text:
\section{The discriminant and Lagrangian cobordisms} \cntrs
\label{s:disc-lcob}

This section provides the proofs of Theorem~\ref{t:cob} and a
generalization of Corollary~\ref{c:del_1=del_2}.

\pbred{In what follows Lagrangian cobordisms $V$ will be generally
  assumed to be connected. In contrast, their boundaries $\partial V$
  are allowed to have several connected components.}

We begin with:
\begin{proof}[Proof of Theorem~\ref{t:cob}]
   Before going into the details of the proof, here is the rationale
   behind it. To the Lagrangian cobordism $V$ we can associate a
   (relative) quantum homology $QH(V, \partial V)$ which has a quantum
   product. The quantum product on $QH(V, \partial V)$ is related to
   the quantum products for the ends of $V$ via a quantum connectant
   $\delta: QH(V, \partial V) \longrightarrow QH(\partial V) =
   \oplus_{i=1}^r QH(L_i)$. This makes it possible to find relations
   between the products on the quantum homologies $QH(L_i)$ of
   different ends of $V$ and the quantum product on $QH(V, \partial
   V)$. In particular this gives the desired relation between the
   discriminants of the different ends.

   We now turn to the details of the proof. We will use here several
   versions of the pearl complex and its homology (also called
   Lagrangian quantum homology) both for Lagrangian cobordisms as well
   as for their ends.  We refer the reader to~\cite{Bi-Co:Yasha-fest,
     Bi-Co:rigidity, Bi-Co:lagtop} for the foundations of the theory
   in the case of closed Lagrangians and to~\S5 of~\cite{Bi-Co:cob1}
   in the case of cobordisms.

   Throughout this proof we will work with $\mathbb{Q}$ as the base
   field and with $\Lambda = \mathbb{Q}[t^{-1}, t]$ or $\Lambda^{+} =
   \mathbb{Q}[t]$ as coefficient rings. We denote by $\mathcal{C}$ and
   $\mathcal{C}^+$ the pearl complexes with coefficients in $\Lambda$
   and $\Lambda^+$ respectively, and by $QH$ and $Q^+H$ their
   homologies. The latter is sometimes called the positive Lagrangian
   quantum homology.

   Before we go on, a small remark regarding the coefficients is in
   order. Throughout this proof we grade the variable $t \in \Lambda$
   as $|t|=-N_V$. This is the standard grading for $QH(V)$ and $QH(V,
   \partial V)$ and their positive versions. We use the same
   coefficient rings (and grading) also for $QH(L_i)$ and its positive
   version. This is possible since $N_V | N_{L_i}$, hence our ring
   $\Lambda^+$ is an extension of the corresponding ring in which the
   degree of $t$ is $-N_{L_i}$.

   Recall that (for any Lagrangian submanifold) the positive quantum
   homology $Q^+H$ admits a natural map $Q^+H \longrightarrow QH$
   induced by the inclusion $\mathcal{C}^+ \longrightarrow
   \mathcal{C}$. Again, for degree reasons the induced map in homology
   is an isomorphism in degree $0$ and surjective in degree $1$:
   \begin{equation} \label{eq:Q^+H-QH} Q^+H_0 \xrightarrow{\; \; \cong
        \; \;} QH_0, \quad Q^+H_1 \rrightarrow QH_1.
   \end{equation}
   In fact, the last map is an isomorphism whenever the minimal Maslov
   number is $> 2$. We also have $Q^+H_n(K) \cong H_n(K)$ for every
   $n$-dimensional Lagrangian submanifold $K$.

   Coming back to the proof of the theorem, we first claim there is a
   commutative diagram
   \begin{equation} \label{eq:diag-QH-H}
      \begin{CD}
         Q^+H_1(V) @>j_Q>>Q^+H_1(V, \partial V) @ > \delta >>
         Q^+H_0(\partial V) @>i_Q>> Q^+H_0(V) \\
         @V s VV @V s  VV @V s VV @V s VV \\
         H_1(V) @>j>> H_1(V, \partial V) @ > \partial >> H_0(\partial
         V) @>i>> H_0(V)\\
         @. @VVV @VVV @. \\
         @. 0 @. 0 @.
      \end{CD}
   \end{equation}
   with exact rows and columns. The second row of the diagram is the
   classical homology sequence for the pair $(V, \partial V)$ with
   $\partial$ being the connecting homomorphism (we use $\mathbb{Q}$
   coefficients here). The first row is its quantum homology analogue,
   and we remark that the quantum connectant $\delta$ is
   multiplicative with respect to the quantum product (see~\S5
   of~\cite{Bi-Co:cob1} and~\cite{BSing:msc}). The vertical maps $s$
   come from the following general exact sequence of chain complexes:
   \begin{equation} \label{eq:ses-s} 0 \longrightarrow t \mathcal{C}^+
      \xrightarrow{\;\, \iota \;\,} \mathcal{C}^+ \xrightarrow{\;\, s
          \; \, } CM \longrightarrow 0,
   \end{equation}
   where $CM$ stand for the Morse complex (defined using the same
   Morse function and metric as used for the pearl complex, but with
   coefficient in $\mathbb{Q}$ rather than $\Lambda^+$). The second
   map in this exact sequence, $s: \mathcal{C}^+ \longrightarrow CM$,
   is induced by $t \mapsto 0$ (i.e. it sends a pearly chain to its
   classical part, omitting the $t$'s), and $\iota$ stand for the
   inclusion. We now explain why the two middle $s$ maps
   in~\eqref{eq:diag-QH-H} are surjective. We start with the third $s$
   map (i.e. the one before the rightmost $s$). We have:
   \begin{equation} \label{eq:H_0(del-V)}
      H_0(\partial V) = \bigoplus_{i=1}^r H_0(L_i), \quad
      Q^+H_0(\partial V) = \bigoplus_{i=1}^r Q^+H_0(L_i).
   \end{equation}
   Next, note that the composition of $s: Q^+H_0(L_i) \longrightarrow
   H_0(L_i)$ with the inclusion $H_0(L_i) \subset H_0(L_i; \Lambda^+)$
   coincides with the augmentation $\widetilde{\epsilon}_{L_i} :
   Q^+H_0(L_i) \longrightarrow H_0(L; \Lambda^+)$. The fact that $s$
   is surjective now follows easily from~\S\ref{sbsb:Q=QH_n}
   and~\eqref{eq:Q^+H-QH}.

   The surjectivity of the second to the left $s$ map requires a
   different argument.  Consider the chain complex $\mathcal{D}_* = (t
   \mathcal{C}^+)_*$, viewed as a subcomplex of $\mathcal{C}^+$. In
   view of the exact sequence~\eqref{eq:ses-s} the surjectivity of the
   second to the left $s$ map in~\eqref{eq:diag-QH-H} would follow if
   we show that $H_0(\mathcal{D})=0$. To this end consider the
   following filtration $\mathcal{F}_{\bullet}\mathcal{D}$ of
   $\mathcal{D}$ by subcomplexes, defined by:
   \begin{align*}
      & \mathcal{F}_m \mathcal{D} := t^{-m} \mathcal{D} = t^{-m+1}
      \mathcal{C}^+ \quad \forall \, m \leq 0, \\
      & \mathcal{F}_k \mathcal{D} := \mathcal{D} \quad \forall \, k
      \geq 0.
   \end{align*}
   Note that this filtration is very similar to the one described
   in~\S\ref{sb:spec-seq} only that here it is applied to the complex
   $\mathcal{D}$ rather than to $\mathcal{C}$.

   A simple calculation (similar to the one in~\S\ref{sb:spec-seq})
   shows that the first page of the spectral sequence associated to
   this filtration satisfies:
   \begin{align*}
      & E^1_{p,q} \cong t^{-p+1} H_{{\scriptscriptstyle
          p+q+N_V-pN_V}}(V, \partial V) \quad
      \forall \, p \leq 0, \\
      & E^1_{p,q} = 0 \quad \forall \, p \geq 1.
   \end{align*}
   It follows from the assumption of the theorem that for all $p,q$
   with $p+q=0$ we have $E^1_{p,q}=0$, hence also
   $E^{\infty}_{p,q}=0$. Since this spectral sequence converges to
   $H_*(\mathcal{D})$ this implies that $H_0(\mathcal{D})=0$. This
   completes the proof of the surjectivity of the second to the left
   $s$ map in~\eqref{eq:diag-QH-H}.

   We proceed now with the proof of the theorem, based on the
   diagram~\eqref{eq:diag-QH-H} and its properties. We first remark
   that due to the assumptions of the theorem the number of ends of
   $V$ must be $r \geq 2$. Indeed, by the results of~\cite{Bi-Co:cob1}
   if a Lagrangian submanifold $L_1$ is Lagrangian null-cobordant
   (i.e. there exists a monotone Lagrangian cobordism $V$ with only
   one end being $L_1$) then $HF(L_1,L_1)=0$, in contrast with the
   assumption that $L_1$ satisfies condition $(3)$ of
   Assumption~$\mathscr{L}$. We therefore assume from now on that $r
   \geq 2$.

   Denote by $p_i \in H_0(L_i) \subset H_0(\partial V)$ the class
   corresponding to a point in $L_i$. Let $\alpha_2, \ldots, \alpha_r
   \in H_1(V, \partial V)$ be classes with $\partial \alpha_i = p_1 -
   p_i$. Choose lifts $\overline{p}_i \in Q^+H_0(\partial V)$ of the
   $p_i$'s under the map $s$ as well as lifts $\overline{\alpha}_2,
   \ldots, \overline{\alpha}_r \in Q^+H_1(V,\partial V)$ of $\alpha_2,
   \ldots, \alpha_r$. Denote by $e_V \in Q^+H_{n+1}(V, \partial V)$
   the unity and by $e_{L_i} \in Q^+H_n(L_i)$ the unities
   corresponding to the $L_i$'s. Note that $\delta(e_V) = e_{L_1} +
   \cdots + e_{L_r}$. Finally, put $\nu = n / N_V$. (Recall that
   $N_{L_i} | n$ by assumption, and since $N_V | N_{L_i}$ we have $N_V
   | n$.) Since the Lagrangians $L_i$ satisfy conditions~(1)~--~(3) of
   Assumption~$\mathscr{L}$ and in view of~\S\ref{sbsb:Q=QH_n}, we
   have:
   $$Q^+H_0(\partial V)
   \cong QH_0(\partial V) = \mathbb{Q} \overline{p}_1 \oplus \cdots
   \oplus \mathbb{Q} \overline{p}_r \oplus \mathbb{Q} e_{L_1} t^{\nu}
   \oplus \cdots \oplus \mathbb{Q} e_{L_r} t^{\nu}.$$

   \begin{prop} \label{p:basis-delta} $\dim_{\mathbb{Q}}
      (\textnormal{image\,} \delta) = r$. Moreover, for every choice
      of $\alpha_i$'s and $\overline{\alpha}_i$'s the elements
      $$\delta(\overline{\alpha}_2), \ldots, \delta(\overline{\alpha}_r),
      (e_{L_1} + \cdots + e_{L_r}) t^{\nu}$$ form a basis (over
      $\mathbb{Q}$) of the vector space $\textnormal{image\,}{\delta}
      \subset QH_0(\partial V)$.
   \end{prop}
   We defer the proof of the lemma and continue with the proof of our
   theorem.

   Denote by $\mathcal{B} \subset Q^+H_1(V, \partial V)$ the kernel of
   $\delta : Q^+H_1(V, \partial V) \longrightarrow Q^+H_0(\partial
   V)$. By Proposition~\ref{p:basis-delta} the elements
   $$\overline{\alpha}_2, \ldots, \overline{\alpha}_r, e_V t^{\nu}$$
   induce a basis for the vector space $Q^+H_1(V, \partial V) /
   \mathcal{B}$.

   We now continue by proving that $\Delta_{L_1} = \Delta_{L_2}$. The
   other equalities follow by the same recipe. Using the preceding
   basis we can write:
   \begin{equation} \label{eq:a2*a2-delta-1}
      \begin{aligned} & \overline{\alpha}_2 *
         \overline{\alpha}_2 = \sum_{j=2}^r \xi_j \overline{\alpha}_j
         t^{\nu} + Bt^{\nu} + \rho e_V t^{2 \nu}, \\
         & \delta(\overline{\alpha}_2) = \overline{p}_1 - \overline{p}_2 +
         \sum_{k=1}^r a_k e_{L_k} t^{\nu},
      \end{aligned}
   \end{equation}
   for some $\xi_j, a_k, \rho \in \mathbb{Q}$ and $B \in \mathcal{B}$.
   For the first equality we have used the fact that
   $\overline{\alpha}_2 * \overline{\alpha}_2 \in Q^+H_{1-n}(V,
   \partial V) \cong t^{\nu} Q^+H_1(V, \partial V)$.

   We will also need a similar equality to the second one
   in~\eqref{eq:a2*a2-delta-1}, but for $\delta(\overline{\alpha}_i)$:
   \begin{equation} \label{eq:delta-alpha-i}
      \delta(\overline{\alpha}_i) = \overline{p}_1 - \overline{p}_i +
      \sum_{k=1}^r a_k^{(i)} e_{L_k} t^{\nu}, \quad \forall \, 2 \leq
      i \leq r,
   \end{equation}
   where $a_k^{(i)} \in \mathbb{Q}$. (Note that according to our
   notation $a_k = a_k^{(2)}$.)

   At this point we need to separate the arguments to the cases $r
   \geq 3$ and $r=2$. (As we have already remarked, $r=1$ is
   impossible under the assumptions of the theorem.) We assume first
   that $r \geq 3$. The case $r=2$ will be treated after that.

   We now perform a little change in the basis and the choice of the
   lift $\overline{p}_i$ as follows:
   \begin{align*}
      & \overline{\alpha}_2 \longrightarrow \overline{\alpha}_2 - a_3
      e_V t^{\nu}, \quad \overline{\alpha}_i \longrightarrow
      \overline{\alpha}_i \quad
      \forall i \geq 3, \\
      & \overline{p}_1 \longrightarrow \overline{p}_1 +
      (a_1-a_3)e_{L_1} t^{\nu}, \quad \overline{p}_2 \longrightarrow
      \overline{p}_2 - (a_2-a_3) e_{L_2} t^{\nu}, \quad \overline{p}_i
      \longrightarrow \overline{p}_i \quad \forall i \geq 3.
   \end{align*}
   To simplify notation we continue to denote the new basis elements
   by $\overline{\alpha}_i$ and similarly for the $\overline{p}_i$'s.
   By abuse of notation we also continue to denote the new
   coefficients $a_k$, $a_k^{(i)}$, $\xi_j$ and $\rho$ resulting from
   the basis change by the same symbols, and similarly for the term $B
   \in \mathcal{B}$. The outcome of the basis change is that now the
   second equality in~\eqref{eq:a2*a2-delta-1} becomes:
   \begin{equation} \label{eq:delta-alpha-2-new}
      \delta(\overline{\alpha}_2) = \overline{p}_1 - \overline{p}_2 +
      \sum_{k=4}^r a_k e_{L_k} t^{\nu}.
   \end{equation}
   (Of course, if $r=3$ then the third term in the last equation is
   void.)  We now use the fact that $\delta$ is multiplicative
   (see~\cite{Bi-Co:cob1}):
   \begin{equation} \label{eq:delta-alpha-2*alpha-2}
      \delta(\overline{\alpha}_2 * \overline{\alpha}_2) =
      \delta(\overline{\alpha}_2) * \delta(\overline{\alpha}_2) =
      \overline{p}_1^{*2} + \overline{p}_2^{*2} + \sum_{k=4}^r a_k^2
      e_{L_k} t^{2 \nu}.
   \end{equation}
   We now express $\overline{p}_1^{*2} \in Q^+H_{-n}(L_1) \cong
   t^{\nu}Q^+H_{0}(L_1)$ in terms of the basis $\{ \overline{p}_1
   t^\nu, e_{L_1}t^{2\nu}\}$ and similarly for $\overline{p}_2^{*2}$:
   $$\overline{p}_1^{*2} = \sigma_1 \overline{p}_1 t^{\nu} +
   \tau_1 e_{L_1}t^{2\nu}, \qquad \overline{p}_2^{*2} = \sigma_2
   \overline{p}_2 t^{\nu} + \tau_2 e_{L_2}t^{2\nu},$$ where $\sigma_1,
   \sigma_2 \in \mathbb{Q}$ and $\tau_1, \tau_2 \in \mathbb{Q}$. (In
   fact, by choosing the $\alpha_i$'s, $\overline{\alpha}_i$'s and
   $\overline{p}_i$'s carefully, over $\mathbb{Z}$, the coefficients
   $\sigma_1, \sigma_2, \tau_1, \tau_2$ will in fact be in
   $\mathbb{Z}$, but we will not need that.)  Substituting this
   into~\eqref{eq:delta-alpha-2*alpha-2} we obtain:
   \begin{equation} \label{eq:delta-alpha-2*alpha-2-II}
      \delta(\overline{\alpha}_2 * \overline{\alpha}_2) = \sigma_1
      \overline{p}_1 t^{\nu} + \sigma_2 \overline{p}_2 t^{\nu} +
      \tau_1 e_{L_1}t^{2\nu} + \tau_2 e_{L_2}t^{2\nu} + \sum_{k=4}^r
      a_k^2 e_{L_k} t^{2 \nu}.
   \end{equation}
   Applying $\delta$ to the first equality in~\eqref{eq:a2*a2-delta-1}
   and using~\eqref{eq:delta-alpha-2-new}
   and~\eqref{eq:delta-alpha-2*alpha-2-II} we obtain:
   \begin{align*}
      & \xi_2\Bigl(\overline{p}_1 - \overline{p}_2 + \sum_{k=4}^r a_k
      e_{L_k} t^{\nu}\Bigr) t^{\nu} + \sum_{i=3}^r \xi_i
      \Bigl(\overline{p}_1 - \overline{p}_i + \sum_{q=1}^r a_q^{(i)}
      e_{L_q} t^{\nu}\Bigr) t^{\nu} + \rho (e_{L_1} + \cdots +
      e_{L_r})t^{2 \nu} \\
      & = \sigma_1 \overline{p}_1 t^{\nu} + \sigma_2 \overline{p}_2
      t^{\nu} + \tau_1 e_{L_1}t^{2\nu} + \tau_2 e_{L_2}t^{2\nu} +
      \sum_{k=4}^r a_k^2 e_{L_k} t^{2 \nu}.
   \end{align*}
   Comparing the coefficients of $\overline{p}_3, \ldots,
   \overline{p}_r$ we deduce that $\xi_3 = \cdots = \xi_r = 0$. The
   last equation thus becomes:
   \begin{equation} \label{eq:coeffs-xi=0}
      \begin{aligned}
         & \xi_2\Bigl(\overline{p}_1 - \overline{p}_2 + \sum_{k=4}^r
         a_k e_{L_k} t^{\nu}\Bigr) t^{\nu} + \rho (e_{L_1} + \cdots +
         e_{L_r})t^{2 \nu} \\
         & = \sigma_1 \overline{p}_1 t^{\nu} + \sigma_2 \overline{p}_2
         t^{\nu} + \tau_1 e_{L_1}t^{2\nu} + \tau_2 e_{L_2}t^{2\nu} +
         \sum_{k=4}^r a_k^2 e_{L_k} t^{2 \nu}.
      \end{aligned}
   \end{equation}
   Comparing the coefficients of $\overline{e}_3$ on both sides
   of~\eqref{eq:coeffs-xi=0} (recall that $r \geq 3$) we deduce that
   $\rho=0$. It easily follows now that $\tau_1 = \tau_2 = 0$ and that
   $\sigma_1 = \xi_2 = -\sigma_2$. By the definition of the
   discriminant it follows that
   $$\Delta_{L_1} = \sigma_1^2 = \sigma_2^2 = \Delta_{L_2}.$$
   Note that the relation between our $\sigma_i$'s and $\tau_i$'s and
   the notation used in~\S\ref{sb:discr-intro} and
   in~\S\ref{sbsb:Q=QH_n} is $\sigma_1 = \sigma_1(p_1,
   \overline{p}_1)$, $\sigma_2 = \sigma_2(p_2, \overline{p}_2)$ and
   similarly for $\tau_1, \tau_2$.  Finally we remark that since
   $\Delta_{L_1} = \sigma_1^2 \in \mathbb{Z}$ we must have $\sigma_1
   \in \mathbb{Z}$, hence $\Delta_{L_1}$ is a perfect square.

   We now turn to the case $r=2$. In that case we can
   write~\eqref{eq:a2*a2-delta-1} as
   \begin{equation} \label{eq:a2*a2-delta-2}
      \begin{aligned} & \overline{\alpha}_2 *
         \overline{\alpha}_2 = \xi \overline{\alpha}_2
         t^{\nu} + Bt^{\nu} + \rho e_V t^{2 \nu}, \\
         & \delta(\overline{\alpha}_2) = \overline{p}_1 - \overline{p}_2 +
         a_1 e_{L_1} t^{\nu} + a_2 e_{L_2} t^{\nu},
      \end{aligned}
   \end{equation}
   By an obvious basis change (among $\overline{p}_1$,
   $\overline{p}_2$) we may assume that $a_1 = a_2 = 0$. Then the
   identity $\delta(\overline{\alpha}_2 * \overline{\alpha}_2) =
   \delta(\overline{\alpha}_2) * \delta(\overline{\alpha}_2)$ becomes:
   $$\xi(\overline{p}_1 - \overline{p}_2)t^{\nu} +
   \rho(e_{L_1} + e_{L_2})t^{2 \nu} = \sigma_1 \overline{p}_1 t^{\nu}
   + \sigma_2 \overline{p}_2 t^{\nu} + \tau_1 e_{L_1} t^{2\nu} +
   \tau_2 e_{L_2} t^{2\nu}.$$ It follows immediately that
   $\sigma_1=-\sigma_2$ and $\tau_1 = \tau_2$. Consequently
   $\Delta_{L_1} = \Delta_{L_2}$.

   To complete the proof of the theorem it remains to prove
   Proposition~\ref{p:basis-delta}. For this purpose we will need the
   following Lemma.
   \begin{lem} \label{l:div-t} Let $j \geq 0$ and consider the
      connecting homomorphism $$\delta: Q^+H_{1+jN_V}(V, \partial V)
      \longrightarrow Q^+H_{jN_V}(\partial V).$$ Let $\eta \in
      Q^+H_{1+jN_V}(V,
      \partial V)$ and assume that $\delta(\eta)$ is divisible by $t$.
      Then $\eta$ is also divisible by $t$.
   \end{lem}
   \begin{proof}[Proof of the lemma]
      The connecting homomorphism $\delta$ is part of the following diagram:
      \begin{equation} \label{eq:diag-QH-H-2}
         \begin{CD}
            @. Q^+H_{1+jN_V}(V, \partial V) @ > \delta >>
            Q^+H_{jN_V}(\partial V) \\
            @. @V s  VV @V s VV \\
            H_{1+jN_V}(V) @>j>> H_{1+jN_V}(V, \partial V) @ > \partial >>
            H_{jN_V}(\partial
            V) \\
         \end{CD}
      \end{equation}
      where the vertical $s$-maps are induced by~\eqref{eq:ses-s}.
      Since $\delta(\eta)$ is divisible by $t$ we have $s
      (\delta(\eta))=0$ hence $\partial (s(\eta))=0$. By assumption
      $H_{1+jN_V}(V)=0$ hence the bottom map $\partial$ is injective,
      and therefore we have $s(\eta)=0$. Looking again
      at~\eqref{eq:ses-s} it follows that $$\eta \in
      \textnormal{image\,} \Bigl( H_{1+jN_V}(t\mathcal{C}^+)
      \xrightarrow{\;\; \iota_* \;\;} Q^+H_{1+jN_V}(V,
      \partial V) \Bigr),$$ where $\mathcal{C}^+$ stands for the
      positive pearl complex of $(V,
      \partial V)$. But $$H_{1+jN_V}(t \mathcal{C}^+) \cong
      tQ^+H_{1+(j+1)N_V}(V,\partial V)$$ via an isomorphism for which
      $\iota_*$ becomes the inclusion $$tQ^+H_{1+(j+1)N_V}(V,\partial
      V) \subset Q^+H_{1+jN_V}(V, \partial V).$$ This proves that
      $\eta$ is divisible by $t$.
   \end{proof}

   We are finally in position to prove the preceding proposition.
   \begin{proof}[Proof of Proposition~\ref{p:basis-delta}]
      Note that $$\{\overline{p}_1, \delta(\overline{\alpha}_2),
      \ldots, \delta(\overline{\alpha}_r), \delta(e_V)t^{\nu}, e_{L_2}
      t^{\nu}, \ldots, e_{L_r} t^{\nu}\}$$ is a basis for
      $Q^+H_0(\partial V)$ (recall that $\delta(e_V) = e_{L_1} +
      \cdots + e_{L_r}$).  Therefore it is enough to show that the
      subspace of $Q^+H_0(\partial V)$ generated by $\overline{p}_1,
      e_{L_2} t^{\nu}, \ldots, e_{L_r} t^{\nu}$ has trivial
      intersection with $\textnormal{image\,}(\delta)$.

      Let $\gamma = c \overline{p}_1 + \sum_{j=2}^r b_j e_{L_j}
      t^{\nu}$, where $c, b_j \in \mathbb{Q}$ and assume that $\gamma
      = \delta(\beta)$ for some $\beta \in Q^+H_1(V, \partial V)$.  We
      have $s(\gamma) = c p_1$, where the map $s$ is the third
      vertical map from diagram~\eqref{eq:diag-QH-H}. It follows from
      that diagram that $\partial (s(\beta))= c p_1$. But this is
      possible only if $c=0$ since $p_1 \not\in
      \textnormal{image\,}(\partial)$.

      Thus $\gamma = \sum_{j=2}^r b_j e_{L_j} t^{\nu}$ and we have to
      show that $\gamma=0$.  Recall that $\gamma = \delta(\beta)$. We
      claim that $\beta$ is divisible by $t^{\nu}$, i.e. there exists
      $\beta' \in Q^+H_{n+1}(V,\partial V)$ such that $\beta = t^{\nu}
      \beta'$. To prove this we first note that $\gamma$ is divisible
      by $t$. By Lemma~\ref{l:div-t}, $\beta$ is also divisible by
      $t$. Thus there exists $\beta_1 \in Q^+H_{1+N_V}(V, \partial V)$
      with $\beta = t \beta_1$. In particular $\delta(\beta_1) =
      \sum_{j=2}^r b_j e_{L_j} t^{\nu-1}$.  Continuing by induction, using
      Lemma~\ref{l:div-t} repeatedly, we obtain elements $\beta_j \in
      Q^+H_{1+jN_V}(V, \partial V)$ with $t \beta_{j+1} = \beta_j$ for
      every $1 \leq j \leq \nu-1$. Take $\beta' = \beta_{\nu}$.

      It follows that $t^{\nu} \delta(\beta') = \sum_{j=2}^r b_j
      e_{L_j} t^{\nu}$ for some $\beta' \in Q^+H_{n+1}(V, \partial
      V)$. As $Q^+H_{n+1}(V, \partial V) = \mathbb{Q} e_V$ we have
      $\beta' = a e_V$ for some $a \in \mathbb{Q}$. But $\delta(e_V) =
      e_{L_1} + \cdots +e_{L_r}$ hence $a(e_{L_1} + \cdots +
      e_{L_r})t^{\nu} = (\sum_{j=2}^r b_j e_{L_j}) t^{\nu}$. Since by
      condition~(3) of Assumption~$\mathscr{L}$ the element $e_{L_1}
      \in Q^+H_n(\partial V)$ is not torsion (over $\Lambda^+$), it
      follows that $a=0$.  Consequently $b_2 = \cdots = b_r =0$ and so
      $\gamma = 0$. This concludes the proof of
      Proposition~\ref{p:basis-delta}.
   \end{proof}

   Having proved Proposition~\ref{p:basis-delta}, the proof of
   Theorem~\ref{t:cob} is now complete.
\end{proof}

\subsection{Lagrangians intersecting at one point} \label{sb:lag-1-pt}
\cntrsb

We start with a stronger version of Corollary~\ref{c:del_1=del_2}
from~\S\ref{sb:discr-intro}.

\begin{cor} \label{c:L1-L2-1pt} Let $(M, \omega)$ be a monotone
   symplectic manifold. Let $L_1, L_2 \subset M$ be two Lagrangian
   submanifolds that satisfy conditions $(1)$ -- $(3)$ of Assumption
   $\mathscr{L}$ and such that $N_{L_1} = N_{L_2}$. Denote by $N =
   N_{L_i}$ their mutual minimal Maslov number and assume further
   that:
   \begin{enumerate}
     \item $H_{1+jN}(L_1)=H_{1+jN}(L_2)=0$ for every $j$;
     \item $H_{jN-1}(L_1) = H_{jN-1}(L_2)=0$ for every $j$;
     \item either $\pi_1(L_1 \cup L_2) \to \pi_1(M)$ is injective, or
      $\pi_1(L_i) \to \pi_1(M)$ is trivial for $i=1,2$.
   \end{enumerate}
   Finally, suppose that $L_1$ and $L_2$ intersect transversely at
   exactly one point. Then $$\Delta_{L_1} = \Delta_{L_2}$$ and
   moreover this number is a perfect square.
\end{cor}
Note that if $L_1$, $L_2$ are even dimensional Lagrangian spheres then
conditions~(1)~--~(3) of Corollary~\ref{c:L1-L2-1pt} are obviously
satisfied, hence Corollary~\ref{c:del_1=del_2} follows from
Corollary~\ref{c:L1-L2-1pt}.

We now turn to the proof of Corollary~\ref{c:L1-L2-1pt}. We will need
the following Proposition.

\begin{prop} \label{p:surgery} Let $L_1, L_2 \subset (M, \omega)$ be
   two Lagrangian submanifolds intersecting transversely at one point.
   Then there exists a Lagrangian cobordism $V \subset \mathbb{R}^2
   \times M$ with three ends, corresponding to $L_1$, $L_2$ and $L_1
   \# L_2$ and such that $V$ has the homotopy type of $L_1 \vee L_2$.
   If $L_1$ and $L_2$ are monotone with the same minimal Maslov number
   $N$ and they satisfy assumption (3) from
   Corollary~\ref{c:L1-L2-1pt} then $V$ is also monotone with minimal
   Maslov number $N_V = N$.  Moreover, if $L_1$ and $L_2$ are spin
   then $V$ admits a spin structure that extends those of $L_1$ and
   $L_2$.
\end{prop}

Before proving this proposition we show how to deduce
Corollary~\ref{c:L1-L2-1pt} from it.

\begin{proof}[Proof of Corollary~\ref{c:L1-L2-1pt}]
   Consider the Lagrangian cobordism provided by
   Proposition~\ref{p:surgery}. Since $V$ is homotopy equivalent to
   $L_1 \vee L_2$ and $L_i$ satisfy assumptions (1) and~(2) of
   Corollary~\ref{c:L1-L2-1pt} a simple calculation shows that
   $$H_{jN}(V, \partial V) = 0, \quad H_{1+jN}(V) = 0, \quad \forall j.$$
   The result now follows immediately from Theorem~\ref{t:cob}.
\end{proof}

We now turn to the proof of the Proposition.

\begin{proof}[Proof of Proposition~\ref{p:surgery}]
   The proof is based on a version of the Polterovich Lagrangian
   surgery~\cite{Po:surgery} adapted to the case of
   cobordisms~\cite{Bi-Co:cob1}.  We briefly outline those parts of
   the construction that are relevant here. More details can be found
   in~\cite{Bi-Co:cob1}.

   Consider two plane curves $\gamma_1, \gamma_2$ as in
   Figure~\ref{f:gamma-12}.
   \begin{figure}[htbp]
      \begin{center}
         \epsfig{file=pic-1.eps, width=0.5\linewidth}
      \end{center}
      \caption{\label{f:gamma-12}}
   \end{figure}
   Consider the Lagrangian submanifolds $\gamma_1 \times L_1, \gamma_2
   \times L_2 \subset \mathbb{R}^2 \times M$. The surgery construction
   from~\cite{Bi-Co:cob1} produces a Lagrangian cobordism $V \subset
   \mathbb{R} \times M$ with two negative ends which coincide with
   negative ends of $\gamma_i \times L_i$ and with whose positive end
   looks like the positive end of $\gamma_3 \times (L_1 \# L_2)$,
   where the curve $\gamma_3$ is depicted in Figure~\ref{f:pi-V} and
   $L_1 \# L_2$ stands for the Polterovich surgery (in $M$) of $L_1$
   and $L_2$ (which coincides with the connected sum of the $L_i$'s
   because they intersect transversely at exactly one point). The
   projection of $V$ to $\mathbb{R}^2$ is depicted in
   Figure~\ref{f:pi-V}.
   \begin{figure}[htbp]
      \begin{center}
         \epsfig{file=pic-2.eps, width=0.85\linewidth}
      \end{center}
      \caption{\label{f:pi-V}}
   \end{figure}

   Next we determine the topology of $V$.  Consider the curves
   $\widetilde{\gamma}_1, \widetilde{\gamma}_2$ (which are extensions
   of the $\gamma_i$'s to curves with positive ends as in
   Figure~\ref{f:gamma-tilde-12}.)
   \begin{figure}[htbp]
      \begin{center}
         \epsfig{file=pic-3.eps, width=0.7\linewidth}
      \end{center}
      \caption{\label{f:gamma-tilde-12}}
   \end{figure}
   Consider the Polterovich surgery $W = (\widetilde{\gamma}_1 \times
   L_1) \# (\widetilde{\gamma}_2 \times L_2) \subset \mathbb{R}^2
   \times M$ (note that the latter two Lagrangians also intersect
   transversely at a single point). See Figure~\ref{f:cob-W}.
   \begin{figure}[htbp]
      \begin{center}
         \epsfig{file=pic-4.eps, width=0.85\linewidth}
      \end{center}
      \caption{\label{f:cob-W}}
   \end{figure}
   \begin{figure}[htbp]
      \begin{center}
         \epsfig{file=pic-5.eps, width=0.85\linewidth}
      \end{center}
      \caption{\label{f:V_0-strip-S}}
   \end{figure}

   Denote by $\pi: \mathbb{R}^2 \times M \longrightarrow \mathbb{R}^2$
   the projection, and by $S \subset \mathbb{R}^2$ the strip depicted
   in Figure~\ref{f:V_0-strip-S}. Put $V_0 = W \cap \pi^{-1}(S)$.
   According to~\cite{Bi-Co:cob1}, $V_0$ is a manifold with boundary,
   with two obvious boundary components corresponding to the $L_i$'s
   and a third boundary component which is $W \cap \pi^{-1}(0)$. The
   latter is exactly the Polterovich surgery $L_1 \# L_2$.  Moreover
   $V_0$ is homotopy equivalent to $V$ (in fact $V_0 \subset V$ and is
   a deformation retract of $V$). A straightforward calculation shows
   that there is an embedding $L_1 \vee L_2 \subset V_0$ and moreover
   that $L_1 \vee L_2$ is a deformation retract of $V_0$. (In fact,
   one can show that $V_0$ is diffeomorphic to the boundary connected
   sum of $[0,1] \times L_1$ and $[0,1] \times L_2$, where the
   connected sum occurs among the boundary components $\{1\} \times
   L_i$, $i=1,2$.)

   The statement on monotonicity follows from the Seifert - Van~Kampen
   theorem (see also~\cite{Bi-Co:cob1}).

   Assume now that $L_1, L_2$ are spin. Then $\widetilde{\gamma}_1
   \times L_1$ and $\widetilde{\gamma}_2 \times L_2$ are also spin,
   with a spin structure extending those of the ends. Recall that the
   connected sum of spin manifolds is also
   spin~\cite{La-Mi:spin-geom}. Thus $W = (\widetilde{\gamma}_1 \times
   L_1) \# (\widetilde{\gamma}_2 \times L_2)$ is spin too and by
   standard arguments it follows that the spin structure on $W$ can be
   chosen so that it extends those given on the ends. By restriction
   we obtain a spin structure on $V_0 \subset W$ and consequently also
   the desired one on $V$.
\end{proof}

config: arxiv


TITLE: $Re(\frac{f}{g})\geq 0$ on the boundary of disk implies $f$ and $g$ have same number of zeros in the disk
QUESTION [0 upvotes]: I came across the following problem while solving some question paper. I have not been able to make any progress on it.
“Let $f$ and $g$ be holomorphic in a neighbourhood of unit disk. Assume that $f$ and $g$ have no zeros on the boundary, and $Re(\frac{f}{g})\geq 0$ on the boundary. Show that $f$ and $g$ have same number of zeros in the disk.”
I do not understand how to proceed. I thought of using Rouche’s theorem but I do not know how to obtain $|f+g|<|f|+|g|$ on the boundary from the given condition.
I am not sure but I was also thinking if I could get from the above condition that $\frac{f}{g}$ extends as holomorphic function on the disk and then show that it must be a constant. But, I also feel that it would be a little strong thing to conclude as this would mean that $f$ and $g$ have the same zeros.
Any hint would be highly appreciated.

REPLY [3 votes]: The principal branch of $\log$ is analytic on $\Bbb{C}-(-\infty,0]$
$\frac{f(z)}{g(z)}$ is analytic and $ \in \Bbb{C}-(-\infty,0]$ for $|z| \in (1-\epsilon,1+\epsilon)$
thus $h(z)=\log \frac{f(z)}{g(z)}$ is analytic for $|z| \in (1-\epsilon,1+\epsilon)$ so that $$0=\int_{|z|=1} h'(z)dz=\int_{|z|=1} (\frac{f'(z)}{f(z)}-\frac{g'(z)}{g(z)})dz$$
The proof of Rouche's theorem is based on the same argument.


TITLE: Understanding angle-preserving definition
QUESTION [4 upvotes]: My book (Real and complex analysis, by Rudin) gives the following definition:
Let $A(z) = \frac z{|z|}$. Then we say $f$ preserves angles at $z_0$ if $$\lim_{r \to 0}e^{-i\theta} A[f(z_0 + re^{i\theta} )- f(z_0)] \ \ \ \ (r > 0)$$ exists and it is independent of $\theta$.
It then adds

The requirements is that fr any two rays $L'$ and $L''$ starting at
  $z_0$, the angles which their images $f(L')$ and $f(L'')$ make at
  $f(z_0)$ is the same as that made by $L'$ and $L''$ in size as well
  orientation

I am trying to understand why the definition says that. Does anyone know of a cool geometric interpretation?
My thoughts
$A(z)$ returns a complex number on the unit circle whose argument is the argument of $z$. So $A(z - w)$ basically is a measure of the angle between $z$ and $w$.
Now in the definition I take the angle between $f(z_0 + re^{i \theta})$ and $f(z_0)$, and divide it by $e^{i \theta}$ (meaning I rotate by $-\theta$ our number). Since the original angle between $z_0$ and $z_0 + re^{i \theta}$ is $\theta$, if I want angles to be preserved, I would like the result to have an argument of $0$, right? Why is instead required the limit to exists and be independent of $\theta$?
Moreover, the reason why we take the limit is because we only care about local properties of the function, right?
Finally, why was it defined this way? I mean one could be naive and define it like $$\lim_{r \to 0} \arg(f(z_0 + re^{i\theta}) - f(z_0)) = \theta$$. What's the reason this isn't a good definition for our purposes?

REPLY [5 votes]: Your thougts are almost spot on, save for a little detail: $A(z-w)$ does not measure the argument between $z$ and $w$, it actually measures the argument of $z$ after being translated by $w$ (which is not the same thing as there is no rotation involved in translating something by $w$, I recommend you make a drawing to convince yourself of this fact).
So in fact $A(f(z + r e^{i\theta}) - f(z))$ measures the argument of $f(z+re^{i\theta})$ when we regard $f(z)$ as the origin.
And I guess your alternative definition is correct, but the one in the book suggests a more concrete analytical treatment.


TITLE: Temperature of Neon Light
QUESTION [0 upvotes]: I recently installed a neon light in a configuration such that it is resting on a book shelf and near some books. I would like to know if this stupid/if it is going to start a fire if I leave it on without watching it. I'm also curious about how to calculate the temperature of the bulb by physical principles.
To be clear, I'm not going to leave the bulb on when I am not present and/or not awake until I've measured the equilibrium temperature and it is sufficiently low.
Parameters: the bulb is a 5 feet long. The outer radius is approximately 1.5cm. I don't know how thick the glass is. The bulb contains argon and mercury and emits a pink color. It presumably is made of glass and has a coating. By touch, the cathode/anode (? -- I think this is the terminology) is the hottest part. These nodes are enclosed in a plastic insulator. I can touch the lamp and the ends without burning myself after a few hours of it being on, which I am taking (correctly?) to be the equilibrium temperature. The lamp is at STP, defined as: in a room at atmospheric pressure and temperature, out of direct sunlight, and with ambient temperature unlikely to exceed 110 degrees F/45 degrees C at any time. (yes, I am American).
I can make more measurements or determine them with some amount of money/materials/effort if I have a sense of what the important measurements are.
So: the questions are:
(1) Practical: Should I worry about starting a fire
(2) Theoretical: How do I calculate the equilibrium temperature of the bulb given the voltage, current, number density of the gas, ratio of the two species, absorptive properties of the coating, length and inner+outer radius of the cylindrical glass bulb, and pressure and temperature of the surroundings. Or any other parameters I might not be thinking of.
(3) Characteristic: What are the characteristic values of these parameters that I should use.
Progress so far: the Internet tells me that Neon lights are a plasma with electron temperature between $10^5$K and $10^6$K, and plasma density between $10^{12}$ and $10^{20}$ in units of $cm^{-3}$. I don't know what this means. I would appreciate a reference.

REPLY [1 votes]: here is a simplified way to estimate whether or not you are in trouble.
on the box which encloses the high-voltage power supply for the neon tube, there will be a power rating for the box which will say something like "120VAC 1.5A input" possibly accompanied by something else which reads like "5000VDC 30ma output MAX". these numbers refer to the input power consumed by the power supply from the AC mains (120VAC x 1.5A = 180 watts) and the maximum amount of power it can deliver to the neon tube (5000VDC x 0.030A = 150 watts, slightly less than the input power due to losses in the power supply).
Assume now that the neon tube is drawing the full rated output power, 150 watts in this example, to power up the tube, whose surface area you can calculate knowing its length and diameter. Assume further that all that electrical power is dissipated as heat in the tube (a worst-case assumption). Now you can calculate the worst-case power density in watts per square centimeter which the tube will produce as it sits upon your shelf.
Finally, we assume that half the tube output will be radiated away from the shelf into the room and half will be absorbed by the shelf. This represents the heat load which the shelf must dissipate else it will ignite. Now you must answer the final question: If the shelf, which we assume is wood, is presented with a heat input of x watts/square centimeter, will it begin to burn?
I can tell you now that the answer will be No, based on experience, and that the tube and shelf will be safe: because the wattage produced will be insufficient to raise the temperature of the wood to the ignition point.


TITLE: Example of inner product as matrix multiplication
QUESTION [1 upvotes]: Can anyone please give an example for the following:
If I take $B=\{[1\ 0],[0\ 1]\} \implies M=I$. If $x=[1 \ 2], y=[3 \ 4]$, then $\langle x,y\rangle =11$. $[y]^H=[3 \ 4]^T$. But $[3 \ 4]^T M=[3 \ 4]^TI=[3 \ 4]^T \ne \langle x,y\rangle$? What am I doing wrong here?

REPLY [1 votes]: You've calculated $y^H M$ instead of $y^H Mx$.


text:
(*  Title:      Well-Quasi-Orders
    Author:     Christian Sternagel <c.sternagel@gmail.com>
    Maintainer: Christian Sternagel
    License:    LGPL
*)

section \<open>Instances of Well-Quasi-Orders\<close>

theory Wqo_Instances
imports Kruskal
begin

subsection \<open>The Option Type is Well-Quasi-Ordered\<close>

instantiation option :: (wqo) wqo
begin
definition "x \<le> y \<longleftrightarrow> option_le (\<le>) x y"
definition "(x :: 'a option) < y \<longleftrightarrow> x \<le> y \<and> \<not> (y \<le> x)"

instance
  by (rule class.wqo.of_class.intro)
     (auto simp: less_eq_option_def [abs_def] less_option_def [abs_def])
end

subsection \<open>The Sum Type is Well-Quasi-Ordered\<close>

instantiation sum :: (wqo, wqo) wqo
begin
definition "x \<le> y \<longleftrightarrow> sum_le (\<le>) (\<le>) x y"
definition "(x :: 'a + 'b) < y \<longleftrightarrow> x \<le> y \<and> \<not> (y \<le> x)"

instance
  by (rule class.wqo.of_class.intro)
     (auto simp: less_eq_sum_def [abs_def] less_sum_def [abs_def])
end

subsection \<open>Pairs are Well-Quasi-Ordered\<close>

text \<open>If types @{typ "'a"} and @{typ "'b"} are well-quasi-ordered by \<open>P\<close>
and \<open>Q\<close>, then pairs of type @{typ "'a \<times> 'b"} are well-quasi-ordered by
the pointwise combination of \<open>P\<close> and \<open>Q\<close>.\<close>

instantiation prod :: (wqo, wqo) wqo
begin
definition "p \<le> q \<longleftrightarrow> prod_le (\<le>) (\<le>) p q"
definition "(p :: 'a \<times> 'b) < q \<longleftrightarrow> p \<le> q \<and> \<not> (q \<le> p)"

instance
  by (rule class.wqo.of_class.intro)
     (auto simp: less_eq_prod_def [abs_def] less_prod_def [abs_def])
end

subsection \<open>Lists are Well-Quasi-Ordered\<close>

text \<open>If the type @{typ "'a"} is well-quasi-ordered by \<open>P\<close>, then
lists of type @{typ "'a list"} are well-quasi-ordered by the homeomorphic
embedding relation.\<close>

instantiation list :: (wqo) wqo
begin
definition "xs \<le> ys \<longleftrightarrow> list_emb (\<le>) xs ys"
definition "(xs :: 'a list) < ys \<longleftrightarrow> xs \<le> ys \<and> \<not> (ys \<le> xs)"

instance
  by (rule class.wqo.of_class.intro)
     (auto simp: less_eq_list_def [abs_def] less_list_def [abs_def])
end

end

subset_name: curated


TITLE: Classical $\mathcal{A}$ quantifier: prove that $(\mathcal{A} \forall)_{1}$ relations and $\Sigma_1^1$ relations have the same expressive strength..
QUESTION [3 upvotes]: A quantifier on $\mathbb{N}$ is a set $Q$ such that $\emptyset \subsetneq Q \subsetneq P(\mathbb{N})$ and $Q$ is closed upwards, i.e. if $X \subseteq Y \subseteq \mathbb{N}$ and $X \in Q$, then $Y \in Q$.
Given a quantifier $Q$ on $\mathbb{N}$ we can expand a language $L$ to the language $L(Q)$ which has the additional symbols $Q$ and $\breve{Q}$, so that if $\phi$ is a formula of $L(Q)$, then so is $Q x \phi(x)$ and $\breve{Q} x \phi(x)$ with the following interpretation:
$$
Q x \phi(x) \text { is true } \Leftrightarrow\{x: \phi(x)\} \in Q.
$$
Suppose we also have a coding scheme in the language, such that we can code finite sequences of numbers as one natural number. $n = \langle n_1,\ldots,n_m\rangle$ is a single natural number and we can decode this too, thus we have that $(n)_i = n_i$. Now we define the classical $\mathcal{A}$ quantifier as $$
\mathcal{A}=\left\{Y \subseteq \mathbb{N}:\left(\exists x_{1} \exists x_{2} \ldots\right)(\forall n)\left(\left\langle x_{1}, x_{2}, \ldots, x_{n}\right\rangle \in Y\right)\right\}
.$$
We say that a second-order relation $\varphi(\vec{x}, X)$ is $(\mathcal{A} \forall)_{1}$ on $\mathbb{N}$ if it is definable by a formula of the language $L(\mathcal{A})$ of the form $(\mathcal{A} y)(\forall z) \chi(\vec{x}, y, z, X)$ where $\chi$ only has bounded quantification and $\vec{x}$ and $X$ are free variables, $X$ being second-order.
In "Recursion, induction and quantifiers" in theorem 3.16 and subsequent notes, P. Kolaitis treats $(\mathcal{A} \forall)_{1}$ relations as if they are $\Sigma^1_1$ relations. $\Sigma_1^1$ relations are defineable by a formula of the form $\exists Y \forall z \psi(\vec{x}, Y, z, X)$, here $\exists Y$ is second-order quantification.
How does one prove that these do in fact play the same role? That for every $\Sigma_1^1$ relation $\varphi$ there exists a $(\mathcal{A} \forall)_{1}$ relations $\varphi'$ such that $$ \varphi(\vec{x},X) \Leftrightarrow \varphi'(\vec{x},X).$$ And vice versa? Or are they different in a subtle way?
I have already found that chapter 25 of Jech contains similar ideas by proving an normal form of $\Sigma_1^1$ sets through trees but this does not seem to generalise to the result I am searching for.

REPLY [2 votes]: Here's one direction (I'll add the other later) - namely, that every $(\mathcal{A}\forall_1)$-relation is $\Sigma^1_1$:
We have $$(\mathcal{A}y)(\forall z)\chi(\overline{x},y,z,X)\quad\equiv\quad \{y: \forall z\chi(\overline{x},y,z,X)\}\in\mathcal{A}$$ by the general definition of quantifiers. Now the right hand side can be rewritten, via the definition of $\mathcal{A}$ specifically, as $$\exists (y_i)_{i\in\mathbb{N}}\forall n(\langle y_1,...,y_n\rangle\in\{y: \forall z\chi(\overline{x},y,z,X)\}),$$ which itself can be rewritten as $$\exists (y_i)_{i\in\mathbb{n}}\forall n(\forall z\chi(\overline{x}, \langle y_1,...,y_n\rangle, z, X)).$$ This is clearly $\Sigma^1_1$, so every $(\mathcal{A}\forall_1)$-on-$\mathbb{N}$ formula is $\Sigma^1_1$.


TITLE: a characterisation of proper maps via ultrafilters
QUESTION [2 upvotes]: Let $B$ be a topological space.
Call a subset $A\subset B$ ultrafilter-like iff $A$ is dense in $B$ and
each decomposition $A=A_1\cup A_2$,
into the union of two open subsets extends to a decomposition of
$B=B_1\cup B_2$, $A_1\subset B_1$ and $A_2\subset B_2$ into the union of two open subsets $B_1$ and $B_2$.
Is the following true? Let $f:X\rightarrow Y$ be a proper map
and $A\subset B$ is ultrafilter-like.
For any  $g:A\rightarrow X$ and $h:B\rightarrow Y$  such that
$f(g(a))=h(a)$ for each $a\in A$, there is a map $g':B\rightarrow X$
extending $g$ such that $f(g'(b))=h(b)$ for each $b\in B$.
Is this true if $Y$ is a point (and thus $X$ is an arbitrary
quasi-compact space).
This is true provided $B=A\cup \{\omega\}$
where $\omega$ is closed. If  $A$ is discrete, then the neighbourhoods of $\omega$
define an ultrafilter on $A$ and the property becomes
the definition of a proper map via ultrafilters (see Bourbaki, General Topology, I\S10.2, Theorem I).
 If $A$ is not necessarily discrete, then there is an ultrafilter $F$ on $A$
such that the neighbourhoods of $\omega$ are $F$-big,
and the same arguments works. (To fix the terminology: by a proper map
I mean a closed map such that the preimage of any point is quasi-compact).

REPLY [1 votes]: Let $C$ be a connected Tychonoff space and $a,b\in \beta C\setminus C$ be two distinct points. Let $X=\beta C$, $Y=X/\{a,b\}$ be the quotient space and $f:X\to Y$ be the quotient map. It is clear that $f$ is perfect and hence proper.
Let $B=Y$, $A=B\setminus \{q(a)\}$, $h:B\to Y$ be the identity map, $g=q^{-1}|A:A\to X\setminus\{a,b\}\subset X$ be the homeomorphism of $A=Y\setminus\{q(a)\}$ onto $X\setminus\{a,b\}$. It is easy to see that $f\circ g=h|A$. On the other hand, it can be shown that $g$ admits no continuous extension to a map $g':B\to X$ such that $f\circ g'=h$.
The space $g(A)$ is connected sinse it contains a dense connected subspace $C$. Then the space $A$ is connected as well (being homeomorphic to $g(A)$). The connectedness of $A$ implies that $A$ is ultrafilter-like in $B$ (being connected the space $A$ admits no non-trivial partitions into two open sets).
So, to get a sensible answer, we should assume that the space $A$ is disconnected, or better (strongly) zero-dimensional. In this case there is a hope for the positove answer since ultrafilter-likenes of $A$ in $B$ should imply that the identity embedding $A\to\beta A$ extends to an embedding of $B$ into $\beta(A)$.


TITLE: finite rank operators are dense in trace class operators and Hilbert–Schmidt operators
QUESTION [2 upvotes]: In many reference books,the proofs were not given,I hope someone can give me hints on how to construct finite rank operators to approximate trace class operators and  Hilbert–Schmidt operators.Thnak you in advance!

REPLY [1 votes]: Rolands answer already gives the idea of how these finite rank operators are constructed. The following argument can be looked up in the proof of Proposition 16.2 / 16.3 from "Introduction to Functional Analysis" by Meise & Vogt (1997).
Any trace-class or Hilbert-Schmidt operator is compact (by construction; in the case of Meise/Vogt) and every compact operator $A\in\mathcal K(\mathcal H)$ on any Hilbert space $\mathcal H$ has a singular value decomposition $A=\sum_{n=1}^\infty s_n(A)\langle \cdot,e_n\rangle f_n$ where $(e_n)_{n\in\mathbb N},(f_n)_{n\in\mathbb N}$ are orthonormal systems in $\mathcal H$ and $(s_n(A))_{n\in\mathbb N}$ is a decreasing null sequence in $[0,\infty)$. Define the sequence of finite-rank operators $A_m:=\sum_{n=1}^m s_n(A)\langle \cdot,e_n\rangle f_n$. From Pythagoras's theorem it follows then that for each $x\in\mathcal H$ and $m\in\mathbb N$
$$
\Vert (A-A_m)x\Vert^2=\Big\Vert \sum_{n=m+1}^\infty s_n(A)\langle x,e_n\rangle f_n\Big\Vert^2\overset{s_n\in[0,\infty)}=\sum_{n=m+1}^\infty s_n(A)^2|\langle x,e_n\rangle|^2 \underbrace{\Vert f_n\Vert^2}_{=1}.
$$
Now since $(s_n(A))_{n\in\mathbb N}$ is decreasing, Bessel's inequality yields
$$
\Vert (A-A_m)x\Vert^2=\sum_{n=m+1}^\infty s_n(A)^2|\langle x,e_n\rangle|^2\leq (s_{m+1}(A))^2\sum_{n=m+1}^\infty |\langle x,e_n\rangle|^2\leq (s_{m+1}(A))^2\Vert x\Vert^2.
$$
Since $x\in\mathcal H$ was arbitrary, taking the square root on this inequality by the definition of the operator norm gives $\Vert A-A_m\Vert\leq s_{m+1}(A)$ for any $m\in\mathbb N$. The fact that $(s_n(A))_{n\in\mathbb N}$ is null sequence thus shows $\lim_{m\to\infty}\Vert A-A_m\Vert=0$ and concludes this argument.


TITLE: Bibliography request: Entropy for continued fractions
QUESTION [9 upvotes]: Given a strictly positive real number $x$ we set $e(x)=\log(1+x)$ if $x$ is an integer and
$$e(x)=\log(1+x)+\frac{1+\lbrace x\rbrace}{1+x}\left(e(1/\lbrace x\rbrace)-\log(1+\lbrace x\rbrace)\right)$$
otherwise (i.e. if the fractional part $\lbrace x\rbrace$ of $x$ is strictly positive).
This defines $e(x)$ if $x$ is rational and the recursive definition of $e(x)$ yields its value as a (perhaps infinite) limit if $x$ is irrational.
The real number $e(x)$ can be interpreted as an entropy (or information content) as follows: Consider the segment $[-1,x]$
endowed with the uniform probability law. Apply to the positive and to the negative part of this segment iteratedly the Euclidean
Algorithm (removing from the longer piece a segment of length
the length of the shorter piece). This yields a partition of $[-1,x]$
into finitely many segments if $x$ is rational and into infinitely many segments accumulating at the origin otherwise. The function $e(x)$ is simply a normalized entropy of this partition
($e(x)$ is essentially the expected amount of information on the position of a random point on $[-1,x]$ which can be inferred when only knowing the interval containing the random point).
We illustrate this with $x=5/19$. We start with the interval $[-1,5/19]$. Since only proportions matter, we replace it by $[-19,5]$ in order to work over $\mathbb Z$. The positive interval $[0,5]$ is shorter than the negative interval $[-19,0]$. We therefore cut a piece of length $5$ from the negative interval in order to get $[-14,5]$. Iteration yields then $[-9,5]$ and $[-4,5]$. At this point the negative interval is shorter and we get therefore $[-4,1]$. The positive interval $[0,1]$ is now shortest leading to $[-3,1],[-2,1],[-1,1]$ and finally $[
0,1]$. We get therefore the partition
of $[-19,5]$ into intervals $[-19,-14],[-14,-9],[-9,-4],[-4,-3],[-3,-2],[-2,-1],[-1,0],[0,1],[1,5]$. Endowing the interval $[-19,5]$ with the uniform probability measure $\mu$ (with probability of an interval given by its length divided by the total length $19+5=24$)
the topological entropy of this partition is given by $$e(5/19)=-\frac{1}{24}(5\log(5/24)+5\log(5/24)+5\log(5/24)+5\cdot 1\log(1/24)+4\log(4/24))$$ and gives the expected amount of information (up to $\log(2)$ when working with bits, I guess) encoded by the partition.
A few easy facts:
$e(x)=e(1/x)$ (obvious from the definition).
$e(x)$ is discontinous at every rational value (it is in fact unbounded in the neighbourhood of a rational value) but should be continuous on a set of full measure. $e(x)$ is infinite at
some irrational values with very huge coefficients in their
continued fraction expansion.
The graph of $e(x)$ is a beautiful fractal.
The function $x\longmapsto e(x)$ is probably locally integrable.
It would be somewhat surprising if nobody has ever looked at this notion.
Are there paper, references etc. on this notion?

REPLY [0 votes]: Below is a graph of $e(x)$. Precisely, the $y$-value associated with the given $x$-value is $y = e(x/70009)$.


TITLE: Is matrix algebra a special case of Grassmann (exterior) algebra, and if so what is more general case?
QUESTION [2 upvotes]: Just a little question. I only recently heard about Grassmann algebra while reading a book on the history of vector algebra and quaternions.
I still don't understand what does exterior product mean exactly, and what is its most general case (according to the book I've read, Grassmann was trying to describe every possible algebraic system).
But this thing about matrices is easy to understand. Since the product of $m  \times n $ and $n  \times m $ matrices is $m  \times m $ matrix, I can see that.
$$
[a_1, a_2, a_3] \begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix} = a_1 b_1+a_2 b_2+a_3 b_3
$$
Which is $1 \times 1$ matrix, i.e. a number, a dot product of two vectors. So this operation lowers the rank of a tensor by one.
But in the other case
$$
\begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix} [a_1, a_2, a_3] = \begin{bmatrix}a_1 b_1 & a_2 b_1 & a_3 b_1 \\a_1 b_2 & a_2 b_2 & a_3 b_2 \\a_1 b_3 & a_2 b_3 & a_3 b_3 \end{bmatrix}
$$
So this operation elevates the rank of a tensor by one, creates a square matrix from two vectors.
So is this a special case of exterior and interior products? And if it what is the more general case?

REPLY [1 votes]: The second product you give is equivalent to a tensor product; if it were an exterior product instead, the resulting matrix would be antisymmetric.  Also notice that the trace of that matrix is the inner product.  So the inner product can be identified with the trace, and the exterior product could be identified with the antisymmetric part of this matrix, but there's still a symmetric, traceless piece.
That is, the tensor product captures both the inner product and the exterior product, but it also clearly has more stuff.
Exterior algebra can be described completely in terms of tensor algebra, but not vice versa.


TITLE: How much information is lost in this binary sequence?
QUESTION [0 upvotes]: Given a sequence of $X= y_1, y_2 , ...,y_m$ binary data points. That means $y_i = 0 \text{ or } y_i=1$ for $i\in \{1,...,m\}$
Say that we select $n<m$ datapoints, 'damage' our file such that we are not certain about the outcomes of those $n$ points. How much information is now lost?

For example, say we have a sequence 000000. Then we damage this file
such that we are not certain about the first 2 values any more (the
file now could be 000000, 010000, 100000 or 110000
Does this mean that we lose $\frac{2^2}{2^6}=4/64=$ 6,25% information?

My question is: In general, how much information is lost when we are not certain about $n$ chosen outcomes of the sequence. Alternatively asked: how much information do the $m-n$ datapoints contain when compared to the original $m$ datapoints?
And what about if the $m$ datapoints are not binary, but for example can take $3$, $4$ or $r\in \mathbb{N}$ values?
Im looking for a clear, comprehensive answer, in a general formula. Thank you :-)

REPLY [1 votes]: Each entry $Y_k\in\{0,1\}$ of your vector $X=(Y_1,\ldots ,Y_m)$ incorporates $1$ bit of information; therefore a "measurement" of $X$ is worth $m$ bits. This statement implicitly assumes that the $Y_k$ are independent and assume the value $0$ or $1$ with probability ${1\over2}$ each, in other words: that all $2^m$ possible words $X$ are equiprobable.
If $n$ of the outcomes $y_k$ are corrupted, and you know the affected set $J\subset[m]$, then the information loss is $n$ bits, since such a measurement $x$ does not specify a point in $\{0,1\}^m$ exactly, but admits $2^n$ points compatible with $x$.
When it is only known that $|J|=n$, but $J$ is unspecified otherwise, then things are much more complicated.
When $2$ is replaced by some $r>1$ in these arguments the information gained or lost per "correct digit" amounts to $\log_2 r$ bits.


TITLE: How can I calculate cohomology?
QUESTION [0 upvotes]: I know what homology means and I know what (de Rham) cohomology means. But I can’t actually calculate cohomology of a given space (while I can do it for homology)
My question is that is there a relationship between homology and cohomology? (I mean can I derive something about cohomology is I know its homology?) If not, how can I calculate cohomology?

REPLY [1 votes]: de Rham cohomology is only defined for smooth manifolds. It it is known that it agrees with singular cohomology on smooth manifolds.
Now you can apply the universal coefficient theorem for cohomology. See for example https://en.wikipedia.org/wiki/Universal_coefficient_theorem or any textbook on algebraic topology. This theorem establishes a short exact sequence
$$0 \to \text{Ext}(H_{n−1}(X);\mathbb{Z})\to H^n(X) \to \text{Hom}(H_n(X);\mathbb{Z}) \to 0$$
In many cases you can compute $H^n(X)$ using this sequence. However, in general it is an extension problem with a non-unique solution.


TITLE: Showing the correctness of inequality with $e$
QUESTION [0 upvotes]: Is true that always for every $x$ and $n, n\not=0$ this inequality is true:
$$\left(1-\frac{1}{n}\right)^{x} \leq e^{-x/n}\;\;?$$
I have the doubt about the values of $n$ which are very close to $0$, but I don't know how to prove it or show counterexample that this inequality is wrong?

REPLY [2 votes]: The function $f\colon t\mapsto \log(1-t)$ is concave $\left(f''(t)=\frac{-1}{(1-t)^2}\right)$ so the curve of $f$ is below its tangent line at the point $t=0$ with equation $y=-t$ hence
$$\log(1-t)\le -t\quad \forall t<1$$
hence we have
$$\log\left(1-\frac 1 n\right)\le -\frac 1 n$$
now multiply by $x\ge0$ and apply the exponantial function you get the result.


TITLE: find the minimum value for the following expression $\max \{ {x_{i_1}, x_{i_2}-x_{i_1},x_{i_3}-x_{i_2} ... x_{i_n}-x_{i_{n-1}}, 1-x_{i_n} }\}$
QUESTION [3 upvotes]: Given the numbers $0<x_1<x_2<...<x_{n^2}<1$,
For each sub group of n of those elements: $x_{i_1}<x_{i_2}<...<x_{i_n}$
we will look at the value of the following expression : $\max \{ {x_{i_1}, x_{i_2}-x_{i_1},x_{i_3}-x_{i_2} ... x_{i_n}-x_{i_{n-1}}, 1-x_{i_n}  }\}$
What would be the most efficient possible algorithm that will find the minimal value of that expression that a sub group containing $n$ variables could achieve?

REPLY [2 votes]: I have an algorithm for this, but I haven't been able to prove that it's optimal (in my experience, proving strong lower bounds for computational problems is typically very hard), so maybe this isn't the answer you're looking for.
Let's rephrase the problem: given numbers $n < m$ and $x_0 < x_1 < \cdots < x_{m-1} < x_m$, we want to find
$$\min_{0 < i_1 < i_2 < \cdots < i_n < m} \max \{x_{i_1} - x_0, x_{i_2} - x_{i_1}, \dots, x_{i_n} - x_{i_{n-1}}, x_m - x_{i_n}\}.$$
We can think of this as the problem of "cutting" the interval $[x_0, x_m]$ into $n+1$ blocks, where we are only allowed to cut in the prescribed places $x_i$, and we want to minimize the largest resulting block size.
Define $A(k, r)$ as the minimum largest block size resulting from cutting $[x_0, x_k]$ into $r+1$ blocks, i.e.
$$A(k, r) = \min_{0 < i_1 < i_2 < \cdots < i_r < k} \max \{x_{i_1} - x_0, x_{i_2} - x_{i_1}, \dots, x_{i_r} - x_{i_{r-1}}, x_k - x_{i_r}\}$$
so we want to find $A(m, n)$.
For $r > 1$, it is not hard to see that
$$A(k, r) = \min_{r \leq j < k} \max\{A(j, r-1), x_k - x_j\}$$
since
\begin{align*}
A(k, r)
&= \min_{r \leq i_r < k} \min_{0 < i_1 < \cdots < i_r} \max\{x_{i_1} - x_0, x_{i_2} - x_{i_1}, \dots, x_{i_r} - x_{i_{r-1}}, x_k - x_{i_r}\} \\
&= \min_{r \leq i_r < k} \max \left\{\min_{0 < i_1 < \cdots < i_r} \max\{x_{i_1} - x_0, x_{i_2} - x_{i_1}, \dots, x_{i_r} - x_{i_{r-1}}\},  x_k - x_{i_r} \right\} \\
&= \min_{r \leq i_r < k} \max \{A(i_r, r-1),  x_k - x_{i_r}\} \\
\end{align*}
This gives a pretty direct dynamic programming approach:
set all A(k, 1) = x_k - x_0
for r in {2, ..., n}:
    for k in {r+1, ..., m}:
        iterate over all j in {r, ..., k-1} to find the smallest max(A(j, r-1), x_k - x_j)
        set A(k, r) to this value

which clearly runs in $O(nm^2)$ steps. However, we can actually get this down to $O(nm \log m)$ steps by using binary search, as described below.
The important fact is that in the computation of
$$A(k, r) = \min_{r \leq j < k} \max\{A(j, r-1), x_k - x_j\},$$
$x_k - x_j$ is a strictly decreasing function of $j$, while maybe less obviously, $A(j, r-1)$ is in fact an increasing function of $j$. To show this last part, note that $A(k, r)$ can also be defined as the minimum largest block size when we cut $[x_0, x_k]$ into at most $r+1$ blocks (that is, with at most $r$ cuts): making fewer cuts cannot decrease the largest block size. Then for $j' < j$, if we have a choice of at most $r-1$ cuts of $[x_0, x_j]$, this gives a choice of at most $r-1$ cuts of $[x_0, x_{j'}]$ (namely those cuts lying in the smaller interval), and the largest block size cannot increase, so it follows that $A(j', r-1) \leq A(j, r-1)$.
Thus, if we consider $f(j) = \max\{A(j, r-1), x_k - x_j\}$ as a function of $j \in \{r, \dots, k-1\}$, we can break $\{r, \dots, k-1\}$ into the two regions $\{r, \dots, t-1\}$ where $x_k - x_j$ is larger, and $\{t, \dots, k-1\}$ where $A(j, r-1)$ is larger (though one of these may be empty), so that $f$ is decreasing on the first region, and increasing on the second region. This means the minimum must occur at either $t-1$ or $t$, and one of these must be the point at which $g(j) = A(j, r-1) - (x_k - x_j)$ (a strictly increasing function) is closest to 0.
Now, since $g(j)$ is strictly increasing, we can find the index $j_0$ where $g(j)$ is closest to $0$ by binary searching for $0$ in the list $\{g(r), g(r+1), \dots, g(k-1)\}$. Importantly, we never explicitly construct this list, we only construct the values $g(j)$ when we query them, to save time, so this takes $O(\log m)$ steps. Then the index $j^*$ which minimizes $f(j)$ is among $j_0-1, j_0, j_0+1$, so we set $A(k, r)$ to be the smallest of $f(j_0-1), f(j_0), f(j_0+1)$. Since we take $O(\log m)$ steps to compute each $A(k, r)$, it takes $O(nm\log m)$ steps to compute all $A(k, r)$, and in particular, $A(m, n)$.
Thus in the relevant case, where $m = n^2 + 1$, and $x_0 = 0$, $x_m = 1$, the algorithm takes $O(n(n^2 + 1)\log(n^2 + 1)) = O(n^3 \log n)$ steps.


text:
\begin{document}

\title{A Positivity-Preserving Limiting Strategy for Locally-Implicit Lax-Wendroff Discontinuous Galerkin Methods}

\titlerunning{A Positivity-Preserving Limiting Strategy for Locally-Implicit Lax-Wendroff DG}

\author{Camille Felton \and
        Mariana Harris \and
        Caleb Logemann \and
        Stefan Nelson \and
        Ian Pelakh \and
        James A. Rossmanith\footnote{Corresponding author}}
\institute{Camille Felton \at
           University of Wisconsin--Platteville, Department of Mathematics, 435 Gardner Hall, 1 University Plaza,
           Platteville, WI 53818, USA, \email{\url{camillefelton129@gmail.com}} \\
           \and
           Mariana Harris \at
           Instituto Tecnol\'ogico Aut\'onomo de M\'exico,
           R\'io Hondo 1, Progreso Tizap\'an,
           01080 Alvaro Obreg\'on, CDMX, Mexico,
           \email{\url{harris-94@hotmail.com}} \\
           \and
           Caleb Logemann \at
           Iowa State University,
           Department of Mathematics,
           411 Morrill Road,
           Ames, IA 50011-2104, USA,
           \email{\url{logemann@iastate.edu}} \\
           \and
           Stefan Nelson \at
           Minnesota State University Moorhead,
           Department of Mathematics,
           1104 7th Ave South,
           Moorhead, MN 56563, USA,
           \email{\url{nelsonste@mnstate.edu}} \\
           \and
           Ian Pelakh \at
           University of Florida,
           Department of Mathematics,
           1400 Stadium Rd,
           Gainesville, FL 32611, USA
           \email{\url{ipelakh@ufl.edu}} \\
           \and
           James A. Rossmanith \at
           Iowa State University,
           Department of Mathematics,
           411 Morrill Road,
           Ames, IA 50011-2104, USA,
           \email{\url{rossmani@iastate.edu}}
}

\date{Received: date / Accepted: date}

\maketitle

\begin{abstract}

Nonlinear hyperbolic conservation laws admit singular solutions such as shockwaves (discontinuities in conserved variables), rarefaction waves (discontinuities in derivatives), and vacuum states (loss of strong hyperbolicity). When ostensibly high-order numerical methods are applied in such solution regimes, unphysical oscillations present themselves that can lead to large errors and a breakdown of the numerical simulation. In this work we develop a new Lax-Wendroff discontinuous Galerkin (LxW-DG) method with a limiting strategy that keeps the solution non-oscillatory and positivity-preserving for relevant variables, such as height in the shallow water equations and density and pressure in the compressible Euler equations. The proposed LxW-DG scheme updates the solution over each time-step with a locally-implicit predictor followed by an explicit corrector. The locally-implicit prediction phase is formulated in terms of primitive variables, which greatly simplifies the solver. The resulting system of nonlinear algebraic equations are approximately solved via a Picard iteration, where the number of iterations is equal to the order of accuracy of the method. The correction phase is an explicit evaluation formulated in terms of conservative variables in order to guarantee numerical conservation. In order to achieve full positivity-preservation, limiting is required in both the prediction and correction steps. The resulting scheme is applied to several standard test cases for the shallow water and compressible Euler equations. All of the presented examples are written in a freely available open-source Python code.

\keywords{discontinuous Galerkin \and
          Lax-Wendroff \and
          shallow water \and
          compressible Euler \and
          positivity-preserving \and
          hyperbolic conservation laws}

\subclass{65M12 \and 65M60 \and 35L65}

\end{abstract}

\input{intro.tex}
\input{equations.tex}
\input{numerical_method.tex}
\input{limiters.tex}
\input{python_code.tex}
\input{numerical_examples.tex}
\input{conclusions.tex}

\section*{Acknowledgments}
This research was carried out as part of the 2017 Summer REU (Research Experience for Undergraduates) Program at Iowa State University in Ames, Iowa. We would like to thank Iowa State University for their hospitality and the National Science Foundation for funding the REU program under the following grant: NSF Grant DMS--1457443.
Additionally, JAR was supported in part by NSF Grant DMS--1620128. We also thank the developers of Python
(\url{python.org}), Matplotlib (\url{matplotlib.org}), and PyPy (\url{pypy.org}) for developing the excellent open-source software tools that allow research projects such as this one to remain fully open-source.

\bibliographystyle{plain}


TITLE: Induced electric field in circular wire around solenoid
QUESTION [1 upvotes]: I have a stationary solenoid of radius $a$ and length $L$ with $n$ windings per unit length. There is a time varying current in the wire $I = kt$, with $k$ a constant. A conducting wire with radius $r$, mass $m$ and resistance $R$ is placed around the wire centred around the solenoid and is free to move.
I am at first interested in the induced electric field. However I am a bit confused.
I know that $$\oint_C \mathbf{E} \cdot d \mathbf{l} = - \frac{d \phi}{dt}$$
For my integral, we are used to choosing circular path, but I am unsure about what radius to choose for this? Shall I choose a radius $r$ or radius $a$? Similarly, what radius do I choose to use for the area that comes into play with the $\phi$?
And what is the difference between the cases $r \geq a$ and $r<a$?
Thanks :)

REPLY [1 votes]: Inside the solenoid:
The field is $$B(t)=\mu_0ni(t)$$
So for $r<a$ the flux is $$\phi_B=\mu_0ni(t)*\pi r^2$$
$$\oint E\,ds=\frac{d\phi_B}{dt}$$
$$E2\pi r=\mu_0nk\pi r^2$$
$$E=\frac{\mu_0nkr}{2}$$
Outside the solenoid:
$$B=0$$
So the magnetic flux is only due to the field inside the solenoid:
$$\phi_B=\mu_0ni(t)*\pi a^2$$
$$\oint E\,ds=\frac{d\phi_B}{dt}$$
$$E2\pi r=\mu_0nk\pi a^2$$
$$E=\frac{\mu_0nk a^2}{2r}$$


TITLE: What should be the minimum value of the function $f(x)= \frac{1}{\sin(x)}$, where $0 \leq x < 1$
QUESTION [1 upvotes]: $$f(x)=\frac{1}{\sin(x)}$$  where $$(.)$$ represents the fractional part of function.
I have computed for the maximum value and that tends to infinity but I cannot find the lower bound of the range of the function. A little guidance is appreciated.

REPLY [2 votes]: The function does not have a minimum on $0 \leq x <1$ because it is not a closed interval.  We can only say that the infimum is $\lim_{x \rightarrow 1}\frac{1}{\sin(x)} = \frac{1}{\sin(1)},$ since $\sin(\cdot)$ is an increasing function on this interval.  The fractional part of $x$ is the same as $x$.
This explanation works even if you restrict $x$ to be rational.  Given any $\epsilon>0$, I can find a rational $x<1$ for which $|1/\sin(1)-1/\sin(x)|<\epsilon$.


TITLE: Maximum number of circle packing into a rectangle
QUESTION [8 upvotes]: I'm asked to pack the maximum number of 10m^2 circle into a 257 x 157m rectangle. After a lot of research, I found out that there are no optimal solution. So, i try to pack as many as possible (taking this website as reference):
1) First, I tried to place them in rectangular pattern:

I had the width 257/d (diameter) -> I got about 72.024 --> So along the width, i can place 72 circle.
I had the height 157/d (diameter) -> I got about 43.999 --> So along the height, i can place 43 circle.

--> That means in this case, i can fit in 43*72= 3096 circles
2) Then I try triangular pattern, which can fit more circles, 3575 circles. However, I find my math calculation kinda inefficient, long, and not correct in any other cases.
So my question is: Did I calculate it in a correct way? Are there any other more effective calculation methods?
Because in later question, it asks me to find the area of the circle to so that we get the maximum profit. Giving the profit of each circle is: P(a) = 200 - 200/a (a is the area of the circle)

REPLY [6 votes]: Consider the following diagram of a triangular packing:

If the circles have radius $r$, then each pair of horizontal red lines is a distance $r$ apart, and they're a distance $r$ from the edges. Each pair of vertical blue lines is a distance $r \sqrt 3$ apart, and they're still a distance $r$ from the edges.
So if you want the triangular packing to have $m$ circles in each column, and $n$ columns, then the rectangle must be at least $(2m+1) \cdot r$ units tall and $(2 + (n-1)\sqrt3) \cdot r$ units long. (Also, if the rectangle is only $2m \cdot r$ units tall, we can alternate columns with $m$ and $m-1$ circles.)
If the rectangle is $257 \times 157$ and the radius of a circle is $\sqrt{\frac{10}{\pi}}$, then:

If we make $257$ the vertical dimension, then the rectangle is a bit over $144 \cdot r$ units tall, and a bit over $(2 + 49\sqrt3) \cdot r$ units wide. So we can arrange the circles in $50$ columns that alternate between $72$ and $71$ circles, for $25 \cdot 72 + 25 \cdot 71 = 3575$ circles.
If we make $157$ the vertical dimension, then the rectangle is a bit over $87 \cdot r$ units tall, and a bit over $(2 + 82\sqrt3) \cdot r$ units wide. So we can arrange the circles in $83$ columns that have $43$ circles each, for $3569$ circles.

We pick the first option, which gave us the greater value.
If you want to know the exact number of circles that can fit, there is nothing better to be done than this calculation. But you can estimate the number of circles that will fit by knowing that the limiting density of the triangular packing is $\frac{\pi}{2\sqrt 3}$.
The $257 \times 157$ rectangle has area $40349$, but at most a $\frac{\pi}{2\sqrt 3}$ fraction of that area can be used: at most area $\frac{40349 \pi}{2\sqrt 3} \approx 36592.5$. If all circles have area $10$, then at most $3659$ circles can fit in that area. As you can see, this is an overestimate, because we aren't using the space around the edges of the packing as efficiently as possible.


TITLE: Prove that $\left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| \le \sqrt{2n}$.
QUESTION [2 upvotes]: For all positive integers $n$, prove that $$\large \left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| \le \sqrt{2n}$$

This is the last problem of a book I (wrongly) stole from the shelves from our class. And there definitely were shortcuts taken.
I also don't really understand the last part $$ [\cdots ]\le \frac{m - 2}{2} + \frac{n}{m} < \frac{\sqrt{2n} - 1}{2} + \sqrt{\frac{n}{2}} < 2n$$. Could someone please explain?
Moreover, the answer is lengthy, at least to me. So if you have any other solution that is shorter, please answer below and I would be appreciated.

REPLY [0 votes]: Let $m = \lfloor\sqrt{2n} + 1\rfloor$. We have that $$\left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right|$$
equals the absolute value of
$$\underbrace{\left(\left\{\frac{n}{1}\right\} + \left\{\frac{n}{2}\right\} - \left\{\frac{n}{3}\right\} + \cdots - (-1)^{m - 1}\left\{\frac{n}{m - 1}\right\}\right)}_{A}$$
$$ - (-1)^m\underbrace{\left(\frac{n}{m} + \frac{n}{m + 1} - \frac{n}{m + 2} + \cdots + (-1)^{n - m}\frac{m}{n}\right)}_{B}$$
$$ + (-1)^m\underbrace{\left(\left\lfloor\frac{n}{m}\right\rfloor + \left\lfloor\frac{n}{m + 1}\right\rfloor - \left\lfloor\frac{n}{m + 2}\right\rfloor + \cdots + (-1)^{n - m}\left\lfloor\frac{n}{n}\right\rfloor\right)}_{C}$$
Going through each term, we have that
$$ - \left(\left\{\dfrac{n}{2}\right\} + \left\{\dfrac{n}{4}\right\} + \left\{\dfrac{n}{6}\right\} + \cdots\right) \le A \le \left\{\dfrac{n}{1}\right\} + \left\{\dfrac{n}{3}\right\} + \left\{\dfrac{n}{5}\right\} + \cdots$$
with the expressions on the left- and right-hand side having $\left\lfloor\dfrac{m - 1}{2}\right\rfloor$ and $\left\lfloor\dfrac{m}{2}\right\rfloor$ respectively.
(It should be also noted that $\left\{\dfrac{n}{1}\right\} = 0$.)
For all $\left\{\dfrac{n}{p}\right\}$ $(p \in \mathbb Z^+, 1 \le p \le m)$, we have that $\left\{\dfrac{n}{p}\right\} \le \dfrac{p - 1}{p} \le \dfrac{m - 2}{m - 1}$
$$\implies |A| \le \left\lfloor\frac{m - 1}{2}\right\rfloor \cdot \dfrac{m - 2}{m - 1} \le \dfrac{m - 2}{2}$$
It is also evident that $$0 \le \left(\frac{m}{n} - \frac{n}{m + 1}\right) + \left(\frac{n}{m + 2} - \frac{n}{m + 3}\right) + \cdots $$
$$ = B = \frac{m}{n} - \left(\frac{n}{m + 1} - \frac{n}{m + 2}\right) - \cdots \le \dfrac{n}{m}$$
and $$0 \le \left(\left\lfloor\frac{m}{n}\right\rfloor - \left\lfloor\frac{n}{m + 1}\right\rfloor\right) + \left(\left\lfloor\frac{n}{m + 2}\right\rfloor - \left\lfloor\frac{n}{m + 3}\right\rfloor\right) + \cdots $$
$$ = B = \left\lfloor\frac{m}{n}\right\rfloor - \left(\left\lfloor\frac{n}{m + 1}\right\rfloor - \left\lfloor\frac{n}{m + 2}\right\rfloor\right) - \cdots \le \left\lfloor\dfrac{n}{m}\right\rfloor \le \dfrac{n}{m}$$
$$\implies \left|\left\{\frac{n}{1}\right\} - \left\{\frac{n}{2}\right\} + \left\{\frac{n}{3}\right\} - \cdots - (-1)^n\left\{\frac{n}{n}\right\}\right| = |A - (-1)^mB + (-1)^mC|$$
$$ \le \frac{m - 2}{2} + \frac{n}{m} < \frac{\sqrt{2n} - 1}{2} + \sqrt{\frac{n}{2}} < 2n$$


TITLE: Interpretation of vector mesons in QCD
QUESTION [1 upvotes]: It is well-known that scalar mesons are interpreted as pseudogoldstone bosons which is connected with spontaneous broken $SU(3) \times SU(3)$ symmetry to $SU(3) \times SU(3) / SU(3)_{chiral}$.
Is there similar explanation for vector mesons? I understand that broken symmetry corresponds to existense of particles with spin zero, but I expect that vector mesons can also be explained through something like that, because they correspond to irreducible representation of $SU(3)$ group (as well as scalar mesons).

REPLY [1 votes]: A note on your question: the chiral SU(3)×SU(3) is dynamically broken by QCD to just the flavor ("eightfold way") SU(3). Your coset expression would be meaningful if the subscript were "vector" not "chiral", and it could serve as a summary of the 8 SSBroken axial generators of the full chiral starting group, not the surviving ones. So you might be asking for an explanation of the non-existent.
Indeed, 8 pseudoscalar mesons are the pseudogoldstone bosons corresponding to these 8 broken generators. And, indeed, glorious speculation has been proffered on vector mesons being, effectively, SSB gauge fields notably in the reference by Bando, Kugo, et al.
However, consider that any other mesons, scalars, tensors, etc., do not admit of such interpretations. If you are seeking a grand reductionist picture for all mesons in chiral symmetry breaking, it appears to be lacking---so far!


TITLE: Isolated zeros on closure of a domain
QUESTION [10 upvotes]: Let $f$ be an analytic function on the open unit disk domain $D$.  Suppose also that $f$ is bounded.
Since $f$ is bounded I believe that $f$ can be continuously extended to the closed unit disk.
I know that the zeros of $f$ in the open disk $D$ are isolated.   Are the zeros of $f$ in the closed unit disk also necessarily isolated?

REPLY [5 votes]: Jonas Meyer's answer goes much deeper, but let me say the following as well.
Even if your (nonzero -- you didn't say that, but of course you meant it!) function does extend continuously to the closed unit disk it need not have isolated zeros.  Indeed, suppose that $f$ has infinitely many zeros in the open unit disk.  Then the zero set of the (putative) extension of $f$ to the closed unit disk is an infinite subset of a compact space, so must have an accumulation point: i.e., there will be at least one point on the boundary which is a nonisolated zero.
In fact, by the Weierstrass Factorization Theorem, the zero set of an analytic function on the open unit disk can be any discrete subset without accumulation points in the open disk.  You can choose such a set to have the entire boundary $|z| = 1$ of the disk contained in its closure.  We conclude that there exists a nonzero analytic function $f$ on the open unit disk such that -- if it admits a continuous extension to the closed unit disk -- this extension is identically zero on the boundary.
I would be interested to know if this construction can be made unconditional, i.e., whether any analytic function with such a zero set can be extended continuously to the boundary.


TITLE: Adjoint of Complex Matrix
QUESTION [0 upvotes]: Assume A is a matrix with complex entries. Prove that all diagonal elements of
the matrix (A*)A are real where A* is the adjoint of A.
Where do I start here?
Thanks in advanced!

REPLY [2 votes]: Let $A \in \mathbb{C}^{m \times n}$ have matrix elements $(A_{ij})$ where $i$ denotes row position and $j$ denotes column position. Then we have that the adjoint $A^* \in \mathbb{C}^{n \times m}$ has elements $(\bar A_{ji})$, with $\bar A_{ji}$ denoting the complex conjugate of $A_{ji}$, and $i$ again denoting row position and $j$ column position.
So from the definition of matrix multiplication, we have the formula for a matrix element $x_{ij}$ of $A^*A$:

 $$x_{ij} = \sum_{k=1}^m \bar A_{ki} A_{kj}$$

When we calculate a diagonal element $x_{ii}$ with this formula, we obtain

 $$x_{ii} = \sum_{k=1}^m \bar A_{ki} A_{ki} = \sum_{k=1}^m |A_{ki}|^2$$


TITLE: Convexity with conditional probabilities
QUESTION [0 upvotes]: Consider the random variables $X$, $Y$, and $Z$ supported on $\mathbb{R}^3$ and with joint distribution $P_{XYZ}$.
Can I say anything about $P_{XY|Z}(x,y|z)\log P_{XY|Z}(x,y|z)$ being convex or concave on $z$?

REPLY [0 votes]: In general, no, not without a lot of additional information about the functional form of $P_{XY|Z}(x,y|z)$.


TITLE: Handling halogen lamps
QUESTION [7 upvotes]: When consulting manuals, electricians, online sources, etc., they always instruct you to handle halogen light bulbs with gloves.
The "explanation" that usually accompanies this statement is that oils and salts from a bare hand can "react" with the surrounding glass/quartz, owing to the fact that halogen bulbs are much hotter than ordinary ones. These reactions would cause weak spots in the quartz (or in some versions of the story, the filament), decreasing the lifetime of the bulb.
Is there any merit to this explanation? What are then the precise mechanisms involved in the bulb's degradation when the bulb is touched by a dirty monkey finger?

REPLY [2 votes]: The interesting point in halogen bulbs is that they managge to not, over time, cover the inside of the bulb glass with condensed vapour of the filament (1).
Therefore, the bulb can be smaller, as it's not needed to spread that metal over a large area.
Now, when touching the bulb, a fingerprint is left, which contains various carbon compounds. When switching on the bulb, it carbonizes, leaving slight dark residues (2).
Even if they are barely visible, they cover a significant fraction of the surface area, and make that area somewhat less translucent. That causes a reduction of light output over the livetime of the bulb,
adding up to a significant loss in overall efficiency of the halogen bulb.

On the alternative explanations:
The idea that somehow temperature differences are created, and cause a problem by weakening mechanical stability makes no sense - the kind of glass a halogen bulb is made of, quarz glas, is very stable under heat differences, as it changed it's volume only very little, compared to other glass.
(1) Instead, that vapour reacts with the halogen, to a compound that will subsequently split again into metal and halogen on the surface of the filament.
(2) I assume these carbon residues are long-lasting or permanent because the bulb surface is not hot enough to just burn them off.


text:
\section{Introduction}\label{sec:intro}
Persistent (co)homology has been the subject of extensive study in applied topology. Roughly speaking it is a homology theory for filtrations or filtered spaces. A landmark result in applied topology is that persistent (co)homology and more importantly persistence diagram are stable with respect to perturbations of the input filtration.  The classical result states:
\begin{theorem}[\cite{CohEdeHar2007}] Let $X$ be a triangulable space with continuous tame functions $f,g : X \rightarrow \R$. Then the persistence diagrams $\Dgm(f)$ and $\Dgm(g)$ for their sublevel set filtrations satisfy
$$d_B(\Dgm(f),\Dgm(g)) \leq ||f -g||_\infty$$
 where $d_B(\cdot) $ represents the bottleneck distance.
\end{theorem}
This result has been generalized to algebraic~\cite{bauer2014induced} and categorical settings~\cite{bubenik2015metrics}, with recent work strongly aimed at multiparameter and more general settings, particularly where classical notions of persistence diagrams do not exist.

Here we study the $p$-Wasserstein stability of persistence diagrams for $1\leq p \leq \infty$. This has been far less studied,  with existing results almost exclusively in terms of the classical results relating \emph{interleaving distances} between filtrations and the $\infty$-Wasserstein distance,  i.e.  bottleneck distance. Upper bounds on the $p$-Wasserstein distances are less common and often relying on bottleneck stability and providing pessimistic bounds. Furthermore, there is often a requirement of $p$ being sufficiently large.

Wasserstein distance for small values of $p$, i.e. $p=1,2$ rather than $p=\infty$, are important for a number of reasons. From the applications side, the 2-Wasserstein distance on persistence diagrams has been found to be much more effective than bottleneck distance. In stochastic topology, the bottleneck distance provides bounds which are often too coarse, as they are most often controlled by extreme points. $p$-Wasserstein stability provides a new tool with which to study the topology of probabilistic phenomena.

One of the main difficulties in establishing a $p$-Wasserstein bound is that \emph{interleaving}, a key tool in persistence stability results, cannot be applied.  Indeed, at first there does not seem to be a natural algebraic description of Wasserstein distance in the same way as for bottleneck distance.

However, here we take a fundamentally different approach to proving $p$-Wasserstein stability which at its core focuses on a cellular $p$-Wasserstein stability theorem. The proof exploits the local correspondences between coordinates of the points in the persistence diagram with critical cells in a filtration over a cellular complex. This cellular $p$-Wasserstein stability theorem then can be modified and applied to a variety of settings to prove a range of stability theorems.

In contrast, the stability of linear representations of persistent homology are usually stated of upper bounds in terms of the $1$-Wasserstein distance, for which the pre-existing stability results for $p$-Wasserstein distances cannot be applied. An important corollary of this paper is that we also get stability results for these linear representations.

Using the cellular stability theorem as inspiration, we also develop the algebraic theory to deal with Wasserstein distance for persistence modules.  The description of the distance is different to the notion of interleaving, being closer to the single morphism characterization from \cite{bauer2014induced}. Our generalized description recovers this result in the special case of bottleneck distance.

Using this theory, we prove the stabilty results in a more general setting, albeit with more technically involved proofs. We also show a Minkowski-type of result connecting the norms of a persistence modules, also called total persistence, in a short exact sequence. The upper bound follows naturally from our study of the Wasserstein distance, but surprisingly, there is a lower bound as well, giving a geometric result for group extensions of persistence modules for all $p\geq 1$.

In summary, the main contributions of this paper are:
\begin{enumerate}

  \item A cellular $p$-Wasserstein stability theorem, which also provides a simplified proof of bottleneck stability of diagrams for finite complexes.
  \item We apply the above theorem to produce stability theorems for a number of applications, including grey-scale images, persistent homology transforms, and Vietoris-Rips filtrations. As the 2 -Wasserstein distance is widely used in applications, this addresses a significant gap in the applied topology literature.
  \item We give an algebraic formulation of the $p$-Wasserstein distance and prove stability which applies to  pointwise finite dimensional modules with an additional technical condition,
  \item Upper and lower bounds relating the $p$-Wasserstein distance of diagrams of modules in a short exact sequence,
  \item Sufficient algebraic conditions which ensure that our algebraic definition yields a psuedo-distance without depending on the structure of persistence modules.
\end{enumerate}
The paper can be thought of as split into two main parts. After introducing the relevant preliminaries, the first part of the paper is focused on results which are immediately relevant for applications in topological data analysis. We present a number of examples which illustrate the inherent instabilities in persistence diagrams, followed by general cellular stability theorem.  We then discuss applications to the important cases of image analysis,  persistent homology transforms, Vietoris-Rips filtrations, and other topological summaries. The results are focused on explicit bounds for finite cases, which are the main interest in applications.
The results on Vietoris-Rips filtrations are interesting in their own right with a critical dependence on the underlying geometry.

The second part of the paper provides an algebraic perspective on Wasserstein stability. This includes a formulation which we show is equivalent to the Wasserstein distance for persistence diagrams but which applies to a wider class of modules. The main idea can be thought of the algebriac generalization of the cellular stability proof. This section is not required for the applications and so can be skipped by readers more interested in the practical implications of the results. Likewise, the algebraically minded reader need not go through the applications of the cellular stability theorem.


TITLE: Given n integers, show that there is a subset whose sum is divisible by n.
QUESTION [0 upvotes]: I’ve tried this.
Let the integers be $a_1, a_2,...,a_n$.
On dividing any number by $n$, we have $n$ possible remainders which are $0,1,2,...n-1$. Let these $n$ remainders be $n$ pigeonholes.
From the $n$ integers available, we can form $2^n - 1$ possible subsets, assuming each subset must contain at least one element. Each subset is assumed to be a pigeon.
$2^n - 1$ is always greater than or equal to $n$, for all values of $n$ greater than or equal to 1.
So, the pigeonhole of the remainder 0, must contain at least one element.
That element is divisible by $n$.
Now, when it contains more than one element, the sum of such elements will also be divisible by n, since the individual numbers are divisible by n.
I don’t think this is the right approach though. Any corrections and other methods will be appreciated.
Edit: This approach is completely wrong, as is my previous understanding of the pigeonhole principle. Answers and a link to a similar question are in the comments.

REPLY [3 votes]: This part is incorrect:

$2^n-1$ is always greater than or equal to $n$, for all values of $n$ greater than or equal to $1$. So, the pigeonhole of the remainder $0$, must contain at least one element.

That's not what the pigeonhole principle states:

In mathematics, the pigeonhole principle states that if $n$ items are put into $m$ containers, with $n > m$, then at least one container must contain more than one item.

Solution:
Lets examine the following subsets:
$\{a_1\}, \{a_1, a_2\}, \{a_1, a_2, a_3\}, ..., \{a_1, a_2, a_3\, ... , a_N\}$ and their sums' remainders. If any remainder is $0$ then we've solved the problem. Otherwise we have $N-1$ different remainders $(1, 2, 3, ..., N-1)$ and $N$ different subsets. By the pigeonhole principle we know that at least $2$ of those have the same remainder. Let those $2$ be $\{a_1, a_2, a_3\, ... , a_i\}$ and $\{a_1, a_2, a_3\, ... , a_j\}$ where $i < j$ and their remainder is $r$. We know that the sum of the subset $\{a_1, a_2, a_3\, ... , a_j\}$ is $\{a_{i+1}, , ... , a_j\}$ = $\{a_1, a_2, a_3\, ... , a_j\} - \{a_1, a_2, a_3\, ... , a_i\}$. Therefore the remainder is $r-r=0$ and it is divisible by $N$.


text:
\section{Relation is Symmetric iff Inverse is Symmetric}
Tags: Symmetric Relations

\begin{theorem}
A [[Definition:Relation|relation]] $\mathcal R$ is [[Definition:Symmetric Relation|symmetric]] {{iff}} its [[Definition:Inverse Relation|inverse]] $\mathcal R^{-1}$ is also [[Definition:Symmetric Relation|symmetric]].
\end{theorem}

\begin{proof}
Let $\mathcal R$ be [[Definition:Symmetric Relation|symmetric]].
Then from [[Relation equals Inverse iff Symmetric]]:
:$\mathcal R = \mathcal R^{-1}$
The result follows.
{{qed}}
\end{proof}

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