text:
\section{Introduction}

How are two Leavitt path algebras related if the underlying graphs are related by the so called Cuntz splice?
This is arguably one of the most important open problems in the theory of graph algebras.
More concretely, the question is whether the Cuntz splice preserves either the Morita equivalence class or the isomorphism class of the Leavitt path algebras considered as either rings, algebras, or $*$-algebras.
In this paper, we provide the first answer to any of these questions:
We show that there cannot exist a $*$-isomorphism between the Leavitt path algebras over the ring of integers induced by the graphs $E_2$ and $E_{2-}$ shown below.

\begin{center}
\begin{tikzpicture}[shorten >=1pt]

\tikzset{vertex/.style = {shape=circle,draw,minimum size=2em}}
\tikzset{edge/.style = {->,> = latex'}}

\node at (0,0) {$E_2$};

\node[vertex] (u) at (2,0) {$u$};

\draw[edge] (u) to[loop, out=120, in=60, looseness=8] node [above] {$a$} (u);
\draw[edge] (u) to[loop, out=240, in=300, looseness=8] node [below] {$b$} (u);

\end{tikzpicture}
\end{center}

\begin{center}
\begin{tikzpicture}[shorten >=1pt]

\tikzset{vertex/.style = {shape=circle,draw,minimum size=2em}}
\tikzset{edge/.style = {->,> = latex'}}

\node at (0,0) {$E_{2-}$};

\node[vertex] (u) at  (2,0) {$u$};
\node[vertex] (v1) at  (4,0) {$v_1$};
\node[vertex] (v2) at  (6,0) {$v_2$};

\draw[edge] (u) to[loop, out=120, in=60, looseness=8] node [above] {$f_1$} (u);
\draw[edge] (u) to[loop, out=240, in=300, looseness=8] node [below] {$f_2$} (u);
\draw[edge] (u) to[bend left] node [midway, above] {$d_1$} (v1);

\draw[edge] (v1) to[loop, out=120, in=60, looseness=8] node[above] {$e_1$} (v1);
\draw[edge] (v1) to[bend left] node[midway, above] {$e_2$} (v2);
\draw[edge] (v1) to[bend left] node[midway, below] {$d_2$} (u);

\draw[edge] (v2) to[loop, out=30, in=-30, looseness=8] node[right] {$e_4$} (v2);
\draw[edge] (v2) to[bend left] node[below] {$e_3$} (v1);

\end{tikzpicture}
\end{center}

\noindent Some authors prefer the name $R_2$ (``rose with two petals'') for the graph we call $E_2$.

The interest in isomorphism and Morita equivalence of Leavitt path algebras of graphs related by the Cuntz splice stems from a desire to understand whether Leavitt path algebras can be classified through algebraic analogues of the celebrated results of R\o rdam (\cite{RordamClassificationOfCK}) and Kirchberg--Phillips (\cite{KirchbergClassification, KirchbergPhillipsEmbedding, PhillipsClassification}).
We now give a short overview of some of the historic developments in the field.

To each irreducible non-permutation matrix $A \in M_n(\mathbb{N})$, one can associate a shift of finite type, $\mathcal{X}_A$.
For such shift spaces, the Bowen--Franks invariant consisting of the Bowen--Franks group $BF(A) = \mathbb{Z}^n / (I-A)\mathbb{Z}^n$ and the sign of the determinant of $I - A$ is a complete invariant of flow equivalence \cite{BowenFranks, FranksFlowEquivalence, ParrySullivan}.
To each non-permutation matrix $A \in M_n(\mathbb{N})$, one can also associate  a $C^*$-algebra $\mathcal{O}_A$, called the Cuntz--Krieger algebra, and the stable isomorphism class of $\mathcal{O}_A$ is a flow invariant for $\mathcal{X}_A$.
In fact, $K_0(\mathcal{O}_A) \cong BF(A)$.
Considering the flow classification of $\mathcal{X}_A$, one might wonder if $\mathcal{O}_A$ also contains enough information to determine the other half of the Bowen--Franks invariant, $\sgn(\det(I-A))$. On its own, $\mathcal{O}_A$ does not: R{\o}rdam proved that simple Cuntz--Krieger algebras are classified up to stable isomorphism by their $K_0$-groups \cite{RordamClassificationOfCK}.

Leavitt path algebras were introduced independently in \cite{AbramsArandaPino05} and \cite{AraMorenoPardo} as universal algebras over some field $K$ with relations given by the underlying graph.
They are purely algebraic analogues of graph $C^*$-algebras, which, in turn, are generalizations of Cuntz--Krieger algebras, see for instance \cite{BatesPaskRaeburnSzymanski, EnomotoWatatani, FowlerLacaRaeburn, KumjianPaskRaeburn, KumjianPaskRaebrunRenault, RaeburnBook}.
More generally, work of Tomforde shows how to define a Leavitt path algebra over any unital, commutative ring \cite{TomfordeLeavittOverRing}.
A good overview of the theory of Leavitt path algebras can be found in the survey paper \cite{AbramsFirstDecade}.

In \cite{AbramsLoulyPardoSmith}, it is shown that if $\mathcal{X}_A$ is flow equivalent to $\mathcal{X}_B$ for finite irreducible non-permutation matrices $A$ and $B$, then $L_K(E_A)$ is Morita equivalent to $L_K(E_B)$ for any field $K$.
It is also shown that the $K_0$-group of $L_K(A)$ coincides with the Bowen--Franks group of $A$, so that a Leavitt path algebra is classified by its $K_0$-group and the sign of the determinant.
However, it is an open problem whether the sign of the determinant is an isomorphism invariant, and this leaves the theory of Leavitt path algebras in a situation very similar to the situation in the theory of Cuntz--Krieger algebras prior to R\o rdam's classification result.

For the graphs $E_2$ and $E_{2,-}$ defined above, it is straightforward to check that the associated Bowen--Franks groups are trivial.
However, the signs of the determinants are different, and hence, the associated shifts of finite type are not flow equivalent.
The graph $E_{2-}$ is constructed from $E_2$ by attaching two extra vertices and the associated edges.
This gluing operation is called ``performing a Cuntz splice'', and it has played an important role in the study Cuntz--Krieger algebras, graph $C^*$-algebras, and Leavitt path algebras \cite{AbramsLoulyPardoSmith, ERRS, RestorffClassificationOfCK, RordamClassificationOfCK, RuizTomforde,  SorensenGeometricClassification}.
It serves to flip the sign of the determinant while keeping the Bowen--Franks group fixed.
The graph $E_2$ is the simplest possible example of a graph where this operation can be performed without obviously changing Morita equivalence class.

In R{\o}rdam's work, it was shown that $C^*(E_2) \cong C^*(E_{2-})$, and this was then used to derive the general non-reliance on the sign of the determinant \cite{RordamClassificationOfCK}.
If we wish to pursue a similar approach to the classification of Leavitt path algebras and understand whether there can exists an algebraic Kirchberg--Phillips theorem in this setting, then we must know whether the Leavitt path algebras associated to $E_2$ and $E_{2-}$ are isomorphic.

Our main results are Theorem \ref{thm: algebraic matsumoto matui}, Theorem \ref{thm: isomorphism same sign}, and Corollary \ref{cor: only diagonal projections}.
All these results concern a subalgebra of $L_{R}(E)$ called the diagonal, see Definition \ref{def: diagonal}.
Specifically, Theorem \ref{thm: algebraic matsumoto matui} states that for a subring $R$ of $\mathbb{C}$ that is closed under complex conjugation and graphs $E$ and $F$ satisfying standard assumptions, there can only exist a diagonal preserving $*$-isomorphism between $L_R(E)$ and $L_R(F)$ if the Bowen--Franks determinants are equal.
This is a partial algebraic analogue of a result of Matsumoto and Matui (\cite{MatsumotoMatui}) and it is derived directly from their result.
Theorem \ref{thm: isomorphism same sign} shows that if all projections in $L_{R}(F)$ are elements of the diagonal, then $L_{R}(E)$ and $L_{R}(F)$ can only be isomorphic as $*$-algebras if their Bowen--Franks determinants are equal.
Corollary \ref{cor: only diagonal projections} shows that all projections of $L_{\mathbb{Z}}(E)$ are elements of the diagonal if $E$ is a finite graph.

In the initial version of this paper, we only proved that all projections in $L_{\mathbb{Z}}(E_2)$ are diagonal, which was sufficient to let us conclude that $L_{\mathbb{Z}}(E_2)$ and $L_{\mathbb{Z}}(E_{2-})$ are not $*$-isomorphic.
Our arguments were based on a complete description of the unitaries in $L_{\mathbb{Z}}(E_2)$ given in \cite{BS}.
Since then, Chris Smith has shown us how to extend the description of the unitaries in $L_{\mathbb{Z}}(E_2)$ from \cite{BS} to cover all finite graphs, see Proposition \ref{prop: standard form unitaries}, and we have realized how to extend this from $\mathbb{Z}$ to slightly more general rings.
Therefore, we can show that all projections in $L_{R}(E)$ are diagonal whenever $E$ is  a finite graph and $R$ is a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup\ (see Definition \ref{def:PropertyX}).
In particular, this holds if $R = \mathbb{Z}$.

The three main results combine to show that if $E,F$ are finite, strongly connected, essential and non-trival graphs (see Definition \ref{def: graph concepts}), if $R$ is a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup\ and if $L_{R}(E)$ and $L_{R}(F)$ are $*$-isomorphic, then $\sgn(I - A_E) = \sgn(I - A_F)$ (Corollary \ref{cor: iso implies same sign}).
Possible applications of the results to questions concerning Morita equivalence and Leavitt path algebras over fields are discussed in Remarks \ref{remark:fields} and \ref{remark:morita}.

Since the initial version of this paper appeared on the arXiv, there have been some interesting developments related to the problem discussed here.
In \cite[Corollary 4.8]{AraBosaHazratSims} the authors use groupoid techniques to show that when $R$ is an integral domain, there can be no diagonal preserving \emph{ring} isomorphism from $L_{2,R}$ to $L_{2-,R}$.
In \cite[Corollary 4.7]{CarlsenRuizSims}, it is shown that $L_{2,\mathbb{Z}}$ to $L_{2-,\mathbb{Z}}$ are not stably isomorphic as $*$-algebras, in a suitable sense.

\section{Preliminaries}

\begin{definition}
A graph $E = (E^0, E^1, r, s)$ is a four tuple where $E^0$ is the vertex set, $E^1$ the edge set, and $r,s \colon E^1 \to E^0$ are the range and source maps.
\end{definition}

\noindent Given a graph $E$, we denote by $A_E$ its adjacency matrix, i.e.\ the $E^0 \times E^0$ matrix with $A(v,w) = |s^{-1}(v) \cap r^{-1}(w)|$.
The following graph concepts will be needed below.

\begin{definition}[{\cite{KumjianPaskRaeburn}}]
A graph $E$ satisfies Condition (L) if every cycle in $E$ has an exit.
\end{definition}

\begin{definition} \label{def: graph concepts}
A graph $E$ is said to be:
\begin{itemize}
	\item \emph{strongly connected} if there is a path between any two vertices,
	\item \emph{essential} if it has no sinks or sources, and
	\item \emph{trivial} if it is a single cycle with no other edges or vertices.
\end{itemize}
\end{definition}

\noindent Note that if $E$ is an essential and strongly connected graph then $A_E$ will be an irreducible matrix with no zero rows or columns.
Furthermore, if $E$ is non-trivial then $A_E$ will not be a permutation matrix.

For a graph $E$, $E^n$ will denote the set of \emph{paths} of length $n$, and $E^* = \cup_{n \in \mathbb{N}} E^n$ will denote the collection of all finite paths. The range and source maps are extended to these sets in the natural way. Similarly, $E^\infty = \{ e_1 e_2 \cdots \mid e_i \in E^1 \textnormal{ and } r(e_i) = s(e_{i+1}) \}$ will denote the set of infinite paths.
In the case of $E_2$, $E_2^\ast = \{a,b\}^* $ and $E_2^\infty = \{a,b\}^\mathbb{N}$. The \emph{cylinder set} of $\alpha \in E^\ast$ is defined to be
\begin{displaymath}
\mathcal{Z}(\alpha) = \{ \alpha \xi \mid \xi \in E^\infty \textnormal{ and } \alpha \xi \in E^\infty\}.
\end{displaymath}
For $\alpha, \beta \in E^*$, $\mathcal{Z}(\alpha)$ and $\mathcal{Z}(\beta)$ are disjoint if and only if neither $\alpha$ nor $\beta$ extends the other.

To each graph $E$, we associate a graph $C^*$-algebra and a collection of Leavitt path algebras as defined below.

\begin{definition}[{\cite[Definition 1]{FowlerLacaRaeburn}}]
Let $E$ be a graph.
The graph $C^*$-algebra of $E$, $C^*(E)$, is the universal $C^*$-algebra generated by mutually orthogonal projections $\{ p_v \mid v \in E^0 \}$ and partial isometries $\{ s_e \mid e \in E^1 \}$ subject to the relations
\begin{enumerate}[(i)]
 \item $s_e^* s_f = 0$, if $e \neq f$,
 \item $s_e^* s_e = p_{r(e)}$,
 \item $s_e s_e^* \leq p_{s(e)}$, and,
 \item $p_v = \sum_{e \in s^{-1}(v)} s_e s_e^*$, if $s^{-1}(v)$ is finite and nonempty.
\end{enumerate}

\end{definition}

\noindent Readers unfamiliar with the subject should be warned that there are two competing conventions for the definition of $C^*(E)$. This dichotomy stems from a necessary asymmetry between sinks and sources in the defining relations.
Raeburn's monograph \cite{RaeburnBook} uses the other possible convention.

\begin{definition}[{\cite[Definitions 2.4 and 3.4]{TomfordeLeavittOverRing}}] \label{def:lpa}
Let $R$ be a commutative ring with unit and let $E$ be a graph.
The Leavitt path algebra of $E$ over $R$ is the universal $R$-algebra generated by pairwise orthogonal idempotents $\{ v \mid v \in E^0\}$ and elements $\{e, e^* \mid e \in E^1\}$ satisfying
\begin{enumerate}[(i)]
 \item $e^* f = 0$, if $e \neq f$,
 \item $e^* e = r(e)$,
 \item $s(e) e = e = e r(e)$,
 \item $e^* s(e) = e^* = r(e) e^*$, and,
 \item $v = \sum_{e \in s^{-1}(v)} e e^*$, if $s^{-1}(v)$ is finite and nonempty. \label{item:summation_relation}
\end{enumerate}
\end{definition}

By \cite[Proposition 3.4]{TomfordeLeavittOverRing}, $L_R(E) = \operatorname{span}_{R}\{ \alpha \beta^* \mid \alpha, \beta \in E^*, r(\alpha) = r(\beta)\}$.
For any unital, commutative ring $R$, we can extend the map $\alpha \beta^* \mapsto \beta \alpha^*$ to a \emph{linear} involution on $L_R(E)$.
If $R$ is a subring of $\mathbb{C}$ that is closed under complex conjugation, we will
extend $\alpha \beta^* \mapsto \beta \alpha^*$ to a \emph{conjugate linear} involution on $L_{R}(E)$ instead.
Throughout this paper, we are mainly interested in the latter case.

Since we consider $L_{R}(E)$ as a $*$-algebra, we prefer to work with self-adjoint idempotents rather than just idempotents.
Following the name conventions of $C^*$-algebras, we call these elements projections.

\begin{definition}
An element $p \in L_{R}(E)$ is called a \emph{projection} if $p = p^2 = p^*$.
\end{definition}

Each graph algebra contains a distinguished subalgebra called the diagonal:

\begin{definition} \label{def: diagonal}
Let $E$ be a graph and $R$ a commutative ring with unit.
We define
\[
  \mathcal{D}(C^*(E)) = \overline{\newspan}_\C\{ s_\alpha s_\alpha^* \mid \alpha \in E^* \},
\]
and
\[
 \mathcal{D}(L_R(E)) = \operatorname{span}_{R}\{ \alpha \alpha^* \mid \alpha \in E^* \}.
\]
We refer to these sets as the diagonal of $C^*(E)$ and $L_R(E)$, respectively.
\end{definition}

\noindent It is shown in \cite[Theorem 5.2]{HopenwasserPetersPower} and \cite[Theorem 3.7]{NagyReznikoff}) that $\mathcal{D}(C^*(E))$ is a MASA (maximal abelian sub-algebra) in $C^*(E)$ if $E$ satisfies Condition (L).
We note that $\mathcal{D}(L_R(E))$ is generated by projections.
It has recently been shown that if $E$ satisfies condition (L) then $\mathcal{D}(L_{R}(E))$ is maximal abelian in $L_R(E)$ (see \cite[Proposition 3.12 and Theorem 3.9]{GilNasr} and \cite[Lemma 3.13]{AraBosaHazratSims}).

We are interested in subrings of $\mathbb{C}$ that behave like $\mathbb{Z}$ in the following very specific way.

\begin{definition}[Essentially unique partition of the unit] \label{def:PropertyX}
A unital (same unit) subring $R \subset \mathbb{C}$ closed under complex conjugation has an \emph{essentially unique partition of the unit} if whenever $\lambda_1, \lambda_2, \ldots, \lambda_n \in R$ satisfy
\[
  \sum_{i=1}^n \lambda_i \overline{\lambda_i} = \sum_{i=1}^n |\lambda_i|^2 = 1,
\]
then all but one of the $\lambda_i$ is zero.
\end{definition}

\begin{example}
The following is an incomplete list of subrings of $\mathbb{C}$ with an essentially unique partition of the unit.
\begin{itemize}
 \item $\mathbb{Z}$.
 \item $\mathbb{Z} + \sqrt{p}\mathbb{Z}$, for $p$ a prime.
 \item The Gaussian integers, $\mathbb{Z} + i\mathbb{Z}$.
 \item The ring $\mathbb{Z}[\pi]$, $\mathbb{Z} + \pi\mathbb{Z} + (\pi^2)\mathbb{Z} + \cdots$.
\end{itemize}
\end{example}

\begin{remark} \label{rmk:PropX_nohalf}
Suppose $R \subset \mathbb{C}$ has an essentially unique partition of the unit.
For each $n \neq 1$, $\frac{1}{n}\notin R$ since
\[
  \sum_{k=1}^{n^2} \left( \frac{1}{n} \right)^2 = 1.
\]
Note that this means that no subfield of $\mathbb{C}$ has an essentially unique partition of the unit.
\end{remark}

\section{Matsumoto and Matui's Result for Graph Algebras}

In \cite{MatsumotoMatui}, Matsumoto and Matui show that under standard assumptions on matrices $A$ and $B$, there can only exist a diagonal preserving isomorphism between the Cuntz--Krieger algebras (see Definition \ref{def: Cuntz Krieger Algebra}) $\mathcal{O}_A$ and $\mathcal{O}_B$ if $\sgn(\det (I - A)) = \sgn( \det(I - B))$.
In this section, we will translate this part of Matsumoto and Matui's spectacular result into the world of graph algebras. In Section \ref{subsection:cstar}, the result is translated into a statement about graph $C^*$-algebras and in Section \ref{subsection:leavitt}, this is used to prove an algebraic analogue for Leavitt path algebras over subrings of $\mathbb{C}$. Both results are straightforward adaptations of those presented in \cite{MatsumotoMatui}.

\subsection{Graph $C^*$-algebras}\label{subsection:cstar}

First, recall the definition of a Cuntz--Krieger algebra and its diagonal (originally from \cite{CuntzKrieger}).

\begin{definition} \label{def: Cuntz Krieger Algebra}
Let $A$ be an $N \times N$ matrix with entries in $\{0,1\}$.
The Cuntz--Krieger algebra $\mathcal{O}_A$ is the universal $C^*$-algebra generated by partial isometries $S_1, S_2, \ldots, S_N$ satisfying the relations
\begin{itemize}
 \item $\sum_{i=1}^N S_i S_i^* = 1$, and,
 \item $S_i^* S_i = \sum_{j=1}^N A(i,j) S_j S_j^*$,
\end{itemize}
where $A(i,j)$ denotes the $ij$'th entry of $A$.
The diagonal of $\mathcal{O}_A$ is
\[
  \mathcal{D}(\mathcal{O}_A) = \overline{\newspan}_\C\{ S_{i_1} S_{i_2} \ldots S_{i_k} S_{i_k}^* \ldots S_{i_1}^* \mid k \in \mathbb{N}, i_1, i_2, \ldots i_k \in \{1,2,\ldots, N\}  \}.
\]
\end{definition}

We observe that for two indicies $i,j$ we have
\[
  S_i S_j = S_i S_i^* S_i S_j = S_i \left( \sum_{k=1}^N A(i,k) S_{k} S_{k}^* \right) S_j = A(i,j) S_i S_j.
\]
Hence, $S_i S_j = 0$ if $A(i,j) = 0$.

Condition (L) for graphs was designed to be an equivalent of Cuntz and Krieger's Condition (I) (\cite{CuntzKrieger}).
We recall that a $\{0,1\}$ matrix will satisfy Condition (I) if it is not a permutation matrix.

Graph $C^*$-algebras were originally defined to be generalizations of Cuntz--Krieger algebras, and there are strong connections between the two concepts.
We have been unable to find a reference for the following lemma, but it is certainly well-known to researchers in the field.

\begin{lemma} \label{lem:CKisGraph}
Let $A$ be an irreducible $N \times N$ matrix with entries in $\{0,1\}$ that satisfies Condition (I). Let $E_A$ be the graph with $E_{A}^0 = \{1,2, \ldots, N\}$, $E_{A}^1 = \{ [ij] \mid A(i,j) = 1 \}$ and range and source maps given by
\[
  r([ij]) = j \quad \quad \text{ and } \quad \quad s([ij]) = i.
\]
There exists an isomorphism $\phi \colon C^*(E_A) \to \mathcal{O}_{A}$ such that $\phi(\mathcal{D}(C^*(E_A))) = \mathcal{D}(\mathcal{O}_A)$.
\end{lemma}
\begin{proof}
Define $\phi$ on generators by
\begin{align*}
 \phi(p_{i}) &= S_i S_i^*, \\
 \phi(s_{[ij]}) &= S_i S_j S_j^*.
\end{align*}
By (the proof of) \cite[Proposition 4.1]{MannRaeburnSutherland}, $\phi$ is an isomorphism.

Given a path $\alpha = [i_1 i_2] [i_2 i_3] \ldots [i_{k-1} i_k]$ in $E_A$ we see that
\begin{align*}
  \phi(s_\alpha) &= (S_{i_1} S_{i_2} S_{i_2}^*) (S_{i_2} S_{i_3} S_{i_3}^*) \cdots (S_{i_{k-1}} S_{i_k} S_{i_k}^*) \\
	       &= S_{i_1} (S_{i_2} S_{i_2}^* S_{i_2}) (S_{i_3} S_{i_3}^* S_{i_3}) \cdots (S_{i_{k-1}} S_{i_{k-1}}^* S_{i_{k-1}}) S_{i_k} S_{i_k}^* \\
	       &= S_{i_1} S_{i_2} \cdots S_{i_k} S_{i_k}^*.
\end{align*}
Hence
\[
  \phi(s_\alpha s_\alpha^*) = S_{i_1} S_{i_2} \cdots S_{i_k} S_{i_k}^* S_{i_k} S_{i_k}^* S_{i_{k-1}}^* \cdots S_{i_1}^* = S_{i_1} S_{i_2} \cdots S_{i_k} S_{i_k}^* S_{i_{k-1}}^* \cdots S_{i_1}^*.
\]
From this, it follows that $\phi(\mathcal{D}(C^*(E_A))) \subseteq \mathcal{D}(\mathcal{O}_A)$.
Since $S_i S_j \neq 0$ if and only if $[ij]$ is an edge in $E_A$, the computation above  also shows that $\phi$ maps $\mathcal{D}(C^*(E_A))$ onto $\mathcal{D}(\mathcal{O}_A)$.
\end{proof}

\begin{theorem}[{cf.\ \cite[Theorem 3.6]{MatsumotoMatui}}] \label{thm: matsumoto matui for graphs}
Let $E,F$ be finite, essential, strongly connected graphs that satisfy Condition (L).
If there exists an isomorphism $\phi \colon C^*(E) \to C^*(F)$ such that $\phi(\mathcal{D}(C^*(E))) = \mathcal{D}(C^*(F))$, then
\[
  \sgn(\det (I - A_E)) = \sgn( \det(I - A_F)).
\]
\end{theorem}

\begin{proof}
In order to apply the result from \cite{MatsumotoMatui}, out-splitting will be used to replace $E$ and $F$ by graphs without multiple edges (see \cite[Section 3]{BatesPask}).
For each $v \in E^0$, partition $s^{-1}(v)$ into singleton sets. Note that these partitions are proper, and let $E_1$ be the graph obtained from the corresponding out-splitting.
Note that since $E$ is essential, strongly connected and satisfies Condition (L), $E_1$ will be essential, strongly connected and will satisfy Condtion (L).
Analogously, let $F_1$ be the graph obtained from $F$ by such a complete out-splitting.
By \cite[Corollary 6.2]{BrownloweCarlsenWhittaker}, there exist diagonal preserving isomorphisms $C^*(E) \to C^*(E_1)$ and $C^*(F) \to C^*(F_1)$.
Hence, there also exists a diagonal preserving isomorphism from $C^*(E_1)$ to $C^*(F_1)$.

Since $E_1$ and $F_1$ have no multiple edges, satisfy Condition (L), are essential, and strongly connected, $A_{E_1}$ and $A_{F_1}$ are  irreducible $\{0,1\}$ matrices that satisfy Condition (I).
By Lemma \ref{lem:CKisGraph}, $C^*(E_1) \cong \mathcal{O}_{A_{E_1}}$ and $C^*(F_1) \cong \mathcal{O}_{A_{F_1}}$ in a diagonal preserving way, so the argument above guarantees the existence of a diagonal preserving isomophism from $C^*(\mathcal{O}_{A_{E_1}})$ to $C^*(\mathcal{O}_{A_{F_1}})$, and hence, it follows from \cite[Theorem 3.6]{MatsumotoMatui} that $\sgn(\det (I - A_{E_1})) = \sgn( \det(I - A_{F_1}))$.
For edge shifts, out-splitting produces a shift space conjugate to the original (see e.g.\ \cite[Theorem 2.4.10]{LindMarcus}). Since flow equivalence is a coarser equivalence relation than conjugacy, this implies that the edge shift of $E$ is flow equivalent to the edge shift of $E_1$. Clearly, the same relation exists between $F$ and $F_1$. By \cite{ParrySullivan}, the sign of the determinant is an invariant of flow equivalence, so
\[
 \sgn(\det (I - A_E)) = \sgn(\det (I - A_{E_1})) = \sgn( \det(I - A_{F_1})) = \sgn( \det(I - A_F)). \qedhere
\]
\end{proof}

\subsection{Leavitt path algebras over certain subrings of $\mathbb{C}$}\label{subsection:leavitt}

In this section, we will prove an algebraic analogue of Theorem \ref{thm: matsumoto matui for graphs}, i.e.\ a weak algebraic version of Matsumoto and Matui's result.
We will look only at rings $R$ that are subrings of $\mathbb{C}$ closed under complex conjugation, and we will equip $L_R(E)$ with a \emph{conjugate} linear involution.
In the following, $\mathbb{Z}$ will be the main example of such a ring.
First, we formalize the connection between $L_{R}(E)$ and $C^*(E)$.

\begin{lemma} \label{lem: LZE into CE}
Let $E$ be a graph and let $R$ be a subring of $\mathbb{C}$ closed under complex conjugation.
The map $\iota_E \colon L_{R}(E) \to C^*(E)$
given on generators by

\begin{align*}
  \iota_E(v) &= p_v, \quad \text{ for } v \in E^0, \\
  \iota_E(e) &= s_e, \quad \text{ for } e \in E^1.
\end{align*}
extends to a $*$-algebra embedding.
\end{lemma}
\begin{proof}
By the universal property of $L_{R}(E)$,
$\iota_E$ extends to a $*$-homomorphism.
It is injective by the Graded Uniqueness Theorem (\cite[Theorem 5.3]{TomfordeLeavittOverRing}), see \cite[Theorem 7.3]{TomfordeUniquenessTheorems} for details on how to apply the Graded Uniqueness Theorem.
\end{proof}

We now provide a version of \cite[Theorem 4.4]{AbramsTomforde} for Leavitt path algebras over subrings of $\mathbb{C}$
(the result in \cite{AbramsTomforde} is for Leavitt path algebras over $\mathbb{C}$).
This version of the result additionally tracks the image of the diagonal.

\begin{lemma} \label{lem: extending diagonal iso}
Let $R$ be a subring of $\mathbb{C}$ closed under complex conjugation and
let $E,F$ be finite graphs that satisfy Condition (L).
If $\phi \colon L_R(E) \to L_R(F)$
is a $*$-isomorphism such that $\phi(\mathcal{D}(L_R(E))) \subseteq \mathcal{D}(L_R(F))$,
then $\phi$ extends to an isomorphism $\overline{\phi} \colon C^*(E) \to C^*(F)$ such that $\overline{\phi}(\mathcal{D}(C^*(E))) = \mathcal{D}(C^*(F))$.
\end{lemma}
\begin{proof}
As in the proof of \cite[Theorem 4.4]{AbramsTomforde}, $\phi$ extends to an isomorphism $\overline{\phi} \colon C^*(E) \to C^*(F)$ satisfying $\overline{\phi} \circ \iota_{E} = \iota_F \circ \phi$.
Hence,
\begin{align*}
  \overline{\phi}(\mathcal{D}(C^*(E)))	&= \overline{\operatorname{span}}_\mathbb{C}\left\{\overline{\phi}(s_\alpha s_\alpha^*) \mid \alpha \in E^*\right\} \\
				&= \overline{\operatorname{span}}_\mathbb{C}\left\{\overline{\phi}(\iota_E(\alpha \alpha^*)) \mid \alpha \in E^*\right\} \\
				&= \overline{\operatorname{span}}_\mathbb{C}\left\{\iota_F(\phi(\alpha \alpha^*)) \mid \alpha \in E^*\right\} \\
				&= \overline{\operatorname{span}}_\mathbb{C}\left\{\iota_F(\phi(\operatorname{span}_R\{\alpha \alpha^* \mid \alpha \in E^* \} ))\right\} \\
				&= \overline{\operatorname{span}}_\mathbb{C}\left\{\iota_F(\phi(\mathcal{D}(L_R(E))))\right\} \\
				&\subseteq \overline{\operatorname{span}}_\mathbb{C}\left\{\iota_F(\mathcal{D}(L_R(F)))\right\} \\
				&=  \overline{\operatorname{span}}_\mathbb{C}\{ \operatorname{span}_R \{ s_\alpha s_\alpha^* \mid \alpha \in F^* \}\} \\
				&= \overline{\operatorname{span}}_\mathbb{C}\{ s_\alpha s_\alpha^* \mid \alpha \in F^* \} \\
				&= \mathcal{D}(C^*(F)).
\end{align*}
Since $\overline{\phi}$ is an isomorphism and $\mathcal{D}(C^*(E))$ is a MASA in $C^*(E)$,
$\overline{\phi}(\mathcal{D}(C^*(E)))$ is a MASA in $C^*(F)$.
As $\mathcal{D}(C^*(F))$ is also a MASA in $C^*(F)$, it follows that $\overline{\phi}(\mathcal{D}(C^*(E))) = \mathcal{D}(C^*(F))$.
\end{proof}

\begin{theorem} \label{thm: algebraic matsumoto matui}
Let $E,F$ be finite, essential, strongly connected graphs that satisfy Condition (L), and let $R$ be a subring of $\mathbb{C}$ closed under complex conjugation.
If $\phi \colon L_R(E) \to L_R(F)$
is a $*$-isomorphism such that $\phi(\mathcal{D}(L_R(E))) \subseteq \mathcal{D}(L_R(F))$,
then
\[
  \sgn(\det (I - A_E)) = \sgn( \det(I - A_F)).
\]
\end{theorem}
\begin{proof}
By Lemma \ref{lem: extending diagonal iso}, $\phi$ extends to a diagonal preserving isomorphism $\overline{\phi} \colon C^*(E) \to C^*(F)$, so the result follows from Theorem \ref{thm: matsumoto matui for graphs}.
\end{proof}

\section{Unitaries in $L_{R}(E)$}
In this section, we show that all unitaries in $L_{R}(E)$ can be written in a certain standard from when $E$ is a finite graph and $R$ is a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup\ (see Proposition \ref{prop: standard form unitaries}).
This is similar to what is done in \cite{BS} for unitaries in $L_{\mathbb{Z}}(E_2)$, and the authors are grateful to Chris Smith for showing them how to generalize the arguments given there.
A particularly interesting feature of Smith's argument is that it eschews the presentation of $L_{\mathbb{Z}}(E)$ as endomorphisms on an uncountably generated free module used in \cite{BS} in favor of techniques based on the grading of Leavitt path algebras.
We generalize Smith's argument to cover all rings that have an essentially unique partition of the unit.

We first note a simple way to find unitaries in $L_{R}(E)$.

\begin{lemma} \label{lem: orthogonal paths}
Let $E$ be a finite graph and let $R$ be a unital commutative ring with characteristic $0$.
If $\alpha_1, \alpha_2, \ldots, \alpha_n \in E^*$ satisfy
\[
 \sum_{i=1}^n \alpha_i \alpha_i^* = 1,
\]
then $\alpha_i^* \alpha_j = 0$ for $i \neq j$.
\end{lemma}
\begin{proof}
By relabeling, we can assume that $|\alpha_1| \geq |\alpha_2| \geq \cdots \geq |\alpha_n|$.
We see that
\[
	\alpha_1 = 1 \alpha_1 = \left( \sum_{i=1}^n \alpha_i \alpha_i^* \right) \alpha_1 = \sum_{i=1}^n \alpha_i \alpha_i^* \alpha_1.
\]
The first term in the sum on the right hand side is $\alpha_1$, and since $\alpha_1$ has maximal length among the $\alpha_i$, each subsequent term is either a real path or equals $0$.
Since the real paths are linearly independent (\cite[Proposition 4.9]{TomfordeLeavittOverRing}) and $R$ has characteristic $0$, it follows that $ \alpha_i \alpha_i^* \alpha_1 = 0$, for each $i \neq 1$.
Hence, $\alpha_i^* \alpha_1 = 0$.
By taking adjoints, $\alpha_1^* \alpha_i = 0$ for $i = 2, 3, \ldots, n$.

Let $k = 2, 3, \ldots, n$ be given, and assume that $\alpha_i^* \alpha_k = 0$ for all $i < k$.
Then
\[
	\alpha_k = 1 \alpha_k = \sum_{i=1}^n \alpha_i \alpha_i^* \alpha_k = \sum_{i=k}^n \alpha_i \alpha_i^* \alpha_k.
\]
Since $\alpha_k$ has maximal length among the $\alpha_i$ appearing in the sum on the right hand side, we see as above that $\alpha_i^* \alpha_k = 0 = \alpha_k^* \alpha_i$ for $i = k+1, k+2, \ldots, n$.
By induction, $\alpha_i^* \alpha_j = 0$ all $i \neq j$.
\end{proof}

This gives us a simple way to find unitaries in $L_{R}(E)$:

\begin{lemma} \label{lem: certain unitaries}
Let $E$ be a finite graph and let $R$ be a unital commutative ring with characteristic $0$.
Suppose that $\alpha_i, \beta_i \in E^*$ and $\lambda_i \in R$, $i=1,2,\ldots,n$, are such that
\begin{enumerate}
	\item $\sum_{i=1}^n \alpha_i \alpha_i^* = 1$, \label{item:certain unitaries 1}
	\item $\sum_{i=1}^n \beta_i \beta_i^* = 1$, \label{item:certain unitaries 2}
\end{enumerate}
and (if $L_{R}(E)$ is equipped with a linear involution)
\begin{enumerate}
\setcounter{enumi}{2}
\item $\lambda_i^2 = 1$ for all $i$  \label{item:certain unitaries 3}
\end{enumerate}
or (if $L_{R}(E)$ is equipped with a conjugate linear involution)
\begin{enumerate}
\setcounter{enumi}{2}
\item $\lambda_i \overline{\lambda_i} = 1$ for all $i$. \label{item:certain unitaries 3'}
\end{enumerate}
Then
\[
	u = \sum_{i=1}^n \lambda_i \alpha_i \beta_i^*,
\]
is a unitary.
\end{lemma}
\begin{proof}
This follows from a simple computation of $uu^*$ and $u^*u$ using Lemma \ref{lem: orthogonal paths}.
\end{proof}

In general, not all unitaries in $L_{R}(E)$ will have the form described in Lemma \ref{lem: certain unitaries}.
However, in the special case of $R = \mathbb{Z}$ all unitaries do in fact have this form.
Before proving this result, we note the following simple consequence of the defining relations of a Leavitt path algebra.

\begin{remark} \label{rmk: extend paths}
Let $E$ be a finite graph.
Fix a vertex $v \in E^0$, some natural number $m$, and define
\[
  X_{v,m} = \{ \gamma \in E^* \mid s(\gamma) = v \text{ and either } |\gamma| = m \text{ or } |\gamma| < m \text{ and } r(\gamma) \text{ is a sink} \}.
\]
A straightforward induction argument using (\ref{item:summation_relation}) from Definition \ref{def:lpa} shows that
\[
 \sum_{\gamma \in X_{v,m}} \gamma \gamma^* = v.
\]
Consider
\[
 x = \sum_{i=1}^n \lambda_i \alpha_i \beta_i^*  \in L_{R}(E),
\]
with $\lambda_i \in R$ and $r(\alpha_i) = r(\beta_i)$, and let $k = \max\{ |\beta_i| \}$.
Then
\begin{align*}
  x &= \sum_{i=1}^n \lambda_i \alpha_i \beta_i^* = \sum_{i=1}^n \lambda_i \alpha_i r(\alpha_i) \beta_i^* \\
    &= \sum_{i=1}^n \lambda_i \alpha_i \left( \sum_{\gamma \in X_{r(\alpha_i),k-|\beta_i|}} \gamma \gamma^* \right) \beta_i^* \\
    &= \sum_{i=1}^n \sum_{\gamma \in X_{r(\alpha_i),k-|\beta_i|}} \lambda_i \alpha_i \gamma (\beta_i \gamma)^*.
\end{align*}
Re-indexing the last sum, we see that
\[
  x = \sum_{j=1}^l \hat{\lambda}_j \mu_j \nu_j^*,
\]
where each $\mu_j$ extends an $\alpha_i$, each $\nu_j$ extends a $\beta_i$, and all the $\nu_j$ have length $k$ or are shorter but end at a sink.
In particular, if $\nu_j \neq \nu_i$ then $\nu_j^* \nu_i = 0$.
\end{remark}

We are now ready to show that all unitaries in $L_{R}(E)$ can be written in the form from Lemma \ref{lem: certain unitaries} when $R$ is a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
The following proposition is due to Chris Smith in the case where $R = \mathbb{Z}$.

\begin{proposition} \label{prop: standard form unitaries}
Let $E$ be a finite graph, let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup, and let $u \in L_{R}(E)$ be a unitary.
Then there exist paths $\alpha_i, \beta_i$ such that
\[
	u = \sum_{i=1}^n \lambda_i \alpha_i \beta_i^*,
\]
with
\begin{enumerate}
 \item $\sum_{i=1}^n \alpha_i \alpha_i^* = 1$, \label{item:standard form alpha}
 \item $\sum_{i=1}^n \beta_i \beta_i^* = 1$, and, \label{item:standard form beta}
 \item $|\lambda_i| = 1$. \label{item:standard form lambda}
\end{enumerate}
\end{proposition}
\begin{proof}
By Remark \ref{rmk: extend paths},
\[
	u = \sum_{i=1}^n \lambda_i \alpha_i \beta_i^*
\]
for some $\lambda_i \in R$
and paths $\alpha_i, \beta_i$ with $r(\alpha_i) = r(\beta_i)$ such that the $\beta_i$ are all of some fixed length $t$ or shorter but ending in a sink.
By combining like terms and dropping terms where $\lambda_i = 0$, we may assume that the pairs $(\alpha_i, \beta_i)$ are distinct and that each $\lambda_i \neq 0$.

We will now show that the $\beta_i$ are distinct and incidentally that $\lambda_i = \pm 1$ for each $i$.
Fix an index $1 \leq l \leq n$.
We have
\begin{align*}
  r(\beta_l) &= \beta_l^* \beta_l = \beta_l^* u^*u \beta_l = \beta_l^* \left( \left(\sum_{i=1}^n \lambda_i \beta_i \alpha_i^* \right) \left( \sum_{j=1}^n \overline{\lambda_j} \alpha_j \beta_j^* \right) \right) \beta_l \\
    &= \beta_l^* \left( \sum_{i,j= 1}^n \lambda_i \overline{\lambda_j} \beta_i \alpha_i^* \alpha_j \beta_j^* \right) \beta_l
    = \sum_{i,j= 1}^n \lambda_i \overline{\lambda_j} \beta_l^* \beta_i \alpha_i^* \alpha_j \beta_j^* \beta_l.
\end{align*}
By construction of the $\beta_i$, we see that $\beta_i^* \beta_j = 0$ if $\beta_i \neq \beta_j$.
To ease notation, let $L = \{i \mid \beta_i = \beta_l\}$.
Continuing the computation we get
\begin{align} \label{eqn: r(beta_l)}
 r(\beta_l) &= \sum_{i,j \in L} \lambda_i \overline{\lambda_j} \beta_l^* \beta_l \alpha_i^* \alpha_j \beta_l^* \beta_l = \sum_{i,j \in L} \lambda_i \overline{\lambda_j} \alpha_i^* \alpha_j,
\end{align}
where the last equality follows from $\beta_l^* \beta_l = r(\beta_l) = r(\alpha_i)$ for $i \in L$.

Recall that all Leavitt path algebras are $\mathbb{Z}$-graded, that vertex projections are homogeneous of degree $0$, and elements of the form $\mu \nu^*$ are homogeneous of degree $|\mu| - |\nu|$ (\cite[Definition 4.5 and Proposition 4.7]{TomfordeLeavittOverRing}).
Since the left hand side of (\ref{eqn: r(beta_l)}) is homogeneous of degree $0$ the right hand side must be so too.
Hence, we can discard any element not of degree $0$ from the sum.
Since the pairs $(\alpha_i, \beta_i)$ are distinct, and since we only sum over indices $i,j$ with $\beta_i = \beta_l = \beta_j$, we see that the $\alpha_i$ appearing in the sum must be distinct.
Thus, for $i \neq j$ the term $\alpha_i^* \alpha_j$ is either $0$ or homogeneous of degree $|\alpha_i| - |\alpha_j|$ (note that if $|\alpha_i| = |\alpha_j|$ then $\alpha_i^* \alpha_j = 0$ since the paths are distinct).
Thus, we get
\[
 r(\beta_l) = \sum_{i \in L} |\lambda_i|^2  \alpha_i^* \alpha_i.
\]
Using that $\alpha_i^* \alpha_i = r(\alpha_i) = r(\beta_l)$, it follows that
\[

 r(\beta_l) = \left( \sum_{i \in L} |\lambda_i|^2 \right) r(\beta_l).
\]
Since each $\lambda_i$ is non-zero and $R$ has an essentially unique partition of the unit, this equality can only hold if $|L| = 1$ and $|\lambda_l| = 1$.
Hence, all the $\beta_i$ are distinct and $|\lambda_i| = 1$ for each $i$, i.e.\ (\ref{item:standard form lambda}) holds.

Since the $\beta_i$ are distinct, $\beta_i^* \beta_j = 0$ if $i\neq j$, and as always $\beta_i^* \beta_i = r(\beta_i)$.
Thus,
\begin{align*}
  1 &= uu^* = \left( \sum_{i=1}^n \lambda_i \alpha_i \beta_i^* \right) \left( \sum_{j=1}^n \overline{\lambda_i} \beta_j \alpha_j^* \right) \\
    &= \sum_{i=1}^n |\lambda_i|^2 \alpha_i \alpha_i^* = \sum_{i=1}^n \alpha_i \alpha_i^*.
\end{align*}
That is, (\ref{item:standard form alpha}) holds.
By Lemma \ref{lem: orthogonal paths}, (\ref{item:standard form alpha}) implies $\alpha_i^* \alpha_j = 0$ if $i \neq j$, so
\[
  1 = u^* u = \left( \sum_{j=1}^n \lambda_i \beta_j \alpha_j^* \right) \left( \sum_{i=1}^n \overline{\lambda_i} \alpha_i \beta_i^* \right) = \sum_{i=1}^n \beta_i \beta_i^*.
\]
Hence, (\ref{item:standard form beta}) also holds.
\end{proof}

In the case of a finite graph with no sinks, we can give a completely geometric description of when a set of paths $\alpha_1, \alpha_2, \ldots, \alpha_n$ satisfy that the projections $\alpha_i \alpha_i^*$ sum to $1$.

\begin{lemma} \label{lem: disjoint cylinder sets}
Let $E$ be a finite graph with no sinks, let $R$ be a unital commutative ring with characteristic $0$, and let $\alpha_i$, $i=1,2,\ldots,n$, be paths in $E$.
The following are equivalent
\begin{enumerate}
 \item $\bigsqcup_{i=1}^n \mathcal{Z}(\alpha_i) = E^{\infty}$.\label{item:cylinders}
 \item $\sum_{i=1}^n \alpha_i \alpha_i^* = 1$.\label{item:sum}
\end{enumerate}
\end{lemma}
\begin{proof}

By repeated applications of (\ref{item:summation_relation}) from Definition \ref{def:lpa}, it follows that $(\ref{item:cylinders})$ implies $(\ref{item:sum})$.
Suppose now that $(\ref{item:sum})$ holds.
We first show that $\cup_i \mathcal{Z}(\alpha_i) = E^{\infty}$.
Suppose for contradiction that there is some $\mu \in E^{\infty}$ that is not in $\sqcup_i \mathcal{Z}(\alpha_i)$, and let $\nu$ be the initial segment of $\mu$ of length $\max{|\alpha_i|}$.
Then $\alpha_i^* \nu = 0$ for all $i$ and therefore
\[
 \nu = 1\nu = \left(\sum_{i=1}^n \alpha_i \alpha_i^* \right) \nu = \sum_{i=1}^n \alpha_i \alpha_i^* \nu = 0,
\]
a contradiction.
By Lemma \ref{lem: orthogonal paths}, we have $\alpha_i^* \alpha_j = 0$ for $i \neq j$.
This implies that no $\alpha_i$ is an initial segment of any $\alpha_j$, and hence that the cylinder sets are disjoint.
\end{proof}

\section{Projections in $L_{R}(E)$}

The aim of this section is to study the projections in $L_{R}(E)$
for a finite graph $E$ and $R$ a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
Specifically, it will be proved that all projections in $L_{R}(E)$
are elements of the diagonal.

\begin{definition}\label{def: diagonal projections}
Let $E$ be a graph.
A projection $p \in L_R(E)$ is said to be \emph{diagonal} if $p \in \mathcal{D}(L_{R}(E))$.
\end{definition}

\noindent The following Example gives a basic illustration of a Leavitt path algebra where all projections are diagonal.

\begin{example}
Consider the graph

\begin{center}
\begin{tikzpicture}[shorten >=1pt]

\tikzset{vertex/.style = {shape=circle,draw,minimum size=2em}}
\tikzset{edge/.style = {->,> = latex'}}

\node at (0,0) {$F_n$};

\node[vertex] (u1) at (2,0) {$u_1$};
\node[vertex] (u2) at (4,0) {$u_2$};
\node (u3) at (6,0) {$\cdots$};
\node[vertex] (un) at (8,0) {$u_n$};

\draw[edge] (u1) to (u2);
\draw[edge] (u2) to (u3);
\draw[edge] (u3) to (un);

\end{tikzpicture}
\end{center}

\noindent It is well known that $L_{\mathbb{Z}}(F_n) \cong M_n(\mathbb{Z})$, and furthermore that the diagonal is exactly the diagonal matrices.
We claim that $L_{\mathbb{Z}}(F_n)$ only has diagonal projections.

Suppose
\[
  P = \begin{pmatrix}
       a_{11} & a_{12} & \cdots & a_{1n} \\
       a_{21} & a_{22} & \cdots & a_{2n} \\
       \vdots & \vdots & \ddots & \vdots \\
       a_{n1} & a_{n2} & \cdots & a_{nn}
      \end{pmatrix}
\]
is a projection in $M_n(\mathbb{Z})$.
The relations $P = P^*$ and $P = P^2$ imply that $P = PP^T$, so for every $i$ we have
\[
  a_{ii} = \sum_{j=1}^n a_{ij}^2.
\]
This can only be satisfied if $P$ is a diagonal matrix with entries in $\{0,1\}$.
\end{example}

The following two lemmas will pave the way for the proof that all projections in $L_{\mathbb{Z}}(E)$ are diagonal.

\begin{lemma} \label{lem: doubling}
Let $E$ be a graph and let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
If $x,y \in L_{R}(E)$
satisfy $2x = 2y$ then $x = y$.
\end{lemma}
\begin{proof}
Let $\iota_E \colon L_{R}(E) \to C^*(E)$
be the embedding from Lemma \ref{lem: LZE into CE}.
We have
\[
  2\iota_E(x) = \iota_E(2x) = \iota_E(2y) = 2\iota_E(y),
\]
so since $2$ is invertible in $\mathbb{C}$, we get $\iota_E(x) = \iota_E(y)$.
Because $\iota_E$ is an embedding, $x = y$.
\end{proof}

\begin{remark} \label{rmk: add tail}
Let $E$ be a graph, $R$ a commutative unital ring.
Using a process called ``adding tails'' (see \cite{BatesPaskRaeburnSzymanski}),  we can find a graph $F$ with no sinks such that $L_{R}(E)$ embeds into $L_{R}(F)$ as $*$-algebras, furthermore this embedding maps $E^*$ into $F^*$.

For Leavitt path algebras over fields this is proved in \cite[Section 5]{AbramsArandaPino08}, where infinite emitters are also ``desingularized'' .
We are only interested in removing sinks, so we preform a partial desingularization, we are also interested in working over general rings, not just fields, however the arguments for constructing the embedding given in \cite{AbramsArandaPino08} works perfectly well in this setting, so we only review them briefly.
We define $F$ as follows
\begin{align*}
  F^0 &= E^0 \sqcup \{ w_{i} \mid i \in \mathbb{N}, w \in E^0, w \text{ is a sink} \}, \\
  F^1 &= E^1 \sqcup \{ e^{w}_i \mid i \in \mathbb{N}, w \in E^0, w \text{ is a sink} \},
\end{align*}
we let the range and source maps extend those of $E$ and define $r(e^{w}_i) = e^{w}_{i+1}$ and
\[
  s(e^{w}_i) = \begin{cases}
                w_{i-1}, & \text{if } i > 1 \\
                w, &  \text{otherwise}
               \end{cases}.
\]
Now we can define a $*$-homomorphism $\phi \colon L_{R}(E) \to L_{R}(F)$ by stipulating that $\phi(v) = v$ and $\phi(e) = e$ for all $v \in E^0$ and $e \in E^1$.
By the Graded Uniqueness Theorem (\cite[Theorem 5.3]{TomfordeLeavittOverRing}) $\phi$ is injective.
\end{remark}

\begin{lemma} \label{lem: sum of paths}
Let $E$ be a graph, let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup, let $x \in L_{R}(E)$ and let $\lambda \in R$.
If $\alpha, \beta \in E^*$ are such that
\[
  2 x = \alpha + \lambda \beta,
\]
then $\alpha = \beta$.
\end{lemma}
\begin{proof}
By Remark \ref{rmk: add tail}, we can assume that $E$ has no sinks.
We may write
\[
 x = \sum_{i=1}^n \lambda_i \mu_i \nu_i^*,
\]
where $\lambda_i \in R$
$\mu_i, \nu_i \in E^*$.
Pick a path $\gamma$ based at $r(\alpha)$ such that $|\gamma|$ is greater than $\max\{ |\nu_i| \}$.
Then $x \gamma$ is a polynomial in real edges,
\[
  x \gamma = \sum_{k=1}^m \hat{\lambda}_k \xi_k,
\]
where $\hat{\lambda}_k \in R$
and $\xi_k \in E^*$.
Thus,
\[
  \sum_{k=1}^m 2 \hat{\lambda}_k \xi_k = 2x\gamma = \alpha \gamma + \lambda \beta \gamma.
\]
The real paths form a linearly independent set in $L_{R}(E)$ (\cite[Proposition 4.9]{TomfordeLeavittOverRing}), so if $\alpha \gamma \neq \beta \gamma$, it follows from the above equation that there is some subset $I \subseteq \{1,2,\ldots,m\}$ such that
\[
 1 = \sum_{k \in I} 2 \hat{\lambda}_k = 2 \left( \sum_{k \in I} \hat{\lambda}_k \right),
\]
which contradicts that $\frac{1}{2}\notin R$ (Remark \ref{rmk:PropX_nohalf}).
Therefore, we must have $\alpha \gamma = \beta \gamma$.
Since $s(\gamma) = r(\alpha)$, we have $\alpha \gamma \neq 0$ and hence we must have $\beta \gamma \neq 0$ so $s(\gamma) = r(\beta)$.
Thus $\alpha = \beta$.
\end{proof}

\begin{theorem} \label{thm: projections look nice}
Let $E$ be a finite graph and let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
If $p \in L_{R}(E)$
is a projection then
\[
  p = \sum_{i=1}^n \beta_i \beta_i^*,
\]
for some paths $\beta_i \in E^*$ with $\beta_i^* \beta_j = 0$ for $i \neq j$.
\end{theorem}
\begin{proof}
Put $u = 2p - 1$.
Then $u$ is a self-adjoint unitary, i.e.\ $u = u^*$ and $u^2 = 1$.
By Proposition \ref{prop: standard form unitaries}, we can write
\[
  u = \sum_{i=1}^n \lambda_i \alpha_i \beta_i^*,
\]
where $|\lambda_i| = 1$,
$\alpha_i, \beta_i \in E^*$ and $\sum_{i=1}^n \alpha_i \alpha_i^* = 1 = \sum_{i=1}^n \beta_i \beta_i^*$.
Thus,
\[
  2p = 1 + u = \sum_{i=1}^n \beta_i \beta_i^* + \sum_{i=1}^n \lambda_i \alpha_i \beta_i^*
     = \sum_{i=1}^n \left( \beta_i + \lambda_i \alpha_i \right) \beta_i^*.
\]

Let $k \in \{1,2,\ldots, n\}$ be given.
By Lemma \ref{lem: orthogonal paths} $\beta_i^* \beta_k = 0 $ when $i \neq k$, so
\[
  2p\beta_k = \beta_k + \lambda_k \alpha_k.
\]
By Lemma \ref{lem: sum of paths}, this implies that $\alpha_k = \beta_k$.
So
\[
  u = \sum_{i=1}^n \lambda_i \beta_i \beta_i^*.
\]
We then get
\[
  \beta_k = \left( u^2 \right) \beta_k = \left( \sum_{i=1}^n \lambda_i \beta_i \beta_i^* \right) \left( \sum_{j=1}^n \lambda_j \beta_j \beta_j^* \right) \beta_k = \left( \sum_{i=1}^n \lambda_i^2 \beta_i \beta_i^* \right) \beta_k = \lambda_k^2 \beta_k.
\]
Since the real paths are linearly independent (\cite[Proposition 4.9]{TomfordeLeavittOverRing}), we get that $\lambda_k = \pm 1$.
It follows that
\[
  2 p = \sum_{i=1}^n \varepsilon_i \beta_i \beta_i^*,
\]
where $\varepsilon_i \in \{0,2\}$.
An application of Lemma \ref{lem: doubling} completes the proof.
\end{proof}

\noindent The following result is an immediate consequence of the preceding theorem.

\begin{corollary} \label{cor: only diagonal projections}
Let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
If $E$ is a finite graph, then $L_{R}(E)$ only has diagonal projections.
\end{corollary}

\noindent
It is not clear that this result only holds when $R$ is a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup, however it is worth noting that the result certainly requires some restriction on the ring of coefficients.
For instance,
\[
  p = \frac{1}{2}\left( aa^* + ab^* + ba^* + bb^*\right)
\]
is non-diagonal projection in $L_{2,\mathbb{C}}$.

\section{Conclusions}

\begin{proposition}\label{prop: only dynamical homomorphisms}
Let $E,F$ be graphs, let $R$ be a subring of $\mathbb{C}$ closed under complex conjugation, and let $\phi \colon L_{R}(E) \to L_{R}(F)$ be a $*$-homomorphism.
If $L_{R}(F)$ only has diagonal projections, then $\phi(\mathcal{D}(L_{R}(E))) \subseteq \mathcal{D}(L_{R}(F))$.
\end{proposition}
\begin{proof}
Let $\alpha \in E^*$ be given.
Since $\alpha \alpha^*$ is a projection in $L_{R}(E)$ and since $\phi$ is a $*$-homomorphism, $\phi(\alpha \alpha^*)$ is a projection.
By the assumption on $L_{R}(F)$, it follows that $\phi(\alpha \alpha^*) \in \mathcal{D}(L_{R}(F))$.
By linearity of $\phi$ and the fact that $\mathcal{D}(L_{R}(F))$ is an algebra, we get that $\phi(\mathcal{D}(L_{R}(E))) \subseteq \mathcal{D}(L_{R}(F))$.
\end{proof}

\begin{remark}
The requirement that $L_{R}(F)$ only has diagonal projections is clearly crucial for the result.
For instance, there exist $*$-automorphisms of $L_{2,\mathbb{C}}$ that do not preserve the diagonal.
As an example of this, consider the unitary
\[
  u = \frac{1}{\sqrt{2}} \left( aa^* - ab^* + ba^* + bb^* \right),
\]
and define an automorphism $\psi$ of $L_{2,\mathbb{C}}$ by $\psi(x) = u x u^*$.
Then
\[
  \psi(aa^*) = \frac{1}{2}\left( aa^* + ab^* + ba^* + bb^*\right),
\]
so $\psi$ does not preserve the diagonal.
This $u$ was constructed by mapping the unitary that rotates by $45$ degrees from $M_2(\mathbb{C})$ into $L_{2,\mathbb{C}}$ using the map discussed in \cite[Example 5.13]{BS}.
\end{remark}

\begin{theorem} \label{thm: isomorphism same sign}
Let $E,F$ be finite, essential, strongly connected graphs that satisfy Condition (L), and let $R$ be a subring of $\mathbb{C}$ closed under complex conjugation.
Assume furthermore that $L_{R}(F)$ only has diagonal projections.
If $L_{R}(E)$ is $*$-isomorphic to $L_{R}(F)$ then
\[
  \sgn(\det (I - A_E)) = \sgn( \det(I - A_F)).
\]
\end{theorem}
\begin{proof}
Suppose $\phi \colon L_{R}(E) \to L_{R}(F)$
is a $*$-isomorphism.
By Proposition \ref{prop: only dynamical homomorphisms}, we must have $\phi(\mathcal{D}(L_{R}(E))) \subseteq \mathcal{D}(L_{R}(F))$, so by Theorem \ref{thm: algebraic matsumoto matui},
\[
  \sgn(\det (I - A_E)) = \sgn( \det(I - A_F)). \qedhere
\]
\end{proof}

\begin{corollary} \label{cor: iso implies same sign}
Let $E,F$ be finite, essential, strongly connected graphs that satisfy Condition (L) \and let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
If $L_{R}(E)$ is $*$-isomorphic to $L_{R}(F)$ then
\[
  \sgn(\det (I - A_E)) = \sgn( \det(I - A_F)).
\]
\end{corollary}
\begin{proof}
This an immediate consequence of Theorem \ref{thm: isomorphism same sign}, since all projections in $L_{R}(E)$ are diagonal by Corollary \ref{cor: only diagonal projections}.
\end{proof}

\begin{corollary} \label{cor: main result}
Let $R$ be a unital, commutative subring of $\C$ closed under complex conjugation that has an \eup.
Then $L_{2,R}$ is not $*$-isomorphic to $L_{2-,R}$.
In particular, $L_{2,\mathbb{Z}}$ is not $*$-isomorphic to $L_{2-,\mathbb{Z}}$.
\end{corollary}
\begin{proof}
Recall that by definition $L_{2,R} = L_{R}(E_2)$ and $L_{2-,R} = L_{R}(E_{2-})$.
We see that the adjacency matrices are
\[
  A_{E_2} = \begin{pmatrix}
            2
           \end{pmatrix},
\]
and
\[
  A_{E_{2-}} = \begin{pmatrix}
                2 & 1 & 0 \\
                1 & 1 & 1 \\
                0 & 1 & 1
               \end{pmatrix}.
\]
Hence, $\det(I - A_{E_2}) = -1$ and $\det(I - A_{E_{2-}}) = 1$, so by Corollary \ref{cor: iso implies same sign}, $L_{2,\mathbb{Z}} \not \cong L_{2-,\mathbb{Z}}$
as $*$-algebras.
\end{proof}

Although Corollary \ref{cor: main result} does not apply to fields, we can use it to say something about the situation for fields.

\begin{remark}\label{remark:fields}
Suppose $K$ is a field with characteristic $0$.
Then $L_{2,\mathbb{Z}}$ will embed into $L_{2,K}$, and $L_{2-,\mathbb{Z}}$ will embed into $L_{2-,K}$.
Assume that there exists a $*$-isomorphism $\phi \colon  L_{2-,K} \to  L_{2,K}$ which restricts to a $*$-homomorphism from $L_{2-,\mathbb{Z}}$ to $L_{2,\mathbb{Z}}$.
The restriction is also injective, so by Corollary \ref{cor: main result}, it cannot be surjective.

One now wonders how common it is for $*$-isomorphisms between Leavitt path algebras over $K$ to restrict to $*$-homomorphisms between the corresponding algebras over $\mathbb{Z}$.
In general, homomorphisms will not restrict in this way. For instance, each $z \in \mathbb{C}$ of modulus one defines a \emph{gauge automomorphism} $\gamma_z \colon L_{2,\mathbb{C}} \to L_{2,\mathbb{C}}$ given by $\gamma_z(a) = za$ and $\gamma_z(b) = zb$ which clearly does not restrict to an automorphism of $L_{2,\mathbb{Z}}$.
However, in many cases where homomorphisms are written down explicitly, the field of coefficients has not been used, and such homomorphisms must clearly restrict to homomorphisms over $\mathbb{Z}$.
For some examples of this phenomenon, see for instance \cite{AbramsAnhPardo, AbramsLoulyPardoSmith, BS, RuizTomforde}.
In fact, at the end of \cite{TomfordeLeavittOverRing}, Tomforde suggests that Leavitt path algebras over $\mathbb{Z}$ may provide the key to understanding this non-dependence on the field of coefficients.

Our result leaves two ways for $L_{2-,K}$ and $L_{2,K}$ to be $*$-isomorphic: Either the isomorphism makes explicit use of the field, so that it does not induce a $*$-homomorphism from $L_{2-,\mathbb{Z}}$ to $L_{2,\mathbb{Z}}$.
Or, if the isomorphism does induce such a $*$-homomorphism, then this induced map cannot be surjective.
Either way, a $*$-isomorphism between $L_{2-,K}$ and $L_{2,K}$, if it exists, will have to be a fairly complicated map.
\end{remark}

We conclude with a few remarks on the possibility that $L_{2,R}$ and $L_{2-,R}$ are Morita equivalent.

\begin{remark
\label{remark:morita}
Let $K$ be a field.
If $L_{2,K}$ and $L_{2-,K}$ are Morita equivalent, they must be isomorphic as rings (this follows from the argument given in the proof of $(2)$ implies $(1)$ in \cite[Proposition 10.4]{RuizTomforde}).
It is also worth noting that there are no known examples of Leavitt path algebras that are isomorphic as rings, but not as $*$-algebras.
See \cite{AbramsTomforde} for a discussion of how various notions of isomorphism of Leavitt path algebras are related to isomorphism of graph $C^*$-algebras.

When working over a commutative, unital ring $R$, the connection between Morita equivalence and ring isomorphism of $L_{2,R}$ and $L_{2-,R}$ is less clear.
If $R$ is a regular supercoherent ring, then both $L_{2,R}$ and $L_{2-,R}$ have trivial algebraic $K$-theory \cite{AraBrustengaCortinas}.
So at least this does not provide an obvious proof that they are not Morita equivalent.
We note that the class of regular supercoherent rings covers the class of Noetherian regular rings which in turn contains the class of principal ideal domains.

To go from a Morita equivalence of $L_{2,K}$ and $L_{2-,K}$ to a ring isomorphism, one uses that $L_{2,K}$ is simple.
In particular, the description of the $K_0$-group of a simple purely infinite ring given in \cite[Corollary 2.2]{AraGoodearlPardo} is crucial.
However, if the ring $R$ has non-trivial ideals, then $L_{2,R}$ will clearly not be simple.
Therefore, it is unclear to the authors if Morita equivalence of $L_{2,R}$ and $L_{2-,R}$ implies ring isomorphism.
\end{remark}

\section*{Acknowledgements}
As mentioned above, the proof of Proposition \ref{prop: standard form unitaries} is due to Chris Smith when the ring is $\mathbb{Z}$, and the authors are very grateful to him for sharing this result.
After the second named author gave talk at the 2015 Norwegian Operator Algebras Meeting about the case $R = \mathbb{Z}$, emphasizing the property of $\mathbb{Z}$ used in the proof (having an essentially unique partition of the unit), it was pointed out to the authors by Christian Skau and Lars Tuset that many rings have this property.
The authors thank them for this valuable insight.
Additionally, the authors would like to thank Gene Abrams, Mark Tomforde, and Efren Ruiz for their comments on an early draft of this paper.

The authors' collaboration began at the conference \emph{Classification of $C^*$-algebras, flow equivalence of shift spaces, and graph and Leavitt path algebras} hosted at the University of Louisiana at Lafayette, and the authors are grateful to the organizers and to the National Science Foundation for providing this opportunity.
This work was supported by VILLUM FONDEN through the experimental mathematics network at the University of Copenhagen and by the Danish National Research Foundation through the Centre for Symmetry and Deformation (DNRF92).

redpajama_set_name: RedPajamaArXiv


text:
Q: spawn ENOENT error with NodeJS I get an error using GM() on my remote server, I think i get an issue when I try to write the file... It works perfectly on my local server...
My code :
gm()
.in('-page', '+0+0')
.in('/var/www/myapp/public/images/instabox.jpg')
.in('-page', '+10+10')
.in(image)
.mosaic()
.minify()
.write(newFileName, function (err) { // newfilename = '/var/www/myapp/public/uploads/mydir/image.jpg'
  if (!err) console.log('done');
  if (err) console.log(err);
  callback();
});

My error :
{ [Error: spawn ENOENT] code: 'ENOENT', errno: 'ENOENT', syscall: 'spawn' }

Response of the issue:

I put back GM on my remote server and it was the issue, GM was not
  well installed...

redpajama_set_name: RedPajamaStackExchange


text:
Q: How to rewrite $x-y=\frac{x}{y}$ so that it become $y=$ (something...)? For example, $x+y=x\times y$ is easy to express as $y=\frac{x}{x-1}$, how about $x-y=\frac{x}{y}$?
I tried multiply both sides by $y$ and become
$y^2-xy+x=0$
but up to this step I don't know how to continue, can anyone help?

A: Solve for y in quadratic equation
$y^2-xy+x=0$
Using quadratic formula.

A: It cannot be done, because, as you correctly noticed, the expression (if $y\neq 0$) is equivalent to the expression
$$y^2 - xy + x = 0$$
This expression can have, for a fixed value of $x$, two solutions, one solution, or even no real solution.
For example, if $x=1$, then there is no real value of $y$ that satisfies the equality, but if $x=-1$, there are two such values of $y$.


text:
The Most Dangerous Job in New York
Construction accidents are up over 220% in NYC, and workers are calling Cellino & Barnes to help hold negligent employers accountable for safety violations that led to injuries
NEW YORK – Construction is now the deadliest job in New York City.
According to the Bureau of Labor Statistics, the construction industry has more fatal work-related accidents than any other industry in New York City. In addition, many construction sites continue to ignore safety regulations, putting workers at risk.
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Accidents Up 221%
In the past year, more than a dozen workers died in construction accidents, including three workers in just the past month.
Overall, statistics show that construction accidents are up over 220-percent, compared to just five years ago. Falls were the top cause of these occupational injuries, representing nearly 60-percent of all construction accidents in New York City.
1 in 4 Work Sites Not In Compliance
The Department of Buildings says there are about 45,000 active construction sites across New York City. However, about 25-percent of them are not in compliance with safety regulations.
That's over 10,000 sites that are putting workers at risk.
In response to these violations, the Department of Buildings stated that inspectors will be working 12-hour shifts, and contractors that are not in compliance with today's safety regulations will face steep fines.
Cellino & Barnes Helping Hard-Working New Yorkers
All construction accidents are preventable if workers are supplied with the modern safety equipment and training.
Sadly, many contractors will place profits and deadlines above their workers' safety, and it's unacceptable.
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After any construction accident, workers are urged to contact Cellino & Barnes for a FREE confidential consultation, or by calling us at (800) 888-8888.
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redpajama_set_name: RedPajamaCommonCrawl


The 328d model will have 250 horsepower and 280 pounds of torque. Then, the 340i will have a 3.0-liter six-cylinder engine. It will be able to produce 326 horsepower and 332 pounds of torque.
Tags: 2020 BMW 4 Series Grand Coupe, 2020 BMW 4 Series Redesign, 2020 BMW 4 Series Convertible, 2020 BMW 4 Series Coupe.


Just a quick post today to wish everyone a happy and safe Easter. I'm taking a few days off to recharge the batteries and spend time with family and friends. I'm sure there will also be some chocolate consumption! I hope you're doing the same... I'll be back on Tuesday but first, here's a little bit of Easter moodboard inspiration. Enjoy!
Your moodboards feel so cosy... love them !


text:
Ukrainian crisis - demonstration against military drafting in the Lower Carpathians
Budapest, Friday August 8, 2014 - On Friday, Jobbik and Sixty-Four Counties Youth Movement held a demonstration in front of the Embassy of Ukraine in Budapest, protesting against military drafting and the settlement of refugees in the Lower Carpathians.
Jobbik's vice president István Szávay pointed out in his speech that the Hungarian community of the Lower Carpathians had nothing to do with the "mindless bloodshed" going on a thousand kilometres away. He added that several Hungarians were drafted nonetheless, treated as volunteers against their will, while the Hungarian government did nothing to protect them.
"This is not a Hungarian war, and neither is it the war of any other nation in the Lower Carpathians", so we must speak out for the Hungarian, Ruthenian and Ukrainian families alike – he stated.
István Szávay demanded that conscripts drafted in the Lower Carpathians not be deployed in the East, but be allowed to do their military service locally. He also condemned the settlement of war-zone refugees in the areas populated by Hungarians or other ethnic minority groups. Anatoliy Sava, a councilman of the World Congress of Ruthenians in the Lower Carpathians addressed the demonstrators in the Ruthenian language. "Now it is time for Ruthenians and Hungarians to realize the results of the regional and local referenda of the Lower Carpathian region and the Berehove district held in 1991.
Ruthenian people hope that Hungary will not allow such a genocide to take place in the Lower Carpathians as the one in Eastern Ukraine. Ruthenians have been living together with Hungarians for a thousand years and this is the token of peace and a better future in the Lower Carpathians. Ruthenians do believe in it" - concluded Anatoliy Sava.
In his speech, László Toroczkai, the mayor of Ásotthalom and the founder of Sixty-Four Counties Youth Movement described the Ukrainian events as "clearly a top-down" crisis, commenting that the secret services of the US and its allies were "busy working" when the conflict broke out.
Toroczkai reminded the demonstrators that the news of the Lower Carpathians are quite similar to those of the Yugoslav Wars in the Southern Regions in the early 90s. Just like back then, they want to draft Hungarian men into a war they have nothing to do with. He pointed out that the Romanian president and other Romanian politicians did make a clear stand for protecting the interests of ethnic Romanians in the Ukraine, while the Hungarian government failed to do so for the Hungarians and Ruthenians of the Lower Carpathians.
Sources: MTI (Hungarian News Agency), Alfahir.hu, Kuruc.info


Do you tire often and easily?
Standard Weight for a given height).
Is your waist / hip ratio > 0.9?
If you have answered in the affirmative to any of these questions, it is high-time you begin on some exercise to Keep Fit!
Fitness is LOST if you exercise for two days or less in a week!
Fitness is MAINTAINED if you exercise three days a week!
Fitness is IMPROVED if you exercise six days a week!
Fat people have an internal chemistry that adapts to a low-calorie intake.
Fat people gain weight when they overindulge!
Start Exercising and Keep Fit!
Small Frame- Subtract 2 to 3 Kg.
Heavy Frame- Add 2 to 3 Kg.
Is losing weight or gaining weight your problem?
· Diets do not work for only losing weight.
· Losing weight is not the basic problem, the problem is gaining weight.
· Fat people gain weight easily and quickly.
· A fat person's body adjusts to the making of excess fat.
· Yes, people who have excess fat should go on a diet to get rid of the fat and change their internal chemistry through regular exercise so that fewer calories are converted to fat.
It is not right to say that the reason for becoming overweight is either eating too much or exercising too little or both. It is mainly due to internal body chemistry.
We are not merely passive reservoirs or tanks where input is more calories and output is less calories. We are in fact, active metabolizing machines, each different in some way and each handling calories differently.
60 - 70% of the energy that the muscles need when one is resting is supplied by fat.
Storage of fat is a natural body function.
Fat people have a greater propensity for storing fat and less than the normal propensity for burning it.
Fitness is a vicious cycle - the more the fat one has, the body chemistry or metabolism changes to favour the build up of more fat.
What is the maximum fat content for a fit person?
Your weight = fat + lean body mass.
The more fat you have, the better you will be able to float in a swimming pool! But it is of no good to your health.
In contrast to fat, lean muscle and bone do not float.
If weighed underwater, it is the skinny people who weigh more. Dense is beautiful like gold.
People with 25% fat can float easily in a swimming pool.
People with 20% fat can float easily while breathing.
People with 15% fat will sink very slowly even with a chest-full of air.
People with 13% fat readily sink even in salt water.
Larger bones and more muscle can carry more fat without exceeding the limits.
One may become over-fat without getting overweight.
As we grow, our muscles gradually become less dense less lean and more fatty, due to reducing physical activities.
Work is not exercise. Daily routine work never stresses more than 50% of the muscles. Unused 50% of the muscle can waste away giving way to fat in the long run.
One can make no realistic determination of own fatness by referring to weight.
Overweight is tolerable, but over-fat is not.
If one has left active sports-life, after taking to a job, then the muscles are slowly but surely converting to fat, i.e. getting fatter but not heavier.
To avoid gaining fat, start exercising! Like food, exercise should become an essential item in our daily routine.
The correct weight of a person depends on several factors, namely, the size of the bones, width of the shoulder, age, height and body type.
But the 'correct weight' is really quite ambiguous.
A person's correct weight' changes with the change in lean mass.
If one's weight is 60 Kg. then the maximum fat content should be 9Kg. for a male (15%) and 13.2 Kg. for female (22%). The difference should be the lean body mass (LBM) for efficient working.
Fat is calorie, so the fat people cannot consume calories. LBM burns up the calories.
Therefore, for ideal maximum weight, we have to start with the active metabolizing tissue we have i.e. in LBM and calculate the amount of fat we could add on to our LBM.
Exercise increases the lean body mass.
It isn't so much excess fat, as the lack of athletically trained muscles, that is the cause of obesity.
The ultimate cure for obesity is exercise.
Brisk walk for 30-40 minutes daily is the most efficient way of removing fat for people at 35 plus. One can take a glass or two of water early in the morning, but certainly not tea or coffee before the walk.
The walk has to be continuous and steady.
The heart beat per minute should be increased to the "training rate", which takes about eight minutes of warm-up.
Then walk at that rate for twelve minutes.
Finally cool down the heart rate for next ten minutes.
Stretching exercises in the beginning help to get warmed up before walking.
How hard should one exercise?
As you work your muscles, the oxygen demand goes up and the heart beats faster.
The older you are, the slower your heart beats during maximum exercise.
For regular, everyday, efficient exercise we should work only hard enough to make our hearts go at 80% of the maximum for our age.
If you are terribly out of shape at forty years range, you can easily drive your pulse to 146 by just moderate walking.
If you are in a better shape, you might have to walk briskly or jog to get 146 beats per minute.
If you are lean and athletic, you may have to run at a fast pace to reach the same heart rate.
Never combine your walking with others.
Determine your correct exercise "Training Heart Rate" from the table and continue the steady exercise for twelve minutes. Don't forget to warm-up and cool down.
How to determine your correct Heart Rate?
First check your resting heart rate.
An average heart beats seventy two times per minute. Take your pulse as soon as you wake up.
If your morning heart rate is above 85 or erratic, see the doctor before starting any exercise.
This is the fastest your heart can beat for your age.
Do not exercise at this rate !!!!!!!
e.g.: A forty-year-old has a resting rate of 70. His training heart rate is 70 + 65% x (180-70) = 142.0. When this person exercises, he should maintain a heart rate of 142-144 beats per minute.
When you exercise, check your pulse immediately at the end of the exercise. Count it for 6 seconds and multiply by 10. If it is going 10 or more beats slower than your calculated training rate, your exercise is not hard enough. If it is going 10 or more beats faster, slow down your speed. Remember we are working with time, not racing against it.
Check your Measurement now !
Is Intense Exercise better than Moderate Exercise?
Intense exercise will overheat the muscles, doing more damage than good if the body isn't ready for that.
If one wants fast improvement, one should rather exercise moderately for a longer period.
In intense exercise one is also likely to lose muscle, if not combined with a proper diet.
After the age of forty, it is recommended that exercise be changed on alternate days to involve all the muscles in the system and to avoid exercising the same muscles often.
Give time for the muscles to repair between exercises.
If one is in a hurry to get in shape "Exercise longer not harder" and Never exceed your training heart rate!
How to take an exercise pulse?
Your pulse is a much more accurate measurement of your health than your weight.
Men's average resting pulse is about 72 and for women it is about 80.
As one becomes physically fit, the resting pulse rate drops.
You can find your pulse on the thumb side of the wrist or on the side of the neck.
Once you have found the pulse, count it for exactly 6 seconds and multiply by ten.
One can take the pulse several times during the day to get the average resting pulse.
If you exercise correctly and long enough, the heart muscles will get stronger and it will soon pump slower, while pumping more blood at each stroke.
If you want to take your training pulse, stop and do a quick 6second count. When you first start any exercise stop many times to check the exercise pulse until you know exactly how hard to do the exercise to get the correct pulse rate.
Many people begin an exercise programme, only to give up after a few weeks.
Reason is the selection of an exercise not suited for themselves and over exercise by running/jogging etc.
Walking is an excellent exercise for all ages. It is not as strenuous to the knees and joints as jogging.
Just be sure the walk is brisk enough to maintain your heart rate at your training heart rate (for 35 plus, it is around 140.
If you feel brisk walking is not enough to drive your pulse to the training heart rate, try to carry some weight like a back-pack.
Jogging! Running is recommended for young and fit people.
A stress electro - cardiogram is recommended for thirty five plus before starting jogging/running.
Your fat loss (in turn weight loss) will be faster in jogging / running.
Stationary cycling and swimming is also considered a good exercise for reducing fat and also for a change of routine brisk walk.
A valuable criterion of physical fitness is heart recovery rate after exercise which is an important health indicator.
Heart rate slows down after the exercise in two stages.
First there is a sharp drop from 140 to 100 BPM (beats per minute) within a minute. It stays there for few minutes (say 2 to 3 minutes) and then gradually drops to original resting heart rate.
The first drop in heart rate, immediately after the exercise is most meaningful in judging the health condition.
Recovery rate can be calculated by taking the pulse for 6 seconds immediately at the end of an exercise and then again for 6 seconds exactly one minute later.
Body weight index (BWI) is a very easy method of ascertaining obesity.
Divide your weight (W) in Kg. by the square of your height in meters, i.e.
Measure the waist and hip with a measuring tape.
Calculate the waist to hip ratio.
For men an index of < 0.9 is considered normal.
For women an index of < 0.85 is considered normal.
There are several ways of having the Body Composition assessed - underwater weighing, skin-fold measurements, ultrasound testing, electrical impedance, to name a few.
But now there is a simple chart based on thousands of these tests that allows you to get an accurate estimate of your body fat percentage. To use the chart, locate your body weight on the Weight scale and draw a straight line from there to your waist size (measured at your belly button) on the opposite scale. Your estimated percentage of body fat is the point at which the line the diagonal scale in the middle of the chart.
What changes - weight or shape?
Unless one is quite a lot over-fat, there will be little, if any, reduction in total weight. In fact, one may gain weight.
The fat will be converted into muscle and muscle is much heavier than fat. What does change is the SHAPE.
- muscular fat is removed and the muscle will go back to its original long, lean shape.
from big pear to a little pear, change will be less.
Muscle loss will occur, if the exercise is too intense.
Exercise at the proper heart rate.
Muscle loss will occur, if there is insufficient time allowed for recovery.
Age 30 - 50 Wait 24 hours before next exercise period and switch exercise day-to-day.
Do not exercise when systemically ill; the body needs protein to repair sick tissues.
Muscle loss will occur. if dietary protein is inadequate or imbalanced.
Absorb 60 gms. of protein in a day. Be sure the diet is balanced, when carbo hydrate is low, protein will be used for the making of glucose instead of tissue repair.
Muscle and muscle enzymes can increase when a person exercises properly on a day-to-day basis.
If one exercises too long and strenuously, the body will not be able to re-synthesize the protein lost, by the next session.
The energy demand will always take precedence over protein bio-synthesis.
Is it possible to reduce fat from particular spots?
Most people feel fat concentrates in specific areas and that it can be got rid of with super exercise. But this is not true.
In women, the fat concentrates on the hips and thighs and in men, around the midsection.
Passive spot reducers include belly-belts and rollers in health clubs and in other methods they use heated belts, sweat-suits etc.
Heat and pressure drive the water out of the tissues and they lose inches but within hours the tissues get back to original shape.
Most of the time the weight loss will actually be water loss from the body.
People who try to do sit-ups, leg-rises and donkey-kicks beyond their capacity, with the aim of spot reduction, work their poor muscles to death.
When large sets of muscles of the body are exercised, fat will be drawn from all parts of the body to meet the energy requirement.
Work and Exercise - do they produce same results?
Very few calories are used during any exercise.
But exercise changes the metabolic rate, increases the amount of muscle, raises the level of calorie-consuming enzymes inside the muscle and increases the burning of fats.
Sustained exercise at 80% of maximum heart rate is very efficient at bringing about the changes.
Physical work is a form of exercise but it is not sufficient to make the changes in our body.
The work one does may cause the heart to beat faster, but work rarely sustains long enough to get any benefits.
Work is non-aerobic like weight-lifting or sprinting. It is too low or too high in intensity or too short in duration.
Work usually makes demands on only one set of muscles but aerobic walking summons all the muscles of the body, including the heart muscles.
Metabolically, even the arms get conditioned during aerobic walking.
Aerobic walking will get you in condition for work, but work will not get you in condition for exercise.
"Insensible Exercise' is unconscious muscular activity -carried out without being aware of it.
Fit people often get involved in exercise without sensing that they are doing exercise at all.
Though twenty minutes of jogging burns only 180 calories equivalent to one glass of milk, exercise changes the body so that we burn more calories throughout the day during routine work.
Physically fit people have slightly elevated metabolism which burns more calories at rest than in fat people.
Normally, fat people tend to be more inactive than fit ones and unconsciously find ways to get the same job done with less movement.
Fit people are inclined towards insensible exercise and fat people towards sensing their exercise.
Exercise for a few minutes a day has far-reaching effects by altering our insensible exercise for the rest of the day.
Fat, carbohydrate and protein are the only sources of calories in the diet.
The various weight-loss diets consist of an endless list of manipulations of just these three foods.
Once digested and in the blood stream, the body readily converts one to another.
The liver is so sensitive to the body's needs of all the three and it starts converting very quickly if one eats an unbalanced diet.
Fat is never converted into protein or carbohydrate, whereas protein and carbohydrate can be converted to fat.
The only thing the body can do with fat is burn it up in the muscle.
Protein can be converted into glucose which can be used by muscle. The brain mainly uses two-thirds of the glucose in the blood.
If one endangers blood glucose supply in any way (over exercising) the powerful body function will convert protein into glucose.
Liver holds stored glycogen and converts protein into glucose.
High protein diet to reduce fat has the danger of overburdening the kidney.
Similarly over exercising instead of building muscles will break the muscles into glucose for blood sugar.
What happens when one eats too much protein?
One cannot repair any more tissue than that which needs repair.
When amino acids are not needed, they are sent to the liver where they are de-aminated and then converted into fat.
The process of de-amination is stressful to the body.
During de-amination the nitrogen that is released from amino acids is quickly converted into ammonia.
Ammonia is very toxic to the body and in turn is changed to urea.
Urea is also toxic and to eliminate it from the body, it must be diluted into urine.
If the protein content is about 12 to 13% of the total calorie intake, the body can very easily get rid of the urea.
But if it is more, then one needs enormous amounts of water to dilute it. The body will take from its own tissue and kidney has to work overtime to get rid of the urea.
In high protein diet the weight loss is only water loss. Not reduction of fat.
One can get enough protein in the diet along with a good balance of carbohydrates and vitamins.
The thumb rule is to eat two servings (3 ounces) of protein-rich split peas, dried beans, lentils etc.
For non-vegetarians, low fat meats such as chicken and fish may be recommended. Best avoided for proven reasons.
In addition, have two servings (one cup each) of non-fat or low fat milk or a milk substitute such as curd, yogurt or cheese.
Balance these high protein foods with carbohydrates by eating four servings of fruits or vegetables each day and four servings of rice and cereals.
One serving is just 3/4 cup filled with small size pieces.
It is almost impossible not to get fat in foods we eat which is very much present in nuts and other seeds we take.
The essence of the whole exercise is to have a well balanced diet.
So eat well and exercise longer.
Do muscles burn fat or glucose?
Muscles use mainly the fat for energy.
Nearly 70% of the energy need of muscles is met by fat and only 30% by glucose.
We burn fat all the time even while we are not exercising.
Fat can be burned, metabolized or oxidized and there is no better source of energy for your body than fat.
Fat is such an efficient, beautifully stored energy source which we never run out of even though we burn it all the time.
Glucose is like a twig, it burns easily but fat is like a log; it is hard to get started but once started it burns for a long time, giving off lots of heat. Continuous walk for thirty minutes plays an important role here.
Glucose is good for quick energy which is required during a sprint or short distance running etc.
Fat people should exercise moderately but for a longer period to burn the fat.
All living things, including plants, have to store a certain amount of food for the days when they cannot find or make food.
Carbohydrates are stored only by plants which are stationary and all mobile animals store in the form of fat. Fat contains twice as many calories per gram as carbohydrates.
Fat can be packed into a small space and the fat cell can contain up to 85% fat and 15% is used by water-based life function of the cell.
Carbohydrate stored in the form of glycogen needs larger space and glycogen occupies only 15% of the cell space. So a mobile creature is far better off with this marvellous invention called fat.
Almost everything you eat can be converted into fat and fat people are particularly efficient at doing this.
The ability to store food is like having a good bank balance.
Body visualizes fat as a physiological safety mechanism.
15 % fat for men and 20% for the women of their total weight are the optimum limits for efficient and effective working.
Is it advisable to reduce weight by fasting?
If the body is stressed by fasting, it will try to hoard extra fat when it is made available.
Fasting to reduce weight will encourage the body to become fatter.
Fasting once in a week or a fortnight or a month will certainly help the body to cleanse the toxins and give total rest to the digestive system.
Never make the mistake of eating one meal a day (23 hour fast). But spread those calories over the day in five to six meals.
Otherwise when you go off the diet, the body is set to gain heavy weight.
So exercise is the only way to knock off the fat from all over the body and stay healthy. All hats off to slim people! But you need exercise to keep fit.
Recommended percentage of daily calories for good health is Carbohydrates 65%, Fat 23% and Protein 12%.
Starch in rice, wheat etc., is only the beginning of the complex carbohydrates.
But there is cellulose, hemicellulose, and lignifl which are only partly digested. These form the roughage.
Eating more roughage means absorbing more carbohydrates.
All carbohydrates after intestine digestion become simple carbohydrate called glucose. Some carbohydrates breaks down very quickly, some very slowly and some (like cellulose) not at all.
Complex carbohydrates will become blood glucose very slowly while the simple sugars such as maltose, lactose in milk and table sugar will become blood glucose almost immediately.
Avoid simple carbohydrates because it causes our blood glucose levels to fluctuate too quickly.
Adding bran to the food to increase the roughage should be avoided.
Eat whole wheat, unspoiled rice, ragi etc., unrefined conditions close to the way it is grown.
Only then you get all the carbohydrates, vitamins and minerals associated with it.
Benefits of eating whole grain without polishing (cooked properly, of course).
a. Calorific value is decreased by the roughage.
b. Roughage is hard to chew, thus decreasing the tendency to overeat.
c. Roughage gives some protective effect to the digestive tract.
d. One can eat the vitamins with their food as nature intended.
e. The simple sugars give taste but are delayed in their entrance to blood stream.
Chew the food very well and feel the sweetness of rice, chappathi etc. Bon Appetite with Carbohydrates and balanced protein and fat.
Overall percentage of fat consumption is increased in developed countries mainly due to high-grade meat.
Fat in India is mainly consumed in the form of edible oils used for frying foods.
Milk products including sweets taken daily can add excess fat in the long run.
Unsaturated fats, although they decrease blood cholesterol, can increase the cholesterol in the arterial valves, artery walls and other body tissues.
It would be best to decrease all fat intake.
Again I emphasize no fat at all because you have enough fat hidden inside your food.
Eat very less meat and meat products - better to be a vegetarian. Decrease the use of fat in cooking - Indians from the North to specially note.
Oils, margarines and fat that we use in food preparation is similar to refined sugar in food value.
Ground nut or coconut has reasonable food value, but when we squeeze the oil out, we leave most of the nutrition behind and are left with pure fat.
These fats and oils are almost pure grease and we don't need them.
Unfortunately, sugars and fats make food tasty.
"if you want to stay fit and healthy, do not indulge excessively in sweets and fats. "
How energy is produced inside a Muscle?
Glucose from carbohydrates, fatty acids from fats, and amino-acids from proteins are burned inside muscle cells to get energy.
Cellular burning is more like taking a building apart! It is a careful procedure requiring special tools and detonators at every step. In our body, those tools are called enzymes.
Literally, hundreds of enzymes are needed in each cell and each one is quite different from the other.
Each enzyme is made of protein.
They are large, complex molecules that cannot pass through the wall of a cell.
So, the only way enzymes increase in a muscle is when the DNA makes more enzymes inside a cell.
This is called enzyme biosynthesis and takes place only if you eat adequately, if your cells aren't sick and if you exercise to stimulate the DNA to go to work.
DNA (de-oxyribonucleic acid) contains the blue print of our total body and dictates the cells how to behave, what to manufacture, what to seek, what to avoid etc. DNA can be compared to an architect.
The RNA (Ribo Nucleic acid) can be compared to the builder who constructs the house.
Millions of cells die each second and millions are born and this process continues till the end except for the brain cells.
The cells manufacture more than 600 enzymes. Cellular enzymes build complex hormones and disease fighting antibodies.
Exercise alters muscle metabolism in such a way that fit people can eat more without making more fat.
Muscle is unique in its ability to produce sudden bursts of energy.
Brain cells use twice as many calories during intense thinking as during sleep.
But muscle cells may increase their energy demand fifty-fold over a split second depending on rest or intense physical work.
Nutrients that have filtered into the cells from the blood will supply the source of energy.
However only in the muscle cells one can find enzymeS competent to metabolize the nutrients so quickly.
Muscle has special enzymes enabling it to burn up tremendous amount of calories in short periods.
Muscle is the only tissue with enzymes that are specialised for sudden increase in calorie burning. This alone is sufficient check to keep our muscles in good tone.
Fatness is a Vicious circle. As one gets fat, the body changes to favour getting all the more fatter.
Eating carbohydrates causes a rise in blood sugar, followed by a rise in blood insulin, which the pancreas produces to handle the sugar.
Since fat persons tissues are insensitive to their own insulin, these responses are abnormally high.
All body cells are impervious to glucose until insulin comes along and stimulates them to open certain pores, which then allows the entrance of glucose.
As one gets fat, one's muscles get out of shape. They lose their ability to respond to insulin. This means glucose can enter the muscle cells very slowly.
Since muscles account for 30-50% of the body, blood glucose levels remain elevated longer after a meal.
This means extra glucose is circulating around the blood stream looking for some cells to enter.
So what kind of cells does a fat person have in abundance? Yes!! Fat cells! Sugar is dumped to fat cells.
When the muscle cells reject the glucose, the fat tissue becomes a glucose sink.
You can ask "So what?" Glucose becomes the skeleton for fat formation.
The paradox then is that the person with the least need for more fat, makes it the fastest.


text:
ASIO To Sydney Sheikh: Deport Yourself, Please
Tom Allard
ASIO has asked a prominent Sydney Shiite cleric, Sheik Mansour Leghaei, to "voluntarily deport" himself in the latest, bizarre twist in a lengthy bid by the spy agency to enforce an adverse security assessment against him.
Sheikh Mansour, who came to Australia from Iran and is a father of four, has hosted Attorney-General Philip Ruddock at his Imam Husain Islamic Centre at Earlwood and been the recipient of a glowing reference from Labor frontbenchers Robert McClelland and Anthony Albanese.
He has also been widely praised for his moderate teachings and leading role in setting up inter-faith dialogues.
But ASIO has accused him of being a spy for Iran. It also confiscated a notebook from him which it said contains jihadist material.
According to ASIO, parts of the book discussed a holy war to be fought with "infidels" who "do not accept the Koran as the book of heaven", and gave advice on spying.
However, Sheikh Mansour has said the notebook was translated incorrectly and has denied he is, or was, an Iranian spy.
Sheikh Mansour and his legal team have never been able to see the full detail of the security assessment made against him that required the Department of Immigration to revoke his visa.
He had appealed to the Federal Court in 2005, but it ruled that national security considerations overruled any notion of procedural fairness or natural justice.
Last week the High Court declined an application for him to appeal the decision.
As Sheikh Mansour has four dependants who are Australian residents, including one child who was born in Australia, he said that meant he could not be forcibly deported. That led to the Australian Government Solicitor, acting for ASIO, to ask Sheikh Mansour this week to voluntarily leave the country.
"That's the dilemma they have," Sheikh Mansour said yesterday. "I don't intend to go. I have asked the community and they don't want me to go."
ASIO yesterday declined to comment.
← Fogh Ahead For Denmark
Dutch Float 'Migrant Prison' Scheme →


text:
Q: Mockito.When method not able to return mocked object when the method is called I see lot of posts on this question but still not able to solve my problem.
when(queryEngineService.getRelationshipWithResources(anyString(), refEq(filterDto), anyString(), anyString(),
                anyInt(), anyInt(), anyString())).thenReturn(tenantResponseDto);
TenantResponseDto dto = queryEngineService.getRelationshipWithResources(anyString(), refEq(filterDto),
                anyString(), anyString(), anyInt(), anyInt(), anyString());

When i debug the object. I'm not able to see the mock data in it. is there anything which I am missing?
Below one is my TestClass:
    public class QueryEngineAPITest extends AbstractTest {

    QueryEngineService queryEngineService = Mockito.mock(QueryEngineService.class);
    TenantResponseDto tenantResponseDto;

    private QueryEngineDto queryEngineDto;
    private RelationshipFilterDto filterDto;
    @Before
    public void setUp() throws Exception {
        super.setUp();
        prepareMockData();

    }

    void prepareMockData() {
        tenantResponseDto = new TenantResponseDto();
        int i = 0;
        queryEngineDto = new QueryEngineDto();
        List<Map<String, Object>> mockDataList = new ArrayList<Map<String, Object>>();
        while (i < 10) {
            Map<String, Object> mockDataMap = new HashMap<String, Object>();
            mockDataMap.put("APPLICATION", "AWS");
            mockDataMap.put("ENVIRONMENT_NAME", "SIGMA LABS" + i);
            mockDataList.add(mockDataMap);
            i++;
        }
        queryEngineDto.setQueryData(mockDataList);
        tenantResponseDto.setResponseAsQueryEngineDto(queryEngineDto);
tenantResponseDto.setResponseCount(Long.valueOf(queryEngineDto.getQueryData().size()));

    filterDto = new RelationshipFilterDto();
    filterDto.setBefore("");
}

@Test
public void getRelationShipsTest() throws Exception {
    when(queryEngineService.getRelationshipWithResources(eq(""), refEq(filterDto), eq(""), eq(""), eq(1), eq(1),
            eq(""))).thenReturn(tenantResponseDto);
    TenantResponseDto dto = queryEngineService.getRelationshipWithResources(eq(""), refEq(filterDto), eq(""),
            eq(""), eq(1), eq(1), eq(""));
    assertFalse(dto.getResponseAsQueryEngineDto().getQueryData().isEmpty());
    assertEquals(Long.valueOf(dto.getResponseAsQueryEngineDto().getQueryData().size()), Long.valueOf(10));

}

A: Your mocked method with arguments doesn't match method you are calling
when(queryEngineService.getRelationshipWithResources(eq(""), refEq(filterDto), eq(""), eq(""), eq(1), eq(1),eq(""))).thenReturn(tenantResponseDto);
TenantResponseDto dto = queryEngineService.getRelationshipWithResources(eq(""), refEq(filterDto), eq(""),
                eq(""), eq(1), eq(1), eq(""));

You need to change second line to:
TenantResponseDto dto = queryEngineService.getRelationshipWithResources("",filterDto, "", "", 1, 1, "");

eq and refEq are called argument matchers, when you want to run test you need to provide arguments as they are, not in argument matcher


text:
Formula 1 Lewis Hamilton
Lewis Hamilton Net Worth, Childhood and Royal Ties
Posted on October 11, 2018 November 2, 2018 Author Jordan Ewanss	Comments Off on Lewis Hamilton Net Worth, Childhood and Royal Ties
Lewis Hamilton is one of the most popular Formula One drivers and definitely the most successful driver which came from Britain. Hamilton has had a fruitful career so far and he was crowned the F1 World Champion four times over the course of his career.
Hamilton recoded wins at both the German and Hungarian Grand Prix and due to his massive success, he has become of the richest drivers in the world. When he was four, he couldn't even imagine that he would become one of the best in this beautiful sport.
source:hellomagazine.com
It was then when his father Anthony bought him a remote-control car. Lewis started competing when he was just a kid and he finished second in the national BRCA championship. He said of that time: "I was racing these remote-controlled cars and winning club championships against adults". It was a sign for his father. He then purchased Lewis' first go-kart for Christmas at the age of six fueling his passion for motorsports.
Lewis Hamilton Net Worth
According to Celebrity Net Worth website, the popular F1 driver has the estimated net worth of $285 million. Yearly, Hamilton earns around $50 million. When he joined Mercedes' racing team in 2012, he signed a contract worth $100 million. And let us not forget about the money Hamilton has earned through advertising sponsorships with global brands such as Tommy Hilfiger, Puma and L'Oreal.
Racing Gear Sponsorships
Just by looking at Hamilton's racing gear, he has a $148 million worth of sponsors. Some of the names are Energy, Bose, Union Bank of Switzerland, Epson and the two we have already mentioned – Tommy Hilfiger and Puma. They all pay millions to the driver for their logo to be featured on Lewis' white polo top.
Family influence
Hamilton has mentioned several times that his father had a vast influence on him and the story when he received his first remote-controlled car. About six months ago, Lewis shared a cute video of him as a kid and his journey into racing. You can see him compete in his go-kart race when he was eight. At first it was just something that bonded father and son but it appeared to be much bigger.
source:edition.cnn.com
Lewis said: "To be honest, it started out as a hobby, something for my Dad and I to do together. At one point, my dad had four jobs just to keep us karting. He gave up everything for me. I wouldn't be a four-time world champion if it wasn't for my dad."
Early Ambitions
When he was a kid, Lewis talked about his future career: "Imagine being in a Formula One car, that must be very powerful that." When he was nine, Hamilton met McLaren boss Ross Dennis and he told him: "Hi, I'm Lewis Hamilton. I won the British championship and one day I want to be racing your cars." Dennis gave him the autograph, writing: "Phone me in nine years, we'll sort something out then." Three years has passed since that moment when Lewis became one of the youngest drivers who was signed to the McLaren Driver Development Support programme.
source:dailymail.co.uk
Hamilton and the Royal Family
Hamilton has impressed every man and woman living in Britain and the royal family was no exception. Following his victory at the F1 World Championship in 2014, Prince Harry said: "Lewis thank you very much for not making the British public sweat. You are an absolute legend. Well done mate."
source:telegraph.co.uk
He even had lunch with the Queen and he recalls: "She is a sweet woman, and we talked about how she spends her weekends, houses and music. She is really cool."
Tagged Childhood and Royal Ties, F1, Formula 1, Lewis Hamilton childhood, Lewis Hamilton girlfriend, Lewis Hamilton Net Worth, Lewis Hamilton Net Worth 2018
20 Women Drivers to Participate in the New Female-Only Racing Series
Posted on October 17, 2018 October 17, 2018 Author Jordan Ewanss
The new W series, for female-only drivers, aims at bringing women to Formula One sometimes in the future. The series is supported by non-other than David Coulthard, the British legend of Formula One.
Formula 1 Race
Top 7 Driver Bans in Formula One
Posted on September 3, 2018 September 6, 2018 Author Jordan Ewanss
Formula One is the most superior segment of automobile racing and the vehicles that are used here are the fastest single-seated cars on our planet. It is a real joy to watch the professional drivers steer through the bends and speed up on the straight line.
Tesla Technology Is Possible Thanks to Formula 1
Posted on February 7, 2019 February 7, 2019 Author Jordan Ewanss
According to Paddy Lowe, the Chief Technical Officer at Williams Martini Racing, the rate of development of Formula 1 surpasses the one in other major industries. Likewise, it has assisted in developing advances in electric motoring.
2019 Mercedes-Benz EQC Is Finally Here
26 Amazing F1 Racecar Concepts