TITLE: Complete atomless Boolean algebras with abelian automorphism group
QUESTION [5 upvotes]: Is there any example of a complete atomless Boolean algebra with a non-trivial abelian automorphism group?
This is equivalent, by Stone duality, to asking for an extremally disconnected compact Hausdorff space with no isolated points, having abelian homeomorphism group. Note that no such example can be metrizable.
I only know of examples of rigid complete Boolean algebras, i.e., those having trivial automorphism groups. For examples, see

Jónsson, B., A Boolean Algebra Without Proper Automorphisms, Proc. AMS, 1951.

Another rigid example coming from forcing appears in

McAloon, K., Consistency Results About Ordinal Definability, Ann. Math. Log, 1971.

REPLY [4 votes]: Don Monk's answer confirms my expectation; let me now also provide a more precise statement and a proof:

If $A$ is a Boolean algebra with a trivial automorphism group, then the automorphism group of $A\times A$ is canonically isomorphic to $(A,+)$, acting by $a\cdot(b,c)=((1-a)b+ac,(1-a)c+ab)$.

By Stone duality, this amounts to proving: if $X$ is a Stone space (totally disconnected, Hausdorff compact topological space) with trivial self-homeomorphism group, then any self-homeomorphism of $X\times\{-1,1\}$ is given by $(x,t)\mapsto (x,u(x)t)$ where $u$ is a continuous function $X\to\{-1,1\}$.
To prove the latter, let us first check that for any two clopen subsets $Y,Z$ of $X$ and homeomorphism $h:Y\to Z$, we have $Y=Z$ and $h$ is the identity. Indeed, otherwise there exists $y\in Y$ such that $h(y)\neq y$. By passing to smaller clopen subsets, we can then assume that $Y$ and $Z$ are disjoint. Then we can extend $h$ to a self-homeomorphism of $X$, as equal to $h^{-1}$ on $Z$ and identity elsewhere. (Note that in this part, we use that $X$ is Hausdorff with a basis of clopen subsets, but compactness does not play any role).
Then it's easy to conclude. Let $h$ be a self-homeomorphism of $X\times\{-1,1\}$. Call "component" the two subsets $X\times\{t\}$. We find a finite clopen partition of $X$ such that for each part $Y$ of the partition, each of $Y\times\{t\}$ is mapped into a single component. Then by the previous fact on partial homeomorphism, we have $h(y,t)=(y,s(y,t))$ for some $s(y,t)\in\{-1,1\}$ (which is constant when $y$ varies in $Y$). By injectivity, we have $s(y,1)=-s(y,-1)$. So we can write $s(y,t)=u(y)t$. $\Box$
Similarly, for a finite set $F$, the automorphism group of $X\times F$ consists of the $(x,t)\mapsto (x,\sigma(x)(t))$, where $\sigma$ ranges over the continuous functions $X\to\mathfrak{S}(F)$. The corresponding Boolean algebra action can be described accordingly.


TITLE: Rouche’s theorem Application right half plane
QUESTION [1 upvotes]: Show that there are two roots of $(z^4+2)e^z+z+1=0$ which lies in the right half plane (that is, Rez > 0).
I have been stuck in this problem for a week, and I  only know that $|e^z|$>1
Can somebody give me some hints? Thanks a lot!

REPLY [2 votes]: Along the imaginary axis, $|(z^4 + 2)e^z| > |z+1|$ (note that $z^4$ is real and positive in this case), while on a large enough semicircle centered at the origin going into the right half-plane, we clearly have $|(z^4 + 2)e^z| > |z+1|$. Using that semicircle together with its diameter, Rouche's theorem implies that $(z^4 + 2)e^z + z + 1$ has as many zeroes within this semicircle as does $(z^4 + 2)e^z$, which clearly has two. Since we can choose the semicircle to be as large as we want, there can be only two zeroes in the right half-plane.


TITLE: Identities of the spectrum of a ring
QUESTION [1 upvotes]: Let $R$ be a commutative ring, then the spectrum of $R$ is a topological space:
$$Spec(R)=\{ P:P \;\text{is a prime ideal of R} \}$$
where the close sets are
$$V_{I}=\{P \in Spec(R) \;\text{such that} \; I \subset P\}$$
then we have that
$$V_I = \bigcap_{f\in I}V_f $$

So I want to prove that

$V_{\Sigma I_{\alpha}}=\bigcap _{\alpha}V_{I_{\alpha}}$
$V_{I_1 \cap I_2}=V_{I_1} \cup V_{I_2}$

But I don't know how to proceed in the first one. For the second one I used the distributive property of  union and intersection to get:
$$V_{I_1} \cup V_{I_2}= \bigcap_{f \in I_1} V_{f} \cup \bigcap _{f_1 \in I_2} V_{f_1}=\bigcap_{f_1 \in I_2 } \bigcup_{f \in I_1} V_{f_1} \cap V_f$$
but I don't know how to compare it or get from this $V_{I_1 \cap I_2}$
So, can someone help me to prove this please?

Edition

For 1) I've got this
$\Rightarrow]$ Since we know that $\cup I_k \subset \Sigma I_k $ then the result follows.
$\Leftarrow]$ I don't know how to get the other contention because how can you embed a bigger set into a small one?
For 2) I've got this
$\Leftarrow]$ Since $I_1 \cap I_2 \subset I_1$ and  $I_1 \cap I_2 \subset I_2$
then the result follows.
$\Rightarrow]$ The same problem as before.

REPLY [1 votes]: I'll use $V(I)$ to denote what you mean as $V_I$ to clean up the notation.

$\sum I_\alpha$ is the smallest ideal containing all of the ideals $I_\alpha$. therefore, $P$ contains the sum if and only if $P$ contains each $I_\alpha$.
If $I$ or $J$ is contained in $P$, then $P$ certainly contains $I \cap J$. Conversely, suppose $I \cap J \subseteq P$. Suppose, WLOG, that $I$ is not contained in $P$ though. Then there is some $x \in I - P$. But for any $y \in J$, $xy \in I \cap J$ and so $xy \in P$. Thus, $y \in P$ so $J \subseteq P$ as desired.


TITLE: Can energy be defined as the most fundamental particle which exists in different forms as protons, electrons etc
QUESTION [0 upvotes]: What we see all the different forms of matter around is just a form of energy. Why can't this energy be the most fundamental particle. Given dust can turn into a star and then emit all sorts of radiation loosing energy over a period of time either collapse to worst case of black hole or burst as a supernova. The dust and gas would again collapse again at some point of time and start the cycle all over again. Why can't energy be then the fundamental particle?

REPLY [3 votes]: Students have a tendancy to think of the energy as a thing in it's own right - possibly as a result of watching too many episodes of Star Trek. As various people have commented, energy is a property of a system not an object in it's own right.
You can see this very easily. Take two objects of mass M coming towards each other at speed $v$. What is the kinetic energy of the system? If you're in the centre of mass frame you see two objects both moving at speed $v$, so the total kinetic energy is $mv^2$. But suppose you're sitting on one of the objects. Now you see your object has zero KE because it's stationary while the other object is moving at $2v$, so the total KE is $2mv^2$, which is different from the previous result.
So energy is frame dependant unlike for example the number of particles, which is frame independant (we'll gloss over the Unruh effect here!).


TITLE: Difference between cofinal map and cofinal
QUESTION [2 upvotes]: In this note I read, the cofinal map is defined as,

Let $\mathbb D´$ and $\mathbb D$ be two directed sets, $f: \mathbb D´ \to \mathbb D$ is a cofinal map, if $M$ is cofinal in $\mathbb D´$, then $f[M]$ is cofinal in $\mathbb D$.

What´s the difference between the aformentioned definition with merely defining as $f[\mathbb D´]$ is
cofinal in $\mathbb D$?
I think they´re supposed to be different, but it seems to me the second, the weaker one, implies the first.

REPLY [2 votes]: The two definitions are not equivalent. E.g., take $\mathbb{D}' = \mathbb{D} = \mathbb{N}$ with the usual ordering and define $f(n)$ to be $n$ if $n$ is even and to be $0$ if $n$ is odd. Then $f[\mathbb{D}]$ comprises the even numbers and is cofinal (so $f$ satisfies your proposed simpler definition). However, the set $O$ of odd numbers is cofinal, but $f[O] = \{0\}$ is not (so $f$ does not satisfy the definition you quoted from the note).


TITLE: My proof that there are primitive roots modulo $p^2$
QUESTION [0 upvotes]: Let $p$ be a prime number. I'd like to prove that there are primitive roots modulo $p^2$. Could someone check this argument?

Note that if $r\in\mathbb Z$ is a primitive root modulo $p^2$, it must be a primitive root modulo $p$ as well, since it must have powers congruent to $1, 2, 3, ... p$ modulo $p^2$ and therefore those powers will also be congruent to $1, 2, 3, ... p$ modulo $p$.
So, we restrict our search to primitive roots modulo $p$. Let $r$ be such a root, and let's look for one of order $\phi(p^2)=p(p-1)$. Let $n$ be the order of $r$ modulo $p^2$. $n$ must divide $\phi(p^2)=p(p-1)$. But it must also be a multiple of $\phi(p)=p-1$, since $r^n\equiv 1$ modulo $p^2$ and therefore also modulo $p$. This implies that $n/(p-1)$ divides $p$ and thus either $n=p(p-1)$ or $n=(p-1)$.
So we just need to find out if all values of $r$ are of order $p-1$. Well, if $n$ is the order of $r$, then $r+p$ is another primitive root modulo $p$ and thus another candidate in our search. But $(r+p)^n\equiv r^n + pr^{n-1}\equiv1 + pr^{n-1}$ modulo $p^2$. But the $pr^{n-1}$ can't be zero because $r^{n-1}$ is a unit modulo $p^2$. So $(r+p)^n\not\equiv 1$, and in particular $r$ and $r+n$ have different orders.
So, if there is even one value of $r$ of order $p-1$, then there is another that isn't (namely $r+p$), and is therefore of order $p(p-1)$, and therefore is a primitive root. Of course, if there aren't any values of $r$ of order $p-1$, then they're all of order $p(p-1)$, so we still reach the desired result.

REPLY [1 votes]: There is actually a serious flaw in this proof which I didn't spot. In the second paragraph I want to show that $r$ and $r+p$ are always of different orders. I write:
$$(r+p)^n\equiv r^n+pr^{n-1} \text{ modulo $p^2$}$$
I got the binomial formula wrong - it should be
$$(r+p)^n\equiv r^n+npr^{n-1} \text{ modulo $p^2$}$$
Thus we also need $n$ to be coprime with $p$ in order to conclude. In our case we really just want this argument to work when $n=p-1$, so we're fine, although we can't show the more general statement (which might be false) that the orders of $r$ and $r+p$ are always different.
This flaw becomes apparent when we try to generalize this argument to show that if there are primitive roots modulo $p^n$, there are primitive roots modulo $p^{n+1}$, in which case the argument fails if $p=2$, basically because to go from $2^2$ to $2^3$ we need to take $n=\phi(4)$, which is not coprime to $2$ at all.


TITLE: How to prove two closed intervals are equinumerous?
QUESTION [1 upvotes]: Prove that $[0,1] \approx [-\pi,e^2]$
(The notation $\approx$ is used to denote equinumerous)
I know to prove two closed intervals are equinumerous I need to show a bijective function that will map from one interval to the other but I cannot seem to find the correct mapping.

REPLY [0 votes]: Hint: $[-\pi, e^2] \cong [0, e^2 + \pi]$
Edit: I don't understand why I received a downvote so here's the full solution: The linear function from $[0,1] \to [0, e^2 + \pi]$ is just multiplication by $e^2 + \pi$ and is clearly a bijection. We then compose this with the map which subtracts $\pi$. This composition is a composition of bijections and is therefore a bijection.


TITLE: Conditional PDF of X, given random variable Y based on the PDF of X
QUESTION [0 upvotes]: X is a random variable with PDF: $F_X(x)=\frac{1}{2}sin(x)$ on the interval $[0,\pi]$
If $X\in[0,\frac{\pi}{3}],Y=3$
If $X\in(\frac{\pi}{3},\frac{2\pi}{3}],Y=2$
If $X\in(\frac{2\pi}{3},\pi],Y=1$
What is the conditional probability density function of X given Y =2?
$f_{X|Y}(x|Y=2)=?$
I tried starting by calculating $P(Y=2)$, but could not figure out how to do that from the way thay Y is defined

REPLY [1 votes]: Since $F_X$ is a probability density function (not CDF as is usually indicated by use of capital lettering): $$\begin{align}\mathsf P(Y{=}2)~&=~\mathsf P(\tfrac\pi 3{\lt} X{\leqslant}\tfrac{2\pi}{3})\\[1ex]&=~\int_{\pi/3}^{2\pi/3} F_X(x)~\mathrm d x\end{align}$$And so using Bayes' Rule: $$F_X(x\mid Y{=}2)=\dfrac{F_X(x)\mathbf 1_{\pi/3<x\leqslant 2\pi/3}}{\mathsf P(Y=2)}$$


text:
\begin{document}
\footnotetext[1]{Corresponding author, helmut.laenger@tuwien.ac.at}
\footnotetext[2]{Support of the research of the authors by the Austrian Science Fund (FWF), project I~4579-N, and the Czech Science Foundation (GA\v CR), project 20-09869L, entitled ``The many facets of orthomodularity'', as well as by \"OAD, project CZ~02/2019, entitled ``Function algebras and ordered structures related to logic and data fusion'', and, concerning the first author, by IGA, project P\v rF~2020~014, is gratefully acknowledged.}
\maketitle
\begin{abstract}
The concept of a Kleene algebra (sometimes also called Kleene lattice) was already generalized by the first author for non-distributive lattices under the name pseudo-Kleene algebra. We extend these concepts to posets and show how (pseudo-)Kleene posets can be characterized by identities and implications of assigned commutative meet-directoids. Moreover, we prove that the Dedekind-MacNeille completion of a pseudo-Kleene poset is a pseudo-Kleene algebra and that the Dedekind-MacNeille completion of a finite Kleene poset is a Kleene algebra. Further, we introduce the concept of a strict (pseudo-)Kleene poset and show that under an additional assumption a strict Kleene poset can be organized into a residuated structure. Finally, we prove by using the so-called twist construction that every poset can be embedded into a pseudo-Kleene poset in some natural way.
\end{abstract}

{\bf AMS Subject Classification:} 06A11, 06D30, 03G25

{\bf Keywords:} (pseudo-)Kleene algebra, (pseudo-)Kleene poset, strong pseudo-Kleene \\
poset, strict (pseudo-)Kleene poset, commutative meet-directoid, Dedekind-\\
-MacNeille completion, twist construction

\section{Introduction}

Kleene algebras, or in another terminology Kleene lattices (see \cite{Ci} and \cite K), are special cases of De Morgan algebras, i.e.\ distributive lattices with an antitone involution satisfying the so-called {\em normality condition}, i.e.\ the identity $x\wedge x'\leq y\vee y'$. These algebras were also called {\em normal i-lattices} (see \cite K) or quasi-Boolean algebras (by A.~Bialynicki and H.~Rasiova). They are important models in the field of logic since they generalize Boolean algebras, \L ukasiewicz algebras and Post algebras. The name ``Kleene algebra'' was introduced by R.~Cignoli (\cite{Ci}). The case when the underlying lattice need not be distributive was treated by the first author in \cite{Ch} under the name pseudo-Kleene algebras.

In some propsitional logics the identification of disjunction with lattice join $\vee$ turns out to be problematic. For example, the logic of quantum mechanics was originally modeled by orthomodular lattices (which are special pseudo-Kleene algebras) and later on by orthomodular posets in which the existence of the join $x\vee y$ of two elements $x$ and $y$ is guaranteed only in the case when these elements are orthogonal to each other which means that $x\leq y'$ (or, equivalently, $y\leq x'$). Similar problems may occur also in other models of non-classical logics. Hence the question arises whether results obtained for Kleene algebras or pseudo-Kleene algebras can be generalized to posets with an antitone involution satisfying some condition analogous to normality. In this paper we solve this problem by investigating so-called Kleene posets, pseudo-Kleene posets and strong pseudo-Kleene posets. We believe that these may be successfully applied in the algebraic axiomatization of several non-classical logics. Moreover, we introduce some kind of residuation which may be applied in fuzzy logic.

For the reader's convenience, we recall several concepts concerning posets.

Let $\mathbf P=(P,\leq)$ be a poset, $a,b\in P$ and $A,B\subseteq P$. We write $a\parallel b$ if $a$ and $b$ are incomparable and we extend $\leq$ to subsets by defining
\[
A\leq B\text{ if and only if }x\leq y\text{ for all }x\in A\text{ and }y\in B.
\]
Instead of $\{a\}\leq B$ and $A\leq\{b\}$ we also write $a\leq B$ and $A\leq b$, respectively. Analogous notations are used for the reverse order $\geq$. Moreover, we define
\begin{align*}
L(A) & :=\{x\in P\mid x\leq A\}, \\
U(A) & :=\{x\in P\mid A\leq x\}.
\end{align*}
Instead of $L(A\cup B)$, $L(\{a\}\cup B)$, $L(A\cup\{b\})$ and $L(\{a,b\})$ we also write $L(A,B)$, $L(a,B)$, $L(A,b)$ and $L(a,b)$, respectively. Analogous notations are used for $U$. Instead of $L(U(A))$ we also write $LU(A)$. Analogously, we proceed in similar cases. Sometimes we identify singletons with their unique element, so we often write $L(a,b)=0$ and $U(a,b)=1$ instead of $L(a,b)=\{0\}$ and $U(a,b)=\{1\}$, respectively. The {\em poset} $\mathbf P$ is called {\em downward directed} if $L(x,y)\neq\emptyset$ for all $x,y\in P$. The {\em poset} $\mathbf P$ is called {\em bounded} if it has a least element $0$ and a greatest element $1$. This fact will be expressed by notation $(P,\leq,0,1)$. The {\em poset} $\mathbf P$ is called {\em distributive} if it satisfies one of the following equivalent identities:
\begin{align*}
 L(U(x,y),z) & \approx LU(L(x,z),L(y,z)), \\
UL(U(x,y),z) & \approx U(L(x,z),L(y,z)), \\
 U(L(x,y),z) & \approx UL(U(x,z),U(y,z)), \\
LU(L(x,y),z) & \approx L(U(x,z),U(y,z)).
\end{align*}
In fact, the inclusions
\begin{align*}
LU(L(x,z),L(y,z)) & \subseteq L(U(x,y),z), \\
UL(U(x,z),U(y,z)) & \subseteq U(L(x,y),z)
\end{align*}
hold in every poset. Hence, to check distributivity, we need only to confirm one of the converse inclusions. A unary operation $'$ on $P$ is called
\begin{itemize}
\item {\em antitone} if, for all $x,y\in P$, $x\leq y$ implies $y'\leq x'$,
\item an {\em involution} if it satisfies the identity $x''\approx x$.
\end{itemize}
For $A\subseteq P$ we define $A':=\{x'\mid x\in A\}$. If the poset is bounded and distributive, we can prove the following property of an antitone involution.

\begin{lemma}
Let $(P,\leq,{}',0,1)$ be a bounded distributive poset with an antitone involution and $a,b\in P$ with $a\leq b$ and $L(b,a')=\{0\}$. Then the following hold:
\begin{align*}
L(a,a') & =L(b,b')=\{0\}, \\
U(a,a') & =U(b,b')=\{1\}.
\end{align*}
\end{lemma}

\begin{proof}
We have
\begin{align*}
L(a,a') & =LUL(a,a')=LU(L(a,a'),0)=LU(L(a,a'),L(b,a'))=L(U(a,b),a')= \\
        & =L(U(b),a')=L(b,a')=\{0\}, \\
L(b,b') & =LUL(b',b)=LU(0,L(b',b))=LU(L(a',b),L(b',b))=L(U(a',b'),b)= \\
        & =L(U(a'),b)=L(a',b)=\{0\}, \\
U(a,a') & =(L(a',a))'=\{0\}'=\{1\}, \\
U(b,b') & =(L(b',b))'=\{0\}'=\{1\}.
\end{align*}
\end{proof}

\section{Kleene posets and pseudo-Kleene posets}

Now we define our main concepts.

\begin{definition}
A {\em pseudo-Kleene poset} is a poset $\mathbf P=(P,\leq,{}')$ with an antitone involution satisfying
\begin{enumerate}
\item[{\rm(K)}] $L(x,x')\leq U(y,y')$ for all $x,y\in P$.
\end{enumerate}
An element $a$ of $P$ is called a {\em fixed point} of $\mathbf P$ if $a'=a$. By a {\em Kleene poset} we mean a distributive pseudo-Kleene poset.
\end{definition}

\begin{lemma}
Let $\mathbf P=(P,\leq,{}')$ be a pseudo-Kleene poset. Then $\mathbf P$ has at most one fixed point.
\end{lemma}

\begin{proof}
If $a$ and $b$ are fixed points of $\mathbf P$ then
\begin{align*}
L(a) & =L(a,a)=L(a,a')\leq U(b,b')=U(b,b)=U(b), \\
L(b) & =L(b,b)=L(b,b')\leq U(a,a')=U(a,a)=U(a)
\end{align*}
and hence $a\leq b$ and $b\leq a$, i.e.\ $a=b$.
\end{proof}

\begin{example}\label{ex2}
The poset visualized in Figure~1 is a Kleene poset which is not a lattice:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(4,8)
\put(2,1){\circle*{.3}}
\put(1,3){\circle*{.3}}
\put(3,3){\circle*{.3}}
\put(1,5){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(2,7){\circle*{.3}}
\put(1,3){\line(0,1)2}
\put(1,3){\line(1,1)2}
\put(1,3){\line(1,-2)1}
\put(3,3){\line(-1,-2)1}
\put(3,3){\line(-1,1)2}
\put(3,3){\line(0,1)2}
\put(2,7){\line(-1,-2)1}
\put(2,7){\line(1,-2)1}
\put(1.85,.25){$0$}
\put(.3,2.85){$a$}
\put(3.4,2.85){$b$}
\put(.3,4.85){$b'$}
\put(3.4,4.85){$a'$}
\put(1.3,7.4){$1=0'$}
\put(1.2,-.75){{\rm Fig.\ 1}}
\end{picture}
\end{center}
\vspace*{4mm}
\end{example}

\begin{example}
The poset $(P,\leq,{}')$ visualized in Figure~2 is a pseudo-Kleene poset which is neither a lattice nor a Kleene poset since
\[
L(U(a,c),b)=L(1,b)=L(b)\neq\{0\}=L(P)=LU(0)=LU(0,0)=LU(L(a,b),L(c,b)).
\]
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(8,8)
\put(4,1){\circle*{.3}}
\put(3,3){\circle*{.3}}
\put(5,3){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(5,5){\circle*{.3}}
\put(4,7){\circle*{.3}}
\put(1,4){\circle*{.3}}
\put(7,4){\circle*{.3}}
\put(3,3){\line(0,1)2}
\put(3,3){\line(1,1)2}
\put(3,3){\line(1,-2)1}
\put(5,3){\line(-1,-2)1}
\put(5,3){\line(-1,1)2}
\put(5,3){\line(0,1)2}
\put(4,7){\line(-1,-2)1}
\put(4,7){\line(1,-2)1}
\put(4,1){\line(-1,1)3}
\put(4,1){\line(1,1)3}
\put(4,7){\line(-1,-1)3}
\put(4,7){\line(1,-1)3}
\put(3.85,.25){$0$}
\put(2.3,2.85){$a$}
\put(5.4,2.85){$b$}
\put(.3,3.85){$c$}
\put(7.4,3.85){$c'$}
\put(2.3,4.85){$b'$}
\put(5.4,4.85){$a'$}
\put(3.3,7.4){$1=0'$}
\put(3.2,-.75){{\rm Fig.\ 2}}
\end{picture}
\end{center}
\vspace*{4mm}
\end{example}

Recall that a {\em pseudo-Kleene algebra} (see \cite{Ch}) is a lattice $(L,\vee,\wedge,{}')$ with an antitone involution satisfying the identity
\[
x\wedge x'\leq y\vee y'\text{ for all }x,y\in L.
\]
A {\em Kleene algebra} (see \cite K) is a distributive pseudo-Kleene algebra. Observe that a lattice with an antitone involution is a (pseudo-)Kleene poset if and only if it is a (pseudo-)Kleene algebra. Hence, our concepts defined above are appropriate generalizations of pseudo-Kleene algebras and Kleene algebras as will be shown in the next section.

\begin{example}
The lattice visualized in Figure~3 is a pseudo-Kleene algebra which is not a Kleene algebra since the lattice is not distributive:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(4,10)
\put(2,1){\circle*{.3}}
\put(1,3){\circle*{.3}}
\put(3,3){\circle*{.3}}
\put(1,5){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(1,7){\circle*{.3}}
\put(3,7){\circle*{.3}}
\put(2,9){\circle*{.3}}
\put(1,3){\line(0,1)4}
\put(1,3){\line(1,2)2}
\put(1,3){\line(1,-2)1}
\put(3,3){\line(-1,-2)1}
\put(3,3){\line(0,1)4}
\put(2,9){\line(-1,-2)1}
\put(2,9){\line(1,-2)1}
\put(1.85,.25){$0$}
\put(.3,2.85){$a$}
\put(3.4,2.85){$b$}
\put(.3,4.85){$c$}
\put(3.4,4.85){$c'$}
\put(.3,6.85){$b'$}
\put(3.4,6.85){$a'$}
\put(1.3,9.4){$1=0'$}
\put(1.2,-.75){{\rm Fig.\ 3}}
\end{picture}
\end{center}
\vspace*{4mm}
\end{example}

\begin{example}
The lattice visualized in Figure~4 is a Kleene algebra:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(4,10)
\put(2,1){\circle*{.3}}
\put(2,3){\circle*{.3}}
\put(1,5){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(2,7){\circle*{.3}}
\put(2,9){\circle*{.3}}
\put(2,3){\line(0,-1)2}
\put(2,3){\line(-1,2)1}
\put(2,3){\line(1,2)1}
\put(2,7){\line(-1,-2)1}
\put(2,7){\line(1,-2)1}
\put(2,7){\line(0,1)2}
\put(1.85,.25){$0$}
\put(2.4,2.85){$a$}
\put(.3,4.85){$b$}
\put(3.4,4.85){$b'$}
\put(2.4,6.85){$a'$}
\put(1.3,9.4){$1=0'$}
\put(1.2,-.75){{\rm Fig.\ 4}}
\end{picture}
\end{center}
\vspace*{4mm}
\end{example}

\section{Dedekind-MacNeille completion}

In what follows we investigate the question for which posets $\mathbf P$ with an antitone involution their Dedekind-MacNeille completion $\BDM(\mathbf P)$ is either a pseudo-Kleene algebra or a Kleene algebra.

Let $\mathbf P=(P,\leq,{}')$ be a poset with an antitone involution. Define
\begin{align*}
 \DM(\mathbf P) & :=\{L(A)\mid A\subseteq P\}, \\
            A^* & :=L(A')\text{ for all }A\in\DM(\mathbf P), \\
\BDM(\mathbf P) & :=(\DM(\mathbf P),\subseteq,^*)
\end{align*}
Then $\BDM(\mathbf P)$ is a complete lattice with an antitone involution, called the {\em Dedekind-MacNeille completion} of $\mathbf P$. That $^*$ is an antitone involution on $(\DM(\mathbf P),\subseteq)$ can be seen as follows. Let $A,B\in\DM(\mathbf P)$. If $A\subseteq B$ then $A'\subseteq B'$ and hence $B^*=L(B')\subseteq L(A')=A^*$. Moreover, $A^{**}=L((L(A'))')=LU(A)=A$. We have
\begin{align*}
 (L(A))^* & =L((L(A))')=LU(A')\text{ for all }A\subseteq P, \\
  A\vee B & =LU(A,B)\text{ for all }A,B\in\DM(\mathbf P), \\
A\wedge B & =A\cap B\text{ for all }A,B\in\DM(\mathbf P).
\end{align*}

\begin{theorem}\label{th6}
Let $\mathbf P=(P,\leq,{}')$ be a poset with an antitone involution. Then $\BDM(\mathbf P)$ is a pseudo-Kleene algebra if and only if $\mathbf P$ is a pseudo-Kleene poset.
\end{theorem}

\begin{proof}
Assume that $\mathbf P$ is a pseudo-Kleene poset. Then the following are equivalent:
\begin{align*}
                   & \BDM(\mathbf P)\text{ is a pseudo-Kleene algebra}, \\
       C\wedge C^* & \subseteq D\vee D^*\text{ for all }C,D\in\DM(\mathbf P), \\
L(A)\wedge(L(A))^* & \subseteq L(B)\vee(L(B))^*\text{ for all }A,B\subseteq P, \\
 L(A)\cap L(U(A')) & \subseteq LU(L(B),LU(B'))\text{ for all }A,B\subseteq P, \\
        L(A,U(A')) & \subseteq L(UL(B)\cap ULU(B'))\text{ for all }A,B\subseteq P, \\
        L(A,U(A')) & \subseteq L(UL(B)\cap U(B'))\text{ for all }A,B\subseteq P, \\
        L(A,U(A')) & \subseteq LU(L(B),B')\text{ for all }A,B\subseteq P, \\
        L(A,U(A')) & \leq U(L(B),B')\text{ for all }A,B\subseteq P.
\end{align*}
Now let $A,B$ be fixed subsets of $P$. Then
\begin{align*}
L(A,U(A')) & =L(A)\cap LU(A')=\bigcup_{x\in L(A)}L(x)\cap\bigcap_{y\in U(A')}L(y)= \\
& =\bigcup_{x\in L(A)}(L(x)\cap\bigcap_{y\in U(A')}L(y))\subseteq\bigcup_{x\in L(A)}(L(x)\cap L(x'))=\bigcup_{x\in L(A)}L(x,x'), \\
U(L(B),B') & =UL(B)\cap U(B')=\bigcap_{x\in L(B)}U(x)\cap\bigcup_{y\in U(B')}U(y)= \\
& =\bigcup_{y\in U(B')}(\bigcap_{x\in L(B)}U(x)\cap U(y))\subseteq\bigcup_{y\in U(B')}(U(y')\cap U(y))=\bigcup_{y\in U(B')}U(y,y')
\end{align*}
and
\[
\bigcup_{x\in L(A)}L(x,x')\leq\bigcup_{y\in U(B')}U(y,y').
\]
Hence $\BDM(\mathbf P)$ is a pseudo-Kleene algebra provided $\mathbf P$ is a pseudo-Kleene poset. The converse is evident.
\end{proof}

\begin{theorem}
If $\mathbf P=(P,\leq,{}')$ is a finite poset with an antitone involution then $\BDM(\mathbf P)$ is a Kleene algebra if and only if $\mathbf P$ is a Kleene poset.
\end{theorem}

\begin{proof}
According to a result by M.~Ern\'e (\cite E), the Dedekind-MacNeille completion of a finite distributive poset is a distributive lattice. Hence, by Theorem~\ref{th6} the Dedekind-MacNeille completion of a finite Kleene poset is a Kleene algebra. The converse is evident.
\end{proof}

The Dedekind-MacNeille completion of the Kleene poset from Example~\ref{ex2} is visualized in Figure~5:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(4,10)
\put(2,1){\circle*{.3}}
\put(1,3){\circle*{.3}}
\put(3,3){\circle*{.3}}
\put(2,5){\circle*{.3}}
\put(1,7){\circle*{.3}}
\put(3,7){\circle*{.3}}
\put(2,9){\circle*{.3}}
\put(1,3){\line(1,2)2}
\put(1,3){\line(1,-2)1}
\put(3,3){\line(-1,2)2}
\put(3,3){\line(-1,-2)1}
\put(2,9){\line(-1,-2)1}
\put(2,9){\line(1,-2)1}
\put(1.85,.25){$0$}
\put(.3,2.85){$a$}
\put(3.4,2.85){$b$}
\put(2.4,4.85){$c=c'$}
\put(.3,6.85){$b'$}
\put(3.4,6.85){$a'$}
\put(1.3,9.4){$1=0'$}
\put(1.2,-.75){{\rm Fig.\ 5}}
\end{picture}
\end{center}
\vspace*{4mm}

\section{A representation by commutative meet-directoids}

Now we recall the concept of a commutative meet-directoid. We will use it for the characterization of pseudo-Kleene posets, Kleene posets, strong pseudo-Kleene posets and strict pseudo-Kleene posets. The advantage of this approach is that we characterize properties of posets by means of identities and quasiidentities of algebras. Hence, one can use algebraic tools for their investigation.

A {\em commutative meet-directoid} (see \cite{CL11} and \cite{JQ}) is a groupoid $\mathbf D=(D,\sqcap)$ satisfying the following identities:
\begin{align*}
                   x\sqcap x & \approx x\text{ (idempotency)}, \\
                   x\sqcap y & \approx y\sqcap x\text{ (commutativity)}, \\
(x\sqcap(y\sqcap z))\sqcap z & \approx x\sqcap(y\sqcap z)\text{ (weak associativity)}.
\end{align*}
If $\mathbf P=(P,\leq)$ is a downward directed poset, if we define $x\sqcap y:=x\wedge y$ for comparable $x,y\in P$ and if we put for $x\sqcap y=y\sqcap x$ an arbitrary element of $L(x,y)$ if $x,y\in P$ are incomparable, then $\mathbb D(\mathbf P):=(P,\sqcap)$ is a commutative meet-directoid which is called a {\em meet-directoid assigned} to $\mathbf P$. Conversely, if $\mathbf D=(D,\sqcap)$ is a commutative meet-directoid and we define
\[
x\leq y\text{ if and only if }x\sqcap y=x
\]
for all $x,y\in D$ then $\mathbb P(\mathbf D):=(D,\leq)$ is a downward directed poset, the so-called {\em poset induced} by $\mathbf D$. Though the assignment $\mathbf P\mapsto\mathbb D(\mathbf P)$ is not unique, we have $\mathbb P(\mathbb D(\mathbf P))=\mathbf P$ for every downward directed poset $\mathbf P$. Sometimes we consider posets and commutative meet-directoids together with a unary operation. Let $(D,\sqcap,{}')$ be a commutative meet-directoid $(D,\sqcap,{}')$ with an antitone involution. W e define
\[
x\sqcup y:=(x'\sqcap y')'\text{ for all }x,y\in D.
\]
Then $\sqcup$ is also idempotent, commutative and weakly associative and we have for all $x,y\in D$
\begin{align*}
x\sqcup y & =x\vee y\text{ if }x,y\text{ are }comparable, \\
x\sqcup y & =y\sqcup x\in U(x,y)\text{ if }x\parallel y, \\
x\sqcap y & =x\text{ if and only if }x\sqcup y=y, \\
     L(x) & =\{z\sqcap x\mid z\in P\}, \\
     U(x) & =\{z\sqcup x\mid z\in P\}, \\
   L(x,y) & =\{(z\sqcap x)\sqcap(z\sqcap y)\mid z\in P\}, \\
   U(x,y) & =\{(z\sqcup x)\sqcap(z\sqcup y)\mid z\in P\}.
\end{align*}

Posets with an antitone involution can be characterized in the language of commutative meet-directoids by identities as follows.

\begin{lemma}\label{lem1}
Let $\mathbf P=(P,\leq,{}')$ be a downward directed poset with a unary operation and $\mathbb D(\mathbf P)$ an assigned meet-directoid. Then $\mathbf P$ is a poset with an antitone involution if and only if $\mathbb D(\mathbf P)$ satisfies the identities
\begin{enumerate}[{\rm(1)}]
\item $x''\approx x$,
\item $(x\sqcap y)'\sqcap y'\approx y'$.
\end{enumerate}
\end{lemma}

\begin{proof}
Condition (1) is evident by definition. Further, the following are equivalent:
\begin{align*}
& (2), \\
& x'\sqcap y'=y'\text{ for all }x,y\in P\text{ with }x\leq y, \\
& y'\leq x'\text{ for all }x,y\in P\text{ with }x\leq y, \\
& '\text{ is antitone}.
\end{align*}
\end{proof}

Now we characterize pseudo-Kleene posets by identities of an assigned commutative meet-directoid.

\begin{theorem}\label{th1}
Let $\mathbf P=(P,\leq,{}')$ be a downward directed poset with a unary operation and $\mathbb D(\mathbf P)$ an assigned meet-directoid. Then $\mathbf P$ is a pseudo-Kleene poset if and only if $\mathbb D(\mathbf P)$ satisfies identities {\rm(1)} -- {\rm(3)}:
\begin{enumerate}
\item[{\rm(3)}] $(z\sqcap x)\sqcap(z\sqcap x')\leq(w\sqcup y)\sqcup(w\sqcup y')$.
\end{enumerate}
\end{theorem}

\begin{proof}
It is easy to check that (3) is equivalent to $L(x,x')\leq U(y,y')$ for all $x,y\in P$. Applying Lemma~\ref{lem1} completes the proof.
\end{proof}

In order to characterize Kleene posets in a similar way we need to capture distributivity of posets in the language of commutative meet-directoids.

\begin{theorem}\label{th3}
Let $\mathbf P=(P,\leq,{}')$ be a downward directed poset with a unary operation and $\mathbb D(\mathbf P)$ an assigned meet-directoid. Then $\mathbf P$ is a Kleene poset if and only if $\mathbb D(\mathbf P)$ satisfies identities {\rm(1)} -- {\rm(3)} and implication {\rm(4)}:
\begin{enumerate}
\item[{\rm(4)}] $w\sqcap((t\sqcup x)\sqcup(t\sqcup y))=w\sqcap z=w$ and $s\sqcup((t\sqcap x)\sqcap(t\sqcap z))=s\sqcup((t\sqcap y)\sqcap(t\sqcap z))=s$ for all $t\in P$ imply $w\leq s$.
\end{enumerate}
\end{theorem}

\begin{proof}
Since
\begin{align*}
& U(x,y)=\{(t\sqcup x)\sqcup(t\sqcup y)\mid t\in P\}, \\
& w\sqcap u=w\text{ is equivalent to }w\in L(u),
\end{align*}
$w\sqcap((t\sqcup x)\sqcup(t\sqcup y))=w\sqcap z=w$ is equivalent to $w\in L(U(x,y),z)$. Further, since
\begin{align*}
& L(x,z)=\{(t\sqcap x)\sqcap(t\sqcap z)\mid t\in P\}, \\
& L(y,z)=\{(t\sqcap y)\sqcap(t\sqcap z)\mid t\in P\}, \\
& s\sqcup u=s\text{ is equivalent to }s\in U(u),
\end{align*}
$s\sqcup((t\sqcap x)\sqcap(t\sqcap z))=s\sqcup((t\sqcap y)\sqcap(t\sqcap z))=s$ is equivalent to $s\in U(L(x,z),L(y,z))$. Hence the following are equivalent:
\begin{align*}
& (4), \\
& w\in L(U(x,y),z)\text{ and }s\in U(L(x,z),L(y,z))\text{ imply }w\leq s, \\
& L(U(x,y),z)\subseteq LU(L(x,z),L(y,z)), \\
& \mathbf P\text{ is distributive}.
\end{align*}
Applying Theorem~\ref{th1} completes the proof.
\end{proof}

Let us note that the class of all directoids assigned to downward directed pseudo-Kleene posets forms a variety due to Theorem~\ref{th1}. As shown in \cite{CL11} and \cite{CKL}, every variety of directoids with an antitone involution is congruence distributive.

\begin{definition}
A {\em strong pseudo-Kleene poset} is a poset $(P,\leq,{}')$ with an antitone involution satisfying
\begin{enumerate}
\item[{\rm(S)}] $x\parallel y$ implies $L(x,x')=L(y,y')$.
\end{enumerate}
\end{definition}

\begin{lemma}\label{lem2}
Let $\mathbf P=(P,\leq,{}')$ be a strong pseudo-Kleene poset. Then $\mathbf P$ is a pseudo-Kleene poset.
\end{lemma}

\begin{proof}
Let $a,b\in P$. \\
If $a\leq b$ then $L(a,a')\leq a\leq b\leq U(b,b')$. \\
If $b\leq a$ then $a'\leq b'$ and hence $L(a,a')\leq a'\leq b'\leq U(b,b')$. \\
If $a\parallel b$ then $L(a,a')=L(b,b')\leq U(b,b')$.
\end{proof}

\begin{example}\label{ex1}
The poset visualized in Figure~6 is a strong pseudo-Kleene poset which is not a lattice and hence not a pseudo-Kleene algebra:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(8,16)
\put(4,1){\circle*{.3}}
\put(4,3){\circle*{.3}}
\put(4,5){\circle*{.3}}
\put(1,7){\circle*{.3}}
\put(3,7){\circle*{.3}}
\put(5,7){\circle*{.3}}
\put(7,7){\circle*{.3}}
\put(1,9){\circle*{.3}}
\put(3,9){\circle*{.3}}
\put(5,9){\circle*{.3}}
\put(7,9){\circle*{.3}}
\put(4,11){\circle*{.3}}
\put(4,13){\circle*{.3}}
\put(4,15){\circle*{.3}}
\put(4,5){\line(0,-1)4}
\put(4,5){\line(-3,2)3}
\put(4,5){\line(-1,2)1}
\put(4,5){\line(1,2)1}
\put(4,5){\line(3,2)3}
\put(1,7){\line(0,1)2}
\put(1,7){\line(1,1)2}
\put(3,7){\line(-1,1)2}
\put(3,7){\line(0,1)2}
\put(5,7){\line(0,1)2}
\put(5,7){\line(1,1)2}
\put(7,7){\line(-1,1)2}
\put(7,7){\line(0,1)2}
\put(4,11){\line(-3,-2)3}
\put(4,11){\line(-1,-2)1}
\put(4,11){\line(1,-2)1}
\put(4,11){\line(3,-2)3}
\put(4,11){\line(0,1)4}
\put(3.85,.25){$0$}
\put(4.4,2.85){$a$}
\put(4.4,4.85){$b$}
\put(.3,6.85){$c$}
\put(3.4,6.85){$d$}
\put(4.3,6.85){$e$}
\put(7.4,6.85){$f$}
\put(.3,8.85){$f'$}
\put(3.4,8.85){$e'$}
\put(4.3,8.85){$d'$}
\put(7.4,8.85){$c'$}
\put(4.4,10.85){$b'$}
\put(4.4,12.85){$a'$}
\put(3.3,15.4){$1=0'$}
\put(3.2,-.75){{\rm Fig.\ 6}}
\end{picture}
\end{center}
\vspace*{4mm}
\end{example}

We are going to determine the class of directoids assigned to strong pseudo-Kleene posets.

\begin{theorem}\label{th2}
Let $\mathbf P=(P,\leq,{}')$ be a downward directed poset with a unary operation and $\mathbb D(\mathbf P)$ an assigned meet-directoid. Then $\mathbf P$ is a strong pseudo-Kleene poset if and only if $\mathbb D(\mathbf P)$ satisfies identities {\rm(1)} and {\rm(2)} and implication {\rm(5)}:
\begin{enumerate}
\item[{\rm(5)}] $x\neq x\sqcap y\neq y$ and $x\sqcap z=x'\sqcap z=z$ imply $y\sqcap z=y'\sqcap z=z$.
\end{enumerate}
\end{theorem}

\begin{proof}
Since
\begin{align*}
& x\sqcap y=x\text{ is equivalent to }x\leq y, \\
& x\sqcap y=y\text{ is equivalent to }y\leq x
\end{align*}
we have that $x\neq x\sqcap y\neq y$ is equivalent to $x\parallel y$. Further, since
\begin{align*}
& x\sqcap z=x'\sqcap z=z\text{ is equivalent to }z\in L(x,x'), \\
& y\sqcap z=y'\sqcap z=z\text{ is equivalent to }z\in L(y,y'),
\end{align*}
the following are equivalent:
\begin{align*}
& (5), \\
& x\parallel y\text{ and }z\in L(x,x')\text{ imply }z\in L(y,y'), \\
& x\parallel y\text{ implies }L(x,x')\subseteq L(y,y'), \\
& x\parallel y\text{ implies }L(x,x')=L(y,y').
\end{align*}
Applying Lemma~\ref{lem1} completes the proof.
\end{proof}

The following concept will be used in the sequel.

\begin{definition}
A {\em strict pseudo-Kleene poset} is a bounded poset $(P,\leq,{}')$ with an antitone involution satisfying
\[
x,y\neq0,1\text{ implies }L(x,x')=L(y,y').
\]
A {\em strict Kleene poset} is a distributive strict pseudo-Kleene poset.
\end{definition}

Obviously, every strict pseudo-Kleene poset is a strong pseudo-Kleene poset and hence a pseudo-Kleene poset according to Lemma~\ref{lem2}, but not conversely (see Example~\ref{ex1} where $L(a,a')=a\neq b=L(b,b')$).

Of course, every Boolean poset, i.e.\ every bounded distributive poset where the antitone involution which is a complementation (i.e.\ $L(x,x')\approx\{0\}$ and $U(x,x')=\{1\}$), is a strict Kleene poset. In the next example we show a strict Kleene poset which is not Boolean.

\begin{example}
The poset visualized in Figure~7 is a strict Kleene poset which is not a lattice and hence not a Kleene algebra:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(8,14)
\put(4,1){\circle*{.3}}
\put(4,3){\circle*{.3}}
\put(1,5){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(5,5){\circle*{.3}}
\put(7,5){\circle*{.3}}
\put(1,7){\circle*{.3}}
\put(7,7){\circle*{.3}}
\put(1,9){\circle*{.3}}
\put(3,9){\circle*{.3}}
\put(5,9){\circle*{.3}}
\put(7,9){\circle*{.3}}
\put(4,11){\circle*{.3}}
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\put(4,3){\line(0,-1)2}
\put(4,3){\line(-3,2)3}
\put(4,3){\line(-1,2)1}
\put(4,3){\line(1,2)1}
\put(4,3){\line(3,2)3}
\put(4,11){\line(-3,-2)3}
\put(4,11){\line(-1,-2)1}
\put(4,11){\line(1,-2)1}
\put(4,11){\line(3,-2)3}
\put(1,5){\line(0,1)4}
\put(7,5){\line(0,1)4}
\put(1,7){\line(1,1)2}
\put(1,5){\line(1,1)4}
\put(3,5){\line(1,1)4}
\put(5,5){\line(1,1)2}
\put(3,5){\line(-1,1)2}
\put(5,5){\line(-1,1)4}
\put(7,5){\line(-1,1)4}
\put(7,7){\line(-1,1)2}
\put(4,11){\line(0,1)2}
\put(3.85,.25){$0$}
\put(4.4,2.85){$a$}
\put(.3,4.85){$b$}
\put(2.3,4.85){$c$}
\put(5.4,4.85){$d$}
\put(7.4,4.85){$e$}
\put(.3,6.85){$f$}
\put(7.4,6.85){$f'$}
\put(2.3,8.85){$d'$}
\put(.3,8.85){$e'$}
\put(5.4,8.85){$c'$}
\put(7.4,8.85){$b'$}
\put(4.4,10.85){$a'$}
\put(3.3,13.4){$1=0'$}
\put(3.2,-.75){{\rm Fig.\ 7}}
\end{picture}
\end{center}
\vspace*{4mm}
\end{example}

Analogously as above, also strict pseudo-Kleene posets and strict Kleene posets can be characterized by means of properties of assigned meet-directoids.

\begin{theorem}
Let $\mathbf P=(P,\leq,{}')$ be a downward directed poset with a unary operation and $\mathbb D(\mathbf P)$ an assigned meet-directoid. Then the following hold:
\begin{enumerate}[{\rm(i)}]
\item $\mathbf P$ is a strict pseudo-Kleene poset if and only if $\mathbb D(\mathbf P)$ satisfies identities {\rm(1)} and {\rm(2)} and implication {\rm(6)}:
\begin{enumerate}
\item[{\rm(6)}] $x,y\notin\{0,1\}$ and $x\sqcap z=x'\sqcap z=z$ imply $y\sqcap z=y'\sqcap z=z$.
\end{enumerate}
\item $\mathbf P$ is a strict Kleene poset if and only if $\mathbb D(\mathbf P)$ satisfies identities {\rm(1)} and {\rm(2)} and implications {\rm(4)} and {\rm(6)}.
\end{enumerate}
\end{theorem}

\begin{proof}
\
\begin{enumerate}[(i)]
\item The following are equivalent:
\begin{align*}
& (6), \\
& x,y\notin\{0,1\}\text{ and }z\in L(x,x')\text{ imply }z\in L(y,y'), \\
& x,y\notin\{0,1\}\text{ implies }L(x,x')\subseteq L(y,y'), \\
& x,y\notin\{0,1\}\text{ implies }L(x,x')=L(y,y').
\end{align*}
Applying Lemma~\ref{lem1} completes the proof.
\item This follows from (i) and the proof of Theorem~\ref{th3}.
\end{enumerate}
\end{proof}

\section{Residuation in Kleene posets}

In the study of non-classical logics we prefer to have a logical connective implication which enables deduction, i.e.\ to derive new propositions from given ones. The question arises how to define implication in the logic based on pseudo-Kleene posets and Kleene posets. Usually implication is considered to be well-behaved if it is related with conjunction by means of adjointness.

In the following we will study residuation in strict pseudo-Kleene posets. Contrary to the case of lattices, $\odot$ and $\rightarrow$ cannot be binary operations, they are only operators, i.e.\ mappings from $P^2$ to $2^P$. We extend $\odot$ to $(2^P)^2$ by defining
\[
A\odot B:=\bigcap_{x\in A,y\in B}(x\odot y)
\]
for all $A,B\subseteq P$.

\begin{definition}
A {\em Kleene residuated poset} is an ordered six-tuple $(P,\leq,\odot,\rightarrow,0,1)$ \\
where $(P,\leq,0,1)$ is a bounded strict pseudo-Kleene poset and $\odot$ and $\rightarrow$ are mappings from $P^2$ to $2^P$ satisfying the following conditions for all $x,y,z\in P$:
\begin{itemize}
\item $x\odot y\approx y\odot x$,
\item $x\odot1\approx1\odot x\approx L(x)$,
\item $(x\odot y)\odot z\approx x\odot(y\odot z)$,
\item $x\odot y\leq z$ if and only if $x\leq y\rightarrow z$ {\rm(}adjointness{\rm)}.
\end{itemize}
\end{definition}

Let $(P,\leq,{}')$ be a poset with an antitone involution. Define mappings $\odot$ and $\rightarrow$ from $P^2$ to $2^P$ as follows:
\begin{enumerate}
\item[(R)] $\quad x\odot y:=\left\{
\begin{array}{ll}
0      & \text{if }x\leq y', \\
L(x,y) & \text{otherwise}
\end{array}
\right.
\quad\quad\quad x\rightarrow y:=\left\{
\begin{array}{ll}
1       & \text{if }x\leq y, \\
U(x',y) & \text{otherwise}
\end{array}
\right.$
\end{enumerate}

\begin{theorem}\label{th4}
Let $(P,\leq,{}',0,1)$ be a bounded strict Kleene poset satisfying
\begin{enumerate}
\item[{\rm(7)}] $L(x,y)\neq0$ for all $x,y\in P\setminus\{0\}$
\end{enumerate}
and let $\odot$ and $\rightarrow$ be defined by {\rm(R)}. Then $(P,\leq,\odot,\rightarrow,0,1)$ is a Kleene residuated poset.
\end{theorem}

\begin{proof}
Let $a,b,c\in P$. One can easily see that $x\odot0\approx0\odot x\approx0$. Since $0\in L(x,y)$ for all $x,y\in P$ and therefore $0\in x\odot y$ for all $x,y\in P$, we have $0\in a\odot b$ and $0\in b\odot c$ and hence $(a\odot b)\odot c=0=a\odot(b\odot c)$ proving associativity of $\odot$. Because $a\leq b'$ is equivalent to $b\leq a'$ and, moreover, $L(a,b)=L(b,a)$, $\odot$ is commutative. Further,
\begin{align*}
& \text{if }a=0\text{ then }a\odot1=0=L(0)=L(a), \\
& \text{if }a\neq0\text{ then }a\odot1=L(a,1)=L(a).
\end{align*}
The rest follows by commutativity of $\odot$. We consider the following cases:
\begin{itemize}
\item $a\leq b'$ and $b\leq c$. \\
Then $a\odot b=0\leq c$ and $a\leq1=b\rightarrow c$.
\item $a\leq b'$ and $b\not\leq c$. \\
Then $a\odot b=0\leq c$ and $a\leq b'\leq U(b',c)=b\rightarrow c$.
\item $a\not\leq b'$ and $b\leq c$. \\
Then $a\odot b=L(a,b)\leq b\leq c$ and $a\leq1=b\rightarrow c$.
\item $a\not\leq b'$, $b\not\leq c$ and $a=1$. \\
Then $a\odot b=L(a,b)=L(1,b)=L(b)\not\leq c$. Moreover, $b,c'\neq0$ and therefore $b\rightarrow c=U(b',c)=(L(b,c'))'\neq0'=1$ according to (7) whence $a=1\not\leq b\rightarrow c$.
\item $a\not\leq b'$, $b\not\leq c$ and $b=1$. \\
Then the following are equivalent:
\begin{align*}
a\odot b & \leq c, \\
  L(a,b) & \leq c, \\
  L(a,1) & \leq c, \\
    L(a) & \leq c, \\
       a & \leq c, \\
       a & \leq U(c), \\
       a & \leq U(0,c), \\
       a & \leq U(b',c), \\
       a & \leq b\rightarrow c.
\end{align*}
\item $a\not\leq b'$, $b\not\leq c$ and $c=0$. \\
Then $a,b\neq0$ and therefore $a\odot b=L(a,b)\neq0$ according to (7) whence $a\odot b\not\leq c$. Moreover, $a\not\leq U(b')=U(b',0)=U(b',c)=b\rightarrow c$.
\item $a\not\leq b'$, $b\not\leq c$, $a,b\neq1$ and $c\neq0$. \\
Then $a,b,c\neq0,1$. If $a\odot b\leq c$ then
\begin{align*}
b\rightarrow c & =U(b',c)\subseteq U(b',a\odot b)=U(b',L(a,b))=UL(U(b',a),U(b',b))= \\
               & =UL(U(b',a),U(a',a))\subseteq ULU(a)=U(a)
\end{align*}
and hence $a\leq b\rightarrow c$. If, conversely, $a\leq b\rightarrow c$ then
\begin{align*}
a\odot b & =L(a,b)\subseteq L(b\rightarrow c,b)=L(U(b',c),b)=LU(L(b',b),L(c,b))= \\
         & =LU(L(c',c),L(c,b))\subseteq LUL(c)=L(c)
\end{align*}
and hence $a\odot b\leq c$.
\end{itemize}
This shows that in any case $a\odot b\leq c$ is equivalent to $a\leq b\rightarrow c$.
\end{proof}

\begin{remark}
It seems to be impossible to generalize Theorem~\ref{th4} in such a way that the assumption of distributivity is replaced by the weaker assumption of modularity. If, e.g., $L(x,y)$ and $U(x',y)$ in {\rm(R)} are replaced by $L(U(x,y'),y)$ and $U(x',L(x,y))$, respectively, then adjointness cannot be proved in the last case considered in the proof of Theorem~\ref{th4}.
\end{remark}

As it is usual in logics satisfying the double negation law, the connectives conjunction (i.e.\ $\odot$) and implication (i.e.\ $\rightarrow$) can be derived one by the other by means of involution.

\begin{theorem}
Let $\mathbf P=(P,\leq,{}',0,1)$ be a bounded poset with an antitone involution, $\odot$ and $\rightarrow$ defined by {\rm(R)} and $a,b\in P$. Then the following hold:
\begin{enumerate}[{\rm(i)}]
\item $a\odot b=(a\rightarrow b')'$,
\item $a\rightarrow b=(a\odot b')'$.
\end{enumerate}
If, moreover, $\mathbf P$ satisfies {\rm(7)} then
\begin{enumerate}
\item[{\rm(iii)}] $a\odot b=0$ if and only if $a\leq b'$,
\item[{\rm(iv)}] $a\rightarrow b=1$ if and only if $a\leq b$.
\end{enumerate}
If, moreover, $\mathbf P$ is a strict Kleene poset {\rm(}not necessarily satisfying {\rm(7))} then
\begin{enumerate}
\item[{\rm(v)}] If $a\leq b$ and $L(a',b)=0$ then $a=b$.
\end{enumerate}
\end{theorem}

\begin{proof}
\
\begin{enumerate}[(i)]
\item
\begin{align*}
& \text{If }a\leq b'\text{ then }a\odot b=0=1'=(a\rightarrow b')', \\
& \text{if }a\not\leq b'\text{ then }a\odot b=L(a,b)=(U(a',b'))'=(a\rightarrow b')',
\end{align*}
\item According to (i) we have $a\rightarrow b=(a\rightarrow b'')''=(a\odot b')'$.
\item
\begin{align*}
& \text{If }a\leq b'\text{ then }a\odot b=0\text{ by definition}, \\
& \text{if }a\not\leq b'\text{ then }a,b\neq0\text{ and hence }a\odot b=L(a,b)\neq0\text{ according to (7)}.
\end{align*}
\item According to (ii) and (iii) the following are equivalent:
\begin{align*}
a\rightarrow b & =1, \\
  (a\odot b')' & =1, \\
     a\odot b' & =0, \\
             a & \leq b.
\end{align*}
\item
\begin{align*}
& \text{If }a=0\text{ then }L(b)=L(1,b)=L(a',b)=0\text{ and hence }a=0=b, \\
& \text{if }a=1\text{ then }b=1\text{ because of }a\leq b\text{ and hence }a=1=b, \\
& \text{if }b=0\text{ then }a=0\text{ because of }a\leq b\text{ and hence }a=1=b, \\
& \text{if }b=1\text{ then }L(a')=L(a',1)=L(a',b)=0\text{ and hence }a'=0,\text{ i.e.\ }a=1=b, \\
& \text{if }a,b\neq0,1\text{ then }U(a)=U(0,a)=U(L(a',b),a)=UL(U(a',a),U(b,a))= \\
& \quad =UL(U(b',b),U(b))=ULU(b)=U(b)\text{ which implies }a=b.
\end{align*}
\end{enumerate}
\end{proof}

\section{Twist construction}

Now we show how to construct pseudo-Kleene posets and Kleene posets from posets and distributive posets, respectively. We embed an arbitrary given poset into a pseudo-Kleene poset by using the so-called twist construction known already for distributive lattices.

For an arbitrary poset $\mathbf Q=(Q,\leq)$ and an arbitrary element $a$ of $Q$ we define
\[
P_a(\mathbf Q):=\{(x,y)\in Q^2\mid L(x,y)\leq a\leq U(x,y)\}.
\]
Let $p_1$ and $p_2$ denote the first and second projection from $P_a(\mathbf Q)$ to $Q$, respectively. In $P_a(\mathbf Q)$ we introduce a binary relation $\leq$ and a unary operation $'$ as follows:
\begin{align*}
(x,y)\leq(z,v) & :\Leftrightarrow x\leq z\text{ and }y\geq v, \\
        (x,y)' & :=(y,x).
\end{align*}
Put $\mathbb P_a(\mathbf Q):=(P_a(\mathbf Q),\leq,{}')$.

\begin{theorem}
Let $\mathbf Q=(Q,\leq)$ be a poset and $a\in Q$. Then the following hold:
\begin{enumerate}[{\rm(i)}]
\item $\mathbb P_a(\mathbf Q)$ is a pseudo-Kleene poset with fixed point $(a,a)$,
\item the mapping $x\mapsto(x,a)$ is an embedding of $\mathbf Q$ into $\mathbb P_a(\mathbf Q)$,
\item $\mathbf Q$ is distributive if and only if $\mathbb P_a(\mathbf Q)$ is a Kleene poset.
\end{enumerate}
\end{theorem}

\begin{proof}
Let $(b,c),(d,e),(f,g)\in P_a(\mathbf Q)$ and $h,i\in Q$. It is easy to show that
\begin{align*}
L(A) & =L(p_1(A))\times U(p_2(A)), \\
U(A) & =U(p_1(A))\times L(p_2(A))
\end{align*}
for all $A\subseteq P_a(\mathbf Q)$.
\begin{enumerate}[(i)]
\item Clearly, $\mathbb P_a(\mathbf Q)$ is a poset with an antitone involution and fixed point $(a,a)$. According to (i),
\[
L((b,c),(c,b))=L(b,c)\times U(b,c)\leq(a,a)\leq U(d,e)\times L(d,e)=U((d,e),(e,d)).
\]
\item We have $(h,a),(i,a)\in P_a(\mathbf Q)$. Moreover, $(h,a)\leq(i,a)$ if and only if $h\leq i$.
\item Using (i), we obtain
\begin{align*}
          L(U((b,c),(d,e)),(f,g)) & \approx L(U(b,d)\times L(c,e),(f,g))\approx \\
                                  & \approx L(U(b,d),f)\times U(L(c,e),g), \\
LU(L((b,c),(f,g)),L((d,e),(f,g))) & \approx LU(L(b,f)\times U(c,g),L(d,f)\times U(e,g))\approx \\
                                  & \approx L(U(L(b,f),L(d,f))\times L(U(c,g),U(e,g)))\approx \\
                                  & \approx LU(L(b,f),L(d,f))\times UL(U(c,g),U(e,g)).
\end{align*}
\end{enumerate}
\end{proof}

It is well-known (see e.g.\ \cite K) that for an arbitrary distributive lattice $\mathbf Q$ the constructed poset $\mathbb P_a(\mathbf Q)$ is a Kleene algebra. We have shown that this construction can be extended to arbitrary posets.

\begin{example}
If $\mathbf Q$ is the poset $(Q,\leq)$ visualized in Figure~8:
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(4,6)
\put(2,1){\circle*{.3}}
\put(2,3){\circle*{.3}}
\put(1,5){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(2,3){\line(0,-1)2}
\put(2,3){\line(-1,2)1}
\put(2,3){\line(1,2)1}
\put(1.85,.25){$0$}
\put(2.4,2.85){$a$}
\put(.85,5.4){$b$}
\put(2.85,5.4){$c$}
\put(1.2,-.75){{\rm Fig.\ 8}}
\end{picture}
\end{center}
\vspace*{4mm}
then $\mathbb P_a(\mathbf Q)$ is the pseudo-Kleene poset shown in Figure~9
\vspace*{-2mm}
\begin{center}
\setlength{\unitlength}{7mm}
\begin{picture}(6,10)
\put(1,1){\circle*{.3}}
\put(5,1){\circle*{.3}}
\put(1,3){\circle*{.3}}
\put(3,3){\circle*{.3}}
\put(5,3){\circle*{.3}}
\put(1,5){\circle*{.3}}
\put(3,5){\circle*{.3}}
\put(5,5){\circle*{.3}}
\put(1,7){\circle*{.3}}
\put(3,7){\circle*{.3}}
\put(5,7){\circle*{.3}}
\put(1,9){\circle*{.3}}
\put(5,9){\circle*{.3}}
\put(1,1){\line(0,1)8}
\put(1,1){\line(1,1)2}
\put(5,1){\line(-1,1)2}
\put(5,1){\line(0,1)8}
\put(3,5){\line(-1,-1)2}
\put(3,5){\line(-1,1)2}
\put(3,5){\line(0,1)2}
\put(3,5){\line(1,1)2}
\put(3,5){\line(1,-1)2}
\put(3,5){\line(0,-1)2}
\put(1,9){\line(1,-1)2}
\put(5,9){\line(-1,-1)2}
\put(.35,.25){$(0,c)$}
\put(4.35,.25){$(0,b)$}
\put(2.35,1.9){$(0,a)$}
\put(-.6,2.85){$(a,c)$}
\put(-.6,4.85){$(b,c)$}
\put(5.3,2.85){$(a,b)$}
\put(5.3,4.85){$(c,b)$}
\put(3.3,4.85){$(a,a)$}
\put(-.6,6.85){$(b,a)$}
\put(5.3,6.85){$(c,a)$}
\put(2.35,7.75){$(a,0)$}
\put(.35,9.4){$(b,0)$}
\put(4.35,9.4){$(c,0)$}
\put(2.2,-.75){{\rm Fig.\ 9}}
\end{picture}
\end{center}
\vspace*{4mm}
which is neither a pseudo-Kleene algebra nor a Kleene poset since
\begin{align*}
L(U((a,c),(0,a)),(a,b)) & =L(U((a,a)),(a,b))=L((a,a),(a,b))=\{(0,b),(a,b)\}\neq \\
                        & \neq\{(0,b)\}=L(P_a(\mathbf Q)\setminus\{(0,c),(a,c),(b,c)\})=LU((0,b))= \\
                        & =LU(L((a,c),(a,b)),L((0,a),(a,b))).
\end{align*}
\end{example}

config: arxiv


text:
\begin{document}
\newtheorem*{thm}{Theorem}

\title{Abelian fixed point free endomorphisms and the Yang-Baxter equation}
\author{Alan Koch}
\address{Department of Mathematics, Agnes Scott College, 141 E. College Ave., Decatur, GA 30030, USA \\ akoch@agnesscott.edu}

\author{Laura Stordy}
\address{School of Mathematics, Georgia Institute of Technology, 686 Cherry St NW, Atlanta, GA 30332, USA \\ lstordy3@gatech.edu }

\author{Paul J.~Truman}
\address{School of Computing and Mathematics, Keele University, Staffordshire, ST5 5BG, UK \\ p.j.truman@keele.ac.uk }
\date{\today       }

\begin{abstract} We obtain a simple family of solutions to the set-theoretic Yang-Baxter equation, one which depends only on considering special endomorphisms of a finite group. We show how such an endomorphism gives rise to two non-degenerate solutions to the Yang-Baxter equation, solutions which are inverse to each other. We give concrete examples using dihedral, alternating, symmetric, and metacyclic groups.
	\end{abstract}

\maketitle
\section{Introduction}

In the span of just a few years, Chen-Ning Yang \cite{Yang67}, working in theoretical physics, and Rodney Baxter \cite{Baxter72}, working in statistical mechanics, independently developed an equation which is now commonly referred to as the {\it Yang-Baxter equation}. The Yang-Baxter equation has been found to have applications in many areas of mathematics and physics, including quantum groups \cite{LambeRadford97}, knots \cite{Jimbo89a,Jimbo94}, analysis of integrable systems \cite{Jimbo89b}, and quantum computing \cite{Chen12}.

The Yang-Baxter equation describes a relationship between functions, namely \[(R\otimes 1)(1\otimes R)(R\otimes 1) = (1 \otimes R)(R\otimes 1)(1\otimes R),\] where $R\in \End(V\otimes V)$ for some finite dimensional vector space $V$. Interesting solutions of this equation are difficult to find, so an easier problem was introduced by Drin'feld  in \cite{Drinfeld92}: that of finding {\it set-theoretic solutions}, where $V$ is replaced by a finite set $B$, and tensor product replaced by the cross product. Given such a set-theoretic solution on $B$, one obtains a solution for a vector space with $B$ as a basis.

Many interesting example of set-theoretic solutions start to appear when $B$ carries some additional structure. For example, if $B=(B,\cdot)$ is an abelian group then Rump \cite{Rump07} showed how to construct solutions to the Yang-Baxter equation by giving $B$ the structure of a {\it left brace}, where $(B,\cdot)$ is endowed with another binary operation $\circ$ such that $(B,\circ)$ is a group and a certain relation holds. The corresponding solutions to the equation were both {\it non-degenerate} (see section \ref{YBEsec}) and {\it involutive} (i.e., self-inverse).

Later, the concept of left brace was generalized by Guarnieri and Vendramin \cite{GuarnieriVendramin17} to the case where $(B,\cdot)$ is nonabelian, obtaining a {\it (left) skew brace}. Skew braces correspond to non-degenerate solutions of the Yang-Baxter equations which are not non-involutive, though one can easily compute the inverse to any solution \cite{KochTruman20b}.

Thus, identifying set-theoretic solutions to the Yang-Baxter equation reduces to finding skew braces. But skew braces may not be easy to find: they involve putting two  (potentially) different group structures on $B$ that interact with each other in a precise way. A breakthrough was obtained in 2016 by Bachiller \cite{Bachiller16}, who observed a correspondence between skew braces and Hopf-Galois structures on Galois field extensions, a subject which had been studied for 30 years by numerous people--see, e.g., \cite{Kohl98}, \cite{CarnahanChilds99}, \cite{Childs00}, \cite{Byott04b}. The classification of Hopf-Galois structures on a Galois extension with Galois group $ G $ follows from a theorem of Greither and Pareigis, and amounts to finding regular subgroups $N$ of $\Perm(G)$, satisfying a certain stability condition. Given $N\le\Perm(G)$ as above, a skew brace can be constructed where one of the underlying groups is isomorphic to $N$, the other to $G$.

But Hopf-Galois theory still involves two groups interacting in a very precise way. In 2013, Childs \cite{Childs13} developed a technique using {\it fixed point free abelian  endomorphisms} (definition \ref{fpf} below) of a group $G$ to construct a certain class  of suitable regular subgroups, hence of Hopf-Galois structures.

In this work, we use Childs's theory to construct explicit solutions to the Yang-Baxter equation. If $\psi:G\to G$ is a fixed point free abelian endomorphism, then one can obtain a regular subgroup of $\Perm(G)$. Here, we construct the skew  brace $\B_{\psi}$ associated to $\psi$, and then we find a solution to the Yang-Baxter equation based on $\B_{\psi}$. We then use the theory of opposite braces in \cite{KochTruman20} to develop a second solution, namely, the inverse to the first solution. In short, we prove the following.

\begin{thm}[Theorem \ref{main}]
	Let $G$ be a finite group.
	\begin{enumerate}
		\item Let $\psi:G\to G$ be a fixed point free abelian endomorphism. Let $R_{\psi},R'_{\psi}:G\times G \to G \times G$ be given by
		\begin{align*}
		R_{\psi}(g,h)&=\left(\psi(g^{-1})h\psi(g),\psi(hg^{-1})h^{-1}\psi(g)g\psi(g^{-1})h\psi(gh^{-1}\right))\\
		R_{\psi}'(g,h)&=\left(g\psi(g^{-1})h\psi(g)g^{-1},\psi(h)g\psi(h^{-1})\right)
		\end{align*}
		Then:
		\begin{enumerate}
			\item Both $R_{\psi}$ and $R'_{\psi} $ are non-degenerate, set-theoretic solutions to the Yang-Baxter equation.
			\item The functions $R_{\psi}$ and $R'_{\psi}$ are inverse to each other.
		\end{enumerate}
		\item Two fixed point free abelian endomorphisms $\psi_1,\psi_2:G\to G$ give {\it equivalent} solutions to the Yang-Baxter equation $R_{\psi_1},\;R_{\psi_2}$ if and only if $\psi_2=\varphi\psi_1\varphi^{-1}$ for some $\varphi\in\Aut(G)$.
	\end{enumerate}
\end{thm}

In section \ref{bracesec}, we define precisely the concepts of {\it type} of solution and {\it equivalence} of solutions.

The significance of this result lies in its simplicity. Although the proof presented here uses the Greither-Pareigis correspondence for Hopf-Galois structures on separable field extensions and the theory of skew braces, the reader can generate pairs of solutions to the Yang-Baxter equation without a deep understanding of these concepts: one simply needs to be familiar with elementary group theory.

We start by describing, in greater detail, the Yang-Baxter equation. Then, we give an overview of Greither-Pareigis theory, which we then connect to skew braces which, in turn, give us set-theoretic solutions to the Yang-Baxter equation. Next, we discuss fixed-point free abelian endomorphisms on a finite group $G$ and we explicitly construct a brace corresponding to such an endomorphism. We then prove the main result above (with some minor differences in notation) in Section 5. Finally, we provide examples in which $G$ is dihedral, alternating, symmetric, or metacyclic.

Unless otherwise specified, we assume all groups (and braces) are finite.

\section{The Yang-Baxter Equation and its Solutions}\label{YBEsec}

Let $V$ be a finite dimensional vector space over some field $K$. A {\it solution} to the Yang-Baxter equation is a $K$-linear map $R:V\otimes V\to V\otimes V$ such that
\[(R\otimes I)(I\otimes R)(R\otimes I)=(I\otimes R)(R\otimes I)(I\otimes R): V\otimes V \otimes V\to V\otimes V \otimes V,\]
where $I:V\to V$ is the identity map. If we view $R\in M_{n^2}(K)$ where $\dim_K V=n$ then $R$ is what is called an {\it $R$-matrix}. As a way to develop non-trivial examples of solutions, in \cite{Drinfeld92}, Drin'feld suggests considering set-theoretic solutions to the YBE, that is, a set $B$ and a function $ R: $ $B\times B\to B\times B$ such that
\[(R\times \mathrm{id})(\mathrm{id}\times R)(R\times \mathrm{id}) = (\mathrm{id}\times R)(R\times \mathrm{id})(\mathrm{id}\times R)\]
holds, where $\mathrm{id}$ is the identity map on $B$. If $B$ is finite, we let $V_B$ be the $K$-vector space with basis $B$; then a set theoretic solution on $B$ gives a vector space solution on $V_B$.

A set-theoretic solution $R:B\times B\to B\times B$ is said to be {\it non-degenerate} if the functions $f_y,g_x:B\to B$ given by $ R(x,y)=(f_y(x),g_x(y))$ are invertible. As above, we say $R$ is {\it involutive} if $\mathrm{id}$.

In the work being presented here, the set $B$ will carry the structure of a (finite)  group, say $G$, and the set-theoretic solutions will rely heavily on the group operation. Working with $G$ will allow us to identify $V_G$ naturally with $K[G]$, thereby naturally obtaining $R$-matrices on group algebras.

\section{Regular Subgroups, Braces, and Solutions to the YBE}\label{bracesec}

In this section, we introduce many of the structures needed to prove theorem \ref{main}. We start with the definition of regular $G$-stable subgroups of $\Perm(G)$, which give rise to skew braces, which in turn give rise to solutions to the Yang-Baxter equation.

Let $G$ be a group, and denote by $\Perm(G)$ the group of all permutations of $G$. We say a subgroup $N$ of $\Perm(G)$ is {\it regular} if the action of $N$ on $G$ is free and transitive, that is
\begin{enumerate}
	\item for $\eta\in N, g\in G$ we have $\eta[g]=g$ if and only if $\eta=1_N$; and
	\item for $g,h\in G$ there exists an $\eta\in N$ such that $\eta[g]=h$.
\end{enumerate}

If $N\le\Perm(G)$ is regular then $|N|=|G|$. Additionally, if $N\le \Perm(G)$ with $|N|=|G|$ and either of the above two conditions are satisfied, then $N$ is a regular subgroup. Two fundamental examples of regular subgroups are the images of the left regular representation $\lambda: G \hookrightarrow \Perm(G)$ and the right regular representation $\rho:G\hookrightarrow \Perm(G)$.

In general, it is not necessarily the case that $N\cong G$, however in the examples we will consider here these groups will in fact be isomorphic.

The subgroup $\lambda(G)\le\Perm(G)$ acts on $\Perm(G)$ by conjugation. We say $N\le\Perm(G)$ is {\it $G$-stable} if this action restricts to an action on $N$. In other words, for all $\eta\in N,g\in G$ we have
\[^g\eta:=\lambda(g)\eta\lambda(g^{-1})\in N.\]

Certainly, $\lambda(G)$ is $G$-stable, and since $\lambda(G)$ and $\rho(G)$ commute in $\Perm(G)$ we see that $\rho(G)$ is $G$-stable as well.

Subgroups of $\Perm(G)$ which are regular and $G$-stable are essential in identifying Hopf-Galois structures on Galois extensions $L/K$ with $\Gal(L/K)\cong G$. This was first discovered by Greither and Pareigis in \cite{GreitherPareigis87}, but will not be needed here.

We will now introduce the notion of a skew (left) brace, and connect it to the regular, $G$-stable subgroups above.

A {\it skew left brace} is a triple $\B=(B,\cdot,\circ)$ where $(B,\cdot)$ and $(B,\circ)$ are groups; and for all $a,b,c\in B$ the following condition holds:
\[a\circ(b\cdot c) = (a\circ b)\cdot a^{-1} \cdot (a\circ c),\]
where $a^{-1}$ is the inverse to $a$ in $(B,\cdot)$. We call this condition the {\it brace relation}.

For simplicity, we adopt the following conventions.
\begin{enumerate}
	\item We will refer to $\B$ simply as a {\it brace}. The term ``brace'' was first used by Rump in \cite{Rump07} to describe a triple $(B,\cdot,\circ)$ satisfying the brace relation where $(B,\cdot)$ is abelian. The notion of brace was generalized 10 years later by Guarnieri and Vendramin \cite{GuarnieriVendramin17}.
	\item When no confusion will arise we will suppress the dot and write $ab=a\cdot b$.
	\item While $(B,\cdot)$ and $(B,\circ)$ share the same identity element, their inverses are different in general: we will denote the inverse to $a\in(B,\circ)$ by $\overline{a}$.
\end{enumerate}

The reader should be made aware that there is no universal notation for braces at this point. Not only are the symbols for the two operation not standard, the order in which they appear in the triple is not standard as well.

Simple examples of braces include $(G,\cdot, \cdot)$, called the {\it trivial brace on $G$}, and $(G,\cdot,\cdot')$ where $(G,\cdot')$ is the opposite group to $(G,\cdot)$, giving the {\it almost trivial brace on $G$}.

As with regular, $G$-stable subgroups of $\Perm(G)$, we do not require $(B,\cdot)\cong (B,\circ)$, however this paper will only deal with braces satisfying this property.

A regular, $G$-stable subgroup $N\le \Perm(G)$ gives rise to a brace as follows. The map $\varkappa:N\to G$ given by $\varkappa(\eta)=\eta(1_G)$ is a bijection. Define  $\B(N)=(N,\cdot,\circ)$, where $\cdot$ is the usual group operation on $N$, and
\[\eta\circ \pi = \varkappa^{-1}(\varkappa(\eta)\ast_G \varkappa(\pi)),\;\eta,\pi\in N.\]

It is well-known that every brace $(B,\cdot,\circ)$ arises from a regular, $G$-stable subgroup $N\le\Perm(G)$ with $G\cong (B,\circ)$ \cite[Prop. 6.1]{Bachiller16}. In general, the correspondence is not one-to-one \cite[Appendix A]{GuarnieriVendramin17}.

Braces give set-theoretic solutions to the Yang-Baxter equation. For $\B=(B,\cdot,\circ)$ a brace, define $R_{\B}:B\times B\to B\times B$ by
\[R_{\B}=(a^{-1}(a\circ b),\overline{a^{-1}(a\circ b)}\circ a \circ b) \label{YBE}]\tag{1}.\]
Then $R_{\B}$ is a non-degenerate, set-theoretic solution to the YBE.

\begin{definition}
	Let $\B=(B,\cdot,\circ)$ be a brace. Then $R_{\B}$ is said to be a solution of {\it type $(N,G)$} if $N$ and $G$ are abstract groups with $N\cong(B,\cdot)$ and $G\cong (B,\circ)$. If $N=G$ then we say $R_{\B}$ is a solution of {\it type $G$}.
\end{definition}

If $(B,\cdot)$ is abelian, then $R_{\B}$ is involutive, i.e., $R_{\B}$ is self-inverse. On the other hand, if $(B,\cdot)$ is non-abelian, two of the authors in \cite[Th. 4.1]{KochTruman20} construct the inverse to $R_{\B}$ through the use of opposite braces. The inverse to $R_{\B}$, denoted $R'_{\B}$, is simply
\[R'_{\B}=((a\circ b)a^{-1},\overline{(a\circ b)a^{-1}}\circ a \circ b). \tag{2}\label{YBEopp}\]
 In this case, $R'_{\B}$ is a solution of the same type as $R_{\B}$. Notice that $R'_{\B}= R_{\B'}$, i.e., it is the solution obtained from \ref{YBE} using the opposite brace to $\B$.

It should be pointed out that braces can give additional set-theoretic solutions. For example, suppose $\phi:\B\to \B$ is a brace isomorphism, i.e., a bijective map which preserves both operations, and $R$ is any solution to the Yang-Baxter equation with underlying set $B$. Then $R_{\phi}:=R(\phi\times\phi):B \times B\to B\times B$ is given by
\begin{align*}
R_{\phi} (a,b) &= (\phi(a)^{-1}(\phi(a)\circ \phi(b)),\overline{\phi(a)^{-1}(\phi(a)\circ \phi(b))}\circ \phi(a) \circ \phi(b))\\
 &= (\phi(a^{-1}(a\circ b)),\phi(\overline{a^{-1}(a\circ b)}\circ a \circ b)),
\end{align*}
so $R(\phi\times\phi)=(\phi\times\phi)R$ and the Yang-Baxter equation is clearly satisfied. However, if one is interested in set-theoretic solutions as a means to find vector space solutions, then $R_{\phi}$ and $R$ can be thought of as being the same solution under a change of basis. Thus if $N_1$ and $N_2$ are regular, $G$-stable subgroups which produce the same brace (up to isomorphism), we think of them as giving {\it equivalent} solutions to the Yang-Baxter equation.

\section{Fixed-Point Free Abelian Endomorphisms and Braces}

The purpose of this section is to explicitly show how to obtain a brace from a fixed point free abelian endomorphism.

We start with a formal definition of what it means for a an endomorphism to be fixed point free abelian. The term is self-explanatory, however since these endomorphisms are central to this work we wish to be explicit.

\begin{definition}\label{fpf}
	Let $G$ be a group. An endomorphism $\psi:G\to G$ is said to be {\it fixed point free abelian} if both of the following hold:
	\begin{enumerate}
		\item If $\psi(g)=g$ then $g=1_G$. (Fixed point free)
		\item The image $\psi(G)$ is abelian. (Abelian)
	\end{enumerate}
\end{definition}

Observe that if $\psi$ is an abelian endomorphism, then $\psi(ghg^{-1})=\psi(h)$ for all $g,h\in G$. That is, $\psi$ is constant on conjugacy classes. We will use this fact extensively in section \ref{exSec} when constructing these maps.

For a given group $G$, we will denote by $\FPF(G)$ the set of all fixed point free abelian endomorphisms of $ G $. If $\psi_1,\psi_2\in\FPF(G)$ their composition is typically not fixed point free, so $\FPF(G)$ should be viewed as simply a pointed set, base point $\psi_0$.

\begin{example}\label{triv0}
	Let $G$ be any group. Then the trivial map $\psi_0:G \to G,\; \psi_0(g)=1_G$ is fixed point free abelian.
\end{example}

\begin{example}\label{D40}
		Let $G=D_4=\gen{r,s:r^4=s^2=rsrs=1_G}$ be the dihedral group of order $8$. Define $\psi:G\to G$ by $\psi(r) = \psi(s) = rs$. Since $\psi(G)=\gen{rs}$ this map is clearly abelian. Since the image of $\psi$ contains only one nontrivial point, the only possible fixed point is $rs$; since $\psi(rs)=(rs)^2=1_G$ we get that $\psi$ is fixed point free, hence $\psi\in\FPF(G)$.

		This example is well-known: see \cite[p. 1264, (6)]{Childs13}.
\end{example}

In \cite{Childs13} Childs constructs, from $\psi\in\FPF(G)$, a regular $G$-stable subgroup of $\Perm(G)$, namely $N=\{\lambda(g)\rho(\psi(g)):g\in G\}$. Explicitly,
\[\eta_g[h]=gh\psi(g^{-1}) {\mbox{ for all $ g,h \in G $}}.\]
The reader can check that $N$ is regular, and that $^k\eta_g = \eta_{kgk^{-1}}$ for all $g,k\in G$.

Different fixed point free endomorphisms can yield the same group $N$. Indeed,  $\psi_1$, $\psi_2\in\FPF(G)$ produce the same regular, $G$-stable subgroup of $\Perm(G)$ if and only if there is a $\zeta\in\FPF(G)$ with $\zeta(G)\le Z(G)$ and such that
\[ \psi_2(g)=\psi_1(g\zeta(g^{-1}))\zeta(g),\;g\in G.\]
In particular, $\psi\in\FPF(G)$ gives the same regular, $G$-stable subgroup of $\Perm(G)$ as $\psi_0$--namely, $\lambda(G)\le\Perm(G)$--if and only if $\psi(G)\le Z(G)$.
 See \cite[Th. 2]{Childs13} for details.

Thus, $N$ gives rise to a brace $\B_{\psi}=(N,\cdot,\circ)$ where the dot operation is the usual operation in $N$, which itself is inherited from $G$: since $\lambda(G)$ and $\rho(G) $ commute in $\Perm(G)$ we have
\[\eta_g\eta_h = \lambda(g)\rho(\psi(g))\lambda(h)\rho(\psi(h)) = \lambda(g)\lambda(h)\rho(\psi(g))\rho(\psi(h))=\lambda(gh)\rho(\psi(gh))=\eta_{gh}.\]

 We will now explicitly compute the circle operation, completing the description of $\B_{\psi}$. Note that the bijection $\varkappa:N\to G$ is $\varkappa[\eta_g] = \eta_{g}[1_G] = g\psi(g^{-1})$. For $g,h\in G$ we have

\begin{align*}
\eta_g\circ \eta_h &= \varkappa^{-1}(\varkappa(\eta_g)\ast_G \varkappa(\eta_h))\\
&= \varkappa^{-1}\big(g\psi(g^{-1})h\psi(h^{-1})\big).
\end{align*}

On the other hand, since $\psi$ is constant on conjugacy classes,
\begin{align*}
\varkappa\big(\eta_{g\psi(g^{-1})h\psi(g)}\big) &= g\psi(g^{-1})h\psi(g)\psi\big((g\psi(g^{-1})h\psi(g))^{-1}\big)\\
&=g\psi(g^{-1})h\psi(g)\psi(h^{-1}g^{-1}) \\
&=g\psi(g^{-1})h\psi(h^{-1}),
\end{align*}
hence
\[\eta_g\circ \eta_h = \eta_{g\psi(g^{-1})h\psi(g)}. \]

At this point we remark that the map $g\mapsto \eta_g:G\to (N,\cdot)$ is in fact an isomorphism. This allows us to think of the underlying set of our brace as $G$, obtaining:
\begin{proposition}
	Let $\psi:G \to G$ be a fixed point free abelian endomorphism. Then $\B_{\psi}=(G,\cdot,\circ)$ is a brace, where
	\[g\circ h = g\psi(g^{-1})h\psi(g),\;g,h\in G.\]
\end{proposition}

As mentioned earlier, the two groups in a brace need not be isomorphic, however for braces constructed from fixed point free abelian maps, it is necessary that they are.

\begin{corollary}
	Let $\psi\in\FPF(G)$, and let $B_{\psi}=(G,\cdot,\circ)$ be its corresponding brace. Then $(G,\circ)\cong (G,\cdot)$.
\end{corollary}

\begin{proof}
	Define $\phi:(G,\circ)\to (G,\cdot)$ by $\phi(g)=g\psi(g^{-1})$. We claim that $\phi$ is an isomorphism.  Indeed, we have
	\begin{align*}
	\phi(g\circ h)&= \phi( g\psi(g^{-1})h\psi(g))\\
	&= g\psi(g^{-1})h\psi(g)\psi(\psi(g^{-1})h^{-1}\psi(g)g^{-1})\\
	&=g\psi(g^{-1})h\psi(h^{-1})\tag{$\psi$ is abelian}\\
	&=\phi(g)\phi(h)
	\end{align*}
	so $\phi$ is a homomorphism. If $g\in \ker \phi$ then $\phi(g)=g\psi(g^{-1})=1_G$, from which it follows that $g=\psi(g)$. As $\psi$ is fixed point free, this means $g=1_G$.
Thus, $\phi$ is injective, hence an isomorphism.
\end{proof}

Direct calculation establishes the following, which will be needed in the sequel.
\begin{corollary}\label{thecor}
	Let $\psi\in\FPF(G)$. Then for all $g\in G$ we have
	\[\overline{g}=\psi(g)g^{-1}\psi(g^{-1}).\]
\end{corollary}

From \cite[\S 5]{KochTruman20b}, the group of automorphisms $\Aut(G)$ acts on $\FPF(G)$ via conjugation, i.e., $\varphi\psi\varphi^{-1}\in\FPF(G)$ whenever $\varphi\in\Aut(G),\;\psi\in\FPF(G)$. Furthermore, the braces formed by $\psi$ and $\varphi\psi\varphi^{-1}$ are isomorphic: in fact, any regular subgroup of $\Perm(G)$ which gives a brace isomorphic to the one determined by $\psi$ is of this form.

\begin{definition}
	We say $\psi_1,\;\psi_2\in\FPF(G)$ are {\it brace equivalent} if $\psi_2=\varphi\psi_1\varphi^{-1}$ for some $\varphi\in\Aut(G)$.
\end{definition}

\begin{example}\label{triv}
	Let $G$ be any group, and let $\psi_0:G\to G$ be the trivial map as in example \ref{triv0}. Then
	\[g \circ h = g \psi_0(g^{-1})h\psi_0(g)=gh\]
	and $\B_{\psi}=(G,\cdot,\cdot)$ is the trivial brace on $G$.

	Notice that for any $\varphi\in\Aut(G)$ we have
	\[\varphi^{-1}\psi_0\varphi(g) = \varphi^{-1}(1_G)=1_G\]
	so $\varphi^{-1}\psi_0\varphi=\psi_0$. Thus, no nontrivial $\psi\in\FPF(G)$ is equivalent to $\psi_0$.
\end{example}

\begin{example}\label{D4}
	Let $G=D_4=\gen{r,s:r^4=s^2=rsrs=1_G}$ be the dihedral group of order $ 8 $?, and let $\psi_1\in\FPF(G)$ be $\psi_1(r) = \psi_1(s) = rs$ as in example \ref{D40}. The circle operation in $\B_{\psi_1}$ is given by
	\begin{align*}
	r^i\circ r^j &= r^i(rs)^ir^j(rs)^i=r^{i+(-1)^ij}\\
	r^i\circ r^js &= r^i(rs)^ir^js(rs)^i=r^{-i+(-1)^ij}s\\
	r^is\circ r^j&= r^i{s}(rs)^{i+1}r^j(rs)^{i+1} = r^{i+(-1)^ij}s \\
	r^is\circ r^js &= r^is(rs)^{i+1}r^js(rs)^{i+1}=r^{2-i+(-1)^ij}.\\
	\end{align*}
\end{example}

\begin{example}\label{other D4}
Let $G=D_4$ as above, and let $\psi_2 : G \to G$ be the endomorphism given by $\psi_2(r)=r^2s$, $\psi_2(s)=1$. Since $\psi_2(G)=\gen{r^2s}$ the map is abelian, and since $\psi(r^2s) = 1_G \ne r^2s$ we get $\psi_2\in\FPF(G)$. The circle operation in ${\B_{\psi_2}}$ is given by
\begin{align*}
r^i \circ r^j &= r^i(r^2s)^ir^j(r^2s)^i = r^{i + (-1)^ij}\\
r^i \circ r^js &= r^i(r^2s)^ir^js(r^2s)^i = r^{i + (-1)^ij}s\\
r^is \circ r^j &= r^is(r^2s)^ir^j(r^2s)^i = r^{i-(-1)^ij}s\\
r^is \circ r^js &= r^is(r^2s)^ir^js(r^2s)^i = r^{i - (-1)^ij}.
\end{align*}

Now let $\varphi\in\Aut(G)$ be given by $\varphi(r)=r,\;\varphi(s)=r^3s$. Then
\begin{align*}
\varphi^{-1}\psi_1\varphi(r) &= \varphi^{-1}\psi_1(r) = \varphi^{-1}(rs) = r^2s =\psi_2(r)\\
\varphi^{-1}\psi_1\varphi(s) &= \varphi^{-1}\psi_1(r^3s) = \varphi^{-1}(1_G) = 1_G =\psi_2(s),
\end{align*}
so $\psi_2$ is equivalent to the $\psi_1$ found in example \ref{D4}.
\end{example}

\section{Proof Of The Main Result}

We are now well-positioned to prove our main theorem.

\begin{theorem}\label{main}
	Let $G$ be a finite group.
	\begin{enumerate}
		\item Let $\psi\in\FPF(G)$. Let $R_{\psi},R'_{\psi}:G\times G \to G \times G$ be given by
		\begin{align*}
		R_{\psi}(g,h)&=\left(\psi(g^{-1})h\psi(g),\psi(hg^{-1})h^{-1}\psi(g)g\psi(g^{-1})h\psi(gh^{-1}\right))\\
		R_{\psi}'(g,h)&=\left(g\psi(g^{-1})h\psi(g)g^{-1},\psi(h)g\psi(h^{-1})\right).

		\end{align*}
		Then:
		\begin{enumerate}
			\item Both $R_{\psi}$ and $R'_{\psi}$ are non-degenerate, set-theoretic solutions to the Yang-Baxter equation of type $G$.
			\item The functions $R_{\psi}$  and $R'_{\psi}$ are inverse to each other.
		\end{enumerate}
	\item The maps $\psi_1,\psi_2\in\FPF(G)$ give equivalent pairs of solutions to the Yang-Baxter equation $R_{\psi_1},\;R_{\psi_2}$ if and only if $\psi_2=\varphi\psi_1\varphi^{-1}$ for some $\varphi\in\Aut(G)$.
	\end{enumerate}
\end{theorem}

\begin{proof}
	Any $\psi\in\FPF(G)$ gives rise to a regular, $G$-stable subgroup $N_{\psi}$ of $\Perm(G)$, which in turn corresponds to the brace $\B_{\psi}=(G,\cdot,\circ)$ from before. The solution $R_{\psi}$ can be readily computed from equation (\ref{YBE}):
	\begin{align*}
	R_{\psi}(g,h) &=\big(g^{-1}(g\circ h),\overline{g^{-1}(g\circ h)}\circ g \circ h\big)\\
	&=\big(g^{-1}g\psi(g^{-1})h\psi(g),\overline{g^{-1}g\psi(g^{-1})h\psi(g)}\circ g \circ h\big) \tag{cor. \ref{thecor}} \\
	&=\left(\psi(g^{-1})h\psi(g),(\psi(\psi(g^{-1})h\psi(g)))(\psi(g^{-1})h^{-1}\psi(g))\psi(\psi(g^{-1})h^{-1}\psi(g))\circ g \circ h\right).
	\end{align*}
	The first component now agrees with the statement of the theorem. For the second component,
	\begin{align*}
     &(\psi(\psi(g^{-1})h\psi(g)))(\psi(g^{-1})h^{-1}\psi(g))\psi(\psi(g^{-1})h^{-1}\psi(g))\circ g \circ h\\&=\psi(h)\psi(g^{-1})h^{-1}\psi(g)\psi(h^{-1})\circ g\psi(g^{-1})h\psi(g)\\
     &=(\psi(hg^{-1})h^{-1}\psi(gh^{-1})\psi\big(\psi(gh^{-1})h\psi(hg^{-1})\big) g\psi(g^{-1})h\psi(g) \psi\big(\psi(hg^{-1})h^{-1}\psi(gh^{-1})\big)\\
     &=\psi(hg^{-1})h^{-1}\psi(gh^{-1})\psi(h)g\psi(g^{-1})h\psi(g)\psi(h^{-1})\\
     &=\psi(h\psi(h^{-1}))g\psi(g^{-1})h\psi(g\psi(h))
	\end{align*}
	as desired.

The solution $R_{\psi}'$ can be computed using the theory of opposites as in \cite{KochTruman20}. Given a brace $\B=(B,\cdot,\circ)$ with $(B,\cdot)$ nonabelian one can construct an opposite brace $\B'=(B,\cdot',\circ)$ where $a\cdot' b = b\cdot a$ for all $a,b\in B$. One then obtains a second solution to the YBE, namely
\[(g,h)\mapsto\big((g\circ h)g^{-1},\overline{(g\circ h)g^{-1}}\circ g \circ h\big).\]
One can show that the expression above reduces to $R_{\psi}'$ as in the statement of the theorem; that $R_{\psi}'$ is the inverse to $R_{\psi}$ can be found in \cite[Th. 4.1]{KochTruman20}.

Finally, to show (2) we use the fact that two regular, $G$-stable subgroups are brace equivalent if and only if their opposites are brace equivalent \cite[Cor. 7.2]{KochTruman20b}.
\end{proof}

\begin{example}\label{trivYBE}
	For any nonabelian group $G$, let $\psi:G\to G$ be the trivial map as in example \ref{triv}. Then we get the trivial brace on $G$, and
	\begin{align*}
	R_{\psi}(g,h) &= \left(\psi(g^{-1})h\psi(g),\psi(hg^{-1})h^{-1}\psi(g)g\psi(g^{-1})h\psi(gh^{-1}\right)) = (h,h^{-1}gh)\\
	R_{\psi}'(g,h) &= \left(g\psi(g^{-1})h\psi(g)g^{-1},\psi(h)g\psi(h^{-1})\right)=(ghg^{-1},g).
	\end{align*}
\end{example}

\begin{example}
	As in example \ref{D4}, let $G=D_4=\gen{r,s:r^4=s^2=rsrs=1_G},\;\psi(r) = \psi(s) = rs$. Using theorem \ref{main} we see, e.g., $R_{\psi}(rs,r^2) = (r^2,rs)$. More generally,
	the values for the solution $R_{\psi}$ are given in the following table:

	\[\begin{array}{c|cccccccc}      & 1        & r          & r^2        & r^3        & s           & rs        & r^2s         & r^3s        \\\hline
1    & (1,1)    & (r, 1)     & (r^2, 1)   & (r^3, 1)   & (s,1)       & (rs,1)    & (r^2s,1)     & (r^3s,1)    \\
r    & (1,r)    & (r^3,r^3)  & (r^2,r)    & (r,r^3)    & (r^2s,r)    & (rs,r^3)  & (s,r)        & (r^3s,r^3)  \\
r^2  & (1,r^2)  & (r,r^2)    & (r^2,r^2)  & (r^3,r^2)  & (s,r^2)     & (rs,r^2)  & (r^2s,r^2)   & (r^3s,r^2)  \\
r^3  & (1,r^3)  & (r^3,r)    & (r^2,r^3)  & (r,r)      & (r^2s,r^3)  & (rs,r)    & (s,r^3)      & (r^3s,r)    \\
s    & (1,s)    & (r^3,s)    & (r^2,s)    & (r,s)      & (r^2s,r^2s) & (rs,r^2s) & (s,r^2s)     & (r^3s,r^2s) \\
rs   & (1,rs)   & (r,r^3s)   & (r^2,rs)   & (r^3,r^3s) & (s,r^3s)    & (rs,rs)   & (r^2s, r^3s) & (r^3s,rs)   \\
r^2s & (1,r^2s) & (r^3,r^2s) & (r^2,r^2s) & (r,r^2s)   & (r^2s,r^3s) & (rs,s)    & (s,s)        & (r^3s,s)    \\
r^3s & (1,r^3s) & (r,rs)     & (r^2,r^3s) & (r^3,rs)   & (s,rs)      & (rs,r^3s) & (r^2s,rs)    & (r^3s,r^3s) \end{array}\]
A table for $R'_{\psi}$ can be similarly constructed using the formula as in theorem \ref{main}; alternatively one could compute it by explicitly constructing the inverse to the table of values above.
\end{example}

\section{Examples}\label{exSec}

We finish with examples, where we explicitly construct the solutions to the Yang-Baxter equation whose underlying sets are some well-known groups. We have tried to strike a balance between explicitness and simplicity of results.

Sometimes, the use of fixed point free abelian endomorphisms will find {\it all} solutions to the Yang-Baxter equation with underlying group $G$. We explore this topic in the examples below.

\subsection{Abelian Groups}
The theory presented above gives useful results when $G$ is nonabelian. In the case where $G$ is abelian,
\[g\circ h = g\psi(g^{-1})h\psi(g) = gh\]
giving only the trivial brace. The only solution to the Yang-Baxter equation we construct is
\[R_{\psi}(a,b)=(b,a)\]
as $R'_{\psi}=R_{\psi}$ when $G$ is abelian.

By \cite{Byott96}, our construction will yield all possible solutions to the Yang-Baxter equation if and only if $\gcd(|G|,\phi(|G|))=1$, where $\phi$ is the Euler totient function. Such a group is necessarily cyclic.

\subsection{The Dihedral Groups $D_n$}

Generalizing the notation from example \ref{D4}, write $G=D_n=\gen{r,s:r^n=s^2=rsrs=1_G}$. Classification of $\FPF(D_n)$ appears in \cite[\S 5]{Childs13}. Here, we will give the Yang-Baxter solutions given by those maps.

Suppose first that $n$ is odd. Then $\FPF(D_n)$ consists only of the trivial map. Indeed, if $\psi\in\FPF(D_n)$ is nontrivial then $\ker\psi=\gen{r}$. Additionally, $\psi(s)=r^is$ for some $0\le i \le n-1$. But then $\psi(r^is)=r^is$ is a fixed point, which is a contradiction.

Thus, the two solutions to the Yang-Baxter equation we obtain when $n$ is odd are the two from example \ref{trivYBE}.

By the main theorem in \cite{Kohl20}, the number of regular, $G$-stable subgroups of $D_n$ isomorphic to $n$ for $n$ odd is equal to the number of self-inverse elements in $\mathbb{Z}_n^{\times}$, from which it follows that our process obtains all such subgroups if and only if $p$ is a power of an odd prime.

Now suppose that $ n $ is even, and write $ n=2m $.
Each element of $\FPF(D_n)$ which does not give the trivial regular subgroup $\lambda(G)$ is in at least one of the following four forms.
\begin{enumerate}
	\item $\alpha_i(r)=r^{2i}s,\;\alpha_i(s) = 1,\;0\le i\le m-1$;
	\item $\beta_i(r)=r^{2i+m}s,\;\beta_i(s) = r^m,\;0\le i\le m-1$;
	\item $\gamma_i(r)=r^{2i+1}s,\;\gamma_i(s) = r^{2i+1}s,\;0\le i\le m-1$;
	\item $\delta_i(r)=r^{2i+1+m}s,\;\delta_i(s) = r^{2i+1}s,\;0\le i\le m-1$.
\end{enumerate}

For $0\le i,j \le n-1,\;\gcd(i,n)=1$ let $\varphi_{i,j}:D_n\to D_n$ be given by $\varphi_{i,j}(r)=r^i,\;\varphi_{i,j}(s)=r^js$. Then $\Aut(D_n)=\{\varphi_{i,j}\}$. If we set $i=1$ we have
\begin{align*}
\varphi_{1,j}^{-1}\alpha_0\varphi_{1,j}(r) &= \varphi_{1,j}^{-1}\alpha_0(r)=\varphi_{1,j}^{-1}(s)=r^{-j}s\\
\varphi_{1,j}^{-1}\alpha_0\varphi_{1,j}(s) &= \varphi_{1,j}^{-1}\alpha_0(r^js)=\varphi_{1,j}^{-1}(s^j),\\
\end{align*}
hence if $j$ is even we get $\varphi_{1,j}^{-1}\alpha_0\varphi_{1,j}=\alpha_{j-2}$, and if $j$ is odd we get $\varphi_{1,j}^{-1}\alpha_0\varphi_{1,j}(s)=r^{-j}s$ and so $\varphi_{1,j}^{-1}\alpha_0\varphi_{1,j}=\gamma_{(-j-1/2)}$. By allowing $j$ to vary we see that $\{\alpha_i,\gamma_j\}$ are all equivalent.

Using $\varphi_{1,j}$ it can also be shown that $\{\beta_i,\delta_j\}$ are all brace equivalent in a very similar manner. Furthermore, in the case $m$ is even the map $\zeta: D_n\to D_n$ given by $\zeta(r)=r^m,\;\zeta(s)=r^m$ is a fixed point free endomorphism whose image is the center of $D_n$. Additionally,
\begin{align*}
\alpha_0(r\zeta(r^{-1}))\zeta(r) &= \alpha_0(r^{1+m})r^m = s^{1+m}r^m=r^ms=\beta_0(r)\\
\alpha_0(s\zeta(s^{-1}))\zeta(s) &= \alpha_0(r^ms)r^m = s^{m}r^m = r^m=\beta_0(s),
\end{align*}
so there is only one nontrivial class.

We have
\begin{align*}
R_{\alpha_0}(r^is^j,r^ks^{\ell})&=(s^ir^ks^{\ell},s^{k+i+\ell}r^{-k}s^ir^is^{i+j}r^ks^{j+k+\ell})\\
R_{\alpha_0}'(r^is^j,r^ks^{\ell})&=(r^is^{i+k}r^ks^{i+\ell+j}r^i,s^kr^is^{j+k})
\end{align*}
and for $m$ odd we have, since $r^m\in Z(D_n)$,
\begin{align*}
R_{\beta_0}(r^is^j,r^ks^{\ell})&=(s^ir^ks^{i+\ell},s^{i+k+\ell}r^{-k}s^ir^is^{i+j}r^ks^{i+k+\ell})\\
R'_{\beta_0}(r^is^j,r^ks^{\ell})&=(r^is^{i+j}r^ks^{i+j+{\ell}}r^{-i},s^kr^is^{j+k}).
\end{align*}

As mentioned above, in \cite{Kohl20}, Kohl enumerates all regular, $G$-stable subgroups of $ \Perm(D_{n}) $ isomorphic to $D_n$. Since a regular, $G$-stable subgroup arising from $\psi\in\FPF(G)$ never gives the same brace as one arising from a regular, $G$-stable subgroup which does {\it not} come from such a $\psi$, this allows us to determine when we have found all the Yang-Baxter solutions of dihedral type. More precisely:

\begin{proposition}
	With the constructions above, we obtain every Yang-Baxter solution of dihedral type if and only if one of the following holds:
	\begin{enumerate}
		\item $n=p^k$ for some prime $p$ and positive integer $k$;
		\item $n=2p^k$ for some odd prime $p$ and positive integer $k$;
	\end{enumerate}
\end{proposition}
\begin{proof}
	The main result of \cite{Kohl20} indicates that the number $\mathscr{R}(D_n)$ of regular, $G$-stable subgroups of $D_n$ isomorphic to $D_n$ is given by the following formula:
	\[ \mathscr{R}(D_n)= |\Upsilon_n|\cdot \begin{cases} 1 & v_2(n)=0\\ 2m+1 & v_2(n)=1 \\ m+1 & v_2(n)=2 \\ m+2 & v_2(n)>2\end{cases},\]
	where $v_2(n)$ is the $2$-adic valuation of $n$, $n=2m$ where appropriate, and $\Upsilon_n$ is the subgroup of self-inverse elements of $\mathbb{Z}_n^{\times}$. Note that $\Upsilon_n$ is nontrivial for $n>3$, and $|\Upsilon_n|=2$ if and only if $\mathbb{Z}_n^{\times}$ is cyclic, which holds if and only if $n=1,2,p^k$ or $2p^k$ for some odd prime $p$ and $k\ge 1$.

	On the other hand, in \cite{Childs13} Childs counts the number of regular, $G$-stable subgroups arising from fixed point free abelian endomorphisms on $D_n$. For our purposes, since the theory of opposites allow us to obtain {\it two} regular, $G$-stable subgroups from each $\psi$, we get that the number of subgroups  $\mathscr{S}(D_n)$ we obtain is
	\[\mathscr{S}(D_n) = \begin{cases} 2 & v_2(n)=0 \\ 4m+2 & v_2(n)=1 \\ 8m+2 & v_2(n)>1 \end{cases}.\]

	By comparing these, we see that if $n$ is odd we will get every regular, $G$-stable subgroup (hence, every Yang-Baxter solution) if and only if $|\Upsilon_n|=2$, i.e., $n$ is a power of a prime. If $v_2(n)=1$ we again require $|\Upsilon_n|=2$; since $n\ge 3$ is even this means $n=2p^k$ for some $p$. If $v_2(n)=2$ then we would need $(m+2)|\Upsilon_n|=4m+2$. Since $|\Upsilon_n|$ is necessarily a nontrivial power of $2$ and $4m+2\equiv 2 \pmod 4$ we see that we cannot get all regular, $G$-stable subgroups in this manner. Similarly, if $v_2(n)>2$ then we would need $(m+2)|\Upsilon_n|=8m+2$ which cannot happen.
\end{proof}

\subsection{The Alternating Group $A_4$}

It is well-known that $G=A_4$ can be expressed as a semidirect product $V\rtimes C_3$ where $V$ is the (unique) Sylow $2$-subgroup of $A_4$ and $C_3$ is generated by a $3$-cycle, say $x$. Then
\[A_4=\{vx^i: 0\le i\le 2, v\in V\}.\]
Let $\psi\in\FPF(G)$ be nontrivial. Then, for some $v\in V$ we have either $\psi(v)=1_G$ or $\psi(v)$ is an element of order $2$, say $\psi(v)=w$. But $v$ and $w$ are conjugate in $A_4$, hence $\psi(w)=\psi(v)=w$, a contradiction. Thus, $\psi(v)=1_G$ for all $v\in V$.

Since $\psi$ is assumed to be nontrivial, $\psi(x)$ is an element of order $3$ which is not conjugate to $x$. There are two conjugacy classes for the $3$-cycles in $A_4$, and every $3$-cycle is in a different class than its inverse. Thus each conjugacy class has four elements, hence there are four possibilities for $\psi(x)$. One such example is $\psi(x)=x^2$. The reader can verify that all four choices give valid fixed-point free abelian endomorphisms.

We omit the technical details above because each of the nontrivial elements of $\FPF(G)$ give the same brace. To see this, suppose $\psi_1(x)=y$ and $\psi_2(x)=z$. Necessarily, $y$ and $z$ are in the same conjugacy class, so there exists an inner automorphism of $ G $ carrying $\psi_1(x)$ to $\psi_2(x)$.

Including the trivial map, we have
\begin{proposition}
	There are five fixed point free abelian maps on $A_4$ and two brace equivalence classes of fixed point free abelian maps.
\end{proposition}

If we use $\psi(x)=x^2$ the solutions obtained are
\begin{align*}
R_{\psi}(v_1x^i,v_2x^j) &= (x^iv_2x^{j-i},x^iv_2x^{-i}v_1x^{-i}v_2x^{-i-j})\\
R_{\psi}'(x^iv_1,x^jv_2) &= (v_1x^{-i}v_2x^{i-j}v_1,x^{-j}v_1x^{i+j})
\end{align*}
where $v_1,v_2\in V$.

As $\Perm(A_4)$ has $10$ regular, $G$-stable subgroups \cite[Th. 7]{CarnahanChilds99}, we see that the solutions presented above are all of the solutions of type $A_4$.

The situation changes drastically for $A_5$, as we shall see.
\subsection{The Alternating Groups $A_n,n\ge 5$}

It is easy to compute $\FPF(A_n)$ for $n\ge 5$. In fact, we generalize, computing $\FPF(G)$ in the case $G$ is a simple group. This is well known and can be shown by direct computation or by considering Hopf-Galois structures on Galois extensions with simple Galois group \cite{Byott04c}.

\begin{proposition}
	Suppose $G$ is a nonabelian simple group. Then $\FPF(G)$ consists of only the trivial map.
\end{proposition}
\begin{proof}
	If $\psi:G\to G$ is any homomorphism, then $\ker\psi=\{1_G\}$ or $\ker\psi=G$ since $\ker\psi\triangleleft G$. If $\ker\psi=\{1_G\}$ then $\psi(G)=G$ and hence $\psi$ cannot be abelian. Thus $\psi$ is trivial.
\end{proof}

Thus the only solutions to the Yang-Baxter equation obtained via fixed-point-free abelian maps are the solutions found in example \ref{trivYBE}. Unlike in the abelian case, we do get multiple solutions. Trivially, this gives us all the Yang-Baxter equation solutions of type $G$ for $G$ a nonabelian simple group.
\subsection{The Symmetric Groups $S_n,n\ge 5$}

Let $\psi:S_n\to S_n$ be a homomorphism. In a manner similar to the previous section, $\ker \psi=\{1_G\},A_n$, or $S_n$. If we assume $\psi$ is nontrivial and fixed point free, then $\ker \psi=A_n$. Now for $\sigma,\pi\in S_n\setminus A_n$ we have $\sigma\pi^{-1}\in A_n$, hence
\[\psi(\pi) = \psi(\pi\sigma^{-1})\psi(\sigma) = \psi(\sigma),\]
so $\psi$ is constant on odd permutations. Furthermore, $\psi(\sigma^2)=1_G$, hence a nontrivial $\psi$ sends odd permutations to some element of order $2$, say $\tau$. Finally, if $\tau$ is odd then $\psi(\tau)=\tau$ and we have a fixed point. In fact, we have:

\begin{proposition}
	Let $A_n^{(2)}$ be the largest subgroup of $A_n$ of exponent two. Then there is a bijection between $A_n^{(2)}$ and $\FPF(S_n)$ given by $\tau\mapsto\psi_{\tau}$, where
	\[\psi_{\tau}(\sigma)=\begin{cases} 1 & \sigma\in A_n \\ \tau & \sigma\notin A_n \end{cases}.\]
	Furthermore, $\psi_{\tau_1}$ and $\psi_{\tau_2}$ are brace equivalent if and only if $\tau_1$ and $\tau_2$ have the same cycle structure.
\end{proposition}

\begin{proof}
	From the discussion above it is clear that any $\psi\in\FPF(G)$ must be of this form, and it is routine to show that $\psi_{\tau}\in\FPF(G)$ for any $\tau\in A_n$ with $\tau^2=1_G$. Any $\tau_1,\;\tau_2\in A_n$ with $\tau_1^2=\tau_2^2=1_G$ are conjugate if and only if they have the same cycle structure. Now, if $\tau_2=\xi\tau_1\xi^{-1},\;\xi\in S_n$ then the map $C(\xi):S_n\to S_n$ given by conjugation by $\xi$ is an automorphism. If $\sigma\in S_n$ then
	\[C(\xi)\psi_1C(\xi)^{-1}(\sigma)= C(\xi)\psi_1(\xi^{-1}\sigma\xi)=C(\xi)(1_G)=1_G=\psi_2(\sigma)\]
 if $\sigma$ is even; and 	\begin{align*}
	C(\xi)\psi_1C(\xi)^{-1}(\sigma) &= C(\xi)\psi_1(\xi^{-1}\sigma\xi) \\
	&= C(\xi)\tau_1\\
	&= \xi\tau_1\xi^{-1}\\
	&=\tau_2\\
	&=\psi_2(\sigma)
	\end{align*}
	if $\sigma\in S_n$ is odd. Thus if $\tau_1,\;\tau_2$ have the same cycle structure then the corresponding maps $\psi_1$ and $\psi_2$ are brace equivalent.

	Conversely, for $n\ne 6$ every automorphism of $S_n$ is inner, hence these account for all the brace equivalences. Finally, if $n=6$ every element of order $2$ has the same cycle type, so the result follows.
\end{proof}

\begin{corollary}
	Let $\tau \in A_n$ have order at most two. Then
	\[R_{\tau}(\sigma,\pi)=\begin{cases}
	(\pi,\pi^{-1}\sigma\pi) & \sigma,\pi\in A_n\\
	(\pi,\tau\pi^{-1}\sigma\pi\tau) & \sigma\in A_n,\;\pi\notin A_n\\
	(\tau\pi\tau,\tau\pi^{-1}\tau\sigma\tau\pi\tau) & \sigma\notin A_n,\;\pi\in A_n\\
	(\tau\pi\tau,\pi^{-1}\tau\sigma\tau\pi) & \sigma,\pi\notin A_n\\
	\end{cases},\;
	R'_{\tau}(\sigma,\pi)=\begin{cases}
	(\sigma\pi\sigma^{-1},\sigma)& \sigma,\pi\in A_n\\
	(\sigma\pi\sigma^{-1},\tau\sigma\tau) & \sigma\in A_n,\;\pi\notin A_n\\
	(\sigma\tau\pi\tau\sigma^{-1},\sigma) & \sigma\notin A_n,\;\pi\in A_n\\
	(\sigma\tau\pi\tau\sigma^{-1},\tau\sigma\tau) & \sigma,\pi\notin A_n\\
	\end{cases}
	\]
	are solutions to the Yang-Baxter equation.
\end{corollary}

Here, we get all of the solutions of type $S_n$. This can be seen from \cite[Th. 5]{CarnahanChilds99}, where the authors find two families of regular, $G$-stable subgroups of $\Perm(G)$, each parameterized by elements of order $2$ in $S_n$.

\subsection{The Family of Nonabelian Metacyclic Groups $M_{p,q}$}

For our last set of examples we consider the simplest nonabelian metacyclic groups. Let $p,q$ be primes with $p\equiv 1\pmod q$, and define
\[M_{p,q}=\gen{s,t:s^p=t^q=1_G,\;tst^{-1}=s^d},\] where $d$ is chosen to have order $ q $ modulo $ p $. (Note that the group is independent of the choice of $d$: changing its value merely changes the presentation.) By Sylow theory, the Sylow $p$-subgroup $\gen{s}$ is normal in $M_{p,q}$, however since $sts^{-1} \notin\gen{t}$
 the Sylow $q$-subgroups are not normal in $M_{p,q}$. Thus, if $\psi$ is any endomorphism of $M_{p,q}$, then $\ker\psi=\{1_G\},\;\gen{s}$, or $M_{p,q}$. Thus for any $\psi$ fixed point free and abelian, $\ker \psi=\gen{s}$ and hence $\psi(s)=1_G$.

It therefore suffices to determine $\psi(t)$. Since the Sylow $p$-subgroup is normal, an element $s^it^j$ will have order $q$ if and only if $j\ne 0$. Let us write \[\psi(t)=s^i t^j,\;0\le i \le p-1,\;1\le j\le q-1.\]
Then $\psi$ respects the relations in the presentation of $M_{p,q}$, hence is an endomorphism. Since $\psi(M_{p,q})=\gen{s^it^j}\cong C_q$ the map is abelian. Suppose $\psi(s^kt^{\ell})=s^kt^{\ell}$. Then
\[(s^it^j)^{\ell} = s^kt^{\ell}.\]
For this to hold we require $j\ell\equiv \ell\pmod q$, so either $j=1$ or $\ell=0$. If $\ell=0$ then $\psi(s^k)=s^k$, so $k=0$ and we have the trivial element. If $j=1$ then $\psi(s^it)=s^it$ and we have a fixed point. Thus:
\begin{proposition}
	If $\psi\in \FPF(M_{p,q})$ is nontrivial, then $\psi=\psi_{i,j}$, where
	\[\psi_{i,j}(s)=1,\;\psi_{i,j}(t)=s^it^j,\;0\le i \le p-1, 2\le j \le q-1.\]
\end{proposition}
It follows from \cite[Lemma 8.13]{KochTruman20b} that $\psi_{i_1,j_1}$ is brace equivalent to $\psi_{i_2,j_2}$ if and only if $i_1=i_2$. From this we obtain
\begin{corollary}
	The set $\{\psi_j:=\psi_{0,j}:2\le j \le q-1\}$ is a complete set of nontrivial brace classes of fixed point free abelian maps.
\end{corollary}

The endomorphism $\psi_j$ produces the following solutions to the Yang-Baxter equation:
\begin{align*}
R_{\psi_j}(s^kt^{\ell}, s^mt^n) &= (
t^{-j{\ell}}s^mt^{n+j{\ell}}, t^{j(n-{\ell})}(s^mt^n)^{-1}t^{jl}s^kt^{{\ell}-j{\ell}}s^mt^{n+j({\ell}-n)})\\
R'_{\psi_j}(s^kt^{\ell}, s^mt^n) &= (s^kt^{{\ell}-j{\ell}}s^mt^{n+j{\ell}}(s^kt^{\ell})^{-1}, t^{jn}s^kt^{{\ell}-jn}).
\end{align*}

One can see that we get all solutions of metacyclic type by using the regular, $G$-stable classification from \cite{Byott04b}; alternatively, see \cite[\S 8]{KochTruman20b}, which explicitly shows this.

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\end{document}


TITLE: Evaluating integral using invalid substitution
QUESTION [1 upvotes]: I was trying to show that for suitable t:
$$
2\pi(1+t/(\sqrt{(1-t)(3-t)})=\sum_{0}^{\infty}(t^n\int_0^{2\pi}1/(2-cos(\theta))^nd\theta
$$
By uniqueness this is clearly the Taylor series about $0$ and the form of $c_n $ suggests the use of Cauchy's formula so we get $c_n=\frac{1}{2i\pi}\int_{\gamma(0,r)}\frac{2\pi(1+t/(\sqrt{(1-t)(3-t)})}{t^{n+1}}dt $ for $n\geq1$
$$
=-i\int_{\gamma(0,r)}\frac{1}{t^{n}\sqrt{(1-t)(3-t)}}dt
$$
as $ \int_{\gamma(0,r)}\frac{1}{t^{n+1}}dt=0 \forall n\geq 1$
Curiously the substitution $t=2-cos(\theta)$ for $\theta\in [0,2\pi]$ gives
$$
\frac{dt}{d\theta}=sin(\theta)=\sqrt{1-cos(\theta)^2}=\sqrt{-3+4t-t^2}=i\sqrt{(1-t)(3-t)}
$$
$$
\therefore c_n=\int_0^{2\pi} \frac{1}{(2-cos(\theta))^n}d\theta
$$
and so solves the problem, except for the small issue that it is not a valid substitution.
I was wondering if anyone has any insight into why this gives the correct answer and if there are other integrals where invalid substitutions can give the correct answer.

REPLY [1 votes]: Here's an explanation for the invalid substitution.
First, we note that
$$f(z)=\frac1{z^n\sqrt{(1-z)(3-z)}}$$
is a well-defined holomorphic function on $\mathbb C_\infty\setminus(\{0\}\cup[1,3])$ (choose a branch for $\sqrt{(1-z)(3-z)}$), where $\mathbb C_\infty$ is the Riemann sphere $\mathbb C\cup\{\infty\}$
For $0<r,\epsilon<1/4$, as you did, you can draw a circle $C_r(0)$ around $0$. Moreover, you can picture a contour $\Gamma$ around $[1,3]$ such that for each point $z\in\Gamma$, the minimal distance between $z$ and $[1,3]$ is $\epsilon$. You can easily see that $\Gamma$ is made up of two straight line segments and two half-circles.
By an appropriate choice of orientation, $\int_{C_r(0)}f(z)dz=\int_\Gamma f(z)dz$. It's because that $f$ is holomorphic. (It's important that $f$ is holomorphic at $\infty$. If you don't want to utilize this term, you can draw a large $C_R(0)$ and get $\int_{C_r(0)}-\int_\Gamma=\int_{C_R(0)}$, and let $R\to\infty$)
Now let $\epsilon\to0^+$ and $\int_\Gamma$ tends to the value you calculated by the invalid substitution.


TITLE: Does Shilov's Linear Algebra cover SVD?
QUESTION [1 upvotes]: I'm teaching myself Georgi E. Shilov's Linear Algebra, and have just finished chapter 6 except the final two starred sections. So far, I haven't seen Singular Value Decomposition (SVD). I skimmed the rest of the book, and looked it up in the Index, but had no luck. I'd like to quit studying the book prematurely if it turns out not to cover the subject. You information is valuable. Thank you!

REPLY [1 votes]: No, it doesn't. SVD had yet to become a textbook topic in the English speaking world when Shilov's book was first published (1971). I think it entered the textbook scene by virtues of Gilbert Strang's Linear Algebra and Its Applications and Golub and Van Loan's Matrix Computations, which were published at later times.


TITLE: Does $g$ always normalize $H \cap H^g$?
QUESTION [0 upvotes]: Let $G$ be any group, $H$ a subgroup, $g$ an element of $G$ and $H^g:=g^{-1}Hg$ the conjugate of $H$ by $g$. Is it true that $$g\in N_G(H\cap H^g)?$$

REPLY [3 votes]: No.
As a counterexample take $G$ to be the split extension of an elementary abelian group of order $16$ by a cyclic group of order $4$ with this action
$$
G=\left\langle x_1,x_2,x_3,x_4, g\mid x_i^2=1, x_i x_j=x_j x_i,   g^4=1, x_i^g=x_{i+1 \mod 4}\right\rangle
$$
and let $H=\langle x_1,x_2\rangle$.
Then $H\cap H^g=\langle x_2\rangle$ is not normalised by $g$.