text:
\begin{document}

\title[Calabi-Yau hypersurfaces in weighted projective spaces]{Mirror symmetry
for quasi-smooth Calabi-Yau hypersurfaces in weighted projective spaces}

\author[Victor Batyrev]{Victor Batyrev}
\address{Mathematisches Institut, Universit\"at T\"ubingen,
Auf der Morgenstelle 10, 72076 T\"ubingen, Germany}
\email{victor.batyrev@uni-tuebingen.de}

\author[Karin Schaller]{Karin Schaller}
\address{Mathematisches Institut, Freie Universit\"at Berlin,
Arnimallee 3, 14195 Berlin, Germany}
\email{karin.schaller@fu-berlin.de}

\begin{abstract}
We consider a $d$-dimensional  well-formed weighted projective
space $\P(\w)$ as a toric variety associated with a fan $\Sigma(\w)$ in
$N_{\w} \otimes \R$ whose $1$-dimensional cones are spanned  by primitive
vectors $v_0, v_1, \ldots, v_d \in N_{\w}$ generating a lattice
$N_{\w}$ and satisfying the linear relation $\sum_i w_i v_i =0$.
For any fixed dimension $d$, there exist only finitely
many weight vectors $\w = (w_0, \ldots, w_d)$ such that $ \P(\w)$
contains a quasi-smooth Calabi-Yau hypersurface $X_w$
defined by a transverse weighted homogeneous polynomial
$W$ of degree $w = \sum_{i=0}^d w_i$.
Using a formula of Vafa for the orbifold Euler number
$\chi_{\rm orb}(X_w)$, we show
that for any quasi-smooth Calabi-Yau hypersurface $X_w$
the number $(-1)^{d-1}\chi_{\rm orb}(X_w)$ equals
the stringy Euler number  $\chi_{\rm str}(X_{\w}^*)$ of
Calabi-Yau compactifications $X_{\w}^*$ of
affine toric hypersurfaces $Z_{\w}$ defined by
non-degenerate Laurent polynomials $f_{\w} \in \C[N_{\w}]$
with Newton polytope $\conv(\{v_0, \ldots, v_d\})$.
In the moduli space of Laurent polynomials $f_{\w}$
there always exists a special point $f_{\w}^0$ defining  a mirror $X_{\w}^*$
with a $\Z/w\Z$-symmetry group such that $X_{\w}^*$ is birational to a
quotient of a Fermat hypersurface via a Shioda map.
\end{abstract}

\maketitle
\thispagestyle{empty}

\section{Introduction}

Many topologically different
examples of smooth Calabi-Yau threefolds and many evidences
for mirror symmetry were obtained from quasi-smooth
Calabi-Yau hypersurfaces $X_w$  in $4$-dimensional
weighted projective spaces
$\P(w_0,w_1,w_2,w_3,w_4)$ defined by weighted homogeneous polynomials
$W \in \C[z_0, \ldots, z_4]$ of degree $w = \sum_{i=0}^4 w_i$
such that the differential
$dW$ vanishes exactly in the origin $0 \in \C^5$
\cite{CLS90,GP90,CdlOGP91,CdlOK95}.
The Hodge numbers of two $d$-dimensional smooth Calabi-Yau varieties $V$ and $V^*$ that
are mirror symmetric to each other must satisfy the equalities
\[ h^{p,q}(V) = h^{d-p,q}(V^*)  \]
for all $0 \leq p,q \leq d$ \cite{Wit92}. In particular, the Euler number
$\chi = \sum_{p,q} (-1)^{p+q} h^{p,q}$ must satisfy the equality
\[ \chi(V) =  (-1)^d  \chi(V^*).\]
Unfortunately, quasi-smooth Calabi-Yau hypersurfaces
 $X_w \subset \P(w_0,w_1,w_2,w_3,w_4)$ defined
by weighted homo\-ge\-neous polynomials $W$ are usually singular.
The first ma\-the\-ma\-tical verifications of the above mentioned equalities for
Hodge and Euler numbers were based on the existence of
crepant desingularizations $\rho: Y \to X_w$
that allow to replace $X_w$ by
smooth Calabi-Yau threefolds  $Y$ \cite{Roa90}.
Note that crepant desingularizations  $\rho: Y \to X_w$
of quasi-smooth Calabi-Yau hypersurfaces $X_w \subset
\P(w_0, w_1, \ldots, w_d)$ do not exist in general if  $d \geq 5$.

\smallskip
Let us recall some definitions and facts on
$d$-dimensional weighted projective spaces $\P(\w)$
and quasi-smooth Calabi-Yau hypersurfaces in
$\P(\w)$, where $\w \defeq (w_0, w_1, \ldots, w_d)$.

\begin{dfn}
A weighted projective space
$\P(\w) $ is a quotient of
$\C^{d+1} \setminus \{0 \}$ by the $\C^*$-action  $\lambda \cdot(a_0, a_1, \ldots, a_d)
=(\lambda^{w_0} a_0, \lambda^{w_1} a_1, \ldots, \lambda^{w_d} a_d)$
defined by the {\em weight vector} $\w=(w_0, w_1, \ldots, w_d) \in \Z^{d+1}_{>0}$,
whose coordinates $w_i$ are called {\em weights}.
A weighted projective space $\P(\w)$ is called {\em well-formed} if
\[\ggT(w_0, \ldots, w_{i-1}, w_{i+1}, \ldots, w_d) =1 \;\; \forall
 i \in I \defeq \{0,1, \ldots, d\}. \]
In this paper, we consider only well-formed weighted projective spaces $\P(\w)$.
A weighted homogeneous polynomial $W \in \C[z_0, z_1, \ldots,
z_d]$ of degree $k$ is characterized by the condition
\[ W(\lambda^{w_0}z_0, \lambda^{w_1}z_1, \ldots, \lambda^{w_d}z_d)
= \lambda^k \cdot W(z_0, z_1, \ldots, z_d). \]
Moreover, a weighted homogeneous polynomial $W$ of degree $k$ and the hypersurface
$X_k \defeq \{ W({\bf z}) = 0 \}  \subset \P(\w)$ of degree $k$
are called {\em transverse} if the common zeros of all partial derivations
\[ \partial W/\partial z_i = 0 \;\;  ( 0 \leq i \leq d)\] is the point $0 \in \C^{d+1}$.
A weight vector $\w= (w_0, w_1 , \ldots, w_d) \in \Z_{>0}^{d+1}$ is called {\em transverse}
if there exists at least one transverse Calabi-Yau
hypersurface $X_w \subset \P(\w)$ of degree
$w = \sum_i w_i$. Another name, more often used by mathematicians,
for transverse Calabi-Yau hypersurfaces is {\em quasi-smooth} Calabi-Yau hypersurfaces.
The quasi-smoothness (or transversality) condition
ensures that the hypersurface $X_w$ has no singularities in addition to those coming
from the singularities of the ambient space \cite{KS98a}, where
the only singularities of weighted projective spaces are
cyclic quotient singularities \cite[Definition 11.4.5]{CLS11}.
\end{dfn}

\begin{dfn} A weight vector $\w \in \Z^{d+1}_{>0}$
has the {\em $\text{IP}$-property} if
\[ \Delta(W) \defeq \conv(  \{ (u_0, u_1, \ldots, u_d) \in \Z^{d+1}_{\geq 0}
\, \vert \, \sum_{i=0}^d w_i u_i = w \} ) \subseteq \R^{d+1} \]
is a $d$-dimensional lattice polytope containing the lattice point
$(1,1,\ldots, 1) \in \Z^{d+1}$ in its interior.
\end{dfn}

By \cite[Lemma 2]{Ska96}, any transverse weight vector
$\w \in \Z^{d+1}_{>0}$ has the $\text{IP}$-property.
For any fixed dimension $d$, there exist only finitely many
transverse weight vectors and only finitely many
weight vectors $\w \in \Z_{> 0}^{d+1}$ with $\text{IP}$-property.
The numbers $T(d)$ and $\text{IP}(d)$ are known for $d \leq 5$:

\begin{itemize}
\item[-] $T(2) = \text{IP}(2) =3$:  $\{ \w \} = \{(1,1,1), (1,1,2), (1,2,3) \}$;
\item[-]  $T(3) = \text{IP}(3) = 95$   \cite{Rei80};
\item[-]  $T(4) = 7,\!555$ \cite{KS92,KS94},
$\text{IP}(4) =184,\!026$ \cite{Ska96};
\item[-]  $T(5) = 1,\!100,\!055$
\cite{LSW99,BK16},
$\text{IP}(5) =322,\! 383,\! 760,\! 930 $ \cite{SchS19}.
\end{itemize}

The following statement is a general result of
Artebani, Comparin, and Guilbot \cite[Theorem 1]{ACG16}
in the particular case of weighted projective spaces:

\begin{prop}
Let $\w = (w_0, w_1, \ldots, w_d)$ be a weight vector
with $\text{IP}$-property. Then a general hypersurface
$X_w \subset \P(\w)$ of degree $w = \sum_i w_i$ is
a Calabi-Yau variety.
\end{prop}

\smallskip
The present paper is inspired
by mirror symmetry and the following formula of Vafa for
the orbifold Euler number  $\chi_{\rm orb}(X_w)$ of
quasi-smooth Calabi-Yau hypersurfaces
$X_w  \subset P(\w)=\P(w_0,w_1, \ldots ,w_d)$
of degree $w= \sum_{i=0}^d w_i$ for arbitrary $d\geq 2$:
\begin{align} \label{vafa0}
\chi_{\rm orb}(X_w) = \frac{1}{w}
\sum_{l,r=0}^{w-1} \, \, \prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}
\left(1 - \frac{1}{q_i} \right),
\end{align}
where $q_i \defeq  \frac{w_i}{w}$ for all $i \in I = \{0,1,\ldots,d\}$ and one assumes
$\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z} \left( 1 - \frac{1}{q_i} \right) =1$
if $lq_i,rq_i \not\in \Z$  for all $i \in I$.
It is remarkable that Formula \eqref{vafa0} appeared without using
algebraic geometry as the {\em Witten index} ${\rm Tr}(-1)^F$ of the $N=2$
superconformal Landau-Ginzburg field theory defined by a weighted homogeneous superpotential $W$
of degree $w$ \cite[Formula (28)]{Vaf89}.

\smallskip
A first general mathematical ($K$-theoretic) interpretation
of Vafa's Formula \eqref{vafa0} was obtained by Ono and Roan:

\begin{theopar}[{\cite[Theorem 1.1]{OR93}}]  \label{K-th}
Let $S^{2d+1} \subseteq \C^{d+1} \setminus \{0 \}$ be
the unit sphere. Consider the compact smooth $(2d-1)$-dimensional real manifold
 $S_w \defeq S^{2d+1} \cap \{ W = 0\}$  together with the
 $S^1$-fibration $S_w \to X_w$ which is the restriction of the Seifert $S^1$-fibration
$S^{2d+1} \to \P(w_0,w_1, \ldots ,w_d)$ to a quasi-smooth Calabi-Yau hypersurface
$X_w$ of degree $w = \sum_{i=0}^d w_i$.
Then the $S^1$-equivariant $K$-groups $K^i_{S^1}(S_w)$  $(i=0,1)$
have finite rank and
\[ {\rm rank}\,  K^0_{S^1}(S_w) -  {\rm rank}\,  K^1_{S^1}(S_w) =
\frac{1}{w} \sum_{l,r=0}^{w-1} \, \, \prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}
\left( 1 - \frac{1}{q_i} \right). \]
In particular,
the right side of this equation is an integer.
\end{theopar}

Quasi-smooth Calabi-Yau hypersurfaces
$X_w \subset \P(\w)=\P(w_0,w_1, \ldots ,w_d)$ can be considered
as orbifolds, \emph{i.e.}, as geometric objects that are locally quotients
of smooth manifolds by finite group actions. The
orbifold cohomology theory developed by Chen and Ruan \cite{CR04,ALR07}
allows to define for any projective orbifold $V$ certain vector spaces
$H^{p,q}_{CR}(V)$, whose dimensions $h^{p,q}_{\rm orb}(V)$
are called {\em orbifold Hodge numbers}. In particular, one obtains
\[ \chi_{\rm orb}(V) \defeq \sum_{p,q} (-1)^{p+q} h^{p,q}_{\rm orb}(V). \]
Another mathematical approach to orbifold Hodge and orbifold Euler
numbers was suggested  in \cite{BD96,Bat98} via so-called
{\em stringy Hodge numbers}
$h^{p,q}_{\rm str}(X)$ that were defined for arbitrary
projective algebraic varieties $X$ with at worst
Gorenstein canonical singularities.
We note that such singularities do not necessarily
admit a local orbifold structure.
The mathematical definition of stringy Hodge numbers $h^{p,q}_{\rm str}(X)$
uses an arbitrary desingularization $\rho : Y \to X$. Their definition
immediately implies the equalities
$h^{p,q}_{\rm str}(X) =h^{p,q}(Y)$
if the desingularization  $\rho$ is crepant \cite{Bat98}.

\smallskip
Our consideration of the orbifold Euler number
is motivated by the combinatorial mirror duality
for Calabi-Yau hypersurfaces $X$ in $d$-dimensional
Gorenstein toric Fano varieties  \cite{Bat94}.
This mirror duality is based on the
polar duality $\Delta \leftrightarrow \Delta^*$ between $d$-dimensional
reflexive lattice polytopes $\Delta$ and
$\Delta^*$. Recall that a $d$-dimensional lattice polytope
$\Delta \subseteq M_\R \cong \R^d$
is called {\em reflexive} if it contains the lattice point
$0 \in M$ in its interior and the {\em dual polytope}
\[ \Delta^* \defeq \{ y \in N_\R \, \vert\,
\langle x, y \rangle \geq -1 \;\; \forall x \in \Delta \} \subseteq N_\R \]
is also a lattice polytope. If $\Delta $ is reflexive, then
$\Delta^*$ is also reflexive and $(\Delta^*)^* = \Delta$.
The polar duality induces a natural
one-to-one correspondence $\theta \leftrightarrow \theta^*$
between $k$-dimensional faces $\theta \preceq\ \Delta$ of $\Delta$ and
$(d-k-1)$-dimensional faces $\theta^* \preceq \Delta^*$ of $\Delta^*$.
If $X$ is a general Calabi-Yau hypersurface in the Gorenstein toric Fano variety
$\P_\Delta$ associated with the normal fan $\Sigma_{\tope}$ of $\Delta$,
then the {\em stringy Euler number}
of $X$ can be computed by the following combinatorial
formula \cite[Cor. 7.10]{BD96}:
\begin{align}  \label{comb-Euler}
\chi_{\rm str}(X)= \sum_{k=1}^{d-2} (-1)^{k-1} \sum_{ \theta \preceq \Delta \atop \dim(\theta) = k}
\Vol_k (\theta) \cdot \Vol_{d-k-1} (\theta^*),
\end{align}
where $\Vol_{\dim(\cdot)}(\cdot) \in \N$ denotes the normalized volume (cf. Definition \ref{normvol}).
Using Formula \eqref{comb-Euler} and the bijection $\theta \leftrightarrow \theta^*$,
 one immediately obtains the equality
\begin{align*}
\chi_{\rm str}(X) = (-1)^{d-1} \chi_{\rm str}(X^*),
\end{align*}
where $X^*$ denotes the mirror of $X$ obtained as a
Calabi-Yau hypersurface in the Gorenstein toric Fano
variety $\P_{\Delta^*}$. Moreover, by combinatorial
methods one can show the following equalities of stringy Hodge numbers
\[h^{p,q}_{\rm str}(X) = h^{d-1-p,q}_{\rm str}(X^*) \;\; (0 \leq p,q \leq d-1) \]
predicted by mirror duality \cite{BB96}.

\begin{rem}
The following combinatorial property of the polar duality
for reflexive polytopes is crucial for the combinatorial  Mirror Construction \ref{conj-mirr}:

\begin{itemize}
\item[] The set of  generators of  $1$-dimensional cones $\sigma$ in the normal fan
$\Sigma_\Delta \subseteq N_\R$ of a reflexive polytope $\Delta \subseteq M_\R$
is the set of vertices of the dual reflexive polytope
$\Delta^* \subseteq N_\R$.
\end{itemize}
\end{rem}

We consider a well-formed weighted projective space  $\P(\w)$ as
a projective toric variety corresponding to a $d$-dimensional
simplicial fan $\Sigma(\w)$ in $N_{\w} \otimes \R$ whose $1$-dimensional
cones are spanned by primitive lattice vectors
$v_0, v_1, \ldots, v_d \in N_{\w}$ generating the lattice $N_{\w}$ and satisfying the linear
relation $\sum_{i=0}^d w_i v_i =0$. In this way, the lattice
$N_{\w}$ can be identified with the quotient lattice $\Z^{d+1}/\Z\w$.

 \smallskip
 If $\w \in \Z_{>0}^{d+1}$ is a weight vector having the $\text{IP}$-property, then
 the $d$-dimensional lattice polytope
 $\Delta'(W) \defeq  \Delta(W) - (1, \ldots, 1) \subseteq M_{\w} \otimes \R$
 contains the origin $0 \in M_{\w}$ in its interior.
We define the {\em spanning fan}  $\Sigma^{\vee}(\w)$ of $\Delta'(W)$ as
 the fan  in $M_{\w} \otimes \R$  consisting  of all cones $\R_{\geq 0}\theta'$
 over faces $\theta' \preceq \Delta'(W)$ of  the lattice polytope $\Delta'(W)$.
 By the theory of toric varieties \cite{CLS11},  the fan  $\Sigma^{\vee}(\w)$ defines
 a $d$-dimensional $\Q$-Gorenstein toric Fano variety $\P^{\vee}(\w)$, which is a
compactification of the algebraic torus $\T_{\w}$ with the lattice of
characters $N_{\w}$, \emph{i.e.},
 \[\T_{\w} \defeq \{(x_0,x_1,\ldots,x_d) \in  (\C^*)^{d+1} \, \vert \,  \prod_{i=0}^d x_i^{w_i} =1 \}
\subset  (\C^*)^{d+1}. \]

Using several results from \cite{Bat17},
we prove the following main statement of our paper:

\begin{theo} \label{vafemirrortest}
Let  $\w = (w_0, w_1, \ldots, w_d)$ be a weight vector with $\text{IP}$-property
and $Z_{\w} \subset \T_{\w}$ a non-degenerate affine
hypersurface defined by a Laurent
polynomial $f_{\w}$ with Newton polytope
 $\Delta_{\w}^*= \conv(\{v_0, \ldots, v_d\}) \subseteq N_{\w} \otimes \R$.
Then the Zariski closure of $Z_{\w}$
in the $\Q$-Gorenstein toric Fano variety $\P^{\vee}({\w})$
is a Calabi-Yau hypersurface $X_{\w}^*$ and
\[  \chi_{\rm str}(X^*_{\w})  =  (-1)^{d-1} \frac{1}{w} \sum_{l,r=0}^{w-1} \, \,
\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z} \left( 1 - \frac{1}{q_i} \right), \]
where $q_i = \frac{w_i}{w}$ $($$i \in I$$)$.
In particular, if $\w$ is a transverse weight vector, \emph{i.e.}, there exists
a quasi-smooth Calabi-Yau hypersurface
$X_w \subset \P(\w)$ of degree $w = \sum_i w_i$, then
\[ \chi_{\rm str}(X_{\w}^*) = (-1)^{d-1} \chi_{\rm orb}(X_w). \]
\end{theo}

The last statement in Theorem \ref{vafemirrortest}
supports the following mirror construction:

\begin{constr} \label{conj-mirr}
Let $\w \in \Z_{>0}^{d+1}$ be an arbitrary transverse
weight vector. Then mirrors of quasi-smooth Calabi-Yau hypersurfaces
$X_w\subset \P(\w)$ can be obtained as Calabi-Yau compactifications
$X_{\w}^*$ of non-degenerate affine hypersurfaces $Z_{\w} \subset \T_{\w}$
with Newton polytope $\Delta_{\w}^* = \conv(\{v_0, v_1, \ldots, v_d\})$.
\end{constr}

\begin{rem}
If the weighted projective space
$\P(\w)=\P(w_0, w_1, \ldots, w_d)$ is not Gorenstein, then
the lattice simplex $\Delta_{\w}^*$ is not reflexive and its
dual simplex
\[\Delta_{\w} =  \{ (u_0, u_1, \ldots, u_d) \in \R^{d+1}_{\geq 0}
\, \vert \, \sum_{i=0}^d w_i u_i = w \}  - (1,\ldots, 1)\]
is a rational simplex containing
the lattice polytope $\Delta'(W) \subseteq M_{\w} \otimes \R$. It
follows from \cite[Theorem 2]{ACG16} that if a $\Q$-Gorenstein
weighted projective space $\P(\w)$ contains
a Calabi-Yau hypersurface $X_w$ defined by an
{\em invertible} weighted homogeneous polynomial
$W$, then the lattice simplex $\Delta_{\w}^*$ is the Newton polytope
of the Berglund-H\"ubsch-Krawitz mirror of $X_w \subset \P(\w)$
\cite[Section 5]{CR18}, \cite{Kra10,CR11,Bor13}.
\end{rem}

\begin{rem}
Note that a weighted projective space $\P(\w)= \P(w_0,w_1, \ldots ,w_d)$
is a Gorenstein if and only if
each weight $w_i \; ( 0 \leq i \leq d)$ divides the degree
$w = \sum_{i=0}^d w_i$. In the latter case, one can
choose a  quasi-smooth Calabi-Yau hypersurface  $X_w$
defined by the following weighted homogeneous polynomial
of {\em Fermat-type}:
\[ W= z_0^{w/w_0} + z_1^{w/w_1} + \cdots + z_d^{w/w_d}
\in \C[z_0, z_1, \ldots, z_d]. \]
In this case, the two simplices $\Delta_{\w}^*$ and
\[ \Delta_{\w} = \Delta'(W) =\Delta(W) - (1, \ldots, 1) \]
are reflexive and Mirror Construction \ref{conj-mirr}
coincides with the orbifoldizing mirror
construction of Greene-Plesser \cite{GP90}.
\end{rem}

\begin{rem}
The lattice simplex $\Delta_{\w}^* = \conv(\{v_0, v_1, \ldots, v_d\})$
is the Newton polytope of the special non-degenerate Laurent polynomial
$f_{\w}^0({\bf t}) \defeq \sum_{i=0}^d {\bf t}^{v_i}  \in \C[N_w]$
defining the affine hypersurface
\[Z_{\w}^0 = \T_{\w} \cap
 \{(x_0,x_1,\ldots,x_d) \in (\C^*)^{d+1}   \,  \vert \, \sum_{i=0}^d x_i =0  \}
\subset \T_{\w},\]
\emph{i.e.},  $f^0_{\w}$ is the restriction of the linear function
$x_0 + x_1 + \ldots + x_d: (\C^*)^{d+1} \to \C $
to the $d$-dimensional subtorus $\T_{\w} \subset  (\C^*)^{d+1}$.
Note that the Laurent polynomial $f_{\w}^0$ appears in the Givental-Hori-Vafa mirror
construction for the weighted projective space $\P(\w)$ \cite[Section 4.4]{CR18}, \cite{HV00}.
The restriction  of the projection
\[ (\C^*)^{d+1} \to (\C^*)^d, \;\; (x_0, x_1, \ldots, x_d) \mapsto
(x_1/x_0, \ldots, x_d/x_0) \]
to the affine hypersurface
$Z_{\w}^0 \subset \T_{\w} \subset  (\C^*)^{d+1}$
defines an unramified cyclic  Galois covering
$\gamma_w : Z_{\w}^0 \to Z_0$ of order $w$, where
\[ Z_0 \defeq \{ (y_1, \ldots, y_d) \in (\C^*)^d \, \vert \,
\sum_{i=1}^d y_i = -1 \} \subset (\C^*)^d. \]
Therefore, the affine hypersurface $Z_{\w}^0$ can be obtained
as a $(\Z/w\Z)^{d-1}$-quotient of the affine
Fermat hypersurface
\[ F_w \defeq  \{ (y_1, \ldots, y_d) \in (\C^*)^d \, \vert \,
\sum_{i=1}^d y_i^w = -1 \} \subset (\C^*)^d. \]
Thus, the Givental-Hori-Vafa polynomial $f^0_{\w}$ defines a special
"Fermat-type" point in the moduli space
of mirrors of quasi-smooth
Calabi-Yau hypersurfaces $X_w \subset \P(\w)$.
The Givental-Hori-Vafa mirrors  $Z^0_{\w}$ were considered by Kelly \cite{Kel13} via
so-called {\em Shioda maps} in the Berglund-H\"ubsch-Krawitz mirror constructions
for Calabi-Yau hypersurfaces $X_w \subset \P(\w)$ defined by invertible
polynomials $W$.
\end{rem}

\begin{expar}[{\cite[Example 53]{CR18}}]
Let   $\w =(w_0, w_1, \ldots, w_d)$ be a weight vector
with $w_0 =1$. Then   $v_0 = - \sum_{i=1}^d w_i$ and
the lattice vectors $v_1, \ldots, v_d \in \Z^d$ form
a $\Z$-basis of $N_{\w}$. Thus, the Givental-Hori-Vafa
polynomial $f^0_{\w} \in \C[N_{\w}]$
can be written  in the form
\[ f^0_{\w}({\bf t}) = \sum_{i=0}^d {\bf t}^{v_i} =
\frac{1}{t_1^{w_1} \cdot \ldots \cdot t_d^{w_d}}+  t_1 + \ldots + t_d. \]
\end{expar}

We note that we can not expect Mirror Construction \ref{conj-mirr}
working for Calabi-Yau hypersurfaces $X_w  \subset \P(\w)$
that are not quasi-smooth. The following Examples \ref{not-a-mirror}
and \ref{EulerQ} were proposed by Harald Skarke:

\begin{expar}[{\cite[Example 2]{KS92}}]  \label{not-a-mirror}
Consider the $\text{IP}$-weight vector  $\w =(1,1,6,14,21)$.
One can show that  the weight vector $\w$
is not transverse. The $4$-dimensional lattice polytope
$\Delta'(W) \subset  \Delta_{\w}$
is reflexive \cite[Lemma 1]{Ska96}.
Therefore, general Calabi-Yau hypersurfaces $X_{43} \subset \P(\w)$
are birational to smooth Calabi-Yau
threefolds $Y$ with  Hodge numbers
$h^{1,1}(Y) = 21$, $h^{2,1}(Y) = 273$,
and Euler number $\chi(Y) = - 504$   \cite{Bat94}.

\smallskip
On the other hand, by Theorem \ref{vafemirrortest},
the Givental-Hori-Vafa hypersurface
$Z_{\w}^0 \subset (\C^*)^4$ defined by the equation
\[ \frac{1}{t_1 t_2^6 t_3^{14} t_4^{21}} + t_1 +t_2 + t_3 + t_4 = 0  \]
is birational to a $3$-dimensional Calabi-Yau variety $X^*_{\w}$
with  stringy Euler number
\[\chi_{\rm str}(X^*_{\w}) = - \frac{1}{43}
\sum_{l,r=0}^{42} \prod_{0 \leq i \leq 4 \atop lq_i, rq_i \in \Z}
\left( 1 - \frac{1}{q_i} \right) =     506  \neq 504 =
 - \chi_{\rm str}(X_{43}) = -\chi(Y).\]
Therefore, $X^*_{\w}$ is not mirror of $X_{43}$.
\end{expar}

\begin{ex}\label{EulerQ}
The weight vector  $\w =(1,1,2,4,5)$ has {\rm IP}-property.
Since the  $4$-dimensional lattice polytope
$\Delta'(W) \subset  \Delta_{\w}$ is reflexive,
a general Calabi-Yau hypersurface $X_{13} \subset \P(\w)$
is birational to a smooth Calabi-Yau
threefold. On the other hand,
by Theorem \ref{vafemirrortest},
the Givental-Hori-Vafa hypersurface
$Z_{\w}^0 \subset (\C^*)^4$ defined by the equation
\[ \frac{1}{t_1 t_2^2 t_3^{4} t_4^{5}} + t_1 +t_2 + t_3 + t_4 = 0  \]
is birational to a $3$-dimensional Calabi-Yau variety $X^*_{\w}$
with  stringy Euler number
\[\chi_{\rm str}(X^*_{w}) = - \frac{1}{13}
\sum_{l,r=0}^{12} \prod_{0 \leq i \leq 4 \atop lq_i, rq_i \in \Z}
\left( 1 - \frac{1}{q_i} \right) =   \frac{1032}{5} \not\in \Z. \]
Therefore, the Calabi-Yau variety $X^*_{\w}$ cannot have a  Landau-Ginzburg
description and  $X^*_{\w}$ has no
mirror at all.
\end{ex}

\section{Stringy Euler numbers of toric Calabi-Yau hypersurfaces}

Let $M \cong \Z^d$ be a lattice of rank $d$, $N \defeq \Hom(M, \Z)$
its dual lattice, and
\[\langle \cdot, \cdot \rangle : M \times N \to \Z\]
the natural pairing. Consider a $d$-dimensional lattice polytope
$\tope \subseteq M_\R \defeq M \otimes_{\Z} \R \cong \R^d$ and
denote by $\Sigma_{\tope}$ the normal fan of $\tope$ in the dual real vector
space $N_\R \defeq N \otimes_{\Z} \R$. There exists a natural
one-to-one correspondence $\theta \leftrightarrow \sigma_{\theta}$  between
$k$-dimensional faces $\theta \preceq \tope$ and $(d-k)$-dimensional
cones $\sigma_{\theta} \in \Sigma_{\tope}$, where
\[ \sigma_{\theta}= \{ y \in N_\R\, | \,
\min_{x \in \Delta} \langle x, y \rangle  =   \langle x', y \rangle  \; \; \forall x' \in \theta \}. \]

\smallskip
We are interested in $d$-dimensional
Newton polytopes $\Delta \subseteq M_\R$ of Laurent polynomials $f_{\tope} \in \C[M]$
defining non-degenerate affine hypersurfaces
\[Z_{\tope}  \subset \T \defeq \Hom(M,\C^*) \cong (\C^*)^d\]
that are birational to {\em Calabi-Yau varieties}
$X$, \emph{i.e.}, normal projective algebraic varieties with at worst
canonical singularities and trivial canonical class.
By \cite[Theorem on page 41]{Kho78}, such a Newton
polytope $\Delta$ must be a {\em canonical Fano polytope},
\emph{i.e.}, a lattice polytope containing exactly one interior lattice
point that we can assume to be $0 \in M$.

\smallskip
We review two results from \cite{Bat17}.
The first is a combinatorial
characterization of $d$-dimensional canonical  Fano polytopes $\tope$
such that $Z_{\tope}$ is birational to a
Calabi-Yau hypersurface $X$:

\begin{theopar}[{\cite[Theorem 2.23]{Bat17}}]  \label{Bat17223}
Let $\Delta \subseteq M_\R$ be a $d$-dimensional canonical Fano polytope
and $\tope^* = \{ y \in N_\R\; \vert \; \langle x, y \rangle \geq -1 \;\; \forall
x \in \tope \}$ its rational dual polytope. We set
\[[\tope^*] \defeq \conv( \tope^* \cap \; N) \subseteq N_\R \]
to be the convex hull  of all lattice points in $\tope^*$.
Then a non-degenerate affine hypersurface $Z_{\tope} \subset \T$
is birational to a Calabi-Yau variety $X$ if and only if $[\tope^*]$
is also a $d$-dimensional canonical Fano polytope.
\end{theopar}

\begin{rem} \label{ifpart}
The "if"-part in Theorem \ref{Bat17223} was independently proven by
Artebani, Comparin, and Guilbot \cite[Theorem 1]{ACG16}.
Moreover, they proved that a canonical Calabi-Yau model $X$
of the hypersurface $Z_\Delta \subset \T$ can be explicitly
constructed, \emph{e.g.},  as the Zariski closure
of $Z_\Delta$ in the toric $\Q$-Fano variety
associated with the spanning fan of the canonical Fano polytope
$[\Delta^*] \subseteq N_\R$.
\end{rem}

\begin{rem}
Canonical Fano polytopes $\tope \subseteq M_\R$
satisfying the criterion  in Theorem \ref{Bat17223}
can be equivalently characterized by the condition
that the Fine interior of $\tope$ is exactly its interior
lattice point
$\{0\}$ \cite[Section 2]{Bat17}, \cite{BKS19}.
\end{rem}

\begin{rem}
A $d$-dimensional canonical Fano polytope $\Delta$ is called
{\em pseudoreflexive} if it satisfies the condition
\[ [[\Delta^*]^*] = \Delta.\]
It is easy to see that if $\Delta$ is pseudoreflexive, then
$[\Delta^*]$ is also pseudoreflexive. Thus, pseudoreflexive
polytopes satisfy the combinatorial duality $\Delta \leftrightarrow [\Delta^*]$,
which was proposed by Mavlytov as a generalization of
the polar duality for reflexive polytopes \cite{Mav11}. We note
that any pseudoreflexive polytope $\Delta$ of dimension
$d \leq 4$ is reflexive. A simple example of
a $5$-dimensional pseudoreflexive polytope $\Delta$
that is not reflexive can be obtained
as the Newton polytope of a general $4$-dimensional Calabi-Yau hypersurface
of degree $7$ in the weighted projective space $\P(1,1,1,1,1,2)$ \cite{KS98b}.
It was expected by Mavlyutov that if  $(\Delta, [\Delta^*])$ is
a pair of pseudoreflexive polytopes, then
Calabi-Yau hypersurfaces in toric
$\Q$-Fano varieties corresponding to rational
polytopes $\Delta^*$ and $[\Delta^*]^*$
are mirror symmetric to each other.  For Calabi-Yau
hypersurfaces of degree $7$ in $\P(1,1,1,1,1,2)$ this
is true, but in general this is false \cite[Section 5]{Bat17}.
\end{rem}

\begin{rem} \label{almostpseudo}
Let $\Delta$ be an arbitrary canonical Fano polytope satisfying the
condition in Theorem \ref{Bat17223},
\emph{i.e.}, $[\Delta^*]$ is canonical. In general, $\Delta$
need not to be pseudoreflexive.  However,
the canonical polytopes $[\Delta^*]$
and $[[\Delta^*]^*]$ are always pseudoreflexive. For this
reason, we call a canonical Fano polytope $\Delta$
{\em almost pseudoreflexive} if
$[\Delta^*]$ is also a canonical Fano polytope.
The name is motivated by the fact that in this case
$[[\Delta^*]^*]$ is the smallest pseudoreflexive
polytope containing $\Delta$. So it is natural to call $[[\Delta^*]^*]$
the {\em pseudoreflexive hull} of $\Delta$. Since
pseudoreflexive polytopes $\Delta$ of dimension $3$ and $4$
are reflexive, we will use
the expressions  {\em almost reflexive} and {\em reflexive hull}
instead of {\em almost pseudoreflexive} and {\em pseudoreflexive hull},
respectively.
\end{rem}

\begin{rem}
Recall some classification results concerning
special classes of canonical Fano polytopes:
\begin{itemize}
\item Any $2$-dimensional
canonical Fano polytope is reflexive.
There exist exactly 16 of them.
\item All $3$-dimensional canonical Fano polytopes
are classified by Kasprzyk \cite{Kas10}. There exist exactly  $665,\!599$
$3$-dimensional almost reflexive polytopes among all $674,\!688$
$3$-dimensional canonical Fano polytopes. This list extends the classification of
all $4,\!319$ reflexive $3$-dimensional polytopes obtained by Kreuzer and
Skarke \cite{KS98a}.
\item All $4$-dimensional reflexive polytopes are classified by
Kreuzer and Skarke \cite{KS00}. This list consists of $473,\!800,\!776$ reflexive
polytopes, but the list of all $4$-dimensional
almost reflexive polytopes is still unknown.
\end{itemize}
\end{rem}

The second result from \cite{Bat17} is a combinatorial formula
for the stringy Euler number  $\chi_{\rm str} (X)$ of a  Calabi-Yau
variety $X$ generalizing Formula \eqref{comb-Euler}
for reflexive polytopes.
We recall a general definition for the stringy Euler number
of a $d$-dimensional normal projective $\Q$-Gorenstein variety
$X$ with at worst log-terminal singularities:

\begin{dfn}
Let $\rho :Y \to X$ be a log-desingularization of $X$ whose
exceptional locus is the union of smooth irreducible divisors $D_1,
\ldots, D_s$ with only simple normal crossing and
\[ K_Y = \rho^* K_X + \sum_{i =1}^s a_i D_i \] for some rational numbers
$a_i > -1$ $(1 \leq i \leq s)$. We set  $D_\emptyset \defeq Y$ and
\[D_J \defeq \bigcap_{j \in J} D_j, \; \;\; \emptyset \neq
J \subseteq I\defeq \{1, \ldots, s\}.\]
Then the {\em stringy Euler number} $\chi_{\rm str}(X)$
is defined to be
\[ \chi_{\rm str}(X) \defeq \sum_{ \emptyset \subseteq J \subseteq I}
\chi(D_J) \prod_{j \in J} \left(  \frac{-a_j}{a_j +1} \right),  \]
where $\chi(D_J)$ denotes the usual Euler number of
the smooth projective variety $D_J$ \cite{Bat98}.
\end{dfn}

\begin{rem}
It is important to note that the stringy Euler number is
independent on the log-desingularization $\rho: Y \to X$
and $\chi_{\rm str}(X)$ equals the usual Euler number of $Y$
if $\rho: Y \to X$ is crepant. More generally, if two
projective algebraic varieties $X$ and $X'$ are $K$-equivalent,
then $\chi_{\rm str}(X) = \chi_{\rm str}(X')$. In particular,
the stringy Euler number of a given birational class
of algebraic varieties of non-negative Kodaira dimension is well-defined.
\end{rem}

In addition, we will need the notion of the
normalized volume of a rational polytope:

\begin{dfn} \label{normvol}
Let $\Delta \subseteq N_{\R}$ be a $d$-dimensional {\em rational polytope}, \emph{i.e.},
$\Delta$ has vertices in $N_{\Q} \defeq N \otimes \Q$.
Then  the positive rational number
\[\Vol_d(\Delta) \defeq \frac{1}{l^d} \Vol_d(l \Delta)= \frac{1}{l^d} \cdot d!\cdot \vol_d (l \Delta)\]
is called the {\em normalized volume} of $\Delta$,
where $l$ is a positive integer such that $l\Delta$ is
a lattice polytope and $\vol_d(\cdot)$ denotes the $d$-dimensional volume of
$\Delta$ with respect to the lattice $N$. Similarly, one obtains
the normalized volume $\Vol_k(\theta) \defeq
\frac{1}{l^k} \Vol_k(l\theta)$ with respect to the sublattice $\text{span}(\theta) \cap N$
for a $k$-dimensional rational face
$\theta \preceq \Delta$.
\end{dfn}

\begin{theopar}[{\cite[Theorem 4.11]{Bat17}}] \label{Bat17411}
Assume that a $d$-dimensional canonical Fano polytope $\tope \subseteq M_\R$
satisfies the condition in Theorem \ref{Bat17223}, \emph{i.e.},
a non-degenerate affine hypersurface $Z_\Delta \subset \T$
is birational to a Calabi-Yau variety $X$.
Then the stringy Euler number of $X$ can be computed
by the following combinatorial formula
\begin{align*}
\chi_{\rm str} (X)  = \sum_{k=1}^{d} (-1)^{k-1}  \sum_{ \theta \preceq \Delta \atop \dim (\theta) = k}
\Vol_k (\theta) \cdot \Vol_{d-k} (\sigma_\theta \cap \Delta^*),
\end{align*}
where $\sigma_\theta \cap \Delta^*$ is the $(d-k)$-dimensional
pyramid with vertex $0 \in N$ over the $(d-k-1)$-dimensional dual face $\theta^* \preceq \Delta^*$
of the dual rational polytope $\Delta^*$.
\end{theopar}

\begin{ex}
Let $\tope \subseteq M_\R$ be a $3$-dimensional
almost reflexive polytope. Then $X$
is a $K3$-surface with  $\chi_{\rm str} (X)=24$. In this case, the
combinatorial formula in Theorem
\ref{Bat17411} is equivalent to the identity
\[ 24 = \Vol_3(\Delta) - \sum_{ \theta \prec \Delta \atop
\dim (\theta) = 2} \frac{1}{n_\theta} \cdot
\Vol_2(\theta)  + \sum_{ \theta \prec \Delta \atop
\dim (\theta) = 1} \Vol_1(\theta) \cdot \Vol_1(\theta^*), \]
where $n_\theta$ denotes the integral distance
between a facet $\theta \preceq \Delta$ and $0 \in M$
\cite[Theorem A]{BS19}.
\end{ex}

\begin{ex}
Let  $\tope \subseteq M_\R$ be a $d$-dimensional reflexive polytope. Then
the dual polytope $\tope^*$ is also a reflexive polytope.
Therefore, we obtain
\[ \Vol_{d-k}(\sigma_\theta \cap \Delta^*) =
\Vol_{d-k-1}(\theta^*)\]
for any $k$-dimensional face $\theta \preceq \tope$ of $\tope$ $(1 \leq k \leq d-2)$.
Together with the equality
$ \Vol_d(\Delta) = \sum_{ \theta \preceq \Delta \atop \dim (\theta) = d-1}
\Vol_{d-1} (\theta)$, Theorem \ref{Bat17411} implies the already discussed
combinatorial Formula \eqref{comb-Euler}.
\end{ex}

\section{Calabi-Yau hypersurfaces in weighted projective spaces}

Let $(w_0, w_1, \ldots, w_d) \in \Z_{>0}^{d+1}$ be a
well-formed weight vector.
In this section, we apply Theorems \ref{Bat17223} and \ref{Bat17411}
to the special  $d$-dimensional lattice simplex
\[\tope \defeq  \tope_{\w}^* = \conv(\{v_0, v_1, \ldots, v_d\}) \subseteq  N_{\w} \otimes \R,\]
where $v_0, \ldots, v_d$ generate the lattice
$M  \defeq N_{\w} $ and satisfy the linear relation $\sum_{i=0}^d w_i v_i = 0$.
Consider the lattice $\widetilde{M} \defeq M \oplus \Z$ together with $d+1$
linearly independent lattice vectors $\widetilde{v}_i \defeq (v_i,1) \in \widetilde{M}$
($0 \leq  i \leq d$). Set $M'\defeq \langle  \widetilde{v}_i \rangle_{0 \leq i \leq d}$ and
identify $M'$ with $\Z^{d+1}$ via the basis $\widetilde{v}_i$ ($0 \leq  i \leq d$).
Since $M$ is generated by ${v}_i$ ($0 \leq  i \leq d$), the lattice $\widetilde{M}$
is generated by $\widetilde{v}_i $ together with the lattice vector
$\widetilde{v} \defeq (0, 1) \in M \oplus \Z= \widetilde{M}$ and
\[ \widetilde{v} = \sum_{i=0}^d \frac{w_i}{w} \cdot \widetilde{v}_i. \]
So the quotient $\widetilde{M} / M'$ is a cyclic
group of order $w$ generated by $\widetilde{v} + M'$ and we can write
\[ \widetilde{M} =\Z \left( \frac{w_0}{w},  \frac{w_1}{w},
\ldots,  \frac{w_d}{w}\right) +  \Z^{d+1} =
\Z( q_0, q_1, \ldots, q_d) + \Z^{d+1}. \]
In certain cases, it will be convenient to use the multiplicative
description of $\widetilde{M} / M'$ by the cyclic group
$G \defeq \langle g \rangle  = \{g^s \, |\, s \in \Z/w\Z \} \subseteq  SL(d+1, \C)$
generated by the diagonal matrix
\[ g \defeq {\rm diag}(g_0, g_1, \ldots, g_d)
\defeq {\rm diag}(e^{2\pi i q_0}, e^{2\pi i q_1}, \ldots, e^{2\pi i q_d}). \]
In the language of  toric varieties, the $(d+1)$-dimensional
simplicial cone $\R_{\geq 0}(1, \Delta) \subseteq \widetilde{M}_\R$ describes the
Gorenstein cyclic quotient singularity $\C^{d+1}/G$ considered
by Corti and Golyshev \cite{CG11}.

\begin{dfn}
For any non-empty subset $J \subseteq I$, we define two
$\vert J \vert$-dimensional sublattices
\[ M_J' \defeq \sum_{j \in J} \Z \widetilde{v_j} \; \; \text{ and } \;\;
\widetilde{M}_J \defeq \widetilde{M} \cap (M_J' \otimes \Q), \]
and the subgroup
\[G_J\defeq \left\{ g^s \in G \, \middle \vert \,
g_j^s =1 \;\; \forall j \in J\right\} \subseteq G.\]
\end{dfn}

Each $k$-dimensional face $\theta \preceq \tope =\tope^*_w$
is a $k$-dimensional simplex
\[ \theta_J \defeq \conv(v_{i_0}, v_{i_1}, \ldots, v_{i_k}) \]
determined by a $(k+1)$-element subset
$J \defeq \{i_0, i_1, \ldots, i_k \} \subseteq I = \{0, 1,\ldots, d \}$.
We set $\widetilde{\theta}_J$ to be
the $(k+1)$-dimensional simplex
\[\widetilde{\theta}_J \defeq \conv(0,\widetilde{v}_{i_0},
\widetilde{v}_{i_1}, \ldots, \widetilde{v}_{i_k}) .  \]

\begin{prop} \label{volume}
Let $\emptyset \neq J \subseteq I$ be a non-empty subset of $I$. Then
\begin{itemize}
\item[{\rm (i)}] $\vert \widetilde{M}_J/ M_J' \vert =
\Vol_{\vert J \vert}(\widetilde{\theta}_J) $;
\item[{\rm (ii)}] $\vert G_J\vert = n_J \defeq \ggT (w,  w_j \, \vert \, j \in J )$;
\item[{\rm (iii)}] $\Vol_{\vert J \vert-1}(\theta_J)  =
n_{\overline{J}}= \vert G_{\overline{J}} \vert,$
where $\overline{J} \defeq I \setminus J$.
\end{itemize}
\end{prop}

\begin{proof}
{\rm (i)}
The normalized volume
$\Vol_{\vert J \vert}(\widetilde{\theta}_J)$ of $\widetilde{\theta}_J$ equals
the index of the sublattice $M'_J$ generated by the lattice vectors
$\widetilde{v}_j \;( j \in J)$ in the $|J|$-dimensional lattice
$\widetilde{M}_J = \widetilde{M} \cap (M_J' \otimes \Q)$.

\smallskip
{\rm (ii)}
Assume without loss of generality $J = \{0,1,\ldots,k \}$. We
set $u \defeq w/n_J \in \N$. Then one has  $u q_j = w_j/n_J  \in \Z$ for all
$j \in J $, \emph{i.e.}, $g^u \in G_J$.
This implies the inclusion $\langle g^u
\rangle \subseteq G_J$ and the inequality $n_J =  |\langle g^u
\rangle| \leq |G_J|$. In order to prove the opposite inclusion
$G_J \subseteq \langle g^u
\rangle$, we set $u_j\defeq w_j/n_J \in \N$
for $j \in J$ and assume that  $g^s \in G_J$.
Then $su_j/u=sq_j \in \Z$ for all $j \in J$,
\emph{i.e.}, $u$ divides $su_j$ for all $j \in J$.
Since $\ggT(u, u_0, u_1, \ldots, u_k) =1$,  there
exist integers $a$ and $a_0, \ldots,a_k$ such that
\[ a u + \sum_{j \in J} a_j u_j = 1. \]
Therefore, $u$ divides  $s = asu +  \sum_{j \in J} a_j s u_j$,
\emph{i.e.}, $g^s \in  \langle g^u
\rangle$.

\smallskip
{\rm (iii)}
Note that the quotient $\widetilde{M}_J/ M_J'$
can be considered as a subgroup of the cyclic group
$\widetilde{M}/M' \cong \Z/w\Z$. Indeed, if we take the
homomorphism $\varphi\, :\,  \widetilde{M}_J \to \widetilde{M}/M'$
obtained from the embedding of $\widetilde{M}_J$ into $\widetilde{M}$,  then
the kernel of $\varphi$ is
$\widetilde{M}_J \cap M' = \widetilde{M} \cap (M_J' \otimes \Q) \cap M' = M_J'$
because $(M_J' \otimes \Q) \cap M' =  M_J'$. Furthermore, an element
$s(\widetilde{v} + M') \in \widetilde{M}/M'$ belongs to the
subgroup $\widetilde{M}_J/ M_J'$ if and only if
the coefficients $s q_i $ in the equation
\[s \widetilde{v}  = \sum_{i =0}^d s q_i   \widetilde{v}_i \]
are integers for all $i \not\in J$.
By Proposition \ref{volume} (ii), the latter happens if and only if
$g^s \in G_{\overline{J}}$. This shows that  $\widetilde{M}_J/ M'_J \cong
G_{\overline{J}}$ and $\Vol_{k}(\theta_J) = \Vol_{k+1}(\widetilde{\theta}_{J}) =n_{\overline{J}}$.
\end{proof}

For the lattice simplex
$\Delta = \Delta_{\w}^*= \conv(\{v_0, v_1, \ldots, v_d\})$ the dual polytope $\tope^*$ is
the  $d$-dimensional rational simplex
\[ \Delta^* = \tope_{\w} = \{ (u_0, u_1, \ldots, u_d) \in \R_{\geq 0}^{d+1} \, \vert \,
\sum_{i=0}^d w_i u_i = w \}  - (1, 1,\ldots, 1).  \]
Let $J \subseteq I=\{0,1,\ldots,d\}$ be a (k+1)-element subset of $I$.
Then we denote by $\sigma_J \defeq \sigma_{\theta_J}$ the normal cone
in the normal fan $\Sigma_{\tope}= \{\sigma_{J'} \, \vert \, \emptyset \subseteq J' \subseteq I\}$
corresponding to the face $\theta_J \preceq \tope$
of the lattice simplex $\tope$ with $\dim(\sigma_J) = d - \dim(\theta_J) = d-k$,
\emph{i.e.}, $\sigma_J$ is the cone generated by all inward-pointing facet normals of
facets containing the face $\theta_J \preceq \tope$.

\begin{prop} \label{v-normal}
Let $J \subseteq I=\{0,1,\ldots,d\}$ be a $(k+1)$-element subset of $I$. Then
the normalized volume $\Vol_{d-k}(\sigma_J \cap \tope^*) $ of
the rational polytope $\sigma_J \cap \tope^*$ is given by
\[\Vol_{d-k}(\sigma_J \cap \tope^*) =
\frac{n_{\overline{J}}}{w} \prod_{i \in \overline{J}} \frac{1}{q_i} , \]
where $\overline{J} = I \setminus J$, $q_i = \frac{w_i}{w}$ $(i \in I)$,
and $n_{\overline{J}} = \ggT(w,w_i \, \vert \, i \in \overline{J})$.
\end{prop}

\begin{proof}
\begin{figure}[t!]
	\centering
	\scalebox{0.6}{
	\includegraphics[width=\textwidth]{VF1}}
	\caption[]{{\bf Illustration of Proposition \ref{v-normal}.}
	Grey coloured $2$-dimensional rational simplex $\tope'$ with unique interior lattice point $(1,1,1) \in \Z^3$
	and  vertices $\frac{1}{q_i}$ $(i \in I = \{0,1,2\})$ together with a $1$-dimensional dotted face
	$\theta_J' \preceq \tope'$, \emph{i.e.}, $J =\{0\} \subseteq I$ and $d=2$.
	Light grey coloured shifted dual simplex $\tope^* = \tope' - (1,1,1)$ with unique interior lattice point
	$(0,0,0) \in \Z^3$ together with a $1$-dimensional dotted shifted face $\theta_J^* \preceq \tope^*$.
	The grey shaded associated $2$-dimensional normal cone $\sigma_J$ with two dotted rays corresponding
	to the (rational) vertices of $\theta_J^*$, where $\theta_J = \{v_0\} \preceq \Delta$ is a vertex.
	Moreover, the area $\sigma_J \cap \tope^*$ is crosshatched grey.}
	\label{fig:VF1}
\end{figure}

Consider the $d$-dimensional rational simplex
\[\tope' \defeq  \{ (u_0, u_1, \ldots, u_d) \in \R^{d+1} \, \vert \, u_i \geq 0 \; \; \forall i \in I,
w_0 u_0 + w_1 u_1 + \ldots + w_d u_d = w\}.\]
Then the shifted simplex $\tope' - (1,1,\ldots, 1)$ is the dual simplex
$\tope^* \subseteq N_\R$ of $\Delta$ because
\[ \tope^*= \{ (u_0, u_1, \ldots, u_d) \in \R^{d+1} \, \vert \, u_i \geq -1\; \;
\forall i \in I,  w_0 u_0 + w_1 u_1 + \ldots + w_d u_d = 0 \} \]
(Figure \ref{fig:VF1}).
For simplicity, we assume without loss of generality
$J = \{0,1,\ldots,k \}$, \emph{i.e.}, $\overline{J} = \{k+1, \ldots, d \}$ and
denote by $\{e_0, e_1, \ldots, e_d\}$ the standard basis of $\R^{d+1}$.
The dual rational face $\theta_J^* \preceq \tope^* $ of $\tope^*$ as well as the
shifted dual rational face
$\theta_J' \defeq \theta_J^*+(1, 1,\ldots, 1)  \preceq \tope^* +(1, 1,\ldots, 1) =\tope'$
of the shifted simplex $\tope'$ have dimension $d-k-1$.
Moreover, $\tope'$ has vertices $\frac{1}{q_i} e_i$ ($i \in I$) and
$\theta_J^*+(1, 1,\ldots, 1) \preceq \tope'$ has vertices
$ \frac{1}{q_i} e_i  \;\; (i \in \overline{J})$ (Figure \ref{fig:VF1}).

\begin{figure}
	\centering
	\scalebox{.4}{
	\includegraphics[width=\linewidth]{VF2}}
	\caption[]{{\bf Illustration for Proposition \ref{v-normal}.}  Shaded faces are occluded.
	Grey coloured $2$-dimensional rational simplex $\tope'$ with unique interior lattice point $(1,1,1) \in \Z^3$
	and vertices $\frac{1}{q_i}$ $(i \in I = \{0,1,2\})$ together with a $1$-dimensional dotted face
	$\theta_J' \preceq \tope'$ with vertices $\frac{1}{q_i}$ $(i \in \overline{J})$, \emph{i.e.},
	$J =\{0\} \subseteq I$, $\overline{J} = \{1,2\}$, and $d=2$. The $2$-dimensional grey crosshatched
	pyramid $\Pi_{\overline{J}}$ with vertex $(0,0,0) \in \Z^3$ and basis $\theta_J'$ .}
	\label{fig:VF2}
\end{figure}

\smallskip
For $\vert J \vert = \vert I \vert = d+1$, \emph{i.e.}, $k=d$ and $\overline{J}= \emptyset$, we obtain
$\theta_J = \tope$, $\sigma_J = \{0\}$, and
\[\Vol_{d-k}(\sigma_J \cap \tope^*)
=\Vol_{0}(\{0\})  =1
=\frac{w}{w} \prod_{i \in \emptyset} \frac{1}{q_i}
= \frac{n_{\overline{J}}}{w} \prod_{i \in \overline{J}} \frac{1}{q_i}.\]

For $\vert J \vert = d$, \emph{i.e.}, $k=d-1$ and $\overline{J}=\{w_d\}$, the associated
dual rational face $\theta_J^* \preceq \tope^*$ has dimension $0$,
\emph{i.e.}, $\theta_J^* \defeq \{v\}$ is a (rational) vertex and $P_v \defeq \conv(\{(0,0,\ldots,0),v\})$
a $1$-dimensional (rational) polytope. Therefore,
\begin{align*}
\Vol_{d-k}(\sigma_J \cap \tope^*)
&=\Vol_{1}(P_v)
=\frac{\ggT(w,w_d)}{w_d}
= \frac{n_{\overline{J}}}{w} \prod_{i \in \overline{J}} \frac{1}{q_i}.
\end{align*}

Let $k\leq d-2$, \emph{i.e.}, $\vert J \vert \leq d-1$.
Moreover, let $ \Pi_{\overline{J}}$ be the pyramid with vertex $0 \in \R^{d+1}$ over
the shifted dual face $\theta_J' \preceq \tope'$, \emph{i.e.},
$\Pi_{\overline{J}} = \conv(\{0,  \frac{1}{q_i} e_i  \, \vert \, i \in \overline{J}\})$ with
$\dim(\Pi_{\overline{J}})= d-k$ and
\[ \Vol_{d-k}( \Pi_{\overline{J}}) = \prod_{i  \in \overline{J}}  \frac{1}{q_i}  \]
(Figure \ref{fig:VF2}).
The pyramid  $\Pi_{\overline{J}}$ is contained in the $(d-k)$-dimensional subspace
generated by $e_{k+1},\ldots, e_d$.  The basis of the pyramid is the simplex
$\theta_J'$ that belongs to a hyperplane in the affine linear subspace defined by the equation
\[ w_{k+1} u_{k+1} + \ldots + w_d u_d = w. \]
The integral distance between this hyperplane and the origin
equals $w/n_{\overline{J}}$, where $n_{\overline{J}} = \ggT(w, w_{k+1}, \ldots, w_d)$.
Therefore, we obtain
\[  \Vol_{d-k }(\sigma_J \cap \tope^*)= \Vol_{d-k-1}(\theta_J^*) =\Vol_{d-k-1 }(\theta_J') =
\Vol_{d-k}( \Pi_{\overline{J}}) \cdot \frac{n_{\overline{J}}}{w} =
\frac{n_{\overline{J}}}{w} \prod_{i \in \overline{J}}  \frac{1}{q_i}. \]
\end{proof}

\begin{theo} \label{can-strEuler}
Let  $\w = (w_0, w_1, \ldots, w_d)$ be a weight vector with $\text{IP}$-property
and $Z_{\w} \subset \T_{\w}$ a non-degenerate affine
hypersurface defined by a Laurent
polynomial $f_{\w}$ with Newton polytope $\Delta_{\w}^*$.
Then the Zariski closure of $Z_{\w}$
in the $\Q$-Gorenstein toric Fano variety $\P^{\vee}({\w})$
is a $(d-1)$-dimensional Calabi-Yau variety $X_{\w}^*$ and
\[ \chi_{\rm str}(X_{\w}^*) = \frac{1}{w} \sum_{\emptyset \subseteq J \subseteq I \atop |J| \geq 2}
(-1)^{|J|} {n_{\overline{J}}^2} \prod_{i \in \overline{J}} \frac{1}{q_i},\]
where $\overline{J} = I \setminus J$ and $q_i=\frac{w_i}{w}$ $(i \in I)$.
\end{theo}

\begin{proof}
By assumption, $\tope(W)^\circ \cap \Z^{d+1} =(1,1,\ldots, 1)$, \emph{i.e.},
the lattice polytope $[\tope^*] =[(\tope_{\w}^*)^*]=\tope(W)-(1,1,\ldots,1)=\Delta'(W)$ is
a $d$-dimensional canonical Fano polytope.
By Theorem \ref{Bat17223} and Remark \ref{ifpart},
the Zariski closure of the non-degenerate hypersurface $Z_{\w} \subset \T_{\w}$
in the toric variety $\P^{\vee}(\w)$ associated with the spanning fan $\Sigma^{\vee}(\w)$
of the canonical Fano polytope $[\tope^*]$ is a Calabi-Yau variety $X_{\w}^*$.
Therefore, we are able to apply Theorem  \ref{Bat17411} to compute
the stringy Euler number $\chi_{\rm str}(X_{\w}^*)$ of $X_{\w}^*$ via
\begin{align*}
\chi_{\rm str} (X_{\w}^*)  = \sum_{k=1}^{d} (-1)^{k-1}
\sum_{ \theta \preceq \Delta \atop \dim (\theta) = k}
\Vol_k (\theta) \cdot \Vol_{d-k} (\sigma_\theta \cap \Delta^*).
\end{align*}
Moreover, the $k$-dimensional faces of $\tope$ are
simplices $\theta_J = \conv(\{v_j \, \vert \, j \in J\})$ pa\-ra\-me\-tri\-zed
by subsets $J \subseteq I$ with $\vert J \vert =k+1$,
where $k \geq 1$ if and only if $\vert J \vert \geq 2$.

\smallskip
The  normalized volumes $\Vol_{\vert J \vert -1 }(\theta_J)$ and
$\Vol_{d-\vert J \vert +1}(\sigma_J \cap \tope^*)$
have been computed  for every subset $J \subseteq I$
with $\vert J \vert \geq 2$
in Proposition \ref{volume} (iii) and Proposition \ref{v-normal}, respectively.
By substitution, we obtain
\begin{align*}
\chi_{\str}(X_{\w}^*)
&= \sum_{\vert J \vert \geq 2}  (-1)^{\vert J \vert -2} \Vol_{\vert J \vert -1 }(\theta_J)
\cdot \Vol_{d-\vert J \vert +1}(\sigma_J \cap \tope^*)\\
&=\sum_{\emptyset \subseteq J \subseteq I \atop |J| \geq 2}
(-1)^{|J|} \, n_{\overline{J}} \cdot \frac{ n_{\overline{J}} }{w}
\prod_{i \in {\overline{J}}} \frac{1}{q_i}
= \frac{1}{w}\sum_{\emptyset \subseteq J \subseteq I \atop |J| \geq 2}
(-1)^{|J|} \, n_{\overline{J}}^2  \prod_{i \in {\overline{J}}}
\frac{1}{q_i} .
\end{align*}
\end{proof}

\section{Proof of Theorem \ref{vafemirrortest}}

Let $(w_0, w_1, \ldots, w_d) \in \Z_{>0}^{d+1}$ be a
well-formed weight vector. We consider  the cyclic group
$G =  \langle g \rangle = \{ g^s\, | \, s \in \Z/w\Z \}  \subseteq SL(d+1, \C)$,
whose generator $g = {\rm diag}(g_0, g_1, \ldots, g_d)$ acts linearly on $\C^{d+1}$
by the  diagonal matrix
\[{\rm diag}(e^{2\pi i q_0}, e^{2\pi i q_1}, \ldots, e^{2\pi i q_d}).\]

\smallskip
We aim at a combinatorial version of
Vafa's Formula \eqref{vafa0}. This version has been proven
in \cite[Formula (4.1), page 255]{GRY91}
for Calabi-Yau hypersurfaces in $4$-dimensional
weighted projective spaces:

\begin{theo} \label{vafa-re}
Let $(w_0, w_1, \ldots, w_d) \in \Z_{>0}^{d+1}$ be as above. Then
\[ \frac{1}{w}  \sum_{l,r=0}^{w-1}
\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}
\left(1 - \frac{1}{q_i} \right)
= \frac{1}{w}  \sum_{\emptyset \subseteq
J \subseteq I \atop |J| \leq d-1} (-1)^{|J|}   \, n_J^2  \prod_{j \in J}
\frac{1}{q_j}. \]
\end{theo}

\begin{proof}
Obviously, we have
\[\frac{1}{w}  \sum_{l,r=0}^{w-1}
\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}
\left( 1 - \frac{1}{q_i} \right)
= \frac{1}{|G|}  \sum_{(g^l,g^r) \in G^2}
\prod_{0 \leq i \leq d \atop g_i^l=g_i^r =1}
\left(1 - \frac{1}{q_i} \right).\]
First consider the trivial pair $(l,r)=(0,0) \in (\Z/w\Z)^2$. Then
$g_i^l = g^r_i =1$ for all $i \in I$ and we obtain the product
\[ \prod_{i=0}^d
\left( 1 - \frac{1}{q_i} \right) =
\sum_{\emptyset \subseteq
J \subseteq I}  \prod_{j \in J}
\left( - \frac{1}{q_j} \right) = \sum_{\emptyset \subseteq
J \subseteq I}
(-1)^{|J|} \prod_{j \in J}
\frac{1}{q_j}  \]
that appears as one summand in Vafa's Formula \eqref{vafa0}.
For $s \in \Z/w\Z$, we define the subset $J_s \subseteq I$:
\[J_s \defeq  \{ i \in I \, \vert \, g_i^s =1\}. \]
Then for any pair $(g^l,g^r) \in G^2$, we obtain
\[ \prod_{0 \leq i \leq d \atop g_i^l = g_i^r =1}
\left(1 - \frac{1}{q_i} \right) =
\prod_{ i \in J_l \cap J_r} \left( 1 - \frac{1}{q_i} \right) =
\sum_{\emptyset \subseteq
J \subseteq J_l \cap J_r}
(-1)^{|J|} \prod_{j \in J} \frac{1}{q_j}  \]
and
\begin{align*}
\frac{1}{|G|}  \sum_{(g^l, g^r) \in G^2}
\prod_{0 \leq i \leq d \atop g_i^l =g_i^r =1}
\left( 1 - \frac{1}{q_i} \right)
&= \frac{1}{|G|}  \sum_{(g^l,g^r) \in G^2} \left(
\sum_{ \emptyset \subseteq
J \subseteq J_l \cap J_r }  (-1)^{|J|} \prod_{j \in J} \frac{1}{q_j} \right) \\
& = \frac{1}{|G|} \sum_{\emptyset \subseteq
J \subseteq I} \left(
\sum_{(g^l, g^r) \in G_J^2}   (-1)^{|J|}  \prod_{j \in J}
\frac{1}{q_j} \right)   \\
&= \frac{1}{|G|}  \sum_{\emptyset \subseteq
J \subseteq I} (-1)^{|J|}  \, n_J^2  \prod_{j \in J}
\frac{1}{q_j}  ,
\end{align*}
where the last equality holds by Proposition \ref{volume} (ii).
Moreover, we note that $n_J =1$ if $|J| \in \{ d, d+1 \}$, since we assumed
\[\ggT (w_0, \ldots, w_{i-1}, w_{i+1}, \ldots, w_d) = 1 \; \; \forall i \in I. \]
Using the equality $\sum_{i =0}^d q_i = 1$, we obtain
\[ (-1)^{d+1}  \prod_{i=0}^d \frac{1}{q_i}
+ (-1)^d \sum_{i=0}^d \prod_{j \in I \atop j \neq i}   \frac{1}{q_j} =0 \]
and this implies
\[ \frac{1}{w}  \sum_{l,r=0}^{w-1}
\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}
\left(1 - \frac{1}{q_i} \right)  =  \frac{1}{w}  \sum_{\emptyset \subseteq
J \subseteq I \atop |J| \leq d-1} (-1)^{|J|}  \, n_J^2  \prod_{j \in J} \frac{1}{q_j}. \]
\end{proof}

\begin{expar}[{\cite[page 1182]{Vaf89}}]
Let $X_{15} \subset \P(1,2,3,4,5)$ be the quasi-smooth hypersurface of degree $15$
defined by the weighted homogeneous polynomial
$W= z_4^3+z_4z_1^5+z_2^5+z_2z_3^3+z_0^{15}$.
Then $q_0= \frac{1}{15}$, $q_1= \frac{2}{15}$, $q_2= \frac{1}{5}$, $q_3= \frac{4}{15}$,
$q_4= \frac{1}{3}$, and
\begin{align*}
\chi_{\rm orb}(X_w)
&=  \frac{1}{15} \sum_{l,r=0}^{14} \, \, \prod_{0 \leq i \leq 4 \atop lq_i,rq_i \in \Z}
\left( 1 - \frac{1}{q_i} \right)
= \frac{1}{15}  \sum_{\emptyset \subseteq
J \subseteq I \atop |J| \leq 3} (-1)^{|J|}   \, n_J^2  \prod_{j \in J}
\frac{1}{q_j} \\
& = \frac{1}{15}\cdot \left(225 - \frac{585}{4} + \frac{3375}{8} - \frac{19125}{8} \right) = -126.
\end{align*}
\end{expar}

Now we prove the main theorem of our paper:

\begin{proof}[Proof of Theorem \ref{vafemirrortest}]
By assumption, $\w = (w_0, w_1, \ldots, w_d) $ is an arbitrary
weight vector with ${\rm IP}$-property and $Z_{\w} \subset \T_{\w}$ a
non-degenerate affine hypersurface
defined by a Laurent polynomial $f_{\w}$ with Newton polytope
$\Delta_{\w}^*$. Applying Theorem \ref{can-strEuler}, the Zariski closure of $Z_{\w}$
 in the $\Q$-Gorenstein toric Fano variety $\P^{\vee}({\w})$
is a Calabi-Yau variety $X_{\w}^*$ and
\begin{align*}
\chi_{\str}(X_{\w}^*) = \frac{1}{w}\sum_{\emptyset \subseteq J \subseteq I \atop |J| \geq 2}
(-1)^{|J|} \, n_{\overline{J}}^2  \prod_{i \in {\overline{J}}}
\frac{1}{q_i} .
\end{align*}
Since $|J| + |\overline{J}| = |I| = d+1$, we obtain
\[ \chi_{\str}(X_{\w}^*) =
(-1)^{d-1}    \frac{1}{w} \sum_{\emptyset \subseteq J \subseteq I \atop |\overline{J}| \leq d-1}
(-1)^{|\overline{J}|} \, n_{\overline{J}}^2  \prod_{i \in {\overline{J}}}  \frac{1}{q_i}
=(-1)^{d-1}    \frac{1}{w}  \sum_{\emptyset \subseteq J \subseteq I \atop |J| \leq d-1}
(-1)^{|J|}   n_J^2  \prod_{j \in J}
\frac{1}{q_j} \]
and Theorem \ref{vafa-re} implies
\[ \chi_{\str}(X_{\w}^*)  = (-1)^{d-1}  \frac{1}{w}
\sum_{\emptyset \subseteq J \subseteq I \atop |J| \leq d-1}
(-1)^{|J|}   n_J^2  \prod_{j \in J}  \frac{1}{q_j}
= (-1)^{d-1}  \frac{1}{w}  \sum_{l,r=0}^{w-1}
\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}
\left( 1 - \frac{1}{q_i} \right).\]

\smallskip
If  $\w= (w_0, w_1, \ldots, w_d)$ is a transverse weight vector,
Vafa's Formula \eqref{vafa0} yields
\[\chi_{\rm orb}(X_w) =  \frac{1}{w}  \sum_{l,r=0}^{w-1}
\prod_{0 \leq i \leq d \atop lq_i,rq_i \in \Z}\left( 1 - \frac{1}{q_i} \right). \]
Since $\w$ has
${\rm IP}$-property
\cite[Lemma 2]{Ska96},
a combination of the last two equations implies
\[  \chi_{\rm str}(X_{\w}^*) = (-1)^{d-1} \chi_{\rm orb}(X_w). \]
\end{proof}


TITLE: Will sticking two high-coercivity neodymium magnets together of opposite polarity and equal strength demagnetize them?
QUESTION [0 upvotes]: Say I have two neodymium magnets of equal strength. If I stick them together in a way they dont like, will they lose strength over time (days or years)? Or is a much larger magnetic field needed?
What if they are placed 1-2mm apart? Will the magnetic field be different enough that I dont have to worry?

REPLY [0 votes]: To switch a magnet, a stronger magnetic force needs to be applied to it.  How strong that field needs to be in order to flip the magnet depends on the material.
For example, a N40 NIB (Neodymium Iron Boron) needs a magnetic forces of 1000kA/m.  Most other magnets will flip with less Coercive force/


text:
\begin{document}

\subsection{Tensor-train and hierarchical Tucker  format}
\label{subsection:TT_and_HT_format}

The tensor-train format was first introduced to the numerical analysis community in~\cite{Oseledets2009}. It is also known in other areas as \emph{matrix product states}~\cite{Ostlund1995,White1992} or as \emph{linear tensor network}~\cite{Loan2008}.

A tensor of dimension $d$ in the tensor-train format is represented as $d-1$ contractions of $d$ so-called \emph{core tensors}.
Each core tensor represents one mode and is connected to its forward and backward neighbors. The core tensors themselves are of dimension three, or two in case of the first and the last mode.
For a given tensor-train representation, its rank is defined as the tuple of all mode sizes corresponding to edges.~\cref{figure:TT} shows an illustrative example.
\begin{figure}[tbhp]
	\centering

	\includegraphics[width = 0.25\textwidth]{tikz/tensor_train_network}

	\caption{\footnotesize{Tensor-train representation of a tensor with dimension $d = 4$,  mode sizes all equal to $n$ and representation rank $\mathbf{r}=\left(r,r,r\right)$.}}
	\label{figure:TT}
\end{figure}
Instead of the original tensor one needs to store the core tensors. Thus, the required storage cost for a tensor-train representation with dimension $d$,  mode sizes all equal to $n$ and representation rank component-wise bounded by $r$ is in $\mathcal{O}\left( d n  r^2 \right)$.
In our case the dimension of the distribution tensors is equal to the number $d$ of automata, and the mode sizes are given by the number of states $n_{\mu}$ in each automaton.
Note that in general the representation rank can depend on the ordering of the modes/automata, cf.~\cite{Grasedyck2011,Oseledets2009}.
\\

The hierarchical Tucker format was introduced in~\cite{Hackbusch2009} and further analyzed in~\cite{Grasedyck2010}.
As the name suggests, the hierarchical Tucker format is based on a hierarchical binary tree structure defining the arrangement of the auxiliary tensors.
Similar to the tensor-train format there are auxiliary tensors each of which represents one mode of the original tensor. These two-dimensional tensors, the so-called \emph{frames}, build the leaves of the binary tree. The two-dimensional tensor at the root of the tree and the three-dimensional tensors at all other levels are called \emph{transfer tensors}. For a given hierarchical Tucker representation, its rank $\mathbf{r}$ is defined as the tuple of all mode sizes corresponding to the edges of the tree, i.e., $\mathbf{r}= \left(r_t\right)_{t \in \mathcal{T}}$ where $\mathcal{T}$ denotes the given tree.~\cref{figure:HT} shows an illustrative example.
\begin{figure}[tbhp]
	\centering

	\includegraphics[width = 0.25\textwidth]{tikz/hierarchical_tucker_network}

	\caption{\footnotesize{Hierarchical Tucker representation of a tensor with dimension $d = 4$, mode sizes all equal to $n$ and representation rank $\mathbf{r} = \left(r\right)_{t \in \mathcal{T}}$ using a balanced tree $\mathcal{T}$.}}
	\label{figure:HT}
\end{figure}
Note that in the hierarchical Tucker format we have many options for the structure of the binary tree and that the rank depends on our choice, cf.~\cite{Grasedyck2011}.
Instead of the original tensor the frames and transfer tensors have to be stored. Thus, the storage cost for a hierarchical Tucker representation of a $d$-dimensional tensor with maximum mode size $n$ and a representation rank component-wise bounded by $r$ is in $\mathcal{O}\left( d n r + d  r^3 \right)$.
In our case the dimension of the distribution tensors is again equal to the number $d$ of automata, and the mode sizes are the number of states $n_{\mu}$ for each automaton.
\\

Thus, in both the tensor-train and the hierarchical Tucker format, the storage cost grows only linearly in the dimension $d$. This allows for efficient storage and prevents the state-space explosion provided we have low ranks.
It is easily possible to convert a CP representation of rank $r$ to a tensor-train or a hierarchical Tucker representation, where all rank components are bounded by $r$, cf.~\cite{Hackbusch2012}.
Similar to the CP format both formats provide arithmetic operations like addition, application of operators, or scalar products.
Since the storage cost increases with the square or cube of the rank and arithmetic operations also increase the rank, we need a way to truncate a tensor to lower rank in an error-controlled way.
In tree tensor networks we use a generalization of the singular value decomposition to truncate in a quasi-optimal way, see~\cite{Oseledets2009} for the tensor-train and~\cite{Grasedyck2010} for the hierarchical Tucker format.
There it is shown that the error in the Frobenius norm caused by the truncation of a tensor $\mathcal{B} \in \R^{\I}$ to rank $\mathbf{r} = \left(r_{t}\right)_{t \in \mathcal{T}}$ for a given tree $\mathcal{T}$ is bounded by
\begin{align}
	\label{eq:truncation_error}
	\Vert\mathcal{B} - \tau_{\mathbf{r}}(\mathcal{B}) \Vert^2 \leq \sum\limits_{t \in\mathcal{T}} \sum\limits_{m > r_{t}} \sigma_{t, m}^2 \leq~ C d ~\Vert\mathcal{B} - \mathcal{B}^{\operatorname{best}} \Vert^2
\end{align}
where $\tau_{\mathbf{r}}$ denotes the truncation operator, $\sigma_{t, m}$ is the $m$-th singular value of the matricization $\mathcal{B}^{\left(t\right)}$, $\mathcal{B}^{\operatorname{best}}$ is a best rank-$\mathbf{r}$ approximation, and $C < 2$ is a small constant.
According to~\cref{eq:truncation_error}, a steep decay in the singular values of the corresponding matricization leads to a small relative truncation error, i.e., the tensor can be well approximated with one of low rank.

Inspired by~\cite{Hackbusch2012},~\cref{tab:tensor_arithmetic} summarizes some arithmetic operations and their cost in the tensor-train and the hierarchical Tucker format, see~\cite{Grasedyck2010,Hackbusch2012,Oseledets2009}.~\Cref{tab:tensor_arithmetic} shows that using those low-rank tensor formats reduces the exponential computational cost in the dimension $d$ to linear cost.
\begin{table}[tbhp]
	\captionsetup{justification=centering}
	{\footnotesize
			\caption{\footnotesize \\ Operations and their cost in tensor-train $\left(\TT\right)$ and hierarchical Tucker $\left(\HT\right)$  format for $d$-dimensional tensors with maximum mode size $n$ and rank component-wise bounded by $r$.}
			\label{tab:tensor_arithmetic}
		\begin{center}
			\begin{tabular}{lll}
				\hline
				Operation & Cost in $\TT$ & Cost in $\HT$
				\\
				\hline
				Storage & $\mathcal{O}( d n r^2)$ &$\mathcal{O}( d r^3 + d n r)$
				\\
				Addition & $\mathcal{O}( d n r^3)$  & $\mathcal{O}( d n  r^2 + d r^4)$
				\\
				Evaluation & $\mathcal{O}( d r ^2)$  &  $\mathcal{O}( d r^3 )$
				\\
				Scalar product& $\mathcal{O}( d n r^3)$& $\mathcal{O}( d n r^2 + d r^4)$
				\\
				Operator application & $\mathcal{O}( d n^2 r^4)$ & $\mathcal{O}( d n^2 r)$
				\\
				Truncation  &  $\mathcal{O}( d n r^3 )$ & $\mathcal{O}( d n r^2 + d r^4 )$
				\\
				\hline
			\end{tabular}
	\end{center}
	}
\end{table}
Since the hierarchical Tucker format is based on a binary tree, it is possible to do many of the arithmetic operations and especially the costly truncation in parallel level by level of the tree as described in~\cite{Grasedyck2017}.
Thus, for a balanced tree and dimension $d$ the runtime can often be reduced from $\mathcal{O}\left(d\right)$ to $\mathcal{O}\left(\log(d)\right)$, cf.~\cite{Grasedyck2017}.
\end{document}

config: arxiv


TITLE: What is the difference between thermodynamic free energies and the Landau free energy?
QUESTION [7 upvotes]: How and why is the Landau free energy any different from thermodynamic free energies?
It is written on page 140 of Nigel Goldenfeld's book Lectures on Phase Transitions and The Renormalization Group that

The Landau free energy has dimensions of energy, and is related to, but, as we will see, is not identical with the Gibbs free energy of the system.

The explanation in section 5.6 is quite elaborate and too complicated. Please help me with a simple understanding of why the Landau free energy is not the Helmholtz free energy or the Gibbs free energy, and how it is related to the thermodynamic free energies.

REPLY [10 votes]: The Landau free energy, also called the Landau-Ginzburg Hamiltonian, is treated in an adhoc and rather confusing manner in a lot of textbooks. But in the modern view, it has a simple interpretation as an effective Hamiltonian attained by integrating out degrees of freedom.
Suppose we have a spin system, such as an Ising magnet. We can describe the state of the system by a magnetization field $\phi(x)$, noting that this field doesn't make sense if we examine length scales smaller than the lattice spacing $a$. We can write a sum over all spin states by an integral over field configurations, as long as the integral is cut off at the distance scale $a$.
If the Hamiltonian is $H[\phi]$, then the thermodynamic free energy $F$ obeys
$$Z = e^{-\beta F} = \int_{\Delta x\, >\, a} \mathcal{D}\phi \, e^{-\beta H[\phi]}$$
which is just a rewording of the standard identity $F = - k_B T \log Z$. In the Wilsonian view, the thermodynamic free energy is acquired by integrating out all microscopic degrees of freedom. The result only depends on macroscopic quantities like temperature, pressure, and the external field. This is useful because the entire point of thermodynamics is to ignore the microscopic details and focus on macroscopic quantities that are easy to measure. For example, using just the function $F$, we can determine the equilibrium magnetization by minimizing it.
Now, the Landau free energy $H_L$ satisfies
$$Z = \int_{\Delta x \, > \, b} \mathcal{D}\phi \, e^{-\beta H_L[\phi]}$$
where $b$ is a mesoscopic distance scale, larger than $a$ but still much smaller than a macroscopic length. In the Wilsonian view, the Landau free energy is the effective Hamiltonian acquired by integrating out degrees of freedom on lengthscales $a < x < b$. The point of the Landau free energy is that it represents a compromise between the completely microscopic $H$, which has too much detail to be useful, and the completely macroscopic $F$, which tells us nothing about, e.g. position dependence. Like $H$, $H_L$ is a functional, but it's a functional of "fewer variables".
The above explains why $H_L$ can be called a Hamiltonian, but why is it also called a free energy? Usually, the starting point for applying Landau theory is the saddle point approximation, which states that typical equilibrium field configurations minimize $H_L$. Since we're minimizing $H_L$, we're treating it like we would a free energy, which is why it's sometimes called the Landau free energy.
But why is this valid? You definitely can't get the right answer to any thermodynamic question by minimizing $H$, because it doesn't take into account thermal effects; you instead have to minimize $F$. Minimizing $H_L$ gives the right answer precisely when thermal effects are negligible on distance scales greater than $b$. This is true when $b$ is much greater than the system's correlation length $\xi$, which is why Landau theory does such a good job, and usually not true at a critical point where $\xi$ diverges, which is why Landau theory fails to describe continuous phase transitions.


TITLE: What's the error in this argument that Fin$\le_m$Inf
QUESTION [2 upvotes]: There must be an error in the following argument since Fin  is not many-one reducible to Inf, I can't seem to find it. Here it is informally (I hope it's straightforward and not confusing):
Take any infinite r.e. set like $K$  and consider the following:
We will enumerate members of $W_{e}$  and $K$  (simultaneously). We start with $A^0=\emptyset$  and at each stage $i$ given $A^i$ if we've found a new element in $K$ then we add it to $A^{i+1}$.  If at this stage we also find a new element in $W_{e}$ then  we let $A^{i+1}=\emptyset$ otherwise let $A^{i+1}=A^i\cup$ whatever new element we found in $K$ (if one was found). If we found a new element of $W_{e}$  at stage $i$, then  we begin enumerating $K$ from scratch at the next stage, otherwise we keep enumerating $K$ where we left off. Regardless of what happens at stage $i$ we keep enumerating $W_{e}$ where we left off in the next stage.
At the end let  $W_{g(e)}=\bigcup_{i<\omega}A^i$
What will then happen is:
1) if $W_{e}$  is finite then at some point no new elements of $W_{e}$  will be found and so we will be able to place all elements of $K$  into $W_{g(e)}$  without $W_{e}$  hindering us. So in the end we will have $W_{g(e)}=K$,  which means $W_{g(e)}$  will be infinite.
2) if $W_{e}$  is infinite, then we will always be finding new elements of $W_{e}$  and we will be put in a loop and never be able keep anything inside of $W_{g(e)}$.  So in the end we will have $W_{g(e)}=\emptyset$,  which means $W_{g(e)}$  will be finite.
Any help is appreciated, thanks.

REPLY [1 votes]: When you say that you are trying to make $W_{g(e)}$, you are giving a computable enumeration of a set that should be c.e. in the end. A c.e. enumeration can only put elements into a set, but can not take it out. In your construction, "Each time a new element is found in $W_e$ we clear out everything we have in $W_{g(e)}$" This involves removing elements from your enumeration.
Such a construction of a set which merely has an computable approximation where elements come in and out are not necessarily c.e. sets. There is a general notion of a $n$-c.e. set. These are sets that possess a computable approximation where for each element $x$, it can come in and out of the approximation at most $n$ times. In this sense, the 1-c.e. sets are the ordinary c.e. sets. It is very easy to show that there is a 2-c.e. set that is not c.e.


TITLE: Khan Academy, Example: Computing Partial Derivative
QUESTION [1 upvotes]: Struggling in following this problem and it's solution.
Problem:
$f(x, 2) = 8x^2$
Solution:
$\frac{d}{dx}f(x, 2)=\frac{d}{dx}(8x^2)=16x$
$x=3$
$16(3)=48$
I got to the part where it's $8x^2$ - but then assumed the solution would be $8 * 3^2 = 72$
Guess what I am asking is why was the exponent moved down and multiplied instead of being used as an actual exponent?

REPLY [0 votes]: $f(x,2)=8x^2$ is the function that you want to take the derivative.
We compute the derivative and found
$$\frac{d}{dx} f(x,2) = \frac{d}{dx}(8x^2) = 8(2)x^{2-1}=16x$$
Since you want to compute the derivative, we evaluate the derivative at $x=3$.
We have for a monomial, $x^n$,
$$\frac{d}{dx}x^n = nx^{n-1}$$
If you are interested in the value of a function, $f$, we compute $f(x)$.  If we are interested in the derivative, we compute $f'(x) = \frac{d}{dx}f(x)$.


TITLE: How long can the Enterprise's shields hold?
QUESTION [0 upvotes]: The Enterprise is under attack from a Romulan Bird of Prey that is firing phasers which do (say: 33.333) units of damage per minute.
The Enterprise's shields are rated to resist say (100) units of damage before being completely depleted.
However, as the shields are depleted, the resistance drops proportionally: At 100% they absorb 100% of damage, while at 75% they absorb 75% of damage where the remaining (25%) damage is sustained by the Enterprise's hull.
so a one-minute attack which does 33.333 units of damage would reduce the shields from 100% to 75% (absorbing 25 damage), and the hull will sustain the remaining 8.333 damage.
(I think that's right, anyway.)
For the next shot, the shields will start at 75%, meaning greater damage to the hull, but how much?
More importantly, how would I create a general formula which can calculate hull damage for any shield rating/level and any attack power?

REPLY [1 votes]: I don't understand the initial condition.  It seems like the shields start at $100\%$ efficiency, so they would never be depleted.  I think you need to start them at something close to $100\%$, but not exactly.
If we let $y(t)$ be the resistance of the shields at time $t$, where $0\leq y(t)\leq 1$ then the equation is $$y'= -k(1-y)\tag1$$ where $k$ represents the power of the phasers.  In the example, $k=\frac13.$  When $y=\frac34$, we get $$y'=-\frac13\left(1-\frac34\right)=-\frac1{12},$$ so that the resistance decreases by $\frac1{12},$ as in the example.
The general solution to $(1)$ is $$y=ae^{kt}+1,\tag2$$ where $a$ is a constant to be determined from the initial condition.  That is, $a=1-y(0)$.  If we set $y(0)=1$ as you say, then of course, $a=0$, and the shields stay at full resistance always.
If the shields start at $99.99\%$ efficiency, then we get $$y(t)=1-.0001e^{t/3}$$ when $k=\frac13.$


TITLE: Energy to momentum
QUESTION [5 upvotes]: Is there anyway to convert energy to motion IN SPACE? Let's say a satellite collects electric energy from sun using solar panel. Is it possible to convert it to Linear motion? The only way I know to change linear motion in space is by throwing stuff out (Ions, burned propellants etc).

REPLY [1 votes]: You're right, the only way to change linear motion in space is by throwing stuff out, and why is that a problem?
Instead consider, most obviously, a nuclear submarine which any number of people think somehow moves directly on nuclear energy. Really?
It would be just as accurate to brand the critter a "steam submarine" or by the same token to rename the earlier railway "steam" trains as "coal" or "wood" engines.
Either actually turns its energy into power through steam boilers, whether the boilers happen to source their heat energy from nuclear reactors or coal or wood fires.
In principle, yes, your craft uses solar energy to generate electricity, uses that to boil water then exhausts steam to create motive force.
No few writers pay lip-service to that principle by referring to "reaction mass" but how is that more useful than referring to nuclear or solar energy in the first place?
Please remember, none of that says anything about efficiency or sustainability.


TITLE: Unbiased estimator of square of covariance
QUESTION [0 upvotes]: Let $X$ and $Y$ be two correlated random variables. The covariance is:
$\mu_{1,1} = E[(X-E[X])(Y-E[Y])]$
And the unbiased estimator:
$\hat{\mu}_{1,1} = \frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})$
According to some numerical experiments I've run, $\hat{\mu}_{1,1}^2$ is not an unbiased estimator of $\mu_{1,1}^2$. So, apparently I need the "multivariate polyache" (see p.11-12 of this book), but I can't find an implementation.
Any suggestions?
Thank you!

REPLY [1 votes]: I'll just go ahead and answer my own question since I've found the answer in the Mathematica documentation:
An unbiased estimator of the square of covariance is the multivariate polyache $h_{\left \{ \left \{ 1,1 \right \},\left \{ 1,1 \right \} \right \} }$

Where
$S_{r,s} = \sum_{i=1}^n x_i^r y_i^s$
MultiPolyH[P] is a Mathematica function I've written (details bellow)

Just to have it documented, here is a more general answer:
For $X_1,X_2,\dots,X_d$ random variables, an unbiased estimator of the product $\prod_{j=1}^m\mu_{p_{j,1},p_{j,2},...,p_{j,d}}$, where
$\mu_{p_{j,1},p_{j,2},...,p_{j,d}} = E \left[ (X_1-\bar{X}_1)^{p_{j,1}}(X_2-\bar{X}_2)^{p_{j,2}}\dots(X_d-\bar{X}_d)^{p_{j,d}} \right]$
is the multivariate polyache $h_{ P }$, where
$P = \left \{ \left \{ p_{1,1},p_{1,2},...,p_{1,d} \right \} ,\left \{  p_{2,1},p_{2,2},...,p_{2,d}\right \}, \dots \left \{ p_{m,1},p_{m,2},...,p_{m,d} \right \}\right \}$
The following function implemented in Mathematica computes $h_P$.
MultiPolyH[P_] :=
FullSimplify[
    MomentConvert[
        Product[CentralMoment[P[[i]]], {i, 1, Length[P]}],
        "UnbiasedEstimator", "PowerSymmetricPolynomial"]
    /.{PowerSymmetricPolynomial[0] -> n},
    Assumptions -> Element[n, Integers]]

For example, the product of variances $\mu_{2,0}\mu_{0,2}$ can be estimated by:

Note that the Mathematica function also works for $d=1$ and $m=1$. For example the third central moment:


TITLE: What am I doing wrong; evaluating integrals?
QUESTION [0 upvotes]: $$f(x) = \dfrac{2x}{2-3x^2}$$
$$- \dfrac{1}{3} d (2-3x^2) = 2x$$
$$ \dfrac{2x}{2-3x^2} dx = -\dfrac{1}{3} d (2-3x^2) \cdot \dfrac{1}{2-3x^2} = - \dfrac{1}{3} du \cdot u^{-1} = -\dfrac{1}{3} u^{-1} \cdot du$$
And here I'm stuck, since you can't divide by 0.. What am I doing incorrectly here? I've applied the method my book gave me perfectly as far as I know.

REPLY [0 votes]: All your steps are correct. But what it seems you're not aware of is that
$$\int{\frac{1}{x}}\space{dx}=\ln|x|+C$$
This is an exception to the power rule.


TITLE: Finding numbers that double when you switch the first and last digit
QUESTION [1 upvotes]: So, I was watching this video by MindYourDecisions about finding the smallest number that doubles when you move the last digit to become the first digit. Actually, I just saw the title of the video, and I found it interesting, so I began to work on a solution. But, in fact, I had misinterpreted the question posed by the video.
I was looking for a number that doubles when you switch the first and last digits of a number. I started off my search by writing equations like:
$$ 2(10a + b) = 10b + a $$
This simplifies to:
$$ 19a = 8b$$
Since $a$ and $b$ should be integers between 0 and 9 inclusive, I figured out that there are no solutions amongst two-digit numbers.
Next, I applied this technique to three and four digit numbers, getting this equation for three digit numbers (using the convention that $a$ is the first digit, $b$ is the second, and so on):
$$ 199a + 10b = 98c $$     And for four-digit numbers,
$$ 1999a + 100b + 10c = 998d $$
Next, I made tables varying the value of $a$ and the last digit between 0 and 9, looking for numbers that were close enough so that the smaller terms in the middle could make up the difference.
I couldn't find any solutions; however, I was able to find near misses such as 37, 397, 3997, or generally, a 3 followed by a string of 9s followed by a 7, which all differ by just one.
I am wondering if there actually are any solutions to this problem, and if not, how you would go about proving that.
Thank you!

REPLY [0 votes]: Assume x > 0.
Let x == a0 + 10a1 +...+ an10^n where 0 <= ai <= 9, n >=0,  the usual decimal expansion.
Then 2x == an + 10a1+...+a010^n, so x == (an-a0) + (a0-an)10^n since x == 2x - x.
Clearly, a0 != an and an is even.
If a0 < an, then 0 < (an-a0) < 10 and -10^n > (a0-an) so x < 0. Contradiction.
So a0 > an.
Also, if a0 < 5, x == 2x implies an == 2a0, which contradicts the above.
So a0 == 6,7,8 or 9.
Thus an == 2,4,6 or 8, the modulus 10 of 2*a0.
So candidates are 2?6, 4?7, 6?8 or 8?9, where ? is some fixed string of digits.
But 2*(2?6) < 6?2, 2*(4?7) > 7?4, 2*(6?8) > 8&6 and 2*(8?9) > 9?8.
So there is no solution.


TITLE: Irreducible representation of the product of two groups and tensor product
QUESTION [6 upvotes]: Let $G_1, G_2$ be two lie groups, $V$ be a finite dimensional (continuous) irreducible complex representation of $G_1 \times G_2$, must $V \cong V_1 \otimes V_2$ for some irreducible representation $V_i$ of $G_i$?
If $G_i$ are compact, this is true by Peter-Weyl theorem.

REPLY [9 votes]: If the field is $\mathbb C$, There are many ways of seeing this. For $i=1,2$ we may replace $G_i$ by its Zariski closure in $GL(V_i)$ without changing the hypotheses or the conclusion. But if an algebraic subgroup $G\subset GL(V)$ is irreducible, then it is reductive (the unipotent radical will have a fixed space which is G invariant and hence zero).
We now use the fact that the  reductive group $G_i$  has a Zariski dense compact group $K_i$; we may thus replace $G_i$ by the compact $K_i$ where you have accepted the result.


TITLE: Planar bipartite graph
QUESTION [4 upvotes]: I know that the number of edges for a face in a bipartite is at least 4. Clearly, we cannot have 3, because we will need to move back to the other side to have face. But, I'm having difficulty in imagining a face of a planar bipartite graph. Where can I find an example of it?
On the other hand, I have found the following equality and inequality, but I couldn't grasp exactly what they are saying, if someone can elaborate for me them, I would be glad.
$$2 |E| = \sum_{i \geq 4} i F_i$$ where $i$ is the number of edges of the face $F_i$ and $F_i$ is the number of faces with $i$ edges. Clearly this is somehow consequence of the handshaking lemma, but why multiplying number of edges of a face with the faces give us this result? Shouldn't we use the degree of each vertex for handshaking lemma?
The second inequality is this one
$$4 \sum_{i \geq 4} F_i \leq \sum_{i \geq 4} i F_i$$
$$4 |F| \leq 2|E|$$
I mean, the second part is form the above equation, but first part the 4 confuses me. Is it because we need at least 4 edges for a face in bipartite graph? If yes, why do we need it actually in this inequality? Second question, why does this inequality hold? A clarification would be highly welcomed.

REPLY [2 votes]: Here’s a simple example of a planar bipartite graph:
           a
          / \
         /   \
        b--c--d
         \   /
          \ /
           e

It has three vertices, $a,c$, and $e$, in one part and two, $b$ and $d$, in the other. It has three faces: one is the triangular region bounded by the cycle $abcd$, another is the triangular region bounded by the cycle $bcde$, and the third is the external face bounded by the cycle $abed$. Each of these faces has $4$ edges and $4$ vertices.
To see why $$2|E|=\sum_{i\ge 4}iF_i\;,$$ note that each edge of a planar graph is adjacent to two faces, one on each side of it. Suppose that the graph has faces $f_1,\ldots,f_n$, and face $f_k$ has $e_k$ edges for $k=1,\ldots,n$. Then $e_1+\ldots+e_n$, the sum of the numbers of edges of the faces, counts each edge twice, once for each of the two faces adjacent to it. Thus, $e_1+\ldots+e_n=2|E|$. But $e_1+\ldots+d_n$ is the same thing as $\sum_{i\ge 4}iF_i$. $F_i$ is the number of faces with $i$ edges, so $iF_i$ is the total number of edges for all of the faces with $i$ edges, and when we sum over $i$, we get the total number of edges of all of the faces of the graph. The degrees of the individual vertices don’t enter into it.
In the example above, $F_4=3$, and $F_i=0$ for $i\ne 4$, so $\sum_{i\ge 4}iF_i=4\cdot3=12=2\cdot6$, which is as it should be, since there are $6$ edges.
The see why the inequality
$$4\sum_{i\ge 4}F_i\le\sum_{i\ge 4}iF_i$$
holds, notice that if $i\ge 4$, then $4F_i\le iF_i$. Since we’re summing only over values of $i$ greater than or equal to $4$, we therefore have
$$\sum_{i\ge 4}4F_i\le\sum_{i\ge 4}iF_i\;:$$
each term on the left is less than or equal to the corresponding term on the right, so the sum on the left must be less than or equal to the sum on the right. And of course
$$4\sum_{i\ge 4}F_i=\sum_{i\ge 4}4F_i\;.$$
Finally, $\sum_{i\ge 4}F_i$ is simply $|F|$, the total number of faces, since every face has at least $4$ edges: we’re just adding up the number with $4$ edges, the number with $5$ edges, and so on. Thus, $4\sum_{i\ge 4}F_i=4|F|$. Combining this with the inequality and the original equation, we see that $4|F|\le 2|E|$ (and hence $2|F|\le|E|$: the number of edges of a planar bipartite graph is at least twice the number of faces).
In the example above $|F|=3$ and $|E|=6$, so in this case the inequality is actually an equality: the number of edges is exactly twice the number of faces.


TITLE: If $f,g:X \to Y$ are continuous and $Y$ is $T_2$, then $\{x \in X\,|\,f(x)=g(x)\}$ is closed
QUESTION [8 upvotes]: I'd like to know if the following proof is valid.  The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.
"Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions.  Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."
Let $x \in X$ be such that $f(x) \ne g(x)$.  Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$.  Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$.  Then $U \cap f(C)=V \cap f(C)=\varnothing$.  Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$.  Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.
Thanks.

REPLY [1 votes]: Different argument (assuming you did the exercise in the link before):
If $f,g:X\to Y$ are continuous and $Y$ is Hausdorff then by a closed graph theorem we have that $G_f=\{(x,f(x)):x\in X\}$ and $G_g=\{(x,g(x)) : x\in X\}$ are closed in $X\times Y$; therefore, $G_g\cap G_f = \{(x,f(x) ):x\in X , f(x)=g(x)\}$ is closed as well.  But $F:X\to X\times Y$ given by $F(x)=(x,f(x))$ is continuous , whence $F^{-1}(G_f\cap G_g)=\{x : f(x)=g(x)\}$ is closed in $X$.


TITLE: constructive Serre classes
QUESTION [9 upvotes]: A Serre class (of abelian groups) is a class of abelian groups closed under subgroups, quotients, and extensions.  For instance, finitely generated groups and finite groups are both Serre classes.
However, in constructive mathematics, these are no longer examples, at least not with the usual definition of "finite" (= in bijection with $\{0,1,\dots,n\}$ for some $n\in\mathbb{N}$).  In particular, finite sets are not closed under subsets and quotients, so there is no reason that finite groups should be either.
There are other weaker constructive notions of "finite", some of which are described here: subfinite, finitely indexed, subfinitely indexed.  It seems that subfinitely indexed sets — the subquotients of finite sets — are closed under subsets, quotients, and finite products, so that the subfinitely indexed groups should be a Serre class even constructively.  Classically, of course, all subfinitely indexed sets are finite.
My original question was:

Is there a Serre class of abelian groups in constructive mathematics which reduces classically to the finitely generated ones?

As pointed out by Ingo, this has a trivial and uninteresting answer; the real question is

Is there a description of the Serre envelope of the class of finitely generated abelian groups which is more explicit than the trivial inductive one?

REPLY [4 votes]: Any subclass $\mathcal{C}$ of an abelian category determines a smallest Serre class containing it, by iteratively adding (the zero object and) the object $Y$ for any exact sequence $X \to Y \to Z$ where $X$ and $Z$ are objects of the previous stage. Note the missing zeros at the ends of the sequence; alternatively, one can iteratively add the zero object, subobjects, quotients, and extensions (objects $Y$ such that there is a short exact sequence $0 \to X \to Y \to Z \to 0$ with $X$ and $Z$ previously added).
Anyway, this construction can in particular be performed with the class $\mathcal{C}$ of finitely generated abelian groups. Classically, its closure will coincide with $\mathcal{C}$, as $\mathcal{C}$ is already a Serre class.
However, this construction only answers the question in a technical sense, since it describes the sought-after class in rather abstract terms. The construction thus only shows that the question could profitably be reformulated in order to better capture its spirit: "Is there a description of the Serre envelope of the class of finitely generated abelian groups which is more explicit than the trivial inductive one?"


TITLE: Proof verification of $\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$ using $\epsilon$-$\delta$
QUESTION [4 upvotes]: I am trying to prove that
$$\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$$
and this is my approach.
Let $\epsilon>0,\delta>0\,$ so that
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|<\epsilon\quad\text{whenever}\quad-2<x<-2+\delta$$
We have that
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}$$
Let $\delta<1$
$$-2<x<-2+\delta<-1\implies-3<x-1<-2\implies2<|x-1|<3$$
so
$$|x-1|\sqrt{x+2}\leq3\sqrt{x+2}<\epsilon$$
if we pick $\,\delta\leq\min\{1,\epsilon^2/9\}.$
Would this be correct?

REPLY [3 votes]: Seems fine, just write it in the forward direction after working backward.
Given $\epsilon >0$, we let $\delta = \min\{ 1, \frac{\epsilon^2}{9}\}$,
then if $-2 < x < -2 + \delta$, we have $-3<x-1< -2+\delta \implies  2<|x-1|<3$.
hence,
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=\left|\frac{(x-1)(x+2)}{\sqrt{x+2}}\right|=|x-1|\sqrt{x+2}<3\sqrt{x+2} \le 3\sqrt{\frac{\epsilon^2}{9}}=\epsilon$$


TITLE: Proving that the line joining $(at_1^2,2at_1),(at_2^2,2at_2)$ passes through a fixed point based on given conditions on $t_1,t_2$
QUESTION [2 upvotes]: Problem:If $t_1$ and $t_2$ are roots of the equation $t^2+kt+1=0$ , where $k$ is an arbitrary constant. Then prove that the line joining the points $(at_1^2,2at_1),(at_2^2,2at_2)$ always passes through a fixed point.Also find that point.
I have done this question by a fairly difficult procedure(at least I think so) ..by finding the equation of the line passing through the given points as
$$y(t_2+t_1)-2at_1t_2-2at_1^2=2x-2at_1^2$$
And then using the relation from the quadratic equation given and writing it as $$(x+a)+k(\frac{y}{2})=0$$ which is the equation of family of lines and so which gives the fixed point as $(-a,0)$.
But actually I found this question in exercises of parabola and so I was thinking if it was possible to solve this question without this family of lines and all that (that is by using geometrical properties of parabola)...I have been thinking about this ...and I could only notice that since $(at^2,2at)$ is the parametric equation of a general point on a parabola $y^2=4ax$ so the points $(at_1^2,2at_1),(at_2^2,2at_2)$ can be thought of as two given points on parabola $y^2=4ax$ where $t_1t_2=1$ but I am not able to get that fixed point (which apparently is a point on the directrix of parabola $y^2=4ax$ from the above method that is $(-a,0)$)  based on the above information by using the geometrical properties of parabola.
Any help is appreciated on this.

REPLY [1 votes]: Tgts at ends Parb
With some reference to the above:
When the roots are solved, the tangent drawn at these particular slope pairs all cut on the directrix when the chord between $(t_1- t_2)$ points of tangency passes through the focus.
EDIT 1
There appears to be an incorrect tangent/slope supplying equation to start with, at the outset:
$$ t^2 + k t + 1 = 0 $$
whereas the correct equation for the given parametrization ought to be
$$ t^2 + k t - 1 = 0 $$
EDIT 3:
( It is now conceded the former sign is correct for all rays through (-a,0) )
By change of sign in the constant term we have two different "foci"
The sum of roots at either end of a parabola can be arbitrary, but since tangent at either end of focal chord are perpendicular ( meeting on directrix but that is besides the point) their product of slopes must be necessarily $-1$ and not $+1$ as we can see directly from quadratic equation coefficients.
This property of parabola is easy to verify and shall not do it here.
Taking correct sign means chaging sign of $ a, a\rightarrow -a $
$$ (x- a)+k(\frac{y}{2})=0 $$
EDIT 2:
( To find fixed point by C discriminant , partially differentiate w.r.t. k, y=0, and so (x=a, y=0) is the fixed point).
which is the equation of all focal rays passing through a fixed point, viz., the focus of parabola.
(BTW, It will be also an interesting problem to see how the (red) directrix line is fixed)
I am giving a sketch for confirmatory reference.


TITLE: Find the maximum possible area of a certain right triangle
QUESTION [2 upvotes]: I want to find the maximum possible area of a right triangle with hypotenuse $=10$.
My approach so far: let $x,y$ be the lengths of the two sides adjacent to the right angle; then $$100=x^2+y^2$$
Area $=\frac{xy}{2}$, so by the substitution method I got equality. But my result is wrong (probably I made a mistake in the equality), could someone show what to do?

REPLY [0 votes]: The area of triangle $ABC$ is just $\Delta ABO + \Delta BOC$. If $\angle BOC = \theta$, the area is:
$$\frac{1}{2} r^2 \sin \theta + \frac{1}{2}r^2 \sin(180 - \theta) = \frac{1}{2}r^2 \big( \sin \theta + \sin(180 - \theta) \big) = \frac{1}{2}r^2 (2 \sin \theta) = r^2 \sin \theta$$
where we have used the identity $\sin \theta = \sin(180 - \theta)$.
The maximum value of $\sin \theta$ is $1$, so the maximum area is $r^2$. In this case, $r = 5$ so the maximum area is $25$.
This occurs when $\theta = \frac{\pi}{2}$ or $90º$, so $\Delta ABC$ must in fact be isoceles.

Alternatively, if $A = (-5,0)$, $B = (5 \cos t, 5 \sin t)$, $C = (5,0)$, then the area of $\Delta ABC$ using the shoelace formula is:
$$
    \begin{vmatrix}
    5 & 0  \\
    5 \cos t & 5 \sin t \\
    -5 & 0 \\
    5 & 0 \\
    \end{vmatrix}
$$
so the area is: $$\frac{1}{2} \big((25 \sin t + 0 + 0) - (0 + -25 \sin t + 0) \big) = 25 \sin t$$
which takes the maximum value $25(1)$ when $t = \frac{\pi}{2}$, again.
The area of triangle $ABC$ is can also be computed using determinants:
$$
    \begin{vmatrix}
    5 & 0 & 1 \\
    5 \cos t & 5 \sin t & 1 \\
    -5 & 0 & 1 \\
    \end{vmatrix}
$$
which when computed, gives:
$$\frac{1}{2} \left((25 \sin t + 0 + 0) - (-25 \sin t + 0 + 0) \right) = 25 \sin t.$$