TITLE: Solve for $z$ in $3z + \dfrac{z}{1 + i} = 10 - 4i$
QUESTION [0 upvotes]: $3z + \dfrac{z}{1 + i} = 10 - 4i$
Can someone please tell me how to solve this problem? I've tried multiplying both sides by $1+i$ but I just get a really complex equation. Please help!

REPLY [0 votes]: Multiply both sides by $(1+i)$ then you have
$$3z(1+i)+z=(10-4i)(1+i)=10+10i-4i+4=14+6i$$ so $$z(3+3i+1)=z(4+3i)=14+6i$$ and thus $$z=\frac{14+6i}{4+3i}=\frac{14+6i}{4+3i}\times\frac{4-3i}{4-3i}=\frac{74-18i}{25}.$$

REPLY [0 votes]: $\frac{1}{1+i}$ is just a constant, so you have: $$z(3+\frac{1}{1+i})=10-4i\implies z=\frac{10-4i}{3+\frac{1}{1+i}}$$
Use that $$\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(a+bi)(c-di)}{c^2+d^2}$$ to rationalise the complex fractions

REPLY [0 votes]: We have that
$$3z + \dfrac{z}{1 + i} = 10 - 4i \iff 3\bar z + \dfrac{\bar z}{1 - i} = 10 + 4i$$
and summing/subtracting up
$$
\begin{cases}
6\Re(z)+\Re(z)-\Im(z)=20\\\\
6i\Im(z)+i\Im(z)-i\Re(z)=-8i
\end{cases}
\implies
\begin{cases}
7x-y=20\\\\
7y-x=-8
\end{cases}
\implies x=\frac{74}{25},\, y=-\frac{18}{25}$$


TITLE: What is the standard deviation of the sample average (sample size = 16) from a Laplace(0,1) population?
QUESTION [1 upvotes]: I've been running some simulations, and it seems clear to me that the Sample Mean $\overline{X}$ from a Laplace$(0,1)$ population is distributed normal with mean $0$.  But I need to come up with the standard deviation and I'm stumped.
Specifically, I am looking at the sample mean (from a sample of size 16) from the Laplace(0,1) population.  How could I find the standard deviation?

REPLY [0 votes]: Suppose that $X_1,X_2,\dots,X_n$ are independent random variables, each with Laplace distribution, parameters $0$ and $1$. Let $Y=X_1+\cdots +X_n$. It looks as if you want the standard deviation of $\frac{Y}{n}$. We calculate the variance.
The random variable $Y$ has $\chi^2$ distribution. However, we do not need to know that to calculate the variance. By independence, the variance of $Y$ is the sum of the variances of the $X_i$. It follows that the variance of $\frac{Y}{n}$ is $\frac{1}{n}$ times the variance of (any) $X_i$.
For the variance of $X_1$, since the mean is $0$, we need to calculate
$$\int_{-\infty}^\infty \frac{x^2}{2}e^{-|x|}\,dx.$$
By symmetry this is
$$\int_0^\infty x^2e^{-x}\,dx.$$
The integral can be calculated using two integrations by parts. Alternately, it is closely related to the variance of the exponential with parameter $1$.
You should end up concluding that $\frac{Y}{n}$ has variance $\frac{2}{n}$.


TITLE: What is R in the following question
QUESTION [1 upvotes]: I'm stuck on the following question:

Show that the function $f(x) = x − \ln(x^2 + 1)$ is increasing in $\mathbf{R}$

Can anyone explain what is the meaning of $\mathbf{R}$ in this context?
Thanks

REPLY [0 votes]: It is probably referring to $\mathbb{R}$, the set or real numbers, but of course this depends on context. A letter can be used for a reference to any mathematical object, so I would check if there are more details somewhere else in the book/exercise paper.


TITLE: n-bucket water puzzle
QUESTION [3 upvotes]: Inspired by this post. I am thinking of a general case.
Reference to read.
Suppose $N,L_1,L_2,p,q$ are integers, and $N L_1 = pq$. with $p\ge 2$  and $L_1>L_2$.
Now you have $N+1$ buckets,  the first $N$ buckets are full of water, each of these buckets can hold $L_1$ liters of water, the left one is a small bucket which is empty at first, the small bucket can hold $L_2$ liters of water.
Now the problem is I want to split the $N L_1$ liters of water to $p$ people evenly, everybody can get $q$ liters of water.
Question: in what situation $(L_1,L_2,N,p,q)$ we can split the water evenly, in what case we cannot.

For $p=1$ case, that means all the water will go to one people, that is easy.
For the original problem setting, if we change the capacity of the small bucket, let $(L_1,L_2,N,p,q)=(8,L_2,2,4,4)$, we can see that

when $L_2 = 1$, it is easy to do, we just get $1$ liter of water every time.
when $L_2 = 2$, it also easy, we just get $2$ liters of water every time.
$L_2 =3$ is done in the post.
$L_2=4$ is also easy as the first two.
I can prove when $L_2 = 6,7$, we cannot succeed.

I guess, if $L_2>q$, then we cannot succeed. I am trying to cook up a theory on this, but very limited. Since to find out a proper way to split is kind of NP hard.
Conjecture 1:
$gcd(L_1,L_2)\mid q$ and $L_2\le q$ are necessary(not sufficient) to find a solution.

REPLY [1 votes]: I think you mean $\gcd(L_1,L_2) \mid q$? It is clear that if $\gcd(L_1,L_2) \nmid q$, then no solution can exist, as all the amounts of water moved around are divisible by $\gcd(L_1, L_2)$. I shall also prove that if $L_2>q$, no solution can exist.
Indeed suppose $L_2>q$, and we assume without loss of generality that a solution exists. There are 2 possible operations.

Choose a non-empty bucket $A$, and a bucket $B$ which isn't full, and pouring water from $A$ to $B$ till either $A$ is empty or $B$ is full.
Choose a non-empty bucket $A$, and give all of its contents to 1 of the $p$ people.

Let $m$ be the number of buckets with $>q$ litres of water in them, at that point in time. (so $m$ changes as we make moves) Note that $m$ can decrease by at most 1 at each turn. Since we start at $m=n$ (all big buckets have $L_1>q$), and end at $m=0$ (all water distributed), there must be a move taking us from $m=1$ to $m=0$. Suppose that just before that move, bucket $P$ is the only bucket with $>q$ litres of water. In order for $m$ to become $0$, we must perform an operation involving $P$. It is clear that we cannot perform operation $2$, as that would lead to one of the $p$ people getting $>q$ litres of water. Thus we must have performed operation $1$, by pouring water from $P$ to another bucket $Q$. If we pour till $P$ is empty, then now bucket $Q$ has $>q$ litres of water, contradicting $m=0$. If we pour till $Q$ is full, then obviously bucket $Q$ has $\geq L_2>q$ litres of water, contradicting $m=0$.
Thus there can never be a move bringing us from $m=1$ to $m=0$, so no solution exists when $L_2>q$.
Now let us consider the special case $N=1$, so $L_1=pq$, $p \geq 2$. We have a solution if and only if $L_2 \mid q$. It is easy to see that a solution exists when $L_2 \mid q$ (just keep pouring water into the small bucket for distribution.
For the converse, note that if $L_2 \nmid q$, we can never finish distributing $q$ litres of water by just giving $L_2$ litres of water at a time. If we call a transfer of water using operation $2$ which transferred $\not =L_2$ litres as a special transfer, then we must have at least $p \geq 2$ special transfers.
We show that we can treat all special transfers as from operation $2$ applied to the big bucket. Note that the only way for the small bucket to be partially full at any point in time is by pouring water from the big bucket till the big bucket is empty (If we perform operation $2$ on the small bucket it would be empty) But in that case, if we had performed operation $2$ on the small bucket to get a special transfer, we can equivalently pour all the water in the small bucket into the big bucket and perform operation $2$ on the big bucket instead.
Consider the situation right after we had made the first special transfer. Then by the above argument we can treat it as having resulted from operation $2$ on the big bucket, so the big bucket is now empty. Consider the situation just before the special transfer. Then at that time, the big bucket has some water, so by above the small bucket must be either empty or full by the above argument. Now let's return to the point in time after the special transfer. Now the big bucket is empty and the small bucket is either empty ($0$ litres) or full ($L_2$ litres). It is clearly now impossible to make any more special transfers, so at most $1$ special transfer has been made, contradicting the fact that we need at least $p \geq 2$ such transfers.


TITLE: Is $\sin z$ a square of a holomorphic function?
QUESTION [2 upvotes]: Is there a holomorphic function $f$ satisfying $(f(z))^2=\sin(z)$ on a nbhd of 0? I know that for these kind of problems it is common to use a logarithm, in this case $g(z):=\exp(L(\sin(z)/2))$ where $L$ is the logarithm. I assume that such a $g$ cannot exist because of the zeros of $\sin z$.

REPLY [0 votes]: It appears that by $f^2(z)$ you meant $(f(z))^2,$ not $f(f(z)).$ If the latter, I think the answer may be "yes". (Google the phrase "What you have to do twice in order to sin". I think that was the title of a talk someone gave.)
As $z$ approaches $0$ along the positive $x$-axis, $\dfrac d {dz} \sqrt{\sin z}$ approaches $+\infty$ and $\dfrac d {dz} (-\sqrt{\sin z}\,)$ approaches $-\infty.$ If there were a function holomorphic in some neighborhood of $0$ whose square is $\sin z$ within that neighborhood, then on intervals of the form $[0,\varepsilon)$ it would agree with either $+\sqrt{\sin z}$ or $-\sqrt{\sin z}.$ So its derivative could not approach any particular complex number. But the derivative of the proposed holomorphic function would have to approach the value of that function at $0.$ So the answer is no.


TITLE: Prove $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers
QUESTION [0 upvotes]: I am trying to prove by mathematical induction  $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers $n$.
So far I have:

Solved in the first case possible - $1$
Assumed the function to be true for all positive integers
Subbed in $k+1$ for $n$ leaving me with the following
$$((k+1)^2+1)(k+1)!=(k+1)(k+2)!$$

I am unsure where to go from here? Any help would be greatly appreciated!

REPLY [0 votes]: The key point is form the induction hypotesis
$$2\times1!+5\times2!+10\times3!+…+(k^2+1)k!=k(k+1)!$$
is that
$$2\times1!+\ldots+((k+1)^2+1)(k+1)!\stackrel{Ind. Hyp.}=k(k+1)!+((k+1)^2+1)(k+1)!\stackrel{?}=(k+1)(k+2)!$$
and the latter is true indeed
$$k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)!(k^2+3k+2)=(k+1)!(k+1)(k+2)=(k+1)(k+2)!$$


TITLE: Solutions Problem
QUESTION [1 upvotes]: I think this problem would be incomplete? Do I need to know how much % juice graph juice has?
If I did, would I just have to solve this equation?
$$20x+?(8-x)=40(8)$$

REPLY [0 votes]: I would take graph (probably grape) juice to be $100\%$ juice.  You should specify what $x$ represents.  It looks like the number of liters of punch.  In that case your equation is correct.


text:
\begin{document}
\title{Cardinal $\bf p$ and a theorem of Pelczynski}
\author{M. V. Matveev\\
{\small Department of Mathematics, University of California, Davis}\\
{\small Davis CA 95616, USA (address valid till June 30, 2000}\\
{\small e-mail: misha$\underline{\mbox{ }}$matveev@hotmail.com
}}
\date{}
\maketitle

\noindent
{\small{\bf Abstract.}
Are two compactifications of $\omega$ homeomorphic if their remainders
are homeomorphic?
For metrizable compactifications the question was answered
affirmatively by Pelzcynski.
Can the same happen for some non-metrizable remainders?
We consider the case when the remainder is $D^\tau$
for some uncountable $\tau$.
We show that the answer is affirmative if $\tau<\bf p$
and negative if $\tau=\bf c$.
We prove that every isomorphism between two subalgebras of $\Pf$
is generated by a permutation of $\omega$ provided these
subalgebras have independent basis of cardinality fewer than $\bf p$.
Also we consider some special dense countable subsets in $D^\tau$.
}

\medskip
\noindent
{\bf AMS subject Classification:} Primary 54D35; Secondary 54D65,
54A20, 03E10, 06E05

\medskip
\noindent
{\bf Keywords:} compactification, remainder, Boolean algebra,
independent, cardinal $\bf p$, cardinal $\bf b$,
sequentially separable.

\section{Introduction}

In 1965 Pelczynski proved the following:

\begin{theorem}\label{ofPel}
{\em \cite{Pel}}
Let $b_1\omega$ and $b_2\omega$ be two metrizable compactifications
of $\omega$.
If $b_1\omega\setminus\omega$ and $b_2\omega\setminus\omega$
are homeomorphic, then $b_1\omega$ and $b_2\omega$ are homeomorphic.

Moreover, every continuous mapping
$f:b_1\omega\setminus\omega\to b_2\omega\setminus\omega$
can be extended to a continuous mapping $F:b_1\omega\to b_2\omega$
so that $F(\omega)\subset\omega$ and
if $f$ is surjective, then so is $F$.
\end{theorem}

(See also Terasawa's paper \cite{Terasawa} for an alternative proof
and related results).

In this paper we are interested in the question whether
Pelczynski's theorem can be extended to some non-metrizable compacta.
We obtain the following particular results:

\begin{theorem}\label{positive}
If $\tau<\bf p$, then all compactifications of $\omega$
with the remaider homeomorphic to $D^\tau$ are homeomorphic
to each other.
\end{theorem}

\begin{theorem}\label{negative}
There are $2^{\bf c}$ pairwise non-homeomorphic compactifications
of $\omega$ with the remainder homeomorphic to $D^{\bf c}$.
\end{theorem}

(The reader is refered to \cite{vD} or \cite{Vau}
for the definition of cardinal $\bf p$
and other small cardinals and $^*$-terminology).
The following natural questions remain open.

\begin{question}{\em
Can one construct within ZFC a non-metrizable compact space $X$
such that all compactifications of $\omega$ with the remainder
homeomorphic to $X$ are homeomorphic to each other?
}
\end{question}

\begin{question}\label{two}
{\em
For which cardinal numbers $k$, $1<k<2^{{\bf c}}$, is there a compact $X$
such that there are exactly $k$ many pairwise non-homeomorphic
compactifications of $\omega$ with the remainder homeomorphic to $X$?
}
\end{question}

Restriction $<2^{{\bf c}}$ in Question~\ref{two} is natural:
one cannot construct more than $2^{{\bf c}}$ pairwise non-homeomorphic
compactifications of $\omega$.

In the  proof of Theorem~\ref{positive} we essentially use the fact
that the Boolean algebra of clopen subsets of $D^\tau$
has independent basis. So it is not clear how to extend Theorem~\ref{positive}
to other compacta of small weight.

The basic idea of the proof of Theorem~\ref{negative}
is the notion of sequential separability (or, rather, its various negations)
which we discuss in Section~\ref{SeqSep}.

An interesting question was rised by R. Williams in the review of
Pelczynski's paper (MR 32 $\#$4659): can the assignment $f\to F$
in Theorem~1 be made functorial (i.e. such that $FG$
always corresponds to $fg$)?

\section{Proof of Theorem 2}

$\Pf$ denotes the Boolean algebra of classes of subsets of $\omega$
modulo the ideal of finite sets.
A family $\cal A$ of subsets of $\omega$
is independent provided for every
$n,m\in\omega$ and every distinct $A_1,\dots, A_{n+m}\in \cal A$ the
intersection
$A_1\cap\dots\cap A_n\cap(\omega\setminus A_{n+1})\cap(\omega\setminus
A_{n+m})$ is nonempty.
In the same sense we speak about the independence of subfamilies of $\Pf$.
We say that an isomorphism $f: {\cal A}\to{\cal B}$ between two
subalgebras
${\cal A},{\cal B}\subset\Pf$ is generated by a permutation $\pi$ of
$\omega$ provided
for every $a\in\cal A$ and every $A\in a$, $\pi(A)\in f(a)$.

It is clear that any subalgebras of $\Pf$ having independent basis of
the same cardinality are isomprphic.

\begin{lemma}\label{BAInd}
Let $f:{\cal A}\to\cal B$ be an isomorphism between subalgebras
${\cal A},{\cal B}\subset\Pf$.
If $|{\cal A}|<\bf p$ and $\cal A$ has an independent basis
then $f$ can be generated by a permutation of $\omega$.
\end{lemma}

{\bf Proof:}
Denote $\tau=|{\cal A}|$. We can assume that $\tau\geq\omega$.
Let ${\cal A}_0$ be an independent basis for $\cal A$.
Then $|{\cal A}_0|=\tau$ and we can write
${\cal A}_0=\{A_\alpha:\alpha<\tau\}$.
It is clear that $\{f(A_\alpha):\alpha<\tau\}$ is an
independent basis for $\cal B$.
For each $\alpha<\tau$, choose $a_\alpha\in A_\alpha$ and $b_\alpha\in
f(A_\alpha)$.
For $n\in\omega$ and $\alpha<\tau$, put
$$
S_{n,\alpha}=\left\{\begin{array}{l}
b_\alpha,\mbox{ if }n\in a_\alpha\\
\omega\setminus b_\alpha\mbox{, otherwise}.
\end{array}\right.
$$
Then the family
${\cal S}_n=\{S_{n,\alpha}:\alpha<\tau\}$
has sfip and, since $\tau<\bf p$, a pseudointersection, say $H_n$.

Next, for each $\alpha<\tau$ we define a function $h_\alpha:\omega\to
\omega$ by
$h_\alpha(n)=\min\{m: H_n\setminus\{0,\dots,m-1\}\subset S_{n,\alpha}\}$.
Since $\tau<{\bf p}\leq{\bf b}$, there is a strictly increasing function
$g_0:\omega\to\omega$ such that $g_0\geq^* h_\alpha$ for all $\alpha<\tau$.
Define inductively a strictly increasing function $\pi_0:\omega\to\omega$
such that $\pi_0\geq g_0$ and $\pi_0(n)\in H_n$ for all $n$.
Then $\pi_0$ is an injection of $\omega$ to $\omega$ such that
$\pi_0(a_\alpha)\subseteq^* b_\alpha$
and $\pi_0(\omega\setminus a_\alpha)\subseteq^* \omega\setminus b_\alpha$
for all $\alpha<\tau$.
Working simetrically, we obtain an injection $\pi_1:\omega\to\omega$
such that $\pi_1(b_\alpha)\subseteq^* a_\alpha$
and $\pi_1(\omega\setminus b_\alpha)\subseteq^* \omega\setminus a_\alpha$
for all $\alpha\in\tau$.
By the usual Cantor-Bernstein argument
(see, for example \cite{KM}, Theorem~5.5.2) there are decompositions
$\omega=M_1\cup M_2$ and $\omega= N_1\cup N_2$ such that
$\pi_0(M_1)=N_1$ and $\pi_1(N_2)=M_2$.
Then the function $\pi$ defined by the role
$$
\pi(n)=\left\{\begin{array}{l}
\pi_0(n),\mbox{ if }n\in M_1\\
\pi_1^{-1}(n),\mbox{ if }n\in M_2
\end{array}\right.
$$
is a permutation of $\omega$ and satisfies the condition
$\pi(a_\alpha)=^* b_\alpha$ for all $\alpha<\tau$.
So $f$ is generated by $\pi$.
\qed

\bigskip

For a space $X$, let $CO(X)$ denote the family of all clopen sets in $X$.
The next lemma follows directly from normality and compactness.

\begin{lemma}\label{omext}
Let $b\omega$ be a compactification of $\omega$ with a zero-dimensional
remainder $X=b\omega\setminus\omega$.
Then there is a (unique up to $^*$)
mapping that
assigns to each $Y\in CO(X)$ a subset $g(Y)\subset\omega$
so that the set $Y'=Y\cup g(Y)$ is clopen in $b\omega$.
\end{lemma}

Now we prove Theorem~\ref{positive}.
Let $b_1\omega=(\omega\cup D^\tau,{\cal T}_1)$ and
$b_2\omega=(\omega\cup D^\tau,{\cal T}_2)$
be two compactifications of $\omega$ with remainder $D^\tau$.
Further, let $g_1,g_2:CO(D^\tau)\to {\cal P}(\omega)$
be mappings from Lemma~\ref{omext} corresponding to $b_1$ and $b_2$.
We will consider them as mappings to $\Pf$.
Note that ${\cal C}=\{\{f\in D^\tau:f(\alpha)=0\}:\alpha<\tau\}$
is an independent basis basis for the Boolean algebra $CO(D^\tau$.
Then the subalgebras $\cal A$ and $\cal B$ of $\Pf$
generated by $g_1({\cal C})$ and $g_2({\cal C})$
also have independent basis and cardinality $\tau$.
It is clear that they are isomorphic.
By Lemma~\ref{BAInd} this isomorphism is generated
by some permutation, say $\pi$, of $\omega$.
Then the mapping $F:b_1\omega\to b_2\omega$
defined by the formula
$$
F(x)=\left\{\begin{array}{l}
x,\mbox{ if }x\in D^\tau\\
\pi(x),\mbox{ if }x\in\omega
\end{array}\right.
$$
is a homeomorphism.
\qed

\section{Sequential separability, its negations\\
and variations of cardinal $\bf p$}\label{SeqSep}

A space $X$ is {\em sequentially separable} provided there is
a countable subspace $Y\subset X$ sucht that every point of $X$ is the limit
of a sequence from $Y$ ($Y$ is said to be {\em sequentially dense} in $X$).
Wilansky noted that the product of $\bf c$ sequentially separable
spaces need not be sequentially separable and asked if the product
of fewer than $\bf c$ sequentially separable spaces is
sequentially separable \cite{Wilansky}.
Tall showed \cite{Tall} that
the answer to this question depends on additional set-theoretic
assumptions.
In fact, Tall proved that MA($k$) for $\sigma$-centered posets
implies that every product of fewer than $k$ sequentially separable
spaces is sequentially separable.
Similar results were obtained in \cite{FK}.
Further, Bell proved (\cite{Bell}, see also \cite{Vau}, p.~201)
that ${\bf p}={\bf m}_{\sigma-\mbox{{\small centered}}}$ where
${\bf m}_{\sigma-\mbox{{\small centered}}}$ is the minimal cardinal $k$
such that ``MA($k$) for $\sigma$-centered posets'' fails.
So now Tall's theorem can be restated as follows.

\begin{theorem}
Every product of fewer than $\bf p$ sequentially separable spaces
is sequentially separable.
\end{theorem}

\begin{corollary}\label{trivial}
$D^\tau$ is sequentially separable for all $\tau<\bf p$.
\end{corollary}

This particular case of Tall's theorem can be derived also from
the inequality ${\bf p}={\bf p}_{\bf p}={\bf p}_\chi$
(\cite{vD}, Theorem~6.2) where
${\bf p}_{\bf p}=\min\{k:D^k$  is not subsequential$\}$
and
${\bf p}_\chi=\min\{k:$ there exists a regular non-subsequential
space of character $k\}$
(subsequential, in van~Douwen's terminology, means that
if $x\in\overline{A}$ and $A$ is countable, then there is a sequence
converging from $A$ to $x$).
So $D^k$ with $k<\bf p$ is better than just sequentially separable:
every dense countable subspace of $D^k$ is sequentially dense
in it.\footnote{
By the way, ``countable'' in this statement is essential.
Indeed, a $\Sigma$-product is dense but not sequentially dense
in $D^k$, moreover, it is sequentially closed.
}

Consider the following cardinal numbers.

\bigskip\noindent
${\bf p}_1=\min\{\tau:$ there is a dense countable $Y\subset D^\tau$
which is not sequentially

\hskip 1cm dense$\}$,

\bigskip\noindent
${\bf p}_2=\min\{\tau: D^\tau$ is not sequentially separable$\}$,

\bigskip\noindent
${\bf p}_3=\min\{\tau:$ there is a dense countable $Y\subset D^\tau$
such that no nontrivial

\hskip 1cm sequence from $Y$ converges in $D^\tau\}$.

\bigskip
Then ${\bf p}\leq{\bf p}_1\leq{\bf p}_2\leq{\bf c}$
and ${\bf p}_1\leq{\bf p}_3\leq{\bf c}$.
The inequality ${\bf p}\leq{\bf p}_1$
follows from Corollary~\ref{trivial};
the inequality ${\bf p}_3\leq{\bf c}$
follows from Theorem~\ref{badsubspaces} below.
The rest is trivial.

Further, cardinals ${\bf p}_1$
and ${\bf p}_3$ can be expressed in a purely set-theoretic way.
Indeed, it is easy to see that

\bigskip\noindent
${\bf p}_1=\min\{\tau:$ there is an independent family
$\cal A$ of subsets of $\omega$ such

\hskip 1cm that $|{\cal A}|=\tau$ and $\cal A$ does not have infinite
pseudointersection$\}$.

\bigskip
Further, let $\cal A$ and $\cal B$ be families of subsets of $\omega$.
We say that $\cal B$ is a {\it partial inversement} of $\cal A$
if there is a bijection $h:{\cal A}\to{\cal B}$
such that for each $A\in\cal A$ the set $h(A)$ equals either $A$ or
$\omega\setminus A$.
Then it is clear that

\bigskip\noindent
${\bf p}_3=\min\{\tau$: there is an independent family $\cal A$ of subsets
of $\omega$ such that

\hskip 1cm $|{\cal A}|=\tau$ and no partial inversement of $\cal A$
has nonempty

\hskip 1cm pseudointersection$\}$.

\bigskip
Peter Nyikos noted \cite{Nyik} that in fact ${\bf p}_1=\bf p$.
Nyikos' proof goes as follows: let $\{I_\alpha:\alpha<{\bf p}\}$
be an independent family of subsets of $\omega$
and let $\{F_\alpha:\alpha<{\bf p}\}$ be a free filter on
$\omega$ with sfip but no infinite pseudointersection.
Then $\{I_\alpha\times F_\alpha:\alpha<{\bf p}\}$
is a ${\bf p}_1$-witnessing independent family of subsets
of $\omega\times\omega$.

Rothberger proved in \cite{Rot} that $D^{\omega_1}$
is sequentially separable iff there is a Q-set of cardinality
$\omega_1$.
In \cite{Lo} this result is extended as follows: $D^\tau$ is sequentially separable iff
there is a Q-set of cardinality $\tau$;
in other words, ${\bf p}_2=\bf q$ where $\bf q$ is the minimal
cardinal $\tau$ such that there is no Q-set of cardinality $\tau$
(cardinal $\bf q$ was introduced in \cite{GN}).
Since the existence of Q-sets of cardinality $\geq\bf p$
is consistent with ZFC (a model is constructed in \cite{FM} in which
there is a Q-set of cardinality $\omega_2$ and a non-Q-set of cardinality
$\omega_1$
it is consistent also that ${\bf p}_1<{\bf p}_2$.
The question which of strict inequalities
${\bf p}_1<{\bf p}_3<{\bf c}$
can be consistently true remains open
as well as the relationship between ${\bf p}_2$ and ${\bf p}_3$
is not clear.

\begin{theorem}\label{badsubspaces}
There is a family ${\cal Y}=\{Y_\alpha:\alpha<2^{\bf c}\}$
of dense countable subspaces $Y_\alpha\subset D^{\bf c}$
such that:

(1) no nontrivial sequence from $Y_\alpha$ converges in $D^{\bf c}$
and

(2) $Y_\alpha$ and $Y_{\alpha'}$ are disjoint and non-homeomorphic
as soon as $\alpha\neq\alpha'$.
\end{theorem}

(Trivial means eventially constant).

{\bf Proof:}
For convinience, we replace $\bf c$ with $\Bbb R$.
Two filters on $\omega$ have the same type if there is a permutation
of $\omega$ which transforms one of them into the other.
It is well known that there are $2^{\bf c}$ types of ultrafilters on
$\omega$.
We construct $Y_\alpha$ by induction.
Suppose $\alpha<2^{\Bbb R}$ and $Y_\beta$ have been constructed for all
$\beta<\alpha$.
Denote $M_\alpha=\cup\{Y_\beta:\beta<\alpha\}$.
Then $|M_\alpha|<2^{\bf c}$.
Further, denote ${\cal F}_\alpha$ the set of all types of filters
of neighbourhoods of $x$ in $Z$ where $x\in Z\subset Y_\beta$
for some $\beta<\alpha$
(if $\alpha=0$ then $M_\alpha=\emptyset$ and ${\cal F}_\alpha=\emptyset$).
Then $|{\cal F}_\alpha|<2^{\bf c}$ and thus we can pick
an ultrafilter $u_\alpha$ not in ${\cal F}_\alpha$.
Now we need a lemma.

\begin{lemma}
If $M\subset D^{\Bbb R}$ and $|M|<2^{\bf c}$,
then there is an embedding $i:\beta\omega\to D^{\Bbb R}$
such that $i(\beta\omega)\cap M=\emptyset$.
\end{lemma}

{\bf Proof of lemma:} We have $|pr_{{\Bbb R}^+}(M)|<2^{\bf c}=|D^{{\Bbb R}^+}|$
(we denote ${\Bbb R}^+=\{x\in{\Bbb R}: x\geq 0\}$ and
${\Bbb R}^-=\{x\in{\Bbb R}: x<0\}$).
Pick a point $z\in D^{{\Bbb R}^+}\setminus pr_{{\Bbb R}^+}M$.
Denote $Z=\{x\in D^{{\Bbb R}}:pr_{{\Bbb R}^+}(x)=z\}$.
Then $Z\cap M=\emptyset$ and $Z$ is homeomorphic to $D^{\bf c}$
and thus it contains a copy of every zero-dimensional space
of weight $\leq c$, in particular, a copy of $\beta\omega$.
\qed

Now we go on with the proof of the theorem.
By lemma, there is an embedding $i_\alpha:\beta\omega\to D^{\Bbb R}$
such that $i_\alpha(\beta\omega)\cap M_\alpha=\emptyset$.
Denote $m_\alpha=i_\alpha(u_\alpha)$.
Our $Y_\alpha$ will be of the form
$Y_\alpha=i_\alpha(\omega)\cup\{m_\alpha\}\cup Z_\alpha$
where $Z_\alpha$ is yet to be constructed.
This will be made in several steps.

First, consider the set $T$ of all such functions $f\in D^{\Bbb R}$ that $f$:

- takes each of the two values, $0$ and $1$, $\bf c$ times
on the set ${\Bbb R}^-$,

- has only finitely many points of discontinuity;

- all points of discontinuity are in $\Bbb Q$.

\noindent
It is easy to see that $T$ is countable and dense in $D^{\Bbb R}$.

Next, enumerate all infinite, nowhere dense subsets of $T$ as
$\{N_\gamma:\gamma<{\bf c}\}$.
For every $\gamma<{\bf c}$ fix a decomposition
$N_\gamma=N_\gamma^0\cup N_\gamma^1$ where both
$N_\gamma^0$ and $N_\gamma^1$ are infinite.
Also, enumerate the set ${\Bbb R}^+$ as
${\Bbb R}^+=\{q_\gamma:\gamma<{\bf c}\}$.
For every $t\in T$ we define $\tilde{t}\in D^{\Bbb R}$
as follows: for every $r\in\Bbb R$ put
$$
\tilde{t}(r)=\left\{\begin{array}{l}
0,\mbox{ if }r=q_\gamma\mbox{ and }t\in N_\gamma^0\\
1,\mbox{ if }r=q_\gamma\mbox{ and }t\in N_\gamma^1\\
t(r),\mbox{ otherwise.}
\end{array}\right.
$$
Put $\tilde{T}=\{\tilde{t}:t\in T\}$.
Then $\tilde{T}$ is countable.

We claim that $\tilde{T}$ is dense in $D^{\Bbb R}$.
Let $O=O_{r_1\dots r_n}^{i_1\dots i_n}=\{f\in D^{\Bbb R}:
f(r_1)=i_1,\dots,f(r_n)=i_n\}$ be a basic open set in $D^{\Bbb R}$.
Some of $r_1,\dots, r_n$, say
$r_{k_1}=q_{\gamma_1},\dots, r_{k_s}=q_{\gamma_s}$ may be in
${\Bbb R}^+$.
Note that the set $T\setminus(N_{\gamma_1}\cup\dots\cup N_{\gamma_s})$
is dense in $D^{\Bbb R}$.
Pick $t\in(T\setminus(N_{\gamma_1}\cup\dots\cup N_{\gamma_s}))\cap O$.
But then also $\tilde{t}\in O$ since $t$ and $\tilde{t}$ take the same
values at coordinates $q_{\gamma_1}$,..., $q_{\gamma_s}$.
So $\tilde{T}$ is dense in $D^{\Bbb R}$.

Now we claim that $\tilde{T}$ does not contain non-trivial sequences
converging in $D^{\Bbb R}$.
Suppose that $\xi$ is a non-trivial sequence from $T$ converging
in $D^{\Bbb R}$.
Then the set $\tilde{S}=\{\xi(n):n\in\omega\}$
is infinite, and hence so is the set
$S=\{t\in T:\tilde{t}\in\tilde{S}\}$.
Further, $S$ contains an infinite nowhere dense set, say $N$.
But then $N=n_\gamma$ for some $\gamma<\bf c$,
and we have $\tilde{t}(\gamma)=0$ for all $t\in N_\gamma^0$ and
$\tilde{t}(\gamma)=1$ for all $t\in N_\gamma^1$.
This means that the sequence $\xi$ contains both infinitely many
elements taking value $0$ at coordinate $\gamma$ and
infinitely many elements taking value $1$ at coordinate $\gamma$.
Hence $\xi$ cannot converge.

Also we note that every point from $\tilde{T}$ takes each of the two
values, $0$ and $1$, $\bf c$ many times.
Indeed, $t$ and $\tilde{t}$ do not differ on the coordinates from
${\Bbb R}^-$.

Every permutation $\pi:{\Bbb R}\to\Bbb R$ induces
an autohomeomorphism $h_\pi:D^{\Bbb R}\to D^{\Bbb R}$
by the formula $(h_\pi(x))(r)=x(\pi^{-1}(r))$.
It is clear that $h_\pi(\tilde{T})$ has all properties of $\tilde{T}$
(is dense in $D^{\Bbb R}$, countable, and does not contain
nontrivial sequences converging in $D^{\Bbb R}$).
It remains to find such a permutation $\pi$ that $h_\pi(\tilde{T})$
does not intersect $M_\alpha$.
By induction it is easy to find for all $t\in T$ subsets
$A_t, B_t\subset\Bbb R$ such that:

$|A_t|=|B_t|=\bf c$ for every $t\in T$,

$A_t\cap B_t=\emptyset$ for every $t\in T$,

$A_t\cap B_{t'}=\emptyset=A_t\cap A_{t'}$ whenever $t\neq t'$,

$\tilde{t}(a)=0$ for all $t\in T$ and $a\in A_t$,

$\tilde{t}(b)=1$ for all $t\in T$ and $b\in B_t$.

\noindent
Let $t\in T$. Denote ${\cal P}_t$
the set of all such permutations $\pi$ of $\Bbb R$
that $\pi(A_t)=B_t$, $\pi(B_t)=A_t$ and $\pi(a)=a$
for all $a\not\in A_t\cup B_t$.
It is clear that $|{\cal P}_t|=2^{\bf c}$.
Further, if $\pi,\pi'\in{\cal P}_t$ and $\pi\neq\pi'$, then
$pr_{A_t\cup B_t}(h_\pi(\tilde{t}))\neq pr_{A_t\cup B_t}(h_{\pi'}(\tilde{t}))$.
So $|\{pr_{A_t\cup B_t}(h_\pi(\tilde{t})):\pi\in{\cal P}_t\}|=2^{\bf c}$,
and since $|M_\alpha|<2^{\bf c}$
there is a $\pi_t\in{\cal P}_t$ such that
$pr_{A_t\cup B_t}\left(h_{\pi_t}(\tilde{t})\right)\not\in
pr_{A_t\cup B_t}(M_\alpha)$ ($*$).
The function $\pi:{\Bbb R}\to{\Bbb R}$ defined by formula
$$
\pi(a)=\left\{\begin{array}{l}
\pi_t(a),\mbox{ if }a\in A_t\cup B_t\mbox{ for some }t\in T\\
a,\mbox{ otherwise}
\end{array}\right.
$$
is a permutation.
It follows from ($*$) that
$\pi(\tilde{T})\cap M_\alpha=\emptyset$.

Put $Z_\alpha=\pi(\tilde{T})$.
Then $Z_\alpha\cap M_\alpha=\emptyset$,
hence $Y_\alpha\cap M_\alpha=\emptyset$,
hence $Y_\alpha\cap Y_\beta=\emptyset$ for all $\beta<\alpha$.

The fact that $Y_\alpha$ is not homeomorphic to $Y_\beta$
for all $\beta<\alpha$ follows fro the inclusion
$Y_\alpha\supset i_\alpha(\omega)\cup \{m_\alpha\}$.

Last, if $Y_\alpha$ would contain a nontrivial sequence converging
in $D^{\Bbb R}$, then this sequence would contain either
a nontrivial subsequene from $Z_\alpha$, or one from
$i_\alpha(\omega)$.
The first is impossible by the construction of $Z_\alpha$,
the second - by the properties of \v{C}ech-stone compactification.
\qed

\section{Proof of Theorem 3 and more}

Let $Y\subset D^{\bf c}$ be one of $Y_\alpha$ from Theorem~\ref{badsubspaces}.
The following is a restriction of Tkachuk's construction from \cite{Tka}.
Fix a decomposition $\omega=\cup\{\omega_y:y\in Y\}$
where $\omega_y$ is infinite for every $y\in Y$.
Put $b_Y\tilde{\omega}=
(D^{\bf c}\times\{\omega\})\cup\tilde{\omega}\subset
D^{\bf c}\times(\omega+1)$
where $\tilde{\omega}=\cup\{\{y\}\times\omega_y:y\in Y\}$.
Then $b_Y\tilde{\omega}$ is compact; $\tilde{\omega}$ is countable,
dense in $b_Y\tilde{\omega}$ and consists of isolated points.
So $b_Y\tilde{\omega}$ can be considered as a compactification of $\omega$
with the remainder homeomorphic to $D^{\bf c}$.
Further, it is easy to see that
$\{x\in D^{\bf c}:\mbox{ there is a sequence converging from }
\tilde{\omega}\mbox{ to }x\}=Y$.
This fact and Theorem~\ref{badsubspaces} imply that
$b_{Y_\alpha}\tilde{\omega}$ and $b_{Y_{\alpha'}}\tilde{\omega}$
are not homeomorphic whenever $\alpha\neq\alpha'$. \qed

\medskip
We conclude the paper with the following modification of Theorem~3.

\begin{theorem}
There are infinitely many pairwise non-homeomorphic
compactifications of $\omega$
with the remainder homeomorphic to $D^{{\bf p}_3}$.
\end{theorem}

{\bf Proof:}
Recall that the Alexandroff Duplicat $AD(X)$ of a topological space $X$
is the set $X\times\{0,1\}$ topologized by declaring the points of
$X\times\{1\}$ to be isolated while basic neighbourhoods of points of
$X\times\{0\}$ take the form $(U\times\{0,1\})\setminus\{(x,1)\}$
where $x\in X$ and $U$ is a neighbourhood of $x$ in $X$.
For $Y\subset X$ denote $AD_Y(X)=(X\times\{0\})\cup(Y\times\{1\})$
with the topology inhereted from $AD(X)$.
Let $Y\subset D^{{\bf p}_3}$ be dense, countable and
such that no non-trivial sequence
from $Y$ converges in $D^{{\bf p}_3}$.
Let $n\in\Bbb N$.
Choose $n$ distinct points $p_1,\dots,p_n\in D^{{\bf p}_3}$.
Further, let $A_1=N_1\cup\{a_1\}$,..., $A_n=N_n\cup\{a_n\}$
be $n$ disjoint copies of the convergent sequence
(the limit points are $a_1,\dots, a_n$).
Denote $b_{nY}\omega$ the quotient space of the discrete sum
$A_1\cup\dots\cup A_n\cup AD_Y D^{{\bf p}_3}$ obtained by
identifying $a_i$ with $(p_i,0)$ for all $1\leq i\leq n$.
Then $b_{nY}\omega$ can be considered as a compactification of
$N_1\cup\dots\cup N_n\cup(Y\times\{1\})\sim\omega$
with the remainder $D^{{\bf p}_3}\times\{0\}\sim D^{{\bf p}_3}$.
It is clear that exactly $n$ points in $D^{{\bf p}_3}$ are limits
of sequences from $\omega$.
So compactifications $b_{nY}\omega$ are not homeomorphic
to each other for distinct $n$.
\qed

\bigskip
{\bf Asknowlegement.}
The paper was written when the author was visiting University of
California, Davis. The author is grately indebted to colleagues
form UC Davis for their hospitality.

Also the author expresses his gratitude to Ronnie Levy for
numerous usefull discussions and to Peter Nyikos and Joseph Lo
for usefull discussions during Spring Topological Conference in March 2000.

config: arxiv


text:
\begin{document}
\bibstyle{plain}

\maketitle

\begin{abstract}
In this paper a new geometric characterization of the $n$th symmetric product of a curve is given. Specifically, we assume that there exists a chain of smooth subvarieties $V_i$ of dimension $i$, such that $V_i$ is an ample divisor in $V_{i+1}$ and its intersection product with $V_1$ is one. That the Albanese dimension of $V_2$ is $2$ and the genus of $V_1$ is equal to the irregularity of the variety. We prove that in this case the variety is isomorphic to the symmetric product of a curve.

{\parskip7pt
\noindent \textbf{Keywords}: Symmetric product, curve, irregular variety.}
\end{abstract}

\let\thefootnote\relax\footnote{The author has been partially supported by the Proyecto de Investigaci\'on MTM2012-38122-C03-02.}
\let\thefootnote\relax\footnote{2010MSC Codes: 14J30, 14J35, 14J40}
\let\thefootnote\relax\footnote{University of Copenhagen. Contact e-mail: meritxell@math.ku.dk. Phone: +4535337598}

\section{Introduction}

The aim of this paper is to give a new characterization of the $n$th symmetric product of a curve. Following the ideas introduced in the articles \cite{CCM} and \cite{MPP1} we prove a characterization of the $n$th symmetric product of a curve by the existence of a chain of subvarieties with certain properties. This generalizes the $2$-dimensional case proved in the mentioned references.

Let $C$ be a smooth complex projective curve of genus $g$. For an integer $n\geq 1$, the $n$th symmetric product of $C$ is the quotient of the Cartesian product by the action of the $n$th symmetric group. The action of $S_n$ on $C\times \cdots \times C$ is by permutation of the factors. It is well known that $C^{(n)}$ is a smooth and projective variety of dimension $n$ which parametrizes the effective degree $n$ divisors on $C$. Equivalently, it parametrizes the unordered $n$-tuples of points of $C$.

Symmetric products of curves play a very important role both in the theory of algebraic curves and in the theory of higher dimensional algebraic varieties. In the first topic, they are exploited by Brill-Noether theory to study special divisors on curves. Moreover, the $i$th symmetric product determines the cuve $C$. In the second topic, they are particularly simple examples of irregular varieties in any dimension.

The square symmetric product of a curve can be described in a very precise geometric way, simply by the existence of a divisor with certain numerical properties.

\begin{thm}(\cite{MPP1}) \label{mpp}
Let $S$ be a smooth surface of general type with irregularity $q$ containing a $1$-connected divisor $D$ such that $p_a(D)=q$ and $D^2>0$. Then the minimal model of $S$ is either
\begin{enumerate}
\item the product of two curves of genus $g_1,\,g_2\geq 2$ ($g_1+g_2=q$) or
\item the symmetric product $C^{(2)}$, where $C$ is a smooth curve of genus $q$, and $C^2=1$.
\end{enumerate}
Furthermore, if $D$ is $2$-connected, only the second case occurs.
\end{thm}

We remark that in the proof of this theorem the authors use the characterization of $C^{(2)}$ given in \cite{CCM}. Namely, that $C^{(2)}$ is the only minimal algebraic surface with irregularity $q$ that is covered by curves of genus $q$ and self-intersection $1$. These are the coordinate curves $C_{2,P}$, $P\in C$, that parametrize the degree two divisors in $C$ which contain the point $P$.

In general, given a point $P\in C$, we define the divisor $C_{n,P}$ of $C^{(n)}$ as
\[
C_{n,P}=\left\{P+\mathcal{Q}\ |\ \mathcal{Q}\in C^{(n-1)}\right\}.
\]
That is, $C_{n,P}$ is the image of the inclusion map $i_P: C^{(n-1)}\rightarrow C^{(n)}$ with $i_P(\mathcal{Q})=P+\mathcal{Q}$. The divisor $C_{n,P}$ is ample in $C^{(n)}$ (see \cite[Pag. 247]{Poli}) and isomorphic to $C^{(n-1)}$.

The numerical equivalence class of $C_{n,P}$ is independent of $P$, and so, when talking about numerical classes, the subindex $P$ will not be significant. We will call these divisors the \textbf{coordinate divisors}. When $n=2$, they are the usual \textbf{coordinate curves} in $C^{(2)}$. These coordinate divisors form a $1$-dimensional family, $\mathcal{C}$, of algebraically equivalent divisors in $C^{(n)}$ (not linearly equivalent). Moreover, its numerical class determines the family. That is, if an effective divisor of $C^{(n)}$ is numerically equivalent to $C_{n,P}$ then it belongs to the family $\mathcal{C}$ (see \cite{CS}).

The main result in this paper is the following theorem characterizing symmetric products of curves:

\begin{thm}\label{charCn}
Let $X$ be a smooth projective variety of dimension $n$. Assume that there exists a chain of inclusions
\[
X=V_n\supset V_{n-1} \supset \cdots \supset V_2 \supset V_1=C
\]
such that
\begin{enumerate}
\item $V_i$ is a smooth irreducible variety with $dim(V_i)=i$.
\item For $i<n$, $V_i$ is an ample divisor in $V_{i+1}$.
\item $V_i\cdot C=1$ inside $V_{i+1}$.
\item The Albanese dimension of $V_2$ is $2$.
\item $q(X)=g(C)$.
\end{enumerate}

Then $X\cong C^{(n)}$. Moreover, $V_i\cong C^{(i)}$ and it is a coordinate divisor inside $V_{i+1}$ for $i<n$.
\end{thm}

We prove this result by induction on the dimension of the variety. The $2$-dimensional step is a consequence of Theorem \ref{mpp}. To prove the induction step, we observe first that the $Pic^0$ varieties of the elements in the chain are isomorphic. Using these isomorphisms and generic vanishing results, we find a one dimensional algebraic family of dicisors which are birational to the $C^{(n-1)}$. This family allows us to construct a birational map between our variety and $C^{(n)}$. The image of this family by the morphism is the family $\mathcal{C}$ of the coordinate divisors. Finally, we deduce that the map is an isomorphism.

\noindent \textbf{Acknowledgments} The author thanks Miguel Angel Barja and Joan Carles Naranjo for the multiple discussions and invaluable help on the development of this article. Many thanks also to the Universitat de Barcelona for the pre-doctoral grant APIF and their hospitality afterwards.

\textbf{Notation:} We work over the complex numbers. All varieties considered are projective and irreducible. For a smooth variety $X$ we denote by $q(X)=h^0(X, \Omega^1_X)$ its irregularity. The Albanese dimension of a variety is the dimension of its image by the Albanese morphism.

\section{Proof of the main theorem}

First, we remind some results that are useful for the proof of Theorem \ref{charCn}.

\begin{lemma}\label{injrest}
Let $X$ be an algebraic variety of dimension $n\geq 3$ and let $D$ be an ample effective reduced divisor. Then, the restriction map $Pic^0(X)\rightarrow Pic^0(D)$ is injective.
\end{lemma}

\begin{proof}
By the Lefschetz Theorem for Picard Groups (\cite{Laz}), we have that the restriction morphism $Pic^0(X)\rightarrow Pic^0(D)$ has trivial kernel.
\end{proof}

We remind some results on generic vanishing theory. The main objects of interest are the cohomological support loci.

\begin{defi}
Let $X$ be an irregular (smooth) variety of dimension $d$. The cohomological support loci of $\mathcal{O}_X$ are the algebraic sets
\[
V^i(X)=V^i(X, \mathcal{O}_X)=\{\eta\in Pic^0(X)\ |\ h^i (X, \mathcal{O}_X \otimes \eta)\neq 0\},
\]
where $i = 1,\dots, d$.
\end{defi}

The main result about the structure of the cohomological support loci was proved by Green and Lazarsfeld, with an important addition due to Simpson (the fact that the translations are given by torsion elements).
\begin{thm}\label{gv1}(\cite{GL1},\cite{GL2}, \cite{Simp})
Let $X$ be an irregular variety of dimension $d$, then $V^i(X)$ is formed by translates of subtorus of $Pic^0(X)$ by torsion elements. Moreover,
\[
codim_{Pic^0(X)} V^i(X)\geq dim\,a (X)-i
\]
where $a(X)$ is the image of $X$ by its Albanese morphism.

In particular, $h^i (X,L) = 0$ for general $L\in Pic^0(X)$ and $i < dim\,a(X)$.
\end{thm}

\vspace{5pt}
We define the index of a family of divisors:

\begin{defi}
Given an irreducible family $\mathcal{D}\subset B \times X$, with dimension $1$ ($dimB=1$), of effective divisors in a projective variety $X$, the \textbf{index} $i=i(\mathcal{D})$ of $\mathcal{D}$ is the degree of the projection, $p_2: \mathcal{D} \rightarrow X$. Equivalently, it is the number of divisors of $\mathcal{D}$ containing the general point of $X$.
\end{defi}

Notice that the family of coordinate divisors in $C^{(n)}$ has index $n$.
\vspace{5pt}

Now, we have all the necessary tools to prove our main theorem.

\begin{proof}[Proof of Theorem \ref{charCn}]
We prove the theorem by induction. First, we observe that since $C\subset V_2$ is an irreducible smooth curve it is $2$-connected. Moreover, its self-intersection is one and hence, following the proof of Theorem \ref{mpp}, we deduce that $S:=V_2$ is birational to $C^{(2)}$. Furthermore, since the divisor $C$ is ample in $S$, in fact $S\cong C^{(2)}$, because any exceptional divisor would have intersection product $0$ with $C$. Hence, the case $n=2$ is already known. By the proof of Theorem \ref{mpp} (see \cite{MPP1}, Proposition 4.3) we have that there exists a $1$-dimensional family in $Pic^0(S)$
\[
\mathcal{W}:=\{\tilde{\eta} \in Pic^0(S)\ |\ h^0(S, \mathcal{O}_S(C)\otimes\tilde{\eta})=1\}.
\]
It is the image of
\[
W_1(C)=\{\eta\in Pic^0(C)\ |\ h^0(C,\mathcal{O}_C(C)\otimes \eta)=1\}
\]
by the isomorphism $Pic^0(S)\cong Pic^0(C)$ given by the restriction map. That is, we consider $W_1(C)$ as the image of $C$ by the natural map $C\rightarrow Pic^0(C)$ defined as $p\rightarrow \mathcal{O}_C(p-C|_C)$.

Furthermore, $\mathcal{C}=\{C_{\eta},\ \eta\in \mathcal{W}\}$ is the family of coordinate curves in $C^{(2)}$, where $C_{\eta}$ is the curve such that $\mathcal{O}_S(C_{\eta})=\mathcal{O}_S(C)\otimes \eta$.

We observe that since $V_1=C$ is algebraically equivalent to a coordinate curve in $V_2\cong C^{(2)}$, it is, in fact, a coordinate curve and thus $0\in \mathcal{W}$.

We assume now that $n\geq 3$ and that the result is proven for all $dim(X)\leq n-1$. We are going to prove the theorem for $dim(X)=n$. We consider $S:=V_2$ for the inductive process.

Since $V_i$ is ample in $V_{i+1}$ and $q(X)=g(C)$, by Lemma \ref{injrest} we obtain the following chain of isomorphisms given by the restriction maps:
\[
Pic^0(X)\cong Pic^0(V_{n-1})\cong \dots \cong Pic^0(V_i)\cong \cdots \cong Pic^0(S)\cong Pic^0(C).
\]

We add the following statement to the inductive process:

\begin{itemize}
\item  The image of $\mathcal{W}$ in $Pic^0(V_i)$ by this chain of isomorphisms parametrizes the family of coordinate divisors in $V_{i}\cong C^{(i)}$ for $i<n$.
\end{itemize}

We remind that by the induction hypothesis, $V_{i-1}$ is a coordinate divisor in $V_i\cong C^{(i)}$ for all $i<n$.

In what follows we denote by $N$ the divisor associated to the line bundle $\mathcal{O}_{V_{n-1}}(V_{n-1}|_{V_{n-1}})$.

\textbf{Claim:} \emph{There exists $\alpha \in Pic^0(X)$ such that
\[
\alpha|_{V_{n-1}}=\mathcal{O}_{V_{n-1}}(V_{n-2}-N)
\]
and $V_{n-1}^n=1$.}

Consider $\mathcal{O}_S(V_{n-2}|_S-N|_S)$. We observe that $V_{n-1}\cong C^{(n-1)}$ and $S\cong C^{(2)}$ with the inclusion $S\hookrightarrow V_{n-2}$ given by a point in $C^{(n-4)}$ (when $n=3$, $V_{n-2}$ is just $C$). Therefore, $V_{n-2}|_S$ is algebraically a coordinate curve $C_{2,Q}$ in $S\cong C^{(2)}$.

Moreover, $N|_S\cdot C=V_{n-1}|_S \cdot C=1$ and hence
\[
(V_{n-2}|_S-N|_S)\cdot C=(C_{2,Q}-V_{n-1}|_S)\cdot C=0
\]
and
\[
\begin{array}{c}
(V_{n-2}|_S-N|_S)^2=(C_{2,Q}-V_{n-1}|_S)^2=\\
C_{2,Q}^2-2C_{2,Q}\cdot V_{n-1}|_S + (V_{n-1}|_S)^2 =-1+(V_{n-1}|_S)^2\geq 0
\end{array}
\]
because $V_{n-1}$ is ample in $V_n$.

Since $C$ is ample in $S$, by the Hodge index Theorem, we deduce that $V_{n-2}|_S-N|_S$ is numerically trivial. In fact, it is algebraically trivial, because there is no torsion in $H^2(C^{(2)}, \mathbb{Z})$ (see \cite{MD}).

By the Lefschetz Theorem for Picard Groups applied to the chain of $V_i$'s we have that the restriction map gives an injective morphism $Pic(V_{n-1}) \hookrightarrow Pic(S)$. Then, from the isomorphism $Pic^0(V_{n-1})\cong Pic^0(S)$ and
\[
\mathcal{O}_S(V_{n-2}|_S-N|_S)\in Pic^0(S)
\]
we deduce that
\[
\mathcal{O}_{V_{n-1}}(V_{n-2}-N)\in Pic^0(V_{n-1}).
\]
Consequently, by the isomorphism between the $Pic^0$'s, there exists an $\alpha$ as claimed.

Finally, since $V_{n-2}$ and $N$ are numerically equivalent, we obtain that $1=V_{n-2}^{n-1}=N^{n-1}=(V_{n-1}|_{V_{n-1}})^{n-1}=V_{n-1}^n$. $\Diamond$

Now, consider the exact sequence
\begin{equation}\label{exactsequence}
0\rightarrow \mathcal{O}_{X} \rightarrow \mathcal{O}_{X}({V_{n-1}}) \rightarrow \mathcal{O}_{V_{n-1}}({V_{n-1}}) \rightarrow 0.
\end{equation}

Let $\mathcal{W}_n$ be the image of $\mathcal{W}$ by the isomorphism $Pic^0(V_{n-1})\cong Pic^0(S)$ and $\eta \in \mathcal{W}_n$ a general element. We tensor \eqref{exactsequence} with $\alpha\otimes \eta$ and get
\[
0 \rightarrow  \alpha\otimes \eta \rightarrow \alpha\otimes \eta \otimes \mathcal{O}_{X}({V_{n-1}}) \rightarrow \mathcal{O}_{V_{n-1}}(V_{n-2})\otimes\eta|_{V_{n-1}} \rightarrow 0.
\]
We take cohomology and obtain
\[
\begin{array}{ll}
0 \rightarrow  & H^0({X},\alpha\otimes \eta) \rightarrow H^0({X},\alpha\otimes \eta \otimes \mathcal{O}_{X}({V_{n-1}})) \rightarrow \\[4pt]
&H^0({V_{n-1}}, \mathcal{O}_{V_{n-1}}(V_{n-2})\otimes\eta|_{V_{n-1}}) \rightarrow H^1({X}, \alpha\otimes \eta) \rightarrow \dots
\end{array}
\]

First of all, we observe that $H^0({X},\alpha\otimes \eta)=0$ since $\alpha\otimes \eta \in Pic^0({X})$ is non trivial.

Second, we notice that the image of the Albanese morphism of $X$ has dimension greater or equal than two. Indeed, we know that the image of the Albanese morphism of $S\cong C^{(2)}$ is a two dimensional subvarity of $Alb(S)=J(C)$. By the identification of $Pic^{0}$'s, this subvariety of $J(C)$ lives inside the image of the Albanese morphism of $X$, hence, it is of dimension at least two. We can apply generic vanishing results and deduce that $V^1(X)=\{\varsigma\in Pic^0(X)\ |\ h^1(X, \varsigma)>0\}$ is the union of finitely many translates of proper abelian subvarieties.

Furthermore, we know that $W_1(C)$ generates $Pic^0(C)$, hence, its image by the identification $Pic^0(X)\cong Pic^0(C)$ generates $Pic^0(X)$. When we translate it by a fixed element $\alpha\in Pic^0(X)$ it still generates, so by the generic vanishing results, it cannot be contained in $V^1(X)$. Hence, for a general $\eta\in \mathcal{W}_n$ we obtain that $\alpha \otimes \eta \notin V^1(X)$ and thus $H^1(X, \alpha \otimes \eta)=0$.

Therefore, for $\eta\in \mathcal{W}_n$ general we have that
\[
h^0(X,\alpha\otimes \eta \otimes \mathcal{O}_X(V_{n-1}))=h^0(V_{n-1}, \mathcal{O}_{V_{n-1}}(V_{n-2})\otimes\eta|_{V_{n-1}})=1>0.
\]
And by semicontinuity, $h^0(X, \alpha \otimes \eta \otimes \mathcal{O}_{X}({V_{n-1}}))>0$ for all $\eta\in \mathcal{W}_n$.

Thus, we have a $1$-dimensional family, $\mathcal{D}$ in ${X}$, of effective divisors algebraically equivalent to ${V_{n-1}}$. Let $H_{\eta}$ denote the effective divisor in $X$ such that $\mathcal{O}_X(H_{\eta})=\mathcal{O}_X(V_{n-1})\otimes \alpha \otimes \eta$. We observe that:
\begin{itemize}
\item $V_{n-1}\cdot H_{\eta} =(V_{n-2})_{\eta}$. Indeed,
\[
\mathcal{O}_{V_{n-1}}(H_{\eta})=\mathcal{O}_{V_{n-1}}(N)\otimes \alpha|_{V_{n-1}}\otimes \eta=\mathcal{O}_{V_{n-1}}({V_{n-2}})\otimes \eta
\]
where we consider $\eta \in Pic^0(V_{n})$ or $Pic^0(V_{n-1})$ indistinctively by the isomorphism given by the restriction map.

\item Since $H_{\eta}$ is algebraically equivalent to $V_{n-2}$, we have that $H_{\eta}$ is ample and $H_{\eta}^n=1$. Hence, $Pic^0(X) \cong Pic^0(H_{\eta})\cong Pic^0(C)$. In particular, when $H_{\eta}$ is smooth, $q(H_{\eta})=g(C)$.

\item If $H_{\eta}$ is smooth, since $V_{n-1}\cdot H_{\eta} =(V_{n-2})_{\eta}\cong C^{(n-2)}$, we can apply the induction hypothesis to $H_{\eta}$ and deduce that $H_{\eta} \cong C^{(n-1)}$. In addition we obtain that in $Pic^0(H_{\eta})$ there is a $1$-dimensional family $\{\varsigma \in Pic^0(H_{\eta})\ |\ h^0(H_{\eta}, \mathcal{O}_{H_{\eta}}((V_{n-2})_{\eta})\otimes \varsigma)>0\}$ which is the image of $\mathcal{W}$ via the identification $Pic^0(H_{\eta})\cong Pic^0(C)$.
\end{itemize}

Assume for a moment that $H_{\eta}$ is smooth for $\eta$ generic.

Since $C^{(i)}$ deforms in an algebraic family only as the $i$th symmetric product of a curve, we deduce that the general element in $\mathcal{D}$ is birational to $C^{(n-1)}$. Moreover, since $V_{n-1}\cdot H_{\eta} = (C^{n-2})_{\eta}$ we obtain that the restriction of $\mathcal{D}$ to a general divisor in the family is the family of coordinate divisors in $H_{\eta}\approx C^{(n-1)}$.

Consequently, since the index of the family of coordinate divisors on $C^{(n-1)}$ is $n-1$, we deduce that the index of $\mathcal{D}$ in $X$ is $n$. Indeed, given a general point in $H_{\eta}$, we have $n-1$ other elements of $\mathcal{D}$ containing it, that together with $H_{\eta}$ are a total of $n$ elements of the family.

Next, we see that indeed $X\cong C^{(n)}$.

Let $Q\in X$ be a general point and let $H_1, \dots, H_n$ be the divisors in $\mathcal{D}$ containing $Q$. Let $D_1=V_{n-1}\cdot H_1$, then $D_1$ is a coordinate divisor in $V_{n-1}\cong C^{(n-1)}$, hence, it is of the form $C^{(n-2)}+{P_1}$, for certain $P_1\in C$. In a similar way, $H_i\cdot V_{n-1}=C^{(n-2)}+{P_i}$. Thus, we have a birational map
\[
\begin{matrix}
X & \dashrightarrow & C^{(n)} \\
Q & \dashrightarrow & P_1+\dots+ P_n.
\end{matrix}
\]

In fact, $X\cong C^{(n)}$. Indeed, any curve contracted by the birational map would have product $0$ with $V_{n-1}$, which is not possible since $V_{n-1}$ is ample in $X$. Observe finally that if $V_{n-1}\cdot H_{\eta}= C^{(n-2)}+{P}$, then $H_{\eta}=C_{n,P}$, the coordinate divisor with base point $P$, and hence, $\mathcal{W}_n$ parametrizes the coordinate divisors in $C^{(n)}$.

Finally, we study the possible singularities of the hypersurfaces $H_{\eta}$ to prove that indeed the general one is smooth.

First, a divisor $H_{\eta}$ does not contain a curve of singularities. Otherwise, since $V_{n-1}$ is ample, this curve would cut $V_{n-1}$ in a point, and then $(C^{(n-2)})_{\eta}$ should be singular, contradicting our hypothesis. Hence, each $H_{\eta}$ has at most a finite number of singularities.

Second, the possible singularities do not deform with the divisors in the family. Otherwise, there would be some curves $\{B_i\}$ such that the intersection point of $B_i$ and $H_{\eta}$ would be a singular point of $H_{\eta}$. Since $V_{n-1}$ is ample, a curve $B_i$ would intersect $V_{n-1}$ in a point $P\in (C^{(n-2)})_{\eta}$ for a certain $\eta$, and then $(C^{(n-2)})_{\eta}$ should be singular.

Third, there is no base curve for the family $\mathcal{D}$. Otherwise, this curve would intersect $V_{n-1}$ and then the family $\mathcal{D}_{n-2}$ of coordinate divisors in $C^{(n-1)}$ should have a base point.

Finally, there is no singularity $Q$ common to all $H_{\eta}\in \mathcal{D}$. Otherwise, the point $Q$ would be a base point of the family and all varieties would have multiplicity at least two at this point, therefore, $V_{n-1}^n\geq 2$, contradicting $V_{n-1}^n=1$.

Therefore, not all elements $H_{\eta}\in \mathcal{D}$ are singular, in fact, the general one is smooth, and those singular have at most isolated singularities.
\end{proof}

From the theorem we deduce the following result, less general but simpler in its hypothesis.

\begin{cor}
Let $X$ be a smooth projective variety of dimension $n$. Assume that there exists a divisor $D$ isomorphic to $C^{(n-1)}$ such that, if $C$ denotes a coordinate curve in $D\cong C^{(n-1)}$, then $D\cdot C=1$. Assume also that $q(X)=g(C)$. Then $X\cong C^{(n)}$.
\end{cor}

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\end{document}


TITLE: Relation between line bundles over $\mathbb{P}^1$ in geometry and topology
QUESTION [3 upvotes]: For an algebraic geometry assignment I recently had to prove that the isomorphism classes of line bundles over $\mathbb{P}_k^n$ for any field $k$ are the powers of the tautological line bundle, or in other words the Picard group is given by Pic($\mathbb{P}_k^n) \cong \mathbb{Z}$.  I was trying to consider simple examples to reconcile this with my intuition about line bundles from algebraic topology.
Taking $k = \mathbb{R}$ and $n = 1$, we have Pic($\mathbb{RP}^1) \cong \mathbb{Z}$.  Naively, I might have thought that since $\mathbb{RP}^1 \cong S^1$, there would be only the two isomorphism classes of (topological) line bundles over the circle.  I was wondering: what are the relations between line bundles in algebraic geometry and topology in this case, and why the discrepancy?
I figure maybe the real numbers are not a very well behaved field in algebraic geometry. If $k = \mathbb{C}$ and we again take $n = 1$, do the classifications of bundles match up on $\mathbb{CP}^1 \cong S^2$ in the geometric and topological contexts?

REPLY [8 votes]: You need to be careful what kind of $\mathrm{Pic}$ you are talking about!
The point is that if $X/\mathbb{R}$ is a finite type $\mathbb{C}$-scheme then $X(\mathbb{R})$ (resp. $X(\mathbb{C})$) is a real (resp. complex) manifold (the former being only 'locally a manifold' depending on what sort of axioms of a point-set topological nature you impose). This then allows you to define maps
$$\mathrm{Pic}_\mathrm{alg}(X)\to \mathrm{Pic}_\mathrm{smooth}(X(\mathbb{R}))\to\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{R}))\qquad (1)$$
and maps
$$\mathrm{Pic}_\mathrm{alg}(X_\mathbb{C})\to \mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))\to\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))\qquad (2)$$
But, in general, these maps needn't all be isomorphisms!
For example if $M$ is a smooth real manifold then there is actually an isomorphism
$$\mathrm{Pic}_\mathrm{smooth}(M)\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{cont.}}(M)\cong H^1_\mathrm{sing}(M,\mathbb{Z}/2\mathbb{Z})$$
The latter isomorphism comes from the fact that the classifying space of continuous real line bundles is $\mathbb{RP}^\infty$ which is a $K(\mathbb{Z}/2\mathbb{Z},1)$. The former isomorphism can be thought about in two ways:

The fact that we have smooth approximation for maps $M\to \mathbb{RP}^\infty$
The fact that smooth real line bundles should be classified by $\mathcal{O}_M^\times$. There is then a SES
$$0\to \mathcal{O}_M\to \mathcal{O}_M^\times\to \underline{\mathbb{Z}/2\mathbb{Z}}\to 0$$
of sheaves and the fact that $H^1(M,\mathcal{O}_M)=H^2(M,\mathcal{O}_M)=0$ because the sheaves $\mathcal{O}_M$ are so-called 'fine' (and so are acyclic).

From this we see that we can refine $(1)$ to
$$\mathrm{Pic}_\mathrm{alg}(X)\to \mathrm{Pic}_\mathrm{smooth}(X(\mathbb{R}))\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{R}))$$
but this former map needn't be an isomorphism. As you pointed out, if $X=\mathbb{P}^1_\mathbb{R}$ then this fomer map is
$$\mathrm{Pic}_\mathrm{alg.}(\mathbb{P}^1_\mathbb{R})\cong \mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}= \mathrm{Pic}_{\mathrm{smooth}}(\mathbb{RP}^1)$$
where explicitly the map takes $\mathcal{O}(n)$ to the trivial bundle if $n$ is even and the Mobius bundle if $n$ is odd! The point being that while the algebraic structures on $\mathcal{O}$ and $\mathcal{O}(2n)$ (as well as $\mathcal{O}(1)$ and $\mathcal{O}(2n+1)$) are not algebraically equivalent, they are smoothly equivalently. Try it for yourself with overlap maps $x$ and $x^3$!
In the $\mathbb{C}$-case if you assume that $X$ is, in addition, proper then you actually get an isomorphism
$$\mathrm{Pic}_{\mathrm{alg.}}(X)\xrightarrow{\approx}\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))$$
this follows from Serre's GAGA results. Moreover, we can describe the map $\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C}))\to \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))$ quite nicely. Namely, the complex continuous line bundles on $X(\mathbb{C})$ can be described as $H^2_\mathrm{sing}(X,\mathbb{C})$. Again, the reason for this is that the classifying space of complex line bundles is $\mathbb{CP}^\infty$ which is a $K(\mathbb{Z},2)$. The holomorphic line bundles on $X(\mathbb{C})$ are classified by $H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}^\times)$. The connection between the two is given by the exponential sequence
$$0\to \underline{\mathbb{Z}}\to \mathcal{O}_{X(\mathbb{C})}\to\mathcal{O}_{X(\mathbb{C})}^\times\to 0$$
where the second map is $f\mapsto \exp(2\pi i f)$. Taking the long exact sequence in cohomology we get (part of the ) long exact exponential sequence
$$H^1_\mathrm{sing}(X,\mathbb{Z})\to H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})\to H^1(X(\mathbb{C}),\mathcal{O}_X^\times)\to H^2_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z})\to H^2(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})$$
And, in fact, the natural diagram
$$\begin{matrix}\mathrm{Pic}_{\mathrm{hol.}}(X(\mathbb{C})) & \to & \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C})\\ \downarrow & & \downarrow\\ H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}^\times) & \to & H^2_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z})\end{matrix}$$
commutes with the vertical maps being isomorphisms. This map is called the Chern class of an algebraic/holomorphic line bundle.
So, if $X$ is a smooth projective (geometrically) connected curve of genus $g$ we see that the map
$$\mathrm{Pic}_{\mathrm{alg.}}(X(\mathbb{C}))\to \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))$$
is surjective with kernel a quotient of a vector space of dimension $g$. So, if $X=\mathbb{P}^1_\mathbb{R}$ the kernel is trivial and we get the desired isomorphism
$$\mathrm{Pic}_{\mathrm{alg.}}(\mathbb{P}^1_\mathbb{C})\cong \mathrm{Pic}_{\mathrm{cont.}}(X(\mathbb{C}))\cong H^2(\mathbb{CP}^1,\mathbb{Z})\cong \mathbb{Z}$$
In fact, what's generally true is that if $X/\mathbb{R}$ is a smooth projective (geometrically) connected curve then the (portion of )the long exact exponental sequence can be written
$$\begin{matrix}H^1_\mathrm{sing}(X(\mathbb{C}),\mathbb{Z}) & \to & H^1(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})}) & \to & \mathrm{Pic}(X_\mathbb{C}) & \to & H^2(X(\mathbb{C}),\mathbb{Z}) & \to & H^2(X(\mathbb{C}),\mathcal{O}_{X(\mathbb{C})})\\ \downarrow & & \downarrow & & & &\downarrow & & \downarrow\\ \mathbb{Z}^{2g} & &\mathbb{C}^g & & &  & \mathbb{Z} & & 0\end{matrix}$$
And, if you take it on faith (this is the beginning of Hodge theory!) that $\mathbb{Z}^{2g}$ is embedded into $\mathbb{C}^g$ as a lattice, then we see that we get a short exact sequence
$$0\to \mathbb{C}^g/\mathbb{Z}^{2g}\to \mathrm{Pic}(X_\mathbb{C})\to \mathbb{Z}\to 0$$
so that $\mathrm{Pic}(X_\mathbb{C})$ looks like a disconnected complex Lie group with component group $\mathbb{Z}$ and identity component an abelian variety (i.e. a compact complex Lie group). This identity component is called the Jacobian of $X_\mathbb{C}$ and is denoted $\mathrm{Jac}(X_\mathbb{C})$. The map from
$$\mathrm{Pic}(X_\mathbb{C})\to \mathbb{Z}=\pi_0(\mathrm{Pic}(X_\mathbb{C}))$$
is just the degree map. Of course, since $\mathbb{Z}$ is projective and discrete this sequence non-canonically splits to give you that $\mathrm{Pic}(X_\mathbb{C})\cong \mathrm{Jac}(X_\mathbb{C})\times\mathbb{Z}$.
For example, if you take $X=E$ an elliptic curve, then it turns out that $\mathrm{Jac}(E_\mathbb{C})\cong E_\mathbb{C}$!
All of this then starts the fascinating journey into the Albanese variety/Picard scheme of a smooth proper scheme!

The last thing I'll say that is that, in some sense, the algebraic bundles on $X/\mathbb{R}$ smooth projective are much more closely related to the holomorphic bundles on $X(\mathbb{C})$ than the continuous bundles on $X(\mathbb{R})$! In fact, there's the so-called 'Picard-Brauer sequence' which contains the terms
$$0\to \mathrm{Pic}(X)\to \mathrm{Pic}(X_\mathbb{C})^{\mathrm{Gal}(\mathbb{C}/\mathbb{R})}\to \mathrm{Br}(\mathbb{R})(\cong\mathbb{Z}/2\mathbb{Z})$$
From this we see that the algebraic bundles on $X$ embed into the algebraic bundles on $X_\mathbb{C}$ (which is equal to the holomorphic bundles on $X(\mathbb{C})$) and that up to a $\mathbb{Z}/2\mathbb{Z}$-term they exactly hit the Galois invariants of the algebraic bundles on $X_\mathbb{C}$.
In the case of $X=\mathbb{P}^1_\mathbb{R}$ this sequence is not very interesting. It looks like
$$0\to \mathbb{Z}\to\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$$
where the map $\mathbb{Z}\to\mathbb{Z}$ is an isomorphism and the map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ is trivial.
But, if instead of $\mathbb{P}^1_\mathbb{R}$ you took it's only non-trivial twist $X:=V(x^2+y^2+z^2)\subseteq\mathbb{P}^2_\mathbb{R}$ then your sequence actually looks like
$$0\to 2\mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}\to 0$$
The point being that if you take any degree $2$-point on $X$ then $\mathrm{Pic}(X)\cong \{\mathcal{O}(np):n\in\mathbb{Z}\}$. But, since $p$ is a point of degree $2$ when you base change to $\mathbb{C}$ you get that $p$ splits into two points--$q_0$ and its Galois conjugate $\sigma(q_0)$ so that $\mathcal{O}(p)$ maps to $\mathcal{O}(q_0)\otimes \mathcal{O}(\sigma(q_0))\cong \mathcal{O}(2)$.