TITLE: Number of $n$-bead binary necklaces [OEIS-A000013]
QUESTION [2 upvotes]: I tried to obtain the number of $n$-bead binary necklaces from my program written in C++. Then, one formula came up when I looked up the number to see if my thought is correct.

Number of $n$-bead binary necklaces with beads of $2$ colors where the colors may be swapped but turning over is not allowed
$$a(n)=\sum_{d|n}\frac{\phi(2d) \, 2^\frac{n}{d}}{2n}$$ for $n>0$.

Source: A000013
However, although I see the number I obtained from my program was correct, I could not grasp why the formula holds.
Thus, I would appreciate it if someone could explain its derivation to me.

REPLY [0 votes]: As was suggested in a comment, the Burnside's lemma is possibly the simplest way to deal with similar problems, since you are in fact looking for the number of orbits of an action of $\Pi_n\times\Sigma_2$ on $A_n=\{0,1\}^n$, with $\pi$ and $\sigma$ being rotation and inversion operators:
$$
(\pi a)_k=a_{(k+1)\bmod n},\quad (\sigma a)_k=1-a_k,\quad  k=0\dots n.
$$
There are altogether $2n$ elements in this action group with general form $\pi^i\sigma^j$ ($i=0\dots n-1;\ j=0,1)$, the element $\pi^0\sigma^0$ being the identity. It remains to find the number of the elements of $A_n$ fixed by each of the $2n$ operations.
The elements fixed by the rotation $\pi^j$ are well known. One can imagine that a necklace is a train consisting of $n/(n,j)$ identical railcars with length $(n,j)$. Clearly the necklace coincides with itself by any rotation $\pi^{k(n,j)}$, particularly by $\pi^j$ and $\pi^n=\pi^0$. As the content of a railcar can be arbitrary the number of fixed elements is $2^{(n,j)}$ for binary necklaces. If the number of railcars $n/(n,j)$ is odd this is the end of story as $\sigma$ cannot come in the game. This changes however if the number is even. In this case the number of fixed elements is doubled since the trains consisting both of identical as well as of alternating "inverted" cars are the fixed elements. It is important to observe that the number $j/(n,j)$ (the number of cars staying between $0$ and $j$) is  always odd in the considered case.
Thus we have by Burnside's lemma for the number of orbits $N$:
$$\begin{align}
N(n)&=\frac1{2n}\left[\sum_{1\le i\le n}^{n/(n,i)=0\bmod 2}2^{(n,i)+1}
+\sum_{1\le i\le n}^{n/(n,i)=1\bmod 2}2^{(n,i)}\right]\\
&=\frac1{2n}\sum_{k|n}\left[\sum_{1\le i\le n}^{(n,i)=k \atop n/k=0\bmod 2}2^{k+1}
+\sum_{1\le i\le n}^{(n,i)=k \atop n/k=1\bmod 2}2^{k}\right]\\
&=\frac1{2n}\left[\sum_{k|n}^{n/k=0\bmod 2}2^{k+1}\sum_{1\le i\le n}^{(n,i)=k}1
+\sum_{k|n}^{n/k=1\bmod 2}2^{k}\sum_{1\le i\le n}^{(n,i)=k}1\right]\\
&=\frac1{2n}\left[\sum_{k|n}^{n/k=0\bmod 2}2^{k+1}\phi(n/k)
+\sum_{k|n}^{n/k=1\bmod 2}2^{k}\phi(n/k)\right]\\
&=\frac1{2n}\left[\sum_{d|n}^{d=0\bmod 2} 2^{n/d}\cdot2\phi(d)
+\sum_{d|n}^{d=1\bmod 2}2^{n/d}\cdot\phi(d)\right]\\
&=\frac1{2n}\sum_{d|n} 2^{n/d}\phi(2d),
\end{align}
$$
where the last equality follows from already mentioned in comments property:
$$
\phi(2d)=\begin{cases}
2\phi(d),& d=0\bmod 2\\
\phi(d),& d=1\bmod 2\\
\end{cases}.
$$


TITLE: Evaluate $\int^4_0 g(x) \text { d}x$
QUESTION [1 upvotes]: Evaluate $\int^4_0 g(x) \text { d}x$

Breaking this up into separate integrals, I got:
$$\int^4_0 g(x) \text { d}x = \int^1_0 x \text { d}x + \int^2_1 (x-1) \text { d}x + \int^3_2 (x-2) \text { d}x + \int^4_3 (x-3) \text { d}x$$
$$ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2$$
However, is this correct? I ask because we have closed/open intervals and so I'm not sure how integration behaves on closed/open intervals.

REPLY [2 votes]: What you wrote is correct. The value of an integral of a function is not altered by changing the value of the function at an isolated point.

REPLY [1 votes]: You are most right.
As a result of its definition, the riemann integral does not change its value if you change a function in a finite set of points, because you can just choose a partition to have those troublesome points as boundaries of the intervals of the partition. Likewise, integrating  $f $ over the four intervals $[a, b]; [a, b); (a, b]; (a, b) $ yields the same result.


TITLE: If $X_t$ is independent of $\mathcal G$ for all $t\geq 0$, does $X_{\sigma }$ will be independent of $\mathcal G$ for a stopping time $\sigma $?
QUESTION [3 upvotes]: Let $(X_t)$ be such that $X_t$ is independent of a $\sigma -$algebra $\mathcal G$ for all $t$. Let $\sigma $ a stopping time (We can suppose that $\sigma <\infty $ a.s.). Is $X_\sigma $ independent of $\mathcal G$ ? I think it is, but I don't know how to prove it. What I could prove so far is that if $(X_t)$ is continuous, $$\mathbb E[f(X_{\sigma })\boldsymbol 1_A]=\mathbb E[f(X_\sigma )]\mathbb P(A)\tag{1}$$
for all $f$ continuous and bounded, and all $A\in \mathcal G$ using the fact that there is a decreasing sequence $(\sigma _n)$ s.t. $\sigma _n\searrow\sigma $ and $\sigma _n$ taking finitely many value. But I couldn't prove $(1)$ if $f$ is only Borel-measurable or if $(X_t)$ is not continuous.
Can someone tell me if the result is true (at least if $(X_t)$ is continuous), and how to prove $(1)$ for $f$ measurable, but not necessarily continuous.

REPLY [2 votes]: The claim is not true.
Hint: Let $(B_t^{(1)}, B_t^{(2)})$ be a two-dimensional Brownian motion starting at $(0,1)$. Let $\mathcal{F}_t$ and $\mathcal{G}_t$ be the filtrations generated by $B^{(1)}$ and $B^{(2)}$, respectively, $\mathcal{F} = \mathcal{F}_\infty$ and $\mathcal{G} = \mathcal{G}_\infty$. Set $\sigma = \inf \{ t > 0 \, : \, B_t^{(2)} = 0 \}$. Then, by construction, $B_t^{(1)}$ is independent of $\mathcal{G}$.
Can you show $B_\sigma^{(1)}$ is not independent of $\mathcal{G}$?


TITLE: A Conway problem on identifying groups
QUESTION [7 upvotes]: I first saw this on the Missouri State problem page and it has never been solved there.
Consider the group generated by a,b,c, and d subject to the relations
ab = c, bc = d, cd = a, and da = b
Using the first relation, the second relation becomes bab = d. Using this expression and the first relation, we obtain
abbab = a and baba = b
Taking the second relation above and multiplying both sides on the left by a^-1b^-1 and on the right by a^-1, we have b = a^-2. Now first relation above becomes aa^-2a^-2aa^-2 = a or a^-4 = a, hence i = a^5. Therefore our group consists of the five elements i, a, a^2, a^3, a^4. The other elements can be expressed in terms of a as follows: b = a^3, c = a^4, and d = a^2.
Finally, we get to his month's problems. How many elements are there in the groups given by the following generators and relations?
* Generators:a,b,c
  Relations: ab = c, bc = a, ca = b

* Generators:a,b,c,d,e
  Relations: ab = c, bc = d, cd = e, de = a, ea = b

* Generators:a,b,c,d,e,f
  Relations: ab = c, bc = d, cd = e, de = f, ef = a, fa = b

Source: John H. Conway
I recognize the first one as the quaternions but have made no progress on the other two.

REPLY [5 votes]: For your second group, we can write it as:
$$ \langle a,b\ |\ babab^2ab=a, ab^2aba=b\rangle. $$
The first relation gives $baba=ab^{-1}a^{-1}b^{-2}$, and plugging that into the second relation gives $ab(ab^{-1}a^{-1}b^{-2})=b$, or $aba=b^3ab$.  Plugging that again into the second relation gives $ab^2(b^3ab)=b$, or $ab^5a=1$, so $b^5=a^{-2}$.  Thus $a^2$ is central; combining the first and second relation gives $bab^2=a^2$, and plugging that into the first relation gives $ba^4b=a$.  Since $a^2$ is central, this is the same as saying $b^2=a^{-3}$. This is enough to abelianize your group, so it was abelian to begin with; a simple check then shows it is $C_{11}$.
It is easy to see the last group is infinite: the presentation can be given as
$$ \langle a,b\ |\ b^2ab^2abab^2ab, abab^2aba\rangle;$$
Quotienting out by the normal subgroup $\langle a^2,b^2\rangle$ (you can check it is normal), gives the group $C_2\ast C_2$, the infinite dihedral group.


TITLE: Continuity of the operation in a topological group.
QUESTION [1 upvotes]: Consider a topological group $\mathrm{D},$ whose group operation I shall denote multiplicatively. The following result must be true but I fail to find a proof of it (except in the commutative case, which is not what I need). Consider a permutation $\sigma$ of $1, \ldots, p.$

Is the mapping $\mathrm{D}^p \to \mathrm{D}$ given by $(x_1, \ldots, x_p) \mapsto x_{\sigma(1)} \cdots x_{\sigma(p)}$ continuous with respect to the product topology on $\mathrm{D}^p$ induced by that of
  $\mathrm{D}$?

Again, in the commutative case $\sigma$ has no effect, thus one can write
$$(x_1 \cdots x_p) (y_1 \cdots y_p)^{-1} = (x_1 y_1^{-1}) \cdots (x_p y_p^{-1});$$
one can consider then a neighbourhood $\mathrm{T}$ of the neutral element and another neighbourhood $\mathrm{K}$ (of the neutral element again) such that $\underbrace{\mathrm{K} \cdots \mathrm{K}}_{n \text{ times}} \subset \mathrm{T}$, then it suffices to take $x_i \in y_i \mathrm{K}$ for $1 \leq i \leq p$. I ask now, can this proof be modified slightly to allocate the noncommutative case?

REPLY [2 votes]: The permutation is largely irrelevant; your map is just the
composition of the permutation-of-coordinates map $D^p\to D^p$
(which is a homeomorphism) with the multiplication map
$\mu_p:(x_1,\ldots,x_p)\to x_1\cdots x_p$.
The continuity of $\mu_p$ is an induction on $p$. For $p=2$
this is part of the definition of topological group. In general
if $\mu_p$ is continuous, so is $(x_1,\ldots,x_{p+1})\mapsto
x_1\cdots x_p$. Clearly $(x_1,\ldots,x_{p+1})\mapsto x_{p+1}$
is continuous. From the continuity of multiplication,
$\mu_{p+1}:(x_1,\ldots,x_{p+1})\to (x_1\cdots x_p)x_{p+1}$ is continuous.


TITLE: Hopf algebra: Identity under convolution
QUESTION [5 upvotes]: In Hopf algebra texts, it is usually stated that  $1=\eta\epsilon\in$Hom($H^C,H^A$) is the identity under convolution.
$\eta$ is the unit, $\epsilon$ is the counit.
My question is, is that a definition, or can it be proved?
Sincere thanks for any help.
(Do let me know if you need any clarification on the above notations.)

REPLY [9 votes]: It can be proved using the definition of (co-)associativity and the (co-)unit.
Let $\mathbb K$ be a field. Let $(A,m)$ be a associative $\mathbb K$-algebra with unit $\eta: \mathbb K \to A$ and let $(C,\Delta)$ be a coassociative $\mathbb K$-coalgebra with counit $\varepsilon: C \to \mathbb K$.
The convolution $\star: \operatorname{Hom}(C,A) \times \operatorname{Hom}(C,A) \to \operatorname{Hom}(C,A)$ is defined by
$$f \star g := m \circ (f \otimes g) \circ \Delta.$$
Let $\mathbf 1 := \eta(1_{\mathbb K})$, then from the definition of the unit follows
$$m(\mathbf 1 \otimes a) = a \quad \text{for all } a\in A.$$
Furthermore,
$$(\varepsilon \otimes \operatorname{id}) \circ \Delta = 1_{\mathbb K} \otimes \operatorname{id},$$
by definition of counit. Using this we show
$$\eta \varepsilon \star f = f \star \eta\varepsilon = f \quad \text{for all } f \in\operatorname{Hom}(C,A).$$
For all $c \in C$ we have
\begin{align*}
(\eta\varepsilon \star f)(c) &= (m \circ (\eta \varepsilon \otimes f) \circ \Delta)(c)\\
&=(m \circ (\eta \otimes f) \circ (\varepsilon \otimes \operatorname{id}) \circ \Delta)(c)\\
&= (m \circ (\eta \otimes f))(1_{\mathbb K} \otimes c)\\
&= m(\mathbf 1 \otimes f(c))\\
&= f(c),
\end{align*}
hence $\eta\varepsilon \star f = f$. Similarly one shows $f \star \eta\varepsilon = f$.


TITLE: How to find equations of motion when potential is given by inverse-square?
QUESTION [0 upvotes]: When potential is $U=-\dfrac{a}{r^2}$ ($a>0$), how can I find $r=r(\phi)$?
I'm trying to solve this problem during several hours.
From $E=T+U$, and constant angular momentum $L$, I can get the integral equation $\phi=\phi(r)$, but this integral is too complicated to calculate (or I don't know how to integrate though I've tried a few hours) because $U$ is proportional to $r^2$, I think.
So, I got equation of motion about $r$ from Lagrangian, which is $$\dfrac{d^2r}{dt^2}+\dfrac{1}{r^3}=0$$
I think this is easier than first one, but this is non-linear ODE and I didn't find the solution. Could you let me know this?

REPLY [2 votes]: With a potential which is proportional to the inverse square of the radius, you can write the energy as
$$E=\frac{1}{2}m \dot{r}^2 +\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$
where $L$ is the angular momentum
$$L=m r^2 \dot{\phi}$$
You can rewrite the energy in the following way
$$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \dot{\phi}   \right)^2+\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$
Now substitute $\dot{\phi}$ using the definition of the angular momentum
$$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \frac{L}{mr^2}   \right)^2+\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$
and change variable setting $u=1/r$. You obtain
$$E=\frac{L^2}{2m}\left( \frac{du}{d\phi} \right)^2+\left( \frac{L^2}{2m}-a\right)u^2$$
By setting $dE/d\phi=0$ (energy is conserved) you obtain a motion equation of the form
$$\frac{d^2u}{d\phi^2}+\left(1-\frac{2ma}{L^2} \right)u=0$$
If $2ma/L^2<1$ this is an harmonic oscillator, and the general solution is
$$u=\frac{1}{r} = A\cos \left( k \phi + B \right)$$ with
$$k=\sqrt{1-\frac{2ma}{L^2}}$$ and $A$,$B$ arbitrary constants.
This describes an orbit that winds $1/k$ times around the attraction centre and then escape to the infinity. If $2ma/L^2>1$ the potential is larger that the centrifugal barrier. The solution of the equation can be written as
$$u=\frac{1}{r} = A\cosh \left( k \phi + B \right)$$ and the particle falls in the centre of attraction (with an infinite number of windings).
The particular case $2ma/L^2=1$ correspond to
$$u=\frac{1}{r} = A \phi + B$$
In this case the potential cancels exactly the centrifugal barrier, and the particle falls on the centre of attraction coming from the infinity.
There is an interesting geometrical interpretation of these results when $k$ is real. If you introduce
$$\tilde{L} = L \sqrt{1-\frac{2ma}{L^2}}$$
you can rewrite the energy as
$$E=\frac{1}{2}m \dot{r}^2 + \frac{\tilde{L}^2}{2mr^2}$$
and
$$\tilde{L} = mr^2 \dot{\theta}$$
where the new "angular variable" is
$$\theta= \sqrt{1-\frac{2ma}{L^2}} \phi$$
$E$ and $\tilde{L}$ seems to be the energy and the angular momentum of a free particle, in polar coordinates. The trajectories will be simply lines.
However a complete orbit around the centre $\Delta \phi=2\pi$ is given by
$$\Delta \theta = 2\pi \sqrt{1-\frac{2ma}{L^2}}$$
so you must cut the plane at $\theta= \Delta \theta$ and identify the border with $\theta=0$ (in other words, you obtain a cone). The "line" will wind around the attraction centre several times, if the factor $k=\sqrt{1-\frac{2ma}{L^2}}$ is small, accordingly with the analytic solution.


TITLE: Conditional expectation formula
QUESTION [0 upvotes]: can I get an explanation why the following is the formula for conditional expectation? $$ E(X|Y)=\int_{-\infty}^\infty xf_{X|Y}(x|y)dx$$

REPLY [1 votes]: Regards @matt , I may not give full detail. Hope this will be useful.
-In the discrete case, the expectation $E(X)$ is the sum of each element (represents an event) in the random variable Sample Space $S$ multiplied by its corresponding probability.
$$  S = \{ X, \: X = e_{1}, \: \: e_{2}, \:\: , .... \}  $$
$ e_{i} $ is a representation of the event (in a number).
$$ E(X) = e_{1} P(X=e_{1}) + e_{2} P(X=e_{2}) + ....  $$
Now consider about Conditional Probability and with sets of events $X$ and $Y$ : the notation $X|Y$ , or $X|Y$ in particular means we are considering the problem only within a Sample Space $S_{Y}$. We only consider events in $X$ that are subsets of $Y$, so $Y$ becomes the sample space. So :
$$ E(X|Y) = e_{1} P(X=e_{1}|Y) + e_{2} P(X=e_{2}|Y_{2}) + ....  $$
-In the continuous case, we cannot calculate probability at a point, but only probability density. So we use integral instead of summation and get
$$ E(X|Y=e_{y}) = \int x f_{X|Y}(x, e_{y}) dx $$
with the definition : $$f_{X|Y}(x,e_{y}) = \frac{f(x,e_{y})}{f_{Y}(e_{y})} $$
so the ratio (or in continuous case : 'relative density') is with respect to $f_{Y}(e_{y})$. So we see the events in $X$ that are coincide with the event $\{Y=e_{y}\}$.

Note that $f_{Y}(e_{y})$ is not the probability $P(Y = e_{y})$, but it does 'approximate' it as an average. $$ f_{Y}(e_{y}) = \frac{F_{Y}(e_{y}+ \delta y) - F_{Y}(e_{y})}{ \delta y} $$ Similarly for the joint pdf $f(x,e_{y})$.
Notice also the definition of conditional pdf $f_{X|Y}(x,e_{y})$ looks similar with the conditional discrete case :
$$ f_{X|Y}(x,e_{y}) = \frac{f(x,e_{y})}{f_{Y}(e_{y})}, \: \: \: \: \: P(X|Y) = \frac{P(X \cap Y)}{P(Y)} $$

Thanks, all the best.


TITLE: Field $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$
QUESTION [1 upvotes]: Am I right to say that the field $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$ is an algebraic extension of $\mathbb{Q}$?
Because $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt{3})( \sqrt{-1})=\mathbb{Q}(\sqrt{3}, \sqrt{-1})$.
Thanks.

REPLY [6 votes]: HINT $\rm\ K =\: \mathbb Q(\sqrt 3, \sqrt{-1})\:$ is a $4$-dimensional vector space over $\rm\mathbb Q\:,\:$ viz. $\rm\: K =\: \mathbb Q\langle 1,\:\sqrt 3,\:\sqrt{-1},\:\sqrt{-3}\rangle\:.$
Hence $\rm\:\alpha\in K\ \Rightarrow\ 1,\ \alpha,\ \alpha^2,\ \alpha^3,\ \alpha^4\:$ are linearly dependent over $\rm\:\mathbb Q\:.\:$ This dependence relation yields a nonzero polynomial $\rm\:f(x)\in \mathbb Q[x]\:$ of degree $\le 4\:$ such that $\rm\:f(\alpha)=0\:.$


text:
\begin{document}

\title{Higher Kiss terms}
\author{ Andrew Moorhead}

\address[Andrew Moorhead]{
  University of Kansas
  Dept.\ of Electrical Engineering and Computer Science;
  Eaton Hall;
  Lawrence, KS 66044;
  U.S.A.}
\email[Andrew Moorhead]{apmoorhead@gmail.com}

\thanks{Andrew Moorhead has received funding from the European Research Council
(ERC) under the European Unions Horizon 2020 research and
innovation programme (grant agreement No 771005).
}

\begin{abstract}
We show that the modular term condition higher commutator is equal to the modular hypercommutator. As a consequence, we arrive at a new proof that HC8 holds for modular varieties. Next, we develop a procedure for a modular variety for producing the higher dimensional congruences that characterize the hypercommutator. This procedure allows us to demonstrate that every modular variety has an infinite sequence of what we call higher dimensional Kiss terms. We use these results to extend the scope of a theorem of Opr{\v s}al from permutable varieties to modular varieties.
\end{abstract}
\maketitle

\section{Introduction}

The topic of this paper is the higher commutator. First defined by Bulatov in \cite{buldef}, the higher commutator is a congruence lattice operation that is a higher arity generalization of the binary commutator. Our overarching goal is to determine the extent to which the well-developed theory of the binary commutator (see for instance \cite{FM}, \cite{gumm}, and \cite{kearneskiss}) can be found as a low-dimensional case inside of a general multidimensional commutator theory. The results of this article demonstrate that there is indeed a higher dimensional commutator theory for modular varieties which extends the essential parts of Chapter 4 of \cite{FM} along with the properties of the Kiss term defined in \cite{threeremarks}.

The first results of this kind were proved by Aichinger and Mudrinski \cite{aichmud}. They demonstrate that, for congruence permutable varieties, many properties of the binary modular commutator are true also of the higher commutator. They also show that the higher commutator for a permutable variety satisfies an inequality relating nested commutator terms, which they refer to as HC8.

Opr{\v s}al gives alternative proofs of these properties for permutable varieties in \cite{orsalrel}. His development of the $(n)$-ary commutator flows from the properties of $(2^n)$-ary relations in permutable varieties, usually called matrices, that satisfy higher dimensional versions of reflexivity and symmetry. He also defines a sequence of terms, each derived from a Mal'cev operation, which may be used along with knowledge of the higher commutator to recursively generate all of the matrices for an algebra in a permutable variety. He goes on to deduce that the collection of clones on a set sharing congruences, higher commutators, and a Mal'cev operation has a greatest element.

In this article we show that the same kind of result is true for modular varieties, except that the relations we consider are also required to satisfy a higher dimensional version of transitivity and Day terms fill the role of a Mal'cev operation. This is the first of two articles that expand on the theory developed in \cite{taylorsupnil}. There, we define a new operation which we call the hypercommutator. The term condition higher commutator is defined with respect to what we call a higher dimensional tolerence, while the hypercommutator is defined with respect to what we call a higher dimensional congruence.

The structure of this article is as follows. In Section \ref{sec:hyper=tc}, we show that the modular term condition higher commutator is equal to the modular hypercommutator. We derive as a consequence an alternative proof of HC8 for modular varieties, first shown to hold by Wires in \cite{Wires}. In Section \ref{sec:hdkt} we define the higher dimensional Kiss terms and establish some of their basic properties, and in Section \ref{sec:application} we extend Opr{\v s}al's Theorem 1.3 of \cite{orsalrel} from permutable varieties to modular varieties. We rely on the notation and theory developed in \cite{taylorsupnil} and cite results from \cite{FM}, so the reader should have both of these on hand.

\section{The hypercommutator for a modular variety}\label{sec:hyper=tc}

\subsection{A relational characterization of the term condition modular higher commutator}
\cite{taylorsupnil}

In this section we argue that the term condition commutator is equal to the hypercommutator in a modular variety. This can be used to produce alternative proofs of HC8 for modular varieties and that the modular commutator is equal to the modular two-term commutator. The proof combines machinery from \cite{moorheadHC} and \cite{taylorsupnil}. To make the presentation consistent, we will first build a more general version of the machinery from \cite{moorheadHC} using the notation from \cite{taylorsupnil}.

\begin{defn}\label{def:modularshiftrotations}
Let $\var$ be a modular variety with Day terms $m_0, \dots, m_{k}$ (see Theorem 2.2 of \cite{FM}). Let $\A \in \var$ and $n \geq 2$. For each $ e \in k+1$ and $ i \neq j \in n$, we define the \textbf{$e$th shift rotation at $(i,j)$} as
\begin{align*}
\rot^e_{i,j}: A^{2^n } &\to A^{2^n}\\
			\gamma &\mapsto m_e\left( \refl_j^1(\gamma). \gamma, \refl_i^0(\gamma), \refl_j^1(\refl_i^0(\gamma))\right)
\end{align*}
\end{defn}
While the notation for a shift rotation used here does not specify the dimension of the relation on which it acts, the dimension should be clear from the context (see \emph{(4)} of Lemma \ref{lem:basicrotationproperties}).
\begin{lem}[cf.\ 3.1 of \cite{moorheadHC}]\label{lem:basicrotationproperties} Let $\var$ be a modular variety with Day terms $m_0, \dots, m_{k}$. Let $\A \in \var$, $\delta \in \Con(\A)$, $n \geq 2$, and $ i\neq j \in n$. The following hold:

\begin{enumerate}

\item If $R$ is an $(n)$-dimensional tolerance of $\A$, then
$
\rot^e_{i,j} : R \to R,
$
for all $e \in k+1$.

\item Let $\gamma \in A^{2^n}$, $f \in 2^{n \setminus \{i, j \}}$, and suppose the $(i)$-supporting line of  $\squares_{i,j}(\gamma)_f$ is a $\delta$-pair. The $(i)$-pivot line of $\squares_{i,j}(\gamma)_f$ is a $\delta$-pair if and only if the $(j)$-pivot line of $\squares_{i,j}(\rot^e_{i,j}(\gamma))_f$ is a $\delta$-pair for every $e \in k+1$.

\item If $\gamma \in A^{2^n} $ is such that every $(i)$-supporting line of $\gamma$ is a $\delta$-pair, then every $(j)$-supporting line of $\rot^e_{i,j}(\gamma)$ is a $\delta$-pair for every $e \in k+1$. The same statement holds if `supporting' is replaced with `cross section'.

\item For  $ i \neq j \neq l \in n$,
\[\rot_{i,j}^e(\gamma) =
\glue_{\{l\}} \left( \rot^{e}_{i,j}( \faces_{l}^0(\gamma)), \rot^{e}_{i,j}( \faces_{l}^1(\gamma)) \right).
\]
\end{enumerate}

\end{lem}
\begin{proof} Assume that $R$ is an $(n)$-dimensional tolerance of $\A$ and take $\gamma \in R$. Because $R$ is $(n)$-reflexive, it contains  $\refl_j^1(\gamma), \refl_i^0(\gamma)$, and $\refl^1_j(\refl_i^0(\gamma))$. Because $R$ is closed under the operations of $\A$, it follows that \emph{(1)} holds.

To see that the remaining properties hold, notice that if $\gamma \in A^{2^n} $ is such that
\begin{align*}
\squares_{i,j}(\gamma) &=
\left\{
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$c_f$}
      -- ++(1,0)  node (ne) {$d_f$}
      -- ++(0,-1) node (se) {$b_f$}
      -- ++(-1,0) node (sw) {$a_f$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

\end{tikzpicture}
: f \in 2^{n \setminus \{ i,j \}}
\right\}, \text{ then for each $e \in k+1$}\\
\squares_{i,j}(\rot^e_{i,j}(\gamma)) &=
\left\{
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$c_f= m_e(c_f,c_f, c_f, c_f)$}
      -- ++(3,0)  node (ne) {$m_e(d_f,d_f, c_f, c_f)$}
      -- ++(0,-1) node (se) {$m_e(d_f,b_f, a_f, c_f)$}
      -- ++(-3,0) node (sw) {$c_f= m_e(c_f,a_f, a_f, c_f)$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

\end{tikzpicture}
: f \in 2^{n \setminus \{ i,j \}}
\right\}.
\end{align*}
Notice that we have applied one of Day's identities to conclude that the $(j)$-supporting line of any $(i,j)$-cross section square of any $\rot_{i,j}^e(\gamma)$ is a constant pair. Using Lemma 2.3 from \cite{FM}, the reader may now easily verify that \emph{(2)} and \emph{(3)} hold.

The final property \emph{(4)} asserts that taking a shift rotation of an $(n)$-cube $\gamma$ commutes with projecting $\gamma$ onto a lower dimensional face, provided the image of the projection still depends on the coordinates $i$ and $j$. This is a consequence of Definition \ref{def:modularshiftrotations}.

\end{proof}

We have the following immediate consequence of the properties of the shift rotations.

\begin{prop}[cf.\ 3.2 of \cite{moorheadHC} ]\label{prop:modularcentralityissymmetric}
Let $\var$ be a modular variety, $\A \in \var$, and $n \geq 2$. Let $R$ be an $(n)$-dimensional tolerance of $\A$ and let $\delta \in \Con(\A)$. If $R$ has $(\delta, j)$-centrality for some $j \in n$, then $R$ has $(\delta, i)$-centrality for each $i \in n$.

\end{prop}

\begin{proof}
Suppose that the assumptions hold and that $R$ has $(\delta, i)$-centrality. Let $ j \in n $ and take $\gamma \in R$ such that every every $(i)$-supporting line of $\gamma$ is a $\delta$-pair. If $i = j$ then it follows from our assumptions that the $(i)$-pivot line of $\gamma$ is a $\delta$-pair. If $i \neq j$ we apply \emph{(3)} then \emph{(2)} of Lemma \ref{lem:basicrotationproperties} and conclude that the $(i)$-pivot line of $\gamma$ is a $\delta$-pair.
\end{proof}

In view of Proposition \ref{prop:modularcentralityissymmetric}, we may now assert with no chance for confusion that an $(n)$-dimensional tolerance of an algebra $\A$ belonging to a modular variety has $\delta$-centrality for some $\delta \in \Con(\A)$.

As in both \cite{moorheadHC} and \cite{taylorsupnil}, we need to consider certain compositions of shift rotations and will use finite trees to index these compositions. Suppose that $\var$ is a modular variety with Day terms $m_0, \dots, m_k$ and let $n \geq 2$. Set

\[
\mathbb{D}_n = \langle  (k+1)^{ < n}; \leq \rangle,
\]
where $ (k+1)^{ < n} = \Union \{ (k+1)^i : i \in n \}$  and two sequences $d_1$, $d_2$ are $\leq$-related when $d_1 \subseteq d_2$. Note that $\mathbb{D}_n$ has the empty sequence $\emptyset$ as its root. For $\A \in \var$ and $\gamma \in A^{2^n}$, set $\gamma^\emptyset = \gamma$. We recursively define $\gamma^d = \rot_{i,i+1}^{d_i}(\gamma^c)$, where $d =(d_0, \dots , d_i) \in \mathbb{D}_{n}$ is non-empty and $c$ is the predecessor of $d$.

\begin{lem}[cf.\ 4.1 of \cite{moorheadHC}]\label{lem:treesuccessorrotations}
Let $\var$ be a modular variety with Day terms $m_0, \dots, m_k$. Let $\A \in \var$ and $R \leq \A^{2^n}$ be an $(n)$-dimensional tolerance for some $n\geq 2$. If $d \in \mathbb{D}_{n}$ is a tuple of length $i \in n$, then
\begin{enumerate}

\medskip
\item
$\gamma^d \in R$, and
\medskip

\item
if $f \in 2^{n \setminus \{i\}}$ satisfies $f(j) = 0$ for some $j\in i$, then the $(i)$-cross-section line of $\gamma^d$ at $f$
is a constant pair.
\end{enumerate}

\end{lem}
\begin{proof}
The result is trivially true for $\gamma^{\emptyset} = \gamma$. Suppose that it holds for $c \in \mathbb{D}_{n}$ of length $i$ and let $d=(d_0, \dots, d_i)$ be a successor of $c$. Set
$
\gamma^d = \rot^{d_i}_{i,i+1}(\gamma^c).
$
Notice that \emph{(1)} of Lemma \ref{lem:basicrotationproperties} guarantees that $\gamma^d \in R$, which establishes \emph{(1)} of this lemma.

Now let $f \in 2^{n\setminus{i+1}}$ be such that $f(j) = 0$ for some $j \in i+1$, and let $f^*$ be the restriction of $f$ to the set $n\setminus{\{i, i+1\}}$.  There are two cases to consider.

\begin{enumerate}
\item[Case 1:] Suppose $j \notin i$, in which case $f(i)=0$. If
\[
\squares_{i,i+1}(\gamma^c)_{f^*} = \Square[a][e][b][d],
\]
then it follows from the proof of Lemma \ref{lem:basicrotationproperties} that
\[
\squares_{i,i+1}(\gamma^d)_{f^*} =
 \begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$e= m_{d_i}(e,e, e, e)$}
      -- ++(3,0)  node (ne) {$m_{d_i}(d,d, e, e)$}
      -- ++(0,-1) node (se) {$m_{d_i}(d,b, a, e)$}
      -- ++(-3,0) node (sw) {$e= m_{d_i}(e,a, a, e)$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

\end{tikzpicture}.
\]
The left column of the above square is equal to the $(i+1)$-cross-section line of $\gamma^d$ at $f$ and it is a constant pair, as claimed.
\item[Case 2:] Suppose $ j \in i$. In this case we apply the inductive assumption that \emph{(2)} holds for $\gamma^c$ and conclude that
\[
\squares_{i, i+1}(\gamma^c)_{f^*} = \Square[a][e][a][e], \text{ so it follows that}
\]

\[
\squares_{i,i+1}(\gamma^d)_{f^*} =
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$e= m_{d_i}(e,e, e, e)$}
      -- ++(3,0)  node (ne) {$m_{d_i}(e,e, e, e)=e$}
      -- ++(0,-1) node (se) {$m_{d_i}(e,a, a, e)=e$}
      -- ++(-3,0) node (sw) {$e= m_{d_i}(e,a, a, e)$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

\end{tikzpicture}.
\]
The $(i+1)$-cross-section line of $\gamma^d$ at $f$ is either the left column or right column of the above square, and each of these columns is a constant pair. This finishes the proof.
\end{enumerate}
\end{proof}

\begin{prop}[cf.\ 4.7 of \cite{taylorsupnil}]\label{prop:tchascentrality}
Let $\var$ be a modular variety with Day terms $m_0, \dots, m_k$. Let $\A \in \var$, $n \geq 2$, $\delta \in \Con(\A)$. Let $l \in n$. If $R \leq \A^{2^n}$ is an $(n)$-dimensional tolerance of $\A$ that has $\delta$-centrality, then $R^{\circ_l}$ is also an $(n)$-dimensional tolerance of $\A$ that has $\delta$-centrality.
\end{prop}

\begin{proof}
Before we embark on the proof, we point out that the argument we give is an amalgamation of the proof that the modular commutator distributes over arbitrary joins (see 5.1 of \cite{moorheadHC}) and the proof of the `perpendicular stage' (see 4.7 of \cite{taylorsupnil}). Each of those proofs is accompanied by a picture that the reader may find useful for visualizing the argument.

Without loss it suffices to consider the case when $l=n-1$. We noted in Lemma 2.9 of \cite{taylorsupnil} that $R^{\circ_{n-1}} $ is an $(n)$-dimensional tolerance of $\A$, so it remains to verify that $R^{\circ_{n-1}}$ has $\delta$-centrality whenever $R$ has $\delta$-centrality. By Proposition \ref{prop:modularcentralityissymmetric}, it is enough to show that $R^{\circ_{n-1}}$ has $(\delta, 0)$-centrality.

So, let $\gamma \in R^{\circ_{n-1}}$ be such that every $(0)$-supporting line is a $\delta$-pair. Our task is to show that the $(0)$-pivot line of $\gamma$ is also a $\delta$-pair. By the definition of $R^{\circ_{n-1}}$, there are $\mu_0, \dots, \mu_{s-1} \in A^{2^{n-1}}$ so that
\begin{enumerate}
\item $\faces_{n-1}^0(\gamma) = \mu_0$,
\item $ \faces_{n-1}^1(\gamma) = \mu_{s-1}$, and
\item $\glue_{\{n-1\}}(\langle \mu_r, \mu_{r+1} \rangle ) \in R$, for each
$r \in s-1$.
\end{enumerate}

\begin{claim}\label{claim:perpclaim1}

Take $d \in \mathbb{D}_{n-1}$ to be a leaf. The sequence
$
(\mu_0)^d, \dots, (\mu_{s-1})^d \in A^{2^{n-1}}
$
satisfies
\begin{enumerate}
\item for all $r \in s$, each $(n-2)$-supporting line of $(\mu_r)^d$ is a $\delta$-pair,
\item every $(n-2)$-cross section line of $(\mu_0)^d $ is a $\delta$-pair, and
\item $\glue_{\{n-1\}}(\langle (\mu_r)^d, (\mu_{r+1})^d \rangle) \in R$,
 for all
$r \in s-1$.

\end{enumerate}
\end{claim}

\begin{claimproof}
Suppose $d = (d_0, \dots, d_{n-3})$. The first property of the claim follows from \emph{(2)} of Lemma \ref{lem:treesuccessorrotations} and the fact that $\delta$ contains all constant pairs.
To show the second property of the claim, we proceed by induction over the branch in $\mathbb{D}_{n-3}$ determined by $d = (d_0, \dots, d_{n-3})$. We assume $\mu_0 = \faces_{n-1}^0(\gamma)$ and that every $(0)$-supporting line of $\gamma$ is a $\delta$-pair. It follows that every $(0)$-cross section line of $\mu_0 = (\mu_0)^\emptyset$ is a $\delta$-pair, which establishes the basis of the induction. Now let $(d_0, \dots, d_i)$ be an ancestor of $d$ and suppose that every $(i+1)$-cross section line of $(\mu_0)^{(d_0, \dots, d_i)}$ is a $\delta$-pair. Now \emph{(3)} of Lemma \ref{lem:basicrotationproperties} shows that every $(i+2)$-cross section line of $(\mu_0)^{(d_0, \dots, d_{i+1})}$ is a $\delta$-pair. It follows that every $(n-2)$-cross section line of $(\mu_0)^d$ is a $\delta$-pair, as claimed.

A similar induction using \emph{(1)} and \emph{(4)} of Lemma \ref{lem:basicrotationproperties} establishes the third property of the claim.

\end{claimproof}

\begin{claim}\label{claim:perpclaim2}
If $d \in \mathbb{D}_{n-1}$ is a leaf, then the $(n-2)$-pivot line of $(\mu_{r+1})^d$ is a $\delta$-pair for all $r \in s-1$.
\end{claim}
\begin{claimproof}
The claim follows by induction on $r \in  s$. The claim holds for $r=0$ by \emph{(2)} of Claim  \ref{claim:perpclaim1}. Suppose the claim holds for $\mu_r$ for $r \in s-2$. This assumption along with \emph{(1)} and \emph{(3)} of Claim \ref{claim:perpclaim1} show that every $(n-2)$-supporting line of
\[
\glue_{\{n-1\}}(\langle (\mu_{r+1})^d, (\mu_{r+2})^d \rangle) \in R
\]
is a $\delta$-pair. We now apply the assumption that $R$ has $\delta$-centrality and conclude that the $(n-2)$-pivot line of this cube is also a $\delta$-pair. Because
\[\glue_{\{n-1\}}(\langle (\mu_{r+1})^d, (\mu_{r+2})^d \rangle)
\] and $(\mu_{r+2})^d$ have the same $(n-2)$-pivot line, the claim is proved.
\end{claimproof}

\begin{claim}\label{claim:perpclaim3}
Let $c  \in \mathbb{D}_{n-1}$ be a tuple of length $z \in n-1$. The $(z)$-pivot line of $(\mu_{s-1})^c$ is a $\delta$-pair. In particular, the $(0)$-pivot line of
$(\mu_{s-1})^\emptyset = \mu_{s-1}$ is a $\delta$-pair.
\end{claim}
\begin{claimproof}

We proceed by an induction from the leaves of $\mathbb{D}_{n-1}$ to its root. The basis has been established by Claim \ref{claim:perpclaim2}. Suppose that the claim holds for all tuples of length $z +1 \in n-1$ belonging to $\mathbb{D}_{n-1}$ and let $c \in \mathbb{D}_{n-1}$ be a tuple of length $z$. Our assumption that every $(0)$-supporting line of $\gamma$ is a $\delta$-pair implies that every $(0)$-supporting line of $\mu_{s-1}$ is a $\delta$-pair. An induction using \emph{(3)} of Lemma \ref{lem:basicrotationproperties} shows that every $(z)$-supporting line of $(\mu_{s-1})^c$ is a $\delta$-pair. In particular, the $(z)$-supporting line of the $(z,z+1))$-pivot square of $(\mu_{s-1})^c$ is a $\delta$-pair. The inductive assumption that the claim holds for all $d \in \mathbb{D}_{n-1}$ of length $z+1$ implies that the $(z+1)$-pivot line of $\rot_{z,z+1}^{e}((\mu_{s-1})^c)$ is a $\delta$-pair for every $e \in k+1$. It follows that \emph{(2)} of Lemma \ref{lem:basicrotationproperties} may be applied, and we conclude that the $(z)$-pivot line of $(\mu_{s-1})^c$ is a $\delta$-pair.

\end{claimproof}

The $(0)$-pivot lines of $\gamma$ and $\mu_{s-1}$ are the same, and the conclusion of Claim \ref{claim:perpclaim3} is that the $(0)$-pivot line of $\mu_{s-1}$ is a $\delta$-pair. This is what we wanted to show, so the proof is finished.

\end{proof}

\begin{thm}\label{thm:tc=hhcinmodular}
If $\var$ is a modular variety, then the $(n)$-ary term condition commutator for $\var$ is equal to the $(n)$-ary hypercommutator for $\var$.
\end{thm}

\begin{proof}
Let $\var$ be modular and let $\A \in \var$. Take $n \geq 2$ and pick some $(\theta_0, \dots, \theta_{n-1}) \in \Con(\A)^n$. Our task is to show that

\[
[\theta_0, \dots, \theta_{n-1}]_{TC} = [ \theta_0, \dots, \theta_{n-1}]_{H}.
\]
The hypercommutator is always bigger than the term condition commutator, so it suffices to show that $\Delta(\theta_0, \dots,\theta_{n-1})$ has $\delta$-centrality, where $\delta = [\theta_0, \dots, \theta_{n-1}]_{TC}$ (see 2.17 of \cite{taylorsupnil}). This follows from an induction using Proposition \ref{prop:tchascentrality}. Indeed,

\[
\Delta(\theta_0, \dots, \theta_{n-1}) = \tc(M(\theta_0, \dots, \theta_{n-1})),
\]
and $M(\theta_0, \dots, \theta_{n-1})$ has $\delta$-centrality by definition.
\end{proof}

The theory of the hypercommutator for any variety includes HC8 as a true statement (we denote this fact by HHC8, see 4.13 of \cite{taylorsupnil}), so the following corollary is immediate.

\begin{cor}
HC8 is a true statement belonging to the theory of the modular higher commutator.
\end{cor}

\subsection{The $\Delta$ relation in a modular variety}

In this section we develop a method for generating the $\Delta$ relation from lower dimensional $\Delta$ relations in the case that the algebra under investigation belongs to a modular variety. As we will see later, this characterization of $\Delta$ is important for the computations that demonstrate the existence of the `higher dimensional' Kiss terms for a modular variety.

We first look at the relationship between the definition of a $(2)$-dimensional congruence and the development of the binary modular commutator in \cite{FM}. For this paragraph let $\A$ be an algebra belonging to a modular variety and let $\alpha, \beta \in \Con(\A)$. In Chapter 4 the authors define the relation $\Delta_{\beta, \alpha}$ to be the least congruence of $\beta$ that collapses any two diagonal pairs whose values are $\alpha$-related. The authors use $2\times2$ matrices to represent pairs in this relation, with the convention that rows or columns determine $\alpha$ or $\beta$ pairs, respectively. The reader is asked to prove in exercise 5 of Chapter 4 that $\Delta_{\beta, \alpha}$ is a $(2)$-dimensional congruence (we are translating the statement using our language). This means that, in the modular setting, the $\Delta_{\beta, \alpha}$ defined in \cite{FM} is the same relation the $\Delta(\alpha, \beta)$ defined in \cite{taylorsupnil}.

For us the notion of a $(2)$-dimensional congruence is taken as a definition, so the ideas in \cite{FM} can be turned around to give sufficient conditions for a $(2)$-dimensional tolerance to be a $(2)$-dimensional congruence in a modular variety. Informally, the next lemma asserts that in a modular variety, any $(2)$-dimensional tolerance that is `almost' a $(2)$-dimensional congruence is a $(2)$-dimensional congruence.

\begin{lem}\label{lem:almost2transitivegivestransitive}
Let $\var$ be a modular variety and let $\A \in \var$. Let $k \neq l \in \nat$ and suppose that $R \leq \A^{2^{\{k,l\}}}$ is a $(2)$-dimensional tolerance of $\A$. If $\faces_k(R) \in \Con(\faces_k^0(R))$ and $\faces_k^0(R) \in \Con(\A)$, then $R$ is a $(2)$-dimensional congruence.
\end{lem}

\begin{proof}
Before we begin, we remark that this proof is essentially identical to the solution of exercise 5 of Chapter 4 of \cite{FM}. Because we have weakened the assumptions and changed the terminology, we reproduce the argument here for the convenience of the reader.

In this proof every labeled square belonging to $R$ is oriented so that the $k$-direction is horizontal and the $l$-direction is vertical. Our goal is to show that $\faces_l(R)$ is a transitive relation. We begin by establishing the following special case.

\begin{claim}\label{claim:shiftingclaim}
If
$
\Square[a][c][a][d][.6], \Square[b][a][b][a][.6] \in R,
$
then $\Square[b][c][b][d][.6] \in R$.
\end{claim}
\begin{claimproof}
We denote by $\theta$ the congruence $\faces_k^0(R)$. The assumptions of the claim guarantee that $a,b,c$ and $d$ belong to the same $\theta$-class, so in particular
\[
\LinePic[b][c][.6], \LinePic[b][d][.6] \in \theta.
\]
To demonstrate that these two pairs are related by $\faces_k(R)$, it suffices to apply the Shifting Lemma (see 2.4 of \cite{FM}) to the following diagram, where the $\eta_i$'s are projection kernels.

\[
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$\LinePic[b][c][.6]$}
      -- ++(2,0)  node (ne) {$\LinePic[a][c][.6]$}
      -- ++(0,-1.5) node (se) {$\LinePic[a][d][.6]$}
      -- ++(-2,0) node (sw) {$\LinePic[b][d][.6]$};
    \draw (nw) -- node[above] {$\eta_0$}(ne) -- (se) -- (sw) -- node[right] {$\eta_1$} (nw);

    \draw (ne) edge[bend left] node[right] {$\faces_k(R)$} (se);

    \end{tikzpicture}
\]

\end{claimproof}
Moving on to the general case, suppose that
$
\Square[u][x][v][y][.6], \Square[r][u][s][v][.6] \in R.
$
Suppose that $\var$ has Day terms $m_0, \dots, m_n$. We want to show that $\Square[r][x][s][y][.6] \in R$. By 2.3 of \cite{FM}, it is enough to see that
\[
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$m_e(x,x,y,y)$}
      -- ++(2.2,0)  node (ne) {$m_e(x,x,y,y)$}
      -- ++(0,-1) node (se) {$m_e(r,u,v,s)$}
      -- ++(-2.2,0) node (sw) {$m_e(r,r,s,s)$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);
    \end{tikzpicture}
    \in R,
\]
for every $e \in n+1$. The assumptions imply that both
\[
\LinePic[u][u][.6] \faces_k(R) \LinePic[v][v][.6]
\qquad \text{and} \qquad
\LinePic[s][v][.6] \faces_k(R) \LinePic[r][u][.6],
\]
and it follows that
\[
\LinePic[m_e(r,r,s,s)][m_e(u,u,v,v)][.6]
\faces_k(R)
\LinePic[m_e(r,r,r,r)][m_e(u,u,u,u)][.6]
=
\LinePic[r][u][.6]
=
\LinePic[m_e(r,v,v,r)][m_e(u,v,v,u)][.6]
\faces_k(R)
\LinePic[m_e(r,u,v,s)][m_e(u,u,v,v)][.6].
\]
Because $\faces_k(R)$ is transitive, we have shown that
\newline
$
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$m_e(u,u,v,v)$}
      -- ++(2.2,0)  node (ne) {$m_e(u,u,v,v)$}
      -- ++(0,-1) node (se) {$m_e(r,u,v,s)$}
      -- ++(-2.2,0) node (sw) {$m_e(r,r,s,s)$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);
    \end{tikzpicture} \in R.
$ Furthermore, it is easy to see that
\newline
$
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$m_e(x,x,y,y)$}
      -- ++(2.2,0)  node (ne) {$m_e(x,x,y,y)$}
      -- ++(0,-1) node (se) {$m_e(u,u,v,v)$}
      -- ++(-2.2,0) node (sw) {$m_e(u,u,v,v)$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);
    \end{tikzpicture}
    \in R,
$
and the lemma follows by applying Claim \ref{claim:shiftingclaim}.

\end{proof}

We now argue that the previous lemma is actually a special case of a more general principle. Informally, this principle guarantees that in a modular variety, every $(n)$-dimensional tolerance that is `almost' an $(n)$-dimensional congruence is an $(n)$-dimensional congruence. More precisely, we have the following proposition.

\begin{prop}\label{prop:almostnconisncon}
Let $\var$ be a modular variety and let $\A \in \var$. Let $n \geq 2$ and suppose that $R \leq \A^{2^n}$ is an $(n)$-dimensional tolerance of $\A$ with the property that $\faces_i^0(R)$ is a $(n-1)$-dimensional congruence for each $i \in n$. If there exists $k\in n$ such that $\faces_k(R) \in \Con(\faces_k^0(R))$, then $R$ is an $(n)$-dimensional congruence.
\end{prop}

\begin{proof}
Because $R$ is an $(n)$-dimensional tolerance, we just need to show that $R$ is $(n)$-transitive. Let $k$ be as in the proposition statement. We assume that $\faces_k(R)$ is transitive, so let $l \neq k \in n$. We use Lemma \ref{lem:almost2transitivegivestransitive} to deduce that $R$ is transitive in the $l$-direction from the transitivity of $R$ in the $k$-direction. We first ask the reader to verify the following easy claim.

\begin{claim}\label{claim:sufficienttochecksquares}
Let $s\neq t \in n$.
 The relation $\faces_s(R) $ is transitive if and only if $\faces_s(\cut_{\{s,t\}}(R))$ is transitive. Similarly, $\faces_t(R) $ is transitive if and only if $\faces_t(\cut_{\{s,t\}}(R))$ is transitive.
\end{claim}

Now, a typical element of
$\cut_{\{k,l\}}(R)$ looks like

\[
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$\cut_{\{k,l\}}(\gamma)_{\{\langle k,0 \rangle, \langle l,1\rangle \}}$}
      -- ++(7,0)  node (ne) {$\cut_{\{k,l\}}(\gamma)_{\{\langle k,1 \rangle, \langle l,1\rangle \}}$}
      -- ++(0,-2) node (se) {$\cut_{\{k,l\}}(\gamma)_{\{\langle k,1 \rangle, \langle l,0\rangle \}}$}
      -- ++(-7,0) node (sw) {$\cut_{\{k,l\}}(\gamma)_{\{\langle k,0 \rangle, \langle l,0\rangle \}}$};
      \node at (3.5,-1) {$\leftarrow \faces_k(\cut_{\{k,l\}}(R)) \rightarrow$};
    \draw (nw) --  (ne) -- (se) -- node[below] {$\faces_k(\faces_l^0(R))$} (sw) -- node[left] {$\faces_l(\faces_k^0(R))$}(nw);
    \end{tikzpicture}.
\]
We can apply Lemma \ref{lem:almost2transitivegivestransitive}, with the algebra under consideration set to
\newline
$\cut_{\{k,l\}}(R)_{\{\langle k, 0 \rangle , \langle l, 0 \rangle\}}$ and the $(2)$-dimensional relation under consideration set to $\cut_{\{k,l\}}(R)$. Claim \ref{claim:sufficienttochecksquares} ensures that $\faces_k(\cut_{\{k,l\}}(R))$ is transitive. It is similarly straightforward to see that the other assumptions of the lemma are satisfied, so we conclude that $\faces_l(\cut_{\{k,l\}}(R))$ is transitive.

\end{proof}

\begin{thm}\label{thm:deltaofdeltaisdelta}
Let $\var$ be a modular variety. Let $\A \in \var$, $S \subseteq \nat$ with $|S| \geq 3$, and $\{\theta_k\}_{k \in S} \in \Con(\A)^S$. Let $Q \subseteq S$. If for each $i \in Q$ we set
\begin{align*}
\alpha_i &= \faces_i
\bigg(
\Delta
\left(
\{\theta_j\}_{j \in \{i\} \union S \setminus Q}
\right)
\bigg), \text{ then}\\
\Delta
\left(
\{ \theta_k \}_{k \in S}
\right)
 &= \glue_Q
\bigg(
\Delta
\left(
\{\alpha_i\}_{ i \in Q}
\right)
\bigg).
\end{align*}

\end{thm}

\begin{proof}
We should first explain how the theorem statement makes sense. First, notice that for each $i \in Q$,
\[
\alpha_i \in
\Con
\bigg(
\Delta
\left(
\{\theta_j\}_{j \in  S \setminus Q}
\right)
\bigg).
\]
It follows that
\begin{align*}
\Delta(\{\alpha_i\}_{i \in Q}) &\subseteq
\left(
\Delta
\left(
\{\theta_j\}_{j \in  S \setminus Q}
\right)
\right)^Q
\subseteq (A^{2^{S \setminus Q}})^Q.
\end{align*}
Hence, $\Delta(\{\alpha_i\}_{i \in Q})$
is a subset of the domain of the mapping $\glue_{Q}$ and theorem's assertion contains no formal conflict. We also notice that if $|Q| \in \{ 0, 1, |S| \}$, then the theorem is trivial, and so we assume that we are not dealing with one of these cases.

Set
$\Delta = \Delta
\left(
\{ \theta_k \}_{k \in S}
\right)$ and
$\Delta' = \glue_Q
\bigg(
\Delta
\left(
\{\alpha_i\}_{ i \in Q}
\right)
\bigg).
$
The result will follow after we notice that
\[
M
\left(
\{\theta_k\}_{k \in S}
\right)
\subseteq
\Delta'
\subseteq \Delta
\]
and then demonstrate that $\Delta'$ is a $(|S|)$-dimensional congruence. Denote by $\cube(x)$ the vertex labeled cube with every vertex labeled by $x$. Note that the dimension of this cube depends on the context in which it is used.

To show that
$M
\left(
\{\theta_k\}_{k \in S}
\right)
\subseteq
\Delta'
$,
it is enough to see that $\Delta'$ contains its generating set. Let $k \in S$ and $\langle x,y \rangle \in \theta_k$. We have that

\[
\cut_Q(\cube_k(x,y)) =
\begin{cases}
\cube_k( \cube(x), \cube(y)) &\text{ if } k \in Q \\
\cube(\cube_k(x,y)) &\text{ if } k \in S \setminus Q.
\end{cases}
\]
In either case, $\cut_Q(\cube_k(x,y))$ is a generator of $\Delta\left( \{\alpha_i\})_{i \in Q} \right)$, hence $\cube_k(x,y) \in \Delta'$.

To show that $\Delta' \subseteq \Delta$, we will demonstrate that
$\glue_Q(M(\{\alpha_i\}_{i \in Q})) \subseteq \Delta$. This is adequate, because it has the consequence that
\[
\Delta' = \glue_Q(\tc
\left(
M(\{\alpha_i\}_{i \in Q})
\right)
\subseteq \tc
\left(
\glue_Q(
M(\{\alpha_i\}_{i \in Q})
)
\right)
\subseteq \Delta .
\]
Because term operations commute with $\glue_Q$, we need only show that $\glue_Q(\mu) \in \Delta$, where $\mu$ is a generator of $M(\{\alpha_i\}_{i \in Q})$.
So, let $i \in Q$ and suppose $\mu = \cube_i( \langle \zeta, \eta \rangle)$, where $\zeta, \eta \in \Delta(\{\theta_j\}_{j \in S \setminus Q})$ and $\langle \zeta, \eta \rangle \in \alpha_i$. Now, Lemma 2.14 of \cite{taylorsupnil} indicates that $\cut_{Q\setminus \{i\}}(\Delta)_\textbf{1} = \Delta(\{\theta_j\}_{j \in \{i\} \union S\setminus Q})$. There is therefore a $\gamma \in \Delta$ satisfying
$
\cut_{Q\setminus{\{i\}}}(\gamma)_\textbf{1}= \glue_{\{i\}}(\langle \zeta, \eta\rangle).
$
Applying Corollary 2.5 of \cite{taylorsupnil} to this situation produces a $\tau$ such that
\[
\cut_{Q\setminus \{i\}}(\tau)_f = \glue_{\{i\}}(\langle \zeta, \eta\rangle)
\]
holds for every $f \in 2^{Q\setminus \{i\}}$. This exactly means that $\cut_Q(\gamma) = \mu$, or put another way, that $\glue_Q(\mu) = \gamma \in \Delta$.

Now we apply Proposition \ref{prop:almostnconisncon} to show that $\Delta'$ is a $(|S|)$-dimensional congruence. Let us establish that the conditions are satisfied. We first show that $\Delta'$ is a $(|S|)$-dimensional tolerance. First, we note that $\Delta' \subseteq \tc(\glue_Q(M(\{\alpha_i\}_{i\in Q})))$, so it is enough to establish that $\glue_Q(M(\{\alpha_i\}_{i\in Q}))$ is a $(|S|)$-dimensional tolerance (this follows from Lemma 2.9 of \cite{taylorsupnil}).

Let $\gamma \in \Delta'$, $k \in S$ and $j \in 2$. We want to show that $\refl^j_k(\gamma), \sym_k(\gamma) \in \glue_Q(M(\{\alpha_i\}_{i\in Q}))$. It is sufficient to check this for those $\gamma = \glue_Q(\mu)$ for $\mu$ a generator. The case when $k \in Q$ is obvious, so suppose $k \in S\setminus Q$. Let $i \in Q$ and $\langle \zeta, \eta \rangle \in \alpha_i$ determine a generator $\mu =\cube_i(\zeta, \eta)$. It follows from the definition of $\alpha_i$ that $\langle \refl^j_k(\zeta), \refl^j_k(\eta) \rangle, \langle \sym_k(\zeta), \sym_k(\eta) \rangle  \in \alpha_i$, therefore $\mu^{r_j} = \cube_i(\refl^j_k(\zeta), \refl^j_k(\eta))$ and $\mu^s = \cube_i(\sym_k(\zeta), \sym_k(\eta) )$ are also generators. It is easy to see that
$
\refl_k^j(\gamma) = \glue_Q(\mu^{r_j} )
$ and
$
\sym(\gamma) = \glue_Q(\mu^s)$.

Recall that we are assuming that $|Q| \geq 2$. Therefore, there exists $i \in Q$. Evidently, $\faces_i(\Delta')$ is a congruence. So, to finish the proof we need to show that
$
\faces_k^0(\Delta')
$
is a $(|S|-1)$-dimensional congruence, for each $k \in S$. Here we proceed by induction, supposing that the theorem we are proving is true for indexing sets with cardinality strictly less than $S$.

Take $k \in S$.
If $k \in Q$, we compute

\begin{align*}
\faces_k^0(\Delta') &=
\faces_k^0(\glue_Q(\Delta( \{\alpha_i\}_{i \in Q}))) \\
&= \glue_{Q \setminus \{k\}}(\faces_k^0(\Delta(\{\alpha_i\}_{i \in Q})))\\
&= \glue_{Q \setminus \{k\}}(\Delta(\{\alpha_i\}_{i \in Q\setminus \{k\}}))\\
&=_{\text{(inductive assumption)}} \Delta(\{\theta_i\}_{i \in Q \setminus \{k\}}).
\end{align*}
Now suppose $k \in S \setminus Q$.  For each $i \in Q$, set
$\beta_i =
\faces_i
\bigg(
\Delta
\left(
\{\theta_j\}_{j \in \{i\} \union S \setminus
(Q \union \{k\})}
\right)
\bigg)
$
Abusing notation slightly, we have that $\beta_i = \faces_k^0(\alpha_i)$. A straightforward induction on the number of transitive closures taken over $Q$ of $\glue_Q(M(\{\alpha_i\}_{i\in Q}))$ shows that
\begin{align*}
\faces_k^0(\Delta') &= \faces_k^0(\glue_Q(\Delta( \{\alpha_i\}_{i \in Q})))\\
&= \glue_Q(\Delta( \{\faces_k^0(\alpha_i)\}_{i \in Q}))\\
&= \glue_Q(\Delta( \{\beta_i\}_{i \in Q}))\\
&=_{\text{(inductive assumption)}} \Delta(\{\theta_i\}_{i \in Q \setminus \{k\}}).
\end{align*}

\end{proof}

\section{Higher dimensional Kiss terms}\label{sec:hdkt}

In \cite{threeremarks}, Kiss demonstrates that every modular variety $\var$ has a certain $4$-ary term which one may use to obtain from an arbitrary element of $\rect(\alpha, \beta)$ an element of $\Delta(\alpha, \beta)$, for any $\A \in \var$ and $\alpha, \beta \in\Con(\A)$. A Kiss term therefore has a close connection to the behaviour of the modular binary commutator. In this section we derive from the Kiss term a sequence of `higher dimensional' Kiss terms, each of which has a close connection to the behavior of a higher arity commutator.

\begin{defn}[see 3.3 of \cite{threeremarks}]\label{def:4differenceterm}
A four variable term $q_2$ is said to be a $(2)$-dimensional Kiss term for a variety $\var$ if
\begin{enumerate}
\item $q_2(x,x,y,y) \approx q_2(x,y,x,y) \approx y$ holds in $\var$, and
\item for all $\A \in \var$ and $\alpha, \beta \in \Con(\A)$, if
$
\Square[a][c][b][d][.5], \Square[a][c][b][d'][.5] \in \rect(\alpha, \beta),
$
then $\langle q_2(a,b,c,d), q_2(a,b,c,d') \rangle \in [ \alpha, \beta]_{TC} $.
\end{enumerate}
\end{defn}

The following proposition states an important feature of a $(2)$-dimensional Kiss term for a modular variety.
\begin{prop}[see 3.8 of \cite{threeremarks}]\label{prop:twokisscompletes}
Let $\var$ be a modular variety with a $(2)$-dimensional Kiss term $q_2$. Let $\A \in \var$ and $\alpha, \beta \in \Con(\A)$. If $\Square[a][c][b][d][.5] \in \rect(\alpha, \beta)$, then
\[
\Square[a][c][b][q_2(a,b,c,d)] \in \Delta(\alpha, \beta).
\]
\end{prop}

Before we go any further, we point out again that the definitions of $\Delta(\alpha, \beta)$ given in \cite{FM} and \cite{taylorsupnil} coincide in the binary modular case. Our use of the higher dimensional Kiss terms makes frequent use of the fact that $\Delta(\alpha, \beta)$ is a $(2)$-dimensional congruence. In particular, the construction relies on the following lemma.

\begin{lem}\label{lem:twokissmasterlemma}
Let $\var$ be a modular variety with a $(2)$-dimensional Kiss term $q_2$. Let $\A \in \var$ and let $\alpha, \beta \in \Con(\A)$. If $x,y,u,v,x',y',v' \in A$ are such that
\[
\Square[x][u][y][v][.6] \in \rect(\alpha, \beta)
\qquad \text{and} \qquad
\Square[x][u][y'][v'][.6] \in \Delta(\alpha, \beta),
\]

\[
\text{then }
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$u$}
      -- ++(1.5,0)  node (ne) {$q_2(y',y,v',v)$}
      -- ++(0,-1) node (se) {$y$}
      -- ++(-1.5,0) node (sw) {$x$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

    \end{tikzpicture}
\in \Delta(\alpha, \beta).
\]
\end{lem}

\begin{proof}
The assumptions imply that
\[
\Square[y'][v'][y][v][.6] \in \rect(\alpha, \beta),
\]
so it follows from Proposition \ref{prop:twokisscompletes} that
\[
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$v'$}
      -- ++(1.5,0)  node (ne) {$p(y',y,v',v)$}
      -- ++(0,-1) node (se) {$y$}
      -- ++(-1.5,0) node (sw) {$y'$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

    \end{tikzpicture}
\in \Delta(\alpha, \beta).
\]
Because $\Delta(\alpha, \beta)$ is $(2)$-transitive, the result follows.
\end{proof}

\begin{defn}[Higher dimensional Kiss terms]\label{def:higherkissterms}
Let $\var$ be a modular variety with a $(2)$-dimensional Kiss term $q_2$. For each $n \geq 3$, recursively define a $2^n$-ary term $q_n$ as
\[
q_n \coloneqq q_2 \big(
q_{n-1}(x_0, \dots, x_{2^{n-1}-1}), x_{2^{n-1}-1},
q_{n-1}( x_{2^{n-1}}, \dots, x_{2^n -1}) , x_{2^n -1}
\big)
\]
\end{defn}

We order the functions belonging to $2^n$ colexicographically in what follows. That is, we write $f < g$ if $f_i < g_i$, where $i$ is the greatest input on which $f$ and $g$ differ.
The remainder of this section is devoted to proving the following theorem.

\begin{thm}\label{thm:higherkisscompletes}
Let $\var$ be a modular variety with higher dimensional Kiss terms $q_n$ for $n \geq 2$. Let $\A \in \var$, $n \geq 2$, and $(\theta_0, \dots, \theta_{n-1}) \in \Con(\A)^n$. For $\gamma \in A^{2^n}$, let $\widehat{\gamma} \in A^{2^n}$ be defined as
\[
\widehat{\gamma}_f =
\begin{cases}\gamma_f \text{ if } f \neq \textbf{1}\\
q_n(\gamma_\textbf{0}, \dots, \gamma_f, \dots , \gamma_\textbf{1}) \text{ if }
f = \textbf{1},
\end{cases}
\]
where the $\gamma_f$ are ordered according to the colexicographical ordering on $2^n$.

If $\gamma \in A^{2^n}$ has the property that $\faces_i^j(\gamma) \in
\Delta
\left( \{\theta_k\}_{k \in n\setminus \{i\}}
\right) $ for every $i \in n$ and $j \in 2$, then $\widehat{\gamma} \in \Delta(\theta_0, \dots, \theta_{n-1})$.
\end{thm}

\begin{proof}[Proof when $n=3$]

Notice that the $n=2$ case of Theorem \ref{thm:higherkisscompletes} is handled by Proposition \ref{prop:twokisscompletes}. The proof of this special case is included to illustrate the proof of the general case. So, suppose that we are in an $n=3$ situation and that

\[
\gamma  = \Cube[a][c][b][d][e][g][f][h] \in A^{2^3}
\]
satisfies the condition in the theorem statement.

Set $\beta_0 = \faces_0(\Delta(\theta_0, \theta_2))$ and $\beta_1= \faces_1(\Delta(\theta_1, \theta_2))$. Our assumption about $\gamma$ implies that

\[
\cut_{\{0,1\}} (\gamma) =
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.02em}, scale=1.2 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (1nwA) {$c$}
      -- ++(1.3,0)  node (1neA) {$d$}
      -- ++(0,-1) node (1seA) {$b$}
      -- ++(-1.3,0) node (1swA) {$a$};
    \path (.5,-.5)   node (0nwA) {$g$}
      -- ++(1.3,0)  node (0neA) {$h$}
      -- ++(0,-1) node (0seA) {$f$}
      -- ++(-1.3,0) node (0swA) {$e$};
     \draw (0nwA) -- node (c) {} (1nwA);
     \draw (0swA) -- node (d) {}(1swA);
     \draw (0neA) -- node (a) {} (1neA);
     \draw (0seA) -- node (b) {} (1seA);
     \draw (a) -- (b) -- (d) -- (c) -- (a);
    \end{tikzpicture}
\in \rect(\beta_0, \beta_1).
\]
We apply Lemma \ref{lem:twokissmasterlemma} and obtain
$
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.02em}, scale=1.2 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (1nwA) {$c$}
      -- ++(1.3,0)  node (1neA) {$q_2(a,b,c,d)$}
      -- ++(0,-1) node (1seA) {$b$}
      -- ++(-1.3,0) node (1swA) {$a$};
    \path (.5,-.5)   node (0nwA) {$g$}
      -- ++(1.3,0)  node (0neA) {$q_2(e,f,g,h)$}
      -- ++(0,-1) node (0seA) {$f$}
      -- ++(-1.3,0) node (0swA) {$e$};
     \draw (0nwA) -- node (c) {} (1nwA);
     \draw (0swA) -- node (d) {}(1swA);
     \draw (0neA) -- node (a) {} (1neA);
     \draw (0seA) -- node (b) {} (1seA);
     \draw (a) -- (b) -- (d) -- (c) -- (a);
    \end{tikzpicture}
\in \Delta(\beta_0, \beta_1).
$
Theorem \ref{thm:deltaofdeltaisdelta} indicates that
$
\glue_{\{0,1\}}\left(\Delta(\beta_0, \beta_1)\right) =
\Delta(\theta_0, \theta_1, \theta_2),
$
so we conclude that
\[
\gamma' =
\Cube[a][c][b][q_2(a,b,c,d)][e][g][f][q_2(e,f,g,h)][1.2]
\in \Delta(\theta_0, \theta_1, \theta_2).
\]

Set $\alpha_1= \faces_1\left(\Delta(\theta_0, \theta_1)\right)$ and $\alpha_2= \faces_2\left(\Delta(\theta_0,\theta_2)\right)$. Applying the mapping $\cut_{\{1,2\}}$ to $\gamma$ and $\gamma'$ respectively produces the following labeled squares, each belonging to the universe of the indicated algebra:
\[
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.02em}, scale=1.2 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (1nwA) {$c$}
      -- ++(1.3,0)  node (1neA) {$d$}
      -- ++(0,-1) node (1seA) {$b$}
      -- ++(-1.3,0) node (1swA) {$a$};
    \path (.5,-.5)   node (0nwA) {$g$}
      -- ++(1.3,0)  node (0neA) {$h$}
      -- ++(0,-1) node (0seA) {$f$}
      -- ++(-1.3,0) node (0swA) {$e$};
    \node at (.65,0) (a) {\bf{y}};
    \node at (.65,-1) (b) {\bf{x}};
    \node at (1.15,-.5) (c) {\bf{v}};
    \node at (1.15,-1.5) (d) {\bf{u}};
     \draw (0nwA) -- (c) -- (0neA);
     \draw (0swA) -- (d) -- (0seA);
     \draw (1nwA) -- (a) -- (1neA);
     \draw (1swA) -- (b) -- (1seA);
     \draw (a) -- (b) -- (d) -- (c) -- (a);
    \end{tikzpicture}
\in \rect
\left(
\alpha_1, \alpha_2
\right)
\text{ and }
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.02em}, scale=1.2 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (1nwA) {$c$}
      -- ++(1.6,0)  node (1neA) {$q_2(a,b,c,d)$}
      -- ++(0,-1) node (1seA) {$b$}
      -- ++(-1.6,0) node (1swA) {$a$};
    \path (.5,-.5)   node (0nwA) {$g$}
      -- ++(1.6,0)  node (0neA) {$q_2(e,f,g,h)$}
      -- ++(0,-1) node (0seA) {$f$}
      -- ++(-1.6,0) node (0swA) {$e$};
    \node at (.7,0) (a) {\bf{y'}};
    \node at (.7,-1) (b) {\bf{x}};
    \node at (1.2,-.5) (c) {\bf{v'}};
    \node at (1.2,-1.5) (d) {\bf{u}};
     \draw (0nwA) -- (c) -- (0neA);
     \draw (0swA) -- (d) -- (0seA);
     \draw (1nwA) -- (a) -- (1neA);
     \draw (1swA) -- (b) -- (1seA);
     \draw (a) -- (b) -- (d) -- (c) -- (a);
    \end{tikzpicture}
\in \Delta(\alpha_1, \alpha_2).
\]

Indeed, the truth of the first membership follows from our assumption about $\gamma$ and the truth of the second membership follows from Theorem \ref{thm:deltaofdeltaisdelta}. The vertex labels of the above labeled squares have been named so that we may apply Lemma \ref{lem:twokissmasterlemma}, and we conclude that
\[
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.02em}, scale=1.2 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,.1)   node (1nwA) {$c$}
      -- ++(3,0)  node (1neA) {$d$}
      -- ++(0,-1) node (1seA) {$b$}
      -- ++(-3,0) node (1swA) {$a$};
    \path (.5,-.4)   node (0nwA) {$q_2(g,g,g,g)=g$}
      -- ++(3,0)  node (0neA) {$q_3(a,b,c,d,e,f,g,h)$}
      -- ++(0,-1) node (0seA) {$f$}
      -- ++(-3,0) node (0swA) {$e$};
    \node at (1.5,.1) (a) {};
    \node at (1.5,-.9) (b) {};
    \node at (2,-.4) (c) {};
    \node at (2,-1.4) (d) {};
     \draw (0nwA) -- (c) -- (0neA);
     \draw (0swA) -- (d) -- (0seA);
     \draw (1nwA) -- (a) -- (1neA);
     \draw (1swA) -- (b) -- (1seA);
     \draw (a) -- (b) -- (d) -- (c) -- (a);
    \end{tikzpicture}
 \in \Delta (\alpha_1, \alpha_2 ).
\]
A final application of Theorem \ref{thm:deltaofdeltaisdelta} finishes the proof of Theorem \ref{thm:higherkisscompletes} for the case $n=3$.

\end{proof}

\begin{proof}[Proof of general case]
Let $n \geq 3$ and suppose that Theorem \ref{thm:higherkisscompletes} holds for $q_{n-1}$. Let $\gamma \in A^{2^n}$ satisfy the conditions of the theorem statement. For each $i \in n-1$, set $\beta_i = \faces_i(\Delta(\theta_i, \theta_{n-1}))$. Let $i \in n-1$. We assume that
$
\faces_i^j(\gamma) \in \Delta\left( \{\theta_k\}_{k \in n\setminus \{i\}} \right),
$
for each $j \in 2$.
We apply the $\lines_{n-1}$ map to each side of this and conclude that
\[
\lines_{n-1}\left(
\faces_i^j(\gamma)
\right)
=
\faces_i^j
\left(
\lines_{n-1}(\gamma)
\right)
\in \lines_{n-1}
\left(
\Delta\left( \{\theta_k\}_{k \in n\setminus \{i\}} \right)
\right).
\]
We now use Theorem \ref{thm:deltaofdeltaisdelta} and the fact that $\glue_{n-1}$ is the inverse of $\lines_{n-1}$ to see that
\begin{align*}
\lines_{n-1}
\left(
\Delta\left( \{\theta_k\}_{k \in n\setminus \{i\}} \right)
\right)
&= \lines_{n-1}
\left(
\glue_{n-1}
\left(
\Delta
\left( \{\beta_k\}_{k \in n-1\setminus \{i\}}
\right)
\right)
\right)\\
&=
\Delta
\left( \{\beta_k\}_{k \in n-1\setminus \{i\}}
\right).
\end{align*}
Set $\mu = \lines_{n-1}(\gamma)$. We have just shown that $\mu$ satisfies the assumption of the theorem we are proving for the lower dimension of $n-1$, and we inductively assume the theorem to hold in this case. There therefore exists $\widehat{\mu}$ as in the theorem statement such that
$
\widehat{\mu} \in \Delta
\left( \{\beta_k\}_{k \in n-1}
\right).
$
Set $\gamma' = \glue_{n-1}(\widehat{\mu})$. We apply Theorem \ref{thm:deltaofdeltaisdelta} once again and conclude that
$
\gamma' \in \Delta(\theta_0, \dots, \theta_{n-1}).
$

It follows from the definition of $\widehat{\mu}$ that
\[
\gamma'_f=
\begin{cases}
 q_{n-1}
\big(
\underbrace{
\faces_{n-1}^0(\gamma)_\textbf{0}, \dots, \faces_{n-1}^0(\gamma)_g, \dots,
\faces_{n-1}^0(\gamma)_\textbf{1}
}_{\text{colex ordering on $g \in 2^{n-1}$}}
\big)
\text{ if } f = (1, \dots, 1,0), \\
 q_{n-1}
\big(
\underbrace{
\faces_{n-1}^1(\gamma)_\textbf{0}, \dots, \faces_{n-1}^0(\gamma)_g, \dots,
\faces_{n-1}^1(\gamma)_\textbf{1}
}_{\text{colex ordering on $g \in 2^{n-1}$}}
\big)
\text{ if } f=(1, \dots, 1,1), \\
\gamma_f \text{ otherwise.}
\end{cases}
\]
Set $\zeta = \cut_{\{n-2, n-1\}}(\gamma) $ and $\zeta' = \cut_{\{n-2,n-1\}}(\gamma').$
We now have

\[
\zeta= \Square[\textbf{x}][\textbf{u}][\textbf{y}][\textbf{v}]
\text{ and }
\zeta' = \Square[\textbf{x}][\textbf{u}][\textbf{y'}][\textbf{v'}]
\]
where
\begin{align*}
\textbf{x} &= \cut_{\{n-2, n-1\}}(\gamma)_{\{\langle n-2, 0\rangle, \langle n-1, 0 \rangle \}},
 \\
\textbf{y} &= \cut_{\{n-2, n-1\}}(\gamma)_{\{\langle n-2, 1\rangle, \langle n-1, 0 \rangle \}},
&&\textbf{y'} = \cut_{\{n-2, n-1\}}(\gamma')_{\{\langle n-2, 1\rangle, \langle n-1, 0 \rangle \}}, \\
\textbf{u} &= \cut_{\{n-2, n-1\}}(\gamma)_{\{\langle n-2, 0\rangle, \langle n-1, 1 \rangle \}},
 \\
\textbf{v} &= \cut_{\{n-2, n-1\}}(\gamma)_{\{\langle n-2, 1\rangle, \langle n-1, 1 \rangle \}}.
 &&\textbf{v'}= \cut_{\{n-2, n-1\}}(\gamma')_{\{\langle n-2, 1\rangle, \langle n-1, 1 \rangle \}}. \\
\end{align*}
Our goal is to apply Lemma \ref{lem:twokissmasterlemma}, with the algebra under consideration set to $\Delta(\theta_0, \dots, \theta_{n-3})$ and the two congruences under consideration set to
\[
\alpha_{n-2} = \faces_{n-2}(\Delta(\{\theta_j\}_{j\in n\setminus \{n-1\}}))
\text{ and }
\alpha_{n-1} = \faces_{n-1}(\Delta(\{\theta_j\}_{j\in n\setminus \{n-2\}})).
\] Again, the assumption we made of $\gamma$ implies that $\zeta \in \rect(\alpha_{n-2}, \alpha_{n-1})$, while Theorem \ref{thm:deltaofdeltaisdelta} implies that $\zeta' \in \Delta(\alpha_{n-2}, \alpha_{n-1})$. Having established the conditions of Lemma \ref{lem:twokissmasterlemma}, we conclude that
\[
\epsilon=
\begin{tikzpicture}
    [ baseline=(center.base), font=\small,
      every node/.style={inner sep=0.25em}, scale=1 ]
    \node at (0.5,-0.5) (center) {\phantom{$\cdot$}};
    \path (0,0)   node (nw) {$\textbf{u}$}
      -- ++(1.5,0)  node (ne) {$q_2(\textbf{y'}, \textbf{y}, \textbf{v'}, \textbf{v})$}
      -- ++(0,-1) node (se) {$\textbf{y}$}
      -- ++(-1.5,0) node (sw) {$\textbf{x}$};
    \draw (nw) -- (ne) -- (se) -- (sw) -- (nw);

    \end{tikzpicture}
\in \Delta(\alpha_{n-2}, \alpha_{n-2}).
\]

\end{proof}

It is now straightforward to check that, for $g \in 2^{n-2}$,
\[
q_2(\textbf{y'}, \textbf{y}, \textbf{v'}, \textbf{v})_g =
q_2(\textbf{y'}_g, \textbf{y}_g, \textbf{v'}_g, \textbf{v}_g)=
\begin{cases}
\textbf{v}_g &\text{ if } g\neq \textbf{1}\\
q_n(\underbrace{\gamma_\textbf{0}, \dots, \gamma_f, \dots, \gamma_{\textbf{1}}}_{\text{colex ordering on $f \in 2^{n}$}})
&\text{ if } g  =\textbf{1}.
\end{cases}
\]
Therefore, we set $\widehat{\gamma} = \glue_{\{n-2, n-1\}}(\epsilon)$. It is similarly straightforward to check that $\widehat{\gamma}$ has the property claimed by the theorem, so the proof is finished.

\section{An application of higher dimensional Kiss terms}\label{sec:application}
In this section we show that, for an algebra $\A$ belonging to a modular variety, the collection of all $\Delta(\theta_0, \dots, \theta_{n-1})$ for all $n\geq 2$ and $(\theta_0, \dots, \theta_{n-1}) \in \Con(\A)^n$ is completely determined by the higher commutator. We established in \cite{taylorsupnil} that this collection of relations completely determines the hypercommutator for $\A$. In view of Theorem \ref{thm:tc=hhcinmodular}, this means that the term condition higher commutator and the collection of all $\Delta$ relations completely determine each other for an algebra with Day terms. After this is established we will follow \cite{orsalrel} and conclude that any collection of clones on some set that share congruences, higher commutators, and Day terms has a greatest element.

We begin with a lemma which, we wish to note, could be used to provide a cleaner proof of parts of Theorem 4.10 from \cite{taylorsupnil}. Recall from \cite{taylorsupnil} that for an algebra $A$, $n\geq 2$, and $x,y \in A$, the $(n)$-dimensional commutator cube for the pair $\langle x, y\rangle$ is the labeled cube belonging to $A^{2^n}$ such that the pivot vertex is labeled by $y$, while the other vertices are labeled by $x$. This cube is denoted $\com_n(x,y)$.

\begin{lem}\label{lem:deltashifting}
Let $A$ be an algebra, $n\geq 1$, and $R \subseteq A^{2^n}$ an $(n)$-dimensional congruence. Let $\gamma \in R$. If $\com_n(\gamma_\textbf{1}, q) \in R$ for some $q \in A$, then $\widehat{\gamma} \in R$, where
\[
\widehat{\gamma}_f =
\begin{cases}
q &\text{ if } f = \textbf{1}\\
\gamma_f &\text{ otherwise}.
\end{cases}
\]
\end{lem}

\begin{proof}
The proof proceeds by induction. The case when $n=1$ is an obvious application of the fact that a $(1)$-dimensional congruence is a transitive relation. So, suppose that $n\geq 2$ and that the lemma holds for $n-1$. Let $R \leq A^{2^n}$, $\gamma \in R$, and suppose that there is a $q \in A$ such that $\com_n(\gamma_{\textbf{1}}, q) \in R$. Suppose that the $(n-1)$-pivot line of $\gamma$ is the pair $\langle z, \gamma_\textbf{1} \rangle$. It follows from Corollary 2.5 of \cite{taylorsupnil} that $\cube_{n-1}(z, \gamma_\textbf{1}) \in R$. Because $R$ is $(n)$-transitive, we conclude that
\begin{align*}
\mu &= \glue_{n-1}
\left(
\langle
\faces_{n-1}^0(\cube_{n-1}(z,\gamma_\textbf{1})),
\faces_{n-1}^1(\com_n(\gamma_\textbf{1},q))
\rangle
\right)\\
&= \glue_{n-1}(\langle \cube(z) , \com_{n-1}(\gamma_\textbf{1}, q) \rangle) \in R.
\end{align*}
We can describe $\mu$ more concretely, where for $f \in 2^n$
\[
\mu_f =
\begin{cases}
z &\text{ if } f_{n-1}= 0\\
\gamma_\textbf{1} &\text{ if } f_{n-1} =1 \text{ and } f\neq \textbf{1}\\
q &\text{ if } f = \textbf{1}.
\end{cases}
\]

It follows from Lemma 2.4 of \cite{taylorsupnil} that $\lines_{n-1}(R)$ is a $(n-1)$-dimensional congruence of $\A^2$. Notice that
$\lines_{n-1}(\mu) = \com_{n-1}(\langle z, \gamma_\textbf{1} \rangle, \langle z, q \rangle )$. So, we can apply inductive assumption to get that $ \epsilon \in \lines_{n-1}(R)$, where, for $g \in n-1$, we have that
\[
\epsilon_g=
\begin{cases}
\lines_{n-1}(\gamma)_g &\text{ if } g \neq \textbf{1}, \\
\langle z, q \rangle &\text{ if } g = \textbf{1}.
\end{cases}
\]
To finish the proof, set $\widehat{\gamma} = \glue_{\{n-1\}}(\epsilon)$.

\end{proof}

\begin{prop}\label{prop:characterizeDelta}
Let $\A$ be an algebra belonging to a modular variety $\var$. Let $n\geq 2$, $q_n$ be an $(n)$-dimensional Kiss term for $\var$, and $(\theta_0, \dots, \theta_{n-1}) \in \Con(\A)^n$. Then, $\gamma \in \Delta(\theta_0, \dots, \theta_{n-1})$ if and only if
\begin{enumerate}
\item $\faces_i^j(\gamma) \in
\Delta
\left( \{\theta_k\}_{k \in n\setminus \{i\}}
\right) $ for every $i \in n$ and $j \in 2$, and
\item $\langle \gamma_\textbf{1} , q_n(\gamma_\textbf{0}, \dots, \gamma_f, \dots, \gamma_{\textbf{1}}) \rangle \in [\theta_0, \dots, \theta_{n-1}]$.
\end{enumerate}
\end{prop}

\begin{proof}
Suppose that $\gamma \in \Delta(\theta_0, \dots, \theta_{n-1})$. It follows from Lemma 2.14 of \cite{taylorsupnil} that $\gamma$ satisfies \emph{(1)}, which also means that $\gamma$ is within the scope of Theorem \ref{thm:higherkisscompletes}. Let $\widehat{\gamma}$ be as in the conclusion of Theorem \ref{thm:higherkisscompletes}.
It follows from the $(n)$-symmetry and $(n)$-transitivity of $\Delta(\theta_0, \dots, \theta_{n-1})$ that
\[\mu = \glue_{\{n-1\}}
\left(
\langle \faces_{n-1}^1(\gamma), \faces_{n-1}^1(\widehat{\gamma}) \rangle
\right)
\in \Delta(\theta_0, \dots, \theta_{n-1}).
\]
Because the pair $\langle \gamma_\textbf{1} , q_n(\gamma_\textbf{0}, \dots, \gamma_f, \dots, \gamma_{\textbf{1}} )\rangle$ is $(n-1)$-supported by $\mu$, we may apply Theorem 4.10 of \cite{taylorsupnil} along with \ref{thm:tc=hhcinmodular} and conclude that $\langle \gamma_\textbf{1} , q_n(\gamma_\textbf{0}, \dots, \gamma_f, \dots, \gamma_{\textbf{1}} )\rangle \in [\theta_0, \dots, \theta_{n-1}]$. This establishes \emph{(2)}.

Now, suppose that \emph{(1)} and \emph{(2)} hold. We may again apply Theorem \ref{thm:higherkisscompletes} and obtain $\widehat{\gamma} \in \Delta(\theta_0, \dots, \theta_{n-1})$. Next, we use \emph{(2)} along with Theorem 4.10 of \cite{taylorsupnil} to conclude that $\com_n( q_n(\gamma_\textbf{0}, \dots, \gamma_f, \dots, \gamma_{\textbf{1}}) ,\gamma_\textbf{1} ) \in \Delta(\theta_0, \dots, \theta_{n-1})$. We apply Lemma \ref{lem:deltashifting} and conclude that $\gamma \in \Delta(\theta_0, \dots, \theta_{n-1})$.
\end{proof}

In what follows we need to distinguish between commutators computed in different algebras, provided it makes sense to do so. Let $\A$ and $\mathbb{B}$ be algebras with a common universe and suppose that $\{\theta_i\}_{i \in S} \subseteq \Con(\A)\cap \Con(\mathbb{B})$ for some $S \finsub \nat$. We denote the $\Delta(\{\theta_i\}_{i \in S})$ and $[\{\theta_i\}_{i \in S}]$ computed in either $ \A$ or $\mathbb{B}$ with a subscript, i.e.\ $\Delta_\A( \{\theta_i\}_{i \in S})$ and $[\{\theta_i\}_{i \in S}]_\A$ are these objects computed in $\A$, while $\Delta_\mathbb{B}( \{\theta_i\}_{i \in S})$ and $[\{\theta_i\}_{i \in S}]_\mathbb{B}$ are computed in $\mathbb{B}$. We make no effort to distinguish between the term condition and hypercommutator here, because each algebra we work with belongs to a modular variety.

\begin{cor}[cf.\ Corollary 4.3 of \cite{orsalrel}]\label{cor:shareddelta}
If $\A$ and $\mathbb{B}$ are algebras that share a universe, Day terms $p_0, \dots, p_k$, and congruences $\{\theta_i\}_{i \in S}$ for some $S \finsub \nat$, then
$[\{\theta_i\}_{i \in T}]_\A = [\{\theta_i\}_{i \in T}]_\mathbb{B}$ for all $T \subseteq S$
if and only if
$\Delta_\A( \{\theta_i\}_{i \in S}) = \Delta_\mathbb{B}( \{\theta_i\}_{i \in S})$.
\end{cor}

\begin{proof}
Notice that, because $\A$ and $\mathbb{B}$ share a choice of Day terms, they also share a sequence $q_2, \dots, q_n, \dots $ of higher dimensional Kiss terms. For the forward implication we proceed by induction on the size of $S$. The statement is trivial if $|S| =1$. Let $n\geq 2$ and assume it holds for $|S| =n-1$. Take $\gamma \in \Delta_\A(\{\theta_i\}_{i \in S})$. Then $\gamma$ satisfies \emph{(1)} and \emph{(2)} of Proposition \ref{prop:characterizeDelta}. Applying the inductive assumption, the assumption that $[\{\theta_i\}_{i \in T}]_\A = [\{\theta_i\}_{i \in T}]_\mathbb{B}$, and the fact that these algebras share the term $q_n$ allows us to conclude that $\gamma$ satisfies \emph{(1)} and \emph{(2)}, now for the algebra $\mathbb{B}$. This shows that $\gamma \in \Delta_\mathbb{B}(\{\theta_i\}_{i \in S})$. So, $\Delta_\A(\{\theta_i\}_{i \in S}) \subseteq \Delta_\mathbb{B}(\{\theta_i\}_{i \in S})$. The other containment is proved in an identical manner.

The other direction follows from Lemma 2.14 and Theorem 4.10 of \cite{taylorsupnil}.
\end{proof}

\begin{thm}[cf.\ Theorem 1.3 of \cite{orsalrel}]\label{thm:comparecommutators}
Let $\A$ be an algebra with Day terms $m_0, \dots, m_k$. Suppose $S \finsub \nat$ and let $\{\theta_i\}_{i \in S} \subseteq \Con(\A)$. There exists a greatest clone $\mathcal{C}$ on $A$ satisfying
\begin{enumerate}
\item $m_0, \dots, m_k \in \mathcal{C} $,
\item $\mathcal{C}$ preserves $\{\theta_i\}_{i \in S}$, and
\item $[\{\theta_i\}_{i \in T}]_\A = [\{\theta_i\}_{i \in T}]_{\langle A; \mathcal{C}\rangle}$ for every $T \subseteq S$.
\end{enumerate}
\end{thm}

\begin{proof}
Let $\mathcal{C}$ be the collection of all polymorphisms of $\Delta_\A(\{\theta_i\}_{i \in S}$. We first show that \emph{(1)}, \emph{(2)} and \emph{(3)} hold. It is immediate that \emph{(1)} holds. It follows from Lemma 2.14 of \cite{taylorsupnil} that $\Delta_\A(\{\theta_i\}_{i \in T})$ is definable from $\Delta_\A(\{\theta_i\}_{i \in S})$ by a positive primitive formula, for every $T\subseteq S$. Therefore, $\mathcal{C}$ preserves all of these lower dimensional relations as well. In particular, \emph{(2)} holds. Because $\clo(\A) \subseteq \mathcal{C}$, we have that $\Delta_\A(\{\theta_i\}_{i \in S}) \subseteq \Delta_{\langle A; \mathcal{C}\rangle } (\{\theta_i\}_{i \in S})$. Because $\Delta_\A (\{\theta_i\}_{i \in S})$ is preserved by $\mathcal{C}$ and is a higher dimensional equivalence relation containing the generators of $ \Delta_{\langle A; \mathcal{C}\rangle } (\{\theta_i\}_{i \in S})$, we see that the two relations are equal. We apply Corollary \ref{cor:shareddelta} and conclude that \emph{(3)} holds.

Now we show that $\mathcal{C}$ is the greatest such clone. Suppose that $\mathcal{D}$ also satisfies \emph{(1)}, \emph{(2)}, and \emph{(3)}. We apply Corollary \ref{cor:shareddelta} to the algebras $\langle A; \mathcal{C}\rangle$ and $\langle A; \mathcal{D} \rangle$ and conclude that $\Delta_{\langle A; \mathcal{C}\rangle } (\{\theta_i\}_{i \in S}) = \Delta_{\langle A; \mathcal{D}\rangle } (\{\theta_i\}_{i \in S})$. Therefore, every operation belonging to $\mathcal{D}$ preserves $\Delta_\A(\{\theta_i\}_{i \in S})$, hence $\mathcal{D} \subseteq \mathcal{C}$.
\end{proof}

\begin{cor}[cf.\ Corollary 1.4 of \cite{orsalrel}]
Let $\A$ be an algebra that has Day terms $m_0, \dots, m_k$. There exists a largest clone $\mathcal{C}$ on $A$ satisfying
\begin{enumerate}
\item $p_0, \dots, p_k \in \mathcal{C}$
\item $\langle A; \mathcal{C} \rangle$ and $\A$ have the same congruences, and
\item $\langle A; \mathcal{C} \rangle$ and $\A$ have the same higher commutator operations.
\end{enumerate}
\end{cor}
\begin{proof}
The clone $\mathcal{C}$ is the intersection of all possible clones guaranteed by Theorem \ref{thm:comparecommutators}.
\end{proof}

\bibliographystyle{amsplain}
\bibliography{refs.bib}
\begin{center}
  \rule{0.61803\textwidth}{0.1ex}
\end{center}
\subjclass{MSC 08A40 (08A05, 08B05)}
\end{document}

config: arxiv


TITLE: Does Michelson Morley experiment assume that the Earth drags aether?
QUESTION [1 upvotes]: My knowledge is that the purpose of Michelson Morley experiment was to measure the velocity of Earth with respect to aether. It was assumed that the Earth does not "drag" the aether or interfere with it, as it was shown by measures on stellar aberration by Bradley.
Nevertheless, considering the red light ray in picture, in order to hit the mirror this ray must travel at the same velocity of earth horizontally, otherwise it would surely miss the mirror.

So does this mean that in the  description of the experiment it is assumed that the Earth drags the aether? Otherwise why should the ray have the horizontal velocity of Earth (which allows it to hit the mirror) instead of just going on a straight line and miss the mirror?

REPLY [3 votes]: The light ray only need to be aimed towards the future position of the mirror. Let $\Delta t$ be the time it takes for the wave front to move from the first splitter mirror to the second mirror (horizontal on your diagram). The latter has moved by $v\Delta t$ and therefore light has to travel a distance $\sqrt{L^2 + (v\Delta t)^2}$ to reach it, and that distance shall be equal to $c\Delta t$ as light propagates at the speed $c$ in all directions in the Aether frame where your diagram is drawn, which yield
$$\Delta t = \frac{L}{\sqrt{c^2-v^2}},$$
which is half of $T_t$ on your diagram, as it should be.
But the light ray is not purposely "aimed toward the future position of the mirror". It is only the incoming light ray reflecting by the splitter mirror. Naively applying the reflection law "incident angle = reflected angle", we would expect the reflected ray to go vertically on your diagram, so what is going on? In fact, this reflection law is not valid for a moving mirror. So first, let's see what is the reflection law in the case you wondered about. The following diagram is the same as yours with extra information relevant to the question at hand.

The light rays are in red. The angle between the mirror plane and the speed of the lab frame with respect to the Aether is $\varphi=\pi/4$; the incident angle is $\alpha=\pi/4$; the reflected angle is $\beta=\pi/4+\theta$. Then the argument I developed in my first paragraph implies that
$$\sin\theta=\frac{v\Delta t}{c\Delta t}=\frac{v}{c}.$$
Since $v/c \ll 1$, the angle $\theta$ is very small and therefore $\sin\theta \approx\theta$, hence putting everything together
$$\beta = \frac{\pi}{4} + \frac{v}{c}.\tag{1}$$
This is the reflection law in this configuration. It is a special case of the general law. Let me update the above diagram of mine to present the general case.

The reflection law valid when $v/c \ll 1$ is
$$\beta = \alpha + 2\frac{v}{c}\sin\varphi\sin\alpha.\tag{2}.$$
As you can see, with $\alpha=\varphi=\pi/4$, eqn. (2) gives indeed eqn. (1).
The last step is then to derive (2). There is an excellent paper Gju04 working out the law of reflection for any speed $v$ by using Huygens' principle. Another paper of the same author Gju04bis derives it using Fermat's principle. It should be noted that Einstein did derive the formula in his seminal relativity paper!
[Gju04] Aleksandar Gjurchinovski. Reflection of light from a uniformly moving mirror. American Journal of Physics, 72(10):1316–1324, 2004.
[Gju04bis] Aleksandar Gjurchinovski. Einstein’s mirror and fermat’s principle of least time. American Journal of Physics, 72(10):1325–1327, 2004.


TITLE: Product of finite order elements in a group
QUESTION [2 upvotes]: Let $G$ be a group. Let $a,b\in G$ be of finite order.
Prove or disprove:
(1) If $ab$ has finite order, then $ba$ has finite order.
(2) If $ab$ has finite order, then $a^{-1}b^{-1}$ has finite order.
I do not see both (1) and (2) has to be true. However, I could not give a counter example. I would guess the counterexample must be some kind of non-abelian group.

REPLY [0 votes]: @Timbuc, you still have to prove (2) of the original OP. To this end we note that ord($a$)=ord($a^{-1}$), that is, the order of an inverse of an element is the same as the order of the element in question. Since $(ab)^{-1}=b^{-1}a^{-1}$, one can now apply (1).


TITLE: Calculating Magnetic Field Strength from FFT Amplitude
QUESTION [1 upvotes]: Briefly on the exposition; I'm an undergraduate assistant to a professor at UMass Amherst. We contribute to the Muon g-2 experiment in Fermilab, designing and optimizing the magnetic-measurement equipment. As you might imagine I utilize the Fourier Transform often to analyze data. The data I'm analyzing is photo-current which comes from a laser of known power and is modulated by a magnetic field. The modulation by the magnetic field perturbs the plane of polarization in a well-defined manner (Faraday Rotation), resulting in a modulation of the intensity of photo-current and thus the voltage measured in our DAQ Assist box.
I use LabView to analyze the data and thus use a proprietary "virtual instrument" in the program to do the FFT for me (obviously as the sheer amount of data would be overwhelming!). I understand the concept of an FFT, it being a mathematical tool to take a signal and decompose it into it's frequency domain and measure the strength of each frequency directly by their amplitude, as they contribute to the resultant "summed" signal. However what I need to understand is how to translate the amplitude of one of the frequencies (say 60 hz) into the amplitude for the magnetic field. In other words my professor wants me to be able to tell the magnetic field strength from the amplitudes present on our power spectrum. Now I have a few questions...
i) Can this be done by simply analyzing one frequency and it's amplitude. My intuition says no as I would imagine the same strength magnetic field could produce different profiles for the power spectrum. However the way my professor puts it he makes it seem like he wants me to look and see "oh our 60 hz signal is at 1x10^-4, so that corresponds to a field of x Tesla"
Well, I guess I don't have any more questions but if anyone could explain to me (or show me) how to go about calculating the strength of the magnetic field from the power spectrum that would be incredible.

REPLY [1 votes]: Relating the power in a frequency spectrum can be both straightforward and not depending on how the signal varies with the magnetic field. Multiplying the original signal by a constant scales the Fourier transform by an equal amount, so doubling the magnetic field strength should double the amplitude of every frequency. However, that assumes you are directly measuring the field.
I'm going to make some assumptions about your setup and show a possible analysis.

The laser power is constant.
The two polarizers (one magnetically modulated and one constant) are crossed so that zero magnetic field means zero light is transmitted.
The rotation of the polarizing angle in the magnetically modulated polarizer is proportional to the magnetic field strength.
The laser shines through both polarizers before hitting the laser power meter.

The amplitude of light let through the polarizers when the angle between them is $\theta$ off perpendicular is
$$A = a\sin(\theta).$$
The measuring device actually detects power, so the signal is
$$P = a^2\sin^2(\theta),$$
where $a^2$ is the total input laser power. The angle $\theta(t)$ is given by $kB(t),$ where $k$ is some constant empirically derived from the polarizer material and dimensions and $B(t) = B_0\sin(\omega t)$ is the magnetic field strength parallel to the light. As a function of time, the instantaneous power is
$$P = a^2\sin^2(kB_0\sin(\omega t)).$$
Assuming the angle change $\theta$ is small, we can say that $\sin\theta \approx \theta.$
$$P = a^2(kB_0\sin(\omega t))^2 = a^2k^2B_0^2\sin^2(\omega t).$$
From the double-angle trig formulas, we know that a $\sin^2$ signal is really a sinusoid at twice the original frequency with a DC offset. So, the laser power transmitted varies at twice the frequency of the magnet. The amplitude of any frequency of the oscillation will be proportional to the square of the maximum magnetic field.
Getting the magnetic field from this can be complicated since you need to know $k$ from the varying polarizer, the total laser power, and how that laser power varies over time. You also need to check how the FFT of your software deals with multiplicative constants ($1/2\pi$, $1$, $1/\sqrt{2\pi}$, etc.).
If everything was ideal, you would measure the amplitude of the power oscillations at $2\omega,$ since every other frequency would be from noise, imperfections in the function generator waveform, and other uninteresting sources.


TITLE: Evaluate this limit in terms of f
QUESTION [4 upvotes]: I want to evaluate the following limit:
$$\lim_{d\to x} \dfrac{\dfrac{2x}{f'(x)}+f(x)-f(d)-\dfrac{x^2-d^2}{f(x)-f(d)}}{2\left(\dfrac{d-x}{f(x)-f(d)}+\dfrac{1}{f'(x)}\right)}$$
I tried L'hopital's rule but it just keeps getting worse and worse.
I got this limit by wondering about circles fitting on a curve at a point $(x,f(x))$ and this limit is the $x$ coordinate of the circles center, in terms of a dummy point $d$.

REPLY [2 votes]: This isn't pretty, but it works:
Multiply the top and bottom by $f'(x)(f(x)-f(d))$ to clear denominators, giving:
$$
\lim_{d \to x} \frac{2x(f(x)-f(d))+f'(x)(f(x)-f(d))^2-f'(x)(x^2-d^2)}{2(f(x)-f(d)-f'(x)(x-d))}
$$
By adding $(x-d)^2f'(x)-(x-d)^2f'(x)$ to the numerator, this becomes
$$
\lim_{d \to x} \frac{2x(f(x)-f(d))-2xf'(x)(x-d) + f'(x)(f(x)-f(d))^2+f'(x)(x-d)^2}{2(f(x)-f(d)-f'(x)(x-d))}
$$
$$
 = x + \lim_{d \to x} \frac{f'(x)(f(x)-f(d))^2+f'(x)(x-d)^2}{2(f(x)-f(d)-f'(x)(x-d))}
$$
Now use L'Hopital's Rule, differentiating with respect to $d$, to give:
$$
=x+ \lim_{d \to x} \frac{-2f'(x)f'(d)(f(x)-f(d))-2f'(x)(x-d)}{2f'(x)-2f'(d)}
$$
This is still an indeterminate form, so cancel the $2$'s, and use L'Hopital's Rule again:
$$
=x + \lim_{d \to x} \frac{f'(x)f'(d)^2-f'(x)f''(d)(f(x)-f(d)) + f'(x)}{-f''(d)}
$$
$$
=x - \lim_{d \to x} \frac{f'(x)f'(d)^2+f'(x)}{f''(d)}
$$
$$
=x - \frac{f'(x)+(f'(x))^3}{f''(x)}
$$
As long as $f''(x) \neq 0$.


TITLE: Expected time of sequence getting typed when the letters are typed randomly
QUESTION [5 upvotes]: This question has been asked in an examination of the Indian  Statistical Institute, Kolkata for second year Master of Statistics students in the subject Martingale Theory.

Q. Mr.Trump  decides to post a random message on the  Facebook   and
  starts typing a random sequence $\{U_k\}_{k\geq1}$ of letters such that
  they are chosen independently and uniformly  from the $26$ English
  alphabets. Find out the expected time of the first appearance of the
  word "COVFEFE". We may assume that Trump has his caps lock on so
  that only upper case letters are typed. Assume further that the letters
  are typed at the rate of one letter per second.

I have no idea as to how to proceed. I will be grateful for any help.

REPLY [1 votes]: I'm a little late to the party, but here's my solution:
Let's say we have a word $w$ in an alphabet of size $b$, and we have an infinite stream of random letters. let's say we have a gremlin that runs along the stream up until the first occurrence of the word, and then starts from scratch at the next letter. for example, if $w=aba$ then given the stream $aababaababb$ the gremlin will find $a{\bf aba}baababb$ first, then start over from the next letter and find the next occurence of $w$, $a{\bf aba}ba{\bf aba}bb$. note the gremlin missed a copy of $w$ because it overlapped a copy that it did find.
we can think of the stream as a sequence of "gremlin runs", which are the runs from the point the gremlin starts at to the end of the next occurrence of $w$. for example, in the example the runs are $a{\bf aba}$, $ba{\bf aba}$, with a trailing $bb$ (one can assume there are no trails in the infinite stream case). all the runs are completely independent and have the exact same distribution. if we denote by $L$ the expected length of a gremlin run (which is the quantity we are trying to find), and denote by $\mu$ the density of the occurrences of $w$ that the gremlin finds, we get $L=\frac{1}{\mu}$. (the density is the ratio between how many occurrences the gremlin found to how many letters he read, when taken to infinity.)
whenever our gremlin found the word, it is possible that it missed another instance of $w$ sharing $k$ of the same letters, exactly when the first and last $k$ letters of the word are the same. (for example, for $aba$ the first $a$ is the same as the last $a$ so it's true for $k=1$, so it's possible the gremlin would find ${\bf aba}ba$ and miss the second copy). in that case, there are $|w|-k$ more letters to chose, so the chance of having a missed copy of $w$ sharing $k$ letters is exactly $b^{|w|-k}$.
denote $c_k=1$ if the first and last $k$ letters of $w$ are the same for $1\le k<|w|$ and $c_k=0$ otherwise. also denote $\rho=\sum_{k=1}^{|w|-1}c_k b^{k-|w|}$. so, the density of missed occurrences of $w$ is exactly $\rho\mu$, and the density of occurrences of $w$ is $(1+\rho)\mu$. that's just plainly $b^{-|w|}$, so $\mu=\frac{b^{-|w|}}{1+\rho}$.
therefore, $L=\frac{1}{\mu}=(1+\rho)b^{|w|}=\sum_{k=1}^{|w|-1}c_k b^{k}+b^{|w|}$. if you will, the first and last $|w|$ letters of $w$ are just the word itself, so $c_{|w|}=1$, and $L=\sum_{k=1}^{|w|}c_k b^{k}$.
this would explain why the "covfefe" example gave such a nice answer - $26^7$.
formal sidenote: the strong law of large numbers formally justifies that the densities exist with probability 1 and that everything that was claimed holds with probability 1.


TITLE: Boolean Algebra: Prove associate law for products
QUESTION [0 upvotes]: I'm working through some old papers for an exam coming up, and I found this question that I cannot seem to answer, and I don't have the memos.
Could someone please help me out?
The question:
Prove the associative law for products: For any elements a, b, and c in a Boolean
algebra,
$$(a ∗ b) ∗ c = a ∗ (b ∗ c).$$
Begin your proof by letting $L = (a ∗ b) ∗ c$ and
$R = a ∗ (b ∗ c)$, and first show that $a + L = a + R$, and then that $a′ + L = a′ + R$.
You may only use the axioms of Definition 1.2, and then also idempotency, boundedness, and absorption.
And Definition 1.2:

Commutativity:
$a+b=b+a$
$a*b=b*a$
Distributivity:
$a + (b ∗ c) = (a + b) ∗ (a + c)$
$a ∗ (b + c) = (a ∗ b) + (a ∗ c)$
Identities:
$a + 0 = a$
$a * 1 = a$
Complements:
$a + a' = 1$
$a *a' = 0$

Any help would be greatly appreciated

REPLY [1 votes]: Let us calculate those terms:
\begin{align}
a + L
&= a + ( (a * b) * c )\\
&= (a + (a * b)) * (a + c)\tag{distributivity}\\
&= a * (a + c)\tag{absortion}\\
&= a.\tag{absortion}
\end{align}
\begin{align}
a + R &= a + ( a * (b*c) )\\
&= a. \tag{absortion}
\end{align}
Thus, $a+L=a+R$.
\begin{align}
a' + R
&= a'+ (a * (b*c))\\
&= (a' + a) * (a' + (b*c))\tag{distributivity}\\
&= 1 * ((a' + b) * (a' + c))\tag{distributivity, complement}\\
&= (a' + b) * (a' + c).\tag{identities}
\end{align}
\begin{align}
a' + L
&= a' + ((a*b)*c)\\
&= (a' + (a*b)) * (a' + c)\tag{distributivity}\\
&= ((a' + a) * (a' + b))*(a' + c)\tag{distributivity}\\
&= ( 1 * (a' + b) ) * (a' + c)\tag{complement}\\
&= (a' + b) * (a' + c),
\end{align}
and so $a' + R = a' +L$.
Now given that $a+L=a+R$ and $a' + R = a' +L$,
\begin{align}
L
&= (a * a') + L \\
&= (a + L) * (a' + L)  \\
&= (a + R) * (a' + R)\\
&= (a * a') + R\\
&= R,
\end{align}
where by now, I suppose you can justify the above equalities.


TITLE: How are complex sound waves combined?
QUESTION [3 upvotes]: Audio is often explained by single frequencies. Typically this is a sound wave:
plot sin(x) * 2 from 0 to 10

However we usually deal with more complex sounds, more specific various frequencies and amplitudes. Varying amplitudes within same frequency etc. in the same time frame.
How are such sound patterns visualized? Say we have the previous tone + this one:
plot sin(5x) * 2 from 0 to 10

Would this be the end-result? Would a wave like this travel trough the air?
plot (sin(5x) + sin(x)) * 2 from 0 to 10

Even a mono track can have a rather detailed sound picture of say a philharmonic playing some composition. Would such a recording (or live observation for that matter – but that would give a different picture I guess as one would have multiple sound sources) be a complexified × ten folds version of the two sine waves?

My end goal is to better understand PCM audio and how digital audio works. A starting point is to better understand the physics behind audio. Then again a most of what I find deals with one frequency samples and the like. (I'm likely missing some terminology.)

REPLY [3 votes]: Yes, you have the right idea.  You will want to learn about Fourier analysis, which lets you take a complicated-looking waveform like your third figure and analyze it to say "this is two sine waves, frequencies 1 and 5, equal amplitudes, zero relative phase."
I like to think of a piano as an inverse Fourier transform machine: you push the keys to tell the piano "please generate frequencies C, E, and G, with the C having larger amplitude than the others" and the piano makes the air vibrate for you.  Your auditory system then does the ordinary Fourier transformation: with ear training, you can take those vibrations and say "Oh, a major triad, with a strong root."