text:
\begin{document}

\title[On essentially large divisors]{On essentially large divisors}

\author{Gordon Heier}
\author{Min Ru}

\address{Department of Mathematics\\University of Houston\\4800 Calhoun Road, Houston, TX 77204\\USA}

\email{heier@math.uh.edu}
\email{minru@math.uh.edu}

\subjclass[2000]{11G35, 11G50, 14C20, 14G40, 32H30}

\begin{abstract}
Motivated by the classical Theorems of Picard and Siegel and their generalizations, we define the notion of an {\it essentially large} effective divisor and derive some of its geometric and arithmetic consequences. We then prove that on a nonsingular projective variety $X$ whose codimension is no greater than $\dim X-2$, every effective divisor with $\dim X +2$ or more components in general position is essentially large.
\end{abstract}

\maketitle

\section{Introduction}
In \cite{cz_cr}, Corvaja and Zannier found an innovative way of using Schmidt's Subspace Theorem to give a new proof of the classical Theorem of Siegel on integral points on affine varieties. They subsequently expanded their approach to obtain certain results on integral points in higher dimensions (\cite{cz_crelle}, \cite{cz_annals}, \cite{cz_ajm}, \cite{cz_imrn}). The approach was translated to Nevanlinna theory in \cite{ru_ajm}.\par
Evertse and Ferretti (\cite{ferretti_compositio}, \cite{ef_imrn}) used similar (yet more general) arguments to obtain diophantine inequalities on projective varieties. Their approach is largely based on Mumford's degree of contact. In Nevanlinna theory, this approach was used by Ru \cite{ru_annals} to establish a Second Main Theorem for holomorphic curves into projective varieties intersecting hypersurfaces.\par
As stated in \cite[p.\ 609]{levin_annals}, the article \cite{cz_cr} motivated Levin to make the following Definition. Its significance is due to the hyperbolicity-type and Mordell-type properties enjoyed by the complements of large divisors. We let $K$ denote either a number field or $\CC$.
\begin{definition}[{\cite[Definition 8.1]{levin_annals}}]
Let $D$ be an effective divisor on a nonsingular projective variety $X$ defined over $K$. Then $D$ is said to be {\it very large} if for every $P\in D(\bar K)$, there exists a basis $B$ of the finite-dimensional vector space
$$L(D)=\{f \text{\ rational function on\ } X| \divisor(f) \geq -D\}$$
such that $\ord_E \prod_{f\in B} f > 0$ for every irreducible component $E$ of $D$ with $P\in E(\bar K)$. Moreover, an effective divisor is said to be {\it large} if it has the same support as some very large divisor.
\end{definition}
Recall that for an effective divisor $D$ as in the above Definition, the finite-dimensional vector spaces $L(D)$ and $H^0(X,\OO_X(D))$ are isomorphic via
\begin{equation}H^0(X,\OO_X(D))\to L(D),\quad s\mapsto \frac s {s_D},\label{h0_isom}\end{equation}
where $s_D$ is a section of $\OO_X(D)$ with $\divisor (s) =D$.\par
For a divisor $D=\sum_{i=1}^{r} D_i$ with $D_i$ effective (but not necessarily irreducible), Levin's main sufficient criterion for very largeness is the following.
\begin{lemma}[{\cite[Lemma 9.1]{levin_annals}}]\label{Levin_9_1} Let $D=\sum_{i=1}^{r} D_i$ be a divisor with each $D_i$ effective (but not necessarily irreducible) on a nonsingular projective variety $X$, all defined over $K$. For $P\in D(\bar K)$, let $D_P=\sum_{\{i:P\in D_i\}}D_i$. Also, for integers $m,n$, let
$$f_P(m,n)=h^0(X,\OO_X(nD-mD_P))-h^0(X,\OO_X(nD-(m+1)D_P)).$$
If there exists $n>0$ such that $\sum_{m=0}^\infty (m-n)f_P(m,n) > 0$ for all $P\in D(\bar K)$, then $nD$ is very large.
\end{lemma}
The short proof provided in \cite{levin_annals} is based on the filtration argument introduced in \cite{cz_cr}. The main idea is to define a filtration $$V_j=H^0(X,\OO_X(nD-jD_P)) \quad (j=0,1,2,3,\ldots)$$
of $H^0(X,\OO_X(nD))$ and to choose a basis $f_1,\ldots ,f_{h^0(X,\OO_X(nD))}$ of $L(nD)$ according to this filtration, beginning with the last nonzero subspace. Since $\dim V_j/V_{j+1}=f_P(j,n)$, we get
$$\ord_E \prod_{i=1}^{h^0(X,\OO_X(nD))} f_i \geq (\ord_E D)\sum_{m=0}^{\infty} (m-n)f_P(m,n) > 0.$$\par
The main result on large divisors in \cite{levin_annals} is the following.
\begin{theorem}[\cite{levin_annals}]\label{levin_2mq_thm}
Let $X$ be a $q$-dimensional nonsingular projective variety, defined over $K$. Let $D=\sum_{i=1}^{r} D_i$ be a divisor, also defined over $K$, with each $D_i$ effective (but not necessarily irreducible) and big and nef. Moreover, assume that every irreducible component of $D$ is nonsingular and that the intersection of any $m+1$ distinct $D_i$ is empty over $\bar K$. If $r>2mq$, then $D$ is large.
\end{theorem}
In the proof of this Theorem given in \cite{levin_annals}, the difference $h^0(X,\OO_X(nD-mD_P))-h^0(X,\OO_X(nD-(m+1)D_P))$ is bounded from above based on the $H^0$-part of the corresponding long exact cohomology sequence. As remarked by Levin, using the double filtration argument from \cite[Lemma 3.2]{cz_annals}, the factor of $2$ in the lower bound on $r$ can be removed.\par
It would of course be interesting to decrease the lower bound on $r$ in Theorem \ref{levin_2mq_thm}. In light of the classical Theorems of Picard and Siegel and their generalizations, and some of the standard conjectures in hyperbolicity theory, one might suspect that when $m=q$, the lower bound of $2mq$ can be replaced by $q+1$. However, it is quite clear that the sufficient criteria in \cite{levin_annals} cannot be used to prove this. To be precise, we make the following Remark.
\begin{remark}\label{P_q_suff_crit_remark}
In the situation of Lemma \ref{Levin_9_1}, let $X=\PP^q$. Let $D$ be the sum of $r$ general smooth hypersurfaces of degree $d$. Then
$$\sum_{m=0}^\infty (m-n)f_P(m,n) > 0\quad (n \text{\ sufficiently large})$$
if and only if $r> q^2+q$.
\end{remark}
We leave the proof of Remark \ref{P_q_suff_crit_remark} as an exercise to the reader. Note that since $\Pic(\PP^q)$ is generated by $\OO_{\PP^q}(1)$ and since $h^0(\PP^q,\OO_{\PP^q}(\ell))={q+\ell\choose \ell}$, the dimensions of all the $H^0$'s involved can be computed without problem.\par
We now turn to the new results presented in this note. The starting point is our definition of {\it essentially (very) large} divisors as follows.
\begin{definition}\label{ess_v_l_def}
Let $D$ be an effective divisor on a nonsingular projective variety $X$, all defined over $K$. Then we define $D$ to be {\it essentially very large} if there exists a linear subspace $V$ of $L(D)$ such that for every $P\in D(\bar K)$, there exists a basis $B$ of $V$ such that $\ord_E \prod_{f\in B} f > 0$ for every irreducible component $E$ of $D$ with $P\in E(\bar K)$. Moreover, we define an effective divisor to be {\it essentially large} if it has the same support as some essentially very large divisor.
\end{definition}
This is obviously a weakening of the notion of large divisors, which requires $V=L(D)$. However, our definition seems technically easier to handle, while it retains the important function theoretic and number theoretic consequences of the original definition of large divisors derived in \cite{levin_annals}. In fact, in Section \ref{section_nevan}, we prove new inequalities of Second Main Theorem-type and Schmidt Subspace Theorem-type for essentially large divisors that are actually stronger than the consequences for large divisors derived in \cite{levin_annals}. Since the statements of our Theorems require some (standard) definitions, we simply refer the reader to Theorem \ref{our_smt} and Theorem \ref{our_schmidt} at this point of the Introduction.\par
As a matter of convention, we will always assume that the constant function $1$ is an element of the basis $B$, although it does not contribute to the vanishing of the product.
\begin{remark}
The reader might wonder how essential very largeness fits in with the standard notions of ``size'' for a divisor in algebraic geometry. In this paper, we will not make any statements in this respect, other than the trivial observation that a basis for $V$ yields a nonconstant rational map to projective space whose locus of indeterminacy (i.e., the base locus of $V$ understood as a subspace of $H^0(X,\OO_X(D))$) is contained in $D$.
\end{remark}
In a second line of inquiry, beginning in Section \ref{section_ess_v_l_proofs}, it is clearly interesting to ask for (sharp) sufficient criteria for a divisor to be essentially large. In this direction, we first discovered that the filtration method from \cite{cz_ajm} (see also \cite{ru_ajm}) yields the expected sharp bound (even for the original version of largeness) in the case of $X=\PP^q$ as stated in the following Theorem. (Cf. our comments in Remark \ref{P_q_suff_crit_remark}.)
\begin{theorem}\label{thm_1}
Let $q\geq 1$ and $r\geq q+2$ be integers. On $\PP^q$, let $D=\sum_{i=1}^r D_i$ be a divisor defined over $K$, where each $D_i$ is a hypersurface (not necessarily irreducible or reduced). Assume that the $D_i$ are in general position, i.e., the intersection of any $q+1$ of them is empty over $\bar K$. Then $D$ is large.
\end{theorem}
In the case of nonsingular  projective varieties different from projective space, our result is the following.
\begin{theorem}\label{thm_2}
Let $q\geq 1$ and $r\geq q+2$ be integers. Let $X\subseteq \PP^\ell$ be a nonsingular projective variety of dimension $q$ defined over $K$. Let $D=\sum_{i=1}^r D_i$ be a divisor on $X$ such that each $D_i$ is defined by the restriction to $X$ of a homogeneous polynomial of degree $d_i$ in $K[X_0,\ldots,X_\ell]$. Finally, assume that the $D_i$ are in general position on $X$, i.e., the intersection of $X$ with any $q+1$ of them is empty over $\bar K$. Then $D$ is essentially large.
\end{theorem}
While the assumptions in Theorem \ref{thm_2} are certainly restrictive, they correspond exactly to the current state-of-the-art for generalized geometric versions of the Second Main Theorem (see \cite{ru_annals}).\par
It seemed to be an interesting problem to find a broad class of projective varieties to which Theorem \ref{thm_2} can actually be applied. This is treated in Section \ref{section_alg_geom}, where we obtain the following Theorem, which is repeated as Corollary \ref{cor}.
\begin{theorem}\label{cor_intro}
Let $q\geq 1$ and $r\geq q+2$ be integers. Let $X\subseteq \PP^\ell$ be a nonsingular projective variety of dimension $q$, defined over $K$. Assume that $2q - \ell \geq 2$ holds. Let $D=\sum_{i=1}^r D_i$ be an effective divisor on $X$ defined over $K$ such that the $D_i$ are in general position. Then $D$ is essentially large.
\end{theorem}
In closing this Introduction, we conjecture that the bound on the number of components in Theorem \ref{cor_intro} is generally the correct one, regardless of the codimension of $X$:
\begin{conjecture}\label{conjecture}
Let $q\geq 1$ and $r\geq q+2$ be integers. Let $X\subseteq \PP^\ell$ be a nonsingular projective variety of dimension $q$, defined over $K$. Let $D=\sum_{i=1}^r D_i$ be an effective divisor, defined over $K$, on $X$ such that the $D_i$ are big and in general position. Then $D$ is essentially large.
\end{conjecture}
\section{The geometric and arithmetic properties of essentially large divisors}\label{section_nevan}
\subsection{The geometric case}\label{geom_case}
We first derive the inequality of Second Main Theorem-type for essentially very large divisors. We begin by recalling some standard definitions from Nevanlinna Theory. \par
Let $g: {\CC} \rightarrow {\PP}^m$ be a
holomorphic map.
Let $g=[g_0, \dots, g_m]$ be a reduced representative of $g$,
where $g_0, \dots, g_m$ are  entire functions
on  ${\CC}$ and have no
common zeros. The Nevanlinna-Cartan characteristic function (or the
height function) $T_g(r)$
is defined by
$$ T_g(r) = {1\over 2\pi}\int_0^{2\pi}\log  \max_{j=0,\ldots,m}
|g_j(re^{\sqrt{-1}\theta})| d\theta.$$
The above definition is independent, up to an additive constant, of the choice of
the reduced representation of $g$. \par
For a divisor $D$ on a projective variety $X$, represented by a local defining function $\rho$, and a holomorphic map
$g:\CC\to X$,
the counting function is defined as
$$N_g(r,D)=\int_0^ r {n_g(t,D)-n_g(0,D)\over t} dt + n_g(0,D)\log r,$$
where $n_g(t,D)$ is the number of zeros of $\rho\circ g$ inside
$|z|<t$ (counting multiplicities), and $n_g(0,D)=\lim_{t\to 0^+}n_g(t,D)$.\par
By the Jensen formula, we also have
$$N_g(r,D)=\sum_{a\in D_r\setminus \{0\}} \ord_a(\rho\circ g)\log
\left|\frac r a \right|+
\ord_0(\rho\circ g) \log r.$$\par
The following generalized version of Cartan's Second Main Theorem (see \cite{ru_gen_form_smt_tams}, \cite{vojta_smt_ajm_1997}) will be the basis of the proof of Theorem \ref{our_smt}. Note that by $\|\ \|$ we mean the (coefficient-wise) maximum norm in this Subsection.
\begin{theorem}\label{gen_smt}
Let $f=[f_0:\ldots:f_m]:\CC\to \PP^m$ be a holomorphic map whose image is not contained  in a proper linear subspace. Let $H_1,\ldots,H_q$ be arbitrary hyperplanes in $\PP^m$. Let $L_j$, $1\leq j \leq q$, be linear forms defining $H_1,\ldots,H_q$. Then, for every $\varepsilon>0$,
$$\int_0^{2\pi}\max_K \log \prod_{j\in K} \frac{\|f(r e^{\sqrt{-1}\theta})\|\|L_j\|}{|L_j(f(re^{\sqrt{-1}\theta})|}\frac{d\theta}{2\pi}.\leq. (m+1+\varepsilon)T_f(r),$$
where ``$.\leq.$'' means that the inequality holds for all $r$ outside of a set $\Gamma$ with finite Lebesgue measure, and the maximum is taken over all subsets $K$ of $\{1,\ldots,q\}$ such that the linear forms $L_j$, $j\in K$, are linearly independent.
\end{theorem}
To prove our result, we need a general formula for the height $T_g(r)$
when  $g=[g_0:\ldots:g_m]$, where $g_0, \dots, g_m$ are meromorphic
functions (i.e., it is not necessarily the {\it reduced} representation
of $g$). According to \cite[p. 202]{lang_book}, such a formula reads
\begin{align*}T_g(r)=&\int_0^{2\pi} \log \max_{j=0,\ldots,m}|g_j(re^{\sqrt{-1}\theta})|\frac{d\theta}{2\pi}-\log\max_{j\in A} |c_{g_j}|\\
&+\sum_{a\in D_r\setminus{0}}\max_{j=0,\ldots,m}(-\ord_a(g_j))\log\left|\frac r a \right|+\max_{j=0,\ldots,m}(-\ord_0(g_j))\log |r|,
\end{align*}
where $\log\max_{j\in A} |c_{g_j}|$ is a correction term that makes
the definition independent of multiplication by a nowhere zero
holomorphic function (see \cite{lang_book} for full details). \par
Our main result in this Subsection is the following inequality of Second Main Theorem-type. According to the subsequent Corollary \ref{hyp_cor}, it can be understood from the geometric point of view as a quantitative version of the Chern hyperbolicity of the complement of an essentially large divisor.
\begin{theorem}\label{our_smt}
Let $D$ be an essentially very large divisor on a nonsingular complex projective variety $X$. Let $V$ be a linear subspace of $L(D)$ as in Definition \ref{ess_v_l_def}. Let $\phi_0,\ldots\phi_m$ be an arbitrary basis for $V$. Let $f:\CC\to X$ be an algebraically nondegenerate holomorphic map. Let $\Phi$ be the rational map $[\phi_0:\ldots:\phi_m]:X\to \PP^m$, and write
$$F=\Phi\circ f: \CC \to \PP^m.$$
When $E_1,\ldots, E_\ell$ denote the irreducible components of $D$, let
$$M=\max\{-\ord_{E_i}(\phi_j)\ |i=1,\ldots,\ell,\ j=0,\ldots,m\}.$$
Then, for every $\epsilon>0$,
$$T_F(r) .\leq. (M(m+1)+1+\epsilon) N_f(r,D),$$
where ``$.\leq.$'' means that the inequality holds for all $r$ outside of a set $\Gamma$ with finite Lebesgue measure.
\end{theorem}
Note that since $\Phi$ is nonconstant and $f$ is algebraically nondegenerate, the map $F$ also is nonconstant.\par

\begin{corollary} \label{hyp_cor} Let $D$ be an essentially large divisor on a
  nonsingular complex projective variety $X$. Then every holomorphic
  map $f: \CC \to X \setminus D$ must be algebraically degenerate,
  i.e., the image of $f$ must be contained in a proper subvariety of
  $X$. In other words, $X \setminus D$ is Chern hyperbolic (aka quasi Brody hyperbolic).
\end{corollary}
\begin{proof}[Proof of Corollary \ref{hyp_cor}] Since the statement of the Corollary only refers to the support of $D$, we can assume that $D$ is essentially {\it very} large. Assume that
$f: \CC \to X\setminus D$ is algebraically nondegenerate.  Since $f$ omits $D$,  $N_f(r,D)=0$. So the above inequality implies that $T_F(r)$ is bounded outside of a set of finite Lebesgue measure. However, since $F$ is nonconstant, this is false and gives a contradiction.
\end{proof}
\begin{proof}[Proof of Theorem \ref{our_smt}]
It is easy to see (\cite[Remark 8.2]{levin_annals}) that there exists a finite set $J$ of elements in $V$ such that for every $P\in D$ there exists a subset $I\subset J$ that is a basis of $V$ with $\ord_{E}\prod_{f\in I} f>0$ when $E$ is a component of $D$ with $P\in E$. Let $J'$ be the set of linear forms $L$ in $m+1$ variables such that $L\circ\Phi\in J$.\par
According to Theorem \ref{gen_smt},
$$\int_0^{2\pi}\max_I \log \prod_{L\in I} \frac{\|F(r e^{\sqrt{-1}\theta})\|\|L\|}{|L(F(re^{\sqrt{-1}\theta}))|}\frac{d\theta}{2\pi}.\leq. (m+1+\varepsilon)T_F(r),$$
where the maximum is taken over subsets $I\subset J'$ such that $I$ consists of exactly $m+1$ independent linear forms.\par
Since the left-hand side is independent of the choice of the representation of $F$, we can rewrite the above inequality as
$$\int_0^{2\pi}\max_I \log \prod_{L\in I} \frac{\max_{0 \leq j \leq
    m}|\phi_j(f(re^{\sqrt{-1}\theta}))|}
{|L(F(re^{\sqrt{-1}\theta}))|}\frac{d\theta}{2\pi}.\leq.
(m+1+\varepsilon)T_F(r) +O(1),$$
where $O(1)$ is introduced by dropping the factor $\|L\|$ after assuming that a fixed choice of coefficients has been made.\par
Next, we observe that since $D$ is essentially very large, outside of the support of the divisor $D$, we have
\begin{equation}\label{ess_v_l_boundedness}
\max_j |\phi_j(f(re^{\sqrt{-1}\theta}))|^{\frac 1 M} \min_I \prod_{L\in I} {|L(F(re^{\sqrt{-1}\theta}))|} \leq  C,
\end{equation}
where $C$ is some constant. The reason is that we can cover $X$ by a finite number of open sets on each of which $|\phi_j\min_I (\prod_{L\in I} {|L\circ \Phi |)^M}|$ is bounded for each $j=0,\ldots,m$. Namely, for $P\in X$, define $U_P$ as follows. If $P\not \in D$, take a small neighborhood $U_P$ of $P$ such that its closure is disjoint from the support of $D$. Since the $\phi_j$ have no poles outside of the support of $D$, the boundedness on $U_P$ is obvious. If $P\in D$, take $U_P$ to be a small neighborhood of $P$ with the property that $Q\in E\cap U_P$ implies $P\in E$ for all components $E$ of $D$. Consequently, by the essential very largeness, there exists $I_0$ such that $ \ord_E (\prod_{L\in I_0} {L\circ \Phi )}$ is a positive integer and $\ord_E\left(\phi_j\min_I (\prod_{L\in I} {|L\circ \Phi |)^M}\right)\geq 0$ for all $j=0,\ldots,m$. Perhaps after shrinking $U_P$, this implies that $|\phi_j\min_I (\prod_{L\in I} {|L\circ \Phi |)^M|}$ is bounded on $U_P$ for every $j=0,\ldots,m$. Finally, since $X$ is compact, we can cover it with a finite number of open sets of the form $U_P$.\par
Applying $\log$ to both sides of \eqref{ess_v_l_boundedness} yields
$$\log \max_j |\phi_j(f(re^{\sqrt{-1}\theta}))|^{\frac 1 M}
+  \log \min_I \prod_{L\in I} {|L(F(re^{\sqrt{-1}\theta}))|} \leq  \log C.$$
Therefore,
$$\log \max_j
|\phi_j(f(re^{\sqrt{-1}\theta}))|^{m+1+\frac 1 M}
\leq \max_I \log \prod_{L\in I}\frac{\max_j |\phi_j(f(re^{\sqrt{-1}\theta}))|}
{{|L(F(re^{\sqrt{-1}\theta}))|}} +\log C.$$
Thus
 \begin{gather*}
\int_0^{2\pi} \log \max_j |\phi_j(f(re^{\sqrt{-1}\theta}))| d\theta
 \\ = \frac {M}{M(m+1)+1}
\int_0^{2\pi}\log  \max_j |\phi_j(f(re^{\sqrt{-1}\theta}))|^{m+1+\frac{1}{M}} d\theta \\
\leq  \frac {M}{M(m+1)+1}  \left(  \int_0^{2\pi}
\max_I \log {\prod_{L\in I} \frac{\max_j |\phi_j(f(re^{\sqrt{-1}\theta}))|}  {|L(F(re^{\sqrt{-1}\theta}))|}}  d\theta +\log C \right) \\
.\leq.   \frac {M(m+1+\epsilon)}{M(m+1)+1}
T_F(r)  +O(1).
\end{gather*}
Recall that from Lang's formula \cite[p. 202]{lang_book},
\begin{align*}
T_F(r)=&\int_0^{2\pi} \log
\max_j|\phi_j(f)(re^{\sqrt{-1}\theta})|\frac{d\theta}{2\pi}-
\log\max_{j\in K} |c_{\phi_j(f)}|\\
&+\sum_{a\in D_r\setminus{0}}\max_{j=0,\ldots,m}(-\ord_a(\phi_j(f)))\log\left|\frac r a \right|\\ &+\max_{j=0,\ldots,m}(-\ord_0(\phi_j(f)))\log |r|.
\end{align*}
Thus,
\begin{align*}
T_F(r)  .\leq.&  \frac{M(m+1+\epsilon)}{M(m+1)+1} T_F(r)
+ \sum_{a\in
  D_r\setminus{0}}\max_{j=0,\ldots,m}(-\ord_a(\phi_j(f)))\log\left|\frac
  r a \right|\\&+\max_{j=0,\ldots,m}(-\ord_0(\phi_j(f)))\log |r|  +O(1).
\end{align*}
Recall that due the isomorphism \eqref{h0_isom}, there are sections $s_0,\ldots , s_m\in H^0(X,\OO_X(D))$ such that $\phi_j=\frac{s_j}{s_D}$. Therefore, for all $a\in D_r$,
\begin{align*}
\max_{j} (-\ord_a(F_j))&=\max_{j}(-\ord_a(\phi_j\circ f))\\
&=\max_{j}\left(-\ord_a\left(\frac{s_j}{s_D}\circ
    f\right)\right)\\&=\ord_a(s_D\circ
f)-\min_{j}\ord_a(s_j\circ f) \\&\leq \ord_a(s_D\circ f).
\end{align*}
Consequently, we obtain
\begin{align*}  \frac {1-\epsilon}{M(m+1)+1} T_F(r)
.\leq. &\sum_{a\in D_r\setminus\{0\}}\ord_a(s_D\circ f)\log\left|\frac r a \right|\\ &+\ord_0(s_D\circ f)\log r+O(1)\\
=&N_f(r,D) +O(1).
\end{align*}\par
Since $T_F(r)$ is unbounded for $r\to \infty$ outside of a set of finite Lebesgue measure, we can, after perhaps enlarging $\Gamma$, continue to estimate as follows, where $\delta$ is any positive number.
\begin{align*}  &N_f(r,D) +O(1)\\
.\leq.& N_f(r,D)+\delta T_F(r).
\end{align*}
We can now subtract $\delta T_F(r)$ from both sides of the inequality and then divide both sides by $\frac {1-\epsilon}{M(m+1)+1}-\delta$. Choosing small enough $\epsilon$ and $\delta$ (depending on $\epsilon$) concludes the proof.
\end{proof}
\subsection{The arithmetic case}\label{arithm_case}
We now derive an inequality of Schmidt Subspace Theorem-type for essentially large divisors. We again begin by recalling some standard definitions.\par
Let $k$ be a number field of degree $d$. Denote by $M_k$ the set
of places (equivalence classes of absolute values) of $k$ and
write $M^{\infty}_k$ for the set of
archimedean places of $k$.  In an archimedean class $\upsilon$, we choose the absolute value $|~~|_{\upsilon}$ such that
$|~~|_{\upsilon}
= |~~|$ on $\QQ$ (the standard absolute value). For a nonarchimedean class $\upsilon\in M_k\setminus M^{\infty}_k$, we let $|p|_{\upsilon} = p^{-1}$ if $\upsilon$ lies above the
rational
prime $p$. \par
Denote by $k_{\upsilon}$
the completion of $k$ with respect to $\upsilon$ and by
$d_{\upsilon} = [k_{\upsilon}
: {\QQ}_{\upsilon}]$ the local degree. Let $||~~||_{\upsilon} =
|~~|_{\upsilon}^{d_{\upsilon}/d}.$
We have the product formula: for every $x\in k^*$,
$$\prod_{\upsilon\in M_k} \|x\|_{\upsilon}=1.$$
 For $Q=[x_0, \dots, x_m] \in \PP_k^{m}$, let
 $\|Q\|_{\upsilon} =
\max _{1 \leq i \leq m}\|x_i\|_{\upsilon}$. Moreover, define the  logarithmic  height of
$Q$ by
$$h(Q) =\sum_{\upsilon \in {M_k}} \log \|{Q}\|_{\upsilon}.$$
By the product formula, its definition is independent of the choice of the
representations. \par
We now come to the notion of a Weil function. Let $X$ be a projective nonsingular variety and $D$ be an effective divisor on $X$, both defined over
a number field $k$. Extend $\|~~\|_{\upsilon}$ to an
absolute value on the algebraic closure $\bar{k}_{\upsilon}$ for all $\upsilon\in M_k$. Then a
{\it (local) Weil function for $D$ relative to $\upsilon$} is a function
$\lambda_{D, \upsilon}: X(\bar{k}_{\upsilon})\setminus  \mbox{Supp}(D)\rightarrow {\RR}$
such that if $D$ is represented locally by $(f)$
on an open set $U$, then
$$\lambda_{D, \upsilon}(P): =-\log\|f(P)\|_{\upsilon}+\alpha(P),$$
where $\alpha(P)$ is a continuous function on $U({\bar k}_{\upsilon}).$
We sometimes think of $\lambda_{D, \upsilon}$ as a function of
$X(k)\setminus \mbox{Supp}(D)$ or $X(\bar{k})\setminus \mbox{Supp}(D)$ by simply choosing an embedding
${\bar k}\rightarrow \bar{k}_{\upsilon}.$\par
Given a finite set $S \subset M_k$ containing all archimedean places, define the {\it proximity function}
$m_S(x, D)$ by, for $x \in X({\bar k})\setminus \mbox{Supp}(D) $,
\begin{equation}m_S(x, D) = {1\over [k:{\QQ}]} \sum_{\upsilon \in S}  \lambda_{D, \upsilon}(x).
\end{equation}
The {\it counting function} $N_S(x, D)$ is defined by
\begin{equation}N_S(x, D) = {1\over [k:{\QQ}]} \sum_{\upsilon \not\in S}   \lambda_{D, \upsilon}(x).
\end{equation}
Note that the sum above is still a finite sum, since the terms all vanish except for finitely
many.

\begin{definition} \label{int_pt_def} A subset $\Sigma$ of $X({\bar k})\setminus \mbox{Supp}(D)$ is
$(D, S)$-{\it integral} if there is a Weil function
$\lambda_D$ for $D$ and constants $c_{\upsilon}\in {\RR}$ for all
$\upsilon\not\in S$ such that $c_{\upsilon}=0$ for almost all $c_{\upsilon}=0$
 and $\lambda_{D, \upsilon}(P)\leq c_{\upsilon}$ for all
$\upsilon\not\in S$ and $P\in \Sigma$. In other words, a subset $\Sigma$ of $X({\bar k})\setminus \mbox{Supp}(D)$ is
$(D, S)$-{\it integral} if and only if $N_S(x, D)$ is bounded over $\Sigma$.
\end{definition}\par

We recall the following generalized version of Schmidt's Subspace Theorem from \cite{vojta_ajm_87}.
\begin{theorem}  \label{schmidt} Let $k$ be a number field
with its set of canonical places $M_k$. Let $S \subset M_k$ be a finite set containing
all archimedean places.  Let $H_1, \dots, H_m$ be hyperplanes in
$\PP^n$ defined over ${\bar k}$ with corresponding
Weil functions $\lambda_{H_1}, \dots, \lambda_{H_m}$.
Then there exists a finite union of hyperplanes $Z$, depending only
on $H_1, \dots, H_m$ (and not $k, S$), such that for
any $\epsilon>0$,
$$\sum_{\upsilon \in S} \max_I \sum_{i\in I}
\lambda_{H_i, \upsilon}(P) \leq (n+1+\epsilon)h(P)$$
holds for all but finitely many $P\in {\PP}^n\setminus Z$, where the maximum is taken over subsets $I \subset \{1, \dots, m\}$ such that the linear forms defining $H_i, i\in I,$ are linearly independent.
\end{theorem}\par
Our main result in this Subsection is the following inequality of Schmidt Subspace Theorem-type for essentially large divisors.
\begin{theorem} \label{our_schmidt} Let $D$ be an essentially very large divisor on a nonsingular  projective variety $X$, both defined over a number field $k$.  Let $V$ be a linear subspace of $L(D)$ as in Definition \ref{ess_v_l_def}. Let $\phi_0,\ldots\phi_m$ be an arbitrary basis for $V$ and let $\Phi$ be the rational map $[\phi_0:\ldots:\phi_m]:X\to \PP^m$. Let $S\subset M_k$ be a finite set of places containing all archimedean ones. Then for every $\epsilon >0$,
$$h(\Phi(P)).\leq.  (M(m+1)+1+\epsilon)N_S(P, D),$$
where ``$.\leq.$'' means that the inequality holds for all $P\in X(k)$ outside a Zariski closed
variety $Z$ of $X$.
\end{theorem}
\begin{proof}
Let $I$ be defined as in the proof of Theorem \ref{our_smt}. Due to arguments analogous to those that gave the boundedness \eqref{ess_v_l_boundedness}, we have for $P\in X(k)$ not contained in the support of $D$:
$$ \sum_{\upsilon \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon}^{m+1+\frac 1 M}
\leq \sum_{\upsilon \in S} \max_I \log \prod_{L\in I}\frac{\max_j
 \|\phi_j(P)\|_{\upsilon}} {\|L(\Phi(P))\|_{\upsilon}} +\log C,$$
for some constant $C$. In the sequel, we will denote abuse and denote all constants with the same symbol $C$.\par
By Theorem \ref{schmidt},  for all $P\in X(k)$ outside a Zariski closed
variety  $Z$ of $X$,
$$\sum_{\upsilon \in S} \max_I \log \prod_{L\in I}\frac{\max_j
 \|\phi_j(P))\|_{\upsilon}} {\|L(\Phi(P))\|_{\upsilon}}
.\leq.  (m+1+\epsilon)
h(\Phi(P)).$$
Hence
$$
\sum_{\upsilon \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon}^{m+1+\frac 1 M} .\leq. (m+1+\epsilon)
h(\Phi(P))+\log C.$$
Now,
$$h(\Phi(P)) =\sum_{\upsilon \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon}+ \sum_{\upsilon\not \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon}.$$\par
We again use the fact that, due to the isomorphism \eqref{h0_isom}, there are sections
$s_0,\ldots , s_m\in H^0(X,\OO_X(D))$ such that $\phi_j=\frac{s_j}{s_D}$, $j=0,\ldots,m$, where $s_D$ is a section of $\OO_X(D)$ with $\divisor (s) =D$. Moreover, we think of a section $s$ of $\OO_X(D)$ as given by a collection $(s_\beta)$ of functions pertaining to an open cover $(U_\beta)$ such that $s_\beta=h_{\beta\gamma}s_\gamma$ on $U_\beta\cap U_\gamma$, where $h_{\beta\gamma}$ are the nowhere vanishing transition functions of $\OO_X(D)$.\par
If $P\in U_\beta(\bar{k}_{\upsilon})$, then we can write
$$\|\phi_j(P)\|_{\upsilon}=\frac{\|s_{j,\beta}(P)\|_{\upsilon}}{\|s_{D,\beta}(P)\|_{\upsilon}}$$
independently of $\beta$.
Thus
$$
 \sum_{\upsilon \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon} - \sum_{\upsilon\not \in S} \log {\|s_{D,\beta}(P)\|_{\upsilon}
\over   \max_j
\|s_{j,\beta}(P)\|_{\upsilon}}
= h(\Phi(P)).$$
From the definition of the Weil-function $ \lambda_{D, \upsilon} (P)$,
since $D$ is represented by $s_{D, \beta}$ on $U_\beta$,
$$\lambda_{D, \upsilon} (P) =-\log \|s_{D, \beta}(P)\|_{\upsilon}+\alpha(P),$$
where $\alpha$ is continuous on $U_\beta(\bar{k}_{\upsilon})$.
Since $\log  \max_j \|s_{j, \beta}(P)\|_{\upsilon}$ is bounded from above on $U_\beta(\bar{k}_{\upsilon})$, there exists a constant $C_\beta$ such that
\begin{align*}
- \log {\|s_{D, \beta}(P)\|_{\upsilon}
\over   \max_j
\|s_{j, \beta}(P)\|_{\upsilon}}
=&-\log \|s_{D, \beta}(P)\|_{\upsilon} + \log  \max_j \|s_{j, \beta}(P)\|_{\upsilon}\\
\leq&  -\log \|s_{D, \beta}(P)\|_{\upsilon}+\alpha(P)+C_\beta\\=&\lambda_{D, \upsilon} (P) +C_\beta .
\end{align*}
Therefore
$$\sum_{\upsilon \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon} +  \sum_{\upsilon\not \in S}  \lambda_{D, \upsilon} (P) +C
\geq  h(\Phi(P)).$$
So we have
 \begin{align*}
 \left(m+1+{1\over M}\right) h(\Phi(P))
 \leq&
\sum_{\upsilon \in S} \max_j
\log \|\phi_j(P)\|_{\upsilon}^{m+1+\frac 1 M} \\
&+\left(m+1+{1\over M}\right)\sum_{\upsilon\not \in S} \lambda_{D, \upsilon} (P) +C\\
.\leq. & (m+1+\epsilon)
h(\Phi(P))+ \left(m+1+{1\over M}\right)
\sum_{\upsilon\not \in S} \lambda_{D, \upsilon} (P) \\
&+C.
\end{align*}
By Northcott's Theorem, for any $\delta>0$, $\{Q\in \PP^m | h(Q)< \frac C \delta \}$ is a finite set. Thus, after possibly enlarging $Z$, we can continue to estimate
\begin{align*}
 &(m+1+\epsilon)  h(\Phi(P))+ \left(m+1+{1\over M}\right)
\sum_{\upsilon\not \in S} \lambda_{D, \upsilon} (P) +C\\
.\leq.& (m+1+\epsilon+\delta)  h(\Phi(P))+ \left(m+1+{1\over M}\right)
\sum_{\upsilon\not \in S} \lambda_{D, \upsilon} (P).
\end{align*}
After rearranging the inequality and choosing $\epsilon$ and $\delta$ (depending on $\epsilon$) small, we obtain the Theorem.
\end{proof}
In strict analogy to Corollary \ref{hyp_cor}, we obtain the following
\begin{corollary} Let $D$ be an essentially large divisor on a nonsingular projective variety $X$, both defined over a number field $k$. Let $S\subset M_k$ be a finite set of places containing all archimedean ones. Then any set of $(D,S)$-integral points is contained in a proper subvariety of $X$. In other words, $X\setminus D$ is quasi Mordellic.
 \end{corollary}
\begin{proof}
We can assume without loss of generality that $D$ is essentially {\it very} large. Let $\Sigma$ be a set of $(D,S)$-integral points. According to Definition \ref{int_pt_def}, $N_S(x,D)$ is bounded over $\Sigma$. So, by Theorem \ref{our_schmidt}, $h(\Phi(P))$ is bounded for all $P\in \Sigma$ outside a Zariski closed variety $Z$ of $X$. This means that, due to Northcott's Theorem, the set $\{\Phi(P)|P\in (\Sigma \setminus Z)(k)\}$ is finite. Since $\Phi$ is a nonconstant morphism outside of $D$, the preimage of this finite set is a proper subvariety $Z_1$ of $X$. We have shown that $\Sigma\subset Z\cup Z_1$, which concludes the proof.
 \end{proof}
\section{Proofs of the sharp sufficient criteria for essentially large divisors}\label{section_ess_v_l_proofs}
As was stated in the Introduction, the filtration method from \cite{cz_ajm}, \cite{ru_ajm} can be used to prove the sharp bound for the number of components of a large divisor in the case of $X=\PP^q$ (Theorem \ref{thm_1}). We now give the proof.
\begin{proof}[Proof of Theorem \ref{thm_1}]
The number field case being completely analogous (see \cite{cz_ajm}), we only prove the case when $K=\CC$.\par
Let $Q_1,\ldots,Q_r$ be homogeneous polynomials of degree $d_i$ such that $D_i=\{Q_i=0\}$. Let $P\in D$. After replacing each $Q_i$ by $Q_i^{d/d_i}$, where $d$ is the least common multiple of the $d_i$'s, we can assume that all $Q_i$ have the same degree $d$.  Without loss of generality, assume that precisely $q$ of the $D_i$ contain the point $P$, and let $\{\gamma_1,\ldots,\gamma_q\}$ be the unique set of distinct elements of $\{Q_1,\ldots,Q_r\}$ such that $P\in \cap_{i=1}^q\{\gamma_i=0\}$.\par
Let $V_N$ be the vector space of homogeneous polynomials of degree $N$ in $\CC[X_0,\ldots,X_q]$. With the lexicographical ordering on the $q$-tuples $\vec{i}=(i_1,\ldots,i_q)$ with $\sigma(\vec{i}):=\sum_{j=1}^q i_j\leq N/d$, we obtain a filtration on $V_N$ by letting
$$W_{\vec{i}}=\sum_{\vec{e}\geq \vec{i}}\gamma_1^{e_1}\ldots\gamma_q^{e_q}V_{N-d\sigma(\vec{e})}.$$
Note that clearly $W_{\vec{0}}=V_N$ and $W_{\vec{i}}\supseteq W_{\vec{i}'}$ for $\vec{i'}\geq \vec{i}$.
Combining Lemma 2.3 and Lemma 3.1 of \cite{cz_ajm}, we note the following Lemma for our situation.
\begin{lemma}\label{hilbert_lemma}
There exists an integer $N_0$ (depending only on $\gamma_1,\ldots,\gamma_q$) such that for all integers $N> N_0$ and for all $\vec{i}$ with $d\sigma(\vec{i}) < N-N_0$:
$$\Delta_{\vec{i}}:=\dim W_{\vec{i}}/W_{\vec{i'}}=d^q.$$
\end{lemma}
Now choose a basis $\psi_1,\ldots,\psi_m$ ($m={N+q\choose q}$) for $V_N$ with respect to the above filtration. For all $\nu=1,\ldots,m$, write
$$\psi_\nu=\gamma_1^{i_1}\ldots\gamma_q^{i_q}\gamma^{(\nu)}$$
for some $\gamma^{(\nu)}\in V_{N-d\sigma(\vec{i})}$ according to its place in the filtration.\par
For $\tilde N\in \NN$, let $N=\tilde N\cdot r \cdot d$ and write
$$H^0(\PP^q,\OO_{\PP^q}(\tilde N D))=H^0(\PP^q,\OO_{\PP^q}(\tilde N\cdot r \cdot d))=H^0(\PP^q,\OO_{\PP^q}(N))=V_N.$$
For $\nu=1,\ldots,m$, let
$$f_\nu=\frac{\psi_\nu}{Q_1^{\tilde N}\ldots Q_r^{\tilde N}}.$$
These rational functions clearly form a basis for $L(\tilde N D)$. To conclude the proof of the Theorem, we show that they satisfy
 $\ord_{E} \prod_{\nu=1}^{m} f_\nu > 0$ for any component $E$ in $D$ with $P\in E$. We assume that $E$ is contained in $D_{j_0}$. Then
$$
\frac 1 {\ord_E D} \ord_{E} \prod_{\nu=1}^{m} f_\nu = \left(\sum_{\vec{i}}\Delta_{\vec{i}}i_{j_0}\right)-\tilde N {N+q\choose q}.$$
Since the number of nonnegative integer $m$-tuples with
sum $\leq t$ is equal to the number of nonnegative integer
$(m+1)$-tuples with sum exactly $t$ which is
${t+m\choose m}$, and since the sum below is independent of
$j$, we have that, for
$N$ divisible by $d$ and for every $j$,
\begin{align*}
\sum_{\vec{i}} i_j&={1\over q+1}\sum_{\hat{\vec{i}}} \sum_{j=1}^{q+1}
i_j ={1\over q+1} \sum_{\hat{\vec{i}}}  {N\over d}\\
&={1\over q+1} {N/d+q\choose q}{N\over d}
={N^{q+1}\over d^{q+1}(q+1)!}+O(N^q),
\end{align*}
where $\displaystyle{\sum_{\hat{\vec{i}}}}$ is taken over all nonnegative integer $(q+1)$-tuples with sum exactly $N/d$.
Combining it with Lemma \ref{hilbert_lemma},
we have, for every $1 \leq j_0 \leq q$,
$$\sum_{\vec{i}}\Delta_{\vec{i}}i_{j_0} ={N^{q+1}\over d(q+1)!}+O(N^q).$$
Thus,
\begin{align*}
\frac 1 {\ord_E D} \ord_{E} \prod_{\nu=1}^{m} f_\nu &= \left(\sum_{\vec{i}}\Delta_{\vec{i}}i_{j_0}\right)
-\tilde N {N+q\choose q}\\&=\left(\frac{N^{q+1}}{d\cdot (q+1)!}+O(N^q)\right)- \frac N{dr} \left(\frac {N^q} {q!}+O(N^{q-1})\right)\\
&=\left(\frac{1}{q+1}-\frac{1}{r}\right) \frac{N^{q+1}}{d\cdot q!} + O(N^q).
\end{align*}
For $N$ sufficiently large, this is positive if $r \geq q+2$.
\end{proof}\par
\begin{remark} We remark that under the assumptions of Theorem \ref{thm_1}, the paper \cite{ru_ajm}, using the same methods from \cite{cz_ajm}, obtained a sharp Second Main Theorem that is stronger than Theorem \ref{our_smt}. The same can be said for Theorem \ref{thm_2} and the paper \cite{ru_annals}, which is in turn based on \cite{ferretti_compositio}, \cite{ef_imrn}. On the other hand, Theorem \ref{our_smt} does not require these assumptions, so it is still a new result.\end{remark}

We continue with the proof of Theorem \ref{thm_2}, based on the methods of \cite{ferretti_compositio}, \cite{ef_imrn} and  \cite{ru_annals}.
\begin{proof}[Proof of Theorem \ref{thm_2}]
The number field case being completely analogous (see \cite{ferretti_compositio}), we only prove the case when $K=\CC$.\par
Let $Q_1,\ldots,Q_r$ be homogeneous polynomials such that $D_i=\{Q_i=0\}\cap X$.
As above, we can assume that the degrees of the homogeneous polynomials $Q_i$ equal $d$ for all $i$. Let
$$\varphi: X\to \PP^{r-1}, \quad x\mapsto [Q_1(x),\ldots,Q_r(x)].$$
Let $Y:=\varphi(X)$. By the general position assumption, $\varphi$ is a finite holomorphic map $X\to Y$. \par
On $\PP^{r-1}$, we have for all $N\in \NN$ a short exact sequence
$$0\to \II_Y(N)\to \OO_{\PP^{r-1}}(N)\to \OO_Y(N)\to 0.$$
The beginning of the corresponding long exact sequence reads
$$0\to H^0(\PP^{r-1},\II_Y(N))\to H^0(\PP^{r-1},\OO_{\PP^{r-1}}(N))\stackrel {\tau}{\to} H^0(Y,\OO_Y(N)),$$
where $\tau$ denotes the restriction map. We let
\begin{align*}
W_N&:=\image(\tau)\\& \cong H^0(\PP^{r-1},\OO_{\PP^{r-1}}(N))/ \ker(\tau)\\ &\cong H^0(\PP^{r-1},\OO_{\PP^{r-1}}(N))/H^0(\PP^{r-1},\II_Y(N))\\ &\cong\CC[Y_0,\ldots,Y_{r-1}]_N/(I_Y)_N,
\end{align*}
where $(I_Y)_N$ denotes the set of those homogeneous polynomials of degree $N$ vanishing on $Y$.
Let $V_N:=\varphi^*(W_N)\subseteq H^0(X,\OO_X(ND))$.\par
Since $\varphi:X\to Y$ is a finite surjective holomorphic map,
\begin{align*}
\dim(V_N)=\dim (W_N)=\dim \CC[Y_0,\ldots,Y_{r-1}]_N/(I_Y)_N=H_Y(N),
\end{align*}
where $H_Y(N)$ is the Hilbert function of $Y$.\par
Let $P\in D$. Without loss of generality, we assume again that there are distinct
$Q_{i_1},\ldots,Q_{i_q} \in \{Q_1,\ldots,Q_r\}$ such that $P\in \cap_{j=1}^q\{Q_{i_j}=0\}$.\par
Let $\vec{c}=(c_1, \dots, c_r)$ be the
$r$-vector whose $i_j$-th entry ($1 \leq j \leq q$) is $1$ and
elsewhere is $0$.
Let $\vec{y}^{\vec{a}^{(1)}},\ldots,\vec{y}^{\vec{a}^{(H_Y(N))}}$ be monomials such that their equivalence classes in $\CC[Y_0,\ldots,Y_{r-1}]_N/(I_Y)_N$ give a basis and such that
$$S_Y(N,\vec{c})=\sum_{i=1}^{H_Y(N)} \vec{a}^{(i)} \bullet \vec{c},$$
where $S_Y(N,\vec{c})$ is the $N$-th Hilbert weight and the bullet denotes the usual dot product. Recall that the  $N$-th Hilbert weight
 is given by
$$S_Y(N,\vec{c})=\max \sum_{i=1}^{H_Y(N)} \vec{a}^{(i)} \bullet \vec{c},$$
where the maximum is taken over  all sets of monomials
  $\vec{y}^{\vec{a}^{(1)}},\ldots,\vec{y}^{\vec{a}^{(H_Y(N))}}$ whose residue class modulo
$I_Y$ form a basis of $\CC[Y_0,\ldots,Y_{r-1}]_N/(I_Y)_N$.
For $\nu=1,\ldots,H_Y(N)$, and $N$ a multiple of $r$, let
$$f_\nu = \frac{Q_1^{a_1^{(\nu)}}\ldots Q_r^{a_r^{(\nu)}}}{Q_1^{N/r}\ldots Q_r^{N/r}}|_X.$$
These functions form a basis for $V_N$ understood as a subspace of $L(ND)$. To conclude the proof of the Theorem, we show that they satisfy
 $\ord_{E} \prod_{\nu=1}^{H_Y(N)} f_\nu > 0$ for any component $E$ in
 $D$ with $P\in E$. We assume that $E$ is contained in $D_{j_0}$. \par
We recall two basic lemmas in \cite{ru_annals}:
\begin{lemma} Let $X \subset {\PP}^{N}_{\CC}$
be an algebraic variety of
dimension $n$ and degree $\bigtriangleup$. Let $m>\bigtriangleup$
be an integer and
let   ${\vec c}=(c_0, \dots, c_N) \in {\Bbb R}^{N+1}_{\ge 0}$.
Then
$$ {1\over mH_X(m)}S_X(m, {\vec c})
\ge {1\over (n+1)\bigtriangleup} e_X({\vec c}) - {(2n+1)\bigtriangleup
\over m}\cdot
\left(\max_{i=0, \dots, N}c_i\right),$$
where ${1\over (n+1)\bigtriangleup} e_X({\vec c})$ is the normalized Chow weight of $X$ with respect to $\vec{c}$.
\end{lemma}
We will not give a definition of the normalized Chow weight here (see \cite{ru_annals}). For our purposes, all we need is the following estimate from below.
 \begin{lemma}
Let $Y$ be a subvariety of ${\Bbb P}^{q-1}_{\CC}$ of dimension $n$ and degree
$\bigtriangleup$. Let ${\vec c}=(c_1, \dots, c_q)$ be a tuple of
positive reals.
Let $\{i_0, \dots, i_n\}$ be a subset of $\{1, \dots, q\}$ such that
$$Y\cap \{y_{i_0}=0, \dots, y_{i_n}=0\}=\emptyset. $$
Then
$$e_Y({\vec c})\ge (c_{i_0}+\cdots +c_{i_n})\cdot \bigtriangleup. $$
\end{lemma}
We now continue our proof. With our chosen $\vec{c}$ and $\vec{a}^{(i)}$, using Lemmas 3.3
and 3.4 (notice the condition that $Q_1, \dots, Q_r$ are in general position), and the symmetry property of the $\vec{a}^{(1)},\ldots,\vec{a}^{(H_Y(N))}$ (\cite[p. 61]{mumford_enseign}),
\begin{align*}
{1\over \ord_E D}\ord_E \prod_{\nu=1}^{H_Y(N)} f_\nu &= \left(\sum_{\nu=1}^{H_Y(N)} a_{j_0}^{(\nu)}\right)-\frac{N}{r}H_Y(N)\\
&=\frac 1 q \left(\sum_{\nu=1}^{H_Y(N)}  \vec{a}^{(\nu)} \bullet \vec{c}\right)- \frac{N}{r}H_Y(N)\\
&=\frac 1 q S_Y(N,\vec{c})- \frac{N}{r}H_Y(N)\\
&\ge\frac 1 q \frac 1 {q+1} N H_Y(N)(\sum_{j=1}^q  c_{i_j})-O(H_Y(N))- \frac{N}{r}H_Y(N)\\
&=\frac 1 {q+1} N H_Y(N) - \frac{N}{r}H_Y(N)-O(H_Y(N))\\
&=(\frac 1 {q+1} - \frac{1}{r})NH_Y(N)- O(H_Y(N)).\\
\end{align*}
For $N$ sufficiently large, this is positive if $r \geq q+2$.
\end{proof}

\section{Essentially large divisors on projective varieties of small codimension}\label{section_alg_geom}
Our ultimate (and so far unreached) goal is of course to find a (sharp) lower bound for the number of (say ample) components in general position necessary for a divisor on a nonsingular projective variety to be essentially large (cf.\ Conjecture \ref{conjecture}). For a general nonsingular projective variety, it is clearly not true that all effective divisors are cut out by hypersurfaces in the ambient projective space, as assumed in Theorem \ref{thm_2}. In this final Section, we exhibit a class of nonsingular projective varieties  for which this is the case and to which Theorem \ref{thm_2} can thus be applied. Consequently, the above-mentioned ultimate goal has been achieved for these varieties.\par
We refer the reader to \cite{hartshorne_bull_ams} or \cite[Section 3.2]{PAGI} for more on the algebraic geometry behind this question, which is essentially the problem of extending the Lefschetz Hyperplane Theorem to varieties that are not complete intersections. In particular, the theorems of Barth \cite{barth_ajm}, Larsen \cite{larsen_invent}, and, in the case of a general ground field of characteristic zero, Ogus \cite{ogus_annals} apply. For our purposes, we simply state the following Proposition, which is an immediate consequence of these theorems.
\begin{proposition} Let $X\subset \PP^\ell$ be a nonsingular projective variety of dimension $q$, defined over $K$. If $2q - \ell \geq 2$, then restriction yields an isomorphism
$$\Pic(\PP^\ell)\stackrel{\cong}{\to} \Pic(X).$$
\end{proposition}\par
Since $\Pic(\PP^\ell)=\ZZ$, this Proposition gives the following Corollary to Theorem \ref{thm_2}.
\begin{corollary}\label{cor}
Let $q\geq 1$ and $r\geq q+2$ be integers. Let $X\subseteq \PP^\ell$ be a nonsingular projective variety of dimension $q$, defined over $K$. Assume that $2q - \ell \geq 2$ holds. Let $D=\sum_{i=1}^r D_i$ be an effective divisor on $X$ defined over $K$ such that the $D_i$ are in general position. Then $D$ is essentially large.
\end{corollary}
We conclude by pointing out that the condition $2q - \ell \geq 2$ is not that restrictive. It is well-known that, by general linear projections, any nonsingular projective variety of dimension $q$ can be embedded in $\PP^\ell$ with $\ell=2q+1$. Thus, for a generic nonsingular projective variety, one can always find an embedding with $2q-\ell=-1$, which is quite close to the above condition. In particular, Corollary \ref{cor} applies to many interesting special projective varieties such as hypersurfaces, appropriate complete intersections, and certain Grassmannians, such as $G(2,5)$ embedded into $\PP^{9}$ under the Pl\"ucker embedding. Note that the latter is not a complete intersection due to B\'ezout's Theorem, because its degree is (the prime number) $5$, while it is not contained in any hyperplane.


TITLE: Nontrivial critical exponents in exactly solvable models?
QUESTION [5 upvotes]: Are there any exactly solvable models in statistical mechanics that are known to have critical exponents different from those in mean-field theory, apart from the two-dimensional Ising model? I wonder about this because most easily solved models are either mean-field or do not exhibit a phase transition (Ising chain).

REPLY [1 votes]: If you want to expand your search to all critical exponents, I would recommend the following article:
Géza Ódor, Universality classes in nonequilibrium lattice systems, REVIEWS OF MODERN PHYSICS, VOLUME 76, JULY 2004
This contains lots of different systems that can be (are) modeled with lattices (e.g. Ising, percolation and interface growth). The contents list is actually pretty good starting point for simple listed classification. Some of them are exactly solvable, however most are not (try searching the word 'exactly' in the document).
URL: http://link.aps.org/doi/10.1103/RevModPhys.76.663
DOI: 10.1103/RevModPhys.76.663
There is more on this topic in this question: Where can I find a good classification for phase transitions?


TITLE: Proof of half angle identity
QUESTION [0 upvotes]: Remember one of the step in proving it is to replace theta with 1/2 A. I am wondering how could you replace a variable to prove the identity. A is not the same as theta. Plus aren't we trying to find half theta?

REPLY [1 votes]: Presumably, you are asking to prove the half-angle identities for cosine and sine given the double-angle formulas for cosine:
$$\begin{align}
\cos 2 \theta &= 2 \cos^2 \theta - 1 &(1) \\
&= 1 - 2 \sin^2\theta &(2)
\end{align}$$
The identities show that trig functions of "some angle" (shown as "$\theta$") on the right are equal to a trig function of "twice that angle" ("$2\theta$") on the left; as identities, these relations hold no matter what "some angle" is.
Being clever, we can replace "some angle" with "half some other angle", as in $\theta \to \alpha/2$, whereupon "twice that angle" becomes simply "the other angle" ($2\theta \to 2(\alpha/2) = \alpha$), and we have:
$$\begin{align}
\cos\alpha &= 2 \cos^2\frac{\alpha}{2} - 1 &(1^\prime) \\
&= 1 - 2 \sin^2\frac{\alpha}{2} &(2^\prime)
\end{align}$$
You can then solve these equations for $\cos^2\frac{\alpha}{2}$ and $\sin^2\frac{\alpha}{2}$ in terms of $\cos\alpha$.
(And then, if having the identities in terms of "$\alpha$" bothers you, you can change the name of the angle to "$\theta$", so that you have $\cos^2\frac{\theta}{2}$ and $\sin^2\frac{\theta}{2}$ in terms of $\cos\theta$. I suspect that the instructions tell you to use $\alpha$ to avoid some confusion with replacing $\theta \to \theta/2$ in the earlier step.)


TITLE: A simple algebra that is not semisimple
QUESTION [2 upvotes]: I was said that can exists simple algebras $A$ that are not semi-simples in the following sense:

$A$ is simple, i.e. doesn't have non trivial ideals;
$A$ is not semi-simple, i.e. is not a semi-simple module over itself.

Does anybody has an example?
Thanks in advance

REPLY [7 votes]: Yes, the Weyl algebra $A = \mathbb{C}[x, \partial]/(\partial x - x \partial - 1)$ is a standard example. Proving that it's simple but not semisimple is a nice exercise.


TITLE: Check on Summations
QUESTION [1 upvotes]: I'm completing some time complexity analysis on loops which requires me to represent psuedocode loops as summations. I have two summations that I believe I have done to a degree of correctness, but am unsure. Here's the initial pseudocode for the first:
for i=1 to n:
    for j=1 to n:
        if i+j<n:
            for k=1 to n:
                sum++

I take this to mean the following:
$\sum_{i=1}^n\sum_{j<n-i}^n\sum_{k=1}^n 1$
$\sum_{i=1}^n\sum_{j<n-i}^nn = \sum_{i=1}^n n\sum_{j<n-i}^n$
$\sum_{i=1}^n n(n-(n-i)+1) = n\sum_{i=1}^n 1-i = n(\sum_{i=1}^n 1 - \sum_{i=1}^n i)$
$n(n-\frac{n(n+1)}{2}) = n^2-\frac{n^3}{2}+\frac{n^2}{2}$
So the overall time complexity of this loop (in the worst case) is $O(n^3)$. Is this accurate? I'm concern that the upper bound for my second summation is incorrect.
The second psuedocode is this:
for i=1 to n:
    for k=1 to n:
        sum++
        k=k/2

This psuedocode contains no tricky 'if' statement, but the $k=\frac{k}{2}$ confuses me a bit. Here's the summation I have:
$\sum_{i=1}^n\sum_{k=1}^n 1+\frac{k}{2} = \sum_{i=1}^n 1 + \frac{1}{2}\sum_{k=1}^n k$
$\sum_{i=1}^n n+\frac{n(n+1)}{4} = \sum_{i=1}^n n+ \sum_{i=1}^n\frac{n(n+1)}{4} = n\sum_{i=1}^n 1 +  \frac{n^2}{4}\sum_{i=1}^n 1 + \frac{n}{4}\sum_{i=1}^n 1$
$n^2+\frac{n^3}{4} + \frac{n^2}{4}$
So the time complexity is $O(n^3)$. Again, I'm not sure if I handled that $k=\frac{k}{2}$ properly, so feedback would be appreciated!

REPLY [1 votes]: In the first case the correct formula (for $n \geq 3$) is:
$$\begin{align}
\sum_{i=1}^{n-2}\sum_{j=1}^{n-i-1} n &= \sum_{i=1}^{n-2} n(n-i-1) \\
&= n^2(n-2) - n\frac{(n-2)(n-1)}{2} - n(n-2) \\
&= \frac{(2n^3 - 4n^2) - (n^3 - 3n^2 + 2n) - (2n^2 - 4n)}{2} \\
&= \frac{n^3 - 3n^2 + 2n}{2} \\
&= \frac{n(n-1)(n-2)}{2}
\end{align}$$
The $n-2$ in the upper limit of the sum is because values of $i \geq n-1$ never end up being computed in the second sum.
In the second case, I don't think the internal loop will ever finish.
If we simulate what's happening on the third for loop we get:
k=1      % begin of the for loop
k <= n   % we enter the for loop
sum++    % computed, presumably now sum=1 (there were no sum=0 before)
k = k/2  % now k=0.5
k++      % in the end of the for loop, so k=1.5
k <= n   % we continue inside the for loop (assuming n >=2)
sum++    % computed, presumably now sum=2
k = k/2  % now k=0.75
k++      % in the end of the for loop, so k=1.75
k <= n   % we continue inside the for loop (assuming n >=2)
         % and so on for a long time
sum++    % computed, now sum is very large
k = k/2  % now k=0.999999...
k++      % in the end of the for loop, so k=1.99999...
k <= n   % we continue inside the for loop (assuming n >=2)
         % and the loop will continue forever

Therefore, you'll get $k=1$, $k=0.5$, $k=1.5$, $k=0.75$, ... approaching an infinite loop that with $k$ alternating between just under $k=1$ and just under $k=2$.
Are you sure it's not something else inside the internal loop? Perhaps sum += k/2?


TITLE: Strength of BTEE
QUESTION [3 upvotes]: What is the consistency strength of Basic Theory of Elementary Embeddings (BTEE) from The spectrum of elementrary embeddings j : V → V by Paul Corazza?
BTEE uses the language of $(V,∈,j)$ and asserts:
ZFC (without separation and replacement for formulas using $j$)
(schema) $j$ a is nontrivial elementary embedding of $V$ into $V$ (for formulas without $j$)
critical point: the least ordinal moved by $j$ exists.
Without critical point, the theory would be conservative over ZFC.  With full replacement (and given the axiom of choice), the theory would be inconsistent; and there is a hierarchy between the two extremes.
BTEE suffices for a basic theory of $j$.  For every natural number $n$, BTEE proves that $κ = \mathrm{crit}(j)$ is $n$-ineffable and totally indescribable (see the linked paper).  However, the only upper bound on the strength of BTEE in the paper is $ω$-Erdős.
This question is in Q/A format.

REPLY [2 votes]: BTEE is conservative over the stationary reflection principle (SRP), i.e. ZFC + (schema) {there is $n$-subtle cardinal}$_{n∈\mathbb{N}}$. Using $n$-ineffable in the schema is equivalent.
Note that we get conservativity and not just equiconsistency, and also that (without full conservativity) SRP shows up in a number of other results including in a basic axiomatization of reflective cardinals of finite orders (an extension of the language of set theory), and (separately) in Harvey Friedman's natural arithmetic statements independent of ZFC.
As noted above, for every $n$, BTEE implies existence of $n$-subtle cardinals.
For the converse, given a proposition $A$, consistency of ZFC + $A$ + (schema) {there is $n$-subtle cardinal}$_{n∈\mathbb{N}}$ implies existence of a model $M$ of ZFC + $A$ + $n$-subtle cardinal for a nonstandard number $n$, plus existence of $λ$ with $V_λ^M≺M$.  Fix such an $M$ and an $M$-well-ordering $<^*$ of $V_λ^M$, and (in $M$; $n$ is nonstandard outside of $M$) use an $n$-subtle cardinal $<λ$ to produce an $n$-tuple $(κ_1,...,κ_n)$ of good indiscernibles for $(V_λ^M, ∈^M, <^*)$.  ("good" is a technical term here and is why we need the $n$-subtle cardinal; it will ensure that $j$ does not move $M$-ordinals below $κ_1$.)  Let $M'$ be the $(V_λ^M, ∈^M, <^*)$ Skolem hull of the first $ω$ of these indiscernibles (as computed in $V$ so all Gödel numbers are standard).  $(M',∈^M,j)$ satisfies BTEE + $A$ where $j(κ_i)=κ_{i+1}$ and $j$ is extended to all sets in $M'$ using the Skolem hull.
Adding induction (and especially transfinite induction) would significantly increase the strength.  For the strength of existence of a well-founded model of BTEE, see this question.
$n$-huge cardinals
Nonstandard models can also be used to show that WA0 (bounded quantifier Wholeness Axiom, which consists of BTEE and separation for bounded quantifier $j$-formulas) is conservative over ZFC + (schema) {there is $n$-huge $κ$ with $V_κ≺_{Σ_n}V$}$_{n∈\mathbb{N}}$ .  (In turn, this is $Σ_2^V$ conservative over ZFC + (schema) {there is $n$-huge $κ$}$_{n∈\mathbb{N}}$ .)
One direction is proved in the linked paper.  For the converse, given a proposition $A$, consistency of ZFC + $A$ + (schema) {there is $n$-huge $κ$ with $V_κ≺_{Σ_n}V$}$_{n∈\mathbb{N}}$ implies existence of a model $M$ of ZFC + $A$ + $n$-huge cardinal $κ$ for a nonstandard number $n$, and with $V_κ^M≺M$.  Starting with such $M$ and an $n$-huge embedding $j$ in $M$ with $κ = \mathrm{crit}(j)$, let $M' = \{x∈M: ∃m<ω \,\, x ∈^M j^m(V_κ^M)\}$.  (Note that we use $ω$ rather than $ω^M$.) $(M', ∈^M, j)$ satisfies WA0 + $A$.


text:
\begin{document}

\title{Connective eccentric index of some graph operations}

\author[nd]{Nilanjan De}
\ead{de.nilanjan@rediffmail.com}
\author[ap]{Anita Pal}
\ead{anita.buie@gmail.com}
\author[man]{Sk. Md. Abu Nayeem\corref{cor1}}
\ead{nayeem.math@aliah.ac.in}

\address[nd]{Department of
Basic Sciences and Humanities (Mathematics),\\ Calcutta Institute of Engineering and Management, Kolkata, India.}
\address[ap]{Department of
Mathematics, National Institute of Technology, Durgapur, India.}
\address[man]{Department of Mathematics, Aliah University,\\ DN - 20, Sector -  V, Salt Lake, Kolkata - 700 091, India.}
\cortext[cor1]{Corresponding Author.}

\begin{abstract}
The connective eccentric index of a graph is a topological index involving degrees and eccentricities of vertices of the graph. In this paper, we have studied the connective eccentric index for double graph and double cover. Also we give the connective eccentric
index for some graph operations such as joins, symmetric difference, disjunction and
splice of graphs.

\medskip
\noindent \textsl{MSC (2010):} Primary: 05C35; Secondary: 05C07, 05C40\\
\end{abstract}
\begin{keyword}
Eccentricity, degree, graph invariant, connective eccentric
index, graph operations.
\end{keyword}

\maketitle

\section{Introduction}
Let $G$ be a simple connected graph with vertex set $V(G)$ and edge set $E(G)$. Let
$n$ and $m$ be the number of vertices and edges of $G$
respectively. We denote the degree of a vertex $v$ of $G$ by $d_G(v)$. For $u,v \in V(G)$, the distance between $u$
and $v$ is defined as the length of a shortest path between $u$
and $v$. For a given vertex $v$ of $G$, the eccentricity $\varepsilon_G(v)$ is the largest distance from $v$ to any other vertices  of $G$. The sum of eccentricities of all the
vertices of $G$ is denoted by $\theta(G)$ \cite{dan04}. If any vertex $v \in V(G)$ is
adjacent to all the other vertices of $G$ then $v$ is called a
well-connected vertex. Thus, if $v \in V(G)$ is a well-connected vertex,
then $\varepsilon_G(v) = 1$. For example, all the vertices of a complete graph are
well-connected.

Recently, a number of topological indices involving vertex degree
and eccentricity were subject to a lot of mathematical as well as chemical studies.
A topological index of this type, introduced by Gupta et al. \cite{gup00}, was named as the connective eccentric index and was defined as $$C^\xi(G) = \sum\limits_{v \in V(G)}\frac{d_G(v)}{\varepsilon_G(v)}.$$ M. Ghorbani \cite{gho11} gave some bounds of connective eccentricity index and also computed this index for two infinite classes of dendrimers. De \cite{de12} reported some bounds for this index in terms of some graph invariants such as maximum and minimum degree, radius, diameter, first Zagreb index and first Zagreb eccentricity index, etc. In \cite{gho12a}, Ghorbani and Malekjani computed the eccentric connectivity index and the
connective eccentric index of an infinite family of fullerenes. In \cite{yu13}, Yu and
Feng also derived some upper or lower bounds for the connective eccentric index
and investigated the maximal and the minimal values of connective eccentricity
index among all $n$-vertex graphs with fixed number of pendent vertices.

In \cite{gho11}, Ghorbani showed that for a vertex transitive graph $G$, the
connective eccentric index is given by $$C^\xi(G) = \sum_{j = 1}^t |{A_j}|\frac{d_G(x_i)}{\varepsilon_G(x_i)},$$ where ${A_1},{A_2},\ldots,{A_t}$ are the orbits of $Aut(G)$ under its natural action on $V(G)$ and $x_i \in {A_i}, 1 \le i \le t$. In
particular, if $G$ is a regular graph, then ${C^\xi}(G) = \d\frac{n\delta}{r(G)}$, where $n$ is the number of vertices of $G$, which is a $\delta$-regular graph and $r(G)$ is the radius of $G$. Let, ${K_n},{C_n},{Q_m},{\Pi_m},{A_m}$ denote the complete graph with $n$
vertices, the cycle on $n$ vertices, $m$-dimensional hypercube,
$m$-sided prism and the $m$-sided antiprism respectively. It can be easily verified that the explicit formulae for the connective eccentric index of ${K_n},{C_n},{Q_m},{\Pi_m},{A_m}$ are as follows.

\begin{prop}
${C^\xi }({K_n}) = n(n - 1).$
\end{prop}

\begin{prop}
${C^\xi }({C_n}) = \frac{2n}{\lfloor n/2 \rfloor}.$
\end{prop}

\begin{prop}
${C^\xi }({Q_m}) = {2^m}.$
\end{prop}

\begin{prop}
${C^\xi }({\Pi _m}) = \left\{ \begin{array}{ll}
\frac{12m}{m + 2},&\mbox{when } m \mbox{ is even}\\[1mm]
\frac{12m}{m + 1},&\mbox{when } m \mbox{ is odd}.
\end{array}\right.$
\end{prop}

\begin{prop}
${C^\xi }({A_m}) = \left\{ \begin{array}{ll}
16,&\mbox{when } m \mbox{ is even}\\[1mm]
\frac{16m}{m + 1},&\mbox{when } m \mbox{ is odd}.
\end{array}\right.$
\end{prop}

Several studies on different topological indices related to graph operations of different
kinds are available in the literature \cite{ash11, gho12b, hos09, kha08, kha09}.

In this paper, first we calculate connective eccentric index of double graph and double cover
and hence the explicit formulae for the connective eccentric indices of join,
symmetric difference, disjunction and splice of graphs are obtained. For the
definitions and different results on graph operations, such as join, symmetric
difference, disjunction, etc., readers are referred to the book of Imrich and Klav\v{z}ar \cite{imr00}.

\section{Main Results}
In this section, first we define and then compute eccentric connectivity index of
double graph and double cover graph.

\subsection{Connective eccentric index of double graph and double cover}
Let us denote the double graph of a graph $G$ by $G^*$, which is constructed from two copies of
$G$ in the following manner \cite{alo86, hua12}. Let the vertex set of $G$ be $V(G) = \left\{
v_1, v_2, \ldots, v_n\right\}$, and the vertices of
${G^*}$ are given by the two sets $X = \left\{x_1, x_2, \ldots, x_n\right\}$ and $Y =
\left\{y_1, y_2, \ldots, y_n\right\}$. Thus for each vertex ${v_i} \in
V(G)$, there are two vertices ${x_i}$ and ${y_i}$ in $V(G^*)$. The double
graph $G^*$ includes the initial edge set of each copies of $G$, and for any edge
${v_i}{v_j} \in E(G)$, two more edges ${x_i}{y_j}$ and ${x_j}{y_i}$ are added.

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[place/.style={circle,draw=black!50,fill=black!20,
inner sep=0pt, minimum size = 5.2mm}]
{\footnotesize
\node[place] (1) {$v_1$};
\node[place] (2) [right=of 1] {$v_2$}
edge (1);
\node[place] (3) [right=of 2] {$v_3$}
edge (2);}
\end{tikzpicture}
\hspace{3cm}
\begin{tikzpicture}[place/.style={circle,draw=black!50,fill=black!20,
inner sep=0pt, minimum size = 5.2mm}]
{\footnotesize
\node[place] (1) {$x_1$};
\node[place] (2) [right=of 1] {$x_2$}
edge (1);
\node[place] (3) [right=of 2] {$x_3$}
edge (2);
\node[place] (4) [below=of 1] {$y_1$}
edge (2);
\node[place] (5) [below=of 2] {$y_2$}
edge (1) edge (3) edge (4);
\node[place] (6) [below=of 3] {$y_3$}
edge (2) edge (5);}
\end{tikzpicture}
\caption{The graph $P_3$ and its double graph $P_3^*$.}
\end{center}
\end{figure}

\begin{thm} The connective eccentric index of the double graph
${G^*}$ is given by ${C^\xi}(G^*) = 4{C^\xi}(G) - 2(n - 1){\left\|n - 1\right\|_G}$, where, ${\left\|n - 1\right\|_G}$ is the number of vertices with degree $n-1$, i.e., of eccentricity
one.
\end{thm}

\n\textit{Proof.} From the construction of double graph, it is clear that
$d_{G^*}(x_i) = d_{G^*}(y_i) = 2d_G(v_i)$, where $v_i \in V(G)$ and $x_i, y_i \in V(G^*)$ are the corresponding clone vertices of $v_i$. Also we can write $\varepsilon_{G^*}(x_i) = \varepsilon_{G^*}(y_i) = \varepsilon_G(v_i)$, when $\varepsilon_G(v_i) \ge 2$ and $\varepsilon_{G^*}(x_i) = \varepsilon_{G^*}(y_i) = \varepsilon_G(v_i) +
1$ when $\varepsilon_G(v_i) = 1$.

Thus the connective eccentric index of double graph $G^*$ is

\begin{eqnarray*}C^\xi(G^*)& = &\sum_{i = 1}^n\frac{d_{G^*}(x_i)}{\varepsilon_{G^*}(x_i)} + \sum_{i=1}^n \frac{d_{G^*}(y_i)}{\varepsilon_{G^*}(y_i)}\\
& = & 2\left[\sum_{\varepsilon_G(v_i) \ge 2}\frac{2d_G(v_i)}{\varepsilon_G(v_i)}+\sum_{\varepsilon_G(v_i)=1}\frac{2d_G(v_i)}{\varepsilon_G(v_i) + 1}\right]\\
&=&4\left[\sum_{\varepsilon_G(v_i)\ge 2}\frac{d_G(v_i)}{\varepsilon_G(v_i)}+\sum_{\varepsilon_G(v_i)=1}\frac{d_G(v_i)}{2}\right]\\
&=& 4\left[\sum_{\varepsilon_G(v_i)\ge 2}\frac{d_G(v_i)}{\varepsilon_G(v_i)}+ \sum_{\varepsilon_G(v_i)= 1}\frac{d_G(v_i)}{1}\right] - 2\sum_{\varepsilon_G(v_i)=1}d_G(v_i)\\
&=& 4C^\xi(G) - 2(n - 1){\left\|n - 1\right\|_G}.
\end{eqnarray*}
\qed

\bigskip
Let $G=(V,E)$ be a simple connected graph with $V = \left\{
{{v_1},{v_2},...,{v_n}} \right\}$. The extended double cover of $G$, denoted by
${G^{**}}$ is the bipartite graph with bipartition $(X,Y)$ where $X =
\left\{ {x_1},{x_2},...,{x_n}\right\}$ and $Y = \left\{
{y_1},{y_2},...,{y_n}\right\}$ in which ${x_i}$ and ${y_j}$ are adjacent if
and only if either ${v_i}$ and ${v_j}$ are adjacent in $G$ or $i = j$. For example,
the extended double cover of the complete graph is the complete bipartite graph.
This construction of the extended double cover was introduced by Alon \cite{alo86} in
1986.

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[place/.style={circle,draw=black!50,fill=black!20,
inner sep=0pt, minimum size = 5.2mm}]
{\footnotesize
\node[place] (1) {$v_1$};
\node[place] (2) [right=of 1] {$v_2$}
edge (1);
\node[place] (3) [right=of 2] {$v_3$}
edge (2);}
\end{tikzpicture}
\hspace{3cm}
\begin{tikzpicture}[place/.style={circle,draw=black!50,fill=black!20,
inner sep=0pt, minimum size = 5.2mm}]
{\footnotesize
\node[place] (1) {$x_1$};
\node[place] (2) [right=of 1] {$x_2$};
\node[place] (3) [right=of 2] {$x_3$};
\node[place] (4) [below=of 1] {$y_1$}
edge (1) edge (2);
\node[place] (5) [below=of 2] {$y_2$}
edge (1) edge (2) edge (3);
\node[place] (6) [below=of 3] {$y_3$}
edge (2) edge (3);}
\end{tikzpicture}
\caption{The graph $P_3$ and its extended double cover $P_3^{**}$.}
\end{center}
\end{figure}

\begin{thm}
The connective eccentric index of the extended
double cover ${G^*}^*$ satisfies the inequality
\[
{C^\xi }({G^{**}}) \le 2{C^\xi }(G) - (n - 2){\left\| {n - 1} \right\|_G}.
\]
\end{thm}

\n\textit{Proof.} If $G$ is a graph with $n$ vertices and
$m$ edges then, from definition, the extended double cover graph ${G^{**}}$
consists of $2n$ vertices and $(n+2m)$ edges and $d_{G^{**}}(x_i) = d_{G^{**}}(y_i) = d_G(v_i) + 1$ and $\varepsilon _{G^*}(x_i) = \varepsilon _{G^*}(y_i) = \varepsilon_G(v_i) + 1$ for $i=1,2,\ldots,n$.

Thus the connective eccentric index of extended double cover graph ${G^{**}}$ is
given by
\begin{eqnarray*}
C^\xi(G^{**})& = &\sum_{i = 1}^n
\frac{d_{G^{**}}(x_i)}{\varepsilon_{G^{**}}(x_i)} +
\sum_{i = 1}^n\frac{d_{G^*}(y_i)}{\varepsilon_{G^*}(y_i)}\\
&=& 2\left[\sum_{\varepsilon_G(v_i) \ge 2}
\frac{d_G(v_i) + 1}{\varepsilon_G(v_i) + 1} +
\sum_{\varepsilon_G(v_i) = 1}\frac{d_G(v_i) + 1}{2}
\right]\\
&\le & 2\sum_{\varepsilon_G(v_i) \ge 1}\frac{d_G(v_i)}{\varepsilon_G(v_i)} + \sum_{\varepsilon_G(v_i) = 1}\left\{d_G(v_i) + 1\right\}\\
&= & 2\left[\sum_{\varepsilon_G(v_i)\ge 1}\frac{d_G(v_i)}{\varepsilon_G(v_i)} +  \sum_{\varepsilon_G(v_i) = 1}d_G(v_i)\right] - \sum_{\varepsilon_G(v_i) = 1}d_G(v_i) +\sum_{\varepsilon_G(v_i) = 1} 1\\
& = & 2C^\xi(G) - (n - 1){\left\|n - 1\right\|_G} + {\left\|n - 1\right\|_G}\\
&=& 2C^\xi(G) - (n - 2){\left\|n - 1\right\|_G}.
\end{eqnarray*}
\qed

\bigskip

Now some exact formula for the eccentric connectivity index of joins, symmetric
difference, disjunction, splice graphs are presented.

\subsection{Join}
The join $G_1 + G_2$ of two graphs $G_1$ and $G_2$ with disjoint vertex
sets $V_1, V_2$ and edge sets $E_1, E_2$, is the graph union $G_1 \cup G_2$ together with all the edges joining ${V_1}$ and ${V_2}$, i.e., $G_1 +G_2$ consists of the
vertex set $V_1\cup V_2$ and edge set $E_1\cup E_2 \cup \{ xy:x \in V_1, y \in V_2\}$.

\begin{thm} Let ${G_1}$ and ${G_2}$ be two graphs without
well-connected vertices. Then,
$
{C^\xi }({G_1} + {G_2}) = |E({G_1})| + |E({G_2})| + |V({G_1})||V({G_2})|.
$
\end{thm}

\n\textit{Proof.} Let $n_1$ and $n_2$ be the numbers of vertices in $G_1$ and $G_2$ respectively. Thus, $|V({G_1} + {G_2})| = |{V_1}| + |{V_2}| = {n_1} + {n_2}$. For vertices $u \in {V_1}$ and $v \in {V_2}$, it holds that $d_{G_1+G_2}(u) = d_{G_1}(u) + n_2$ and $d_{G_1+G_2}(v) =
d_{G_2}(v) + n_1$. Since none of ${G_1}$ and ${G_2}$ contains
well connected vertices, then for every $u \in V({G_1} + {G_2}), \varepsilon_{G_1+G_2}(u) = 2$. So, from definition of connective eccentric index, we have
\begin{eqnarray*}
C^\xi(G_1+G_2)& = &\sum_{v\in V(G_1)}\frac{d_{G_1+G_2}(v)}{\varepsilon_{G_1+G_2}(v)}+ \sum_{v\in
V(G_2)}\frac{d_{G_1+G_2}(v)}{\varepsilon_{G_1+G_2}(v)}\\
&= &\frac{1}{2}\left[\sum_{v \in V(G_1)}\left\{d_{G_1}(v)+n_2 \right\}+\sum_{v\in V(G_2)}\left\{d_{G_2}(v)+n_1\right\}\right]
\end{eqnarray*}
from where the desired result follows.
\qed

\bigskip
Now we generalize the above result for $n$ disjoint graphs $G_1,G_2,\ldots,{G_n}$.

\begin{thm} Let ${G_1},{G_2},\ldots,{G_n}$ be $n$ graphs with disjoint vertex sets
${V_i} = V({G_i})$ and edge sets ${E_i} = E({G_i}), 1 \le i \le n$, without
well-connected vertices. Then,
\[
{C^\xi }({G_1} + {G_2} + ... + {G_n}) = \sum\limits_{i = 1}^n {|E({G_i})|}  +
\frac{1}{2}{\left\{ {\sum\limits_{i = 1}^n {|V({G_i})|} } \right\}^2} -
\frac{1}{2}\sum\limits_{i = 1}^n {|V({G_i}){|^2}}.
\]

\end{thm}

\n\textit{Proof.} From definition of join, we have
$
d_{G_1 + G_2 + ... + G_n}(v_j) = d_{G_i}(v_j) + |V(G_1 +G_2 + ... + G_n)| - |V(G_i)|$ and since none of $G_1, G_2, ..., G_n$ contains any well connected
vertex, we have $\varepsilon_{G_1 + G_2 + ... + G_n}(v_j) = 2$ for all $v_j\in V(G_i)$.

Thus the connective eccentric index of $G_1+ G_2+ ... + G_n$ is given by
\begin{eqnarray*}
C^\xi(G_1 + G_2 + \ldots + G_n) &=& \sum_{i = 1}^n\sum_{j
= 1}^{|V({G_i})|}\frac{d_{G_1 + G_2 + ... +G_n}(v_j)}{\varepsilon_{G_1 + G_2 + ... +G_n}(v_j)}\\
&=& \sum_{i = 1}^n\sum_{j = 1}^{|V(G_i)|}\frac{d_{G_i}(v_j) + \left|V(G_1 + G_2 + ... + G_n)\right| - \left|V(G_i)\right|}{2} \\
& = & \frac{1}{2}\bigg[|E(G_1)| + |E(G_2)| + ... +|E(G_n)|\\
&&+ |V(G_1+ G_2+... + G_n)|\left\{|V(G_1)| + |V(G_2)| + ... \right.\\
&&\left. + |V(G_n)|\right\}\left. - \left\{ |V(G_1)|^2 + V(G_2)|^2 +... + |V(G_n)|^2\right\}
\right]
\end{eqnarray*}
from where the desired result follows.
\qed

\bigskip
The following corollaries are direct consequences of the theorem.

\begin{cor}
If $nG$ denotes the join of $n$ copies of $G$, then
\[
{C^\xi }(nG) = n|E(G)| + \frac{{n(n - 1)}}{2}|V(G){|^2}.
\]
\end{cor}

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[place/.style={circle,draw=black!50,fill=black!20,
inner sep=0pt, minimum size = 5.2mm}, bend angle=15]
\draw (1.8,0) ellipse (30mm and 6mm);
\draw (1.8,-1.5) ellipse (30mm and 6mm);
\draw (1.8,-4.5) ellipse (30mm and 6mm);
\draw (2.8,0) node{\ldots};
\draw (2.8,-1.5) node{\ldots};
\draw (1.8,-3) node{\vdots \hspace{1cm}\vdots \hspace{1cm}\vdots};
\draw (2.8,-4.5) node{\ldots};
\draw (5,0) node{\footnotesize $V_1$};
\draw (5,-1.5) node{\footnotesize $V_2$};
\draw (5,-4.5) node{\footnotesize $V_n$};
{\tiny
\node[place] (1) {$v_{11}$};
\node[place] (2) [right=of 1] {$v_{12}$};
\node[place] (3) [right=of 2, xshift=1cm] {$v_{1m_1}$};
\node[place] (4) [below=of 1] {$v_{21}$}
edge (1) edge (2) edge (3);
\node[place] (5) [below=of 2] {$v_{22}$}
edge (1) edge (2) edge (3);
\node[place] (6) [below=of 3] {$v_{2m_2}$}
edge (1) edge (2) edge (3);
\node[place] (7) [below=of 1, yshift=-3cm]{$v_{n1}$}
edge[bend left] (1) edge (2) edge (3) edge (4) edge (5) edge (6);
\node[place] (8) [below=of 2, yshift=-3cm] {$v_{n2}$}
edge (1) edge[bend left] (2) edge (3) edge (4) edge (5) edge (6);
\node[place] (9) [below=of 3, yshift=-3cm] {$v_{nm_n}$}
edge[bend angle=10, bend left] (1) edge (2) edge[bend left] (3) edge (4) edge (5) edge (6);
}
\end{tikzpicture}
\caption{The complete $n$-partite graph.}
\end{center}
\end{figure}

\begin{cor} Let $G = K_{m_1, m_2, ..., m_n}$ be the complete
$n$-partite graph having $v = |V(G)|$ number of vertices. Here the vertex
set can be partitioned into $n$ subsets ${V_1},{V_2},...,{V_n}$ such that ${V_1}
\cup {V_2} \cup ... \cup {V_n} = V(G)$ and ${V_i} \cap {V_j} = \phi,i \ne j, 1 \le i,j \le n$. From definition of join it is clear
that $G = K_{m_1,m_2,...,m_n}$ is the join of $n$ empty graphs with
${m_1},{m_2},...,{m_n}$  number of vertices. Then the
connective eccentric index of $G = K_{m_1, m_2,..., m_n}$ is given by
\[
{C^\xi }(K_{m_1, m_2, ..., m_n}) = \sum_{{1 \le i\le n\atop 1\le j \le n}\atop i \ne j}
{{m_i}{m_j}}.
\]
\end{cor}

\subsection{Symmetric difference}
Let ${G_1}$ and ${G_2}$ be two graphs with vertex sets $V({G_1})$ and $V({G_2})$ and
edge sets $E({G_1})$ and $E({G_2})$. Then the symmetric difference of
${G_1}$ and ${G_2}$, denoted by ${G_1} \oplus {G_2}$ is the graph with vertex set
$V({G_1}) \times V({G_2})$ in which any two vertices $({u_1},{u_2})$ is adjacent
to $({v_1},{v_2})$ whenever ${u_1}$ is adjacent to ${v_1}$ in $G_1$ or
${u_2}$ is adjacent to ${v_2}$ in $G_2$, but not both. From definition
of symmetric difference, the degree of a vertex $({u_1},{u_2})$ of ${G_1} \oplus
{G_2}$ is given by \cite{imr00}
\[
d_{G_1\oplus G_2}({u_1},{u_2}) = |V({G_2})|{d_{{G_1}}}({u_1}) +
|V({G_1})|{d_{{G_2}}}({u_2}) - 2{d_{{G_1}}}({u_1}){d_{{G_2}}}({u_2}).
\]

\begin{thm} The connective eccentric index of the symmetric
difference ${G_1} \oplus {G_2}$ of two graphs ${G_1}$ and ${G_2}$ is given by
\[
{C^\xi }({G_1} \oplus {G_2}) = |E({G_1})||V({G_2}){|^2} +
|E({G_2})||V({G_1}){|^2} - 4|E({G_1})||E({G_2})|
\]
where, none of ${G_1}$ and ${G_2}$ contains well-connected vertices.
\end{thm}

\n\textit{Proof.} Since the distance between any two vertices of a symmetric
difference cannot exceed two, if none of the components contains well-connected
vertices, the eccentricity of all vertices is constant and equal to two, i.e., ${\varepsilon _{{G_1} \oplus {G_2}}}({u_1},{u_2}) = 2$, for all
vertices $({u_1},{u_2})$ \cite{ash11, gho12b}.

Thus the connective eccentric index of symmetric difference ${G_1}
\oplus {G_2}$ of two graphs ${G_1}$ and ${G_2}$ is given by
\begin{eqnarray*}
C^\xi(G_1\oplus G_2)&=& \sum_{(u_1,u_2) \in V(G_1\oplus G_2)}\frac{d_{G_1\oplus G_2}(u_1,u_2)}{\varepsilon_{G_1\oplus G_2}(u_1,u_2)}\\
&=& \frac{1}{2}\sum_{u_1\in V(G_1)}\sum_{u_2\in V(G_2)}\left[ |V(G_2)|d_{G_1}(u_1) + |V(G_1)|d_{G_2}(u_2) -2d_{G_1}(u_1)d_{G_2}(u_2) \right]\\
&=& \frac{1}{2}\left[ |V(G_2)|^2\sum_{u_1 \in V(G_1)}d_{G_1}(u_1) + |V(G_1)|^2\sum_{u_2\in V(G_2)}d_{G_2}(u_2)\right.\\
&&\left.  - 2\sum_{u_1\in V(G_1)}d_{G_1}(u_1)\sum_{u_2\in V(G_2)}d_{G_2}(u_2)\right]
\end{eqnarray*}
from where the desired result follows.
\qed

\subsection{Disjunction}
The disjunction ${G_1} \vee {G_2}$of two graphs ${G_1}$and ${G_2}$is the graph
with vertex set $V({G_1}) \times V({G_2})$ in which $({u_1},{u_2})$ is adjacent to
 $({v_1},{v_2})$ whenever ${u_1}$ is adjacent with ${v_1}$ in $G_1$
or ${u_2}$ is adjacent with ${v_2}$ in $G_2$. Obviously, the degree of a
vertex $({u_1},{u_2})$ of ${G_1} \vee {G_2}$ is given by \cite{ash11, gho12b}
\[
{d_{({G_1} \vee {G_2})}}({u_1},{u_2}) = \left[ {|V({G_2})|{d_{{G_1}}}({u_1}) +
|V({G_1})|{d_{{G_2}}}({v_2}) - {d_{{G_1}}}({u_1}){d_{{G_2}}}({v_2})} \right].
\]

\begin{thm}The connective eccentric index of the disjunction ${G_1}
\vee {G_2}$ of two graphs${G_1}$ and ${G_2}$ is given by
\[
{C^\xi }({G_1} \vee {G_2}) = |E({G_1})||V({G_2}){|^2} + |E({G_2})||V({G_1}){|^2}
- 2|E({G_1})||E({G_2})|
\]
where none of ${G_1}$ and ${G_2}$ contains well-connected vertices.
\end{thm}

\n\textit{Proof.} Since the distance between any two vertices of a disjunction
cannot exceed two, if none of the components contains well-connected vertices,
the eccentricity of all vertices is constant and equal to two \cite{ash11, gho12b}. Then the
connective eccentric index of the disjunction ${G_1} \vee {G_2}$ of two
graphs ${G_1}$ and ${G_2}$ is computed as
\begin{eqnarray*}
C^\xi(G_1\vee G_2) &=& \sum_{(u_i,v_j) \in V(G_1\vee G_2)}\frac{d_{G_1\vee G_2}(u_1,u_2)}{\varepsilon_{G_1\vee G_2}(u_1,u_2)}\\
&=& \frac{1}{2}\sum\limits_{{u_1} \in V({G_1})} {\sum\limits_{{v_2} \in V({G_2})}
{\left[ {|V({G_2})|{d_{{G_1}}}({u_1}) + |V({G_1})|{d_{{G_2}}}({u_2}) -
{d_{{G_1}}}({u_1}){d_{{G_2}}}({u_2})} \right]} }
\end{eqnarray*}
from where the desired result follows.
\qed

\subsection{Splice}
Let ${G_1} = ({V_1},{E_1})$ and ${G_2} = ({V_2},{E_2})$ are two graphs ${V_1} \cap
{V_2} = \phi $. Let ${v_1} \in {V_1}$ and ${v_2} \in {V_2}$ be two given vertices
of ${G_1}$ and ${G_1}$ respectively. A splice of ${G_1}$ and ${G_1}$ at the
vertices ${v_1}$ and ${v_2}$ is obtained by identifying the vertices ${v_1}$
and ${v_2}$ in the union of ${G_1}$ and ${G_1}$ and is denoted
by $S({G_1},{G_2},{v_1},{v_2})$ \cite{dos05}. Different topological indices of the splice
graphs have already been computed \cite{sha13}. Let $u \in V(S)$, then ${\varepsilon
_1}(u)$ denotes the eccentricity of $u$ as a vertex of ${G_1}, {\varepsilon
_2}(u)$ denotes the eccentricity of $u$ as a vertex of ${G_2}$ and
$\varepsilon (u)$ denotes the eccentricity of $u$ as a vertex of $G$.
 Let ${d_S}(x)$ be the degree of the vertex $x \in V(S)$. Then the connective
eccentric index of $S({G_1},{G_2},{v_1},{v_2})$ is computed as follows.

\begin{thm} The connective eccentric index of splice of  ${G_1}$
and ${G_2}$ is given by
\[
{C^\xi }(S) = \sum\limits_{x \in {V_1}} {\frac{{{d_S}(x)}}{{max\left\{
{{d_{{G_1}}}(x,{u_1}) + \varepsilon ({u_2})\varepsilon (x)} \right\}}} + }
\sum\limits_{y \in {V_2}} {\frac{{{d_S}(y)}}{{max\left\{ {{d_{{G_2}}}(y,{u_2}) +
{\varepsilon _1}({u_1})\varepsilon (y)} \right\}}}}.
\]
\end{thm}

\n\textit{Proof.} For any vertex $x \in {V_1}$, Sharafdini and Gutman \cite{sha13} showed
that
\[
\varepsilon (x) = \max\left\{ {{d_{{G_1}}}(x,{u_1}) + {\varepsilon
_2}({u_2}){\varepsilon _1}(x)} \right\}.
\]
Similarly, for any vertex $y \in {V_2}$,
\[
\varepsilon (y) = \max\left\{ {{d_{{G_2}}}(y,{u_2}) + {\varepsilon _1}({u_1})\varepsilon (y)} \right\}.
\]
Thus, the desired result follows from the definition of connective eccentric index.
\qed

config: arxiv


TITLE: Is the set discrete?
QUESTION [1 upvotes]: Is the set $S=\{(m+\frac{1}{2^{|p|}},n+\frac{1}{2^{|q|}}):m,n,p,q\in \mathbb{Z}$} discrete in $\mathbb{R}^2$?
I'm not  getting how shall I check discrete here?

REPLY [3 votes]: Hint: Take a particular point $P=\left(m+\frac{1}{2^p}, n+\frac{1}{2^q}\right)$.  We want to show that the point is an isolated point. To do this, show that our set contains no point other than $P$ within distance $\min\left(\frac{1}{2^{p+2}},\frac{1}{2^{q+2}}\right)$ of $P$.


TITLE: How to calculate an object breaking point or force require to break it?
QUESTION [1 upvotes]: for example, i have a 10 ft rope that has a strength to hold 2000lbs.
when I attach the rope to an object of 1500lb and pull it toward me, (keep it simple with no friction apply) the rope should not be damaged
Now I wrap the rope around the object and have 2 persons identity with the same amount of force pulling toward to themselves. does each person share equal amount of 2000lbs ? which law of physics define this?

REPLY [1 votes]: You have asked at least two different questions here. I will answer the one in the title of your post.
To solve for when the rope will break, you need to know three things: 1) the magnitude of the load, 2) the cross-sectional area of the rope, and 3) the yield strength of the rope material.
The yield strength is quoted in terms of pounds of load per square inch of area at the breaking point. So you solve for (magnitude of load)/(cross sectional area of rope). If this number is greater than the yield strength, the rope breaks.


TITLE: Can you learn Algebra and Calculus at the same time?
QUESTION [0 upvotes]: So I'm trying to catch up to Calc as soon as possible and I have an exam about differential/integral calc in a couple months. I'm not really sharp in Algebra yet, covering Algebra 1 and about to start Algebra 2. Would it be possible to learn algebra 2 and co-learn calc as well? Would like to hear if someone had experience with that.
Thank you.

REPLY [3 votes]: I would say "No." If you don't already know algebra,  you can't possibly learn calculus. And algebra itself isn't really enough. You really should have exposure to limits and analytic geometry, which typically are introduced in a precalculus course.
So I would buckle down and concentrate on mastering algebra first. Hit calculus after that (or after precalculus following algebra) -- you will understand it more deeply and more quickly if you have the proper tools in your toolkit.


TITLE: Question about functional analysis, bounded set
QUESTION [0 upvotes]: Let X be a normed space, Z⊆X,  and for all $f \in X^*$ set $\{ x:  x \in \mathbb{R} ~,~ \exists z \in Z ~~with~~  x=f(z) \}$ is bounded, prove that Z is bounded.
I was thinking of using the Theorem (Bounded linear functionals) on page 223 of the book kreyszig introductory functional analysis with applications, but I could do the demonstration, if anyone has any ideas I would appreciate it.

REPLY [0 votes]: For fixed $z$, let $x_{z}^{\ast\ast}\in X^{\ast\ast}$ be such that $x_{z}^{\ast\ast}(f)=f(z)$, so for fixed $f\in X^{\ast}$, $|x_{z}^{\ast\ast}(f)|\leq M_{f}$ for all $z$ and hence by Uniform Boundedness Principle we have $\|x_{z}^{\ast\ast}\|\leq M$ for all $z$, then for each $z$, choose an $f$ such that $\|z\|=|f(z)|$ and $\|f\|=1$, then $\|z\|\leq\|x_{z}^{\ast\ast}\|\|f\|\leq M$.


TITLE: Converges Uniformly Example
QUESTION [0 upvotes]: Folland's Real Analysis/Measure Theory textbook on page 61 states that, $n \chi_{[0,1/n]}$ converges uniformly to 0.
I cannot see how this can be true as I thought that
$\sup|f_n(x)-f(x)|$=$\sup |n \chi_{[0,1/n]}-0|=n$ which goes to $
\infty$ as $n$ approaches $\infty$, so it cannot converge uniformly to $0$. Am I mistaken or is there a typo in the textbook? I looked at the errata but I don't see anything listed. Thanks.

REPLY [2 votes]: On pg 61 of Folland, of which this is example (iii)
It is worded like

In (i), (ii), (iii), $f_n \rightarrow 0$ uniformly, pointwise, and a.e. respectively.

So for this third example, it is just meant a.e convergence. Uniform convergence applies to the (i) example, which is $n^{-1}\chi_{(0,n)} $.


TITLE: Frequency plot with photoelectric current and anode potential
QUESTION [0 upvotes]: Keep your eyes near 0V anode potential - here you can see that the light with maximum frequency has more photoelectric current and that the light with minimum frequency has less photoelectric current but how is it possible?
When we have two different frequencies of light $\nu_1$ and $\nu_2$ ($\nu_2>\nu_1$) and having same intensity of light then number of photons per unit time per unit area decrease for 2nd and increase for 1st.. Which means photoelectric current must have a low value for 2nd and a high value for 1st - Which is the opposite of the graph.

REPLY [0 votes]: The photoelectric current depends in complicated ways on frequency. The figure is wrong for real photocathodes.
But the idea is probably that this is a theoretical material with a quantum yield of 100 % for all frequencies above the threshold, and that the different curves are for the same photon flux.


TITLE: Nullspace that spans $\mathbb{R}^n$?
QUESTION [5 upvotes]: My professor said that if for a $n \times n$ matrix $A$, $\text{null}(A) = \mathbb{R}^n$, then $A = 0_{n}$. Why is this true? I understand what its saying - if everything times this matrix is zero, then the matrix has to be zero.
The intuition is simple enough with numbers, but could someone explain why this is true with matrices? Thanks

REPLY [4 votes]: If $A \neq 0$ then some column of $A$ is not the $0$ (column) vector. Say the $i$th column of $A$ is not the $0$ vector. Then $$A e_i = i\text{th column of } A$$
where $e_i$ is the (column) vector with $1$ in the $i$th positions and zero otherwise. Why does this imply that null$(A)$ isn't all of $\Bbb{R}^n$?

REPLY [2 votes]: A linear transformation can be visualized as taking a grid over space and skewing/distorting it. The grid lines all have to stay lines, and parallel lines remain parallel (i.e. it respects vector addition and multiplication), but that's it. So "stretch the z-axis by 2" and "rotate around the y axis by 30$^\circ$" are valid linear transformations, represented by particular matrices.
In particular, squashing $\mathbb{R}^n$ onto some subspace is also a linear transformation (e.g. "squish everything down into the xz plane"). If we end up with a subspace, then that means that some other subspace (in this case the y-axis) got mapped to the origin. When this happens we say that this subspace is part of the nullspace of the matrix. So $\operatorname{null}(A)=\mathbb{R}^n$ is saying "this matrix squishes everything to the origin". The zero matrix multiplies everything by zero, so it squishes everything to the origin. What your professor has proved is that the only matrix that does this is the zero matrix: the "squish everything to zero" property is enough to pin it down, because nothing else can do this.
In particular, any matrix with a nonzero entry in its $n^{th}$ column can be multiplied by the basis vector $(0,0,0,...1...0)$ with $1$ at position $n$, which will then produce this column vector as output (which by hypothesis is nonzero). So any nonzero matrix leaves something unsquished, and the conclusion follows.


TITLE: Understanding limit points for topological space $X$
QUESTION [0 upvotes]: Let $X$ be a topological space and $A \subset X$. We say $x \in X$ is a ${\bf limit \; point}$ of $A$ if every neighborhood of $x$ intersects $A$ in a point different from $x$.
This can be rephrased as follows: Let $L$ be set of limit points of $A$. $x \in L$ means that for every $U$ open in $X$, $(U \setminus \{ x \} ) \cap A \neq \varnothing$.

Consider now $X$ with discrete topology. That is, let $\tau_X = \mathcal{P}(X)$. For simplicity, say $X = \{a,b,c\}$. For example, consider  $A = \{b,c\}$. Observe that this has no limit point as the open $\{ a\}$ fails to intersect $\{b,c\}$ and the other points intersect $A$ on those points violating definition of limit point. Therefore, every subset of $X$ has no limit point... Does this generalize?
Another example, Consider $\mathbb{R}$ with standard topology and let $A = [0,1)$. Isnt every point of $A$ here a limit point of $A$ ? Indeed, take any $ 0 \leq x < 1 $ and choose some $U$ open so that $x \in U $ and clearly $U \cap [0,1) = \varnothing $.
Am I understanding these concepts correctly ?

REPLY [1 votes]: It is indeed true that in any discrete space $(X, \mathcal{T}=\mathscr{P}(X))$ no subset $A$ of $X$ has a limit point, as for any $x$, $U=\{x\}$ is open and $U\setminus\{x\} = \emptyset$. So in that sense your observation generalises.
Note that the definition of limit point is "for every open set $U$ that contains $x$, we have that $(U\setminus \{x\}) \cap A \neq \emptyset$. (Otherwise only dense sets can have limit points, it's about the points "close to $x$")
In $\Bbb R$ in the standard topology things are different. $A=[0,1)$ has all points of $[0,1]$ as its limit points. If $x \in [0,1]$ and $U$ is open containing $x$, we know $(x-\delta, x+\delta) \subseteq U$ and we always have points of $[0,1)$ to the left or right of $x$ (tto the right only if $x=0$, and to the left only if $x=1$, but at least one of $x-\frac{\delta}{2}$ or $x+\frac{\delta}{2}$ is a point in $(U\setminus \{x\}) \cap [0,1)$ if $x\in [0,1]$. If $x>1$, $U=(1, x+1)$ is an open set that contains $x$ and misses $A$ entirely. Same for $U=(x-1,0)$ if
$x<0$ so points outside $[0,1]$ are not limit points of $[0,1)$.
Now check that in the same topology $\Bbb Z$ has no limit points at all.


TITLE: Failing to understand some basic idea behind differentiation
QUESTION [2 upvotes]: I just discovered I must have some big holes in my knowledge of basic calculus, and this is scary honestly.
I have to compute some derivatives of the solution of a dynamical system:
\begin{equation*}
\frac{\text d y(t)}{\text dt} = f(t,y(t)),\quad y(t_0) = y_0,\quad t_0\leq t\leq T.
\end{equation*}
Say that I have to compute derivatives with respect to $t$. Clearly, $\dfrac{\text d y(t)}{\text dt}$ is given.

I want to compute $\dfrac{\text d y(t)}{\text du}$ with $u<t$.
I write:
\begin{equation*}
y(t) = y(u) +\int_u^t f(\tau,y(\tau))\text d \tau\therefore\dfrac{\text d y(t)}{\text du}=\dfrac{\text d y(u)}{\text du}+\dfrac{\text d}{\text du}\int_u^t f(\tau,y(\tau))\text d \tau=f(u,y(u))+?
\end{equation*}
The question mark stays for the fact that I have some uncertainties in how to compute the derivative of the integral by Leibniz rule. I will not report here all my doubts, I could fill pages.

I assume $\dfrac{\text d y(t)}{\text du}=0$ with $u>t$ for physical reasons (how can future influence past?), but is it actually true? If I unwind all the computation I will get by chain rule some terms like $\dfrac{\text d t}{\text du}$. Intuitively, it should be zero, but since $\dfrac{\text d t}{\text du}=\left(\dfrac{\text d u}{\text dt}\right)^{-1}$, then I would set it to 1.

the specific case $\dfrac{\text d y(t)}{\text d t_0}$ is the funniest. I get different results when computing it as
\begin{equation*}
\dfrac{\text d y(t)}{\text d t_0} = \dfrac{\text d}{\text d t_0}\left(y_0+\int_{t_0}^t f(\tau,y(\tau))\text d\tau\right) = -f(t_0,y_0)
\end{equation*}
or
\begin{equation*}
\dfrac{\text d y(t)}{\text d t_0} = \dfrac{\text d y(t)}{\text d t}\dfrac{\text d t}{\text d t_0} = f(t,y(t))
\end{equation*}

what about $\dfrac{\text d y(u)}{\text d y(t)}$ with $t<u$, by Leibniz rule? I obtain different results writing $y(u)=y(t)+\int_t^u f(\tau,y(\tau))\text d\tau$ or $y(u)=y_0+\int_0^u f(\tau,y(\tau))\text d\tau$

$\dfrac{\text d y(u)}{\text d y(t)}$ with $t>u$ would be 0 for physical reasons, or the inverse of what results in point 4, by algebra of differentials.

How would you solve these doubts? I think I don't get completely the meaning of derivative ..

REPLY [3 votes]: Premise. As already pointed out in the comments, I believe your confusion stems from a somewhat excessive use of the Leibniz notation for the derivative, rather than a poor understanding of the derivative itself. While the notation $\rm{d}y/\rm{d}t$ is undeniably intuitive and `agile' for some things it is not the most precise notation possible. In particular, it makes a complete mess of the point at which the derivative is being evaluated, making it impossible to distinguish between the independent variable and some fixed point. I suggest we ditch it completely, and instead write $y'(t)$ for the derivative of $y$ at the point $t$. With this in mind, let me address your points one by one
1) If $u,t\in[t_0,T]$ and $u<t$, it's certainly true that
$$y(t)=y(u)+\int_u^tf(\tau,y(\tau))\,\rm{d}\tau,$$
but it's not at all clear what $\rm{d}y(t)/\rm{d}u$ even means. If you mean the derivative of $y$ evaluated at the point $u$, i.e. $y'(u)$, then this is just $f(u,y(u))$. Otherwise, you could say $t$ is now fixed and $u$ is your independent variable (this effectively reduces your interval to $[t_0,t]$). Now that the meaning of all symbols has been clarified, you can take a derivative of both sides of your equation: the LHS vanishes because $y(t)$ is now a constant, whereas the RHS reads
$$y'(u) - f(u,y(u)) = y'(u) - y'(u) = 0,$$
i.e. you get the trivial identity $0=0$, as you should.
2) Your writing in this point corroborates the idea that to you $\rm{d}y(t)/\rm{d}u$ must stand for $y'(u)$. Let me point out, however, that you have absolutely no reason to assume $y'(u)$ vanishes for $u$ greater than some fixed $t\in[t_0,T]$, unless the function $f$ dictates so. For instance if you take $f(t,u(t))=A$ where $A$ is a non-zero constant this is false, but the equation $y'(t)=f(t,y(t))$ makes perfect sense regardless (and it's easily solved). The equation you solve at the end of your post is another example. This is because while we use differential equations to describe the time evolution of physical systems, they don't have to describe them. Thus if you want a system to be causal, that's a condition you have to impose on the system, otherwise it may well be false.
3) Here is where you can see clearly how the Leibniz notation falls short—and, at the same time, how useful it is. The inverse function is defined by the identity
$y(t)=\eta \Longleftrightarrow y^{-1}(\eta)=t,$
then under some very reasonable circumstances it holds
$$(y^{-1})'(\eta) = \frac{1}{y'(y^{-1}(\eta))} = \frac{1}{y'(t)}.$$
Notice that the LHS is evaluated at a different point than the RHS, but if we forget about this and use the Leibniz notation (which does not specify the evaluation point anyway) the identity above reads
$$\frac{\rm{d}y^{-1}(\eta)}{\rm{d}\eta} = \frac{\rm{d}t}{\rm{d}\eta} = \frac{1}{\rm{d}u/\rm{d}t}.$$
Now we may formally write this as $$\frac{\rm{d}t}{\rm{d} y}=\left(\frac{\rm{d} y}{\rm{d}t}\right)^{-1},$$
so long as we understand this is just a notational shorthand and not a full-fledged identity. You should now be able to see why the correct identity is
$$y'(t_0) = f(t_0,y_0),$$
and not
$$y'(t_0) = -f(t_0,y_0).$$
4,5) I believe my previous answers address these points as well. In particular, there is no "algebra of differentials", only formal differential identities which mock specific theorems, and as such they should be taken with a grain of salt.