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\begin{document}

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\title[ Cohomological dimension ]
 { Cohomological dimension with respect to the linked ideals}

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     \author[M. jahangiri]{Maryam jahangiri$^1$}
     \author[kh. sayyari]{khadijeh sayyari$^2$}

\address{$^{1, 2}$ Faculty of Mathematical Sciences and Computer,
Kharazmi University, Tehran, Iran.}

\email {jahangiri@khu.ac.ir, std-sayyari@khu.ac.ir}

\keywords{Linkage of ideals,  local cohomology modules}

 \subjclass[2010]{13C40, 13D45.}

\begin{abstract}

  Let $R$ be a commutative Noetherian ring.
 Using the new concept of linkage of ideals over a module, we show that if $\fa$ is an ideal of $R$ which is
 linked by the ideal $I$, then $\cd(\fa,R) \in \{ \grad \fa, \cd(\fa, H^{\grad \fa}_ {\fc} (R)) + \grad \fa\}, $
  where $\fc : = \bigcap_{\fp \in \Ass \frac{R}{I}- V(\fa)}\fp$. Also, it is shown that
for every ideal $\fb$ which
is geometrically linked with $\fa,$ $\cd(\fa, H^{\grad \fb}_ {\fb} (R))$  does not depend on $\fb$.
\end{abstract}

\maketitle

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\section{Introduction}

Let $R$ be a commutative
Noetherian ring, $\fa$   be an ideal  of $R$ and
$M$ be an $R$-module. For each $i\in \Z$, $H^i_{\fa} (M)$ denotes the $i$-th local cohomology module of $M$ with respect to $\fa$ (our terminology on local cohomology modules comes from \cite{BS}). Vanishing of these modules is an important problem in this topic and it attracts lots of interest, see for example \cite{L} and \cite{O}.
One of the most various invariants in local cohomology theory is
the cohomological dimension of   $M$ with respect to the
ideal $\fa$, i.e. $$\cd(\fa , M) : = \Sup \{ i \in \mathbb{N}_0|
H^i_{\fa} (M) \neq 0 \}.$$

In this paper, we consider the cohomological dimension of $M$ with respect to the "linked ideals" over it.

Following \cite{PS}, two proper ideals $\fa$ and $\fb$ in a Cohen-Macaulay
local ring $R$ is said to be linked if there is a regular sequence
$\underline{\fx}$ in their intersection such that $\fa =
(\underline{\fx}) :_R \fb$ and $\fb = (\underline{\fx}) :_R \fa$.
In a recent paper, \cite{JS}, the authors
introduced the concept of linkage of ideals over a module and
studied some of its basic properties. Let   $\fa$ and $\fb$ be two non-zero ideals of $R$ and
$M$ denotes a non-zero finitely generated $R$-module. Assume that
$\fa M\neq M\neq \fb M$  and let $I\subseteq \fa \cap \fb$ be an
ideal generating by an $M$-regular sequence. Then the ideals $\fa$
and $\fb$ are said to be linked by $I$ over $M$, denoted by
$\fa\sim_{(I;M)}\fb$, if $\fb M = IM:_M\fa$ and $\fa M = IM:_M\fb $.

  This concept is the classical concept of linkage of ideals in
\cite{PS}, where $M= R.$ Note that
these two concepts do not coincide \cite[2.6]{JS} although, in some cases they do
(e.g. Example \cite[2.4]{JS}).  We can also characterize linked ideals over $R,$ see \cite[2.7]{JS2}.

As an application of this generalization, one may characterize Cohen-Macaulay modules in terms of the type of linked ideals over it, see \cite[3.5]{JS2}.

  In this paper, we consider the above
generalization of linkage of ideals over a module and, among other
things, study the cohomological dimension of an $R$-module $M$ with
respect to the ideals which are linked over $M$. In particular, in
 Theorem \ref{t5} we show that if $\fa$ is an ideal of $R$ which is
 linked by $I$ over $M$, then $$\cd(\fa,M) \in \{ \grad_M \fa, \cd(\fa, H^{\grad_M \fa}_ {\fc} (M)) + \grad_M \fa\},
 $$
  where $\fc : = \bigcap_{\fp \in \Ass \frac{M}{IM}- V(\fa)}\fp$.

  And
in Corollary \ref{c2}  it is shown that
for every ideal $\fb$ which
is geometrically linked with $\fa$ over $M$, $\cd(\fa, H^{\grad_M \fb}_ {\fb} (M))$ is constant and does not depend on $\fb$.

Also, we show that if $\cd(\fb, R)< \dim(R)$ for any linked ideal  $\fb$ over $R$, then $\cd(\fa, R)< \dim(R)$
  for any ideal  $\fa$ (Corollary \ref{c1}).

Throughout the paper, $R$ denotes a commutative Noetherian ring with $1\neq 0$, $\fa$ and $\fb$ are two non-zero proper ideals
of $R$ and $M$ denotes a non-zero finitely generated $R$-module.

\section{ Cohomological dimension}

The cohomological dimension of an $R$-module $X$ with respect to $\fa$ is defined
by $$\cd(\fa , X) : = \Sup \{ i \in \mathbb{N}_0| H^i_{\fa} (X) \neq 0 \}.$$ It is a significant invariant in
local cohomology theory and attracts lots of interest, see for example \cite{L} and \cite{O}. In this section, we study
this invariant via "linkage". We begin by the definition of our main tool.

\begin{defn}\label{F1}

 \emph{Assume that $\fa M\neq M\neq \fb M$
 and let $I\subseteq \fa \cap \fb$ be an ideal generated by an $M$-regular sequence.
  Then we say that the ideals $\fa$ and $\fb$ are linked by $I$ over $M$,
  denoted $\fa\sim_{(I;M)}\fb$, if $\fb M = IM:_M\fa$ and $\fa M = IM:_M\fb $.
  The ideals $\fa$ and $\fb$ are said to be geometrically linked by $I$ over $M$ if $\fa M \cap \fb M =
  IM$. Also, we say that the ideal $\fa$ is linked over $M$ if there
  exist ideals $\fb$ and $I$ of $R$ such that $\fa\sim_{(I;M)}\fb$.
  $\fa$ is $M$-selflinked by $I$ if $\fa\sim_{(I;M)}\fa$.
  Note that in the case where $M = R$, this concept is the classical concept of linkage of ideals in \cite{PS}.}
\end{defn}
The following lemma, which will be used in the next proposition, finds some relations between local cohomology modules of $M$ with respect to ideals which are linked over $M.$

\begin{lem}\label{l07}

 Assume that $I$ is an ideal of $R$ such that $\fa\sim_{(I;M)}\fb$. Then
  \begin{itemize}
    \item [ (i) ] $\sqrt{I + \Ann M} = \sqrt{(\fa \cap \fb) + \Ann M}$. In particular, $H^i_{\fa \cap \fb} (M) \cong H^i_ I (M)$, for all $i$.
    \item [ (ii) ]Let $I=0$. Then, $\sqrt{0:_{\frac{R}{\Ann M}}\fa}= \sqrt{\frac{\Ann \fa M}{\Ann M}}= \sqrt{\frac{\fb + \Ann_R M}{\Ann_R M}}.$ Therefore, $H^ i _{\Ann_R M :_R \fa}(M)\cong H^ i _{\Ann \fa M}(M)\cong H^ i _{\fb}(M)$. In other words, if $M$ is faithful, then $H^ i _{\fb}(M)\cong H^ i _{0:_R \fa}(M)$.
  \end{itemize}
 \end{lem}

\begin{proof}
\begin{itemize}
  \item [ (i) ] Let $r \in\fa \fb$. By the assumption, $rM \subseteq IM $. Therefore, in view of \cite[2.1]{M}, there exist an integer $n$ and $b_1 ,..., b_n \in I$ such that $(r^n + r^ {n-1}b_1 + ...+ b_n)M= 0$. This implies that $r+\Ann M \in \sqrt{\frac{I + \Ann M}{\Ann M}} $, as desired. Now, the result follows using \cite[4.2.1]{BS}.
  \item [ (ii) ] Let $r \in \Ann \fa M$. Then,
  by the assumption,  $rM \subseteq \fb M $ and using similar argument  in part (i) one can see that  $r+ \Ann M \in \sqrt{\frac{\fb + \Ann M}{\Ann M}}.$ Also, via the fact
  that $\fa \fb \subseteq \Ann M,$ $\frac{\fb + \Ann M}{\Ann M} \subseteq 0:_{\frac{R}{\Ann M}}\fa.$ This proves
  the desired equalities. Now, the isomorphisms between local cohomology modules follows using \cite [4.2.1]{BS}.

\end{itemize}

\end{proof}
\begin{prop}\label{t8}

 Let $I$ be an ideal of $R$ such that $\fa\sim_{(I;M)}\fb$ and set $t: = \grad_M I$. Then $ \cd (\fa + \fb, M) \leq  \Max \{ \cd(\fa, M), \cd(\fb, M), t+1 \}.$ Moreover, if $ \cd (\fa + \fb, M) \geq t+1,$ e.g. $\fa$ and $\fb$ are geometrically linked over $M$, then the equality holds.
 \end{prop}

\begin{proof}
Note that $H^i_ I (M) = 0$ for all $i \neq t$, by \cite[1.3.9 and 3.3.1]{BS}. Now, the result follows from \ref{l07} and using the Mayer-Vietoris sequence

\begin{eqnarray*}... \longrightarrow H^i_ {\fa \cap \fb} (M) \longrightarrow H^{i+1}_ {\fa + \fb} (M)\longrightarrow H^{i+1}_ {\fa} (M) \oplus H^{i+1}_ {\fb} (M) \longrightarrow H^{i+1}_ {\fa\cap \fb}(M)\longrightarrow ... \textbf{.}\end{eqnarray*}

\end{proof}
The following corollary, which is immediate by the above
proposition, shows that, in spite of \cite[21.22]{e}, parts of an
$R$-regular sequence can not be linked over $R.$
\begin{cor}\label{c11}

Let $(R, \fm)$ be local and $x_1, ... , x_n \in \fm$ be an $R$-regular sequence, where $n \geq 4$. Then $(x_{i_1}, ... , x_{i_j})\nsim (x_{i_{j+1}}, ... , x_{i_{2j}})$, for all $1 < j \leq [\frac{n}{2}]$ and any permutation $(i_1,...,i_{2j}) $ of $\{ 1,..., 2j\}$ .
\end{cor}

Let $M \neq \fa M$. It is well-known,  by \cite[1.3.9]{BS}, that $\grad_M \fa \leq \cd(\fa, M).$ Then $M$ is said to be relative Cohen-Macaulay with respect to $\fa$ if $$\cd(\fa,M)= \grad_M \fa.$$

In the following proposition we compute the cohomological dimension of an $R$-module $M$ with respect to $\fa$ in two cases.
\begin{prop}\label{l5}

Let $I$ be an ideal of $R$ generating by an $M$-regular sequence of length $t$ and $\fa\sim_{(I;M)}\fb$.
\begin{itemize}
  \item [(i)] If $M$ is relative Cohen-Macaulay with respect to $\fa + \fb,$ then $H^i_{\fa} (M) =0$ for all $i \notin \{\grad_M \fa,\grad_M \fa+\fb \}.$
  \item [(ii)] If $I=0,$ then $\cd(\fa,M)= \cd(\fa, \frac{M}{\fb M}).$
\end{itemize}

 \end{prop}

\begin{proof}
\begin{itemize}
  \item [(i)] Using the assumption, $H^i_ {\fa \cap \fb} (M)= 0$ for all $i \neq t.$ Now, the result follows from the isomorphism $H^i_{\fa+\fb} (M) \cong H^i_{\fa} (M) \bigoplus H^i_{\fb} (M)$, for all $i>t+1$, and the surjective map $$H^{t+1}_ {\fa+\fb} (M)\longrightarrow H^{t+1}_ {\fa} (M) \oplus H^{t+1}_ {\fb} (M) \longrightarrow 0,$$ which are deduced by the Mayer-Vietoris sequence.
  \item [(ii)] It follows from the fact that $\fb M$ is $\fa$-torsion.
\end{itemize}

\end{proof}
The following lemma will be used in the rest of the paper.
\begin{lem}\label{l1}
Let $I$ be a proper ideal of $R$ such that $\fa\sim_{(I;M)}\fb$. Then, $\frac{M}{\fa M}$ can be embedded in finite copies of $\frac{M}{IM}$.
\end{lem}

\begin{proof}
Assume that $F \rightarrow \frac{R}{I} \rightarrow \frac{R}{\fb}
\rightarrow 0 $ is a free resolution of $\frac{R}{\fb}$ as
$\frac{R}{I}$-module. Then, using $^*:= \Hom_{\frac{R}{I}} (- ,
\frac{M}{IM})$, we get the exact sequence
 $0 \rightarrow (\frac{R}{\fb})^* \rightarrow (\frac{R}{I})^* \stackrel{f}{\rightarrow} F^*$,
 where $\frac{M}{\fa M}\cong Im (f) \subseteq F^* \cong \oplus \frac{M}{IM} $.

\end{proof}
The next theorem, which is our main result, provides a formula for $\cd(\fa,M)$ in the case where $\fa$ is linked over $M$.
\begin{thm}\label{t5}

Let $I$ be an ideal of $R$ generating by an $M$-regular sequence such that
 $\Ass \frac{M}{IM} = \Min Ass \frac{M}{IM}$ and $\fa$ is linked by $I$
 over $M.$ Then $$\cd(\fa,M) \in \{ \grad_M \fa, \cd(\fa, H^{\grad_M \fa}_ {\fc} (M)) + \grad_M \fa\}, $$
  where $\fc : = \bigcap_{\fp \in \Ass \frac{M}{IM}- V(\fa)}\fp$.
 \end{thm}

\begin{proof}
Note that, by \ref{l1}, $\Ass \frac{M}{\fa M}\subseteq \Ass \frac{M}{IM}$. Set $t:= \grad_M \fa$. Without loss of generality, we may assume that $\cd(\fa,M) \neq t$. Hence, there exists $\fp \in \Ass \frac{M}{IM}- V(\fa)$, else, $\sqrt{I + \Ann M}= \sqrt{\fa + \Ann M}$ which implies that $\cd(\fa,M)= t$. We claim that \begin{equation}\label{e2} \grad_M(\fa +\fc)> t.\end{equation} Suppose the contrary. So, there exist $\fp \in \Ass \frac{M}{IM}$ and $\fq \in \Ass \frac{R}{\fc}$ such that $\fa +\fq \subseteq \fp$. By the assumption, $\fp = \fq$ which is a contradiction to the structure of $\fc$.

          Let $A:= \{\fp|\fp \in\Ass \frac{M}{IM}\cap V(\fa)\}.$ Then, in view of \ref{l1},
          $$\sqrt{\fa+ \Ann M} = \bigcap_{\fp\in \Min Ass \frac{M}{\fa M}}\fp \supseteq \bigcap_{\fp\in A}\fp.$$
          On the other hand, let $\fp\in \Min A.$ Then, there exists $\fq\in \Min Ass \frac{M}{\fa M}$ such that $\fq\subseteq \fp.$ Hence,
          again by \ref{l1}, $\fq\in A$ and, by the structure of $\fp,$ $\fq=
          \fp.$ Therefore,
\begin{equation}\label{e1}
            \sqrt{\fa+ \Ann M} = \bigcap_{\fp\in A}\fp.
          \end{equation}
          Whence, using (\ref{e1}), it follows that $$\sqrt{I + \Ann M}= \bigcap_{\fp \in \Ass \frac{M}{IM}}\fp = \bigcap_{\fp \in \Ass \frac{M}{\fa M}}\fp \cap \fc =\sqrt{\fa \cap \fc+ \Ann M}.$$ Now, in view of (\ref{e2}), we have the following Mayer-Vietoris sequence \begin{equation}\label{e8} 0 \longrightarrow H^t_ {\fa} (M) \oplus H^t_ {\fc} (M) \longrightarrow H^t_ I (M)\longrightarrow N \longrightarrow 0\end{equation} for some $\fa$-torsion $R$-module $N$. Applying $\Gamma_{\fa}(-)$ on (\ref{e8}), we get the exact sequence $$  0 \rightarrow H^t_ {\fa} (M) \oplus \Gamma_{\fa}(H^t_ {\fc} (M)) \rightarrow \Gamma_{\fa}(H^t_ I (M))\rightarrow N \stackrel{f}\rightarrow H^1_{\fa}(H^t_ {\fc} (M)) \rightarrow H^1_{\fa}(H^t_ I (M))\rightarrow 0$$ and the isomorphism $$ H^i_{\fa}(H^t_ {\fc} (M)) \cong H^i_{\fa}(H^t_ I (M)), \textrm{for all }i>1.$$ Also, using \cite[3.4]{SA}, we have $H^{i+t}_{\fa}(M) \cong H^i_{\fa}(H^t_ I (M))$, for all $i\in \mathbb{N}_0$. This implies that

$$H^i_ {\fa} (M) \left\lbrace
           \begin{array}{c l}
              \cong H^{i-t}_{\fa}(H^t_ {\fc}(M)\ \ & \text{ \ \ $i>t+1$,}\\
              \cong \frac{H^1_{\fa}(H^t_ {\fc}(M)}{im(f)}\ \ & \text{ \ \ $i=t+1$,}\\
              \neq 0\ \   & \text{   \ \ ${i=t}$,}\\
              0\ \   & \text{   \ \ $\textrm{otherwise}$.}
           \end{array}
        \right.$$\\%

Now, the result follows from the above isomorphisms.

\end{proof}

The following corollary, which follows from the above theorem,
provides a precise formula for $\cd(\fa,M)$ in the case where $\fa$
is geometrically linked over $M$ and shows how far $\cd(\fa,M)$ is from $\grad_M \fa.$ Note that by \cite[1.3.9]{BS}, $\grad_M \fa \leq \cd(\fa, M).$
\begin{cor}\label{c3}
Let $I$ be an ideal of $R$ generating by an $M$-regular sequence and $\fa$ and $\fb$ be geometrically linked by $I$ over $M$. Also, assume that $M$ is not relative Cohen-Macaulay with respect to $\fa.$ Then $$\cd(\fa,M) = \cd(\fa, H^{\grad_M \fa}_ {\fb} (M)) + \grad_M \fa.$$
\end{cor}
\begin{proof}
First, we show that \begin{equation}\label{e3}
  \{\fp|\fp \in \Ass \frac{M}{IM}- V(\fa)\}= \{\fp|\fp \in \Ass \frac{M}{IM}\cap V(\fb)\}.
\end{equation}
Let $\fp \in \Ass \frac{M}{IM}- V(\fa).$ Then, by \ref{l07}(i),
$\fp\in  V(\fb).$ On the other hand, if $ \fp\in \Ass
\frac{M}{IM}\cap V(\fb)$ then $\fp\nsupseteqq \fa,$ else, $$
0:_{\frac{M}{IM}}\fa\cap0:_{\frac{M}{IM}}\fb =
0:_{\frac{M}{IM}}\fa+\fb \neq 0,$$ which is a contradiction.

Now, in view of \cite[2.8(iii)]{JS},
$$\sqrt{\fb + \Ann M} =\bigcap_{\fp \in \Ass \frac{M}{IM}\cap V(\fb)}\fp.$$ This, in conjunction with (\ref{e3}),
 implies that $$\sqrt{\fb + \Ann M} =\bigcap_{\fp \in \Ass \frac{M}{IM}- V(\fa)}\fp.$$ Now, the result follows using similar argument as used in the proof of theorem \ref{t5}.
\end{proof}

\begin{prop}\label{t}
  Let $I$ be an ideal of $R$ generating by an $M$-regular sequence and $\fa$ and $\fb$ be geometrically linked by $I$ over $M$. Then $\grad_M \fa + \fb = \grad_M I + 1.$
\end{prop}

\begin{proof}
  Assume that $\fa$ and $\fb$ are geometrically linked by some $M$-regular sequence $I$ of length $t$ over $M.$ Then $\fa $ and $\fb$ are geometrically linked by zero over $\frac{M}{I M}.$ In view of the fact that $\grad _M (\fa +\fb) = \grad _{\frac{M}{IM}} (\fa +\fb)+ t,$ one can replace $M$ by $\frac{M}{I M}$ and assume that $I =0.$  Hence, it is enough to show that $\grad_M \fa + \fb = 1.$

   Let $\fp \in V(\fa + \fb).$ By \cite[2.12]{JS}, $\fa R_{\fp}$ and $\fb R_{\fp}$ are geometrically linked by zero over $M_{\fp}.$ In conjunction with the facts that $\grad_M \fa + \fb \leq \grad_{M_{\fp}} (\fa + \fb)R_{\fp}$ and $\grad_M \fa + \fb \geq 1,$ it is enough to show that $\grad_{M_{\fp}} (\fa + \fb)R_{\fp} = 1.$ Therefore, one may assume that $R$ is local.

   On the contrary, assume that $\grad_M \fa + \fb > 1.$ Using the long exact sequence $$0 \rightarrow \Hom (\frac{R}{\fa+\fb}, M)\rightarrow \Hom (\frac{R}{\fa}, M) \oplus \Hom (\frac{R}{\fb}, M) \rightarrow \Hom (\frac{R}{\fa\cap\fb}, M)\rightarrow \Ext_R^1(\frac{R}{\fa+\fb}, M)$$ and the assumption $(\fa \cap\fb)M\subseteq \fa M \cap\fb M = 0$, we get $$M \cong \fa M \oplus \fb M.$$ By the fact that $\fa \fb M=0$ we get, for any $i>0,$ $\fb^i M = \fb M.$ Then, by krull theorem, $\fb M =0$ and so $\fa M = M,$ which is a contradiction.
\end{proof}

\begin{rem}\label{e5}
\indent
\begin{itemize}
\item [ (i) ]
An ideal can be linked with more than one ideal. As an example, let $R$ be local and $x , y, z$ be an $M$-regular sequence. Then, $Rx$ is  geometrically linked with $Ry$ and $Rz$ over $M$.
\item[ (ii) ]
 Let $\fa$ and $\fb$ be geometrically linked by $I$ over $R$. Then, by \cite[2.4]{JS2}, $\sqrt{\fa}$ is linked by $I$. In particular, $\sqrt{\fa}$ is linked by every ideal $I'\subset \sqrt{\fa}$ which is generated by a maximal $R$-seqence in $\fa.$ Indeed, by \cite[2.8]{JS} and \cite[Theorem 1]{He},

 \begin{eqnarray*}
 \Ass \frac{R}{\fa} = V(\fa) \cap\Ass\frac{R}{I} = \Ass H^{\grad \fa}_ {\fa} (R)&=& \Ass\Hom_R(\frac{R}{\fa},\frac{R}{I'})\\ &=& V(\fa) \cap\Ass\frac{R}{I'},
 \end{eqnarray*}
which, in view of \cite[2.4]{JS2}, implies that $\sqrt{\fa}$ is linked by $I'.$
 \end{itemize}
\end{rem}
 The following corollary shows that for all ideals $\fb$ which are geometrically linked with $\fa$ over $M$, $\cd(\fa, H^{\grad_M \fb}_ {\fb} (M))$ is constant.��
\begin{cor}\label{c2}
  Let $\fa$ be  linked over $M$. Then, for every ideal $\fb$ which is geometrically linked with $\fa$ over $M$, $\cd(\fa, H^{\grad_M \fb}_ {\fb} (M))$ is constant. In particular,
  $$\cd(\fa, H^{\grad_M \fb}_ {\fb} (M))=\left\lbrace
           \begin{array}{c l}
               1,   &  ��\text{   M is relative Cohen-Macaulay}�\\�
                � &  �\text{with respect to} \ \fa,\\
               \cd(\fa, M)- \grad_M \fa,   & \text{   \textrm{otherwise.}}
           \end{array}
        \right.$$\\%

 \end{cor}

\begin{proof}
Assume that $M$ is relative Cohen-Macaulay with respect to $\fa$ and $\fa$ and $\fb$ are geometrically linked by some $M$-regular sequence $I$ of length $t$ over $M.$ Then, by \cite[2.8]{JS}, we have the following Mayer-Vietoris sequence \begin{equation}\label{e.8} 0 \longrightarrow H^t_ {\fa} (M) \oplus H^t_ {\fb} (M) \longrightarrow H^t_ I (M)\longrightarrow N \longrightarrow 0\end{equation} for some $\fa$-torsion $R$-module $N$. Applying $\Gamma_{\fa}(-)$ on (\ref{e.8}), we get the exact sequence $$ H^{i-1}_ {\fa}( N) \rightarrow H^i_{\fa}(H^t_ {\fb} (M)) \rightarrow H^i_{\fa}(H^t_ I (M)),$$ for $i>1.$ Now, by \cite[3.4]{SA} and the assumption, we get $H^i_{\fa}(H^t_ {\fb} (M))= 0,$ for $i>1.$ On the other hand, again by \cite[2.8]{JS} , $\Gamma_{\fa}(H^t_ {\fb} (M))= 0.$

Therefore, using the convergence of spectral sequences $$H^i_{\fa}(H^j_ {\fb} (M)) \Rightarrow_i H^{i+j}_ {\fa+\fb} (M)$$ and the assumption, we get $H^1_{\fa}(H^t_ {\fb} (M))\cong H^{t+1}_ {\fa+\fb} (M).$ Now, by \ref{t}, $H^1_{\fa}(H^t_ {\fb} (M))\neq 0$ and $\cd(\fa, H^{\grad_M \fb}_ {\fb} (M))= 1.$

In the case where $M$ is not relative Cohen-Macaulay with respect to $\fa,$ the result follows from \ref{c3}.

\end{proof}
\begin{con}

 Assume that $I$ is an ideal of $R$ which is generated by an $M$-regular sequence. We define the set $$ S_{(I;M)} : = \{ \fa \triangleleft R | I \subsetneqq \fa,  \fa = IM:_R IM:_M \fa  \}.$$
$S_{(I;R)}$ actually contains of all linked ideals by $I$.
\end{con}

The following proposition, which is needed in the next two items, shows that any ideal $\fa$ with $\fa M \neq M$ can be embedded in a redical ideal $\fa'$ of $S_{(I;M)}$ for some $I.$
\begin{prop}\label{t7}

Assume that $\fa M \neq M$. Then,
\begin{itemize}
 \item [(i)] There exists an ideal $I$, generating by an $M$-regular sequence, such that $\fa$ can be embedded in a radical element $\fa'$ of $S_{(I;M)}$ with $\grad_M \fa' = \grad_M \fa= :t$. Also, $\fa'$ can be chosen to be the smallest radical ideal with this property.

 \item [(ii)] Let $\fa'$ be as in (i). Then $\Ass H^t_{\fa} (M) = \Ass \frac{R}{\fa'}$. In particular, $\fa'= \bigcap _{\fp \in \Ass H^t_{\fa} (M)} \fp$ and it is independent of the choice of the ideal $I.$
\end{itemize}
\end{prop}

\begin{proof}
\begin{itemize}
 \item [(i)] Let $x_1,...,x_t\in\fa$ be an $M$-regular sequence such that $\fa \subseteq Z_R (\frac{M}{(x_1,...,x_t)M})$. Replacing $(x_1,...,x_t^2)$ with $(x_1,...,x_t)$, we may assume, in addition, that $x_1,...,x_t\neq\fa$ and it is not a prime ideal. Set $\Lambda: = \{\fp | \fp \in \Ass (\frac{M}{(x_1,...,x_t)M}), \fa \subseteq \fp\}$ and $\fa' = \cap_{\fp \in \Lambda} \fp $. Then, setting $I:= (x_1,...,x_t)$ and using \cite[2.4]{JS2}, $\fa\subseteq\fa'$ is a radical ideal of $S_{(I;M)}.$

Assume that there exists a radical ideal $\fc\in S_{(I;M)}$ such that $\fa \subseteq \fc$. Hence, by \cite[2.4]{JS2}(iii), $\Ass\frac{R}{\fc} \subseteq \Ass\frac{M}{IM} \cap V(\fa) =\Lambda$. Therefore, $\fa' \subseteq \fc$.

\item [(ii)] By \cite[Theorem 1]{He} and \cite[1.4]{SW}, we have $$\Ass H^t_{\fa} (M) = \Ass \Ext^t_R(\frac{R}{\fa},M).$$ Also, using the above notations and in view of \cite[1.2.4 and 1.2.27]{BH},$$\Ass \Ext^t_R(\frac{R}{\fa},M)= \Ass\Hom_R(\frac{R}{\fa},\frac{M}{IM})= V(\fa) \cap\Ass\frac{M}{IM}= \Ass \frac{R}{\fa'_I}.$$ Now, by the above equalities, we have $\fa'= \bigcap _{\fp \in \Ass H^t_{\fa} (M)} \fp.$

\end{itemize}
\end{proof}

\begin{cor}\label{c4}
 Let $\fa$ be a linked ideal over $M$. Then, $\sqrt{\fa + \Ann M} = \bigcap _{\fp \in \Ass H^t_{\fa} (M)} \fp.$
\end{cor}
\begin{proof}
  It follows from \cite[2.4]{JS2}(v) and the above proposition.
\end{proof}
The following theorem provides some conditions in order to have $cd(\fa, M)< \dim M.$
\begin{thm}\label{t1}

 Let $(R,\fm)$ be local and $\underline{\fx} = x_1,...,x_t$ be an
 $M$-regular sequence of length $t$. Assume that $H^{\dim M}_{\fp} (M) =0$ for all $\fp \in \Ass_R \frac{M}{(\underline{\fx})M}$. Then, $H^{\dim M}_{\fa} (M) =0$ for any ideal $\fa \supseteq (\underline{\fx})$ with $\grad_M \fa = t.$
 \end{thm}

\begin{proof}

 Let $n := \dim M$ and $\fb \in S_{((\underline{\fx});M)}$ be a radical ideal. Then, by \cite[2.4]{JS2}, $\fb = \cap^l_{i=1} \fp_i$ for some $l\in\mathbb{N}$ and $\fp_1,...,\fp_l\in \Ass \frac{M}{(\underline{\fx})M}$. By the assumption, $H^n_ {\fb}(M)=0$ when $l= 1$. In the case $l>1$, set $\fc = \cap^l_{i=2} \fp_i$. Then, using the Mayer-Vietoris sequence

$$... \longrightarrow H^n_ {\fp_1} (M) \oplus H^n_ {\fc} (M) \longrightarrow H^n_ {\fb}(M)\longrightarrow 0$$ and the inductive hypothesis, we have $H^n_ {\fb} (M) = 0$.

Now, let $\fa \supseteq (\underline{\fx})$ be an ideal with $\grad_M \fa = t$. Then, by \ref{t7}, there exists a radical ideal $\fb\in S_{((\underline{\fx});M)}$ such that $\fa \subseteq \fb$. Let $\fb = \fa +(y_1,...,y_m).$ Then, using induction on $m$ and \cite[8.12]{BS}, it is straight forward to see that there exists an onto homomorphism $H^n_ {\fb} (M) \longrightarrow H^n_ {\fa}(M)\longrightarrow 0,$ and the result follows.

\end{proof}

\begin{rem}\label{r2}

Let the situations be as in the above theorem and assume, in
addition, that $(R,\fm)$ is complete. Let $\fa \supseteq
(\underline{\fx})$ be an ideal with $\grad_M \fa = t.$ Then, the \LH
\ shows that $\fa$ can not
 be coprimary with a member of $\Assh M$,i.e. there is no $\fp \in \Assh M$ with $\sqrt{\fa + \fp}= \fm$.
 \end{rem}

\begin{cor}\label{c1}
There is a linked ideal $\fb$ over
$R$ such that $H^{\dim R}_{\fb} (R) \neq 0.$
\end{cor}

\bibliographystyle{amsplain}


TITLE: What's the 3D equivalent of a 2D rotation?
QUESTION [1 upvotes]: In 2 dimensional vector space, a rotation can be described as a rotation around an axis pointing along a third basis vector.
What would the equivalent operation look like in 3 dimensions? i.e. a "rotation" around an "axis" defined by a fourth basis vector?

REPLY [1 votes]: Yes, in 3D (and only 3D) all rotations can still be described by an (oriented) axis $a$, and the angle of rotation $\theta$ (measured counterclockwise in a plane orthogonal to and oriented consistently with $a$, say) about that axis. This is the so-called axis-angle representation of a rotation.
The difference here is that the axis of rotation is just an ordinary vector in $\mathbb{R}^3$, unlike in 2D where the "axis of rotation" is a vector pointing out of the plane.


TITLE: Solving for streamlines from numerical velocity field
QUESTION [8 upvotes]: Say I have a given numerical velocity field in two dimensions, (u,v). I am trying to find the streamlines from this data set at a particular contour level and I thus have to solve the differential equation
$$
dy/dx = v/u=g(x,y)
$$
I can rewrite the equation to
$$
dy = g(x_i,y_i)dx
$$
The subscript $i$ denotes that $g$ is given at discrete points. Do I solve this equation simply by choosing a $y_i$ and then summing for all $x_i$?

REPLY [3 votes]: It's easy to spoil decent content, therefore: never change a (bounty) winning answer. Instead, it will be referred to as the
first answer
to the question. There are two unsatisfactory issues with that first answer, though:

The software accomplishing all this is [ available for free, but it is ] far from optimal.
In two respects: (1) too many zeroes are stored and (2) there is a lot of computation with those zeroes.
The example chosen in the first answer does very much bear the flavor of Finite Differences.
I would like to present another example, more Finite Element like, for the sheer purpose of demonstrating the power of the Least Squares method for unstructured grids.

So there are two reasons for devising a second answer.
Ideal flow past a circular cylinder (called Potential flow around a circular cylinder in
Wikipedia)
has been selected as the demonstration problem here and now:

Instead of taking mean values, known velocities are evaluated at the midpoints of the line segments,
as is shown in the picture on the left:
$$
u_{12} = u\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \quad ; \quad
v_{12} = v\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
$$ $$
d\psi = -v\,dx + u\,dy \quad \Longrightarrow \quad
\int_{(1)}^{(2)} d\psi = -\int_{(1)}^{(2)}v\, dx + \int_{(1)}^{(2)}u\, dy
\quad \Longrightarrow \\ \psi_2-\psi_1 = -v_{12}(x_2-x_1) + u_{12}(y_2-y_1)
\quad \mbox{: numerically}
$$
Velocity and stream function values are given by
Wikipedia
in polar coordinates. Which must be converted into Cartesian coordinates for our purpose.
The velocities are input for the program, the stream function is for comparison of the numerical
outcome with the exact solution eventually:
$$
\begin{cases}
V_{r}=\left(1-{R^{2}/r^{2}}\right)\cos(\theta)\\
V_{\theta}=-\left(1+R^{2}/r^{2}\right)\sin(\theta)\end{cases}
\quad \Longrightarrow \\ \begin{cases}
u(x,y) = \left[1-R^2(x^2-y^2)/(x^2+y^2)^2\right]\\
v(x,y) = -R^2.2xy/(x^2+y^2)^2\end{cases}\\
\psi =\left(r-\frac{R^2}{r}\right)\sin(\theta) = \left(1-\frac{R^2}{x^2+y^2}\right)y
$$
Another unsatisfactory issue in the first answer is inefficiency of the code in space & time:
too many zeroes are stored and there is a lot of computation with those zeroes. This can be greatly improved.
There are two main stages involved with the final equations: assembly and solving.
Assembly. In the first answer, $M \times N$ overdetermined equations $A$ are defined first,
where in our case $M = 180$ and $N = 100$. While forming the system matrix $S=A^TA$ we thus have $180\times 100^2$
operations, which is ridiculous because $A$ is almost empty: only $2$ of the $100$ entries in a row are non-zero!
It's easy to improve this by defining $A$ as being an $1 \times 2$ array and forming $A^TA$ as an elementary matrix $E$,
corresponding with just one equation; $r$ is the right hand side of it.
$$
E = A^TA = \begin{bmatrix} -1 \\ +1 \end{bmatrix} \begin{bmatrix} -1 & +1 \end{bmatrix} =
\begin{bmatrix} +1 & -1 \\ -1 & +1 \end{bmatrix} \\
r = \begin{bmatrix} -1 \\ +1 \end{bmatrix}\left[-v_{12}(x_2-x_1) + u_{12}(y_2-y_1)\right]
$$
The elementary matrices $E$ and loads $r$ must subsequently be assembled into the system matrix $S$ and global load vector $b$.
Which happens to be exactly the same procedure as is common with the conventional finite element method.
The name Least Squares Finite Element Method is justified herewith.
Solving. Instead of defining a full $100\times 100$ system matrix $S$, it's far more efficient to define
it as a band matrix. That will save us a factor $10$ of space in the first place.
But there are a few other properties that are interesting:
$$ \begin{cases}
S_{ij} = \sum_k A_{ki}A_{kj} \quad \Longrightarrow \quad S_{ij}=S_{ji} \\
S_{ii} = \sum_k A_{ki}A_{ki} \quad \Longrightarrow \quad S_{ii} > 0 \\
\left(\sum_k A_{ki}A_{kj}\right)^2 \le \sum_k A_{ki}^2\cdot\sum_k A_{kj}^2 \quad \Longrightarrow \quad
\left| S_{ij}\right| \le \sqrt{S_{ii} S_{jj}} \end{cases}
$$
The last property by Schwarz inequality. Symmetry will save us another factor $2$ in space, but more importantly, the
computation time for solving the equations will be reduced with a factor $100$ or so. The last two properties show that
the system matrix $S$ is positive definite and diagonally dominant, meaning that all pivots are on the main diagonal.
All of the improvements have been implemented in the updated version of the software, labeled as "Second answer" of the
question in MSE publications / references 2016.
There is still another issue with the streamline (= streamfunction isoline) plot, as shown in the above picture on the right.
Solving for the parameters of a
Quadrilateral Interpolation,
as would be needed with the algorithm for defining an isoline plot, is virtually impossible with quadrilaterals that
are not paralellograms. Therefore we have subdivided each of our quads into four
triangles. Thanks to an algorithm by Paul Bourke,
defining isolines with triangles is a matter of routine.
The end-result is shown in the above picture on the right. Needless to say that there is a match with the exact solution for the streamline function.


text:
(*
 * This is an autogenerated file. Do not edit.
 * It was created using instantiate_laws.py.
 * It checks whether the other autogenerated files are up-to-date.
 *)

theory Check_Autogenerated_Files
  (* These imports are not actually needed, but in jEdit, they will conveniently trigger a re-execution of the checking code below upon changes. *)
  imports Laws_Classical Laws_Quantum Laws_Complement_Quantum
begin

ML \<open>
let
  fun check kind file expected = let
    val content = File.read (Path.append (Resources.master_directory \<^theory>) (Path.basic file))
    val hash = SHA1.digest content |> SHA1.rep
    in
      if hash = expected then () else
      error (kind ^ " file " ^ file ^ " has changed.\nPlease run \"python3 instantiate_laws.py\" to recreated autogenerated files.\nExpected SHA1 hash " ^ expected ^ ", got " ^ hash)
    end
in
  check "Source" "Axioms_Classical.thy" "f4a0dac97bed23ec5b7c4cbf779f8eb2a12aa488";
  check "Source" "Axioms_Quantum.thy" "b1ac4a827c2b943202c03611176d3c723119d8e1";
  check "Source" "Laws.thy" "37803f67bcda2df6bf7abe2417d3bf49e6317dcd";
  check "Source" "Laws_Complement.thy" "6065101853fa432ca4bd6fd3113315e856e21ecb";
  check "Generated" "Laws_Classical.thy" "051bc83ae9bd061fba08bfa29468a3421817068b";
  check "Generated" "Laws_Complement_Quantum.thy" "c58ba1680287643d1c3f9b2b97d9784db8e1dd84";
  check "Generated" "Laws_Quantum.thy" "36d05e686993e4f9b7c5bf57639d840b3e9b2e47"
end
\<close>

end

subset_name: curated


TITLE: Does string substitution have a definition, similar to the one for string homomorphism in terms of monoid morphism of the free monoid?
QUESTION [1 upvotes]: string homomorphism is defined in formal language theory as:

A string homomorphism (often referred to simply as a homomorphism in formal language theory) is a string substitution such that each character is replaced by a single string. That is,  f(a)=s , where s is a string, for each character a.

and has a clearer and equivalent algebraic definition:

String homomorphisms are monoid morphisms on the free monoid, preserving the empty string and the binary operation of string concatenation.

The first definition shows that in formal language theory, string homomorphism is defined as a special case of substitution, which is defined as:

Let L be a language, and let Σ be its alphabet. A string substitution or simply a substitution is a mapping f that maps characters in Σ to languages (possibly in a different alphabet).

Does string substitution have a clearer and equivalent definition, similar to the one for string homomorphism in terms of monoid morphism of the free monoid?
If yes, does that equivalent definition  of substitution still consider homomorphism as a special case?
Thanks.

REPLY [1 votes]: Yes. Let $A$ and $B$ be alphabets. Then $A^*$ and $B^*$ are (free) monoids, but ${\cal P}(B^*)$ is also a (non-free) monoid under the usual product of languages. Now, a substitution is a monoid morphism from $A^*$ to ${\cal P}(B^*)$.
Homomorphisms are a special case if you identify $B^*$ as the submonoid of ${\cal P}(B^*)$ consisting of the singletons.


TITLE: A Dirichlet Convolution involving $\mu(n)$ and $\log n$
QUESTION [8 upvotes]: The following arithmetic identity holds:
\begin{align}
\Lambda(n) = \sum_{d \mid n} \mu(d) \log \frac{n}{d}
\end{align}
where $\mu(n)$ is the Moebius function and $\Lambda(n)$ is the von Mangoldt function. Does the following related Dirichlet convolution simplify to known (or simpler) functions?
\begin{align}
n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}
\end{align}

REPLY [4 votes]: Yes,$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=n \sum_{d \mid n} \frac{\mu(d)}{d} (\log(n)-\ln(d))$$
$$= \sum_{d \mid n} \frac{\mu(d)}{d} n\log(n)-n\frac{\mu(d)}{d}\ln(d)=n\ln(n) \sum_{d \mid n} \frac{\mu(d)}{d} -n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)$$
$$=\ln(n)\phi(n)-n\sum_{d\mid n}\frac{\mu(d)}{d}\ln(d)=\ln(n)\phi(n)+\phi(n)\sum_{p\mid n}\frac{\ln(p)}{p-1}$$
Thus,
$$n \sum_{d \mid n} \frac{\mu(d)}{d} \log \frac{n}{d}=\phi(n)( \ln(n)+\sum_{p\mid n}\frac{\ln(p)}{p-1})$$


TITLE: Derived Distributions: PDF of -ln|X| [answered]
QUESTION [0 upvotes]: I am studying questions for a probability exam. I am stuck on derived distributions.
One of my textbook's questions asks: If $X$ is a random variable uniformly distributed between $-1$ and $1$, find the PDF of $-\ln|x|$.
The solution in the textbook is given above:
I don't understand why they have the initial condition for $y \ge 0$ on the first line. Won't $\ln|X|$ always give a positive number, making $-\ln|X|$ negative, and always mapping $Y$ to a negative number?
Second, I don't understand how they go from $P(X \ge e^{-y}) + P(X \le -e^{-y}) = 1 - e^{-y}$.
I tried using the CDF of the exponential distribution to evaluate this, but I didn't get anywhere.
Any help would be greatly appreciated! Thank you.

REPLY [0 votes]: Firstly, I think you meant the distribution of $X$ to be uniform over $(-1,1)$. So, clearly the absolute value of X will always be a fraction. The logarithm of a fraction is always negative; so, $-ln|X|$ is always a non-negative quantity. Hence $Y = -ln|X|$ is always a non-negative random variable.
Next, if $X$ is uniform over $(-1,1)$, then the cdf of $X$, $$F_X(x) = P[X \le x] = \begin{cases} 0 & \text{for $x \le -1$}
\\ (1+x)/2 & \text{for $-1<x<1$}
\\ 1 & \text{for $x \ge 1$} \end{cases}$$
Use this to find why $P[X \ge e^{-y}] + P[X \le -e^{-y}] = 1 - e^{-y}$.


TITLE: How does the size of the slits (apertures) affect the detected pattern in single and double slit particle interference experiments?
QUESTION [0 upvotes]: If you emit an electromagnetic wave with visible light frequency toward a single or double slit or aperture opening you get an interference pattern that depends on the ratio of the width or radius of the openings and wavelength of the wave in the electromagnetic field.

In cases when you fire single photons like eraser and delayed eraser experiments (whether or not the photons position information is 'exposed' or not) does the detected pattern depend the same way to the relation between the photons wavelength and opening size?
Is there a similar relation to the openings size and all bosonic particles, fermionic particles, atoms and molecules minimum uncertainty wave packets size in double and single slit interference experiments?

REPLY [1 votes]: The angular position of the minima in single-slit diffraction is given by
$$d\sin{\theta_n}=n\lambda$$
so a wider slit width $d$ will cause the diffraction peaks to move closer together.
In double-slit diffraction, the width of the slit is typically taken to be negligible compared to the separation $D$ between the slits. The angular position of the maxima in double-slit interference is given by
$$D\sin{\theta_n}=n\lambda$$
so a wider separation means closer interference peaks. In addition, there is a 180-degree phase shift in the angular position of the interference pattern as compared to single-slit diffraction.
In practice, the slits in a double-slit interference also have a finite width, which means that single-slit diffraction occurs from each of them. Because $D$ is by necessity larger than $d$, the double-slit interference pattern will have many more peaks/radian than the single-slit diffraction, so the single-slit contributions are typically thought of as an "envelope" which gradually modifies the intensity of the double-slit pattern.
Interference effects can be observed in any particle that considers the wall in which the slits are cut to be opaque. So a neutrino beam, which doesn't interact with the wall very much at all, would exhibit essentially no interference pattern.


TITLE: The $\mathbf{F}$-metric induces the weak topology on the set of bounded varifolds
QUESTION [6 upvotes]: Some preliminary definitions and notation:
(1) Given a vector space $\mathbb{V}$, we denote by $G_k(\mathbb{V})$ the $k$-grassmannian of $\mathbb{V}$, i.e. the set of all $k$-dimensional vector subspaces of $\mathbb{V}$;
(2) Given a differential (or riemannian) manifold $M$, we denote by $G_k(M)$ the $k$-grasmannian bundle on $M$, i.e. the bundle with fibers $G_k(T_pM)$;
(3) A $k$-dimensional varifold on $M$ is any Radon measure $V$ on $G_k(M)$;
(4) We denote by $\mathcal{V}_k(M)$ the set of all $k$-dimensional varifolds on $M$;
(5) The mass of a varifold $V\in \mathcal{V}_k(M)$ is defined by $\|V\|=\int_{G_k(M)}dV$.

One can define a topology - the weak topology - on $\mathcal{V}_k(M)$ as follows. Let $J:=\mathcal{C}_c(G_k(M))$ be the set of compactly supported continuous functions $f:G_k(M)\to \mathbb{R}$ and consider the family of functions $\{\varphi_f\,:\, f\in J\}$ where
$$\begin{matrix}
\varphi_f:&\mathcal{V}_k(M)&\to&\mathbb{R}\\
&V&\mapsto&\int_{G_k(M)}f\,dV.
\end{matrix}$$
The weak topology on $\mathcal{V}_k(M)$ is the least topology on $\mathcal{V}_k(M)$ for which every $\varphi_f$ is continuous or, in other words, is the topology on $\mathcal{V}_k(M)$ generated by the subbasis
$$\beta=\{\varphi_f^{-1}(\Omega)\,:\, f\in J,\,\Omega\subset_{op} \mathbb{R}\}.$$

One can ask if the weak topology on $\mathcal{V}_k(M)$ is metrizable. Well, define $\mathbf{F}:\mathcal{V}_k(M)\times \mathcal{V}_k(M)\to [0,+\infty]$ by
$$\mathbf{F}(V,W)=\sup E(V,W),$$
with
$$E(V,W)=\left\{\int_{G_k(M)}f\,dV-\int_{G_k(M)}f\,dW\,:\, f\in J,\,|f|\leq 1,\,\mathrm{Lip}(f)\leq 1\right\}.$$
One verifies that $\mathbf{F}$ has the three axioms of a metric, although it can be $+\infty$. But given any $c\geq 0$, $\mathbf{F}$ is always finite on
$$\mathcal{V}_{k,c}(M):=\{V\in \mathcal{V}_k(M)\,:\, \|V\|\leq c\}.$$
Indeed, for any $f\in J$, $|f|\leq 1$, $\mathrm{Lip}(f)\leq 1$,
$$\int_{G_k(M)}f\,dV-\int_{G_k(M)}f\,dW\leq \int_{G_k(M)}\,dV+\int_{G_k(M)}\,dW=\|V\|+\|W\|\leq 2c.$$
Therefore, $\mathbf{F}$ is a finite genuine metric on $\mathcal{V}_{k,c}(M)$.

I want to prove that:

($\ast$) The weak topology and the $\mathbf{F}$-induced topology coincide on $V_{k,c}(M)$.

This $\mathbf{F}$-metric appears in several papers in the internet (generally talking about minimal surfaces) but always with its definition and the claim ($\ast$) above, followed by some sort of "it is easy to prove" or "one easily verifies".
Is it really that easy? I guess I could prove that $\tau_w\subset \tau_{\mathbf{F}}$ but, given an open $\mathbf{F}$-ball $B_\epsilon(V)$, how do I find a weak-basic open $U$ with $V\in U\subset B_\epsilon(V)$? (such open must be a finite intersection of elements of $\beta$. Which $f_i$'s and real opens $I_i$'s to choose?)

REPLY [2 votes]: I'm pretty sure this is false in general. Consider the degenerate case where $M = \mathbb{R}$ and $k=0$, then $\mathcal{V}_0(\mathbb{R})$ is the space of Radon measures on $\mathbb{R}$, which includes the dirac delta measures $\delta_x : f \mapsto f(x)$. Now note that for any $f$ with compact support:
$$
\lim_{x \to \infty} f(x) = 0
$$
hence $\lim_{x\to\infty} \delta_x = 0$ in the weak*-topology. And yet:
$$
\lim_{x \to \infty} \mathbf{F}(\delta_x, 0) = 1
$$
so the $\delta_x$ do not converge in the $\mathbf{F}$ topology (or equivalently $\mathbf{F}$ is not continuous in the weak*-topology). Restricting to a subspace of measures $V$ with $\|V\| \leq c$ does not help since $\|c \delta_x \| = c$.


TITLE: Know total amount. Must divide in two parts so that one part is 10% from another part.
QUESTION [0 upvotes]: I know total amount.
Must divide the total amount in two parts. Smaller part (amount) is 10% of bigger part (amount) so that smaller part plus bigger part is total amount.
So total is 17091.54
Bigger part is 15537.76
Calculate 10% from 15537.76 and get smaller part 1553.78
15537.76 + 1553.78 = 17091.54
But what is the formula to get either 15537.76 (90%) or 1553.78 (10%) knowing only total 17091.54 (100%)?
If 17091.54 * 0.9 = get 15382.39 (not correct)
As know formula is something like (17091.54 - 15382.39) / ((1-0.9)+0.9). But result is not correct (1709.15)
Update
I can only get result following long way.
17091.54 = 100
17091.54 / 100 = 0.005850849
To get proportion of 15537.76 to 17091.54, divide 15537.76 by 0.005850849 = 90.90909091
Then 100 / 90.90909091 = 1.1
Then 17091.54 / 1.1 = 15537.76
But this seems not the best way.
Please advice what would be better way

REPLY [0 votes]: In the Singapore education system, children are encouraged to solve this sort of problem with model drawing. Previously, I would've immediately resorted to algebra like Claude, but after having taught my 9-year old, I now draw models! :)
It's quite simple. 10% is one-tenth. So represent the small part as 1 unit. The large part is therefore 10 units. The total is 11 units, and that's 17091.54
So 1 unit = 17091.54/11 = 1553.776363636...
Therefore the small unit is 1553.78 and the large unit is 10*1553.7763 = 155377.76
I'm rounding off both answers to 2 decimal places like the total figure. They won't add up to exactly the total because of round-off errors but it'll be close.
I just wanted to illustrate a more "elementary" method than algebra to do this.


TITLE: problems in mathematics
QUESTION [4 upvotes]: Where can i find long and interesting problems related to analysis (or to mathematics in general) in order to prepare for the Phd qualifying exams (outside of usa)
i don't want research problems, but enough difficult for self training and that can be solved in some hours.

REPLY [1 votes]: Most universities have their PhD qualifying exams posted on their websites. For example:
University of Florida
http://gma.math.ufl.edu/past-exams/first-year-analysis/
UCLA
https://www.math.ucla.edu/grad/handbook/quals
Stanford
http://mathematics.stanford.edu/academics/graduate/phd-program/phd-qualifying-exams/
To name a few. I googled "phd qualifying exams math." It appears the page at the University of Florida also has qualifiers for Measure Theory, Complex Analysis, and Functional Analysis.


TITLE: Squares ending in 6: Why do they have this in common?
QUESTION [0 upvotes]: If you remove the last digit of a square number (for example $15^2=(22)5,13^2=(16)9$), why is it that when that number is odd ($16^2=(25)6$) then the final digit is always 6?
I'm aware that due to this there's only a limited number of final digits. But why does only the squares of 4 and 6 have an odd "prefix"
I assume it has something to do with the fact that the single digit squares only 4 and 6 have an odd prefix
1 01
2 04
3 09
4 16 -> odd
5 25
6 36 -> odd
7 49
8 64
9 81

But I don't see the connection.

REPLY [1 votes]: For any real number $~r~$, let $\lfloor r\rfloor$ denote the largest integer $~\leq r$.
For $a \in \Bbb{Z^+}, b \in \{0,1,2,\cdots,9\}$, I claim that
$$\left\lfloor\frac{(10a + b)^2}{10}\right\rfloor$$
is odd if and only if
$$\left\lfloor \frac{b^2}{10}\right\rfloor$$
is odd.
Proof
$$\left\lfloor\frac{(10a + b)^2}{10}\right\rfloor $$
$$= \left\lfloor\frac{(100a^2 + 20ab + b^2}{10}\right\rfloor \tag1 $$
$$ = 10a^2 + 2ab + \left\lfloor \frac{b^2}{10}\right\rfloor. \tag2 $$
The expressions in (1) and (2) above are equivalent because of the following principle:
if $P \in \Bbb{Z}$ and $r \in \Bbb{R}$, then
$$\left\lfloor P + r\right\rfloor = P + \lfloor r \rfloor.$$
Further, in (2) above, each of the first two terms is an even number.  Therefore, the overall sum will be an even number if and only if the 3rd term is even.

With the claim established, the remainder of the answer to the posted question is supplied by the analysis already given by the OP (i.e. original poster).
That is, the OP has shown that for $b \in \{0,1,\cdots,9\}$, the only instances when
$$\left\lfloor \frac{b^2}{10}\right\rfloor$$
is odd are when $b \in \{4,6\}.$  Further, these are (also) the only instances where the rightmost (i.e. one's) digit of $b^2$ is $(6)$.


text:
\begin{document}
	\title[(Filtered) cohomological rigidity of Bott towers]{(Filtered) cohomological rigidity of Bott towers}
	\date{\today}
	\author[H. Ishida]{Hiroaki Ishida}
	\address{Department of Mathematics, Graduate School of Schience, Osaka City University, Sugimoto, Sumiyoshi-ku, Osaka 558-8585, Japan}
	\email{hiroaki.ishida86@gmail.com}
	\keywords{Bott manifold, Bott tower, cohomological rigidity, toric manifold}
	\subjclass[2000]{57R19, 57R20, 57S25, 14M25}
	\begin{abstract}
		A Bott tower is an iterated $\CP ^1$-bundle over a point, where each $\CP ^1$-bundle
		is the projectivization of a rank $2$ decomposable complex vector bundle. For a Bott tower,
		the \emph{filtered cohomology} is naturally defined.
		We show that isomorphism classes of Bott towers are distinguished by their filtered cohomology rings.
		We even show that any filtered cohomology ring isomorphism between two Bott towers
		is induced by an isomorphism of the Bott towers.
	\end{abstract}

	\maketitle
	\section{Introduction}
		A \emph{Bott tower} of height $n$ is a sequence of $\CP^1$-bundles
		\begin{equation}\label{eq:Bott tower}
			B_n\stackrel{\pi_n}\longrightarrow B_{n-1} \stackrel{\pi_{n-1}}\longrightarrow
			\dots \stackrel{\pi_2}\longrightarrow B_1 \stackrel{\pi_1}\longrightarrow
			B_0=\{\text{a point}\},
		\end{equation}
		where each fibration $\pi _i: B_i\to B_{i-1}$ for $i=1,\dots,n$
		is the projectivization of a rank $2$ decomposable complex vector bundle over $B_{i-1}$.
		A decomposable complex vector bundle is the Whitney sum of line bundles.
		Each $B_i$ is called a \emph{Bott manifold}.
		As is known, a Bott manifold $B_n$ is a smooth projective toric variety.
		We denote the Bott tower \eqref{eq:Bott tower} of height $n$ by
		$B_\bullet =(\{ B_i\} _{i=0}^n , \{\pi _i\} _{i=1}^n )$.

		Two Bott towers $B_\bullet = (\{ B_i\} _{i=0}^n , \{\pi _i\} _{i=1}^n )$ and
		$B'_\bullet =(\{ B'_i\} _{i=0}^n , \{\pi '_i\} _{i=1}^n )$ are isomorphic if there is
		a collection of diffeomorphisms $\varphi _\bullet = \{ \varphi _i: B_i \to B'_i\} _{i=0}^n$
		such that the following diagram is commutative:
			\begin{equation*}
				\xymatrix{
					B_n \ar[r]^{\pi _n} \ar[d]^{\varphi _n} & B_{n-1} \ar[r]^{\pi _{n-1}} \ar[d]^{\varphi _{n-1}}
					& \dots \ar[r]^{\pi _2} & B_1 \ar[r]^{\pi _1} \ar[d]^{\varphi _1} & B_0 \ar[d]^{\varphi _0}\\
					B'_n \ar[r]^{\pi '_n} & B'_{n-1} \ar[r]^{\pi '_{n-1}}
					& \dots \ar[r]^{\pi '_2} & B'_1 \ar[r]^{\pi '_1} & B'_0 .
					}
			\end{equation*}

		Since $\pi _i: B_i \to B_{i-1}$ has a cross section,
		the induced homomorphism $\pi _i^* : \mathrm{H}^*(B_{i-1}) \to \mathrm{H}^*(B_i)$ is injective for each $i$.
		Therefore the cohomology ring $\mathrm{H}^*(B_{i-1})$ can be regarded as a subring of $\mathrm{H}^*(B_i)$ via $\pi _i^*$.
		We define a filtered graded $\Z$-algebra $\FH (B_\bullet )$ by
		\begin{itemize}
			\item For $i \geq n$, $\mathrm{F}_i\mathrm{H}^*( B_\bullet) := \mathrm{H}^*(B_n)$.
			\item For $0 \leq j \leq n-1$,
				$\mathrm{F}_j\mathrm{H}^*( B_\bullet ):=
				\pi _n^*\circ \pi _{n-1}^* \circ \dots \circ \pi _{j+1}^*(\mathrm{H}^*(B_j))$.
		\end{itemize}
		We call $\FH (B_\bullet )$ the \emph{filtered cohomology ring} of the Bott tower $B_\bullet$.

		The following is our main result:
		\begin{Theo}\label{theorem1.1}
			Isomorphism classes of Bott towers are distinguished by their filtered cohomology rings.
			Moreover, any filtered cohomology ring isomorphism between two Bott towers are induced
			by an isomorphism between the Bott towers.
		\end{Theo}
		Our study is motivated by the so-called \emph{cohomological rigidity problem} for toric manifolds.
		A \emph{toric manifold} is a smooth (complete) toric variety. Bott manifolds are examples of toric manifolds.
		The cohomological rigidity problem asks whether topological types or diffeomorphism types of toric manifolds
		are distinguished by their cohomology rings or not. This problem is open and we do not know any counterexamples.

		This paper is organized as follows.	In Section \ref{section2}, we recall some preliminary facts about Bott manifolds.
		In Section \ref{section3}, we see that
		rank $2$ decomposable vector bundles over Bott manifolds are distinguished by their total Chern classes.
		In Section \ref{section4}, we study $\CP ^1$-bundles over Bott manifolds by using the results
		obtained in Section \ref{section3}. In Section \ref{section5}, we give a proof of Theorem \ref{theorem1.1}.

		Throughout this paper, all cohomology groups are taken with $\Z$-coefficient.
		\bigskip \\
		{\bf Acknowledgement}.
		The author would like to thank professor Mikiya Masuda for stimulating discussion about
toric topology.
	\section{Preliminaries}\label{section2}
		In this section, we recall some preliminary facts (see \cite[section 2]{CM} and \cite[section 2]{CMS} for more details).
		For a complex vector bundle $\eta $ over a topological space, we denote its projectivization by $P(\eta )$.

		\begin{Lemm}[Lemma 2.1, \cite{CMS}]\label{lemma2.1}
			Let $B$ be a smooth manifold, $\xi $ be a complex line bundle over $B$ and $\eta $ be a complex vector bundle over $B$.
			Then, the projectivizations $P(\eta )$ and $P(\xi \otimes \eta )$ are isomorphic as bundles.
		\end{Lemm}

		Let $B_{i-1}$ be a Bott manifold and $\eta _i= \xi _i \oplus \xi _i'$ be the Whitney sum of complex line bundles
		$\xi _i$ and $\xi _i'$ over $B_{i-1}$. Since $\xi _i^* \otimes \eta _i = \uC \oplus (\xi _i^* \otimes \xi _i')$, the projectivization $B_i =P(\eta _i)$ is isomorphic to $P (\uC \oplus (\xi _i^*\otimes \xi _i'))$
		as fiber bundles by Lemma \ref{lemma2.1}, where $\uC $ denotes the trivial line bundle over $B_{i-1}$ and $\xi _i^*$ denotes the dual line bundle of $\xi _i$.
		Hence we can assume that $\eta _i$ is the Whitney sum of $\uC $ and a complex line bundle, denoted $\xi _i$ again, in the definition of Bott towers.

		Let $B_\bullet =(\{ B_i\} _{i=0}^n , \{\pi _i\} _{i=1}^n )$ be a Bott tower of height $n$.
		We suppose that each fibration $B_i \to B_{i-1}$ is the projectivization $P(\uC \oplus \xi _i)$.
		Let $\gamma _i$ be the tautological line bundle over $P(\uC \oplus \xi _i) =B_i$.
		By Borel-Hirzebruch formula, the cohomology ring $\mathrm{H}^*(B_i)$ viewed as an $\mathrm{H}^*(B_{i-1})$-algebra via $\pi _i^*$ is of the form
		\begin{equation*}\label{BorelHirzebruch}
			\mathrm{H}^*(B_i) \cong \mathrm{H}^*(B_{i-1})[X]/(X^2-c_1(\xi _i)X)
		\end{equation*}
		where $c_1(\xi _i)$ denotes the first Chern class of $\xi _i$
		and $X$ represents the first Chern class of the tautological line bundle $\gamma _i$.
		Using this formula inductively on $i$, we see that the cohomology ring $\mathrm{H} ^*(B_n)$ is generated by
		\begin{equation*}
			x_i := \pi _n^*\circ \dots \circ \pi _{i+1}^*(c_1(\gamma _i))
		\end{equation*}
		for $i = 1,\dots ,n$. More precisely, $\mathrm{H} ^*(B_n)$ is of the form
		\begin{equation*}
			\mathrm{H} ^*(B_n) = \Z [x_1,\dots ,x_n]/(x_i^2 - \alpha _ix_i ; i=1,\dots ,n)
		\end{equation*}
		as rings,
		where $\alpha _i := \pi _n^*\circ \dots \circ \pi _{i+1}^*(c_1(\xi _i))$ for all $i$.

		With this understood, we may regard $\mathrm{H} ^*(B_k)$ as a subring of $\mathrm{H}^*(B_n)$
		generated by $x_j$'s with $j =1,\dots ,k$. Namely, we have
		\begin{equation*}
			\mathrm{F}_k\mathrm{H}^*(B_\bullet ) = \Z [x_1,\dots ,x_k]/(x_j^2 - \alpha _jx_j ; j=1,\dots ,k).
		\end{equation*}

		A \emph{vanishing pair} is an ordered pair $(z,\overline{z})$ of elements in $\mathrm{H}^2(B_n)$ such that $z\overline{z} = 0$.
		We say that a vanishing pair is \emph{primitive} if both elements are primitive. The following lemma is very helpful for our purpose.
		\begin{Lemm}[Lemma 2.3, \cite{CM}]\label{pvp}
			A primitive vanishing pair is of the form
			\begin{equation*}
				(ax_j+u, \pm (a(x_j-\alpha _j)-u))
			\end{equation*}
			for some $j$, where $a$ is a non-zero integer, $u$ is a linear combination of $x_i$'s with $i < j$, and $u(u+a\alpha _j)=0$.
		\end{Lemm}

	\section{Rank $2$ decomposable vector bundles over Bott manifolds}\label{section3}
		For an element $\alpha \in \mathrm{H}^2(B_n)$, we denote by $\gamma ^\alpha $ the complex line bundle over a Bott manifold $B_n$
		whose first Chern class is $\alpha$.
		The purpose of this section is to prove the following.
		\begin{Theo}\label{theorem3.1}
			Rank $2$ decomposable vector bundles over a Bott manifold
			are distinguished by their total Chern classes.
		\end{Theo}
		We prepare two lemmas.
		\begin{Lemm}\label{lemma3.2}
				$\gamma ^{x_j}\oplus \gamma ^{-x_j+ \alpha _j} \cong \uC \oplus \gamma ^{\alpha _j}$.
			\end{Lemm}
			\begin{proof}
				Since $\gamma ^{x_j}$ is the pull-back bundle of the tautological line bundle over $P(\uC \oplus \xi _j)=B_j$,
				$\gamma ^{x_j}$ is a subbundle of $\uC \oplus \gamma ^{\alpha _j}$ where $\alpha _j = c_1(\xi _j)$. It follows that the first Chern class of the orthogonal complement of $\gamma ^{x_j}$ in $\uC \oplus \gamma ^{\alpha _j}$ (with an Hermitian metric) is
				$ -x_j + \alpha _j$. Since the orthogonal complement is a complex line bundle
				and complex line bundles are classified by their first Chern classes, it must be isomorphic to $\gamma ^{-x_j +\alpha _j}$, proving the lemma.
			\end{proof}
			\begin{Lemm}\label{lemma3.3}
				If $\gamma ^\alpha \oplus \gamma ^\beta \cong \uC \oplus \gamma ^{\alpha + \beta }$,
				then $\gamma ^{a\alpha } \oplus \gamma ^{b\beta } \cong \uC \oplus \gamma ^{a\alpha + b\beta }$ for any integers $a$ and $b$.
			\end{Lemm}
			\begin{proof}
				There is a nowhere zero cross section $f$ of $\gamma ^\alpha \oplus \gamma ^\beta$ since $\gamma ^\alpha \oplus \gamma ^\beta $ contains a trivial complex line bundle as a subbundle by assumption. Let $f_\alpha $ and $f_\beta$ be the projections of $f$ on $\gamma ^\alpha$ and $\gamma ^\beta $, which are cross sections of $\gamma ^\alpha$ and $\gamma ^\beta $ respectively. Their zero loci do not intersect. If $a$ (respectively, $b$) is positive, we define $f_{a\alpha } := f_\alpha ^a = f_\alpha \otimes \dots \otimes f_\alpha$ (respectively, $f_{b\beta } := f_\beta ^b$) which is a cross section of $\gamma ^{a\alpha }$ (respectively, $\gamma ^{b\beta }$).
				Note that there is a map $\varphi _{a\alpha } : \gamma ^{a\alpha } \to \gamma ^{-a\alpha }$
				(respectively, $\varphi _{b\beta} : \gamma ^{b\beta } \to \gamma ^{-b\beta}$) which is an isomorphism as
				real $2$-plane bundles but anti-$\mathbb{C}$-linear on fibers.
				If $a$ (respectively, $b$) is negative, we define a cross section $f_{a\alpha }$ of $\gamma ^{a\alpha }$
				(respectively, $f_{b\beta }$ of $\gamma ^{b\beta }$)
				to be the pull-back of $f_\alpha ^{-a}$ (respectively, $f_\beta ^{-b}$) by $\varphi _{a\alpha } : \gamma ^{a\alpha } \to \gamma ^{-a\alpha}$ (respectively, $\varphi _{b\beta} : \gamma ^{b\beta } \to \gamma ^{-b\beta}$). The pair $(f_{a\alpha }, f_{b\beta })$ determines a nowhere zero cross section of $\gamma ^{a\alpha }\oplus \gamma ^{b\beta}$.
				Thus $\gamma ^{a\alpha } \oplus \gamma ^{b\beta }$ contains a trivial complex line bundle as a subbundle.
				By computing the first Chern class of the orthogonal complement of the trivial subbundle in
				$\gamma ^{a\alpha } \oplus \gamma ^{b\beta }$, we get the lemma.
			\end{proof}
		\begin{Prop}\label{proposition3.4}
			Let $\xi $ and $\xi '$ be complex line bundles over $B_n$. Then, the second Chern class $c_2( \xi \oplus \xi ')$ is zero if and only if $\xi \oplus \xi'$ is isomorphic to $\uC \oplus (\xi \otimes \xi ')$.
		\end{Prop}
		\begin{proof}
			The \lq\lq{if}'' part is clear. So we show the \lq\lq{only if}'' part.

			We assume that $c_2(\xi \oplus \xi ' )= 0$. Then $c_1(\xi )c_1(\xi ') =0$.
			Note that if either $c_1(\xi )=0$ or $c_1(\xi ')=0$, then $\xi $ or $\xi '$ is trivial and Proposition \ref{proposition3.4} holds. Thus, by lemma \ref{lemma3.3} and the above note, it suffices to show that $\xi \oplus \xi ' \cong \C \oplus (\xi \otimes \xi ')$ only when the vanishing pair $(c_1(\xi _i),c_1(\xi _i'))$ is primitive.

			By Lemma \ref{pvp} and \ref{lemma3.3}, we may assume
			\begin{equation*}
				(c_1(\xi ),c_1(\xi '))=(ax_j+u, a(x_j-\alpha _j)-u)
			\end{equation*}
			for some $j$, where $a$ is a non-zero integer, $u$ is a linear combination of $x_i$'s with $i < j$, and $u(u+a\alpha _j) = 0$.
			Let $j$ be the minimal integer such that $c_1(\xi )$ and $c_1(\xi ')$ can be written as linear combinations of $x_k$'s with $k \leq j$.
			We will show that $\xi  \oplus \xi ' \cong \uC \oplus (\xi \otimes \xi ')$ by induction on $j$.
			If $j=1$, then $(c_1(\xi ),c_1(\xi ')) = \pm ( x_1, x_1)$ and we are done by Lemma \ref{lemma3.2} and \ref{lemma3.3}. Now we assume that $\gamma ^\alpha \oplus \gamma ^{\alpha '} \cong \uC \oplus (\gamma ^\alpha \otimes \gamma ^{\alpha '})$ for any vanishing pair $(\alpha , \alpha ')$ such that $\alpha$ and $\alpha '$ can be written as linear combinations of $x_k$'s with $k < j$.
			Then,
			\begin{equation*}
				\begin{split}
					\gamma ^{ax_j+u}\oplus \gamma ^{a(x_j-\alpha _j)-u} & \cong \gamma ^{ax_j} \otimes (\gamma ^u \oplus \gamma ^{-(u+a\alpha _j)})  \\
					& \cong \gamma ^{ax_j} \otimes (\uC \oplus \gamma ^{-a\alpha _j}) \text{ (by hypothesis of induction)} \\
					& \cong \gamma ^{ax_j} \oplus \gamma ^{a(x_j - \alpha _j)}  \\
					& \cong \uC \oplus \gamma ^{a(2x_j -\alpha _j)} \text{ (by Lemma \ref{lemma3.2} and \ref{lemma3.3})}.
				\end{split}
			\end{equation*}
			This completes the induction step, proving the proposition.
		\end{proof}
		\begin{proof}[Proof of Theorem \ref{theorem3.1}]
			Let $\alpha $, $\beta $, $\alpha '$ and $\beta '$ be elements in $\mathrm{H}^2(B_n)$ such that
			\begin{equation*}
				c(\gamma ^\alpha \oplus \gamma ^\beta ) = c(\gamma ^{\alpha '} \oplus \gamma ^{\beta '}),
			\end{equation*}
			that is, $(1+\alpha )(1+\beta )= (1+\alpha ')(1+\beta ')$. Then, we have
			\begin{equation*}
				\beta - \alpha = (\alpha ' -\alpha ) + (\beta ' - \alpha )
			\hspace*{5mm}\text{ and }\hspace*{5mm}
				(\alpha ' -\alpha )(\beta ' -\alpha )=0.
			\end{equation*}
			Therefore, it follows from Proposition \ref{proposition3.4} that
			\begin{equation*}
				\uC \oplus \gamma ^{\beta - \alpha } \cong \gamma ^{\alpha ' -\alpha } \oplus \gamma ^{\beta ' -\alpha }.
			\end{equation*}
			By tensoring the both sides above with $\gamma ^\alpha $,
			we obtain the theorem.
		\end{proof}
	\section{$\CP ^1$-bundles over Bott manifolds}\label{section4}
		In this section, we study $\CP ^1$-bundles over Bott manifolds.
		\begin{Theo}\label{theorem4.1}
			Let $B_n$ be a Bott manifold. Let $\gamma ^\alpha $ and $\gamma ^\beta $ be complex line bundles over $B_n$.
			If $\mathrm{H}^*(P(\uC \oplus \gamma ^\alpha ))$ and $\mathrm{H}^*(P(\uC \oplus \gamma ^\beta ))$ are isomorphic as $\mathrm{H}^*(B_n)$-algebras,
			then $P(\uC \oplus \gamma ^\alpha )$ and $P(\uC \oplus \gamma ^\beta )$ are isomorphic as bundles.
		\end{Theo}
		\begin{proof}
			By Borel-Hirzebruch formula, $\mathrm{H}^*(P(\uC \oplus \gamma ^\alpha ))$ and
			$\mathrm{H}^*(P(\uC \oplus \gamma ^\beta ))$ are of the forms
			\begin{equation*}
				\mathrm{H}^*(P(\uC \oplus \gamma ^\alpha )) = \mathrm{H}^*(B_k)[X]/(X^2-\alpha X)
			\end{equation*}
			and
			\begin{equation*}
				\mathrm{H}^*(P(\uC \oplus \gamma ^\beta )) = \mathrm{H}^*(B_k)[Y]/(Y^2-\beta Y).
			\end{equation*}
			Let $\Phi : \mathrm{H}^*(P(\uC \oplus \gamma ^\beta )) \to \mathrm{H}^*(P(\uC \oplus \gamma ^\alpha ))$ be an $\mathrm{H}^*(B_n)$-algebra isomorphism. We write $\Phi (Y) =: sX+ \alpha '$, $s=\pm 1$, $\alpha ' \in \mathrm{H}^2(B_k)$. Since $\Phi (Y^2 -\beta Y)=0$ and
			$X^2=\alpha X$, we have
			\begin{equation*}
				\begin{split}
					0 &= (sX+\alpha ')^2 - \beta (sX+\alpha ') \\
					&=(\alpha + 2s\alpha ' - s\beta )X + \alpha '(\alpha '-\beta ).
				\end{split}
			\end{equation*}
			Therefore $\alpha = s(-2\alpha '+ \beta )$ and $\alpha '(\alpha '-\beta ) =0$. Thus we obtain
			\begin{equation*}
				\begin{split}
					P(\uC \oplus \gamma ^\alpha ) &= P(\uC \oplus \gamma ^{s(-2\alpha '+\beta )})\\
					&\cong P(\uC \oplus \gamma ^{-2\alpha '+\beta } ) \text{ (by taking complex conjugation when $s=-1$)}\\
					&\cong P(\gamma ^{\alpha '} \oplus \gamma ^{-\alpha '+ \beta }) \text{ (by Lemma \ref{lemma2.1})}\\
					&\cong P(\uC \oplus \gamma ^\beta ) \text{ (by Theorem \ref{theorem3.1})}.
				\end{split}
			\end{equation*}
			This proves the theorem.
		\end{proof}
		\begin{Lemm}\label{lemma4.2}
			The group of $\mathrm{H}^*(B_n)$-algebra automorphisms of $\mathrm{H}^*(P(\uC \oplus \gamma ^\alpha ))$ is of order $2$ and
			generated by the automorphism induced by the bundle map $p : P(\uC \oplus \gamma ^\alpha ) \to P(\uC \oplus \gamma ^\alpha )$
			which assigns a line $\ell $ in $\uC \oplus \gamma ^\alpha$ (with an Hermitian metric) to its orthogonal complement $\ell ^\perp$.
		\end{Lemm}
		\begin{proof}
		Let $\Phi : \mathrm{H}^*(P(\uC \oplus \gamma ^\alpha )) \to \mathrm{H}^*(P(\uC \oplus \gamma ^\alpha ))$ be an $\mathrm{H}^*(B_n)$-algebra automorphism.
		The same argument as in the proof of Theorem \ref{theorem4.1} shows that $\Phi (X) = X$ or $-X +\alpha$.
		Thus the group of $\mathrm{H}^*(B_n)$-algebra automorphisms of $\mathrm{H}^*(P(\uC \oplus \gamma ^\alpha ))$ is of order $2$.
		We have to show that $p^*(X) = -X+\alpha $.
		Note that the total space of the tautological line bundle $\gamma ^X$ over $P(\uC \oplus \gamma ^\alpha )$ is the set
		\begin{equation*}
			\{ (\ell , v) \in P(\uC \oplus \gamma ^\alpha )\times E(\uC \oplus \gamma ^\alpha ); \ell \ni v\}
		\end{equation*}
		and the total space of the pull-back $p^*\gamma ^X$ of $\gamma ^X$ by $p$ is given by
		\begin{equation*}
			\{ (\ell , v) \in P(\uC \oplus \gamma ^\alpha )\times E(\uC \oplus \gamma ^\alpha ); \ell ^\perp \ni v\} ,
		\end{equation*}
		so $p^*\gamma ^X  \oplus \gamma ^X  \cong \pi ^*(\uC \oplus \gamma ^\alpha )$, where $\pi : P(\uC \oplus \gamma ^\alpha ) \to B_n$ is the projection.
		Therefore we obtain $p^*(X) = -X+\alpha $.
		\end{proof}
		The following corollary follows from Theorem \ref{theorem4.1} and Lemma \ref{lemma4.2}.
		\begin{Cor}\label{corollary4.3}
			Let $B_n$ be a Bott manifold. Let $\eta $ and $\eta '$ be
			rank $2$ decomposable vector bundles over $B_n$. Then, their projectivizations $P(\eta )$ and $P(\eta ')$ are isomorphic as bundles
			if and only if
			their cohomology rings are isomorphic as $\mathrm{H}^*(B_n)$-algebras. Moreover, any $\mathrm{H}^*(B_n)$-algebra isomorphism between $\mathrm{H}^*(P(\eta ))$ and $\mathrm{H}^*(P(\eta '))$ is induced by a bundle isomorphism.
		\end{Cor}
	\section{Proof of Theorem \ref{theorem1.1}}\label{section5}
	Let $B_\bullet = (\{ B_i\} _{i=0}^n , \{\pi _i\} _{i=1}^n )$ and $B'_\bullet =(\{ B'_i\} _{i=0}^n , \{\pi '_i\} _{i=1}^n )$
	be Bott towers of height $n$. Let $\Phi _\bullet : \FH (B_\bullet ')  \to \FH (B_\bullet )$ be an isomorphism.
	We will show Theorem 1.1 by induction on height. Assume Theorem \ref{theorem1.1} holds for Bott towers of height $k$.
	For $\Phi _k : \mathrm{F}_k\mathrm{H}^* (B_\bullet ')=\mathrm{H}^*(B_k') \to \mathrm{H}^*(B_k) = \mathrm{F}_k\mathrm{H}^*( B_\bullet )$,
	there is a diffeomorphism $\varphi _k : B_k \to B_k '$ such that $\varphi _k^* = \Phi _k$ by the hypothesis of induction. Then, the pull-back bundle $\varphi _k^*B_{k+1}' \to B_k$ is isomorphic to $B_{k+1} \to B_k$. In fact, there is a commutative diagram
	\begin{equation*}
		\xymatrix{
			\mathrm{H}^*(B_{k+1}) & & \mathrm{H}^*(B_{k+1}') \ar[ll]_{\Phi _{k+1}} \ar[ld]_{\widetilde{\varphi _k}^*} \\
			 & \mathrm{H}^*(\varphi _k^*B_{k+1}') & \\
			\mathrm{H}^*(B_k) \ar[uu]^{\pi _{k+1}^*} \ar[ru]^{(\varphi _k^*\pi _{k+1}')^*} & &
			\mathrm{H}^*(B_k') \ar[ll]^{\varphi _k^* = \Phi _k} \ar[uu]_{\pi _{k+1}'^*}
		}
	\end{equation*}
	where $\widetilde{\varphi _k}^*$  is the homomorphism induced from the bundle map
	$\widetilde{\varphi _k} : \varphi _k^*B_{k+1}' \to B_{k+1}'$ and $(\varphi _k^*\pi _{k+1}')^*$
	is the homomorphism induced from the projection
	$\varphi ^*\pi _{k+1}' : \varphi _k^*B_{k+1}' \to B_k$.
	Since $\widetilde{\varphi _k}$ is a diffeomorphism, the induced homomorphism $\widetilde{\varphi _k}^*$ is a ring isomorphism.
	The composition $\Phi _{k+1}\circ (\widetilde{\varphi _k}^*)^{-1} : \mathrm{H}^*(\varphi _k^*B_{k+1}') \to \mathrm{H}^*(B_{k+1})$
	is an $\mathrm{H}^*(B_k)$-algebra isomorphism. By Corollary \ref{corollary4.3}, there is a bundle isomorphism $b : B_{k+1} \to \varphi _k^*B_{k+1}'$ such that
	$b^* = \Phi _{k+1}\circ (\widetilde{\varphi _k}^*)^{-1}$.

	By the construction of $b$, the bundle map $\varphi _{k+1}:= \widetilde{\varphi _k} \circ b$ satisfies $\varphi _{k+1}^* = \Phi _{k+1}$.
	\qed

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