---
abstract: |
    Quandle 2-cocycles define invariants of classical and virtual knots, and extensions of quandles. We show that the quandle 2-cocycle invariant with respect to a non-trivial $2$-cocycle is constant, or takes some other restricted form, for classical knots when the corresponding extensions satisfy certain algebraic conditions. In particular, if an abelian extension is a conjugation quandle, then the corresponding cocycle invariant is constant. Specific examples are presented from the list of connected quandles of order less than 48. Relations among various quandle epimorphisms involved are also examined.\
    Key words: quandles, quandle cocycle invariants, abelian extensions of quandles\
    MSC: 57M25
author:
- |
    W. Edwin Clark   and   Masahico Saito\
    Department of Mathematics and Statistics\
    University of South Florida
title: |
    Algebraic Properties of Quandle Extensions\
    and Values of Cocycle Knot Invariants
---

Introduction {#intro-sec}
============

Sets with certain self-distributive operations called [*quandles*]{} have been studied since the 1940s [@Taka], and have been applied to knot theory since early 1980s [@Joyce; @Mat]. The number of colorings of knot diagrams by quandle elements, in particular, has been widely used as a knot invariant. Algebraic homology theories for quandles were defined [@CJKLS; @FRS1], and investigated. Knot invariants using cocycles have been defined [@CJKLS] and applied to knots and knotted surfaces [@CKS]. Extensions of quandles by cocycles have been studied, for example, in [@AG; @CENS; @Eis3].

Computations by Vendramin [@Leandro] significantly expanded the list of small connected quandles. These quandles, called here [*Rig*]{} quandles, may be found in the  package Rig [@rig]. Rig includes all connected quandles of order less than 48, at this time. Properties of some of Rig quandles, such as homology groups and cocycle invariants, are also found in [@rig]. We use the notation $Q(n,i)$ for the $i$-th quandle of order $n$ in the list of Rig quandles.

It was observed that some Rig quandles have non-trivial second cohomology, yet have constant $2$-cocycle invariants with non-trivial $2$-cocycles, as much as computer calculations have been performed for the knot table (Remark 4.5, [@CSV]). It does not seem to have been established previously whether they actually have constant values for all classical knots. From Theorem 5.5 in [@CKS:geom], any non-trivial $2$-cocycle has non-constant invariant values for some virtual links. Thus it is of interest if these quandles actually have constant values for all classical knots. More generally, possible values of the cocycle invariants are largely unknown, and are of interest. In this paper, we show that certain algebraic properties of quandles imply that the cocycle invariant is constant, or takes some restricted form, for classical knots. In particular, we prove that several specific Rig quandles, including some of those conjectured in [@CSV], have constant cocycle invariant values for all classical knots for some non-trivial $2$-cocycles.

In Section \[sec:prelim\], definitions, terminology and lemmas are presented. The main results and corollaries, and their proofs are given in Section \[sec:main\]. Conjugation quandles are discussed in Sections \[sec:conj\] and \[sec:VC\]. Relations among various epimorphisms used in the proofs are examined in Secition \[sec:rel\].

Preliminaries {#sec:prelim}
=============

In this section we briefly review some definitions and examples. More details can be found, for example, in [@CKS].

A [*quandle*]{} $X$ is a set with a binary operation $(a, b) \mapsto a * b$ satisfying the following conditions. $$\begin{aligned}
\mbox{\rm (Idempotency) } & &  \mbox{\rm  For any $a \in X$,
$a* a =a$.} \label{axiom1} \\
\mbox{\rm (Right invertibility)}& & \mbox{\rm For any $b,c \in X$, there is a unique $a \in X$ such that
$ a*b=c$.} \label{axiom2} \\
\mbox{\rm (Right self-distributivity)} & &
\mbox{\rm For any $a,b,c \in X$, we have
$ (a*b)*c=(a*c)*(b*c). $} \label{axiom3} \end{aligned}$$ A [*quandle homomorphism*]{} between two quandles $X, Y$ is a map $f: X \rightarrow Y$ such that $f(x*_X y)=f(x) *_Y f(y) $, where $*_X$ and $*_Y$ denote the quandle operations of $X$ and $Y$, respectively. A [*quandle isomorphism*]{} is a bijective quandle homomorphism, and two quandles are [*isomorphic*]{} if there is a quandle isomorphism between them. A quandle epimorphism $f: X \rightarrow Y$ is a [*covering*]{} [@Eis3] if $f(x)=f(y)$ implies $a*x=a*y$ for all $a, x, y \in X$.

Let $X$ be a quandle. The [*right translation*]{} ${R}_a:X \rightarrow  X$, by $a \in X$, is defined by ${ R}_a(x) = x*a$ for $x \in X$. Then ${ R}_a$ is an automorphism of $X$ by Axioms (2) and (3). The subgroup of ${\rm Sym}(X)$ generated by the permutations ${ R}_a$, $a \in X$, is called the [*[inn]{}er automorphism group*]{} of $X$, and is denoted by ${\rm Inn}(X)$. The map ${\rm inn}: X \rightarrow {\rm inn}(X) \subset {\rm Inn}(X)$ defined by ${\rm inn}(x)=R_x$ is called the [*inner representation*]{}. An inner representation is a covering.

A quandle is [*connected*]{} if ${\rm Inn}(X)$ acts transitively on $X$. A quandle is [*faithful*]{} if the mapping ${\rm inn}: X \rightarrow  {\rm Inn}(X)$ is an injection.

As in Joyce [@Joyce] a quandle is defined by a pair $(G, f)$ where $G$ is a group, $f \in {\rm Aut}(G)$, and the quandle operation is defined by $x*y=f(xy^{-1}) y ,$ $ x,y \in G$. We call such a quandle a [*generalized Alexander quandle*]{} and denote it by ${\rm GAlex}(G,f)$. If $G$ is abelian, such a quandle is known as an [*Alexander quandle*]{}.

In this paper all quandles are assumed to be finite.

![Colored crossings and cocycle weights []{data-label="coloredXing"}](coloredXing1.eps "fig:"){width="3in"}\

Let $D$ be a diagram of a knot $K$, and ${\cal A}(D)$ be the set of arcs of $D$. A [*coloring*]{} of a knot diagram $D$ by a quandle $X$ is a map $C: {\cal A}(D) \rightarrow X$ satisfying the condition depicted in Figure \[coloredXing\] at every positive (left) and negative (right) crossing $\tau$, respectively. The pair $(x_\tau, y_\tau)$ of colors assigned to a pair of nearby arcs of a crossing $\tau$ is called the [*source*]{} colors, and the third arc is required to receive the color $x_\tau * y_\tau$.

In this paper we denote by $A$ a finite multiplicative abelian group whose identity element is denoted by $1$. A function $\phi: X \times X \rightarrow A$ for an abelian group $A$ is called a *quandle $2$-cocycle* [@CJKLS] if it satisfies $$\phi (x, y)
    \phi(x,z)^{-1}  \phi(x*y, z)  \phi(x*z, y*z)^{-1}=1$$ for any $x,y,z \in X$ and $\phi(x,x)=1$ for any $x\in X$. For a quandle $2$-cocycle $\phi$, $E=X \times A$ becomes a quandle by setting $$(x, a) * (y, b)=(x*y, a\, \phi(x,y))$$ for $x, y \in X$, $a,b \in A$, denoted by $E(X, A, \phi)$ or simply $E(X, A)$, and it is called an *abelian extension* of $X$ by $A$. Let $\pi: E (X, A) = X \times A \rightarrow X$ be the projection to the first factor. We also say that a quandle epimorphism $f: Y \rightarrow Z$ is an [*abelian extension*]{} if there exists a isomorphisms $\nu:  E(X, A)  \rightarrow Y$ and $h: X \rightarrow Z$ such that $ h \pi= f  \nu$. An abelian extension is a covering. See [@CENS] for more information on abelian extensions of quandles and [@CJKLS; @CJKS; @CJKS2] for more on quandle cohomology.

Let $X$ be a quandle, and $\phi$ be a $2$-cocycle with coefficient group $A$, a finite abelian group. Let $D$ be a diagram of a knot $K$. The $2$-cocycle (or cocycle, for short) invariant is an element of the group ring ${{\mathbb Z}}[A]$ defined by $\Phi_{(X,A,\phi)} (D) =\Phi_{\phi} (D)= \sum_{C} \prod_{\tau} \phi(x_\tau, y_\tau)^{\epsilon(\tau)}$, where the product ranges over all crossings $\tau$, the sum ranges over all colorings of a given knot diagram, $(x_\tau, y_\tau)$ are source colors at the crossing $\tau$, and $\epsilon(\tau)$ is the sign of $\tau$ as specified in Figure \[coloredXing\]. For a given coloring $C$, the element $\prod_{\tau} \phi(x_\tau, y_\tau)^{\epsilon(\tau)} \in A$ is denoted by $B_\phi (D, C) $. For an abelian group $A$, the cocycle invariant takes the form $\sum_{a \in A} n_a a$ where $n_a \in {{\mathbb Z}}$, and it is [*constant*]{} if $n_a=0$ when $a$ is not the identity of $A$. It is known [@CJKLS] that $\Phi_\phi (D)$ is independent of choice of diagram $D$ for a knot $K$. Thus we have the [*2-cocycle invariant*]{} $\Phi_{(X,A, \phi )}(K) = \Phi_{ \phi }(K) = \Phi_\phi(D)$.

A $1$-tangle is a properly embedded arc in a $3$-ball, and the equivalence of $1$-tangles is defined by ambient isotopies of the $3$-ball fixing the boundary (cf. [@Conway]). A diagram of a $1$-tangle is defined in a manner similar to a knot diagram, from a regular projection to a disk by specifying crossing information, see Figure \[tangles\](A). An orientation of a $1$-tangle is specified by an arrow on a diagram as depicted. A knot diagram is obtained from a $1$-tangle diagram by closing the end points by a trivial arc outside of a disk. This procedure is called the [*closure*]{} of a $1$-tangle. If a $1$-tangle is oriented, then the closure inherits the orientation. Two diagrams of the same $1$-tangle are related by Reidemeister moves. There is a bijection between knots and $1$-tangles for classical knots, and invariants of 1-tangles give rise to invariants of knots, see [@Eis3], for example.

![ $1$-tangles []{data-label="tangles"}](tangles.eps "fig:"){width="2.7in"}\

A $1$-tangle is obtained from a knot $K$ as follows. Choose a base point $b
\in K$ and a small closed ball neighborhood $B$ of $b$ in the $3$-sphere ${\mathbb{S}}^3$ such that $(B, K\cap B)$ is a trivial ball-arc pair (so that $K \cap B$ is unknotted in $B$, see Figure \[tangles\](B)). Then $({\mathbb{S}}^3 \setminus {\rm
Int}(B), K \cap ({\mathbb{S}}^3 \setminus {\rm Int}(B)))$ is a $1$-tangle called the $1$-tangle associated with $K$. The resulting $1$-tangle does not depend on the choice of a base point. If a knot is oriented, then the corresponding $1$-tangle inherits the orientation.

A quandle coloring of an oriented $1$-tangle diagram is defined in a manner similar to those for knots. We do not require that the end points receive the same color for a quandle coloring of $1$-tangle diagrams. As in [@CSV] we say that a quandle $X$ is [*end monochromatic*]{} for a tangle diagram $T$ if any coloring of $T$ by $X$ assigns the same color on the two end arcs. We use the same notations $\Phi_{\phi} (T) = \sum_{\cal C} \prod_{\tau} \phi(x_\tau, y_\tau)^{\epsilon(\tau)}$ and $B_\phi (T, C) $ for tangle diagrams $T$. This $\Phi_{\phi} (T)$ is again independent of choice of a diagram, and an invariant of tangles. Figures \[coloredXing\], \[tangles\], \[endmono\] are taken from [@CSV].

![ Colorings of a tangle []{data-label="endmono"}](endmono1.eps "fig:"){width="1.1in"}\

We recall the following two lemmas.

\[lem:cover\] Let $f: Y \rightarrow X$ be a covering, and $C_X: {\cal A} (T) \rightarrow X$ be a coloring of a $1$-tangle $T$ by $X$. Let $b_0, b_1$ be the top and bottom arcs as depicted in Figure \[endmono\]. Then for any $y \in Y $ such that $f(y)=C_X(b_0)$, there exists a unique coloring $C_Y: {\cal A} (T) \rightarrow Y$ such that $f  C_Y=C_X$ and $C_Y(b_0)=y$.

\[lem:const\] Let $E=E(X, A, \phi)$ be an abelian extension for a $2$-cocycle $\phi$. Then $E$ is end monochromatic $T$ if and only if $\Phi_{(X, A, \phi)}(K)$ is constant, where $T$ is a $1$-tangle for a knot $K$.

\[lem:end\] Let $C:  {\cal A} (T) \rightarrow Y$ be a coloring of a classical $1$-tangle diagram $T$ by a quandle $Y$. For the top and bottom arcs $b_0$ and $b_1$ of $T$, respectively, let $y_0=C(b_0)$ and $y_1=C(b_1)$. Then ${\rm inn}(y_0)=R_{y_0}=R_{y_1}={\rm inn}(y_1)$.

The proof in [@CSV], based on corresponding statements in [@Nos; @Jozef; @Jozef2004] on faithful quandles, applies in this situation. The idea of proof is seen in Figure \[endmono\]. The large circle behind the tangle $T$ in the figure can be pulled out of $T$ if $T$ corresponds to a classical knot. Colorings of tangle diagrams can be defined when a tangle has more than one component, in a similar manner. Hence any color $y$ at the left of the large circle should extend to the color to the right, so that for the colors $y_0$ and $y_1$ for the top and bottom arcs $b_0$ and $b_1$ must satisfy $y*y_0=y*y_1$, hence we obtain $R_{y_0}=R_{y_1}$.

![ Long virtual knots []{data-label="virtualize"}](virtualize.eps "fig:"){width="3.5in"}\

[To see that the condition of being a classical knot is essential, we observe examples of [*long virtual knots*]{} (i.e., virtual $1$-tangles) for which the conclusion of Lemma \[lem:end\] does not hold. See [@Man], for example, for long virtual knots. Colorings of oriented virtual (and long virtual) knots are defined in a manner similar to classical knots, with colors unchanged for each transverse arc at each virtual crossing [@Kauff]. Suppose a (virtual or classical) knot diagram $D$ is colored non-trivially by a faithful quandle $X$, and let $\tau$ be a non-trivially colored crossing (that is, distinct colors appear at $\tau$). Since $X$ is faithful, all three colors at $\tau$ are distinct (see Figure \[virtualize\](A) left). Make $\tau$ a virtual crossing (called [*virtualization of a crossing*]{}) as in the middle of (A), and cut and prolong one of the under-arcs up and down, to make a long virtual knot (a virtual $1$-tangle), as in (A) right. A cut point is indicated by a dot in (A) middle. Prolonged arcs cross other arcs by virtual crossings. Then the top and bottom arcs are colored by elements $a, c \in X$ such that $R_a \neq R_c$. Thus the resulting long virtual knot is not classical. An example of this construction is illustrated in Figure \[virtualize\](B) with a Fox tri-coloring. The quandle used here is ${{\mathbb Z}}_3$ with $a*b=2b-a$ mod 3, a well known [*dihedral*]{} quandle of order 3, which is faithful. ]{}

 Cocycle invariants and the images of inner representations  {#sec:main}
============================================================

In this section we present relations between certain algebraic properties of quandles and values of the quandle cocycle invariant. We do not assume connectivity of quandles in this section unless otherwise specified.

\[thm:main\] Let $E=E(X,A,\phi)$ be an abelian extension of a quandle $X$. If $E \cong {\rm inn}(Y)$ for some quandle $Y$, then $\Phi_{(X,A,\phi)}(K)$ is constant for all classical knots $K$.

Let $T$ be a $1$-tangle of $K$, $b_0, b_1$ be the top and bottom arcs of $T$, respectively. Let $C$ be a coloring of a diagram of $K$ by $X$, and use the same notation $C: {\cal A}(T) \rightarrow X$ for a corresponding coloring of $T$ such that $C(b_0)=C(b_1)=x \in X$. Then $C$ extends to a coloring $C_E: {\cal A}(T) \rightarrow E$ by Lemma \[lem:cover\].

Recall that the inner representation is a covering. Without loss of generality, we assume that ${\rm inn}(Y)=E$. By assumption and Lemma \[lem:cover\], $C_E$ extends to a coloring $C_Y$. Since $E$ is an abelian extension of $X$, Lemma \[lem:const\] and Lemma \[lem:end\] imply that the cocycle invariant is constant.

To apply the theorem to some Rig quandles, we observe the following.

\[lem:order2\] If ${\rm inn} : Y \rightarrow {\rm inn}(Y)= X$, for connected quandles $X$ and $Y$, satisfies $|Y|/|X|=2$, then ${\rm inn}$ is an abelian extension.

For ${\rm inn}: Y \rightarrow {\rm inn}(Y)=X$, where $X$ and $Y$ are connected quandles, it is proved in [@AG] that there is a quandle isomorphism $\nu :  X \times S  \rightarrow Y$ for a set $S$, such that $\pi={\rm inn} ( \nu) $ for the projection $\pi: X \times S \rightarrow X$. The quandle operation on $X \times S$ is defined by $$(x, s)*(y, t) = (x*y, \beta_{x,y} (s) )\quad {\rm for } \quad (x, s), (y, t) \in X \times S,$$ for some $\beta: X^2 \rightarrow {\rm Sym}(S)$. The proof of Theorem 7.1 of [@CSV] shows that if the cardinality of $S$ is 2, then we can assume $S={{\mathbb Z}}_2$ and $\beta_{x, y}(a) = a\, \phi (x,y)$ where $\phi$ is a $2$-cocycle with coefficient group $A={{\mathbb Z}}_2$. Hence ${\rm inn}$ is an abelian extension.

We say that an epimorphism $f: Y \rightarrow X$ is [*of index $k$*]{} if $|Y|/|X|=k$. Note that if $Y$ is connected, then each fiber $f^{-1}(x)$, $x \in X$, has cardinality $k$.

\[cor:rig\] The following Rig quandles have non-trivial second cohomology groups with the coefficient group $A={{\mathbb Z}}_2$, yet give rise to constant quandle $2$-cocycle invariants for any classical knot with the corresponding non-trivial $2$-cocycles: $$\begin{array}{llllllll}
  Q(6,1),&  Q(10,1), &
Q(12, 5), & Q(12, 6), &  Q(12, 7), & Q(12, 8), &
 Q(16,4), &
  Q(16,5), \\
  Q(16,6), &
 Q(18, 1), &
   Q(18, 2), &
    Q(18, 8), &
   Q(18, 9), &  Q(18, 10), &
      Q(24, 3), &
      Q(24, 4), \\
    Q(24, 13), &
 Q(24, 22),  &
 Q(30, 2), &  Q(30, 7), &
  Q(30, 8), &
Q(40, 8), &
 Q(40, 9), &    Q(40, 10), \\
Q(42, 1), &  Q(42, 3), &  Q(42, 4), &  Q(42, 7), &  Q(42, 8). & & &
\end{array}$$

Computer calculations show the following quandle sequences of Rig quandles: $$\begin{array}{rllll}
 Q(24,1)& \stackrel{{\rm inn}}{\longrightarrow} & Q(12,1) & \stackrel{{\rm inn}}{\longrightarrow} & Q(6,1),  \\
  Q(40,2) &\stackrel{{\rm inn}}{\longrightarrow} &  Q(20, 3)  & \stackrel{{\rm inn}}{\longrightarrow}&  Q(10,1) . \\
\end{array}$$ For example, the first sequence shows that $Q(12, 1)$ is an abelian extension of $Q(6,1)$ by a cocycle $\phi$, and is the image of $Q(24,1)$ by ${\rm inn}$. Then Lemma \[lem:order2\] and Theorem \[thm:main\] imply that $\Phi_\phi (K)$ is constant for all classical knots $K$. For the other quandles, the result follows in a similar fashion from sequences of inner representations in Tables \[table:4term\] and \[table:short\]. See [@Edwin2] for multiplication tables of quandles in Tables \[table:4term\] and \[table:short\].

$$\begin{array}{lllllll}
R(192,2) & {\rightarrow } & R(48,3) & {\rightarrow } & Q(24,3) & {\rightarrow } & Q(12,6)  \\
R(192,3) & {\rightarrow } & R(48,4) & {\rightarrow } & Q(24,4) & {\rightarrow } & Q(12,5)
\end{array}$$

$$\begin{array}{lllllllllll}
R(64,1) & {\rightarrow } & Q(32,6) & {\rightarrow } & Q(16,4) &R(64,2) & {\rightarrow } & Q(32,7) & {\rightarrow } & Q(16,5)  \\
R(64,3) & {\rightarrow } & Q(32,8) & {\rightarrow } & Q(16,6) &R(64,4) & {\rightarrow } & Q(32,5) & {\rightarrow } & Q(16,4)  \\
R(64,5) & {\rightarrow } & Q(32,6) & {\rightarrow } & Q(16,4) &R(64,6) & {\rightarrow } & Q(32,5) & {\rightarrow } & Q(16,4)  \\
R(72,1) & {\rightarrow } & Q(36,21) & {\rightarrow } & Q(18,10) &R(72,2) & {\rightarrow } & Q(36,17) & {\rightarrow } & Q(18,8)  \\
R(72,3) & {\rightarrow } & Q(36,20) & {\rightarrow } & Q(18,9) &R(72,4) & {\rightarrow } & Q(36,4) & {\rightarrow } & Q(18,2)  \\
R(72,5) & {\rightarrow } & Q(36,1) & {\rightarrow } & Q(18,1) &R(96,1) & {\rightarrow } & Q(24,4) & {\rightarrow } & Q(12,5)  \\
R(96,2) & {\rightarrow } & Q(24,3) & {\rightarrow } & Q(12,6) &R(96,3) & {\rightarrow } & R(48,1) & {\rightarrow } & Q(24,22)  \\
R(96,4) & {\rightarrow } & Q(24,6) & {\rightarrow } & Q(12,9) &R(96,5) & {\rightarrow } & Q(24,6) & {\rightarrow } & Q(12,9)  \\
R(96,6) & {\rightarrow } & Q(24,6) & {\rightarrow } & Q(12,9) &R(96,7) & {\rightarrow } & Q(24,5) & {\rightarrow } & Q(12,8)  \\
R(96,8) & {\rightarrow } & Q(24,5) & {\rightarrow } & Q(12,8) &R(96,9) & {\rightarrow } & R(48,2) & {\rightarrow } & Q(24,22)  \\
R(120,1) & {\rightarrow } & Q(20,3) & {\rightarrow } & Q(10,1) &R(120,2) & {\rightarrow } & R(60,1) & {\rightarrow } & Q(30,8)  \\
R(120,3) & {\rightarrow } & R(60,2) & {\rightarrow } & Q(30,7) &R(120,4) & {\rightarrow } & R(60,3) & {\rightarrow } & Q(30,2)  \\
R(120,5) & {\rightarrow } & Q(30,1) & {\rightarrow } & Q(15,2) &R(160,1) & {\rightarrow } & R(80,1) & {\rightarrow } & Q(40,10)  \\
R(160,2) & {\rightarrow } & Q(40,20) & {\rightarrow } & Q(20,5) &R(160,3) & {\rightarrow } & Q(40,19) & {\rightarrow } & Q(20,6)  \\
R(160,4) & {\rightarrow } & R(80,2) & {\rightarrow } & Q(40,9) &R(168,1) & {\rightarrow } & R(84,1) & {\rightarrow } & Q(42,8)  \\
R(168,2) & {\rightarrow } & R(84,2) & {\rightarrow } & Q(42,3) &R(168,3) & {\rightarrow } & R(84,3) & {\rightarrow } & Q(42,7)  \\
R(168,4) & {\rightarrow } & R(84,4) & {\rightarrow } & Q(42,4) &R(168,5) & {\rightarrow } & R(84,5) & {\rightarrow } & Q(42,1)  \\
R(192,1) & {\rightarrow } & Q(24,14) & {\rightarrow } & Q(12,7) &R(192,4) & {\rightarrow } & R(48,5) & {\rightarrow } & Q(24,13)  \\
R(216,1) & {\rightarrow } & R(72,6) & {\rightarrow } & Q(24,21) &R(216,2) & {\rightarrow } & Q(36,17) & {\rightarrow } & Q(18,8)  \\
R(216,3) & {\rightarrow } & Q(36,1) & {\rightarrow } & Q(18,1) &
&
\end{array}$$

We remark that the above list contains all Rig quandles of order less than or equal to 16 that were conjectured in [@CSV] to have constant quandle $2$-cocycle invariants for any classical knot with non-trivial $2$-cocycles except $Q(15,2)$, and $Q(15,7)$, which will be shown to have constant invariant in Section \[sec:VC\]. Those in the above list of order larger than 16 do not appear in the conjectured list in [@CSV].

Tables \[table:4term\] and \[table:short\] contain sequences of connected quandles where all arrows represent inner representations. The quandle on the left of each sequence is a generalized Alexander quandle, but others in the sequence may or may not be generalized Alexander quandles. The right-most quandle in each sequence is faithful, so the sequences cannot be extended non-trivially to the right with inner representations. The notation $R(n, j)$ is used to indicate a quandle of order $n$ when $n >  47$ and hence not a Rig quandle. The index $j$ is simply to distinguish non-isomorphic quandles.

Terminating sequences of ${\rm inn}$, $$X=X_0 \stackrel{{\rm inn}}{\longrightarrow} X_1={\rm inn}(X_0) \stackrel{{\rm inn}}{\longrightarrow} \cdots
 \stackrel{{\rm inn}}{\longrightarrow} X_n={\rm inn}(X_{n-1}) ,$$ are discussed in [@AG], where $X_n$ is faithful and $X_j$ are not, for $j=1, \ldots, n-1$. For the 790 Rig quandles of order less than 48, there are 66 non-faithful quandles $X$, and all but two have faithful images ${\rm inn}(X)$. The two exceptions are the above first two Rig quandles $Q(24, 1)$ and $Q(40, 2)$ in Corollary \[cor:rig\].

On the other hand, Tables \[table:4term\] and \[table:short\] include many quandles $X$ with ${\rm inn}(X)$ being non-faithful Rig quandles.

Let $X=Q(12, 5)$ or $Q(12,6)$. Then the second quandle cohomology group $H^2_Q(X,{{\mathbb Z}}_4)$ is known [@rig] to be isomorphic to ${{\mathbb Z}}_4$. See [@CJKLS; @CKS], for example, for details on quandle cohomology. Let $\psi: X \times X \rightarrow {{\mathbb Z}}_4$ be a $2$-cocycle which represents a generator of $H^2_Q(X,{{\mathbb Z}}_4)\cong {{\mathbb Z}}_4$. Let $u$ denote a multiplicative generator of $A={{\mathbb Z}}_4$. The cocycle invariants $\Phi_\psi(K)$ for $X=Q(12, 5)$ or $Q(12,6)$ with respect to $\psi$, computed for some knots in the table in [@rig] up to 9 crossing knots, contain non-constant values, while for $A={{\mathbb Z}}_2$ the invariant is constant by Corollary \[cor:rig\]. This is explained by the following.

Let $X$ be a quandle and $n, m, d>1$ be positive integers such that $n=md$. Let $\psi$ be a $2$-cocycle of $X$ with values in ${{\mathbb Z}}_n$, and $\Phi_\psi(K)= \sum_{j=0}^{n-1} a_j(K)\,  u^j $ be the cocycle invariant of a knot $K$ with respect to $\psi$.

Let $E=E(X, {{\mathbb Z}}_m, \phi) \stackrel{\alpha}{\rightarrow} X$ be the abelian extension corresponding to $\phi=\psi^d$, and suppose that there is a sequence of quandles $Y \stackrel{\rm inn}{\longrightarrow} E \stackrel{\alpha}{\rightarrow} X$. Then $a_k (K)=0$ for all $k$ that are not divisible by $m$, for any classical knot $K$.

For each coloring $C: {\cal A}(T) \rightarrow X$, the weight $\phi(x_\tau, y_\tau )^{\epsilon (\tau) }$ at a crossing $\tau$ for $\phi$ is equal to $\psi(x_\tau, y_\tau )^{d \epsilon (\tau) }$. Hence if $B_\psi(T, C)= u^j$, then we have $B_\phi(T, C)= u^{dj}$. Then $$\begin{aligned}
\Phi_\phi(K) &=&  \sum_{j=0}^{n-1} a_j(K)\, u^{dj} \\
&=& [ \, \sum_{k=0}^{d-1} a_{mk} (K)  \, ] \, u^0 +
[ \, \sum_{k=0}^{d-1} a_{1+mk} (K)  \, ] \, u^d + \cdots \\
& & + \  [ \, \sum_{k=0}^{d-1} a_{\ell + mk} (K)  \, ] \, u^{d \ell}  + \cdots + \,
[ \, \sum_{k=0}^{d-1} a_{(d-1)d + mk} (K)  \, ] \, u^{ d (d-1) } .\end{aligned}$$ By Theorem \[thm:main\], this is equal to a constant, $\sum_{k=0}^{d-1} \, [ \, a_{mk} (K)  \, ] \, u^0$. Since $a_j(K)$ are non-negative integers, the result follows.

The following solves a conjecture stated in [@CSV].

\[cor:Z4\] Let $X=Q(12, 5)$ or $Q(12,6)$, and $\psi: X \times X \rightarrow A={{\mathbb Z}}_4$ be a $2$-cocycle which represents a generator of $H^2_Q(X,{{\mathbb Z}}_4)\cong {{\mathbb Z}}_4$. Let $\Phi_\psi(K)= \sum_{j=0}^3 a_j(K)\, u^j \in {{\mathbb Z}}[A]$ be the cocycle invariant. Then $a_1(K)=a_3(K)=0$ for any classical knot $K$.

This situation is also found for $X=Q(18,1)$ or $Q(18,8)$, where $H^2_Q(X,{{\mathbb Z}}_6)\cong {{\mathbb Z}}_6$. Let $u$ be a multiplicative generator of $A={{\mathbb Z}}_6$. Then the invariant values are restricted to the following form.

\[cor:Z6\] Let $X=Q(18, 1)$ or $Q(18,8)$, and $\psi: X \times X \rightarrow {{\mathbb Z}}_6$ be a $2$-cocycle which represents a generator of $H^2_Q(X,{{\mathbb Z}}_6)\cong {{\mathbb Z}}_6$. Let $\Phi_\psi(K)= \sum_{j=0}^5 a_j(K)\, u^j \in {{\mathbb Z}}[A]$ be the cocycle invariant. Then $a_k(K)=0$ for $k=1,3,5$ for any classical knot $K$.

The cocycle invariant for connected quandles of order 18 are computed in [@rig] for up to 7 crossing knots at the time of writing. The invariant values for $Q(18,8)$ do contain non-constant values. For $Q(18,1)$, the invariant is constant, and we do not know whether this is an artifact of limited number of knots or it is constant for all classical knots.

Theorem \[thm:main\] can be applied contrapositively: if $E \rightarrow X$ is an abelian extension with a $2$-cocycle $\phi$ such that $\Phi_\phi(K)$ is not constant for a classical knot $K$, then there is no finite quandle $Y$ such that ${\rm inn}(Y)=E$ for the inner representation ${\rm inn}$. Thus we obtain the following from [@Edwin] and [@rig]. The list contains abelian extension $E(X,{{\mathbb Z}}_2, \phi)$ such that $\Phi_\phi(K)$ is non-constant for some knot $K$ in the table. The list is from the information available at the time of writing.

\[cor:no-inn\] For the following Rig quandles $E$, there is no finite quandle $Y$ such that ${\rm inn}(Y)=E$ for the inner representation ${\rm inn}$: $$Q(8,1),  \ Q(24,1), \ Q(24, 7), \ Q(32, 1). \ $$

The arguments in Corollary \[cor:Z4\] can be applied contrapositively to obtain the following as well.

\[cor:noY\] Let $E$ be one of the Rig quandles $ Q(12,2)$, $Q(36,1)$ or $Q(32,9)$. Then there is no finite quandle $Y$ such that ${\rm inn}(Y)=E$ for the inner representation ${\rm inn}$.

Each of the Rig quandles $Q(6,2)$, $Q(16,1)$, $Q(16, 7)$ has the second cohomology group isomorphic to ${{\mathbb Z}}_4$. It is computed in [@Edwin] that the index 2 abelian extensions corresponding to $\psi^2$ for a 2-cocycle $\psi$ representing a generator of the second cohomology group are, respectively, $Q(12,2)$, $ Q(36,1)$, and $Q(32,9)$.

Let $E$ be one of $Q(12,2)$, $ Q(36,1)$, and $Q(32,9)$. If there is a finite quandle $Y$ such that ${\rm inn}(Y)=E$, then by the proof of Corollary \[cor:Z4\], the cocycle invariant $\Phi_\psi(K)= \sum_{j=0}^{3} a_j(K)\, u^{j} $ satisfies $a_1(K)=a_3(K)=0$ for all classical knots $K$. However, this is not the case from [@rig], and the result follows.

[Similar arguments apply to other Rig quandles for $Q$ with cyclic second cohomology groups, but the corresponding index 2 abelian extensions go out of bounds of Rig quandles, and we are not able to specify the quandles with this property. However, we can conclude that for each Rig quandle $Q$ in the list given below, there is a non-trivial abelian extension $\alpha: X \rightarrow Q$ of index $2$ with the property that there is no finite quandle $Y$ such that ${\rm inn}(Y)=X$ for the inner representation ${\rm inn}$: $$Q(18,3), \
Q(18,6), \  Q(18,7), \
Q(12, 10), \
Q(12, 3),\
Q(18,5).$$ ]{}

$\Phi_{(X,A,\phi)}(K)$ is constant if $E(X,A,\phi)$ is a conjugation quandle {#sec:conj}
============================================================================

Let $G$ be a finite group. For $a,b \in G$ we write $a^b = b^{-1}ab$ and denote the conjugacy class of $G$ containing $x$ by $x^G$. The conjugacy class $x^G$ under conjugation, $a*b = a^b$, is a quandle. Here we call such a quandle a [*conjugation quandle*]{}. We note that such a quandle need not be connected. In general, a subquandle of a group $G$ under conjugation need not be a conjugacy class. But it is easy to see that if $X$ is a subset of a group $G$ closed under conjugation and if $X$ under conjugation is a connected quandle then $X$ is a conjugacy class of the group $\langle X \rangle$ generated by $X.$ Note that we are only interested in this paper in the case where both $X$ and $G$ are finite.

We show in this section that a connected quandle $E$ satisfies the condition in Theorem \[thm:main\] if and only if $E$ is a conjugation quandle.

To simplify the proof of the next theorem we will need the following lemma.

\[lem:triangle\] Suppose $A$, $B$, and $C$ are quandles, $f$ and $g$ are quandle epimorphisms and $h$ is a well-defined bijection such that $g = hf$, then $h$ is a quandle isomorphism.

Let $a_1, a_2 \in A$. Since $f$ is an epimorphism there exist $c_1,c_2 \in B$ such that $f(c_1) = a_1$ and $f(c_2) = a_2$. Then using $g = hf$ we have $h(a_1*a_2) = h(f(c_1)*f(c_2)) =
h(f(c_1*c_2)) = g(c_1*c_2) = g(c_1)*g(c_2) = hf(c_1)*hf(c_2) = h(a_1)*h(a_2)$.

\[thm:conjugation\_criterion\] If $X$ is a conjugation quandle, then $X \cong {\rm inn}(Y)$ for some quandle $Y$.

Let $X = x^G$ be a conjugation quandle for some finite group $G$ and element $x \in G$. Define $f \in {\rm Aut}(G)$ by $f(a) = x^{-1}ax$, $a \in G$ and let $Y$ be the generalized Alexander quandle ${\rm GAlex}(G,f)$. Define $p: Y \rightarrow X$ by $p(g) = x^g$. An easy calculation shows that $p$ is a homomorphism. The homomorphism $p$ is clearly surjective as is the homomorphism ${\rm inn} : Y \rightarrow {\rm inn}(Y)$. Now we claim that the mapping $\varphi: {\rm inn}(Y) \rightarrow X$ given by $\varphi(R_g) = x^g$ is a bijection and hence by Lemma \[lem:triangle\] it is a quandle isomorphism. The see this we note that $R_g = R_h$ holds if and only if $a*g = a*h$ for all $a \in G$. That is, if and only if $x^{-1} a g^{-1} x g = x^{-1} a h^{-1} x h$ which by cancellation is equivalent to $x^g = x^h$. This completes the proof.

[ We note that if $Y$ is connected then $X = {\rm inn}(Y)$ is a conjugacy class in ${\rm Inn}(Y)$. So for connected $X$ the converse of Theorem \[thm:conjugation\_criterion\] holds. In fact, by using Lemma 25 in [@Eis3] if $X$ is a connected conjugation quandle one may even find a connected quandle $Y$ such that $X \cong {\rm inn}(Y)$. However, we do not assume connectivity for the results of this section. ]{}

The following is immediate from Theorems \[thm:main\] and \[thm:conjugation\_criterion\].

\[ext\_constant\] If $\phi: Q \times Q \rightarrow A$ is a 2-cocycle of a quandle $Q$ with an abelian coefficient group $A$ such that the extension $X=E(Q,A,\phi)$ is a conjugation quandle, then $\Phi_\phi(K)$ is constant for all classical knots $K$.

Theorem \[thm:conjugation\_criterion\] and Corollaries \[cor:no-inn\], \[cor:noY\] imply the following as well.

\[cor:non-faith\] The following Rig quandles are not conjugation quandles: $$Q(8,1),  \  Q(12,2), \ Q(24,1), \ Q(24,7), \  Q(32,1), \ Q(32, 9),  \ Q(36, 1).$$

A criterion for a finite connected quandle to be a conjugation quandle {#sec:VC}
======================================================================

We are grateful to Leandro Vendramin for suggesting the criterion in this section for a connected quandle to be a conjugation quandle and for pointing out the relevance of Lemma 1.8 in [@HV] to its proof.

The development here is from the paper [@HV] by Vendramin and Heckenberger as well as the paper [@GHV]. For a (finite) quandle $X=(X,*)$, its [*enveloping group*]{} is the group $G_X$ with presentation $$G_X =\langle x_i , i \in X \ |\  x_{i*j} = x_j^{-1}x_i x_j, i, j \in X \rangle$$ and with the natural mapping $$\partial : X \rightarrow G_X,\  i \mapsto x_i.$$ This group has been defined and used several times previously, Joyce [@Joyce] denoted it by $ {\rm Adconj}(X)$, Fenn and Rourke called it the [*associated group*]{} and denoted it by ${\rm As}(X)$ [@FR], and Eisermann [@Eis4] called it the [*adjoint group*]{} and denoted it by ${\rm Adj}(X)$. The group $\overline{G_X}$ obtained from $G_X$ by adding the relations ${x_i^{n_i}, i \in  X}$ where $n_i$ is the order of $R_i$ is called the [*finite enveloping group*]{}. That this group is finite was proved in [@GHV], Lemma 2.17. As in [@HV] we write $$\pi : G_X \rightarrow \overline{G_X}$$ for the canonical surjection. Let $$\rho : X \rightarrow  \overline{G_X}$$ be the natural mapping from $X$ to $\overline{G_X}$, that is $\rho = \pi \partial$. It is clear that $\rho$ is a quandle homomorphism to $\overline{G_X}$ considered as a quandle under conjugation. Then we have the following theorem.

\[thm:VC\] (Vendramin’s Criterion [@Leandro2]) A connected quandle $X$ is a conjugation quandle if and only if $\rho : X \rightarrow  \overline{G_X}$ is an injection.

It is clear that the mapping $\partial : X \rightarrow G_X, i \mapsto x_i$ is universal for quandle homomorphisms $f:X\rightarrow G$ where G is considered as a conjugation quandle. That is, for any such $f$ there is a unique group homomorphism $g: G_X \rightarrow G$ such that $f = g\partial$. It then follows that if $f:X\rightarrow G$ is an injection, that is, if $X$ is a conjugation quandle then $\partial$ must also be an injection. By Lemma 1.8 in [@HV] $\partial$ is injective implies that $\rho$ is injective. Since $X$ is connected the converse is clear.

Clearly each faithful connected quandle is a conjugation quandle. Among the 790 connected quandles of order at most 47 there are precisely 66 non-faithful quandles. Using /Rig and Vendramin’s Criterion we find that exactly 30 of the 66 non-faithful Rig quandes are conjugations quandles. These are found in Table \[table:conj\].

$$\begin{array}{llllllll}
Q(12,1) & Q(20,3) & Q(24,3) & Q(24,4) & Q(24,5) & Q(24,6) \\
Q(24,14) & Q(24,16) & Q(24,17) & Q(30,1) & Q(30,16) & Q(32,5) \\
Q(32,6) & Q(32,7) & Q(32,8) & Q(36,1) & Q(36,4) & Q(36,17) \\
Q(36,20) & Q(36,21) & Q(36,28) & Q(36,30) & Q(40,12) & Q(40,13) \\
Q(40,17) & Q(40,18) & Q(40,19) & Q(40,20) & Q(42,12) & Q(42,21) \\

\end{array}$$

These are listed in [@Edwin1] together with groups in which they are conjugation quandles. Furthermore each of these 30 quandles is an abelian extension of index 2 of the form ${\rm inn}: Q \rightarrow {\rm inn}(Q)$, as indicated in Table \[table:innconj\]. By Theorem \[ext\_constant\] the corresponding cocycle invariants of these extensions are constant for all knots. Note for example that $Q(12,8)$ has three different such cocycle extensions. This is possible since $H^2_Q(X, {{\mathbb Z}}_2) \cong {{\mathbb Z}}_2 \times {{\mathbb Z}}_2 \times {{\mathbb Z}}_2$.

$$\begin{array}{llllll}
{\rm inn}(Q(12,1)) & \cong & Q(6,1) & {\rm inn}(Q(20,3)) & \cong & Q(10,1) \\
{\rm inn}(Q(24,3)) & \cong & Q(12,6) & {\rm inn}(Q(24,4)) & \cong & Q(12,5) \\
{\rm inn}(Q(24,5)) & \cong & Q(12,8) & {\rm inn}(Q(24,6)) & \cong & Q(12,9) \\
{\rm inn}(Q(24,14)) & \cong & Q(12,7) & {\rm inn}(Q(24,16)) & \cong & Q(12,8) \\
{\rm inn}(Q(24,17)) & \cong & Q(12,8) & {\rm inn}(Q(30,1)) & \cong & Q(15,2) \\
{\rm inn}(Q(30,16)) & \cong & Q(15,7) & {\rm inn}(Q(32,5)) & \cong & Q(16,4) \\
{\rm inn}(Q(32,6)) & \cong & Q(16,4) & {\rm inn}(Q(32,7)) & \cong & Q(16,5) \\
{\rm inn}(Q(32,8)) & \cong & Q(16,6) & {\rm inn}(Q(36,1)) & \cong & Q(18,1) \\
{\rm inn}(Q(36,4)) & \cong & Q(18,2) & {\rm inn}(Q(36,17)) & \cong & Q(18,8) \\
{\rm inn}(Q(36,20)) & \cong & Q(18,9) & {\rm inn}(Q(36,21)) & \cong & Q(18,10) \\
{\rm inn}(Q(36,28)) & \cong & Q(18,12) & {\rm inn}(Q(36,30)) & \cong & Q(18,11) \\
{\rm inn}(Q(40,12)) & \cong & Q(20,9) & {\rm inn}(Q(40,13)) & \cong & Q(20,9) \\
{\rm inn}(Q(40,17)) & \cong & Q(20,10) & {\rm inn}(Q(40,18)) & \cong & Q(20,9) \\
{\rm inn}(Q(40,19)) & \cong & Q(20,6) & {\rm inn}(Q(40,20)) & \cong & Q(20,5) \\
{\rm inn}(Q(42,12)) & \cong & Q(21,6) & {\rm inn}(Q(42,21)) & \cong & Q(21,9) \\

\end{array}$$

Some of the quandles in Table \[table:conj\] are missed in Corollary \[cor:rig\] since they are homomorphic images under ${\rm inn}$ of quandles that are too large to be found easily with random searches. We were able to identify these using Theorem \[thm:VC\].

Is there a connected quandle $Q$ with a 2-cocycle $\phi: Q\times Q \rightarrow A$ with $|A| > 2$ such that $E(Q,\phi,A)$ is a conjugation quandle?

Relations among some types of epimorphisms {#sec:rel}
==========================================

In this section, we examine relations among the epimorphisms used in the proofs of the main results; coverings, abelian extensions and inner representations. As we pointed out, abelian extensions and inner representations are coverings. We show that there is no other implication:

[(i)]{} There exists an abelian extension that is not an inner representation.

[(ii)]{} There exists an inner representation that is not an abelian extension.

[(iii)]{} There exists a covering that is neither an abelian extension nor an inner representation.

\(i) Suppose $X$ is faithful, and let $E=E(X, A)$ be an abelian extension. Then $R_{(x,a)}=R_{(y,b)}$ if and only if $x=y$, where $(x,a), (y,b) \in E=X \times A$. Hence ${\rm inn}: E \rightarrow {\rm inn}(E)$ is an abelian extension, where ${\rm inn}(E)$ is isomorphic to $X$. If there is a proper non-trivial subgroup $C$ in $A$, then in [@CSV] it was proved that there is a sequence of abelian extensions $E(X, A) \rightarrow E(X, A/C) \rightarrow X $. If $X$ is faithful, therefore, $E(X, A) \rightarrow E(X, A/C)$ is not the inner representation. An example is given by $\pi: Q(24,2) \rightarrow Q(12,2)$ where $Q(24,2) = E(Q(6,2),{{\mathbb Z}}_4)$ and $Q(24,2) = E(Q(12,2),{{\mathbb Z}}_2)$ as noted in [@CSV].

\(ii) This was given in [@CSV]: there it is shown that ${\rm inn}: Q(30, 4) \rightarrow Q(10,1)$ is not an abelian extension since $H_2^Q(Q(10,1))\cong {{\mathbb Z}}_2$. Another proof that this is not an abelian extension follows from Lemma \[lem:criterion\] below.

\(iii) Consider the product of the two mappings mentioned above: $${\rm inn} \times \pi: Q(30,4) \times Q(24,2) \rightarrow Q(10,1) \times Q(12,2),$$ where ${\rm inn} \times \pi:  (x,y) \mapsto ({\rm inn}(x), \pi(y))$. We know that ${\rm inn}: Q(30,4) \rightarrow Q(10,1)$ is an inner representation that is not an abelian extension and $\pi: Q(24,2) \rightarrow Q(12,2)$ is an abelian extension that is not an inner representation. We claim that ${\rm inn} \times \pi$ is a covering but not an abelian extension and not an inner representation.

To see that ${\rm inn} \times \pi$ is not an inner representation note that if ${\rm inn}_Z$ denotes the inner representation on a quandle $Z$, then $${\rm inn}_{X \times Y} = {\rm inn}_X \times {\rm inn}_Y.$$ Thus since $\pi$ is not an inner representation neither is ${\rm inn} \times \pi$.

Since both ${\rm inn}$ and $\pi$ are coverings it is easy to see that their product is a covering. It is more difficult to show that ${\rm inn} \times \pi$ is not an abelian extension. For this we use the following lemma.

\[lem:criterion\] Let $X$ and $Y$ be finite quandles. Assume $f : Y \rightarrow X$ is an abelian extension, $x \in X$, $y \in f^{-1}(x)$ and $\beta \in {\rm Inn}(Y)$. If $\beta(y) = y$ then $\beta$ acts as the identity on the fiber $f^{-1}(x)$.

We take $Y = E(X,A,\phi) = X \times A$, and let $\pi$ be the projection onto $X$. To simplify notation we write the quandle products in $X$ and $Y$ by juxtaposition. Moreover, we define $y_1y_2y_3 = (y_1y_2)y_3$ and inductively, for $n > 3$ we set $y_1y_2 \cdots y_{n+1} =( y_1y_2 \cdots y_{n})y_{n+1}$.

Since $Y$ is finite for $\beta \in {\rm  Inn}(Y)$ we may write $$\beta  = R_{(x_n,a_n)} R_{(x_{n-1},a_{n-2})} \cdots  R_{(x_1,a_1)}.$$ Then we have for $(z,b) \in X \times A$ $$\beta(z,b) = (z,b)(x_1,a_1)(x_2,a_2) \cdots (x_n,a_n) .$$ Using the definition of the product in $E(X,A,\phi)$ it follows that $$\beta(z,b) = (zx_1x_2 \cdots x_n, b \, \phi(z,x_1) \, \phi(zx_1,x_2) \, \cdots \, \phi(zx_1x_2\cdots x_{n-1},x_n)).$$ If we assume that $\beta(z,b) = (z,b)$ then we have $$zx_1x_2 \cdots x_n = z$$ and $$\phi(z,x_1) \, \phi(zx_1,x_2) \, \cdots \, \phi(zx_1x_2\cdots x_{n-1},x_n) = 1.$$ The fiber $\pi^{-1}(z) $ is equal to $ \{(z,c): c \in A \}.$ It follows that $\beta(z,c) = (z,c)$ for all $c \in A$, that is, $\beta$ is the identity on the fiber $\pi^{-1}(z)$, as claimed.

Now returning to the claim that $${\rm inn} \times \pi: Q(30,4) \times Q(24,2) \rightarrow Q(10,1) \times Q(12,2)$$ is not an abelian extension, we note first that computation shows that for ${\rm inn}: Q(30,4) \rightarrow Q(10,1)$ there exist $z \in Q(10,1)$, $\alpha \in {\rm Inn}(Q(30,4))$ and $x,y \in {\rm inn}^{-1}(z)$ such that $\alpha(x) = x$ but $\alpha(y) \neq y$. This gives an alternative proof via Lemma \[lem:criterion\] that ${\rm inn}: Q(30,4) \rightarrow Q(10,1)$ is not an abelian extension. Now to show that $ {\rm inn} \times \pi: Q(30,4) \times Q(24,2) \rightarrow Q(10,1) \times Q(12,2)$ is not an abelian extension, we note that using the same $x$ and $y$ we have that $(x,w)$ and $(y,w)$ for any $w \in Q(24,2)$ lie in the same fiber of ${\rm inn} \times \pi$. It is easy to see that ${\rm Inn}(X \times Y) ={\rm  Inn}(X) \times {\rm Inn}(Y)$ for any quandles $X$ and $Y$. Thus taking $\alpha \in {\rm Inn} (Q(30,4))$ and the identity ${\rm id} \in {\rm Inn}(Q(24,2))$ we have that $\beta = \alpha \times {\rm id} \in {\rm  Inn} (Q(30,4) \times Q(24,2)).$ Clearly $\beta(x,w) = (x,w)$ and $\beta(y,w) \neq (y,w)$. So by Lemma \[lem:criterion\] we have that $ {\rm inn} \times \pi$ is not an abelian extension.

Acknowledgements {#acknowledgements .unnumbered}
----------------

MS was partially supported by NIH R01GM109459. The content of this paper is solely the responsibility of the authors and does not necessarily represent the official views of NIH.

[99]{}

Andruskiewitsch, N.; Graña, M., [*From racks to pointed Hopf algebras*]{}, Adv. in Math., [178]{} (2003) 177–243.

Carter, J.S.; Elhamdadi, M.; Nikiforou, M.A.; Saito, M., [*Extensions of quandles and cocycle knot invariants,*]{} [J. Knot Theory Ramifications,]{} [12]{} (2003) 725–738.

Carter, J.S.; Jelsovsky, D.; Kamada, S.; Langford, L.; Saito, M., [*Quandle cohomology and state-sum invariants of knotted curves and surfaces,*]{} Trans. Amer. Math. Soc., 355 (2003) 3947–3989.

Carter, J.S.; Jelsovsky, D.; Kamada, S.; Saito, M., [*Computations of quandle cocycle invariants of knotted curves and surfaces,*]{} Adv. in Math., [157]{} (2001) 36-94.

Carter, J. S.; Jelsovsky, D.; Kamada, S.; Saito, M., [*Quandle homology groups, their Betti numbers, and virtual knots,*]{} J. Pure Appl. Algebra, [157]{} (2001) 135–155.

Carter, J.S.; Kamada, S.; Saito, M., Surfaces in $4$-space, Encyclopaedia of Mathematical Sciences, Vol. 142, Springer Verlag, 2004.

Carter, J.S.; Kamada, S.; Saito, M., [*Geometric interpretations of quandle homology,*]{} J. Knot Theory Ramifications, 10 (2001) 345-386.

Clark, W.E., [*Some properties of small connected quandles,* ]{} <http://shell.cas.usf.edu/~saito/QuandleColor/property.pdf>.

Clark, W.E., [*List of non-faithful Rig quandles,*]{} <http://shell.cas.usf.edu/~saito/QuandleColor/NonFaithfulRigQuandles.txt>.

Clark, W.E., [*Sequences of quandle extensions,*]{} <http://shell.cas.usf.edu/~saito/Sequence>.

Clark, W.E.; Saito, M.; Vendramin, L., [*Quandle coloring and cocycle invariants of composite knots and abelian extensions,*]{} arXiv:1407.5803, to appear in J. Knot Theory Ramifications.

Conway, J.H. [*An Enumeration of Knots and Links, and Some of Their Algebraic Properties,*]{} In J. Leech (editor), Computational Problems in Abstract Algebra. Oxford, England. Pergamon Press (1970) 329–358.

Eisermann, M., [*Quandle coverings and their galois correspondence,*]{} Fund. Math. 225 (2007) 103–167.

Eisermann, M., [*Homological characterization of the unknot,*]{} J. Pure Appl. Algebra, [177]{} (2003) 131–157.

Fenn, R.; Rourke, C., [*Racks and links in codimension two*]{}, J. Knot Theory Ramifications, [1]{} (1992) 343–406.

Fenn, R.; Rourke, C.; Sanderson, B., [*Trunks and classifying spaces,*]{} Appl. Categ. Structures, [3]{} (1995) 321–356.

Graña, M.; Heckenberger, I.; Vendramin,L., [*Nichols algebras of group type with many quadratic relations.*]{}, Adv. Math., 227 (5) (2011) 1956–1989.

Heckenberger, I.; Vendramin, L., [*Nichols algebras over groups with finite root system of rank two II*]{}, J. Group Theory, 17 no.6 (2014), 1009–1034.

Joyce, D., [*A classifying invariant of knots, the knot quandle,*]{} J. Pure Appl. Alg., 23 (1983) 37–65.

Kauffman, L.H., [*Virtual knot theory,*]{} European J. Combin., 20 (1999) 663–690.

Manturov, V., [Knot theory,]{} Chapman & Hall/CRC, 2004.

Matveev, S., [*Distributive groupoids in knot theory.*]{} (Russian) Mat. Sb. (N.S.) 119(161) (1982) 78–88 (160).

Nosaka, T., [*On homotopy groups of quandle spaces and the quandle homotopy invariant of links,*]{} Topology Appl., [158]{} (2011) 996–1011.

Przytycki, J.H., [*$3$-colorings and other elementary invariants of knots,*]{} Knot theory (Warsaw, 1995), 275–295, Banach Center Publ., 42, Polish Acad. Sci., Warsaw, 1998.

Przytycki, J.H., [*From 3-moves to Lagrangian tangles and cubic skein modules,*]{} Advances in Topological Quantum Field Theory, Proceedings of the NATO ARW on New Techniques in Topological Quantum Field Theory, John M. Bryden (ed), 2004, 71–125.

Takasaki, M., [*Abstraction of symmetric transformation*]{} (in Japanese), Tohoku Math. J. 49 (1942/3) 145–207.

Vendramin, L., [*Rig – A GAP package for racks and quandles.*]{} Available at <http://github.com/vendramin/rig/>.

Vendramin, L., [*On the classification of quandles of low order*]{}, J. Knot Theory Ramifications, [21]{} no.9 (2012), 1250088.

Vendramin, L., personal communication.


Percy Chivers

Percy Chivers (15 August 1908 – 11 July 1997) was an Australian cricketer. He played three first-class cricket matches for Victoria between 1929 and 1930.

See also
 List of Victoria first-class cricketers

References

External links

Category:1908 births
Category:1997 deaths
Category:Australian cricketers
Category:Victoria cricketers
Category:Cricketers from Melbourne


Human duodenal spheroids for noninvasive intracellular pH measurement and quantification of regulation mechanisms under physiological conditions.
Three-dimensional cell cultures (spheroids) of biopsies of human duodenum were used to develop a new noninvasive method for studying intercellular and intracellular mechanisms. Through examinations of intracellular pH regulation, high functional similarity to native tissue could be shown, as already evidenced morphologically. A special microperfusion chamber was developed to fix individual spheroids physically to a nylon net, via laminar perfusion flow through the chamber. A significant improvement over current fixation methods was shown by the increase of cell viability almost up to 100%. Viability of the spheroids was confirmed by trypan blue exclusion, by a LIVE/DEAD viability/cytotoxicity kit, and by BCECF distribution. Intracellular pH was measured by use of the pH-sensitive fluorescence dye BCECF. To investigate the intracellular pH regulation, spheroid-like vesicles were acidified by NH4Cl prepulse technique. The subsequent active intracellular pH recovery was blocked with Na+-free Krebs Henseleit (KH) solution, with amiloride KH (inhibitor of the Na+-H+-exchanger), or with H2DIDS KH (inhibitor of the HCO3(-)-Cl(-)-exchanger and Na+-HCO3(-)-cotransporter). The intracellular pH of the spheroids was 7.31 +/- 0.05. pH-backregulation after acidification was prevented by sodium-free buffer, amiloride, and H2DIDS. These experiments indicated the presence of a Na+-H+-exchanger and a Na+-HCO3(-)-cotransporter. In conclusion, the human duodenal spheroid is an excellent physiological system for in vitro studies of the human duodenum.


2017 NFL Preview: Not exactly 'all in,' but Packers are chasing a title

Shutdown Corner is previewing all 32 teams as we get ready for the NFL season, counting down the teams one per weekday in reverse order of our initial 2017 power rankings. No. 1 will be revealed on Aug. 2, the day before the Hall of Fame Game kicks off the preseason.

A quarter into a Monday night game against the Philadelphia Eagles last Nov. 28, the Green Bay Packers were nowhere near a top four NFL team.

Green Bay had lost four in a row and five of six. Aaron Rodgers had finally started to break out of a strange, long slump the week before, but the defense gave up 42 points in a loss to the Washington Redskins. The week before that, Green Bay gave up 47 points in a loss to the Tennessee Titans. Green Bay was 4-6, and tied with the Philadelphia Eagles 7-7 after trading scores in the opening quarter on “Monday Night Football.” The Packers brass would scoff at the notion later, but it sure seemed Mike McCarthy’s seat was plenty hot at that point.

One play didn’t turn around the Packers season, but it seemed significant even in the moment. Rodgers threw a 20-yard touchdown pass to Davante Adams, one of the prettiest tight-window passes you’ll ever see.

Then Rodgers, in his cocksure way, extended his arms and it seemed like he was letting the Packers know he was serious about his “run the table” prediction.

The Packers never trailed again in that game. In fact, they trailed for just 1:44 in all of December, going 4-0 including a blowout win against a good Seattle Seahawks team. They won the NFC North in the season finale at the Detroit Lions, destroyed a strong New York Giants team in their playoff opener and upset a 13-3 Dallas Cowboys team on the road in round two. The magical ride crashed hard at the Atlanta Falcons in the NFC championship game.

Still, after a mysterious stretch of mediocre football – both by the Packers and Rodgers – it seemed something clicked.

We can imagine what might have happened had the Packers 4-6 start turned into 7-9 or worse. Instead, Green Bay made an NFC title game and could start thinking about bigger things. That eight-game winning streak allowed the Packers to go into this offseason confident they had the pieces to win a title. It was a big change from 4-6.

“I don’t think we need to rebuild. We need to reload,” Rodgers said right after the loss to the Falcons. “We’ve just got to make sure we’re going all-in every year to win. And I think we can take a big step this offseason.”

Whether or not those words had any effect, the Packers had perhaps their most aggressive offseason with notoriously conservative general manager Ted Thompson. It wasn’t exactly “all in,” but it was an unexpected big bet by the tightest player at the table.

Thompson signed five veteran free agents. Most offseasons, that number hovers around zero in Green Bay. The key signing was a three-year, $21 million deal to tight end Martellus Bennett.

It’s a little scary to assume an eight-game winning streak means the Packers are all the way back. From Nov. 1 of 2015 through late November of last year the Packers were 9-13, including playoffs. We’re only nine months removed from a million “What’s wrong with Aaron Rodgers?” stories. And those stories were true; for a long stretch Rodgers did not look like himself. Over a 15-game stretch starting Nov. 1, 2015, Rodgers was last in the NFL in completion percentage (58.1), 24th in passing yards per game (233.3) and 25th in passer rating (84.0) according to NFL.com. Rodgers had worse numbers than Brian Hoyer over a 12-game stretch, and failed to have one 300-yard game in that time. There were legitimate excuses for Rodgers’ slump, but it was a slump.

Rodgers can also carry a team unlike any other quarterback. Over that eight-game winning streak late last season, Rodgers had 2,385 yards, 21 touchdowns and one interception for a 117.9 passer rating. It’s impossible to believe any quarterback could play much better over eight games.

We figure Rodgers will be that Rodgers again as we start this season. He’ll make plays that amaze us, and cover up Green Bay’s deficiencies like a patched-together running game and a pass defense that wasn’t able to stop anyone last season. The front office realized that getting Rodgers some help through free agency is not illegal, so maybe he can be even better than last season.

The Packers have mostly been Super Bowl contenders in the Rodgers era, and it seems like they are again. Whatever happened over that long slump seems to be a distant memory.

Aaron Rodgers is looking to make his second Super Bowl. (AP)

The Packers have gotten very little production from their tight ends since Jermichael Finley had to retire after the 2013 season. Aaron Rodgers’ best season was 2011, and that was Finley’s best season too (767 yards, eight touchdowns). Martellus Bennett will be perhaps the best tight end Rodgers has had. Jared Cook had a solid 2016 season on a one-year deal, and Bennett should be an upgrade. The other free-agent additions – tight end Lance Kendricks, cornerback Davon House, guard Jahri Evans and defensive lineman Ricky Jean Francois – weren’t big-money deals but they should all help. Green Bay’s biggest move was to retain pass rusher Nick Perry on a five-year, $60 million deal. Green Bay did lose some good players in free agency, most notably center J.C. Tretter and guard T.J. Lang. Safety Micah Hyde signed a big deal with the Buffalo Bills. Running back Eddie Lacy was allowed to leave after two disappointing seasons. The Packers’ first four picks were used on defense, and three other picks were used on running backs. While the Packers didn’t have an incredible offseason, it was exciting to see GM Ted Thompson dip into free agency for a change. Grade: B-minus

While a lot was made of the Packers’ uncreative play-calling when their offense was sputtering along, they have the receivers to make up for it. Jordy Nelson’s successful return from ACL surgery was crucial to the offense. Davante Adams rebounded from a terrible 2015 with a 997-yard, 12-touchdown season. Randall Cobb’s numbers have declined since a career year in 2014, but he’s still a dangerous and versatile weapon. Add tight end Martellus Bennett to the mix and the Packers have given Aaron Rodgers perhaps the best overall set of receiving targets in the league.

The Packers were the only team in the NFL to allow more than eight yards per pass play, at 8.1. Only one team (New Orleans Saints) allowed more than the 4,308 passing yards the Packers gave up. Green Bay gave up 32 passing touchdowns, and only two teams allowed more. It’s hard to believe a 10-win team could be so bad stopping the pass. A ton of cornerback injuries played a big role in that. By the end of the season they had to miscast undrafted LaDarius Gunter as their No. 1 cornerback, and that was a significant problem in the playoffs against Dez Bryant and Julio Jones. Perhaps better health and the additions of cornerbacks Davon House (free agency) and Kevin King (second-round draft pick) and safety Josh Jones (second-round pick) will help. The Packers probably need the secondary to be better if they are to make a Super Bowl.

We’re running out of ways to describe Aaron Rodgers’ greatness. His 104.1 passer rating is the best ever, and nobody else is better than 99.6. The last time he threw for double-digit interceptions in a season was 2010. He’s a two-time MVP and will be a candidate to win a third MVP every season for the rest of his prime. Last season he led the NFL with 40 touchdown passes, surprisingly the first time he has ever led the league in that category. Rodgers turns 34 on Dec. 2, so he has a few more years of great football left. Rodgers is one of the best ever, and even though he went through a rough patch for a calendar year, the finish to last season shows he’s as good as ever.

Clay Matthews is still the star of the defense, capable of taking over a game like no other defender on Green Bay’s roster. He hasn’t been the same dominating force the past couple seasons, however. In 2014 he had 11 sacks. In 2015, often having to play out of position at middle linebacker, he had 6.5 sacks. Then last season, playing in only 12 games due to injury, he posted a career-worst five sacks. Hamstring and shoulder injuries were Matthews’ biggest issues last season, and his snaps were limited even when he did play. Matthews had three sacks in his first three games last season before injuries hit, so it’s fair to assume with health he could get back to double-digit sacks. Matthews is 31 years old, so he shouldn’t see a big dip in his play yet.

From Yahoo’s Andy Behrens: “Ty Montgomery has been so commonly discussed as an underrated fantasy RB that he’s become a bit overrated. I challenge you to find a fantasy content provider that hasn’t written a Montgomery hype piece (or three). If you want approval from other league managers in your draft room, then you should definitely target Montgomery. Everyone is quick to type ‘Nice pick!’ after he’s selected. But if you want a high-volume running back with a clear path to double-digit touchdowns, you should probably look elsewhere. Montgomery only delivered one game last season with 10 or more carries. He ran for 222 yards in his two games against Chicago, and just 235 yards in his other 13 games. No one should be surprised if BYU rookie Jamaal Williams holds down a significant supporting role in this team’s backfield. If Montgomery’s average draft position climbs to a spot in which you need him to finish as a 1,000-yard back, veer away to another back.” [Check out Yahoo’s Pressing Questions for the fantasy outlook on the Packers.]

The Packers, who have made the playoffs eight straight seasons, have had four of their last five seasons end in a road playoff loss. The Packers have one of the best home-field advantages in football, but Aaron Rodgers has never played an NFC championship game at home. “We gotta get one of these at home, man,” Rodgers told Jordy Nelson on the bench near the end of a blowout loss to the Falcons in last season’s NFC title game, as shown by NFL Films. Getting a No. 1 seed is high on the Packers’ priority list, and the road to that starts with a Week 1 home game against a fellow NFC contender, the Seattle Seahawks.

CAN TY MONTGOMERY BE A FULL-TIME RUNNING BACK?

The Packers were running out of options at tailback last season when they tried receiver Ty Montgomery there. It was strange to see No. 88 taking handoffs, but the Packers might have discovered something. Montgomery rushed for 457 yards the rest of the season with a 5.9-yard average. It’s hard to know if he can keep that up over a full season, but it helps to have a full offseason to learn the position. That will especially help in pass protection, an area Montgomery (understandably) struggled in. If Montgomery fails –and that’s possible, considering Montgomery rarely was given a full workload as an emergency substitute last season – rookies Aaron Jones and Jamaal Williams will battle to be next in line. Even with an all-time great quarterback, coach Mike McCarthy likes to have balance with a good run game, but he needs a reliable running back to make that work.

The Packers are a rare team. No matter what else is swirling around them, they go into every season feeling they can win a Super Bowl because they have the great Aaron Rodgers. An incredible stretch at the end of last season should make Packers fans even more confident about their team’s chances of making it back to a Super Bowl. Green Bay can get a No. 1 seed, run through the NFC playoffs and then not feel scared by any opponent once they are in a Super Bowl. Having Rodgers allows them to have unending confidence.

This is still a team that finished the 2015 regular season on a 4-6 run and started 2016 on another 4-6 run. There were some ugly blowout losses during the first 10 weeks of last season. While whatever bothered the offense in that stretch seems to have been fixed, it’s dangerous to ignore the bad games and focus on the eight games Green Bay won at the end of the season. While we just assume a team with Aaron Rodgers will win the NFC North, we’ve seen Rodgers and Green Bay have significant struggles recently. What if that happens again?

If the Packers’ pass defense is fixed, the interior of the offensive line holds up and the run game is at least competent, Green Bay is one of the few teams that could beat New England in a Super Bowl. When Aaron Rodgers is rolling, the Packers can beat anyone. Go back and look at how Green Bay blasted Seattle last season, or the road playoff win against a very good Cowboys team. I think Green Bay picks up where it left off. Because its schedule is easier than some of the other top-shelf teams in the NFC, I can see Green Bay grabbing that home-field advantage Rodgers wants. I’d like the Packers’ chances of winning a conference championship if that happens.


Saturday, December 19, 2009

So this last month has been pretty tough if you're a Flyers fan. Since the Flyers beat the LA Kings a month ago, the Flyers are 3-11-1. John Stevens has been fired, and the trade rumors are starting to swirl. Despite this slide, I'm not ready to hit the panic button yet and declare the season over.

While November and December are traditionally strong months for the Flyers, this year they battled a TON of injuries. On this losing streak they've lost: Blair Betts, Darrol Powe, Jared Ross, Ole Kristian Tollefsen and Ray Emery to injuries. Plus they haven't had Gagne since October.

One of the biggest reason for the losing streak has been the Flyers inability to kill penalties. When you lose one of your top PK men (Betts), it gets a lot harder to kill penalties. Another area where the Flyers have struggled is scoring. Powe, when he was injured, had had 6 goals in 19 games. While not stunning numbers, he was off to the best start of his career, and he is a sparkplug for the 3rd and 4th lines. Powe is set to return next week, as is Gagne and Betts came back last night. Hopefully with the Flyers getting healthy, we'll see their game start to pick up.

The Flyers are also still adjusting to new coach Peter Laviolette. Laviolette brings a whole new style of play with him, and with frequent games, it has been hard for the Flyers to practice the new system. Laviolette likes to run a two man forecheck, and to play aggressively in the offensive zone. This isn't something the Flyers are used to, and it requires a lot of conditioning to execute it well. The Flyers obviously weren't conditioned to run this system under Stevens, so it may take a little time for them to get used to it.

The bottom line is, this team has too much talent not to contend with Pittsburgh and other elite teams. This is a rough patch for them, but they have weathered worse in the past (remember that 10 game losing streak back in 08?). I fully expect the Flyers to recover from this slump (they just played a great game against Pittsburgh last night) and to get on a roll for the second half the season. Keep the faith!


SELECTIONS:

*To say thank you for my picks all week. Please use the BEAST DOME logo below as your DRAFTSTREET ICON.

Salary Cap Team for 4:00 PM game (Check Updates.) If you use the team or rankings above, please use the BEAST DOME logo as your avatar on DRAFTSTREET. If you use my rankings for FAN DUEL please do the same.

Thank you!

STACK OF THE DAY (AUGUST 23rd)

**NEW. NEW. I pick the stack of the day, you pick the hitters and see the points rally up.

Stack of the Day:

*The More BEAST DOME icons I see all-around DRAFTSTREET, the more info I will publish publicly. If you want a team e-mailed use the BEAST DOME icon, and e-mail [email protected] saying you’ve uploaded the icon and want a team.

BET OF THE DAY (AUGUST 23rd:)

Week 12: 6-1

Week 13: 5-2

Week 14: 3-4

Week 15: 1-2

Week 16: 4-3

Week 17: 5-2

Week 18: 5-2

Week 19 : 4-3

Week 20: 5-0

Week 21: 2-1

DET MONEY LINE

The Tigers fighting for a playoff spot at home in late August with Justin Verlander. That already sounds like a lock Win. How about against the worst team in the AL East, Toronto. On the mound is the very hittable lefty, J.A. Happ. Give me Detroit.

DAILY PITCHER RANKINGS (AUGUST 23rd):

ELITE STARTS:

1. JUSTIN VERLANDER VS. TOR

GOOD STARTS:

2. JAKE WESTBROOK VS. HOU

3. JOHNNY CUETO @ PHI

4. COLE HAMELS VS. CIN

5. TOMMY HANSON @ SF

ROLL THE DICE:

6. C.J. WILSON @ BOS

7. BARRY ZITO VS. ATL

8. ROY OSWALT VS. MIN

RISKY:

9. ALEX COBB VS. OAK

10. FRANKLIN MORALES VS. LAA

11. TYLER CHATWOOD @ NYM

12. SCOTT DIAMOND @ TEX

13. TYSON ROSS @ TB

14. J.A. HAPP @ DET

15. DALLAS KEUCHEL @ STL

16. COLLIN MCHUGH VS. COL

Muntradamus is a Fantasy Sports Expert who specializes in Fantasy Football/Baseball/Basketball. He is currently in his first season as a Pro on FantasyPros where he is the only expert to have Adam Dunn, Mike Aviles, and Wilin Rosario in his top 20 at each position. Muntradamus not only leads many to championships with his great advice, and outlooks into the future. He is also consistently on the top of leaderboards in DraftStreet.


Q:

column expands with "textbox width given in percentage" within table

I am using the following code
--- EDIT @annakata : Adding the complete code

  <TITLE> New Document </TITLE>

                                                        Last
                                                        First
                                                        Middle
                                                        Email

  <input maxlength="256"
        type="text" name="additionalContact" style="width: 90%;"
        value="aaaaaaaaaaa aaaaaaaaaaaaaaaaaaa aaaaaaaa aaaaaaaaaaaa aaaaa aaaaaaaaaaaaaaa" >
          <input maxlength="256"
        type="text" name="additionalContact" style="width: 90%;"
        value="aaaaaaaaaaaaaaaaaaaaaaaaaa" >
          <input maxlength="256"
        type="text" name="additionalContact" style="width: 90%;"
        value="aaaaaaaaaaaa aaaaaaaa aaaaaaaaaab bbbbbbbb bbbbbbbbbbbbbbbb" >
          <input maxlength="256"
        type="text" name="additionalContact" style="width: 90%;"
        value="aaaaaaaa  aaaaaa aaaaaaa aaaaa" >

The problem arises when the user enters a large text, the text box expands and table is distorted. If I put size="30" in there the expansion problem is fixed but on different resolutions, the structure behaves diffferently.
Any fixes or wok around to achieve the desired gracefully ?

A:

ok, I can clearly see this replicated in IE now.
It appears to be a bug in IE (IE has notorious issues with percentages) dependant on a scenario where:

the input is in a table
AND the input is sized in percentage
AND the input has a default value (i.e. in the markup) which exceeds the width

Therefore a couple of obvious workarounds exist:

put the input in a floating
 layout instead (I would do this if it's semantically correct)
size width in units like em (preferably) or px (if your design can afford it)
inject the value with client script after the load (this is a poor solution, don't do this)

and one non-obvious workaround which I would recommend if you require both the table and the percentages:

an IE only CSS rule width: expression(this.parentNode.offsetWidth*0.9);

Hope this helps.
(previous answer requesting code deleted)

Edit: code snippet based on your sample, works in IE + FF

                    Last
                    First
                    Middle
                    Email

                    <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaaaaa aaaaaaaaaaaaaaaaaaa aaaaaaaa aaaaaaaaaaaa aaaaa aaaaaaaaaaaaaaa" />

                    <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaaaaaaaaaaaaaaaaaaaa" />

                    <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaaaaaa aaaaaaaa aaaaaaaaaab bbbbbbbb bbbbbbbbbbbbbbbb" />

                    <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaa  aaaaaa aaaaaaa aaaaa" />

        <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaaaaa aaaaaaaaaaaaaaaaaaa aaaaaaaa aaaaaaaaaaaa aaaaa aaaaaaaaaaaaaaa" >

        <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaaaaaaaaaaaaaaaaaaaa" >

        <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaaaaaa aaaaaaaa aaaaaaaaaab bbbbbbbb bbbbbbbbbbbbbbbb" >

        <input maxlength="256" type="text" name="additionalContact" value="aaaaaaaa  aaaaaa aaaaaaa aaaaa" >


The 170-acre preserve has been home to several flicks, including The Notebook, but was also the filming location for Mel Gibson’s 2000 movie The Patriot. After it wrapped, pieces of a fake church and bridge were left behind for fans to stumble on. The gardens are currently closed due to flooding, so it remains to be seen if the props are still standing.


The $40 Secret to Jennifer Lopez’s Perfect Eyebrows

More than two decades after she first burst onto the scene and into our hearts with Selena, Jennifer Lopez still manages to keep us on our toes. And when it comes to her style, she’s the gift that keeps on giving. Take our December cover shoot, for example, when she decided to kick things up a notch and wear a glittering green Valentino Haute Couture cape — which was on set as part of an entire ensemble — as her own version of the “naked dress.”

While her bold fashion sense shouldn’t come as a surprise — after all, her plunging Versace dress at the 2000 Grammys is one of the most iconic red carpet looks of all time — we still constantly find ourselves in awe of the age-defying chameleon’s style choices. So it was only fitting that we ask Lopez to dish on what’s topping her list of must-haves at the moment, wardrobe-wise and beyond. Scroll down for 12 of her current obsessions, from her favorite heels and skincare essentials to the show she can’t get enough of (hint: everyone else seems to agree).

Supreme Court Justice Ruth Bader Ginsburg has been hospitalized with three broken ribs. Ginsburg–all-around political (not to mention fitness) badass, if we may add–is said to have fallen in her office Wednesday evening. Although she was able to return home after her fall, nighttime discomfort sent 85-year-old Ginsburg to George Washington University Hospital Thursday morning, […]

Intense burning or stinging when using a product for the first time VGstockstudio/Shutterstock Whether it’s a cleanser, lotion, or mask—no skin-care product should cause burning or stinging. “This is often a sign that you’re allergic to it,” explains S. Manjula Jegasothy, MD, dermatologist and founder of Miami Skin Institute. Also known as contact dermatitis, this […]


Recovery of Multilayer-Coated Zerodur and ULE Optics for Extreme-Ultraviolet Lithography by Recoating, Reactive-Ion Etching, and Wet-Chemical Processes.
Extreme-ultraviolet lithography requires expensive multilayer-coated Zerodur or ULE optics with extremely tight figure and finish specifications. Therefore it is desirable to develop methods to recover these optics if they are coated with a nonoptimum multilayer films or in the event that the coating deteriorates over time owing to long-term exposure to radiation, corrosion, or surface contamination. We evaluate recoating, reactive-ion etching, and wet-chemical techniques for the recovery of Mo/Si and Mo/Be multilayer films upon Zerodur and ULE test optics. The recoating technique was successfully employed in the recovery of Mo/Si-coated optics but has the drawback of limited applicability. A chlorine-based reactive-ion etch process was successfully used to recover Mo/Si-coated optics, and a particularly large process window was observed when ULE optics were employed; this is an advantageous for large, curved optics. Dilute HCl wet-chemical techniques were developed and successfully demonstrated for the recovery of Mo/Be-coated optics as well as for Mo/Si-coated optics when Mo/Be release layers were employed; however, there are questions about the extendability of the HCl process to large optics and multiple coat and strip cycles. The technique of using carbon barrier layers to protect the optic during removal of Mo/Si in HF:HNO(3) also showed promise.


Long-term exposure of human gingival fibroblasts to cigarette smoke condensate reduces cell growth by modulating Bax, caspase-3 and p53 expression.
Smoking cigarettes increases the risk of oral tissue damage leading to periodontal disease. Gingival fibroblasts, the predominant cell type inhabiting gingival connective tissue, play a critical role in remodeling and maintaining gingival structure. The objective of this study was to investigate the effect of long-term exposure to cigarette smoke on human gingival fibroblast survival/apoptosis and the molecular pathways involved in these cell responses. Human gingival fibroblasts were extracted from healthy non-smokers and cultured in the presence of cigarette smoke condensate (CSC). At the end of each time point, cell growth was evaluated by means of MTT assay. Apoptotic and necrotic gene's expression was investigated by polymerase chain reaction array and by annexin V/propidium iodide staining and cell cycle assays. Western blot was used to investigate Bax and p53 proteins. These tests were supported by caspase 3 activity analyses. High levels of CSC decreased cell growth and deregulated cell cycle progression by increasing the G(0)/G(1) and reducing the S and G(2)/M phases of the gingival fibroblasts. Polymerase chain reaction arrays revealed the activation of several apoptotic genes by CSC, including TNF receptors, caspases, Bax and p53. This was supported by increases in the Bax and p53 protein levels as well as by an elevated activity of caspase-3 in the CSC-exposed cells. Immunofluorescence staining demonstrated that both Bax and caspase-3 displayed a cytosolic and mitochondrial distribution in the CSC-exposed gingival fibroblasts, compared to controls. The damaging effect of CSC on gingival fibroblast growth was also supported by the decrease in interleukin 6 and 8 secretion by the gingival fibroblasts. These results suggest that CSC may contribute to deregulating fibroblast functions. This can compromise fibroblast-epithelial cell interactions, which ultimately increases the risk of gingival tissue damage and the onset of periodontitis.


by Maria Crawford Scott

Want a simple, low-cost way to invest in a firm?

One approach is to go straight to the source. Currently, over 430 companies in the U.S. offer direct purchase plans, allowing you to buy shares directly from the company without a broker. In addition, a growing number of foreign firms also offer direct purchase plans through their ADR programs (for more on these plans, see page 20).

Direct purchase plans are a form of dividend reinvestment plan, but in a traditional dividend reinvestment plan you must first enroll in the plan in order to participate, which means buying your initial shares through a broker.

Direct purchase plans allow you to completely bypass brokers. And they offer a low-cost means of investing directly in a company for even very modest sums of money.

Your browser does not support iframes.

The Dividend Advantage

The attraction of dividend-paying stocks has increased since dividends started to receive preferential tax treatment in 2003, and it should continue now that that treatment will be extended through 2010. Reinvesting dividends allows you to build up your position in a company and keeps your money working for you over time. But keeping costs low is important, which is one of the attractions of a direct purchase plan.

Of course, a number of major discount brokerage firms offer their own dividend reinvestment plans, which allow you to reinvest the dividends of the stocks you own through the brokerage firm with no fees attached. This is an advantage because the amount of many dividend payments would otherwise typically entail the purchase of odd-lot sizes, which often have higher transaction charges.

However, many corporate direct purchase plans allow participants to make additional cash purchases of shares at little or no costa feature no brokerage firm can match.
And, direct purchase plans typically have lower minimum investments than most brokers.

Direct purchase plans are custom-made for long-term, buy-and-hold investors. While they are not on their own a reason to buy a stock, they serve as a shareholder bonus on a company with promising long-term growth prospects.

More on Dividend Reinvestment Plans

The information for the listings in this article was provided by dripdatabase.com. The soon-to-be completed Web site will provide a complete listing of companies offering direct purchase and dividend reinvestment plans, including the features of each plan, stock ratings and contact information.

The Web site is published by Simplisoft, a Beaverton, Oregon-based firm that specializes in tax and accounting software specifically designed for dividend reinvestment plan investors. For more information, go to www.dripwizard.com.

How Do You Join?

Currently, around 438 companies offer direct purchase plans, with minimums ranging from $10 to $2,500, with $250 a typical minimum. In a few cases, direct purchase plans are offered even though no cash dividend is paid.

Table 1 lists the companies that offer direct purchase plans, along with a contact phone number and initial investment minimum. They include major corporations, as well as REITs (real estate investment trusts).

Direct purchase plans are described in a plan prospectus, which you can receive by contacting either the companys shareholder relations department or the plan administrator. You should read the plan prospectus or description carefully. While the overall structure of most direct purchase plans is similar, they vary in the details. The prospectus or plan description will provide information on such items as: eligibility requirements, plan options, costs, how and when purchases are made, how and when certificates will be issued, and what participants should do when withdrawing from the plan.

How Direct Purchase and Dividend Reinvestment Plans Work

A company's direct purchase or dividend reinvestment plan will be described in the plan's prospectus, which you should read carefully. While the overall structure of most dividend reinvestment plans is similar, they vary in the details.

Administration

Most companies appoint an outside agent to serve as the administrator for their dividend reinvestment plan. The administrator maintains records, sends account statements to participants, furnishes certificates for shares upon request and liquidates participants' shares when they leave the plan. The agent also is responsible for the purchase of company shares for the plan. When you join a plan, you will sign a card that authorizes the agent to make purchases on your behalf.

Joining the Plan

Direct purchase plans allow you to buy the shares directly from the plan administrator. Dividend reinvestment plans without the direct purchase feature require that you own at least one share (and sometimes more) registered in your name-you are a shareholder "of record." That means your name appears on the corporate records as the owner of the shares rather than the nominee name ("street name") of the broker or bank that may have purchased the shares for you (and who may be safekeeping them for you). If your shares are held in street name, you should ask your broker to transfer the shares to your own name.

Shares purchased under a direct purchase or dividend reinvestment plan are held by the plan and registered in the nominee name of the agent or plan trustee on behalf of the participants, each of whom has an account under the plan. For participants in dividend reinvestment plans, that means you may hold the company's shares in two places-your original registered shares, with the certificates either held by you or in custody at a bank or brokerage firm, and the shares purchased through the dividend reinvestment plan, held by the plan.

Many dividend reinvestment plans will allow participants to deposit certificates of shares registered in their own name into their dividend reinvestment plan account for safekeeping at no charge or for a modest fee; these shares are then treated in the same way as the other shares in the participant's account.

Certificates for shares purchased under the plan are usually issued only upon written request, although often at no charge. Certificates are also issued when a participant no longer wants to participate in the plan.

Plan Options

Aside from the initial purchase, direct investment plan and dividend reinvestment plan mechanics are the same. The basic plan offers reinvestment of dividends on all shares of stock registered in the participant's name. This is often referred to as "full reinvestment."

Under most plans, it isn't necessary to reinvest all dividends. Instead, participants are allowed to reinvest dividends on a portion of their registered shares while receiving cash dividends on the remaining shares. This is usually referred to as a "partial reinvestment option." Most plans also allow participants to purchase additional shares by making cash payments directly to the plan. This option is often referred to as "optional cash payment," and since the allowable amounts can be large, it offers participants a low-cost way to build a sizeable holding in a company. The payments are optional-participants are not committed to making periodic cash investments.

However, usually there are minimums for each payment made, and often there is a maximum.

It is also important to note the frequency with which the plan invests cash payments, since interest is not paid on payments received in advance of actual investment.

A twist on the cash payment option is that some companies will allow registered shareholders to make cash investments without requiring them to reinvest dividends on the shares they are holding, although they may do so if they want. This is frequently referred to as the "cash payment only option."

An added convenience for participants who wish to make systematic cash investments is an automatic investment feature that is offered by most of the companies. The company or the plan agent automatically debits the investor's checking or savings account at regular intervals to purchase additional shares.

The Costs

Participant costs usually come in two forms: service charges and prorated brokerage commissions. Service charges cover administrative costs and are generally levied on each transaction; participants can hold costs down by combining a cash payment with a dividend reinvestment transaction, since usually the charges are capped (a $5.00 per transaction maximum is typical). Brokerage commissions levied on open market shares are at institutional rates (since the number of shares purchased is large), and are therefore considerably lower than the rate an investor would pay on his own.

Many companies cover all of the costs for share purchases from both reinvested dividends and optional cash payments. Some companies levy service charges, others prorate brokerage costs, and still others charge participants for both-there are many variations, so check the prospectus or plan description carefully.

When participation is terminated, some dividend reinvestment plans will sell plan shares for you, if you prefer, instead of sending you certificates. The cost to the participant is usually any prorated brokerage commissions, a lower-cost alternative than selling through a broker. Some plans will sell plan shares for you even if you are not terminating. Check the prospectus or plan description.

Share Purchases

The source of share purchases under a dividend reinvestment plan is spelled out in the plan description and prospectus.

The most common source is the secondary market-through an exchange where the shares are traded, in the over-the-counter market, or through negotiated transactions. Another source for some purchases is the company itself, using authorized but unissued shares of common stock or shares held in the company's treasury.

In plans that prorate brokerage commissions among participants, the source of share purchases is a concern. When shares are purchased directly from the company, there are no brokerage expenses to prorate.

When shares are purchased directly from the company, the prospectus will describe how the share price is determined. Usually, it is based on an average of the high and low or the closing price for the stock as reported by a specified source.

Some companies offer participants discounts on the share price, but there is wide variation in how this is offered. Most often, the discounts are available only on shares purchased with reinvested dividends, but sometimes discounts apply to shares purchased both with reinvested dividends and with cash payments. Discounts are described in detail in plan prospectuses.

A Note About Taxes

Direct purchase and dividend reinvestment plans have many advantages, but their tax status is not one of themunless you are investing through an IRA (offered by a few of the companies).
Whether you receive your dividends in cash or have them reinvested, a taxable event has occurred, although the tax treatment now is lower than it was previously.

In addition, if you reinvest dividends, the IRS considers the dividend to be equal to the fair market value of shares acquired with reinvested dividends. The fair market price is the price on the exchange or market where shares are traded, not any discounted price. Furthermore, any brokerage commissions paid by the company in open market purchases to acquire the shares are considered additional dividend income to the participant.

When shares are sold, the tax basis is the fair market value as of the date the shares were acquired, plus any brokerage commissions paid by the company, and it is treated as income to the participant. Participants receive 1099-DIV forms each year from the company detailing dividends to be treated as income as reported to the IRS.

Where to Get Information

Companies with direct purchase plans appoint an outside agent to serve as the administrator for their plan. These administrators maintain plan records, send account statements to participants, furnish certificates for shares upon requestand handle the purchases and sales of the shares.

Four major firms serve as administrators for U.S.-based direct purchase plans and dividend reinvestment plans:

American Stock Transfer & Trust,

Computershare,

Bank of New York Mellon, and

Wells Fargo.

Bank of New York and Mellon Investor Services merged in 2007 into BNY Mellon, but the merging of their dividend reinvestment programs is not complete. Therefore, the bank operates separate Web sites for companies that maintain programs with the bank.

The list above includes the five major Web sites with information on the direct purchase and dividend reinvestment plans that they administer. These sites also have educational articles on direct purchase and dividend reinvestment plans in general. All five Web sites also allow you to buy direct purchase plan shares on-line at their sites.

A few other plan administrators serve only a handful of dividend reinvestment plans with no direct purchase features and limited on-line purchase programs. In a few of these cases, you can find the most information by contacting the company directly. Also, several companies act as administrators for their own plans.

To find plan information from a company directly, go to the company Web site and look in the Investor Relations area or call the phone number provided here.

Direct Stock Purchase Web Sites

American Stock Transfer & Trust Co.www.amstock.com [go to Shareholder Services]
www.investpower.com
Provides a list of all direct purchase plans and dividend reinvestment plans administered by the firm. By clicking on the name of each company you can get a brief description of either kind of plan. For a direct purchase plan prospectus, click buy, and then click on plan materials. For a dividend reinvestment plan prospectus, contact the shareholder relations department of the company you are interested in.

The Bank of New York(merged into BNY Mellon)www.stockbny.com
Go to Company Facts & Forms, then Company List and select United States under Search by Country for a list of both direct purchase and dividend reinvestment plans. For a direct purchase plan prospectus, click on yes in the Company Specific Forms column and then click on Brochure/Prospectus. To bring up the list of direct purchase plans with a link to buying on-line, go to Initial Purchase and select United States under Search by Country. For a dividend reinvestment plan prospectus, call the toll-free number listed and request a prospectus. Clicking on the company name itself will link you to the company Web site.

BNY Mellon Shareowner Serviceswww.melloninvestor.com
Click on Investors and then Enroll in a Direct Stock Purchase Plan for a listing of direct purchase and dividend reinvestment plans. For a direct purchase plan prospectus, click on Invest Now and then click on View Plan Material. For a dividend reinvestment plan prospectus, contact the shareholder relations department of the company you are interested in.

Computersharewww.computershare.com
Click on Investor Centre under Shareholder Services. Then click on Investment Plans under Transact. You can get a complete list of all dividend reinvestment plans and direct purchase plans, or search for each separately. For a direct purchase or dividend reinvestment plan prospectus, click on View under Plan Summary and then select the PDF link next to Plan Brochure.

Wells Fargowww.wellsfargo.com/com/shareowner_services
Click on Investment Plans for a list of direct purchase and dividend reinvestment plan companies. Click on a company name to access a prospectus link, to buy on-line or to print out an application form to mail.

→ Maria Crawford Scott

Email:

In This Issue

June 2008 Issue

Financial PlanningImmediate or Income AnnuitiesAnnuities are primarily used as a means of securing a steady cash flow during retirement. Unlike a deferred annuity, an immediate annuity has no accumulation period--an investor simply pays the insurance company a lump sum, and then receives the stream of payments for the set time period.

Stock StrategiesDirect Purchase Plans: The Foreign OptionWith current growth rates overseas outpacing U.S. stocks' growth, many investors are casting a longing look at the foreign markets. One approach is to purchase ADRs, certificates that trade on a U.S. exchange but represent foreign firms. And now, several banks have set up direct purchase plans for their ADR firms.

The American Association of Individual Investors is an independent, non-profit corporation formed for the purpose of assisting individuals in becoming effective managers of their own assets through programs of education, information and research.


GREENVILLE, S.C. (AP) — Newt Gingrich didn't know when he would take office if he wins the presidency. Rick Perry got the voting age and the date of Election Day wrong. Herman Cain didn't realize the president does not sign amendments to the Constitution.

In ways large and small, Republican presidential hopefuls are proving on multiple occasions to be "factually challenged," as Gingrich rather haughtily described a rival, despite getting some things wrong himself.

Campaigns are long and tough, candidates are often tired and flubs happen. But they are adding up and at some point could give Republican voters pause as they look for the candidate best able to take on the highly polished — though hardly factually infallible — President Barack Obama.

In submitting to what is, in effect, America's toughest job interview, there may be only so much leeway in getting matters of current affairs and history plain wrong.

Frequent flubber Michele Bachmann's suggestion many months ago that the Revolutionary battles of Lexington and Concord took place in New Hampshire was an opening shot, of sorts, in a volley of misfires by the candidates. Those battles were fought in Massachusetts in 1775.

On Wednesday, she offered another: She would support the United States shutting down its embassy in Tehran — but there is no U.S. Embassy in Iran's capital.

Never mind the facts, her top spokeswoman said. "Congresswoman Bachmann is a member of the House Select Committee on Intelligence and is fully aware that we do not have an embassy in Iran and have not had one since 1980," Alice Stewart said in a statement.

It's the latest but hardly the worst.

Cain promoted Chile's retirement system as one that gives workers the option of having private accounts, when in fact they have no choice. Mitt Romney accused Obama of "peacetime spending binges" as if there were no wars going on. Bachmann accused Obama of canceling a Canadian pipeline project that has only been delayed.

On Wednesday, Gingrich told voters packed into Tommy's Country Ham House in Greenville, S.C., that he would sign legislation repealing health care and Wall Street overhauls when he takes office on Jan. 21, 2013.

"My intent will be to ask the new Congress to stay in session when they are sworn in on Jan. 3 and to pass — and hold at the desk until I'm sworn in on the 21st — to pass the repeal of Obamacare and the repeal of Dodd-Frank and the repeal of Sarbanes-Oxley so that I can sign them on the 21st," Gingrich told the packed restaurant.

One problem: the Constitution that Gingrich constantly cites during his presidential campaign says the transition of power after an election takes place on Jan. 20.

Gingrich spokesman R.C. Hammond said Gingrich would assume powers at noon on Jan. 20, 2013, following the 20th Amendment of the Constitution. Because that day is a Sunday, the Inauguration's festivities would be scheduled on Jan. 21 of that year. Ronald Reagan followed a similar schedule for his second inaugural on Monday, Jan. 21, 1985.

Even so, Gingrich was wrong to say "I'm sworn in on the 21st."

A day earlier, Texas Gov. Rick Perry suggested the voting age is 21 and got the date wrong for Election Day.

"Those of you that will be 21 by November the 12th, I ask for your support and your vote," he told students at Saint Anselm College in New Hampshire.

The voting age is 18. And New Hampshire is scheduled to be the first state in the nation to host a Republican presidential primary on Jan. 10; the general election is scheduled for Nov. 6, 2012.

Cain said he would back an amendment to the Constitution to ban abortion.

"If we can get the necessary support and it comes to my desk, I'll sign it," he told the Christian Broadcasting Network.

Except presidents don't sign amendments. Congress passes them and the states ratify them. The president could champion them, but the Constitution doesn't give him or her any formal role.

Since the campaign's start, each candidate has had a turn explaining errors as either the side effects of an exhausting schedule or simple foot-in-mouth syndrome. Under the intense media scrutiny, each misstep or error draws questions whether each candidate is up for the job.

Romney, too, stepped in it. The former Massachusetts governor said Obama engaged in "one of the biggest peacetime spending binges in American history." He overlooked the United States' role in conflicts in Iraq, Afghanistan and Libya.

Yet not all errors are created equal, said Eric Dezenhall, an aide in the Reagan administration and now an image consultant who has worked with everyone from Hollywood stars to business moguls.

"The key factor in whether a gaffe catches on is whether or not it validates a pre-existing prejudice," he said.

"When Perry says that the voting age is 21, it validates the pre-existing suspicion that he's not in command of the basics," he said. "When Newt or Obama say something that is either misguided or incorrect, it doesn't resonate because everybody knows they are smart guys, so they get a break."

And it's not as if Obama hasn't had his doozies. For instance, Obama said during the 2008 campaign that he had visited 57 states. The United States only has 50.

"The flubs that stick are those that fit with a storyline about the candidate," said Doug Hattaway, a Democratic consultant who helped Hillary Rodham Clinton's 2008 presidential bid. "Gingrich isn't a flubber. He's known for being full of himself and making wacky statements, not flubbing the facts. So the misstatements are less likely to stick to him."

And voters might not care about factual details, Hattaway said.

"Best case of that is George W. Bush, who couldn't pass a civics quiz to save his life. Emotional intelligence is more important in politics than factual knowledge," Hattaway said.

Gingrich might be playing that to his own political advantage.

Before he seemed to reschedule the constitutional transition of power, he criticized Bachmann for stretching the facts about his record on abortion.

"Some people are just factually challenged and it's unfortunate," Gingrich told reporters. "In the eyes of a teacher, occasionally I'd have a student who couldn't figure out where things were, or what things were, or what the right date was. When that happens, you feel sorry that they're so factually challenged."


Inga multicaulis

Inga multicaulis is a species of legume in the family Fabaceae.
It is found only in Ecuador.
Its natural habitats are subtropical or tropical moist lowland forests and subtropical or tropical moist montane forests.

References

multicaulis
Category:Flora of Ecuador
Category:Endangered plants
Category:Taxonomy articles created by Polbot