text:
\begin{document}

\title{Torsion groups of elliptic curves over quadratic fields $\mathbb{Q}(\sqrt{d}),$ $0<d<100$}

\author{Antonela Trbovi\'c}
\address{Department of Mathematics, University of Zagreb, Bijeni\v{c}ka
cesta 30, 10000 Zagreb, Croatia}
\email{antonela.trbovic@gmail.com}

\begin{abstract}
We prove results towards classifying the possible torsion
subgroups of elliptic curves over quadratic fields $\mathbb{Q}(\sqrt{d})$, where $0<d<100$ is a square-free integer, and obtain a complete classification for 49
out of 60 such fields. Over the remaining
11 quadratic fields, we cannot rule out the possibility of the group $\mathbb{Z}/16\mathbb{Z}$
appearing as a torsion group of an elliptic curve.
\end{abstract}

\subjclass[2000]{Primary 11G05}

\keywords{torsion groups, elliptic curves, quadratic fields.}

\thanks{The author was supported by the QuantiXLie Centre of Excellence, a project co-financed by the Croatian
	Government and European Union through the European Regional Development Fund - the Competitiveness and
	Cohesion Operational Programme (Grant KK.01.1.1.01.0004).
}

\maketitle

\section{Introduction}

For an elliptic curve $E$ defined over a number field $K,$ let $E(K)$ be the set of all $K-$rational points on $E.$ By the Mordell-Weil theorem and the structure theorem for finitely generated abelian groups, we know that $E(K)$ is isomorphic to $E(K)_{tors} \oplus \mathbb{Z}^r,$ where $r$ is a non-negative integer and $E(K)_{tors}$ is the torsion subgroup.

In the case of $K=\mathbb{Q},$ Mazur's theorem \cite{BM} gives us all possible torsion subgroups of $E(\mathbb{Q})$:
$$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=1,...,10,12,$$
$$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=1,2,3,4.$$

We also have a similar result by Kamienny, Kenku and Momose \cite{SK,MAKFM} concerning possible torsion subgroups of elliptic curves defined over any quadratic field, which are the following 26 groups:
$$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=1,...,16,18,$$
$$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=1,...,6,$$
$$\mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/3n\mathbb{Z}, \:\:\: n=1,2,$$
$$\mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}.$$
While this theorem settles the question on what are the possibilities for the torsion subgroup over all quadratic fields, we are interested in what happens when we fix a quadratic field. In order to see what happens over a fixed field, one would have to go through each of the 26 groups mentioned above and check whether that is a possible torsion subgroup or not.

We are going to see why every group mentioned in Mazur's theorem has to appear as a possible torsion subgroup over all quadratic fields, and what happens with the groups $\mathbb{Z}/3\mathbb{Z}\oplus \mathbb{Z}/3n\mathbb{Z}, \: n=1,2, \: \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}.$
For the rest of the groups we will follow the methods described in \cite{SKFN}.

From now on, let $K$ be a fixed quadratic field.
Let $Y_1(m,n)$ be the affine modular curve whose every $K-$rational point corresponds to an isomorphism class of an elliptic curve together with an $m-$torsion point $P_m\in E(K)$ and an $n-$ torsion point $P_n\in E(K)$ such that $P_m$ and $P_n$ generate a subgroup isomorphic to $\mathbb{Z}/m\mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z},$ and let $X_1(m,n)$ be its compactification (the same curve with adjoined cusps). We denote $X_1(1,n)$ by $X_1(n).$

More precisely, what we need to do in order to determine whether an elliptic curve with torsion $\mathbb{Z}/n\mathbb{Z} \oplus  \mathbb{Z}/m\mathbb{Z}$ over $K$ exists, for the rest of the 26 groups, is to determine whether there are $K-$rational points on $X_1(m,n)$ that are not cusps. These modular curves are either elliptic or hyperelliptic.

If the modular curve $X_1(m,n)$ is elliptic, we compute its rank. If the rank is positive, there are infinitely many elliptic curves over $ K $ with the given torsion subgroup, as the number of cusps is finite. If the rank is 0, we have to compute the torsion subgroup and check whether any torsion point corresponds to a $ K- $rational point on the modular curve that is not a cusp.

If the modular curve $X_1(m,n)$ is hyperelliptic, we compute the rank of the Jacobian of the curve. If the rank is 0, we also have to check whether any torsion point arises from a $ K- $rational point on the modular curve that is not a cusp. If the rank is positive, the problem becomes more difficult. More about this can be found in \cite{SKFN}.

\section{Groups from Mazur's theorem and \\ $\mathbb{Z}/3\mathbb{Z}\oplus \mathbb{Z}/3n\mathbb{Z}, \: n=1,2, \: \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$}
\label{sec:two}

In this section, we are going to show that every group mentioned in Mazur's theorem has to appear as a possible torsion group over any quadratic field $K.$

Let $E$ be an elliptic curve and denote by $\rho_{E,n}:\Gal(K/\mathbb{Q}) \rightarrow \GL_2(\mathbb{Z}/n \mathbb{Z})$ the $mod \: n$ Galois representation attached to $E.$

If ${P, P'}$ is a basis for $E[n],$ the subgroup of $E$ of points of order $n,$ and if $P$ is a point of order $n$ in $E(\mathbb{Q}),$ then $\rho_{E,n}(\sigma)=\begin{bmatrix}
    1       & b \\
    0       & d
\end{bmatrix}, \: b, \in\mathbb{Z}/n\mathbb{Z}, d \in (\mathbb{Z}/n\mathbb{Z})^{\times}$  for every $\sigma \in \Gal(K/\mathbb{Q}),$ with respect to the basis $\{P,P'\}.$

We define a subgroup of $\GL_2(\mathbb{Z}/n\mathbb{Z}),$
$$\Gamma_1(n)=\left\{\begin{bmatrix}
1 & b \\ 0 & d
\end{bmatrix}: b \in \mathbb{Z}/n\mathbb{Z}, d \in
(\mathbb{Z}/n\mathbb{Z})^{\times}\right\},$$
which corresponds to $ X_1(n), $ i.e. the $mod \:n $ representations of elliptic curves parameterized by the points on $ X_1(n) $ are elements in $ G(n), $ with an appropriate choice of basis. Similarly, we define a subgroup
 $$\Gamma_1(2,2n)=\left\{\begin{bmatrix}
1 & b \\ 0 & d
\end{bmatrix}: b \in 2\mathbb{Z}/2n\mathbb{Z}, d \in (\mathbb{Z}/2n\mathbb{Z})^{\times}\right\},$$ which corresponds to $ X_1(2,2n), $ in the sense described above.

For any of the groups $\mathbb{Z}/n\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}$ appearing in Mazur's theorem, we
have that the corresponding modular curve $X_1(n)$ or $X_1(2,2n)$,
respectively, is of genus 0. Now, with $ X_G=X_1(n) $ or $ X_G=X_1(2,2n) $ in \cite[Lemma 3.5]{DJZ}, using the same arguments as in the proof
of  the lemma, but taking the base field to be a quadratic field $ K $
instead of $\mathbb{Q}$, we have that there are infinitely many elliptic curves $E$ over $K$
such that $\rho_{E,n}(\Gal (K/\mathbb{Q}))$ is conjugate (not just contained) in $\GL_2(\mathbb{Z}/n\mathbb{Z})$
to $\Gamma_1(n)$ or $\Gamma_1(2,2n)$, respectively, proving our claim.

Now, we will focus on the groups
$$\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/3n\mathbb{Z}, \:\: n=1,2,$$
$$\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}.$$
From the properties of the Weil pairing, we know that
$\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \subset E(K)$ only if $\mathbb{Q}(\zeta_n) \subset K.$

Hence, $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3n\mathbb{Z} \subset E(K),$ $n=1,2,$ only when $K\supset\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z} \subset E(K)$ only if $K \supset \mathbb{Q}(i).$ Moreover, the mentioned groups are the only groups, except the ones from the Mazur's theorem, that appear over $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(i),$ respectively \cite{FN1,FN2}.

\section{Torsion over $\mathbb{Q}(\sqrt{17})$}

We will now demonstrate how to carry out the mentioned methods over the quadratic field $\mathbb{Q}(\sqrt{17}).$ We chose $\mathbb{Q}(\sqrt{17})$ because $17 \equiv 1 \pmod{8},$ and the significance of this relation will be clear later on.

As stated above, all groups from Mazur's theorem are possible torsion subgroups over $\mathbb{Q}(\sqrt{17}),$ and the groups $\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/3n\mathbb{Z}, \: n=1,2,$ $\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/4\mathbb{Z}$ are not.

For the rest of the groups we will follow the methods from \cite{SKFN} described above.

The equations for $X_1(m,n)$ can be found in \cite{HB,FPR}.

All computations in the following propositions will be done in Magma \cite{MAGMA}. Computations for this paper can be found  \href{https://web.math.pmf.unizg.hr/~atrbovi/magma.txt}{here}.

\begin{prop}\label{prop1}
There are infinitely many elliptic curves with torsion $\mathbb{Z}/11\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$
\end{prop}
\begin{proof}
To show this, we have to prove that the modular curve $X_1(11)$ defined over $\mathbb{Q}(\sqrt{17})$ has infinitely many points. It will suffice to see that the rank is positive, since the number of cusps on $X_1(11)$ is finite. For the modular curve $$X_1(11): \:\: y^2-y=x^3-x^2,$$ we compute $$rank(X_1(11)(\mathbb{Q}(\sqrt{17})))=1$$ in Magma. Now we can conclude that there are infinitely many elliptic curves with torsion $\mathbb{Z}/11\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$

We can also compute a generator of the group $X_1(11)(\mathbb{Q}(\sqrt{17}))$ (modulo the torsion subgroup), which is $$\left(\dfrac{1}{8}(-\sqrt{17}+1),\dfrac{1}{16}(\sqrt{17}+7)\right).$$
\end{proof}

\begin{prop} There are infinitely many elliptic curves with torsion $\mathbb{Z}/14\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}
For the modular curve $$X_1(14): \: \: y^2+xy+y=x^3-x$$ we compute $$rank(X_1(14)(\mathbb{Q}(\sqrt{17})))=1.$$ Using similar reasoning to the one in Proposition $\ref{prop1}$ we conclude that there are infinitely many elliptic curves with torsion $\mathbb{Z}/14\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$

Also, one point of infinite order is given by $$\left(\dfrac{1}{2}(\sqrt{17}+3),-\sqrt{17}-5\right).$$
\end{proof}

\begin{prop}
$\mathbb{Z}/15\mathbb{Z}$ cannot be a torsion group of an elliptic curve over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}
For the modular curve $$X_1(15): \:\: y^2+xy+y=x^3+x^2,$$ we compute $$rank(X_1(15)(\mathbb{Q}(\sqrt{17})))=0.$$
Hence, we only have to show that $$Y_1(15)(\mathbb{Q}(\sqrt{17}))=\emptyset,$$ i.e. that there are only cusps in $X_1(15)(\mathbb{Q}(\sqrt{17})).$

The $x-$coordinates of the cusps on $X_1(15)$ satisfy the equation $$x(x+1)(x^4+3x^3+4x^2+2x+1)(x^4-7x^3-6x^2+2x+1)=0.$$
So, the set of all cusps is $$X_1(15)(\mathbb{Q}(\sqrt{17})) \backslash Y_1(15)(\mathbb{Q}(\sqrt{17}))=\{ O, (0,0), (0,-1), (-1,0) \}.$$
We compute $$X_1(15)(\mathbb{Q}(\sqrt{17})) \cong \mathbb{Z}/4\mathbb{Z},$$ so now it is obvious that all points on the modular curve $X_1(15)$ defined over $\mathbb{Q}(\sqrt{17})$ are cusps. Therefore, $Y_1(15)(\mathbb{Q}(\sqrt{17}))=\emptyset,$ so there are no elliptic curves with torsion $\mathbb{Z}/15\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$
\end{proof}

\begin{prop} There are infinitely many elliptic curves with torsion $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/10\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}
For the modular curve $$X_1(2,10): \:\: y^2=x^3+x^2-x,$$ we compute $$rank(X_1(2,10)(\mathbb{Q}(\sqrt{17})))=1.$$
One point of infinite order is $$\left(\sqrt{17}+4,3\sqrt{17}+12\right).$$
\end{proof}

\begin{prop} There are infinitely many elliptic curves with torsion $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/12\mathbb{Z}$ over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}
For the modular curve $$X_1(2,12): \:\: y^2=x^3-x^2+x,$$ we compute $$rank(X_1(2,12)(\mathbb{Q}(\sqrt{17})))=1.$$
A point of infinite order is $$\left(\dfrac{1}{2}(-\sqrt{17}+9),\dfrac{1}{2}(-3\sqrt{17}+19)\right).$$
\end{proof}

Now we have determined whether $\mathbb{Z}/n\mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$ is a possible torsion of an elliptic curve over $\mathbb{Q}(\sqrt{17}),$ for all modular functions $X_1(m,n)$ that are elliptic curves. To determine if the $\mathbb{Z}/n\mathbb{Z}, \:\: n=13, 16, 18,$ are possible torsion groups is somewhat more difficult, since corresponding modular curves are hyperelliptic curves.

The groups $\mathbb{Z}/n\mathbb{Z}, \:\: n=13, 18,$ are generally easier to deal with over quadratic fields, since we have the two following results that can be found in \cite{BBDN, KRUMM}.

\begin{thm} \label{tm1}
If $X_1(13)$ has a point defined over $\mathbb{Q}(\sqrt{d}),$ then:
\begin{enumerate}
   \item $d>0,$
   \item $d\equiv 1 \pmod{8}.$
\end{enumerate}
\end{thm}

\begin{thm} \label{tm2}
If $X_1(18)$ has a point defined over $\mathbb{Q}(\sqrt{d}),$ $d\neq -3,$ then:
\begin{enumerate}
   \item $d>0,$
   \item $d\equiv 1 \pmod{8},$
   \item $d\not\equiv 2 \pmod{3}.$
\end{enumerate}
\end{thm}

Now it becomes clear why we chose the field $\mathbb{Q}(\sqrt{17}),$ as we did not want to rule out the existence of the groups $\mathbb{Z}/n\mathbb{Z}, \:\: n=13, 18,$ as possible torsion subgroups.

\begin{prop} \label{tm3}
$\mathbb{Z}/13\mathbb{Z}$ is a possible torsion over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}
Let $J_1(13)$ be the Jacobian of the hyperelliptic curve
$$X_1(13): \: \: y^2=x^6-2x^5+x^4-2x^3+6x^2-4x+1$$ and let  $J_1^{17}(13)$ be its quadratic twist by 17, which becomes isomorphic to $J_1(13)$ over $\mathbb{Q}(\sqrt{17}).$
We compute $$rank(J_1(13)(\mathbb{Q}))=0,$$ $$rank(J_1^{17}(13)(\mathbb{Q}))=2.$$ Now, we have $$
rank(J_1(13)(\mathbb{Q}(\sqrt{17})))=rank(J_1(13)(\mathbb{Q}))+rank(J_1^{17}(13)(\mathbb{Q}))=2.$$

By searching for points on $X_1(13)(\mathbb{Q}(\sqrt{17}))$ in Magma, we find a point $\left(\dfrac{1}{2}, \dfrac{1}{8}\sqrt{17}\right)$ on the curve.

Since the $x-$coordinates of the cusps on the $X_1(13)$ are the solutions of the equation $$x(x-1)(x^3-4x^2+x+1)=0,$$
the cusps are $$X_1(13)(\mathbb{Q}(\sqrt{17})) \backslash Y_1(13)(\mathbb{Q}(\sqrt{17}))=\{\infty_+, \infty_-,(0,\pm 1),(1,\pm 1)\}.$$ We conclude that the point mentioned above is not a cusp and so the elliptic curve over $\mathbb{Q}(\sqrt{17})$ with a torsion subgroup $\mathbb{Z}/13\mathbb{Z}$ exists.
\end{proof}

Unlike in previous propositions, we do not have infinitely many elliptic curves with torsion $\mathbb{Z}/13\mathbb{Z},$ since by Falting's theorem the modular curve $X_1(13)$ can only have finitely many points over a number field.

\begin{prop}
$\mathbb{Z}/16\mathbb{Z}$ cannot be a torsion group of an elliptic curve over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}
Let $J_1(16)$ be the Jacobian of the hyperelliptic curve $$X_1(16): \: \: y^2 = x(x^2+1)(x^2+2x-1)$$ and let  $J_1^{17}(16)$ be its quadratic twist.

We compute  $$rank(J_1(16)(\mathbb{Q}(\sqrt{17})))=rank(J_1(16)(\mathbb{Q}))+rank(J_1^{17}(16)(\mathbb{Q}))=0.$$

Since the rank is zero, we need to find the cusps in $X_1(16)(\mathbb{Q}(\sqrt{17}))$ and the torsion subgroup of $J_1(16)(\mathbb{Q}(\sqrt{17}))$ in order to determine if there exists a point on the Jacobian that arises from a point on the modular curve that is not a cusp.

As the $x-$coordinates of the cusps satisfy $$x(x -1)(x + 1)(x^2- 2x - 1)(x^2 + 2x-1)=0,$$ the cusps on $X_1(16)(\mathbb{Q}(\sqrt{17}))$ are $$X_1(16)(\mathbb{Q}(\sqrt{17})) \backslash Y_1(16)(\mathbb{Q}(\sqrt{17}))=\{\infty,(0,0),(1,\pm 2),(-1,\pm 2)\}.$$

We also compute $$J_1(16)(\mathbb{Q})_{tors} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/10\mathbb{Z},$$
$$J_1^{17}(16)(\mathbb{Q})_{tors} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.$$

The set of points of odd order on the Jacobian $J_1(16)$ defined over $\mathbb{Q}(\sqrt{17})$ is $$J_1(16)(\mathbb{Q}(\sqrt{17}))_{(2')}\cong J_1(16)(\mathbb{Q})_{(2')} \oplus J_1^{17}(16)(\mathbb{Q})_{(2')}\cong \mathbb{Z}/5\mathbb{Z},$$

and the $2-$torsion subgroup is $$J_1(16)(\mathbb{Q}(\sqrt{17}))_{tors} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.$$

Thus, $J_1(16)(\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/10\mathbb{Z}$ and $J_1(16)(\mathbb{Q}) \cong J_1(16)(\mathbb{Q})_{tors}.$

In Magma, we find 20 divisor classes in Mumford representation \cite{CF},
$$(1, 0, 0), \:\:(x^2 + 2x + 1, 2x, 2), \:\:(x^2 + 2x + 1, -2x, 2),\:\: (x^2 - 2x + 1,
4x - 2, 2), $$ $$(x^2 - 2x + 1, -4x + 2, 2),\:\: (x + 1, 2, 1),\:\: (x + 1, -2, 1),\:\: (x, 0,
1), $$ $$ (x - 1, 2, 1),\:\: (x - 1, -2, 1),\:\: (x^2 + 2x - 1, 0, 2), \:\:(x^2 + x, 2x, 2),\:\:
(x^2 + x, -2x, 2), $$ $$(x^2 - 1, 2x, 2),\:\: (x^2 - 1, -2x, 2), \:\:(x^2 - 1, 2, 2),\:\: (x^2
    - 1, -2, 2), $$ $$(x^2 + 1, 0, 2), \:\:(x^2 - x, 2x, 2),\:\: (x^2 - x, -2x, 2).$$

The first divisor class represents the point at infinity, and for the rest of the divisor classes in Mumford representation, we follow the methods described in \cite{MAGMA} in order to retrieve the point on the Jacobian from its Mumford representation.

For a triple $(a(x),b(x),d)$ in Mumford representation we define $A(x,z)$ as the homogenisation of the polynomial $a(x)$ of degree $d$ and $B(x,z)$ as the homogenisation of the polynomial $b(x)$ of degree $g+1,$ where $g$ is a genus of $X_1(16),$ in this case $g=2$.

Now, by solving the equations
$$ A(x,z)=0, y=B(x,z), $$
we get the points $P_i=(x_i:y_i:1), \: i=1,...,d,$ in projective coordinates. Note that the number of points is exactly $d,$ as the polynomial $A(x,z)$ is of degree $d$.

The point on the Jacobian represented by $(a(x),b(x),d)$ is then the divisor class $$\left[P_1+...+P_d-d\infty\right],$$ if there is a single point $\infty$ at infinity, or  $$\left[P_1+...+P_d-\dfrac{d}{2}(\infty_+ + \infty_-)\right],$$ if there are two points $ \infty_+$ and $\infty_-$ at infinity.

For example, for the point $(x^2+2x+1,2x,2)$ on $J_1(16)(\mathbb{Q}(\sqrt{17}))$ in Mumford representation we have
$$A(x,z)=x^2+2xz+z^2,$$
$$B(x,z)=2xz^2,$$
and $P_1=P_2=(-1:-2:1),$ so we conclude that the point $(x^2+2x+1,2x,2)$ represents the divisor class $\left[(-1:-2:1)+(-1:-2:1)-2\infty\right]$ on the Jacobian $J_1(16)(\mathbb{Q}(\sqrt{17})).$

By doing so for every divisor class in Mumford representation, one can check that all divisor points correspond to the cusps in $X_1(16)(\mathbb{Q}(\sqrt{17})),$ so we conclude that $\mathbb{Z}/16\mathbb{Z}$ cannot be a torsion group of an elliptic curve over $\mathbb{Q}(\sqrt{17}).$
\end{proof}

\begin{prop} \label{tm4}
$\mathbb{Z}/18\mathbb{Z}$ cannot be a torsion group of an elliptic curve over $\mathbb{Q}(\sqrt{17}).$
\end{prop}

\begin{proof}

Let $J_1(18)$ be the Jacobian of the hyperelliptic curve $$X_1(18): \: \: y^2=x^6+2x^5+5x^4+10x^3+10x^2+4x+1$$ and let  $J_1^{17}(18)$ be its quadratic twist. We compute  $$
rank(J_1(18)(\mathbb{Q}(\sqrt{17})))=rank(J_1(18)(\mathbb{Q}))+rank(J_1^{17}(18)(\mathbb{Q}))=0.$$

The $x-$coordinates of the cusps in $X_1(18)$ satisfy the equation $$x(x+1)(x^2+x+1)(x^2-3x-1)=0,$$ so the cusps are $$X_1(18)(\mathbb{Q}(\sqrt{17})) \backslash Y_1(18)(\mathbb{Q}(\sqrt{17}))=\{\infty_+, \infty_-,(0,\pm 1),(-1,\pm 1)\}.$$

On the other hand, we compute $$J_1(18)(\mathbb{Q})_{tors} \cong \mathbb{Z}/21\mathbb{Z},$$
$$J_1^{17}(18)(\mathbb{Q})_{tors} \cong \{ O \}.$$ The set of points of odd order is $$J_1(18)(\mathbb{Q}(\sqrt{17}))_{(2')}\cong J_1(18)(\mathbb{Q})_{(2')} \oplus J_1^{17}(18)(\mathbb{Q})_{(2')}\cong \mathbb{Z}/21\mathbb{Z}.$$
As the polynomial $$f(x)=x^6+2x^5+5x^4+10x^3+10x^2+4x+1$$ has no zeros defined over $\mathbb{Q}(\sqrt{17}),$ we conclude that $J_1(18)(\mathbb{Q}(\sqrt{17}))$ has no points of order 2, and $$J_1(18)(\mathbb{Q}(\sqrt{17}))_{tors} \cong \mathbb{Z}/21\mathbb{Z}.$$
The elements of $J_1(18)(\mathbb{Q}(\sqrt{17}))_{tors}$ in Mumford representation are

$$(1, 0, 0), \:\: (1, x^3 + x^2, 2), \:\: (1, -x^3 - x^2, 2),\:\:    	(x^2 + 2x + 1, x, 2),\:\: (x^2
    + 2x + 1, -x, 2),$$
$$(x^2, 2x + 1, 2), \:\:(x^2, -2x - 1, 2),\:\: (x + 1, x^3, 2),\:\:
	(x + 1, -x^3, 2),$$ $$ (x + 1, x^3 + 2, 2),\:\:  (x + 1, -x^3 - 2, 2),\:\:
	(x, x^3 - 1, 2),\:\: (x, -x^3 + 1, 2),\:\:(x, x^3 + 1, 2),$$
$$(x, -x^3 - 1, 2), \:\:(x^2 + x, 2x + 1, 2),\:\:
	(x^2
    + x, -2x - 1, 2)),\:\: (x^2 + x, 1, 2),$$ $$ (x^2 + x, -1, 2), \:\: (x^2 + x + 1, x - 1,
2), \:\: (x^2 + x + 1, -x + 1, 2), $$
	and one can easily conclude that all of the points correspond to the cusps, so we obtain our result.
\end{proof}

We proved the following theorem:

\begin{thm}\label{tm} The possible torsion subgroups of elliptic curves defined over $\mathbb{Q}(\sqrt{17})$ are the following:
$$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=1,...,14,$$
$$\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=1,...,6.$$
\end{thm}

\begin{remark}
	In \cite[Theorem 2.7.7, 2.7.8]{KRUMM} Krumm found a list of possible quadratic fields $\mathbb{Q}(\sqrt{d})$ over which torsion subgroups $\mathbb{Z}/13\mathbb{Z}$ and $\mathbb{Z}/18\mathbb{Z}$ may appear, for $0<d<1000.$

	In our case, for $0<d<100,$ there are only two such fields, $\mathbb{Q}(\sqrt{17}),$ over which $\mathbb{Z}/13\mathbb{Z}$ appears, and $\mathbb{Q}(\sqrt{33}),$ over which $\mathbb{Z}/18\mathbb{Z}$ appears.

	In this paper we were able to eliminate the rest of the fields for $0<d<100$ using only conditions from \ref{tm1} and \ref{tm2} and methods described in \ref{tm3} and \ref{tm4}.
\end{remark}

\section{Torsion over $\mathbb{Q}(\sqrt{d})$, $0<\lowercase{d}<100$}

For every quadratic field $\mathbb{Q}(\sqrt{d}),$ where $d$ is a non-negative squarefree integer $d\neq 1,$ $0<d<100,$ we found the torsion subgroups appearing by methods similar to the ones described in Theorem \ref{tm}.

Since we know that the groups from the Mazur's theorem appear as torsion subgroups, and the groups $\mathbb{Z}/3\mathbb{Z}\oplus \mathbb{Z}/3n\mathbb{Z}, \: n=1,2, \: \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}$ do not appear as torsion subgroups of elliptic curves over mentioned fields, we give a list of possible torsion subgroups within the groups $$\mathbb{Z}/n\mathbb{Z}, \:\: n=11,13,14, 15, 16, 18, $$ $$\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2n\mathbb{Z}, n=5,6, \:\: $$ i.e. the rest of the possible 26 torsion subgroups over quadratic fields.

Note that the groups mentioned above appear as a torsion subgroup 37, 1, 38, 37, 6-17, 1, 35, 36 times, respectively, over all $\mathbb{Q}(\sqrt{d})$, where $0<d<100.$

We were unable to determine how many times exactly does the group $ \mathbb{Z}/16\mathbb{Z}$ appear as a torsion subgroup because of the following problem: computing (in Magma) the rank of the Jacobian of the modular curve $X_1(16)$ defined over the problematic quadratic fields listed in the table did not give a result, only the lower and the upper bound that were not the same. That was a problem since it is important to know the mentioned rank in order to know which method to use for determining whether that is a possible torsion subgroup or not.

Also, searching for the points on the modular curve $X_1(16)$ over the same fields did not yield a result.

\begin{table}[]
\centering
\caption{List of all possible torsion groups over quadratic fields $\mathbb{Q}(\sqrt{d}),$ for $0<d<100,$ without the groups from the Mazur's theorem}
\label{my-label}
\begin{tabular}{ll}
\hline
\multicolumn{1}{|l|}{Fields}                  & \multicolumn{1}{l|}{Possible torsion subgroups over a given field$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$}                                                                                \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{2}) \:\:\:\:\:\:\:\:\:\:\:\:$}  & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\end{tabular}}                         \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{3})$}  & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}}                 \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{5})$}  & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=15$\end{tabular}}                 \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{6})$}  & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}}                         \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{7})$}  & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}}                     \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{10})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15, 16$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}}                 \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{11})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15$\end{tabular}}                         \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{13})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}}                         \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{14})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}}                 \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{15})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=15, 16$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}}             \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{17})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11,13,14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}}                             \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{19})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}}                         \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{21})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}}                         \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{22})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}}                     \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{23})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}} $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}}                     \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{26})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$
\end{tabular}} \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{29})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}}                     \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{30})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}}                     \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{31})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, \text{ maybe } \mathbb{Z}/16\mathbb{Z} $\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$
\end{tabular}}     \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{33})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15, 18$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}}                 \\ \hline
\multicolumn{1}{|l|}{$\mathbb{Q}(\sqrt{34})$} & \multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}}                 \\ \hline

\end{tabular}
\end{table}

\begin{table}[]
\centering
\label{my-label}
\begin{tabular}{|l|l|}
\hline
Fields                  & Possible torsion subgroups over a given field$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$                                                                                \\ \hline
$\mathbb{Q}(\sqrt{35})\:\:\:\:\:\:\:\:\:\:\:\:$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{37})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{38})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{39})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{41})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15, 16$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{42})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{43})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{46})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{47})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$
\end{tabular}     \\ \hline
$\mathbb{Q}(\sqrt{51})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 16$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{53})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{55})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{57})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{58})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11,15, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{59})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{61})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{62})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$ \end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{65})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{66})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{67})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=15$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{69})$ & \begin{tabular}[c]{@{}l@{}} $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{70})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15, 16$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}
\\ \hline
$\mathbb{Q}(\sqrt{71})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}         \\ \hline

\end{tabular}
\end{table}

\begin{table}[]
\centering
\label{my-label}
\begin{tabular}{|l|l|}
\hline
Fields                  & Possible torsion subgroups over a given field$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$                                                                                \\ \hline
$\mathbb{Q}(\sqrt{73})\:\:\:\:\:\:\:\:\:\:\:\:$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{74})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$
\end{tabular}     \\ \hline
$\mathbb{Q}(\sqrt{77})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{78})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=15, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular} \\ \hline
$\mathbb{Q}(\sqrt{79})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\  $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$ \end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{82})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$ \end{tabular} \\ \hline
$\mathbb{Q}(\sqrt{83})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{85})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=6$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{86})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{87})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$ \end{tabular}     \\ \hline
$\mathbb{Q}(\sqrt{89})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{91})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}             \\ \hline
$\mathbb{Q}(\sqrt{93})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15, 16$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}     \\ \hline
$\mathbb{Q}(\sqrt{94})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11, 14, \text{ maybe } \mathbb{Z}/16\mathbb{Z}$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}         \\ \hline
$\mathbb{Q}(\sqrt{95})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=11$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5,6$\end{tabular}                 \\ \hline
$\mathbb{Q}(\sqrt{97})$ & \begin{tabular}[c]{@{}l@{}}$\mathbb{Z}/n\mathbb{Z}, \:\:\: n=14, 15$\\ $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2n\mathbb{Z}, \:\:\: n=5$\end{tabular}         \\ \hline
\end{tabular}
 \\
\end{table}

\clearpage
\bibliographystyle{unsrt}

config: arxiv


TITLE: Simple non-closed geodesic.
QUESTION [11 upvotes]: In torus there exists simple non-closed geodesic. One example is take the irrational slope curves in $\mathbb{R}^2$ and project it down to torus. Is this thing can happen in closed hyperbolic surface i.e.
Does there exists a simple non-closed geodesic in a closed (i.e. compact without boundary) hyperbolic surface?

REPLY [2 votes]: Let's call $\Sigma$ a closed hyperbolic surface. Each free homotopy class of curves on a hyperbolic surface has exactly one geodesic representative. Conversely, a closed geodesic gives a free homotopy class. Thus (up to basepoint) there is a bijective correspondence
$$\bigg\{ \mbox{ closed geodesics }\bigg\} \longleftrightarrow \pi_1(\Sigma).$$
Since $\pi_1(\Sigma)$ is finitely generated, it is a countable set. In particular, it lifts to a countable set of geodesics in $\mathbb{H}^2$. Any and every geodesic in $\Sigma$ can be obtained by pushing a geodesic of $\mathbb{H}^2$ forward under the covering map, so the set of all geodesics of $\Sigma$ lifts to the set of geodesics of $\mathbb{H}^2$, which is uncountable. So there are many more open geodesics than closed geodesics on $\Sigma$.

Here's a simple example on a non-closed hyperbolic cylinder that illustrates the phenomenon. Let $\Sigma = \mathbb{H}^2/\mathbb{Z}$, where the $\mathbb{Z}$ action is given by $z\mapsto \lambda z$, for some $\lambda > 1$. The single closed geodesic is the image (under quotient) of the $y$-axis. Take any other geodesic which limits to $0$ and push it down under the covering map; it will spiral in from one side of the cylinder and limit to the closed geodesic.


TITLE: Factor Theorem in Unique Factorization Domain
QUESTION [2 upvotes]: To Prove: Suppose $r \in R$ and $P(x)$ a polynomial in $R[x]$ where R is a unique factorization domain. Prove that $x-r$ divides $P(x)$ iff $P(r)=0.$ I am able to prove this if R is a field because I can use the Division Algorithm but I am really stuck on this. Help would be greatly appreciated

REPLY [3 votes]: The Division Algorithm works with no change for polynomials over an integral domain, if we are dividing by a monic polynomial.
Added:
Theorem: Let $D$ be an integral domain with unity, and let $a(x)$, $b(x)$ be polynomials with coefficients in $D$, where $q(x)$ is monic. Then there exist polynomials $q(x)$, $r(x)$ with coefficients in $D$ such that $a(x)=q(x)b(x)+r(x)$, where $r(x)$ is the zero polynomial or has degree less than the degree of $b(x)$.
The proof is basically the same as the proof for $D$ a field. For example we can consider the collection $P$ of all polynomials over $D$ of the form $a(x)-f(x)b(x)$, and let $r(x)$ have smallest degree among these. It is easy to show that if the degree of $r(x)$ is greater than or equal to the degree of $b(x)$, there is a polynomial in $P$ of degree smaller than the degree of $r(x)$. It is at this stage where the  fact that $b(x)$ is monic is needed.


TITLE: Uniform convergence for a summation involving exponentials
QUESTION [0 upvotes]: For $y > 0$, why does the series $\sum_{m = 1}^{\infty} e^{-2\pi my}$ converge uniformly in $y$? Can I just use the following argument? For $m$ large enough, there exists a constant $C$ independent of $y$ such that $e^{-2\pi my} \leq m^{-2}$. Then as $\sum m^{-2} < \infty$,  $\sum_{m = 1}^{\infty} e^{-2\pi my}$ converges uniformly in $y$.

REPLY [2 votes]: Let $g(y)=e^{-2\pi y}$.  Then the partial sum of your series is:
$$f_n(y) = \sum_{m=0}^n g(y)^m = \frac{1-g(y)^{n+1}}{1-g(y)}$$
For $y>0$, $0<g(y)<1$, so this coverges to $f(y)=\frac{1}{1-g(y)}$
Now, to prove uniform convergence, you need for each $\epsilon>0$ an $N$ such that if $n>N$, $\forall y: |f(y)-f_n(y)|<\epsilon$
But $f(y) - f_n(y) = \frac{g(y)^{n+1}}{1-g(y)}$
As  $y\rightarrow 0$,  $g(y)\rightarrow 1$, so $|f(y)-f_n(y)|$ approaches infinity.  In particular, then, for any $n$, there is always a $y$ such that $|f(y)-f_n(y)|>\epsilon$.
Therefore, this doesn't converge uniformly in y.
Note: You can actually show that it cannot be uniformly convergent just by noting that $f(y)$ is unbounded on $(0,+\infty)$, and that each $f_n$ is bounded above (by $n$.)


TITLE: Movement in a grid
QUESTION [1 upvotes]: I am in a middle of a grid of 8X8 . Each box of grid  is a square with side of 80 units. Suppose my current coordinates are 594,422 . How many boxes will i have to move to enter the square that contains the coordinates 594,24 ?
The grid starts from the coordinates 350,20 as shown here.

The answer of the above problem is : I will have to move 5 boxes to capture the coordinate 594,24. But how can i generalize this result ? Any expression that can evaluate this is what i want ? Please also show the logic i.e where does the expression come from .
If my current coordinate is x,y and i have to capture x,q how many boxes will i have to move to do that ?
Getting more general , if my current coordinate is x,y and i have to capture p,q how many boxes will i have to move to do that ?
In this grid x and y increase as shown.

REPLY [2 votes]: You can do this in two steps: First figure out how to get from point coordinates to box coordinates; then just subtract the box coordinates corresponding to the source and destination. To find the box coordinates corresponding to given point coordinates, first subtract the coordinates $350,20$ of the origin. Then the box coordinate is just the result of dividing that by the box size, $80$, with integer division. In a programming language where / applied to integers results in integer division, you could write this as
bx = (x - 350) / 80;
by = (y -  20) / 80;

In mathematical notation, this would be
$$x_b = \left\lfloor\frac{x-350}{80}\right\rfloor\;,$$
$$y_b = \left\lfloor\frac{\,y\,\;-\;\,20\,}{80}\right\rfloor\;,$$
where the floor function $\lfloor\cdot\rfloor$ yields the greatest integer not greater than its argument.


TITLE: How fast can matrix multiplication with/on integers be performed?
QUESTION [1 upvotes]: I would like to know how fast a matrix of integers can be multiplied by another matrix of integers.  My motivation is because I've got a few ideas that seem to make the multiplication possible in around $O(n^2)$, but maybe this has already been accomplished somewhere else.  Most likely I've made a zillion errors or this has been done already...  I guess I'd like to know the best methods available, such as Strassen's algorithm.

REPLY [1 votes]: There is a newer O(n^2.373) algorithm, see Virginia Vassilevska Williams, "Breaking the Coppersmith-Winograd barrier": http://theory.stanford.edu/~virgi/matrixmult-f.pdf (published in 2014).


TITLE: Why is it useful to express PDE solutions as $L^2$-convergent series?
QUESTION [1 upvotes]: The existence of an $L^2$ orthonormal basis consisting of eigenfunctions of a Sturm-Liouville equation helps us to express the solutions of various ODEs and PDEs as infinite series. However, in the general case these infinite series converge to the said solution only in the sense of mean square convergence.
Why is it useful to have series representations of solutions which converge only 'in the mean'? I can think of a few reasons:

because sometimes it may be the best we can do;
because it's a stepping stone to proving stronger types of convergence for particular cases;
because we can be happy that they are 'generally' correct, although they can be atrociously wrong particularly at single (zero-measure) points.

However, I can't help noticing that the existence of $L^2$-convergent series as solutions to equations - and more generally approximations to $L^2$ functions - are celebrated in their own right as a practical, applicable achievement. I would welcome people's thoughts about the direct use of these solutions, especially how the general idea of being 'comfortable with these solutions on average' translates into some kind of practical reliability.
Thank you.

REPLY [1 votes]: In my experience, it seems that the Hilbert-space context (e.g., Plancherel theorems) allows the most decisive assertions, even if they do not respond quite directly to the primordial questions. This slight disconnect is well-known with Fourier series, and with Sturm-Liouville problems.
I remember being shocked, "in my youth", that $L^2$ convergence was not the same as pointwise, or as uniform pointwise. It seemed a hostile act on nature's part.
However, eventually, for example by looking at somewhat modern theory of linear, especially elliptic, PDE, one sees that "$L^2$-differentiation" is entirely tolerable, and, invoking Sobolev's inequalities at moment where necessary, quite reasonable assertions about "classical" differentiability can be cashed-in when necessary.
I learned how simple Sobolev theory can be from G. Folland's Tata notes on PDE. (W. Rudin's "Functional Analysis" seems not to appreciate Sobolev theory.) The simplest possible case, of Fourier series in one variable, is written out in detail sufficient for students who aren't seasoned analysts, in my notes
 Functions on circles .
Even in more serious structured situations, one can hope for a Plancherel theorem for $L^2$. There is no reasonable general expectation for other $L^p$ spaces, really. Thus, $L^2$ Sobolev theory has some universality not shared by other viewpoints.


TITLE: $\int\sin^2(x) \, dx$ - Which one is right?
QUESTION [2 upvotes]: If I enter $\int\sin^2(x)\,dx$ into an integral calculator it returns:
$$\int\sin^2(x)\,dx=-\frac{\sin(2x)-2x}{4}+c$$
If I calculate it with Angular multiples I come to:
$$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{\sin(2x)-2x} 4 +c$$
Here is my calculation
If I calculate it with partial integration and substitution I come to:
$$\int\sin^2(x)\,dx=-\frac{\sin(x)\cos(x)+x} 2 +c$$
Here is my second calculation
If I plot all of the functions they look also different...
So which one is right? Or maybe are all of them right due to the "+c"?
EDIT:
I have added my calculations, maybe you can help me to spot the wrong signs - thanks!

REPLY [1 votes]: With angular multiples,
$\int\sin^2(x)\,dx=\int\frac{1-\cos(2x)}2 \, dx=\frac{x}{2} - \frac{\sin(2x)}4 + c$
Now $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$\frac{x}{2} - \frac{2\sin(x)\cos(x)}4 + c$
$\frac{x}{2} - \frac{\sin(x)\cos(x)}2 + c$
$\frac{- \sin(x)\cos(x) + x}2 + c$
That is with partial integration and substitution.
Again $\frac{x}{2} - \frac{\sin(2x)}4 + c$
$ \frac{- \sin(2x)}4 + \frac{x}{2} + c$
$ - \frac{\sin(2x) - 2x}4 + c$
That is with integral calculator.
So all are same. I think you have either typing mistake in your third solution or mistake in solving.


TITLE: A construction using straightedge and compass
QUESTION [6 upvotes]: Given a circle, it's easy to contruct its center.
The question is: given an ellipse, draw the foci.
I don't know whether it's possible to do this using only straightedge and compass.

REPLY [13 votes]: Let $\lambda$ a given ellipse, let's draw its foci.
See the figure below:

Draw two parallel chords $AB$ and $DC$.
Find out $M$ and $N$ (midpoints of $AB$ and $DC$ respectively).
Draw the straightline $r$ through $M$ and $N$.
Let $\{E, E'\} = r \cap \lambda$, find out $O$ the midpont of $EE'$.
Draw an arc $\mu$ centered at $O$ and radius $AO$, such that $\{A, A'\}= \mu \cap \lambda$.
Draw the straightline $s$ through $A$ and $A'$.
Draw a straightline $t$ such that $O \in t$ and $t\parallel s$.
Draw $u$ (the perpendicular bisector of $AA'$).
Let $\{G, G'\} = u \cap \lambda$, and $\{J, J'\} = t \cap \lambda$, $GG'$ and $JJ'$ are the axes of symmetry of $\lambda$. Now draw an arc $\nu$ centered at $G$ and radius $JO$, such that $\{F1, F2\}= \nu \cap t$. $F1$ and $F2$ are the desired foci of $\lambda$.


TITLE: Lemma 2.4.4 from Kempf's Algebraic Varieties
QUESTION [1 upvotes]: I have a question about the following lemma in the book.

Its proof starts by saying that without loss of generality we can assume that $X$ and $Y$ are affine, and $k[X]$ is a finitely generated module over $k[Y]$. Why is this justified? In particular, since cover by affines is not necessarily disjoint, why does solving it locally suffice?
Furthermore, they follow it with the an assertion that in fact it suffices to consider the case when $X$ is irreducible affine and $Y = f(X)$.  I fail to see why finiteness of morphism is preserved in this case?
Definition of a finite morphism is given as follows: a morphism $f: X \rightarrow Y$ is finite if there is an open cover $Y = \cup_i U_i$by affines $U_i$ whose inverse images are affine and $k[f^{-1}(U_i)]$ is finitely generated as $k[U_i]$ module.
Any help is appreciated. Thanks.

REPLY [2 votes]: The reason we can make these reductions is that finiteness is preserved by base change, and both steps of this reduction (restricting to some subset) can be described by base change. Here's how to make the reduction steps:
First, reducing to $Y$ irreducible: If $Z_2$ is irreducible, then $f(Z_2)$ is also irreducible, and it is contained in an irreducible component $Y_0$ because irreducible components are maximal with respect to inclusion among irreducible subsets. By restricting to $Y_0$, aka base changing along the closed immersion $Y_0\to Y$, we get to the situation $f|_{f^{-1}(Y_0)}: f^{-1}(Y_0)\to Y_0$, where $Y_0$ is irreducible and $f|_{f^{-1}(Y_0)}$ is again finite, being the base change of a finite morphism. Thus it suffices to treat the case where $Y$ is irreducible.
If we reduce to the case where $Y$ is affine, this will get us to the case where $X$ is affine as well, as finite morphisms are affine: the preimage of any affine subset under an affine map is again affine. In order to reduce to the case of $Y$ affine, it suffices to note that we can tell if $f(Z_1)\subset f(Z_2)$ is a proper inclusion or not if $(f(Z_1)\cap U)\subset (f(Z_2)\cap U)$ is proper for every open subset $U$ in some cover of $Y$ (think about why this is - how do you know an inclusion is proper? You find a point in one but not in the other...). As restricting to $U\subset Y$ is the base change by the open immersion $U\to Y$, we see that finiteness (and therefore affineness) is again preserved. As every variety admits an open cover of affine varieties, this means we can reduce to the case of $X,Y$ affine.
By performing these reductions in sequence, we get what we want: $X,Y$ affine, $f$ finite, and $Y$ irreducible.


TITLE: Matrix $M$ such that $M^3 = -M$
QUESTION [1 upvotes]: I couldn't find a similar question.
I want to prove that a matrix $M$ such that $M^3 = -M$ satisfies
$R^3 = ker(M)\oplus ker(M^2+I)$
And that the dimension of $Ker(M^2+I)$ is even.
I know how to prove that $Ker(M)\cap Ker(M^2+I) = {0}$, but then, I don't know how to proceed (to show that $R^3$ is written by elements of the two kernels and its dimension). Any help is welcomed. Thank you

REPLY [2 votes]: Note that $M^3+M = M(M^2+1)$, so $p(X) = X(X^2+1)$ annihilates $M$. Since $X$ and $X^2+1$ are coprime, and in fact $g(X)X+f(X)(X^2+1) = 1$ for $g(X) = -X$ and $f(X) = 1$, any vector can be written as a sum of two vectors $v_1+v_2$ where $v_1 = g(M)Mv$ and $v_2 = g(M)(M^2+1)v$, and certainly $v_1\in \ker(M^2+1)$ while $v_2\in \ker M$.
Moreover, this shows that if $v$ is in both kernels it is zero, for in such case $v_1=v_2=0$. Observe that this proof works whenever a polynomial $p$ annihilates a matrix and you have a decomposition of $p$ into irreducible factors.
To show that the kernel $V$ of $M^2+1$ is of even real dimension, note that such vectors satisfy $M^2v=-v$, so the operator $M : V\to V$ can be used to extend the $\mathbb R$-structure to a $\mathbb C$-structure by declaring that $iv = Mv$ for $v\in V$, and complex vector spaces have even real dimension.


TITLE: Finding and proving $\bigcap\limits_{n=1}^{\infty} \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$
QUESTION [1 upvotes]: Letting $A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$, I am trying to find $\bigcap\limits_{n=1}^{\infty} A_n$. I believe the answer is $\left[0, \frac{1}{3}\right)$. I don't have any intuition for why this is the case other than attempting to prove that the answer is $\left[0, \frac{1}{2}\right)$. Approacing the left and right endpoints as sequences and taking $n \to \infty$ didn't help.
I'm going to attempt to prove this.
Let $x \in \left[0, \frac{1}{3}\right)$. Then $0 \leq x < \frac{1}{3}$. We must show that $x \in A_n$ for all $n$. So fix $n \geq 1$. Certainly $- \frac{1}{n} < 0 \leq x$. Now we show that $x < \frac{n}{2n+1}$. Since $n \geq 1$, $2n \geq 2$ and $2n + 1 \geq 3$, so $\frac{1}{2n+1} \leq \frac{1}{3}$. Certainly, because $n \geq 1$, $\frac{n}{2n+1} \geq \frac{1}{2n+1}$, but that doesn't help, so at this point I'm stuck.
For the reverse direction, let $x \in \bigcap\limits_{n=1}^{\infty} A_n$, so $x \in \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$ for all $n$. So $- \frac{1}{n} < x < \frac{n}{2n+1}$. Outside of taking $n \to \infty$, I don't know how to proceed.
Any help or hints would help.

REPLY [1 votes]: Let $A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$. We want to find $\bigcap_{n=1}^\infty A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$.

Suppose $x\in \bigcap_{n=1}^\infty A_n$. Then, $x\in A_n$ for all $n\ge 1$. That is, $x > -\frac1n$ for all $n\ge 1$, and taking limits as $n\to\infty$ gives you $x\ge  0$. Also, $x < \frac{n}{2n+1}$ for all $n\ge 1$. This is where you have to be careful - simply taking limits as $n\to\infty$ will not help. Here's why:
$$x < \frac{n}{2n+1} = \frac12\cdot \left(1  - \frac{1}{2n+1} \right) \quad (\forall n\ge 1)$$
and $\frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ is an increasing function of $n$. As a result, if $x < \frac12\cdot \left(1  - \frac{1}{2k+1} \right)$ for some $k\in \mathbb N$, then we certainly have $x < \frac12\cdot \left(1  - \frac{1}{2m+1} \right)$ for all $m \ge k$. Thus, put $n=1$ and see that $x < \frac13$. As discussed, $x < \frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ for all $n\ge 1$ is then automatically satisfied. Hence, $x \in [0, \frac13)$.

On the other hand, suppose $x\in [0,\frac13)$. You have already shown $x > -\frac1n$ for all $n\ge 1$. $\frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ is an increasing function of $n \in \mathbb N$, and its minimum (over natural numbers) is attained at $n=1$. In particular, this minimum is $\frac13$. Thus $x < \frac13$ ensures that $x < \frac12\cdot \left(1  - \frac{1}{2n+1} \right)$ for all $n\ge 1$. Consequently, $x\in \bigcap_{n=1}^\infty A_n$, as desired.

Let me know if you have any questions.


text:
\begin{document}

\title{Transparent lattices and their solitary waves}

\author{E. Sadurn\'i}
 \email{sadurni@ifuap.buap.mx}
\affiliation{Instituto de F\'isica, Benem\'erita Universidad Aut\'onoma de Puebla,
Apartado Postal J-48, 72570 Puebla, M\'exico}

\date{\today}

\begin{abstract}
We provide a familiy of transparent tight-binding models with non-trivial potentials and site-dependent hopping parameters. Their feasibility is discussed in electromagnetic resonators, dielectric slabs and quantum-mechanical traps. In the second part of the paper, the arrays are obtained through a generalization of supersymmetric quantum mechanics in discrete variables. The formalism includes a finite-difference Darboux transformation applied to the scattering matrix of a periodic array. A procedure for constructing a hierarchy of discrete hamiltonians is indicated and a particular biparametric family is given. The corresponding potentials and hopping functions are identified as solitary waves, pointing to a discrete spinorial generalization of the Korteweg-deVries family.
\end{abstract}

\pacs{42.50.Md, 42.81.Dp, 04.60.Nc, 11.30.Pb}

\maketitle

\section{ Introduction \label{sec:1}}

There is an increasing interest in the study of tight-binding models and their application to several areas, such as optical waveguides \cite{russell2003, keil2010, foresi1997}, ultracold atoms in optical lattices \cite{oberthaler2006, bloch2005, oberthaler1996} and artificial realizations of condensed matter, e.g. graphene \cite{esslinger2013, kuhl2010, barkhofen2013, bittner2010, bellec2013}. Noteworthy is the area of quantum metamaterials \cite{felbacq2012, luukkonen2009}, where transmission properties can be tailored at the level of quantum degrees of freedom. The subject has reached the domain of supersymmetric models through a recent realization of a Dirac oscillator \cite{franco2013, sadurni2010, longhi2010}, which seems to be the first experimental construction of an $N=2$ supersymmetry. In these applications, technical developments towards the engineering of local potentials and couplings between sites have been reached with several purposes \cite{esslinger2008, christodoulides2008}, even with microwaves \cite{sadurni2013}. A particular example of interest is the perfect transmission of signals through discrete arrays and its associated scattering problem in the presence of local lattice modifications. It is important to mention that transparency has been extensively studied in continuous variables \cite{cooper1995, samsonov2000}, but never in a tight-binding array \cite{berry2008}. In this paper we provide an elegant solution of the problem by extending the well-developed apparatus of supersymmetric quantum mechanics \cite{kac1973, catterall2002} (SUSYQM) to discrete variables.

We shall proceed in the following order: In section \ref{sec:2} we define our problem and state the main result of our work by giving a family of transparent potentials in tight-binding arrays. In subsection \ref{sec:experiment} we study the possibility of implementing such models: matter waves in optical traps \ref{sec:qmwaves}, electromagnetic waves in nanoscopic and mesoscopic arrays \ref{sec:emwaves} and dielectric slabs \ref{sec:slabs} are considered. Section \ref{sec:3} provides the necessary definitions and generalizations for discrete SUSYQM. This section can be read separately and is divided in five parts: \ref{sec:3.0} The factorization method, \ref{sec:3.1} isospectrality and transparency (where we address the problem of discrete-variable Darboux transformations), \ref{sec:3.2} the continuous limit of the theory, \ref{sec:4} a numerical check of the required properties and, in full analogy with traditional SUSYQM, subsection \ref{sec:5} studies a set of discrete solitons given by potentials and hopping parameters.  Section \ref{sec:6} gives a brief conclusion.

\section{A family of models \label{sec:2}}

\begin{figure}[t]
\includegraphics[width=8.6cm]{twochains}
\caption{\label{fig:0} Two asymptotically periodic arrays. Bonds represent couplings $\Delta_n$ and site colors represent potential values $V_n$. Example a) has two identical asymptotic regions indicated by rectangular boxes. In example b), the asymptotically periodic regions do not coincide.}
\end{figure}

\begin{figure*}[t]
\includegraphics[width=10.6cm]{fig1}
\caption{\label{fig:1} Variation of our discrete potential (blue-filled curve) and hopping function (green-filled curve) with respecto to $\gamma$. The minimum of the potential and the maximum of the hopping parameter lie in the same region and suffer a linear displacement to the right when $\gamma$ is increased exponentially (the center is denoted by C).}
\end{figure*}

\begin{figure}[t]
\includegraphics[width=7cm]{rods}
\caption{\label{fig:1.1} Dielectric rods of variable radii. The couplings $\Delta_n$ can be tuned by varying the distance between the cylinders. The optical axis $z$ is parallel to their longitudinal coordinate.}
\end{figure}

\begin{figure}[t]
\includegraphics[width=7cm]{slabs}
\caption{\label{fig:1.2} Dielectric slabs of variable thickness and alternating refraction index. As before, the distance between dark-blue slabs can be used to control $\Delta_n$.}
\end{figure}

Several areas benefit from second quantized tight-binding models with and without particle interactions. In one dimension, and with units such that $\hbar=1$, one typically has

\bea
H_{\mbox{\small quantum}} = \sum_{\< n, m\>} \left[\Delta_{nm} C^{\dagger}_n C_m +  \Delta_{nm}^*  C^{\dagger}_m C_n  \right]
\label{a1}
\eea

where $C_n, C_m^{\dagger}$ are local field operators of bosonic or fermionic nature. The off-diagonal $\Delta_{nm}$'s are hopping parameters and the diagonal $\Delta_{nn}$ represent on-site potentials. The propagation of signals in such models can be viewed as a scattering problem in which one or many particle states with well defined momenta can be ideally prapared at infinity. In the absence of terms of the form $C_n^{\dagger} C_m C_i^{\dagger} C_j$ (dilute matter waves \cite{zoller1998} or photonic crystal waveguides \cite{chien2007}), the lack of particle interactions allow a single-particle treatment of the scattering problem. Therefore, we restrict ourselves to a hamiltonian in first quantization. With the help of Wannier functions $\< x | n \>$ localized around site $n$, we have

\bea
H = \sum_{n} \Delta_n | n \>\<n-1| + \Delta_n^* | n-1 \>\<n| + V_n | n \>\<n|.
\label{a2}
\eea

For simplicity, only nearest neighbors are assumed. One can expand a stationary state $|\psi \>$ as

\bea
|\psi \> = \sum_{n} \psi_n | n \>
\label{a3}
\eea

and from this expansion, the Schr\"odinger equation associated to (\ref{a2}) acquires its typical recurrence form

\bea
\Delta_n \psi_{n-1} + \Delta_{n+1} \psi_{n+1} + V_n \psi_n = E \psi_n.
\label{a4}
\eea

Now we define $H$ to be {\it asymptotically periodic\ }if at the far ends of the array (left and right) we have asymptotically constant couplings and potentials, i.e. $\Delta_{n} \sim \Delta_0$ and $V_n \sim V_0$  if $n >> 1$ . A diagram is shown in figure \ref{fig:0}. The main finding of this work is the existence of a family of potentials and hopping paramaters (or functions, since they depend on site $n$) for which the reflection coefficient of a Bloch wave vanishes for all energies. The most general expression for the biparametric family can be written in terms of continued fractions, as we shall see in section \ref{sec:5}. To this end, we employ the notation

\bea
\left[a_n, b_n; a_{n-1}, b_{n-1}; ... \right] = a_n - \frac{b_n}{a_{n-1} - \frac{b_{n-1}}{\cdots}}.
\label{a6}
\eea

An important case is given by a monoparametric subfamily $\tilde V_n, \tilde \Delta_n$, which can be written in terms of hyperbolic functions

\bea
\tilde V_n &=& V_0  \nonumber \\ &+& \Delta_0\left(\frac{\cosh\left[ n \lambda + \beta \right]}{\cosh\left[ (n-1) \lambda + \beta \right]} - \frac{\cosh\left[ (n+1) \lambda + \beta \right]}{\cosh\left[ n \lambda + \beta \right]}\right), \nonumber \\
\label{a7}
\eea

\bea
\tilde \Delta_n = \Delta_0 \frac{\sqrt{ \cosh \left[ (n-2) \lambda + \beta \right] \cosh \left[ n \lambda + \beta \right] } }{ \cosh \left[ (n-1) \lambda + \beta \right] },\nonumber \\
\label{a8}
\eea

where $\beta$ is a free parameter and the convenient definition
\bea
\lambda = \ahalf \ln \left(\frac{V_0-\sqrt{V_0^2 - 4 \Delta_0^2}}{V_0+\sqrt{V_0^2 - 4 \Delta_0^2}}\right)
\label{a9}
\eea

has been used. Evidently, $|V_0| \geq 2|\Delta_0|$ is a necessary condition and it can always be met by recognizing that $V_0, \Delta_0$ are independent parameters. Some examples are depicted in figure \ref{fig:1} as a function of $\gamma=\mbox{e}^{2\beta}$.

\subsection{Experimental feasibility \label{sec:experiment}}

Tight binding models can be implemented in a variety of settings, ranging from matter waves to electromagnetic waves. The implementation of our model relies strongly on the possibility of reaching physical values for our on-site energies $V_n$ and hopping functions $\Delta_n$. In this respect, it is important to ensure that all the values provided by (\ref{a7}, \ref{a8}) are bounded. It is a simple task to find upper and lower bounds for such functions:

\bea
|\Delta_0| < |\tilde \Delta_n| < |\Delta_0 \cosh (\lambda)|,
\label{b1}
\eea

and
\bea
V_0 + \Delta_0 \left( 1 - \frac{\cosh(3\lambda/2)}{\cosh(\lambda)} \right) < \tilde V_n < V_0 + \Delta_0.
\label{b2}
\eea

These bounds depend entirely on $V_0, \Delta_0$, and there is no value of $\lambda$ in (\ref{a9}) that leads to singular limits for non-trivial configurations.

With these bounds, we are in the position to discuss realistic parameters in available experiments.

\subsubsection{Quantum-mechanical waves \label{sec:qmwaves}} With the aim of understanding wave propagation in solids, artifical crystals have been produced by means of optical traps. Bose-Einstein Condensates (BECs) are produced and held by potentials that, in some applications, can be tuned in frequency and shape. Hubbard models with tailored hopping parameters \cite{struck2012} have been achieved with different purposes, such as the emulation of non-abelian gauge fields in 2d (phase variation of $\Delta$) and the study of localization phenomena, with its corresponding insulator transition as a function of disorder \cite{inguscio2007}. It is important to recall that reasonable on-site energies have values around $|V_n-V_0|<h \times 3.2$kHz, providing thus an interval to play with. The hopping parameters are in turn determined by such potential heights and by the lattice spacing ($830$ nm $<\lambda<1076$ nm for laser traps). Moreover, the minimization of self-interactions (or supression of non-linear terms in the effective Gross-Pitaevskii wave equation) has been achieved in BECs with atoms such as K, Cs and Li, making linear tight-binding models more realistic \cite{roati2007}. We must recognize, however, that full trap tunability depends on the use of lasers at many different frequencies. Well widths and depths can be controlled in this way, but recent applications have only used a few frequencies; for example Ref. \cite{inguscio2007} reports two colors in order to produce non-periodic configurations.

In a different domain of quantum mechanics, we find the so-called quantum metamaterials \cite{felbacq2012}, where it is possible to tune the transmission properties of arrays made of doped nanorods. By illuminating our proposed configuration of resonators, one may be able to switch its properties from full reflectivity (due to a bandgap) to transparency (as in our case).

\subsubsection{Electromagnetic waves \label{sec:emwaves}} We propose the use of dielectric media disposed in convenient configurations (figs. \ref{fig:1.1} and \ref{fig:1.2}), such as those employed in the fabrication of photonic crystals (nanoscopic scales) and microwave resonators (mesoscopic scales). The first possibility is within range, if we recall that carved structures in controlled patterns have been consistently produced for a few decades \cite{yablanovitch1987, little1995}. See \cite{bienstman2003} and, in particular, a so-called {\it stage I coupler\ }in the inset of fig. 5 of the same paper. The main idea is to vary the size, the interdistance and refraction indices of such structures. Realistic parameters can be quoted; for example, in \cite{little1995} we find refraction indices in the range $3.20<n<3.25$ and slab thicknesses $d\sim 1\mu$m. Realizations with nanorods of AlO and GaAs mixtures work with $1.61<n<3.37$, as reported in \cite{bienstman2003}, and with variations of their radii (around $\rho \sim 125$nm) one may further induce shifts in their resonant frequencies. In this way, one may achieve a variable on-site potential in effective tight-binding models as a function of resonator size. The hopping parameters can be tuned by varying the distance between structures such as slabs, rods or coupled waveguides in general. Finally, the values cited above determine the elements of transfer matrices for the propagation of TE and TM modes, e.g. Eq. (1) in Ref. \cite{wang2008}.

For photonic structures carved in Si substrates \cite{wang2008}, the reported permittivities are $\epsilon_a = 2.22$ and $\epsilon_b =1$ (air layers). Frequency gaps can be achieved at central wavelengths $\lambda_{\scriptsize \mbox{center}} \sim 1011$ nm for normal incidence of light, but such values can be tuned down to $350$ nm by varying the angle of incidence, covering the complete optical range. When the structures are built in a non-periodic pattern of a similar size, the corresponding frequency levels tend to be discretely distributed, but they lie in the same (optical) region. The thickness of each barrier or slab can be in the range $95$ nm $<d<459$ nm.

Other realizations with variable couplings using photons in waveguides have been reported \cite{keil2013}. The couplings $\Delta_n$ are tuned again by the interdistance mechanism. Similar settings were produced in \cite{sadurni2013}. An exponential law for $\Delta_n$ as a function of interdistance $d$ was shown in both cases. This dependence can be further refined by modified Bessel functions arising from cylindrical geometries.

In the case of ceramic microwave resonators  (Temex series from E2000 to E7000), different types of spectra have been produced by the method of variable couplings \cite{kuhl2010, franco2013, sadurni2013}. The parameters were tuned by analyzing the level separation of a dimer as a function of the distance between resonators. It is important to mention that the emulation of (\ref{a7}) also requires a mechanism that adjusts the on-site potential. To this end, it is convenient to employ resonators made of different materials, with selected resonant frequencies ranging from 800MHz to 50GHz and various permittivities. Isolated sharp peaks of $\Gamma \sim$2MHz for each resonator provide reasonable spectra in arbitrary arrays. It should be noted that other approaches to vary on-site energies are possible, as indicated in \cite{barkhofen2013}. The spectral gap of Boron Nitride was emulated by breaking a dimer symmetry using antenna couplings, rather than varying the permittivity of the constituents.

\subsubsection{Couplings and on-site potentials for dielectric layers \label{sec:slabs}}

Let us analyze a problem of parallel dielectric slabs, with the aim of extracting the thickness and interdistance dependence of couplings and on-site potentials. We start with a single slab of permittivity $\epsilon_1$ immersed in a medium of permittivity $\epsilon_2$ (cladding). We have $\epsilon_1>\epsilon_2$. If the slab is parallel to the $x$ axis, as in fig. \ref{fig:1.2}, we will have a continuous TE mode $\v E_x$ travelling along $\v k$, which lies in the $y$-$z$ plane. We would like to analyze the behavior of $\v E_x$ along the coordinate $y$, i.e. across media interfaces. Our starting point is the 3d Helmholtz equation ($c=1$)

\bea
\left( \nabla^2 + \epsilon_1 \omega^2 \right) \v E_x = 0, \quad \v r \in \Omega, \nonumber \\
\left( \nabla^2 + \epsilon_2 \omega^2 \right) \v E_x = 0, \quad \v r \notin \Omega.
\label{b3}
\eea

This equation can be transformed to a Schr\"odinger-like 1d equation with a potential well: assuming no confinement along $x$, we reduce (\ref{b3}) to

\bea
\left( -\frac{\partial^2}{\partial y^2} + U_0 \Theta(L/2 - |y|) \right) \phi(y) = E \phi(y)
\label{b4}
\eea

with $E\equiv\epsilon_2 \omega^2 - k_z^2 <0, U_0 \equiv (\epsilon_2 - \epsilon_1 )\omega^2 <0$ and $\v E_x(x,y,z)=\v x \phi(y) e^{ik_z z}$. The solutions $\phi(y)$ are confined along $y$ by the walls of the potential, i.e.

\bea
\phi(y) = \ncal \times \begin{cases}
\exp (-\frac{|y-L/2|}{\lambda}) & \text{($y > L/2$)} \\
\exp (-\frac{|y+L/2|}{\lambda}) & \text{($y < -L/2$)},
\end{cases}
\label{b5}
\eea

and for $|y|<L/2$, $\phi$ is a trigonometric function. The skin depth $\lambda$ is related to the effective energy by the simple relation $\lambda = 1/\sqrt{|E|}$. In order to find $E$ we must solve a well-known transcendental equation; with the variables $\xi_0 \equiv L^2 U_0 / 4, \xi \equiv \sqrt{E+U_0} L / 2$, one has $\sqrt{\xi_0^2-\xi^2} = \xi \tan \xi$, but in the regime of interest only one bound state is needed. Therefore we take $\xi \ll 1$ and solve the previous relation, which leads to the following estimate to lowest order in $U_0$:

\bea
E \approx -\frac{L^2  U_0^2}{4} = -\frac{ \omega^4 L^2  (\epsilon_1-\epsilon_2)^2}{4}
\label{b6}
\eea

and
\bea
\lambda \approx  \frac{4}{\omega^2 L  (\epsilon_1-\epsilon_2)}.
\label{b6.1}
\eea

We are now able to find the couplings of a nearest-neighbor hamiltonian. When a second slab of permittivity $\epsilon_1$, skin depth $\lambda'$ and width $L'$ is centered at $y=d +(L + L')/ 2$, a significant overlap between functions $\phi, \phi'$ appears and the eigenfrequencies of the system are modified. We obtain a non-diagonal element in the effective hamiltonian (\ref{b4}), given by

\bea
\Delta = \int_{-\infty}^{+\infty} dy (\phi H_{\scriptsize \mbox{eff}} \phi'^*) \approx  - \int_{L/2}^{L/2 + d} dy \phi \frac{ d^2 \phi'}{dy^2}.
\label{b7}
\eea

This integral contains exponential tails and it can be carried out easily. It is important to note that the normalization constants of $\phi, \phi'$, i.e. $\ncal, \ncal'$, must be redefined in the interval $[L/2,L/2+d]$. With this in mind, we obtain an exponential dependence

\bea
\Delta(d) &=& \Delta(0) \exp \left( -\frac{(\lambda + \lambda') d}{2 \lambda \lambda'} \right) \nonumber \\ &\times& \frac{\sqrt{\lambda \lambda'} \sinh \left( \frac{(\lambda - \lambda') d}{2 \lambda \lambda'} \right) }{(\lambda-\lambda')\sqrt{\sinh \left( \frac{d}{2 \lambda} \right)\sinh \left( \frac{d}{2 \lambda'} \right)} }.
\label{b8}
\eea

For practical purposes, we can approximate this expression by the exponential alone. On the other hand, the on-site potentials in a tight-bidning approximation are given by the eigenvalues of isolated potential wells $E \sim 1/\lambda^2, E' \sim 1/\lambda'^2$. Finally, we can see that a careful control of the independent parameters $L, L', d$ -- and possibly $\epsilon_1, \epsilon_2$ -- generates couplings and on-site potentials that can be used in arrays of many sites. Given a set of couplings and potentials $\{ \Delta_n, V_n \}$, we obtain
a set of widths and separations $\{ L_n, d_n \}$ through the relations
\bea
 L_n = \frac{2}{\omega^2(\epsilon_1-\epsilon_2)} \sqrt{|V_{\scriptsize \mbox{offset}}-V_n|},
\label{b9}
\eea
\bea
d_n = - \left( \frac{2}{\sqrt{|V_n|} + \sqrt{|V_{n+1}|}} \right) \log \left( \frac{\Delta_n}{\Delta(0)} \right).
\label{b10}
\eea

\section{A path to discrete SUSYQM \label{sec:3}}

In the rest of this paper we establish the mathematical methods that lead to transparency in the context of discrete variables. This shall be done with the help of discrete SUSYQM.

\subsection{The discrete factorization method \label{sec:3.0}}
 The central discretization of the Schr\"odinger equation leads naturally to a nearest-neighbor tight-binding hamiltonian, therefore we focus on such local quantum-mechanical models for our constructions. A generic expression for $H$ in terms of operators can be written as

\bea
H = \Delta(N) T + T^{\dagger} \Delta(N)^{\dagger} + V(N),
\label{e1}
\eea

where $N$ is the site number operator and $T$ is a translation in one unit. These operators satisfy $\left[ F(N), T \right]= T \{F(N+1)-F(N) \}$ for any function $F$. Their action on localized states $|n\>$ is given by $N|n\> = n |n\>$ and $T |n\> = |n+1\>$. The functions $V(N)$ and $\Delta(N)$ represent the on-site potential and the nearest- neighbor hopping function, respectively. Their eigenvalues are $V_n$ and $\Delta_n$, with $V_n$ real. In full analogy with traditional SUSYQM in continuous variables, we propose a factorization scheme of (\ref{e1}) as follows

\bea
H = A^{\dagger} A, \quad A =  F(N)T + G(N).
\label{e2}
\eea

The reconstruction of (\ref{e1}) will be possible if we impose the restrictions $V_n = |G_n|^2 + |F_{n+1}|^2$ and $\Delta_n = G_{n}^* F_n$, as can be verified by applying the product $A^{\dagger} A$ to some state $|n\>$. It is important to recognize that this scheme can be applied only if the potential is positive definite or, without loss of generality, if it is bounded below. This property enables us to consider the existence of a ground state and to further subtract it from the hamiltonian, which is a usual procedure \cite{cooper1995}. We should also point out that given a positive $V_n$ and a complex $\Delta_n$, we may determine $F_n$ and $G_n$ up to phase factors (which can be gauged away trivially in 1d). To this end one has to solve the recurrence $V_n = |F_{n+1}|^2 + |\Delta_n|^2 / |F_n|^2$ for $|F_n|^2$ by the method of continued fractions

\bea
|F_n|^2 = \left[ V_{n-1}, |\Delta_{n-1}|^2;V_{n-2}, |\Delta_{n-2}|^2;... \right]
\label{e3}
\eea

and then reconstruct $G$ with the relation

\bea
|G_n|^2 = \frac{|\Delta_n|^2}{\left[ V_{n-1}, |\Delta_{n-1}|^2;V_{n-2}, |\Delta_{n-2}|^2;... \right]}.
\label{e4}
\eea

Our proposal for the factorization of $H$ is in full correspondence with the continuous case, in view of the analogy $ A_{\mbox{\small discrete}} \leftrightarrow  A_{\mbox{\small continuous}}$, i.e.

\bea
F(N) T + G(N) \longleftrightarrow \frac{d}{dx} + W(x),
\label{e4}
\eea

where $W(x)$ is a superpotential satisfying the Riccati equation \cite{cooper1995}. A clear connection with continuous variables can be given by means of a lattice spacing $a$: The translation of wave functions $\<x-a|n\>=\<x|n+1\>$ motivates the substitutions $T=\exp (- a \cdot d/dx),  x= aN$, which in turn lead to the limit

\bea
F \left(\frac{x}{a} \right)\exp \left(- a \frac{d}{dx} \right) + G \left(\frac{x}{a} \right) \longrightarrow \phi(x) \frac{d}{dx} + \gamma(x).\nonumber \\
\label{e5}
\eea

Here we have imposed $F(x/a) + G(x/a) \rightarrow \gamma(x)$ and $aF(x/a) \rightarrow -\phi(x)$. This limit resembles the usual Darboux operator \cite{darboux1882, manas1997}. Moreover, our discrete $A$ in (\ref{e2}) is a particular case of a series of automorphisms, but we should stress that  $F(N) \neq \mbox{constant}$ is a more general choice and allows more freedom in our models. In connection with singularities in our limits, we should note that $a F(x/a)$ is regular at $a=0$, but $F(x/a)$ is not. The function $G(x/a)$ compensates for the singularity of $F(x/a)$ rendering a finite $\gamma(x)$. We shall come back to this point in connection with the P\"oschl-Teller potential as a plausible continuous limit.

With these considerations, we are ready to construct a discrete superpartner $\tilde H$ with the prescription

\bea
\tilde H = AA^{\dagger}= \tilde \Delta(N) T +  T^{\dagger} \tilde \Delta(N)^{\dagger} + \tilde V(N).
\label{e6}
\eea

The new potential and hopping functions are given by

\bea
\tilde \Delta(N) = G(N-1)^{\dagger} F(N),
\label{e7}
\eea

\bea
\tilde V(N) =   G(N)G(N)^{\dagger} + F(N)F(N)^{\dagger},
\label{e8}
\eea

and their eigenvalues obey the relations $\tilde \Delta_n = G_{n-1}^* F_{n}, \tilde V_n = |G_{n}|^2 + | F_{n}|^2$. Remarkably, the functions which determine the superpartners come in pairs, since both $\Delta$ and $V$ must be modified.

\subsection{Isospectrality and Transparency \label{sec:3.1} }

Our method can be readily applied to bound states as well as scattering solutions. Let $|\psi_k\>$ be a solution of $H|\psi_k\>=E_k|\psi_k\>$. If $\tilde H|\tilde \psi_k\>= \tilde E_k| \tilde \psi_k\>$, we deduce the relations

\bea
|\tilde \psi_{k} \> = \left( E_{k+1} - E_0 \right)^{-1/2} \{ F(N)T + G(N) \} |\psi_{k+1} \>, \nonumber \\
\label{e9}
\eea

\bea
\tilde E_{k} = E_{k+1}.
\label{e10}
\eea

The $N=2$ supersymmetry is realized by defining supercharges with the help of $A, A^{\dagger}$ and the Pauli matrices $\sigma_{\pm}$. We have

\bea
Q=\sigma_- A, \quad Q^{\dagger}=\sigma_+ A^{\dagger}, \quad \hcal = \{ Q, Q^{\dagger} \}.
\label{e10.1}
\eea

where $\hcal$ is the central charge. The aforementioned isospectrality can be used for many purposes, but we are interested now in the transformation properties of the scattering matrix and their relation with transparency.

Let us define the conditions of a scattering problem in discrete variables. We take $H, \tilde H$ asymptotically periodic, such that in the limit $n \rightarrow \pm \infty$ we have $V_n \rightarrow V_{\pm}, \tilde V_n \rightarrow \tilde V_{\pm}, \Delta_n \rightarrow \Delta_{\pm}, \tilde \Delta_n \rightarrow \tilde \Delta_{\pm}$. The solutions with continuous parameter $k$ become Bloch waves: $\< n| \psi_k\> \rightarrow \ncal \mbox{e}^{ikn} $ and similarly for $\< n| \tilde \psi_k\>$. The limit values of $F$ and $G$ can be reconstructed via the relations $|F_{\pm}|^2= \ahalf (V_{\pm} \pm \sqrt{V_{\pm}^2-4\Delta_{\pm}^2})$, where $V_{\pm}^2 \geq 4 \Delta_{\pm}^2$. With this information on the factorization parameters, we are ready to apply the discrete Darboux transformation to the asymptotic form of the wave functions

\bea
\< n |A| \psi_k \> = \left( G_n + \mbox{e}^{ik} F_n \right) \psi_n^{k} \rightarrow  \left( G_{\pm} + \mbox{e}^{ik} F_{\pm} \right) \mbox{e}^{ikn}. \nonumber \\
\label{e11}
\eea

It may happen that the asymptotic regions at $\pm \infty$ have different associated constants, i.e. $V_+ \neq V_-, \Delta_+ \neq \Delta_-$. However, we always have the energy (or dispersion) relation $E = 2 \Delta_- \cos k + V_- = 2 \Delta_+ \cos k' + V_+$ satisfied at both ends of the array. In general $k'\neq k$, and the reflected and transmitted waves have different Bloch quasi-momenta, expressed by the limits

\bea
\<n | \psi_k \> \begin{array}{c} _{\longrightarrow} \\  _{n \rightarrow -\infty} \end{array} \mbox{e}^{ikn} + R \mbox{e}^{-ikn},
\label{e12}
\eea

\bea
\<n | \psi_k \> \begin{array}{c} _{\longrightarrow} \\  _{n \rightarrow +\infty} \end{array}  T \mbox{e}^{ik'n}.
\label{e13}
\eea

The application of the discrete Darboux transformation (\ref{e11}) to (\ref{e12}) and (\ref{e13}) leads to waves of the same energy but with modified transmission and reflection coefficients $\tilde T, \tilde R$. The results are

\bea
\tilde R = \left( \frac{G_- + \mbox{e}^{ik}F_-}{G_- + \mbox{e}^{-ik}F_-} \right) R
\label{e14}
\eea

\bea
\tilde T = \left( \frac{G_+ + \mbox{e}^{ik'}F_+}{G_- + \mbox{e}^{-ik}F_-} \right) T
\label{e15}
\eea

which constitute a generalization of the usual scattering matrix transformations with the replacement $ik \mapsto e^{ik}$. The reality condition for $G_{-}$ and  $F_-$ establishes that $|\tilde R|= |R|$. Moreover, in the problem of transparency $\tilde R$ vanishes if $R=0$.  Let us examine this possibility by proposing $H$ as a {\it free\ }hamiltonian, i.e. a periodic chain. This means that $\Delta_n=\Delta_{\pm} \equiv \Delta_0$ and $V_n=V_{\pm} \equiv V_0$. Now we must find $\tilde V_n$ and $\tilde \Delta_n$ by solving the recurrence

\bea
|G_n|^2 = V_0 - |F_{n+1}|^2 = \frac{|\Delta_0|^2}{ |F_{n}|^2}
\eea

for $|F_n|^2$. With the definitions $\mu_{\pm} = \ahalf (V_{0} \pm \sqrt{V_{0}^2-4\Delta_{0}^2})$ we obtain

\bea
F_n = \sqrt{\frac{\alpha_+ \mu_+ ^{n} + \alpha_- \mu_- ^{n}}{\alpha_+ \mu_+ ^{n-1} + \alpha_- \mu_- ^{n-1}}},
\label{e16}
\eea

\bea
G_n = \Delta_0 \sqrt{\frac{\alpha_+ \mu_+ ^{n-1} + \alpha_- \mu_- ^{n-1}}{\alpha_+ \mu_+ ^{n} + \alpha_- \mu_- ^{n}}},
\label{e17}
\eea

where $\alpha_{\pm}$ are arbitrary constants of the same sign, preserving the reality of $F_n$. The potential and hopping functions are now

\bea
\tilde V_n = V_0  + \frac{\alpha_+ \mu_+ ^{n} + \alpha_- \mu_- ^{n}}{\alpha_+ \mu_+ ^{n-1} + \alpha_- \mu_- ^{n-1}} - \frac{\alpha_+ \mu_+ ^{n+1} + \alpha_- \mu_- ^{n+1}}{\alpha_+ \mu_+ ^{n} + \alpha_- \mu_- ^{n}} \nonumber \\
\label{e18}
\eea

\bea
\tilde \Delta_n = \Delta_0 \frac{\sqrt{ (\alpha_+ \mu_+ ^{n} + \alpha_- \mu_- ^{n})(\alpha_+ \mu_+ ^{n-2} + \alpha_- \mu_- ^{n-2}) } }{ \alpha_+ \mu_+ ^{n-1} + \alpha_- \mu_- ^{n-1} }.\nonumber \\
\label{e19}
\eea

Finally, $\tilde H$ given in (\ref{e6}) is a non-trivial monoparametric familiy enjoying the property of being reflectionless. One can show that the solutions depend only on $\gamma \equiv \alpha_- / \alpha_+$, and that $\tilde V, \tilde \Delta$ can be put in terms of hyperbolic functions $\sinh ( n \ln \mu_{\pm}), \cosh ( n \ln \mu_{\pm})$. In figure \ref{fig:1} we show the behavior of potentials and hopping parameters under the modification of $\gamma$, resulting in a translation of the interaction region or potential well. This is reminiscent of shape invariance in continuous variables, where generalized P\"oschl-Teller or Scarf potentials can be translated at will on the real line (among other operations, such as rescaling). However, such a similarity is to be taken with a grain of salt, since continuous translations of discrete variables yield the same (translated) potentials only if $\nu = \ln(\gamma) / \ln (\mu_+ / \mu_-)$ is an integer. We can distinguish these features in the panels of figure \ref{fig:1}, as the centers of $\tilde V, \tilde \Delta$ move to the right and the functions suffer slight variations for non-integer values of $\nu$.

\begin{figure*}[t]
\includegraphics[width=10.6cm]{fig2}

\caption{\label{fig:2} Numerical solutions of the scattering problem in discrete variable $n$ for a Bloch wave impinging from the left. Each panel shows a different scattering energy $E$. The parameters are $\gamma=\mbox{e}^{-6}$, $V_0 = 6 \Delta_0$. Solid gold: $|\tilde \psi|$, dashed blue:  $\mbox{Re}(\tilde \psi)$,  dashed red:  $\mbox{Im}(\tilde \psi)$, blue-filled curve: $\tilde V$, red-filled curve: $\tilde \Delta$, solid black: $E$. In all cases $|T|^2 = 1$ and the original Bloch wave picks up a phase at the right end of the array.}
\end{figure*}

\subsection{The continuous limit \label{sec:3.2}}

It is important to make contact with the well-known results of transparent potentials in continuous SUSYQM. The limits can be reached by letting $a \rightarrow 0$ as before. In order to recover a Schr\"odinger equation with double derivatives and ground state energy $U_0$, we must impose $x=an$, $\Delta_0 \sim - R_0 / a^2$ with $R_0 >0$ and $V_0 + 2\Delta_0 \sim U_0$. In the process, we note that $T \sim 1 - a\cdot d/dx$ and $(\mu_+ / \mu_-)^n \sim \exp \left(4x\sqrt{R_0/U_0} \right)$. The particular choice $\gamma=1$ leads to a familiar case of hyperbolic superpotentials; we have

\bea
F_n \sim \frac{\sqrt{R_0}}{a} + \sqrt{\frac{U_0}{R_0}} \tanh \left( 2\sqrt{\frac{U_0}{R_0}} x \right),
\label{e21}
\eea

\bea
G_n \sim -\frac{\sqrt{R_0}}{a} + \sqrt{\frac{U_0}{R_0}} \tanh \left( 2\sqrt{\frac{U_0}{R_0}} x \right),
\label{e22}
\eea

and the Darboux operator becomes

\bea
A \sim 2\sqrt{\frac{U_0}{R_0}} \tanh \left( 2\sqrt{\frac{U_0}{R_0}} x \right) + \sqrt{R_0} \frac{d}{dx}.
\label{e23}
\eea

This is the usual operator for the Rosen-Morse superpotential with non-zero ground state $U_0$.

\subsection{A numerical test for transparency \label{sec:4}}
We test the reflectionless property by solving numerically the scattering problem for various energies. A reasonable choice of parameters for the potential and hopping functions is $\gamma=e^{-6}$ (producing strong asymmetry in the potential) and $V_0/\Delta_0 = 6$, localizing the region of interaction in a small portion of a few sites. The numerical solution is reached by imposing a Bloch wave at least at two sites, since (\ref{a4}) is a second order recurrence. For instance, at sites $n=0$ and $n=1$ we have

\bea
\tilde \psi_0 = \mbox{e}^{i \times 0}= 1,\quad \tilde \psi_1 = \mbox{e}^{ik}.
\eea

Such boundary conditions generate $\tilde \psi_{n+1}$ through the recurrence
\bea
\tilde \psi_{n+1} = \frac{ \tilde \Delta_n \tilde \psi_{n-1} +  (E-\tilde V_n)\tilde \psi_n}{\tilde \Delta_{n+1}},
\label{e20}
\eea

provided that $E$ and $k$ are related by $E=V_0 + \Delta_0 \cos k$. We have used four different energies in the scattering regime: $E/\Delta_0 = 7.66, 7.80, 7.91, 7.97$, verifying that the modulus of the {\it transmitted\ }wavefunction recovers the value $1$ in all cases. The results are shown in figure \ref{fig:2}, where the modulus, the real and the imaginary part of $\psi_n$ are displayed. At the right end of the array (25 sites) the wave recovers its modulus and phase factor, but at this region the potential is negligible and the solution will continue to be a Bloch wave propagating to the right if the array is prolonged indefinitely.

\begin{figure*}[t]
\includegraphics[width=10.6cm]{fig3}

\caption{\label{fig:3} The motion of discrete solitonic potentials (blue-filled curve) and solitonic hopping functions (green-filled curve). The parameter $\alpha = 1.5$ produces two solitons (a maximum and a minimum) for each graph. As $\gamma$ increases, their centers $C_1$ and $C_2$ move to the right at different velocities. This is shown progressively from top left to right bottom.}
\end{figure*}

\subsection{Biparametric solitons \label{sec:5}}

The solutions of the Korteweg-deVries (KdV) equation \cite{kruskal1967, kdv1895} are known to be represented by a hierarchy of superpotentials \cite{wang1990}. Such a hierarchy can be obtained through a step-by-step method for generating superpartners. In our discrete case, however, the application of new Darboux transformations can be increasingly challenging. Yet, a simpler strategy to obtain families of solutions consists of finding at once all the superpartners of a given reflectionless problem. We proceed in this direction in what follows. Let us start with a monoparametric transparent problem given by (\ref{e18}) and (\ref{e19}). We now consider a hamiltonian

\bea
H^{(\alpha)} = \Delta^{(\alpha)}(N) T + T^{\dagger} \Delta^{(\alpha)}(N)^{\dagger} + V^{(\alpha)}(N),
\label{e24}
\eea

where $\alpha$ is a new parameter, yet to be determined. The factorization procedure yields the relations

\bea
V^{(\alpha)}_n = |G_n^{(\alpha)}|^2 + |F_{n+1}^{(\alpha)}|^2, \nonumber \\ \Delta^{(\alpha)}_n = \left[G_n^{(\alpha)}\right]^* F_n^{(\alpha)},
\label{e25}
\eea

but another set of recurrences in terms of $\tilde V_n$ and $\tilde \Delta_n$ must be satisfied for the reflectionless problem:
\bea
\tilde V_n = |G_n^{(\alpha)}|^2 + |F_{n}^{(\alpha)}|^2, \nonumber \\ \tilde \Delta_n = \left[G_{n-1}^{(\alpha)}\right]^* F_n^{(\alpha)},
\label{e25.1}
\eea

where (\ref{e18}) and (\ref{e19}) must be substituted in the l.h.s. of (\ref{e25.1}). These relations are sufficient to determine $|F_n^{(\alpha)}|^2$ and $|G_n^{(\alpha)}|^2$. A particular solution is given, of course, by (\ref{e16}) and  (\ref{e17}), but the most general solution of (\ref{e25}) is a continued fraction

\bea
|F_n^{(\alpha)}| = \left[ |\tilde \Delta_n|^2,  \tilde V_{n-1}; ...; |\tilde \Delta_1|^2, \tilde V_0 - \alpha \right], \nonumber \\
|G_n^{(\alpha)}| = \frac{|\tilde \Delta_{n+1}|^2}{\left[ |\tilde \Delta_{n+1}|^2,  \tilde V_{n}; ...; |\tilde \Delta_1|^2, \tilde V_0 - \alpha \right]}.
\label{e26}
\eea

We identify the new parameter $\alpha$ with the initial condition of the recurrence, i.e. $\alpha = |F_0^{(\alpha)}|^2$. The potentials and hopping functions can be reconstructed by means of the relations

\bea
V_n^{(\alpha)} = \tilde V_n + |F_{n}^{(\alpha)}|^2 - |F_{n+1}^{(\alpha)}|^2 ,
\label{e27}
\eea

\bea
\Delta_n^{(\alpha)} = \tilde \Delta_n \frac{F_{n}^{(\alpha)}}{F_{n+1}^{(\alpha)}}.
\label{e28}
\eea

The expressions (\ref{e27}) and (\ref{e28}) represent a biparametric family of transparent potentials and hopping functions. It is worthwhile to investigate their behavior as a function of both $\alpha$ and the original parameter $\gamma$. For example, setting $\alpha=1.5$ takes us to two solitons for each of the functions $V^{(\alpha)}_n, \Delta^{(\alpha)}_n$. For values $\gamma \sim 1$ the solitons are close to each other (see figure \ref{fig:3}). Increasing $\gamma$ exponentially produces their motion with respect to the origin as well as a relative displacement between them. Thus, we have two solitons with two different velocities \cite{wang1990}.

\section{Conclusion and outlook \label{sec:6}}

Lattice design by site and coupling engineering gives rise to many possibilities of which the present paper is an example. A concrete experiment showing transparency --among other properties predicted by SUSYQM-- can be proposed using current technologies, as discussed in section \ref{sec:experiment}. A particular configuration using dielectric slabs was provided in \ref{sec:slabs}. We also recognize that SUSYQM is indeed a powerful method; its application to discrete problems has been demonstrated by finding systems with a desired property. From the mathematical point of view, we have found that some aspects of solitons \cite{lederer2008, nijhoff1995, wang1990} can be reproduced also in tight-binding arrays, motivating further explorations towards discrete spinorial KdV equations. The extension of the present study to 2D lattices seems plausible. Moreover, discrete exactly solvable problems \cite{natig1991, natig2001} and their relation with shape invariant potentials can be explored in this context.

\begin{acknowledgments}
I am grateful to T. H. Seligman for useful comments on the manuscript. Financial support from CONACyT under project CB 2012-180585 is acknowledged.
\end{acknowledgments}

\providecommand{\noopsort}[1]{}\providecommand{\singleletter}[1]{#1}


TITLE: A question on the fibered product of schemes
QUESTION [2 upvotes]: In Hartshorne's "Algebraic Geometry", on p.99, at the end of the proof of Lemma 4.5, the author mentions that in general, if $\xi \in X \times_Y X$ and $p_1(\xi) = p_2(\xi)$, where $p_1$ and $p_2$ are the projection morphisms onto the first and second factors respectively, then it is not true in general that $\xi$ belongs necessarily to $\Delta(X)$, where
$\Delta: X \to X \times_Y X$
is the diagonal morphism.
I haven't thought much about it, but is it easy to construct a counterexample? I ask for examples of schemes $X$ and $Y$, with a morphism $f: X \to Y$, such that there is a point $\xi \in X \times_Y X$ with $p_1(\xi) = p_2(\xi)$, but $\xi \notin \Delta(X)$.
Edit: If we take $X = \mathbb{A}^1_\mathbb{C}$ to be the affine line over $\mathbb{C}$. Then $X \times_\mathbb{C} X$ is the affine plane $\mathbb{A}^2_\mathbb{C}$. Let $\xi$ be the generic point in $\mathbb{A}^2_\mathbb{C}$ of the line $y=2x$. Then is it true that $p_1(\xi) = p_2(\xi)$, and that $\xi \notin \Delta(X)$? Let $f_1: \mathbb{C}[x] \to \mathbb{C}[x,y]$ and $f_2: \mathbb{C}[y] \to \mathbb{C}[x,y]$ be the natural ring homomorphisms, corresponding to the 2 projections. Then $f_i^{-1}(y-2x) = (0)$, for $i=1,2$, which confirms my suspicions, that $p_1(\xi) = p_2(\xi)$. Moreover, $\xi$ does not belong to $\Delta(X)$. Thus I indeed found a counterexample.

REPLY [2 votes]: Let $Y=\operatorname{Spec}(\mathbb R), X=\operatorname{Spec}(\mathbb C)$. Since $X$ consists of one point, $\Delta(X)$ is just one point, and therefore also $p_1(q)=p_2(q)$ for each $q\in X\times_Y X$. Note that
$$X\times_Y X=\operatorname{Spec}(\mathbb C\otimes_{\mathbb R}\mathbb C)=\operatorname{Spec}(\mathbb C\times\mathbb C)=\operatorname{Spec}(\mathbb C)\amalg\operatorname{Spec}(\mathbb C)$$
consists of two points, so $\Delta(X)\neq X\times_Y X$.


TITLE: Hairy Ball Theorem on $\mathbb{S}^2$: a counter-example?
QUESTION [4 upvotes]: Let $(U, (\theta, \varphi))$ be the spherical coordinate chart on the sphere $\mathbb{S}^2$, and consider the vector field on $U$ defined by $\dfrac{\partial}{\partial \varphi}$. Then in this chart, the metric tensor $g_{ij}$ can be expressed as follows:
$$g_{ij} = d\varphi^2 + \sin^2\varphi \, d\theta^2. $$
This means that the vector $\dfrac{\partial}{\partial \varphi} \Bigr|_{p} \in T_p\mathbb{S}^2$ has norm $1$ for any point $p \in U$. But this seems to contradict the Hairy Ball theorem, which says that every smooth vector field on $\mathbb{S}^2$ has to vanish somewhere. What's going on? What is wrong with this logic?
I think the problem is that the spherical coordinate chart on $\mathbb{S}^2$ is not a global chart. So perhaps this vector field defined on $U$ cannot be extended to a smooth vector field on the whole manifold. My question is: is there an intuitive geometric reason why we would expect that this vector field cannot be smoothly extended? Something involving pictures of vector fields would be helpful. Thanks.

REPLY [6 votes]: In the hope a picture is worth a thousand words:


TITLE: Limits with two variables
QUESTION [1 upvotes]: $$
\lim_{(x,y)\to(0.0)} \frac{xy\cos y}{3x^2+y^2}
$$
Approach along the path $x=0$, $y$ not $0$:
$$\frac{0}{3x^2}=0$$
Approach along $x=y$:
$$\frac{x^2\cos x}{4x^2}=\frac{\cos x}{4}$$
Therefore, it shouldn't exist as the limits are not the same. Is this approach / logic right?
(Also, how do I properly type up maths formatting? Is there some easy way to type in stuff like "3x/y^5" and it will automatically render the appropriately coded text?)

REPLY [2 votes]: Your approach is good. As for the second question, you have to use the \frac and $$ operator here.
As an example, you have to put the following code between \$\$ and \$\$
\frac{3x}{y^5}


TITLE: Division of Factorials [binomal coefficients are integers]
QUESTION [46 upvotes]: I have a partition of a positive integer $(p)$. How can I prove that the factorial of $p$ can always be divided by the product of the factorials of the parts?
As a quick example $\frac{9!}{(2!3!4!)} = 1260$ (no remainder), where $9=2+3+4$.
I can nearly see it by looking at factors, but I can't see a way to guarantee it.

REPLY [3 votes]: Let's prove using induction, the special case of two numbers, i.e., the statement that if $p, q$ $\in \mathbb{N}$ then $p!q!|(p+q)!$.
(Assume any new variables introduced below to refer to natural numbers.)
First note that since $(p+1)! = (p+1)p!1!$, the statement is true for $q = 1$ and any $p$ (including $p = 1$). In particular, it is true for $p + q = 1 + 1 = 2$. Let us assume that the statement holds for $p, q$ such that $p + q = n$.
Now, for $p, q$ such that $p + q = n + 1$, we can write
$$(p + q)! = (p + q)(p + q - 1)!$$
$$= p [\underbrace{(p - 1) + q}_n]! + q [\underbrace{p + (q - 1)}_n]!$$
$$= p \underbrace{k_1 (p-1)!q!}_{\text{using induction assumption}} + q \underbrace{k_2 p!(q-1)!}_{\text{using induction assumption}}$$
$$= (k_1 + k_2) p! q!$$. Hence the principle of mathematical induction implies the truth of the statement.
Now it is easy to prove the analogous statement for three numbers, i.e. $p!q!r! | (p + q + r)!$, since (using the statement just proven) $(p + q + r)!$ is divisible by $p! (q+r)!$ and $(q + r)!$ is divisible by $q!r!$.
This can be generalised for any number of parts, by induction.