TITLE: Overconstrained equilibrium binary mixture phase seggregation
QUESTION [1 upvotes]: Assume a binary mixture, with the component $A$ in concentration $c$ and the component $B$ in concentration $1-c$. The total energy $E(c)$ is thus given by a function of one variable $c$ (if we do not take into account the kinetic energy...). Now let's assume that energy has two minima at $c_0$ and $c_1$. The minimization of energy should lead to a phase segreggation.
But : In order for two phases $1$ and $2$ to be at equilibrium, we need two conditions : $\mu_1=\mu_2$ and $\pi_1=\pi_2$. Two equations for 1 variable !! The system is overconstrained and there cannot be any phase segregation. How come the minimization of the energy and basic themodynamics not compatible ?

REPLY [0 votes]: Actually my question was mostly wrong. There are indeed 2 variables : $\phi_1$ of the first phase and $\phi_2$ of the second phase. So we need 2 equations.
However, if the second phase $\phi_2$ is imposed (if we want the second phase to be at some given), then there will be overconstrained, and the potential will have to be "chosen" ad hoc, so that the constraints do not contradict each other.


text:
\section{The Properties of the inhomogeneous LWR model}
To apply the results for hyperbolic systems of conservation laws, we express the inhomogeneity factor $a(x)$ as an additional conservation law, i.e., $a_t=0$. Hence, we can write the inhomogeneous LWR model as
\bqn
U_t+F(U)_x&=&0,\label{system}
\eqn
where $U=(a,\r), F(U)=(0,f(a,\r)),x\in R,t\geq 0$. In this chapter we consider one type of inhomogeneity -- variable number of lanes. The fundamental diagram is thus written as $f(a,\r)=\r\vs(\ra)$, given that all the lanes are of the same condition.

The inhomogeneous LWR model \refe{system} can be linearized as
\bqn
U_t+DF(U) U_x&=&0, \label{linear}
\eqn
where the differential $DF(U)$ of the flux vector $F(U)$ is
\bqn
DF&=&\mat{{cc}0&0\\-\frac {\r^2}{a^2} \vs'(\ra)&\vs(\ra)+\ra\vs'(\ra)}.
\eqn
The two eigenvalues for $DF$ are
\bqn
\l_0=0\qquad \l_1=\vs(\ra)+\ra\vs'(\ra).
\eqn
The corresponding right eigenvectors are
\bqs
\vecr_0=\mat{{c}\vs(\ra)+\ra\vs'(\ra)\\(\ra)^2\vs'(\ra)}
\qquad \vecr_1=\mat{{c}0\\1},
\eqs
and the left eigenvector of $\partial f/\partial \r$ as $\vecl_1=1$.
The system \refe{system} is a non-strictly hyperbolic system, since $\l_1$ may be equal to $\l_0=0$.

We consider a traffic state $U_{\ast}=(a_{\ast},\r_{\ast})$ as critical if
\bqn
\lambda_1(U_{\ast})=0\label{assum_1};
\eqn
i.e., at critical states, the two wave speeds are the same and system \refe{system} is singular.
For a critical traffic state $U_{\ast}$ we also have
\bqn
\frac {\partial}{\partial \r} \l_1(U_{\ast})&=&f_{\r\r}<0 \label{assum_2},
\eqn
and
\bqn
\frac{\partial}{\partial a} f(U_{\ast})&=&-(\ra)^2\vs'(\ra)\big|_{U_{\ast}}=\ra \vs(\ra)\big |_{U_{\ast}}>0. \label{assum_3}
\eqn

A consequence of properties \refe{assum_2} and \refe{assum_3} is that the linearized system \refe{linear} at $U_{\ast}$ has the following normal form
\bqn
\mat{{c} \delta a\\\delta \r}_t+\mat{{cc}0&0\\1&0}\mat{{c} \delta a\\\delta \r}_x&=&0. \label{nlinear}
\eqn
The system \refe{nlinear} has the solution $\delta \r(x,t)=\delta a'(x)t+c$, and the solution goes to infinity as $t$ goes to infinity. Therefore \refe{nlinear} is a linear resonant system, and the original inhomogeneous LWR model \refe{system} is a nonlinear resonant system.

For \refe{system}, the smooth curve $\Gamma$ in $U$-space formed by all critical states $U_{\ast}$ are  named a transition curve. Therefore $\Gamma$ is defined as
\bqs
\Gamma&=&\left\{ U \big | \l_1(U)=0\right\}.
\eqs
Since $\l_1(U)=\vs(\ra)+\ra\vs'(\ra)$, we obtain
\bqn
\Gamma &=&\left\{(a,\r)\big|\ra=\alpha, \m{ where } \alpha \m{ uniquely solves }\vs(\alpha)+\alpha \vs'(\alpha)=0\right\}; \label{trancur}
\eqn
i.e., the transition curve for \refe{system} is a straight line passing through the origin in $U$-space. In \refe{trancur}, $\alpha$ is unique since $f(a,\r)$ is concave in $\r$.

The entropy solutions to a nonlinear resonant system are different from those to a strict hyperbolic system of conservation laws. Isaacson \& Temple (1992\nocite{isaacson_t1992}) proved that solutions to the Riemann problem  for system \refe{system} exist and are unique with the conditions \refe{assum_1}-\refe{assum_3}. Lin et al. (1995\nocite{lin_t_w1995}) also presented solutions to a scalar nonlinear resonant system, which is similar to our system \refe{system} except that $f$ is convex. In the next section we apply those results to solve the Riemann problem for the inhomogeneous LWR model.

config: arxiv


TITLE: Interpreting $\{x~|~x\in\Bbb N_1~\land~x\leq50~\land~(\forall y~(y\in\{2,\dots,x/2\})\implies(x\not\equiv0\pmod y))\}$
QUESTION [0 upvotes]: I am learning universal quantifiers and existential quantifiers and I am still at the beginner level. I found a question in my lecture slides as following
$$\left\{\;x~|~x\in\Bbb N_1~\land~x\leq50~\land~(\;\forall y~[(y\in \{2,\dots,x/2\})\implies(x\not\equiv0\pmod y)]\;)\;\right\}$$
and the answer was given as
$$\{1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}$$
$\Bbb N_1$ means NAT1 which means natural numbers which are starting from $1$ ($1$ inclusive).
Unfortunately, I don't understand how does that answer come. So any help with a step by step explanation is greatly appreciated.

REPLY [1 votes]: Let us decompose the whole expression step by step. First, we have our intial declaration of what type of things we are even looking for to collect in our set

$$\{x~|~x\in\Bbb N_1\}=\{1,2,3,4,\dots\}\tag1$$

So we are looking for natural numbers $x$(starting by $1$). But they have to met some other conditions too, denoted by conjunction $\land$. Second, we have a restriction concerning their size

$$\{x~|~x\in\Bbb N_1~\land~x\leq50\}=\{1,2,3,4,\dots,49,50\}\tag2$$

Now, a new variable comes into play: $y$. In contrast to $x$ its not clearly mentioned that we also have $y\in\Bbb N_1$ as all of our $y$ are defined to belong to the list $2,3,\dots,x/2-1,x/2$, or to put it in a slightly different way: we have $y\in\{2,3,\dots,x/2-1,x/2\}$. We demand that for every $y\in\{2,3,\dots,x/2-1,x/2\}$ it holds that $x\not\equiv0\mod y$ simultaneously. This is encoded by the formula

$$\forall y((y\in\{2,3,\dots,x/2-1,x/2\})\implies(x\not\equiv0\mod y))\tag3$$

The role $\forall y$ plays here is to ensure the aforementioned simultaneity. Translating this into more familiar notions: we want to have that our number $x\in\Bbb N_1$ which is not divisble by any of the preceding numbers up to $x/2$ (we exclude all $y$ such that $x/2\leq y\leq x$ as they trivially fulfill this condition); that are precisely the primes (and $1$)!
Now, take a look at our complete set description

$$M=\{x~|~x\in\Bbb N_1~\land~x\leq50~\land~(\forall y~(y\in \{2,\dots,x/2\})\implies(x\not\equiv0\pmod y))\}$$

We conclude that by $(1)$ and $(2)$ that we are looking for natural numbers $x\leq50$ and by $(3)$ such that they are not divisble by any of the preceding numbers. These are precisely the primes $\leq50$ expanded by $1$. So our solution set is

$$\therefore~M=\{1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}$$


TITLE: Trying to find more information about "Darboux's method/theorem" on coefficients of an analytic function
QUESTION [4 upvotes]: My supervisor briefly showed me a statement of something she called "Darboux's theorem," but I am having trouble finding more information about it on the internet. Here is what I have written down (some details may be incorrect):

Let function $R$ be analytic in a neighborhood about $z=0$ of radius $\rho$. Suppose it has exactly one singularity at distance $\rho$ from $z=0$ (not sure about this sentence). If $R$
 can be written in the form
$$R(z) = \left(1-\frac zp\right)^{-s} G(z) + H(z)$$
where $s \notin \{0, -1, -2,\ldots\}$, where $G$ and $H$ are analytic in neighborhoods about $z=0$ of radius larger than $\rho$, and $G(\rho)\ne 0$, then the coefficient of $z^n$ in the power expansion of $R(z)$ follows this asymptotic description
$$[z^n] R(z) = \rho^{-n} n^{s-1} \frac{G(s)}{\Gamma(s)} \left(1 + O\left(\frac 1n\right)\right)$$

My guess is that it's called "Darboux's method" in English, but still I only have these two links (here and here), that don't have a statement that is exactly the above, although they seem close. Can anyone comment or help me find what I'm looking for? Also any corrections to what I've written above would be appreciated.

REPLY [4 votes]: You should take a look at the book Analytic Combinatorics, the section about analytic method contains an extensive discussion of Darboux's and other related methods.
The book is available here
http://algo.inria.fr/flajolet/Publications/books.html
In general these types of results are known as "Tauberian theorems".


TITLE: how to determine if the following sequence is monotonous
QUESTION [1 upvotes]: I have the following sequence:
$$U_n =\displaystyle \sum_{k=n}^{2n} \frac1k$$
I need to determine if the sequence is increasing or decreasing; to do this I want to calculate $U_{n+1} - U_n$ and know if it is positive or not. However, I do not know how to simplify such an expression containing sigma.

REPLY [0 votes]: If $S_n=\sum_{k=n}^{2n}\frac1k$, then we have
$$\begin{align}
S_{n+1}-S_n&=\sum_{k=n+1}^{2n+2}\frac1k -\sum_{k=n}^{2n}\frac1k\\\\
&=\frac{1}{2n+2}+\frac1{2n+1}-\frac1n
\end{align}$$
Can you finish now?


TITLE: Complicated series converges to $\pi$.
QUESTION [2 upvotes]: How do I get this result?
$$\frac {426880 \sqrt {10005}}{\large \sum_{k = 0}^{\infty}\frac {(6k)!(545140134k + 13591409)}{(k!)^3 (3k)! (-640320)^{3k}}} = \pi$$
It seems formidable.
Context: I came across this when reading a book (in Chinese) on sequences for high school students where the author started to introduce infinite sequences. He listed a few other famous results and then this, as an illustration that this kind of series can be really complicated.

REPLY [4 votes]: This series was obtained by Chudnovsky brothers through the theory of modular forms. To describe their formula let us define functions $P, Q, R$ via the equations
\begin{align}
P(q) &= 1 - 24\left(\frac{q}{1 - q} + \frac{2q^{2}}{1 - q^{2}} + \frac{3q^{3}}{1 - q^{3}} + \cdots\right)\notag\\
Q(q) &= 1 + 240\left(\frac{q}{1 - q} + \frac{2^{3}q^{2}}{1 - q^{2}} + \frac{3^{3}q^{3}}{1 - q^{3}} + \cdots\right)\notag\\
R(q) &= 1 - 504\left(\frac{q}{1 - q} + \frac{2^{5}q^{2}}{1 - q^{2}} + \frac{3^{5}q^{3}}{1 - q^{3}} + \cdots\right)\notag
\end{align}
where $q$ is a real number with $|q| < 1$ (so that the series above are convergent).
Let $n > 3$ be an integer and we write
\begin{align}
P_{n} &= P(-e^{-\pi\sqrt{n}}),\,\, Q_{n} = Q(-e^{-\pi\sqrt{n}}),\,\, R_{n} = R(-e^{-\pi\sqrt{n}})\notag\\
j_{n} &= 1728\frac{Q_{n}^{3}}{Q_{n}^{3} - R_{n}^{2}}\notag\\
b_{n} &= \sqrt{n(1728 - j_{n})}\notag\\
a_{n} &= \frac{b_{n}}{6}\left\{1 - \frac{Q_{n}}{R_{n}}\left(P_{n} - \frac{6}{\pi\sqrt{n}}\right)\right\}\notag
\end{align}
then Chudnovsky brothers establish the following general series for $1/\pi$ $$\boxed{{\displaystyle \frac{1}{\pi} = \frac{1}{\sqrt{-j_{n}}}\sum_{m = 0}^{\infty}\frac{(6m)!}{(3m)!(m!)^{3}}\frac{a_{n} + mb_{n}}{j_{n}^{m}}}}\tag{1}$$ The series in question is obtained by putting $n = 163$. Obtaining the actual numbers given in the series (like $13591409$) from the value of $n = 163$ is very very difficult via hand calculation.
I have shown in my blog post that the above general series $(1)$ can be obtained through Ramanujan's theory described in Ramanujan's most famous paper Modular Equations and Approximations to $\pi$ without using the deep and difficult theory of modular forms. Ramanujan explicitly did not obtain the formula $(1)$ but indicated a general method to obtain series for $1/\pi$ and the series given by Chudnovsky can also be obtained via this method. See my blog post for more details.
Also see one of my answers regarding Ramanujan's series for $1/\pi$.


TITLE: When the loudspeaker is arranged in a plane, I want to calculate the sound pressure level at a certain distance from the front side
QUESTION [1 upvotes]: I would like to calculate the initial sound pressure that must be guaranteed to guarantee 100dB at a distance of 2000m with a frequency of 17kHz.
Next, I would like to know how to arrange a few loudspeakers with an output of 140 dB to ensure the initial sound pressure. Should all loudspeakers have the same phase? Which arrangement structure is better?

REPLY [1 votes]: As already posted in this answer, in order to calculate the sound pressure at a specific point in space, knowing the pressure in some other point, you can use the inverse square law (this is the simplest case). In order to do that you first have to specify what the second point is. You cannot calculate the sound pressure at the position of the source, since the law has a singularity at that point. On top of that it is not physically valid to calculate sound pressure for short distances from the source, since you will be in the near-field where the inverse square law does not hold. The law does hold for short distances only in the ideal case where the source has zero dimensions and so the near-field (which is frequency/wavelength and source dimension dependent) has also zero dimensions. Of course this is not the case in any real source.
Just for completeness, the inverse square law is stated as (in polar coordinates)
$$p(r) = p_{ref}(r_{ref}) \frac{1}{\frac{r}{r_{ref}}} \implies p(r) = p_{ref}(r_{ref}) \frac{r_{ref}}{r}$$
where $r$ is the distance from the source of interest and $r_{ref}$ is the reference distance, at which the pressure $p_{ref}$ is known. Note that when $r$ is less than $r_{ref}$ you get increase in pressure. Also note that there is no square in the law for pressure. This is due to the fact that the initial formulation of the law is for intensity and pressure is a derived quantity.
In addition to geometrical attenuation, you also have to take into account the air absorption which is frequency, temperature and humidity dependent. For this, you can have a look at the linked resources of the aforementioned answer to get some representative tabulated (or extract from plots) values of attenuation per distance for various frequencies and conditions.
Now, regarding the loudspeaker arrangement, according to references in this answer (this is just a reference but there's a plethora of other sources you can search for), the maximum increase in pressure (power/intensity) you can achieve by arranging many sources in a linear arrangement is when all of them have the same phase characteristics (thus achieving coherent summation) in the frequency of interest and are weighted with the same coefficients (effectively, having the same amplitude, which of course can be unity, or else, maximum gain). In this case you achieve an $N$ multiplicative gain from the array compared to the individual element, where $N$ is the number of elements.
Replicating the formula presented in the linked answer to make some notes.
$$Y(\psi) = \frac{1}{N} \frac{\sin \left( N \frac{\psi}{2}\right)}{\sin \left( \frac{\psi}{2}\right)}$$
where $\psi$ is given by $\psi = \frac{2 \pi}{\lambda} d \cos (\theta)$ with $\lambda$ being the wavelength corresponding to the frequency of interest, $d$ the interelement distance (here assumed to be uniform in the array, thus talking about ULAs - Uniform Linear Array -) and $\theta$ the angle of interest. You should look for $\theta = 90^{o}$ (broadside of the array) where the elements will be arranged to "shoot". This way you achieve maximum again, which as stated (and can also be verified with the formula) will be $N$ times that of an individual source. Needless to say that all sources are assumed to be identical.
In case you arrange the sources to "shoot" in the direction parallel to the array axis you will end up with an end-fire array where the treatment is somewhat different and you will have to introduce additional delays to each source in order to achieve coherent summation. I believe (haven't double checked that) that the gain will be the same in this case, which is again $N$ times the amplitude of individual sources.
Please keep in mind that in real life, if there is considerable distance between the elements, geometrical attenuation (inverse square) and air absorption may apply to each source and the above formula may not yield exact results.


TITLE: Solution to Systems of Equations, Orthogonal Polynomials
QUESTION [2 upvotes]: Let $\mu_j=\int_a^b x^jw(x)dx$ be the $j$th moment of the weight distribution $w(x)$. Show that the linear system of equations
$$\left[\begin{matrix}
\mu_0 & \mu_1 & \cdots & \mu_{n-1} \\
\mu_1 & \mu_2 & \cdots & \mu_n \\
\vdots & \vdots & \ddots & \vdots \\
\mu_{n-1} & \mu_n & \cdots & \mu_{2n-2}
\end{matrix}\right]\left[\begin{matrix}c_0\\c_1\\\vdots\\c_{n-1}\end{matrix}\right]=\left[\begin{matrix}\mu_n\\\mu_{n+1}\\\vdots\\\mu_{2n-1}\end{matrix}\right]$$
has as solution the coefficients of a polynomial $x^n-\sum_{j=0}^{n-1} c_jx^j$, which is a member of the family of orthogonal polynomials associated with the weight function $w$.

By "has as solution" does that just mean the $c_i$ here are the same as in the sum?
I've tried expanding this matrix out and solving it but that doesn't really get anywhere.
Is there a trick that I'm not spotting?
Thanks for the help!

REPLY [2 votes]: An important thing to notice is the matrix is a gramian matrix G:
$$G_{i,j}=\int_a^b x^ix^jw(x)dx=(x^i|x^j)$$
where the right hand side denotes the inner product $(f|g)=\int_a^b f(x)g(x)w(x)dx$, on the linear space of polynomials.
Then, reading one line $i$, your system simply states that:
$$\sum_{j=0}^{n-1} (x^i|x^j)c_j=(x^i|x^n)$$
Or
$$(x^i|\sum_{j=0}^{n-1} c_jx^j)=(x^i|x^n)$$
That is, for all $i=0,\dots,n-1$, $(x^i|p_n)=0$, with
$$p_n=x^n-\sum_{j=0}^{n-1}c_jx^j$$
Therefore the $p_n$ form a family of orthogonal polynomials: $p_n$, of degree $n$, is orthogonal to all polynomials of degree $<n$, w.r.t. the inner product $(f|g)=\int_a^b f(x)g(x)w(x)dx$.

One also needs to check this system has actually a solution. For a given $n$, and two vectors $u=(d_0,\dots,d_{n-1})^T$ and $v=(c_0,\dots,c_{n-1})^T$, the product $Gv$ is a vector whose $i$-th row is $(x^i|\sum_{j=0}^{n-1} c_jx^j)$, as shown above. Thus
$$u^TGv=\sum_{i=0}^{n-1} d_i(x^i|\sum_{j=0}^{n-1} c_jx^j)=\left(\sum_{i=0}^{n-1}d_ix^i\right|\left. \sum_{j=0}^{n-1}c_jx^j\right)$$
That is, given two polynomials $P,Q$ of degree at most $n-1$, and their coefficients given bu $u,v$, then $u^TGv=(P|Q)$.
Thus $u^TGu=(P|P)=\int_a^b P^2(x)w(x)dx$, which is $>0$ when $P\neq0$, granted the density function is positive. Thus the matrix $G$ is symmetric definite positive, and it's invertible.


text:
\begin{document}

\begin{abstract}
We give the complete classification of Mukai pairs of dimension $4$ and rank $2$ with Picard number one, that is, pairs $(X,\sE)$ where $X$ is a Fano $4$-fold with Picard number one, and $\sE$ is an ample vector bundle of rank two on $X$ with $c_1(X) = c_1(\sE)$.
Equivalently, the present paper completes the classification of ruled Fano $5$-folds with index two, which was partially done by C.~Novelli and G.~Occhetta in 2007.
\end{abstract}

\maketitle

\section*{Introduction}
\subsection{}
A \emph{Fano manifold} $M$ is, by definition, a smooth projective variety with ample anticanonical divisor $-K_M$.
One of the most important and fundamental invariants for a given Fano manifold $M$ is its \emph{index}, denoted by $r_M$, which is defined as the greatest integer dividing $-K_M$ in the Picard group.
Roughly speaking, the index measures how large the anticanonical divisor is, and philosophically the positivity of anticanonical divisor poses some restriction on the (biregular) structure of $M$.
Indeed, Fano manifolds $M$ with $r_M \geq \dim M -2$ are completely classified in celebrated articles \cite{KO73,Fuj82a,Fuj82b,Muk89} (cf.\ \cite{Mel99,Amb99}).

On the other hand, it appears that, due to lack of knowledge on Calabi-Yau $3$-folds or Fano $4$-folds, the complete classification of Fano $n$-folds with index $n-3$ is far from being complete, particularly when the Picard number is one.
Nevertheless, if $n \geq 5$ and the Picard number is bigger than 1, then we still have some room to attack the problem by studying its contractions, whose existence is promised by the fundamental theorems in Mori theory.
It is known that, if $M$ is a Fano $n$-fold with index $n-3$ and $\rho_M>1$,  then $n\leq 8$  \cite{Wis90}.
Moreover, by works on Fano manifolds with large index  \cite{Wis90,Wis91,PSW92b,Wis93,BW96} (cf.\ \cite{Occ05}), such Fano manifolds are completely classified when $n \geq6$.
Also, in a series of papers \cite{CO06,NO07,CO08}, Chierici, Novelli and Occhetta started to classify Fano $5$-folds with index two and Picard number bigger than one by the above strategy, though the classification is not completed yet.

Among their study, the ruled case \cite{NO07} is of particular interest in the present paper.
In that paper, Novelli and Occhetta classified ruled Fano $5$-folds $M$ with index two with the assumption $\rho_M \geq 3$ (see Remark~\ref{rem:NO} for their treatment in the case $\rho_M=2$).
The purpose of this paper is to complete the classification of ruled Fano $5$-folds with index two by studying the case $\rho_M =2$, which are not treated sufficiently in \cite{NO07}.

\subsection{}
There is another point of view; \emph{classification of Mukai pairs with large rank}.
In 1988, Mukai \cite{Muk88} introduced study of pairs $(X,\sE)$ where $X$ is a Fano manifold, and $\sE$ is an ample vector bundle on $X$ with $c_1(X) = c_1(\sE)$.
In \cite{Kan17}, such a pair is called a \emph{Mukai pair}.
Given a Mukai pair $(X,\sE)$, its rank is defined as the rank of the bundle $\sE$.
This invariant ``rank $\sE$ for a Mukai pair'' is an analogue of ``index for a Fano manifold.''
For example, a Fano manifold $X$ with index $r_X$ gives a Mukai pair $(X, \cO(H_X)^{\oplus r_X})$ of rank $r_X$, where $H_X\coloneqq -K_X/r_X$ is the fundamental divisor of $X$.
Based on the above analogy, Mukai pairs are classified when $\rank \sE \geq \dim X -1$ around 1990s \cite{Fuj92,Pet90,Pet91,YZ90,Wis89b,PSW92b} (cf.\ \cite{Occ05}).

Therefore, it is natural to hope the classification of Mukai pairs with $\rank \sE = \dim X -2$ as a next step.
In \cite{Kan17}, the author classified such pairs when $\dim X \geq 5$.
Note that,  for the case $\rank \sE = \dim X -2$, the smallest possible value of $\dim X$ is three, and the classification in this case is equivalent to the classification of Fano $3$-folds, which is established by Fano, Iskovskih, Shokurov, Fujita, Mori and Mukai (see \cite{IP99} and references therein).
On the other hand, the classification of Mukai pairs with dimension $4$ and rank $2$ is equivalent to the classification of ruled Fano $5$-folds with index two by taking the projectivization $\bP(\sE)$, and, as we mentioned, such a classification is partially done by Novelli and Occhetta.
Thus, when we are dealing with the classification problem of Mukai pairs with $\rank \sE = \dim X-2$, the remaining part to be considered is the case $\dim X=4$, $\rank \sE =2$ and $\rho _X =1$.
The purpose of this paper is to classify Mukai pairs in this missing case, and hence to complete the classification of Mukai pairs of  $\rank \sE = \dim X -2$ in arbitrary dimension.

\begin{theorem}[Classification of Mukai pairs with dimension $4$ and rank $2$]\label{thm:main}
Let $(X,\sE)$ be a Mukai pair of dimension $4$ and rank $2$.
Assume that $\sE$ is indecomposable and $\rho _X =1$.
Then $(X,\sE)$ is isomorphic to either
\[
(V_4, p^*\sS^* _{\bQ^4} (1)) \text{ or } (V_5, \sS ^*_{V_5} (1)).
\]
Here we use the following symbols:
\begin{itemize}
 \item $V_4$ is a quartic del Pezzo $4$-fold obtained as a double cover $p \colon V_4 \to \bQ ^4$ of the hyperquadric of dimension $4$ branched along a smooth divisor $B \in \left| \cO_{\bQ^4}(2) \right|$.
 \item $\sS_{\bQ^4}$ is a spinor bundle on $\bQ^4$.
 \item $V_5$ is a general linear section of the Grassmannian $\Gr (2,5)$ embedded into the projective space $\bP ^9$ via the Pl\"ucker embedding.
 \item $\sS _{V_5}$ is the restriction of the universal subbundle on $\Gr(2,5)$ to $V_5$.
\end{itemize}
\end{theorem}

\begin{remark}[Novelli and Occhetta's result]\label{rem:NO}
The classification problem of pairs as in the assumption of Theorem~\ref{thm:main} was also discussed in Section~8 of Novelli and Occhetta's paper \cite{NO07}.
Essentially, they classified such pairs with the extra assumption $r_X \geq 2$, which is not required in this paper.
Also, since the author could not follow their argument (see Remark~\ref{rem:comp}), we will provide a proof of Theorem~\ref{thm:main} that is independent from their argument.
\end{remark}

\subsection{}
Here we briefly sketch the strategy of the proof and give an outline of this paper.
Let $(X,\sE)$ be a Mukai pair as in Theorem~\ref{thm:main}.
  Then $W \coloneqq \bP(\sE)$ is a Fano $5$-fold with index two and $\rho_W=2$.
 We will denote by $\pi \colon W \to X$ the natural projection.
 Since $\rho _W =2$, it admits another contraction $\varphi \colon W \to Y$.
Note that, in each outcome of Theorem~\ref{thm:main},
\begin{enumerate}
\item the corresponding manifold $Y$ is isomorphic to a projective space $\bP^m$ and
\item each $\pi$-fiber gives a line in $Y \simeq \bP^m$.
\end{enumerate}
In our proof, a key step consists of proving these properties for our Mukai pair $(X,\sE)$.
Then we have a morphism $X \to \Gr (2,m+1)$, which will be proved to satisfy the desired description of the pair $(X,\sE)$.

In Section~\ref{sect:pre}, we gather some results on fundamental extremal contractions, such as scrolls, quadric bundles and blow-ups.
Those results will be applied to $\varphi$ in later sections.

In Section~\ref{sect:RC}, we will study rational curves on a Mukai pair $(X,\sE)$ as in Theorem~\ref{thm:main}.
There we will see that $X$ is covered by rational curves with anticanonical degree $3$.
Such rational curves are called \emph{minimal rational curves}.
For a minimal rational curve $f \colon \bP^1 \to X$, the pull-back $f^*\sE$ is isomorphic to $\cO(2) \oplus \cO (1)$, and thus the pull-back of $\bP(\sE)$ over $\bP^1$ is the Hirzebruch surface $\bF_1$.
Then the minimal section of this ruled surface gives a rational curve on $\bP(\sE)$, which we call the \emph{minimal lift} of the minimal rational curve $f \colon \bP^1 \to X$.

In Section~\ref{sect:stable}, we will see that the bundle $\sE$ is stable, and hence the Bogomolov inequality holds for $\sE$ (Proposition~\ref{prop:Bogomolov}).
This inequality is crucial in our proof of Theorem~\ref{thm:main}, and will be used twice: first to give rough description of the elementary contraction $\varphi$ of $\bP(\sE)$ (Section~\ref{sect:rough}), and second to prove that minimal lifts of minimal rational curves are contracted by $\varphi$ (Propositions~\ref{prop:Ineq} and \ref{prop:comp}).

In Section~\ref{sect:rough}, we will give a rough description of $\varphi$; the morphism $\varphi$ is  either a quadric bundle or a special \Banica scroll (see Definition~\ref{def:sBscroll} and Theorem~\ref{thm:phi}).
Then, in Sections~\ref{sect:quarticDP} and \ref{sect:quinticDP}, we will complete the proof of Theorem~\ref{thm:main} by studying each case.

\medskip
In the subsequent sections, we frequently use the following setting and notations.
\begin{setting}[Mukai pairs of dimension $4$ and rank $2$]\label{set}
$(X,\sE)$ is a Mukai pair of dimension $4$ and rank $2$ such that $\sE$ is indecomposable and $\rho _X =1$.

 We will use the following notations:
\begin{itemize}
 \item $W \coloneqq \bP(\sE)$ is the Grothendieck projectivization of $\sE$.
 \item $\pi \colon W \to X$ is the natural projection.
 \item $\varphi \colon W \to Y$ is the other elementary contraction.
 \item $\xi=\xi_\sE$ is the tautological divisor of the projectivization $\bP(\sE)$.
 \item $H_X$ (resp.\ $H_Y$) is the ample generator of $\Pic(X)$ (resp.\ $\Pic(Y)$).
 \item $r_X$ is the index of $X$.
 \item $l_X$ is the length of $X$.

 \item $R_\pi \subset \NE(X)$ (resp.\ $R_\varphi\subset \NE(X)$) the ray corresponding to $\pi$ (resp.\ $\varphi$).
 \item $\tR \subset \NE(X)$ is the half line $\bR_{\geq0}[\tC]$ spanned by the class $[\tC]$ of minimal lifts of minimal rational curves.
\end{itemize}
\end{setting}

\begin{convention}
 \hfill
\begin{enumerate}
 \item For a morphism $f \colon X \to Y$ and a coherent sheaf $\sF$ on $Y$, we will denote by $\sF|_X$ the pull-back $f^*\sF$ if no confusion arises.
 \item A \emph{rational curve} $C$ on a Fano manifold $X$ is, by definition, a $1$-dimensional closed subvariety whose normalization is a projective line $\bP^1$.
Sometimes the normalization map $f \colon \bP^1 \to X$ itself is also referred to as a rational curve.
\end{enumerate}
\end{convention}

\begin{acknowledgements}
The author wishes to express his gratitude to Professors Shigeru Mukai and Hiromichi Takagi for their valuable comments and suggestions.
He is also grateful to Professor Gianluca Occhetta for his helpful comments on the paper \cite{NO07}.
The author is also grateful to Doctor Takeru Fukuoka for his helpful comments and discussions on quintic del Pezzo manifolds.
He also wishes to thank Professor Hiraku Kawanoue for useful comments on the proof of Proposition~\ref{prop:Ineq}.

Main part of this work was done at the Graduate School of Mathematical Sciences, the University of Tokyo, and this paper was prepared at Research Institute for Mathematical Sciences,
Kyoto University.
\end{acknowledgements}

\section{Preliminaries: fundamental extremal contractions}\label{sect:pre}

Here we briefly review some facts on fundamental extremal contractions.
Let $M$ be a smooth projective variety and $\varphi \colon M \to N$ be its elementary contraction, that is, a contraction associated with a $K_M$-negative extremal ray $R$ of $\cNE(M)$.
Recall that the length $l(R_\varphi)$ of the ray $R$ is defined as the minimum anticanonical degree of rational curves that are contracted by $\varphi$:
\[
l(R)\coloneqq\{\, -K_M\cdot C \mid\, \text{$C$ is a rational curve such that $[C] \in R$} \}.
\]
The following is a fundamental inequality, which gives a bound of dimensions of the exceptional locus and fibers from below in terms of length.
\begin{theorem}[{Ionescu-Wi\'sniewski inequality \cite[Theorem~0.4]{Ion86}, \cite[Theorem~1.1]{Wis91}}]\label{thm:IWineq}
 Let $\varphi \colon M \to N$ be an elementary contraction associated to an extremal ray $R$, $E$ an irreducible component of $\Exc (\varphi)$ and $F$ a fiber contained in $E$.
 Then
 \[
 \dim E + \dim F \geq \dim M + l(R) -1.
 \]
\end{theorem}

Assume that the ray $R$ is supported by a Cartier divisor of the form $K_M+rL$, where $L$ is a $\varphi$-ample Cartier divisor and $r \in \bZ_{>0}$ is a positive integer.
Then $l(R) \geq r$, and thus we have
\[
 \dim E + \dim F \geq \dim M + r -1.
\]
Therefore, if $\varphi$ is of fiber type, then $\dim F \geq r-1$.
Hence, by taking $F$ as a general $\varphi$-fiber, we have
\[
\dim M-\dim N +1\geq r.
\]

\begin{definition}[Scrolls and quadric fibrations]
An elementary contraction $\varphi$ is called an \emph{adjunction theoretic scroll}, or simply a \emph{scroll}, if it is of fiber type and there is a $\varphi$-ample line bundle $L \in \Pic (M)$ such that
\[
K_M + (\dim M -\dim N +1)L
\]
is a supporting divisor for the contraction $\varphi$.

Similarly, the contraction is called a \emph{quadric fibration} if it is of fiber type and there is a $\varphi$-ample line bundle $L \in \Pic (M)$ such that
\[
K_M + (\dim M -\dim N)L
\]
is a supporting divisor for the contraction $\varphi$.
\end{definition}
Note that in the above definition $\varphi$ is assumed to be  elementary.
Also, in each case, a general $\varphi$-fiber is isomorphic to $\bP^n$ or $\bQ^n$ respectively.

The projection of a projectivized vector bundle $\bP(\sF) \to N$ is a typical example of scrolls.
Conversely, Fujita proved that an equidimensional scroll $\varphi \colon M \to N$ is a projection of a projectivized vector bundle \cite[Lemma~2.12]{Fuj87}.
The corresponding result for quadric fibrations is proved by Andreatta, Ballico and Wi\'sniewski:
\begin{theorem}[Characterization of quadric bundles {\cite[Theorem~B]{ABW93}}]\label{thm:Qbundle}
 Let $\varphi \colon M \to N$ be an equidimensional quadric fibration (from a smooth manifold $M$).
 Then $N$ is smooth and $\varphi$ is a quadric bundle, that is,  there exists a vector bundle $\sF$ of rank $\dim M -\dim N +2$ on $N$ such that $M$ is embedded into $\bP(\sF)$ as a divisor of relative degree two.
\end{theorem}

We also need the following characterization of bolw-ups.
\begin{theorem}[Characterization of blow-ups {\cite[Theorem~4.1]{AW93}}]\label{thm:Bl}
Let $r\in \bZ_{>0}$ be a positive integer and $\varphi \colon M \to N$ an elementary birational contraction whose nontrivial $\varphi$-fibers have dimension $r$.
Assume that there is a $\varphi$-ample line bundle $L$ such that  $K_M + rL$ gives a supporting divisor of the contraction $\varphi$.
Then $N$ is smooth and $\varphi$ is a blow-up of N along a smooth subvariety of codimension $r+1$.
\end{theorem}

\subsection{Special \Banica scrolls}
A special kind of scrolls, called \emph{\Banica scrolls}, are introduced in \cite{BW96}.
Here we will briefly recall the definition  and some properties of  \Banica sheaves and \Banica scrolls, based on \cite{BW96,AW93}.

\begin{definition}[\Banica sheaf]
 A \emph{\Banica sheaf} $\sF$ on a normal variety $N$ is a coherent sheaf $\sF$ on $N$ whose projectivization $\bP(\sF)$ is a smooth variety.
\end{definition}
Let $\sF$ be a \Banica sheaf on a normal variety $N$.
Then $M\coloneqq \bP(\sF)$ is a smooth variety.
Let $\varphi \colon M \to N$ be its projection.
Then it is known that the sheaf $\sF$ is reflexive and that the contraction $\varphi$ is an elementary contraction \cite[Lemma~2.2]{BW96}.
Moreover, the contraction $\varphi$ is a scroll.
A scroll obtained in this way is called a \emph{\Banica scroll}.
Conversely, a sufficient condition for a scroll to be a \Banica scroll is given in \cite{AW93,BW96} as follows:

\begin{theorem}[{\cite[Theorem~4.1, Remark~4.12]{AW93}, cf.\ \cite[Proposition~2.5]{BW96}}]\label{thm:Bscroll}
 Let $\varphi \colon M \to N$ be a scroll and  $L \in \Pic (M)$ a $\varphi$-ample line bundle on $M$ that gives a supporting divisor
\[
K_M + (\dim M -\dim N +1)L
\]
for the contraction $\varphi$.

Assume that
\[
\dim F \leq \dim M -\dim N +1
\]
holds for every $\varphi$-fiber $F$.
Then $N$ is smooth and $M \simeq \bP(\varphi_*L)$.
In particular, $\varphi_*L$ is a \Banica sheaf.
\end{theorem}

\begin{definition}[Special \Banica scrolls]\label{def:sBscroll}
 A scroll satisfying the assumptions in Theorem~\ref{thm:Bscroll} is called a \emph{special \Banica scroll}.
\end{definition}

A more geometric description of special \Banica scrolls is presented in \cite{AW93} as follows:
Let $\varphi \colon M \to N$ be a special \Banica scroll and $Z \subset N$ the set of points over which the dimension of fibers is $\dim M - \dim N +1$.
Set $E \coloneqq \varphi^{-1} (Z)$.

\begin{proposition}[{Geometry of special \Banica scrolls \cite[Remark~4.13]{AW93}}]\label{prop:Bscroll}
 Let the notation be as above.
 Then the following hold:
\begin{enumerate}
 \item $Z$ is smooth of codimension $\dim M -\dim N +2$.
 \item The morphism $\varphi|_{E} \colon E \to Z$ is a smooth $\bP^{\dim M -\dim N +1}$-bundle.
 \item We have the following commutative diagram with $\bP^{\dim M -\dim N}$-bundle $\tvarphi$
 \[
 \xymatrix{
 \tM \ar[r]^-{\tvarphi} \ar[d] & \tN \ar[d]\\
 M \ar[r]^-\varphi & N,
 }
 \]
 where $\tM \to M$ is the blow-up of $M$ along $E$, and $\tN \to N$ the blow-up of N along $Z$.
\end{enumerate}
\end{proposition}

\section{Rational curves and Mukai pairs}\label{sect:RC}

 Let $X$ be a smooth Fano manifold.
Recall that a rational curve $f \colon \bP^1 \to X$ is called \emph{free} if the pull-back of the tangent bundle $f^*T_{X}$ is a nef vector bundle on $\bP^1$.
Then the \emph{length} $l_X$ is defined as the minimum anticanonical degree of free rational curves on $X$.
Note that we have $l_X \geq r_X \geq 1$ by definition and that these values are at most $\dim X +1$ by Mori's theorem \cite{Mor79}.
This value $l_X$ is of particular interest because of the following  theorem, which gives a characterization of projective space and hyperquadric in terms of length.
\begin{theorem}[{Characterization of projective space and hyperquadric \cite{CMSB02,Miy04a}, cf.\ \cite{Keb02,DH17}}]\label{thm:PQ}
Let $X$ be a Fano manifold with Picard number one.
If $l_X \geq \dim X$, then $X$ is isomorphic to either a projective space $\bP^n$ or a hyperquadric $\bQ^n$.
\end{theorem}

In this section, we will prove $l_X=3$ for a Mukai pair as in Setting~\ref{set} (Lemma~\ref{lem:lX}).
Thus, we have a family of rational curves on $X$ whose anticanonical degree is $3$.
Further, we will construct minimal liftings of these rational curves (Definition~\ref{def:lift} and Lemma~\ref{lem:lift}).

The following lemma will be frequently used later.
\begin{lemma}[Splitting type of $\sE$]\label{lem:splitting}
 Let $(X,\sE)$ be a Mukai pair of dimension $n$ and rank $r$, and $f \colon \bP^1 \to C \subset X$ a rational curve on $X$.
 Then the following hold:
\begin{enumerate}
 \item \label{lem:splitting1} $f^* \sE \simeq \bigoplus_{i=1}^r \cO(a_i)$ for positive integers $a_i \in \bZ_{>0}$ such that $\sum_{i=1}^r a_i = -K_X \cdot C$.
 \item \label{lem:splitting2} $n+1 \geq l_X \geq r$.
\end{enumerate}
\end{lemma}
\begin{proof}
By Grothendieck's theorem, every vector bundle on $\bP^1$ is isomorphic to a direct sum of line bundles.
Thus, $f^* \sE \simeq \bigoplus_{i=1}^r \cO(a_i)$ for integers $a_i \in \bZ$.
Since $\sE$ is ample and $c_1(\sE)=c_1(X)$, each $a_i$ is positive and $\sum_{i=1}^r a_i = -K_X \cdot C$.
This proves the first assertion.

As we mentioned, $l_X \leq \dim X +1$ by Mori's theorem \cite{Mor79}.
Also, it follows from \ref{lem:splitting1} that $l_X \geq r$.
\end{proof}

 A \emph{family of rational curves} on $X$ is an irreducible component $M$ of the parameter space $\RC ^n (X)$  for the rational curves on $X$.
For an account of the theory of rational curves, we refer the reader to \cite{Kol96}.
Given a family $M$ of rational curves, there is the following diagram, which gives the realization of the family:
\begin{equation}\label{dia:UF}
\xymatrix{
 U \ar[r]^-e \ar[d]^-p & X \\
 M.
}
\end{equation}
Here $p \colon U \to M$ is the universal family of rational curves, which is known to be a smooth $\bP^1$-fibration,  and $e$ is the evaluation map of rational curves.
Roughly speaking, each point $m \in M$ corresponds to a rational curve $C \subset X$, and the map $e|_{p^{-1}(m)} \colon p^{-1}(m) \to X$ gives the normalization map of $C$.
Given a family $M$ of rational curves, its \emph{anticanonical degree} is defined as the anticanonical degree of a rational curve parametrized by $M$.

Recall that a family $M$ of rational curves is called \emph{covering} (resp.\ \emph{dominating}) if the evaluation map $e$ is surjective (resp.\ \ \emph{dominating}), and it is called \emph{unsplitting} if $M$ is proper.
Note that a family of rational curves is dominating if and only if a general rational curve in this family is free \cite[Proposition~1.1]{KMM92a}.

The next proposition is formulated in \cite{Kan17} for $n\geq 5$, while its proof works for a slightly weaker assumption $n\geq4$.
For the proof, we refer the reader to  \cite[Proof of Propositions~1.7 and 1.10]{Kan17}.
\begin{proposition}[{\cite[Propositions~1.7 and 1.10]{Kan17}}]\label{prop:RCn-2}
 Let $(X,\sE)$ be a Mukai pair of dimension $n \geq 4$ and rank $n-2$ with $\rho_X = 1$.
 Then the following hold:
\begin{enumerate}
 \item \label{prop:RCn-21} There exists an unsplit covering family $M$ of rational curves with anticanonical degree $l_X$.
 \item \label{prop:RCn-22} $X$ is chain connected by rational curves in the family $M$, that is, for each pair of points $x,y \in X$, there is a connected chain of rational curves in $M$ which contains both $x$ and $y$.
 \item \label{prop:RCn-23} If $l_X = n-2$, then $\sE$ is a direct sum of line bundles.
\end{enumerate}
\end{proposition}

Now we can prove $l_X =3$ for a pair as in Setting~\ref{set}.
\begin{proposition}\label{lem:lX}
Let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set}.
Then $l_X=3$.
In particular, $r_X =1$ or $3$.
\end{proposition}

\begin{proof}
Note that $5 \geq l_X \geq 2$ by Lemma~\ref{lem:splitting}~\ref{lem:splitting2} and that $l_X\neq2$ by Proposition~\ref{prop:RCn-2}~\ref{prop:RCn-23}.

Assume to the contrary that $l_X  \geq 4 $.
Then $X \simeq \bP^4$ or $\bQ^4$ by Theorem~\ref{thm:PQ}.
It follows from the classification of Fano bundles of rank two on $\bP^n$ or $\bQ^n$ \cite{APW94} that the bundle $\sE$ is decomposable.
This contradicts our assumption that $\sE$ is indecomposable.
Thus $l_X = 3$.
Since the index $r_X$ divides the length $l_X$, we have $r_X =1$ or $3$.
\end{proof}

In what follows, we will fix a family $M$ of rational curves on $X$ as in Proposition~\ref{prop:RCn-2}~\ref{prop:RCn-21} for our Mukai pair $(X,\sE)$ as in Setting~\ref{set}.
A rational curve in this family is called a \emph{minimal rational curve}.
By the above proposition, we have $l_X=3$.
Thus, if $f \colon \bP^1 \to C \subset X$ is a minimal rational curve on $X$, then $f^*\sE \simeq \cO(2) \oplus \cO (1)$ by Lemma~\ref{lem:splitting}~\ref{lem:splitting2}.
Corresponding to the surjection $\cO(2) \oplus \cO (1) \to \cO(1)$, a unique rational curve $\tf \colon \bP^1 \to \tC \subset \bP(\sE)$ exists such that $\xi_{\sE}\cdot\tC =1$ and the following diagram is commutative:
 \[
 \xymatrix{
 & \bP(\sE) \ar[d]^-\pi\\
 \bP^1 \ar[ru]^-\tf \ar[r]^-f & X.
 }
 \]
\begin{definition}[Minimal lifts]\label{def:lift}
 The above constructed rational curve $\tf \colon \bP^1 \to \tC \subset \bP(\sE)$ is called the \emph{minimal lift} of the minimal rational curve $f \colon \bP^1 \to C \subset X$.
\end{definition}

This construction can be globalized as follows:
let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set} and $M$ a family of minimal rational curves on $X$.
Then we have the following diagram as in \eqref{dia:UF}:
\[
\xymatrix{
 U \ar[r]^-e \ar[d]^-p & X \\
 M.
}
\]

\begin{lemma}[Lifting map]\label{lem:lift}
 There exists a unique lift $\te \colon U \to \bP(\sE)$ of the evaluation map $e$ that restricts to the minimal lift on each minimal rational curve.
\end{lemma}
\begin{proof}
Let $\omega_p$ denote the relative canonical bundle of the universal family $p \colon U \to M$.
 Then the bundle $e^*\sE \otimes\omega_p$ is isomorphic to $\cO \oplus \cO (-1)$ on each $p$-fiber $\bP^1$.
 Thus we have the following exact sequence of \emph{vector bundles} on $U$
 \[
 0 \to p^*p_*e^*\sE \otimes \omega_p \to e^*\sE \otimes \omega_p \to Q \to 0,
 \]
 which restricts on each $p$-fiber to the sequence
  \[
 0 \to \cO \to \cO \oplus \cO(-1) \to \cO(-1) \to 0.
 \]
 This exact sequence gives the map $\te$ as desired.
\end{proof}

Finally, we include here a useful lemma.
\begin{lemma}\label{lem:BB}
Let $(X,\sE)$ be a pair as in Setting~\ref{set}.
Then there is no curve $D$ on $X$ such that $\dim\varphi(\pi^{-1}(D))=1$.
\end{lemma}
\begin{proof}
 Assume to the contrary that there is a curve $D$ on $X$ such that
 \[\dim\varphi(\pi^{-1}(D))=1.\]
Since $\rho_X=1$, each minimal rational curve $f \colon \bP^1 \to C \subset X$ on $X$ is numerically proportional to $D$, and hence $\dim \varphi(\pi^{-1}(C)) =1$.
This contradicts to the fact that $f^*\sE \simeq \cO(2) \oplus \cO(1)$.
\end{proof}

\section{Stability and Bogomolov's inequality}\label{sect:stable}
In this section, we will check that the bundle $\sE$ is stable.
\begin{proposition}[Stability of $\sE$]
 Let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set}.
 Then the bundle $\sE$ is stable.
\end{proposition}
As a consequence, we have the Bogomolov inequality for $\sE$, which is a crucial ingredient in our proof of Theorem~\ref{thm:main}.
Roughly speaking the inequality enables us to bound some invariants for the pair $(X,\sE)$.
Accordingly, the boundedness of invariants implies some finiteness results on possibilities of the structure of the pair.

First, we will recall the definition of stability of vector bundles, or torsion free sheaves.
Let $X$ be a smooth projective variety of dimension $n$ with an ample divisor $H_X$.
Given a torsion free sheaf $\sF$ on $X$, the \emph{slope} $\mu (\sF)$ with respect to $H_X$ is defined as
\[
\mu (\sF) \coloneqq \frac{c_1(\sF)\cdot H_X^{n-1}}{\rank \sF}.
\]

A subsheaf $\sF_1 \subset \sF$ is said to be \emph{saturated} if the quotient $\sF/\sF_1$ is torsion free.

\begin{definition}[Stability of vector bundles]
A torsion free sheaf $\sF$ is called \emph{$H_X$-stable} if
\begin{equation}\label{eq:slope}
 \mu(\sF_1)<\mu(\sF)
\end{equation}

holds for every nonzero saturated subsheaf $\sF_1 \subsetneq\sF$.
\end{definition}

In the following, we will restrict our attention to the case of a locally free sheave $\sF$ of rank two.
In this case, the stability of $\sF$ is easy to prove:
it suffices to check \eqref{eq:slope} for any saturated subsheaf $\sF_1$ of rank one.
Since $\sF$ is locally free, such a subsheaf $\sF_1$ is reflexive, and hence it is a line bundle \cite[Propositions~1.1 and 1.9]{Har80}.
Thus the stability of $\sF$ is equivalent to the vanishing
\[
H^0(\sF \otimes \sL^{-1}) =0
\]
for any line bundle $\sL$ with
\[
c_1(\sL)\cdot H_X^{n-1} \geq \frac{c_1(\sF)\cdot H_X^{n-1}}{2}.
\]

Moreover, if $\Pic(X)$ is generated by $\cO(H_X)$, then the above condition is equivalent to
\[
H^0(\sF(-m))=0
\]
for any $m \geq \frac{c_1(\sF)}{2}$.

Let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set}.
 Then, keeping with the above observation, the stability of $\sE$ follows from the next lemma.
\begin{lemma}
  Let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set}.
Then
\[
H^0(\sE(-m))=0
\]
for any $m \geq \frac{r_X}{2}$.

\end{lemma}

\begin{proof}
 Take a section $s \in H^0(\sE(-m))$ for $m \geq r_X /2$ and assume $s \neq 0$ to the contrary.
 Then by restricting to a general minimal rational curve $f \colon \bP^1 \to X$, we also have a nonzero section $f^*s$ of the bundle $f^*(\sE(-m))$, which is isomorphic to
\[
\cO\left(2-\frac{3m}{r_X}\right)\oplus \cO\left(1-\frac{3m}{r_X}\right).
\]

Since the section $f^*s$ is nonzero, we have $2-\frac{3m}{r_X} \geq 0$, or equivalently, $\frac{2r_X}{3} \geq m$.
Note that $r_X=1$ or $3$ (Lemma~\ref{lem:lX}), and also that $m \geq \frac{r_X}{2}$ by our assumption.
Thus this is possible only if $r_X=3$ and $m=2$.
In this case, the bundle $f^*(\sE(-m))$ is isomorphic to $\cO\oplus \cO(-1)$.
Therefore, the section $f^*s$ is nowhere vanishing.
As $X$ is chain connected by  minimal rational curves (Proposition~\ref{prop:RCn-2}~\ref{prop:RCn-22}), the original section $s$ is also nowhere vanishing.
Hence we have an exact sequence
\[
0 \to \cO \to \sE(-2) \to \sE(-2)/\cO \to 0
\]
of \emph{vector bundles} on $X$,
and the quotient $\sE(-2)/\cO$ is isomorphic to $\cO(-1)$.
This exact sequence, however, splits, since $\Ext ^1 (\cO(-1),\cO) =0$ by the Kodaira vanishing theorem.
This contradicts our unsplit assumption on $\sE$.
\end{proof}

As a consequence, the bundle $\sE$ is stable.
Set $\varDelta \coloneqq c_1(\sE)^2-4c_2(\sE)$.
Then the Bogomolov inequality yields the following.
\begin{proposition}[Bogomolov's inequality]\label{prop:Bogomolov}
 Let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set}.
 Then
 \[
 \varDelta\cdot H_X^2 \leq0.
 \]
\end{proposition}
Note that, by the Grothendieck relation on the projective bundle $\bP(\sE)$,
we have
\begin{equation}\label{eq:Grel}
 \pi^*\varDelta = (2\xi_\sE-r_X\pi^*H_X)^2.
\end{equation}

\section{Rough description of the second contraction}\label{sect:rough}

Let $(X,\sE)$ be a pair as in Setting~\ref{set}, $M$ a family of minimal rational curves and $f \colon \bP^1 \to C \subset X$ a minimal rational curve.
Then the numerical equivalence class of the minimal lift $\tC$ does not depend on the choice of $C$;
the class $[\tC]$ is characterized by the numerical conditions $\xi_\sE \cdot [\tC] =1$ and $\pi^*(-K_X)\cdot [\tC] =3$.
\begin{definition}[Half line $\tR$]
 We will denote by $\tR$ the half line $\bR_{\geq0}[\tC]$ spanned by the class $[\tC]$.
\end{definition}

Since the Picard number of $\bP(\sE)$ is two, the nef cone $\NE(\bP(\sE))$ is spanned by two extremal rays; the ray $R_{\pi}$  corresponding to the projection $\pi$, and  the other ray $R_{\varphi}$ that defines the other extremal contraction $\varphi \colon \bP(\sE) \to Y$.
Note that $\tR \subset \NE(\bP(\sE))$:
\begin{figure}[h]
 \begin{picture}(100,100)(-20,0)
  \put(0,0){}
  \put(0,0){\line(1,0){70}}
  \put(0,0){\line(0,1){70}}
  \put(0,0){\line(1,1){60}}
  \put(70,-10){$R_\varphi$}
  \put(-15,70){$R_{\pi}$}
  \put(60,70){$\tR \coloneqq \bR_{\geq0}[\tC]$}
\end{picture}
\caption{$\NE (\bP(\sE))$}
\end{figure}

\begin{remark}[$\tR$ and $R_\varphi$]\label{rem:comp}
\hfill
\begin{enumerate}
\item A priori, it is not clear that the half line $\tR$ actually coincides with the extremal ray $R_\varphi$, or equivalently, that the minimal lifts $\tC$ are contracted by the second extremal contraction $\varphi$, cf. \cite[Lemma~3.1]{PSW92b}, \cite[Theorem~3.2]{Kan17}.
In our proof of Theorem~\ref{thm:main}, an important step consists of the proof of this assertion $\tR = R_{\varphi}$ by using the Bogomolov inequality (see Propositions~\ref{prop:Ineq} and \ref{prop:comp}).

\item In Section~8 of their paper \cite{NO07}, Novelli and Occhetta discussed the classification problem of Mukai pairs $(X,\sE)$ as in Setting~\ref{set} with an extra assumption $r_X \geq 2$.
Their argument proceeds as follows:
take a general element $Y$ of its fundamental linear system $|H_X|$.
By virtue of their assumption $r_X \geq 2$, $Y$ is a smooth Fano variety.
Then, they claimed that the restricted projective bundle $\bP(\sE|_Y)$ is also a Fano variety.
And, by using the classification results on $(Y,\sE|_Y)$, they recovered the structure of original $(X,\sE)$.
However, at least for the author, it is not clear that $\bP(\sE|_Y)$ is a Fano variety, as they claimed;
in fact, it is equivalent to the non-trivial assertion $\tR=R_\varphi$ (this equivalence follows from Proposition~\ref{prop:jumpingfib} below).

\end{enumerate}
\end{remark}

The dual cone of $\NE(\bP(\sE))$ is the nef cone $\Nef(\bP(\sE))$, which is spanned by two rays $\bR_{\geq0}[\pi^*H_X]$ and $\bR_{\geq0}[\varphi^* H_Y]$, where $H_Y$ is a divisor corresponding to the ample  generator of $\Pic(Y)$.
Since $\xi_\sE$ is ample on $\bP(\sE)$, we can find a positive rational number $a \in \bQ_{>0}$ such that $[a\xi_\sE - \pi^*H_X] \in \bR_{\geq0}[\varphi^* H_Y]$.
\begin{figure}[h]
 \begin{picture}(100,70)(-50,0)
  \put(0,0){}
  \put(0,0){\line(1,0){70}}
  \put(0,0){\line(-1,1){55}}
  \put(70,-10){$[\pi^*H_X]$}
  \put(5,50){$\xi_\sE$}
  \put(-100,60){$[a\xi_\sE - \pi^*H_X]$}
  \multiput(0,0)(0,2){30}{\line(0,1){1}}
\end{picture}
\caption{$\Nef (\bP(\sE))$}
\end{figure}

\begin{lemma}[$ar_X \geq3$]\label{lem:arX}
Let the notation be as above.
Then $ar_X \geq3$ with equality holding if and only if $\tR = R_\varphi$
\end{lemma}
\begin{proof}
 Since $a\xi_\sE - \pi^*H_X$ is nef, we have
 \[
( a\xi_\sE - \pi^*H_X) \cdot \tC \geq 0,
 \]
 and the equality holds if and only if the curve $\tC$ is contracted by $\varphi$.
 The assertion is simply a rephrasing of this fact.
\end{proof}

First, by using Theorem~\ref{thm:IWineq} and \cite{Wat14b}, we show that there exists at least one $\varphi$-fiber whose dimension is not small.

\begin{proposition}\label{prop:jumpingfib}
 Let $(X,\sE)$ be a pair as in Setting~\ref{set}.
 Then there exists a $\varphi$-fiber $F$ with $\dim F \geq 2$.
\end{proposition}
\begin{proof}
By \cite[Theorem~1.1]{Wat14b}, the other contraction $\varphi$ is not a smooth $\bP^1$-bundle.

 Assume to the contrary that every $\varphi$-fiber has dimension one or zero.
Then, by Theorem~\ref{thm:IWineq}, the morphism $\varphi$ is of fiber type and every $\varphi$-fiber has dimension one.
Moreover, by adjunction, each general $\varphi$-fiber $F$ is isomorphic to $\bP^1$ and $\cO(\xi_\sE)|_F  \simeq \cO_{\bP^1}(1)$.
This implies that $\varphi$ is a smooth $\bP^1$-bundle \cite[Lemma~2.12]{Fuj87}, and we get a contradiction.
\end{proof}

On the other hand, as a first application of the Bogomolov inequality (Proposition~\ref{prop:Bogomolov}), we can now prove the following.

\begin{proposition}[Bounding the dimension of $\varphi$-fibers]\label{prop:bound}
Let $(X,\sE)$ be a pair as in Setting~\ref{set} and the notation as above.
Then $\dim F \leq 2$ for every $\varphi$-fiber $F$.
\end{proposition}

\begin{proof}
 Assume to the contrary that there is a $\varphi$-fiber $F$ with $\dim F \geq 3$.
 Take $F'$ a closed subvariety of $F$ with dimension three.
 Then $\pi (F')$ is a divisor in $X$ and, since $\rho_X =1$, the divisor $\pi(F')$ is numerically proportional to $H_X$.
 Thus the Bogolmolov inequality and the projection formula yields
 \[
 \pi^*\varDelta \cdot \pi^*H_X \cdot F' \leq 0.
 \]
 Then, by the Grothendieck relation \eqref{eq:Grel} for $\sE$, the above inequality is equivalent to
 \[
 \pi^*(2\xi_\sE -r_X\pi^*H_X)^2 \cdot \pi^*H_X \cdot F' \leq 0.
 \]
 Since $F'$ is contracted to a point by the second contraction $\varphi$, we have $a\xi_\sE \nequiv \pi^*H_X$ on $F'$.
 Thus the inequality gives
 \[
 a(ar_X-2)^2 \xi_\sE ^3 \cdot F' \leq 0.
 \]
 This contradicts the facts that $\xi_\sE$ is ample and $ar_X \geq 3$.
\end{proof}

As a corollary, we have the following.
\begin{corollary}[Length of extremal rays]\label{cor:length}
 Let $(X,\sE)$ be a pair as in Setting~\ref{set}.
 Then the following hold:

\begin{enumerate}
 \item\label{cor:length1} $l(R_\varphi) =2$.
 \item\label{cor:length2} There is a rational curve $C_\varphi$ on $\bP(\sE)$ such that $C_\varphi$ is contracted by $\varphi$ and $\xi_\sE \cdot C_\varphi =1$.
 \item\label{cor:length3} $a$ is an \emph{integer}.
 \item\label{cor:length4} There is an ample divisor $L$ on $\bP(\sE)$ such that $K_{\bP(\sE)} + 2 L$ is a supporting divisor for $\varphi$.
\end{enumerate}
\end{corollary}

\begin{proof}
Note that $l(R_\varphi) \in 2 \bZ$ since $-K_{\bP(\sE)}=2\xi_\sE$.
The first assertion follows from the Ionescu-Wi\'sniewski inequality (Theorem~\ref{thm:IWineq}) and the inequality $\dim F \leq 2$ (Proposition~\ref{prop:bound}).
The second one follows from the definition of $l(R_\varphi)$.
Let $C_\varphi$ be a curve as in \ref{cor:length2}.
Then $(a\xi_\sE - \pi^*H_X)\cdot C_\varphi =0$.
Therefore, $a =\pi^*H_X\cdot C_\varphi \in \bZ $.
This proves the third assertion.
By setting $L \coloneqq (\pi^*H_X\cdot C_\varphi +1)\xi_\sE - \pi^*H_X$, we have the last assertion.
\end{proof}

\begin{theorem}[Rough description of the other contraction]\label{thm:phi}
Let $(X,\sE)$ be a Mukai pair as in Setting~\ref{set}.
Then one of the following holds:
\begin{enumerate}
 \item \label{thm:phi2}
 The morphism $\varphi$ is a quadric bundle over a smooth projective $3$-fold $Y$.
 \item \label{thm:phi3}
 The morphism $\varphi$ is a special \Banica scroll over a smooth projective $4$-fold $Y$.
\end{enumerate}
\end{theorem}
\begin{proof}
Let $E$ be an irreducible component of $\Exc(\varphi)$ and $F$ a fiber contained in $E$.
Then, by the Ionescu-Wi\'sniewski inequality (Theorem~\ref{thm:IWineq}), we have
\[
\dim E + \dim F \geq \dim \bP (\sE) + l(R_\varphi) -1.
\]
Since $\dim F \leq 2$ (Proposition~\ref{prop:bound}) and $l(R_\varphi ) = 2$, the above inequality yields
\[
\dim E + 2 \geq \dim E + \dim F \geq 6.
\]
In particular, the morphism $\varphi$ is either divisorial or of fiber type.
Hence $E = \Exc (\varphi)$.

First assume that $\varphi$ is of fiber type.
Then, since $\dim F \leq 2$, we have $\dim Y = 3$ or $4$.
If $\dim Y = 3$, then $\dim F = 2$ for any $\varphi$-fiber $F$ and hence \ref{thm:phi2} holds by Theorem~\ref{thm:Qbundle}.
On the other hand, if $\dim Y =4$, then $\varphi$ is a scroll and \ref{thm:phi3} holds by Theorem~\ref{thm:Bscroll}.

Next assume that $\varphi$ is divisorial.
Then $\dim F =2$ for any non-trivial fiber $F$.
Then, by Theorem~\ref{thm:Bl}, the morphism $\varphi$ is obtained by blowing up a smooth projective $5$-fold $Y$ along a smooth surface $S \subset Y$.\[
\xymatrix{
&W\ar[ld]_-\pi \ar[rd]^-\varphi &E \coloneqq \Exc(\varphi) \ar@{_{(}-_>}[l] \ar[rd]&\\
X&&Y&S.\ar@{_{(}-_>}[l]
}
\]
Thus $W$ is a smooth Fano $5$-fold with index two that has a birational contraction $\varphi$ as above and admits a contraction $\pi$ of fiber type.

Thus this Fano $5$-fold $W$ fits Case (b) of \cite{CO08}.
One can deduce from their classification that there is no Fano $5$-fold $W$ as above.
Thus $\varphi$ is not birational.
\end{proof}

\begin{remark}
In the above proof, we used the classification result of Case (b) in \cite{CO08}, which is carried out in Section~4 of their paper.
Note that their argument in \cite[Section~4]{CO08}, more precisely, the proof of \cite[Lemma~4.4]{CO08} relies on \cite[Theorem~1.3]{NO07},
which, as we mentioned, seems to contain a gap in the proof.
Nevertheless, their usage of \cite[Theorem~1.3]{NO07} is mild; they only use the fact that the minimum anticanonical degree of rational curves on $X$ is two or three, which we already have proved.
Thus we can avoid a circular argument.
\end{remark}

\section{Case: $\varphi$ is a quadric bundle}\label{sect:quarticDP}
As before, $(X,\sE)$ is a pair as in Setting~\ref{set}.
Then, by Theorem~\ref{thm:phi}, $\varphi$ is either a quadric bundle or a special \Banica scroll.
In this section, we will deal with the case of quadric bundle, and prove the following theorem.
\begin{theorem}\label{thm:quarticDP}
 Let $(X,\sE)$ be a pair as in Setting~\ref{set}.
 Assume that the second contraction $\varphi$ is a quadric bundle.
 Then $(X ,\sE)$ is isomorphic to
 \[
 (V_4, p^*\sS^* _{\bQ^4} (1)),
 \]
 where $V_4$ is a quartic del Pezzo $4$-fold obtained as a double cover $p \colon V_4 \to \bQ ^4$ branched along a smooth divisor $B \in \left| \cO_{\bQ^4}(2) \right|$, and $\sS_{\bQ^4}$ is a spinor bundle on $\bQ^4$.
\end{theorem}

The rest of this section is occupied with the proof of Theorem~\ref{thm:quarticDP}.
First, we will determine the isomorphic class of the image $Y$ of the second contraction $\varphi$.
\begin{lemma}
Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quarticDP}.
Then $Y$ is isomorphic to $\bP^3$.
\end{lemma}
\begin{proof}
Let $F$ be a general $\varphi$-fiber.
Then by taking a base change of the projection $\pi$ by the morphism $\pi|_F$, we have the following diagram:
\begin{equation}\label{diagram:fiber}
\vcenter{
\xymatrix{
  \bP(\sE |_{F}) \ar[d]_-{\pi_{F}}  \ar[r] \ar@(ur,ul)[rr]^-{\theta _{F}}  & \bP(\sE) \ar[d]_-{\pi} \ar[r]_-{{\varphi}} & Y         \\
F \ar[r]^-{\pi|_{F}} & X.                                        & \\
}
}
\end{equation}
The morphism $\theta _F$ is surjective.
Otherwise $\dim  \theta_F (\bP(\sE|_F)) < \bP(\sE|_F)$.
Then, by the bend and break lemma \cite[Chapter II, Theorem~5.4]{Kol96}, there is a curve $D \subset F$ such that $\dim \theta_F (\pi_F^{-1}(D)) =1$.
This contradicts Lemma~\ref{lem:BB}.

By adjunction, $F$ is isomorphic to $\bP^1 \times \bP^1$ and
\[
\cO(\xi_\sE)|_F \simeq \pr_1^*\cO(1) \otimes \pr_2^*\cO(1),
\]
where $\pr_i \colon \bP^1 \times \bP^1 \to \bP^1$ is the projection to the $i$-th factor.

By restricting the relative Euler sequence of the projectivization $\bP(\sE) \to X$, we have the following exact sequence:
\[
0 \to \cO(K_\pi +\xi _\sE)|_F  \to \sE|_F \to \pr_1^*\cO(1) \otimes \pr_2^*\cO(1) \to 0. \]

Since $K_\pi \simeq -2 \xi_\sE+ r_X \pi^*H_X$, the bundle $\cO(K_\pi +\xi _\sE)|_F$ is isomorphic to
\[
\cO(r_X \pi^*H_X)|_F \otimes  \pr_1^*\cO(-1) \otimes \pr_2^*\cO(-1).
\]
Thus the class of the above exact sequence belongs to
\[
\Ext^1(\pr_1^*\cO(1) \otimes \pr_2^*\cO(1),\cO(r_X \pi^*H_X)|_F \otimes  \pr_1^*\cO(-1) \otimes \pr_2^*\cO(-1)),
\]
which is isomorphic to
\[
H^1(K_F\otimes \cO(r_X \pi^*H_X)|_F).
\]
Since $F$ is a $\varphi$-fiber, the map $\pi|_F \colon F \to X$ is finite, and hence the divisor $\pi^*H_X|_F$ is ample.
Thus, by the Kodaira vanishing theorem, the cohomology group $
H^1(K_F\otimes \cO(r_X \pi^*H_X)|_F)$ is trivial, and the above exact sequence splits.
Therefore, $\bP(\sE|_F)$ is a toric variety.
Now the assertion follows from \cite{OW02}.
\end{proof}

Second, we will determine the values $a$, $r_X$ and $H_X^4$.
Recall that $a$ is the integer for which the divisor $a\xi_\sE -\pi^*H_X$ is a supporting divisor of the contraction $\varphi$.
\begin{lemma}
Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quarticDP}.
Then $a=1$, $r_X =3$ and $H_X^4 =4$.
\end{lemma}
\begin{proof}
 We will denote by $H_Y$ the divisor class of a hyperplane in $Y \simeq \bP^3$.
 Since the divisor $a\xi_\sE -\pi^*H_X$ is not a multiple of another divisor, we have
 \[
 a\xi_\sE -\pi^*H_X = \varphi^*H_Y.
 \]
Thus
\[
 (a\xi_\sE -\pi^*H_X)^3\cdot\xi_\sE^2
=2
\text{ and }
(a\xi_\sE -\pi^*H_X)^4=0.
\]
Set
\begin{align*}
 x&\coloneqq H_X^4,\\
 y&\coloneqq c_2(\sE)\cdot H_X^2,\\
 z&\coloneqq c_2(\sE)^2.
\end{align*}
By a direct calculation with the Grothendieck relation
\[
 \xi_\sE^2=\xi_\sE\cdot (r_X\pi^*H_X) - \pi^*c_2(\sE),
\]
 the above equations yield
\begin{align*}
(r_X^4a^3-3r_X^3a^2+3r_X^2a-r_X)x-(3r_X^2a^3-6r_Xa^2+3a)y+a^3z= 2,\\
(r_X^3a^4-4r_X^2a^3+6r_Xa^2-4a)x+(4a^3-2r_Xa^4)y=0,\\
x-(r_X^2a^4-4r_Xa^3+6a^2)y+a^4z= 0.
\end{align*}
Solving this system of linear equations for $(x,y,z)$, we have
\begin{align}
x&=\frac{4a}{(r_Xa-2)^2}, \label{eq:Q1}\\
y &= \frac{2r_X^2a^2-4r_Xa +4}{a(r_Xa-2)^2},\label{eq:Q2}\\
z &=\frac{2r_X^4a^4-12r_X^3a^3+32r_X^2a^2-40r_Xa+20}{a^3(r_Xa-2)^2}.\label{eq:Q3}
\end{align}
Note that, by Lemmas~\ref{lem:lX} and \ref{lem:arX}, we have $r_X=1$ or $3$, and $r_Xa \geq 3$.
Assume $r_X =1$. Then, by \eqref{eq:Q1}, we have
\[
\frac{4a}{(a-2)^2} \in \bZ.
\]
Thus $a = 3$ or $4$.
In each case, however, we have $y = 5/4$ or $10/3$ respectively, which is impossible since $y= c_2(\sE)\cdot H_X^2 \in \bZ$.
Thus we have $r_X =3$.
Then, since $x=\frac{4a}{(3a-2)^2} \in \bZ$, we have $a=1$.
This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:quarticDP}]
Since $\xi_\sE-\pi^*H_X =\varphi^*H_Y$, each $\pi$-fiber maps to a line in $Y \simeq \bP^3$, and, by the universality of Grassmannian variety, we have the following commutative diagram:
\[
\xymatrix{
  \bP(\sE ) \ar[d]_-{\pi} \ar[r] \ar@(ur,ul)[rr]^-{\varphi}  & \bP(\sS ^* _\bQ) \ar[d]_-{\pi_\bQ^4} \ar[r] & \bP^3         \\
X \ar[r]^-{p} & \bQ^4 \simeq \Gr(2,4),& \\
}
\]
with the condition $p^*\sS_{\bQ^4}^* \simeq \sE(-1)$.
In particular $p^*\cO_{\bQ^4}(1) \simeq \cO_X(1)$.

Since $\rho_X=1$, the morphism $p$ is finite and surjective.
Moreover, the covering $p$ is of degree two since $H_X ^4 =4$.
Then, by adjunction, we see that the double covering $p$ is branched along a smooth divisor $B \in \left| \cO_{\bQ^4}(2)\right|$.
This completes the proof.
\end{proof}

\section{Case: $\varphi$ is a special B\u anic\u a scroll}\label{sect:quinticDP}
Let $(X,\sE)$ be a pair as in Setting~\ref{set}.
In this final section, we will deal with the case where the second contraction $\varphi$ is a special \Banica scroll, and thus complete the proof of Theorem~\ref{thm:main}.

\begin{theorem}[Special \Banica scroll]\label{thm:quinticDP}
Let $(X,\sE)$ be a pair as in Setting~\ref{set}.
Assume that $\varphi$ is a special \Banica scroll.
Then the pair $(X,\sE)$ is isomorphic to
\[
(V_5, \sS ^*_{V_5} (1)),
\]
where $V_5$ is a general linear section of the Grassmannian $\Gr (2,5)$ embedded into $\bP ^9$ via the Pl\"ucker embedding, and $\sS _{V_5}$ is the restriction of the universal subbundle on $\Gr(2,5)$ to $V_5$.
\end{theorem}

First, we will set up the notations.
\begin{notation}[$\sF$, $E$, $B$, $d_i$, $\tW$, $\tY$, $\tE$, $\tB$]
 From now on, we will assume that $(X,\sE)$ is a pair as in Theorem~\ref{thm:quinticDP}.
 Recall that $W$ denotes the projectivization $\bP(\sE)$.

Set $\sF \coloneqq \varphi _* \cO(\xi_\sE)$.
Then, by Theorem~\ref{thm:Bscroll}, the sheaf $\sF$ is a \Banica sheaf and the morphism $\varphi$ is the projection of the projectivization $\bP(\sF) \to Y$.
Then general $\varphi$-fibers are isomorphic to $\bP^1$, and the other fibers $\bP^2$;
let
\[
B \coloneqq \{\, y\in Y \mid \varphi^{-1}(y) \simeq \bP^2 \,\}
\]
 be the locus parametrizing jumping fibers and $E \coloneqq \varphi^{-1}(B)$ the preimage of $B$.
 Then, by Proposition~\ref{prop:Bscroll}, $Y$ is smooth, $B$ is a disjoint union of smooth irreducible curves $B_i$ for $i = 1, \dots, m$, and the morphism $E \to B$ is a $\bP^2$-bundle.
Set $E_i \coloneqq \varphi^{-1}(B_i)$.

Let $\tW \to W$ (resp.\ $\tY \to Y$) be the blowing-up of $W$ (resp.\ $Y$) along $E$ (resp.\ $B$), and $\tE_i$ (resp.\ $\tB_i$) the exceptional divisor over $E_i$ (resp.\ $B_i$).
Then, by Proposition~\ref{prop:Bscroll} again, we have the following commutative diagram:
 \[
 \xymatrix{
 &\tE = \coprod \tE_i \ar@{_{(}-_>}[ld] \ar[dd] \ar[rr]^-{\tvarphi_{\tE}=\coprod\tvarphi_{\tE_i}}&&\tB= \coprod \tB_i\ar@{_{(}-_>}[ld] \ar[dd]\\
 \tW \ar[rr]^(0.6){\tvarphi} \ar[dd] && \tY \ar[dd]&\\
 &E = \coprod E_i \ar@{_{(}-_>}[ld] \ar[rr]&  & B= \coprod B_i\ar@{_{(}-_>}[ld]\\
 W \ar[d]^-\pi \ar[rr]^-\varphi && Y &\\
 X&&&
 }
 \]
with the following conditions:
\begin{enumerate}
 \item $\tvarphi \colon \tW \to \tY$ is a $\bP^1$-bundle,
 \item $\tE = \tvarphi^{-1}(\tB)$.
\end{enumerate}

As $Y$ is smooth uniruled projective manifold with Picard number one, it is a Fano manifold with Picard number one.
As usual, $H_Y$ is the ample generator of $\Pic (Y)$, $r_Y$ the index of $Y$ and $l_Y$ the length of $Y$.
Let $d_i \coloneqq H_Y \cdot B_i$ be the degree of $B_i$ with respect to $H_Y$.
Note that $\xi_\sE = \xi_\sF$ on $W$.

\end{notation}

In the following lemma, we will see that the pair $(Y,\sF)$ is nearly a Mukai pair; the sheaf $\sF$ is not locally free, but the pair satisfies conditions in the definition of Mukai pairs.
\begin{lemma}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
Then
\begin{enumerate}
 \item \label{lem:Mukai1}$\sF$ is ample and $c_1(Y) = c_1(\sF)$.
 \item \label{lem:Mukai2}$l_Y \geq 3$.
\end{enumerate}
\end{lemma}
\begin{proof}
 \ref{lem:Mukai1} Recall that $\sF$ is said to be ample if $\xi_{\sF}$ is ample.
 Thus the ampleness of $\sF$ follows from the definition.
 Set $W^0 \coloneqq W \setminus E$ and $Y^0 \coloneqq Y \setminus B$, and let $\varphi^0 \colon W^0 \to Y^0$ be the restriction of the projection.
 Then the canonical bundle formula for projective bundles yields
 \[
 c_1(W^0 ) = 2 \xi_{\sF|_{Y^0}}+(\varphi^0)^* (c_1(Y^0)-c_1(\sF|_{Y^0})).
 \]
 Since
 \[
 c_1(W^0)=2\xi_{\sE}|_{W^0} = 2 \xi_{\sF}|_{W^0}=2 \xi_{\sF|_{Y^0}},
 \]
 we have $c_1 (\sF|_{Y^0}) =c_1(Y^0)$.
 Since $\codim _Y B \geq 2$, we have $c_1(\sF) = c_1 (Y)$.

 \ref{lem:Mukai2} Assume that $l_Y =2$.
Then $Y$ admits a dominating family $M_Y$ of rational curves of anticanonical degree $2$.
Let $g \colon \bP^1 \to C_Y \subset Y$ be a general rational curve in this family $M_Y$.
Then, by \cite[Chapter~II, Proposition~3.7]{Kol96}, the image $C_Y$ does not intersect with $B$.
Thus $g^*\sF$ is a locally free sheaf of rank two on $\bP^1$.
Moreover, by \ref{lem:Mukai1}, the bundle $g^*\sF$ is isomorphic to $\cO(1)\oplus\cO(1)$.
Therefore, by considering the minimal sections of $\bP(g^*\sF) \to \bP^1$, we have an unsplit covering family $\tM_Y$ of rational curves $\tC_Y$ on $W$ such that $\xi_\sF \cdot \tC_Y =1$ (this family is unsplit by the numerical condition $\xi_\sF \cdot \tC_Y =1$).
By \cite[Lemma~2.4]{CO06}, the numerical equivalence class $[\tC_Y]$ spans the extremal ray $R_\pi$.
Thus we have $\dim \pi (\varphi^{-1}(C_Y)) =1$.
This implies $\pi^{-1}(\pi (\varphi^{-1}(C_Y))) = \varphi^{-1}(C_Y)$, and hence $\varphi(\pi^{-1}(\pi (\varphi^{-1}(C_Y)))) = C_Y$.
This contradicts Lemma~\ref{lem:BB}.
\end{proof}

Now we can determine the curves $B_i$.
\begin{lemma}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
Then each curve $B_i$ is isomorphic to $\bP^1$.
\end{lemma}
\begin{proof}
It is enough to see that the first Betti number $b_1(B_i)$ is zero for each $i$.

 By the blow-up formula and the projective bundle formula for Betti numbers (see for example \cite[EXPOSE XVIII]{SGA72}), we have
\begin{align}
 b_k(\tY) &= b_k(Y)+\sum_{i=1}^m b_{k-2} (B_i)+\sum_{i=0}^m b_{k-4} (B_i), \label{eq:betti1}\\
  b_k(\tW) &= b_k(W)+\sum_{i=1}^m b_{k-2} (E_i).\label{eq:betti2}
\end{align}
and
\begin{align}
 b_k(\tW) &= b_k(\tY)+ b_{k-2} (\tY),\label{eq:betti3}\\
  b_k(W) &= b_k(X)+b_{k-2} (X),\label{eq:betti4}\\
  b_k(E_i) &= b_k(B_i)+ b_{k-2} (B_i)+b_{k-4} (B_i)\label{eq:betti5}.
\end{align}
By equations~\eqref{eq:betti2}, \eqref{eq:betti4} and \eqref{eq:betti5}, we have
\[
b_k(\tW) = b_k(X)+b_{k-2} (X) + \sum_{i=1}^m (b_{k-2}(B_i)+ b_{k-4} (B_i)+b_{k-6} (B_i)).
\]
Also, by equations~\eqref{eq:betti1} and \eqref{eq:betti3}, we have
\[
b_k(\tW) = b_k(Y)+b_{k-2} (Y) + \sum_{i=1}^m (b_{k-2}(B_i)+ 2 b_{k-4} (B_i)+b_{k-6} (B_i)).
\]
Hence,
\[
b_k(X)+b_{k-2} (X) = b_k(Y)+b_{k-2} (Y) + \sum_{i=1}^m  b_{k-4}(B_i).
\]
By letting $k=3$, $4$ or $5$, we obtain:
\begin{align*}
 b_3(X) &=b_3(Y),\\
  b_4(X)+b_{2} (X) &= b_4(Y)+b_{2} (Y) + \sum_{i=1}^m  b_{0}(B_i),\\
 b_5(X)+b_{3} (X) &= b_5(Y)+b_{3} (Y) + \sum_{i=1}^m  b_{1}(B_i).
\end{align*}
Note that $b_k(X) = b_{8-k}(X)$ and $b_k(Y) = b_{8-k} (Y)$ by the Poincar\'e duality, and that $b_2(X) = b_2 (Y) =1$.
Thus, by solving the above three equations, we have $b_1(B_i)=0$, $b_3(X) =b_3(Y)$ and $b_4(Y) +m =b_4(X)$, and the assertion follows
\end{proof}

In the following, $\sC_{{\blank/\blank}}$ (resp.\ $\sN_{\blank/\blank}$) denotes the conormal (resp.\ normal) bundle.
\begin{lemma}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
Then $E_i$ is isomorphic to $\bP(\sN_{B_i/Y}(1))$ over $B_i \simeq \bP^1$, and, via this identification, we have
 \[
 \xi_{\sE}|_{E_i} \simeq \xi _{\sN_{B_i/Y}(1)}.
 \]
\end{lemma}
\begin{proof}
In the following square
 \[
 \xymatrix{
  \tE_i \ar[d]_-f \ar[r]^-{\tvarphi_{\tE_i}}&\tB_i \ar[d]^-g\\
 E_i \ar[r]  & B_i,\\
 }
 \]
each $f$-fiber gives a line in a $g$-fiber, and $E_i$ parametrizes all the lines in $g$-fibers.
Since $\tB_{i}\simeq \bP(\sC_{B_i/Y})$, the parameter space of lines in $g$-fibers is isomorphic to the dual projectivization $\bP(\sN_{B_i/Y})$.
Thus we have the first assertion.

Note that $\tvarphi^*\tB_{i} = \tE_i$.
 Thus
\begin{align*}
(\tvarphi_{\tE_i})^*\xi_{\sC_{B_i/Y}} &\simeq  (\tvarphi_{\tE_i})^*(-\tB_i|_{\tB_i}) \\
  &\simeq -\tE_i|_{\tE_i}\\
  &\simeq \xi_{\sC_{E_i/W}}.
\end{align*}

Since $-K_W = 2 \xi_{\sE}$, we have $-K_{\tW} = 2\xi_{\sE}|_{\tW}-\tE$.
Thus, by adjunction,
\[
-K_{\tE_i} = 2\xi_\sE|_{\tE_i} + 2 \xi_{\sC_{E_i/W}}.
\]
Then, we have
\[
-K_{\tE_i} = 2\xi_\sE|_{\tE_i} + 2 \xi_{\sC_{B_i/Y}}|_{\tE_i}.
\]

On the other hand, we have
\[
-K_{\tE_i} = (2\xi_{\sN_{B_i/Y}} + 2 \xi_{\sC_{B_i/Y}} + 2H_{B_i})|_{\tE_i},
\]
where $H_{B_i}$ is the ample generator of $\Pic (B_i)$ (note that $B_i \simeq \bP^1$).

Thus
\[
\xi_\sE|_{\tE_i} = (\xi_{\sN_{B_i/Y}} +  H_{B_i})|_{\tE_i}.
\]
Hence
\[
\xi_\sE|_{E_i} =\xi_{\sN_{B_i/Y}} +  H_{B_i}|_{E_i} = \xi_{\sN_{B_i/Y}(1)}.
\]
\end{proof}

As a second application of Bogomolov's inequality (Proposition~\ref{prop:Bogomolov}), we now prove the following lemma, which enables us to control invariants $a$ and $r_X$.

\begin{proposition}\label{prop:Ineq}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
Then
\[
a < \frac{3r_X+2r_Y+\sqrt{9r_X^2+4r_Yr_X+4r_Y^2}}{2r_Yr_X}.
\]
In particular, one of the following holds:
\begin{enumerate}
 \item $ar_X =3$ and $r_Y\geq2$,
 \item $a \leq 3$, $r_X=3$ and $r_Y =1$,
 \item $a \leq 4$, $r_X=1$ and $r_Y=1$.
\end{enumerate}
\end{proposition}
\begin{proof}

 Since $\pi_{*}(E_i)$ is an effective divisor (possibly zero), we have
 \[
0 \geq E_i\cdot\pi^*H_X\cdot\pi^*\varDelta
 \]
 by the Bogomolov inequality (Proposition~\ref{prop:Bogomolov}) and the projection formula.
 This inequality yields

\begin{align*}
 0 &\geq E_i\cdot\pi^*H_X\cdot\pi^*\varDelta\\
 &= E_i\cdot\pi^*H_X\cdot(2\xi_\sE-r_X\pi^*H_X)^2\\
&= E_i\cdot(a\xi_\sE-\varphi^*H_Y)\cdot((ar_X-2)\xi_\sE-r_X\varphi^*H_Y)^2\\
&= (a\xi_\sE-\varphi^*H_Y)|_{E_i}\cdot((ar_X-2)\xi_\sE-r_X\varphi^*H_Y)^2|_{E_i}\\
&= (a\xi_{\sN_{B_i/Y}}+(a-d_i)H_{B_i}|_{E_i})\cdot((ar_X-2)\xi_{\sN_{B_i/Y}}+(ar_X-d_ir_X-2)H_{B_i}|_{E_i})^2.
\end{align*}

Note that $c_1(\sN_{B_i/Y}) = d_i r_Y -2$, and hence $\xi_{\sN_{B_i/Y}}^3 =d_i r_Y-2$.
Thus, by a straightforward calculation, we have
\begin{align*}
0 &\geq(r_Xa-2)\left((d_ir_Xr_Y+r_X)a^2-(3d_ir_X+2d_ir_Y+2)a+2d_i \right)\\
&=(r_Xa-2)\left(d_i\left(r_Xr_Ya^2-(3r_X+2r_Y)a+2\right) +a\left(r_Xa-2\right) \right).
\end{align*}
Since $ar_X \geq 3$, $a>0$ and $d_i>0$, the above inequality yields the following quadratic inequality for $a$:
\[
0 > r_Xr_Ya^2-(3r_X+2r_Y)a+2.
\]
This gives the inequality as claimed.

Note that we have only finite possibilities for $r_X$ and $r_Y$, and that $ar_X \geq3$.
The rest of the assertion follows from case-by-case analysis.
\end{proof}

\begin{proposition}[$ar_X=3$]\label{prop:comp}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
Then $ar_X =3$, or equivalently $\tR=R_\varphi$.
\end{proposition}
\begin{proof}
 Assume to the contrary $ar_X >3$, or equivalently, $\tR \neq R_\varphi$.
 Then, by Proposition~\ref{prop:Ineq}, one of the following holds:
\begin{enumerate}
 \item $a=3$, $r_X=3$ and $r_Y=1$,
 \item $a=2$, $r_X=3$ and $r_Y=1$,
 \item $a=4$, $r_X=1$ and $r_Y=1$.
\end{enumerate}
In any case, we have $\varphi^*(-K_Y)\cdot \tC = a -\frac{3}{r_X}\leq 2$.
Thus the $\varphi$-image of $\tC$ gives a rational curve on $Y$ whose anticanonical degree is at most two.
Since $l_Y =3$, the family of these rational curves is not dominant.
Hence $\varphi(\te(U))\neq Y$.
This also implies $\te(U) \neq W$.

Since $\pi(\te(U)) =X$, we have $\dim \te(U) \geq 4$.
Therefore, $D \coloneqq \te(U)$ is a divisor on $W$.
Then $\varphi(D)$ is also a divisor on $Y$, and $\varphi^*(\varphi(D)) = D$.
In this case, since $\rho _Y=1$, the divisor $\varphi(D)$ is ample, and hence $\varphi(D)\cap B_i \neq \emptyset $.
Thus $D$ contains a two dimensional fiber $F$ of $\varphi$.
Then, by \cite[Lemma~5.4]{ACO04}, \cite[Lemma~3.2 and Remark~3.3]{Occ06} (see also \cite[Corollary~2.2 and Remark~2.4]{CO06}),
\[
\dim \te(\pi^{-1}(\pi(\te^{-1}(F)))) \geq 4
\]
and
\[
\NE(\te(\pi^{-1}(\pi(\te^{-1}(F)))),W) \subset \langle \tR,R_{\varphi}\rangle.
\]
This contradicts the following lemma \cite[Lemma~1.18]{Kan17}, which is a generalization of \cite[Claim 4.1.1]{PSW92b}.
Note that the original assumption in \cite[Lemma~1.18]{Kan17} is a little bit stronger than that in our case, while the same proof does work.
\end{proof}

\begin{lemma}[{\cite[Lemma~1.18]{Kan17}}]\label{lem:section}
 Let $(X,\sE)$ be a Mukai pair as in \ref{set}.
 Then there is no subvariety $V$ in $W$ with $\dim V \geq 4$ such that
 $\NE (V,W) \subset \langle \tR,R_{\varphi} \rangle$.
\end{lemma}
\begin{proof}
Assume to the contrary that there is a closed subvariety $V$ in $W$ with $\dim V \geq 4$ and
 $\NE (V,W) \subset \langle \tR,R_{\varphi} \rangle$.
By the same proof of \cite[Lemma~1.18]{Kan17}, we can show that the subvariety $V$ gives a section of $\pi$ corresponding to the exact sequence
\[
0 \to \cO(2) \to \sE \to \cO(1) \to 0.
\]
This sequence, however, splits, and we get a contradiction.
\end{proof}

Therefore, by Proposition~\ref{prop:comp}, the minimal lifts $\tC$ are contracted by the second contraction $\varphi$.
Conversely, each line in a $\varphi$-fiber  gives a minimal rational curve on $X$.
Using this fact, we can now prove that the image $Y$ is $\bP^4$.
\begin{proposition}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
 Then $Y \simeq \bP^4$.
\end{proposition}
\begin{proof}
By taking the base change of $\pi$ by the morphism $\pi|_{E_i} \colon E_i \to X$, we have the following diagram:
\[
\xymatrix{
\bar {E_i} \ar@{^{(}-_>}[d]^-{\iota} \ar[rr]&&B_i \ar@{^{(}-_>}[d]\\
  \bP(\sE |_{E_i}) \ar[d]_-{\pi_{E_i}}  \ar[r] \ar@(ur,ul)[rr]^-{\theta _{E_i}}  & W \ar[d]_-{\pi} \ar[r]_-{{\varphi}} & Y         \\
E_i \ar[r]^-{\pi|_{E_i}} & X,                                        & \\
}
\]
where $\bar {E_i}$ is a section of $\pi_{E_i}$ corresponding to the original inclusion $E_i \subset W$.

First, we will prove that $\theta_{E_i}$ is surjective.
It is enough to see that $\pi|_{E_i}$ is generically finite.
Otherwise, $\dim \pi(E_i) =2$ and thus $\pi^{-1}(\pi(E_i))=E_i$.
Fix a line $\ell$ in a $\varphi$-fiber contained in $E_i$.
Then $\varphi(\pi^{-1}(\pi(\ell)))=B_i$ since $\pi^{-1}(\pi(E_i))=E_i$.
This is impossible since ${(\pi|_{\ell})}^*\sE \simeq \cO(2)\oplus\cO(1)$.
Thus $\theta_{E_i}$ is surjective.

Second, we will prove that, for each jumping fiber $F \simeq \bP^2$ of $\varphi$, we have $\sE|_F \simeq \cO(2)\oplus \cO(1)$.
Take a jumping fiber $F\simeq \bP^2$.
Then, by restricting the Euler sequence for $\pi$, we have the following exact sequence:
\[
0 \to \cO(2) \to \pi^*\sE|_F \to \cO(1) \to 0.
\]
The assertion is now clear.

Finally, we will check that $Y \simeq \bP^4$.
Set $q \coloneqq \pi_{E_i} \circ \varphi|_{E_i} \colon  \bP(\sE |_{E_i}) \to B_i$.
Now each $q$-fiber is isomorphic to $\bP_{\bP^2}(\cO(2)\oplus\cO(1))$, and the minimal section of this projectivization $\bP_{\bP^2}(\cO(2)\oplus\cO(1))$ is contracted to a point by $\theta_{E_i}$.
Thus the map $\theta_{E_i}$ factors through a $\bP^3$-bundle $\bP(\sG_i)$ over $B_i$:
\[
\xymatrix{
\bP(\sE|_{E_i}) \ar[d]^-q \ar[r] \ar@(ur,ul)[rr]^-{\theta _{E_i}} & \bP(\sG_i) \ar[dl] \ar[r] & Y \\
B_i.&&
}
\]

Since every vector bundle on $B_i$ splits into a direct sum of line bundles, every projective bundle over $B_i$ is a toric variety.
Thus, by \cite{OW02}, $Y$ is isomorphic to $\bP^4$.
\end{proof}

As $Y \simeq \bP^4$, we now can calculate several invariants of $(X,\sE)$ by relating to those of $Y$.
Set
\begin{align*}
 x&\coloneqq H_X^4,\\
 y&\coloneqq c_2(\sE)\cdot H_X^2,\\
 z&\coloneqq c_2(\sE)^2,\\
 u&\coloneqq c_2(X)\cdot H_X^2,\\
 v&\coloneqq c_2(X)\cdot c_2(\sE).
\end{align*}
\begin{proposition}\label{prop:RR}
 Let $(X,\sE)$ be a pair as in Theorem~\ref{thm:quinticDP}.
Then $a=1$ and $r_X = 3$.
Moreover, we have
\begin{align*}
 y&= \frac{5x-1}{2},\\
 z&= \frac{13x-7}{2},\\
 u&= 2x+12,\\
 v&= \frac{13x+41}{2}.
\end{align*}

\end{proposition}
\begin{proof}
By the the Hirzebruch-Riemann-Roch theorem,
\begin{equation}\label{eq:RR}
 \chi(m(a\xi_\sE-\pi^*H_X))  = [\ch(m(a\xi_\sE-\pi^*H_X))\cdot\td(W)]_{\dim W}.
\end{equation}

Since $\chi(\cO_X) =1$, the fourth Todd class $\td_4(X) $ is the class of a point $[\pt]$.
Thus
\[
\td(X) = 1+ \frac{1}{2}r_X H_X + \frac{1}{12}(r_X^2H_X^2+c_2(X))+\frac{1}{24}r_XH_Xc_2(X)+[\pt].
\]
Also,
\[
\td(T_\pi) = 1+ \frac{1}{2}(-K_\pi) + \frac{1}{12}(-K_\pi)^2-\frac{1}{720}(-K_\pi)^4.
\]
Thus
\begin{align*}
 \td(W) &= \pi^*\td(X)\cdot\td(T_\pi)\\
 &= \pi^*\left( 1+ \frac{1}{2}r_X H_X + \frac{1}{12}(r_X^2H_X^2+c_2(X))+\frac{1}{24}r_XH_Xc_2(X)+[\pt]\right)\\
 &\qquad\cdot\left(1+ \frac{1}{2}(-K_\pi) + \frac{1}{12}(-K_\pi)^2-\frac{1}{720}(-K_\pi)^4\right).
 \end{align*}
 Noting $-K_\pi = 2 \xi-r_X\pi^*H_X$ and the grothendieck relation \[
 \xi_\sE^2=\xi_\sE\cdot (r_X\pi^*H_X) - \pi^*c_2(\sE),
\]
we can calculate the right hand side of equation~\eqref{eq:RR}:
\begin{equation}
 \chi(m(a\xi_\sE-\pi^*H_X))  = f_1m^5+f_2m^4+f_3m^3+f_4m^2+f_5m+1,
 \end{equation}
 where

\begin{align*}
 f_1&=
 \frac{a^5 r^4-5a^4 r^3+10a^3 r^2-10a^2 r+5a}{120}x
-\frac{3a^5 r^2-10a^4 r+10a^3}{120}y
+ \frac{a^5}{120}z,\\

 f_2&=
 \frac{a^4 r^4-4a^3r^3+6a^2r^2-4a r+1}{24}x
-\frac{3a^4r^2-8a^3r+6a^2}{24}y
 +\frac{a^4}{24}z,\\

 f_3&=
 \frac{5 a^3 r^4-15 a^2 r^3+15 a r^2-6r}{72}x
 -\frac{5 a^3 r^2-10 a^2 r+4a}{24}y
 +\frac{a^3}{18}z\\
 &\qquad
 +\frac{a^3 r^2-3a^2 r+3a}{72}u
 -\frac{a^3}{72}v,&\\

 f_4&=
 \frac{a^2 r^4-2a r^3+r^2}{24}x
 -\frac{3a^2 r^2-4a r}{24}y
 +\frac{a^2 r^2-2a r+1}{24}u
 -\frac{a^2}{24}v,\\

 f_5&=
 \frac{a r^4}{180}x
 -\frac{a r^2}{60}y
 -\frac{a}{45}z
 +\frac{2a r^2-3r}{72}u
 -\frac{a}{36}v+a.
\end{align*}

On the other hand, since $Y \simeq \bP^4$, we have
\[
\chi(m(a\xi_\sE-\pi^*H_X))=\chi(mH_Y)=\binom{m+4}{m}.
\]
Thus
\begin{align*}
 &f_1= 0,\\
 &f_2= \frac{1}{24},\\
 &f_3= \frac{5}{12},\\
 &f_4= \frac{35}{24},\\
 &f_5= \frac{25}{12}.
\end{align*}
Note that $(a,r_X) = (1,3)$ or $(3,1)$.
Thus, for each pair $(a,r_X)$, we can solve this system of linear equations for $(x,y,z,u,v)$.

If $a=3$ and $r_X=1$, then the above system of linear equations has no solution.
Thus we have $a=1$ and $r_X=3$.
Now the assertion follows from a straightforward calculation.
\end{proof}
\begin{proof}[Proof of Thoerem~\ref{thm:quarticDP}]
As in the proof of Theorem~\ref{thm:quarticDP}, we have the following commutative diagram:
\[
\xymatrix{
  \bP(\sE ) \ar[d]_-{\pi} \ar[r] \ar@(ur,ul)[rr]^-{\varphi}  & \bP(\sS ^* _{\Gr}) \ar[d]_-{\pi_{\Gr}} \ar[r] & \bP^4         \\
X \ar[r]^-{j} & \Gr(2,5),& \\
}
\]
with the condition $j^*\sS_{\Gr}^* \simeq \sE(-1)$.
In particular $j^*\cO_{\Gr}(1) \simeq \cO_X(1)$.
Thus the map $j$ is finite, and the image $V \coloneqq j(X)$ is a closed subvariety of dimension $4$.

Recall that the Chow group of $\Gr(2,5)$ is generated by the Schubert cycles: fix a complete flag $(0\subset V_1 \subset V_2\subset V_3\subset V_4\subset V_5 =\bC^5 )$ in the vector space $\bC^5$.
Then, for a pair $(a_1, a_2)\in \bZ_{\geq0}$ with $2 \geq a_1 \geq a_2$, the Schubert cycle $\sigma_{(a_1,a_2)}$ is defined as the cycle associated to the closed subset
\[
\Sigma_{(a_1,a_2)} \coloneqq \{\, W \in \Gr(2,5) \mid \dim (V_{3+i-a_i} \cap W)\geq i \,\}.
\]

With this notation, we have
\[
c(\sS^*_{\Gr}) =1+ \sigma _{1,0} +\sigma _{1,1}.
\]
Thus
\[
c(\sE) =1+ 3H_X +j^*(\sigma _{1,1})+2H_X^2.
\]

Since $V = j(X)$ is a closed subvariety of dimension $4$, the class $[V]$ is written as $c\sigma_{1,1} + d\sigma _{2,0}$.
Then $j_*(X)=m[V]= cm\sigma_{1,1} + dm\sigma _{2,0}$, where $m$ is the degree of the map $j$.
Then, we have
\begin{align*}
 x&= H_X^4 = m(3c+2d),\\
 y&= c_2(\sE)\cdot H_X^2 = m(7c+5d),\\
 z&= c_2(\sE)^2= m(17c+12d).
\end{align*}

On the other hand, by Proposition~\ref{prop:RR}, we have
$ y= \frac{5x-1}{2}$ and
$ z= \frac{13x-7}{2}$,
and hence we have $m=1$, $c=1$ and $d=1$.

Since $m=1$, the map $j$ is birational, and hence it is the normalization map of $V$.
Also, since $c=1$ and $d=1$, the subvariety $V$ is rationally equivalent to the codimension two linear section $V_5\subset \Gr(2,5)$ via the Pl\"ucker embedding $\Gr(2,5) \subset \bP^9$.
Moreover, since $H_X^4=5$, we already know that $X$ is a quintic del Pezzo $4$-fold.
Now it is easy to check that $V$ itself is a smooth quintic del Pezzo $4$-fold $V_5\subset \Gr(2,5)$ and the map $j$ is an isomorphism.
This completes the proof.
\end{proof}

\bibliographystyle{amsalpha}
\bibliography{}

\end{document}


TITLE: rewriting equation in terms of $y$
QUESTION [0 upvotes]: From Stewart, Precalculus, 5th ed, P98, Q.45
$$x^2+xy+y^2=4$$
how can I re-write this equation in terms of $y$? I want to put this equation into graphing software but don't know to put $y$ on one side and $x$ on the other.
Any suggestion would be great.

REPLY [1 votes]: You can think of $x$ (or $y$) as a constant, then you get something of the form:
$$ay^2 + by + c = 0.$$
The solution to this $2$. degree equation will give you $y$ as a function of $x$. You will get two solutions (plus/minus square root) which can both be plotted.
More specifically you will get:
$$y^2 + xy + x^2 - 4 = 0$$
Solution is then:
$$y = -\frac{x}{2} \pm \frac{\sqrt{x^2 - 4x^2 + 16}}{2}$$


TITLE: Limit of the sequence $\frac{n}{a^n}$ for a>1
QUESTION [1 upvotes]: In general, the sequence defined by
$$a_n=\frac{n}{c^n} $$
Diverges if $\,c\lt1$, and converges to $0$  if $\,c\gt1$. The proof of divergence for $\,c\lt1$ is quite straight forward, but I'm trying to proof that the sequence $\{a_n\}$ converges to $0$ when $\,c\gt1$ by only using the limit definition for convergence. Other results from limits are allowed, as long as they can be proved with algebra and by using some functions (differentiation or integration is not allowed, while trascendental functions are).
I have been trying to prove that
$$\frac{n}{c^n}$$ is less than some other expresion involving $n$ so a direct relationship with some $\epsilon\gt0$ can be obtained, in order to use the definition of limit. Other ways involve to prove that ${a_n}<b_n$ for some sequence $\{b_n\}$ that is known to be convergent to $0$ but I haven't found the way.  Thanks in advance!

REPLY [1 votes]: Write $c=1+p$ for some $p>0$. Then $(1+p)^{n}\geq 1+np+\dfrac{n(n-1)}{2}p^{2}$ and $\dfrac{n}{c^{n}}\leq\dfrac{n}{1+np+\dfrac{n(n-1)}{2}p^{2}}$. Now taking $n\rightarrow\infty$.


TITLE: is a solid torus homeomorphic to $S^1$?
QUESTION [1 upvotes]: The question is in the title. A solid torus is $S^1 \times D^2$, and while it is homotopic to $S^1$ (since $D^2$ is contractible), I suspect it is not homeomorphic to $S^1$. But I don't know how to prove there is no homoeomorphism from $S^1 \to D^2$ to $S^1$. From cardinality considerations there is no bijective map, let alone homeomorphism, $D^2 \to pt$, but I don't know if a a cardinality argument works for my case of $S^1$ and $S^1 \times D^2$ .., both have cardinality $\aleph_1$, and I don't know how to compare $\aleph_1 \aleph_1$ and $ \aleph_1$...maybe they are equal?

REPLY [3 votes]: Hint: Show that if you remove any two points from $S^1$, you get a disconnected space, but if you remove any two points from $S^1\times D^2$, you get a connected space.

REPLY [1 votes]: A solid torus is a manifold of dimension $3$, whereas $S^1$ is a manifold of dimension $1$.


TITLE: How to prove that a set of functions are linearly dependent?
QUESTION [2 upvotes]: Is $S$ linearly dependent on $\textsf V = \mathcal{F}(\Bbb R,\Bbb R)$ and $S = \{t, e^t ,\sin(t)\}$.
How to prove that a set of functions are linearly dependent?

REPLY [0 votes]: Also you may find if
$$\begin{vmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t)\\ f_1''(t) & f_2''(t) & f_3''(t)\end{vmatrix}= 0,$$ then the three functions are linearly dependent (LD) otherwise they are linearly independent (LI).


TITLE: The disjoint union of covering spaces is again a covering space
QUESTION [0 upvotes]: Let $B$ be an arc-connected and locally arc-connected space. Suppose that $p:E\to B$ and $q: F\to B$ are covering spaces. Let $r:E\sqcup F\to B$ be the function such that $r(x)=p(x)$ for all $x\in E$ and $r(x)=q(x)$ for all $x\in F$
Show that $r:E\sqcup F\to B$ is a covering space.
Take $x\in B$, then there is an open $U$ of $B$ such that $x\in U$ and $p^{-1}(U)=\sqcup_{\alpha\in A}V_{\alpha}$, where $V_{\alpha}\in E$ for all $\alpha\in A$, in addition, there is also an open $W$ of $B$ such that $x\in W$ and $q^{-1}(W)=\sqcup_{\beta\in B}S_{\beta}$, where $S_{\beta}\in F$ for all $\beta\in B$. But I do not know what else to do, could someone help me please? Thank you very much.

REPLY [2 votes]: So, $p^{-1}(U)=\sqcup_\alpha V_\alpha$ with $V_\alpha\subseteq E$ and such that $p|_{V_\alpha}:V_\alpha\to U$ is a homeomorphism.
Restricting the codomains to $U\cap W$, we still obtain homeomorphisms $V_\alpha\cap p^{-1}(W)\to U\cap W$,
and we will explicitly have
$$r^{-1}(U\cap W)=\bigsqcup_\alpha (V_\alpha\cap p^{-1}(W))\ \sqcup\ \bigsqcup_\beta (S_\beta\cap q^{-1}(U))\,.$$


TITLE: Simultaneously conjugate finitely matrices
QUESTION [2 upvotes]: Let $E/F$ be a field extension and $n$ a positive integer. Given two ordered subsets $U=\{A_1,\dots,A_k\}$ and $V=\{B_1,\dots,B_k\}$ of $M_n(F)$. If there is a matrix $P\in GL_n(E)$ such that $PA_iP^{-1}=B_i$ for every $1\le i\le k$, then can we find  an element $P'\in GL_n(F)$ with $P'A_iP'^{-1}=B_i$ for every $i$.
The simplest case with $k=1$ is well-known.
An evidence is as follows. Assume that $F$ is of characteristic $0$. Let $G_1,G_2$ be two finite subgroups of $GL_n(F)$.  If there is $P\in GL_n(E)$ such that $PG_1P^{-1}=G_2$, then there is $P'\in GL_n(F)$ such that $P'G_1P'^{-1}=G_2$.
A proof: Let $\rho_1:G_1\to GL_n(F)$ be the injection, and $\rho_2:G_1\to GL_n(F)$ is defined by $\rho_2(h)=PhP^{-1}$. Each $\rho_i$ is a completely reducible representation of the group $G_1$ by  Maschke's theorem.  Moreover, for every $h\in G_1$, the traces coincide $Tr(\rho_1(h))=Tr(\rho_2(h))$. By Bourbaki lemma (Corollary 3.8 p.650 of Algebra (3rd edition) by S.Lang),
the two representations are isomorphic. That means there is $P'\in GL_n(F)$ such that $PhP^{-1}=P'hP'^{-1}$ for every $h\in G_1$. In particular, $P'G_1P'^{-1}=G_2$.
Any comments are welcome.

REPLY [1 votes]: Yes, if $F$ is infinite.
The set of equations $MA_i=B_iM,i=1,...,k$ is a linear system $CX=0$, where $C$ has entries in $F$ and $X=\pmatrix{m_{11} \cr m_{12}\cr \vdots \cr m_{nn}}\in F^{n^2}$.
Let $V=\{X\in F^{n^2}\mid CX=0\}$ and $V_E=\{X\in E^{n^2}\mid CX=0\}$
Note that an $F$-basis of $V$ is an $E$-basis of $V_E$. Indeed, if $(X_1,\ldots,X_d)$ is an $F$-basis of $V$, the matrix whose columns are $X_1,\ldots,X_d$ has rank $d$. But its also has rank $d$ when viewed as a matrix with coefficients in $E$. Hence $(X_1,\ldots,X_d)$ are also $E$-linearly independent. Now similarly, the rank of $C$ is the same , viewed has a matrix with entries in $F$ or in $E$. Consequently, $\dim_F(V)=\dim_E(V_E)$ and it follows that  $(X_1,\ldots,X_d)$ is an $E$-basis of $V_E$.
So let $(X_1,...,X_d)$ be an $F$-basis of $V$, and let $f=\det(T_1X_1+\cdots+T_d X_d)\in F[T_1,...,T_d]$.
Since $P\in V_E$, $P=t_1X_1+...+t_dX_d, t_j\in E$ by the previous point.  Since $P$ is invertible, $f(t_1,...,t_d)\neq 0$. Hence, $f$ is a nonzero polynomial. Since $F$ is infinite, it cannot be identically zero on $F^d$, meaning that there is an $F$-linear combination of $X_1,...,X_d$  which is invertible. This exactly means that our set of equations has an invertible solution with coefficients in $F$.


TITLE: Generic set that is a proper subgroup
QUESTION [7 upvotes]: For a group $G$ generated by a finite set $S$ we denote by $B_{G,S}(n)$ the ball of radius $n$, that is the set of all elements in $G$ which are expressible as products $x_1x_2\ldots x_n$ where $x_i\in S\cup S^{-1}\cup\{1\}$. One calls the set $Q$ generic in $G$ with respect to $S$ if $$\lim_{n\to\infty}\sup \frac{|Q\cap B_{G,S}(n)|}{|B_{G,S}(n)|}=1.$$
My question is whether there exist a group $G$ and a proper subgroup $H<G$ such that $H$ is a generic subset in $G$ with respect to some finite generating set $S$ of the group $G$.

REPLY [9 votes]: No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B_{G,S}(n) = S^n$. Suppose
$$
  |H\cap S^{n_i}|/|S^{n_i}|\to 1
$$
for some subsequence $(n_i)$. Let $x$ be an element of $S$ not in $H$. Since $H$ and $Hx$ are disjoint we have
$$
  |H\cap S^{n_i}x^{-1}|/|S^{n_i}| = |Hx\cap S^{n_i}|/|S^{n_i}| \to 0.
$$
Thus in particular
$$
  |S^{n_i} \cap S^{n_i} x^{-1}|/|S^{n_i}| \to 0.
$$
But
$$
  S^{n_i-1} x \subset S^{n_i},
$$
so
$$
  S^{n_i-1} \subset S^{n_i} \cap S^{n_i} x^{-1}.
$$
Also $S^{n_i} \subset S^{n_i-1} S$. Thus
$$
  |S^{n_i} \cap S^{n_i} x^{-1}| / |S^{n_i}| \geq |S^{n_i-1}|/|S^{n_i}| \geq 1/|S| \not\to 0,
$$
a contradiction.
But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$? However, we would need to have (1) lots of generators, (2) exponential growth. (If $|S^n|$ has subexponential growth then you have an invariant measure so you can't get bigger than upper density $1/2$.) Something in the free group $F_n$?


TITLE: Is it true that $\sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n} a_i = \sum\limits_{i=1}^{n}a_i$, where $a_i\in \Bbb R$
QUESTION [0 upvotes]: Is it true that $\sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n} a_i = \sum\limits_{i=1}^{n}a_i$ where $a_i\in \Bbb R$ for $i=1,...n$.
I think it is true since $\sum\limits_{i=1}^{n} \left(\sum\limits_{i=1}^{n} a_i\right)=\sum\limits_{i=1}^{n} C =C$, where $C=\sum\limits_{i=1}^{n} a_i$.

I wanted to use Cauchy's inequality in a proof and I was not sure if I could use this double summation.
$$\sum\limits_{i=1}^{n}a_i^2\sum\limits_{i=1}^{n}b_i^2\sum\limits_{i=1}^{n}c_i^2 \geq \left(\sum\limits_{i=1}^{n}a_ib_i\right)^2\sum\limits_{i=1}^{n}c_k^2,$$
then I wanted to add a summation $\sum\limits_{i=1}^{n}$ to the right-hand side
$$\sum\limits_{i=1}^{n}\left(\sum\limits_{i=1}^{n}a_ib_i\right)^2\sum\limits_{i=1}^{n}c_k^2\geq \sum\limits_{i=1}^{n}\sum\limits_{i=1}^{n}a_ib_ic_i = \sum\limits_{i=1}^{n}a_ib_ic_i.$$
Can I do that?

Ok, I can't.

REPLY [2 votes]: It is not true, generally speaking. As Martin's comment points out:
$$
\sum_{i=1}^n\sum_{i=1}^na_i=\sum_{i=1}^n\left(\sum_{i=1}^na_i\right)=\sum_{i=1}^nC=nC=n\sum_{i=1}^na_i
$$
so, no. But, for instance, if $a_i=0\;\forall i$ then the equality you posted will be true. You can ask whether there are sequences $\{a\}_n$ such that your equality is true, or maybe prove that there is no possible $\{a\}_n$ (with at least one nonzero element) that satisfies your above equation.


TITLE: Homology of a space obtained from $S^n$ by attaching a cell $e^{n+1}$ by a map of degree $m$
QUESTION [5 upvotes]: I am trying to understand how to use cellular homology on this simple example: let $X$ be a space obtained from $S^n$ by attaching a cell $e^{n+1}$ by a map of degree $m$. I understand that the generators of the group $H_k(X^{(k)},X^{(k-1)})$ are in one-to-one correspondence with the $k$-cells of $X$, and I am supposing the CW structure of $S^n$ with 2 cells in each dimension. But how do I calculate the $Ker(d_k)$ and $Im(d_k)$, where $d_k$ are the boundary maps in the cellular chain complex? I know about the cellular boundary map formula, but I understand how it works on this example.

REPLY [5 votes]: Chain complex for an easier CW structure.
Well, first it is easier to compute the homology if you use the CW structure on $S^n$ having one $n$-cell and one $0$-cell. The cellular chain complex just looks like $$\mathbb{Z}\to0\to\cdots\to0\to\mathbb{Z}$$ or if $n=1$, then $\mathbb{Z} \overset{0}\to\mathbb{Z}$.
Then if you adjoin one more cell via a degree $m$ map, by definition this gives you a cellular chain complex $$\mathbb{Z} \overset{m}\to\mathbb{Z}\to0\cdots\to0\to\mathbb{Z}.$$ Following the instructions for the cellular boundary formula, as on page 140 of Hatcher, there is really no work to do here since in this case the complement of our cell in dimension $n$ is already just a point.
Chain complex for the CW structure having two cells in each dimension.
If you are really set on using the CW structure with two cells in each dimension, it takes a little more work to find the cellular chain complex for $S^n$. Also there is some amibiguity here in that there is more than one CW structure with two cells in each dimension that produces $S^n$. Of course the chain complexes will all be quasi-isomorphic, but if you do it in an organized way then you will get $$\mathbb{Z}^2 \overset{d_n}\longrightarrow \mathbb{Z}^2 \overset{d_{n-1}}\longrightarrow \cdots \overset{d_1}\longrightarrow\mathbb{Z}^2$$
where each of the $d_i$ are just given by the matrix $\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$. I will describe how to do this below. Once you see this, then attaching the $n+1$ cell via a degree $m$ map should give you
$$\mathbb{Z} \overset{\begin{pmatrix}m\\ m\end{pmatrix}}\longrightarrow\mathbb{Z}^2 \overset{d_n}\longrightarrow \mathbb{Z}^2 \overset{d_{n-1}}\longrightarrow \cdots \overset{d_1}\longrightarrow\mathbb{Z}^2.$$
Computing $d_{n+1}$ with the cellular boundary formula.
To compute that $d_{n+1}$ is as above using the cellular boundary formula, consider the two $n$-cells in $S^n$, one for each hemisphere. The coefficient of $d_{n+1}(e^{n+1})$ on the top hemisphere is obtained by crushing the lower hemisphere and checking the degree of the map $$\partial D^{n+1} \to S^n \overset{\text{crush}}\longrightarrow S^n.$$ By the local degree theorem, this only depends on the size of the preimage at a suitable point - but this is the same as for the original map since all we did was crush the lower hemisphere. So the degree of this map is still $m$. The same logic applies to the coefficient on the cell for the lower hemisphere.
Seeing that $d_i = \begin{pmatrix}1&-1\\1&-1\end{pmatrix}$.
Label the cells in dimension $k$ by $e^k$ and $f^k$, which will be the correct ordered basis for the matrices above. Here is how to define attaching maps so that all chain maps are given by the matrix above.
Consider the $k$-skeleton $S^k$ inside of the plane $\mathbb{R}^{k+1}$ in $\mathbb{R}^{k+2}$ (you should probably take $k=1$ for visualization purposes).
Above $S^k$ in $\mathbb{R}^{k+2}$ is the top hemisphere of $S^{k+1}$ which is isomorphic to $D^{k+1}$ via the projection map. The attaching map is the identity, so as above we have that $d_{k+1}(e_{k+1}) = e_k + f_k$. The lower hemisphere of $S^{k+1}$ is isomorphic to $D^{k+1}$ as well, but the orientation of the boundary is the opposite, relative to $S^k$. To see this, note that the attaching map is the same identifying $e^k$ with the upper hemisphere, reflecting through $\mathbb{R}^{k+1}$, then attaching along the identity map. So, the cellular boundary formula tells you $d_{k+1}(f_{k+1}) = -e_k - f_k$.


TITLE: Prove the boundedness of a bilinear continuous mapping.
QUESTION [8 upvotes]: Let $X,Y,Z$ are Banach spaces and $$B:X\times Y\to Z$$ is bilinear and continuous. Prove that there exists $M<\infty$ such that
$$\lVert B(x,y)\rVert \leq M\lVert x\rVert\lVert y\rVert.$$ Is completeness needed here? The bilinearity of $B$ means for a fixed  $x\in X$,$B_x\colon Y\to Z$ is linear and for a fix $y\in Y$, $B^y:X\to Z$ is linear.
Since $X,Y,Z$ are metrizable by norms, I think what we do is to show that $B$ is bounded. Let $E_1,E_2$ are bounded, then $E_1\subset t_1V_1$ and $E_2\subset t_2V_2$ for all large $t_1,t_2$; put $M=\max\{t_1,t_2\}$, so that $$B(E_1\times E_2)\subset B(t_1V_1\times t_2V_2)\subset MB(V_1\times V_2)\subset MW,$$ in which $V_1,V_2$ are balanced neighborhoods of  origins in $X,Y$ respectively and $W$ is a neighborhood of $0$ in $Z$.
Can someone point out if there is some errors in my thought or give a straightforward path?

REPLY [0 votes]: This post is just a supplement and detailed version of Timon's answer.
I think we may need to change $X_1$ to unit circle. So in this post, $X_1$ will be the unit circle in $X$. And the rest of notations are the same with those in Timon's answer.
Following from Timon's answer, we know that, $\forall x\in X_1,\,y\in Y$, we will have that $|B(x,y)|\le||B_x|| \cdot||y|| \le M \cdot ||y||$, where $M:=\sup_{x\in X_1} ||B_x||<\infty$.
Note that $\forall x\in X_1$, $||x||=1$.
Hence, $\forall x\in X_1,\,y\in Y$, we have that $|B(x,y)|\le M \cdot ||y||\cdot ||x||$
Now for any arbitrary $\alpha\in X,$ we know there exists an $x_0 \in X_1$ and some constant $c$ s.t. $\alpha=c\, x_0.$
Therefore, $$|B(\alpha,y)|=|B(c\,x_0,y)|=|cB(\,x_0,y)|=|c|\cdot|B(\,x_0,y)|\, \le |c|\cdot M \cdot ||y||\cdot ||x_0||$$
$$=M \cdot ||y||\cdot ||c\,x_0||= M \cdot ||y||\cdot ||\alpha||$$
I.e. $|B(\alpha,y)|\le M \cdot ||y||\cdot ||\alpha||,\,\forall \alpha\in X,\,y\in Y$.
$\tag*{$\blacksquare$}$