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Home / India News / For Kashmir, 2016 was a long year of discontent
For years, the Valley had not witnessed the kind of protests that erupted in Kashmir after militant Burhan Wani was killed in an encounter.(AFP)
For Kashmir, 2016 was a long year of discontent
Hizbul militant Burhan Wani's death in an encounter triggered widespread protests in the Valley, a curfew that lasted over 50 days and deaths by pellet guns.
By Toufiq Rashid | Hindustan Times, Srinagar, Srinagar
The Kashmir Valley was savouring Eid after the month-long Ramazan fast when news of a shootout trickled in with the evening summer breeze on July 8.
More news poured in soon after. Burhan Wani, the 22-year-old Hizbul Mujahideen poster boy and face of a resurgent local militancy after nearly two decades, was shot dead. What was dismissed as rumour minutes earlier shook the Valley — and aftershocks are continuing to shake the troubled region.
Summer rain drenched Kashmir days on end, but failed to calm the anger as tens of thousands of people poured into the streets and took the first available transport to Tral — Wani's native, restive town in south Kashmir.
More than half-a-million people gathered for the militant's funeral, according to independent estimates. Much more than the funeral of chief minister and Peoples Democratic Party (PDP) patriarch Mufti Mohammad Sayeed, whose death on January 7 shoved the state into three months of political uncertainty.
The young militant's namaz-e-janaza, or funeral prayer, was offered 40 times to accommodate people. The burial scheduled for the morning happened late in the afternoon. Kashmir has not recorded such public reaction for years. Former chief minister Omar Abdullah was quick to react that "a dead Burhan could be more lethal" because "Kashmir's disaffected got a new icon".
The Valley plunged into violence; angry mobs threw stones and tried to torch police stations and official buildings. Anti-India slogans filled the air.
The authorities panicked — telephone and internet services were snapped. The curfew followed; the longest spell of 51 days without relaxation. The next casualties in the following days and months were thousands of men, women and children. Security forces fired bullets and pellets; the official non-lethal ammunition for mob control. Around 91 people were killed and more than 12,000 wounded as the cycle of death went on for months.
Pellet guns became the brutal, diabolical face of street battles. The non-deadly weapon killed and maimed. About 1,100 people lost sight in one eye or both; some will struggle to live with permanent disabilities. Even homes were not safe. Bursts of pellets pierced through windows, doors and wood walls, taking down children.
The unrest gave relevance to the separatists, who called strikes and shutdowns almost every day. The muezzin's call to prayer followed a hate-filled rant in mosques. The authorities pandered the tirade as an excuse to tighten their clampdown — and fire more pellets at people swelling the protests. Trade came to a standstill, shops were shut; schools opened only on weekends when separatists announced a "normal" day. The state education department announced mass promotion for students from class 1 to 8.
Watch | An interview with Burhan Wani's father
South Kashmir, famous for its apple orchards, was the worst hit as it was the turf of homegrown militants such as Wani. Kashmir's sweetest and crunchiest gift didn't reach fellow countrymen because of the shutdown this apple season.
Wani's death was perhaps the tipping point. But the anger was a reflection of the people's growing cynicism against the PDP when it joined hands with the BJP to rule the state together after a fractured verdict in the 2014 elections. It was perceived as a "betrayal", a rallying point for those harbouring separatist sentiments.
The anger was evident as the Valley didn't shut down for a day when chief minister Sayeed died.
"The Centre never addressed the anti-India sentiment. It simply tried to shut up the people by force … a first-aid kind of approach."
The Valley was surcharged over the government's announcement of making a military campus and a separate colony for rehabilitation of migrant Pandits. A high court bench in Srinagar upheld that the state flag must be unfurled next to the national flag on government buildings, while another bench insisted on enforcing a beef ban. People felt the PDP was helping the BJP to do away with Article 370, the constitutional provision that accords special status to the state.
Kashmir is boiling this winter over alleged reports of government issuing domicile certificates to West Pakistani Refugees, families who migrated during Partition. They remain refugees with no right to vote. But any attempt to change their status is viewed as an affront towards Article 370. The Valley is a dormant volcano. "The Centre never addressed the anti-India sentiment. It simply tried to shut up the people by force … a first-aid kind of approach," said Shiekh Mushtaq, senior journalist.
For more of HT's coverage of the past year, visit http://www.hindustantimes.com/year-ender-2016


My goodness, I don't even know where to begin with these two.
Caution: be aware of surroundings. Side effects include excitement and may induce happy-dancing through tall grass and/or in Starbucks.
Engagement season has been lovely this year, and this session was no exception! It was such a delight to spend a few moments with Ben and Mallory after capturing the proposal moment from the bushes, once again. This is becoming a favorite thing. Congratulations, you two!
Two weeks ago, I had the joy and honor of capturing the moment when a dear friend asked his lady to spend the rest of her life with him. The ensuing moments - and their love - were well worth hiding in bushes.
I am very excited to finally be able to blog this set!


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"Who's the Murderer" is a Chinese web variety show that airs on Mango TV. The show scored an 8.7 out of 10 during season 1, a 8.9 during season 2, a 9.1 during season 3 and season 4 is starting with a 9.5, making it the most popular web show of China.
A murder takes place every episode and the 5 suspects need to find the culprit while trying to prove their innocence to the detective. Only the culprit can lie and only the detective can vote twice. The most voted suspect gets arrested. If the culprit get caught, everyone will receive one gold bar and the detective two. If the murderer escapes, the culprit can take all the gold bars away as prize. Can you find the murderer?
All characters and background of the cases are fictional, but related to each other. All the cases takes place in the same dimension. Some characters like the fictional boyband NZND appears in multiple cases and are often mentioned in other unrelated cases.
The show went viral in China through the case "Terrifying Nursery Rhyme" (S2 Case 07) after reaching 1,000,000,000 views halfway through season 2. "Terrifying Nursery Rhyme" is also known to be the scariest case till this day and was a 4 hour special that was split into two episodes.
"Who's the Murderer" is the Chinese version of the Korean show "Crime Scene", however over time, the show has become a lot different than the original one. Despite being a variety show, the Korean version focussed more on the mystery part while the Chinese version more on comedy, making it more light hearted. However, the Chinese show has a way bigger budget than the Korean version which made it possible to make the cases way more complicated and luxury than "Crime Scene". The case of "Hotel Horrors" (S3 Case 01) took place at a 3000 square meters hotel which the production team worked 25 days on.

redpajama_set_name: RedPajamaCommonCrawl


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Black Lives Matter: a note from Network N
In a world where silence is complicity, businesses and individuals need to deliver concrete steps to show that #BlackLivesMatter.
The racism and injustice that has led to civil unrest across the USA right now, has given us all much pause for thought. As we pay attention to what's happening in the USA it's become increasingly apparent we must focus even harder on what we do here. It is clear that racism and injustice are not exclusively American issues but global problems.
Many of us are looking at how we 'show up', and go beyond a post or tweet to take action, right now, to improve the situation. It's easy to know what to say but harder to know how to do the right thing.
What we do know is that the people in our business care deeply about diversity and inclusion. We know that we want to take an active stance against racism and injustice. We also know we haven't shown up in a way that represents how much we care and it's sad that it's taken these tragic events for us to reflect on that. We all need and can do more, but for us at Network N, here's where we will begin:
With immediate effect we will be taking steps to increase the visibility of BAME and other minority voices within our editorial output. As a first step we will allocate an additional 25% editorial freelance budget to commission content from minority writers. And we commit to reexamining the mix of our editorial voices every quarter.
We will donate $250,000 of advertising space to charities and educational organisations already fighting racism and injustice globally. While we work with our teams to identify the best partners we will in the first instance extend this offer to two global organisations; Amnesty International and Solidar.
We need to be better allies but we don't yet know how. We will host an open allyship forum for our employees, partners, and broader network on a video call. We invite and encourage as many stakeholders as possible to attend. For those that can't we will post the session online and encourage further feedback.
As an industry and individuals, we are often focused on community mental health and actively speak out in ways to eradicate toxicity in our gaming communities. But these issues dwarf our gaming worlds. We may not take all of the right actions at all of the right moments, but now is absolutely the time for us to start.
Tim and James
COO and CEO, Network N.
Network N
Free Steam games


Black vinyl rat with red eyes. Equipped with bump & go mechanism built right in. When rat bumps into something it turns and goes the other way. On/off switch. Requires 3 AA batteries. 12 inches long from nose to tip of tail.


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Halla  ou Hallansaari est une île du golfe de Finlande et un quartier de Kotka en Finlande.

Géographie
L'île de Halla est séparée de Popinniemi par le détroit de Karhusalmi.
Halla est actuellement reliée par un remblai à l'île de Tiutinen, qui est située à son sud.

Références

Voir aussi

Articles connexes

Liens externes

Île de Kotka
Quartier de Kotka

redpajama_set_name: RedPajamaWikipedia


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The History of St. Bartholomew's School
This brief history of the school, by May Yardley, first appeared in the October and November 1993 issues of Sydenham Life. Since this was written, a number of changes have been made to the school - the roll and buildings have both been expanded - but it still makes interesting reading and May Yardley was for many years a governor of the school.
We'd love to hear from anyone who would like to bring this history up-to-date!
It is unusual for an infant and junior school within the State system to be able to trace its history far back into the past. St Bartholomew's Church School can do that, it can trace its history for 178 years back to its beginnings in 1815. Evidence suggests that it began as a 'dame's school' held in a cottage parlour in Lower Sydenham, probably for about a dozen children for whom their parents paid a small weekly sum. It is difficult for us to imagine a time when education was not compulsory in England and the majority of children just played in the streets or were set early to work and so grew up illiterate.
The dame's school was moved in 1832 to Upper Sydenham where had grown a village of larger houses. St Bartholomew's Church was dedicated in 1832 as a daughter church to St Mary's at Lewisham, with a curate in charge, and the school became a church school under the aegis of the new church. A corner site was found in Kirkdale on which a proper school building was erected. The school remained on this site for 136 years until 1968.
Boys and girls were taught separately and on this comparatively small site two houses were also built for a head master and a head mistress. Parents still had to pay small fees. The school must have been poorly built, probably for lack of funds, as according to the school records it was often under repair.
In 1854 the church ceased to be a daughter church of Lewisham and became the parish church of Upper Sydenham. Its first vicar was the Rev Charles English, whose tombstone remains just inside the lower church gate. He was a very enterprising man and in 1867 he had the old school building pulled down and a new one built which was described as being "handsome, suitable and commodious". It was a mixed school for boys and girls.
Many of us remember that building which remained in use for a hundred years until the school moved in 1968. "Commodious" by present day standards it certainly was not; several classrooms had to be reached through other classrooms, the hall was small and so was the playing space, and there was no kitchen.
There was a body of six managers (now, by law, the number was swollen to eighteen) who appointed the head master and the staff. The early log books are still in the school's possession. With all its building limitations, the school was always known for being hardworking, happy and well-conducted, a reputation maintained to this day.
In rebuilding the school in 1867, Charles English probably anticipated the Education Act of 1870 when it became compulsory for all children to attend school. The Sydenham school entered the state system but still retained its close links with the Church, which have happily remained strong until the present day. The official title of the school is St. Bartholomew's Church Voluntary Aided Primary School, and in the order of admission to the school, the first admissions are children of parents who attend the church. As in all dioceses, there is a Southwark Diocesan Board of Education which keeps closely in touch with all its church schools.
A new and lovely site was allotted to the school in the early thirties of this century, but it was not able to enter into its heritage for more than twenty years. The war intervened, during which no civic building was permitted, and immediately after the war the field was commandeered for the erection of a village of "prefabs" for those whose homes had been badly damaged or destroyed by air raids during the war years. So the school waited until at last the site was cleared and the inhabitants re-housed. The long expected building of the new school began. It was thrilling for staff in the old building to watch the rise of a modern school building with much larger and light classrooms, a library, a big hall, and kitchen quarters where school dinners would be cooked, beside the laying of the grass on the field and the making of two netball courts.
The foundation tone was laid by the Countess of Dartmouth and the school opened in 1968 with a service of thanksgiving. It is indeed fortunate in the crowded suburbs of South London to have good playing space, opportunities for growing small gardens and for nature study. A pond has been dug by fathers, with an area of wild growth around it. The school celebrated its twenty-fifth birthday on the present site this summer. (1993)
Under the 1988 Education Act, the governors and head teacher are responsible for the allotment of 80% of the annual budget allocated by Lewisham Education Authority. They are also responsible for covering the requirements of the National Curriculum and, as it is a Church Voluntary Aided School, for 15% of the building maintenance, towards which they ask parents for a contribution.
Because of the pressure from parents for school places, the governors decided to increase the size of the school by adding an extra half-stream of pupils. This means that each year the school will admit forty-five new pupils instead of thirty. We congratulate the governors on their enterprise and courage in raising the money to add the necessary new classrooms to the building. At an outdoor ceremony on 25th June 1993, in the presence of the school, the governors, the parents and many friends, a short service of thanksgiving and seeking God's blessing on the new wing was held. Miss Yardley, a former governor for many years, declared the new wing officially open and unveiled a plaque dedicating it 'Ad Gloriam Dei' and naming it the Yardley Suite.
Proposals will shortly be going in front of the governors to enlarge the staffroom and erect a brick-built administrative block for the Secretary. This will be funded by the Department of Education and, if it receives governors' approval, will be completed by March 1994.
Return from the St Bartholomew's School to Local History
Return to St Bartholomew's Home Page
A Guide to the Eucharist
Mission Action Plan (MAP)
The Wider Community
Our Parish & Location
The Wider Church
West Lewisham Deanery
Southwark Diocese
Children at St. Bartholomew's
Back to Local History
The School Website
© St. Bartholomew's Church Parochial Council, 2014


text:
\section{Introduction}

All simplicial complexes considered in this paper are finite and abstract. All homologies are simplicial
homologies with coefficients in a field $ \ensuremath{\mathbb{F}}$. The vertex set of a simplicial complex $X$ will be denoted by
$V(X)$. For $A \subseteq V(X)$, the induced subcomplex $X[A]$ of $X$ on the vertex set $A$ is defined by $X[A] :=
\{\alpha\in X \, : \, \alpha\subseteq A\}$. A simplicial complex $X$ is said to be a {\em triangulated manifold}
if it triangulates a manifold, i.e., if the geometric carrier $|X|$ of $X$ is a topological manifold. A
triangulated closed $d$-manifold $X$ is said to be {\em $ \ensuremath{\mathbb{F}}$-orientable} if $H_d(X;  \ensuremath{\mathbb{F}}) \neq 0$. So, for a
field $ \ensuremath{\mathbb{F}}$ of characteristic two, any triangulated closed manifold is $ \ensuremath{\mathbb{F}}$-orientable.

Taking his cue from pre-existing notions of tightness in the theory of smooth and polyhedral embedding of
manifolds in Euclidean spaces, K\"{u}hnel \cite{Ku} introduced the following precise notion of tightness of a
simplicial complex with respect to a field.

\begin{definition} \label{def:tight}
{\rm Let $X$ be a simplicial complex and $ \ensuremath{\mathbb{F}}$ be a field. We say that $X$ is} tight with respect to $ \ensuremath{\mathbb{F}}$ {\rm
(in short,} $ \ensuremath{\mathbb{F}}$-tight{\rm ) if (a) $X$ is connected, and (b) for every induced subcomplex $Y$ of $X$, the
$ \ensuremath{\mathbb{F}}$-linear map $H_{\ast}(Y;  \ensuremath{\mathbb{F}}) \to H_{\ast}(X;  \ensuremath{\mathbb{F}})$ (induced by the inclusion map $Y \hookrightarrow X$) is
injective. }
\end{definition}

Recall that, if $X$ is a simplicial complex of dimension $d$, then its {\em face numbers} $f_i(X)$ are  defined
by $f_i(X) := \#\{\alpha\in X \, : \, \dim(\alpha)=i\}, \,  0\leq i\leq d$. For $k\geq 2$, a simplicial complex
$X$ is said to be {\em $k$-neighbourly} if any set of $k$ vertices of $X$ form a face, i.e., if $f_{k-1}(X) =
\binom{f_0(X)}{k}$. A 2-neighbourly simplicial complex is called {\em neighbourly}.

\begin{definition} \label{def:sminimal}
{\rm A simplicial complex $X$ is said to be} strongly minimal {\rm if, for every triangulation $Y$ of the
geometric carrier $|X|$ of $X$, we have $f_i(X) \leq f_i(Y)$ for all $i$, $0\leq i\leq \dim(X)$. }
\end{definition}

Thus, a strongly minimal triangulation of a topological space, if it exists, is the most economical among all
possible triangulations of the space. Unfortunately, there are very few criteria available in the literature
which ensure strong minimality. The notion of tightness is of great importance in combinatorial topology because
of the following tantalizing conjecture \cite{KL}.

\begin{conjecture}[K\"{u}hnel-Lutz] \label{conj:KL}
Every $ \ensuremath{\mathbb{F}}$-tight triangulated closed manifold is strongly minimal.
\end{conjecture}

Intuitively, $ \ensuremath{\mathbb{F}}$-tightness of a triangulated manifold $X$ means that all parts of $X$ are essential in order to
capture the $ \ensuremath{\mathbb{F}}$-homology of the topological space $|X|$. In view of this intuition, Conjecture \ref{conj:KL}
appears to be entirely plausible. However, Example \ref{exam:nsmtight} below shows that this intuition is not
correct for arbitrary simplicial complexes.

\medskip

Recall the following from \cite{MNStacked}.

\begin{definition} \label{def:stacked}
{\rm A triangulated manifold $\Delta$ of dimension $d+1$ is said to be {\em stacked} if all its faces of
codimension (at least) two are in the boundary $\partial\Delta$. A triangulated closed manifold $M$ of dimension
$d$ is said to be {\em stacked} if there is a stacked triangulated manifold $\Delta$ of dimension $d+1$ such that
$M = \partial \Delta$. }
\end{definition}

In particular, a {\em stacked sphere} is a triangulated sphere which may be realized as the boundary of a stacked
triangulated ball.

\begin{definition} \label{def:lstacked}
{\rm A triangulated manifold is said to be {\em locally stacked} if all its vertex links are stacked spheres or
stacked balls.}
\end{definition}

Clearly, all stacked triangulated manifolds are locally stacked, but the converse is false (see Example
\ref{exam:Lutz} below). Due to \cite[Theorem 2.24; case $k=1$]{BDStacked} we have the following.

\begin{proposition}[Bagchi-Datta] \label{prop:BDTh2.24}
Let $M$ be a locally stacked $ \ensuremath{\mathbb{F}}$-orientable, neighbourly, triangulated closed $3$-manifold. Then the following
are equivalent
\begin{enumerate}[{\rm (i)}]
\item $M$ is $ \ensuremath{\mathbb{F}}$-tight,

\item $M$ is stacked, and

\item $\binom{f_0(M)-4}{2} = 10 \beta_1(M;  \ensuremath{\mathbb{F}})$.
\end{enumerate}
\end{proposition}

Thus, all orientable, neighbourly, stacked triangulated closed 3-manifolds are $ \ensuremath{\mathbb{F}}$-tight (for any field $ \ensuremath{\mathbb{F}}$).
Even more, we know from  \cite{BDStellated}

\begin{proposition}[Bagchi-Datta] \label{prop:sminimal}
Every $ \ensuremath{\mathbb{F}}$-tight, locally stacked, triangulated closed manifold is strongly minimal.
\end{proposition}

The first main result of this paper (in Section 4) is a converse of Proposition \ref{prop:BDTh2.24}: if a
triangulated closed 3-manifold is tight with respect to a field of characteristic $\neq 2$, then it must be
(orientable, neighbourly and) stacked. This result answers Question 4.5 of \cite{DM} affirmatively, in the case
of odd characteristic. As a consequence of Proposition \ref{prop:sminimal} it follows that the K\"{u}hnel-Lutz
conjecture is true in a special case, namely, if ${\rm char}( \ensuremath{\mathbb{F}}) \neq 2$ then any $ \ensuremath{\mathbb{F}}$-tight triangulated
closed 3-manifold is strongly minimal.

Let $X_1$ and $X_2$ be two triangulated $d$-manifolds intersecting in a common facet ($d$-face) $\alpha$. That
is, $X_1 \cap X_2 = \bar{\alpha}$, where $\bar{\alpha}$ denotes $\alpha$ together with all of its subfaces. Then
$X_1\#X_2 = (X_1 \cup X_2 ) \setminus \{\alpha\}$ is said to be the {\em connected sum of $X_1$ and $X_2$ along
$\alpha$} (for more general definition, see the end of Section 2). Let $I = I^2_{12}$ be the boundary complex
of the icosahedron. Thus, $I$ is a triangulated 2-sphere on 12 vertices. It is well known that $I$ is the unique
triangulation of $S^2$ in which each vertex is of degree 5. We introduce:

\begin{definition} \label{def:icosian}
{\rm A triangulated $2$-sphere is said to be {\em icosian} if it is a connected sum of finitely many copies of
$I$. A triangulated $3$-manifold is said to be {\em locally icosian} if all its vertex links are icosian.}
\end{definition}

In \cite{SpOdd}, the third author proved the following interesting upper bound theorem for tight triangulations
of odd dimensional manifolds.

\begin{proposition}[Spreer] \label{prop:Spreer}
Let $M$ be an $(\ell- 1)$-connected triangulated closed $(2\ell + 1)$-manifold and $ \ensuremath{\mathbb{F}}$ be a field. If $M$ is
$ \ensuremath{\mathbb{F}}$-tight, then ${\binom{\lfloor\frac{f_0(M)}{2}\rfloor-1}{\ell+1}\binom{\lceil
\frac{f_0(M)}{2}\rceil-1}{\ell+1}}/{\binom{f_0(M)-1}{\ell+1}}\leq \beta_{\ell}(M;  \ensuremath{\mathbb{F}})$.
\end{proposition}

In Section 5, we consider fields of characteristic 2. According to Proposition \ref{prop:BDTh2.24}, every
neighbourly and stacked closed 3-manifold $M$ is $ \ensuremath{\mathbb{Z}}_2$-tight. We do not know whether the converse is true or
not. But in this paper we prove that if $M$ is a $ \ensuremath{\mathbb{Z}}_2$-tight triangulated closed 3-manifold with  $f_0(M) \leq
71$ or $\beta_1(M;  \ensuremath{\mathbb{Z}}_2) \leq 188$, then $M$ must be stacked and neighbourly (and therefore $\binom{f_0(M)-4}{2}
= 10\beta_1(M;  \ensuremath{\mathbb{Z}}_2)$). We also show that, in general, each vertex link of a $ \ensuremath{\mathbb{Z}}_2$-tight closed
3-manifold must be a connected sum of $I$'s and $S^2_4$'s. Further, we prove that any $ \ensuremath{\mathbb{Z}}_2$-tight triangulated
closed 3-manifold $M$ satisfies the following upper bound on $f_0(M)$:
\begin{align}
(f_0(M)-4)(617f_0(M)-3861) \leq 15444\beta_1(M;  \ensuremath{\mathbb{Z}}_2). \nonumber
\end{align}
Equality holds here if and only if $M$ is locally icosian. In conjunction with the fact that $\binom{f_0(M)-4}{2}
= 10\beta_1(M;  \ensuremath{\mathbb{Z}}_2)$ when $f_0(M) \leq 71$, this inequality improves upon the upper bound of Proposition
\ref{prop:Spreer} in case $l=1$. We also prove that, if there is a non-stacked $ \ensuremath{\mathbb{F}}$-tight triangulated
3-manifold $M$, then its integral homology group $H_1(M; \ensuremath{\mathbb{Z}})$ must have an element of order 2. The results of
Section 5 were largely suggested by extensive machine computations using simpcomp \cite{ES}. Altogether, these
results impose severe restrictions on the topology of $3$-manifolds admitting tight triangulations (cf. Corollary
\ref{coro:5.14}).

In Section 6, we present some examples to show that the converses/generalizations of several results proved here
are not true.

\section{Preliminaries on stacked and tight triangulations}

In this section, we gather together a few easy (and mostly known) consequences of tightness. For completeness, we
include their proofs. We shall use\,:

\begin{notation} \label{notation1}
{\rm If $x$ is a vertex of a simplicial complex $X$, then $X^x$ and $X_x$ will denote the antistar and the link
(respectively) of $x$ in $X$. Thus,
\begin{align}
X^x & := \{\alpha\in X \, : \, x \not\in\alpha\} = X[V(X)\setminus\{x\}], \nonumber \\
X_x & := \{\alpha\in X : x\not\in\alpha, \alpha\sqcup\{x\}\in X\}. \nonumber
\end{align}
}
\end{notation}

We denote a face $\{u_1, \dots, u_m\}$ in a simplicial complex by $u_1u_2\cdots u_m$. If $X$ is a simplicial
complex and $a\not\in X$ is an element then the {\em cone} with apex $a$ and base $X$ is the simplicial complex
$X \cup \{\alpha\cup\{a\} \, : \, \alpha\in X\}$ and is denoted by $a\ast X$. For a simplicial complex $X$ of
dimension $d$, and for $0\leq k\leq d$, the {\em $k$-skeleton} ${\rm skel}_k(X)$ is defined by
\begin{align}
{\rm skel}_k(X) & := \{\alpha\in X \, : \, \dim(\alpha) \leq k\}. \nonumber
\end{align}

\begin{lemma} \label{lemma:2.2}
Every $ \ensuremath{\mathbb{F}}$-tight simplicial complex is neighbourly.
\end{lemma}

\begin{proof}
Suppose, if possible, $x\neq y$ are two vertices of an $ \ensuremath{\mathbb{F}}$-tight simplicial complex $X$ such that $xy$ is
not an edge of $X$. Let $Y$ be the induced subcomplex of $X$ on the set $\{x, y\}$. Then $\beta_0(Y;  \ensuremath{\mathbb{F}}) = 2 > 1
= \beta_0(X;  \ensuremath{\mathbb{F}})$, so that $H_0(Y;  \ensuremath{\mathbb{F}}) \to H_0(X;  \ensuremath{\mathbb{F}})$ can not be injective. This is a contradiction since $X$
is $ \ensuremath{\mathbb{F}}$-tight.
\end{proof}

\begin{lemma} \label{lemma:2.3}
Every induced subcomplex of an $ \ensuremath{\mathbb{F}}$-tight simplicial complex is $ \ensuremath{\mathbb{F}}$-tight.
\end{lemma}

\begin{proof}
Let $Y$ be an induced subcomplex of an $ \ensuremath{\mathbb{F}}$-tight simplicial complex $X$. By Lemma \ref{lemma:2.2}, $X$ and
hence also $Y$ are neighbourly. So, $Y$ is connected. Let $Z$ be an induced subcomplex of $Y$. Then $Z$ is an
induced subcomplex of $X$ also. Since the composition of the linear maps $H_{\ast}(Z;  \ensuremath{\mathbb{F}}) \to H_{\ast}(Y;  \ensuremath{\mathbb{F}})
\to H_{\ast}(X;  \ensuremath{\mathbb{F}})$ is injective, the first of them must be injective. Thus, $H_{\ast}(Z;  \ensuremath{\mathbb{F}}) \to H_{\ast}(Y;
 \ensuremath{\mathbb{F}})$ is injective for all induced subcomplexes $Z$ of $Y$. Therefore, $Y$ is $ \ensuremath{\mathbb{F}}$-tight.
\end{proof}

\begin{lemma} \label{lemma:2.4}
If $X$ is an $ \ensuremath{\mathbb{F}}$-tight simplicial complex of dimension $d$, then ${\rm skel}_k(X)$ is $ \ensuremath{\mathbb{F}}$-tight for $1\leq
k\leq d$.
\end{lemma}

\begin{proof}
Since $k\geq 1$ and $X$ is neighbourly by Lemma \ref{lemma:2.2}, it follows that ${\rm skel}_k(X)$ is neighbourly
and hence connected. Let $Y$ be an induced subcomplex of ${\rm skel}_k(X)$. Then $Y = {\rm skel}_k(Z)$, where $Z$
is an induced subcomplex of $X$. Since $X$ is $ \ensuremath{\mathbb{F}}$-tight, it follows that, for $0\leq i\leq k-1$, $H_i(Y) =
H_i(Z) \to H_i(X) = H_i({\rm skel}_k(X))$ is injective. Clearly, $Z_k(Y) \subseteq Z_k({\rm skel}_k(X))$. Since
both $Y$ and ${\rm skel}_k(X)$ are of dimension $\leq k$, $B_k(Y) = 0 = B_k({\rm skel}_k(X))$. These imply that
$H_k(Y) \to H_k({\rm skel}_k(X))$ is injective.
\end{proof}

\begin{lemma} \label{lemma:2.5}
Every $ \ensuremath{\mathbb{F}}$-tight triangulation of a closed manifold is $ \ensuremath{\mathbb{F}}$-orientable.
\end{lemma}

\begin{proof}
If ${\rm char}( \ensuremath{\mathbb{F}})= 2$ then there is nothing to prove. So, assume that ${\rm char}( \ensuremath{\mathbb{F}}) \neq 2$. Let $X$ be an
$ \ensuremath{\mathbb{F}}$-tight triangulated closed  $d$-manifold. We can assume that $d\geq 2$. Since $X$ is a connected
triangulated closed $d$-manifold, it is easy to see ab initio that, for any proper subcomplex $Y$ of $X$, $H_d(Y;
 \ensuremath{\mathbb{F}}) = 0$. Now, fix a vertex $x$ of $X$, and consider the induced subcomplex $X^x$ of $X$ on the complement of
$x$. Then $X = X^x \cup (x \ast X_x)$ and $X^x \cap (x \ast X_x) = X_x$. Since $H_d(X^x;  \ensuremath{\mathbb{F}}) = 0$, $H_{d-1}(x
\ast X_x;  \ensuremath{\mathbb{F}})=0$, $H_{d-1}(X_x;  \ensuremath{\mathbb{F}}) =  \ensuremath{\mathbb{F}}$ and $H_{\ast}(X^x;  \ensuremath{\mathbb{F}}) \to H_{\ast}(X;  \ensuremath{\mathbb{F}})$ is injective, it
follows from the exactness of the Mayer-Vietoris sequence
\begin{align}
\cdots \to & H_d(X^x;  \ensuremath{\mathbb{F}}) \oplus H_d(x \ast X_x;  \ensuremath{\mathbb{F}}) \to H_d(X;  \ensuremath{\mathbb{F}}) \to H_{d-1}(X_x;  \ensuremath{\mathbb{F}})
\nonumber \\
  \to & H_{d-1}(X^x;  \ensuremath{\mathbb{F}}) \oplus H_{d-1}(x \ast X_x;  \ensuremath{\mathbb{F}}) \to H_{d-1}(X;  \ensuremath{\mathbb{F}}) \to \cdots \nonumber
\end{align}
that $H_d(X;  \ensuremath{\mathbb{F}}) =  \ensuremath{\mathbb{F}}$.
\end{proof}

Let $X$ be a simplicial complex of dimension $d$. In \cite{BDStellated} we have defined the sigma vector
$(\sigma_0, \sigma_1, \dots, \sigma_d)$ of $X$ with respect to a field $ \ensuremath{\mathbb{F}}$ as
\begin{align}
\sigma_i = \sigma_i(X;  \ensuremath{\mathbb{F}}) & := \sum_{A\subseteq V(X)} \tilde{\beta}_i(X[A])\left/\binom{f_0(X)}{\#(A)}\right.,
\, 0\leq i \leq d,  \nonumber
\end{align}
and $\sigma_i(X; \ensuremath{\mathbb{F}}) = 0$ for $i>\dim(X)$. In this paper, we introduce the
following normalization of the sigma vector which will be referred to as the {\em sigma-star} vector.
\begin{align}
\sigma_i^{\ast} = \sigma_i^{\ast}(X;  \ensuremath{\mathbb{F}}) & :=  \frac{\sigma_i(X;  \ensuremath{\mathbb{F}})}{1+f_0(X)},  \, \, i \geq 0. \nonumber
\end{align}
(See Theorem \ref{theorem:5.7} below for a justification of this normalization.) In \cite{BaMu} the first author
introduced the mu-vector $(\mu_0, \dots, \mu_d)$ w.r.t. $ \ensuremath{\mathbb{F}}$ of a $d$-dimensional simplicial complex $X$ as
follows.
\begin{align}
\mu_0 = \mu_0(X;  \ensuremath{\mathbb{F}}) & :=  \sum_{x\in V(X)} \frac{1}{1+f_0(X_x)},   \nonumber \\
\mu_i = \mu_i(X;  \ensuremath{\mathbb{F}}) & := \delta_{i1}\mu_0(X;  \ensuremath{\mathbb{F}}) +  \sum_{x\in V(X)} \sigma^{\ast}_{i-1}(X_x;  \ensuremath{\mathbb{F}}),  \, \, 1
\leq i \leq d. \nonumber
\end{align}
(Here $\delta_{i1}$ is Kronecker's symbol. Thus, $\delta_{i1}=1$ if $i=1$, and $=0$ otherwise.) Notice that
$\mu_1(X;  \ensuremath{\mathbb{F}})$ is independent of the field $ \ensuremath{\mathbb{F}}$. Therefore, we will write $\mu_1(X)$ for $\mu_1(X;  \ensuremath{\mathbb{F}})$. We
have\,:

\begin{lemma} \label{lemma:2.6}
Let $M$ be an $ \ensuremath{\mathbb{F}}$-orientable, neighbourly triangulated closed $3$-manifold. Then $\beta_1(M;  \ensuremath{\mathbb{F}}) \leq
\mu_1(M)$. Equality holds here if and only if $M$ is $ \ensuremath{\mathbb{F}}$-tight.
\end{lemma}

\begin{proof}
The inequality (as well as the fact that  equality holds if $M$ is $ \ensuremath{\mathbb{F}}$-tight) is a very special case of
\cite[Corollary 1.8]{BaMu}. Suppose $\mu_1 = \beta_1$. Since $M$ is 2-neighbourly, it also satisfies $\mu_0 = 1 =
\beta_0$. By \cite[Theorem 1.6]{BaMu}, $\mu_2 = \mu_1$ and $\mu_3=\mu_0$. Also, since $M$ is $ \ensuremath{\mathbb{F}}$-orientable,
Poincar\'{e} duality implies $\beta_2=\beta_1$ and $\beta_3 = \beta_0$. Therefore, $\mu_2=\beta_2$ and
$\mu_3=\beta_3$. Hence \cite[Corollary 1.8]{BaMu} implies that $M$ is $ \ensuremath{\mathbb{F}}$-tight.
\end{proof}

Recall that, a triangulated closed $d$-manifold $X$ is called {\em orientable} if $H_d(X;  \ensuremath{\mathbb{Z}}) \neq 0$. It
follows from the universal coefficient theorem that, for a field $ \ensuremath{\mathbb{F}}$ of odd characteristic, a triangulated
closed manifold is $ \ensuremath{\mathbb{F}}$-orientable if and only if it is orientable.

\begin{corollary} \label{coro:2.7}
Let $p$ be a  prime and let $M$ be an orientable triangulated closed $3$-manifold. If $M$ is $ \ensuremath{\mathbb{Z}}_p$-tight but
not $ \ensuremath{\mathbb{Q}}$-tight, then $p$ divides the order of the torsion subgroup of $H_1(M;  \ensuremath{\mathbb{Z}})$.
\end{corollary}

\begin{proof}
The hypothesis and Lemma \ref{lemma:2.6} imply that $\beta_1(M;  \ensuremath{\mathbb{Q}}) < \mu_1(M) = \beta_1 (M;  \ensuremath{\mathbb{Z}}_p)$. Hence the
result follows from the universal coefficient theorem.
\end{proof}

The following lemma is immediate from the definition of tightness.

\begin{lemma} \label{lemma:2.8}
{\rm (a)} A simplicial complex is tight w.r.t. a field of characteristic $p$ if and only if it is $ \ensuremath{\mathbb{Z}}_p$-tight.
{\rm (b)} A simplicial complex is tight w.r.t. a field of characteristic zero if and only if it is $ \ensuremath{\mathbb{Z}}_p$-tight
for all primes $p$.
\end{lemma}

For any non-empty finite set $\alpha$, $\overline{\alpha}$ will denote the simplicial complex whose faces are all
the subsets of $\alpha$. Thus, if $\#(\alpha) = d+1$, $\overline{\alpha}$ is the standard triangulation of the
$d$-ball (namely, it is the face complex of the geometric $d$-simplex). If $\#(\alpha) = d+2$, then the boundary
$\partial \overline{\alpha}$ of $\overline{\alpha}$ is the standard triangulation of the $d$-sphere; it will be
denoted by $S^{\,d}_{d + 2}$. (More generally, $S^{\,d}_n$ usually denotes an $n$-vertex triangulation of the
$d$-sphere.) From the definition of a stacked sphere, one can deduce the following (this also follows from
\cite[Lemmas 4.3 (b) \& 4.8 (b)]{BDLBT}).

\begin{lemma} \label{lemma:stacked_sphere}
A simplicial complex $S$ is a stacked $d$-sphere if and only if $S$ is a connected sum of finitely many copies of
the $(d+2)$-vertex standard sphere $S^{\,d}_{d + 2}$.
\end{lemma}

Let $X$ be a triangulated closed manifold and let $\sigma$ and $\tau$ be facets of $X$. For a bijection $\psi:
\sigma \to \tau$, let $X^\psi$ be the simplicial complex obtained from $X \setminus \{ \sigma, \tau\}$ by
identifying $v$ and $\psi(v)$ for $v \in \sigma$. If ${\rm lk}_X(v) \cap {\rm lk}_X(\psi(v))=\{\emptyset\}$ for
each vertex $v \in \sigma$, then $X^\psi$ is a triangulated manifold. If $\sigma$ and $\tau$ belong to different
connected components, say $\sigma\in X_1$, $\tau \in X_2$ and $X = X_1\sqcup X_2$, then $X^{\psi}$ is said to be the {\em connected sum} of
$X_1$ and $X_2$ and is denoted by $X_1\#_{\psi} X_2$. If $\sigma$ and $\tau$ belong to the same connected
component of $X$, then $X^\psi$ is said to be obtained from $X$ by a {\em combinatorial handle addition}. We know
from \cite{DM}

\begin{proposition}[Datta-Murai] \label{prop:Walkup_class}
Let $\Delta$ be a connected triangulated closed manifold of dimension $d\geq 2$. Then $\Delta$ is stacked if and
only if $\Delta$ can be obtained from a stacked $d$-sphere by a sequence of successive combinatorial handle
additions.
\end{proposition}

\section{Induced surfaces in tight triangulations}

Let $X_x$ and $X^x$ be as in Notation \ref{notation1}. If $x\neq y$ are two vertices of a simplicial complex $X$,
then we shall also use notations such as $X^x_y$ for $(X^x)_y = (X_y)^x$. We say that two vertices in a
simplicial complex $X$ are adjacent (or, that they are neighbours) if they form an edge of $X$. We now
introduce\,:

\begin{notation} \label{notation:3.1}
{\rm If $x\neq y$ are vertices of a simplicial complex $X$, then $c_X(x, y)$ will denote the number of distinct
connected components $K$ of $X^x_y$ such that $x$ is adjacent in $X_y$ with some vertex in $K$.  }
\end{notation}

\begin{lemma} \label{lemma:3.2}
If $X$ is an $ \ensuremath{\mathbb{F}}$-tight simplicial complex then for all $x\in V(X)$, we have $\beta_1(X;  \ensuremath{\mathbb{F}}) =
\tilde{\beta_0}(X_x;  \ensuremath{\mathbb{F}}) + \beta_1(X^x;  \ensuremath{\mathbb{F}})$.
\end{lemma}

\begin{proof}
Clearly, $X = X^x \cup (x \ast X_x)$ and $X^x \cap (x \ast X_x) = X_x$. Therefore, the Mayer-Vietoris theorem
yields the exact sequence (noting that the cone $x\ast X_x$ is homologically trivial)
\begin{align}
H_1(X^x;  \ensuremath{\mathbb{F}}) & \to H_1(X;  \ensuremath{\mathbb{F}}) \to \widetilde{H}_{0}(X_x;  \ensuremath{\mathbb{F}}) \to \widetilde{H}_{0}(X^x;  \ensuremath{\mathbb{F}}). \nonumber
\end{align}
Since $X^x$ is an induced subcomplex of the $ \ensuremath{\mathbb{F}}$-tight complex $X$, the map $H_1(X^x;  \ensuremath{\mathbb{F}}) \to H_1(X;  \ensuremath{\mathbb{F}})$ is
injective. Lemma \ref{lemma:2.2} implies that $X^x$ is connected and hence $\widetilde{H}_{0}(X^x;  \ensuremath{\mathbb{F}}) = 0$. So,
we get the short exact sequence
\begin{align}
0 \to & H_1(X^x;  \ensuremath{\mathbb{F}}) \to H_1(X;  \ensuremath{\mathbb{F}}) \to \widetilde{H}_{0}(X_x;  \ensuremath{\mathbb{F}}) \to 0. \nonumber
\end{align}
Hence the result.
\end{proof}

Lemmas \ref{lemma:2.2} and \ref{lemma:2.3} certainly indicate that tightness is a severe structural constraint on
a simplicial complex. So it is surprising that, beyond these two lemmas, no further structural (combinatorial)
consequence of tightness seems to have been known. The following lemma establishes a strong structural
restriction on the 2-skeleton of an $ \ensuremath{\mathbb{F}}$-tight simplicial complex.

\begin{lemma} \label{lemma:3.3}
Let $X$ be an $ \ensuremath{\mathbb{F}}$-tight simplicial complex for some field $ \ensuremath{\mathbb{F}}$. Then, for any two distinct vertices $x, y$ of
$X$, we have $c_X(x, y) = c_X(y, x)$.
\end{lemma}

\begin{proof}
By Lemma \ref{lemma:3.2}, $\beta_1(X) = \tilde{\beta_0}(X_x) + \beta_1(X^x)$. Since $X^x$ is also tight by Lemma
\ref{lemma:2.3}, applying Lemma \ref{lemma:3.2} to the vertex $y$ of $X^x$, we get $\beta_1(X^x) =
\tilde{\beta_0}(X_y^x) + \beta_1(X^{xy})$. Therefore, $\beta_1(X) = \tilde{\beta_0}(X_x) + \tilde{\beta_0}(X_y^x)
+ \beta_1(X^{xy})$. Interchanging the vertices $x$ and $y$ in this argument yields $\beta_1(X) =
\tilde{\beta_0}(X_y) + \tilde{\beta_0}(X_x^y) + \beta_1(X^{yx})$. Since $X^{xy} = X^{yx}$, we get
\begin{align}
\tilde{\beta_0}(X^x_y) & - \tilde{\beta_0}(X_y) = \tilde{\beta_0}(X_x^y) - \tilde{\beta_0}(X_x). \nonumber
\end{align}
But the two sides of this equation are just one less than $c_X(x, y)$ and $c_X(y, x)$. Hence the result.
\end{proof}

Notice that the {\em graphs} of graph theory are just the simplicial complexes of dimension $\leq 1$. In this
paper, we do not use any non-trivial results from graph theory, but the language and the geometric intuition of
graph theory will be useful. Recall that the {\em degree} of a vertex $v$ (denoted by $\deg(v)$) in a simplicial
complex is the number of edges (1-faces) through $v$. A graph is said to be {\em regular}, if all its vertices
have the same degree. For $n\geq 3$, the {\em cycle of length $n$} (in short, {\em $n$-cycle}) is the unique
connected regular graph of degree two on $n$ vertices. It is the unique $n$-vertex triangulation of the circle
$S^1$. An $n$-cycle with edges $a_1a_2, \dots, a_{n-1}a_n, a_na_1$ will be denoted by $a_1\mbox{-}
a_2\mbox{-} \cdots\mbox{-} a_n\mbox{-} a_1$. For $n\geq 1$, the {\em path of length $n$} (the $n$-path) is the
antistar of a vertex in the $(n+2)$-cycle.
By an {\em induced cycle} (resp., path) in a simplicial complex $X$, we mean an induced subcomplex of $X$
which is a cycle (resp., path). Notice that, in particular, a 3-cycle is induced in $X$ if and only if it does
not bound a triangle (2-face) in $X$. When $n \geq 4$, an $n$-cycle is induced in X if and only if it is induced
in the graph ${\rm skel}_1(X)$. A connected acyclic graph is called a {\em tree}.

\begin{lemma} \label{lemma:3.4}
Let the link of some vertex $x$ in a $2$-dimensional $ \ensuremath{\mathbb{F}}$-tight simplicial complex $X$ be a cycle. Then $X$ is a
triangulation of a closed $2$-manifold.
\end{lemma}

\begin{proof}
Let $C = X_x$ be a cycle. Fix a vertex $y\neq x$ of $X$. It suffices to show that the link $X_y$ is also a cycle.
Note that, since $X$ is neighbourly (Lemma \ref{lemma:2.2}), $y$ is a vertex of $C$. Let $z$ and $w$ be the two
neighbours of $y$ in $C$. It follows that $z$ and $w$ are the only two neighbours of the vertex $x$ in the graph
$X_y$. Therefore, it suffices to show that $X^x_y$ is a path joining $z$ and $w$.

Since $X_x = C$ is a cycle and $X^y_x = C^y$ is a path, they are both connected. So, $c_X(y, x) = 1$. Therefore,
by Lemma \ref{lemma:3.3}, $c_X(x, y) = 1$. That is, the vertices $z$ and $w$ (being the two neighbours of $x$ in
$X_y$) belong to the same component of $X^x_y$. Thus, there is a path in $X^x_y$ joining $z$ to $w$. Let $P$ be a
shortest path in the graph $X^x_y$ joining $z$ to $w$. Then, $P$ is an induced path in $X^x_y$. Take any vertex
$v \neq x, y, z, w$ in $X$. (If there is no such vertex then $X^x_y$ is the edge $zw$, and we are done.) We will
show that $v\in P$.
Look at the induced subcomplex $Y = X^v$ of $X$. Then $Y_x = X^v_x = C^v$ is a path in which $y$ is an interior
vertex. So, $Y^y_x = C^{vy}$ is the disjoint union of two paths. The vertex $z$ belongs to one of these two paths
and $w$ belongs to the other. Therefore, $c_Y(y, x) = 2$. By Lemma \ref{lemma:2.3}, $Y$ is also $ \ensuremath{\mathbb{F}}$-tight. So,
by Lemma \ref{lemma:3.3}, $c_Y(x, y) =2$. That is, the neighbours $z$ and $w$ of $x$ in $Y_y$ belong to different
components of $Y^x_y$. Therefore, $v$ belongs to the path $P$ (or else $P$ would be a path in $Y^x_y$ joining $z$
and $w$). Since $v \neq z, w$ was an arbitrary vertex of $Y^x_y$, this shows that the path $P$ is a spanning path
in $Y^x_y$ (i.e., it passes through all the vertices). Since $P$ is also an induced path in $Y^x_y$, it follows
that $Y^x_y = P$ is a path joining $z$ and $w$.
\end{proof}

Now, the following result is an easy consequence of Lemma \ref{lemma:3.4}. Notice that, in this result, there is
no restriction on the dimension of $M$, and $M$ need not be a triangulated manifold.

\begin{theorem} \label{theorem:3.5}
Let the simplicial complex $M$ be tight with respect to some field. Let $C$ be an induced cycle in the link of a
vertex $x$ in $M$. Then the induced subcomplex of $M$ on the vertex set of the cone $x\ast C$ is a triangulated
closed $2$-manifold.
\end{theorem}

\begin{proof}
Let $Y = M[V(C) \cup\{x\}]$ and let $X = {\rm skel}_2(Y)$. Then, by
Lemmas \ref{lemma:2.3} and \ref{lemma:2.4}, $X$ is $ \ensuremath{\mathbb{F}}$-tight. Clearly, $X$ is two dimensional and $X_x = C$ is
a cycle. Thus, by Lemma \ref{lemma:3.4}, $X$ is a triangulated closed 2-manifold. To complete the proof, it
suffices to show that $Y = X$, i.e., that $\dim(Y) =2$. Suppose, if possible, $\dim(Y) > 2$. Take a 3-face
$\alpha\in Y$. Then the induced subcomplex of $X= {\rm skel}_2(Y)$ on the vertex set $\alpha$ is a 4-vertex
triangulated 2-sphere. Since $X$ is a connected and triangulated closed 2-manifold, it follows that $X = S^2_4$,
and $V(X) = \alpha$. Thus $C$ is the 3-cycle on the vertex set $\alpha\setminus\{x\}$. But, the 2-face
$\alpha\setminus\{x\}$ is in $Y_x\subseteq M_x$. This contradicts the assumption that $C$ is an induced cycle in
the link $M_x$.
\end{proof}

\begin{corollary} \label{coro:3.6}
Let $S$ be the link of some vertex in an $ \ensuremath{\mathbb{F}}$-tight simplicial complex $M$.
\begin{enumerate}[{\rm (a)}]
\item If ${\rm char}( \ensuremath{\mathbb{F}}) = 2$, then $S$ has no induced cycle of length $\equiv 1$ $($mod $3)$.

\item If ${\rm char}( \ensuremath{\mathbb{F}}) \neq 2$, then $S$ has no induced cycle of length $\equiv 0, 1, 4, 5, 7, 8, 9$ or $10$
$($mod $12)$.
\end{enumerate}
\end{corollary}

\begin{proof}
Let $x\in V(M)$ and let $C$ be an induced $n$-cycle in $S = M_x$. By Theorem \ref{theorem:3.5}, the induced
subcomplex $X = M[V(C)\cup\{x\}]$ is an $(n+1)$-vertex triangulated closed 2-manifold. By Lemma \ref{lemma:2.2},
$X$ is neighbourly. So, it has $n+1$ vertices, $n(n+1)/2$ edges and hence $n(n+1)/3$ triangles. So, 3 divides
$n(n+1)$, i.e., $n \not \equiv 1$ (mod 3). This proves part (a).

If ${\rm char}( \ensuremath{\mathbb{F}}) \neq 2$ then, by Lemmas \ref{lemma:2.3} and \ref{lemma:2.5}, $X$ is an orientable
triangulated 2-manifold. So, its Euler characteristic $\chi(X) = (n+1)(6-n)/6$ is an even number. Thus, $n
\not\equiv 0, 1, 4, 5, 7, 8, 9$ or 10 (mod 12). This proves part (b).
\end{proof}

\section{Odd characteristic}

In this section, we prove that any triangulated $3$-manifold is tight with respect to a field of odd
characteristic if and only if it is neighbourly, orientable and stacked. For this, we first need some additional
preliminary results on $2$-sphere triangulations.

\begin{lemma} \label{lemma:4.1}
Let $S$ be a triangulation of $S^{\hspace{.15mm}2}$. If $S$ has no induced cycle of length $\leq 5$ then $S =
S^{\hspace{.15mm}2}_4$.
\end{lemma}

\begin{proof}
Let $x$ be a vertex of minimum degree in $S$. It is well known (and easy to prove) that the minimum degree of any
triangulation of $S^{\hspace{.15mm}2}$ is at most five. So, $\deg(x) = 3, 4$ or 5.

If $\deg(x) = 4$ or 5, then $S_x$ is a 4-cycle or a 5-cycle in $S$, and hence it is not induced. So there are
vertices $y, z$ in $S_x$ such that $yz$ is an edge in $S$ but not in $S_x$. So $x\mbox{-}y\mbox{-}z\mbox{-}x$ is
an induced 3-cycle in $S$, a contradiction. Thus, $\deg(x) = 3$. Say $S_x$ is the 3-cycle $x_1\mbox{-}x_2
\mbox{-}x_3 \mbox{-}x_1$. Since this 3-cycle is not an induced cycle in $S$, it follows that $x_1x_2x_3\in S$.
Then $(x\ast S_x) \cup \{x_1x_2x_3\}=S^{\hspace{.15mm}2}_4$. Since a triangulation of $S^{\hspace{.15mm}2}$ can
not be a proper subcomplex of another triangulation of $S^{\hspace{.15mm}2}$, it follows that
$S=S^{\hspace{.15mm}2}_4$.
\end{proof}

We now introduce\,:

\begin{definition} \label{def:primitive}
{\rm A triangulated $d$-sphere $S$ is said to be {\em primitive} if it can not be written as a connected sum of
two triangulated $d$-spheres.}
\end{definition}

Clearly, every triangulated $d$-sphere is a connected sum of finitely many primitive $d$-spheres. The following
lemma is more or less obvious.

\begin{lemma} \label{lemma:4.3}
Let $S$ be a triangulated $d$-sphere. Then $S$ is primitive if and only if $S$ has no induced subcomplex
isomorphic to $S^{\hspace{.15mm}d-1}_{d+1}$.
\end{lemma}

\begin{proof}
If $S$ is not primitive, then $S = S_1 \# S_2$, where $S_1$ and $S_2$ are triangulated $d$-spheres. Let $\alpha$
be the unique common facet of $S_1$ and $S_2$. Then the boundary of $\overline{\alpha}$ is an induced
$S^{\hspace{.15mm}d-1}_{d+1}$ in $S$.

Conversely, suppose $S$ has an induced $S^{\hspace{.15mm}d-1}_{d+1}$, say with vertex set $\alpha$. This
$S^{\hspace{.15mm}d-1}_{d+1}$ divides $S$ into two triangulated $d$-balls $B_1$ and $B_2$ such that $S = B_1\cup
B_2$, $\partial B_1 = \partial B_2 = B_1\cap B_2 = S^{\hspace{.15mm}d-1}_{d+1}$. Put $S_i = B_i\cup\{\alpha\}$,
$i=1,2$. Then $S_1$, $S_2$ are triangulated $d$-spheres and $S = S_1 \# S_2$. So, $S$ is not primitive.
\end{proof}

The following lemma and definition clarify the meaning of connected sums of several primitive triangulated
spheres.

\begin{notation} \label{notation:c_sum}
{\rm Let $S$ be a triangulation of $S^{\hspace{.15mm}d}$. Put $\mathcal{A}(S) := \{\alpha\subseteq V(S) \, : \,
S[\alpha] \cong S^{\hspace{.15mm}d-1}_{d+1}\}$, $\overline{S} := S \cup \mathcal{A}(S)$, and let $\mathcal{B}(S)$
be the collection of all the induced subcomplexes of the simplicial complex $\overline{S}$ which are primitive
triangulations of $S^{\hspace{.15mm}d}$. Let $\mathcal{T}(S)$ be the graph with vertex set $\mathcal{B}(S)$ such
that $S_1, S_2\in \mathcal{B}(S)$ are adjacent in $\mathcal{T}(S)$ if  $S_1 \cap S_2 = \overline{\alpha}$ for
some $\alpha \in \mathcal{A}(S)$. }
\end{notation}

\begin{lemma} \label{lemma:c_sum}
For any triangulated $d$-sphere $S$ we have the following.
\begin{enumerate}[{\rm (a)}]
\item Any two members of $\mathcal{B}(S)$ have at  most one common facet; if they have a common facet $\alpha$
then $\alpha \in \mathcal{A}(S)$. \item Each member of $\mathcal{A}(S)$ belongs to exactly two members of
$\mathcal{B}(S)$. \item There is a natural bijection from $\mathcal{A}(S)$ onto the set of edges of
$\mathcal{T}(S)$. It is given by $\alpha \mapsto \{S_1, S_2\}$, where, for $\alpha\in \mathcal{A}(S)$, $S_1$ and
$S_2$ are the two members of $\mathcal{B}(S)$ containing $\alpha$. \item The graph $\mathcal{T}(S)$ is a tree. In
consequence, $\#\mathcal{B}(S) = 1 + \#\mathcal{A}(S)$.
\end{enumerate}
\end{lemma}

\begin{proof}
Induction on $m : = 1 + \#\mathcal{A}(S)$. If $m=1$, $\mathcal{A}(S)$ is empty, so that $S$ is primitive by Lemma
\ref{lemma:4.3}. So, let $m >1$, and suppose that the result holds for all smaller values of $m$. In this case,
$\mathcal{A}(S)$ is non-empty. Take $\alpha\in\mathcal{A}(S)$. By the proof of Lemma \ref{lemma:4.3}, there are
triangulated $d$-spheres $S^{\prime}$, $S^{\prime\prime}$ such that $S^{\prime} \cap S^{\prime\prime} =
\overline{\alpha}$, $S = (S^{\prime} \cup S^{\prime\prime}) \setminus\{\alpha\}$. Thus, $S^{\prime},
S^{\prime\prime} \subseteq S \cup\{\alpha\} \subseteq \overline{S}$. Hence it is easy to see that $\mathcal{A}(S)
= \mathcal{A}(S^{\prime}) \sqcup \mathcal{A}(S^{\prime\prime}) \sqcup \{\alpha\}$, $\mathcal{B}(S) =
\mathcal{B}(S^{\prime}) \sqcup \mathcal{B}(S^{\prime\prime})$ and the graph $\mathcal{T}(S)$ is obtained from the
disjoint union of the graphs $\mathcal{T}(S^{\prime})$  and $\mathcal{T}(S^{\prime\prime})$ by adjoining a single
edge from a vertex of $\mathcal{T}(S^{\prime})$ to a vertex of $\mathcal{T}(S^{\prime\prime})$. Since, by
induction hypothesis, the result is valid for $S^{\prime}$ and $S^{\prime\prime}$, its validity for $S$ follows.
This completes the induction.
\end{proof}

\begin{definition} \label{def:c_sum}
{\rm Let $S$ be a triangulated $d$-sphere.  Any tree has a leaf (a vertex of degree one) and the deletion of a
leaf from a non-trivial tree leaves a subtree. Therefore, the members of $\mathcal{B}(S)$ may be ordered as $S_1,
\dots, S_m$ in such a way that, for each $i$, $1\leq i\leq m$, $S_i$ is a leaf of the induced subtree of
$\mathcal{T}(S)$ on $\{S_1, \dots, S_i\}$ (when $m\geq 2$, this ordering is not unique). For any such ordering,
we write $S = S_1\# S_2 \# \cdots \# S_m$, and say that $S$ is the {\em connected sum} of the $S_i$'s. Clearly,
we have $S_1 \# \cdots \# S_m = (S_1 \# \cdots \# S_{m-1}) \# S_m$. }
\end{definition}

As a special case of \cite[Theorem 8.5]{Ka}, Kalai proved a nice characterization of stacked 2-spheres. A
triangulated 2-sphere $S$ is stacked if and only if $S$ has no induced cycle of length $\geq 4$. The following
result is a dramatic improvement on this characterization.

\begin{theorem} \label{theorem:4.7}
Let $S$ be a triangulated $2$-sphere. Then $S$ is stacked if and only if it has no induced cycle of length $4$ or
$5$.
\end{theorem}

\begin{proof}
Write $S = S_1\# S_2\# \cdots\# S_k$, where the $S_i$'s are primitive 2-spheres.

If $S$ is stacked then, by Lemma \ref{lemma:stacked_sphere}, each $S_i$ is a copy of $S^{\hspace{.15mm}2}_4$. Let
$C$ be an induced cycle of length $\geq 4$ in $S$. Since an induced cycle of length $\geq 4$ in a connected sum
of two triangulated manifolds must be an induced cycle in one of the summands, it follows inductively that $C$ is
an induced cycle in one of the $S_i$'s. Since $S^{\hspace{.15mm}2}_4$ has no induced cycle at all, it follows
that $S$ has no induced cycle of length $\geq 4$. This proves the ``only if" part.

For the converse, assume that $S$ has no induced cycle of length 4 or 5. It follows that no $S_i$ has any induced
cycle of length 4 or 5. Being primitive, $S_i$ has no induced cycle of length 3 either (Lemma \ref{lemma:4.3}).
Thus, by Lemma \ref{lemma:4.1}, each $S_i$ is a copy of $S^{\hspace{.15mm}2}_4$. Therefore, by Lemma
\ref{lemma:stacked_sphere}, $S$ is stacked. This proves the ``if" part.
\end{proof}

Now, we are ready to prove one of the main results of this paper.

\begin{theorem} \label{theorem:4.8}
A triangulated closed $3$-manifold $M$ is tight with respect to some field $ \ensuremath{\mathbb{F}}$ with ${\rm char}( \ensuremath{\mathbb{F}}) \neq 2$ if
and only if $M$ is orientable, neighbourly and stacked.
\end{theorem}

\begin{proof}
The ``if" part follows from Proposition \ref{prop:BDTh2.24}. To prove the ``only if" part, let $M$ be an
$ \ensuremath{\mathbb{F}}$-tight triangulated closed 3-manifold, ${\rm char}( \ensuremath{\mathbb{F}}) \neq 2$. By Lemmas \ref{lemma:2.2} and
\ref{lemma:2.5}, $M$ is neighbourly and orientable. It remains to show that $M$ must be stacked.

Corollary \ref{coro:3.6} (b) shows that, for each vertex $x$ of $M$, the vertex link $M_x$ is a triangulated
2-sphere with no induced 4-cycle or 5-cycle. So, Theorem \ref{theorem:4.7} implies that each $M_x$ is a stacked
2-sphere. Thus, $M$ is locally stacked. The result now follows by Proposition \ref{prop:BDTh2.24}.
\end{proof}

\begin{corollary} \label{coro:4.9}
Let $M$ be a triangulated closed $3$-manifold. If $M$ is tight w.r.t. a field $ \ensuremath{\mathbb{F}}$, where ${\rm char}( \ensuremath{\mathbb{F}}) \neq
2$, then $M$ is strongly minimal.
\end{corollary}

\begin{proof}
This result follows from Theorem \ref{theorem:4.8} and Proposition \ref{prop:sminimal}.
\end{proof}

\section{Characteristic two}

By Lemma \ref{lemma:2.8}, a simplicial complex is tight with respect to a field of characteristic two if and only
if it is $ \ensuremath{\mathbb{Z}}_2$-tight. So, without loss of generality, we take $ \ensuremath{\mathbb{F}}=  \ensuremath{\mathbb{Z}}_2$ in this section. However, all the
results apply equally well to arbitrary fields of characteristic two. Here, we characterize the links of
$ \ensuremath{\mathbb{Z}}_2$-tight triangulated $3$-manifolds. This characterization is important since it leads to (a) severe
restrictions on the size and topology of such a triangulation (see Theorem \ref{theorem:5.1} and the tables at
the end of this section) and (b) a polynomial time algorithm to decide tightness of $3$-manifolds which is
described in detail in \cite{BBDSS}.

By Proposition \ref{prop:BDTh2.24}, all stacked and neighbourly triangulated closed 3-manifolds are
$ \ensuremath{\mathbb{Z}}_2$-tight. Here is a partial converse. (In case $M$ is orientable, this result follows from Corollary
\ref{coro:2.7}.)

\begin{theorem} \label{theorem:5.1}
Let $M$ be a $ \ensuremath{\mathbb{Z}}_2$-tight triangulated closed $3$-manifold. If the torsion subgroup of $H_1(M;  \ensuremath{\mathbb{Z}})$ is of odd
order $($possibly trivial$)$, then $M$ is stacked $($and neighbourly$)$.
\end{theorem}

\begin{proof}
If possible, assume that $M$ is not stacked. Then, by Proposition \ref{prop:BDTh2.24}, $M$ is not locally
stacked. So, there exists a vertex $v$ whose link $M_v$ is not a stacked 2-sphere. By Theorem \ref{theorem:4.7}
and Corollary \ref{coro:3.6} (a), $M_v$ has an induced cycle $C$ of length 5. Then, by Theorem \ref{theorem:3.5},
the induced subcomplex $N:= M[\{v\}\cup V(C)]$ of $M$ is a triangulated closed 2-manifold. Since $N$ is an
induced subcomplex of $M$, by Lemmas \ref{lemma:2.2} and \ref{lemma:2.3}, $N$ is neighbourly. Thus, $N$ is a
6-vertex neighbourly triangulated 2-manifold and hence is the 6-vertex triangulation $\mathbb{RP}^2_6$ of
$\mathbb{RP}^2$. Take a non-triangle $abc$ of $N = \mathbb{RP}^2_6$ (there are 10 of them). Then $\alpha= ab+bc+
ca$ can be viewed as an 1-cycle of $N$ with $ \ensuremath{\mathbb{Z}}_2$-coefficient as well as with $ \ensuremath{\mathbb{Z}}$-coefficient. In both view,
it is not a boundary. However $2\alpha$ is a boundary with $ \ensuremath{\mathbb{Z}}$-coefficient. Since the map $H_1(N;  \ensuremath{\mathbb{Z}}_2) \to
H_1(M;  \ensuremath{\mathbb{Z}}_2)$, induced by the inclusion map $N \hookrightarrow M$, is injective, it follows that $[\alpha] \neq
0$ as an element of $H_1(M;  \ensuremath{\mathbb{Z}}_2)$ and hence also as an element of $H_1(M;  \ensuremath{\mathbb{Z}})$. But, $2[\alpha] = 0$ in
$H_1(M;  \ensuremath{\mathbb{Z}})$. So, $[\alpha]$ is an element of order 2 in $H_1(M;  \ensuremath{\mathbb{Z}})$. This is a contradiction to the
assumption on $H_1(M;  \ensuremath{\mathbb{Z}})$.
\end{proof}

\begin{lemma} \label{lemma:5.2}
Let $S$ be a primitive triangulation of $S^{\hspace{.15mm}2}$. Suppose $S\neq S^{\hspace{.15mm}2}_4$ and $S$ has
no induced $4$-cycle. Then,
\begin{enumerate}[{\rm (a)}]
\item All vertex links of $S$ are induced cycles in $S$.
\item Any two adjacent vertices of $S$ have exactly two common neighbours in $S$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $x\in V(S)$. Since $S\neq S^{\hspace{.15mm}2}_4$ is primitive, Lemma \ref{lemma:4.3} implies that $S_x$ can
not be a 3-cycle. Let the vertices $y, z$ of the cycle $S_x$ be neighbours in $S$. Since $S$ is primitive, the
3-cycle $x\mbox{-}y\mbox{-}z\mbox{-} x$ can not be induced in $S$. So, the triangle $xyz$ is in $S$. Therefore,
$y$ and $z$ are neighbours in $S_x$. This implies that $S_x$ is induced in $S$. This proves part (a).

Let $x, y$ be adjacent vertices of $S$, and let $z$ be a common neighbour of $x$ and $y$. Then, as the 3-cycle
$x\mbox{-}y\mbox{-}z\mbox{-}x$ is not induced in $S$, the triangle $xyz$ is in $S$. So, $z$ is the third vertex
of one of the two triangles of $S$ through $xy$. Thus, $x$ and $y$ have exactly two common
neighbours. This proves part (b).
\end{proof}

\begin{lemma} \label{lemma:5.3}
Let $S$ be a primitive triangulation of $S^{\hspace{.15mm}2}$ such that $S$ has no induced cycle of length
$\equiv 1$ $($mod $3)$. Then, either $S= S^{\hspace{.15mm}2}_4$ or all the vertices of $S$ have degree $2$ $($mod
$3)$.
\end{lemma}

\begin{proof}
By assumption and Lemma \ref{lemma:5.2} (a), $V(S) = V_0 \sqcup V_2$, where $V_i$ consists of the vertices of
degree $i$ (mod 3), $i=0,2$. If possible, let $x\in V_0$ and $y\in V_2$ be such that $xy$ is an edge of $S$. By
Lemma \ref{lemma:5.2}, the cycles $S_x$ and $S_y$ are induced in $S$ (of length 0 (mod 3) and 2 (mod 3),
respectively) with exactly two common vertices, say $u$ and $v$. Then $u$ and $v$ are the two neighbours of $x$
in $S_y$ and of $y$ in $S_x$.

\smallskip

\noindent {\em Claim.} No vertex in
$V(S_x)\setminus V(S_y)$ is adjacent to any vertex in $V(S_y)\setminus V(S_x)$.

Indeed, if $a \in S_x$ is a neighbour of $b \in S_y$, then $S$ has the 4-cycle $x\mbox{-}a\mbox{-}b\mbox{-}y
\mbox{-}x$. Since $S$ has no induced 3-cycle or 4-cycle, it follows that one of the triangles $xya$ and $xab$ is
in $S$. Hence either $a$ or $b$ is in $S_x \cap S_y$. This proves the claim.

Therefore, if $C$ is the cycle obtained from $S_x\cup S_y$ by deleting the two vertices $x$, $y$ and the four
edges $xu$, $xv$, $yu$, $yv$, then $C$ is an induced cycle in $S$ of length $\equiv 0+2-4 \equiv 1$ (mod 3). This
is a contradiction. Therefore, no vertex in $V_0$ is adjacent to any vertex in $V_2$. Since $S$ is connected and
$V(S) = V_0\sqcup V_2$, it follows that $V_0 =\emptyset$ or $V_2 = \emptyset$. If $V_2 = \emptyset$ then the
degree of each vertex is 0 (mod 3) and hence $S$ has a vertex $z$ of degree 3. Since $S$ is primitive, $S_z$
bounds a triangle. Then, $S$ contains an $S^{\hspace{.15mm}2}_4$ and hence $S= S^{\hspace{.15mm}2}_4$. Otherwise,
$V(S) = V_2$.
\end{proof}

\begin{theorem} \label{theorem:5.4}
Up to isomorphism, $S^{\hspace{.15mm}2}_4$ and $I^2_{12}$ are the only two primitive triangulations of
$S^{\hspace{.15mm}2}$ with no induced cycle of length $\equiv 1$ $($mod $3)$.
\end{theorem}

\begin{proof}
Clearly, $S^{\hspace{.15mm}2}_4$ has no induced cycle whatsoever. It is easy to see that all the induced cycles
of $I^2_{12}$ are 5-cycles. (Indeed, these are precisely the twelve vertex links.) Thus, these two triangulations
of $S^{\hspace{.15mm}2}$ are primitive with no induced cycle of length 1 (mod 3).

Conversely, let $S$ be a primitive triangulation of $S^{\hspace{.15mm}2}$ with no induced cycle of length 1 (mod
3). Assume $S\neq S^{\hspace{.15mm}2}_4$. By Lemma \ref{lemma:5.3}, the minimum degree of the vertices of $S$ is
five. If all its vertices have degree 5, then $S= I^2_{12}$. So, suppose there is a vertex $u$ with $\deg(u) >
5$. By Lemma \ref{lemma:5.3}, $\deg(u) \geq 8$. So, $S_u$ is a cycle of length $\geq 8$. Therefore, we may choose
vertices $v_1, v_2, w_1, w_2$ in $S_u$ such that $v_1, v_2$ are at distance 2, $w_1, w_2$ are at a distance 2,
and ${\rm dist}(v_i, w_j) \geq 2$, $i, j = 1, 2$, where all the distances are graphical distances measured along
the cycle $S_u$. Let $D := (u\ast S_u) \cup (v_1\ast S_{v_1}) \cup (v_2\ast S_{v_2}) \cup (w_1\ast S_{w_1})\cup
(w_2\ast S_{w_2})$. Figure 1 is a ``picture" of $D$. The boundary of $D$ is the union of six paths in $S$ (drawn
as circular arcs; edges of $S$ are drawn as straight line segments). Since $S$ has no induced cycle of length
$\leq 4$, it is easy to see that these six paths intersect pairwise at most at common end points, and the six end
points marked in Fig. 1 are distinct vertices of $S$. Thus, the boundary of $D$ is a cycle and hence $D$ is a
disc.

\begin{center}
{\includegraphics[height=5cm]{disc.pdf}}

Figure 1: Disc $D$ in the proof of Theorem \ref{theorem:5.4}
\end{center}

Fix an index $i \in \{1, 2\}$. Consider the boundary $C_i$ of the disc $(u\ast S_u) \cup (v_i\ast S_{v_i}) \cup
(w_i\ast S_{w_i})$. Then the vertex set of $C_i$ is the union of the vertex sets of $S_u$, $S_{v_i}$, $S_{w_i}$
minus the three vertices $u, v_i, w_i$. By Lemma \ref{lemma:5.2} (b), $S_u$ and $S_{v_i}$ (as also $S_u$ and
$S_{w_i}$) have exactly two common vertices. Since $D$ is a disc, it follows that $S_{v_i}$ and $S_{w_i}$ have a
unique vertex (namely, u) in common. Also, by Lemma \ref{lemma:5.3}, each of $S_u$, $S_{v_i}$, $S_{w_i}$ have
length 2 (mod 3). Therefore, the inclusion exclusion principle shows that the length of $C_i$ is $\equiv 2+2+2 -
(2+2+1) -3 \equiv 1$ (mod 3). Therefore, $C_i$ is not an induced cycle of $S$. Thus, there is an edge $a_ib_i$ in
$S$ such that $a_i$ and $b_i$ are non-consecutive vertices in the cycle $C_i$. By the proof of Lemma
\ref{lemma:5.3}, no vertex of $S_{v_i}\setminus S_u$ is adjacent in $S$ with any vertex of $S_{u}\setminus
S_{v_i}$; also no vertex of $S_{w_i}\setminus S_u$ is adjacent in $S$ with any vertex of $S_{u}\setminus
S_{w_i}$. Therefore, $a_i \in A^{\hspace{-0.25mm}^\circ}_i$, $b_i\in B^{\hspace{-0.15mm}^\circ}_i$ (see Fig. 1).
That is, $a_i$ is an interior vertex of the path $A_i$ and $b_i$ is an interior vertex of the path $B_i$.

Now, consider the disc $D^{\prime}$ in $S$ complementary to the disc $D$ (i.e., $D^{\prime} := S[V(S) \setminus
\{u, v_1, v_2, w_1, w_2\}]$). Then, $a_1b_1$ and $a_2b_2$ are in $D^{\prime}$. Clearly, $a_1b_1$ separates
$A^{\hspace{-0.25mm}^\circ}_2$ from $B^{\hspace{-0.25mm}^\circ}_2$ in $D^{\prime}$ (i.e.,
$A^{\hspace{-0.25mm}^\circ}_2$ and $B^{\hspace{-0.25mm}^\circ}_2$ are in different connected components of
$|D^{\prime}|\setminus|a_1b_1|$). Therefore, the geometric edges $|a_1b_1|$ and $|a_2b_2|$ intersect at an
interior point. This contradicts the very definition of the geometric realization of a simplicial complex. Thus,
there is no vertex $u$ of $S$ with $\deg(u) > 5$. This completes the proof.
\end{proof}

\begin{corollary} \label{coro:5.5}
For triangulations $S$ of $S^{\hspace{.15mm}2}$, the following conditions are equivalent\,:
\begin{enumerate}[{\rm (a)}]
\item
$S$ has no induced cycle of length $1$ $($mod $3)$,
\item
all the induced cycles of $S$ have length $3$ and $5$, and
\item
$S$ is a connected sum of $S^{\hspace{.15mm}2}_4$'s and $I^2_{12}$'s.
\end{enumerate}
\end{corollary}
(The statement in (c) includes the possibility that $S$ is either icosian or stacked.)

\begin{proof}
Let $S$ be as in (c). It follows that the only possible induced cycles in the summands  are 5-cycles. Therefore,
the only induced cycles in $S$ are 3-cycles and 5-cycles. Thus, (c) $\Rightarrow$ (b). Trivially, (b)
$\Rightarrow$ (a).

Now suppose (a) holds. Write $S=S_1\# S_2\#\cdots\# S_k$, where each $S_i$ is primitive. Since $S$ has no induced
cycle of length 1 (mod 3), it follows that no $S_i$ has an induced cycle of length 1 (mod 3). Therefore, by
Theorem \ref{theorem:5.4}, each $S_i$ is $S^{\hspace{.15mm}2}_4$ or $I^2_{12}$. Hence $S$ is as in (c). Thus, (a)
$\Rightarrow$ (c).
\end{proof}

Now, we can prove the second main result of this paper.

\begin{theorem} \label{theorem:5.6}
Let $M$ be a $ \ensuremath{\mathbb{Z}}_2$-tight triangulation of a closed $3$-manifold. Then each vertex link of $M$ is a connected
sum of $S^{\hspace{.15mm}2}_4$'s and $I^2_{12}$'s.
\end{theorem}

\begin{proof}
This is now immediate from Corollary \ref{coro:3.6} (a) and Corollary \ref{coro:5.5}.
\end{proof}

The following result provides a recursive procedure for the computation of the sigma-star vector.

\begin{theorem} \label{theorem:5.7}
Let $X_1$ and $X_2$ be induced subcomplexes of a simplicial complex $X$ and $ \ensuremath{\mathbb{F}}$ be a field. Suppose $X =
X_1\cup X_2$ and $Y = X_1\cap X_2$. If $Y$ is $k$-neighbourly, $k\geq 2$, then $\sigma_i^{\ast}(X; \ensuremath{\mathbb{F}}) =
\sigma_i^{\ast}(X_1;  \ensuremath{\mathbb{F}}) + \sigma_i^{\ast}(X_2; \ensuremath{\mathbb{F}}) - \sigma_i^{\ast}(Y; \ensuremath{\mathbb{F}})$ for $0\leq i\leq k-2$.
\end{theorem}

\begin{proof}
Let us write $m_1 = f_0(X_1)$, $m_2 = f_0(X_2)$, $m = f_0(Y)$. Thus, $f_0(X) = m_1+m_2-m$. For notational
convenience, we write $\tilde{\beta}_i(A)$ for $\tilde{\beta}_i(X[A];  \ensuremath{\mathbb{F}})$,  $A\subseteq V(X)$. Note that any
subset $A$ of $V(X)$ can be uniquely written as $A = A_1\sqcup B \sqcup A_2$, where $A_1 \subseteq V(X_1)
\setminus V(Y)$, $B \subseteq V(Y)$ and $A_2 \subseteq V(X_2) \setminus V(Y)$. Since $Y$ is $k$-neighbourly, it
follows from the exactness of the Mayer-Vietoris sequence that
\begin{align*}
\tilde{\beta}_i(A_1\sqcup B\sqcup A_2) = \tilde{\beta}_i(A_1\sqcup B) + \tilde{\beta}_i(A_2\sqcup B) -
\tilde{\beta}_i(B) & \mbox{ for } 0\leq i\leq k-2.
\end{align*}
Therefore, we can compute

\begin{align*}
\sigma_i^{\ast}(X;  \ensuremath{\mathbb{F}}) & = \frac{1}{m_1+m_2-m+1}\sum_{A_1, A_2, B}\frac{\tilde{\beta}_i(A_1\sqcup B) +
\tilde{\beta}_i(A_2\sqcup B) -\tilde{\beta}_i(B)}{\binom{m_1+m_2-m}{\#(A_1\sqcup A_2\sqcup B)}} \\
&=  \frac{1}{m_1+m_2-m+1}\left[\sum_{A_1, B}\tilde{\beta}_i(A_1\sqcup B) \sum_{A_2}
\frac{1}{\binom{m_1+m_2-m}{\#(A_1\sqcup B) + \#(A_2)}} \right. \\
& \left. + \sum_{A_2, B}\tilde{\beta}_i(A_2\sqcup B) \sum_{A_1} \frac{1}{\binom{m_1+m_2-m}{\#(A_2\sqcup B)
+ \#(A_1)}} - \sum_{B}\tilde{\beta}_i(B) \sum_{A_1, A_2}
\frac{1}{\binom{m_1+m_2-m}{\#(B) + \#(A_1\sqcup A_2)}}\right]\\
&=  \frac{1}{m_1+m_2-m+1}\left[\sum_{k=0}^{m_1}\sum_{\stackrel{A_1, B}{\#(A_1\sqcup B)=k}}
\hspace{-6mm}\tilde{\beta}_i(A_1 \sqcup B) \sum_{\ell = 0}^{m_2-m}
\frac{\binom{m_2-m}{\ell}}{\binom{m_1+m_2-m}{\ell +k}} \right. \\
& \hspace{45mm} + \sum_{k=0}^{m_2}\sum_{\stackrel{A_2, B}{\#(A_2\sqcup B)=k}}
\hspace{-6mm}\tilde{\beta}_i(A_2 \sqcup B)
\sum_{\ell = 0}^{m_1-m} \frac{\binom{m_1-m}{\ell}}{\binom{m_1+m_2-m}{\ell +k}} \\
 & \hspace{50mm} \left. - \sum_{k=0}^{m}\sum_{\stackrel{B}{\#(B)=k}} \hspace{-3mm}\tilde{\beta}_i(B)
 \sum_{\ell = 0}^{m_1+m_2-2m} \frac{\binom{m_1+m_2-2m}{\ell}}{\binom{m_1+m_2-m}{\ell +k}}\right]\\
 %
&=  \frac{1}{m_1+1}\sum_{k=0}^{m_1}\sum_{\stackrel{A_1, B}{\#(A_1\sqcup B)=k}} \hspace{-5mm}
\frac{\tilde{\beta}_i(A_1\sqcup B)}{\binom{m_1}{k}}  + \frac{1}{m_2+1}\sum_{k=0}^{m_2}\sum_{\stackrel{A_2,
B}{\#(A_2\sqcup B)=k}} \hspace{-5mm}
\frac{\tilde{\beta}_i(A_2 \sqcup B)}{\binom{m_2}{k}} \\
 & \hspace{60mm} - \frac{1}{m+1}\sum_{k=0}^{m}\sum_{\stackrel{B}{\#(B)=k}} \hspace{-3mm}
 \frac{\tilde{\beta}_i(B)}{\binom{m}{k}}\\
& =  \sigma_i^{\ast}(X_1;  \ensuremath{\mathbb{F}}) + \sigma_i^{\ast}(X_2;  \ensuremath{\mathbb{F}}) - \sigma_i^{\ast}(Y;  \ensuremath{\mathbb{F}}).
 \end{align*}
\end{proof}

In the penultimate step of the above computation, we have used the following well known identity to compute the
inner sums. For any three non-negative integers $p, q, r$, we have
\begin{align}
\sum_{i=0}^p \frac{\binom{p}{i}}{\binom{p+q+r}{i+r}} = \frac{p+q+r+1}{q+r+1}\times\frac{1}{\binom{q+r}{r}}.
\nonumber
\end{align}

As a particular case of Theorem \ref{theorem:5.7}, we have a formula for the sigma-star vector of a connected
sum. (Since $\sigma_0^{\ast}(X;  \ensuremath{\mathbb{F}})$ is independent of the field $ \ensuremath{\mathbb{F}}$, we denote it by
$\sigma_0^{\ast}(X)$.)

\begin{corollary} \label{coro:5.8}
For any two triangulated $d$-spheres $S_1$, $S_2$ and any field $ \ensuremath{\mathbb{F}}$, we have
\begin{enumerate}[{\rm (a)}]
\item $\sigma_0^{\ast}(S_1\# S_2) = \sigma_0^{\ast}(S_1) + \sigma_0^{\ast}(S_2) + \frac{1}{d+2}$, and \item
$\sigma_i^{\ast}(S_1\# S_2;  \ensuremath{\mathbb{F}}) = \sigma_i^{\ast}(S_1;  \ensuremath{\mathbb{F}}) + \sigma_i^{\ast}(S_2;  \ensuremath{\mathbb{F}})$ for $1\leq i\leq d-2$.
\end{enumerate}
\end{corollary}

\begin{proof}
Let $S_1\cap S_2 = \overline{\alpha}$, where $\alpha$ is a $d$-face. Then $X : = S_1\# S_2$, $X_1 := S_1\setminus
\{\alpha\}$, $X_2 := S_2\setminus\{\alpha\}$, $Y = \partial\overline{\alpha} = S^{\hspace{.15mm}d-1}_{d +1}$
satisfy the hypothesis of Theorem \ref{theorem:5.7} (with $k=d$), and trivially
$\sigma_0^{\ast}(S^{\hspace{.15mm} d -1}_{d+1}) = -1/(d+2)$, $\sigma_i^{\ast}(S^{\hspace{.15mm}d-1}_{d+1}) = 0$
for $1\leq i\leq d-2$.
\end{proof}

\begin{notation} \label{notation:5.9}
{\rm For $k, \ell\geq 0$; $(k, \ell)\neq (0, 0)$, by $kI^2_{12}\# \ell S^{\hspace{.15mm}2}_4$ we denote a
triangulated $2$-sphere which can be written as a connected sum of $k+\ell$ triangulated 2-spheres, of which $k$
are copies of $I^2_{12}$ and the remaining $\ell$ are copies of $S^{\hspace{.15mm}2}_4$ $($in some order$)$. }
\end{notation}

\begin{corollary} \label{coro:5.10}
Let $k, \ell\geq 0$ and $(k, \ell) \neq (0,0)$. Then $\sigma_0^{\ast}(kI^2_{12}\# \ell S^{\hspace{.15mm}2}_4) =
\frac{617}{1716}k + \frac{1}{20}\ell - \frac{1}{4}$.
\end{corollary}

\begin{proof}
Trivially, $\sigma_0^{\ast}(S^{\hspace{.15mm}2}_{4}) = -1/5$. A computation shows that $\sigma_0^{\ast}
(I^{\hspace{.15mm}2}_{12}) = 47/429$. Thus the result holds when $k+\ell = 1$. The general result follows by an
induction on $k+\ell$, where the induction leap uses Corollary \ref{coro:5.8} (with $d=2$).
\end{proof}

Our next result lists a set of necessary conditions that a $ \ensuremath{\mathbb{Z}}_2$-tight triangulated 3-manifold must satisfy.
Note that the statement of the main result of \cite{BDSS} implies that the inequality in part (a) of Theorem
\ref{theorem:5.11} holds, more generally, for all triangulations of closed 3-manifolds. Equality holds in this
more general setting if and only if the triangulation is neighbourly and locally stacked. Therefore, Proposition
\ref{prop:BDTh2.24} implies that equality holds if and only if the triangulation is $ \ensuremath{\mathbb{Z}}_2$-tight and stacked.
Still we have included this inequality for the sake of completeness, and because its proof arises naturally in the
course of proving the rest of the theorem.

\begin{theorem} \label{theorem:5.11}
Let $M$ be a $ \ensuremath{\mathbb{Z}}_2$-tight triangulated closed $3$-manifold. Then the parameters $n :=f_0(M)$ and $\beta_1 :=
\beta_1(M; \ensuremath{\mathbb{Z}}_2)$ must satisfy the following.
\begin{enumerate}[{\rm (a)}]
\item $(n-4)(n-5) \equiv 20\beta_1$ $($mod $776)$. Also, $(n-4)(n-5) \geq 20\beta_1$, with equality if and only
if $M$ is stacked.
\item  $429(n-4)(n-5) - 776n\lfloor\frac{n-4}{9}\rfloor \leq 8580 \beta_1$. Equality holds here if and only if
each vertex link of $M$ is a triangulated $2$-sphere of the form
$\lfloor\frac{n-4}{9}\rfloor I^2_{12}\# (n-4-9\lfloor\frac{n-4}{9}\rfloor)S^{\hspace{.15mm}2}_4$.
\end{enumerate}
\end{theorem}

\begin{proof}
By Theorem \ref{theorem:5.6}, for each $x\in V(M)$ there are numbers $k(x)$ and $\ell(x)$ such that
\begin{align*}
M_x = k(x)I^2_{12} \# \ell(x) S^{\hspace{0.15mm}2}_4.
\end{align*}
Since $M$ is neighbourly by Lemma \ref{lemma:2.2}, equating the number of vertices in the two sides, we get $n-1
= 9k(x) + 3 + \ell(x)$. Hence $\ell(x) = n-4-9k(x)$, and we have $0\leq k(x)\leq \lfloor\frac{n-4}{9}\rfloor$.
Equality holds in the lower bound if and only if $M_x$ is stacked, and equality holds in the upper bound if and
only if $M_x$ is the connected sum of $\lfloor\frac{n-4}{9}\rfloor$ copies of $I^2_{12}$ and $n-4-
9\lfloor\frac{n-4}{9}\rfloor$ copies of $S^{\hspace{0.15mm}2}_4$. Let us put $k := \sum_{x\in V(M)} k(x)$. Then
$0\leq k\leq n\lfloor\frac{n-4}{9}\rfloor$. The lower bound holds with equality if and only if $M$ is locally
stacked (hence, by Proposition \ref{prop:BDTh2.24}, if and only if $M$ is stacked) and the upper bound holds with
equality if and only if each vertex link of $M$ is the connected sum of $\lfloor\frac{n-4}{9}\rfloor$ copies of
$I^2_{12}$ and $n-4- 9\lfloor\frac{n-4}{9}\rfloor$ copies of $S^{\hspace{0.15mm}2}_4$. Then, by Corollary
\ref{coro:5.10},
\begin{align} \label{eq:1}
\sigma_0^{\ast}(M_x) & = \frac{617}{1716}k(x) + \frac{1}{20}\ell(x) - \frac{1}{4}
= \frac{617}{1716}k(x) + \frac{1}{20}(n-4-9k(x)) - \frac{1}{4} \nonumber \\
& = \frac{1}{20}n-\frac{194}{2145}k(x) - \frac{9}{20}.
\end{align}
Now, Lemma \ref{lemma:2.6} implies $\beta_1 = \mu_1(M) := 1 + \sum_x\sigma_0^{\ast}(M_x)$. Therefore, adding
\eqref{eq:1} over all $x\in V(M)$, we get
\begin{align} \label{eq:2}
429((n-4)(n-5)-20\beta_1)= 776k.
\end{align}
Since 776 is relatively prime to 429, the result follows from \eqref{eq:2} and the above discussion on the
bounds on $k$.
\end{proof}

\begin{corollary}[An upper bound theorem for $ \ensuremath{\mathbb{Z}}_2$-tight 3-manifolds] \label{coro:5.12}
Let $M$ be a $ \ensuremath{\mathbb{Z}}_2$-tight closed $3$-manifold with $n :=f_0(M)$, $\beta_1 := \beta_1(M; \ensuremath{\mathbb{Z}}_2)$. Then $(n-
4)(617n -3861) \leq 15444\beta_1$. Equality holds here if and only if $M$ is locally icosian.
\end{corollary}

\begin{proof}
Since $\lfloor\frac{n-4}{9} \rfloor \leq \frac{n-4}{9}$, Theorem \ref{theorem:5.11} (b) implies that $429(n-4)(n
-5) - 776n(n-4)/9 \leq 8580\beta_1$. This simplifies to the given inequality. Clearly, equality holds here if and
only if $n \equiv 4$ (mod 9) and equality holds in Theorem \ref{theorem:5.11} (b).
\end{proof}

\begin{corollary} \label{coro:5.13}
Let $M$ be a $ \ensuremath{\mathbb{Z}}_2$-tight triangulated closed $3$ manifold. If $f_0(M) \leq 71$ or $\beta_1(M; \ensuremath{\mathbb{Z}}_2) \leq 188$,
then  $M$ is stacked.
\end{corollary}

\begin{proof}
Suppose, if possible, $M$ is not stacked. Then, by Theorem \ref{theorem:5.11} (a), there is an integer $\ell \geq
1$ such that $ (n-4)(n- 5) = 776\ell +20\beta_1$. Hence Corollary \ref{coro:5.12} implies
\begin{align*}
15444(n-4)(n-5) \geq 15444\times 776\ell + 20(617n-3861)(n-4).
\end{align*}
This inequality simplifies to
\begin{align} \label{eq:3}
(n-2)^2 \geq 3861\ell +4.
\end{align}

\noindent {\em Case 1.} $n \leq 71$. Hence, by \eqref{eq:3}, $3861\ell + 4 \leq (71-2)^2 = 4761$. Thus, $\ell=1$.
Therefore, $(n-4)(n-5) = 776 +20\beta_1$ and $(n-2)^2 \geq 3861+4 > 62^2$. Thus, $n\geq 65$. But, $(n-4)(n-5)
\equiv 776 \equiv -4$ (mod 20). Hence $n \equiv 12$ or 17 (mod 20). But, $n\geq 65$. So, $n\geq 72$, a
contradiction. So, the result is true in this case.

\noindent {\em Case 2.} $\beta_1(M; \ensuremath{\mathbb{Z}}_2) \leq 188$. Then, by Corollary \ref{coro:5.12}, $n\leq 73$. Hence
\eqref{eq:3} yields $\ell=1$. Thus $(n-4)(n-5) = 776 + 20 \beta_1$, and therefore $n$ is congruent to 12 or 17
(mod 20). Hence $n \leq 72$, and if $n=72$ then $\beta_1 = 189$. Therefore, $n\leq 71$. The result now follows by
Case 1.
\end{proof}

\begin{corollary} \label{coro:5.14}
Let $M$ be a closed topological $3$-manifold admitting a tight triangulation. If $\beta_1 (M,\mathbb{Z}_2) <
189$, then $M$ is homeomorphic to one of the following manifolds
\begin{align*}
S^3, \, (S^2 \times S^1)^{\#k}, \, (\TPSS)^{\#k},
\end{align*}
where $k = 1, 12, 19, 21, 30, 63, 78, 82, 99, 154, 177$ or $183$.
\end{corollary}

\begin{proof}
Let $X$ be an $n$-vertex tight triangulation of $M$. Then, by Corollary \ref{coro:5.13}, Theorem
\ref{theorem:5.11} (a) and Proposition \ref{prop:Walkup_class}, $M$ is homeomorphic to $S^3$, $(S^2 \times
S^1)^{\#k}$ or $(\TPSS)^{\#k}$, where $(n-4)(n-5) = 20k$. The result follows from this.
\end{proof}

The following table gives a list of small values for the parameters $(n, \beta_1)$ of a locally icosian
$ \ensuremath{\mathbb{Z}}_2$-tight triangulation of a closed 3-manifold. Indeed, the number $n$ of vertices in any such triangulation
must be congruent modulo 15444 to one of the eight values of $n$ listed in this table.

\begin{table}[htbp]
\centering {\small
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
&&&&&&&&\\[-3mm]
$n$ & 1408 & 3865 & 5269 & 8320 &9724&12181&13585& 15448\\[1mm]
\hline
&&&&&&&&\\[-3mm]
$\beta_1$ & 78625 & 595186 & 1106970 &2762081 &3773610&5922778&7367441&9527555 \\[1mm]
\hline
\end{tabular}
} \caption{Small feasible parameters for locally icosian tight 3-manifolds} \label{tbl:icsian.3mfd}
\end{table}

The following tables list the small values for parameters $(n, \beta_1)$ satisfying the conclusion of Theorem
\ref{theorem:5.11}. Table \ref{tbl:Z2tight.3mfd} for $(n, \beta_1)$ with strict inequality in part (b) of the
theorem. Table \ref{tbl:Z2tight.3mfd.c=} for $(n, \beta_1)$ with equality in part (c) of the theorem.
\begin{table}[htbp]
\centering {\small
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
&&&&&&&&&&&&&&&\\[-3mm]
$n$ & 72 & 77 & 92 & 96 &97&101&108& 112&113&116&117&121&128&132&133\\[1mm]
\hline
&&&&&&&&&&&&&&&\\[-3mm]
$\beta_1$ &189&224&344&341&389&388&458&539&511&544&594&601&685&774&748\\[1mm]
\hline
\end{tabular}
} \caption{Small feasible parameters for non-stacked $ \ensuremath{\mathbb{Z}}_2$-tight 3-manifolds} \label{tbl:Z2tight.3mfd}
\end{table}
\begin{table}[ht]
\centering {\small
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
&&&&&&&&&&\\[-3mm]
$n$ &825&1296&1408&1760&1881&1989&2145&2580&3168&3276\\[1mm]
\hline
&&&&&&&&&&\\[-3mm]
$\beta_1$ &26871&66637&78625&123049&140677&157336&183109&264924&399817&427582\\[1mm]
\hline
\end{tabular}
} \caption{Small feasible parameters for $ \ensuremath{\mathbb{Z}}_2$-tight 3-manifolds (with equality in Th. \ref{theorem:5.11} (b))}
\label{tbl:Z2tight.3mfd.c=}
\end{table}

\section{Examples}

In this section, we present some examples which help to put the results of this paper in proper perspective.

The first example shows that a strongly minimal and stacked triangulated closed 3-manifold need not be tight.
So, the converse of Proposition \ref{prop:sminimal} is not true.

\begin{example}[Walkup \cite{Wa}] \label{exam:Walkup}
{\rm Let $\mathcal{J}$ be the pure 4-dimensional simplicial complex with vertex set $ \ensuremath{\mathbb{Z}}_{10} =  \ensuremath{\mathbb{Z}}/10 \ensuremath{\mathbb{Z}}$ and an
automorphism $i\mapsto i+1$ (mod 10). Modulo this automorphism, there is only one representative facet (maximal
face) of $\mathcal{J}$, namely $12345$. The face vector of $\mathcal{J}$ is $(10, 40, 60, 40, 10)$. Each vertex
link of $\mathcal{J}$ is an 8-vertex stacked 3-ball and hence $\mathcal{J}$ is a locally stacked triangulated
4-manifold with boundary. The boundary $\mathcal{K} = \partial\mathcal{J}$ has face vector $(10, 40, 60, 30)$ and
triangulates $S^{\hspace{.15mm}2} \times S^1$. Since each 2-simplex of $\mathcal{J}$ is in the boundary,
$\mathcal{J}$ is stacked and hence $\mathcal{K}$ is a stacked triangulated closed 3-manifold. Since $\mathcal{K}$
is not neighbourly, it is not tight with respect to any field.

Let $Y$ be a triangulation of $S^{\hspace{.15mm}2} \times S^1$. Since the only closed $3$-manifolds with at most
$9$ vertices are $S^{\hspace{.15mm}3}$ and $\TPSS$ (see \cite{BDSbundle}), it follows that $f_0(Y) \geq 10$.
Then, by \cite[Theorem 5.2]{NS}, $f_1(Y) \geq 4f_0(Y) \geq 40$. Since $f_0(Y)- f_1(Y) +f_2(Y) -f_3(Y) = 0$ and
$2f_2(Y) = 4f_3(Y)$, it follows that $f_3(Y) = f_1(Y) -f_0(Y) \geq 4f_0(Y) - f_0(Y) = 3f_0(Y)\geq 30$. Hence
$f_2(Y) = 2f_3(Y) \geq 60$. Thus, $f_i(Y) \geq f_i(\mathcal{K})$ for $0\leq i\leq 3$. Therefore, $\mathcal{K}$ is
strongly minimal.}
\end{example}

The following example shows that a neighbourly and locally stacked triangulated closed 3-manifold need not be
stacked. Equivalently, it need not be tight. Thus, the hypothesis `stacked' in Theorem \ref{theorem:4.8} can not
be relaxed to `locally stacked'.

\begin{example}[Lutz \cite{LuThesis}] \label{exam:Lutz}
{\rm Let $\mathcal{L}$ be the pure 3-dimensional simplicial complex with vertex set $ \ensuremath{\mathbb{Z}}_{10} =  \ensuremath{\mathbb{Z}}/10 \ensuremath{\mathbb{Z}}$ and an
automorphism $i\mapsto i+1$ (mod 10). Modulo this automorphism, a set of representative facets of $\mathcal{L}$
are:
\begin{align}
1236, \, 1237, \, 1257, \, 1368. \nonumber
\end{align}
Then, each vertex link is a 9-vertex stacked 2-sphere and hence $\mathcal{L}$ is a locally stacked, neighbourly,
triangulated closed 3-manifold. It triangulates $S^2 \times S^1$ and hence $\beta_1(\mathcal{L};  \ensuremath{\mathbb{F}}) = 1$ for
any field $ \ensuremath{\mathbb{F}}$. Therefore, by Proposition \ref{prop:BDTh2.24}, $\mathcal{L}$ is not stacked and not tight w.r.t.
any field. Since there are non-neighbourly 10-vertex triangulations of $S^2 \times S^1$ (cf. $\mathcal{K}$ in
Example \ref{exam:Walkup}), $\mathcal{L}$ is not strongly minimal. By Propositions \ref{prop:sminimal} and
\ref{prop:BDTh2.24}, this also shows that $\mathcal{L}$ is not stacked and is not $ \ensuremath{\mathbb{F}}$-tight for any field
$ \ensuremath{\mathbb{F}}$. }
\end{example}

The following example shows that an arbitrary $ \ensuremath{\mathbb{F}}$-tight simplicial complex need not be strongly minimal (we do
not know if it must be minimal in the sense of having the fewest number of vertices among all triangulations of
its geometric carrier).

\begin{example} \label{exam:nsmtight}
{\rm Consider the neighbourly $2$-dimensional simplicial complex $X$ on the vertex set $\{1, 2, 3, 4\}$ whose
maximal faces are $123$, $234$ and $14$. It is easy to see that $X$ is $ \ensuremath{\mathbb{F}}$-tight for any field $ \ensuremath{\mathbb{F}}$. Observe
that $|X|$ is also triangulated by the simplicial complex $Z$ (on the same vertex set) whose maximal faces are
$123$, $14$, $24$. Thus, $X$ is not strongly minimal. Therefore, Conjecture \ref{conj:KL} is not true for
arbitrary simplicial complexes. }
\end{example}

Recall that a $d$-dimensional simplicial complex is a {\em pseudomanifold} if (i) each maximal face is
$d$-dimensional, (ii) each $(d-1)$-face is in at most two $d$-faces, and (iii) for any two $d$-simplices $\alpha$
and $\beta$, there exists a sequence $\alpha=\alpha_1, \dots, \alpha_m=\beta$ of $d$-simplices such that
$\alpha_i \cap\alpha_{i+1}$ is a $(d-1)$-face for $1\leq i\leq m-1$. We now extend the definitions of stackedness
and locally stackedness to pseudomanifolds as follows.

\begin{definition} \label{def:st-pmfd}
{\rm A pseudomanifold $Q$ of dimension $d+1$ is said to be {\em stacked} if all its faces of codimension (at
least) two are in the boundary $\partial Q$. A pseudomanifold $P$ without boundary of dimension $d$ is said to be
{\em stacked} if there is a stacked pseudomanifold $Q$ of dimension $d+1$ such that $P = \partial Q$. A
pseudomanifold is said to be {\em locally stacked} if all its vertex links are stacked pseudomanifolds (with or
without boundaries).}
\end{definition}

\begin{example}[Emch \cite{Em}] \label{exam:3-pmfd}
{\rm Consider the $3$-dimensional pseudomanifold $\mathcal{P}$ with vertex set $\{1, \dots, 8\}$ and the
facet-transitive automorphism group ${\rm PGL}(2,7) = \langle (12345678), (132645),$ $(16)(23)(45)(78)\rangle$.
Modulo this group, a facet representative is $1235$. The link of each vertex is isomorphic to the 7-vertex torus
$T^2_7$. Its face vector is $(8, 28, 56, 28)$.

Since $\mathcal{P}$ is a $3$-neighbourly pseudomanifold with Euler characteristic $8$, it follows that its
integral homologies are torsion free and its vector of Betti numbers is $(1,0,8,1)$. Using its doubly transitive
automorphism group, one may calculate easily that the sigma vector of $T^2_7$ (w.r.t. any field) is $(-1,8,1)$.
Hence the mu-vector of $\mathcal{P}$ (w.r.t. any field) also equals $(1,0,8,1)$. Therefore, \cite[Theorem
1.6]{BaMu} implies that $\mathcal{P}$ is tight w.r.t. all fields.

Now, if $Q$ is a pseudomanifold of dimension 4 such that $\partial Q = \mathcal{P}$ and ${\rm skel}_2(Q) = {\rm
skel}_2(\mathcal{P})$ then for any vertex $v$ of $Q$, we have $\partial(Q_v) = \mathcal{P}_v$ and ${\rm
skel}_1(Q_v) = {\rm skel}_1(\mathcal{P}_v)$. Thus, $Q_v$ is a stacked 3-manifold whose boundary is the 7-vertex
torus $P_v$. But it is easy to see that the 7-vertex torus bounds exactly three (distinct but isomorphic) stacked
3-manifolds, each with seven 3-faces. This implies that $f_4(v\ast Q_v) = f_3(Q_v) = 7$. Then, $f_4(Q) = (8\times
7)/5$, which is not possible. Therefore, $\mathcal{P}$ is not stacked. Thus, Theorems \ref{theorem:4.8} and
\ref{theorem:5.1} are not true for $3$-dimensional pseudomanifolds.

}
\end{example}

The next example disproves a putative generalization of Theorem \ref{theorem:3.5} in which the induced cycle in
the hypothesis is replaced by arbitrary manifolds without boundary.

\begin{example} \label{exam:end}
{\rm Consider the $6$-vertex triangulation $\mathbb{RP}^2_6$ of $\mathbb{RP}^2$ whose facets are $123$, $134$,
$145$, $156$, $126$, $235$, $245$, $246$, $346$, $356$. Let $X = (7\ast \mathbb{RP}^2_6)\cup\{123456\}$. So, $X$
is a 5-dimensional simplicial complex obtained from the cone $7\ast \mathbb{RP}^2_6$ by adding a 5-face. The
simplicial complex $X$ is isomorphic to an induced subcomplex of the 13-vertex triangulation $M$ of $SU(3)/SO(3)$
obtained by Lutz in \cite{LuThesis}. (Indeed, for each vertex $v$ of $M$, $M_v$ contains two induced
$\mathbb{RP}^2_6$'s, say $M_v[A]$ and $M_v[B]$, where $V(M_v) = A\sqcup B$. Both the induced subcomplexes
$M[\{v\}\cup A]$ and $M[\{v\}\cup B]$ are isomorphic to $X$.) Thus, $M$ is $ \ensuremath{\mathbb{Z}}_2$-tight \cite{KL}, $M_v[A] =
\mathbb{RP}^2_6$ is a surface and yet the induced subcomplex $X= M[\{v\}\cup A]$ is not a pseudomanifold. }
\end{example}

\medskip

\noindent {\bf Acknowledgements:} The second and third authors are supported by DIICCSRTE, Australia and DST,
India, under the Australia-India Strategic Research Fund (project \linebreak AISRF06660). The second author is
also supported by the UGC Centre for Advanced Studies.

{\small

redpajama_set_name: RedPajamaArXiv


As many as 50 Tory MPs are being targeted in a coordinated putsch by grassroots activists in a bid to force a no-deal Brexit.
Eurosceptic members of local Conservative associations across the country are connecting via WhatsApp, the encrypted messaging service, in order to orchestrate deselection attempts of their MPs in the hope of replacing them with a hardline Brexiteer.
Efforts are being organised at a time when many Tory associations are holding their Annual General Meetings, with several MPs facing impending votes of no-confidence.
"It's definitely being coordinated," a former Tory minister told i. "I have members raising issues in an attempt to force a no confidence vote that have been successful in other local associations. They are sharing info with each other.
Another ex-Cabinet member who is facing a no confidence vote described it as "very unpleasant". "It is a shame as I am very fond of my association, but Brexit has become so divisive," the Tory said.
While no confidence votes do not automatically lead to an MP being deselected, activists believe they can use a successful confidence bid to force their MP out.
Several Conservatives including Cabinet members are being targeted, such as Greg Clark, who saw a failed no confidence attempt staged against him in Tunbridge Wells last week.
The former universities minister Sam Gyimah is being threatened with a vote by his South Surrey constituency on Friday and Dominic Grieve will face a vote a week later.
It comes just days after Nick Boles, the MP for Grantham and Stamford, quit his local party following a series of run ins with local members who want a harder Brexit.
Mr Boles, who will retain the party whip, said the UK's departure from the EU had "caused the final disintegration of my relationship with my local Conservative association".
i has been told as many as 50 Tory MPs opposed to a no-deal Brexit were at risk of being ousted by their local associations.
Other names believed to be targeted are Philip Lee, MP for Bracknell, George Freeman, who represents Mid-Norfolk, Antoinette Sandbach MP for Eddisbury, Orpington's MP Jo Johnson, Energy Minister Claire Perry and Work and Pensions Secretary Amber Rudd.
It led one hardline Brexiteer to call for a general election.
"If we could oust those soft Brexit and Remain MPs it would change the complexion of the House and we could force through Brexit," the source said.
MPs blamed the pro-Brexit campaign group Leave.EU and in particular Aaron Banks, the former Ukip donor, who has called for former Ukip members to join the Tories.
Leave.EU's website lists Tory MPs opposed to a no-deal Brexit and when their AGMs will be taking place as part of an attempt to "deselect the Tory traitors".
Another backbencher complained that CCHQ was not doing enough to support its MPs against the entryism.
"I've spoken to [Conservative Party Chairman] Brandon Lewis and with the CCHQ but they are not taking this seriously. The party is being shifted further to the right but they are not taking any action," the source said.


text:
import pickle, gzip, os, numpy as np
from itertools import product

from utils import is_verb
from __config__.filepath import AUX_DIR

# Hyperparameters

dataset = 'multicore-5'
D = 400  # Number of dimensions for nouns and verbs (separately)
C_range = [30, 40]
seed_range = [32, 8, 91, 64, 97]

a_range = [0.75, 0.8, 0.9, 1]  # Power that frequencies are raised to under the null hypothesis
k_range = [0]

# Load files

with open(os.path.join(AUX_DIR, '{}-vocab.pkl'.format(dataset)), 'rb') as f:
    pred_name = pickle.load(f)
verb = np.array([is_verb(p) for p in pred_name], dtype='bool')

full_template = os.path.join(AUX_DIR, 'simplevec_card', '{}-{}-{}-{}-{}.pkl.gz')

# Calculate PPMI
def ppmi(vec, a, C):
    # Smooth frequencies
    freq_pred = vec.sum(1) ** a
    freq_context = vec.sum(0) ** a

    # Normalise the smoothed marginal context probabilities (for noun and verb contexts separately)
    freq_context[:D] /= freq_context[:D].sum()
    freq_context[D:] /= freq_context[D:].sum()

    # Calculate joint probabilities (for nouns and verbs separately)
    freq_noun = vec * verb.reshape((-1,1))
    freq_verb = vec * np.invert(verb).reshape((-1,1))
    new = freq_noun / freq_noun.sum() + freq_verb / freq_verb.sum()

    # Take logs
    log_pred = np.log(freq_pred / freq_pred.sum())
    log_context = np.log(freq_context / freq_context.sum() * C)
    new = np.log(new * C)
    # The joint probabilities sum to 1 over just nouns or verbs
    # The marginal probabilities sum to 1 over all preds
    # This will lead to boosted scores

    # Subtract logs
    new -= log_context
    new -= log_pred.reshape((-1,1))

    # Keep positive
    new.clip(0, out=vec)
    return new

for C, seed in product(C_range, seed_range):
    print('C', C, 'seed', seed)

    with gzip.open(full_template.format(dataset, D, C, 'full', seed), 'rb') as f:
        count_vec = pickle.load(f)

    template = full_template.format(dataset, D, C, '{}-{}', seed)

    for a in a_range:
        print('a', a)
        vec = ppmi(count_vec, a, C)

        # Shift and save

        for k in k_range:
            filename = template.format(str(k).replace('.',''),
                                       str(a).replace('.',''))
            if os.path.exists(filename): continue
            print('k', k)
            new = (vec - k).clip(0)
            with gzip.open(filename, 'wb') as f:
                pickle.dump(new, f)

redpajama_set_name: RedPajamaGithub


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