$\pi=u(\omega^{2}\pi)$ for some unit $u$, (4) $-\pi$, for the same reason as
(1), (5) $-\omega\pi$ for the same reason as (2), and (6) $-\omega^{2}\pi$ for
the same reason as (3).

With all of these associates for $\pi$, we can now express each in terms of
$a$ and $b$. Namely, in order, we have

  1. (1)

$a+b\omega$,

  2. (2)

$\omega(a+b\omega)=a\omega+b\omega^{2}=a\omega+b(-1-\omega)=-b+(a-b)\omega$,

  3. (3)

$\omega^{2}(a+b\omega)=a\omega^{2}+b\omega^{3}=a(-1-\omega)+b=(b-a)-a\omega$,

  4. (4)

$-a-b\omega$,

  5. (5)

$-\omega(a+b\omega)=-(-b+(a-b)\omega)=b+(b-a)\omega$,

  6. (6)

$-\omega^{2}(a+b\omega)=-((b-a)-a\omega)=(a-b)+a\omega$.

Among these we must determine the primary associate. Recall that
$N\pi=p=a^{2}-ab+b^{2}$. In this expression, only one of $a$ and $b$ is
divisible by $3$ since $\pi$ is primary. Therefore, we proceed by assuming,
WLOG, that $a\not\equiv 0\pmod{3}$. Then we further assume that $a\equiv
2\pmod{3}$. With these assumptions, we have that $p=a^{2}-ab+b^{2}\implies
p\equiv 1\equiv 2^{2}-2b+b^{2}\pmod{3}\implies-3\equiv b(b-2)\equiv
0\pmod{3}$. We now consider two cases for $b$. If $3|b$, then we have $a\equiv
2\pmod{3}$ and $b\equiv 0\pmod{3}$, such that $\pi\equiv 2\pmod{3}$ and so
$a+b\omega$ is primary. If $b\equiv 2\pmod{3}$, then we must also have
$a\equiv 2\pmod{3}$, so $\pi\equiv 2+2\omega\equiv b+(b-a)\omega\pmod{3}$.
Therefore $b+(b-a)\omega$ is primary.

Now all that remains is to prove uniqueness. Let $a+b\omega$ be primary. This
only occurs when $b\equiv 0\pmod{3}$, so looking at (2), it is clear that
$-b+(a-b)\omega$ cannot be primary. The same is true for (3). For (4), notice
that $b\equiv 0\pmod{3}$ implies that $\pi\equiv-a\equiv-2\equiv 1\pmod{3}$,
so $-a-b\omega$ is not primary. For (5), the argument is identical to (2). For
(6), the argument is identical to (3). ∎

Now we are equipped to state cubic reciprocity.

######  Theorem 3.13 (The Law of Cubic Reciprocity).

Let $\pi_{1}$ and $\pi_{2}$ be primary. Furthermore, let
$N\pi_{1},N\pi_{2}\neq 3$ with $N\pi_{1}\neq N\pi_{2}$. Then

$\chi_{\pi_{1}}(\pi_{2})=\chi_{\pi_{2}}(\pi_{1}).$

In words, if $\pi_{1}$ and $\pi_{2}$ are primary with different norms not
equal to $3$, then $\pi_{1}$ is a cubic residue modulo $\pi_{2}$ if $\pi_{2}$
is a cubic residue modulo $\pi_{1}$, and $\pi_{1}$ is a cubic nonresidue
modulo $\pi_{2}$ if $\pi_{2}$ is a cubic nonresidue modulo $\pi_{1}$.

This theorem requires that we consider 3 scenarios. The first scenario
requires that we consider whether each $\pi$ is rational or not. Namely, when
both $\pi_{1}$ and $\pi_{2}$ are rational, when one of $\pi_{1}$ and $\pi_{2}$
is rational and the other is complex (i.e. in $D$), and when both $\pi_{1}$
and $\pi_{2}$ are complex. In Remark 8, we showed that the first case is
trivial. We will also consider special cases of cubic reciprocity, namely when
the input of the cubic character is either a unit, which we consider in the
first supplement, or a special prime, which we consider in the second
supplement. The first supplement is easily provable, but the second is a
little more difficult.

###  3.4. Supplements to Cubic Reciprocity

We first consider cubic reciprocity when the inputs are the units, namely
$1,\omega,$ and $\omega^{2}$ and their negatives. Clearly, $(-1)^{3}=-1$, so
$-1$ is always a cubic residue modulo any $\pi$ prime, i.e.
$\chi_{\pi}(-1)=1$. By (2) of Proposition 3.9, we have that
$\chi_{\pi}(\omega)=\omega^{\frac{N\pi-1}{3}}$. Therefore, the cubic character
of units can be stated as follows.

######  Theorem 3.14 (First supplement to the Law of Cubic Reciprocity).

Let $\omega$ be a cube root of unity. Then

$\chi_{\pi}(\omega)=\omega^{\frac{N\pi-1}{3}}=\left\\{\begin{array}[]{lll}1&\text{if}\
N\pi\equiv 1\pmod{9},\\\ \omega&\text{if}\ N\pi\equiv 4\pmod{9},\\\
\omega^{2}&\text{if}\ N\pi\equiv 7\pmod{9}.\end{array}\right.$

###### Proof.

This is not difficult to show by considering each case. ∎

Notice that in the identity $1+\omega+\omega^{2}=0$, we can write
$\omega^{2}=-1-\omega$. Therefore the final case requires us to consider the
cubic residue character $\chi_{\pi}(1-\omega)$, where $1-\omega$ is also a
prime in $D$ as shown in Theorem 3.4. The proof is divided into two separate
cases, specifically when the modulus is a rational prime $\pi=q$ and when the
modulus is a non-rational prime $\pi$. The first case is easily considered,
but the second case requires us to consider more about primary elements in
$D$.

######  Theorem 3.15 (Second supplement to the Law of Cubic Reciprocity).

Let $N\pi\neq 3$. If $\pi=q$ is rational, then write $q=3m-1$. If $\pi$ is
primary with $\pi\in D$ then write $\pi=a+b\omega$ and take $a=3m-1$. Then

$\chi_{\pi}(1-\omega)=\omega^{2m}.$

###### Proof of Theorem 3.15 when $\pi=q$ is rational.

This supplement requires us to consider two different cases. The first case is
when $\pi=q$ is a rational prime, and the second is when $\pi=a+b\omega$ is
non-rational. We consider the case where $\pi=q$ is a rational prime in this
proof.

First notice that
$(1-\omega)^{2}=1-2\omega+\omega^{2}=-1-\omega+1-2\omega=-3\omega$. Therefore
by definition of a character,
$\chi_{q}((1-\omega)^{2})=\chi_{q}(-3\omega)=\chi_{q}(-3)\chi_{q}(\omega)$. By
Corollary 3.11, since we know that $\pi=q$ is a rational prime, and since
$\gcd(-3,q)=1$, we can write $\chi_{q}(\overline{-3})=\chi_{q}(-3)=1$. Taking
$\pi=q$ to be a rational prime in Theorem 3.14, and since $N\pi=Nq=q^{2}$, we
can write $\chi_{q}(\omega)=\omega^{(N\pi-1)/3}=\omega^{(q^{2}-1)/3}$. If we
square both sides and notice that $\chi_{q}$ is inherently a cube root of
unity then by Corollary 3.11 we have

$\displaystyle(\chi_{q}(1-\omega)^{2})^{2}=\chi_{q}(1-\omega)^{4}=(\chi_{q}(1-\omega))^{3}\chi_{q}(1-\omega)$
$\displaystyle=\chi_{q}(1-\omega)$
$\displaystyle=\omega^{\frac{2}{3}\cdot\frac{q^{2}-1}{3}}.$

We now evaluate the RHS. Let $q=3m-1$. Then $q^{2}-1=(3m-1)^{2}-1=9m^{2}-6m$.
Therefore the exponent of the RHS is $2/3(9m^{2}-6m)=6m^{2}-4m$. Reducing this
modulo $3$ because we are dealing with all possible powers of $\omega$ \-
namely, the cube roots of unity - we have $6m^{2}-4m\equiv-4m\equiv
2m\pmod{3}$. Substituting, we have $\chi_{q}(1-\omega)=\omega^{2m}$, which is
our desired result. ∎

The second case requires that we investigate some facts about primary elements
in $D$. We begin with the following lemma.

######  Lemma 3.16.

Let $\alpha$ and $\beta$ be two primary elements in $D$. Then $-\alpha\beta$
is primary.

###### Proof.

Let $\alpha=a+b\omega$ and $\beta=c+d\omega$. By definition of primary, we
have $\alpha\equiv 2\pmod{3}$ and $\beta\equiv 2\pmod{3}$. For some
$s,t\in\mathbb{Z}$, let $\alpha=3s+2$ and $\beta=3t+2$. Taking the product, we
have

$\displaystyle-\alpha\beta=-(3s+2)(3t+2)=-(9st+6s+6t+4)$
$\displaystyle\equiv-4\pmod{3}$ $\displaystyle\equiv 2\pmod{3}.$

Therefore $-\alpha\beta$ is primary. ∎

This can in fact be extended to a product of any number primary elements. This
gives us a sort of primary factorization for primary elements in $D$.

######  Corollary 3.17.

Let $\gamma_{1},\gamma_{2},\ldots,\gamma_{n}\in D$ be primary. Then
$(-1)^{n-1}\gamma_{1}\gamma_{2}\cdots\gamma_{n}$ is also primary.

###### Proof.

We prove this with induction. When $n=2$, we have
$(-1)^{1}\gamma_{1}\gamma_{2}$, which is primary by Lemma 3.16. Assume that
this holds for some $n=k$. Then
$(-1)^{k-1}\gamma_{1}\gamma_{2}\cdots\gamma_{k}$ is primary. We want to show
that $(-1)^{k}\gamma_{1}\gamma_{2}\cdots\gamma_{k}\gamma_{k+1}$ is primary.
Notice that

$(-1)^{k}\gamma_{1}\gamma_{2}\cdots\gamma_{k}\gamma_{k+1}=(-1)(-1)^{k-1}\gamma_{1}\gamma_{2}\cdots\gamma_{k}(\gamma_{k+1})=-(\gamma_{k+1})((-1)^{k-1}\gamma_{1}\gamma_{2}\cdots\gamma_{k}).$

We know that $\gamma_{k+1}$ is primary, so by Lemma 3.16, the product is also
primary. ∎

This primary factorization allows us to consider the set of primaries as a
UFD. In other words, if $\gamma\in D$ is some primary element, then we can
write $\gamma=(-1)^{k-1}\gamma_{2}\gamma_{2}\cdots\gamma_{k}$, where, as in
$\mathbb{Z}$, the $\gamma_{i}$s need not be distinct primary primes elements.

###### Proof of Theorem 3.15 when $\pi$ is a non-rational prime.

We can now prove the second case of Theorem 3.15. Let $\pi=a+b\omega$ be a
primary non-rational prime. This is only possible when $a\equiv
2\equiv-1\pmod{3}$ and $b\equiv 0\pmod{3}$. For $m,n\in\mathbb{Z}$, let
$a=3m-1$ and $b=3n$. By definition, $a$ is primary. If $a$ is non-prime, then
Corollary 3.17 asserts that for some sequence of primary primes $a_{i}$, we
can write $a=(-1)^{n-1}a_{1}a_{2}\cdots a_{n}$. WLOG, we assume that $a$ is a
primary rational prime because we can choose any such $a_{i}$ to be our $a$.
By extension, we can say that $a+b$ is also a primary rational prime.
Furthermore, since $a$ and $b$ are nonzero and we assumed $\pi$ to be complex,
it is also true that $\gcd(a,a+b)=\gcd(b,a+b)=\gcd(a,a+bw)$.

Notice that

(1) $\frac{Na-1}{3}=\frac{(3m-1)^{2}-1}{3}\equiv 2m\pmod{3}.$

Furthermore, notice that

(2) $a+b\omega\equiv b\omega\pmod{a}.$

Notice that

$\displaystyle a+b\omega$ $\displaystyle\equiv 0\pmod{\pi}$ $\displaystyle
a-a\omega+a\omega+b\omega$ $\displaystyle\equiv 0\pmod{\pi}$ (3)
$\displaystyle a-a\omega$ $\displaystyle\equiv-(a+b)\omega\pmod{\pi}.$

We have the following. Recall that we defined $N\pi=p=a^{2}-ab+b^{2}$, so

$\displaystyle p$ $\displaystyle=a^{2}-ab+b^{2}$
$\displaystyle=(3m-1)^{2}-(3m-1)(3n)+(3n)^{2}$ $\displaystyle\frac{p-1}{3}$
$\displaystyle=3m^{2}-2m-3mn+3n^{2}$ (4) $\displaystyle\equiv-2m+n\pmod{3}.$

Finally, we have the following.

$\displaystyle a+b$ $\displaystyle\equiv 0\pmod{a+b}$ $\displaystyle
a+b\omega$ $\displaystyle\equiv b\omega-b\pmod{a+b}$ (5) $\displaystyle
a+b\omega$ $\displaystyle\equiv-b(1-\omega)\pmod{a+b}.$

Using these results, we can compute the following. We use $\pi=a+b\omega$.
Since $a$ is a rational primary, by Remark 8, we can write

$\displaystyle\chi_{a+b\omega}(1-\omega)$
$\displaystyle=\chi_{a}(b)\chi_{a+b\omega}(1-\omega)$
$\displaystyle=\chi_{a}(b\omega^{3})\chi_{a+b\omega}(1-\omega)$
$\displaystyle=\chi_{a}(\omega^{2})\chi_{a}(b\omega)\chi_{a+b\omega}(1-\omega).$

By (2), we can expand
$\chi_{a}(\omega)^{2}\chi_{a}(a+b\omega)\chi_{a+b\omega}(1-\omega)=\omega^{\frac{2(Na-1)}{3}}\chi_{a+b\omega}(a)\chi_{a+b\omega}(1-\omega)$.

Combining (1) and (3) and then simplifying, we have

$\displaystyle\omega^{2m}\chi_{a+b\omega}(a(1-\omega))$
$\displaystyle=\omega^{2m}\chi_{a+b\omega}(-(a+b)\omega)$
$\displaystyle=\omega^{2m}\chi_{a+b\omega}(-1)\chi_{a+b\omega}(\omega)\chi_{a+b\omega}(a+b)$
$\displaystyle=\omega^{2m}(1)\omega^{\frac{N\pi-1}{3}}\chi_{a+b\omega}(a+b).$

Applying properties of the cubic character, (4), and (5), we have

$\omega^{2m-2m+n}\chi_{a+b}(a+b\omega)=\omega^{n}\chi_{a+b}(-b(1-\omega))=\omega^{n}\chi_{a+b}(1-\omega).$

It is not difficult to verify the following by evaluating each part
individually and simplifying.

(6) $\frac{2(N(a+b)-1)}{3}\equiv 2(m+n)\pmod{3}.$

Leading toward the final result and recalling that $(1-\omega)^{2}=-3\omega$,
we have

$\displaystyle\chi_{a+b\omega}(1-\omega)=\omega^{n}\chi_{a+b}(1-\omega)=\omega^{n}\chi_{a+b}(1-\omega)^{4}$
$\displaystyle=\omega^{n}(\chi_{a+b}(1-\omega)^{2})^{2}$
$\displaystyle=\omega^{n}\chi_{a+b}(-3\omega)^{2}$
$\displaystyle=\omega^{n}(1)^{2}(1)^{2}\chi_{a+b}(\omega)^{2}$
$\displaystyle=\omega^{n}(\omega^{\frac{N(a+b)-1)}{3}})^{2}.$

By (6), this is equivalent to writing
$\omega^{n}\omega^{2(m+n)}=\omega^{2m+3n}=\omega^{2m}\omega^{3n}=\omega^{2m}$,
which is what we wanted to prove. ∎

In section 3.6, we will also introduce a special supplement of cubic
reciprocity, namely the cubic character of 2 modulo $\pi$. We now proceed to
prove cubic reciprocity.

###  3.5. Proof of Cubic Reciprocity

Before continuing with the proof, we need to make some preliminary statements
regarding $D/\pi D$. We let $\pi\in D$ be a prime such that $N\pi=p\equiv
1\pmod{3}$. We showed earlier that $D/\pi D$ is a finite field with
characteristic $p$, so naturally, it contains the field
$\mathbb{Z}/p\mathbb{Z}$ as well. Both fields have $p$ elements. Therefore, it
is useful to define an isomorphism between $D/\pi D$ and
$\mathbb{Z}/p\mathbb{Z}$. Namely, we have a bijection, where we map residue
classes from $\mathbb{Z}/p\mathbb{Z}$ to their complex counterparts in $D/\pi
D$. In this way, we are mapping the coset of some residue class in
$\mathbb{Z}/p\mathbb{Z}$ to some other coset in $D/\pi D$. This isomorphism
allows us to extend the cubic character $\chi_{\pi}$ to not only $D/\pi D$,
but $\mathbb{Z}/p\mathbb{Z}$ as well. This means that the properties of the
cubic character in $D\pi D$ are also valid in $\mathbb{Z}/p\mathbb{Z}$,
allowing us to consider cubic Gauss sums $g_{a}(\chi_{\pi})$ as well as cubic
Jacobi sums $J(\chi_{\pi},\chi_{\pi})$ on $\mathbb{Z}/p\mathbb{Z}$. This
realization is ultimately what allows us to prove cubic reciprocity, and
explains why we investigated $D/\pi D$ so thoroughly.

Moving on, we need to prove some important properties about the Jacobi sum
that relate directly to cubic reciprocity. We first have the following.

######  Lemma 3.18.

$1^{k}+2^{k}+3^{k}+\cdots+(p-1)^{k}=\sum_{l=1}^{p-1}l^{k}\equiv\left\\{\begin{array}[]{ll}0&\text{if
$p-1\not\equiv 0\pmod{k}$}\\\ -1&\text{if $p-1\equiv
0\pmod{k}$}.\end{array}\right.$

###### Proof.

The proof can follow by considering $[g]$ to be a primitive root of
$(\mathbb{Z}/p\mathbb{Z})^{*}$, and noticing that it is identical to
considering the complete set of representatives
$\\{[0],[1],[2],\ldots,[p-1]\\}$, we can evaluate the sum over the entire
finite field and evaluate each congruence depending on whether $p-1|k$ or
$p-1\nmid k$. ∎

######  Proposition 3.19.

Let $\pi$ be primary. Then

$J(\chi_{\pi},\chi_{\pi})=\pi.$

###### Proof.

Let there exist some other primary number $\pi^{\prime}$ such that
$J(\chi_{\pi},\chi_{\pi})=\pi^{\prime}$. By the definition of $\pi$ we must
have $N\pi=\pi\overline{\pi}=p=\pi^{\prime}\overline{\pi}^{\prime}$, since the
norm of every primary element $\pi$ is $p$. Therefore either
$\pi|\pi^{\prime}$ or $\pi|\overline{\pi}^{\prime}$. However, since all prime
elements are primary, each prime is coprime to every other prime, implying
that we must have either $\pi=\pi^{\prime}$ or $\pi=\overline{\pi}^{\prime}$.
We need to show that the second equation is not possible in order to show that
$\pi$ is unique and that there is only one such primary element.

We begin by writing out the cubic Jacobi sum. Then for some $x$ that runs over
$\mathbb{Z}/p\mathbb{Z}$, we have

$J(\chi_{\pi},\chi_{\pi})=\sum_{x}\chi_{\pi}(x)\chi_{\pi}(1-x).$

By Proposition 3.10, we can rewrite each character so that

$\sum_{x}\chi_{\pi}(x)\chi_{\pi}(1-x)\equiv\sum_{x}x^{\frac{p-1}{3}}(1-x)^{\frac{p-1}{3}}\pmod{\pi}.$

Notice that the degree of this polynomial is $\deg(f(x))=2((p-1)/3)<p-1$.
Obviously, this means that $p-1\nmid\deg(f(x))$. Analogously, by Lemma 3.18,
and since $x$ runs over all elements of $\mathbb{Z}/p\mathbb{Z}$, we can
assert that

$\sum_{x}x^{\frac{p-1}{3}}(1-x)^{\frac{p-1}{3}}\equiv 0\pmod{p}.$

Equivalence modulo $p$ can be extended to equivalence modulo $\pi$. Therefore
we can make the assertion that

$\sum_{x}\chi_{\pi}(x)\chi_{\pi}(1-x)\equiv\sum_{x}x^{\frac{p-1}{3}}(1-x)^{\frac{p-1}{3}}\equiv
J(\chi_{\pi},\chi_{\pi})=\pi^{\prime}\equiv 0\pmod{\pi}.$

By definition of congruence, $\pi|\pi^{\prime}$, which is impossible unless
$\pi=\pi^{\prime}$ since $\pi$ and $\pi^{\prime}$ are both primary. ∎

A simple corollary follows by substituting this result into Corollary 2.13 as
follows.

######  Corollary 3.20.

$g(\chi_{\pi})^{3}=p\pi.$

###### Proof.

By Corollary 2.13, we know that $g(\chi)^{3}=pJ(\chi,\chi)$. Take the
character to be the cubic character. Then by Proposition 3.19, we have
$g(\chi_{\pi})^{3}=p\pi$. ∎

We need a final fact about the Jacobi sum. The significance of the following
result is that it allows us to make the assertion that the Jacobi sum
$J(\chi,\chi)$ is a primary prime in $D$. In fact, since $J(\chi,\chi)$ is
indeed primary, we have $J(\chi,\chi)\overline{J(\chi,\chi)}=p$, i.e.
$J(\chi,\chi)$ has norm $p$. First, we must utilize a fact about algebraic
integers. Let $\Omega$ denote the set of algebraic integers.

######  Lemma 3.21.

Let $\omega_{1},\omega_{2}\in\Omega$ and $p\in\mathbb{Z}$ be prime. Then

$(\omega_{1}+\omega_{2})^{p}\equiv\omega_{1}^{p}+\omega_{2}^{p}\pmod{p}.$

###### Proof.

The proof follows by expanding the LHS with the Binomial Theorem and noticing
that all intermediate terms reduce to $0$ modulo $p$. ∎

In the following, assume that $\pi\in D$ is primary.

######  Proposition 3.22.

If $J(\chi,\chi)=a+b\omega$ where $\omega\in\mathbb{Z}[\omega]$, then
$a\equiv-1\pmod{3}$ and $b\equiv 0\pmod{3}$.

###### Proof.

It is well known that the algebraic integers form a ring. Working with
congruences in $\Omega$ and using Lemma 3.21 as well as the fact that
$\chi(t)$ is a cubic character, we have

$g(\chi)^{3}=\bigg{(}\sum_{t\in\mathbb{F}_{p}}\chi(t)\zeta_{3}^{t}\bigg{)}^{3}\equiv\sum_{t\in\mathbb{F}_{p}}\chi(t)^{3}\zeta_{3}^{3t}\pmod{3}.$

Clearly $\chi(0)=0$, and since $\chi(t)$ is a cubic character, $\chi(t)^{3}=1$
for nonzero $t$. Therefore

$\sum_{t\in\mathbb{F}_{p}}\chi(t)^{3}\zeta_{3}^{3t}=\sum_{t\neq
0}\zeta_{3}^{3t}=\sum_{t\neq 0}e^{2\pi it}=-1.$

By Corollary 3.20, we have

$g(\chi)^{3}=pJ(\chi,\chi)\equiv a+b\omega\equiv-1\pmod{3}.$

Alternatively, consider the conjugate $\overline{\chi}$. By Proposition 3.10,
we have that $\overline{g(\chi)}=g(\overline{\chi})$. Again applying Corollary
3.20, we have

$g(\overline{\chi})^{3}=pJ(\overline{\chi},\overline{\chi})\equiv
a+b\overline{\omega}\equiv-1\pmod{3}.$

Equating these equations modulo $3$, we have

$\displaystyle(a+b\omega)-(a+b\overline{\omega})$
$\displaystyle\equiv-1-(-1)\pmod{3}$ $\displaystyle
b(\omega-\overline{\omega})$ $\displaystyle\equiv 0\pmod{3}$ $\displaystyle
b\bigg{(}\frac{-1+i\sqrt{3}}{2}-\frac{-1-i\sqrt{3}}{2}\bigg{)}$
$\displaystyle\equiv 0\pmod{3}$ $\displaystyle
b\bigg{(}\frac{2i\sqrt{3}}{2}\bigg{)}=bi\sqrt{3}$ $\displaystyle\equiv
0\pmod{3}$ $\displaystyle-3b^{2}$ $\displaystyle\equiv 0\pmod{9}.$

This implies that $3|b$, i.e. $b\equiv 0\pmod{3}$. Therefore $a+b\omega\equiv
a\equiv-1\equiv 2\pmod{3}$. ∎

With all of these results, we are now sufficiently equipped to prove cubic
reciprocity.

###### Proof of The Law of Cubic Reciprocity (Theorem 3.13).

As indicated before, this is a proof by cases. In order to prove the full law,
we need to consider when one of $\pi_{1}$ and $\pi_{2}$ is complex and the
other is rational, and when both are complex.

We first prove the case when one is complex and the other is rational. In
other words, we need to show that if $\pi_{1}=q\equiv 2$ and $\pi_{2}=\pi$ a
primary with $N\pi=p$, then $\chi_{\pi}(q)=\chi_{q}(\pi)$. Begin by taking the
expression in Corollary 3.20 and raising the LHS and RHS to a power of
$(q^{2}-1)/3$. This results in the equivalence
$g(\chi_{\pi})^{((q^{2}-1)/3)\cdot
3}=g(\chi_{\pi})^{q^{2}-1}=(p\pi)^{(q^{2}-1)/3}$. Take this equality modulo
$q$, so by properties of the cubic character, we have
$g(\chi_{\pi})^{q^{2}-1}\equiv\chi_{q}(p\pi)\pmod{q}$. Notice that the RHS of
the congruence can be expressed as $\chi_{q}(p\pi)=\chi_{q}(p)\chi_{q}(\pi)$.
Clearly, since $p$ and $q$ are coprime, by Lemma 3.11, $\chi_{q}(p)=1$.
Therefore we can rewrite the congruence as

$\displaystyle g(\chi_{\pi})^{q^{2}-1}$
$\displaystyle\equiv\chi_{q}(p)\chi_{q}(\pi)\pmod{q}$ $\displaystyle
g(\chi_{\pi})^{q^{2}}$
$\displaystyle\equiv\chi_{q}(\pi)g(\chi_{\pi})\pmod{q}.$

Now we can examine the LHS. If we expand the Gauss sum of the LHS, we have

$g(\chi_{\pi})^{q^{2}}=\bigg{(}\sum_{t\in\mathbb{F}_{p}}\chi_{\pi}(t)\zeta^{t}\bigg{)}^{q^{2}}.$

The intermediate terms of this sum will have some form of $q$ as a factor, so
if we take this modulo $q$, we are left with

$g(\chi_{\pi})^{q^{2}}\equiv\sum_{t\in\mathbb{F}_{p}}\chi_{\pi}(t)^{q^{2}}\zeta^{q^{2}t}\pmod{q}.$

Notice that $q\equiv 2\pmod{3}\implies q^{2}\equiv 1\pmod{3}$. Also, since
$\chi_{\pi}(t)$ is a cube root of 1, meaning that is is a cube root of unity,
we can express the RHS as a Gauss sum so that we can simplify to
$g(\chi_{\pi})^{q^{2}}\equiv g_{q^{2}}(\chi_{\pi})\pmod{q}$. By Lemma 2.6 and
since $\overline{q^{2}}=q^{-2}$ and $\chi_{\pi}(q^{3})=1$, the RHS in fact
becomes
$g_{q^{2}}(\chi_{\pi})=\chi_{\pi}(q^{-2})g(\chi_{\pi})=\chi_{\pi}(q^{-2})\chi_{\pi}(q^{3})g(\chi_{\pi})=\chi_{\pi}(q)g(\chi_{\pi})$.
Thus, setting the two equations equal to each other, we have the expression

$\chi_{\pi}(q)g(\chi_{\pi})\equiv\chi_{q}(\pi)g(\chi_{\pi})\pmod{\pi}.$

Notice that $g(\chi_{\pi})g(\overline{\chi_{\pi}})=\chi_{\pi}(-1)p=p$ by
Corollary 2.11, so multiplying both sides by $g(\overline{\chi_{\pi}})$ we
have

$\displaystyle\chi_{\pi}(q)g(\chi_{\pi})$
$\displaystyle\equiv\chi_{q}(\pi)g(\chi_{\pi})\pmod{q}$
$\displaystyle\chi_{\pi}(q)p$ $\displaystyle\equiv\chi_{q}(\pi)p\pmod{q}$
$\displaystyle\chi_{\pi}(q)$ $\displaystyle\equiv\chi_{q}(\pi)\pmod{q}.$

By Lemma 3.8, this means that $\chi_{\pi}(q)=\chi_{q}(\pi)$.

Now we need to show that this is true when both $\pi_{1}$ and $\pi_{2}$ are
non-rational. In this case, we have that the norm of both must be congruent to
1 modulo 3, i.e. $N\pi_{1}=p_{1}\equiv 1\pmod{3}$ and $N\pi_{2}=p_{2}\equiv
1\pmod{3}$. (Note in some way that this is a result of Lemma 3.4) Let some
$\gamma_{1}=\overline{\pi_{1}}$ and some $\gamma_{2}=\overline{\pi_{2}}$. By
Lemma 3.12, we know that exactly one associate of each $p_{1}$ or $p_{2}$ is
primary, so we call them $\gamma_{1}$ and $\gamma_{2}$. Then
$p_{1}=\pi_{1}\gamma_{1}$ and $p_{2}=\pi_{2}\gamma_{2}$. We approach this
problem in a similar way to when one prime is rational and the other is
complex. Take the expression in Corollary 3.20, and write it as
$g(\chi_{\gamma_{1}})^{3}=p_{1}\gamma_{1}$. Raising the LHS and RHS to a power
of $(N\pi_{2}-1)/3$ or $(p_{2}-1)/3$ and taking the expression modulo
$\pi_{2}$, we obtain

$\displaystyle(g(\chi_{\gamma_{1}})^{3})^{\frac{p_{2}-1}{3}}=g(\chi_{\gamma_{1}})^{\pi_{2}-1}$
$\displaystyle=(p_{1}\gamma_{1})^{\frac{p_{2}-1}{3}}\equiv\chi_{\pi_{2}}(p_{1}\gamma_{1})\pmod{\pi_{2}}$
$\displaystyle g(\chi_{\gamma_{1}})^{p_{2}}g(\chi_{\gamma_{1}})^{-1}$
$\displaystyle\equiv\chi_{\pi_{2}}(p_{1}\gamma_{1})\pmod{\pi_{2}}$
$\displaystyle g(\chi_{\gamma_{1}})^{p_{2}}$ $\displaystyle\equiv
g(\chi_{\gamma_{1}})\chi_{\pi_{2}}(p_{1}\gamma_{1})\pmod{\pi_{2}}.$

We can simplify the LHS. Notice that

$g(\chi_{\gamma_{1}})^{p_{2}}=\bigg{(}\sum_{t\in\mathbb{F}_{p}}\chi_{\gamma_{1}}(t)\zeta^{t}\bigg{)}^{p_{2}}\equiv\sum_{t\in\mathbb{F}_{p}}\chi_{\gamma_{1}}(t)^{p_{2}}\zeta^{p_{2}t}\pmod{\pi_{2}}$

since the intermediate terms are all congruent to 0 modulo $\pi_{2}$, and
disappear after reduction. Notice that this is also a Gauss sum, namely
$g_{p_{2}}(\chi_{\gamma_{1}})$, so $g(\chi_{\gamma_{1}})^{p_{2}}\equiv
g_{p_{2}}(\chi_{\gamma_{1}})\pmod{\pi_{2}}$. Notice that by Lemma 2.6 and
Corollary 3.11, and since $\overline{p_{2}}=p_{2}$, the RHS of this congruence
can be equivalently written as
$g_{p_{2}}(\chi_{\gamma_{1}})=\chi_{\gamma_{1}}(p_{2}^{2})g(\chi_{\gamma_{1}})$.
If we equate this to the expression derived above, then we have

$\chi_{\gamma_{1}}(p_{2}^{2})g(\chi_{\gamma_{1}})\equiv
g(\chi_{\gamma_{1}})\chi_{\pi_{2}}(p_{1}\gamma_{1})\pmod{\pi_{2}}$

Recall again that $g(\chi_{1})\overline{g(\chi_{\gamma_{1}})}=p$, so
multiplying both sides by $\overline{g(\chi_{\gamma_{1}})}$, and by Lemma 3.8
, we have

$\displaystyle\chi_{\gamma_{1}}(p_{2}^{2})p$ $\displaystyle\equiv
p\chi_{\pi_{2}}(p_{1}\gamma_{1})\pmod{\pi_{2}}$
$\displaystyle\chi_{\gamma_{1}}(p_{2}^{2})$
$\displaystyle\equiv\chi_{\pi_{2}}(p_{1}\gamma_{1})\pmod{\pi_{2}}$ (1)
$\displaystyle\chi_{\gamma_{1}}(p_{2}^{2})$
$\displaystyle=\chi_{\pi_{2}}(p_{1}\gamma_{1}).$

We now seek to evaluate the the same thing, but instead using the relation
$g(\chi_{\pi_{2}})^{3}=p_{2}\gamma_{2}$. Raising the LHS and RHS to a power of
$(N\pi_{1}-1)/3$ or $(p_{1}-1)/3$ and taking the expression modulo $\pi_{1}$,
we obtain

$\displaystyle(g(\chi_{\pi_{2}})^{3})^{\frac{p_{1}-1}{3}}$
$\displaystyle=(p_{2}\pi_{2})^{\frac{p_{1}-1}{3}}$ $\displaystyle
g(\chi_{\pi_{2}})^{p_{1}-1}$
$\displaystyle\equiv\chi_{\pi_{1}}(p_{2}\pi_{2})\pmod{\pi_{1}}$ $\displaystyle
g(\chi_{\pi_{2}})^{p_{1}}$
$\displaystyle\equiv\chi_{\pi_{1}}(p_{2}\pi_{2})g(\chi_{\pi_{2}})\pmod{\pi_{1}}.$

Evaluating the LHS of this congruence, notice that by using the same facts
about Gauss sums as before we have

$g(\chi_{\pi_{2}})^{p_{1}}=\bigg{(}\sum_{t\in\mathbb{F}_{p}}\chi_{\pi_{2}}(t)\zeta^{t}\bigg{)}^{p_{1}}\equiv\sum_{t\in\mathbb{F}_{p}}\chi_{\pi_{2}}(t)^{p_{1}}\zeta^{p_{1}t}\pmod{\pi_{1}}.$

This is also a Gauss sum, so we can write
$g(\chi_{\pi_{2}})^{p_{1}}=g_{p_{1}}(\chi_{\pi_{2}})=\chi_{\pi_{2}}(p_{1}^{2})g(\chi_{\pi_{2}})$
since $\overline{p_{1}}=p_{1}$. Equating this to the equation derived above,
we now have

$\displaystyle\chi_{\pi_{2}}(p_{1}^{2})g(\chi_{\pi_{2}})$
$\displaystyle\equiv\chi_{\pi_{1}}(p_{2}\pi_{2})g(\chi_{\pi_{2}})\pmod{\pi_{1}}$
$\displaystyle\chi_{\pi_{2}}(p_{1}^{2})p$
$\displaystyle\equiv\chi_{\pi_{1}}(p_{2}\pi_{2})p\pmod{\pi_{1}}$ (2)
$\displaystyle\chi_{\pi_{2}}(p_{1}^{2})$
$\displaystyle\equiv\chi_{\pi_{1}}(p_{2}\pi_{2})\pmod{\pi_{1}}.$

We have evaluated the cases for both $\pi_{1}$ and $\pi_{2}$, but now we are
interested in relating them. We want to evaluate
$\chi_{\gamma_{1}}(p_{2}^{2})$. Notice that by (1) of Corollary 3.11 again, we
can rewrite this as
$\chi_{\gamma_{1}}(p_{2}^{2})=(\chi_{\gamma_{1}}(p_{2}))^{2}=\overline{\chi_{\gamma_{1}}(p_{2})}$.
Since $\overline{\gamma_{1}}=\overline{\overline{\pi_{1}}}=\pi_{1}$, and since
$\overline{p_{2}}=p_{2}$, we have
$\overline{\chi_{\gamma_{1}}(p_{2})}=\chi_{\pi_{1}}(p_{2})$, so

(3) $\chi_{\gamma_{1}}(p_{2}^{2})=\chi_{\pi_{1}}(p_{2}).$

Now we are able to finish the proof.

We compute the following. We have

(4)
$\chi_{\pi_{1}}(\pi_{2})\chi_{\pi_{2}}(p_{1}\gamma_{1})=\chi_{\pi_{1}}(\pi_{2})\chi_{\gamma_{1}}(p_{2}^{2}),$

which follows by substituting (1). Using (3), we have that

$\chi_{\pi_{1}}(\pi_{2})\chi_{\gamma_{1}}(p_{2}^{2})=\chi_{\pi_{1}}(\pi_{2}\chi_{\pi_{1}}(p_{2})=\chi_{\pi_{1}}(p_{2}\pi_{2}).$

By (2), we can write this as

$\displaystyle\chi_{\pi_{2}}(p_{1}^{2})$
$\displaystyle=\chi_{\pi_{2}}(p_{1}\pi_{1}\gamma_{1})$ (5)
$\displaystyle=\chi_{\pi_{2}}(\pi_{1})\chi_{\pi_{2}}(p_{1}\gamma_{1}).$

Equating the LHS of (4) to (5) and dividing both sides by
$\chi_{\pi_{2}}(p_{1}\gamma_{1})$, we have

$\displaystyle\chi_{\pi_{1}}(\pi_{2})\chi_{\pi_{2}}(p_{1}\gamma_{1})$
$\displaystyle=\chi_{\pi_{2}}(\pi_{1})\chi_{\pi_{2}}(p_{1}\gamma_{1})$
$\displaystyle\chi_{\pi_{1}}(\pi_{2})$
$\displaystyle=\chi_{\pi_{2}}(\pi_{1}),$

which is the statement of cubic reciprocity. ∎

###  3.6. Cubic Character of $2$

Now that we have proven cubic reciprocity, we might be interested in
investigating what values might be cubic residues. The special case that we
will consider in particular is the even prime $2$. We will not prove the final
result as it uses facts about Jacobi sums that lie beyond the scope of this
paper, but a detailed proof may be found in Chapter 10 of [Rou12]. To begin,
we have the following result about special rational primes.

######  Proposition 3.23.

If $q\equiv 2\pmod{3}$ is a rational prime, then every integer is a cubic
residue modulo $q$.

###### Proof.

We begin by assuming that $q\equiv 2\pmod{3}$ is a rational prime. Since $q$
is rational, we can work in the integers modulo $q$, namely
$\mathbb{Z}/q\mathbb{Z}$. We define a group homomorphism
$\phi:(\mathbb{Z}/q\mathbb{Z})^{*}\rightarrow(\mathbb{Z}/q\mathbb{Z})^{*}$
with the mapping $\phi(k)=k^{3}$ for some $k\in(\mathbb{Z}/q\mathbb{Z})^{*}$.
By Theorem 0.3,

$(\mathbb{Z}/q\mathbb{Z})^{*}/\text{Ker}(\phi)\approx\text{Im}(\phi).$

We determine the kernel of $\phi$. Clearly, it is only possible for some
$k\in\text{Ker}(\phi)$ if $k^{3}=1$, i.e. $\phi$ maps $k$ to the identity of
$(\mathbb{Z}/q\mathbb{Z})^{*}$, which is $1$. However, Theorem 2.3 asserts
that the multiplicative group $(\mathbb{Z}/q\mathbb{Z})^{*}$ is cyclic with
order $q-1$. But, $3\nmid q-1$, so naturally the relation $k^{3}=1$ is
possible if and only if $k=1$, as it maps to the identity. Thus
$\text{Ker}(\phi)$ is trivial, in that it only contains one element, so that

$|\text{Im}(\phi)|=|(\mathbb{Z}/q\mathbb{Z})^{*}/\text{Ker}(\phi)|=|(\mathbb{Z}/q\mathbb{Z})^{*}|/1=|(\mathbb{Z}/q\mathbb{Z})^{*}|.$

This satisfies the condition for $\phi$ to be surjective, so due to the
mapping $\phi$ defined earlier, every element of
$(\mathbb{Z}/q\mathbb{Z})^{*}$ is a perfect cube, i.e. every integer is a
cubic residue modulo $q$. ∎

######  Proposition 3.24.

The cubic congruence $x^{3}\equiv 2\pmod{\pi}$ where $\pi\in D$ is prime is
solvable if and only if $\pi\equiv 1\pmod{2}$, i.e. if and only if $a\equiv
1\pmod{2}$ and $b\equiv 0\pmod{2}$ in $\pi=a+b\omega$.

###### Proof.

If $\pi=q$ is some primary rational prime, then Proposition 3.23 asserts that
every integer is a perfect cube, so every integer is a cubic residue modulo
$q$ a primary rational prime. Therefore we need to prove that this result is
true for a non-rational primary prime $\pi$. Let $\pi=a+b\omega$ be a primary
prime. By Theorem 3.13, we have that $\chi_{\pi}(2)=\chi_{2}(\pi)$. We
evaluate the RHS of this equality. We have

$\pi^{\frac{N(2)-1}{3}}=\pi^{(4-1)/3}=\pi\equiv\chi_{2}(\pi)\pmod{2}.$

By Proposition 3.9, $\chi_{2}(\pi)=1$, i.e. $\pi$ is a cubic residue modulo
$2$, if and only if the congruence $x^{3}\equiv\pi\pmod{2}$ is solvable. This
congruence is solvable, however, if and only if $\pi\equiv 1\pmod{2}$, because
otherwise $x$ would have to be complex and thus the congruence unsolvable.
Therefore $\chi_{2}(\pi)=1$ if and only if $\pi\equiv 1\pmod{2}$. Similarly,
we have that $\chi_{\pi}(2)=1$ if and only if $\pi\equiv 1\pmod{2}$. In either
case, we thus the congruence $x^{3}\equiv 2\pmod{\pi}$ is solvable if and only
if $\pi\equiv 1\pmod{2}$. ∎

With this result, we can prove a condition for the solvability of $x^{3}\equiv
2\pmod{p}$, which allows us to characterize the cubic character of $2$. Notice
that Proposition 3.24 asserts that the modulus must be a rational primary
prime such that $a\equiv 1\pmod{2}$ and $b\equiv 0\pmod{2}$. We use this fact
concerning the modulus in the proof of the following due to Gauss.

######  Theorem 3.25.

Let $p\equiv 1\pmod{3}$. Then the congruence $x^{3}\equiv 2\pmod{p}$ is
solvable if and only if there exist $C,D\in\mathbb{Z}$ such that
$p=C^{2}+27D^{2}$.

###### Proof.

The proof may be found in Chapter 10 of [Rou12]. ∎

##  4\. A Brief Survey of Biquadratic Reciprocity

While Carl Friedrich Gauss had published 8 proofs of quadratic reciprocity by
his death, he stated, without proof, cubic and biquadratic reciprocity. Though
Gauss did not provide proofs, he stated that their proofs likely involved
Gauss sums, a new technique for proving higher reciprocity laws. The proof of
cubic reciprocity given in Section 3 is exactly using cubic Gauss sums. In
this section we aim to introduce the fundamentals of the Gaussian integers
$\mathbb{Z}[i]$ and outline the biquadratic reciprocity law.

###  4.1. The Statement of Biquadratic Reciprocity

As explained in the introduction, there are many connections between cubic and
biquadratic reciprocity in that they both utilize finite fields. Throughout
the rest of this section we let $D=\mathbb{Z}[i]$ denote the Gaussian
integers. If some $\alpha\in D$ then $(\alpha)=\alpha D$ is the principal
ideal generated by $\alpha$. This is useful in defining the residue class ring
later. It is well known that $D$ is a Euclidean domain, i.e. there exists a
Euclidean algorithm over $D$. Therefore, an analogue of Euclid’s lemma applies
such that if $\pi\in D$ is some irreducible element and $\alpha,\beta\in D$,
then $\pi|\alpha\beta$ implies that $\pi|\alpha$ or $\pi|\beta$.

There is also a norm function over $D$ so that
$N\alpha=\alpha\overline{\alpha}$. Some $\alpha\in D$ is a unit if and only if
$N\alpha=1$. Suppose that $\alpha=a+bi$ is a unit. Then $\alpha|1$, so for
some $\beta\in D$ we have $\alpha\beta=1$. Taking norms, we have
$N\alpha\beta=N(1)=1$. Now suppose that $N\alpha=1$. Then by the definition of
the norm $a^{2}+b^{2}=1$. This is possible only if either $a$ or $b$ is $0$,
in which case the units of $D$ are $\pm 1$ and $\pm i$.

Let $\pi\in D$ be an irreducible.

######  Theorem 4.1.

The residue class ring $D/\pi D$ is a finite field with $N\pi$ elements.

###### Proof.

The proof of this fundamental result is very similar to Theorem 3.5. In this
proof we use more facts about irreducibles in $D$. ∎

A natural corollary as an analogue to Fermat’s Little Theorem easily follows.

######  Corollary 4.2.

If $\pi\nmid\alpha$ then $\alpha^{N\pi-1}\equiv 1\pmod{\pi}$.

The following is very similar to Proposition 3.7, and the proof is ultimately
identical with consideration for $\mathbb{Z}[i]$ instead.

######  Proposition 4.3.

If $\pi\nmid\alpha$ and the ideal $(\pi)\neq(1+i)$ (the significance of this
is due to the fact that $1+i$ is irreducible in $D$), then there exists a
unique integer $j=0,1,2,3$ such that

$\alpha^{\frac{N\pi-1}{4}}\equiv i^{j}\pmod{\pi}.$

As such, the biquadratic character is defined as follows.

######  Definition 22.

If $\pi$ is irreducible and $N\pi\neq 2$, for $\pi\nmid\alpha$ the biquadratic
character of $\alpha$ is defined as $\chi_{\pi}(\alpha)=i^{j}$ where the value
of $j$ is determined exactly by Proposition 4.3. If $\pi|\alpha$ then we
define $\chi_{\pi}(\alpha)=0$.

With this definition we can state biquadratic reciprocity.

######  Theorem 4.4 (The Law of Biquadratic Reciprocity).

Let $\pi,\lambda\in D$ be relatively prime primary elements. Then

$\chi_{\pi}(\lambda)\chi_{\lambda}(\pi)=(-1)^{\frac{N\lambda-1}{4}\cdot\frac{N\pi-1}{4}}.$

While the statement of biquadratic reciprocity is more complex than cubic
reciprocity, the mechanisms in its proof are fundamentally identical to that
of cubic reciprocity. However, to full prove this result, it is necessary to
prove consistently more results regarding Jacobi sums than we have in this
paper. Full coverage of biquadratic reciprocity and its details can be found
in chapter 9 of [IRR90].

###  4.2. Higher Reciprocity

As mentioned in the introduction, Eisenstein reciprocity was the first
generalized reciprocity law. However, before obtaining generalized
reciprocity, many attempts were made to generalize reciprocity beyond cubic
and biquadratic reciprocity. Gauss (who was the first to state potential
results for higher reciprocity), Jacobi, and Eisenstein made multiple attempts
to no avail. In 1839, Jacobi stated - without proof - special cases of higher
reciprocity for 5th, 8th, and 12th degrees, but he was unable to consider more
general laws; see [Jac39]. The reason for the difficulty in generalizing
reciprocity laws beyond 3rd and 4th degrees is that most sets of complex
integers adjoined with increasing roots of unity fail to contain an analogue
to the Euclidean Algorithm and do not form a UFD. The cases we have considered
in this paper both rely on the fact that $\mathbb{Z}[\omega]$ and
$\mathbb{Z}[i]$ have the aforementioned properties. The existence of these
properties allows us to consider their residue class rings and proceed from
there.

## Acknowledgements

The author would like to thank Dr. Tamar Avineri at the North Carolina School
of Science and Mathematics for being a wonderfully supportive reviewer for
this paper throughout the revision process; the author would also like to
thank Dr. Tamar Avineri for insightful and useful conversations regarding
mathematics and number theory. The author would further like to thank Dr.
Simon Rubinstein-Salzedo for helpful insights and suggestions on numerous
proofs throughout this paper. Finally, the author would like to thank Dr.
Frank Thorne at the University of South Carolina for agreeing to endorse him
on the arXiv.

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# Counting Tripods on the Torus

Jayadev S. Athreya  ,  David Aulicino  and  Harry Richman  Department of
Mathematics, University of Washington, Padelford Hall, Seattle, WA 98195, USA
<EMAIL_ADDRESS>Department of Mathematics, Brooklyn College, Room 1156,
Ingersoll Hall 2900 Bedford Avenue Brooklyn, NY 11210-2889, USA  Department of
Mathematics, The Graduate Center, CUNY 365 Fifth Avenue New York, NY 10016 USA
<EMAIL_ADDRESS>Department of Mathematics, University of
Washington, Padelford Hall, Seattle, WA 98195, USA<EMAIL_ADDRESS>
###### Abstract.

Motivated by the problem of counting finite BPS webs, we count certain
immersed metric graphs, tripods, on the flat torus. Classical Euclidean
geometry turns this into a lattice point counting problem in $\mathbb{C}^{2}$,
and we give an asymptotic counting result using lattice point counting
techniques.

J.S.A. was partially supported by NSF CAREER grant DMS 1559860 and NSF DMS
2003528

D.A. was partially supported by NSF DMS - 1738381 and several PSC-CUNY grants

##  1\. Introduction

Given two points $z,w$ in the plane $\mathbb{C}$, the set of points $p$ that
minimize the sum of distances $|z-p|+|w-p|$ is exactly the line segment
connecting $z$ and $w$. At any point $p$ on this segment, the line segments
between $z$ and $p$ and $w$ and $p$ have angle $2\pi/2=\pi$. Given _three
points_ $z,w,u$ in $\mathbb{C}$ such that the triangle they form has largest
angle at most $2\pi/3$, the _Fermat point_ $p$ minimizes the sum of the
lengths $|z-p|+|w-p|+|u-p|$. A classical result in Euclidean geometry says
that the angles between the line segments connecting $p$ to $z,w,u$ are
$2\pi/3$. We call the configuration of line segments a _tripod_.

An integral lattice point $m+in\in\mathbb{Z}[i]$ is called _primitive_ if $m$
and $n$ are coprime. A classical result states that as $R$ tends to infinity,
the number of primitive points in the circle of radius $R$ centered at the
origin grows like $\frac{R^{2}}{\zeta(2)}$. These points correspond to
embedded closed geodesics in the flat torus $\mathbb{C}/\mathbb{Z}[i]$.
Inspired by the work of [GMN13], we ask for a similar asymptotic for the
number of _immersed_ graphs in flat tori that are the projections of tripods
whose vertices are at points of a unimodular lattice.

Let $\Lambda$ be a lattice in $\mathbb{C}$. Throughout, we will assume that
$\Lambda$ has unit covolume. A primitive vector in the complex plane descends
to a primitive closed trajectory on $\mathbb{T}=\mathbb{C}/\Lambda$ with the
origin marked. Analogously, a tripod on the torus will be an immersed tripod
from the plane such that the Fermat point descends to a point $p\not=0$ and
all three other endpoints of the line segments descend to $0$. More formally,
$\mathcal{I}$ is an isometrically _immersed_ copy in $\mathbb{T}$ of a metric
graph $\mathcal{G}=\mathcal{G}(\ell_{1},\ell_{2},\ell_{3})\subset\mathbb{C}$
(see Figure 1) given by positive parameters $\ell_{1},\ell_{2},\ell_{3}$,

$\mathcal{G}=\\{t:0\leq t\leq\ell_{1}\\}\bigcup\\{te^{2\pi i/3}:0\leq
t\leq\ell_{2}\\}\bigcup\\{te^{4\pi i/3}:0\leq t\leq\ell_{3}\\}.$

The image of the tripod is in fact an immersed copy of an equiangular
$\Theta$-graph; that is, a graph with two vertices with three edges between
them. The vertices are the point $0$ and the tripod point $p$, and it is not
difficult to check that the line segments must meet with angle $2\pi/3$ at
$0$. We will later discuss (see Section 3) that associated to each tripod is a
(minimal) cover of the torus where the tripod becomes embedded. The degree of
the cover is the number of transverse self-intersections of the original
tripod, and the embedded tripod gives a representation of the covering torus
as an equiangular hexagon with parallel sides identified by translation.

$\ell_{1}$$\ell_{2}$$\ell_{3}$$\mathbb{C}$ Figure 1. The graph
$\mathcal{G}(\ell_{1},\ell_{2},\ell_{3})\subset\mathbb{C}$.

######  Definition 1.1.

A tripod $\downY$ consists of a pair
$(\mathcal{G}(\ell_{1},\ell_{2},\ell_{3}),\mathcal{I})$, where
$\mathcal{I}:\mathcal{G}\rightarrow\mathbb{T}$ is an isometric immersion, with
$\mathcal{I}(0)=p$, and

$\mathcal{I}(\ell_{1})=\mathcal{I}(\ell_{2}e^{2\pi
i/3})=\mathcal{I}(\ell_{3}e^{4\pi i/3})=0.$

The _length_ of the tripod is denoted $\ell(\downY)$, and is given by

$\ell(\downY)=\ell_{1}+\ell_{2}+\ell_{3}.$

We say a tripod $\downY$ is _primitive_ if it is not a scaled copy of another
tripod. A primitive tripod is called _reduced_ if the only lattice points
lying on its legs are at its endpoints and the tripod point is not at a
lattice point.

The distinction between primitive and reduced lattices is important for future
work. In the case of lattice points, the concepts are identical: a line
segment from the origin to a lattice point passes through another lattice
point if and only if it is a scaled copy of the vector from the origin to that
other lattice point. However, this is not the case for tripods. In future
work, we plan to analyze tripods on higher genus translation surfaces, which
admit cone points. In the presence of cone points, it is natural to define a
tripod as having endpoints at the cone points and no cone points on either of
the legs or at the tripod point. This is essential because the angle at the
cone point of a translation surface is only well-defined modulo $2\pi$.

Figure 2. A (primitive) tripod $\downY$ drawn in a fundamental domain for
$\mathbb{T}$.

Given a lattice $\Lambda$, let

$N_{\downY}(R,\Lambda):=\\#\\{\downY:\downY\mbox{ primitive with endpoints in
}\Lambda\mbox{ and }\ell(\downY)\leq R\\}.$

Furthermore, let

$N_{red,\downY}(R,\Lambda):=\\#\\{\downY:\downY\mbox{ reduced with endpoints
in }\Lambda\mbox{ and }\ell(\downY)\leq R\\}.$

What is the asymptotic behavior of $N_{\downY}(R,\Lambda)$? Our main result
is:

######  Theorem 1.1.

For all (unit covolume) lattices $\Lambda$ in $\mathbb{C}$,

$\lim_{R\rightarrow\infty}\frac{N_{\downY}(R,\Lambda)}{R^{4}}=\frac{1}{\zeta(4)}\cdot\frac{\sqrt{3}\pi}{24}=\frac{15\sqrt{3}}{4\pi^{3}}.$

We will prove this claim by turning our problem into a problem of counting
pairs $(z,w)\subset\Lambda=\mathbb{Z}+\mathbb{Z}\tau$ satisfying certain
conditions. The $1/\zeta(4)$ term in our theorem arises from the fact that we
will be counting pairs

$z=a+b\tau,\quad w=c+d\tau,\qquad a,b,c,d\in\mathbb{Z}$

with $\gcd(a,b,c,d)=1.$ The term $\frac{\sqrt{3}\pi}{24}$ represents the
volume of a region in $\mathbb{C}^{2}$ in which we will be counting dilations
of sets of points.

To clarify how the concepts of primitive versus reduced tripods affect their
asymptotics, we prove the following two theorems. Let
$N_{nonred,\downY}(R,\Lambda)$ be the count of the non-reduced tripods up to
length $R$.

######  Theorem 1.2.

For almost every lattice $\Lambda$,

$N_{nonred,\downY}(R,\Lambda)=o(R^{4}).$

On the other hand, we show that Theorem 1.2 does not hold for all lattices.

######  Theorem 1.3.

Let $\zeta=e^{2\pi i/6}$. Then

$N_{nonred,\downY}(R,\mathbb{Z}+\mathbb{Z}\zeta)\geq CR^{4}$

for some positive constant $C$, for sufficiently large $R$.

###  1.1. Lifting

Note that if we lift a (primitive) tripod $\downY$ from $\mathbb{T}$ to
$\mathbb{C}$, we obtain a center point $\tilde{p}$ and segments emanating from
$\tilde{p}$ to points in $\Lambda$. We can always choose our lift so that one
of these points is $0$, and we call the other two $z$, $w$ with, say
$\arg(z)<\arg(w)$. To remove ambiguity, we insist that the point $\tilde{p}$
lies in the sector of $\mathbb{C}$ specified by

$0\leq\arg(\tilde{p})<2\pi/3.$

Only certain pairs $(z,w)$ will yield triangles which have inscribed tripods.
The length of the tripod can be computed explicitly in terms of $z$ and $w$,
thus turning our problem into a lattice point counting problem in
$\mathbb{C}^{2}$.

$w=i$$z=1$ Figure 3. Tripod $\downY$ inscribed in triangle $\Delta(0,1,i)$.
The tripod point is $\left(\frac{1}{3+\sqrt{3}},\frac{1}{3+\sqrt{3}}\right)$.

###  1.2. Differentials, saddle connections, and tripods

Our problem is also inspired by the problem of counting _saddle connections_
for _quadratic differentials_. Given a holomorphic quadratic differential $q$
on a compact Riemann surface $X$, there is a singular flat metric associated
to $q$ with conical singularities of angle $(k+2)\pi$ at zeros of order $k$ of
the differential. A _saddle connection_ is a geodesic trajectory connecting
two singular points with no singularities in its interior. Alternatively, one
can think of it as a regular point $p$ on $(X,q)$ with two geodesic rays
$\gamma_{1}$, $\gamma_{2}$ emanating from $p$ with an angle of $\pi$, each
terminating in a singular point. Note that angles of $\pi$ correspond to
different local choices of a square root of $q$. Of course there are
(infinitely) many choices of $p$ in this setting. The problem of counting
saddle connections is very well studied, see [Esk06], for example. Our problem
can be thought of as a problem naturally adapted to the setting of holomorphic
_cubic_ differentials $c$, where it is natural to consider trajectories
emanating from a point at angles of $2\pi/3$, each corresponding to a
different choice of cube root of the differential $c$. More generally, finite
BPS webs arise in the work of [GMN13] and are associated to lists of
differentials of varying orders. This work is motivated by the general problem
of counting finite BPS webs of bounded length.

######  Remark 1.

We remark that recent work of Koziarz-Nguyen [KN21] shows that the leading
term for the normalized asympotics for counting certain types of
triangulations on surfaces is in $\mathbb{Q}\cdot(\sqrt{3}\pi)^{N}$ for an
appropriate power $N$. Nevertheless, we do not see an obvious relation between
the results. There is no primitivity assumption in the work of [KN21], which
accounts for the $\zeta(4)$ factor in our work. However, if the primitivity is
removed, then the $R^{4}$ growth rate of $N_{\downY}(R,\Lambda)$ behaves as
$\sqrt{3}\pi/24\not\in\mathbb{Q}\cdot(\sqrt{3}\pi)^{4}$. So there does appear
to be a fundamental difference in the objects being counted.

### Acknowledgments

We would like to thank Vincent Delecroix and Andy Neitzke for useful
discussions; morally, they are coauthors of this paper. We thank Marty
Lewinter for showing us the connection between tripods and Fermat points. We
would like to thank the Mathematical Sciences Research Institute (MSRI), where
this project started in Fall 2019 at the semester program on Holomorphic
Differentials in Mathematics and Physics.

### Organization

In §2, we explain the Euclidean geometry which allows us to translate the
problem to a lattice point counting problem. In §3, we summarize some nice
properties of tripods, lengths, intersections, and covers. In §4, we state
precisely the lattice point counting problem in $\mathbb{C}^{2}$ and the
lattice point counting results which we use to solve the problem. In §5, we
compute the volume of a region in $\mathbb{C}^{2}$ which gives us the main
term in the asymptotic formula. In §6, we prove Theorems 1.2 and 1.3
concerning non-reduced tripods.

##  2\. Fermat points and Steiner trees

###  2.1. Inscribed Tripods

We recall some beautiful facts from classical Euclidean geometry which are
crucial for our translation of our counting problem to a lattice point
counting problem. The following is due to Toricelli. The problem was posed to
him by Fermat, and published by Toricelli’s student Viviani. An excellent
history of this problem (and the more general Steiner tree problem) can be
found here [BGTZ14].

$A$$B$$D$$C$$E$$F$$P$ Figure 4. Constructing a tripod (aka Steiner tree)
$\downY$ inscribed in a triangle $\Delta(ABC)$.

######  Theorem 2.1.

A triangle $\Delta(ABC)$ contains an inscribed tripod if and only if the
largest angle is at most $2\pi/3$. In this case, the tripod $\downY(ABC)$ is
constructed by constructing equilateral triangles (which do not intersect the
original triangle) on each side, and drawing lines connecting the opposite
vertex of this equilateral triangle to the opposite vertex of the original
triangle. That is,

  * •

build an equilateral triangle $\Delta(ABD)$ on the side $AB$,

  * •

an equilateral triangle $\Delta(ACE)$ on the side $AC$,

  * •

an equilateral triangle $\Delta(BCF)$ on the side $BC$,

  * •

and build the segments $DC$, $BE$, and $AF$.

The three line segments ($DC,BE,AF)$ intersect at a common point, $P$, and the
line segments $AP$, $BP$, $CP$, form the tripod. See Figure 4. Furthermore,
this point $P$ is the unique point which minimizes the quantity

$|AQ|+|BQ|+|CQ|$

over all points $Q$ in the plane. The tripod is known as the _Steiner tree_
associated to the points $A,B,C$ [HRW92]. Moreover, the length of the tripod
is given by

$\displaystyle\ell(\downY(ABC))$ $\displaystyle=|AP|+|BP|+|CP|$
$\displaystyle=|DC|=|BE|=|AF|.$

Applying the above result to the triangle $\Delta(0,z,w)$ for

$z,w\in\mathbb{C}\mbox{ such that }\arg(z)<\arg(w)\mbox{ and
}\Delta(0,z,w)\mbox{ has all angles at most }2\pi/3,$

we obtain the following result.

######  Lemma 2.2.

Let $\downY(z,w)$ denote the tripod inscribed in the triangle $\Delta(0,z,w)$,
and let $p$ denote the tripod point of $\downY(z,w)$. Then

(2.1) $\displaystyle\ell(\downY(z,w))$
$\displaystyle=|e^{i\pi/3}z+e^{-i\pi/3}w|$

and

(2.2) $\arg(p)=\arg(e^{i\pi/3}z+e^{-i\pi/3}w).$

###### Proof.

We write $\downY$ for $\downY(z,w)$. Let $p$ denote the tripod point. Then

$\ell(\downY)=|p|+|w-p|+|z-p|.$

On the other hand, we assume that $\arg(z)<\arg(w)$ and by the fact that the
line segments $p0$, $pz$, and $pw$ are at angles of $2\pi/3$, we have that the
three complex numbers

$e^{i\pi/3}(z-p),\quad p,\quad e^{-i\pi/3}(w-p)$

are parallel (that is, their ratios are real and positive). Therefore, the
magnitude of their sum is the sum of their magnitudes. Since

$1=e^{i\pi/3}+e^{-i\pi/3},$

it follows that

(2.3) $\displaystyle\ell(\downY)$ $\displaystyle=|p|+|w-p|+|z-p|$
$\displaystyle=|p|+|e^{-i\pi/3}(w-p)|+|e^{i\pi/3}(z-p)|$
$\displaystyle=|p+e^{-i\pi/3}(w-p)+e^{i\pi/3}(z-p)|$
$\displaystyle=|e^{-i\pi/3}w+e^{i\pi/3}z|.$

Our earlier observation that the complex numbers
$p,e^{-i\pi/3}(w-p),e^{i\pi/3}(z-p)$ all have the same argument implies that

$\displaystyle\arg(p)$
$\displaystyle=\arg(p+e^{-i\pi/3}(w-p)+e^{i\pi/3}(z-p))$
$\displaystyle=\arg(e^{-i\pi/3}w+e^{i\pi/3}z)$

as claimed. ∎

######  Remark 2.

There are also other ways of defining lengths of tripods. For example, given a
tripod $\downY$ inscribed in the triangle $\Delta(ABC),$ we could define its
_triangle length_ $\ell_{\Delta}$ to be

$\ell_{\Delta}(\downY)=|AB|+|AC|+|BC|.$

###  2.2. Steiner trees

More generally, it could be interesting to consider projections to the torus
of solutions to the _Euclidean Steiner tree problem_ with integer vertices:
given $N$ points in the plane, find the connected embedded graph with minimal
total length with vertices at these points. For two points this is of course
the straight line, and more generally, it is not hard to see that the
minimizer must be a tree.

##  3\. Tripod Properties

In this section, we consider the number of self-intersections of a tripod on a
torus and the number of subregions that a tripod divides a torus into. We
relate these tripod properties to its lengths, defined in the previous
section.

###  3.1. Lattice index and tripod lengths

Given a tripod $\downY$ in a lattice $\Lambda$, we define the lattice index of
$\downY$ in $\Lambda$ as the index $[\Lambda:\Lambda(\downY)]$, where
$\Lambda(\downY)$ is the minimal lattice in $\mathbb{R}^{2}$ which contains
$\downY$ as a tripod. We define the volume of a tripod as the covolume of the
lattice $\Lambda(\downY)$. Recall that the tripod length is given by
$\ell(\downY)=\ell_{1}+\ell_{2}+\ell_{3}$. We define the $L_{2}$ length of a
tripod by

$L_{2}(\downY)=(\ell_{1}^{2}+\ell_{2}^{2}+\ell_{3}^{2})^{1/2}.$

######  Proposition 3.1.

  1. (1)

The tripod volume $vol(\downY)$ is related to the tripod lengths by

$\operatorname{vol}(\downY)=\frac{\sqrt{3}}{4}\left(\ell(\downY)^{2}-L_{2}(\downY)^{2}\right).$

  2. (2)

The lattice index $n(\downY,\Lambda)$ is related to the tripod lengths
$\ell(\downY)$ and $L_{2}(\downY)$ by

$n(\downY,\Lambda)=\operatorname{covol}(\Lambda(\downY))=\frac{\sqrt{3}}{4}\left(\ell(\downY)^{2}-L_{2}(\downY)^{2}\right).$

###### Proof.

Suppose $\downY$ spans the triangle with vertices $0,z,w$. Then the area of a
fundamental domain of $\Lambda(\downY)$ is twice the area of the triangle
$\Delta(0,z,w)$. The tripod dissects the triangle $\Delta(0,z,w)$ into three
subtriangles, each with internal angle $2\pi/3$. By adding up these areas, we
have

$Area(\Delta(0,z,w))=\frac{\sqrt{3}}{4}\left(\ell_{1}\ell_{2}+\ell_{1}\ell_{3}+\ell_{2}\ell_{3}\right).$

Therefore,

$\displaystyle covol(\Lambda(\downY))$
$\displaystyle=\frac{\sqrt{3}}{2}\left(\ell_{1}\ell_{2}+\ell_{1}\ell_{3}+\ell_{2}\ell_{3}\right)=\frac{\sqrt{3}}{4}\left((\ell_{1}+\ell_{2}+\ell_{3})^{2}-(\ell_{1}^{2}+\ell_{2}^{2}+\ell_{3}^{2})\right).$

This proves the first part of the proposition. The second part follows from
the first part by the relation

$n(\downY,\Lambda)=\operatorname{covol}(\Lambda(\downY)).$

∎

###  3.2. Counting intersections and regions

######  Proposition 3.2.

Suppose $\downY$ is a tripod on the torus $\mathbb{T}=\mathbb{C}/\Lambda$ with
only transverse self-intersections, with index $n$. Then the number of self-
intersections of $\downY$ on $\mathbb{T}$ is $n-1$.

###### Proof.

Suppose the tripod $\downY$ spans the triangle with vertices $0,z,w$. Let
$\Lambda(\downY)$ denote the lattice in $\mathbb{C}$ spanned by $z$ and $w$;
we have $\Lambda(\downY)\subset\Lambda$ by assumption that $\downY$ is a
tripod in $\Lambda$.

Let $n=[\Lambda:\Lambda(\downY)]$. If we lift the tripod $\downY$ from
$\mathbb{C}/\Lambda$ to $\mathbb{C}/\Lambda(\downY)$, then the preimage can be
expressed as a union of $n$ tripods $\downY_{1},\downY_{2},\ldots,\downY_{n}$
which are translates of each other. We claim that for each $i\neq j$, the
intersection $\downY_{i}\cap\downY_{j}$ consists of exactly two points in the
covering torus $\mathbb{C}/\Lambda(\downY)$. To verify this claim, first
observe that $\downY_{i}\cap\downY_{j}=\downY_{i}\cap(\downY_{j}+\epsilon)$ as
$\epsilon$ varies over a small neighborhood $U$ of zero in $\mathbb{R}^{2}$,
with the neighborhood chosen such that the intersection remains transverse. By
moving the copy $\downY_{j}$ toward $\downY_{i}$, along a path that keeps the
intersection transverse, we have
$\downY_{i}\cap\downY_{j}=\downY_{i}\cap(\downY_{i}+\epsilon)$. Finally, we
verify that $\downY_{i}\cap(\downY_{i}+\epsilon)=2$ when intersecting a tripod
with a small translate of itself, as demonstrated in Figure 5.

Figure 5. Intersection of $\downY_{i}$ and $\downY_{i}+\epsilon$.

It follows that $\downY_{i}\cap\downY_{j}=2$ as claimed.

From this claim, it follows that $\downY=\cup_{i}\downY_{i}$ has
$2\binom{n}{2}=n(n-1)$ self-intersections on the torus
$\mathbb{C}/\Lambda(\downY)$. The quotient map
$\mathbb{C}/\Lambda(\downY)\to\mathbb{C}/\Lambda$ to the original torus has
degree $n$, so this implies that the tripod on $\mathbb{C}/\Lambda$ has $n-1$
self-intersections. ∎

######  Proposition 3.3.

Suppose $\downY$ is a tripod on the torus $\mathbb{T}=\mathbb{C}/\Lambda$ with
only transverse self-intersections, with index $n$. Then the complement
$\mathbb{T}\setminus\downY$ consists of $n$ connected regions.

###### Proof.

The tripod $\downY$ induces a cell structure on $\mathbb{T}$ as follows. The
vertices ($0$-cells) are the two tripod points of $\downY$ and all self-
intersection points. The edges ($1$-cells) are the segments of $\downY$ after
subdividing along vertices, and the faces ($2$-cells) are the connected
components of $\mathbb{T}\setminus\downY$. Note that the components of
$\mathbb{T}\setminus\downY$ are simply connected, because their lifts to the
universal cover $\mathbb{C}$ are bounded. By Proposition 3.2, the number of
vertices in this cell decomposition is $2+(n-1)=n+1$. To compute the number of
edges, we observe that two vertices have degree $3$ while the other $n-1$
vertices have degree $4$. Thus

$c_{1}=\frac{1}{2}\sum_{v}\deg(v)=\frac{1}{2}(3\cdot 2+4(n-1))=2n+1.$

Finally, we know that the Euler characteristic of $\mathbb{T}$ is zero, so

$c_{2}=c_{1}-c_{0}=2n+1-(n+1)=n,$

as claimed. ∎

###  3.3. Counting tripods by spanning lattice

Given a tripod $\downY$, recall that $\Lambda(\downY)$ denotes the minimal
lattice which contains the endpoints of $\downY$; we call $\Lambda(\downY)$
the spanning lattice of $\downY$.

The association of $\downY$ to $\Lambda(\downY)$ defines a map

$(\text{tripods in $\Lambda$})\to(\text{sublattices of }\Lambda).$

This map is surjective, but not injective. For a fixed sublattice
$\Lambda_{0}\subset\Lambda$, the set of tripods

$\\{\downY:\Lambda(\downY)=\Lambda_{0}\\}$

is finite, but as $\Lambda_{0}$ varies the size of this preimage is unbounded.
In particular, the size of this preimage grows arbitrarily large
asymptotically in proportion to the ratio

$(\text{length of second-shortest vector in }\Lambda_{0})/(\text{length of
shortest vector in }\Lambda_{0}).$

######  Proposition 3.4.

Given a lattice $\Lambda$ in $\mathbb{C}$, there are finitely many tripods
$\downY$ such that $\Lambda(\downY)=\Lambda$.

###### Proof.

Without loss of generality, assume that $\Lambda$ has unit covolume. Say a
triangle with endpoints in $\Lambda$ is a unit triangle if it has area $1/2$.
By Theorem 2.1, it suffices to show that $\Lambda$ contains finitely many unit
triangles with angles at most $2\pi/3$, up to translation so that one triangle
vertex is at the origin.

Suppose $\Delta(0,z,w)$ is a unit triangle with angles at most $2\pi/3$. If
$\theta$ is the angle of the triangle at $0$, then the triangle area satisfies

$\frac{1}{2}|z||w|\sin\theta=\mathrm{area}(\Delta(0,z,w))=\frac{1}{2}.$

Up to translation, we may assume that the largest angle of $\Delta(0,z,w)$ is
at $0$, so that $\theta\geq\pi/3$, and our intial hypothesis is that
$\theta\leq 2\pi/3$. Therefore $\sin\theta\geq\sqrt{3}/2$, which implies

$|z||w|=\frac{1}{\sin\theta}\leq\frac{2}{\sqrt{3}}.$

Let $L=\min\\{|z|:z\in\Lambda,\,z\neq 0\\}$. The above bound implies that $z$
and $w$ lie in the set $\\{z\in\Lambda:|z|\leq\frac{2}{\sqrt{3}L}\\}$ which is
finite. This verifies that there are finitely many unit triangles in $\Lambda$
up to translation with angles at most $2\pi/3$. ∎

##  4\. Lattice point counting

###  4.1. Lifting

We now describe how to turn our counting problem for tripods on the torus into
a lattice point counting problem in $\mathbb{C}^{2}$. Given a tripod $\downY$
on $\mathbb{T}=\mathbb{C}/\Lambda$, we fix a lift to $\mathbb{C}$ by choosing
the center point $\tilde{p}$ to lie in the sector
$0\leq\arg(\tilde{p})<2\pi/3$. The lifted tripod will have one endpoint at
$0$. Denote the other endpoints by $z,w\in\Lambda$, with $\arg(z)<\arg(w)$.
(By $\arg(z)<\arg(w)$, we mean $\arg(z)<\arg(w)<\pi+\arg(z)$.) See Figure 6
for an example of determining a lift.

$w=i$$z=1$$\tilde{p}$ Figure 6. Tripod $\downY$ lifted to $\mathbb{C}$, with
endpoints $0,z,w$. The lift with $\arg(\tilde{p})\in[0,2\pi/3)$ is labelled.

###  4.2. Angle bound

By Theorem 2.1, a necessary and sufficient condition for $0,z,w$ to be
endpoints of a tripod is that

(4.1) the angles of the triangle $\Delta(0,z,w)$ are at most $2\pi/3$.

###  4.3. Length bound

Finally, if we want $\ell(\downY(z,w))\leq R$, we need, by Lemma 2.2,

(4.2) $|ze^{i\pi/3}+we^{-i\pi/3}|\leq R$

Putting (4.1) and (4.2) together, we obtain:

######  Lemma 4.1.

Suppose $\Lambda=\mathbb{Z}+\mathbb{Z}\tau$ is a lattice in
$\mathbb{C}\cong\mathbb{R}^{2}$ with $\text{Im}(\tau)>0$. The number of
primitive tripods $N_{\downY}(R,\Lambda)$ is given by the number of pairs
$(z,w)\in\Lambda^{2}$ satisfying the following conditions:

(4.3) $\begin{gathered}z=a+b\tau,\quad w=c+d\tau,\quad\gcd(a,b,c,d)=1,\\\
\arg(z)<\arg(w),\\\ \Delta(0,z,w)\text{ has all angles }\leq 2\pi/3,\\\
|ze^{i\pi/3}+we^{-i\pi/3}|\leq R\text{ and
}0\leq\arg(e^{i\pi/3}z+e^{-i\pi/3}w)<2\pi/3.\end{gathered}$

The following corollary follows from standard lattice point counting results
[HW08, §24.10].

######  Corollary 4.2.

$\lim_{R\rightarrow\infty}\frac{N_{\downY}(R,\Lambda)}{R^{4}}=\frac{1}{\zeta(4)}\operatorname{vol}(\Omega^{\downY}),$

where

$\Omega^{\downY}=\left\\{(z,w)\in\mathbb{C}^{2}:\begin{gathered}\arg(z)<\arg(w)<\pi+\arg(z),\\\
\Delta(0,z,w)\text{ has all angles }\leq 2\pi/3,\\\
|e^{i\pi/3}z+e^{-i\pi/3}w|\leq 1\text{ and
}0\leq\arg(e^{i\pi/3}z+e^{-i\pi/3}w)<2\pi/3\end{gathered}\right\\}.$

###### Proof.

$N_{\downY}(R,\Lambda)$ counts primitive points in
$\Lambda^{2}=(\mathbb{Z}+\mathbb{Z}\tau)^{2}$ in the dilated set
$R\Omega^{\downY}$. $\Omega^{\downY}$ is a compact region with smooth boundary
(in $\mathbb{C}^{2}\cong\mathbb{R}^{4}$), and so by [HW08, §24.10],
$N_{\downY}(R,\Lambda)$ is asymptotic to

$\frac{1}{\zeta(4)}\operatorname{vol}(\Omega^{\downY})R^{4},$

where the factor of $1/\zeta(4)$ is the probability that a random integer
vector $(x,y,u,v)\in\mathbb{Z}^{4}$ is primitive, that is, $\gcd(x,y,u,v)=1$.
∎

##  5\. Volumes

To finish the proof of Theorem 1.1, we need to compute the volume of
$\Omega^{\downY}$. This is given by:

######  Lemma 5.1.

(5.1)
$\operatorname{vol}(\Omega^{\downY})=\frac{\sqrt{3}}{24}\pi=\frac{1}{4}\cdot\frac{2\pi}{3}\cdot\frac{\sqrt{3}}{4}$

###### Proof of Theorem 1.1.

Combining Corollary 4.2 and Lemma 5.1, we obtain Theorem 1.1.∎

###  5.1. Proof of Lemma 5.1

To prove Lemma 5.1, we will apply the following volume-preserving change of
coordinates. Let $\phi:\mathbb{C}^{2}\to\mathbb{C}^{2}$ be defined by

$\phi\begin{pmatrix}z\\\ w\end{pmatrix}=\begin{pmatrix}1&0\\\
e^{i\pi/3}&e^{-i\pi/3}\end{pmatrix}\begin{pmatrix}z\\\
w\end{pmatrix}=\begin{pmatrix}z\\\ e^{i\pi/3}z+e^{-i\pi/3}w\end{pmatrix}.$

The inverse map $\phi^{-1}$ with $u=e^{i\pi/3}z+e^{-i\pi/3}w$ is

$\displaystyle\phi^{-1}\begin{pmatrix}z\\\
u\end{pmatrix}=\begin{pmatrix}1&0\\\
e^{-i\pi/3}&e^{i\pi/3}\end{pmatrix}\begin{pmatrix}z\\\
u\end{pmatrix}=\begin{pmatrix}z\\\ e^{-i\pi/3}z+e^{i\pi/3}u\end{pmatrix}.$

If $\downY$ is a tripod with endpoints $0$, $z$, and $w$, and tripod point
$p$, then $u=e^{i\pi/3}z+e^{-i\pi/3}w$ is the third vertex of the equilateral
triangle with vertices at $z$ and $w$. We call $u$ the _Toricelli point_ of
the tripod, (see Section 2). By Lemma 2.2, the point $u$ satisfies
$\arg(u)=\arg(p)$ and $|u|=\ell(\downY)=|p|+|z-p|+|w-p|$.

$0$$z$$w$$u$$p$ Figure 7. Tripod with endpoints $0,z,w$, and Toricelli point
$u$.

Recall that $\Omega^{\downY}$ is defined as

$\Omega^{\downY}=\left\\{\begin{pmatrix}z\\\
w\end{pmatrix}\in\mathbb{C}^{2}:\begin{gathered}\arg(z)<\arg(w)<\pi+\arg(z),\\\
\text{triangle }\Delta(0,z,w)\text{ has all angles }\leq 2\pi/3,\\\
|e^{i\pi/3}z+e^{-i\pi/3}w|\leq 1\text{ and
}0\leq\arg(e^{i\pi/3}z+e^{-i\pi/3}w)<2\pi/3\end{gathered}\right\\}$

Its image under $\phi$ is

(5.2) $\phi(\Omega^{\downY})=\left\\{\begin{pmatrix}z\\\
u\end{pmatrix}\in\mathbb{C}^{2}:\begin{gathered}\arg(z)<\arg(e^{-i\pi/3}z+e^{i\pi/3}u)<\pi+\arg(z)\\\
\text{triangle }\Delta(0,z,e^{-i\pi/3}z+e^{i\pi/3}u)\text{ has all angles
}\leq 2\pi/3,\\\ |u|\leq 1,\quad 0\leq\arg(u)<2\pi/3\end{gathered}\right\\}.$

For $u\in\mathbb{C}$ satisfying $|u|\leq 1$ and $0\leq\arg(u)<2\pi/3$, let

$\displaystyle\phi(\Omega^{\downY})_{u}$
$\displaystyle=\\{z\in\mathbb{C}:(z,u)\in\phi(\Omega^{\downY})\\}$
$\displaystyle=\left\\{z\in\mathbb{C}:\begin{gathered}0,\,z,\,\text{and}\,e^{-i\pi/3}z+e^{i\pi/3}u\text{
are endpoints of a tripod}\\\ \text{with
}\arg(z)<\arg(e^{-i\pi/3}z+e^{i\pi/3}u)<\pi+\arg(z)\end{gathered}\right\\}.$

and as a special case

(5.3)
$\phi(\Omega^{\downY})_{1}=\left\\{z\in\mathbb{C}:\begin{gathered}0,\,z,\,\text{and}\,e^{-i\pi/3}z+e^{i\pi/3}\text{
are endpoints of a tripod}\\\ \text{with
}\arg(z)<\arg(e^{-i\pi/3}z+e^{i\pi/3})\end{gathered}\right\\}.$

######  Lemma 5.2.

The region $\phi(\Omega^{\downY})_{1}$ is equal to the equilateral triangle
$\Delta(0,1,e^{-i\pi/3})\subset\mathbb{C}$ (Fig. 9).

###### Proof.

By definition $\phi(\Omega^{\downY})_{1}$ consists of points $z$ such that the
triangle $\Delta(0,z,e^{-i\pi/3}z+e^{i\pi/3})$ has all angles at most
$2\pi/3$. By Theorem 2.1, such a triangle has an inscribed tripod. Using Lemma
2.2 with $w=e^{-i\pi/3}z+e^{i\pi/3}$ and the computation

$e^{i\pi/3}z+e^{-i\pi/3}w=e^{i\pi/3}z+(e^{-2i\pi/3}z+1)=1,$

such a tripod has length $1$ and has its tripod point on the positive real
axis.

Such a tripod has the following description, illustrated in Figure 8. The
center tripod point $p$ lies on the positive real axis, and let $a$ denote its
distance from the origin. The lower endpoint is $z$, and let $b$ denote its
distance from $p$, while the upper endpoint $w$ is at distance $1-a-b$ from
$p$. Each tripod leg has nonnegative length, so we assume $0\leq a,b\leq 1$
and $0\leq a+b\leq 1$.

$a$$b$$1-a-b$$z$$w$$p$ Figure 8. A tripod in $\mathbb{C}$.

In such a tripod, $z=a+be^{-i\pi/3}.$ The points

$\\{z=a+be^{-i\pi/3}:a\geq 0,\,b\geq 0,\,a+b\leq 1\\}$

are exactly those inside the equilateral triangle with endpoints $0$, $1$, and
$e^{-i\pi/3}$.

Figure 9. The unit equilateral triangle $\Delta(0,1,e^{-i\pi/3})$ in
$\mathbb{C}$.

It follows that $\phi(\Omega^{\downY})_{1}=\Delta(0,1,e^{-i\pi/3})$ as
claimed. ∎

######  Lemma 5.3.

The region $\phi(\Omega^{\downY})_{u}$ has area $\frac{\sqrt{3}}{4}|u|^{2}$.

###### Proof.

When $u=1$, the claim follows from Lemma 5.2. For arbitrary $u$, the
association $u\mapsto\phi(\Omega^{\downY})_{u}$ is equivariant under
multiplication, i.e.

$\phi(\Omega^{\downY})_{u}=\\{uz:z\in\phi(\Omega^{\downY})_{1}\\}.$

The region $\phi(\Omega^{\downY})_{u}$ has real dimension $2$, so it follows
that

$\operatorname{vol}(\phi(\Omega^{\downY})_{u})=|u|^{2}\operatorname{vol}(\phi(\Omega^{\downY})_{1})=\frac{\sqrt{3}}{4}|u|^{2}$

as claimed. ∎

###### Proof of Lemma 5.1.

First, note that
$\operatorname{vol}(\Omega^{\downY})=\operatorname{vol}(\phi(\Omega^{\downY}))$
since $\phi$ has unit-magnitude Jacobian:

$|Jac(\phi)|=\left|\det\begin{pmatrix}1&0\\\
e^{i\pi/3}&e^{-i\pi/3}\end{pmatrix}\right|=1.$

Then to compute $\phi(\Omega^{\downY})$: we slice the region according to the
$u$-coordinate.

$\operatorname{vol}(\phi(\Omega^{\downY}))=\int\int_{(z,u)\in\phi(\Omega^{\downY})}dz\,du=\int_{u}\operatorname{vol}(\phi(\Omega^{\downY})_{u})du.$

From the definition of $\phi(\Omega^{\downY})$, it is clear that the region
$\phi(\Omega^{\downY})_{u}$ is nonempty only if

$u=re^{i\theta}\quad\text{where }0\leq r\leq 1\text{ and }0\leq\theta\leq
2\pi/3.$

Using this substitution $u=re^{i\theta}$ and the identity
$\operatorname{vol}(\phi(\Omega^{\downY})_{u})=\frac{\sqrt{3}}{4}|u|^{2}=\frac{\sqrt{3}}{4}r^{2}$
from Lemma 5.3, we have

$\displaystyle\operatorname{vol}(\phi(\Omega^{\downY}))=\int_{\theta=0}^{2\pi/3}\int_{r=0}^{1}\frac{\sqrt{3}}{4}r^{2}\cdot
r\,dr\,d\theta=\frac{\sqrt{3}}{4}\int_{\theta=0}^{2\pi/3}d\theta\int_{r=0}^{1}r^{3}\,dr=\frac{\sqrt{3}}{4}\cdot\frac{2\pi}{3}\cdot\frac{1}{4}=\frac{\sqrt{3}\pi}{24}$

as claimed. ∎

##  6\. Non-reduced Tripods

###### Proof of Thm. 1.2.

We will consider a lattice with a non-reduced tripod and show that the real
and imaginary parts of the complex number determining the lattice are related
by an equation. From this, we conclude that a lattice admitting a non-reduced
tripod must lie in a countable union of positive codimension subset of the
space of lattices.

Consider a lattice $\Lambda$ in $\mathbb{C}$. Consider a tripod in this
lattice with the property that at least one of its legs has a lattice point on
its interior. We consider the following transformations. Let $\ell_{1}$ be a
leg that has a lattice point in its interior. Without loss of generality,
translate the lattice so that the endpoint of $\ell_{1}$ is the origin. Next
observe that rotating a tripod preserves the tripod property. Therefore, we
rotate the lattice $\Lambda$ about the origin to a new lattice such that
$\ell_{1}$ lies in the positive real axis. Next we scale the entire lattice so
that one of its basis vectors is $1$ and we write
$\Lambda^{\prime}=\mathbb{Z}\oplus\tau\mathbb{Z}$, where $\tau=s+it$. We make
no claims that $\Lambda^{\prime}$ has unit covolume. From now on we work
entirely with respect to this tripod in the lattice $\Lambda^{\prime}$.

Let $z$ and $w$ be the other endpoints of the tripod. Then there exist
integers $a_{z},b_{z},a_{w},b_{w}\in\mathbb{Z}$ such that $z=a_{z}+b_{z}\tau$
and $w=a_{w}+b_{w}\tau$. Furthermore, the tripod point is simply given by a
real number $r_{p}$ and the assumption that $\ell_{1}$ has a lattice point on
it implies $r_{p}\geq 1$. The angle conditions on the tripod points imply that

$\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\left(a_{z}+b_{z}\tau-
r_{p}\right)\in\mathbb{R}$

and

$\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\left(a_{w}+b_{w}\tau-
r_{p}\right)\in\mathbb{R}.$

In particular, the imaginary parts of both quantities are $0$. After
multiplying by $2$, this yields the equations

$b_{z}t+\sqrt{3}\left(a_{z}+b_{z}s-r_{p}\right)=0$
$b_{w}t-\sqrt{3}\left(a_{w}+b_{w}s-r_{p}\right)=0.$

It is convenient to define $t=t^{\prime}\sqrt{3}$ to get

(6.1) $\displaystyle b_{z}t^{\prime}+a_{z}+b_{z}s-r_{p}=0$ (6.2)
$\displaystyle b_{w}t^{\prime}-a_{w}-b_{w}s+r_{p}=0.$

Adding these equations yields

$(b_{z}+b_{w})t^{\prime}+(a_{z}-a_{w})+(b_{z}-b_{w})s=0.$

First we consider the case where $b_{z}+b_{w}\not=0$. In this case,
$t^{\prime}=q_{1}+q_{2}s$ for some $q_{1},q_{2}\in\mathbb{Q}$. However, this
implies that $\tau$, which defines the lattice $\Lambda$ is not free to be any
point in the complex plane because its real and complex parts are related by
an equation. This completes the proof in this case.

Next, we consider the case where $b_{z}+b_{w}=0$. We claim that
$b_{z}-b_{w}\not=0$ because otherwise, we would have $b_{z}=b_{w}=0$ and this
would contradict the assumption that $0$, $z$ and $w$ form a tripod in
$\Lambda$. Therefore, we conclude that $s\in\mathbb{Q}$ in which case we have
again reduced to a measure zero subset of the space of lattices. ∎

###  6.1. Many nonreduced tripods

In this section fix $\Lambda=\mathbb{Z}+\mathbb{Z}\zeta$ as the triangular
lattice, where $\zeta=e^{2\pi i/6}$ denotes a sixth root of unity.

###### Proof of Thm. 1.3.

Consider a tripod $\downY$ in $\Lambda$ with endpoints at $0$, $z=a+b\zeta$,
and $w=c+d\zeta$, where $a,b,c,d$ are integers. Assume that $\arg(z)<\arg(w)$.
Let $u$ denote the other vertex of the equilateral triangle with vertices at
$z$ and $w$. the vertex $u$ satisfies

$\displaystyle u$ $\displaystyle=\zeta
z+\zeta^{-1}w=c\zeta^{-1}+d+a\zeta+b\zeta^{2}=c(1-\zeta)+d+a\zeta+b(\zeta-1)$
$\displaystyle=(-b+c+d)+(a+b-c)\zeta.$

This shows that $u\in\Lambda$. The tripod point $p$ of $\downY$ is a positive
real multiple of $u$.

Conversely, if $z,w,u$ are related as above and $z,u\in\Lambda$, then it is
straightforward to check that also $w\in\Lambda$.

Recall that a tripod $\downY$ is nonreduced if its interior contains a lattice
point. If the leg $0p$ contains a lattice point in its interior, then
$0u\supset 0p$ also contains a lattice point so $u$ must be a nonprimitive
lattice point of $\Lambda$. Conversely if $u$ is a nonprimitive lattice point
so that $0u$ contains lattice points in its interior, a sufficient condition
for $0p$ to contain a lattice point in its interior is that

$\ell(0p)>\frac{1}{2}\ell(0u).$

Therefore as a lower bound for the number of nonreduced lattices, we have

$\\#\\{\text{nonreduced }\downY\text{ in
}\Lambda\\}\geq\sum_{\begin{subarray}{c}\text{nonprimitive}\\\
u\in\Lambda\end{subarray}}\\#\left\\{\text{tripods
}\downY:\begin{gathered}\text{tripod point }p\in
0u,\,\ell(0p)>\frac{1}{2}\ell(0u),\\\ 0p\text{ is longest leg of
}\downY\end{gathered}\right\\}.$

The condition that $0p$ is the longest tripod leg is needed to avoid
overcounting on the torus $\mathbb{R}^{2}/\Lambda$. Note that the length of
the tripod $\ell(\downY)$ is equal to the distance $|u|=\ell(0u)$.

Suppose we fix some $u$ with $|u|=R$. Then the set

$\\{\text{tripods }\downY:\begin{gathered}\text{tripod point }p\in
0u\end{gathered}\\}$

is parametrized by choosing $w$ inside the equilateral triangle
$\Delta(0,u,\zeta u)$ with side length $R$, while the set

$\left\\{\text{tripods }\downY:\begin{gathered}\text{tripod point }p\in
0u,\,0p\text{ is longest leg of }\downY\end{gathered}\right\\}$

is parametrized by choosing $w$ inside a subregion of the previous triangle of
one-third size, see Figure 10 (middle). Finally the set

$\left\\{\text{tripods }\downY:\begin{gathered}\text{tripod point }p\in
0u,\,0p\text{ is longest leg of }\downY,\\\
\ell(0p)>\frac{1}{2}\ell(0u)\end{gathered}\right\\}$

is parametrized by choosing $w$ inside the subregion shown in Figure 10
(right), which has one-fourth the size of the original triangle.

Therefore for fixed lattice point $u$, the number of tripods

$\displaystyle\\#\left\\{\text{tripods }\downY:\begin{gathered}\text{tripod
point }p\in 0u,\,\ell(0p)>\frac{1}{2}\ell(0u),\\\ 0p\text{ is longest leg of
}\downY\end{gathered}\right\\}$ $\displaystyle=\\#\\{\text{lattice points in
subregion of }\Delta(0,u,\zeta u)\\}$
$\displaystyle=\frac{\textrm{vol}(\text{triangle
subregion})}{\textrm{covol}(\Lambda)}+O(R)$
$\displaystyle=\frac{\sqrt{3}R^{2}/16}{\sqrt{3}/2}+O(R)$
$\displaystyle=\frac{1}{8}R^{2}+O(R).$

Figure 10. An equilateral triangle and relevant subregions.

Now it remains to sum over the possible choices of $u$. Summing over all
lattice points $u$, we would have

$\displaystyle\sum_{\begin{subarray}{c}u\in\Lambda\\\ |u|\leq
R\end{subarray}}\\#\left\\{\text{tripods }\downY:\begin{gathered}\text{tripod
point }p\in 0u,\,\ell(0p)>\frac{1}{2}\ell(0u),\\\ 0p\text{ is longest leg of
}\downY\end{gathered}\right\\}$
$\displaystyle=\sum_{\begin{subarray}{c}u\in\Lambda\\\ |u|=r\leq
R\end{subarray}}\frac{1}{8}r^{2}+O(r)$
$\displaystyle=\int_{r=0}^{R}\left(\frac{\pi}{4}r^{3}+O(r^{2})\right)dr$
$\displaystyle=\frac{\pi}{16}R^{4}+O(R^{3}).$

If we instead sum over only those $u$ which are nonprimitive lattice points,
our answer changes asymptotically by a constant factor

$1-\zeta(2)^{-1}=\lim_{R\to\infty}\frac{\\#\\{u\in\Lambda:\,|u|\leq
R,\,u\text{ nonprimitive}\\}}{\\#\\{u\in\Lambda,\,|u|\leq
R\\}}=1-\frac{6}{\pi^{2}}\approx 0.392.$

Therefore,

$\displaystyle\\#\\{\text{nonreduced }\downY\text{ in
}\Lambda:\ell(\downY)\leq
R\\}=\left(1-\frac{6}{\pi^{2}}\right)\frac{\pi}{16}R^{4}+O(R^{3}),$

so we may take any postive constant
$C<(1-\frac{6}{\pi^{2}})\frac{\pi}{16}\approx 0.0770$. ∎

Note that the total number of tripods in $\Lambda$ satisfies

$\\#\\{\downY\text{ in }\Lambda:\,\ell(\downY)\leq
R\\}\sim\frac{\pi}{12}R^{4}.$

######  Corollary 6.1.

The number of nonreduced, primitive tripods in $\Lambda$ satisfies

$\\#\\{\downY\text{ nonreduced and primitive}:\ell(\downY)\leq R\\}\geq
CR^{4}$

for some positive constant $C$, for sufficiently large $R$.

###### Proof.

The previous theorem showed that

$\displaystyle\\#\\{\text{nonreduced }\downY\text{ in
}\Lambda:\,\ell(\downY)\leq R\\}$
$\displaystyle\gtrsim\left(1-\frac{6}{\pi^{2}}\right)\frac{\pi}{16}R^{4}$
$\displaystyle\sim\left(1-\frac{6}{\pi^{2}}\right)\frac{3}{4}\\#\\{\text{total
}\downY:\,\ell(\downY)\leq R\\},$

where the constant

$C_{1}=\left(1-\frac{6}{\pi^{2}}\right)\frac{3}{4}\approx 0.294.$

We also know that the number of primitive tripods satisfies

$\displaystyle\\#\\{\text{primitive }\downY:\,\ell(\downY)\leq R\\}$
$\displaystyle\sim\zeta(4)^{-1}\\#\\{\text{total }\downY:\,\ell(\downY)\leq
R\\},$

where

$C_{2}=\zeta(4)^{-1}\approx 0.924.$

Together these bounds imply that

$\displaystyle\\#\\{\text{nonreduced, primitive }\downY:\,\ell(\downY)\leq
R\\}$ $\displaystyle\gtrsim(C_{1}+C_{2}-1)\\#\\{\text{total
}\downY,\,\ell(\downY)\leq R\\}$
$\displaystyle\sim(C_{1}+C_{2}-1)\frac{\pi}{12}R^{4}.$

We may take $C$ to be any positive constant less than
$(C_{1}+C_{2}-1)\frac{\pi}{12}$. ∎

##  Appendix A Numerics

We obtained experimental evidence for Theorem 1.1 using the following
elementary Sage code, which computes the number of tripods in $\mathbb{Z}[i]$
of length at most $R$.

from itertools import productfrom time import timedef is_prim(a,b,c,d):
’’’checks that a+bi and c+di are primitive’’’ return gcd(gcd(a,b),
gcd(c,d))==1def is_longest(a,b,c,d): ’’’checks that the largest angle of the
triangle with vertices at 0, a+bi, c+di is at 0, by checking that the length
of the side opposite 0, whose length is |(a-c) + (b-d)i|, is longest.’’’
return min(a^2+b^2, c^2+d^2) > 2*a*c+2*b*ddef is_positively_oriented(a, b, c,
d): ’’’checks that the triangle with vertices at 0, a+bi, c+di is positively
oriented’’’ return a*d-b*c > 0def is_tripod(a,b,c,d): ’’’checks that the
triangle with vertices at 0, a+bi, c+di admits a tripod’’’ return
is_longest(a, b, c, d) and RR(2*a*c+2*b*d+ sqrt((a^2+b^2)*(c^2+d^2)))>0def
tripod_length_squared(a,b,c, d, R): ’’’computes the length-squared of the
tripod’’’ return RR((a-c)^2 + (b-d)^2 + a*c + b*d + sqrt(3)*(a*d-b*c)) <
RR(R^2)def tripod_counts(R): #guess is 0.20947986097*R^4 =
RR(15*sqrt(3)/(4*pi^3)) R^4 ’’’returns the length of the list of tripods of
length at most $R$’’’ L=[(a,b, c, d) for (a,b,c,d) in product(range(-1.5*R,
1.5*R), repeat=4) if is_prim(a,b,c,d) and is_longest(a,b,c,d) and
is_positively_oriented(a,b,c,d)] return len(L)

For $R=35,$ this yields

tripod_counts(35) = 312488

For comparison,

$\left|\frac{312488}{(35)^{4}}-\frac{15\sqrt{3}}{4\pi^{3}}\right|=0.00124129370635984\ldots.$

We make no claims that this Sage code is particularly efficient.

## References

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