $C$ is real and also the outcome of $dx^H dx$ is real, the operator $\mathcal{R}(\cdot)$ can be ignored and therefore:
$$ \frac{\partial^2}{\partial x^H \partial x}(x^H C x) = 2 C $$
And how about, if $C$ is a constant matrix with property $C = C^H$?

REPLY [1 votes]: If you want a derivative with respect to a random variable, then have a look to the Ito's theory. Now, we consider $x,x^H$ as standard variables; note that they are linked; yet we can write a polynomial $P(x,y)$ in $x,y$ under the (unique) form of a polynomial $Q(z,\overline{z})$ in $z,\overline{z}$. For instance, $P$ is harmonic iff $\dfrac{\partial ^2Q}{\partial z\partial \overline{z}}=0$.
Here your function $f:(x,x^H))\in M_{n,1}\times M_{1,n}\rightarrow x^HCx$ is linear with respect to $x$ and to $x^H$. Then $\dfrac{\partial ^2f}{\partial x\partial x^H}:(h,k)\in M_{n,1}\times M_{1,n}\rightarrow kCh=f(h,k)$.
Remark. I think that you only have a vague idea about these concepts; your formula (valid when $C$ is a hermitian matrix)
$ d(x^H C x) = 2 \mathcal{R}( x^H C dx) $
is a derivative with respect to $x$ only, that is to say that you consider that $x^H$ is a function of $x$ !!!


TITLE: Surjective monoid homomorphism $\text{End}(B)\to \text{End}(A)$ given surjection $g:B\to A$
QUESTION [3 upvotes]: For any set $A\neq\varnothing$ let $\text{End}(A)$ denote the endomorphism monoid, consisting of all functions $f:A\to A$, together with composition. If $A, B\neq \varnothing$ are sets and $g:B\to A$ is a surjection, is there a surjective monoid homomorphism $\varphi:\text{End}(B)\to \text{End}(A)$?

REPLY [7 votes]: This is a cleaner rewrite of my original answer. The answer is no (assuming the surjection is not injective and the smaller set does not have cardinality $1$).
Let $T_A$ be the full transformation monoid on the set $A$.  Then the set $C_A$ of constant maps is the unique minimal two-sided ideal of $T_A$.  Since $C_A$ has the same cardinality as $A$, we have $T_B\cong T_A$ if and only if $A$ and $B$ have the same cardinality.
I claim if $|A|\geq 2$, then the unique minimal non-trivial congruence on $T_A$ is to identify all the constant maps to a zero element (absorbing element).  Assuming $T_A$ acts on the left of $A$, we have that $T_A$ acts faithfully on the left of $C_A$ by essentially the same action.  So any homomorphism that is injective on $C_A$ is injective on $T_A$. On the other hand, if a congruence identifies elements of $C_A$ then the restriction of the congruence to $C_A$ is a system of imprimitivity for the symmetric group $S_A\leq T_A$ acting on the left of $C_A$ which is just the same as its natural action on $A$.  This action is $2$-transitive and hence primitive.  Thus any non-trivial congruence on $T_A$ must collapse $C_A$.
In conclusion, every proper quotient of $T_A$ has an absorbing element and so can only be a $T_X$ if $|X|=1$.  Combined with the fact  that $T_A\cong T_B$ iff $A$ and $B$ have the same cardinality, we get the answer is no.


TITLE: Inner regular measure in Hilbert space
QUESTION [0 upvotes]: let $\mu$ be a probability measure  in a separable hilbert space. Then
  for all $x\in H$, for all $\varepsilon >0$, there exists an $r>0$, such that
  $\mu(B_r(x))>1-\varepsilon$. $B_r(x)$ denotes the closed ball with radius
  $r$.

Proof:
Fix a countable Basis $B_n$ of $H$. Define inductively $E_n=B_n\setminus \cup_{k<n}(B_k)$. $E_n$ are pairwise disjoint and cover $H$, hence $\mu(\cup_n E_n)=1$. By continuity from below, there exists an $N$, with
$$\mu(\cup^N E_n)>1-\varepsilon$$
Take $r=\sup \text{dist}_{e\in \cup^NE_n}(x,e)$.
Question: is this proof correct?
is that just the definition of inner regular measure? is every
Borel probability measure inner regular in Hilbert space?

REPLY [1 votes]: Your proof is incorrect and also a lot more complicated than necessary. Note that if $B = \{B_n: n \in \mathbb{N}\}$ is an orthonormal basis of $H$, as indicated in the comments, then $B$ is a subset of the unit ball of $H$. So just consider $\mu = \delta_{2B_1}$ to see that $\mu(\bigcup_n E_n) = \mu(B) = 0 < 1$.
A simple proof of this fact notices that $H = \bigcup_{r \in \mathbb{N}} B_r(x)$ and so
$$1 = \mu(H) = \lim_{N \to \infty} \mu(\bigcup_{j = 1}^N B_j(x)) = \lim_{N \to \infty} \mu(B_N(x))$$
and so for $N$ large enough $\mu(B_N(x)) > 1- \varepsilon$.
To answer your question about inner regularity, I need to fix first what I will mean by inner regular since different sources I've seen differ.
I will use the definitions found in Bogachev's "measure theory". There, a Borel measure $\mu$ is called inner regular if for every Borel set $A$ we have
$$\mu(A)=\sup\{\mu(K): K\subseteq A, K \text{ is closed}\}$$
and is called tight if
$$\mu(A) = \sup\{\mu(K): K \subseteq A, K \text{ is compact}\}.$$
Note that with this definition, any tight measure is inner regular. However, the property stated is different from inner regularity because inner regularity tells you nothing when you take $A = H$ since $H$ is closed in itself.
Finally, using Bogachev's definitions it is true that every finite Borel measure on a metric space is regular but not every such measure is tight (see e.g. example 7.1.6 and theorem 7.1.7 in the second volume of Bogachev's measure theory).


TITLE: Why has this pattern formed in a photo of my laptop screen, and why does it vary from camera to camera?
QUESTION [0 upvotes]: So I got these weird patterns when I photographed my laptop screen with a phone camera (Realme Narzo 10A). I thought they might have something to do with diffraction. Then, for some reason, when I zoomed in to the pic slightly, the pattern morphed into something else.
What exactly is happening here?
Also, I tried the same thing with an iPhone 12, and the picture there had very few to none of these lines. Why is this so? Is there anything you can conclude about the phone based on these lines?

REPLY [2 votes]: This is a Moiré pattern, it results from the lines in your camera not being perfectly aligned with the lines on the screen, and the different resolutions of the both.


TITLE: What do we call a continuous function that induces a homeomorphism onto its image?
QUESTION [1 upvotes]: I know that an order-homomorphism that induces an isomorphism onto its image is called an order-embedding. So, I was expecting that a continuous function that induces an homeomorphism onto its image would be called a 'continuous embedding.' But, nothing really came up when I googled this phrase.
What's the standard terminology?

REPLY [3 votes]: In the context of general topology such a map is called simply an embedding. You could call it a topological embedding if you wanted to make the domain of discourse absolutely clear, and I’m pretty sure that I’ve also seen homeomorphic embedding.


text:
\section{Mathematical preliminaries}

The \textit{Fourier transform} of a function $f \in \LtR$ is a bounded
operator on $\LtR$ defined as follows:
\begin{equation}
  \mathcal{F} f (\xi) =
    \frac{1}{\sqrt{2 \pi}} \lim_{N \rightarrow \infty} \int_{-N}^N e^{-i x \xi} f(x) dx,
\end{equation}
where the limit is taken in $\LtR$.
We also use the notation $\hat{f}$ for $\mathcal{F} f$.

$\mathcal{F}$ is invertible on $\LtR$, and its bounded inverse, the
\textit{inverse Fourier transform}, is defined as follows:

\begin{equation}
  \mathcal{F}^{-1} g(x) =
    \frac{1}{\sqrt{2 \pi}} \lim_{N \rightarrow \infty} \int_{-N}^N e^{i \xi x} g(\xi) d\xi.
\end{equation}
Consequently,
\begin{equation}
	\mathcal{F}^{-1} = \mathcal{F}^\ast,
\end{equation}
i.e. the Fourier transform is a unitary operator on $\LtR$.

A bounded operator $P$ on a Hilbert space is called
\textit{idempotent}, if
\begin{equation}
  P^2 = P.
\end{equation}

A bounded operator $P$ on a Hilbert space is called an
\textit{orthogonal projection}, if it is idempotent and Hermitian,
i.e.
\begin{equation}
  P^2 = P \quad \text{and} \quad P^\ast = P.
\end{equation}

We denote the \textit{characteristic function} of a set $E \subset \RR$
by $\chi_E$
\begin{equation} \label{equ:defchi}
  \chi_E (x) = \left\{
               \begin{array}{ll}
               1, & \mbox{if $x \in E$},\\
               0, & \mbox{otherwise}.
               \end{array} \right.
\end{equation}
For a fixed set $E \subset \RR$, the mapping $f \mapsto \chi_E f$ is an
orthogonal projection on $\LtR$. We denote this projection also by $\chi_E$.
In particular, we use the notation $\chitau$, when $E = (-\tau,\tau)$,
$\tau > 0$.

For a fixed $\omega > 0$, we define the operator $S_\omega$ as follows
\begin{equation} \label{equ:defSomega}
	S_\omega := \mathcal{F} \chi_{(-\omega,\omega)} \mathcal{F}^\ast.
\end{equation}
The operator $S_\omega$ is also an orthogonal projection on $\LtR$.

The integral kernel of $S_\omega$ is well known, namely
\begin{equation}
  S_\omega f(x) = \int_\RR \frac{\sin \omega(x-y)}{\pi (x-y)} f(y) dy.
\end{equation}
This kernel is computed as follows:
\begin{equation} \label{equ:sinckernel}
  k(x,y) = \frac{1}{2 \pi} \int_{-\omega}^\omega e^{- i x t} e^{i t y} dt
         = \frac{1}{\pi} \frac{\sin \omega(x-y)}{x-y}.
\end{equation}
Due to \eqref{equ:defSomega}, $S_\omega$ is also an orthogonal
projection on $\LtR$.

If $P$ is an orthogonal projection, then
\begin{eqnarray}
  (I - P^\ast)(I - P)
  &=& (I - P)^2 \\
  &=& I - 2P + P^2 \\
  &=& I - P.
\end{eqnarray}
Thus for every vector $f$,
\begin{eqnarray}
  \Vert (I - P) f  \Vert^2
  &=& \langle (I - P^\ast)(I - P) f,f \rangle \\
  &=& \langle (I - P) f,f \rangle. \label{equ:norminnerprod}
\end{eqnarray}

For a bounded operator $T$, $\sigma(T)$ denotes the spectrum of $T$.
We need the following well-known lemma, see \cite[Prop. 6,
p. 16]{bonsall_duncan}.
\begin{lemma} \label{lem:spectrum}
  Let $A$ and $B$ be bounded operators on a Hilbert space. For $\lambda \neq 0$,
  $\lambda \in \sigma (AB)$ if and only if $\lambda \in \sigma (BA)$.
\end{lemma}

The following lemma has a straightforward proof, which is omitted.
\begin{lemma} \label{lem:lambdaPinverse} If $P$ is an idempotent
  bounded operator on a Hilbert space and $\lambda \neq 0,1$, then
  \begin{equation}
    (\lambda I  - P)^{-1} = \frac{1}{\lambda} I + \frac{1}{\lambda (\lambda-1)} P.
  \end{equation}
\end{lemma}


TITLE: Does the limit exist?
QUESTION [1 upvotes]: I had a question about the definition of a limit. I know for a limit to exist the right hand limit must equal the left hand limit but what if the graph of a specific function has the domain from [0,5] and if you wanted to find the limit at x=5 you know the left hand limit exists but the right hand doesn't.
So in this situation does the limit exist or not since only the right-hand limit exists?

REPLY [0 votes]: Check out the usual definition of the limit of a function of a real variable (https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit#Precise_statement).
If $D$ is any subset of $\mathbb{R}$, and $f: D \rightarrow \mathbb{R}$ is a function, and $c \in D$ (more generally $c$ can be a limit point of $D$), then we can talk about whether $\lim\limits_{x \to c} f(x)$ exists.
This stuff about left and right hand limits needing to coincide for the limit to exist only applies when $c \in D$ and $D$ contains an open interval about $c$.
For example, suppose $D = [0,5]$, and $f: D \rightarrow \mathbb{R}$ is the function $f(x) = x^2$.  Then $\lim\limits_{x \to 5} f(x)$ exists and is equal to $25$, because for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $x \in D$ and $|x - 5| < \delta$, then $|x^2 - 25| < \epsilon$.
So in this case, $\lim\limits_{x \to 5^-} f(x)$ and $\lim\limits_{x \to 5} f(x)$ exist, but it makes no sense to talk about $\lim\limits_{x \to 5^+} f(x)$.


TITLE: What *really* happens to atoms in chemical reactions?
QUESTION [9 upvotes]: I am imagining that the occasional oxygen might be floating close enough to a certain calcium atom and take the two electrons, which will ionize them and they will now 'glue'. Maybe the same thing happens in a hydrogen and chlorine, and a covalent bond is smoothly and calmly formed.
I expect these cases to be the exception, because if atoms have speeds of hundreds of meters per second, violent collisions would be the norm for their encounter. The question is: Is it the case that electron transfer occurs in collisions, but the kinetic energy overcomes the electrostatics and the atoms fly apart, and then they eventually find other partners to bond as they wander around? Even in covalent bonds? (It was the latter part that sparked the question, because in covalent 'no complete electron transfer occurs', but I couldn't picture atoms nicely slowing down to share electrons). Or is it always (collision=bonding-and-becoming-molecule-in-one-shot)? Or both?

REPLY [0 votes]: To complete John's answer, chemical reactions happen in matter which is at a certain temperature. All matter follows roughly the black body radiation curve which has an energy distribution for the atoms/molecules that comprise it , given its temperature.

The temperature is connected in statistical thermodynamics with the average kinetic and vibrational/rotational  energies in the sample. One can see that at room temperature (~300K), the radiations emitted by the interacting atoms/molecules is at small  fractions of electron volts. The kinetic energy of atoms and molecules is low enough to allow capture by the available channels. The tails of the distribution with high energy collisions have very small probability, falling exponentially.


TITLE: Zeros and the sign of the function
QUESTION [0 upvotes]: I need help solving this task, if anyone had a similar problem it would help me.
The task is:
Determine the zero and sign of the function
$$
y=1-x+\sqrt{{x^3\over x+3}}
$$
I found that the function intersects $x$ axis in $\frac{5+\sqrt{13}}{2}$ and $y$ in $1$. Is that right ?
How to find sign of the function ?
Thanks in advance!

REPLY [0 votes]: Well, if $y=0$ then
$$
\begin{split}
x - 1 &= \sqrt{\frac{x^3}{x+3}} \\
x^{3/2} &= (x-1)\sqrt{x+3} \\
x^3 &= (x-1)^2(x+3) = x^3+x^2-5x + 3 \\
0 &= x^2-5x + 3 \\
\end{split}
$$
and the quadratic formula applies to give you two solutions.
At $x = 0$, your function is positive, and when you cross a zero (in either direction), it will change sign.
Can you finish?


TITLE: Finding average velocity given curved graph?
QUESTION [5 upvotes]: I know that the average velocity is given by the ratio
$$ \frac{\text{total distance}}{\text{total time}}
$$ It should be easy to figure out from a linear displacement-time graph, but what about when only part of it is curved? Here is the graph I have:
I thought that the total distance on the left part is $(25-5) = 20$, and distance on the right part is $(25 - 10) = 15. 20 +15 = 35 = \text{total distance}$. Total time $= 8s$, so I thought avg velocity should be $35/8 - 4.375$, which is the wrong answer.
Where am I going wrong? How do I find total distance when there's a curved graph?

REPLY [1 votes]: My approach is somewhat different. I would say that the average velocity is equal to the total distance traveled divided by the total time. To that you need to calculate
$$v_{avg}=\frac{1}{8}\int \sqrt{1+\left(\frac{dy}{dx} \right)^2}~dx$$
Since the curve portion is not specified, I assumed it's a power law of the form $y=5+x^n$ with $n=\log 20/\log 4$. At this point numerical methods are needed. I found that
$$v_{avg}\approx\frac{36.2983}{8}=4.5373$$
The other answers seem to suggest that the average velocity is path independent. Do you think that your average speed to go the market is independent of your route?