where the latter is seen to be equivalent to the former due to the relation
$16\Delta^{2}=16c^{2}\ell^{2}-(b^{2}-3c^{2}-a^{2})^{2}$, which follows from
Heron’s formula and the expression for $\ell^{2}$ in (1.3). So as the signed
analogue of the latter identity for $M_{b}$, we take

$M_{b,n}=\frac{4(\bar{s}_{n}-\bar{c}_{n})\bar{\ell}_{n}-(\bar{s}_{n}-\bar{b}_{n})^{2}+3(\bar{s}_{n}-\bar{c}_{n})^{2}+(\bar{s}_{n}-\bar{a}_{n})^{2}}{4\Delta_{n}}.$
(3.33)

######  Lemma 3.6.

The signed median lengths $\bar{k}_{n}$, $\bar{\ell}_{n}$ given by (3.31)
satisfy the relations (3.32) and (3.33) for all $n\in{\mathbb{Z}}$.

  * Proof:

To begin with, we rewrite (3.32) as

$P_{a,n}^{-1}=\frac{4(\bar{s}_{n}-\bar{c}_{n})\bar{k}_{n}+(\bar{s}_{n}-\bar{a}_{n})^{2}-(\bar{s}_{n}-\bar{b}_{n})^{2}-3(\bar{s}_{n}-\bar{c}_{n})^{2}}{4\Delta_{n}},$
(3.34)

and consider the symmetry $n\to-n-4$, as in Lemma 3.2. On the left-hand side
we have $P_{a,-n-4}^{-1}=-M_{b,n}$, while on the right we use the
transformations (3.13) together with the linear relation (3.8), as well as
$\bar{k}_{-n-4}=\bar{\ell}_{n}$ and $\Delta_{-n-4}=\Delta_{n}$, to see that
(up to an overall minus sign on both sides), the relation (3.34) is
transformed to (3.33). Therefore it will be sufficient to prove the above
relation involving $\bar{k}_{n}$ alone, and we can proceed as in the proofs of
Lemma 3.1 and Theorem 3.3, by substituting in the analytic formulae for each
of the terms and regarding it as an identity between elliptic functions of
$z=(n+2){\kappa}$, so that the left-hand side has simple poles at the places
$z\equiv 0,{\omega},{\kappa},{\kappa}+{\omega}\bmod\Lambda$. However, we also
wish to exploit the additional symmetry under shifting by the half-period
${\omega}$, as we have $P_{a,n}\to P_{a,n}$ under this symmetry, and using the
results of our previous calculations we see that overall the right-hand side
of (3.34) is also left invariant by this transformation. Thus it is sufficient
to check only the poles at $z=0,{\kappa}$ as well as one other value where the
relation is finite, and the case $n=1$ (corresponding to $z=3{\kappa}$) where
$P_{a,1}=2/3$ is readily verified. Then since we only have residues at two
simple poles to check, corresponding to the values $n=-2,-1$, we can use the
simplified method with a local parameter ${\epsilon}$, as at the end of the
proof of Theorem 3.3. Using (3.5) and (3.8) we can rewrite (3.34) as

$-u_{n+2}v_{n+2}=\frac{4\left(1+\frac{\bar{b}_{n}}{\bar{a}_{n}}\right)\frac{\bar{k}_{n}}{\bar{a}_{n}}+\left(\frac{\bar{s}_{n}}{\bar{a}_{n}}-1\right)^{2}-\left(1+\frac{\bar{c}_{n}}{\bar{a}_{n}}\right)^{2}-3\left(1+\frac{\bar{b}_{n}}{\bar{a}_{n}}\right)^{2}}{4{\Delta_{n}}/{\bar{a}_{n}^{2}}}.$
(3.35)

On the right-hand side above we can make use of the expressions (3.24), as
well as

$\bar{c}_{n}/\bar{a}_{n}=-f_{n}^{2}f_{n+1}v_{n+2}^{2}v_{n+3}^{2}/f_{n+2}$

and

$\frac{\bar{k}_{n}}{\bar{a}_{n}}=-\frac{f_{n+4}v_{n+2}v_{n+3}^{2}(1-f_{n}f_{n+1}^{2}f_{n+2})}{2f_{n+2}},\qquad\frac{\Delta_{n}}{\bar{a}_{n}^{2}}=f_{n}f_{n+1}f_{n+3}f_{n+4}v_{n+2}^{2}v_{n+3}^{2}.$

Now when $n=-2$, at leading order on the left-hand side we have
$-u_{0}v_{0}\sim{\epsilon}^{-1}$, while on the right-hand side the leading
order in the denominator is $4\Delta_{-2}/\bar{a}_{-2}^{2}\sim
4{\epsilon}^{-4}$. In the numerator,
$\bar{b}_{-2}/\bar{a}_{-2}\sim{\epsilon}^{2}$ and
$\bar{k}_{-2}/\bar{a}_{-2}\sim\frac{1}{2}{\epsilon}^{-3}$, hence the first
term gives the leading order contribution
$4(1+\bar{b}_{-2}/\bar{a}_{-2})\bar{k}_{-2}/\bar{a}_{-2}\sim
2{\epsilon}^{-5}$, but for the difference of squares that follows we need a
correction at next-to-leading order, so that from
$\bar{s}_{-2}/\bar{a}_{-2}={\epsilon}^{-3}+\frac{1}{2}{\epsilon}^{-2}+\cdots$,
$\bar{c}_{-2}/\bar{a}_{-2}={\epsilon}^{-3}-\frac{1}{2}{\epsilon}^{-2}+\cdots$
we find
$(\bar{s}_{-2}/\bar{a}_{-2}-1)^{2}-(\bar{c}_{-2}/\bar{a}_{-2}+1)^{2}\sim
2{\epsilon}^{-5}$, and the final term $-3(1+\bar{b}_{-2}/\bar{a}_{-2})^{2}$
gives a lower order contribution at $O({\epsilon}^{-4})$; thus overall at
leading order the right-hand side gives
$(2{\epsilon}^{-5}+2{\epsilon}^{-5})/(4{\epsilon}^{-4})={\epsilon}^{-1}$, as
required. When $n=-2$, the left-hand side is
$-u_{1}v_{1}\sim-{\epsilon}^{-1}$, while the denominator of the right-hand
side is $\Delta_{-1}/\bar{a}_{-1}^{2}\sim 8{\epsilon}^{-1}$, and the numerator
contains the terms $\bar{b}_{-1}/\bar{a}_{-1}\sim{\epsilon}^{-1}$,
$\bar{k}_{-1}/\bar{a}_{-1}\sim-{\epsilon}^{-1}$,
$\bar{s}_{-1}/\bar{a}_{-1}\sim-4$,
$\bar{c}_{-1}/\bar{a}_{-1}\sim-{\epsilon}^{-1}$, so that overall these combine
to give
$(-4{\epsilon}^{-2}-{\epsilon}^{-2}-3{\epsilon}^{-2})/(8{\epsilon}^{-1})=-{\epsilon}^{-1}$
at leading order, in agreement with the left-hand side. Note also that for the
values $n=-4,-3$ and $0$, the left-hand side is finite while the other side
has removable singularities: some of the terms in the numerator/denominator on
the right-hand side of (3.35) are singular, but overall these cancel to give a
finite value in the limit ${\epsilon}\to 0$; so the relation holds as an
identity between elliptic functions of $z=(n+2){\kappa}$, and in particular
for all $n\in{\mathbb{Z}}$. ∎

Given that $\bar{k}_{n}$ satisfies (3.32), for each $n$ we can compare this
with the corresponding positive Schubert parameter $P_{a}>0$, given by

$P_{a}=\frac{4\Delta}{4ck+a^{2}-b^{2}-3c^{2}}=\frac{4ck-a^{2}+b^{2}+3c^{2}}{4\Delta},$

where we use either the first or the second rational expression above
involving $a,b,c,k$ and $\Delta$, according to whether $P_{a,n}=\pm P_{a}^{\pm
1}$, determined by $n\bmod 7$ as in Corollary 3.5. Now
$\mathrm{sgn}(\Delta_{n})=\mathrm{sgn}(T_{n}T_{n+1}T_{n+3}T_{n+4})$, which
repeats the pattern

$\Delta_{n}:\qquad++--+++$

with period 7, while our previous analysis also showed that, for each $n$,
$\bar{s}_{n}-\bar{c}_{n}=\mathrm{sgn}(\bar{s}_{n}-\bar{c}_{n})\,c$, where the
sign pattern is

$\bar{s}_{n}-\bar{c}_{n}:\qquad++-+--+--+-++-$

with period 14. For the cases $n\equiv 1,2,5,6,0\,(\bmod\,7)$ we see that
$\Delta_{n}=\Delta$, and then according to whether $P_{a,n}>0$ (when $n\equiv
1,5,0$) or $P_{a,n}<0$ (when $n\equiv 2,6$) we can directly compare the right-
hand side of (3.32) with either the first formula for $P_{a}$ above, or
compare $-P_{a,n}^{-1}$ with the second formula above, respectively; then
since the squared terms can always be identified, that is
$a^{2}=(\bar{s}_{n}-\bar{a}_{n})^{2}$ etc., we have
$ck=(\bar{s}_{n}-\bar{c}_{n})\bar{k}_{n}$ in the first case, giving
$\bar{k}_{n}=\mathrm{sgn}(\bar{s}_{n}-\bar{c}_{n})\,k$, but
$ck=-(\bar{s}_{n}-\bar{c}_{n})\bar{k}_{n}$ in the second case, giving
$\bar{k}_{n}=-\mathrm{sgn}(\bar{s}_{n}-\bar{c}_{n})\,k$. However, the cases
$n\equiv 3,4\,(\bmod\,7)$, when $\Delta_{n}=-\Delta$, are different because in
those cases we need to apply the first of the transformations in (3.30) to
ensure that (1.9) holds. So for $n\equiv 3\,(\bmod\,7)$ that means comparing
$P_{a,n}^{-1}$ with the second formula for $P_{a}$ above, yielding
$\bar{k}_{n}=-\mathrm{sgn}(\bar{s}_{n}-\bar{c}_{n})\,k$, and for $n\equiv
4\,(\bmod\,7)$ it requires that $-P_{a,n}$ should be compared with the first
formula for $P_{a}$, hence
$\bar{k}_{n}=\mathrm{sgn}(\bar{s}_{n}-\bar{c}_{n})\,k$. Thus overall we see
that $\bar{k}_{n}$ is related to the median length $k$ by an overall sign,
which varies with period 14 in the following pattern:

$\bar{k}_{n}:\qquad+-++-++-+--+--.$ (3.36)

Similarly, $\bar{\ell}_{n}$ is equal to the median length $\ell$ up to a sign
with pattern

$\bar{\ell}_{n}:\qquad+-++-+--+--+-+,$ (3.37)

and this verifies that the formulae (1.20) hold.

Based on computer experiments, it was observed in [7] that for the triangles
in the main sequence, the pairs $(\theta,\phi)$ corresponding to the
parametrization (1.10), given by (1.11) with $+$ signs in both equations,
cycle through one of five isomorphic plane curves in a pattern that repeats
with period 7. Applying the symmetry $a\leftrightarrow b$,
$k\leftrightarrow\ell$, one obtains an alternative pair of parameters

$\tilde{\theta}=\frac{c-b+2k}{2s},\qquad\tilde{\phi}=\frac{a-c+2\ell}{2s},$
(3.38)

and Buchholz and Rathbun noted that these pairs cycle with period 7 through
the same set of curves but in a different order, namely ${\cal C}_{1},{\cal
C}_{2},{\cal C}_{3},{\cal C}_{4},{\cal C}_{1},{\cal C}_{2},{\cal C}_{5}$. A
more detailed study of the allowed discrete symmetries in [9] showed that by
applying appropriate permutations of $a,b,c,k,\ell$ and changes of sign, one
could also obtain pairs of coordinates on three more (isomorphic) curves
${\cal C}_{6},{\cal C}_{7},{\cal C}_{8}$. However, until now there was no
explanation for the period 7 behaviour with respect to the index $n$ in the
main sequence, which we provide here.

######  Theorem 3.7.

For $n\geq 1$, the rational parameters

$\theta=\frac{c-a+2\ell}{a+b+c},\qquad\phi=\frac{b-c+2k}{a+b+c}$

corresponding to the main sequence of Heron triangles with sides $(a,b,c)$ and
medians $k,\ell$ lie on one of five birationally equivalent plane curves of
genus one, which repeat in a pattern with period 7. The same is true for any
such sequence of parameters obtained from these by the action of the discrete
symmetry group that leaves the equations $4k^{2}=2b^{2}+2c^{2}-a^{2}$,
$4\ell^{2}=2c^{2}+2a^{2}-b^{2}$ invariant.

  * Proof:

We begin by considering the pair of quantities

$\theta_{n}=\frac{\bar{a}_{n}-\bar{c}_{n}+2\bar{\ell}_{n}}{2\bar{s}_{n}},\qquad\phi_{n}=\frac{\bar{c}_{n}-\bar{b}_{n}+2\bar{k}_{n}}{2\bar{s}_{n}},$

obtained by replacing $a,b,c,k,\ell$ by their signed counterparts
$\bar{s}_{n}-\bar{a}_{n},\bar{s}_{n}-\bar{b}_{n},\ldots$ etc. For $n=1$ this
coincides with $\theta,\phi$. Moreover, from the analytic parametrization of
the two Somos-5 sequences, this gives a pair of independent elliptic functions
of $z=(n+2){\kappa}$. Then by a standard result in the theory of elliptic
functions, the pairs $(\theta_{n},\phi_{n})$ lie on a plane curve ${\cal C}$
of genus one, which is birationally equivalent to the original elliptic curve
(this follows from the Riemann-Roch theorem, but there is also an explicit
classical method for constructing the equation of ${\cal C}$, based on the
expressions for $\theta_{n},\phi_{n}$ in terms of $\wp,\wp^{\prime}$ [2]).
Under the change $n\to n+1$, the original pair $(\theta,\phi)$ corresponds to
a different pair of elliptic functions, because each term $a,b,\ldots$ that
appears is related to $\bar{s}_{n}-\bar{a}_{n},\bar{s}_{n}-\bar{b}_{n},\ldots$
by a sign which changes with $n$; so the pair of coordinates for the next
value of $n$ will lie on a different curve, ${\cal C}^{*}$ say. Now all of
these signs repeat with period 14, but in each block of 7 the signs are
flipped with respect to the previous block, and only the ratio of terms
appears in $\theta,\phi$, so their overall pattern of signs repeats with
period 7, hence there are at most 7 different curves repeating with this
period. However, a closer examination of (3.29), (3.36) and (3.37) reveals
that there are really only 5 different pairs of functions appearing in each
block of 7, because all the signs for $n\equiv 1\,(\bmod\,7)$ and $n\equiv
5\,(\bmod\,7)$ are the opposite of each other, and similarly for $n\equiv
2\,(\bmod\,7)$ and $n\equiv 6\,(\bmod\,7)$. The same argument applies to any
pair $(\theta,\phi)$ obtained from this one by the action of the discrete
symmetry group. ∎

In order to show that the side lengths given by (1.19) have greatest common
divisor 1 for all $n\geq 1$, it is necessary to reconsider the orbits of the
QRT map $\varphi$ defined by (3.27), associated with the rational sequences
$(u_{n})$, $(v_{n})$, and examine their reduction modulo a prime $p$, which
corresponds to considering the map over the finite field ${\mathbb{F}}_{p}$.
(For a detailed treatment of QRT maps over finite fields, see [32].) A general
QRT map (over any field) is a birational map
$\varphi:\,{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}\to{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$
given by a composition of two involutions, and as already mentioned, each
orbit lies on a curve belonging to a pencil of biquadratic curves in the
plane; generic curves ${\cal C}$ have genus one, and the orbit gives a
sequence of points $\hat{{\cal P}}_{0}+n{\cal P}\in{\cal C}$, with each
iteration of $\varphi$ corresponding to addition of the point ${\cal P}$ in
the group law of the curve. For the particular case at hand, the curve is
given by (2.37), which has identity element ${\cal O}=(\infty,\infty)$, and
$\iota$ in (2.38) is the elliptic involution that sends any point $\hat{\cal
P}\to-\hat{\cal P}$. Under the QRT map $\varphi=\iota\circ\iota_{h}$, ${\cal
P}=(\infty,0)$ is the point being added at each step, and the orbit
corresponding to the Somos-5 sequence (1.14) via (2.34) is

$\varphi^{n}(u_{0},u_{1})=(u_{n},u_{n+1})=(1,1)+n{\cal P},$ (3.39)

where the initial point $\hat{{\cal P}}_{0}=(u_{0},u_{1})=(1,1)\in{\cal C}$ is
2-torsion, while the orbit associated with the other sequence (1.15) is

$\varphi^{n}(v_{0},v_{1})=(v_{n},v_{n+1})=n{\cal P},$ (3.40)

with initial point ${\cal O}=(v_{0},v_{1})=(\infty,\infty)$.

The above description is valid over ${\mathbb{C}}$ or any subfield where the
orbit is defined, in particular over ${\mathbb{Q}}$, in which case ${\cal P}$
is an infinite order element in the Mordell-Weil group of the curve ${\cal
C}({\mathbb{Q}})$, which is generated by ${\cal P}$ and the 2-torsion point
$\hat{{\cal P}}_{0}=(1,1)$. Under reduction $\bmod\,p$ the same description
holds provided that $p$ is a prime of good reduction, so that the curve ${\cal
C}({\mathbb{F}}_{p})$ is non-singular, with the main difference being that now
the curve has finitely many points satisfying the Hasse-Weil bound
$\left|p+1-\\#{\cal C}({\mathbb{F}}_{p})\right|\leq 2\sqrt{p}$, so ${\cal P}$
has some finite order which we denote by $\mathrm{ord}_{p}({\cal P})$. Hence
it follows that both orbits (3.39) and (3.40) over ${\mathbb{F}}_{p}$ are
periodic with the same period $\mathrm{ord}_{p}({\cal P})$.

For primes of bad reduction, which in this case are 2,3 and 17, the situation
is slightly more complicated. For the primes $p=2,3$, we find that the
singular curve ${\cal C}({\mathbb{F}}_{p})$ is reducible, since $-5\bmod
2=-5\bmod 3=1$, so in both cases we can factorize (2.37) as

$(UV+1)(U+V+1)=0.$ (3.41)

When $p=2$, the two orbits $(u_{n},u_{n+1})$ and $(v_{n},v_{n+1})$ are the
same up to a shift of starting point, repeating the sequence

$(\infty,\infty),(\infty,0),(0,1),(1,1),(1,0),(0,\infty)$

with period $6=2\times(2+1)$, corresponding to the fact that each (genus zero)
irreducible component of (3.41) is isomorphic to
${\mathbb{P}}^{1}({\mathbb{F}}_{2})$, so contains 3 points, and the orbit
alternates between points on each component, i.e.
$(\infty,0),(1,1),(0,\infty)\in\\{\,UV+1=0\,\\}$ and
$(\infty,\infty),(0,1),(1,0)\in\\{\,U+V+1=0\,\\}$. Similarly, when $p=3$ both
orbits have period $8=2\times(3+1)$, repeating the sequence

$(\infty,\infty),(\infty,0),(0,2),(2,1),(1,1),(1,2),(2,0),(0,\infty),$

which again alternates between the two irreducible components of the curve
(3.41). For $p=17$, the curve (2.37) is irreducible and singular: this is a
case of non-split multiplicative reduction [38], so it has genus zero and
contains $\\#{\cal C}({\mathbb{F}}_{17})=p+2=19$ points, and after removing
the singular point $(5,5)$, ${\cal
C}({\mathbb{F}}_{17})\setminus\\{\,(5,5)\,\\}$ has the structure of an abelian
group of order 18. The two orbits both have period 9, corresponding to the two
cosets of the cyclic subgroup of order 9 generated by ${\cal P}$; the
coordinates of the points in these two orbits can be read off from Table 7,
which also includes part of the associated periodic sequence
$(f_{n}\bmod\,17)$ (of period 18).

Although it is not essential to our main argument, we can now offer a brief
explanation for how the periods modulo a prime arise for the corresponding
Somos-5 sequences. Setting $t=\mathrm{ord}_{p}({\cal P})$ and working
$\bmod\,p$, from the relation $u_{n+t}=u_{n}$ we have

$\frac{S_{n-2+t}S_{n+1+t}}{S_{n-1+t}S_{n+t}}=\frac{S_{n-2}S_{n+1}}{S_{n-1}S_{n}}$

for all $n$, and clearly the period of the Somos-5 sequence must be a multiple
of $t$. (We will write everything for the sequence $(S_{n})$, but the same
treatment applies to $(T_{n})$ or any other Somos-5 sequence.) Somos-5
sequences have a 3-parameter group of gauge transformations which leave the
ratios $u_{n}$, and hence the recurrence (1.16), unchanged: one can replace
$S_{n}\to A^{*}_{\pm}(B^{*})^{n}\,S_{n}$, i.e. rescale all even/odd index
terms by an arbitrary non-zero scalar $A^{*}_{\pm}$, and rescale each term by
the powers of another arbitrary quantity $B^{*}\neq 0$; this symmetry group
has a natural interpretation in terms of the quiver that defines the
associated cluster algebra [20]. Hence we see that $S_{n}$ and $S_{n+t}$
should be related by a gauge transformation of this kind.

######  Proposition 3.8.

Under shifting by $t=\mathrm{per}(u_{n}\bmod\,p)$, the period of the
associated QRT map, the Somos-5 sequence reduced $\bmod\,p$ satisfies

$S_{n+t}=A^{*}_{\pm}(B^{*})^{n}\,S_{n},\qquad
A^{*}_{\pm},B^{*}\in{\mathbb{F}}_{p}^{*}$ (3.42)

for even/odd $n$, respectively, where $A^{*}_{+},A^{*}_{-},B^{*}$ are
constants independent of $n$.

  * Proof:

To prove this gauge transformation formula directly, set
$r_{n}=S_{n+t}/S_{n}$, and observe that $u_{n+t}=u_{n}$ for all $n$ implies
that
$r_{n+2}/r_{n}=r_{n+3}/r_{n+1}=r_{n+4}/r_{n+2}=r_{n+5}/r_{n+3}=(B^{*})^{2}$
say, a constant independent of $n$, and this can be solved separately for
even/odd $n$ to give $r_{n}=A^{*}_{\pm}(B^{*})^{n}$. ∎

######  Corollary 3.9.

Let $t$ be the order of the point ${\cal P}\in{\cal C}({\mathbb{F}}_{p})$ if
$p$ is a prime of good reduction (or equivalently, the period of
$(u_{n}\bmod\,p)$, appropriately reinterpreted in the case of bad reduction).
Then the period of the Somos-5 sequence $\bmod\,p$ is a multiple of $t$, given
by

$\mathrm{per}(S_{n}\bmod\,p)=\ell^{*}\,t\leq 2(p-1)t,$ (3.43)

where (up to a possible factor of 2) $\ell^{*}$ denotes the lowest common
multiple of $\mathrm{ord}_{p}\big{(}(B^{*})^{2}\big{)}$ and/or the orders in
${\mathbb{F}}_{p}^{*}$ of one or two other combinations of
$A^{*}_{+},A^{*}_{-},B^{*}$, depending on the parity of $t$.

  * Proof:

Iterating the quasiperiodicity relation (3.42) $j$ times when $t$ is even
yields

$S_{n+jt}=(A^{*}_{\pm})^{j}(B^{*})^{nj+tj(j-1)/2}\,S_{n},$

for even/odd $n$, so requiring $S_{n+jt}=S_{n}$ for all $n$ imposes the
conditions

$(B^{*})^{2j}=1,\qquad(A_{+}^{*})^{j}=(A_{-}^{*}B^{*})^{j}=(B^{*})^{-tj(j-1)/2}=\pm
1,$ (3.44)

where the choice of $j$ satisfying the first condition for a given $B^{*}$
fixes the sign on the right. Then $\ell^{*}$ is the smallest positive value of
$j$ that satisfies all these conditions. An odd value of $\ell^{*}$ is only
possible in the case that $(B^{*})^{2}$, $A_{+}^{*}$ and $A_{-}^{*}B^{*}$ are
all quadratic non-residues in ${\mathbb{F}}_{p}^{*}$, in which case only the
plus sign can occur on the right-hand side above, and

$\ell^{*}=\mathrm{lcm}\big{(}\mathrm{ord}_{p}(A^{*}_{+}),\mathrm{ord}_{p}(A^{*}_{-}B^{*}),\mathrm{ord}_{p}((B^{*})^{2})\big{)},$

but otherwise there must be an even
$\ell^{*}|\mathrm{lcm}\big{(}\mathrm{ord}_{p}(A^{*}_{+}),\mathrm{ord}_{p}(A^{*}_{-}B^{*}),\mathrm{ord}_{p}((B^{*})^{2})\big{)}$,
where it may be possible to divide out some factors of 2 from the first two
terms, depending on the orders of these elements. For odd $t$, iterating
(3.42) $j$ times produces

$S_{n+jt}=(A^{*}_{\pm})^{\left\lfloor\frac{j+1}{2}\right\rfloor}(A^{*}_{\mp})^{\left\lfloor\frac{j}{2}\right\rfloor}(B^{*})^{nj+tj(j-1)/2}\,S_{n},$

and then by requiring $S_{n+jt}=S_{n}$ for all $n$, in the case of odd $j$
(exploiting the freedom to replace $B^{*}\to-B^{*}$ and
$A^{*}_{-}\to-A^{*}_{-}$ simultaneously) this leads to

$A^{*}_{+}=A^{*}_{-},\qquad(B^{*})^{j}=(A^{*}_{+})^{j}=1\implies\ell^{*}=\mathrm{lcm}\big{(}\mathrm{ord}_{p}(A^{*}_{+}),\mathrm{ord}_{p}(B^{*})\big{)},$

corresponding to a situation where $\tilde{K}_{0}=\tilde{K}_{1}$ and all the
terms satisfy the same Somos-4 recurrence (cf. equation (2.5) and [43]). For
an even value of $j$ with $t$ odd, we find instead

$(B^{*})^{j}=1,\qquad(A^{*}_{+}A^{*}_{-}B^{*})^{j/2}=(B^{*})^{\big{(}1-t(j-1)\big{)}j/2}=1,$

so now, given the values of $(B^{*})^{2}$ and $A^{*}_{+}A^{*}_{-}B^{*}$, the
smallest possible value of $j/2$ can be found, yielding

$\ell^{*}=2\,\mathrm{lcm}\big{(}\mathrm{ord}_{p}(A^{*}_{+}A^{*}_{-}B^{*}),\mathrm{ord}_{p}((B^{*})^{2})\big{)}.$
(3.45)

The upper bound on the period comes from Fermat’s little theorem. ∎

An advantage of the relation (3.42) is that it allows the period of the
Somos-5 sequence to be computed from (3.43) without calculating so many terms:
in general it is sufficient to find the minimum $t$ such that
$(u_{0},u_{1})=(u_{t},u_{t+1})$, giving $t=\mathrm{ord}_{p}({\cal P})$, which
requires the $t+5$ adjacent terms $S_{-2},\ldots,S_{t+2}$, and then
$A^{*}_{\pm},B^{*}$ can be obtained from the ratios $S_{t}/S_{0}$,
$S_{t+1}/S_{1}$ and $S_{t+2}/S_{2}$ (provided $S_{0}S_{1}S_{2}\neq 0$,
otherwise one of the ratios $S_{t-1}/S_{-1}$, $S_{t-2}/S_{-2}$ can be used
instead). For example, in the case of the sequence (1.14) taken $\bmod\,23$,
we find that $(u_{n}\bmod\,23)$ repeats the pattern $1,1,2,13,20,3,20,13,2$
with period 9, so $t=\mathrm{ord}_{23}({\cal P})=9$ which is odd, while
$A^{*}_{+}\equiv S_{9}/S_{0}\bmod\,23=S_{10}/S_{1}\bmod\,23=21$ and
$S_{11}/S_{2}\bmod\,23=20$, thus we find $(B^{*})^{2}=20/21=13=6^{2}$ in
${\mathbb{F}}_{23}$, and taking $B^{*}=6$ gives $A^{*}_{-}=21/6=15\neq
A^{*}_{+}$; then $\mathrm{ord}_{23}(13)=11$ and $\mathrm{ord}_{23}(21\times
15\times 6)=\mathrm{ord}_{23}(21^{2})=11$, so $\ell^{*}=22$ by (3.45), hence
$\mathrm{per}(S_{n}\bmod\,23)=9\times 22=198$, in agreement with the value
found by Robinson in [46]. As another example, for the sequence (1.15) taken
$\bmod\,61$, the corresponding terms $(v_{n}\bmod\,61)$ have a repeating
pattern

$\infty,\infty,0,60,7,25,12,34,11,32,2,1,1,2,32,11,34,12,25,7,60,0$

with even period $t=22$, and from the terms $1,1$ in the middle it is apparent
that this is the same as the orbit $(u_{n}\bmod\,61)$ but shifted; thus from
$T_{21}/T_{-1}\bmod\,61=14\equiv A_{-}^{*}(B^{*})^{-1}$,
$T_{23}/T_{1}\bmod\,61=60\equiv A_{-}^{*}B^{*}$,
$T_{24}/T_{2}\bmod\,61=14\equiv A_{+}^{*}(B^{*})^{2}$ we find
$A_{-}^{*}=13=14^{2}$, $A_{+}^{*}=48=14^{-1}$, $B^{*}=14$ in
${\mathbb{F}}_{61}$, and $\mathrm{ord}_{61}(14)=6$, so $\ell^{*}$ divides
$\mathrm{lcm}\big{(}\mathrm{ord}_{61}(14^{-1}),\mathrm{ord}_{61}(14^{3}),\mathrm{ord}_{61}(14^{2})\big{)}$=6,
the smallest value of $j$ satisfying all the requirements in (3.44), hence the
period of the sequence $(T_{n}\bmod\,61)$ is $22\times 6=132$, and this is the
same as the period of the original Somos-5 sequence $(S_{n}\bmod\,61)$ as
found in [46].

We are now ready to finish off the proof of the main theorem.

######  Lemma 3.10.

For all $n\in{\mathbb{Z}}$, the terms of the sequences (1.14) and (1.15) with
the same index are coprime, that is

$\gcd(S_{n},T_{n})=1.$

  * Proof:

Suppose that for some $n$ the terms $S_{n}$ and $T_{n}$ have a common divisor.
Then for the reduced sequences in ${\mathbb{F}}_{p}$ it follows that
$S_{n}\equiv 0\equiv T_{n}$, and so $u_{n}=u_{n+1}=\infty=v_{n}=v_{n+1}$. If
$p$ is a prime of good reduction for the curve ${\cal C}$ given by (2.37),
then in the group law of ${\cal C}({\mathbb{F}}_{p})$ this implies that
$(1,1)+n{\cal P}={\cal O}=n{\cal P}$, which is a contradiction. If $p$ is a
prime of bad reduction, then we have $p=2,3$ or 17, and $p=2$ or $p=3$ are
impossible by Proposition 2.8, while Table 7 shows that 17 is never a divisor
of $S_{n}$, since the sequence $(u_{n}\bmod\,17)$ has period 9 and remains
finite. ∎

Table 7: The analogue of Table 5 in the finite field ${\mathbb{F}}_{17}$. $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---|---|---
$u_{n}$ | $1$ | $1$ | $2$ | $10$ | $15$ | $6$ | $15$ | $10$ | $2$ | $1$
$v_{n}$ | $\infty$ | $\infty$ | $0$ | $16$ | $7$ | $11$ | $7$ | $16$ | $0$ | $\infty$
$f_{n}$ | $\infty$ | $1$ | $16$ | $2$ | $3$ | $9$ | $12$ | $14$ | $6$ | $\infty$

######  Proposition 3.11.

For all $n\in{\mathbb{Z}}$, the signed lengths satisfy
$\gcd(\bar{a}_{n},\bar{b}_{n},\bar{c}_{n})=1$.

  * Proof:

First of all, note that from Lemma 3.1 we have
$\gcd(\bar{s}_{n},\bar{a}_{n},\bar{b}_{n},\bar{c}_{n})=\gcd(\bar{a}_{n},\bar{b}_{n},\bar{c}_{n})$.
Up to an overall sign, each of the signed lengths is a degree six monomial in
the two sets of five adjacent values $S_{n},\ldots,S_{n+4}$ and
$T_{n},\ldots,T_{n+4}$. Now suppose that a prime $p$ is a divisor of
$\bar{c}_{n}$. If $p|S_{n}$, then by Lemma 2.3 it cannot divide
$S_{n+1},S_{n+2}$ or $S_{n+3}$, so if it divides $\bar{b}_{n}$ then it must
divide $T_{n+2}$, thus (by Lemma 2.3 again) it is coprime to
$S_{n+3},S_{n+4},T_{n}$ and $T_{n+1}$, hence it cannot divide $\bar{s}_{n}$.
Similarly, if $p|T_{n+4}$ and it is a divisor of $\bar{a}_{n}$ then, by the
same lemma, it is coprime to $T_{n+1},T_{n+2}$ and $T_{n+3}$, so it must
divide $S_{n+2}$, but in that case again it is coprime to $\bar{s}_{n}$.
Applying the same lemma once more, we see that if $p|S_{n+1}$ then
$p|\bar{b}_{n}$, so if $p|\bar{s}_{n}$ as well then it must divide either
$T_{n}$ or $T_{n+1}$, but the first case is impossible because $p$ must also
divide $\bar{a}_{n}$ by (3.8), yet it has to be coprime to
$S_{n+2},T_{n+1},T_{n+2},T_{n+3}$; so this leaves the second case, which
requires $p|\gcd(S_{n+1},T_{n+1})$, contradicting Lemma 3.10. Finally, if
$p|T_{n+3}$ then an analogous argument leads to $p|\gcd(S_{n+3},T_{n+3})$,
another contradiction. ∎

The formulae (1.19) imply that
$\gcd(a,b,c)=\gcd(\bar{b}_{n}+\bar{c}_{n},\bar{c}_{n}+\bar{a}_{n},\bar{a}_{n}+\bar{b}_{n})$,
and any prime divisor of the latter three linear combinations of the signed
lengths must be a divisor of
$2\bar{a}_{n}=(\bar{c}_{n}+\bar{a}_{n})+(\bar{a}_{n}+\bar{b}_{n})-(\bar{b}_{n}+\bar{c}_{n})$,
and similarly of $2\bar{b}_{n}$ and $2\bar{c}_{n}$, so by the above
proposition the only possible common divisor of $(a,b,c)$ is 2, but
Proposition 2.8 shows that precisely one of
$\bar{b}_{n}+\bar{c}_{n},\bar{c}_{n}+\bar{a}_{n},\bar{a}_{n}+\bar{b}_{n}$ is
even (cf. Table 6), so $\gcd(a,b,c)=1$ and this completes the proof of Theorem
1.2.

##  4 Brahmagupta angles and geometrical identities

By virtue of the fact that there is a 2-isogeny relating the curves (2.37) and
(2.43), there are infinitely many identities between elements of the
associated function fields, which correspond to relations between terms of the
sequences $(f_{n})$ and $(v_{n})$, or equivalently $(u_{n})$, since
$u_{n}=v_{n}f_{n-2}f_{n+1}/(f_{n-1}f_{n})$. Here we present some identities
that arise naturally from the geometry of Heron triangles, which in particular
leads to an appealing way to represent and visualize the triangles in the main
sequence.

We begin by considering Brahmagupta’s construction of rational Heron
triangles, which is based on concatenating two rational Pythagorean right
triangles with a common height of length $2r$, where $a,b$ are the hypotenuses
of the two triangles. So if $d,e$ are the bases of the two triangles, then we
have

$4r^{2}+d^{2}=a^{2},\qquad 4r^{2}+e^{2}=b^{2},\qquad\pm(d\pm e)=c,$ (4.1)

where the choice of sign inside the bracket above depends on whether the two
triangles are joined back-to-back along their height, or are overlapping, and
in the latter case there is an overall sign outside depending on which of $d$
or $e$ is the larger.

The case of two overlapping right triangles with $d>e$ is the situation
relevant to the triangle for $n=1$ in the main sequence, and we denote the
acute angles at the base of each triangle by $\psi_{a},\psi_{b}$,
respectively. These two angles completely determine the Heron triangle, up to
scale, and we will refer to them as the Brahmagupta angles. Then upon
comparing with Fig.1 it is clear that

$\psi_{a}=\pi-{\beta}-{\gamma},\qquad\psi_{b}=\pi-{\gamma}^{\prime}+{\alpha}^{\prime},$

with $P_{a}=\mathrm{cot}({\beta}/2)$, $X_{a}=\mathrm{cot}({\gamma}/2)$ as
before, and $M_{b}=\mathrm{cot}({\alpha}^{\prime}/2)$,
$X_{b}=\mathrm{cot}({\gamma}^{\prime}/2)$. So from
$2r=a\sin\psi_{a}=b\sin\psi_{b}$ and trigonometric identities we can write

$r=a\,\frac{(P_{a}+X_{a})(P_{a}X_{a}-1)}{(P_{a}^{2}+1)(X_{a}^{2}+1)}=b\,\frac{(M_{b}-X_{b})(M_{b}X_{b}+1)}{(M_{b}^{2}+1)(X_{b}^{2}+1)},$
(4.2)

while we have $c=d-e$ with $d=a\cos\psi_{a}$, $e=b\cos\psi_{b}$, and then we
reproduce Brahmagupta’s formulae (1.2) by taking

$p=a\cos^{2}\left(\frac{\beta+{\gamma}}{2}\right),\qquad
q=b\sin^{2}\left(\frac{{\gamma}^{\prime}-{\alpha}^{\prime}}{2}\right).$

Once again we can rewrite the trigonometric functions in terms of the
appropriate Schubert parameters, to obtain

$\frac{p}{a}=\frac{(P_{a}X_{a}-1)^{2}}{(P_{a}^{2}+1)(X_{a}^{2}+1)},\qquad\frac{q}{b}=\frac{(M_{b}-X_{b})^{2}}{(M_{b}^{2}+1)(X_{b}^{2}+1)}$
(4.3)

The above formulae have been written with the appropriate sign choices for the
$(73,51,26)$ triangle, which has Brahmagupta parameters $p=49/13$, $q=588/13$,
$r=210/13$. However, each of the lengths and Schubert parameters in the two
different expressions for $r$ in (4.3) can now be replaced by their signed
counterparts, immediately yielding another relation, which is equivalent to an
identity between elliptic functions.

######  Theorem 4.1.

For all $n\in{\mathbb{Z}}$ the identity

$(\bar{s}_{n}-\bar{a}_{n})\sin\psi_{a,n}=(\bar{s}_{n}-\bar{b}_{n})\sin\psi_{b,n}$

holds for the signed Brahmagupta angles $\psi_{a,n}$, $\psi_{b,n}$, which are
defined in terms of the Schubert parameters with signs by

$\sin\psi_{a,n}=\frac{(P_{a,n}+X_{a,n})(P_{a,n}X_{a,n}-1)}{(P_{a,n}^{2}+1)(X_{a,n}^{2}+1)},\qquad\sin\psi_{b,n}=\frac{(M_{b,n}-X_{b,n})(M_{b,n}X_{b,n}+1)}{(M_{b,n}^{2}+1)(X_{b,n}^{2}+1)}.$
(4.4)

The advantage of using these angles (which are allowed to be negative) instead
of the actual positive acute angles is that their sines are given by fixed
rational functions of the Schubert parameters, hence they define the same
functions on the curve for all $n$. By the same argument as in the proof of
Theorem 3.7, this means that the pairs $(\sin\psi_{a,n},\sin\psi_{b,n})$ lie
on an algebraic curve that is birationally equivalent to ${\cal C}$. The
sequence of pairs of sines of (signed) Brahmagupta angles for $n=1,\ldots,500$
is plotted in Fig.5(a), which shows that this plane curve has self-
intersections, while in Fig.5(b) this is lifted to 3D by plotting the
corresponding value of $u_{n}$ as a third component, showing a space curve
without self-intersections.

(a) Plot of $(\sin\psi_{a,n},\sin\psi_{b,n})$.

(b) 3D plot of $(\sin\psi_{a,n},\sin\psi_{b,n},u_{n})$.

Figure 5: 500 points on the orbit of the Brahmagupta angles for the main
sequence.

As one more example of using elementary geometry to derive identities between
elliptic functions, we start from a formula for the tangent of the bisected
angle opposite side $a$, which is given on the first page of Schubert’s
monograph [47], and in our notation reads

$\tan\left(\frac{{\alpha}+{\beta}}{2}\right)=\frac{(s-b)(s-c)}{\Delta}.$

######  Theorem 4.2.

The squares of the coordinates of the points on the orbit of the QRT map
$\varphi$ associated wth the sequence (1.15) are given as rational functions
of five adjacent quantities $f_{n}$ by

$v_{n+2}^{2}=\frac{f_{n+1}^{2}f_{n+2}^{3}+f_{n+4}}{f_{n+3}(f_{n}^{2}-f_{n+2}f_{n+4})},\qquad
v_{n+3}^{2}=\frac{f_{n+2}^{3}f_{n+3}^{2}-f_{n}}{f_{n+1}(f_{n+4}^{2}+f_{n}f_{n+2})}.$
(4.5)

  * Proof:

In terms of the Schubert parameters, the tangent of the bisected angle for the
$n$th triangle is $(P_{a,n}+M_{a,n})/(P_{a,n}M_{a,n}-1)$, while the right-hand
side can be written in terms of ratios of $S_{n}$ and $T_{n}$ terms, and
ultimately as $f_{n}f_{n+1}f_{n+2}/f_{n+4}$, so that this becomes an identity
for signed quantities. Then substituting for the Schubert parameters from
(3.5) and writing everything in terms of $f_{n}$ and $v_{n}$, this produces
the formula

$\frac{1}{f_{n+4}}=\frac{f_{n+2}f_{n+3}\,v_{n+2}^{2}+1}{f_{n}^{2}f_{n+3}\,v_{n+2}^{2}-f_{n+1}^{2}f_{n+2}^{3}};$

the right-hand side above is a Möbius transformation of $v_{n+2}^{2}$, which
is inverted to obtain the first expression in (4.5). An analogous calculation
starting from the corresponding formula for the tangent of the bisected angle
opposite side $b$ produces the second expression for $v_{n+3}^{2}$. ∎

######  Remark 4.3.

In principle one can use repeated application of the recurrence (2.42) in
(4.5) to eliminate $f_{n+4},f_{n+3},f_{n+2}$ and write $v_{n+2},v_{n+3}$ as
rational functions of $f_{n},f_{n+1}$ only, then use (2.43) to simplify the
resulting formulae; this would be a rather tedious way to verify the 2-isogeny
given by (2.49) in terms of these pairs of coordinates. Since it involves the
product $(v_{n+2}v_{n+3})^{2}$, substituting for the latter with (4.5) and
applying the same method provides an alternative proof of the identity (3.9),
which is equivalent to the linear relation (3.8) for the signed lengths,
although from the point of view of the logical progression of the paper this
is a somewhat circular argument, since Lemma 3.1 was one of the key steps in
showing that the quantities $f_{n}$ describe the geometry of the main sequence
of Heron triangles as claimed.

##  5 Conclusions

Table 8: Prime factors of the Schubert parameters in the sporadic cases.

$n$ | $M_{a}$ | $P_{a}$ | $X_{a}$ | $M_{b}$ | $P_{b}$ | $X_{b}$
---|---|---|---|---|---|---
| $\tfrac{2^{3}\cdot 7\cdot 13}{3\cdot 17}$ | $17$ | $\tfrac{2^{4}\cdot 3}{7\cdot 13}$ | $\tfrac{3\cdot 7\cdot 11}{2^{2}\cdot 5\cdot 13}$ | $\tfrac{11\cdot 13\cdot 17}{2^{2}\cdot 3\cdot 5\cdot 7}$ | $\tfrac{17}{5\cdot 11}$
* | $\tfrac{3^{2}\cdot 5\cdot 31}{2^{2}\cdot 7\cdot 17}$ | $\tfrac{2^{2}\cdot 5\cdot 31}{3^{2}\cdot 17}$ | $\tfrac{3^{2}\cdot 7}{5\cdot 17}$ | $\tfrac{3\cdot 7\cdot 17}{5\cdot 19}$ | $\tfrac{3\cdot 5\cdot 17\cdot 19}{2^{3}\cdot 7\cdot 31}$ | $\tfrac{3\cdot 19\cdot 31}{2^{4}\cdot 5\cdot 17}$
** | $\tfrac{2^{3}\cdot 19\cdot 47}{3^{2}\cdot 11\cdot 23}$ | $\tfrac{3^{2}\cdot 11\cdot 17\cdot 47}{2^{3}\cdot 7\cdot 19\cdot 23}$ | $\tfrac{2\cdot 7\cdot 11\cdot 47}{17\cdot 19\cdot 23}$ | $\tfrac{2^{4}\cdot 3^{2}\cdot 7\cdot 17\cdot 23}{5\cdot 11\cdot 19\cdot 97}$ | $\tfrac{2\cdot 7\cdot 11\cdot 17\cdot 19}{5\cdot 23\cdot 97}$ | $\tfrac{5\cdot 11\cdot 19\cdot 23}{2^{5}\cdot 3^{2}\cdot 97}$
*** | $\tfrac{2^{2}\cdot 3\cdot 17\cdot 43\cdot 59}{5\cdot 11\cdot 13\cdot 19\cdot 23}$ | $\tfrac{2^{3}\cdot 3\cdot 5\cdot 13\cdot 23\cdot 43}{11\cdot 19\cdot 41\cdot 59}$ | $\tfrac{19\cdot 23\cdot 41\cdot 43}{5\cdot 11\cdot 13\cdot 17\cdot 59}$ | $\tfrac{19\cdot 23\cdot 41\cdot 43}{2^{4}\cdot 3\cdot 7\cdot 11\cdot 59}$ | $\tfrac{11\cdot 19\cdot 41\cdot 59}{7\cdot 17\cdot 23\cdot 43}$ | $\tfrac{7\cdot 19\cdot 43\cdot 59}{2^{5}\cdot 3\cdot 11\cdot 17\cdot 23}$

Table 9: Prime factors of the semiperimeter, reduced side lengths and area in
the sporadic cases.

$n$ | $s$ | $s-a$ | $s-b$ | $s-c$ | $\Delta$
---|---|---|---|---|---
* | $3\cdot 7\cdot 13\cdot 17$ | $2^{3}\cdot 5^{2}\cdot 17$ | $3\cdot 7\cdot 13$ | $2^{3}\cdot 11^{2}$ | $2^{3}\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$
* | $2^{3}\cdot 7\cdot 19^{2}$ | $2^{3}\cdot 3^{6}$ | $5^{2}\cdot 7\cdot 31$ | $17^{2}\cdot 31$ | $2^{3}\cdot 3^{3}\cdot 5\cdot 7\cdot 17\cdot 19\cdot 31$
** | $2^{3}\cdot 3^{2}\cdot 11^{2}\cdot 19^{2}$ | $2^{5}\cdot 3^{2}\cdot 5^{2}\cdot 7\cdot 17$ | $23^{2}\cdot 47^{2}$ | $7\cdot 17\cdot 97^{2}$ | $2^{4}\cdot 3^{2}\cdot 5\cdot 7\cdot 11\cdot 17\cdot 19\cdot 23\cdot 47\cdot 97$
*** | $17\cdot 23^{2}\cdot 59^{2}$ | $5^{2}\cdot 7^{2}\cdot 13^{2}\cdot 41$ | $2^{4}\cdot 3\cdot 11^{2}\cdot 43^{2}$ | $2^{4}\cdot 3\cdot 17\cdot 19^{2}\cdot 41$ | $2^{4}\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 41\cdot 43\cdot 59$

We have proved all the empirical observations on the infinite sequence of
Heron triangles with two rational medians found by Buchholz and Rathbun in
[7], and this led us to explicit formulae for the side lengths, rational
medians and the area in terms of the Somos-5 sequences (1.14) and (1.15). The
crux of our proof was to view the underlying elliptic curve as a complex
torus, and construct an analytic embedding of this torus in the Schubert
surface, such that the required algebraic identities for the two sets of
Schubert parameters are valid for all complex values of the argument, and in
particular at an infinite discrete set of points corresponding to the indices
$n\in{\mathbb{Z}}$. The indices $n\geq 1$ provide infinitely many distinct
triangles, while under the involution $n\to-n-4$ the indices $n\leq 0$
correspond to the same set of triangles repeated with reversed orientation via
the symmetry $a\leftrightarrow b$, together with a finite number of singular
values.

It would also be possible to prove Theorem 3.3 by purely algebraic means,
using (2.42) and the 2-isogeny (2.49), but this would require extensive
amounts of computer algebra, without providing much insight into the problem.
Our hope was that the analytical approach would give a better understanding of
the structure of the triangles in the main sequence, and suggest whether this
might allow the sporadic solutions to be extended to other infinite families,
perhaps by writing them in terms of different Somos sequences, as well as
possibly shedding some light on the harder problem of showing why no perfect
triangle exists. However, the detailed features of the solutions in the main
sequence appear to rely on specific arithmetical properties of the group of
rational points on the curve ${\cal C}$, and even the particular numerical
values of the coordinates of the two generators, and we have been unable to
generalize this to produce one or more other infinite families. Nevertheless,
the prime factors of the Schubert parameters and reduced lengths for the
sporadic triangles (see Tables 8 and 9) give tantalizing hints that there
might be other families that these solutions could belong to, only with a
different structure compared with the main sequence. A more extensive computer
search, and the discovery of more new solutions, would provide further
evidence in this direction, or otherwise suggest that these four solutions are
truly sporadic.

Acknowledgments: This research was supported by Fellowship EP/M004333/1 from
the Engineering & Physical Sciences Research Council, UK, and grant
IEC\R3\193024 from the Royal Society. It is a pleasure to thank Chris Athorne
and Claire Gilson for the opportunity to present a preliminary version of this
work at the joint British Mathematical/Applied Mathematics Colloquium in
Glasgow, and to thank Harry Braden for some helpful comments during the ICMS
Integrable Systems Virtual Seminar in April 2021.

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# Radon transforms with small derivatives and distance inequalities for convex
bodies

Julián Haddad  Departamento de Análisis Matemático, Universidad de Sevilla,
Sevilla, España<EMAIL_ADDRESS>and  Alexander Koldobsky  Department of
Mathematics, University of Missouri-Columbia, Columbia MO 65211, USA
<EMAIL_ADDRESS>
###### Abstract.

Generalizing the slicing inequality for functions on convex bodies from [11],
it was proved in [4] that there exists an absolute constant $c$ so that for
any $n\in\mathbb{N}$, any $q\in[0,n-1)$ which is not an odd integer, any
origin-symmetric convex body $K$ of volume one in $\mathbb{R}^{n}$ and any
infinitely smooth probability density $f$ on $K$ we have

$\max_{\xi\in S^{n-1}}\frac{1}{\cos(\pi
q/2)}\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)\geq\left(\frac{c(q+1)}{n}\right)^{\frac{q+1}{2}}.$

Here $\mathcal{R}f(\xi,t)$ is the Radon transform of $f$, and the fractional
derivative of the order $q$ is taken with respect to the variable
$t\in\mathbb{R}$ with fixed $\xi\in S^{n-1}.$ In this note we show that there
exist an origin-symmetric convex body $K$ of volume 1 in $\mathbb{R}^{n}$ and
a continuous probability density $g$ on $K$ so that

$\max_{\xi\in S^{n-1}}\frac{1}{\cos(\pi
q/2)}\mathcal{R}g(\xi,\cdot)_{t}^{(q)}(0)\leq\frac{1}{\sqrt{n}}(c(q+1))^{\frac{q+1}{2}}.$

In the case $q=0$ this was proved in [5, 6], and it was used there to obtain a
lower estimate for the maximal outer volume ratio distance from an arbitrary
origin-symmetric convex body $K$ to the class of intersection bodies. We
extend the latter result to the class $L_{-1-q}^{n}$ of bodies in
$\mathbb{R}^{n}$ that embed in $L_{-1-q}.$ Namely, for every $q\in[0,n)$ there
exists an origin-symmetric convex body $K$ in $\mathbb{R}^{n}$ so that
$d_{\operatorname{ovr}}(K,L_{-1-q}^{n})\geq cn^{\frac{1}{2(q+1)}}.$

##  1\. Introduction

The slicing problem for functions asks for the greatest constant $s_{n},$
depending only on the dimension $n,$ such that for every origin-symmetric
convex body $K$ in $\mathbb{R}^{n}$ of volume 1 and any continuous probability
density $f$ supported in $K$ there exists a hyperplane $H$ passing through the
origin for which

$\int_{H}f\geq s_{n}.$

This problem was put forward in [9, 10, 11] and further investigated in [3,
1]. The case where $f\equiv 1$ is the slicing problem of Bourgain, for which
the best lower bound so far, due to Klartag [7], is $c(\log n)^{-1/2}$.

More precisely, let us define

$s_{n}=\inf_{f,K}\max_{\xi\in S^{n-1}}\int_{K\cap\xi^{\perp}}f,$

where the infimum is taken over the class of all even continuous probability
densities defined on symmetric convex bodies $K$ of volume $1$ in
$\mathbb{R}^{n}$. Here $\xi^{\perp}=\\{x\in\mathbb{R}^{n}:(x,\xi)=0\\}$ is the
central hyperplane perpendicular to $\xi.$

It was proved in [10] that $s_{n}\geq\frac{1}{2}n^{-1/2}.$ A more general
result was proved in [11], as follows.

######  Theorem 1.

Let $K$ be an origin-symmetric convex body in $\mathbb{R}^{n}$, and let $f$ be
an even continuous non-negative function on $K$. Then

$\int_{K}f\leq
d_{\operatorname{ovr}}(K,\mathcal{I}_{n})\frac{n}{n-1}c_{n}|K|_{n}^{1/n}\max_{\xi\in
S^{n-1}}\int_{K\cap\xi^{\perp}}f,$

where $c_{n}=\frac{|B_{2}^{n}|_{n}^{\frac{n-1}{n}}}{|B_{2}^{n-1}|_{n-1}}<1$
and $d_{\operatorname{ovr}}(K,\mathcal{I}_{n})$ is the outer volume ratio
distance from $K$ to the class of intersection bodies $\mathcal{I}_{n}$ (see
Section 2 for the definition).

The outer volume ratio distance to a class of convex bodies $\mathcal{C}$ is
defined by

$d_{\operatorname{ovr}}(K,\mathcal{C})=\inf\left\\{\left(\frac{|D|_{n}}{|K|_{n}}\right)^{1/n}:K\subseteq
D,D\in\mathcal{C}\right\\}.$

Since the class of intersection bodies contains ellipsoids and $K$ is
symmetric, by John’s theorem one always has
$d_{\operatorname{ovr}}(K,\mathcal{I}_{n})\leq\sqrt{n}$. The constant
$d_{\operatorname{ovr}}(K,\mathcal{I}_{n})$ may be smaller for special classes
of bodies $K$. For example, if $K$ is an unconditional convex body then
$d_{\operatorname{ovr}}(K,\mathcal{I}_{n})\leq e$ (see [11]), and if $K$ is
the unit ball of a subspace of $L_{p}([0,1])$ then
$d_{\operatorname{ovr}}(K,\mathcal{I}_{n})\leq\sqrt{p}$ (see [16, 13]).

On the other hand, the upper bound $s_{n}\leq cn^{-1/2}$ was established in
[5] up to a logarithmic term, and the proof was refined in [6] to remove the
logarithmic factor.

######  Theorem 2.

There exists a symmetric convex body $K\subseteq\mathbb{R}^{n}$ and an even
continuous function $f:\mathbb{R}^{n}\to\mathbb{R}$ for which

$\int_{K}f\geq c,\qquad\int_{K\cap\xi^{\perp}}f\leq\frac{c}{\sqrt{n}}\text{
and }\qquad|K|_{n}\leq c^{n}$

for every $\xi\in S^{n-1}$, where $c>0$ is an absolute constant. Moreover, $f$
belongs to the Schwartz space, and we have $\sqrt{n}\mathbb{B}\subseteq K$.

It immediately follows from Theorems 1 and 2 that

$c\sqrt{n}\leq\sup_{K}d_{\operatorname{ovr}}(K,\mathcal{I}_{n})\leq\sqrt{n},$

where $c>0$ is a universal constant, and the supremum is taken over all
origin-symmetric convex bodies $K$ in $\mathbb{R}^{n}.$

By applying a similar strategy to moments of functions on convex sets, it was
shown in [1] (lower bound) and [12] (upper bound) that

$c_{1}\sqrt{\frac{n}{p}}\leq\sup_{K}d_{\operatorname{ovr}}(K,L_{p}^{n})\leq
c_{2}\sqrt{\frac{n}{p}},$

where $c_{1},c_{2}$ are absolute constants, and $L_{p}^{n}$ is the class of
unit balls of $n$-dimensional spaces that embed in $L_{p}([0,1]),\ p\geq 1$.

For a continuous function $f:\mathbb{R}^{n}\to\mathbb{R}$, the Radon transform
of $f$ is defined by

${\mathcal{R}}f(\xi,t)=\int_{\xi^{\perp}+t\xi}f,\qquad\xi\in S^{n-1},\
t\in\mathbb{R}.$

Clearly, Theorems 1 and 2 can be stated as bounds for ${\mathcal{R}}f(\xi,0)$,
and

$s_{n}=\inf_{f,K}\max_{\xi\in S^{n-1}}{\mathcal{R}}(\chi_{K}f)(\xi,0)$

where $\chi_{K}$ is the indicator function of $K$.

We denote by $\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)$ the fractional derivative
of order $q$ of the function $t\mapsto{\mathcal{R}}f(\xi,t)$ at $t=0$ (see
Section 2 for the definition).

The following Theorem is a variant of [4, Theorem 2].

######  Theorem 3.

Suppose $K\subseteq\mathbb{R}^{n}$ is an infinitely smooth origin-symmetric
convex body, $f:\mathbb{R}^{n}\to\mathbb{R}$ is a non-negative even infinitely
smooth function and $q\in(-1,n-1)$ is not an odd integer. Then

(1)
$\int_{K}f\leq\frac{n}{(n-q-1)2^{q}\pi^{\frac{q-1}{2}}\Gamma(\frac{q+1}{2})}|K|_{n}^{\frac{q+1}{n}}\\\
\times d_{\operatorname{ovr}}(K,L_{-1-q}^{n})^{q+1}\max_{\xi\in
S^{n-1}}\frac{1}{\cos(\pi q/2)}\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)$

where $L_{-1-q}^{n}$ is the class of star bodies in $\mathbb{R}^{n}$ that
embed in $L_{-1-q}$ (see Section 2).

Note that Theorem 3, as stated in [4, Theorem 2] requires $f$ to have support
inside $K$. But a quick inspection of the proof shows that this is actually
not necessary, and the proof carries over without modifications.

In this note we adapt the example of Theorem 2 to obtain a function with small
fractional derivatives of the Radon transform, thus complementing Theorem 3.

######  Theorem 4.

Let $q\in[0,n-1),$ $q$ is not an odd integer. There exists an infinitely
smooth, symmetric convex body $D\subseteq\mathbb{R}^{n}$ of volume $1$, and an
even continuous function $g:\mathbb{R}^{n}\to\mathbb{R}$ for which
$\int_{D}g=1$ and

$\max_{\xi\in S^{n-1}}\frac{1}{\cos(\pi
q/2)}\mathcal{R}g(\xi,\cdot)_{t}^{(q)}(0)\leq\frac{1}{\sqrt{n}}(c(q+1))^{\frac{q+1}{2}},\qquad\forall\xi\in
S^{n-1}.$

Note that the body $D$ does not depend on $q$.

The reader should compare Theorem 4 with [4, Formula (7)]. Namely, for every
$f,K$ under the conditions of Theorem 3,

(2) $\max_{\xi\in S^{n-1}}\frac{1}{\cos(\pi
q/2)}\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)\geq\frac{(c(q+1))^{\frac{q+1}{2}}}{(d_{\operatorname{ovr}}(K,L_{-1-q}^{n}))^{q+1}}\geq\left(\frac{c(q+1)}{n}\right)^{\frac{q+1}{2}}.$

When $q=2k-1,\ k\in\mathbb{N},$ we have

(3) $\lim_{q\to 2k-1}\frac{1}{\cos(\pi
q/2)}\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)\\\
=(-1)^{k}(2k-1)!\int_{0}^{\infty}t^{-2k}\left(Rf(\xi,t)-\sum_{j=0}^{k-1}\frac{t^{2j}}{(2j)!}(Rf(\xi,t))_{t}^{(2j)}(0)\right)dt,$

and Theorems 3 and 4 hold with this quantity.

As a consequence of Theorem 4 we obtain

######  Corollary 5.

For $q\in[0,n-1)$ there exists an origin-symmetric convex body $K$ in
$\mathbb{R}^{n}$ such that

$d_{\operatorname{ovr}}(K,L_{-1-q}^{n})\geq c_{1}n^{\frac{1}{2(q+1)}}.$

Combining this estimate with the upper estimate from [4, Propositions 2 and
3], we get

$c_{1}n^{\frac{1}{2(q+1)}}\leq\sup_{K}d_{\operatorname{ovr}}(K,L_{-1-q}^{n})\leq\min\left\\{c_{2}\sqrt{\frac{n\log^{3}(\frac{ne}{q+1})}{q+1}},\sqrt{n}\right\\},$

where $c_{1},c_{2}$ are absolute constants, and the supremum is taken over all
origin-symmetric convex bodies in $\mathbb{R}^{n}.$

##  2\. Notation and Auxiliary Results

A star body is a set of the form

$K=\\{0\\}\cup\\{x\in\mathbb{R}^{n}\setminus\\{0\\}:|x|_{2}\leq\rho_{K}(x/|x|_{2})\\}$

where $\rho_{K}:\mathcal{S}^{n-1}\to(0,\infty)$ is continuous and bounded. The
radial function $\rho_{K}$ is extended to $\mathbb{R}^{n}\setminus\\{0\\}$ as
an homogeneous function of degree $-1$, this is $\rho_{K}(\lambda
x)=\lambda^{-1}\rho_{K}(x)$. We denote the volume of $K$ in $\mathbb{R}^{n}$
by $|K|_{n}.$

The Minkowsky functional of a star body $K$ is $\|x\|_{K}=\rho_{K}(x)^{-1}$. A
convex star body $K$ is called a convex body, and $K$ is said to be infinitely
smooth if $\rho_{K}$ is an infinitely smooth function of the sphere.

The radial metric in the class of star bodies is defined by

$d_{\rho}(K,L)=\sup_{\xi\in S^{n-1}}|\rho_{K}(\xi)-\rho_{L}(\xi)|.$

We use the Fourier transform of distributions and refer the reader to [14] for
details. For $f\in\mathcal{S}$, the Schwartz space of rapidly decreasing
infinitely differentiable functions, the Fourier transform is defined by

$\hat{f}(\xi)=\int_{\mathbb{R}^{n}}f(x)e^{-i(x,\xi)}dx.$

The Fourier transform of a distribution $f$ is defined by
$\langle\hat{f},\varphi\rangle=\langle f,\hat{\varphi}\rangle$ for every
$\varphi\in\mathcal{S}$. With this normalization we have
$\hat{\hat{f}}=(2\pi)^{n}f$ for every even distribution $f$.

A star body $K$ in $\mathbb{R}^{n}$ embeds (isometrically) in $L_{p},\ p>0$ if
there exists a finite Borel measure $\mu$ on $S^{n-1}$ so that

$\|x\|_{K}^{p}=\int_{S^{n-1}}|(x,\xi)|^{p}d\mu(\xi),\qquad\forall
x\in\mathbb{R}^{n}.$

We denote by $L_{p}^{n}$ the set of all such bodies.

Following Lutwak [15], we define the intersection body of a star body $K$ as
the star body $IK$ for which $\rho_{IK}(\xi)=|K\cap\xi^{\perp}|_{n-1}$. A more
general class of bodies was introduced in [8]. Given two star bodies
$K,L\subseteq\mathbb{R}^{n}$ and an integer $k\in[1,n]$ we say that $K$ is the
$k$-intersection body of $L$ if

$|K\cap H^{\perp}|_{k}=|L\cap H|_{n-k},$

for all $n-k$ dimensional subspaces of $\mathbb{R}^{n}$. The class of
$k$-intersection bodies is defined as the closure in the radial metric of
$k$-intersection bodies of star bodies. For $k=1$ we get the class of
intersection bodies ${\mathcal{I}}_{n}.$

We say that a star body $K$ embeds in $L_{-p}$ if
$(\|\cdot\|_{K}^{-p})^{\wedge}$ is a positive distribution, i.e.
$\langle\|\cdot\|_{K}^{-p},\varphi\rangle\geq 0$ for every non-negative test
function $\varphi\in\mathcal{S}.$ By [14, Theorem 6.16], for integers
$k\in[1,n-1]$, $K$ embeds in $L_{-k}$ if and only if it is a $k$-intersection
body (note that 1-intersection body is the usual intersection body of Lutwak).
Also, every origin-symmetric convex body in $\mathbb{R}^{n}$ embeds in
$L_{-n+3};$ see [14, Corollary 4.9]. Note that the unit balls of finite
dimensional subspaces of $L_{p},\ 0<p\leq 2$ embed in $L_{-p}$ for every
$p\in(0,n).$ In particular, all ellipsoids embed in $L_{-p}.$ For details, we
refer the reader to [14, Section 6.3]. We denote the class of bodies in
$\mathbb{R}^{n}$ that embed in $L_{-p}$ by $L_{-p}^{n}.$

For a test function $\varphi:[0,\infty)\to\mathbb{R}$ the fractional
derivative of order $q\in\mathbb{C}$ at $0$, is defined as the analytic
continuation to $q\in\mathbb{C}$ of

$\varphi^{(q)}(0)=\frac{1}{\Gamma(-q)}\int_{0}^{\infty}t^{-1-q}\varphi(t)dt.$

If $\varphi$ is $m$-times continuously differentiable and integrable, then
$\varphi^{(q)}(0)$ can be extended to an analytic function at $-1<\Re(q)<m$.
Moreover, one has

(4)
$\varphi^{(q)}(0)=\frac{1}{\Gamma(-q)}\int_{0}^{1}t^{-1-q}\left(h(t)-h(0)-\cdots-h^{(m-1)}(0)\frac{t^{m-1}}{(m-1)!}\right)dt\\\
+\frac{1}{\Gamma(-q)}\int_{1}^{\infty}t^{-1-q}h(t)dt+\frac{1}{\Gamma(-q)}\sum_{k=0}^{m-1}\frac{h^{(k)(0)}}{k!(k-q)}$

for $-1<\Re(q)<m$, and the definition does not depend on the choice of $m$ as
long as $\varphi$ is $m$-times continuously differentiable.

The Fourier transform of negative powers of the Euclidean norm $|\cdot|_{2}$
in $\mathbb{R}^{n}$ are given by

(5)
$(|\cdot|_{2}^{-q})^{\wedge}(\xi)=\frac{\pi^{n/2}\Gamma{\left(\frac{n-q}{2}\right)}2^{n-q}}{\Gamma{(q/2)}}|\xi|_{2}^{q-n}$

for $q\in(0,n);$ see [2].

######  Definition 6.

For a function $f\in\mathcal{S}$ and $q\in(-n,\infty)$ the fractional
Laplacian of the order $q/2$ is defined by

$(-\Delta)^{{q}/2}f=(|\xi|^{q}\hat{f}(\xi))^{\vee},$

where $\vee$ stands for the inverse Fourier transform.

By (5), for $q\in(0,n)$ we have the representation

(6) $(-\Delta)^{{-q}/2}f=c_{n,q}|\cdot|^{-n+q}*f,$

where

$c_{n,q}=\frac{\pi^{n/2}2^{n-q}\Gamma\left(\frac{n-q}{2}\right)}{(2\pi)^{n}\Gamma\left(\frac{q}{2}\right)}.$

######  Lemma 7.

For $q\in(-1,\infty)$ not an odd integer, the fractional Laplacian and the
Radon transform are related by

$\frac{1}{\cos(\pi
q/2)}\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)={\mathcal{R}}((-\Delta)^{{q}/2}f)(\xi,0)$

###### Proof.

By Lemma 2.23 in [14], if $q\in\mathbb{C}\setminus\\{-1,-3,-5,\ldots\\}$ and
$t,\xi\in\mathbb{R}$,

$(|t|^{-1-q})^{\wedge}(\xi)=2\Gamma(-q)\cos(\pi q/2)|\xi|^{q}$

in the sense of analytic continuation of distributions.

We use the well-known relation between the Radon and Fourier transforms; see
[14, Lemma 2.11]. For every fixed $\xi\in S^{n-1}$ and any Schwartz test
function $f$, the Fourier transform of the function $t\to\mathcal{R}f(\xi,t)$
is the function $z\to\hat{f}(z\xi)$. Now, for $q\in(-1,0)$, we have

(7) $\displaystyle\frac{1}{\cos(\pi q/2)}\mathcal{R}f(\xi,\cdot)_{t}^{(q)}(0)$
$\displaystyle=\frac{1}{\cos(\pi
q/2)}\int_{-\infty}^{\infty}\frac{1}{2\Gamma(-q)}|t|^{-1-q}{\mathcal{R}}f(\xi,t)dt$
(8)
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}|z|^{q}{\mathcal{R}}f(\xi,\cdot)_{t}^{\wedge}(z)dz$
(9)
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}|z\xi|_{2}^{q}\hat{f}(z\xi)dz$
(10)
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}((-\Delta)^{{q}/2}f)^{\wedge}(z\xi)dz$
(11)
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\mathcal{R}}{((-\Delta)^{{q}/2}f)}(\xi,\cdot)_{z}^{\wedge}(z)dz$
(12) $\displaystyle={\mathcal{R}}{((-\Delta)^{{q}/2}f)}(\xi,0).$

By analytic continuation with respect to $q$ we obtain the equality for every
$q\in(-1,\infty),$ not an odd integer. ∎

For the proof of the main results we need two lemmas.

######  Lemma 8.

For positive measurable functions $f,g$, and measurable sets $A,B$,

$\int_{A+B}f*g(x)dx\geq\int_{A}f(x)dx\int_{B}g(x)dx$

###### Proof.

Since the support of $(f\chi_{A})*(g\chi_{B})$ is contained in $A+B$, we have

(13) $\displaystyle\int_{A+B}f*g(x)dx$
$\displaystyle\geq\int_{A+B}(f\chi_{A})*(g\chi_{B})(x)dx$ (14)
$\displaystyle\geq\int_{\mathbb{R}^{n}}(f\chi_{A})*(g\chi_{B})(x)dx$ (15)
$\displaystyle=\int_{A}f(x)dx\int_{B}g(x)dx$

∎

######  Lemma 9.

For $0\leq\mu\leq\lambda$ we have

$\lambda^{\mu}\Gamma(\lambda-\mu)\geq\Gamma(\lambda).$

###### Proof.

The equality holds for $\mu=0$. Since the left-hand side is log-convex with
respect to $\mu$, it suffices to show that its logarithmic derivative at
$\mu=0$ is non-negative. By Jensen’s inequality,

(17)
$\displaystyle\left.\frac{\partial}{\partial\mu}\left[\mu\log(\lambda)+\log\Gamma(\lambda-\mu)\right]\right|_{\mu=0}$
$\displaystyle=\log(\lambda)-\frac{\Gamma^{\prime}(\lambda)}{\Gamma(\lambda)}$
(18)
$\displaystyle=\frac{1}{\Gamma(\lambda)}\left(\int_{0}^{\infty}-\log(t/\lambda)t^{\lambda-1}e^{-t}dt\right)$
(19)
$\displaystyle\geq-\log\left(\frac{1}{\lambda\Gamma(\lambda)}\int_{0}^{\infty}t^{\lambda}e^{-t}dt\right)$
(20) $\displaystyle=0$

where we used that $\Gamma(\lambda+1)=\lambda\Gamma(\lambda)$. ∎

##  3\. Proof of the Main Results

###### Proof of Theorem 4.

Consider $K$ and $f$ from Theorem 2, and take $h=(-\Delta)^{{-q}/2}f$. By
Lemma 8 and equation (6),

(21) $\displaystyle\int_{2K}h$ $\displaystyle\geq
c_{n,q}\int_{\sqrt{n}\mathbb{B}}|x|_{2}^{-n+q}dx\int_{K}fdx$ (22)
$\displaystyle\geq c_{n,q}n\omega_{n}\frac{n^{q/2}}{q}c=cd_{n,q}.$

where
$d_{n,q}=\frac{2^{-q}n^{q/2}\Gamma\left(\frac{n-q}{2}\right)}{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{q}{2}+1\right)}$.

By Lemma 7 and Theorem 2,

(23) $\displaystyle\frac{1}{\cos(\pi
q/2)}\mathcal{R}h(\xi,\cdot)_{t}^{(q)}(0)$ $\displaystyle=\frac{1}{\cos(\pi
q/2)}\mathcal{R}((-\Delta)^{{-q}/2}f)(\xi,\cdot)_{t}^{(q)}(0)$ (24)
$\displaystyle={\mathcal{R}}f(\xi,0)\leq\frac{c}{\sqrt{n}}.$

We consider $D=|K|_{n}^{-1/n}K$ and

$g(x)=\left(\int_{D}h(|2K|_{n}^{1/n}x)dx\right)^{-1}h(|2K|_{n}^{1/n}x),$

so that $\int_{D}g=1$ and $D$ has volume $1$. Combining (21), (23) and
applying Lemma 9, the inequality $|K|_{n}\leq c^{n}$ and Stirling’s formula

(25) $\displaystyle\frac{1}{\cos(\pi
q/2)}\mathcal{R}g(\xi,\cdot)_{t}^{(q)}(0)$
$\displaystyle=c|2K|_{n}^{\frac{1+q}{n}}d_{n,q}^{-1}\frac{1}{\cos(\pi
q/2)}\mathcal{R}h(\xi,\cdot)_{t}^{(q)}(0)$ (26) $\displaystyle\leq
c^{q}\frac{1}{\sqrt{n}}\frac{2^{q}\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{q}{2}+1\right)}{n^{q/2}\Gamma\left(\frac{n-q}{2}\right)}$
(27) $\displaystyle\leq
c^{q}\frac{1}{\sqrt{n}}\Gamma\left(\frac{q}{2}+1\right)$ (28)
$\displaystyle\leq c^{q}\frac{1}{\sqrt{n}}(q+1)^{\frac{q+1}{2}}.$

∎

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