text:
\begin{document}

\title{Low-complexity eigenstates of  a $\pmb{ \nu = 1/3}$ fractional quantum Hall system}
\author{Bruno Nachtergaele}
\affiliation{Department of Mathematics
and Center for Quantum Mathematics and Physics,
University of California, Davis, CA  95616-8633, USA}

\author{Simone Warzel}
\affiliation{Munich Center for Quantum Science and Technology, and
Zentrum Mathematik, TU M\"{u}nchen,
85747 Garching, Germany}

\author{Amanda Young}
\affiliation{Munich Center for Quantum Science and Technology, and
Zentrum Mathematik, TU M\"{u}nchen,
85747 Garching, Germany}

\begin{abstract}
We identify the  the ground-state of a truncated version of Haldane's pseudo-potential Hamiltonian in a thin cylinder geometry as being composed of exponentially many fragmented matrix product states. These states are constructed by lattice tilings and their properties are discussed. We also report on a proof of a spectral gap, which implies the incompressibility of the underlying fractional quantum Hall liquid at maximal filling $ \nu = 1/3 $. Low-energy excitations
and an extensive number of many-body scars at positive energy density, but nevertheless low complexity, are also identified using the concept of tilings.
\end{abstract}

\maketitle

\section{Introduction}

Our ability to make systematic approximations of physically relevant states of quantum many-body systems rests on the relatively low complexity of these states. A useful measure of complexity is given by bipartite entanglement. Although generic states of quantum many-body systems are nearly maximally entangled \cite{hayden:2006}, ground states of Hamiltonians with short-range interactions are expected to generically satisfy an area law for the bipartite entanglement, or an enhanced area law, and this has been proved for a number of important classes of systems \cite{cramer:2006,wolf:2006,hastings:2007}. The exact ground states of frustration free models provided by Matrix Product States (MPS) and Tensor Network States (TNS) \cite{fannes:1992,perez-garcia:2007} make
this low-complexity aspect explicit. The first examples of MPS and TNS were given in \cite{affleck:1987a,affleck:1988}, where they were called Valence Bond Solid states, a name which itself suggested low complexity and limited entanglement. The success of DMRG inspired numerical approaches is also best understood based on its connection with MPS and TNS \cite{cirac:2009,schollwock:2011}.

In this letter, we report on a recent detailed investigation of the ground state, spectral gap, and incompressibility of a popular model for the Fractional Quantum Hall Effect (FQHE) in a $1/3$ filled system: a truncated version of Haldane's pseudo-potential Hamiltonian \cite{PhysRevLett.51.605}, which is defined in the one-dimensional
lattice and describes the thin cylinder regime. On an interval $ \Lambda =[a,b]  $ the Hamiltonian  is \cite{PhysRevLett.54.237,Pokrovsky_1985,Trugman:1985lv,Bergholtz_2006,Nakamura:2012bu}
\begin{equation}\label{eq:Ham}
H_\Lambda =  \sum_{x= a}^{b-2}  n_x n_{x+2} + \kappa \sum_{x= a}^{b-3} q_x^* q_x  ,
\end{equation}
where

$n_x := c_x^* c_x$,

$q_x := c_{x+1}  c_{x+2} -  \lambda \ c_{x}  c_{x+3}$,

and $c_x$ is the annihilation operator for a spinless fermion at site $x$. Equivalently, the model can be regarded as a
spin-$1/2$ chain. The parameters $\kappa$ and $\lambda$ are given in terms of the dimensionless parameter $\alpha=\ell/R$, the ratio
of the magnetic length and the radius of the cylinder geometry of the system: $\kappa = e^{3\alpha^2/2} / 4 $ and $ \lambda = -3 e^{-2\alpha^2} $.
In what follows, we take model parameters $ \kappa > 0 $ and $ \lambda \in \mathbb{C}$ which are arbitrary unless otherwise stated.

The Hamiltonian~\eqref{eq:Ham} is an example of a particle-number $ N = \sum_x n_x $ and dipole-conserving  $ \sum_x x n_x $  model.
Its spectrum is invariant under $ \lambda \to - \lambda $ as is seen by unitarily transforming $ c_x  \to - c_x $ on every 4th site $ x $ of the chain.

In addition to proving a spectral gap, we investigate the excitation spectrum. Using the concept of tilings, which we use to classify an orthogonal basis for the ground state space, we identify invariant subspaces of the Hamiltonian and prove asymptotically sharp bounds on low-lying excitations. Many-body scars of higher energy but bounded Schmidt rank are also constructed using tiling states.

\section{Ground states -- tilings and fragmented MPS}

In the Tao-Thouless (TT) limit $ \lambda = 0 $ \cite{Tao:1983cd}, the ground-state at maximal filling $ \nu = 1/3 $ is given by the tensor state $ \ket{(100)_L} $ of $ 3$-periodic particle configurations which repeat $ L $-times the basic particle pattern (100) indicating the presence of a particle at every third site. A ground-state of \eqref{eq:Ham} with $ |\Lambda | = 3 \, L  $ at maximal filling is obtained by squeezing
\begin{equation}\label{eq:sTTstate}
\vp_L :=   \prod_{k = 0  }^{L-2} \left( 1 + \lambda c_{3k+2}^* c_{3k+3}^* \, c_{3k+4}  c_{3k+1}  \right)  \ket{(100)_L}  .
\end{equation}
Shifting the reference state $ (100)_L $ by 1 or 2 lattice sites produces two other ground states.
The full ground-states space $ \ker H_\Lambda=  \{ \psi | H_\Lambda\psi = 0 \} $ contains exponentially many other states  of lower filling.
The set of particle configurations, which support ground states, is constructed
by concatenating voids written as (0), and monomers (100). In addition, at the left and right boundary of the finite chain some additional configurations are allowed: (11000) at the left boundary and (011), (10), and (1) at the right boundary. Concatenating these basic patterns or {\em tiles} produces a {\em root tiling} and configurations obtained in this way are called {\em root configurations}. Each root tiling $ R $ corresponds a basis vector of $ \ker  H_\Lambda $.
These vectors are superpositions of many-particle configurations, namely all those that correspond to a compatible tiling of the finite chain in which pairs of successive monomers are replaced with dimers, i.e.
\[  (100)(100) \to (011000) . \]
At the right boundary special replacement rules apply: $(100)(10) \to (01100)$ and $(100)(1) \to (0110)$. For every root tiling $R$ we denote by $ \mathcal{D}_\Lambda(R) $ the
compatible tilings with Voids, Monomers, and Dimers (VMD tilings, for short), obtained by making the allowed subsitutions as described above. The ground state vector corresponding to $R$ is defined as:
\begin{equation}\label{eq:VMDstate}
	\psi_\Lambda(R) := \sum_{\pmb{D}  \in  \mathcal{D}_\Lambda(R) } \lambda^{\#( \pmb{D} )}  \ket{  \pmb{\sigma}(\pmb{D}) } .
\end{equation}
The sum extends over the collection of VMD tilings  $ \pmb{D}  \in\cD_\Lambda(R)$. Each VMD tiling $  \pmb{D} $ is associated with unique particle configuration $  \pmb{\sigma}(\pmb{D})  $ and $ \#( \pmb{D} ) $ counts the number of dimers in the tiling.

\begin{theorem}[Ground State Space]
For all $|\Lambda|\geq 8$, the set of $\psi_\Lambda(R)$ labeled by the root tilings $R$ is an orthogonal basis of the ground state space of $H_\Lambda$. The dimension of the ground state space grows as $a^{|\Lambda|}$, with $a\approx 1.47$.
\end{theorem}

For more detailed information on this and other properties of ground state spaces, its dimension, and factorization formulas presented next, see
\cite{nachtergaele:2020}.

A point of interest is the maximal filling of $1/3$. In fact, it follows easily from tiling structure that the largest particle number for a ground state, $N_\Lambda^{\rm max}$, satisfies
\begin{equation}\label{eq:maxNumber}
	 \frac{1}{3} \leq \frac{ N_\Lambda^{\rm max}}{|\Lambda|} \leq  \frac{1}{3} + \frac{4}{3|\Lambda|} .
\end{equation}
Therefore, the maximal filling in the ground state is asymptotically equal to 1/3.

The squeezed TT-state~\eqref{eq:sTTstate} is the state from~\eqref{eq:VMDstate} generated by the root $ R= (100)_L $. This state has Schmidt rank 2 across any bipartition of the lattice, which is most easily seen through the recursion relation
\begin{equation}\label{eq:recrel}
\varphi_{L_1+L_2} = \varphi_{L_1} \otimes  \varphi_{L_2} + \lambda \varphi_{L_1-1} \otimes | 011000 \rangle \otimes \varphi_{L_2-1} .
\end{equation}
The correlation length of this state in the infinite volume limit was estimated in~\cite{Jansen:2009gv,Jansen:2012da} and identified in~\cite{Nakamura:2012bu} as
\[ \xi(\lambda) := 3/\ln \frac{ \sqrt{4|\lambda|^2 +1}+1}{\sqrt{4|\lambda|^2 +1}-1}. \]

Up to  boundary tiles $ B_l $ and $ B_r $, which give rise to states $ \psi_{\partial^l \Lambda}(B_l) $ and  $\psi_{\partial^r \Lambda}(B_r) $,
any state $\psi_\Lambda(R)$
is a tensor product of squeezed TT-states interspersed with
sites where the state is $\ket{0}$. The latter occur where the root tiling has voids $(0) $.
The state can thus be regarded as a concatenation of intervals of length $3L_j$ on which the state is  $\vp_{L_j}$, and intervals in between where the
state is the vacuum, i.e.
\[
\psi_\Lambda(R) = \psi_{\partial^l \Lambda}(B_l) \otimes \varphi_{L_1} \otimes | 0 \rangle \otimes  \dots   | 0 \rangle   \otimes  \varphi_{L_n} \otimes  \psi_{\partial^r \Lambda}(B_r) .
\]
 We call this structure a {\em fragmented matrix product state} (FMPS) since each fragment $ \vp_L $ (as well as the vacuum and boundary states) is an MPS \cite{Nakamura:2012bu}.

One consequence of this factorization property, is the exponential decay of correlations $ \langle \cdot \rangle_{\Lambda,R} $ in each of the states $\psi_\Lambda(R)$ proven in~\cite{nachtergaele:2020}.

\begin{theorem}[Exponential clustering] \label{thm:expcluster}
There are positive constants $C$ and $\xi\leq 2 \xi(\lambda) $ such that for all finite chains $\Lambda$ and root tilings $R$ the correlations in the state $\psi_\Lambda(R)$ satisfy
$$
|\langle AB\rangle_{\Lambda,R} - \langle A\rangle_{\Lambda,R} \langle B\rangle_{\Lambda,R}|\leq C \Vert A\Vert \Vert B\Vert e^{-\frac{d(\supp A, \supp B)}{\xi}} .
$$
\end{theorem}

Another immediate consequence of the FMPS structure and~\eqref{eq:recrel}  is the uniform bound $ 2 $ on the Schmidt rank of $\psi_\Lambda(R)$  across any cut in the chain. (Hence the entanglement entropy is bounded by $\log 2$.)

In view of the results presented next,
one could say that each of these states
has the properties expected of gapped ground states based on general results about exponential clustering~\cite{hastings:2006,nachtergaele:2006a} and areas law in one dimension \cite{hastings:2007,brandao:2015a}.
At the same time, however, the ground state space also contains vectors with extensive entanglement and states with arbitrarily slow decay of correlations,
cf. Sec. 4.3 in \cite{nachtergaele:2020}. Given the large degeneracy of the ground states, this is not surprising, but it does raise an interesting
question about the excited states: is there a basis of low-complexity eigenvectors of the model at all
energies or a particular range of energies? By constructing eigenstates that explain some prominent features of the low-lying energy
spectrum we obtain evidence in support of the existence of a low-complexity basis of excited states.

\section{Spectral gap and incompressibility}

An essential ingredient in the description of the FQHE is the gap in the spectrum above the ground state. In \cite{nachtergaele:2020} we proved the following lower bound for this gap which is uniform in $\Lambda $. The estimate is expressed in terms of the gap for chains of length $L=8,9,10$, which can be easily calculated numerically, and a function $f(|\lambda|^2)$. This function is defined and analyzed in  \cite{nachtergaele:2020}. Its essential properties can be read off from the plot given in Fig.~\ref{fig:f}.

\begin{figure}
\includegraphics[width=.4\textwidth]{f_with_labels.jpg}
\caption{Plot of the function $|\lambda|\mapsto f(|\lambda|^2)$.} \label{fig:f}
\end{figure}

\begin{theorem}[Uniform spectral gap]\label{thm:gap}
	There is a monotone increasing function $ f : [0,\infty) \to  [0,\infty) $ such that for all $ \lambda \in \mathbb{C} $ with the property $  f\left(|\lambda|^2\right)  < 1/3  $ and all $ \kappa > 0 $
	\begin{equation}\label{eq:gaplower}
\gap H_\Lambda \geq \left( \min_{L\in \{8,9,10\}} \gap H_{[1,L]} \right)   \frac{\left(1-  \sqrt{3 f\left(|\lambda|^2\right)}\right)^2}{3}
	\end{equation}
	for any interval $\Lambda$ of length $L\geq 9$.
\end{theorem}
For open boundary conditions, the gap of \eq{eq:Ham} vanishes when $\lambda\to 0$, cf.\ Fig.~\ref{fig:spec}. This is caused by spurious boundary states (which are also discussed in~\cite{PhysRevB.85.155116})
such as the eigenstate with lowest energy $E=\mathcal{O}(|\lambda|^2) $ in the 2-dimensional Krylov space $ \spa \{ H_{\Lambda}^n \ket{11001}\otimes \ket{(0)_{|\Lambda| -5}} \} $.
The bulk gap is not expected to vanish as $\lambda\to 0$. We discuss the presumed lowest energy excitations in the next section. Finite systems with periodic boundary conditions show the behavior of the bulk: the gap does not vanish in the limit  $\lambda\to 0$ (see  Fig.~\ref{fig:spec}).

\begin{figure}
\includegraphics[width=.5\textwidth]{pbc_L12_kappa2648_Eupper_labels.jpg}
\includegraphics[width=.5\textwidth]{openbc_L12_kappa2648_Eupper_labels.jpg}
\caption{Excitation spectrum as a function of $\lambda \in [0,2]$, for a chain of length $12$ with periodic (top) and open (bottom) boundary conditions with a fixed value of $\kappa = 2.648$. The red curve is the variational upper bound of the first excited state given in (\ref{eq:upper_bound}). } \label{fig:spec}
\end{figure}

Denoting by $E_L(N)$ the lowest energy state of $H_\Lambda$ with particle number $N$ on an interval of length $|\Lambda| =L$, the compressibility is defined by
\begin{equation}
\frac{1}{\kappa_L(N)}  :=   L  \,  \frac{E_{L+1}(N) + E_{L-1}(N)- 2 E_L(N)}{(2\pi \ell^2 )^2} \, .
\end{equation}
As a consequence of the gap estimate, for fixed $\lambda$ the compressibility vanishes as $c/L$ in the maximally filled state.

\begin{cor}[Incompressibility]\label{cor:incompress}
	At zero temperature and critical filling, i.e.~$ N =  N_\Lambda^{\rm max} $, the compressibility $ \kappa_{\Lambda}(N) $ vanishes in the thermodynamic limit $  |\Lambda| \to \infty $.
\end{cor}

\section{Excited states}

The lowest neutral excitation in the $ \nu = 1/3 $  FQHE system is predicted to be a bound-state (`magnetoroton') of a quasi-particle of fractional charge $ -1/3 $ with quasi-hole of charge $ 1/3 $ \cite{PhysRevLett.54.581,PhysRevLett.108.256807}.
As suggested by the numerical analysis in \cite{Nakamura:2012bu,wang:2015} as well as Fig.~\ref{fig:spec}, this remains true for the truncated Hamiltonian~\eqref{eq:Ham} in the bulk.
In the TT-limit $\lambda =0$, the first excited energy state is  obtained by introducing a kink $ (010) $ into the list of monomers. To show that a state with such features persists for small $|\lambda|>0$ and $ \kappa > 1 $, we look at the Krylov spaces $\mathcal{K}_{L,R}^{(l)}$ generated by $\vp_L \otimes \ket{(010)_l} \otimes\vp_R$, where $L$ and $R$ count the number of monomers to the left and right of the kink region, in which $ (010) $ is repeated $ l \in \mathbb{N} $ times.
Due to the domain wall created by $ (0) $ on the left, this space factorizes as $ \mathcal{K}_{L,R}^{(l)} = \vp_L \otimes \ket{0} \otimes \mathcal{K}_R^{(l)}  $ where
\begin{equation}\label{eq:krylov}
\mathcal{K}_R^{(l)} = \spa \{ H_\Lambda^n \left(  \ket{ (100)_{l-1}(10) }\otimes \vp_R \right) \, \big| \, n\geq 0\}
\end{equation}
and we identify $\Lambda=[1,3(R+l)-1]$. The choice of $ l\in \mathbb{N}  $ corresponds to shifting the dipole moment. It was noted in~\cite{Nakamura:2012bu} (and~\cite{PhysRevLett.108.256807} for the full model), that the lowest excitation occurs at $ l = 2 $.

It is natural to embed  $\mathcal{K}_R^{(l)}  $ in the space $\mathcal{T}_R^{(l)} $ spanned by the tiling configurations generated from applying the replacement rules to the root tiling $(100)_{l-1}(10)(100)_R$. When $ l = 1 $, the presence of $(10) $ generates one additional tile of length 8 obtained from the rule:
\begin{equation}\label{eq:tilings1}
(10)(100)(100) \to (10)(011000) \to (01101000).
\end{equation}
The deformation is thus localized to the two monomers following (10) and any tiling to the right of this region is a regular monomer-dimer tiling.
For $ l \geq 2 $ we get two additional tiles from the replacement rules:
\begin{equation}\label{eq:tilings2}
(100)(10)(100) \to  (01100)(100) \to (01011000).
\end{equation}

To study the lowest excitation, we use that $\mathcal{T}_R^{(2)}$ is orthogonal to the ground states, so that minimizing the energy $E(\sigma,\tau)$ over $\sigma, \tau \in\mathbb{C}$ of the variational state
\begin{equation}\label{eq:approx_estate}
\Psi_{\sigma,\tau} =   \ket{10010}\otimes \left( \vp_R + \sigma \eta_R \right) + \tau \ket{01100}\otimes  \vp_R  \in \mathcal{T}_R^{(2)},
\end{equation}
produces an upper bound on the lowest excited energy.
Here $\eta_R := -\bar{\lambda} \alpha_{R-1}^{1/2} \ket{100}\otimes \vp_{R-1} +  \alpha_{R-1}^{-1/2}\ket{011000}\otimes \vp_{R-2}$ has norm $ \| \eta_R\| = \| \vp_R \| $ and is orthogonal to $\vp_R= \ket{100}\otimes \vp_{R-1}  + \lambda   \ket{011000}\otimes \vp_{R-2}$ and
\begin{equation}\label{eq:alpha}
\alpha_R := \frac{ \|\vp_{R-1}\|^2}{\|\vp_R\|^2} \stackrel{R\to \infty}{ \longrightarrow}  \frac{2}{1+\sqrt{1+4|\lambda|^2}} .
\end{equation}
Using~\eqref{eq:recrel}, one finds that the Schmidt-rank of  $\psi_{\sigma,\tau}$ is at most $4$ for any bipartition.
For $ \kappa > 1 $ and small $ |\lambda| $, where $ a :=  \alpha_R(1+\kappa\alpha_{R-1}|\lambda|^4) + \kappa |\lambda|^2 < \kappa $,  the variational minimum of the energy is asymptotically attained at
\begin{align*}
\sigma_{\min} & := \frac{\lambda  \alpha_R   \alpha_{R-1}^{1/2} (1-\kappa|\lambda|^2)}{b-a} =  \frac{\lambda}{\kappa-1} \left(1+\mathcal{O}(|\lambda|^2)\right) \\
\tau_{\min}  & := \frac{\kappa \lambda}{c - a}   =  \frac{\kappa \lambda}{\kappa-1} \left(1+\mathcal{O}(|\lambda|^2)\right)
\end{align*}
where $ b := \kappa + \kappa(1+\alpha_R+\alpha_{R-1}) |\lambda|^2 + \alpha_R\alpha_{R-1}|\lambda|^2 $

and $ c := \kappa + \kappa \alpha_R |\lambda|^2 $. This yields:
\begin{align}\label{eq:upper_bound}
\min_{\Psi\in \mathcal{T}_R^{(2)}} & \frac{\langle \Psi , H_{\Lambda} \Psi \rangle}{\| \Psi \|^2}  \leq  \  E(\sigma_{\min} ,\tau_{\min}) \nonumber \\
= & \ a -  \frac{(b-a)  |\sigma_{\min}|^2 + (c-a)   |\tau_{\min}|^2 }{ 1+  | \sigma_{\min}|^2 +   |\tau_{\min}|^2} .
\end{align}
This bound shares the asymptotic parabolic behavior  $ 1 - \frac{2\kappa}{\kappa-1} |\lambda|^2 + \mathcal{O}(|\lambda|^4) $
exhibited by the first exited state in Fig.~\ref{fig:spec} for small $|\lambda| $.

To show the bound in~\eqref{eq:upper_bound} is sharp up to $ \mathcal{O}(|\lambda|^2) $, we note that $\mathcal{T}_R^{(2)}$ is invariant under both $ H_{\Lambda}$ and  its lower bound $H_{[1,11]}$.
The latter is block diagonal in the configuration basis of $\mathcal{T}_R^{(2)}$ with all non-trivial blocks given by one of two matrices, $h_6$ and $h_3$. The lowest eigenvalue belongs to the $6 \times 6$-matrix $h_6$ which is obtained from the action of $H_{[1,11]}$ on the 6 tiling configurations generated from $(100)(10)(100)_2$, see \eqref{eq:tilings1}-\eqref{eq:tilings2}.
The minimal energy $E$ can then be seen to coincide with~\eqref{eq:upper_bound} up to second order $ \mathcal{O}(|\lambda|^2) $, e.g., by using the characteristic polynomial $p(x) = \det[h_6-(x+1+2\kappa|\lambda|^2)\idty]$ to bound
\begin{equation}
\min_{\Psi\in \mathcal{T}_R^{(2)}} \frac{\langle \Psi , H_{\Lambda} \Psi \rangle}{\| \Psi \|^2}  \geq E \geq 1+2\kappa|\lambda|^2-\frac{p(0)}{p'(0)}.
\end{equation}

The same procedure can be applied to asymptotically estimate the minimum of the spectrum of $ H_\Lambda $ on other sectors of quasi-momenta. E.g., on $ \mathcal{T}_R^{(1)}$ one can similarly show that $ \min_{\Psi\in \mathcal{T}_R^{(1)}} \frac{\langle \Psi , H_{\Lambda} \Psi \rangle}{\| \Psi \|^2}  = 1 - \frac{\kappa}{\kappa-1} |\lambda|^2 + \mathcal{O}(|\lambda|^4) $, which is clearly greater than~\eqref{eq:upper_bound}.\\

Bulk excitations of low complexity other than the magnetoroton can be constructed by decoupling a finite-volume excited state that is separated by pair of domain walls from two ground states.
E.g., the eigenvalues in Fig.~\ref{fig:spec} which are constant in $ \lambda $ are constructed by putting a void $ (0) $ on both sides of $(10)(100)$. An excited state with energy $E=1$ is then given by
\begin{equation}
\psi = \vp_{L} \otimes \ket{0101000} \otimes \vp_{R} .
\end{equation}
Similarly, placing voids around each of $N$-pairs $(10)(100)$ and extending the state to the full system by tensoring with appropriate ground states produces an excited state with energy $N$. Such states all have Schmidt-rank~$ 2 $.

Other excited states can similarly be generated as long as their boundary does not resonate with the domain wall. However, the minimal number of voids needed to construct a sufficiently large domain wall can vary. For example, the highest energy state $\ket{1}^{\otimes N}$ on $N$ sites requires one void on the left and three voids on the right:
\[
\psi = \vp_L \otimes \ket{0111\ldots 1000} \otimes \vp_R.
\]
This produces an eigenstate with energy $ N-2+(N-1)\kappa + (N-3)\kappa|\lambda|^2$.

\section{Discussion}

Motivated by the above description of the ground-state and the magnetoroton, we conjecture the existence of a complete set of low-energy eigenstates with low-complexity.
This would be in accordance with the hierarchical construction of excited states in this model~\cite{Bergholtz_2006}; see also \cite{PhysRevLett.51.605,PhysRevLett.52.1583}.

Our construction of strictly local eigenstates at medium of high-range in energies is akin to that of the `frozen states'
in related dipole-conserving models~\cite{Pollmann19b}. For the model in which one drops in~\eqref{eq:Ham} all electrostatic interactions, a comprehensive discussion of such eigenstates is found in~\cite{arXiv:1906.05292v1,Bernevig10.19}.  Such eigenstates are known as many-body scars and have been identified recently in many other systems \cite{PhysRevB.101.174308,PhysRevB.101.195131}. They violate the Eigenfunction Thermalization Hypothesis (ETH):
although they share common global quantum numbers,
expectations of local observables differ. Nevertheless, the model is not integrable nor is it, as far as we know described by Local Integrals of Motion (LIOMs), as in case of many-body localized systems.

The states discussed here are always of low-complexity and hence should be easily experimentally accessible. The description is the guided by the same principle employed for the ground states and in~\cite{Bernevig10.19}.
Using a root-tiling we generate a set of configurations (and hence tensor-product states) via replacement rules that constitutes an invariant subspace of the Hamiltonian.

\begin{acknowledgments}
This work was supported by the DFG (Germany) under EXC-2111--390814868 and by the National Science Foundation (USA) under Grant DMS-1813149.
\end{acknowledgments}

config: arxiv


TITLE: A Hausdorff space such that all of its subspaces are compact.
QUESTION [0 upvotes]: As I understand the definition of a subspace, I can simply take any subset of a topological space and promote it to a subspace by endowing it with the subspace topology. Then in this case every subset of my Hausdorff space, endowed with the subspace topology for whatever topology the space has, is compact. Therefore every subspace is a closed subset. I would like for this generalization to be true since from here I think I can go on to talk about the topology on my Hausdorff space being the discrete topology and the space being finite.

REPLY [0 votes]: Let $A \subset X$. Then by assumption, $A$ (in the subspace topology, what else?)
is compact. A compact subspace in a Hausdorff space is closed. So $A$ is closed.
So all subsets of $X$ are closed, so all subsets have a closed complement as well, so all subsets are open too. Hence $X$ is discrete etc.


TITLE: Ideal convegerence, Ideal of sets, sequence convergerence
QUESTION [1 upvotes]: What is the connection between Ideal convergerence and 'sequence convergerence'
We say that a sequence of reals is $I$-convergerent to x if and only if for each $\epsilon>0$:
$\left\{n \in \omega \left|x_{n}-x \right| \ge \epsilon  \right\} \in I $
classical convergerence: Convergent sequence

REPLY [1 votes]: It seems that this notion was first introduced in the context of statistics/fuzzy logic; in particular, if you define the ideal to be $$I = \{ A \subset \mathbb{N} : \lim_{n\to\infty} \frac{|A \cap \{1,\dots,n\}|}{n} = 0 \},$$ then you get a notion of "statistical convergence", which is basically a notion of convergence that lets you ignore a bunch of terms in the sequence, and see what the "majority" of elements are doing. More specifically, a sequence $\{a_n\}$ statistically converges to $a$ if the "asymptotic density" of the elements that aren't converging to $t$ is $0$ no matter how far you zoom in, i.e. $$\lim_{n\to\infty} \frac{1}{n}|\{n \in \mathbb{N} : |a_n-a| \geq \epsilon \}| = 0 \text{ for all } \epsilon > 0.$$
On the other hand, if you define $I$ to be all finite subsets of $\mathbb{N}$, you can check that this is the usual notion of sequence convergence.
Sources:
http://thales.doa.fmph.uniba.sk/katc/publ/salat_i_cong.pdf
http://arxiv.org/pdf/math/0612179.pdf


TITLE: Show that if $f \neq 0$ and $x < y \Rightarrow f(x) \le \frac {f(y) - f(x)}{y - x } \le f(y)$ then $f$ is differentiable
QUESTION [3 upvotes]: I have the following problem :

$f \neq 0$ satisfies $x < y \Rightarrow f(x) \le \frac {f(y) - f(x)}{y - x } \le f(y)$  then $f$ is differentiable

Try
For $h > 0, f(x) \le \frac {f(x +h) - f(x)}{h} \le f(x + h)$.
By squeeze theorem, $\frac {f(x +h) - f(x)}{h} \le f(x + h) \to f(x)$ as $h \to 0^+$, but this assumes continuity of $f$. I'm stuck at showing $f$ is continuous.

REPLY [2 votes]: $f(x)(y - x) \le f(y) - f(x) \le f(y)(y - x)$.
Thus as $x \to y, f(x) \to f(y).$
From such considerations and that $f$ is ascending,
continuity should be forth coming.


TITLE: Is associated graded algebra $\mathrm{gr}(k[x_1, \ldots, x_n]/I)$ isomorphic as a vector space to $k[x_1, \ldots, x_n]/I$?
QUESTION [0 upvotes]: Let $A=k[x_1, \ldots, x_n]$ be the polynomial ring generated by $x_1, \ldots, x_n$. Let $I$ be an ideal of $k[x_1, \ldots, x_n]$ (it is possible that $I$ is not homogeneous).
The algebra $A$ is a graded algebra: $A = \oplus_{i \ge 0} A_i$, where $A_i$ consists of degree $i$ homogeneous polynomials.
The algebra $A/I=k[x_1, \ldots, x_n]/I$ is a filtered algebra with the filtration $F_i(A/I) = (F_i(A)+I)/I$, where $F_i(A)=\oplus_{j \le i} A_j$.

Is associated graded algebra $$\mathrm{gr}(k[x_1, \ldots, x_n]/I)=\mathrm{gr}(A/I)=\oplus_{i \ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) \cap I)$$ isomorphic as a vector space to $k[x_1, \ldots, x_n]/I$?

Thank you very much.

REPLY [1 votes]: Yes.
Observe that $\mathrm{gr}(A/I)$ is also a filtered algebra, with $F_i(\mathrm{gr}(A/I)) = \oplus_{i=0}^r~\mathrm{gr}(A/I)_i$.
Further observe that $$\dim F_i(\mathrm{gr}(A/I) = \sum_{i=0}^r \dim \mathrm{gr}(A/I)_i = \sum_{i=0}^r (\dim F_i(A/I) - \dim F_{i-1}(A/I)) = \dim F_i(A/I).$$
Therefore $\mathrm{gr}(A/I)$ is isomorphic to $A/I$ as a vector space.


TITLE: Which polynomial equations of higher degree will have a solution formula?
QUESTION [4 upvotes]: A polynomial equation of degree greater than four will in general have no solution formula. But what are some typical cases one should be aware of as a practical person in which there are solutions?

REPLY [3 votes]: Another example that come up fairly often is $a x^{2n} + b x^n + c = 0$


TITLE: On the abstract bootstrap principle in the book "Nonlinear Dispersive Equations" by Terence Tao
QUESTION [3 upvotes]: In Terence Tao's book "Nonlinear Dispersive Equations", he gives the following "Abstract bootstrap priciple":
"Let $I$ be a time interval, and for each $t \in I$ suppose we have two statements, a "hypothesis" $H(t)$ and a "conclusion" $C(t)$. Suppose we can verify the following four assertions:
(a) (Hypothesis implies conclusion) If $H(t)$ is true for some time $t \in I$, then $C(t)$ is also true for that time $t$.
(b) (Conclusion is stronger than hypothesis) Is $C(t)$ is true for some $t \in I$, then $H(t')$ is true for all $t'\in I$ in a neighborhood of $t$.
(c) (Conclusion is closed) If "$t_1, t_2,\ldots$" is a sequence of times in $I$ which converges to another time $t\in I$, and $C(t_n)$ is true for all $t_n$, then $C(t)$ is true.
(d) (Base case) $H(t)$ is true for at least one time $t \in I$.
Then $C(t)$ is true for all $t\in I$."
My question is about (a) and (b), according to his remarks and examples, I noticed that (a) is not the real "Hypothesis implies conclusion" . I mean $H(t)$ is actually not stronger than $C(t)$, that is if we want to deduce $C(t)$ from $H(t)$, we must employ some other hypothesis or conditions in this system which we are investigating, not just $H(t)$ ($H(t)$ is actually not sufficient!). Is this understanding of this principle right? I want to be sure.
Also in his remarks and examples, he makes (b) truely hold, I mean we don't need to exploit another hypothesis to prove that. I also want to ask: can we make (b) also not very exactly? I mean just like above I discussed about (a), we need also introduce some other hypothesis to prove (b) not just $C(t)$. Is that still correct? I consider this because in the proof of this principle, I didn't see any necessity to make the implication of (a) and (b) hold strictly. The main purpose of (a) and (b) in the proof is to ensure $\Omega$ open.
The following appendix is from Tao's book.
Appendix:

REPLY [3 votes]: The abstract bootstrap principle says nothing more that if you can verify (a) - (d), then $C(t)$ is true in all of the interval $I$. It is entirely agnostic to how you can verify that (a)-(d) are true.
You should compare this to the tool of induction, which we can write in the following form:

Induction: Suppose you have a function $P$ mapping the natural numbers $\mathbb{N}$ to $\{0,1\}$ (or boolean true/false values). Let the hypothesis $H(t)$ and $C(t)$ both represent the statement $P(t) = 1$.

(Hypothesis implies conclusion) $H(t)$ implies $C(t)$.
(Conclusion is stronger than hypothesis) $C(t)$ implies $H(t+1)$.
(Conclusion is closed) If $t_1, \ldots$ is a sequence in $\mathbb{N}$ such that $t_i \to t_\infty$, and if $C(t_i)$ are all satisfied, then $C(t_\infty)$ is also satisfied.
(Base case is true) There exists $t_0\in\mathbb{N}$ such that $H(t_0)$ is true.

then $C(t)$ is true for all $t \geq t_0$.

Note that in this presentation of mathematical induction, the first and third items are trivial.
Note further in the statement of mathematical induction: 1, 2, 3, 4 are assumed to hold. Given an arbitrary function $P$ they do not have to hold always. But when they do hold, the conclusion can be reached.

To return to your question:
Do not take the splitting of (a) and (b) too seriously.
Let us return to the case of mathematical induction: You can redefine $H(t)$ to be the statement $P(t) = 1$, while $C(t)$ to be $P(t+1) = 1$. In this formulation, that the conclusion is stronger than hypothesis is trivial, while the hard step becomes proving that hypothesis implies conclusion.
So perhaps a slightly better way to think about the bootstrap principle is the following three-condition description:

Abstract bootstrap principle. Let $C:I \to \{0,1\}$ be a mapping, where $I\subset \mathbb{R}$ is some interval. Suppose we can show that

(Conclusion implies better than conclusion) If $C(t_0) = 1$ for $t_0\in I$, then there exists some $\epsilon > 0$ such that $C(t) = 1$ for $t\in (t_0 - \epsilon,t_0 + \epsilon) \cap I$.
(Conclusion is continuous) If $t_1,\ldots$ is a sequence in $I$ such that $t_i \to t_\infty$. If furthermore $C(t_i) = 1$ we can conclude $C(t_\infty) = 1$.
(Conclusion holds somewhere) There exists some $t_0\in I$ such that $C(t_0) = 1$.

then we can conclude that $C\equiv 1$ on $I$.

(Note that here I have cleverly hidden any notion of "implication". The three conditions, for the purpose of the bootstrap principle, are hypotheses you impose on the function $C$. The bootstrap principle is agnostic to how those hypotheses are satisfied: as long as you are given a function $C$ such that those conditions hold, you obtain the final conclusion. In applying the abstract bootstrap principle you have to of course verify that the three statements hold. The degree to which each one is obvious or universal depends on the precise problem you study and the definition of the function $C$ of course. But quite often both the first and second need to be shown using the properties of the PDE you are examining, while the third you can often get away by postulating it in the initial data.)
This now brings us to the fundamental form of the bootstrap principle, which is nothing more than the continuity principle from analysis and topology.

Continuity principle Let $X$ be a connected topological space. If $Y\subset X$ is a non-empty, open, and closed subset, then $Y = X$.

To derive the bootstrap principle from the continuity principle, we observe that $I$ is a connected topological space with the usual topology. Let $B = \{0,1\}$ with the discrete topology. Let $Y = C^{-1}(1)$. By the first condition $Y$ is open. By the second condition $Y$ is closed. By the third condition $Y$ is nonempty. Hence $Y = X$.

Going back to the splitting of (a) and (b): generally to prove that "conclusion implies better than conclusion" what one does is to

Start with $C(t)$.
Prove (here on the word prove implicitly means that you are using the properties of the system/PDE you are studying!) $C(t) \implies$ some other statement $C_1(t)$.
Prove $C_1(t) \implies C_2(t) \implies\ldots \implies C_k(t)$.
From $C_k(t)$ you can appeal to local existence of your PDE to get that $C$ holds on a neighbourhood of $t$.

Note that from steps 1 - 3 we can immediately conclude that $C_1,\ldots C_k$ all hold on a neighbourhood of $t$.
That is to say: the "circular" structure of the argument makes it not necessary to identify any one of the $C_i$ as the "hypothesis" and any one of the $C_i$ as the "conclusion". So you do not have to force yourself to think in terms of "hypotheses" and "conclusions".


TITLE: a non- empty set A of a Banach space is bounded<=> there exists a real number k such that llxll <= k for every x in A.
QUESTION [0 upvotes]: How to show that, a non- empty set A of a Banach space is bounded<=> there exists a real number k such that  llxll <= k for every x in A.
Note:
A set A is bounded if diameter of A is finite.
Diameter of A= sup { d(x,y), x,y€A}.
For <= part, I have deduced ,
 diameter of A<= 2k by means of triangle inequality and I think this part does not even need the set to be a subset of the Banach space.
But I am having trouble to prove => part.Any help would be appreciated.Thanks in advance.

REPLY [0 votes]: For your other implication choose $y_0\in A$. Then for every $x\in A$ $$\|x\| \leq \|y_0\|+\|x-y_0\|\leq \|y_0\| + k,$$ where $k  = \sup_{x,y\in A}\|x-y\|$. For $\|y_0\|+k =: a$ we have $\|x\|\leq a$, $x\in A$.


text:
\subsection{Supplementary Material for Section \ref{subsec:Precond}}

Here we provide additional examples of Hilbert scales, satisfying Assumption \ref{ass:link-condition2}.

\begin{example}[Covariance Scale]
\label{lem:example-pre}
Let $\cH_K$ be an RKHS with kernel $K$ and covariance operator \eqref{eq:covariance}.
Given $r\geq 0$ let $\cH:=Ran(\cT^r)$ and let $L=\cT^{-1}$. Then by definition \ref{eq:def-norm}  we have for any $h \in \cH$ the
equality $||h||_\cH = ||\cT^{-r}h||_{\cH_K}$. Thus,  for any $a>0$, by \eqref{eq:iso1}
\begin{align*}
||\cS h||_{L^2} &= || \sqrt{\cT}h||_{\cH_K} = ||\cT^{r+\frac{1}{2}} h ||_{\cH} \\
&\leq ||\cT^{r+\frac{1}{2}-a}|| \; ||\cT^a h||_{\cH} \\
&= ||\cT^{r+\frac{1}{2}-a}|| \; ||h||_{-a} \;.
\end{align*}
Thus, \eqref{eq:link1} holds with $\bar m =  ||\cT^{r+\frac{1}{2}-a}||$, provided $0<a\leq r+\frac{1}{2}$.

Moreover, by the same reasoning, for $\cB_s=\cS L^{-s}: \cH \to L^2(\cX, \rho_X)$
\begin{align*}
||\cB_s h||_{L^2} &= ||\cS L^{-s} h||_{L^2} = || \cT^{\frac{1}{2}+r+s}h ||_\cH \leq  || \cT^{\frac{1}{2}+r+s}|| \; ||h||_\cH \;.
\end{align*}
Thus, $\cB_s$ is bounded if $s \geq -(\frac{1}{2}+r)$. This is certainly satisfied if $s\geq -a$.

\end{example}

\[\]

\begin{example}[Example \ref{ex:pre}; Switching between different Diffusion spaces]
Let $L=e^{\frac{1}{2}\Delta}$, $t>0$ and  $\cH:=Ran(e^{-\frac{t}{2}\Delta})$ be the diffusion space, defined in Example \ref{ex:diffusion}.
For $a>0$ we have
\begin{align*}
||\cS h||_{L^2} &= ||e^{-\frac{t}{2}\Delta} h||_\cH \leq || e^{(-\frac{t}{2}+ \frac{a}{2})\Delta}|| \; ||h||_{-a} \;.
\end{align*}
Hence, \eqref{eq:link1} is satisfied with $\bar m = || e^{(-\frac{t}{2}+ \frac{a}{2})\Delta}||  $, being finite if $0<a \leq t$.
In addition,
\begin{align*}
||\cB_s h||_{L^2} &= ||\cS L^{-s} h||_{L^2} = || e^{-\frac{1}{2}(s+t)\Delta} h||_\cH  \;.
\end{align*}
Therefore, $\cB_s$ is finite if $s \geq -a \geq -t$.
\end{example}

\[\]

\begin{example}[Example \ref{ex:pre2}; Switching between different orders of smoothness]

Let $s\leq 0$, $t>0$ and  $\cH_t:=Ran(e^{-\frac{t}{2}\Delta})$ be the diffusion space, defined in Example \ref{ex:diffusion}. For $u\geq 0$ define
\[ L:=e^{\frac{1}{2}\Delta}(Id + \Delta)^{-\frac{u}{2}}\;.\]
Then, for any $h \in \cH_t$ we have $h=e^{-\frac{t}{2}\Delta}g$ for some $g \in L^2(\cM)$ and
\begin{align*}
 L^{-s} h &= e^{-\frac{s}{2}\Delta}(Id + \Delta)^{\frac{su}{2}}e^{-\frac{t}{2}\Delta}g \\
 &=(Id + \Delta)^{\frac{su}{2}}e^{-\frac{1}{2}(t+s)\Delta}g \;.
\end{align*}
Thus, $L^{-s}$ maps from the Gaussian RKHS $\cH_t$ into the larger Sobolev space $H^{-su}(\cM)$ (recall that $s \leq 0$), see Example \ref{ex:sobolev},
provided that $t+s \geq 0$.

Moreover,  for $a>0$, we have
\begin{align*}
||\cS h||_{L^2} &= ||e^{-\frac{t}{2}\Delta} h||_\cH \leq || e^{-\frac{t}{2}\Delta}L^a || \; ||h||_{-a} \;.
\end{align*}
Since
\[ e^{-\frac{t}{2}\Delta}L^a = (Id + \Delta)^{\frac{au}{2}}e^{-\frac{1}{2}(t-a)\Delta} \]
is bounded if $0<a\leq t$, the link condition \eqref{ass:link-condition2} is satisfied under this assumption.
Finally, a similar calculation shows that $\cB_s = \cS L^{-s}$ is bounded if  $0\geq s \geq -a \geq -t$. Thus, Assumption \ref{ass:link-condition2} is satisfied.
\end{example}

\subsection{Probabilistic Bounds}

In this section we collect some basic and well established probability bounds. To this end, let us introduce some notation:
\begin{equation*}
B_{n,\lam}(a,b):= \frac{1}{\sqrt \lam} \paren{ a \frac{2\kappa_s}{n\sqrt \lam } + b \sqrt{\frac{\cN_{\cT_s}(\lam)}{n \lam}} } \;,
\end{equation*}
where $a,b>0$.

\begin{proposition}[\cite{blanchard2019lepskii}, Proposition A.1]
\label{prop:sub-proba2}
Fix $\lam >0$, $s\geq 0$, let $n \in \mbn$ and $\delta \in (0,1)$. Then with
probability $\rho^{\otimes n}$ not less than $1-\delta$ we have
\[  \Upsilon_s (\lam )  := || (\cT_s + \lam)^{-1/2}(\cT_s - \cT_{\bx , s})  ||_{HS} \leq 2\log(8/\delta)\sqrt{\lam } B_{n,\lam}(\kappa_s, \kappa_s) \;. \]
In particular, if  $\lam >0$ satisfies the condition
\[ \cN_{\cT_s}(\lam)\leq n\lam \;, \]
we obtain with probability $\rho^{\otimes n}$ not less than $1-\delta$
\[   \Upsilon_s (\lam ) \leq   C_{\kappa_s}\log(8/\delta)\sqrt{\lam } \;, \]
for some $C_{\kappa_s}<\infty$.
\end{proposition}

\[\]

\begin{proposition}[\cite{blanchard2019lepskii}, Proposition A.1]
\label{prop:sub-proba}
Let $\varphi: \mbr_+ \to \mbr_+$ be nondecreasing and sublinear. Fix $\lam >0$, let $n \in \mbn$ and $\delta \in (0,1)$. Then with
probability $\rho^{\otimes n}$ not less than $1-\delta$ we have
\[ \Xi ^\varphi_{\bx,s}(\lam):=||\varphi(\lam + \cT_s)\varphi(\lam + \cT_{\bx , s})^{-1}|| \leq 4\log^2(8/\delta)\paren{1+B_{n,\lam}(\kappa_s, \kappa_s)}^2 \;. \]
If $\varphi(t)=t^r$ with $r \in [0,1]$, we have with probability $\rho^{\otimes n}$ not less than $1-\delta$ the sharper estimate
\[ \Xi ^r_{\bx,s}(\lam):=||(\lam + \cT_s)^r (\lam + \cT_{\bx , s})^{-r}|| \leq 4^r\log^{2r}(8/\delta)\paren{1+B_{n,\lam}(\kappa_s, \kappa_s)}^{2r} \;.  \]
In particular, if  $\lam >0$ satisfies the condition
\[ \cN_{\cT_s}(\lam)\leq n\lam \;, \]
we obtain with probability $\rho^{\otimes n}$ not less than $1-\delta$
\[ \Xi ^\varphi_{\bx,s}(\lam) \leq  C_{\kappa_s } \log^{2}(8/\delta) \;, \]
for some $ C_{\kappa_s , r} < \infty$ and
\[ \Xi ^r_{\bx,s}(\lam) \leq  C'_{\kappa_s , r} \log^{2r}(8/\delta) \;,\]
for some $ C'_{\kappa_s , r} < \infty$.
\end{proposition}


TITLE: True or False: if $(a_n)^2$ converges and $a_n\le 0$ for every n, then $a_n$ converges.
QUESTION [0 upvotes]: My attempt: if $(a_n)^2$ converges, then it converges to a finite limit, then by the definition of the limit we get that : $|(a_n)^2 - L| < \epsilon $  then I opened up the absolute value and got: $L-\epsilon < a_n^2 < L + \epsilon $. (I really wanted to take a square root and end it here but if any sides of the inequation has a negative value then my proof is false, I tried to come up with a counter example and I couldn't find one too, but I think here I can choose the $\epsilon $ as I like since the limit exists so if I took $\epsilon = \frac {L} 2$, I would have all positive and be able to take square root on all the inequality sides.  I wanna make sure if my proof is correct, and if it is, then why choosing one specific $\epsilon $, holds for a proof.  if it's not any explanation would be pretty much appreciated.  Thanks in advance.

REPLY [1 votes]: You can treat the case $L=0$ separately using $\epsilon'=\sqrt{\epsilon}$, so let assume $L>0$.
WLOG you can set $b_n=-a_n$ to work with all positive quantities and get rid of annoyances with sign and inequalities.
Let set $L=\ell^2$ with $\ell>0$, we get
$$|b_n-\ell|=\dfrac{\overbrace{|{b_n}^2-\ell^2|}^{\to 0}}{\underbrace{|b_n+\ell|}_\text{bounded}}\le \frac 1\ell|{b_n}^2-L|\to 0$$
The proof essentially relies on showing that the denominator is bounded but this is trivial since $|b_n+\ell|=b_n+\ell\ge \ell>0$ using the positivity of $b_n$.
You can try writing the proof directly with $a_n$ but it requires a bit of gymnastic to deal with inequalities properly.


TITLE: How to parametrize a complex curve
QUESTION [1 upvotes]: I just started Complex Analysis course in my Uni! We are at the part of Complex Line Integrals! There is something there saying something about the Parametrization of line integrals and i wanted to know how you can do that! I read the description in my slides but i can't figure it out how to do it!
Can someone help me understand how to do them! Ideally by providing some examples! If you can't can you help me solve this?
C segment from 1 + i to 3
f(z)=iz

REPLY [2 votes]: To calculate an integral over a complex line froma definition (and not using theorems like the Cauchy's theorem) you need a parametrization of the curve.
A paramterization of a straight line from $z_1$ to $z_2$ is
$$ z(t) = z_1 + t (z_2-z_1), \qquad t\in[0,1]$$
Another useful curve (not in your specific problem, just in general) is an arc of a circle. It can be parametrized as
$$ z(t) = z_0 + R e^{it}$$
when going counterclockwise or
$$ z(t) = z_0 + R e^{-it}$$
when going clockwise, where $z_0$ is the center of the circle, $R$ is the radius, and you need to choose the range of $t$ to fit the initial and final points of the arc.
One you have the parametrization, the integral is defined as
$$ \int_\gamma f(z) dz : = \int_{t_1}^{t_2} f(z(t)) \frac{dz}{dt} dt = \int_{t_1}^{t_2} {\rm Re}\big( f(z(t)) \frac{dz}{dt} \big)  dt + i \int_{t_1}^{t_2} {\rm Im}\big( f(z(t)) \frac{dz}{dt} \big) dt$$
If a curve is more complex, but it can be decomposed into straight lines and arcs, you write the integral as sum of the integrals over each segment and calculate them separately. If it involves a more difficult curve, you need a parametrization of this curve or use some theorems about integrals over complex curves.
Taking your specific example, you need a straight line from $1+i$ to $3$. It can be parametrized as $$ z(t) = 1+i+ t (2-i), \qquad t\in[0,1]$$
Therefore we have
$$ \int_\gamma f(z)dz = \int_0^1 iz(t) \frac{dz}{dt} dt = \int_0^1 i\big(1+i+ t (2-i) \big)(2-i) dt$$

REPLY [1 votes]: To parameterize the line segment, do it just like you would do it in $\mathbb{R}^2$: $$z(t) = (1-t)(1+i)+3t,\ 0\leq t\leq 1$$
By definition, $\mathrm{d}z=z'(t)\mathrm{d}t= (-1-i+3)\mathrm{d} t$, so the integral is $$\int_0^1 i((1-t)(1+i)+3t))(2-i)\mathrm{d}t$$


text:
\begin{document}
\title{Additivity of Heegaard genera of bounded surface sums}

\author {Ruifeng Qiu\footnote{Supported by a grant of NSFC (No.
10625102)},  Shicheng Wang\footnote{Supported by a grant of NSFC
(No. 10631060)}, Mingxing Zhang}
\date{}
\maketitle
\begin{abstract}

Let $M$ be a surface sum of 3-manifolds $M_1$ and $M_2$ along a
bounded connected surface $F$ and $\partial_{i}$ be the component of
$\partial M_{i}$ containing $F$. If $M_i$ has a high distance
Heegaard splitting, then any minimal Heegaard splitting of $M$ is
the amalgamation of those of $M^{1}, M^{2}$ and $M^{*}$, where
$M^{i}=M_{i}\setminus
\partial_{i}\times I$, and $M^{*}=\partial_{1}\times I\cup_{F}
\partial_{2}\times I$. Furthermore, once both
$\partial_i\setminus F$ are connected, then
$g(M)=Min\bigl\{g(M_{1})+g(M_{2}), \alpha\bigr\}$, where $\alpha=
g(M_{1})+g(M_{2})+1/2(2\chi(F)+2-\chi(\partial _{1})-\chi(\partial
_{2}))-Max\bigl\{g(\partial _{1}), g(\partial _{2})\bigl\}$; in
particular $g(M)=g(M_{1})+g(M_{2})$ if and only if $\chi(F)\geq
1/2Max\bigl\{\chi(\partial_{1}), \chi(\partial_{2})\bigr\}.$

The proofs rely on Scharlemann-Tomova's theorem.
\end{abstract}

{\bf Keywords}: Heegaard Distance and Genus, Surface Sum, Weakly
incompressible.

AMS Classification: 57M25

\section{Introduction}

 \ \ \ \ \ All surfaces and 3-manifolds in this paper are assumed to be compact and
orientable.

Let $F$ be either a properly embedded  surface in a 3-manifold $M$
or a sub-surface of $\partial M$. If there is an essential simple
closed curve on $F$ which bounds a disk in $M$ or $F$ is a 2-sphere
which bounds a 3-ball in $M$, then we say $F$ is compressible;
otherwise, $F$ is said to be incompressible. If $F$ is an
incompressible surface not parallel to $\partial M$, then $F$ is
said to be essential. If $M$ contains an essential 2-sphere, then
$M$ is said to be reducible. If $\partial M$ is compressible, then
$M$ is said to be $\partial$-reducible.

Let $M$ be a  3-manifold. If there is a closed surface $S$ which
cuts $M$ into two compression bodies $V$ and $W$ with
$S=\partial_{+} W=\partial_{+} V$, then we say $M$ has a Heegaard
splitting, denoted by $M=V\cup_{S} W$; and $S$ is called a Heegaard
surface of $M$. Moreover, if the genus $g(S)$ of $S$ is minimal
among all the Heegaard surfaces of $M$, then $g(S)$ is called the
genus of $M$, denoted by $g(M)$.

Now let $M$ be a 3-manifold, and $F$ be a compact surface in $M$
which cuts $M$ into two 3-manifolds $M_{1}$ and $M_{2}$. Then $M$ is
called a surface sum of $M_{1}$ and $M_{2}$ along $F$, denoted by
$M=M_{1}\cup_{F} M_{2}$. Note that $F\subset\partial M_{i}$ for
$i=1, 2$.  A central topic in Heegaard splitting is to address
relations between $g(M_1)$, $g(M_2)$ and $g(M)$.

Suppose first that $F$ is a closed surface.  Let
$M_{i}=V_{i}\cup_{S_{i}} W_{i}$ be a Heegaard splitting for $i= 1,
2$. Then $M$ has a natural Heegaard splitting called the
amalgamation of $V_{1}\cup_{S_{1}} W_{1}$ and $V_{2}\cup_{S_{2}}
W_{2}$. From this view, $g(M)\leq g(M_{1})+g(M_{2})-g(F)$. If $F$ is
a 2-sphere, the so-called Haken's lemma claimed
$g(M)=g(M_{1})+g(M_{2})$. For $g(F)>0$, there are some examples to
show that it is possible that $g(M)\leq g(M_{1})+g(M_{2})-g(F)-n$
for any given $n>0$,  see [7] and [18]. Philosophically, in such
examples neither the gluing between $M_1$ and $M_2$ along $F$ nor
the Heegaard splitting of $M_i$ are complicated.

Under various different conditions describing  the complicated
gluing maps, the equality $g(M)=g(M_{1})+g(M_{2})-g(F)$ was proved,
see [1], [9], [10] and [19].  By invoke results of Hartshorn [3],
Scharlemann [13] and Scharlemann and Tomova [16], it is just proved
in [6] that $g(M)=g(M_{1})+g(M_{2})-g(F)$ if  $M_1$ and $M_2$ have
high  distance Heegaard splittings, where the distance of a Heegaard
splitting was introduced by Hempel [4].

Suppose that $F$ is a bounded surface. Then  it is easy to see
$g(M)\leq g(M_{1})+g(M_{2})$ (see Lemma 2.1). By the disk version of
Haken's lemma, $g(M)=g(M_{1})+g(M_{2})$ if $F$ is a disk. If $F$ is
an annulus, various results about if $g(M)=g(M_{1})+g(M_{2})$ hold
or not have been given, see [5], [8], [11] and [12].

In this paper we will address the additivity of Heegaard genus of
surface sum of 3-manifolds along a bounded surface $F$ with
$\chi(F)< 0$, which  seems not touched before.

We first fix some notions.  Suppose $P$ (resp. $H$) is a properly
embedded surface (resp. an embedded 3-manifold) in a 3-manifold $M$.
We use $M\setminus P$ (resp. $M\setminus H$) to denote the resulting
manifold obtained by splitting $M$ along $P$ (resp. removing
$\text{int} H$, the interior of $H$).

Let $M=M_{1}\cup_{F} M_{2}$, $\partial_{i}$ be the component of
$\partial M_{i}$ containing $F$, and $\partial_{i}\times [0,1]$ be a
regular neighborhood of $\partial_{i}$ in $M_{i}$ with $\partial
_{i}=\partial _{i}\times\bigl\{0\bigr\}$. We denote by $P^{i}$ the
surface $\partial_{i}\times\bigl\{1\bigr\}$. Let
$M^{i}=M_{i}\setminus \partial_{i}\times [0,1]$ for $i=1, 2$,  and
$M^{*}=\partial_{1}\times [0,1]\cup_{F}
\partial_{2}\times [0,1]$. Then
$M=M^{1}\cup_{P^{1}}M^{*}\cup_{P^{2}}M^{2}$.  \vskip 3mm

{\bf Theorem 1.} \ Let $M$ be  a surface sum of 3-manifolds $M_1$
and $M_2$ along a bounded surface $F$, and $\partial_{i}$ be the
component of $\partial M_{i}$ containing $F$. If $M_{i}$ has a
Heegaard splitting $V_{i}\cup_{S_{i}} W_{i}$ with $d(S_{i})>
2(g(S_{1})+g(S_{2}))$, $i=1,2$,  then any minimal Heegaard splitting
of $M$ is the amalgamation of Heegaard splittings of $M^{1}$,
$M^{2}$ and $M^{*}$ along $\partial_{1}$ and $\partial_{2}$. \vskip
3mm

The proof of Theorem 1  invokes full energy of Scharlemann-Tomova's
deep result (Lemma 2.5).

{\bf Theorem 2.} \ Under the assumptions of Theorem 1, if
$\partial_i \setminus F$ is connected for $i=1,2$,   then
$g(M)=Min\bigl\{g(M_{1})+g(M_{2}), \alpha\bigr\}$, where  $$\alpha=
g(M_{1})+g(M_{2})+1/2(2\chi(F)+2-\chi(\partial _{1})-\chi(\partial
_{2}))-Max\bigl\{g(\partial _{1}), g(\partial _{2})\bigl\}\qquad
(1.1).$$  Furthermore $g(M)=g(M_{1})+g(M_{2})$ if and only if
$\chi(F)\geq 1/2Max\bigl\{\chi(\partial_{1}),
\chi(\partial_{2})\bigr\}.$ \vskip 3mm

{\bf Corollary 3.} \ Under the assumptions of Theorem 1, if $F$ is
annulus, then $g(M)=g(M_{1})+g(M_{2})$.\vskip 3mm

{\bf Remark \ 4.} It is remarkable to compare Theorem 2 with the
main result in [6] for closed surface case.  In Theorem 2

(1) once $\chi(F)\geq 1/2Max\bigl\{\chi(\partial_{1}),
\chi(\partial_{2})\bigr\}$, $\chi(F)$, therefore $g(F)$, itself
plays no role in the result.

(2) once $\chi(F)< 1/2Min\bigl\{\chi(\partial_{1}),
\chi(\partial_{2})\bigr\}$, the contribution of $\chi(F)$ to $g(M)$
is non-trivial linear function. A particular case when
$g(\partial_{1})=g(\partial_{2})$ and $\chi(F)=\chi(\partial_{1})$,
then $g(M)=g(M_{1})+g(M_{2})-g(\partial_{1})+1$.

{\bf Acknowledgements.} The authors would like to thank Tsuyoshi
Kobayashi and Qing Zhou for helpful discussions on this paper.
\section{Distance of Heegaard splitting}

Weakly incompressible surface in 3-manifolds was introduced in [16]:
Let $P$ be a separating connected closed surface in 3-manifold $M$
which cuts $M$ into two 3-manifolds $M_{1}$ and $M_{2}$. Then $P$ is
said to be bicompressible if $P$ is compressible in both $M_{1}$ and
$M_{2}$. $P$ is strongly compressible if there are compressing disks
for $P$ in $M_{1}$ and $M_{2}$ which have disjoint boundaries in
$P$; otherwise $P$ is weakly incompressible.

Let $M$ be a compact orientable 3-manifold, and $M=V\cup_{S} W$ be a
Heegaard splitting. If there are essential disks $B\subset V$ and
$D\subset W$ such that $\partial B=\partial D$ ($\partial
B\cap\partial D=\emptyset$), then $V\cup_{S} W$ is said to be
reducible (weakly reducible). Otherwise, it is said to be
irreducible (strongly irreducible), see [2]. It is easy to see that
a strongly irreducible Heegaard surface is weakly incompressible.

Now let $P$ be a bicompressible surface in an irreducible 3-manifold
$M$. By maximally compressing $P$ in both sides of $P$ and deleting
the possible 2-sphere components, we get a surface sum structure of
$M$ as follow:
$$M=N_1\cup_{F^{P}_{1}}H^{P}_{1}\cup_{P}H^{P}_{2}\cup_{F^{P}_{2}}N_2,$$
where $H^{P}_{i}$ is a compression body with $\partial_{+}
H^{P}_{i}=P$, and  $\partial_{-} H^{P}_{i}=F^{P}_{i}$ is a
collection (may be empty) of incompressible closed surfaces of genus
at least one in $N_{i}$, $i=1,2$. Note that, if $F^{P}_{i}$ is
empty, then $H^{P}_{i}$ is a handlebody and $N_{i}$ is empty. It is
easy to see that if $M$ has boundary, then $F^{P}_{1}$ and
$F^{P}_{2}$ can not be both empty. Moreover if $P$ is weakly
incompressible, then the Heegaard splitting
$H^{P}_{1}\cup_{P}H^{P}_{2}$ is strongly irreducible.

Two weakly incompressible surfaces $P$ and $Q$ are said to be
well-separated in $M$ if $H^{P}_{1}\cup_{P}H^{P}_{2}$ is disjoint
from $H^{Q}_{1}\cup_{P}H^{Q}_{2}$ by isotopy. \vskip 2mm

{\bf Lemma 2.1.} \ Suppose $P$ is a weakly incompressible surface in
$M$. Then each component of $F^{P}_{i}$ is incompressible in $M$ for
$i=1, 2$.

{\bf Proof.} \ By the definition, each component of $F^{P}_{i}$ is
incompressible in $N_{i}$ for $i=1,2$. Since $P$ is a bicompressible
but weakly incompressible, $H^{P}_{1}\cup H^{P}_{2}$ is a
non-trivial strongly Heegaard splitting. By the disc version of
Haken's Lemma, each component of $F^{P}_{i}$ is incompressible in
$H^{P}_{1}\cup H^{P}_{2}$. Q.E.D. \vskip 2mm

{\bf Lemma 2.2.}  \ Let $S$ be a strongly irreducible Heegaard
surface of a 3-manifold $M$, then $M\setminus (H_{1}^{S}\cup_{S}
H_{2}^{S})$ is homeomorphic to $\partial M\times I$. Fruthermore, if
$P$ is a weakly incompressible surface in $M$, then either
$H^{P}_{1}\cup_{P} H^{P}_{2}\subset\partial \times I$ is
homeomorphic to $\partial\times I$ for one component $\partial$ of
$\partial M$, or, $S$ and $P$ are not well-separated.

{\bf Proof.} \ Now $M=V\cup_{S} W$, where $V$ and $W$ are
compression bodies. We may assume that $H_{1}^{S}\subset V$ and
$H_{2}^{P}\subset W$. Now there are essential disks $D_{1}, \ldots,
D_{n}$ in $V$ such that each component of $V \setminus
\cup_{i=1}^{n} D\times [0,1]$ is an I-bundle of a closed surface.
Since $\partial_{-} H_{1}^{S}$ is incompressible in $M$, hence in
$V$, each $D_{i}$ can be isotoped to be disjoint from $\partial_{-}
H_{1}^{S}$. Hence $\partial_{-} H_{1}^{S}\subset V \setminus
\cup_{i=1}^{n} D\times [0,1]$. This means that $M\setminus
(H_{1}^{S}\cup_{S} H_{2}^{S})$ is homeomorphic to $\partial M\times
I$.

Suppose $P$ and $S$ are well-separated, then
$H^{P}_{1}\cup_{P}H^{P}_{2}\subset \text{closed surface $\times
I$}$. Since $F^{P}_{1}\cup F^{P}_{2}$ is incompressible in $M$, each
component of $F^{P}_{i}$ is isotopic to $F\times\bigl\{0\bigr\}$ in
$F\times I$. Hence the lemma holds. Q.E.D. \vskip 2mm

The distance between two essential simple closed curves $\alpha$ and
$\beta$ on a compact surface $P$, denoted by $d(\alpha,\beta)$, is
the smallest integer $n\geq 0$ so that there is a sequence of
essential simple closed curves
$\alpha_{0}=\alpha,\ldots,\alpha_{n}=\beta$ on $P$ such that
$\alpha_{i-1}$ is disjoint from $\alpha_{i}$ for $1\leq i\leq n$.
When $P$ is a  bicompressible surface in a 3-manifold $M$, the
distance of $P$ is $d(P)=Min\bigl\{d(\alpha,\beta)\bigr\}$, where
$\alpha$ bounds a disk in $H^{P}_{1}$ and $\beta$ bounds a disk in
$H^{P}_{2}$. See [4] and [16].

Lemma 2.3 follows from the definitions and the main result in
[15]:\vskip 3mm

{\bf Lemma 2.3.} \ (1) \ If $M=V\cup_{S}W$ is a reducible Heegaard
splitting, then $d(S)=0$.

(2) \ If $M=V\cup_{S} W$ is a weakly reducible Heegaard splitting,
then $d(S)\leq 1$.

(3) \ If $M=V\cup_{S} W$ is a non-trivial and $\partial$-reducible
Heegaard splitting, then $d(S)\leq 1$.

(4) \ If $C$ is a closed surface, and $V\cup_{S} W$ is a non-trivial
Heegaard splitting of $C\times I$, then $d(S)\leq 2$ [15]. \vskip
2mm

{\bf Lemma 2.4 ([3], [13]).} \   Let $M=V\cup_{S} W$ be a Heegaard
splitting, and $P$ be an incompressible surface in $M$. Then either
$P$ can be isotoped to be disjoint from $S$ or $d(S)\leq 2-\chi(P)$.
\vskip 3mm

{\bf Lemma 2.5 ([16]).} \ Let $P$ and $Q$ be bicompressible but
weakly incompressible connected closed separating surfaces in a
3-manifold $M$. Then either

(1) \ $P$ and $Q$ are well-separated, or

(2) \ $P$ and $Q$ are isotopic, or

(3) \ $d(P)\leq 2g(Q)$. \vskip 3mm

{\bf Lemma 2.6 ([16]).} \ Let $M=V\cup_{S} W$ be a Heegaard
splitting of a 3-manifold $M$. If $d(S)>2g(S)$, then $M$ has the
unique minimal Heegaard spliting $V\cup_{S}W$ up to isotopy. \vskip
3mm

\section{The proof of Theorem 1}

Let $M=M_{1}\cup_{F} M_{2}$, and $F$ be a bounded surface. Then
$M=M^{1}\cup_{P^{1}}M^{*}\cup_{P^{2}} M^{2}$, where  $M^{1}, M^{2},
P^{1}, P^{2},
\partial_{1},
\partial_{2}$ are defined in Section 1. \vskip 3mm

{\bf Lemma 3.1.} \   $g(M)\leq g(M_{1})+g(M_{2})$.

{\bf Proof.} \ Let $M_{i}=V_{i}\cup_{S_{i}} W_{i}$ be a Heegaard
splitting of $M_{i}$ such that $F\subset
\partial_{i}\subset \partial_{-} W_{i}$. Now let $\gamma_{i}$ be a
unknotted arc in $W_{i}$ such that $\partial_{1} \gamma_{i}
\subset\partial_{+} W_{i}$,
$\partial_{2}\gamma_{1}=\partial_{2}\gamma_{2}\subset intF$. Let
$N(\gamma_{1}\cup\gamma_{2})$ be a regular neighborhood of
$\gamma_{1}\cup\gamma_{2}$ in $W_{1}\cup_{F} W_{2}$. Let
$V=V_{1}\cup N(\gamma_{1}\cup\gamma_{2})\cup V_{2}$, and $W$ be the
closure of $(W_{1}\cup_{F} W_{2})\setminus
N(\gamma_{1}\cup\gamma_{2})$. Then $V\cup_{S} W$ is a Heegaard
splitting of $M$, where $S=\partial_{+} V=\partial_{+} W$. Note that
$g(S)=g(S_{1})+g(S_{2})$. Hence the lemma holds. See also [17].
Q.E.D. \vskip 2mm

{\bf Lemma 3.2.} \  If $M_{i}$ has a Heegaard splitting
$V_{i}\cup_{S_{i}} W_{i}$ with $d(S_{i})> 2(g(S_{1})+g(S_{2}))$,
$i=1,2$, then any minimal Heegaard splitting of $M$ is irreducible
and weakly reducible.

{\bf Proof.} \ Since $d(S_{i})>2(g(S_{1})+g(S_{2}))$ for $i=1,2$,
$M_i$ is irreducible and $F$ is an essential surface in $M_i$ by
Lemma 2.3 (1) and (2). Then it follows that $M=M_1\cup _F M_2$ is
irreducible and $F$ is an essential surface in $M$. Furthermore,
$M_{i}$ is not homeomorphic to an I-bundle of a closed surface by
Lemma 2.3 (4).

Let $M=V\cup_{S} W$ be a minimal Heegaard splitting of $M$. Since
$M$ is  irreducible and $V\cup_{S} W$ is minimal, $V\cup_{S} W$ is
irreducible. By Lemma 2.6, $g(M_{i})=g(S_{i})$. By Lemma 3.1,
$g(S)\leq g(S_{1})+g(S_{2})$.

Now suppose $V\cup_{S} W$ is strongly irreducible. Since $S_{i}$ is
separating in $M$,  $S_{i}$ is bicompressible but weakly
incompressible in $M$. By Lemma 2.5, either $S$ and $S_{1}$ are
well-separated, or $S$ and $S_{1}$ are isotopic, or $d(S_{1})\leq
2g(S)$. Since $S_1$ is a Heegaard surface of $M_1$, we have that
$H^{S_{1}}_{1}\cup_{S_{1}}H^{S_{1}}_{2}$ is homeomorphic to $M_{1}$,
which not a product as we just proved. By Lemma 2.2, $S$ and $S_{1}$
are not well-separated. Since $S$ is a Heegaard surface of $M$ and
$S_{1}$ is a Heegaard surface of $M_{1}$,  $S$ is not isotopic to
$S_{1}$. Since $d(S_{1})>2(g(S_{1})+g(S_{2}))$ and $g(S)\leq
g(S_{1})+g(S_{2})$, $d(S_{1})>2g(S)$, a contradiction. Q.E.D. \vskip
3mm

{\bf The proof of Theorem 1.} \ Under the assumptions of Theorem 1,
by Lemma 3.2, any minimal Heegaard splitting of $M$ is irreducible
and weakly reducible. Let $M=V\cup_{S} W$ be a minimal Heegaard
splitting, then $V\cup _{S} W$ has a thin position as
$$V\cup_{S}W=(V_{1}^{'}\cup_{S_{1}^{'}}
W_{1}^{'})\cup_{F_{1}}\ldots\cup_{F_{n-1}}(V_{n}^{'}\cup_{S_{n}^{'}}
W_{n}^{'}) \ \ \ (*)$$ where $n\geq 2$, and each component of
$F_{1},\ldots, F_{n-1}$ is an incompressible closed surface in $M$,
and each $V_{i}^{'}\cup_{S_{i}^{'}} W_{i}^{'}$ consists of a active
component which is a non-trivial strong irreducible Heegaard
splitting and possible some product components each of which is a
trivial Heegaard splitting of an I-bundle of a closed surface. See
[14]. \vskip 3mm

Since $d(S_{i})>2(g(S_{1})+g(S_{2}))$ for $i=1,2$, $M$ is
irreducible and $F$ is an essential surface in $M$ by Lemma 2.3.
Furthermore, $M_{i}$ is not homeomorphic to an I-bundle of a closed
surface.

{\bf Claim 1.} \  $F_{i}$, $1\leq i\leq n-1$, can be isotoped so
that $F_{i}\subset M^{*}$.

{\bf Proof.} \ Suppose that $F_{i}\cap (M^{1}\cup
M^{2})\neq\emptyset$ for some $1\leq i\leq n-1$. We may assume that
$F_{i}\cap M^{1}\neq\emptyset$, and $F_{i}\cap M^{1}$ is
incompressible moreover. Note that $g(F_{i})\leq g(S)\leq
g(S_{1})+g(S_{2})$.
 Now $\chi(F_{i}\cap M^{1})\geq \chi(F_{i}\cap M_{1})\geq \chi(S)= 2-g(S) \ge
2-2(g(S_{1})+g(S_{2}))$, we have $d(S_1) > 2- \chi(F_i\cap M^{1})$.
By Lemma 2.4, $F_{i}$ can be isotoped to be disjoint from $S_{1}$.
Hence $F_{i}\cap M^{1}$ lies in one of $V_{1}$ and $W_{1}$ which
contains $F$.  It follows $F_i$ can be further isotoped  to be
disjoint from $M^{1}$. Q.E.D. (Claim 1) \vskip 3mm

 {\bf Claim 2.} \ There exists a component of $\bigcup_{1\leq i\leq n-1} F_{i}$ isotopic to $P^{1}$ (resp.
 $P^{2}$).

{\bf Proof.} \ Suppose that each component of $\bigcup_{1\leq i\leq
n-1} F_{i}$ is not isotopic to $P^{1}$. By Claim 1, $M^{1}\subset
M_{i}^{'}=V_{i}^{'}\cup_{S_{i}^{'}} W_{i}^{'}$ for some $1\leq i\leq
n$.

If $S_1$ is contained in the product component of
$M_{i}^{'}=V_{i}^{'}\cup_{S_{i}^{'}} W_{i}^{'}$, by the same
argument in the proof of Lemma 2.2,
$H^{S_{1}}_{1}\cup_{S_{1}}H^{S_{1}}_{2}$ is a Heegaard splitting of
an I-bundle of a closed surface. That is $M_1$ is an I-bundle of a
closed surface, it is a contradiction.

Now suppose $S_1$ is contained in the active component of
$M_{i}^{'}=V_{i}^{'}\cup_{S_{i}^{'}} W_{i}^{'}$. We also use
$M_{i}^{'}=V_{i}^{'}\cup_{S_{i}^{'}} W_{i}^{'}$ to denote the active
component. Now $S_{i}^{'}$ and $S_{1}$ are both weakly
incompressible in $M_{i}^{'}$ and
$H^{S_{1}}_{1}\cup_{S_{1}}H^{S_{1}}_{2}$ is not product. By Lemma
2.2, $S_{i}^{'}$ and $S_{1}$ are not well separated. By Lemma 3.1,
$g(S_{i}^{'})\leq g(S)\leq g(M_{1})+g(M_{2})$. By Lemma 2.6,
$g(M_{1})+g(M_{2})=g(S_{1})+g(S_{2})$. Hence
$2g(S_{i}^{'})<d(S_{1})$.

By Lemma 2.5, $S_{i}^{'}$ and $S_{1}$ are isotopic in $M_{i}^{'}$.

It follows $S_1$ is a strongly irreducible Heegaard surface of
$M_i'$. Then by Lemma 2.2 $M_i'\setminus
(H^{S_{1}}_{1}\cup_{S_{1}}H^{S_{1}}_{2})$ is $\partial M_i'\times
I$. By Lemma 2.2 we may assume that
$H^{S_{1}}_{1}\cup_{S_{1}}H^{S_{1}}_{2}=M^{1}$. Hence one component
of $\partial M_{i}^{'}$ is isotopic to $P^{1}$.

Similarly, there exists a component of $\bigcup_{1\leq i\leq n-1}
F_{i}$ isotopic $P^{2}$. Q.E.D. (Claim 2) \vskip 3mm

By Claim 2, Theorem 1 holds. Q.E.D. \vskip 3mm

\section{The proof of Theorem 2}

{\bf Lemma 4.1.} \ Let $N=P_{1}\times I\cup_{F} P_{2}\times I$ be
the surface sum of $P_{1}\times I$ and $P_{2}\times I$ along $F$,
where $P_{1}$ and $P_{2}$ are orientable closed surfaces, and $F$ is
a connected bounded surface in both $P_{1}\times\bigl\{0\bigr\}$ and
$P_{2}\times\bigl\{0\bigr\}$.

(1) \ If both $P_{1}\times\bigl\{0\bigr\}\setminus F$ and
$P_{2}\times\bigl\{0\bigr\}\setminus F$ are connected, then
$g(N)=Min\bigl\{g(P_{1})+g(P_{2}), \alpha\bigr\}$, where
$$\alpha=1/2(2\chi(F)+2-\chi(P_{1})-\chi(P_{2}))+Min\bigl\{g(P_{1}),
g(P_{2})\bigr\}.$$

(2) \ If $F$ is an annulus, then $g(N)=g(P_{1})+g(P_{2})$.

{\bf Proof.} \ We first prove (1).

Since both $P_{1}\times\bigl\{0\bigr\} \setminus F$ and
$P_{2}\times\bigl\{0\bigr\}\setminus F$ are connected, $N$ contains
three boundary components $P_{1}\times\bigl\{1\bigr\}$,
$P_{2}\times\bigl\{1\bigr\}$, and
$P^{*}=(P_{1}\times\bigl\{0\bigr\}\setminus F)\cup
(P_{2}\times\bigl\{0\bigr\}\setminus F)$. See Figure 4.1. Hence

$$g(N)\geq
Min\bigl\{g(P_{1})+g(P_{2}), g(P_{1})+g(P^{*}),
g(P_{2})+g(P^{*})\bigr\}.$$

\begin{center}
\psfrag{a}{$P^*$} \psfrag{c}{$P_1\times I$} \psfrag{d}{$P_2\times
I$} \psfrag{f}{$F$} \psfrag{i}{} \psfrag{j}{} \psfrag{g}{$P_1\times
1$} \psfrag{h}{$P_2\times 1$}
\includegraphics[height=2.5in]{figure3_1.eps}
\centerline{Figure 4.1}
\end{center}

It is easy to see that $N$ is homeomorphic to both $P_{1}\times
I\cup_{P_{1}\times\bigl\{0\bigr\}\setminus F} P^{*}\times I$ and
$P^{*}\times I\cup_{P_{2}\times\bigl\{0\bigr\} \setminus F}
P_{2}\times I$. See Figure 4.2.

\begin{center}
\psfrag{a}{$P^*\times I$}\psfrag{b}{$P_1\times I$} \psfrag{f}{$F$}
\includegraphics[height=2in]{figure3_2.eps}
\centerline{Figure 4.2}
\end{center}

By Lemma 3.1, $$g(N)\leq Min\bigl\{g(P_{1})+g(P_{2}),
g(P_{1})+g(P^{*}), g(P_{2})+g(P^{*})\bigr\}.$$ Now
$$g(N)=Min\bigl\{g(P_{1})+g(P_{2}),
g(P_{1})+g(P^{*}), g(P_{2})+g(P^{*})\bigr\}.$$

Since $P^{*}=(P_{1}\times\bigl\{0\bigr\}\setminus F)\cup
(P_{2}\times\bigl\{0\bigr\}\setminus F)$,
$$2-2g(P^{*})=\chi(P^{*})=\chi(P_{1})+\chi(P_{2})-2\chi(F),$$ and
$$g(P^{*})=1/2(2\chi(F)+2-\chi(P_{1})-\chi(P_{2})).$$
Hence (1) holds.

Now we prove (2).

Suppose now that $F$ is an annulus.  Now there are three cases:

Case 1. \  Both $P_{1}\times\bigl\{0\bigr\}\setminus F$ and
$P_{2}\times\bigl\{0\bigr\}\setminus F$ are connected.

Now $N$ contains three boundary components
$P_{1}\times\bigl\{1\bigr\}$, $P_{2}\times\bigl\{1\bigr\}$ and
$P^{*}=(P_{1}\times\bigl\{0\bigr\}\setminus F)\cup
(P_{2}\times\bigl\{0\bigr\}\setminus F)$. Since $F$ is an annulus,
$g(P^{*})\geq g(P_{1}), g(P_{2})$. By the argument in (1), (2)
holds.

Case 2. One of  $P_{1}\times\bigl\{0\bigr\}\setminus F$ and
$P_{2}\times\bigl\{0\bigr\}\setminus F$ is connected while the other
is non-connected.

The argument is the same with the one in Case 1.

Case 3. \  Both $P_{1}\times\bigl\{0\bigr\}\setminus F$ and
$P_{2}\times\bigl\{0\bigr\}\setminus F$ are non-connected.

Now we denote by $F_{i}^{1}$ and $F_{i}^{2}$ the two components of
$P_{i}\times\bigl\{0\bigr\}\setminus F$. We may assume that
$\partial F_{1}^{j}=\partial F_{2}^{j}$. Then $N$ contains four
boundary components $P_{1}\times\bigl\{1\bigr\}$,
$P_{2}\times\bigl\{1\bigr\}$, $F^{1}=F_{1}^{1}\cup F_{2}^{1}$ and
$F^{2}=F_{1}^{2}\cup F_{2}^{2}$. In this case,
$g(F^{1})+g(F^{2})=g(P_{1})+g(P_{2})$. Hence (2) holds. Q.E.D.
\vskip 3mm

{\bf The proof of Theorem 2.} \ Recalling the definitions of $M^{i},
M^{*}$ defined in Section 1. Since $\partial_{i}$ is separating in
$M$ for $i=1, 2$, and $M^{i}$ is homeomorphic to $M_{i}$ for
$i=1,2$, by Theorem 1,
$$g(M)=g(M_{1})+g(M_{2})+g(M^{*})-g(\partial_{1})-g(\partial_{2}).$$
By Lemma 4.1, $g(M^{*})=Min\bigl\{g(\partial_{1})+g(\partial_{2}),
\alpha\bigr\}$, where
$$\alpha=1/2(2\chi(F)+2-\chi(\partial_{1})-\chi(\partial_{2}))+Min\bigl\{g(\partial_{1}),
g(\partial_{2})\bigr\}.$$ Hence $g(M)=Min\bigl\{g(M_{1})+g(M_{2}),
\alpha\bigr\}$, where  $$\alpha=
g(M_{1})+g(M_{2})+1/2(2\chi(F)+2-\chi(\partial _{1})-\chi(\partial
_{2}))-Max\bigl\{g(\partial _{1}), g(\partial _{2})\bigl\}.$$

It is easy to see that  $g(M)=g(M_{1})+g(M_{2})$ if and only if
$\chi(F)\geq 1/2Max\bigl\{\chi(\partial_{1}),
\chi(\partial_{2})\bigr\}.$ Q.E.D.\vskip 3mm

{\bf Corollary 3.} \ The proof follows immediately from Theorem 1
and Lemma 4.1 (2). Q.E.D.\vskip 3mm

\vskip 10mm

{\bf Reference}\vskip 5mm

[1] \ D. Bachman, S. Schleimer and E. Sedgwick, Sweepouts of
amalgamated 3-manifolds, Algebr. Geom. Topol. 6(2006) 171-194.

[2] \ A. Casson and C. McA Gordon, Reducing Heegaard splittings,
Topology Appl. 27(1987) 275-283.

[3] \ K. Hartshorn, Heegaard splittings of Haken manifolds have
bounded distance, Pacific J. Math. 204(2002) 61-75.

[4] \ J. Hempel, 3-manifolds as viewed from the curve complex,
Topology 40(2001) 631-657.

[5] \ T. Kobayashi, A construction of arbitrarily high degeneration
of tunnel numbers of knots under connected sum, J. Knot Theory
Ramifications 3(1994) 179-186.

[6] \ T. Kobayashi and R. Qiu, The amalgamation of high distance
Heegaard splittings is always efficient, Math. Ann. 341(2008)
707-715.

[7] \ T. Kobayashi, R. Qiu, Y. Rieck and S. Wang, Separating
incompressible surfaces and stabilizations of Heegaard splittings,
Math. Proc. Cambridge Philos. Soc. 137(2004) 633-643.

[8] \ T. Kobayashi and Y. Rieck, Heegaard genus of the connected sum
of $m$-small knots, Commu. Anal. Geom. 14(2006), 1037-1077.

[9] \ M. Lackenby, The Heegaard genus of amalgamated 3-manifolds,
Geom. Dedicata 109(2004) 139-145.

[10] \ Tao Li, On the Heegaard splittings of amalgamated
3-manifolds, Geom. Topol. Monographs 12(2007) 157-190.

[11] \ K. Morimoto, On the super additivity of tunnel number of
knots, Math. Ann. 317(2000), 489-508.

[12] \ R. Qiu, K. Du, J. Ma and M. Zhang, Distance and the Heegaard
genera of annular 3-manifolds, Preprint.

[13] \ M. Scharlemann, Proximity in the curve complex: boundary
reduction and bicompressible surfaces, Pacific J. Math. 228(2006),
325-348.

[14] M. Scharlemann and A. Thompson, Thin position for 3-manifolds.
Geometric Topology (Haifa, 1992) 231-238, Contemp. Math., 164, Amer.
Math. Soc., Providence, RI, 1994.

[15] \ M. Scharlemann and A. Thompson, Heegaard splittings of ${\rm
Surfaces}\times I$ are standard, Math. Ann., 295(1993), 549-564.

[16] \ M. Scharlemann and M. Tomova, Alternate Heegaard genus bounds
distance, Geom. Topol. 10(2006) 593-617.

[17] \ J. Schultens, Additivity of tunnel number for small knots,
Comment. Math. Helv. 75(2000) 353-367.

[18] \ J. Schultens and R. Weidman, Destabilizing amalgamated
Heegaard splittings, Geom. Topol. Monographs 12(2007) 319-334.

[19] \ J. Souto, The Heegaard genus and distance in curve complex,
Preprint. \vskip 5mm

Department of Applied Mathematics, Dalian University of Technology,
Dalian, China

qiurf@dlut.edu.cn \vskip 5mm

School of Mathematics, Peking University, Beijing, China

wangsc@math.pku.edu.cn \vskip 5mm

Department of Applied Mathematics, Dalian University of Technology,
Dalian, China

zhangmx@dlut.edu.cn

\end{document}


TITLE: Find the equation of the plane that goes through the origin and is parallel to the lines
QUESTION [0 upvotes]: Find the equation of the plane that goes through the origin and is parallel to the lines
$$\mathbf{R_1}=3\boldsymbol{\hat{\imath}}+3\boldsymbol{\hat{\jmath}}-\boldsymbol{\hat k} +s(\boldsymbol{\hat \imath}-\boldsymbol{\hat \jmath}-2\boldsymbol{\hat k}),\quad \mathbf{R_2}=4\boldsymbol{\hat\imath}-5\boldsymbol{\hat\jmath}-8\boldsymbol{\hat k}+t(3\boldsymbol{\hat\imath}+7\boldsymbol{\hat\jmath}-6\boldsymbol{\hat k})$$
Additionally, show that one line is on the plane and one is not on the plane
I need some help solving these questions.
So we have two lines that are parallel to the planes and therefore the normal of the plane is perpendicular to the lines
However, How can I find the normal and the plane only with this given information?
Many thanks in advanced

REPLY [0 votes]: The normal vector to the plane, which is, as you said, perpendicular to the two lines, you can get by doing to vector product of the two directing vector of each line.
Then you know one point of your plan, so it should be easy to get the equation of the plane.


TITLE: Conditional probability - 8 tosses of a coin
QUESTION [1 upvotes]: We throw a coin 8 times. What is the probability of getting the same number of heads and tails, if on the last three tosses of a coin we got tails?

REPLY [3 votes]: Getting same number of heads and tails means getting $4$ heads and $4$ tails. You already got $3$ tails. So, the problem is reduced to getting $4$ heads and $1$ tail in the first five tosses. That is nothing but $5C_4$$*$$1\over 2^5$ $= $$5\over 32$

REPLY [1 votes]: Let $x = $ the number of tails that remain, and $y = $ the total number of heads, then: $3+x = y, 3+x+y = 8$. Thus $ x = 1, y = 4$, Thus the remaining one tail can occur in any of the first $5$ tosses. So the probability is: $5\cdot \left(\dfrac{1}{2}\right)^5$


TITLE: From a conditional probability to an indefinite integral? Bayesian Information Criterion
QUESTION [0 upvotes]: I am trying to understand the Bayesian Information Criterion (BIC) using this article. At page 2 the following equality is given:
$$P(y|M_1) = \int f(y|\theta_i)g_i(\theta_i)d\theta_i$$
with

$y$ : observed data $y1, . . . , yn$
$M_i$ : a candidate model
$P(y|Mi)$ : marginal likelihood of the model $M_i$ given the data
$θ_i$: vector of parameters in the model $M_i$
$g_i(θ_i)$ : the prior density of the parameters $θ_i$
$f(y|θ_i)$ : the density of the data given the parameters $θ_i$

I do not understand the presented equality. I have tried to understand it for a couple of hours, but I was not even able to figure out the exact (conceptual) meaning of the product $f(y|\theta_i)g_i(\theta_i)$, let alone its integration. I know I am suppose to give info on the things I have tried to do, but given that this is not a practical problem I do not know what else to say.
Thanks in advance.

REPLY [0 votes]: The integral is a definite integral across the whole domain of $y$. This is not specified in the paper but it is specified in other books that treat the same topic.
The integral $\int f(y|\theta_i)g_i(\theta_i)$ reflects the probability of $y$ provided our model parameters $\theta_i$ weighted by our prior belief in the model: $g_i(\theta_i)$. So basically we define $P(y|M_1)$ as the probability of $y$ under the parameters of the model weighted by the uncertainty of the model. Which is in deed a very Bayesian approach.


TITLE: $P$ is a change of basis matrix from basis $B$ to basis $C$. Given $P$ and $C$ find $B$
QUESTION [1 upvotes]: $$
P=\begin{pmatrix}1&1&0\\0&3&2\\0&1&1\end{pmatrix}
$$ is a change of basis matrix from  basis $B$ to basis $C$.
  $$
C=\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}
$$
  Find $B$.

The way I understand this is that $C=PBP^{-1}$, then $B=P^{-1}CP$. So I need to find $P^{-1}$ which is:
$$
P^{-1}=\begin{pmatrix}1&-1&2\\0&1&-2\\0&-1&3\end{pmatrix}
$$
However when I compute $B=P^{-1}CP$ my solution the result is wrong when compared with the correct solution.
The correct solution also finds the same matrix $P^{-1}$ but after that it is as follows:
let $B=(b_1,b_2,b_3), C=(c_1,c_2,c_3)$ then:
$$
b_1=c_1=\begin{pmatrix}1\\1\\1\end{pmatrix}\\
b_2=-c_1+c_2-v_3=\begin{pmatrix}-1\\0\\-1\end{pmatrix}\\
b_3=2c_1-2c_2+3c_3=\begin{pmatrix}3\\0\\2\end{pmatrix}
$$
What's wrong with my solution?

REPLY [1 votes]: To put it shortly, $C=BP$ because of how the change of basis matrix is defined as well as how matrices $B$ and $C$ represent their respective bases.
I'll include the long version below, however.

For bases $B=\{\vec b_1,...,\vec b_n\}$ and $C=\{\vec c_1,...,\vec c_n\}$, the change of basis matrix, $P$, is defined such that
$$(\forall j)\,\,\vec c_j=\sum_{i=1}^nP_{ij}\vec b_i$$
Let $[\vec v]_A$ represent the components of some vector $\vec v$ in some basis $A=\{\vec a_1,...,\vec a_n\}$, and ${([\vec v]_A)}_i$ represent the $i$th component. Then
$$\vec v=\sum_{i=1}^n{([\vec v]_A)}_i\vec a_i$$
$\implies$
$$(\forall j)\,\,\sum_{k=1}^n{([\vec c_j]_E)}_k\vec e_k=\sum_{i=1}^nP_{ij}\sum_{k=1}^n{([\vec b_i]_E)}_k\vec e_k$$
Note that (what I assume to be) your book lets matrices $B=([\vec b_1]_E\,|\,...\,|\,[\vec b_n]_E)$ and $C=([\vec c_1]_E|...|[\vec c_n]_E)$ (where '$|$' concatenates vectors and $E=\{\vec e_1,...,\vec e_n\}$ is the standard basis).
Using that, we get
$$(\forall j)\,\,\sum_{k=1}^nC_{kj}\vec e_k=\sum_{i=1}^nP_{ij}\sum_{k=1}^nB_{ki}\vec e_k$$
$$=\sum_{k=1}^n\left(\sum_{i=1}^nB_{ki}P_{ij}\right)\vec e_k$$
$\implies$ (using ${(AB)}_{ij}=\sum_kA_{ik}B_{kj}$ from matrix multiplication)
$$(\forall j)\,\,\sum_{k=1}^nC_{kj}\vec e_k=\sum_{k=1}^n{(BP)}_{kj}\vec e_k$$
As $E$ is a basis, and so the vectors $\vec e_1,...,\vec e_n$ are linearly independent, we can conclude
$$(\forall j,k)\,\,C_{kj}={(BP)}_{kj}$$
$\implies$
$$C=BP$$


TITLE: A problem from Dennis Zill's book
QUESTION [0 upvotes]: I have the following exercises from Dennis Zill's book, anyone can give me some help to resolve this exercises, because with my classmate, How I find the solution of this problem?, Regards!
"A cantilever beam of length L is embedded at its right end, and a horizontal tensile force of P pounds is applied to its free left end. When the origin is taken at its free end, the deflection y(x) of the beam can be shown to satisfy the differential equation $$Ely''=Py-w(x)\dfrac{x}{2}$$ Find the deflection of the cantilever beam if $w(x)=w_{0}x$, $0<x<L$, and $y(0)=0$, $y'(L)=0$".

REPLY [0 votes]: Write $y''-\frac{P}{El}y=\frac{-w(x)}{El}\frac{x}{2}$ to see that you have a linear, non-homogeneous second order differential equation which can be solved by taking
$y=y_h+y_p$ where $y_h$ is the general solution to the homogeneous case and $y_p$ is any other particular solution to the original equation.
Before substituting $w(x)=w_0x$ though, you need to be careful because as it stands, $w$ is not even defined at $x=0$ or $x=L$. You need to extend $w$ so that it is defined at these points. This is easy of course.
The real problem here is that the conditions you are given are NOT initial conditions; one of them is a boundary condition. Therefore, the theorem on uniqueness of solution does not apply, which means that any solution you do get may not be the only one, if indeed you happen to find one.
I imagine Zill deals with these difficulties in his text.


TITLE: Monty Hall Application
QUESTION [3 upvotes]: Driver A comes to a 3 way path junction but is not sure which one to take.
Just as he decides to take path 1, a cyclist came by and told driver A all he knows is that he is going on path 3 which would bring him to town.
Driver A is not heading to town but the seaside. Should he change to path 2?
Driver B encounters the cyclist before he can decide which path to take to the village.
Is his chance less than driver A because he cannot switch?

REPLY [2 votes]: If we presume that the seaside and town are different directions, each driver has $\frac 12$ chance going on path $1$ or $2$ and $0$ chance going on path $3$.  This is not a Monty Hall situation because the cyclist is not reacting to the driver's choice.


TITLE: If $R^*(R+S)\equiv S^*(R+S)$, then $R^*\equiv S^*$
QUESTION [0 upvotes]: If $R^*(R+S)\equiv S^*(R+S)$, then $R^*\equiv S^*$
I have to prove whether this is true or false.
Right now I think it's true, because I can't really think of a counter example. This is where I got:
$R^*(R+S)\equiv S^*(R+S)$, then $R^*\equiv S^*$ means
$$L(R^*)L(R+S)= L(S^*)L(R+S)$$
Where $(*)$ is the Kleene star and $L(R)$ means language denoted by $R$.
Don't really know if we can "cross out" on languages, since the $L(R+S)$ in common on both sides, our expression becomes
$$L(R^*)= L(S^*)$$
which means $R^*\equiv S^*$
would this be correct? The only step I am unsure about is when I "cross out" $L(R+S)$ since it is common on both sides.

REPLY [1 votes]: Suppose $R$ is $\{\epsilon\}$ and $S$ is $\{a\}^*$.
Now $R^* = R$ and $S^* = SS = S$.
Also $(R+T) = RT = T$ for any $T$.
So $R^*(R+S) = S^*(R+S) = S$, but $R^* \ne S^*$


TITLE: Why standard model uses Lie groups like $SU(2)$ and not $SL(2,\mathbb{C})$?
QUESTION [1 upvotes]: First of all, the question is written in section $2)$. Also, I known that the $SU(2)$ group do not appears "alone" in standard model, rather, inside the Glashow-Salam-Weinberg model.

1) Introduction
The heuristic picture of the mathematical structure of standard model (SM) lies on Lie group theory. Moreover, SM is a big gauge theory and therefore uses the technology of fiber bundles.
1.1) Bundles and Gauge theory

Given a manifold $\mathcal{M}$ (a spacetime) and a Lie group $G$ (a gauge group), we can readly construct another manifold using lie group a $G$: the principal bundles $P_{G}$.
Once you constructed the $P_{G}$, you can stablish Ehresmann connections and therefore the connection $1$-form $A$: the gauge field (in fact the one can put the gauge field in spacetime using the local connection $1$-form $A_{s} = s^{*}A$. The $s$ is precisely a choose of the local gauge).
Given the $P_{G}$, the implementation of matter fields $\Phi$ in spacetime (spinors, scalars, vectors and tensors) lies on another bundle called Associated bundle: $A_{P_{G}}$. Its definition is given by the quotient:
$$A_{P_{G}} := P_{G} \times_{\rho}V = \frac{P_{G} \times V}{G}, \tag{1}$$
where $\rho$ is the representation map between groups: $\rho: G \to GL(V)$ and $V$ is a vector space. Also, $\Phi$ are sections of $A_{P_{G}}$;
Inside $A_{P_{G}}$ the one can define, in a formal way, our beloved gauge covariant derivatives that acts (locally) on a matter field $\Phi$ as:
$$D_{\mu}\Phi = \partial_{\mu}\Phi + \rho_{*}(A_{s}(X))\Phi \tag{2}.$$

Where $X$ is a vector field and the map $\rho_{*}$ is the representation map acting on Lie algebra elements that follows the diagram:

With the exponential map, $\mathrm{exp}$, you can "translate" the technology of standard Lie group theory, into lie algebra representations as:
$\require{AMScd}$
\begin{CD}
\mathfrak{g}@>{\rho_{*}}>>  \hspace{0.4cm}\mathrm{End}(V)\\
@V{\mathrm{exp}}VV @VV{\mathrm{exp}}V\\
G @>{\rho}>> GL(V)
\end{CD}

2) My Question
The section $1.1)$ shows mathematical structures that are highly dependent on Lie groups, Lie algebras and its representations. Also, complex representations of $SU(2)$ represent non-relativistic spinors and representations of $SL(2,\mathbb{C})$ represent relativistic spinors $(*)$.
Therefore, my question is: why do we use groups like $SU(2)$ to represent gauge symmetry, instead of groups like $SL(2, \mathbb{C})$?

Another possible way to ask the question:

Knowing the steatment $(*)$, we realize that $SU(2)$ represent non-relativistic fields and $SL(2,\mathbb{C})$ represent relativistic fields. Since the standard model is a relativistic theory shouldn't be better to deal with "things" that represent relativistic behaviour?

REPLY [3 votes]: A gauge transformation acts on the wave function and does not transform it in space and time but in an abstract internal space. The corresponding gauge covariant derivative has Lorentz indices that ensure local Lorentz invariance.
https://en.wikipedia.org/wiki/Gauge_covariant_derivative#Gauge_theory


TITLE: Ask some matrix eigenvalue inequalities.
QUESTION [5 upvotes]: Let  $  \begin{bmatrix}
A& B   \\\\ B^*  &C
\end{bmatrix}$ be positive semidefinite, $A,C$ are of size $n\times n$.
Are the following plausible inequalities true? I have seen a lot of similar results, but for the following inequalities, I cannot locate them in the literature or find that they have been pointed out to be false.
1 $$\quad \sum\limits_{i=1}^k\lambda_i\begin{bmatrix}
A& B   \\\\ B^*  &C
\end{bmatrix}\le \sum\limits_{i=1}^k\left(\lambda_i(A)+\lambda_i(C)\right)\quad, $$
where $1\le k\le n$.
2 Modified
$$\quad \prod\limits_{i=1}^{2k}\lambda_i\begin{bmatrix}
A& B   \\\\ B^*  &C
\end{bmatrix}\le \prod\limits_{i=1}^k \lambda_i(A)\lambda_i(C) \quad, $$
where $1\le k\le n$.
3   $$ 2\lambda_i^{1/2}(B^*B)\le \lambda_i(A+C),$$ where $1\le k\le n$.
Here, $\lambda_i(\cdot)$ means the $i$th largest eigenvalue of $\cdot\quad$. Any references or counterexamples are appreciated.

REPLY [17 votes]: Item 1 is true.  This is part of Problem 22 (b) in Section 3.5 of Horn and Johnson [HJ94], which states that for Ky Fan norm ||⋅|| (and in fact for any unitarily invariant norm) and a positive semidefinite block matrix $\begin{pmatrix}A & B \\ B^* & C\end{pmatrix}$, it holds that $\left\|\begin{pmatrix}A & B \\ B^* & C\end{pmatrix}\right\| \le \left\|\begin{pmatrix}A & 0 \\ 0 & 0\end{pmatrix}\right\| + \left\|\begin{pmatrix}0 & 0 \\ 0 & C\end{pmatrix}\right\|$.
([Aud06] contains a proof of a slight generalization of this inequality among other results.)
Item 2 in the original question is false by considering the case where A=C=I/2, B=0, and k=1, where I is the identity matrix.  (Did you mean to square the left-hand side?)
Modified item 2 is false; see Willie Wong’s comment on this answer.
Item 3 is false.  A simple counterexample is n=2, i=1,
$A=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$,
$B=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$,
$C=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$.
Then $2\sqrt{\lambda_1(B^*B)}=2$ but λ1(A+C)=1.
References
[Aud06] Koenraad M. R. Audenaert.  A norm compression inequality for block partitioned positive semidefinite matrices.  Linear Algebra and its Applications, 413(1):155–176, Feb. 2006.  http://dx.doi.org/10.1016/j.laa.2005.08.017
[HJ94] Roger A. Horn, Charles R. Johnson.  Topics in Matrix Analysis.  Cambridge University Press, 1994.


TITLE: Prove that the number of conjugacy classes of a finite group $G$ is given by $ k(G) = \frac{1}{|G|}\sum_{g \in G}{\left|C(g)\right|} $
QUESTION [1 upvotes]: This is really bothering me. I feel like I have the right idea, but I can't get it to work.
The Statement of the Problem:
Prove that the number of conjugacy classes of a finite group $G$ is given by
$$ k(G) = \frac{1}{|G|}\sum_{g \in G}{\left|C(g)\right|} \qquad (*) $$
using the fact that $\left|G/C(g)\right|$ is the size of the conjugacy class of $g$.
Where I Am:
Ok. Since the conjugacy classes partition $G$, we have that
$$ |G| = \sum_{g \in G}{\left|G/C(g)\right|}. $$
Furthermore, we can represent the order of $G$ as the average size of its conjugacy classes multiplied by the number of conjugacy classes (this is to bring $k(G)$ into the picture):
$$ |G| = k(G) \cdot \frac{1}{k(G)} \sum_{g \in G}{\left|G/C(g)\right|}. $$
And, since each $C(g)$ is a subgroup of $G$, we can use Lagrange's Theorem to get
$$ \sum_{g \in G}{\left|G/C(g)\right|} = \sum_{g \in G}{\left|G\right|/\left|C(g)\right|} = \frac{\left|G \right|}{\sum_{g \in G}{\left| C(g) \right|}} $$
since the order of $G$ is obviously a constant.
So, I feel like everything I've stated is valid, and yet these can't be put together to get $(*)$, for some reason. Any help here would be appreciated.

REPLY [3 votes]: Let $\mathcal{O}(g)$ denote the conjugacy class of $g\in G$. Define
$$
f_{gh}=\begin{cases}1&\mbox{if }ghg^{-1}=h\\
0&\mbox{otherwise.}\end{cases}
$$
Now,
$$\sum_g f_{gh}=|C(h)|.$$
Since $|C(h)|=|C(h')|$ for $h'\in\mathcal{O}(h)$, we have
$$\sum_{h'\in\mathcal{O}(h)}\sum_g f_{gh'}=|\mathcal{O}(h)||C(h)|=|G|$$
Let $\Gamma$ be a set of representative of distinct conjugacy classes in $G$, so $|\Gamma|=k(G)$. Then,
\begin{align*}
\sum_{h\in G}|C(h)|&=\sum_{g,h\in G}f_{gh}\\
&=\sum_{h\in\Gamma}\left(\sum_{h'\in\mathcal{O}(h)}\sum_{g\in G}f_{gh'}\right)\\
&=\sum_{h\in \Gamma}|G|\\
&=k(G)|G|
\end{align*}
This proves the claim.


TITLE: Show that $\int\limits_2^{+\infty}\frac{\sin{x}}{x\ln{x}}\, \rm dx$ is conditionally convergent
QUESTION [1 upvotes]: The integral $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x\ln{x}}\,dx$ is conditionally convergent.
I know that $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x}\, dx$
is conditionally convergent and $ {\forall}p > 1$,
$\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x^p}\, dx$ is absolute convergent,
but $\ln{x}$ is between $x$ and $x^p$, so how to prove that $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x\ln{x}}\, dx$ is conditionally convergent?

REPLY [1 votes]: integrate by parts as Lord Shark said.
Since Wu does not know this method, I put it here for future readers of the question.
Integration by parts shows: for $M > 2$,
$$
\int_2^M\frac{\sin x}{x \log x} dx = \frac{\cos 2}{2 \log 2} - \frac{\cos M}{M \log M}
-\int_2^M \frac{1+\log x}{(x\log x)^2}\;\cos x\;dx
$$
Now
$$
\left|\frac{\cos M}{M\log M}\right| \le \frac{1}{M}\qquad\text{so}\qquad
\lim_{M\to \infty}\frac{\cos M}{M\log M} = 0.
$$
Next,
$$
\left|\frac{1+\log x}{(x\log x)^2}\;\cos x\right| \le \frac{2}{x^2}
\qquad\text{and}\qquad \int_2^\infty\frac{1}{x^2}\;dx\text{ converges}
$$
so
$$
\int_2^\infty \frac{1+\log x}{(x\log x)^2}\;\cos x\;dx\quad\text{converges}
$$
Combining them, we get
$$
\int_2^\infty\frac{\sin x}{x \log x} dx\qquad\text{converges}
$$


TITLE: a fixed point theorem
QUESTION [2 upvotes]: A hiker starts to climb up from base B to the summit S on sat 6am one day,spends the night at S and starts to climb down at 6am the next day.Prove that there is a point on the path B-S (there is only one path connecting B-S) which hiker crosses at the same time both days.

REPLY [4 votes]: I don't know a formal proof, but here is a way to think about it. Imagine that, instead of one man going up one day and coming down the next, there are two men, one going up and the other coming down on the same day, both starting at 6am. It is easy to see that they must cross paths, but this is really the same problem. Hope this helps.

REPLY [3 votes]: Do you know the intermediate value theorem? It states that if $f:[a,b] \rightarrow \mathbb{R}$ is a continuous function satisfying $f(a) = -1, f(b)=1$, then for any $x \in [-1,1]$ , there exists $c \in [a, b]$ such that $f(c) = x$.
Apply this to a suitable function describing your hiker's movement, and let $x=0$.


TITLE: Find $n$ such that there are $11$ non-negative integral solutions to $12x+13y =n$
QUESTION [1 upvotes]: What should be the value of $n$, so that $12x+13y = n$ has 11 non-negative integer solutions?

As it is a Diophantine equation, so we check whether the solution exists? it exists if $gcd(12,13)|n$ that is $1|n$ and hence integer solutions to $12x+13y = n$ exist.
To find the solutions explicitly, we write $1 $ as the linear combinations of $12$ and $13$ that is $1 = 12(1) + 13(-1)$, next we write $n = 12(n) + 13(-n)$ so that the integer solutions to $12x+13y = n$ is $x = n + t(12)$ and $y = -n + t(13)$. To check for non-negative integral solutions, $x \geq 0$ and $y \geq 0$ so that $t \geq -\frac{n^2}{12}$ and $t \geq \frac{n^2}{13}$ so $t \geq \frac{n^2}{13}$. But now I am struggling how to relate this  to the 11 number of non-negative integral solutions?.Any help?

REPLY [0 votes]: $12x + 13y = n$
$12(-n)+13(n)=n$ implies $x(t)=-n+13t$ and $y(t)=n-12t$
$x(t) \ge 0 \implies t \ge \dfrac{n}{13}$
$y(t) \ge 0 \implies t \le \dfrac{n}{12}$
So $\dfrac{n}{13} \le t \le \dfrac{n}{12}$
The existence of $11$ solutions implies
$$\text{$\dfrac{n}{13} \le t_0$ and  $t_0 + 10 \le \dfrac{n}{12}$} \tag{1}$$
That is,
$$\dfrac{n}{13} \le t_0 \le \dfrac{n-120}{12}\tag{2}$$
 for some integer $t_0$.
We find $12n \le 156t_0 \le 13n - 1560$
Or $$0 \le 1560t_0-12n \le n - 1560 \tag{3}$$
So $n \ge 1560$.
For $n = 1560$, condition $(3)$ becomes $0 \le 1560t_0-12n \le 0$.
So $t_0 = 120$.
It follows that $12x+13y = 1560$ has solutions,
$\{(0, 120), (13, 108), (26, 96), (39, 84), (52, 72), (65, 60), (78, 48), (91, 36), (104, 24), (117, 12), (130, 0)\}$
That is to say, $n = 1560$.


TITLE: Showing $F_X \cong F_Y\implies |X| = |Y|$
QUESTION [0 upvotes]: Lately I've been studying free groups, I'm a layman on the subject but I came across a step in the demonstration that I couldn't move forward. I know the question seems to be good but: If $F_X \cong F_Y$ implies $|X| = |Y|$.
In the proof the author constructs a quotient group $F_X/N_1$ where $N_1 = \langle x^2 : x \in F_X \rangle$ and $F_Y/N_2$ where $N_2= \langle y^2 : y \in F_Y \rangle$.
My question is: If $F_X \cong F_Y$ and $N_1 \cong N_2$, then $F_X/N_1 \cong F_Y/N_2$? If so, how did you naturally perceive $N_1 \cong N_2$?

REPLY [1 votes]: This has nothing to do with free groups.
So the key isn't just that the $N_1\cong N_2,$ but that the isomorphism $\phi:F_X\to F_Y$ restricts to an isomorphism $\phi_{|N_1}:N_1\to N_2.$
Definition: If $G$ is a group, define $N(G)=\langle g^2\mid g\in G\rangle.$
Claim 1: $N(G)$ is a normal subgroup.
Proof: The follows since $gg_1^2g^{-1}=\left(gg_1g^{-1}\right)$.
If $h\in N(G)$ then $h=g_1^2g_2^2\cdots g_k^2$ for some sequence $g_1,\dots,g_k\in G.$ If $g\in G,$ then $$ghg^{-1}=\left(gg_1g^{-1}\right)^2\left(gg_2g^{-1}\right)^2\cdots \left(gg_kg^{-1}\right)^2\in N(G).$$

Claim 2: If $\phi:G_1\to G_2$ is an isomorphism, then $\phi(N(G_1))= N(G_2).$
Proof: Let $\phi:G_1\to G_2$ be an isomorphism, so $\phi^{-1}$ is also an isomorphism. If $h\in N(G_1),$ then $h=g_1^2g_2^2\cdots g_k^2$ and $\phi(h)=\phi(g_1)^2\phi(g_2)^2\cdots \phi(g_k)^2\in N(G_2).$ Likewise, if $h\in N(G_2),$ $\phi^{-1}(h)\in N(G_1).$ So $\phi,$ restricted to $N(G_1),$ is an isomorphism between $N(G_1)$ and $N(G_2).$

Claim 3: If $\phi$ is an isomorphism between $G_1$ and $G_2,$ and $N_1$ is a normal subgroup of $G_1,$ then $\phi(N_1)$ is a normal subgroup of $G_2,$ isomorphic to $N_1,$ and $G_1/N_1\cong G_2/\phi(N_1).$
I'll leave this last claim to you.

At heart this is all due to the fact that an isomorphism can be thought of as a renaming of the elements of a group. Since the definition of $N(G)$ doesn't care about the names of the elements, the process of computing $N(G)$ and $G/N(G)$ do not depend on the names.


TITLE: What effect does amplitude have on EM waves?
QUESTION [2 upvotes]: There have been previous questions like this regarding light/photons specifically. My question is what the effect of the amplitude is on, say, a gamma wave? It would change its energy, as $E = h f$. Maybe amplitude doesn't even exist with an EM wave.

REPLY [2 votes]: The electric and magnetic field strength ($\bf{E}$ and $\bf{B}$)
is a concept of classical electrodynamics.
It emerges from quantum electrodynamics
as a collective effect only when there are many photons.
When considering only a few photons or even a single photon
(like when you have gamma rays) it doesn't make sense
to talk about electric and magnetic field or their amplitude.
The energy of a single photon is $E=hf$ (where $f$
is the frequency of the EM wave).
Therefore, the amplitude of electric and magnetic field
has no effect on the energy of the single photons.
Instead, the amplitude effects the number of photons.
The energy density of the EM field (i.e. energy per volume) is
$$\rho=\frac{\epsilon_0}{2}\bf{E}^2 + \frac{1}{2\mu_0}\bf{B}^2.$$
Therefore in an EM wave the number $N$ of photons within a volume
can be calculated from the volume integral
$$\begin{align}
N hf &= \int_V \rho\ dV \\
 &= \int_V \left( \frac{\epsilon_0}{2}\bf{E}^2 + \frac{1}{2\mu_0}\bf{B}^2 \right)dV \\
 &= \frac{1}{2} \left( \frac{\epsilon_0}{2}\hat{\bf{E}}^2 + \frac{1}{2\mu_0}\hat{\bf{B}}^2 \right) V
\end{align}$$
where $\hat{\bf{E}}$ and $\hat{\bf{B}}$ are the amplitudes of the EM wave.


TITLE: Are the transformation matrices the relation between basis $\beta$ and $\beta'$?
QUESTION [1 upvotes]: Are the transformation matrices the relation between basis $\beta$ and $\beta'$?
We are told to get the coordinates of vector  $v = (1,2,1)$ in both basis $\beta$ and $\beta'$, and find what the relation between the two basis is:
$$\beta = \{(1,1,-1),(0,-1,2),(2,0,1)\}$$
$$\beta' = \{(1,0,-1),(1,0,1),(0,1,1)\}$$
I get that the coordinates of the vector are:
$$M_\beta(v)= \left(
\begin{matrix}
9\\
7\\
-4\\
\end{matrix}\right) ,
M_\beta'(v)= \left(
\begin{matrix}
1\\
0\\
2\\
\end{matrix}\right)$$
And I get both transformation matrices:
$$M_{\beta}^{\beta'}= \left(
\begin{matrix}
3/2&-3/2&1/2\\
-1/2&3/2&3/2\\
1&-1&0\\
\end{matrix}\right) , M_{\beta'}^{\beta}= \left(
\begin{matrix}
-3&1&6\\
-3&1&5\\
2&0&-3\\
\end{matrix}\right)$$
I've tested that both transformation matrices work and also that I get $I_3$ when multiplying both of them:
$$M_\beta^{\beta'}.M_{\beta'}^{\beta} = I_3\\
M_\beta^{\beta'}.M_\beta(v) = M_{\beta'}(v)\\
M_{\beta'}^{\beta}.M_{\beta'}(v) = M_\beta(v)$$
Can I say that the transformation matrices are the relation between $\beta$ and $\beta'$?
Hope this is not a stupid question...

REPLY [0 votes]: I suppose so.
Note that if you form the matrix consisting of the columns of $\mathcal B$ ($\mathcal B'$) you get the change of basis matrix from $\mathcal B$ (respectively $\mathcal B'$) to the standard basis.   Thus we should have:$$[I]_{\mathcal B}^{\mathcal B'}=[I]_S^{\mathcal B'}[I]_{\mathcal B}^S=\begin {pmatrix}1&1&0\\0&0&1\\-1&1&1\end{pmatrix}^{-1}\cdot \begin{pmatrix}1&0&2\\1&-1&0\\-1&2&1\end{pmatrix}$$. This should be your $[M]_{\mathcal B}^{\mathcal B'}$.  Indeed it is.


TITLE: On notation of metrics.
QUESTION [1 upvotes]: I tried to read a book on Holomorphic Dynamics, and I found that $dp = \text{something}$ is a metric. In which sense an expression of type "infinitesimal" can be a metric. I only understand metrics as distances. Can you tell me where to read the context of such metrics?

REPLY [0 votes]: I suppose that the book that you mentioned is concerned with metric tensors on Riemannian manifolds. They do induce a metric in the usual sense of the word. For instance, the usual metric in $\mathbb{R}^n$ can be obtained by this process.