$\tau\in\\{0,1\\}^{m}$ with the property that for all questions $x,y,z\leq m$
there is some mindset $\mu_{i}$ such that
$\tau(x)=\mu_{i}(x),\tau(y)=\mu_{i}(y)$ and $\tau(z)=\mu_{i}(z)$, but $\tau$
itself is not a mindset.

Let $x,y,z\leq m$ such that there are as few mindsets $\mu_{i}$ as possible
with $\tau(x)=\mu_{i}(x),\tau(y)=\mu_{i}(y)$ and $\tau(z)=\mu_{i}(z)$ and
suppose without loss of generality that $\mu_{1}$ does so. Since every
partition of the mindsets is induced by some question, there is a question
$x^{\prime}$ such that $\mu_{i}(x^{\prime})=\mu_{i}(x)$ for all $i\neq 1$ and
$\mu_{1}(x^{\prime})\neq\mu_{1}(x^{\prime})$.

Now if $\tau(x^{\prime})\neq\mu_{1}(x^{\prime})$, then the triple
$x^{\prime},y,z$ would contradict the fact that minimally many mindsets
answered $x,y,z$ the same way as $\tau$.

Hence we may assume that $\tau(x^{\prime})=\mu_{1}(x^{\prime})$. Now, since
$\tau\neq\mu_{1}$ there is some question $w\leq m$ with
$\tau(w)\neq\mu_{1}(w)$. But then there is no mindset $\mu_{i}$ which answered
the triple $x,x^{\prime},w$ in the same way as $\tau$, which likewise
contradicts the choice of $x,y,z$. ∎

We will prove the two directions of this statement separately, starting with
showing that, with high probability, every mindset gives rise to a tangle:

###### Lemma 9 (Mindsets give tangles).

Let $k,m$ be fixed and let $\alpha=a/n$ be the agreement parameter as fraction
of $n$. If $p~{}<~{}(n-ka)/(3n)$, then the probability that there is one of
the $k$ mindsets which does not induce a tangle is bounded above by

$km\exp\left(-\frac{2n}{9k}(1-k\alpha-3p)^{2}\right),$

which tends to $0$ as $n$ goes to $\infty$.

For more questions and more mindsets it gets less likely that every ground
truth mindset induces a tangle. The larger the number of questions $m$ we
consider the more likely that for one specific question, many (significantly
more than $p\cdot n_{i}$) persons do not agree with their ground truth
mindset. In this case the data might not represent the ground truth anymore,
so we might not be able to recover it. If we consider a larger number of
mindsets, we increase the number of possibilities to make an error. Also the
number of persons per mindset decreases and thus for fixed $\alpha$ will get
inconsistent if they become too small. It is intuitive that the probability of
mindsets inducing tangles also decreases with increasing noise $p$. For a
large number of persons $n$ answering the questionnaire the probability of a
statistical oilier decreases and so does the probability of mindsets not
inducing tangles.

###### Proof.

If some mindset $\mu_{i}$ is not a tangle then it contains a triple of
questions whose intersection has size at most $a$. Hence by the pigeon-hole
principle for at least one of these three questions at most $(2n_{i}+a)/3$ of
the $n_{i}$ persons in $V_{i}$ answered that question as $\mu_{i}$ does.

Each person in $V_{i}$ answers a question as $\mu_{i}$ does with probability
$(1-p)$. We can apply Hoeffding’s tail bound on the binomial distribution with
success percentage $(1-p)$ to find that

$\displaystyle\mathbf{P}[X\leq\frac{2n_{i}+a}{3}]$
$\displaystyle\leq\exp\left(-2\frac{(\frac{2n_{i}+a}{3}-(1-p)n_{i})^{2}}{n_{i}}\right)=\exp\left(-\frac{2n}{9k}(1-k\alpha-3p)^{2}\right)\,,$

where $X$ is the random variable counting the number of persons in $V_{i}$
that answer a fixed question as $\mu_{i}$ does. Note that we used
$a=n\alpha=kn_{i}\alpha$ here.

Since there are $k$ mindsets and $m$ questions, by the union bound we obtain
that the probability of the event that there is some question for which few
persons answer following their mindset is at most

$km\exp\left(-\frac{2n}{9k}(1-k\alpha-3p)^{2}\right)\,.$

Consequently so is the probability that some mindset is not a tangle. ∎

###### Lemma 10 (Only tangles from mindsets).

If $a=\alpha n$ and $p<a/n$, then, given Assumption 7, the probability that
there is a tangle which is not a mindset is bounded above by
$km\exp\left(-2(\alpha-p)^{2}n/k\right)$.

Increasing the number $m$ of questions or the number $k$ of mindsets makes it
more likely that there are tangles which do not completely agree with one of
the mindsets. Like for Lemma 9, the larger the number $m$ of questions or the
number $k$ of mindsets we consider, the more likely it is that for one
specific question and one mindset, significantly more than $p\cdot n_{i}$
persons from that mindset do not agree with the ground truth of that mindset.
This may then result in the existence of a tangle which agrees with one of the
mindsets on all but this question, where the tangle chooses the opposite
orientation, and thus corresponds to no existing mindset. For increasing $k$
this is especially true, since increasing $k$ with fixed $n$ results in
smaller values of $n_{i}$.

###### Proof.

By Assumption 7, for a tangle $\tau$ which does not correspond to a mindset
there need to exist $x_{1},x_{2},x_{3}$ such that for $y_{i}=\tau(x_{1})$ we
have that for $A_{1}\coloneqq A_{x_{1}}^{y_{1}}$, $A_{2}\coloneqq
A_{x_{2}}^{y_{2}}$ and $A_{3}\coloneqq A_{x_{3}}^{y_{3}}$ that $|A_{1}\cap
A_{2}\cap A_{3}|\geq a$ but no mindset $\mu$ incorporate $A_{1},A_{2},A_{3}$,
i.e. there is no $\mu$ with $\mu(x_{i})=y_{i}\;\forall i\in\\{1,2,3\\}$.

So suppose that there are such $A_{1},A_{2},A_{3}$ with $|A_{1}\cap A_{2}\cap
A_{3}|>a_{Q}$. Then there is, by the pigeon-hole principle, a mindset
$\mu_{i}$ for which at least $a/k$ many persons from $V_{i}$ lie in $A_{1}\cap
A_{2}\cap A_{3}$. If this mindset does not incorporate this triple, then it
does not incorporate one of the questions, say $\mu_{i}(x_{1})\neq y_{1}$.

We are going to compute the probability that there is some such $A_{x}^{y}$
and a mindset $\mu_{i}$ where more than $n_{i}-a_{Q}/k$ many persons from
$V_{i}$ lie in $A_{x}^{y}$, but $\mu_{i}(x)\neq y$. Each person in $V_{i}$
answers a question as $\mu_{i}$ does with probability $(1-p)$. Since it
follows from $p\leq\alpha$ that $n_{i}-a/k\leq(1-p)n_{i}$, we can apply
Hoeffding’s tail bound on the binomial distribution with success probability
$(1-p)$ to find that

$\displaystyle\mathbf{P}[X\leq
n_{i}-\frac{a}{k}]\leq\exp\left(-2\frac{(n_{i}-\frac{a}{k}-(1-p)n_{i})^{2}}{n_{i}}\right)=\exp\left(-\frac{2n}{k}(\alpha-p)^{2}\right)\,,$

where $X$ is the random variable counting the number of persons in $V_{i}$
that answer a fixed question as $\mu_{i}$ does. Note that we used
$a_{Q}=n\alpha_{Q}=kn_{i}\alpha_{Q}$ here.

Since there are $k$ mindsets and $m$ questions, by the union bound we obtain
that the probability of the event that there is some such $A_{x}^{y}$ and
$\mu_{i}$ is at most

$mk\exp\left(-\frac{2n}{k}(p-\alpha_{Q})^{2}\right)\,.$

Consequently the probability that there exists a triple $A_{1},A_{2},A_{3}$ as
above is also at most this value since – as argued above – the $A_{1}$ in such
a triple is such an $A_{x}^{y}$, witnessed by $\mu_{i}$. ∎

Proof of Theorem 2.  It is easy to see that in the setting of Theorem 2 the
requirements of the Lemmas 9 and 10 are satisfied.

—————————————————————————————

###  III.2 Stochastic Block Model

We will show Theorem 3 in two stages. The first step is to calculate the
maximum value of $\Psi$ for which the two orientations of $\mathcal{P}_{\Psi}$
induced by $V_{1}$ and $V_{2}$ respectively are indeed tangles.

For this we shall compute the minimum cost of a _forbidden triple_ in these
orientations i.e. a triple which violates $a$-consistency, that is a triple
$A,B,C$ of sides with the property that $\left\lvert A\cap B\cap
C\right\rvert<a$. This we will do as follows: suppose that $A,B,C$ is a
forbidden triple in the orientation given by $V_{1}$. We will sequentially
manipulate an existing such triple, while only reducing its cots, until we
obtain a triple of a specific form, whose cost is easy to calculate.

The following lemma says that (for a certain range of parameters) the cost of
cutting through both blocks is larger than cutting through only one block and
moving the other block entirely to one side of the cut.

###### Lemma 11 (Not cutting through a block reduces the cost).

Let $P=\\{A,A^{\complement}\\}$ be a cut, and let $\alpha_{i}=|A\cap
V_{i}|/|V_{i}|$ and $\beta_{i}=|A^{\complement}\cap
V_{i}|/|V_{i}|=1-\alpha_{i}$. If $\alpha_{2}\geq(1-2\alpha_{1})q/p$, then
${c(\\{A\cup V_{2},A^{\complement}\smallsetminus V_{2}\\})\leq c(P)}$.
Similarly, if $\beta_{2}\geq(1-2\beta_{1})q/p$, then $c(\\{A\smallsetminus
V_{2},A^{\complement}\cup V_{2}\\})\leq c(P)$.

In words: Given a cut $P$ which cuts through one of the blocks, say with a
fraction $\alpha_{2}$, if $\alpha_{2}>(1-2\alpha_{1})q/p$, then we can prove
that the cut that results from moving all these $\alpha_{2}n$ many points to
the other side is cheaper.

###### Proof.

For $\alpha_{2}=(1-2\alpha_{1})q/p$ we have by (5), that

$\displaystyle c(P)=$
$\displaystyle\frac{n^{2}}{4}\left(p(\alpha_{1}-\alpha_{1}^{2}+\alpha_{2}-\alpha_{2}^{2})+q(\alpha_{1}+\alpha_{2}-2\alpha_{1}\alpha_{2})\right)$
$\displaystyle=$
$\displaystyle\frac{n^{2}}{4}\left(p(\alpha_{1}-\alpha_{1}^{2})+q(1-\alpha_{1})\right)=c(\\{A\cup
V_{2},A^{\complement}\smallsetminus V_{2}\\})$

Now if we consider (5), for fixed $n,p,q,\alpha_{1}$ as a real-valued function
$f$ mapping $\alpha_{2}$ to $c(\\{A,A^{\complement}\\})$, then this is a
quadratic function with a unique maximum. Thus by $f(1)=f((1-2\alpha_{1})q/p)$
we have $f(x)\geq f(1)$ for every $(1-2\alpha_{1})q/p\leq x\leq 1$, i.e. for
$\alpha_{2}\geq(1-2\alpha_{1})q/p$ we have that $c(\\{A\cup
V_{2},A^{\complement}\smallsetminus V_{2}\\})\leq c(P)$.

The argument for $\beta_{2}\geq(1-2\beta_{1})q/p$ is analogous. ∎

We will now compute the lowest possible cost of a forbidden triple $A,B,C$ of
which $B$ and $C$ both completely contain $V_{2}$ but $A$ does not. We will
later use Lemma 11 to see that the minimum cost forbidden triples are of this
form.

###### Lemma 12 (Bound on the costs of a forbidden triple not cutting through
a block).

For $q/p\leq 1/2$, if $A,B,C$ are such that they each contain more than half
of $V_{1}$, their intersection has size $\leq a$, and $V_{2}$ is a subset of
$A^{\complement}$, $B$, and $C$, then the minimum possible value of
$\max\\{c(\\{A,A^{\complement}\\}),c(\\{B,B^{\complement}\\}),c(\\{C,C^{\complement}\\})\\}$
is

$\frac{n^{2}}{4}p\left(\frac{1}{3}(1+r-x)(1+r)-(\frac{1+r-x}{3})^{2}\right),\text{
where }r=\frac{q}{p},\,x=\frac{2a}{n}.$

with some such triple attaining this bound (up to discretization).

###### Proof.

We first calculate the cost $\\{A,A^{\complement}\\}$ would have if
$|A^{\complement}\cap V_{1}|=|V_{1}|/2$. This works out to

$\displaystyle c(\\{A,A^{\complement}\\})$
$\displaystyle=\frac{n^{2}}{4}\left(p(1-1^{2}+\frac{1}{2}-\frac{1}{2}^{2})+q(1+\frac{1}{2}-2\cdot
1\cdot\frac{1}{2})\right)=\frac{n^{2}}{4}p\left(\frac{1}{4}+\frac{r}{2}\right).$

We observe for later that

$\left(\frac{1}{3}(1+r-x)(1+r)-(\frac{1}{3}(1+r-x))^{2}\right)\leq\left(\frac{2}{9}(1+r)^{2}\right)\leq\frac{1}{4}+\frac{r}{2}$

for $0<r\leq 1/2$. Moreover, by the calculations in the proof of Lemma 11, for
fixed $p,q,n$ and $\alpha_{1}=0$ the quadratic function given by (5) has its
maximum at

$\frac{1+\frac{(1-2\alpha_{1})q}{p}}{2}>\frac{1}{2}.$

Thus this function is monotone in the interval between $0$ and $1/2$.
Therefore, if $A,B,C$ would be chosen such that
$\max\\{c(\\{A,A^{\complement}\\}),c(\\{B,B^{\complement}\\}),c(\\{C,C^{\complement}\\})\\}$
is smaller than $n^{2}p\left(\alpha(1+r)-\alpha^{2})\right)/4$, both
$|B^{\complement}\cap V_{1}|/|V_{1}|$ and $|C^{\complement}\cap
V_{1}|/|V_{1}|$ need to be smaller than $\alpha$. Therefore, in order for
$A,B,C$ to be a forbidden triple, $|A^{\complement}\cap V_{1}|/|V_{1}|$ needs
to be larger than $(\alpha-r)|V_{1}|$. Since the function given by (5) is
quadratic, and, by (III.2), its value at $1/2$ is larger than its value at
$(\alpha-r)$, this would imply that the cost of $\\{A,A^{\complement}\\}$ is
larger than $n^{2}p\left(\alpha(1+r)-\alpha^{2})\right)/4$, contradicting the
assumption. ∎

Proof of Theorem 3.  We first show that $\tau_{\Psi}^{1}$, and similarly
$\tau_{\Psi}^{2}$ are indeed tangles.

For this, suppose for a contradiction that $A,B,C\in\tau_{\Psi}^{1}$ such that
$|A\cap B\cap C|<a$ and such that the maximum over the cost of the three is as
small as possible. Then, by pigeon hole principle one of $A,B,C$ must contain
at least $(|V_{2}|-a)/3$ vertices from $V_{2}$.

By Lemma 11, we may now suppose without loss of generality,
that$A^{\complement}\supseteq V_{2}$, say. Again by Lemma 11, we may
additionally suppose that $B^{\complement}\cap V_{2}=\emptyset$,
$C^{\complement}\cap V_{2}=\emptyset$. Thus by Lemma 12, the maximal cost of
$A,B,C$ is minimized by

$\frac{n^{2}}{4}p\left(\frac{1}{3}(1+r-x)(1+r)-(\frac{1+r-x}{3})^{2}\right)\,.$

The argument for $\tau_{\Psi}^{2}$ is analogous.

Moreover, since the cost of the cut $\\{V_{1},V_{2}\\}$ is $qn^{2}/4<\Psi$,
$\tau_{\Psi}^{1}$ and $\tau_{\Psi}^{2}$ are distinct as
$V_{1}\in\tau_{\Psi}^{1}$ and $V_{2}\in\tau_{\Psi}^{2}$.

It remains to show that $\tau_{\Psi}^{1}$ and $\tau_{\Psi}^{2}$ are the only
tangles.

So suppose for a contradiction that there is an orientation $\tau$ of
$\mathcal{P}_{\Psi}$ which is a tangle, but
$\tau\notin\\{\tau_{\Psi}^{1},\tau_{\Psi}^{2}\\}$. Then there is some
$A\in\tau$ such that $|A^{\complement}\cap V_{1}|\geq|A\cap V_{1}|$, suppose
that $|A\cap V_{1}|$ is as small as possible; we will show that $|A\cap
V_{1}|=0$.

If $A\cap V_{2}\neq\emptyset$, it is easy to see that the costs of the cut
$\\{A^{\complement}\cup V_{2},A\smallsetminus V_{2}\\}$ are at most the cost
of the cut $\\{A^{\complement},A\\}$. If $\tau$ contains $A\smallsetminus
V_{2}$ we could chose the cut $\\{A^{\complement}\cup V_{2},A\smallsetminus
V_{2}\\}$ instead of $\\{A^{\complement},A\\}$, so suppose that $\tau$
contains $A^{\complement}\cup V_{2}$.

As the costs of the cut $\\{A\cap V_{2},A^{\complement}\cup V_{1}\\}$ are
again at most the costs of $\\{A,A^{\complement}\\}$, our tangle needs to
contain either $A\cap V_{2}$ or $A^{\complement}\cup V_{1}$. However the
latter is not possible as $(A^{\complement}\cup V_{1})\cap(A^{\complement}\cup
V_{2})\cap A=\emptyset$ contradicting the consistency of $\tau$. Thus $A\cap
V_{2}\in\tau$ which already verifies that $\tau$ contains a set disjoint from
$V_{1}$.

On the other hand, if $A\cap V_{2}=\emptyset$ and there is is some $v\in A\cap
V_{1}$ we can consider the cut
$\\{A^{\complement}\cup\\{v\\}\,,\,A\smallsetminus\\{v\\}\\}$. Since $p\geq
2q$ the cost of this cut is, as $A\subseteq V_{1}$, strictly lower than that
of $\\{A,A^{\complement}\\}$, hence $\tau$ contains an orientation of this
cut. Since $\tau$ is a tangle with $a\geq 2$, and
$A\cap(A^{\complement}\cup\\{v\\})=\\{v\\}$, $\tau$ cannot contain
$A^{\complement}\cup\\{v\\}$ and must therefore contain
$A\smallsetminus\\{v\\}$. However, $A\smallsetminus\\{v\\}$ now contradicts
the choice of $A$.

Thus $\tau$ contains an $A$ disjoint from $V_{1}$, and similarly, as
$\tau\neq\tau_{\Psi}^{2}$, the tangle $\tau$ contains an $B$ disjoint from
$V_{2}$. But since $A\cap B=\emptyset$ this contradicts the assertion that
$\tau$ is a tangle.

To prove Theorem 4 we introduce the concept of locally minimal cuts. A cut
$\\{A,A^{\complement}\\}\in\mathcal{P}$ is a _local minimum_ if moving any
single $v\in V$ to the other side does not decrease the cost.

In generality for tangles for $a\geq 2$ every cut which is of minimum cost
such that a given pair of tangles disagrees on it is a local minimum.

However, all these local minimum cuts need to respect the blocks.

###### Lemma 13 (Local minimum cuts do not cut through the blocks).

Independently of the choice of parameters, as long as $p>0$, every local
minimum cut respects the blocks, that is if $\\{A,A^{\complement}\\}$ is a
local minimum cut, then for $i=1,2$ either $V_{i}\subseteq A$ or
$V_{i}\subseteq A^{\complement}$.

###### Proof.

Suppose that $\\{A,A^{\complement}\\}$ is a local minimum cut but that,
without loss of generality, both $V_{1}\cap A$ and $V_{1}\cap A^{\complement}$
are non-empty. Pick some arbitrary $v\in V_{1}\cap A$ and $v^{\prime}\in
V_{1}\cap A^{\complement}$.

Since $\\{A,A^{\complement}\\}$ is locally minimal, the cut
$\\{A\smallsetminus\\{v\\},A^{\complement}\cup\\{v\\}\\}$ has at least the
cost of $\\{A,A^{\complement}\\}$, hence, as
$w(v,v)=w(v^{\prime},v^{\prime})=0$, we have

$\sum_{x\in A}w(v,x)\geq\sum_{x\in A^{\complement}}w(v,x)\quad\text{and,
similarly,}\quad\sum_{x\in A}w(v^{\prime},x)\leq\sum_{x\in
A^{\complement}}w(v^{\prime},x)\,.$

However, since $v,v^{\prime}$ both lie in $V_{1}$, we have that
$w(v,x)=w(v^{\prime},x)$ for all $x\neq v,v^{\prime}$, thus:

$\displaystyle\sum_{x\in A}w(v,x)\geq$ $\displaystyle\sum_{x\in
A^{\complement}}w(v,x)=\sum_{x\in
A^{\complement}}w(v^{\prime},x)+w(v,v^{\prime})\geq\sum_{x\in
A}w(v^{\prime},x)+w(v,v^{\prime})\geq\sum_{x\in A}w(v,x)+2w(v,v^{\prime})$

Which is a contradiction as, $w(v,v^{\prime})=p>0$. ∎

Proof of Theorem 4.  Suppose that there are two tangles. Then there exists a
lowest cost cut on which they disagree. This cut is a local minimum since
$a\geq 2$. By Lemma 13 the only local minimum cuts are $\\{\emptyset,V\\}$ and
$\\{V_{1},V_{2}\\}$. So the cut in question must be $\\{V_{1},V_{2}\\}$. As
this cut has cost $qn^{2}/4$, our two tangles have to be
$\mathcal{P}_{\Psi}$-tangles for some $\Psi\geq qn^{2}/4$.

Let $\tau$ be the tangle with $V_{1}\in\tau$. Pick any two disjoint
$X_{1},X_{2}\subseteq V_{1}$ with $|X_{1}|=|X_{2}|=\lfloor|V_{1}|/2\rfloor$.
We have that

$c(\\{X_{1},X_{1}^{\complement}\\})=c(\\{X_{2},X_{2}^{\complement}\\})=p\left\lfloor\frac{n}{4}\right\rfloor\left\lceil\frac{n}{4}\right\rceil+q\left\lfloor\frac{n}{4}\right\rfloor\frac{n}{2}\leq
p\frac{n^{2}}{16}+q\frac{n^{2}}{8}\leq q\frac{n^{2}}{4}.$

So $\tau$ contains one of each $X_{1}$ or $X_{1}^{\complement}$ and $X_{2}$ or
$X_{2}^{\complement}$. As $|V_{1}\cap X_{1}^{\complement}\cap
X_{2}^{\complement}|\leq 1<a$, the tangle $\tau$ cannot contain both
$X_{1}^{\complement}$ and $X_{2}^{\complement}$.

Without loss of generality we may assume that $X_{1}\in\tau$ and that $X_{1}$
is of minimum size such that $X_{1}\in\tau$. Now for any $x\in X_{1}$ the cut
$\\{X_{1}\smallsetminus\\{x\\},(X_{1}\smallsetminus\\{x\\})^{\complement}\\}$
has lower cost than $\\{X_{1},X_{1}^{\complement}\\}$. Thus by the minimal
choice of $X_{1}$ the tangle $\tau$ contains
$(X_{1}\smallsetminus\\{x\\})^{\complement}$, but
$\left\lvert(X_{1}\smallsetminus\\{x\\})^{\complement}\cap
X_{1}\right\rvert=\left\lvert\\{x\\}\right\rvert=1<a,$ which contradicts our
assumption that $\tau$ is a tangle.

####  III.2.1 Identifying tangles from random cuts is hard

Note that all our results for this model rely on the fact that
$\mathcal{P}_{\Psi}$ contains _all_ cuts of $G$ up to cost $\Psi$ rather than
just a sample of those cuts. In practice $\mathcal{P}_{\Psi}$ might consist of
many cuts, and so one usually would try to perform some sampling strategy to
obtain a ‘sensible’ subset of these cuts. The following result shows that
sampling from $\mathcal{P}_{\Psi}$ uniformly at random is not a valid sampling
strategy for this purpose, as one would still be required to draw a sample of
size exponential in $n$.

Observe that by Theorem 3 the blocks define $\mathcal{P}_{\Psi}$-tangles for
_any_ collection $\mathcal{P}$ of cuts. Thus, in order for the two blocks to
define distinct tangles it suffices for the sampled set of cuts to contain a
single cut which _distinguishes_ them – that is one cut
$\\{A,A^{\complement}\\}$ such that more than half of $V_{1}$ lies in $A$ and
more than half of $V_{2}$ lies in $A^{\complement}$. Unfortunately, the number
of these _good_ cuts is exponentially small compared to the number of cuts up
to any given cost $\Psi$:

###### Theorem 14 (Number of silly cuts).

Let $p,q$ be fixed and let $\Psi,a$ be chosen dependent on $n$ such that the
orientations induced by the blocks are distinct $\mathcal{P}_{\Psi}$-tangles
for $\mathcal{P}_{\Psi}$ the set of all cuts up to cost $\Psi$, then
asymptotically the number of cuts not distinguishing these tangles is
exponentially larger than the number of cuts distinguishing those tangles, for
$n$ going to infinity.

This theorem implies that, if sampling cuts from $\mathcal{P}_{\Psi}$
uniformly at random, one would need to sample exponentially many of the cuts
in order for the two blocks to define distinct tangles, since our sample needs
to include one cut for this which distinguishes the blocks. We will use the
remainder of this section to prove the theorem.

We can construct all good partitions $\\{A,A^{\complement}\\}$ as follows:
Starting with the partition $\\{V_{1},V_{2}\\}$, we pick any subsets
$G_{1}\subseteq V_{1}$ of size $g_{1}|V_{1}|<|V_{1}|/2$ and $G_{2}\subseteq
V_{2}$ of size $g_{2}|V_{2}|<|V_{2}|/2$ and let
$\\{A,A^{\complement}\\}=\\{(V_{1}\smallsetminus G_{1})\cup
G_{2}\,,\,(V_{2}\smallsetminus G_{2})\cup G_{1}\\}$. We observe that the cost
of $\\{A,A^{\complement}\\}$ depends only on $g_{1}$ and $g_{2}$. For fixed
$g_{1}$ and $g_{2}$ there are exactly
$\binom{|V_{1}|}{g_{1}n}\cdot\binom{|V_{2}|}{g_{2}n}$ many distinct cuts
realizing this $g_{1}$ and $g_{2}$. In this case we say that the good cut
$\\{A,A^{\complement}\\}$ _corresponds to_ $(g_{1},g_{2})$.

Similarly, we can construct all bad partitions $\\{C,C^{\complement}\\}$ as
follows: Starting with the partition $\\{\emptyset,V_{1}\cup V_{2}\\}$, we
pick subsets $B_{1}\subseteq V_{1}$ of size $b_{1}|V_{1}|<|V_{1}|/2$ and
$B_{2}\subseteq V_{2}$ of size $b_{2}|V_{2}|<|V_{2}|/2$ and let
$\\{C,C^{\complement}\\}=\\{B_{1}\cup B_{2},(V_{1}\cup
V_{2})\smallsetminus(B_{1}\cup B_{2})\\}$. We observe that again, the cost of
$\\{C,C^{\complement}\\}$ depends only on $b_{1}$ and $b_{2}$. Moreover for
fixed $b_{1}$ and $b_{2}$ there are exactly
$\binom{|V_{1}|}{b_{1}n}\cdot\binom{|V_{2}|}{b_{2}n}$ many distinct cuts
realizing this $b_{1}$ and $b_{2}$. In this case we say that the bad cut
$\\{C,C^{\complement}\\}$ _corresponds to_ $(b_{1},b_{2})$.

Therefore if we can show that for $b_{1}=g_{1}<1/2$ and $b_{2}=g_{2}<1/2$ the
cost of $\\{A,A^{\complement}\\}$ as constructed above is always as least as
big as the cost of $\\{C,C^{\complement}\\}$, this would imply that there are
at least as many bad partitions as good ones, since we can construct an
injective function from the set of good into the set of bad partitions of cost
$<k$. The cost of a bad cut for $b_{1},b_{2}$ is

$\frac{n^{2}}{4}\left(p(b_{1}-b_{1}^{2}+b_{2}-b_{2}^{2})+q(b_{1}+b_{2}-2b_{1}b_{2})\right)$

and the cost of a good cut for $g_{1}$ and $g_{2}$ is

$\displaystyle\frac{n^{2}}{4}\left(p(g_{1}-g_{1}^{2}+(1-g_{2})-(1-g_{2})^{2})+q(g_{1}+(1-g_{2})-2g_{1}(1-g_{2}))\right)$
$\displaystyle=$
$\displaystyle\frac{n^{2}}{4}\left(p(g_{1}-g_{1}^{2}+g_{2}-g_{2}^{2})+q(1-g_{1}-g_{2}+2g_{1}g_{2})\right)\,.$

However, for $b_{1},b_{2}<1/2$ we have that

$2b_{1}+2b_{2}-4b_{1}b_{2}<1$

and thus

$b_{1}+b_{2}-2b_{1}b_{2}<1-b_{1}-b_{2}+2b_{1}b_{2}\,,$

i.e. for $g_{1}=b_{1}$ and $g_{2}=b_{2}$ the cost of $\\{C,C^{\complement}\\}$
is indeed at most the cost of $\\{A,A^{\complement}\\}$. Therefore up to any
given cost there are at least as many bad cuts as there are good ones.

Moreover, we can show a little bit more: For $\delta=1+q/p$ we can show that,
under the condition that the cost of a good cut corresponding to
$(g_{1},g_{2})$ is at most $p\left(2\left(1+\frac{q}{p}\right)^{2}/9\right)$
(for larger $\Psi$, $\tau_{k}^{1}$ is not a tangle by Lemma 12), then also the
cost of a bad cut corresponding to $(\delta g_{1},\delta g_{2})$ is at most
the cost of a good cut corresponding to $(g_{1},g_{2})$.

Thus in this case we have, given a fixed $\Psi$ and the set $A_{\Psi}$ of all
pairs $(g_{1},g_{2})$ for which the cost of a good cut corresponding to
$(g_{1},g_{2})$ is at most $\Psi$, that there are exactly

$\sum_{\begin{subarray}{c}(i,j)\\\ (\frac{i}{|V_{1}|},\frac{j}{|V_{2}|})\in
A_{\Psi}\end{subarray}}\binom{|V_{1}|}{i}\cdot\binom{|V_{2}|}{j}$

good cuts. However, there are at least

$\sum_{\begin{subarray}{c}(i,j)\\\
(\frac{i}{\delta|V_{2}|},\frac{j}{\delta|V_{2}|})\in
A_{\Psi}\end{subarray}}\binom{|V_{1}|}{i}\cdot\binom{|V_{2}|}{j}$

bad cuts. Using these numbers it can be shown that for fixed $\delta>1$ the
number of bad cuts grows exponentially faster than then number of good cuts,
as $n$ goes to infinity.

###  III.3 Feature based data

The following computations are in expectation, which we incorporate by
assuming two things. First, we assume that the cost function behaves like the
density function of the marginal distributions. Concretely, we assume the
following:

######  Assumption 15.

The cost function $c$ is such that if $(A_{j,i},A_{j,i}^{\complement})$ and
$(A_{k,l},A_{k,l}^{\complement})$ are two axis-parallel bipartitions obtained
from Algorithm 3, then $c(\\{A_{j,i},A_{j,i}^{\complement}\\})\leq
c(\\{A_{k,l},A_{k,l}^{\complement}\\})$ if and only if the density at
$x_{j,i}$ of the marginal distribution on the $j$-axis is smaller than the
density at $x_{k,l}$ of the marginal distribution on the $k$-axis.

Secondly, we assume that the intersection of any three sides of bipartitions
contains the fraction of the points that is expected from the density
functions, i.e. we assume the following:

######  Assumption 16.

Given three bipartitions
$\\{A,A^{\complement}\\},\\{B,B^{\complement}\\},\\{C,C^{\complement}\\}$ in
$\mathcal{P}=\bigcup_{j}\mathcal{P}_{j}$, where $A=\\{v\in V\mid v_{i}<a\\}$,
$B=\\{v\in V\mid v_{j}<b\\}$, $C=\\{v\in V\mid v_{k}<c\\}$ for some
$i,j,k\in\\{1,\dots,d\\}$ we have that the intersection of three sides of the
bipartitions contains as many points as we would expect, i.e. as large as the
integral of the density over the intersection of the induced half-spaces.

This means that we can use the quantiles of normal distributions in our
arguments. In particular, we will use that for $\mathcal{N}(0,\sigma^{2})$
less than 16% of the points are below $-\sigma$.

Let us say that a cut $\\{A_{j,i},A_{j,i}^{\complement}\\}$ in
$\mathcal{P}_{j}$ is a _local minimum_ if and only if

$c(\\{A_{j,i-1},A_{j,i-1}^{\complement}\\})>c(\\{A_{j,i},A_{j,i}^{\complement}\\})<c(\\{A_{j,i+1},A_{j,i+1}^{\complement}\\}).$

If we are given Assumption 15, we immediately obtain the following from the
fact that the marginal densities of a Gaussian have exactly one local mimimum:

###### Observation 17.

If Assumption 15 holds, then there is, in every dimension, at most one
separation which is a local minimum.

We prove Theorem 5 by showing that, for the smallest possible order for which
we find a local minimum, this local minimum is oriented differently by $\mu$
and $\nu$ and, on the other hand that the order is small enough such that the
orientations induced by $\mu$ and $\nu$ cannot contain a triple of
bipartitions with too small intersection.

###### Lemma 18.

If $\tau,\tau^{\prime}$ are distinct tangles, then there is a local minimum
which is oriented differently by $\tau$ and $\tau^{\prime}$

###### Proof.

Let $\\{A_{j,i},A_{j,i}^{\complement}\\}$ be a cut of minimal possible cost
distinguishing $\tau$ and $\tau^{\prime}$, say $A_{j,i}\in\tau$ and
$A_{j,i}^{\complement}\in\tau^{\prime}$. If
$\\{A_{j,i},A_{j,i}^{\complement}\\}$ is not a local minimum, say because
$\\{A_{j,i-1},A_{j,i-1}^{\complement}\\}$ has lower cost (the other case is
analogous), then both $\tau$ and $\tau^{\prime}$ would need to orient
$\\{A_{j,i-1},A_{j,i-1}^{\complement}\\}$ similarly, and thus, since one of
the two contains $A_{j,i}^{\complement}$, they would both need to contain
$A_{j,i-1}^{\complement}$ by our consistency condition. However, since
$\left\lvert A_{j,i}\cap A_{j,i-1}^{\complement}\right\rvert<a$ this
contradicts the consistency of the tangle $\tau$. ∎

###### Lemma 19.

There exists a local minimum in $\mathcal{P}_{j}$ if and only if
$\left\lvert\mu_{j}-\nu_{j}\right\rvert>2\sigma$. Moreover if
$(A_{j,i},A_{j,i}^{\complement})$ is a local minimum, then either
$\nu_{j}<x_{j,i}<\mu_{j}$, or $\mu_{j}<x_{j,i}<\nu_{j}$.

###### Proof.

There is a local minimum in $\mathcal{P}_{j}$ if and only if the density
function has a local minimum by Assumption 15. This is the case precisely if
$\left\lvert\mu_{j}-\nu_{j}\right\rvert>2\sigma$, see Helguerro (1904),
Schilling et al. (2002). The ‘Moreover’ part is then also immediate, since the
density function is monotone on the intervals $(-\infty,\mu_{j})$ and
$(\nu_{j},\infty)$. ∎

Proof of Theorem 5.  We consider the set of all cuts in
$\bigcup\mathcal{P}_{j}$ up to and including the cost of
$\\{A_{j,i},A_{j,i}^{\complement}\\}$, that is,
$\\{A_{j,i},A_{j,i}^{\complement}\\}$ is the only local minimum that we
consider and it is also the most expensive cut that we consider.

Since the $j$-axis is the only axis on which we consider a local minimum, each
cut which we are considering except $(A_{j,i},A_{j,i}^{\complement})$ needs to
have $\mu$ and $\nu$ on the same side, which we call the ‘middle’.

We define $\tau_{\mu}$ by orienting $\\{A_{j,i},A_{j,i}^{\complement}\\}$ as
$A_{j,i}$, and all other cuts towards this middle.

Let us now show that this orientation is consistent, i.e., a tangle. Observe
that, by Assumption 16, each side in $\tau_{\mu}$ contains at least $n/2$
points. Further if we consider some three sides of cuts in $\tau_{\mu}$ which
are along the same axis, then at most two of them are relevant to the
intersection, since one of the sides will always be a superset of one of the
others. So, considering three sides of any cuts in $\tau_{\mu}$, we need to
consider just two cases:

_Case I: All cuts are along distinct axes._ Each of the three sides is
expected to contain at least $n/2$ points, as observed. As the distributions
along the distinct axes are independent, we thus have that the intersection of
the three contains at least $n/8>a$ points by Assumption 16.

_Case II: Two cuts
$\\{A_{k,l},A_{k,l}^{\complement}\\},\\{A_{k,l^{\prime}},A_{k,l^{\prime}}^{\complement}\\}$
are along the same axis, the third is along another axis._ We first determine
the size of the intersection of the cuts along the same axis $k$. We claim
that one of the two is chosen at distance more than $\sigma$ from $\mu_{k}$.
Indeed, if one of the two cuts equals $\\{A_{j,i},A_{j,i}^{\complement}\\}$
then this is true by the assumption of this theorem. Otherwise we claim that
for one of the two cuts the point $x_{k,l}$ or $x_{k^{\prime},^{\prime}l}$ has
lower distance from $\nu_{k}$, respectively $\nu_{k^{\prime}}$, than from
$\mu_{k}$, respectively $\mu_{k^{\prime}}$. Indeed, if for both of the two
cuts the points $x_{k,l}$ and $x_{k^{\prime},^{\prime}l}$ would have larger
distance from $\nu_{k}$ and $\nu_{k^{\prime}}$ than from $\mu_{k}$ and
$\mu_{k^{\prime}}$, respectively, then the size of the intersection of the
sides of the two cuts contained in $\tau_{\mu}$ would be equal to the size of
the smaller of the two respective sides, hence we do not need to consider this
triplet of cuts. Therefore, indeed we have for one of the two cuts, say for
$\\{A_{k,l},A_{k,l}^{\complement}\\}$, that $x_{k,l}$ has smaller distance
from $\nu_{k}$ than from $\mu_{k}$. Now if $x_{k,l}$ would have distance less
than $\sigma$ from $\mu_{k}$, it would also need to have distance less than
$\sigma$ from $\nu_{k}$, but this implies that this cut has higher costs than
$\\{A_{j,i},A_{j,i}^{\complement}\\}$, by Assumption 15, as $x_{j,i}$ has
distance at least $\sigma$ from both, $\mu_{j}$ and $\nu_{j}$.

Thus, indeed one of the two cuts, say $\\{A_{k,l},A_{k,l}^{\complement}\\}$,
satisfies $\left\lvert x_{k,l}-\mu_{k}\right\rvert>\sigma$.

By Assumption 16, in expectation, at least 84% of the points belonging to
$\mu$ – that is at least $0.42n$ points – lie on the same side of
$\\{A_{k,l},A_{k,l}^{\complement}\\}$ as $\tau_{\mu}$ chooses. At least half
of the points belonging to $\mu$ – that is at least $0.25n$ points – lie on
the chosen side of the other cut along the same axis,
$\\{A_{k,l^{\prime}},A_{k,l^{\prime}}^{\complement}\\}$. Thus in their
intersection there are expected to be at least $0.17n$ points.

The third cut is along an independent axis and, by the same argument as in
Case I, at most halves this number. So there are already at least
$0.085n>n/12>a$ points in the intersection, alone from the points belonging to
$\mu$.

To prove Theorem 6 we will use the following simplification: Given a sampled
cut $\\{A_{j,i},A_{j,i}^{\complement}\\}$ there exists a whole interval in
$\mathbb{R}$ in which we can chose an $x_{j,i}$ such that $A_{j,i}=\\{v\in
V\mid v_{j}<x_{j,i}\\}$. To simplify the arguments, we will assume that our
Assumptions 15 and 16 hold, regardless of how we have chosen $x_{j,i}$ for our
given cut. This gives us the technical possibility to consider the cut
$\\{B_{j,x},B_{j,x}^{\complement}\\}$ sampled at $x\in\mathbb{R}$, i.e.
$B_{j,x}:=\\{v\in V\mid v_{j}<x\\}$, for any $x\in\mathbb{R}$ and any
dimension $j$. The result of this technicality is, that we do not have to take
into account that, while we know that this cut is contained in
$\mathcal{P}_{j}$, we may have put it into that set for a slightly different
$x$. This simplifies the arguments a lot. Of course, as a formal statement
this is an unrealistic assumption, but as the number of points tends to
$\infty$, we will come closer to this assumption, and so we expect that the
consequence of the theorem still holds.

Proof of Theorem 6.  Suppose there is a tangle $\tau$ which points neither to
$\mu$ nor to $\nu$. Let us suppose that $j$ is chosen such that
$\left\lvert\mu_{j}-\nu_{j}\right\rvert$ is maximal. The general strategy of
this proof is as follows: We will show that there are two local minima cuts in
$\tau$, one along the $j$-axis and one along another axis, such that one of
the two points away from $\mu$, and the other points away from $\nu$. The cost
of these local minima cuts will allow us to find a set of three cuts which
$\tau$ needs to orient a certain way, but which together violate the
consistency condition on $\tau$, due to the quantiles of the gaussian
distribution. This will result in a contradiction to the assumption that
$\tau$ is a tangle.

So let us first deal with the task of finding these local minima cuts.

As $\tau$ does not point towards $\mu$, there exist a cut
$\\{B_{i,x},B_{i,x}^{\complement}\\}$ whose orientation in $\tau$ points away
from $\mu$, say $B_{i,x}^{\complement}\in\tau$ but $\mu_{i}<x$. We want to
show that in this case we may suppose that
$\\{B_{i,x},B_{i,x}^{\complement}\\}$ is a local minimum.

So suppose that we cannot choose $\\{B_{i,x},B_{i,x}^{\complement}\\}$ as a
local minimum. We claim that there is an $x^{\prime}\geq x$ and some
$\varepsilon>0$ such that $B_{i,x^{\prime}}^{\complement}\in\tau$ and
$\\{B_{i,x^{\prime}+\varepsilon^{\prime}},B_{i,x^{\prime}+\varepsilon^{\prime}}^{\complement}\\}$
is not oriented by $\tau$ for any $\varepsilon^{\prime}<\varepsilon$. If we do
not find such an $x^{\prime}$ and $\varepsilon$, then $\tau$ would need to
orient every $\\{B_{i,x^{\prime}},B_{i,x^{\prime}}^{\complement}\\}$ for any
$x^{\prime}>x$. However, since $B_{i,x}^{\complement}\in\tau$ we can conclude
by the consistency condition that also $B_{i,x^{\prime}}^{\complement}\in\tau$
for all $x^{\prime}>x$ where $\left\lvert x^{\prime}-x\right\rvert$ is small
enough. Applying this argument inductively, we can conclude that in fact
$B_{i,x^{\prime}}^{\complement}\in\tau$ for all $x^{\prime}>x$. But, for large
enough $x^{\prime}$ we have that $\left\lvert
B_{i,x^{\prime}}^{\complement}\right\rvert<a$ which is a contradiction to the
fact that $\tau$ is a tangle.

Thus we may suppose without loss of generality that our point $x$ with
$B_{i,x}^{\complement}\in\tau$ but $\mu\in B_{i,x}$ is chosen such that there
is some $\varepsilon>0$ with the property that the cut
$\\{B_{i,x+\varepsilon^{\prime}},B_{i,x+\varepsilon^{\prime}}^{\complement}\\}$
is not oriented by $\tau$ for any $\varepsilon^{\prime}<\varepsilon$. This
means that $x$ was chosen as large as possible with the property that
$\\{B_{i,x},B_{i,x}^{\complement}\\}$ is still oriented differently by $\tau$
and $\mu$.

Now this choice of $x$ implies that $\mu_{i}<x\leq\nu_{i}$ as otherwise, if
$\mu_{i}\leq\nu_{i}<x$, the cut
$\\{B_{i,x+\varepsilon},B_{i,x+\varepsilon}^{\complement}\\}$ would, for any
$\varepsilon>0$, have lower costs than $\\{B_{i,x},B_{i,x}^{\complement}\\}$,
contradicting the choice of $x$. Now, if
$\left\lvert\mu_{i}-\nu_{i}\right\rvert>2\sigma$, for
$x^{\prime}=\mu_{i}+\frac{\left\lvert\mu_{i}-\nu_{i}\right\rvert}{2}$ the cut
$\\{B_{i,x^{\prime}},B_{i,x^{\prime}}^{\complement}\\}$ is a local minimum and
it is easy to see that $\tau$ contains $B_{i,x^{\prime}}^{\complement}$ and
thus this is a local minimum cut of the type that we assumed does not exists.
Hence we have that $\left\lvert\mu_{i}-\nu_{i}\right\rvert\leq 2\sigma$.

Thus the density function along the $i$-axis has a unique local maximum
between $\mu_{i}$ and $\nu_{i}$, and the cost of
$\\{B_{i,\mu_{i}},B_{i,\mu_{i}}^{\complement}\\}$ is less than the cost of
$\\{B_{i,x},B_{i,x}^{\complement}\\}$. Now for any $y\in\mathbb{R}$, the cut
$\\{B_{j,y},B_{j,y}^{\complement}\\}$ along the $j$-axis (recall that $j$ was
chosen such that $\left\lvert\mu_{j}-\nu_{j}\right\rvert$ is maximal) has
lower cost than $\\{B_{i,\mu_{i}},B_{i,\mu_{i}}^{\complement}\\}$ as the cost
of such a cut is maximal if $y$ equals either $\mu_{j}$ or $\nu_{j}$ and in
that case we still have that $\left\lvert
y-\nu_{j}\right\rvert>2\sigma>\left\lvert\mu_{i}-\nu_{i}\right\rvert$ or
$\left\lvert
y-\mu_{j}\right\rvert>2\sigma>\left\lvert\mu_{i}-\nu_{i}\right\rvert$.
However, this implies that $\tau$ needs to orient
$\\{B_{j,y},B_{j,y}^{\complement}\\}$ for every $y\in\mathbb{R}$ which is not
possible without violating the consistency condition. Hence the assumption
that we do not have a local minimum cut in $\tau$ pointing away from $\mu$
results in a contradiction to the fact that $\tau$ is a tangle.

So, there indeed needs to be a local minimum cut in $\tau$ which points away
from $\mu$. Similarly, we find a local minimum cut in $\tau$ which points away
from $\nu$. As $j$ was chosen such that
$\left\lvert\mu_{j}-\nu_{j}\right\rvert$ is maximal, $\tau$ in particular
needs to orient the locally minimal cut $\\{B_{j,l},B_{j,l}^{\complement}\\}$
in dimension $j$, since this cut is the local minimum cut of lowest possible
cost. So let us suppose wlog. that $\tau$ orients this cut the same way as
$\mu$, i.e. $\tau$ contains $B_{j,l}$ and $\mu_{j}<l<\nu_{j}$.

Further let $\\{B_{i,x},B_{i,x}^{\complement}\\}$ be a local minimum cut that
is oriented differently by $\tau$ and $\mu$, so that
$B_{i,x}^{\complement}\in\tau$ but $\mu_{i}<x$.

Since $\\{B_{i,x},B_{i,x}^{\complement}\\}$ is a local minimum, we know that
$\left\lvert x-\mu_{i}\right\rvert>\sigma$, as $\left\lvert
x-\mu_{i}\right\rvert=\left\lvert x-\nu_{i}\right\rvert$ and
$\left\lvert\mu_{i}-\nu_{i}\right\rvert>2\sigma$ by the fact that there is a
local minimum cut along the $i$-axis. Moreover, we can consider the cut
$\\{B_{j,z},B_{j,z}^{\complement}\\}$ corresponding to the point
$z:=\mu_{j}-\left\lvert x-\mu_{i}\right\rvert$ in dimension $j$, which, by
Assumption 15, is also oriented by $\tau$ since the density along the $j$-axis
at $z$ is less than the density along the $i$-axis at $x$. Since $\tau$ is a
tangle, it needs to be the case that $B_{j,z}^{\complement}$ is contained in
$\tau$: If $B_{j,z}\in\tau$ this contradicts consistency, as $\tau$, by
Assumption 15, also orients all the cuts $\\{B_{j,y},B_{j,y}^{\complement}\\}$
for $y\leq z$.

Furthermore let us consider the point $z^{\prime}:=\mu_{j}+\left\lvert
x-\mu_{i}\right\rvert$. As $\left\lvert
x-\mu_{i}\right\rvert<\frac{\left\lvert\nu_{j}-\mu_{j}\right\rvert}{2}$ we
have that $\left\lvert z^{\prime}-\mu_{j}\right\rvert=\left\lvert
x-\mu_{i}\right\rvert$ and $\left\lvert
z^{\prime}-\nu_{j}\right\rvert\geq\left\lvert x-\nu_{i}\right\rvert$, and thus
Assumption 15 again ensures that the cut
$\\{B_{j,z^{\prime}},B_{j,z^{\prime}}^{\complement}\\}$ in dimension $j$
corresponding to $z^{\prime}$ has lower cost than
$\\{B_{i,x},B_{i,x}^{\complement}\\}$ and is thus oriented by $\tau$. Since
$B_{j,l}\in\tau$ it needs to be the case that $B_{j,z^{\prime}}\in\tau$ as
otherwise, $\tau$ would by Assumption 15 also need to orient all the cuts
$\\{B_{j,y},B_{j,y}^{\complement}\\}$ for $z^{\prime}\leq y\leq l$ which would
then contradict consistency.

We can now use Assumption 16 to calculate how many points are contained in
$B_{j,z}^{\complement}\cap B_{j,z^{\prime}}\cap B_{i,x}^{\complement}$. Let us
first consider the points from $\mu$. Let us denote the fraction of points
from $\mu$ contained in $B_{i,x}^{\complement}$ as $p$, so $p=\left\lvert
V_{\mu}\cap B_{i,x}^{\complement}\right\rvert\frac{2}{n}$. As $\left\lvert
x-\mu_{i}\right\rvert>\sigma$, we have that $p<0.16$.

By the choice of $z$ we have that $B_{j,z}^{\complement}$ contains
$(1-p)\frac{n}{2}$ points from $\mu$ and similarly, $B_{j,z^{\prime}}$
contains a fraction of $(1-p)$ of the points from $\mu$, i.e.
$(1-p)\frac{n}{2}$ points.

So in total, by Assumption 16, the set $B_{j,z}^{\complement}\cap
B_{j,z^{\prime}}\cap B_{i,x}^{\complement}$ contains $p(1-p)^{2}\frac{n}{2}$
points from $\mu$.

For the points from $\nu$ we observe that $B_{i,x}^{\complement}$ contains, by
symmetry and the choice of $p$ exactly $(1-p)\frac{n}{2}$ points from $\nu$.

$B_{j,z^{\prime}}$ contains fewer points from $\nu$ then $B_{j,l}$ and
$B_{j,l}$ contains at most a fraction of $q$ of the points from $\nu$. So,
again using Assumption 16, we can bound the total number of points contained
in $B_{j,z}^{\complement}\cap B_{j,z^{\prime}}\cap B_{i,x}^{\complement}$ by
$\frac{n}{2}(p(1-p)^{2}+q(1-p))$. As $p\leq 0.16$ and $q<p$, this is maximized
for $p=0.16$, where we obtain a value of about $n\cdot(0.42q+0.056)$, hence if
$a>n\cdot(0.42q+0.056)$, $\tau$ cannot be a tangle, as claimed.

### Questions in the Narcissistic Personality Inventory Dataset

|  statement A |  statement B
---|---|---
1 |  I have a natural talent for influencing people. |  I am not good at influencing people.
2 |  Modesty doesn’t become me. |  I am essentially a modest person.
3 |  I would do almost anything on a dare. |  I tend to be a fairly cautious person.
4 |  When people compliment me I sometimes get embarrassed. |  I know that I am good because everybody keeps telling me so.
5 |  The thought of ruling the world frightens the hell out of me. |  If I ruled the world it would be a better place.
6 |  I can usually talk my way out of anything. |  I try to accept the consequences of my behavior.
7 |  I prefer to blend in with the crowd. |  I like to be the center of attention.
8 |  I will be a success. |  I am not too concerned about success.
9 |  I am no better or worse than most people. |  I think I am a special person.
10 |  I am not sure if I would make a good leader. |  I see myself as a good leader.
11 |  I am assertive. |  I wish I were more assertive.
12 |  I like to have authority over other people. |  I don’t mind following orders.
13 |  I find it easy to manipulate people. |  I don’t like it when I find myself manipulating people.
14 |  I insist upon getting the respect that is due me. |  I usually get the respect that I deserve.
15 |  I don’t particularly like to show off my body. |  I like to show off my body.
16 |  I can read people like a book. |  People are sometimes hard to understand.
17 |  If I feel competent I am willing to take responsibility for making decisions. |  I like to take responsibility for making decisions.
18 |  I just want to be reasonably happy. |  I want to amount to something in the eyes of the world.
19 |  My body is nothing special. |  I like to look at my body.
20 |  I try not to be a show off. |  I will usually show off if I get the chance.
|  statement A |  statement B
---|---|---
21 |  I always know what I am doing. |  Sometimes I am not sure of what I am doing.
22 |  I sometimes depend on people to get things done. |  I rarely depend on anyone else to get things done.
23 |  Sometimes I tell good stories. |  Everybody likes to hear my stories.
24 |  I expect a great deal from other people. |  I like to do things for other people.
25 |  I will never be satisfied until I get all that I deserve. |  I take my satisfactions as they come.
26 |  Compliments embarrass me. |  I like to be complimented.
27 |  I have a strong will to power. |  Power for its own sake doesn’t interest me.
28 |  I don’t care about new fads and fashions. |  I like to start new fads and fashions.
29 |  I like to look at myself in the mirror. |  I am not particularly interested in looking at myself in the mirror.
30 |  I really like to be the center of attention. |  It makes me uncomfortable to be the center of attention.
31 |  I can live my life in any way I want to. |  People can’t always live their lives in terms of what they want.
32 |  Being an authority doesn’t mean that much to me. |  People always seem to recognize my authority.
33 |  I would prefer to be a leader. |  It makes little difference to me whether I am a leader or not.
34 |  I am going to be a great person. |  I hope I am going to be successful.
35 |  People sometimes believe what I tell them. |  I can make anybody believe anything I want them to.
36 |  I am a born leader. |  Leadership is a quality that takes a long time to develop.
37 |  I wish somebody would someday write my biography. |  I don’t like people to pry into my life for any reason.
38 |  I get upset when people don’t notice how I look when I go out in public. |  I don’t mind blending into the crowd when I go out in public.
39 |  I am more capable than other people. |  There is a lot that I can learn from other people.
40 |  I am much like everybody else. |  I am an extraordinary person.


ab,ab.legacy,braket,nabla.legacy,op.legacy

# Description of Molecular Chirality and its Analysis via High Harmonic
Generation

Akihito Kato  Department of Physics and Electronics, Osaka Metropolitan
University, 1-1 Naka-ku, Sakai, Osaka 599-8531, Japan  Nobuhiko Yokoshi
Department of Physics and Electronics, Osaka Metropolitan University, 1-1
Naka-ku, Sakai, Osaka 599-8531, Japan

###### Abstract

To clarify the microscopic origin of chirality-induced optical effect, we
develop an analytical method that extracts the chiral part of the Hamiltonian
of molecular electronic states. We demonstrate this method in a model chiral
molecule consisting of helically stacked two $N$ regular polygons, and compare
it with numerical calculation for chiral discrimination via high-harmonic
generation (HHG) of the same molecule. The discrimination signal here is the
difference between the harmonic intensity from the bicircular laser field and
that from its reflected laser field, resulting in a pseudoscalar quantity that
reflects the chirality of the molecule. As a result, we find that, to obtain
the large discrimination signal via HHG difference, the increase of capacity
to generate the polar quantity is more advantageous than maximizing the
conversion efficiency via the optimization of molecular chirality. We
speculate this criteria may be extended to other optical and current-induced
processes relevant to the chiral molecules and materials.

##  I Introduction

Chirality, a structural property of an object that lacks reflection and
inversion symmetries, is ubiquitous in nature from single molecules, molecular
assemblies, crystals to living organisms [1]. Much effort has been devoted
toward understanding the functionality inherent in the chiral systems. In
crystals, electronic and phononic band splittings are originated from the
chiral symmetry breaking [2, 3, 4, 5, 6, 7, 8], which has invigorated
possibilities in transporting and converting their angular momentums (AMs) [9,
10, 11]. In particular, properties of the AM in the chiral systems and its
relevance to chirality-induced spin selectivity (CISS) [12, 13, 14, 15] effect
have been increasingly focused on. Thus, as well as elucidating the
characteristics of chiral materials and establishing efficient means of
asymmetric synthesis, discriminating the chirality is also the indispensable
problem.

Recent studies have demonstrated that the chirality in quantum systems can be
represented as the electric toroidal monopole (ETM) [16, 17, 18, 19]. This is
anti-symmetric with respect to the inversion and reflection and has the time-
reversal symmetry, and thus is a generalization of the definition of the
chirality by Kelvin and Barron [20]. Because also Hamiltonian of chiral
systems should include the ETM, we can formally separate it into the symmetric
and anti-symmetric part. The anti-symmetric part is defined by the subtraction
of the Hamiltonians with the opposite chiralities, which belongs to the
different Hilbert spaces composed of the different electronic basis set. Then,
its direct evaluation has a major difficulty. One of the main results of this
work is that we develop a method to enables the direct subtraction of the
matrix elements, and thus the identification of the chirality.

Using the above identification method, we consider the chiral discrimination
signal by light as one example. The chiral systems asymmetrically interact
with the chiral light, which can be utilized to optically discriminate the
chirality. In circular dichroism (CD), the differential absorption spectrum
between the clockwise and the counterclockwise circular polarization of light,
which correspond to the spin angular momentum (SAM), is measured. The measured
spectrum is the pseudoscalar quantity, i.e., each enantiomer has the opposite
sign from the other. The ability to extract the pseudoscalar (more generally
the pseudotensor) spectrum can be a clear manifestation of the chiral
discrimination. Therefore, it is natural to explore criteria from a
microscopic point of view on how to extract a large pseudoscalar spectrum.

Here, we focus on a nonlinear optical process called high harmonic generation
(HHG), which is the light emission with frequencies multiple integers of the
injected one. As a consequence of the SAM conservation, the atoms and
molecules cannot emit the high-harmonics from the light beams with the single
circular polarization and requires the bicircular field, the combination of
both circular polarizing fields [21, 22, 23]. This bicircular field can break
the reflection symmetry, and thus, can be chiral. Therefore, the chiral
discriminating HHGs using the bicircular field have been studied [24, 25, 26,
27, 28, 29], wherein the HHG spectrum measurements have demonstrated that its
Kuhn $g$-factor is pseudoscalar. We demonstrate our developed identification
method in a model chiral molecule consisting of helically stacked two $N$
regular polygons, and compare it with numerical calculation within so-called
three-step model [30, 31]. It is found that the increase of the polar quantity
is more advantageous than maximizing the conversion efficiency of the
molecular chirality to obtain the large difference between the harmonic
intensity from the bicircular laser field and that from its reflected laser
field.

The remaining part of this paper is organized as follows: In Section II, we
introduce the model chiral molecule and its the anti-symmetric part is deduced
in Section III. In Section IV, we present the theory of the HHG basisd on the
three step model and prove that the Kuhn $g$-factor becomes pseudoscalar using
the symmetry argument. In Section V, numerical results of the chiral
discriminating HHGs are presented and its correspondence to the chiral part of
the Hamiltonian are analyzed. Section VI is devoted to the concluding remarks.

##  II Model Chiral Molecule

Figure 1: (a) Illustration of the model chiral molecule that consists of two
$N=3$ regular polygons. The grey and green lines display the intralayer and
interlayer bonds. (b) Top view of our model. Regular polygons are helically
stacked with the twist angle $\beta$.

As a model chiral system, we consider a helical stacking of two $N$ regular
polygon molecules (see Fig. 1 for $N=3$) characterized by the radius $\rho$,
the interlayer distance $h$, and the twist angle $\beta$. The position of the
$j$th atom ($j=1,\ldots,N$) in the $k$th layer ($k=1,2$) is written as
$\bm{R}_{kj}(\beta)=(\rho\cos 2\theta_{kj},\rho\sin
2\theta_{kj},(k-1)h)^{\top}$ with $\theta_{kj}=(k-1)\beta/2+(j-1)\alpha/2$ and
$\alpha=2\pi/N$ being the rotation angle. To break the reflection symmetry,
the twist angle have to satisfy $0<\vab{\beta}<\alpha/2$. Chirality reversal
corresponds to the inversion of the twist angle, $\beta\to-\beta$, under the
action of the reflection operator $\hat{\sigma}$ with respect to the
$xz$-plane, $\hat{\sigma}\bm{R}_{kj}(\beta)=\bm{R}_{kj}(-\beta)$.

Based on the Slater–Koster method [32, 33], we construct the tight-binding
Hamiltonian $\hat{H}(\beta)$ using the electronic basis states
$\\{\ket*{\phi_{kj}^{\nu}(\beta)}\\}$ that consist of $2\mathrm{s}$ ($\nu=0$)
, $2\mathrm{p}_{z}$ ($\nu=z$), and $2\mathrm{p}_{\pm}=2\mathrm{p}_{x}\pm
i2\mathrm{p}_{y}$ ($\nu=\pm$) orbitals from each atom, where
$\ket*{\phi_{kj}^{\nu}(\beta)}$ represents the orbital $\nu$ centered at
$\bm{R}_{kj}(\beta)$,
$\innerproduct*{\bm{r}}{\phi_{kj}^{\nu}(\beta)}=\phi^{\nu}(\bm{r}-\bm{R}_{kj}(\beta))$.
We consider the electronic bonds between the adjacent atoms in the same layer
(intralayer) and the ones with the same $j$ in the different layers
(interlayer). The latter bond is characterized by the helical angle
$\gamma=\arctan(h/2\rho\sin(\beta/2))$ and is necessary to endow the chirality
with the electronic state. The Slater–Koster overlaps for the intralayer bond
are denoted by $V_{\lambda\nu}^{s}$ ($s=\sigma$ and $\pi$ for parallel with
and perpendicular to the bonding), which is parametrized by $V$, and those for
the interlayer bond are denoted by $W_{\lambda\nu}^{s}$, which is parametrized
by $W$. For the constructed Hamiltonian and its detailed derivation, see
Appendix. A.

The groundstate $\ket{\phi_{\mathrm{GS}}(\beta)}$ is the eigenstate of
$\hat{H}(\beta)$, and hence, belong to the irreducible representation (IRREP)
of the point group $\mathrm{C}_{N}$, because $\hat{H}(\beta)$ possess the
$N$-fold rotational symmetry. Each eigenstate $\ket*{\phi_{m}(\beta)}$ is
classified by its rotational quantum number or the pseudo AM [9, 34], $m$,
through the relation
$\hat{C}_{N}\ket{\phi_{m}(\beta)}=e^{-im\alpha}\ket{\phi_{m}(\beta)}$ with
$\hat{C}_{N}$ being the $N$-fold rotational operator. For $N$-odd and even
cases, the pseudo AM takes the value $m=0,\pm 1,\ldots,\pm(N-1)/2$ and
$m=0,\pm 1,\ldots,\pm(N-1)/2,N/2$, respectively. The time-reversal symmetry
enforces the degeneracy of $\ket*{\phi_{m}(\beta)}$ and
$\ket*{\phi_{-m}(\beta)}$. Thus, the groundstate becomes
$\ket{\phi_{\mathrm{GS}}(\beta)}=(\ket*{\phi_{m}(\beta)}+\ket*{\phi_{-m}(\beta)})/\sqrt{2}$.

##  III Description of Molecular Chirality

To discuss the correspondence between the molecular chirality and the Kuhn
$g$-factor, we extract the anti-symmetric part of the Hamiltonian with respect
to the reflection operation. This part is formally written as
$\hat{G}_{0}(\beta)=[\hat{H}(\beta)-\hat{H}(-\beta)]/2$, which satisfies
$\hat{\sigma}\hat{G}_{0}(\beta)\hat{\sigma}^{\dagger}=-\hat{G}_{0}(\beta)$.
Because the matrix element
$\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda}(\beta^{\prime})}{\hat{H}(\beta)}{\phi_{lj}^{\nu}(\beta^{\prime})}$
is well-defined only for the $\beta=\beta^{\prime}$ cases, the direct
subtraction of the Hamiltonians with the opposite $\beta$ requires the
transformation from $\ket*{\phi_{kj}^{\nu}(0)}$ to
$\ket*{\phi_{kj}^{\nu}(\pm\beta)}$. For that purpose, we first translate the
orbital center from $\bm{R}_{kj}(\beta)$ to $\bm{R}_{kj}(0)$ as

$\displaystyle\phi^{\nu}(\bm{r}-\bm{R}_{kj}(0))$
$\displaystyle=\exp\ab[-\frac{i}{\hbar}\bm{\hat{p}}\cdot(\bm{R}_{kj}(0)-\bm{R}_{kj}(\beta))]$
$\displaystyle\quad\times\phi^{\nu}(\bm{r}-\bm{R}_{kj}(\beta))$
$\displaystyle=\hat{T}_{kj}(\beta)\phi^{\nu}(\bm{r}-\bm{R}_{kj}(\beta)),$ (1)

where $\hat{\bm{p}}=-i\hbar\bm{\nabla}$ is the momentum operator. This
procedure is similar to pullback or push forward mapping in the Lie derivative
[35]. Expressing the orbital as the product of the radial distribution and the
spherical harmonics, the achiral basis with $\beta=0$ can be written as a
linear combination of the chiral basis states,

$\ket*{\phi_{kj}^{\nu}(0)}=\hat{T}_{kj}(\beta)\ket*{\phi_{kj}^{\nu}(\beta)}=\sum_{\nu^{\prime}}g_{kj}^{\nu\nu^{\prime}}(\beta)\ket*{\phi_{kj}^{\nu^{\prime}}(\beta)}$
(2)

with $g_{kj}^{\nu\nu^{\prime}}(\beta)\in\mathbb{C}$ (see Appendix. B for
details). This allows us to expand the Hamiltonians with the achiral basis
states as
$\hat{H}(\beta)=\sum_{klij\lambda\nu}\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda}(0)}{\hat{H}(\beta)}{\phi_{lj}^{\nu}(0)}\outerproduct*{\phi_{ki}^{\lambda}(0)}{\phi_{lj}^{\nu}(0)}$
with

$\displaystyle\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda}(0)}{\hat{H}(\beta)}{\phi_{lj}^{\nu}(0)}$
$\displaystyle=\sum_{\lambda^{\prime}\nu^{\prime}}(g_{ki}^{\lambda\lambda^{\prime}}(\beta))^{\ast}g_{lj}^{\nu\nu^{\prime}}(\beta)\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda^{\prime}}(\beta)}{\hat{H}(\beta)}{\phi_{lj}^{\nu^{\prime}}(\beta)},$
(3)

This procedure also applies to $\hat{H}(-\beta)$, and hence, allows us to
directly evaluate the anti-symmetric term $\hat{G}_{0}(\beta)$ by simply
subtracting the matrix elements. For the small $\beta$, $\hat{G}_{0}(\beta)$
is decomposed into two terms as

$\hat{G}_{0}(\beta)=\hat{G}_{0}^{\mathrm{coup}}(\beta)+\hat{G}_{0}^{\mathrm{pos}}(\beta)+\mathcal{O}(\beta^{2}),$
(4)

where $\hat{G}_{0}^{\mathrm{coup}}(\beta)$ depends on $\beta$ from the
interlayer bonds described by the helical angle $\gamma$ and
$\hat{G}_{0}^{\mathrm{pos}}(\beta)$ depends on $\beta$ from the atomic
position itself combined with the $\gamma$-independent interlayer coupling
elements. The former is expressed as

$\innerproduct*{[}{[}3]{\phi_{2j}^{\lambda}(0)}{\hat{G}_{0}^{\mathrm{coup}}(\beta)}{\phi_{1j}^{\nu}(0)}=\frac{i}{\sqrt{2}}\frac{\rho\beta}{h}\begin{pmatrix}O&M\\\
-M^{\dagger}&O\end{pmatrix},$ (5)

where $\nu$ is taken in order $\\{0,z,+,-\\}$. The $2\times 2$ matrix $M$ is
defined by $M_{11}=-M_{12}=W_{sp}^{\sigma}$ and
$M_{21}=-M_{22}=W_{pp}^{\sigma}-W_{pp}^{\pi}$. Thus, the chirality appears as
the imaginary hopping, which is proportional to the radius $\rho$ and
inversely to the interlayer distance $h$. This feature is consistent with the
fact that $\rho/h$ corresponds to the ratio between the curvature and the
torsion in helices. The latter is expressed as

$\innerproduct*{[}{[}3]{\phi_{2j}^{\lambda}(0)}{\hat{G}_{0}^{\mathrm{pos}}(\beta)}{\phi_{1j}^{\nu}(0)}=\frac{\zeta\rho\beta}{\sqrt{2}}\begin{pmatrix}0&0&u_{1}&u_{1}^{\ast}\\\
0&0&-u_{1}&-u_{1}^{\ast}\\\ -u_{2}^{\ast}&0&0&0\\\ u_{2}&0&0&0\end{pmatrix},$
(6)

with $u_{1}=-ie^{-i(j-1)\alpha}W_{ss}^{\sigma}/\sqrt{3}$,
$u_{2}=\sqrt{3}e^{-i(j-1)\alpha}W_{pp}^{\pi}/2$, and $\zeta$ being the
exponent of the radial component of the orbitals. Hereafter, we compare this
result with the numerical calculation results in Sec. V.

##  IV Chiral Discriminating High-Harmonic Generation

###  IV.1 Three step model

Typical spectra and behaviors of HHG for molecular systems can be successfully
explained by the three-step model [30, 31]: The electron is first tunnel-
ionized, which is further accelerated by the laser field. Finally, the
electron recombines with its parent ion emitting the high-harmonics. The
intensity of the $n$th harmonics is given by
$I_{n}=(n\omega)^{4}\vab*{\bm{\mu}(n\omega)}^{2}=(n\omega)^{4}\vab*{\int_{0}^{\infty}\difc{t}{}\,\bm{\mu}(t)e^{in\omega
t}}^{2}$ with $\bm{\mu}$ being the electric dipole moment. In the quantum
mechanical formulation of the three step model, the electron–photon
interaction is treated within the electric dipole approximation. Furthermore,
assuming that only the single active electron participates in the HHG process,
the depletion of the ground state can be neglected, and the nuclear positions
are fixed (the frozen nuclear approximation), the electric dipole moment at
time $t$ is represented as [31, 36, 37]

$\displaystyle\bm{\mu}(t)$
$\displaystyle=\imaginary\int_{0}^{t}\difc{\tau}{}\,C(\tau)\bm{d}^{\ast}(\bm{p}_{\mathrm{st}}-e\bm{A}(t))e^{-\frac{i}{\hbar}S(\bm{p}_{\mathrm{st}},t,t-\tau)}$
$\displaystyle\times\bm{d}(\bm{p}_{\mathrm{st}}-e\bm{A}(t-\tau))\cdot\bm{E}(t-\tau),$
(7)

with $C(\tau)=\frac{2}{\hbar}\ab(\frac{2\pi
m_{\mathrm{e}}\hbar}{\eta+i\tau})^{3/2}$, the vector potential $\bm{A}$, the
electric field $\bm{E}=-\difcp{\bm{A}}{t}$, the electron charge $e$, and the
electron mass $m_{\mathrm{e}}$. Here, $\eta$ and
$\bm{p}_{\mathrm{st}}=\int_{t-\tau}^{t}\difc{t^{\prime}}{}\,e\bm{A}(t^{\prime})/\tau$
denote the regularization constant and the stationary momentum, respectively,
both of which come from the saddle point approximation in the momentum space
integral. During the acceleration process, the strong field approximation is
invoked, wherein the interaction between the ionized electron and the nuclei
is neglected, yielding the propagator described by the action
$S(\bm{p},t,t-\tau)=\int_{t-\tau}^{t}\difc{t^{\prime}}{}[(\bm{p}-e\bm{A}(t^{\prime}))^{2}/2m_{\mathrm{e}}+I_{\mathrm{p}}]$
with $I_{\mathrm{p}}$ being the ionization potential. The ionization amplitude
is defined as
$\bm{d}(\bm{p})=\innerproduct*{[}{[}3]{\bm{p}}{e\hat{\bm{r}}}{\phi_{\mathrm{GS}}}$,
where $\hat{\bm{r}}$ is the electron position operator. In this formulation,
the chiral effect is contained in $\bm{d}$ and $\bm{d}^{\ast}$ through the
dependence on $\ket{\phi_{\mathrm{GS}}(\beta)}$. In our model, where the
groundstate is given by
$\ket*{\phi_{\mathrm{GS}}(\beta)}=(\ket*{\phi_{m}(\beta)}+\ket*{\phi_{-m}(\beta)})/\sqrt{2}$,
$\bm{\mu}(t)$ is a sum of the IRREP-resolved electric dipoles,
$\bm{\mu}(t)=[\bm{\mu}_{m}(t)+\bm{\mu}_{-m}(t)]/2$, where $\bm{\mu}_{m}(t)$ is
obtained by replacing $\ket{\phi_{\mathrm{GS}}(\beta)}$ in Eq. (7) with
$\ket*{\phi_{m}(\beta)}$, because the ionized electron returns to its original
state or orbital, and hence, the mixing of the states with the different $m$
can be neglected.

###  IV.2 Pseudoscalar property of HHG signal

The chiral discrimination is quantified by the Kuhn $g$-factor [38], which is
generalized to the HHG signals as defined below. To prove that the Kuhn
$g$-factor is a pseudoscalar quantity, we assume that the molecular rotational
axis is fixed and parallel with the propagation direction of the light beams.
This assumption affects the signals quantitatively, but is not the
prerequisite for the pseudoscalar nature as exemplified by the numerical [28]
and experimental [24, 25] studies for randomly oriented molecules. Then, the
vector potential for the bicircular field is given by

$\bm{A}(t)=A_{+}\bm{e}_{+}e^{-ir_{+}\omega t}+A_{-}\bm{e}_{-}e^{-ir_{-}\omega
t}+\mathrm{c.c.},$ (8)

where $\bm{e}_{\pm}=\frac{1}{\sqrt{2}}(1,\pm i,0)^{\top}$ are the polarization
vectors and $A_{\pm}$ is the corresponding amplitudes with the fundamental
frequency $\omega$ and $r_{\pm}\in\mathbb{N}$. This is transformed under the
reflection operation to $\bar{\bm{A}}(t)=\hat{\sigma}\bm{A}(t)$ obtained by
interchanging the polarization vectors, $\bm{e}_{\pm}\to\bm{e}_{\mp}$. In the
cases of $A_{+}\neq A_{-}$ or $r_{+}\neq r_{-}$, the inequality
$\bar{\bm{A}}(t)\neq\bm{A}(t)$ holds, and thus, the bicircular field is
chiral. The reflection operation also acts as
$\hat{\sigma}\ket*{\phi_{m}(\beta)}\equiv\ket*{\phi_{-m}(-\beta)}$, which
straghtforwardly leads to
$\bm{\mu}(n\omega,\bm{A},\beta)=\hat{\sigma}\bm{\mu}(n\omega,\bar{\bm{A}},-\beta)$,
and thus, $I_{n}(\bm{A},\beta)=I_{n}(\bar{\bm{A}},-\beta)$. Now, it is clear
that the Kuhn $g$-factor

$g(\beta)\coloneqq
2\frac{I_{n}(\bm{A},\beta)-I_{n}(\bar{\bm{A}},\beta)}{I_{n}(\bm{A},\beta)+I_{n}(\bar{\bm{A}},\beta)},$
(9)

is pseudoscalar, $g(\beta)=-g(-\beta)$, for all harmonic orders $n$. We also
introduce the unnormalized $g$-factor, which we call $\tilde{g}$-factor, as
$\tilde{g}(\beta)\coloneqq I_{n}(\bm{A},\beta)-I_{n}(\bar{\bm{A}},\beta)$ to
discuss the correspondence between the molecular chirality and the HHG signal.
The $\tilde{g}$-factor is also the pseudoscalar.

##  V Results

To calculate the electric dipole moment Eq. (7), we performed the numerical
integration using the Gauss–Kronrod quadrature method [39] with the quad
precision arithmetics. Throughout the calculation, we set to $\omega
A_{+}=\quantity[per-mode=symbol]{3d10}{\per}$, $\omega A_{-}=\quantity[per-
mode=symbol]{4d10}{\per}$, $r_{+}=1$, $r_{-}=2$, and $\omega=\quantity{1}{}$
for the laser field, and $\rho=\quantity{0.256}{}$, $V=\quantity{0.5}{}$,
$W=\quantity{0.2}{}$, and $I_{\mathrm{p}}=\quantity{14}{}$ for the chiral
molecule. Hereafter, we focus on the $N=3$ with $m=\pm 1$ case only, but the
qualitative consistency was also checked for the $N=4$ with $m=\pm 1$ and the
$N=6$ with $m=\pm 1$ and $\pm 2$ cases.

Figure 2: (a) Calculated $g$-factors for $\beta=0.1\,\alpha$ (blue line) and
$\beta=-0.1\,\alpha$ (orange line). Each enantiomer is interconverted to the
other by inverting $\beta$. (b) Twist angle, $\beta$, dependence of the
$g$-factor. The interlayer distance is fixed to $h=\quantity{0.6}{}$. Figure
3: (a) Calculated $\tilde{g}$-factors for various values of the interlayer
distance, $h$. (b) The $\tilde{g}$-factor decomposition into the
$x+y$-component, $\vab*{\mu_{x}(n\omega)}^{2}+\vab*{\mu_{y}(n\omega)}^{2}$,
and the $z$-component, $\vab*{\mu_{z}(n\omega)}^{2}$, for the harmonic orders
$n=28$ and $34$. Twist angle is fixed to $\beta=0.005\,\alpha$.

Figure 2(a) depicts the $g$-factor for each enantiomer of our model chiral
molecule. The pseudoscalar property is confirmed for all harmonic orders $n$,
because the $g$-factor is reversed by inverting the twist angle, $\beta$. The
$g$-factor has a value in $-2\leq g\leq 2$ and its increasing is advantageous
for the chiral discrimination. In Fig. 2(a), this value can exceeds $1.0$,
which is much larger than typical values of the CD. This is because the CD is
contributed from the magnetic dipole moment in the light–matter interaction,
the magnitude of which is sufficiently smaller than that of the electric
dipole moment.

Next, we examine the correspondence between the molecular chirality of Eqs.
(5), (6) and the $g$-factor. Hereafter, we focus on the small $\beta$ region,
where the Hamiltonian is approximated as
$\hat{H}(\beta)\approx\hat{H}(0)+\hat{G}_{0}^{\mathrm{coup}}(\beta)+\hat{G}_{0}^{\mathrm{pos}}(\beta)$.
In Fig. 2(b), we present the $g$-factors for the various values of $\beta$,
where the $g$-factor monotonically increases as $\beta$ is increased. Because
the twist angle does not change the molecular size, which is the another
element to affect the signal, this feature can be attributed to the linear
$\beta$-dependence of $\hat{G}_{0}^{\mathrm{coup}}(\beta)$ and
$\hat{G}_{0}^{\mathrm{pos}}(\beta)$, and thus, the increase of the molecualr
chirality.

To analyze the correspondance without consideration of the effects from
$\hat{G}_{0}^{\mathrm{pos}}(\beta)$, we consider the $h$-dependence of the
chiral discriminating signals, because $\hat{G}_{0}^{\mathrm{pos}}(\beta)$ has
no $h$-dependence. To remove further the contribution from the symmetric part
of the Hamiltonian of the denominator in Eq. (8), we present the
$\tilde{g}$-factors for various values of $h$ in Fig. 3(a), where the peak
intensities are a monotonic increasing function of $h$. For the peaks of
$n=28$ and $34$, we decompose this into
$\vab*{\mu_{x}(n\omega)}^{2}+\vab*{\mu_{y}(n\omega)}^{2}$ and
$\vab*{\mu_{z}(n\omega)}^{2}$ depicted in Fig. 3(b), where the $x+y$-component
is negligibly small, and thus, the $z$-component dominantly contributes to the
$\tilde{g}$-factor. We further found that this $z$-component scales
quadratically with $h$. To unveil the origin of this, the ionization amplitude
$\bm{d}(\bm{p})$ is expanded with the basis states as

$\bm{d}(\bm{p})=\sum_{kj\nu}b_{kj}^{\nu}e^{-i\bm{p}\cdot\bm{R}_{kj}/\hbar}\ab(\bm{R}_{kj}+i\hbar\difcp*{}{\bm{p}})\tilde{\phi}^{\nu}(\bm{p}),$
(10)

where $b_{kj}^{\nu}$ is the linear coefficient for the basis
$\ket*{\phi_{kj}^{\nu}}$ and $\tilde{\phi}^{\nu}(\bm{p})$ is the momentum
representation of $\phi^{\nu}$. Eqs. (7) and (10) shows that the quadratic
dependence is originated from the $z$-component, $(\bm{R}_{kj})_{z}=(k-1)h$,
of the first term of Eq. (10) for the recombination process, $\bm{d}^{\ast}$,
and not the ionization process. This is because the ionization amplitude
emerges as the form $\bm{d}\cdot\bm{E}$ and does not depend on this term due
to no $z$-component of $\bm{E}(\tau)$. Therefore, the chiral discriminating
HHG can be interpreted as the quantification of the longitudinal dipole moment
at the recombination converted from the transverse dipole moment in the
ionization process. Because increasing the interlayer distance, $h$, decreases
$\hat{G}_{0}^{\mathrm{coup}}(\beta)$, simultaneously increasing the molecular
size, this result also indicates that the enhancement of the electric dipole
moment by enlarging the molecular size contributes dominantly to the
$\tilde{g}$-factor rather than increasing the molecular chirality,
$\hat{G}_{0}^{\mathrm{coup}}(\beta)$. This may be because the $z$-component of
Eq. (10) is proportional to $h$ and unbounded, whereas the chiral
contributions from $b_{kj}^{\nu}$ and $\bm{R}_{kj}$ are bounded by the
normalization condition of the wavefunction and the helical angle $\gamma$
appeared as $\sin\gamma$ and $\cos\gamma$, respectively. What should be noted
here is that this $z$-component is contributed only from the
$2\mathrm{p}_{z}$-orbitals, which are involved only in the chiral cases.
Indeed, in the achiral case with $\beta=0$, the $2\mathrm{p}_{z}$-orbital
belong to the different IRREP from the other orbitals, and thus, they are
orthogonal.

##  VI Concluding Remarks

In this study, we have developed the analytical method that extracts the anti-
symmetric part of the Hamiltonian that describes the chirality. Our approach
is based on the translation of the orbital center of each electronic basis
from the achiral system to the chiral system. This allows us to expand the
chiral Hamiltonians with the achiral basis states, enabling the direct
subtraction of the matrix elements of the different chirality. As a
demonstration of this method, we consider the model chiral molecule that
consists of helically stacked two $N$ regular polygons and investigate the
correspondence between the chiral part of our model and the Kuhn $g$-factor in
HHG. Numerical results shows that the $g$-factor is the pseudoscalar quantity
reflecting the chirality of the molecule. Calculated values are much larger
than the typical values of the CD and slightly larger than that reported in
Ref. [25] of the gaseous molecules. Extension of our result to the randomly
oriented molecules may give the quantitatively consistent result with the
experiments. Numerical results also reveal that the enhancement of the
electric dipole moment dominantly contributes to increasing the spectra even
when this enhancement is accompanied by reducing the molecular chirality. This
finding may have the following implication: The chiral molecules and materials
have an ability to interconvert the polar and axial quantities. This
inverconversion works efficiently for the strong chiral systems. However, to
obtain the large axial quantity from the polar one, increasing capacity to
generate the axial quantity can be effective (and vice versa). We demonstrated
this for the light-induced electric dipole moment in the HHG, and therefore,
expect that this validity can be extended to other optical processes.
Furthermore, this may also be relevant to the effects induced by the
metal–molecule interaction with or without the charge injection in the chiral
systems, especially the CISS, wherein the interrelation between the charge
redistribution and the formation of the anti-parallel spin pair [40, 41, 42]
is hypothesized as one of the keys to understand the CISS mechanism [43, 15].

Our method developed in this paper is also applicable to other chiral systems
such as helices and crystals. Generalization to the systems with the
spin–orbit, electron–phonon, and electron–photon interactions will be of great
interest in elucidating the microscopic insight into the AM exchange among
them [44, 45, 46], which is left for future studies.

###### Acknowledgements.

This work was supported by the JSPS KAKENHI Grant No. 21H05019, No. 22K04863,
and No. 22H05132.

##  Appendix A Construction of tight-binding Hamiltonian

In this Appendix, the tight-binding Hamiltonian of the model chiral molecule
introduced in Sec. II is constructed using the Slater–Koster method following
Refs. [32, 33]. For notational simplicity, we omit $\beta$ to specify the
chirality from the description.

The matrix element of the Hamiltonian is written as

$H_{ki,lj}^{\lambda\nu}\coloneqq\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda}}{\hat{H}}{\phi_{lj}^{\nu}}=E_{ki,lj}^{\lambda\nu}\exp\ab(i\xi_{ki,lj}^{\lambda\nu})$
(11)

where $E_{ki,lj}^{\lambda\nu}$ is the overlap integral and
$\xi_{ki,lj}^{\lambda\nu}$ is the phase factor that is determined to maintain
the molecular symmetry. Let $\hat{\bm{n}}(\nu_{ki})$ denote the unit vector
along the orbital $\ket*{\phi_{ki}^{\nu}}$. Then, the overlap integral can be
expressed as

$\displaystyle E_{ki,lj}^{\lambda\nu}$
$\displaystyle=\aab*{\hat{\bm{n}}(\lambda_{ki}),\hat{\bm{n}}(\nu_{lj})}V_{\lambda\nu}^{\pi}$
$\displaystyle+\aab*{\hat{\bm{R}}_{lj,ki},\hat{\bm{n}}(\lambda_{ki})}\aab*{\hat{\bm{R}}_{lj,ki},\hat{\bm{n}}(\nu_{lj})}\ab(V_{\lambda\nu}^{\sigma}-V_{\lambda\nu}^{\pi})$
(12)

where $V_{\lambda\nu}^{\sigma}$ and $V_{\lambda\nu}^{\pi}$ are the
Slater–Koster overlaps and $\bm{R}_{lj,ki}=\bm{R}_{lj}-\bm{R}_{ki}$ is the
bond vector. The normalized bond vector,
$\hat{\bm{R}}_{lj,ki}\equiv\bm{R}_{lj,ki}/\vab*{\bm{R}_{lj,ki}}$, is given by

$\displaystyle\hat{\bm{R}}_{lj,ki}$
$\displaystyle=-\sin(\theta_{lj}+\theta_{ki})\cos\gamma_{lj,ki}\bm{e}_{x}$
$\displaystyle\quad+\cos(\theta_{lj}+\theta_{ki})\cos\gamma_{lj,ki}\bm{e}_{y}$
$\displaystyle\quad+\sin\gamma_{lj,ki}\bm{e}_{z}$ (13)

with $\theta_{ki}=[(k-1)\beta+(i-1)\alpha]/2$,
$\theta_{lj,ki}=\theta_{lj}-\theta_{ki}$ and
$\gamma_{lj,ki}=\arctan[(l-k)h/2\rho\sin\theta_{lj,ki}]$. With these
notations, $\hat{\bm{n}}(\nu_{ki})$ is written as
$\hat{\bm{n}}(s_{ki})=\hat{\bm{R}}_{lj,ki}$,
$\hat{\bm{n}}(s_{lj})=-\hat{\bm{R}}_{lj,ki}$, $\hat{\bm{n}}(x_{ki})=\cos
2\theta_{ki}\bm{e}_{x}+\sin 2\theta_{ki}\bm{e}_{y}$,
$\hat{\bm{n}}(y_{ki})=-\sin 2\theta_{ki}\bm{e}_{x}+\cos
2\theta_{ki}\bm{e}_{y}$, and $\hat{\bm{n}}(z_{ki})=\bm{e}_{z}$. Hence, we can
obtain the analytical expressions of the overlap integrals,
$E_{ki,lj}^{\lambda\nu}$, by directly evaluating Eq. (12). The details of the
calculation is given in Refs. [33]. Finally, we can obtain the matrix elements
for the intralayer bond as

$\displaystyle H_{kj,kj\pm 1}^{ss}$ $\displaystyle=-V_{ss}^{\sigma}$ (14)
$\displaystyle H_{kjkj\pm 1}^{zz}$ $\displaystyle=V_{pp}^{\pi}$ (15)
$\displaystyle H_{kj,kj\pm 1}^{zs}$ $\displaystyle=H_{kj,kj\pm 1}^{sz}=0$ (16)
$\displaystyle H_{kj,kj\pm 1}^{s\eta}$ $\displaystyle=-H_{kj,kj\pm 1}^{\eta
s}=\pm\frac{i\eta}{\sqrt{2}}e^{\pm i\eta\pi/2}V_{sp}^{\sigma}$ (17)
$\displaystyle H_{kj,kj\pm 1}^{z\eta}$ $\displaystyle=H_{kj,kj\pm 1}^{\eta
z}=0$ (18) $\displaystyle H_{kj,kj\pm 1}^{\eta\eta}$
$\displaystyle=\frac{1}{2}e^{\mp
i\eta\alpha}\ab(V_{pp}^{\sigma}+V_{pp}^{\pi})$ (19) $\displaystyle H_{kj,kj\pm
1}^{\eta\bar{\eta}}$
$\displaystyle=-\frac{1}{2}\ab(V_{pp}^{\sigma}-V_{pp}^{\pi})$ (20)

and for the interlayer bond as

$\displaystyle H_{kj,k\pm 1j}^{ss}$ $\displaystyle=-W_{ss}^{\sigma}$ (21)
$\displaystyle H_{kj,k\pm 1j}^{zz}$
$\displaystyle=W_{pp}^{\pi}+\sin^{2}\gamma\ab(W_{pp}^{\sigma}-W_{pp}^{\pi})$
(22) $\displaystyle H_{kj,k\pm 1j}^{zs}$ $\displaystyle=H_{kj,k\pm
1j}^{sz}=\pm\sin\gamma W_{sp}^{\sigma}$ (23) $\displaystyle H_{kj,k\pm
1j}^{s\eta}$ $\displaystyle=\ab(H_{kj,k\pm 1j}^{\eta
s})^{\ast}=\pm\frac{i\eta}{\sqrt{2}}e^{i(j-1)\eta\alpha}\cos\gamma
W_{sp}^{\sigma}$ (24) $\displaystyle H_{kj,k\pm 1j}^{z\eta}$
$\displaystyle=\ab(H_{kj,k\pm 1j}^{\eta z})^{\ast}$
$\displaystyle=\frac{i\eta}{2\sqrt{2}}e^{i(j-1)\eta\alpha}\sin
2\gamma\ab(W_{pp}^{\sigma}-W_{pp}^{\pi})$ (25) $\displaystyle H_{kj,k\pm
1j}^{\eta\eta}$
$\displaystyle=W_{pp}^{\pi}+\frac{1}{2}\cos^{2}\gamma\ab(W_{pp}^{\sigma}-W_{pp}^{\pi})$
(26) $\displaystyle H_{kj,k\pm 1j}^{\eta\bar{\eta}}$
$\displaystyle=-\frac{1}{2}\cos^{2}\gamma\ab(W_{pp}^{\sigma}-W_{pp}^{\pi}).$
(27)

In the above, the unknown parameters are the Slater–Koster overlap constants.
To determine these, we take the carbon atom as a reference [32, 33] and set to
$V_{ss}^{\sigma}=-1.40\,V$, $V_{sp}^{\sigma}=1.84\,V$,
$V_{pp}^{\sigma}=3.24\,V$, $V_{pp}^{\pi}=-0.81\,V$, and
$W_{ss}^{\sigma}=-1.40\,W$, $W_{sp}^{\sigma}=1.84\,W$,
$W_{pp}^{\sigma}=3.24\,W$, $W_{pp}^{\pi}=-0.81\,W$.

##  Appendix B Extraction of the anti-symmetric part of the Hamiltonian

In this Appendix, we present a method to evaluate the anti-symmetric part of
the Hamilonian, and identify the key parameters describing the chirality of
the system.

The Hamiltonian is separated formally into the symmetric and anti-symmetric
terms with respect to the reflection operation as

$\hat{H}(\beta)=\hat{Q}_{0}(\beta)+\hat{G}_{0}(\beta),$ (28)

where $\hat{Q}_{0}(\beta)=[\hat{H}(\beta)+\hat{H}(-\beta)]/2$ and
$\hat{G}_{0}(\beta)=[\hat{H}(\beta)-\hat{H}(-\beta)]/2$ satisfy
$\hat{\sigma}\hat{Q}_{0}(\beta)\hat{\sigma}^{\dagger}=\hat{Q}_{0}(\beta)$ and
$\hat{\sigma}\hat{G}_{0}(\beta)\hat{\sigma}^{\dagger}=-\hat{G}_{0}(\beta)$,
respectively. The anti-symmetric term $\hat{G}_{0}$ is defined as the
subtraction of the Hamiltonians, each of which has the different chirality and
is expanded in terms of the chirality-dependent electronic basis states.
Because the vanishing twist angle, $\beta=0$, corresponds to the achiral
system, and using the fact that the chirality enters as the atomic positions
in the orbital center, eash basis is related to the achiral basis as

$\displaystyle\phi^{\nu}(\bm{r}-\bm{R}_{kj}(0))$
$\displaystyle=\exp\ab(-\frac{i}{\hbar}\bm{\hat{p}}\cdot(\bm{R}_{kj}(0)-\bm{R}_{kj}(\beta)))$
$\displaystyle\times\phi^{\nu}(\bm{r}-\bm{R}_{kj}(\beta))$
$\displaystyle=\hat{T}_{kj}(\beta)\phi^{\nu}(\bm{r}-\bm{R}_{kj}(\beta))$ (29)

where $\hat{\bm{p}}=-i\hbar\gradient$ is the momentum operator. The exponent
is expressed as

$\displaystyle-\frac{i}{\hbar}\hat{\bm{p}}\cdot(\bm{R}_{kj}(0)-\bm{R}_{kj}(\beta))$
$\displaystyle=i\rho\sin\ab(\frac{k-1}{2}\beta)\ab[e^{i\kappa_{kj}}\ab(\partial_{x}-i\partial_{y})-e^{-i\kappa_{kj}}\ab(\partial_{x}+i\partial_{y})].$
(30)

with $\kappa_{kj}=(k-1)\beta/2+(j-1)\alpha$. Because the nuclei are fixed, the
differential operator for $\bm{r}$ is equivalent to that for
$\bm{r}^{\prime}=\bm{r}-\bm{R}_{kj}(\beta)=(x^{\prime},y^{\prime},z^{\prime})^{\top}$,
$\partial_{a}=\partial_{a^{\prime}}$ ($a=x,y,z$). By expressing the orbitals
as
$\phi^{\nu}(\bm{r}^{\prime})=R_{n}(r^{\prime})Y_{l}^{m}(\theta^{\prime},\varphi^{\prime})$,
where $R_{n}$ and $Y_{l}^{m}$ are the radial distributions and the spherical
harmonics with $n,l,m$ being the quantum number corresponding to the orbital
$\nu$, the differential operator acts as

$\displaystyle\ab(\partial_{x^{\prime}}\pm i\partial_{y^{\prime}})\phi^{\nu}$
$\displaystyle=e^{\pm
i\varphi^{\prime}}\left[\sin\theta^{\prime}Y_{l}^{m}\difcp{R_{n}}{r^{\prime}}\right.$
$\displaystyle\left.+\frac{R_{n}}{r^{\prime}}\ab(\cos\theta^{\prime}\difcp{}{\theta^{\prime}}\pm
i\frac{1}{\sin\theta^{\prime}}\difcp{}{\varphi^{\prime}})Y_{l}^{m}\right].$
(31)

Assuming that the Hamiltonian is spanded by the $2\mathrm{s}$,
$2\mathrm{p}_{z}$, and $2\mathrm{p}_{\pm}$ orbitals, and hence, the
contribution from the other is negligibly small, the action of the
differential operator becomes

$\displaystyle\ab(\partial_{x}\pm i\partial_{y})\phi^{0}(\bm{r}^{\prime})$
$\displaystyle=-\sqrt{\frac{2}{3}}\zeta\phi^{\pm}(\bm{r}^{\prime})$ (32)
$\displaystyle\ab(\partial_{x}\pm i\partial_{y})\phi^{z}(\bm{r}^{\prime})$
$\displaystyle=0$ (33) $\displaystyle\ab(\partial_{x}\pm
i\partial_{y})\phi^{\pm}(\bm{r}^{\prime})$ $\displaystyle=0$ (34)
$\displaystyle\ab(\partial_{x}\pm i\partial_{y})\phi^{\mp}(\bm{r}^{\prime})$
$\displaystyle=-\sqrt{\frac{3}{2}}\zeta\phi^{0}(\bm{r}^{\prime}),$ (35)

with $\zeta$ being the exponent of the radial distbribution, $R_{n}(r)\propto
r^{n-1}e^{-\zeta r}$.

What is necessary is to represent the action of $\hat{T}_{kj}(\beta)$ on the
orbital space $\\{\phi^{\nu}(\bm{r}^{\prime})\\}_{\nu}$, namely, to determine
the coefficients that satisfies
$\hat{T}_{kj}(\beta)\phi^{\nu}(\bm{r}^{\prime})=\sum_{\nu^{\prime}}g_{kj}^{\nu\nu^{\prime}}(\beta)\phi^{\nu^{\prime}}(\bm{r}^{\prime})$
with $g_{kj}^{\nu\nu^{\prime}}(\beta)\in\mathbb{C}$. For that purpose, the
eigenstates of the operator
$i(e^{i\kappa_{kj}}(\partial_{x}-i\partial_{y})-e^{-i\kappa_{kj}}(\partial_{x}+i\partial_{y}))$
as the linear coupling of
$\bm{\phi}^{\prime}=(\phi^{0}(\bm{r}^{\prime}),\phi^{+}(\bm{r}^{\prime}),\phi^{-}(\bm{r}^{\prime}))^{\top}$
needs to be constructed. The orbital $\phi^{z}(\bm{r}^{\prime})$ does not
couple to the other orbital and itself form the eigenstate. From Eq. (35), the
action of the above operator is written as

$\displaystyle
i(e^{i\kappa_{kj}}(\partial_{x}-i\partial_{y})-e^{-i\kappa_{kj}}(\partial_{x}+i\partial_{y}))\bm{\phi}^{\prime}$
$\displaystyle=-\sqrt{\frac{2}{3}}\zeta\begin{pmatrix}0&-ie^{-i\kappa_{kj}}&ie^{i\kappa_{kj}}\\\
\frac{3}{2}ie^{i\kappa_{kj}}&0&0\\\
-\frac{3}{2}ie^{-i\kappa_{kj}}&0&0\end{pmatrix}\bm{\phi}^{\prime}.$ (36)

Hence, we have to diagonalize the matrix in the right hand side of Eq. (36).
This can be straightforwardly done, resulting to the eigenvalues
$\epsilon_{0}=0$ and $\epsilon_{\pm}=\mp\sqrt{2}\zeta$ with the corresponding
eigenstates

$\displaystyle\psi_{0}$
$\displaystyle=\frac{1}{\sqrt{2}}\ab(e^{i\kappa_{kj}}\phi^{+}+e^{-i\kappa_{kj}}\phi^{-}),$
(37) $\displaystyle\psi_{\pm}$ $\displaystyle=\mp
i\sqrt{\frac{2}{5}}\phi^{0}+\sqrt{\frac{3}{10}}\ab(e^{i\kappa_{kj}}\phi^{+}-e^{-i\kappa_{kj}}\phi^{-}),$
(38)

respectively. Inversely, each orbital can be expressed in terms of the
eigenstates as

$\displaystyle\phi^{0}$
$\displaystyle=\sqrt{\frac{5}{8}}i\ab(\psi_{+}-\psi_{-})$ (39)
$\displaystyle\phi^{\pm}$ $\displaystyle=e^{\mp
i\kappa_{kj}}\ab[\frac{1}{\sqrt{2}}\psi_{0}\pm\sqrt{\frac{5}{24}}\ab(\psi_{+}+\psi_{-})]$
(40)

Using these results, the action of $\hat{T}_{kj}(\beta)$ on the orbitals
$\bm{\phi}=(\phi^{0}(\bm{r}^{\prime}),\phi^{z}(\bm{r}^{\prime}),\phi^{+}(\bm{r}^{\prime}),\phi^{-}(\bm{r}^{\prime}))^{\top}$
is calculated and summarized as
$\hat{T}_{kj}(\beta)\bm{\phi}=g_{kj}(\beta)\bm{\phi}$ with

$\displaystyle g_{kj}(\beta)$
$\displaystyle=\begin{pmatrix}c_{k}&0&-\frac{\sqrt{3}}{2}s_{k}e^{i\kappa_{kj}}&\frac{\sqrt{3}}{2}s_{k}e^{-i\kappa_{kj}}\\\
0&1&0&0\\\
\frac{i}{\sqrt{3}}s_{k}e^{-i\kappa_{kj}}&0&\frac{1+c_{k}}{2}&\frac{1-c_{k}}{2}e^{-i2\kappa_{kj}}\\\
-\frac{i}{\sqrt{3}}s_{k}e^{i\kappa_{kj}}&0&\frac{1-c_{k}}{2}e^{i2\kappa_{kj}}&\frac{1+c_{k}}{2}\end{pmatrix}$
(41)

with $s_{k}=\sinh\tau_{k}$, $c_{k}=\cosh\tau_{k}$, and
$\tau_{k}=\sqrt{2}\zeta\rho\sin(\frac{k-1}{2}\beta)$.

Now, let us calculate the anti-symmetric part of the Hamiltonian,
$\hat{G}_{0}(\beta)=[\hat{H}(\beta)-\hat{H}(-\beta)]/2$ expanded in terms of
the achiral basis states. The matrix element is given by

$\displaystyle\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda}(0)}{\hat{G}_{0}(\beta)}{\phi_{lj}^{\nu}(0)}$
$\displaystyle=\frac{1}{2}\sum_{\lambda^{\prime}\nu^{\prime}}\left[(g_{ki}^{\lambda\lambda^{\prime}}(\beta))^{\ast}g_{lj}^{\nu\nu^{\prime}}(\beta)\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda^{\prime}}(\beta)}{\hat{H}(\beta)}{\phi_{lj}^{\nu^{\prime}}(\beta)}\right.$
$\displaystyle\left.-(g_{ki}^{\lambda\lambda^{\prime}}(-\beta))^{\ast}g_{lj}^{\nu\nu^{\prime}}(-\beta)\innerproduct*{[}{[}3]{\phi_{ki}^{\lambda^{\prime}}(-\beta)}{\hat{H}(-\beta)}{\phi_{lj}^{\nu^{\prime}}(-\beta)}\right].$
(42)

Using Eq. (41), the right hand side of the above equation can be calculated,
which has the matrix form in the orbital space

$\innerproduct*{[}{[}3]{\phi_{2j}^{\lambda}(0)}{\hat{G}_{0}(\beta)}{\phi_{1j}^{\nu}(0)}=\begin{pmatrix}f_{1}&0&f_{3}&f_{3}^{\ast}\\\
f_{2}&0&f_{4}&f_{4}^{\ast}\\\ f_{5}+f_{6}&f_{7}&f_{8}&-f_{8}^{\ast}+f_{9}\\\
f_{5}^{\ast}-f_{6}^{\ast}&f_{7}^{\ast}&-f_{8}+f_{9}^{\ast}&f_{8}^{\ast}\\\
\end{pmatrix},$ (43)

where its elements are defined as

$\displaystyle f_{1}$
$\displaystyle=-i\sqrt{\frac{3}{2}}\imaginary(a_{1})W_{sp}^{\sigma}$ (44)
$\displaystyle f_{2}$
$\displaystyle=-i\sqrt{\frac{3}{2}}\imaginary(a_{1})\sin\gamma(W_{pp}^{\sigma}-W_{pp}^{\pi})$
(45) $\displaystyle f_{3}$
$\displaystyle=-\frac{i}{\sqrt{3}}a_{2}W_{ss}^{\sigma}+\frac{i}{\sqrt{2}}a_{3}W_{sp}^{\sigma}$
(46) $\displaystyle f_{4}$ $\displaystyle=\frac{i}{\sqrt{3}}a_{2}\sin\gamma
W_{ss}^{\sigma}+\frac{i}{\sqrt{2}}a_{3}\sin\gamma(W_{pp}^{\sigma}-W_{pp}^{\pi})$
(47) $\displaystyle f_{5}$
$\displaystyle=-\frac{i}{\sqrt{2}}\cosh\tau\cos\gamma W_{sp}^{\sigma}$ (48)
$\displaystyle f_{6}$
$\displaystyle=-\frac{\sqrt{3}}{2}\ab(a_{2}^{\ast}W_{pp}^{\pi}+\real(a_{2})\cos^{2}\gamma(W_{pp}^{\sigma}-W_{pp}^{\pi}))$
(49) $\displaystyle f_{7}$
$\displaystyle=-\frac{i}{\sqrt{2}}\sin\gamma\cos\gamma(W_{pp}^{\sigma}-W_{pp}^{\pi})$
(50) $\displaystyle f_{8}$
$\displaystyle=\frac{i}{\sqrt{6}}a_{1}W_{sp}^{\sigma}-\frac{i}{2}a_{4}\cos^{2}\gamma(W_{pp}^{\sigma}-W_{pp}^{\pi})$
(51) $\displaystyle f_{9}$ $\displaystyle=-ia_{4}^{\ast}W_{pp}^{\pi}$ (52)

with

$\displaystyle a_{1}$
$\displaystyle=e^{-i(j-1)\alpha}\sinh\tau\sin\frac{\beta}{2}\cos\gamma$ (53)
$\displaystyle a_{2}$
$\displaystyle=e^{-i(j-1)\alpha}\sinh\tau\cos\frac{\beta}{2}$ (54)
$\displaystyle a_{3}$
$\displaystyle=\cosh^{2}\frac{\tau}{2}+e^{-2i(j-1)\alpha}\sinh^{2}\frac{\tau}{2}\cos\beta$
(55) $\displaystyle a_{4}$
$\displaystyle=e^{-2i(j-1)\alpha}\sinh^{2}\frac{\tau}{2}\sin\beta$ (56)

and $\tau=\tau_{2}=\sqrt{2}\zeta\rho\sin(\beta/2)$. In the above elements, we
can distinguish two origins of $\beta$-dependence, which are the atomic
position through the dependence on $j$ and the interlayer bond through the
dependence on the helical angle $\gamma$. These two contributions are decouped
in the first order component of $\beta$. Therefore, for the sufficiently small
$\beta$ region, the elements from the former and the latter origins are
expressed as $f_{i}^{\mathrm{pos}}$ and $f_{i}^{\mathrm{coup}}$, respectively
and are written as
$f_{1}^{\mathrm{pos}}=f_{2}^{\mathrm{pos}}=f_{5}^{\mathrm{pos}}=f_{7}^{\mathrm{pos}}=f_{8}^{\mathrm{pos}}=f_{9}^{\mathrm{pos}}=0$,

$\displaystyle f_{3}^{\mathrm{pos}}$
$\displaystyle=-\frac{i}{\sqrt{6}}\zeta\rho\beta
e^{-i(j-1)\alpha}W_{ss}^{\sigma},$ (57) $\displaystyle f_{4}^{\mathrm{pos}}$
$\displaystyle=\frac{i}{\sqrt{6}}\zeta\rho\beta
e^{-i(j-1)\alpha}W_{ss}^{\sigma},$ (58) $\displaystyle f_{6}^{\mathrm{pos}}$
$\displaystyle=-\sqrt{\frac{3}{8}}\zeta\rho\beta
e^{i(j-1)\alpha}W_{pp}^{\pi},$ (59)

$f_{1}^{\mathrm{coup}}=f_{2}^{\mathrm{coup}}=f_{6}^{\mathrm{coup}}=f_{8}^{\mathrm{coup}}=f_{9}^{\mathrm{coup}}=0$,

$\displaystyle f_{3}^{\mathrm{coup}}$
$\displaystyle=\frac{i}{\sqrt{2}}\frac{\rho}{h}\beta W_{sp}^{\sigma},$ (60)
$\displaystyle f_{4}^{\mathrm{coup}}$
$\displaystyle=\frac{i}{\sqrt{2}}\frac{\rho}{h}\beta(W_{pp}^{\sigma}-W_{pp}^{\pi}),$
(61) $\displaystyle f_{5}^{\mathrm{coup}}$
$\displaystyle=-\frac{i}{\sqrt{2}}\frac{\rho}{h}\beta W_{sp}^{\sigma},$ (62)
$\displaystyle f_{7}^{\mathrm{coup}}$
$\displaystyle=-\frac{i}{\sqrt{2}}\frac{\rho}{h}\beta(W_{pp}^{\sigma}-W_{pp}^{\pi}).$
(63)

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