TITLE: Chain of inequlities in Complex variables
QUESTION [0 upvotes]: I am having difficulty understanding the following inequalities which is part of a solution to a problem:
Suppose
\begin{align}
|z-1| &< 1/2\\
|z+1|&< 5/2\\
|z|&> 1/2\\
|\sqrt{z^2+3}| &> 3/2
\end{align}
Then we get
$$
\frac{1}{\lvert\sqrt{z^2+3}\rvert+2}<\frac{1}{\frac{3}{2}+2}
$$
and
$$
\frac{1}{\lvert z\rvert+\frac{1}{2}\lvert\sqrt{z^2+3}\rvert} < \frac{1}{\frac{1}{2}+\frac{3}{4}}
$$

REPLY [0 votes]: Shouldn't there be moduli symbols around the denominators in the third line?
If you mean $\displaystyle \frac{1}{| \sqrt{z^2+3}|+2}< \frac{1}{\frac{3}{2} + 2}$ then this is clear because $0 < A < B$ implies that $\displaystyle \frac{1}{A}>\frac{1}{B}$ along with the basic theory of inequalities. The other is proved similarly provided one is asked to show that $\displaystyle \frac{1}{| \frac{1}{2}\sqrt{z^2+3}|+|z|}< \frac{1}{\frac{3}{4} + \frac{1}{2}}$. In either case you don't need the other two inequalities.


TITLE: System of equations modulo $n$
QUESTION [1 upvotes]: Suppose we have the following system of equations module $n$
\begin{align*}
2x-y & \equiv 1\mod{n}\\
4x+3y & \equiv 2 \mod{n}
\end{align*}
Determine the integers $n$ for which the system has solutions.
I have posted my solution as an answer below.

REPLY [2 votes]: Let's derive the answer using general methods. First triangularize:  $ $ eliminate $\,y\,$ by substituting $\, y\equiv 2x\!-\!1\,$ into the second equation, yielding the following equivalent system
$$\begin{align} y\,-\,2x&\equiv -1\!\!\!\pmod{\!n}\\ 10x&\equiv \ \ 5\!\!\pmod{\!n}\end{align}\qquad$$
This is solvable $\!\iff\! 10x\equiv 5\pmod{\!n}\,$ is solvable $\!\iff\! \color{#c00}{\gcd(10,n)\mid 5},\,$ by this criterion.
If $\ 2\mid n\ $ then $\,\color{#90f}{10x}\equiv \color{#0a0}5\pmod{\!n}\,$ persists $\!\bmod 2,\,$ reducing to $\,\color{#90f}0\equiv \color{#0a0}1\pmod{\!2},\,$ contradiction.
So $\,2\nmid n,\,$ so Euclid $\,\Rightarrow\, \color{#c00}{\gcd(10,n)} = \gcd(5,n)\,$ which does $\rm\color{#c00}{divide\ 5},\,$ confirming solvability.
Remark $\ $ Likely your solution arose by eliminating $\,x\,$ to obtain
$$\begin{align}  2x-y&\equiv 1\!\!\pmod{\!n}\\ 5y&\equiv 0\!\!\pmod{\!n}\end{align}\qquad$$
then noticing an "obvious" solution $\,y\equiv 0\,\Rightarrow\, x\equiv 1/2,\,$ which exists by $\,n\,$ is odd (vs. opposite order above, where we notice an "obvious" solution $\,x\equiv 1/2\,\Rightarrow\, y\equiv 0).$


TITLE: Self induction: Why is induced voltage smaller than the applied voltage?
QUESTION [4 upvotes]: Imagine a circuit with a voltage source, a switch and an inductivity all connected in series.
First, the switch is open and there's no current and no magnetic field around. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductivity. Lenz's Law tells us that the induced voltage from the inductivity will always be such that the change in the magnetic flux is reduced. And we have Faraday's Law which tells us that the induced voltage is equal to the change in the magnetic flux.
What is the theoretical reason why the change in magnetic flux must be smaller than the applied voltage? Why does for t -> infinity always run a current trough the inductivity as it would be just a resistor? Is it just because the wires have a resistance? What would then happen when taking superconductors instead of wires? Is it the inner resistivity of the voltage source then?

REPLY [1 votes]: Well there are many questions in your question and I'll try to address them one by one.
As you may know,
$L* di/dt = V (t)$
one important thing to note about inductors is that unless there is an impulsive excitation(that is, the input is bounded), it's current can not make instantaneous jumps.
Therefore, when you apply the voltage, it will still have a 0 current, it will have a high slope, but 0 current nonetheless at $0^+$ where $0^+$ denotes the instant right after 0.
Ok, so using $L* di/dt = V (t)$ and basic first order circuit analysis coupled with a bit of differential equations, you get the current equation:
$i(t) = V/R *(1 - e^ {-tR/L})$
R is the total thevenin resistance in the circuit. This includes the internal resistance of the inductor. This is the Zero state response of the inductor and it does not take into account any form of initial condition, however it is not hard to implement that into the formula. As you can see, the current will only reach its final value as t -> infinity.
If you assume the inductor circuit has zero resistance, it would be a short circuit between the terminals of the voltage source. If you assume there is an output resistance of voltage source, you can separate it from the voltage source and use in the calculation as R.
Change in the magnetic flux is basically the self induced EMF. The differential equation basically points out to the fact that most things in the inductor occur in an exponential decay kind-of fashion. I hope this answers at least some of your questions. Let me know in the comments if anything is not clear.
Edit: Just to be clear * is multiplication in my answer, not convolution or something like that.


TITLE: Image of corresponding operator to specific tensor of type $(1,1)$
QUESTION [0 upvotes]: Suppose $V$ is a finite-dimensional vector space and let $v\in V$ and $\varphi\in V^*$. Find the image of operator which corresponds to the tensor $v\otimes \varphi$.
My approach: Since $V\cong V^{**}$ in canonical way then we can consider vector $v$ as a tensor of type $(0,1)$. And $\varphi$ is tensor of type $(1,0)$. Then $v\otimes \varphi\in T_1^1(V)$, where by $T_p^q(V)$ I mean the linear space of all tensors of type $(p,q)$.
Also we know that there is canonical isomorphism between $L(V)$ and $T_1^1(V)$, where by $L(V)$ I denoted the linear space of all operators on $V$. This canonical isomorphism is given by $\pi:L(V)\to T_1^1(V)$ such that $f\mapsto \pi_f$ s.t. $\pi_f(u,\xi)=\xi(f(u))$ for all $u\in V, \xi\in V^*$.
By above remark it follows that there is an operator $f$ on $V$ s.t. $\pi_f=v\otimes \varphi$. It means that for all $u\in V, \xi\in V^*$ we have $\xi(f(u))=\xi(v)\varphi(u)$.
Let $y\in \operatorname {Im}f$ then $y=f(x),x\in V$. Then $\xi(f(x))=\xi(v)\varphi(x)$ and let $\alpha:=\varphi(x)$ then $\xi(f(x)-\alpha v)=0$ for all $\xi in V^*$ then it means that $f(x)=\alpha v$ i.e. $y=\alpha v\in \langle v\rangle $, i.e. $\operatorname {Im}f\subseteq \langle v\rangle$.
However the converse inclusion is true if $\varphi \neq 0$ because if $\varphi=0$ then $v\otimes \varphi=0$  and corresponding operator $f$ is zero operator. But the image of such operator is $\{0\}$ which may not be $\langle v\rangle$.
Assume that $\varphi\neq 0$ then $\exists x\in V$ s.t. $\varphi(x)\neq 0$. Then $\xi(f(x))=\varphi(x)\xi(v)$ and let $\lambda=\varphi(x)\neq 0$ and then $\xi(f(x)-\lambda v)=0$ for all $\xi \in V^*$ which implies that $f(x)=\lambda v$ then $v=f(\lambda^{-1}x)$. Hence $v\in \operatorname{Im}f$ then $\langle v\rangle \subseteq \operatorname{Im}f$.
In summary, if $\varphi\neq 0$ then $\operatorname{Im}f=\langle v\rangle$.
Is this reasoning correct? Would be very grateful for any comments, please.

REPLY [1 votes]: Yes, your reasoning is correct. In fact, note that the inverse of the isomorphism $L(V) \to T_1^{~1}(V)$ you wrote is the unique linear map $T_1^{~1}(V) \to L(V)$ characterized by $$T_1^{~1}(V) \ni v\otimes \varphi \mapsto \big(x\mapsto\varphi(x)v  \big)\in L(V), $$and $x \mapsto \varphi(x)v$ has rank $1$ if $\varphi$ and $v$ are non-zero, and it is the zero operator if either $\varphi$ or $v$ is zero.


TITLE: Nested Radicals which is approximately 1.75
QUESTION [2 upvotes]: We know that from using Ramanujan formula, we have :
$$3 = \sqrt{1+2\sqrt{1+\sqrt{3+\cdots}}}$$
Now, suppose I define the following sequence :
$$ x_1 = 1\\
x_2 = \sqrt{1+\sqrt{2}}\\
\vdots\\
x_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots}}}
$$
I have checked numerically the the value for $x_n$ is approximately $1.7$ but it does not equate to $\sqrt3$.
Can anyone help me to solve the exact value of this sequence?

REPLY [4 votes]: The value to which the sequence converges is called the Nested radical constant, no closed-form expression is known for this constant.


TITLE: A commutative Banach algebra with an abundance of discountinuous functions
QUESTION [2 upvotes]: Let $A$ be  the  algebra of all bounded functions from $[0,\;1]$ to  $\mathbb{C}$.
For $f\in A,\;$ $\omega_{f}$ is the  standard oscillation  function.. Each of the following two (equivalent) norms on $A$, defines  a  Banach  algebra  structure on  $A$.
$$\parallel f \parallel=\parallel f\parallel_{\infty}+ \parallel \omega_{f}\parallel_{\infty}$$ or $$\parallel f \parallel=\parallel f\parallel_{\infty}+ \int_{[0,\;1]} \omega_{f}(x)dx$$  The  later is  well defined, since the  oscillation  function is  a bounded measurable function.
Questions:

Let $X$ be the Gelfand spectrum of $A$. What is  the  topological  structure of this  compact nonmetrizable disconnected Haussdorf space, precisely ?   Is it homeomorphic  to a  known  space?

Assume that $B$ and  $C$ are two $C^{*}$-algebras which are  embedded isometrically into $A$. Must their $C^{*}$-tensor product be embedded in $A$, too?

Is there a $C^{*}$-norm on $A$ which is equivalent to the above norms? In particular is  $A$, semi-simple?

REPLY [2 votes]: This is an answer to Q2, because as pointed out by Yemon, Q1 and Q3 follow easily from the fact that this algebra is isomorphic to $C(\beta [0,1]_d)$. It is enough to prove then that if two compact spaces $X$ and $Y$ are continuous images of $\beta \kappa$ for some cardinal $\kappa$, then so is $X\times Y$.
Let $X$ be a compact Hausdorff space. Let $D$ be a dense subset of $X$ endowed with the discrete topology. The identity map
$$\iota\colon D \to X$$
is continuous. By the universal property of the Stone–Čech functor $\beta$, $\iota$ extends (uniquely) to a continuous map
$$\beta \iota \colon \beta D \to X.$$
Since continuous images of compact spaces are compact, $\beta \iota$ is surjective. Hence we arrive at the following conclusion.
Conclusion. Let $X$ be a compact Hausdorff space with density character $\kappa$. Then $X$ is a continuous image of $\beta \kappa$. Consequently, since the product of two compact spaces with character $\kappa$ is compact and has density character $\kappa$, if $X$ are $Y$ are continuous, Hausdorff images of $\beta \kappa$, then so is $X\times Y$.


TITLE: Centroid for a rectangular section inclined at an angle theta
QUESTION [1 upvotes]: How to find Centroid for a rectangular section inclined at an angle theta? Is there any general formula available?

REPLY [1 votes]: The location of the centroid of a body doesn't change with rotation . If you are working in a coordinate system, then only the coordinates of the centroid change (not with respect to the body but with respect to axes). Thus,in a rectangle centroid $(x,y)$ is intersection of two diagonals and if the rectangle is rotated by $\theta$ then  new coordinates$(x',y')$ are given by system of equations $$x'=x\cos\theta-y\sin\theta$$ and $$y'=x\sin\theta+y\cos\theta$$


TITLE: If $A^3B = BA^3$, then $AB = BA$.
QUESTION [6 upvotes]: Let $A$ be a Hermitian matrix. Suppose there exists a matrix $B$ such that $A^3B = BA^3$. Show that $AB = BA$.

I was trying to use the fact that since $A$ is Hermitian, there exists a unitary matrix $U$ such that $UDU^* = A$, thus
$UD^3U^* B = BUD^3U^*$,
so $UD^3U^* B - BUD^3U^* = 0$, $D^3U^* BU - U^*BUD^3= 0$ (multiplying both sides by $U^*$ and $U$.
Let $U^* BU = C$, then we have $D^3C = C D^3$, but I get stuck from here...

REPLY [7 votes]: Suppose the diagonal entries of of $D$ are $\lambda_1,\lambda_2,\ldots, \lambda_n$.  Let $p(x)$ be an interpolating polynomial such that $p(\lambda_k^3)=\lambda_k$ for each $k$.  Then you can show that $p(A^3)=A$ (e.g. because this holds when restricted to an eigenbasis), and it is clear that $p(A^3)$ commutes with $B$.
This relies on the fact that real cube roots are unique, so that $p$ can be defined as specified.  You can see by example that this is false for $A^2$, because $A^2$ might have repeated eigenvalues coming from distinct eigenvalues of $A$.  Concretely, this would be the case with $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$.


TITLE: Let $z = f(y(x), x)$. Show that $\frac{dz}{dx} = \frac{\partial z}{\partial x} y + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x}$
QUESTION [0 upvotes]: Hello I am trying to show that $$\frac{dz}{dx} = \frac{\partial z}{\partial x} y + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x},$$ where $z = f(y(x), x)$.
As per my intuition, I went with the total derivative of $z$ with respect to $x$, but I get $$\frac{dz}{dx} = \frac{\partial z}{\delta x} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial x}.$$
Could someone please point out the concept I am missing that is causing no $y$ along with $\dfrac{\partial z}{\partial x}$ in the first term of $\dfrac{dz}{dx}$?
The derivative appears in https://www.cs.cornell.edu/home/kleinber/aer15-prediction.pdf

REPLY [0 votes]: In the linked paper, there are parentheses: $(Y)$. I think it does not mean a product, but an evaluation: after computing the partial derivative, you have to evaluate at $(X_0,Y)$ which requires prediction of $Y$, as discussed later in the paper.


TITLE: Gram determinant
QUESTION [2 upvotes]: How to prove that
$$\sqrt{\Gamma(\vec{a},\vec{b},\vec{c})}=|(\vec{a},\vec{b},\vec{c})|,$$
where $\Gamma(\vec{a},\vec{b},\vec{c})= \left | \begin{array} {ccc} \vec{a} \cdot \vec{a} & \vec{b} \cdot \vec{a} & \vec{c} \cdot \vec{a} \\
\vec{a} \cdot \vec{b} & \vec{b} \cdot \vec{b} & \vec{c} \cdot \vec{b} \\
\vec{a} \cdot \vec{c} & \vec{b} \cdot \vec{c} & \vec{c} \cdot \vec{c}
\end{array} \right | $.
Thanks!

REPLY [0 votes]: Presumably $\vec{a}$ and the like are column vectors. Your $\Gamma$ is just the matrix product $A^\ast A$, where $A=(\vec{a},\vec{b},\vec{c})$ and $A^\ast$ denotes the conjugate transpose of $A$ (or simply the transpose of $A$ if $\vec{a},\vec{b},\vec{c}$ are real). Using the properties that $\det(XY)=\det(X)\det(Y)$ and $\det(X^\ast)=\overline{\det(X)}$, we see that $\det(\Gamma)=\overline{\det(A)}\det(A)=|\det(A)|^2$ and hence the correct statement should be:
$$\sqrt{\det(\Gamma)}=|\det(A)|,$$
where the pair of strokes $|\cdot|$ here denotes the modulus of a complex number (or absolute value of a real number). Without taking the modulus, your statement is wrong. A counterexample is given by $\vec{a}=(-1,0,0)^\top,\ \vec{b}=(0,-1,0)^\top$ and $\vec{c}=(0,0,-1)^\top$.


text:
\begin{document}
\maketitle

\begin{abstract}
We establish Bochner-type formulas for operators related
to $CR$ automorphisms and spherical $CR$ structures. From
such formulas, we draw conclusions about rigidity by making
assumptions on the Tanaka-Webster curvature and torsion.
\end{abstract}

\section{Statement of results}
It is now clear (e.g. [CL1], [CL2], [CT], [Rum]) that certain distinguished
second-order partial differential operators and their fourth-order
``Laplacians'' play
important roles in the study of three-dimensional $CR$ geometry. In
this paper we will establish Bochner-type formulas for operators
related to $CR$ automorphisms and spherical $CR$ structures. From
such formulas, we can draw conclusions about rigidity by making
assumptions on the so-called Tanaka-Webster curvature and torsion.

To be precise, let $(M,J,\theta )$ be a smooth, closed (compact without
boundary) 3-dimensional strictly pseudoconvex pseudohermitian manifold.
(see, e.g. [Web],[Le1]) Here $J$ denotes a $CR$ structure and $\theta$
is a contact form, i.e. a nonvanishing real 1-form defining the underlying
contact structure. Associated to $(M,J,\theta )$, we have a canonical affine
connection and notions of Tanaka-Webster scalar curvature and torsion,
denoted $R$ and $A$ (a tensor with coefficient $A_{11}$ or ${A_1}^{\bar 1}$)
respectively. ([Le1],[Tan],[Web])

Let $T$ be the characteristic vector field of $\theta$ defined by
${\theta}(T)=1, {\mathcal{L}}_{T}{\theta}=0$. (some authors call $T$ the
Reeb vector field) By choosing suitable complex vector fields $Z_{1},Z_{\bar 1}
$ such that $JZ_{1}=iZ_{1}, JZ_{\bar 1}=-iZ_{\bar 1}$, we form a (unitary)
frame $\{ Z_{1},Z_{\bar 1},T \}$. ($h_{1{\bar 1}}=1$) The covariant derivatives
are taken with respect to this frame and indicated by ${_{,1}},
{_{,{\bar 1}}},{_{,0}}$ and so on. Let $Aut_{0}(J)$ denote the identity
component of the $CR$ automorphism group with respect to $J$.

Now we can state our first result.

\medskip

$\bf {Theorem \: A}$. Let $(M,J,\theta )$ be a smooth, connected, closed
3-dimensional strictly pseudoconvex pseudohermitian manifold.

(a) Suppose $R < 0, {\sqrt 3}R_{,0}-2Im(A_{11,{\bar 1}{\bar 1}}) > 0$.

\noindent Then $Aut_{0}(J)$ consists of only the identity diffeomorphism.

(b) Suppose $R < 0, {\sqrt 3}R_{,0}-2Im(A_{11,{\bar 1}{\bar 1}}) = 0$

\noindent Then $dim Aut_{0}(J){\leq}1$.

\medskip

We remark that the torsion $A=0$ implies the second condition in (b) holds
due to the Bianchi identity: $R_{,0}=A_{11,{\bar 1}{\bar 1}}+
A_{{\bar 1}{\bar 1},11}$. And the result is compatible with Prop.4.8(a) in
[CT]. On the other hand, we do not know any examples satisfying curvature
conditions in (a).

The idea of proving $Theorem \: A$ goes as follows. Consider a certain
second-order linear partial differential operator ${\mathcal D}_J$ and its ``Laplacian'' ${\mathcal D}_J
^{\star}{\mathcal D}_J$ acting on functions. Here ${\mathcal D}_J^{\star}$
denotes the adjoint of ${\mathcal D}_J$.  The kernel of ${\mathcal D}_J$
parametrizes infinitesimal $CR$ automorphisms. (see [CL1] where ${\mathcal D}_J$ was denoted $B_J^{'}$) Establish a suitable Bochner-type formula for
$<{\mathcal D}_J^{\star}{\mathcal D}_{J}f,f>$ by using commutation relations
and integration by parts repeatedly. (see section 2 for details) $Theorem\: A$
then follows easily from the final formula. Note that the kernel of
${\mathcal D}_J^{\star}$ parametrizes the infinitesimal slice in the study
of $CR$ moduli spaces. ([CL2])

Next we consider deformation of spherical $CR$ structures. A $CR$ structure
or $CR$ manifold is called spherical if it is locally $CR$-equivalent to
the unit sphere with the standard $CR$ structure. (e.g. [BS],[CL1])
In dimension 3 it can be characterized quantitatively by the vanishing of
a certain fourth-order partial differential operator, the so-called Cartan
(curvature) tensor, denoted $Q_J$. (while in higher dimensions, Chern's
curvature tensor plays the similar role ([CM]), which is of second order)

A spherical $CR$ structure $J$ is called rigid if there is no infinitesimal
deformation up to diffeomorphisms, i.e. for any smooth family of spherical
$CR$ structures $J_{(t)}$ on the base manifold $M$ with $J_{(0)}=J$,
$d/dt|_{t=0}J_{(t)}$ equals ${\mathcal L}_{X}J$ for some vector field $X$ of
$M$.

To study the rigidity of spherical $CR$ structures, we consider
the linearization of $Q_J$ plus a symmetry-breaking term provided by
${\mathcal D}_{J}{\mathcal D}_J^{\star}$. In section 3 we work out a
Bochner-type formula and analyze it to obtain pointwise conditions for
$J$ to be rigid.

\medskip

$\bf {Theorem \: B}$. Let $(M,J)$ be a smooth, closed, spherical $CR$ 3-manifold.
Suppose there is a contact form $\theta$ such that $R > 0$, (3.11) and (3.12)
hold. Then $J$ is rigid.

\medskip

Note that conditions (a),(b) in $Theorem \: A$ and (3.11),(3.12) in $Theorem \: B$ are independent of positive constant multiples of $\theta$. When the torsion
vanishes, we have the simplified expression as follows.

\medskip

$Corollary \: C$. Let $(M,J)$ be a smooth, closed, spherical $CR$ 3-manifold.
Suppose there is a contact form $\theta$ such that the torsion $A=0$ and

$$
R > 0, 4R(5R^{2}+3{\bigtriangleup}_{b}R)-3|{\nabla}_{b}R|_{\theta}^{2}>0
$$

\noindent Then $J$ is rigid.

\medskip

The sublaplacian and subgradient operators ${\bigtriangleup}_{b},{\nabla}_{b}$
acting on (smooth) functions are defined by
${\bigtriangleup}_{b}f=-{f_{,1}}^{1}
-{f_{,{\bar 1}}}^{\bar 1}$ and ${\nabla}_{b}f={f_{,}}^{1}Z_{1}+{f_{,}}^{\bar 1}
Z_{\bar 1}$ respectively. (cf. [Le1] or [Che]) Also we define $|{\nabla}_{b}f|_{\theta}^{2}=2f_{,1}{f_{,}}^{1}$ for real f.

Observe that $A=0$ and $R$ being
a positive constant satisfy conditions in $Corollary \, C$. In this case the
universal cover of $(M,J)$ must be compact by Rumin's pseudohermitian version
of Myers' theorem ([Rum]), and hence $CR$-equivalent to the standard $S^{3}$. It follows that the fundamental group $\Gamma$ of
$M$ is finite. Hence the group cohomology $H^{1}({\Gamma},{\mathcal G})$ in
deformation theory (p.232 in [BS]) vanishes. So in this special case our
result is compatible with the result obtained by ``Lie theoretical'' argument.
Note that a small perturbation of $A=0$ and $R$ being a positive constant
still satisfies the conditions in $Theorem \: B$.

\medskip

$\bf{Acknowledgment}$. This work was carried out during the author's
visit at Harvard University. He would therefore like to thank the members
of the Mathematics Department, and especially Professor Shing-Tung Yau,
for their hospitality during his stay.

\bigskip

\section{Proof of Theorem A}

Let $\{ {\theta}^{1},{\theta}^{\bar 1},{\theta}\}$ be the coframe dual to
the ``unitary'' frame $\{ Z_{1},Z_{\bar 1},T \}$. (with $h^{1{\bar 1}}=
h_{1{\bar 1}}=1$ in mind, hereafter, we'll write tensors with only lower
indices) Recall ([CL2] or [CL1]) that ${\mathcal{D}}_{J}f= 2Re[(f_{,11}+
iA_{11}f){\theta}^{1}{\otimes}Z_{\bar 1}]$ and the adjoint operator
${\mathcal D}_J^{\star}$ acts on a deformation tensor $E= 2Re(E_{11}{\theta}^
{1}{\otimes}Z_{\bar 1})$ by

\begin{equation}
{\mathcal D}_J^{\star}E=E_{11,{\bar 1}{\bar 1}}+iA_{11}E_{{\bar 1}{\bar 1}}+
conjugate.
\end{equation}

Also the generalized Folland-Stein operator $L_{\alpha}$ is defined by
$L_{\alpha}f={\bigtriangleup}_{b}f+i{\alpha}f_{,0}$ for a function f.
By Lemma 2.1 in [CL2], we have

\begin{equation}
{\mathcal D}_J^{\star}{\mathcal D}_{J}=(1/2)L_{\alpha}^{\star}L_{\alpha}
+{\mathcal O}_{2}
\end{equation}

\noindent with ${\alpha}=i{\sqrt 3}$, where ${\mathcal O}_{2}$
is an operator of
weight ${\leq}2$. In the rest of this section, we'll look into the details
of ${\mathcal O}_{2}$. (Note that $L_{\alpha}$ in the leading term of (2.2)
is subelliptic. This implies the existence of the ``infinitesimal slice
decompositions'' in the study of $CR$ moduli spaces without knowing
details of the lower-weight term ([CL2]))

A direct computation shows that for a real-valued function f, we have

\begin{eqnarray}
&&{\mathcal D}_J^{\star}{\mathcal D}_{J}f=f_{,11{\bar 1}{\bar 1}}+f_{,{\bar 1}{\bar 1}11}
+2Re[2iA_{11}f_{,{\bar 1}{\bar 1}} \\
&&+2iA_{11,{\bar 1}}f_{,{\bar 1}}+
iA_{11,{\bar 1}{\bar 1}}f+|A_{11}|^{2}f]. \nonumber
\end{eqnarray}

We'll frequently use the following commutation relations.

\medskip

$\bf {Lemma \: 2.1}$(Ricci identities in pseudohermitian geometry) Let $c_I$
be a coefficient of some tensor with multi-indices $I$. Suppose $I$ consists
of only $1$ and $\bar 1$, and ${\alpha}$=(\# of $1$ in $I$)-(\# of $\bar 1$ in
$I$).Then

\begin{equation}
c_{I,1{\bar 1}}-c_{I,{\bar 1}1}=ic_{I,0}+{\alpha}c_{I}R
\end{equation}

\begin{equation}
c_{I,01}-c_{I,10}=c_{I,{\bar 1}}A_{11}-{\alpha}c_{I}A_{11,{\bar 1}}
\end{equation}

\begin{equation}
c_{I,0{\bar 1}}-c_{I,{\bar 1}0}=c_{I,1}A_{{\bar 1}{\bar 1}}
+{\alpha}c_{I}A_{{\bar 1}{\bar 1},1}
\end{equation}

\noindent
(this lemma generalizes Lemma 2.3 in [Le2] for the three-dimensional case)

\medskip

By using (2.4),(2.5),(2.6) repeatedly, we obtain

\begin{equation}
{\frac{1}{2}}L_{\alpha}^{\star}L_{\alpha}f=f_{,11{\bar 1}{\bar 1}}
+f_{,{\bar 1}{\bar 1}11}
+2Re[({\sqrt 3}-i)(A_{{\bar 1}{\bar 1}}f_{,1})_{,1}-(Rf_{,1})_{,{\bar 1}}]
\end{equation}

\noindent with the choice of ${\alpha}=i{\sqrt 3}$ eliminating
terms having covariant
derivative in the direction $T$. Comparing (2.3) with (2.7) and taking $L^{2}$
-inner product with $f$ give

\begin{eqnarray}
&{\|}{\mathcal D}_{J}f{\|}^{2}&={\frac{1}{2}}{\|}L_{i{\sqrt 3}}f{\|}^{2}
-{\int_M}R|{\nabla}_{b}f|_{\theta}^{2}dv_{\theta} \\
&& +2{\int_M}[Re({\frac{{\sqrt 3}+i}{2}}A_{11,{\bar 1}{\bar 1}})+|A_{11}|^{2}]
f^{2}dv_{\theta} \nonumber \\
&& +2{\int_M}Re[(-i-{\sqrt 3})A_{{\bar 1}{\bar 1}}f_{,11}]f
dv_{\theta} \nonumber
\end{eqnarray}

Here the volume form $dv_{\theta}={\theta}{\wedge}d{\theta}$. With respect to
$dv_{\theta}$, we have the divergence theorem and hence integration by parts
in calculus of pseudohermitian geometry. ([Le2],[Che]) For instance,

\begin{eqnarray*}
&{\int_M}A_{{\bar 1}{\bar 1},1}f_{,1}{\bar f}dv_{\theta} & ={\frac{1}{2}}
{\int_M}A_{{\bar 1}{\bar 1},1}(f^{2})_{,1}dv_{\theta}{\:}(f{\:}being{\:}real)\\
 & & =-{\frac{1}{2}}{\int_M}A_{{\bar 1}{\bar 1},11}f^{2}dv_{\theta}
{\:}(integration{\:}by{\:}parts)
\end{eqnarray*}

\noindent was used in deducing (2.8). Now suppose ${\phi}_t{\in}Aut_{0}(J)$ is
a smooth family of $CR$ automorphisms with ${\phi}_{0}=identity$. Then
$X={\frac{d}{dt}}|_{t=0}{\phi}_{t}{\in}$ Lie $Aut_{0}(J)$ (=Lie algebra of
$Aut_{0}(J)$) is an infinitesimal $CR$ automorphism, in particular, an
infinitesimal contact automorphism. According to Lemma 3.4 and 3.5 in
[CL1], $X=X_f$ is determined by a function $f=-{\theta}(X)$ and satisfies
the relation: ${\mathcal L}_{X}J=2{\mathcal D}_{J}f$. Since ${\mathcal L}_{X}J=0$, we
get

\begin{equation}
0={\mathcal D}_{J}f = f_{,11}+iA_{11}f,
\end{equation}

\noindent and hence

\begin{equation}
A_{{\bar 1}{\bar 1}}f_{,11} = -i|A_{11}|^{2}f.
\end{equation}

Substituting (2.9), (2.10) in (2.8), we finally obtain

\begin{eqnarray}
& 0 & ={\|}L_{i{\sqrt 3}}f{\|}^{2}-2{\int_M}R|{\nabla}_{b}f|_{\theta}^{2}
dv_{\theta}  \\
& & +{\int_M}[{\sqrt 3}R_{,0}+i(A_{11,{\bar 1}{\bar 1}}-A_{{\bar 1}{\bar 1}
,11})]f^{2}dv_{\theta} \nonumber
\end{eqnarray}

Now it is easy to see from (2.11) that the condition in (a) of $Theorem{\:}A$
implies $f=0$. Therefore $X=X_{f}=0$. For (b), the condition implies
${\nabla}_{b}f=0$. So $f_{,0}=0$ by (2.4). Thus $f$ is constant since $M$
is connected. It follows that $dim(Aut_{0}(J))=dim(Lie{\,}Aut_{0}(J))
{\leq}1$.

\bigskip

\section{Proof of Theorem B}
\setcounter{equation}{0}

 Recall ([CL1]) that the Cartan tensor $Q_{J}=iQ_{11}{\theta}^{1}{\otimes}
Z_{\bar 1}-iQ_{{\bar 1}{\bar 1}}{\theta}^{\bar 1}{\otimes}Z_{1}$ where

$$
Q_{11}={\frac{1}{6}}R_{,11}+\frac{i}{2}RA_{11}-A_{11,0}-\frac{2i}{3}A_{11,
{\bar 1}1}
$$

\noindent and $J$ is spherical if and only if $Q_{J}=0$. (note that we have lowered
indices using $h_{1{\bar 1}}=1$; also $Q_J$ changes ``tensorially'' when
we make a different choice of contact form) Let ${\tilde J}_{(t)}$ be
a smooth family of spherical $CR$ structures with ${\tilde J}_{(0)}=J$.
By a theorem of Gray ([Gra] or [Ham]), there exists a smooth family of
diffeomorphisms $\phi_t$ with $\phi_{0}=identity$ so that for all $t$,
$J_{(t)}={\phi_t^{\star}}{\tilde J}_{(t)}$ has the same underlying
contact structure as $J$ does. Write infinitesimal deformation

$$
\frac{d}{dt}|_{t=0}J_{(t)}= 2E= 4Re(E_{11}{\theta^1}{\otimes}Z_{{\bar 1}})
$$

\noindent and compute $DQ_{J}(2E)={\partial_t}Q_{J_{(t)}}|_{t=0}$ as we did in [CL1].
There appears a ``bad'' term $E_{{\bar 1}{\bar 1},1111}$ in the formula,
so we add a ``symmetry-breaking'' term ${\mathcal D}_{J}{\mathcal D}_J^{\star}
E$ to cancel it. The final formula including terms of lower weights reads

\begin{eqnarray}
& &-DQ_{J}(2E)+\frac{1}{6}{\mathcal D}_{J}{\mathcal D}_J^{\star}
E  =  2Re\{ \frac{1}{3}E_{11,{\bar 1}{\bar 1}11}\\
& &-E_{11,00}
-\frac{2i}{3}E_{11,0{\bar 1}1}
  +  \frac{i}{3}(A_{11}E_{{\bar 1}{\bar 1}})_{,11} \nonumber \\
& &-\frac{1}{6}E_{11}
R_{,1{\bar 1}}+\frac{1}{6}E_{11,{\bar 1}}R_{,1}-\frac{1}{6}
(E_{11}R_{,{\bar 1}})_{,1} \nonumber \\
 & & + \frac{1}{2} A_{11}(iE_{11,{\bar 1}{\bar 1}}-iE_{{\bar 1}{\bar 1},11}
-A_{11}E_{{\bar 1}{\bar 1}}-A_{{\bar 1}{\bar 1}}E_{11})+\frac{i}{2}RE_{11,0}
\nonumber \\
 & & +2A_{11}(A_{11}E_{{\bar 1}{\bar 1}}+A_{{\bar 1}{\bar 1}}E_{11})+
\frac{2i}{3}E_{11}A_{11,{\bar 1}{\bar 1}}-\frac{2i}{3}E_{11,{\bar 1}}
A_{11,{\bar 1}}-\frac{2i}{3}(E_{{\bar 1}{\bar 1}}A_{11,1})_{,1}
\nonumber \\
 & & -\frac{4i}{3}(E_{{\bar 1}{\bar 1},1}A_{11})_{,1}+\frac{i}{6}A_{11}
(E_{11,{\bar 1}{\bar 1}}+E_{{\bar 1}{\bar 1},11}+iA_{11}E_{{\bar 1}{\bar 1}}
-iA_{{\bar 1}{\bar 1}}E_{11})\} {\theta^1}{\otimes}Z_{\bar 1} \nonumber
\end{eqnarray}

The right-hand side of (3.1) can be written as $\frac{1}{12}L_{\alpha}^{\star}
L_{\alpha}E+{\mathcal O}_{2}(E)$ with ${\mathcal O}_{2}$ being an operator
of weight $\leq 2$ and ${\alpha}=4+i{\sqrt 3}$. Since $L_{\alpha}$ is
subelliptic, the above expression was used in [CL1] to show the short-
time solution of a certain regularized evolution equation. Using Lemma 2.1
repeatedly, we can write the highest-weight term of (3.1) as follows:

\begin{eqnarray}
&& \frac{1}{3}E_{11,{\bar 1}{\bar 1}11}-E_{11,00}-\frac{2i}{3}E_{11,0{\bar 1}1}=\frac{1}{3}E_{11,{\bar 1}1{\bar 1}1}-E_{11,00}-iE_{11,0{\bar 1}1}\\
&& +\frac{i}{3}(E_{11,1}A_{{\bar 1}{\bar 1}})_{,1}+\frac{2i}{3}
(E_{11}A_{{\bar 1}{\bar 1},1})_{,1}-\frac{1}{3}(RE_{11,{\bar 1}})_{,1}
\nonumber
\end{eqnarray}

On the other hand, we compute

\begin{eqnarray}
&&{\int_M}R|E_{11,1}|^{2}dv_{\theta} = -{\int_M}(RE_{11,1{\bar 1}}
E_{{\bar 1}{\bar 1}}+R_{,{\bar 1}}E_{11,1}E_{{\bar 1}{\bar 1}})dv_{\theta}
\\
&&(by{\:}integration{\:}by{\:}parts)  \nonumber \\
&& =-{\int_M}[R(E_{11,{\bar 1}1}+iE_{11,0}+2RE_{11})E_{{\bar 1}{\bar 1}}
+R_{,{\bar 1}}E_{11,1}E_{{\bar 1}{\bar 1}}]dv_{\theta}
{\:}(by{\:}(2.4)) \nonumber
\end{eqnarray}

To see how we treat (3.1) in general, we first deal with the torsion$=0$
case. By the local slice theorem ([CL2]), there exists a smooth family of
contact diffeomorphisms $\psi_t$ with ${\psi_0}=identity$ so that
$J_{(t)}^{'}={\psi}_{t}^{\star}J_{(t)}$ lies in the local slice passing
through $J$. Since the infinitesimal slice at $J$ is parametrized by the
kernel of ${\mathcal D}_J^{\star}$, we have ${\mathcal D}_J^{\star}E=0$
for $2E=\frac{d}{dt}|_{t=0}J_{(t)}^{'}$.

Now applying (3.1) to such a deformation tensor $E$: ${\mathcal D}_J^{\star}
E=0$, substituting (3.2) in (3.1), and taking $L^2$-inner product with $E$,
we obtain

\begin{eqnarray}
& 0 & =<-DQ_{J}(2E),E>{\:}(since{\:}Q_{J_{(t)}^{'}}=0) \\
& & ={\int_M}2Re\{ \frac{1}{3}|E_{11,{\bar 1}1}|^{2}+|E_{11,0}|^{2}
-iE_{11,0}E_{{\bar 1}{\bar 1},1{\bar 1}}+\frac{1}{6}R|E_{11,1}|^{2}
\nonumber \\
&& +\frac{2i}{3}RE_{11,0}E_{{\bar 1}{\bar 1}}
 -\frac{1}{6}RE_{11,{\bar 1}1}E_{{\bar 1}{\bar 1}}+\frac{1}{6}
R_{,{\bar 1}}E_{11,1}E_{{\bar 1}{\bar 1}} \nonumber \\
&& +[\frac{11}{6}R_{,{\bar 1}1}-
2R_{,1{\bar 1}}+\frac{1}{3}R^{2}]|E_{11}|^{2}\} dv_{\theta}
\nonumber
\end{eqnarray}

\noindent by putting $A_{11}=0$, using (3.3) and integration by parts
repeatedly.

Note that we reduce (3.4) to the formula (6.3) in [CL1] for $R$ being a
constant $\hat R$. Furthermore, if ${\hat R}>0$, the right-hand side of
(3.4) is a positive definite quadratic hermitian form in $E_{11,{\bar 1}1}$,
$iE_{11,0}$, and ${\hat R}E_{11}$.

It follows that $E=0$ and $0=\frac{d}{dt}|_{t=0}J_{(t)}^{'}=
\frac{d}{dt}|_{t=0}({\phi_t}{\circ}{\psi_t})^
{\star}{\tilde J}_{(t)}={\mathcal L}_{X}J+\frac{d}{dt}|_{t=0}{\tilde J}_{(t)}$
where the vector field $X=\frac{d}{dt}|_{t=0}({\phi_t}{\circ}{\psi_t})$.
So $J$ is rigid. For general $R$, we require that the integrand in (3.4)
is a (pointwise) positive definite quadratic hermitian form in
$E_{11,{\bar 1}1}$, $iE_{11,0}$, $E_{11,1}$, $E_{11}$.
Now the conditions in $Corollary{\:}C$ can be deduced from basic linear
algebra.

When the torsion does not vanish, the formula for $<-DQ_{J}(2E),E>$ with
${\mathcal D}_J^{\star}E=0$ reads

\begin{eqnarray}
& & 0=<-DQ_{J}(2E),E>={\int_M}\{ \frac{2}{3}|E_{11,{\bar 1}1}|^{2}+
2|E_{11,0}|^{2} \\
& & +\frac{1}{3}R|E_{11,1}|^{2}
+ [\frac{2}{3}R^{2}+\frac{1}{6}{\bigtriangleup}_{b}R+6|A_{11}|^{2}
+\frac{8i}{3}(A_{11,{\bar 1}{\bar 1}}-A_{{\bar 1}{\bar 1},11})]|E_{11}|^{2}
\nonumber \\
& & +2Re[-iE_{11,0}E_{{\bar 1}{\bar 1},1{\bar 1}}
+\frac{2i}{3}RE_{11,0}E_{{\bar 1}{\bar 1}}-\frac{1}{6}RE_{11,{\bar 1}1}
E_{{\bar 1}{\bar 1}} \nonumber \\
& & +(\frac{1}{6}R_{,{\bar 1}}-2iA_{{\bar 1}{\bar 1},1})
E_{11,1}E_{{\bar 1}{\bar 1}}-\frac{2i}{3}A_{11,1}E_{{\bar 1}{\bar 1},1}
E_{{\bar 1}{\bar 1}}-\frac{5i}{3}A_{11}E_{11,{\bar 1}}
E_{{\bar 1}{\bar 1},{\bar 1}}]\} dv_{\theta} \nonumber
\end{eqnarray}

There are non-cross terms like $A_{11,1}E_{{\bar 1}{\bar 1},1}E_{{\bar 1}{\bar 1}}$ and $A_{11}E_{11,{\bar 1}}E_{{\bar 1}{\bar 1},{\bar 1}}$ in (3.5).
We need an inequality to deal with  $A_{11}E_{11,{\bar 1}}E_{{\bar 1}{\bar 1},{\bar 1}}$.

\medskip

$\bf{Lemma \: 3.1}$. Let $\lambda$, $\rho$ be real numbers. Then

\begin{eqnarray}
2{\lambda}{\rho}{\int_M}R|E_{11,{\bar 1}}|^{2}dv_{\theta} & \leq &
{\lambda}^{2}{\int_M}|E_{11,{\bar 1}1}|^{2}dv_{\theta}+
{\rho}^{2}{\int_M}R^{2}|E_{11}|^{2}dv_{\theta} \\
 & & -{\lambda}{\rho}{\int_M}(R_{,1}E_{11,{\bar 1}}E_{{\bar 1}{\bar 1}}+
R_{,{\bar 1}}E_{{\bar 1}{\bar 1},1}E_{11})dv_{\theta}
\nonumber
\end{eqnarray}

For the term $A_{11,1}E_{{\bar 1}{\bar 1},1}E_{{\bar 1}{\bar 1}}$, we use
the following estimate:

\begin{equation}
2Re(-\frac{2i}{3}A_{11,1}E_{{\bar 1}{\bar 1},1}E_{{\bar 1}{\bar 1}})
{\geq}-\frac{2}{3}|A_{11,1}|^{\frac{2}{3}}|E_{{\bar 1}{\bar 1},1}|^{2}
-\frac{2}{3}|A_{11,1}|^{\frac{4}{3}}|E_{11}|^{2}.
\end{equation}

\noindent (To deduce (3.7), replace $a$, $b$ by ${\omega}a$, ${\omega}^{2}b$ in the
basic inequality $2Re(ab){\geq}-|a|^{2}-|b|^{2}$ with
$a=-iE_{{\bar 1}{\bar 1},1}$, $b=E_{{\bar 1}{\bar 1}}$, and ${\omega}^{3}=
A_{11,1}$)

The reason for taking fractional exponents in (3.7) is to make our conditions
invariant under the scale change of contact form by a positive constant
multiple as we'll see later. Take a small amount of ${\int_M}
|E_{11,{\bar 1}1}|^{2}dv_{\theta}$ and ${\int_M}R^{2}|E_{11}|^{2}
dv_{\theta}$ to gain the term ${\int_M}R|E_{11,{\bar 1}}|^{2}dv_{\theta}$
by (3.6) in the right-hand side of (3.5) while keeping the quadratic
hermitian form in $E_{11,{\bar 1}1}$, $iE_{11,0}$, $E_{11,1}$,
$E_{11,{\bar 1}}$, $E_{11}$ positive definite at least for $R=constant>0$,
$A_{11}=0$. For instance, we can take ${\lambda}={\rho}=\frac{1}{4}$ in
(3.6) and then use it and (3.7) in estimating the right-hand side of (3.5).
The final result reads

\begin{eqnarray}
0&\geq&{\int_M}\{ \frac{29}{48}|E_{11,{\bar 1}1}|^{2}+2|E_{11,0}|^{2}
+\frac{1}{3}R|E_{11,1}|^{2}\\
 &  & +(\frac{1}{8}R-\frac{2}{3}|A_{11,1}|^{\frac{2}{3}})|E_{11,{\bar 1}}|^
{2}+[\frac{29}{48}R^{2}+\frac{11}{48}{\bigtriangleup}_{b}R
\nonumber \\
 &  & +6|A_{11}|^{2}+\frac{8i}{3}(A_{11,{\bar 1}{\bar 1}}-A_{{\bar 1}{\bar 1},
11})-\frac{2}{3}|A_{11,1}|^{\frac{4}{3}}]|E_{11}|^{2}
\nonumber \\
 & & +2Re[-iE_{11,0}E_{{\bar 1}{\bar 1},1{\bar 1}}+\frac{2}{3}iRE_{11,0}
E_{{\bar 1}{\bar 1}}-\frac{1}{6}RE_{11,{\bar 1}1}E_{{\bar 1}{\bar 1}}
\nonumber \\
 & & +(\frac{5}{48}R_{,{\bar 1}}-2iA_{{\bar 1}{\bar 1},1})E_{11,1}
E_{{\bar 1}{\bar 1}}-\frac{5i}{3}A_{11}E_{11,{\bar 1}}E_{{\bar 1}{\bar 1},
{\bar 1}}]\} dv_{\theta}
\nonumber
\end{eqnarray}

The integrand in (3.8) is a quadratic hermitian form in $E_{11,{\bar 1}1}$,
$iE_{11,0}$, $E_{11,1}$, $E_{11,{\bar 1}}$, and $E_{11}$. By basic
linear algebra it is positive definite if and only if $R>0$,

\begin{equation}
{\left| \begin{array}{cccc}
\frac{29}{48} & -1 & 0 & 0 \\
-1 & 2 & 0 & 0 \\
0 & 0 & \frac{1}{3}R &\frac{5i}{3}A_{{\bar 1}{\bar 1}} \\
0 & 0 & \frac{-5i}{3}A_{11} & \frac{1}{8}R-\frac{2}{3}|A_{11,1}|^
{\frac{2}{3}}
\end{array}\right|}>0,\: and
\end{equation}

\begin{equation}
\left| \begin{array}{ccccc}
\frac{29}{48} & -1 & 0 & 0 & -\frac{1}{6}R \\
-1 & 2 & 0 & 0 & \frac{2}{3}R \\
0 & 0 & \frac{1}{3}R & \frac{5i}{3}A_{{\bar 1}{\bar 1}} &
\frac{5}{48}R_{,{\bar 1}}-2iA_{{\bar 1}{\bar 1},1} \\
0 & 0 & \frac{-5i}{3}A_{11} & \frac{1}{8}R-\frac{2}{3}|A_{11,1}|^
{\frac{2}{3}} & 0 \\
-\frac{1}{6}R & \frac{2}{3}R & \frac{5}{48}R_{,1}+2iA_{11,{\bar 1}} &
0 & \mathcal{M}
\end{array} \right|
\end{equation}

\noindent is larger than $0$.
Here $\mathcal{M}=\frac{29}{48}R^{2}+\frac{11}{48}{\bigtriangleup}
_{b}R+6|A_{11}|^{2}-\frac{2}{3}|A_{11,1}|^{\frac{4}{3}}+
\frac{8i}{3}(A_{11,{\bar 1}{\bar 1}}-A_{{\bar 1}{\bar 1},11})$.
A straightforward computation shows that (3.9) is equivalent to

\begin{equation}
\frac{3}{8}R^{2}-2R|A_{11,1}|^{\frac{2}{3}}-25|A_{11}|^{2}>0
\end{equation}

\noindent while (3.10) is equivalent to

\begin{eqnarray}
&& (\frac{3}{8}R^{2}-2R|A_{11,1}|^{\frac{2}{3}}-25|A_{11}|^{2})\{
\frac{83}{3456}R^{2}\\
&& +\frac{55}{1152}{\bigtriangleup}_{b}R+\frac{5}{4}|A_{11}|^{2}
-\frac{5}{36}|A_{11,1}|^{\frac{4}{3}}+\frac{5i}{9}
(A_{11,{\bar 1}{\bar 1}}-A_{{\bar 1}{\bar 1},11})\}
\nonumber \\
&& -\frac{15}{8}(\frac{1}{8}R-\frac{2}{3}|A_{11,1}|^{\frac{2}{3}})
|\frac{5}{48}R_{,{\bar 1}}-2iA_{{\bar 1}{\bar 1},1}|^{2}>0
\nonumber
\end{eqnarray}

Observe that if $\theta$ changes by a positive constant multiple
$k$, $R$ and $A_{11}$ change by multiplying $k^{-1}$ while
$A_{11,1}$,$A_{{\bar 1}{\bar 1},1}$, and $R_{,{\bar 1}}$ change
by multiplying $k^{-\frac{3}{2}}$. Similarly $A_{11,{\bar 1}{\bar 1}}
$, $A_{{\bar 1}{\bar 1},11}$, and ${\bigtriangleup}_{b}R$ change
by multiplying $k^{-2}$. So the conditions (3.11),(3.12) are
invariant under the change of contact form by a positive
constant multiple. Now $Theorem \: B$ follows from (3.8) under
the conditions (3.11),(3.12).

config: arxiv


TITLE: Is the quantization of gravity necessary for a quantum theory of gravity? Part II
QUESTION [7 upvotes]: (At the suggestion of the user markovchain, I have decided to take a very large edit/addition to the original question, and ask it as a separate question altogether.)
Here it is:
I have since thought about this more, and I have come up with an extension to the original question. The answers already given have convinced me that we can't just leave the metric as it is in GR untouched, but at the same time, I'm not convinced we have to quantize the metric in the way that the other forces have been quantized. In some sense, gravity isn't a force like the other three are, and so to treat them all on the same footing seems a bit strange to me. For example, how do we know something like non-commutative geometry cannot be used to construct a quantum theory of gravity. Quantum field theory on curved non-commutative space-time? Is this also a dead end?

REPLY [4 votes]: Quantum field theory on curved non-communative spacetime is exactly what is used in the "semiclassical gravity" approach, an early example of which is Hawking's original derivation of the Hawking radiation effect.  The limitation of this approach should be obvious--when the Hawking radiation ends up having a mass comparable to the original black hole, how can you trust the result?  The Hawking radiation has mass and energy, too, and obviously, this should be factored into the result, but the semiclassical problem explicitly ignores it.  (and if you try to do an iterative approach, factoring in the Hawking radiation as a source, and calculating the result, you quickly run into a LOT of complexity)
Non-communative geometry is an active area of research and a potential solution, albeit one chosen by a minority of researchers


TITLE: Surjective closed morphism of schemes induces inequality of dimensions
QUESTION [4 upvotes]: Let $f: Y\to X$ be a surjective closed morphism of schemes. Prove
  $\dim Y\geq \dim X$.

I was made to believe that this should be easy, but I somehow did not manage to prove this. Using the fact that $f$ is closed and surjective, it is easy to reduce this to the case where both $X$ and $Y$ are irreducible. Now, if both were (locally) of finite type over a field, then the inequality would follow from the fact that the morphism induces an extension of the function fields. In the general case, however, I did not manage to produce a proof.
Clearly it would be enough to prove the following lemma:

Let $f:Y\to X$ be a closed morphism of schemes. Then for any $y\in Y$ and any generization $x_0$ of $x=f(y)$ there is a generization $y_0$ of $y$ with $f(y_0)=x_0$.

This lemma is true in the case where $f$ is an open morphism, but I don't know whether it is true in the closed case as well. At any rate, I also failed to prove this so far.
Any input on this problem would be highly appreciated, especially as I suspect that I am making things unneccessarily complicated right now.

REPLY [0 votes]: EEDIT: The accepted answer has been edited and it looks pretty good now. If that answer is completely clear to you, then there is no need for you to read anything below.
For what is mentioned as the point-irreducible closed subset correspondence below (which is also mentioned as "the key fact" above in the accepted answer), I mean the following proposition in Görtz & Wedhorn, Algebraic Geometry I.

Proposition 3.23. Let $X$ be a scheme. The mapping
$$ X\to \{Z\subset X ; Z \text{ is closed and irreducible}\}:x\mapsto \overline{\{x\}}$$
is a bijection, i.e. every irreducible closed subset contains a unique generic point.

EDIT:
For what is below, let $f:X\to Y$ be a closed surjective morphism (sorry to reverse the $X$ and $Y$, but the version of the problem I met is in this notation).
I just noticed that one can reduce to the case where both $X$ and $Y$ are irreducible using the point-irreducible closed subset correspondence: it suffices to show for each irreducible subset $\overline{\{y’\}}=Y’\subset Y$ that $\dim f^{-1}(Y’)\geq \dim Y’ $, and by choosing $x’\in X$ such that $f(x’)=y’$ we need only show $\dim\overline{\{ x’\}}\geq \dim \overline{\{y’\}}$. It is now that the case where $\dim X=0$ becomes trivial, and the induction step in the first paragraph of the accepted answer works through. Anyway, I shall post this answer here as a complement.

There is a gap in the currently accepted answer , which I and everyone I can ask failed to figure out how to fix (EEDIT: the gap has been removed), so I decide to post this answer, despite that this question is already three-year old. The instructor of mine told me the following solution.
Let $Y_0\subsetneq \cdots\subsetneq Y_n$ be any chain of irreducible closed subsets of $Y$, the claim is that there exists a chain of irreducible closed subsets of $X$, $X_0\subsetneqq\cdots\subsetneq X_n$, such that $f(X_i)=Y_i$ for each $i$.
Since irreducible closed subsets of schemes are corresponded bijectively to the closures of single point, there is $Y_i=\overline{\{y_i\}}$ for some $y_i\in Y$ for each $i$. Since $f$ is closed, there is $f(\overline{\{x\}})=\overline{\{f(x)\}}$ for any $x\in X$ ($f(\overline{\{x\}})\supset\overline{\{f(x)\}}$ is clear by closeness of $f$; note that $x\in f^{-1}(\overline{\{f(x)\}})$, $\overline{\{x\}}\subset f^{-1}(\overline{\{f(x)\}}) $, hence the converse is also clear).
Now choose by surjectivity $x_n\in X$ such that $f(x_n)=y_n$, it suffices to find inductively that given $x_i\in X$ such that $f(x_i)=y_i$ (consequentially $f(\overline{\{x_i\}})=Y_i$), $x_{i-1}\in \overline{\{x_i\}}$ such that $f(x_{i-1})=y_{i-1}$, which implies that $f(\overline{\{x_{i-1}\}})=Y_{i-1}$ and hence $\overline{\{x_{i-1}\}}\subsetneq \overline{\{x_i\}} $ by surjectivity of $f$. Since $y_{i-1}\in \overline{\{y_i\}}=f(\overline{\{x_i\}})$, clearly such $x_{i-1}$ exists, so we are done.
This solution is based on the key fact that an irreducible closed subset of scheme is exactly the closure of a point. I don’t believe this statement is correct for general topological spaces since the irreducible component could have wild behavior, but I haven’t come up with a counterexample yet.


TITLE: Triple Integral bounds question
QUESTION [0 upvotes]: For the solid bounded by $x=0 , x=2, z=y, z=y-1, z=0, z=4$ I am looking at the $yz-plane$ and setting up a triple integral. I get $\int_{0}^{4} \int_{z+1}^{z} \int_{0}^{2} dxdydz$ However, I was wondering if I did this correctly because $z+1$  is greater than $z$, but when I drew the projection on the yz plane, it looked like z=y-1 is under z=y. Any help appreciated.

REPLY [1 votes]: With reference to the following sketch

the set up should be
$$\int_0^2 dx \int_0^4 dz \int_z^{z+1}dy$$


TITLE: Signed measure of subsets of zero-measured sets?
QUESTION [2 upvotes]: Hey, I'm reading the proof of proposition 6.8 in Big Rudin, pg. 120. I'll just mention where my confusion lies: Suppose K is an arbitrary measure (complex, real, whatever). In the proof of property (f), Rudin says that if K(A) = 0, and F is a subset of A, then K(F) = 0. Rudin has forced that by construction in chapter 1 for positive measures, but why should it work for complex or signed measures? Does there exist two sets A,B such that A is a subset of B, measure of B is 0, but measure of A is non-zero (negative?), for some arbitrary measure?
EDIT: Rudin's prop 6.8, part (f) says: given two arbitrary measures $\lambda_1$, $\lambda_2$, a positive measure $\mu$. If $\lambda_1 \ll \mu$, and $\lambda_2 \perp \mu$, then $\lambda_1 \perp \lambda_2$.
Also, $\ll$ and $\perp$ denote absolute continuity and mutual singularity, respectively.

REPLY [2 votes]: Observe that $\mu$ is positive. Since $\mu$ is positive and $\mu \perp \lambda_{2}$, the latter is concentrated on a $\mu$-null set $A$. But as $\lambda_{1} \ll \mu$ and $\mu$ is positive, we conclude for all measurable $E \subset A$ that $\mu(E) = 0$, hence $\lambda_{1}(E) = 0$ by absolute continuity, and thus $\lambda_{1}$ is concentrated on the complement of $A$. In other words, $\lambda_{1} \perp \lambda_{2}$.


TITLE: Biology: How to find the probability of randomly generating multiple, sequentially identical sets
QUESTION [5 upvotes]: If I randomly generate a substring (example "ATGCAGC") with equal probability (1/X where X=4) for each digit with length (L) digits: What is the formula for finding the probability (P) of randomly generating that sequence (T) times, given a total string length (N)?
Example:
Given "ATGCAGC" string length L=7, number of possible characters X=4 with equal probability of being randomly generated 1/X.
In a case where N characters are generated, what is the probability that an exact substring with length L will occur T times?
If I have randomly, sequentially generated N=7000 characters, what is the probability that any exact substring length L=7 "ATGCAGC" will occur T=2,3,4... times?
P is my dependent variable. L, T, N, X are independent.
In terms of dice:
Example: If I sequentially roll a X=6 sided die N=7000 times: What is the P=probability I will roll the die sequentially the same (1,4,6,5,3,2,3) with sequence length L=7 for T=2 sequentially identical occurrences in the N=7000 sequential rolls of a single die?
What is the probability in 7000 rolls I will have any 2 runs of 7 throws that have an exact sequential match? Example: (1,4,6,5,3,2,3 on rolls 201-207)  and (1,4,6,5,3,2,3) on rolls 5001-5007. It could be any number of (T) occurrences, on any roll numbers in (N) total die rolls.
I am specifically solving for the probability, given any values for the independent variables. Overlapping or non-overlapping substrings or both are great.
My question is related to (How many times will a consecutive sequence of throws randomly appear if I throw a four-sided die N times?)

REPLY [1 votes]: The question of a probability of a pattern repeating exactly $T$ times does not have a simple formula.
I have obtained the probability for the pattern ``ATGCAGC''
A directed graph was used to keep track of the patterns.
For $T=1$ itself it's big:
\begin{align*}
  A_1 &= \left(\begin{array}{rrrrrrrrrrrrrr}
3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
3 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0
\end{array}\right)
\end{align*}
for $T=2$,
\begin{align*}
  A_2 &= \left(\begin{array}{rrrrrrrrrrrrrrrrrrrrr}
3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
3 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0
\end{array}\right)
\end{align*}
That pattern of matrix continues for $T=3,4,\ldots$, and the entries of interest are the last 7 entries
of the first row in $A^{7000}$. Adding those and dividing by $4^{7000}$ gives the required probabilities.
Below are the probabilities obtained on computing:
\begin{align*}
  \begin{array}{|l|c|}\hline
T & P \\\hline
1 & 0.278730925452149 \\
2 & 0.059428898295207 \\
3 & 0.008431600808656 \\
4 & 0.000895516555326\\ \hline
\end{array}
\end{align*}


TITLE: Bounded operator on a non-empty set
QUESTION [3 upvotes]: Let $S$ be non-empty set, and let $X$ be the vector space of bounded functions on $S$, subject only to the condition that it be a Banach space when $X$ is supplied with the supremum-norm. Suppose $f:S \to \mathbb{F}$ is a function such that $fg \in X$ for all $g \in X$. Then the multiplication operator $M_f:X \to X$, defined by $M_f(g)=fg(g \in X$) is bounded.
I do not ask for the proof of this theorem. I just wonder if we assume $X\neq \lbrace0\rbrace$,
*) Is it true that $f$ is necessarily bounded? That would certainly explain why $M_f$ is bounded.
*)Is the theorem still true if $X$ is not required to be complete?

REPLY [2 votes]: Well, I don't think there's much to ponder here. Assume that $f$ is bounded. This inequality holds in total generality:
$$
|f(s)g(s)|\le \sup_{s\in S}|f(s)|\sup_{\sigma\in S} |g(\sigma)|.$$
Using the $\|\cdot\|_\infty$ notation we have
$$
\|M_f(g)\|_\infty \le \|f\|_{\infty}\|g\|_{\infty},$$
which means precisely that the linear operator $M_f\colon X\to X$ is bounded and that its operator norm is smaller or equal to $\|f\|_{\infty}$.
For all of this to make sense we only need that $X$ be a normed space, which is always the case for any set $S$. (Actually, $X$ is always Banach, but that is not needed at this stage). Notice that, if $S\ne \varnothing$, you cannot have $X=\{0\}$. Indeed, if $s\in S$, you can define the functions
$$\tag{1}
\lambda_s(\sigma)=\begin{cases} \lambda, &\sigma=s\\ 0, & \sigma \ne s,\end{cases}$$
where $\lambda\in \mathbb R$ is arbitrary. Since $\lambda_s\in X$, the space $X$ is strictly bigger than $\{0\}$.
Finally, you also ask whether the boundedness of $f$ is a necessary condition for the boundedness of $M_f$. The answer is affirmative. If $f\colon S\to \mathbb R$ is an unbounded function then there exists a sequence $s_n\in S$ such that $|f(s_n)|\to \infty$. Then
$$
\|M_f(1_{s_n})\|_\infty \to \infty, $$
(where $1_{s_n}$ is defined in (1) with $\lambda=1$) and $\|1_{s_n}\|_\infty=1$. So $M_f$ maps a bounded sequence to an unbounded one, which means that $M_f$ is not bounded.


TITLE: Is there a Birkhoff-like theorem for axially symmetric gravity fields?
QUESTION [2 upvotes]: As we know, to derive the Schwarzschild solution in GR we want to determine the unknown functions $a,b$ (sometimes $c$) of
$$(ds)^2 = a(r,t)\;dt^2 - b(r,t)\; dr^2 - r^2d\Omega^2 $$
or
$$ (ds)^2 = a(r,t)\;dt^2 + 2b(r,t)\;drdt - c(r,t)\; dr^2 - r^2d\Omega^2  $$
where $d\Omega^2 = d\theta^2 + \sin^2 \theta \; d\phi^2 \; $ is the usual line element of the 2-sphere. When applied to the empty space field equations where $G_{ab} \equiv 0$, Birkhoff's theorem says any static, spherically symmetric solution to the vacuum equations of general relativity is necessarily the Schwarzschild solution up to a coordinate transformation. That basically wraps up looking for exterior spherically symmetric solutions in GR. The next logical step is then to look for interior solutions which is a whole different ball game.
I'm interested in considering axially symmetric spacetimes in GR. The usual approach is to look at Weyl metrics (as mentioned in the comments below) and go for broke.
What I am confused about is trying to consider one of the following line element
$$ (ds)^2 = a(r,\theta)\;dt^2 + 2b(r,\theta)\;drdt - c(r,\theta)\; dr^2 - r^2d\Omega^2. $$
The above line element is axially symmetric also correct? In terms of polar spherical coordinates it is symmetric about the polar angle coordinate.
In terms of a spacetime that models a non-spherically symmetric body for e.g. a planet, I would have thought that this line element is more suitable than the general Weyl metric. However, I have never seen a method beginning with the above line element.
Question
Is the Weyl solution to axially symmetric gravitational fields as Schwarzschild solutions are to spherically symmetric gravitational fields?
 In other words, is there a Birkhoff-type theorem for axially symmetric solutions to the field equations of GR?

REPLY [1 votes]: Many exact solutions to EFE (Einistein's field equations) with varying degrees of symmetry have been found.
This Wikipedia article and references therein are excellent introduction to this field.
As a book reference, the standard one is:
H. Stephani, D. Kramer, M. MacCallum, C. Hoenselaers, E. Herlt, “Exact Solutions of Einstein’s Field Equations: 2nd Edition”, (2003), Cambridge University Press.


TITLE: Inertial mass and gravitational mass of 5 dimensional stars
QUESTION [6 upvotes]: Consider the following metric which is 5 dimensional (2-parameter) spherically symmetric Kaluza-Klein solution
$$-\left(\frac{1-m/r}{1+m/r}\right)^{2/\alpha}dt^2+(1+\frac{m}{r})^4\left(\frac{1-m/r}{1+m/r}\right)^{2(\alpha-\beta-1)/\alpha}(dr^2+r^2d\Omega^2)+\left(\frac{1-m/r}{1+m/r}\right)^{2\beta/\alpha}dx_5^2$$
where $x_5$ is the periodic fifth coordinate. In this paper (page 15) I read

the inertial mass of the star can be determined by the asymptotic behavior of this metric, assuming no interior singularities, and is equal to $M_{in}=\frac{(1+\beta)m}{\alpha}$

What should I understand as "inertial mass" of a star here? How can I see that this formula represents inertial mass?
In the same page, just a little below it says

the gravitational mass of the stars can be determined by the asymptotic form of $g_{00}$ and is given by $M_g=m/\alpha$

In this case, what should I understand as "gravitational mass" here? Again, how can I see that this formula represents gravitational mass?

REPLY [5 votes]: Are they identifying inertial mass with the ADM mass?
There is an old style way of writing $g_{00}$ of a stationary metric, far from the matter source, as $-\left(1 - \frac{1}{2}\phi({\vec x})\right)$, where $\phi$ is the potential function for the metric, and then you can monopole expand this and identify the numerator of the $\frac{1}{r}$ term of the expansion with the "mass" of the body.${}^{1}$.  This interpretation would explain the rest of their statements (as well as the dependence of the "gravitational mass" on $\alpha$ but not $\beta$).  This is difficult to answer without more context, though.
${}^{1}$Myself, this metric is simple enough that I"d just look for geodesics with $\dot x_{5} = 0$, and do ordinary physics to find the "gravitational mass" of the object.

REPLY [2 votes]: Disclaimer: while I have a good grasp of GR fundamentals, it is not my area of expertise.
The gravitational mass is distinct from the inertial mass as follows. The gravitational mass defines how strongly the body curves space-time, qualitatively it answers the question "how strong is the gravitational force from this object?". The inertial mass defines how the body reacts to forces, qualitatively it answers the question "how will the body react to the application of a given force".
The metric you give is approximated to first order near $r\rightarrow\infty$ by:
$${\rm d}s^2 \approx -\left(1-\frac{4m}{\alpha r}\right){\rm d}t^2 + \left(1+\frac{4(\beta+1)m}{\alpha r}\right)\left({\rm d}r^2+r^2{\rm d}\Omega^2\right) + \left(1-\frac{4\beta m}{\alpha r}\right){\rm d}{x_5}^2$$
For the gravitational mass, there is a similarity with the Schwarzschild metric. Begin with the approximate (at $r\rightarrow\infty$) $g_{00}$:
$$g_{00} \approx -\left(1-\frac{4m}{\alpha r}\right)$$
Compare this to the Schwarzschild solution in isotropic coordinates, also expanded at $r\rightarrow\infty$:
$$ds^2_{\rm Scwarzschild} \approx -\left(1-\frac{2M}{r}\right){\rm d}t^2 + \left(1+\frac{2M}{r}\right)\left({\rm d}r^2+r^2{\rm d}\Omega^2\right)$$
$$g_{00,\rm Schwarzschild} = -\left(1-\frac{2M}{r}\right)$$
Notice that $M\leftrightarrow 2m/\alpha$. I'm not sure why there's a difference of a factor of $2$, but I think it may have to do with what Jerry Schirmer mentioned in his answer, with $g_{00} = -\left(1-\frac{1}{2}\phi(\vec{x})\right)$; they do mention (about the Schwarzschild metric):

[...] and the gravitational potential is $\frac{1}{2}(g_{00}+1)=\frac{M}{R}$.

For the inertial mass, there's a $4(\beta+1)m/\alpha r$ (the inertial mass they quote, and again a factor of $2$) in the approximation to the metric that I gave where you'd expect to find a $2M$ in the Schwarzschild case, this time in the spatial coordinate term.
If anyone can explain that factor of $2$, I'm curious...


TITLE: How to show $e_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}$ is standard orthogonal basis of $L^2[0,2\pi]$?
QUESTION [1 upvotes]: I find the conclusion in a book, but without proof. It say that  $e_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}$ is standard orthogonal basis of $L^2[0,2\pi]$. I can prove that
$$
\lVert e_n \rVert =1
$$
and
$$
(e_n,e_m)=\delta_{nm}
$$
But I don't know how to show it is a basis, namely, for any $u\in L^2[0,2\pi]$, the $u$ can be presented as
$$
u=\sum\limits_{i=1}^\infty  (u, e_i) e_i.
$$
I know it is equal to prove the Parseval equation, namely
$$
\lVert u \rVert^2=\sum\limits_{i=1}^\infty |(x,e_i)|^2.
$$
Or equal to prove
$$
\{e_n\}^\bot=\{\theta\}.
$$
But no one of them I can prove.

REPLY [1 votes]: Assume that the orthogonal complement is non-zero, containing $f$.
Think to it as a $2\pi$-periodic function. The shifts of $f$ are in the orthogonal complement too. For the right choice of $\phi$ continuous we get that $g(t) = \int_0^{2\pi} f(t-u)\phi(u)du$ is non-zero, continuous, and orthogonal to every $e^{int}$.
But if $g(a)\ne 0$ then for $m$ large enough $\int_0^{2\pi} g(t) \left(\frac{1+\cos(t-a)}2\right)^mdt$ can't be $0$, contradicting that $g$ is orthogonal to the exponentials.


TITLE: Why is every continuously differentiable function with a uniform bounded derivative lipschitz continuous
QUESTION [0 upvotes]: I only know how to prove this for functions on a convex set by using the mean value theorem, but is this also true for this general case when nothing is said about the domain of the function besides the fact that it is a subset of $\mathbb{R}^n$?

REPLY [4 votes]: The function $f(x) = \operatorname{sgn} x$ defined on the domain $D = \mathbb{R} \setminus \{0\}$ is differentiable at every point of $D$ and the derivative is continuous  and bounded (it's identically zero), but it is not Lipschitz.

REPLY [2 votes]: If the domain is not connected the assertion is incorrect.  Define a function on the union of intervals $[2n, 2n+1]$ to be equal to $n^2$ on the corresponding interval $[2n, 2n+1]$.  This is continuously differentiable but not Lipschitz.


TITLE: images of group homomorhpisms
QUESTION [0 upvotes]: I want to know the possible images of group homomorphisms that map
$D_{10} \rightarrow G$, where $G$ is some arbitrary group.
Things I know: $\frac{G}{ker(f)} \cong Im(f)$;
$D_{10}$ normal subgroups are $\{e\}$, $<\sigma>$ and $D_{10}$.
I do not know how to connect all this information, so please help!

REPLY [1 votes]: You're basically there! Suppose $\theta:D_{10}\to G$ is a group homomorphism. Since we're only interested in the image of $\theta$, we might as well assume that $\theta$ is surjective, so that $G=\mathrm{Im}(\theta) $. Then by the first isomorphism theorem, $$G\cong D_{10}/\ker\theta$$which isn't quite what you stated.
Hence, if $G$ is the image of a such a homomorphism, then we must have $G\cong D_{10}/N$ for some normal subgroup $N$.
The converse is also true: if $N$ is a normal subgroup, then the map$$D_{10}\to D_{10}/N\\g\mapsto gN$$is a homomorphism.
Hence the possible images are exactly $D_{10}/N$ where $N$ is normal in $D_{10}$.


text:
\begin{document}
\title[Self-adjoint Dirac type Hamiltonians in one space dimension with a mass jump]{Self-adjoint Dirac type Hamiltonians in one space dimension with a mass jump.}

\author{L. A. Gonz\'{a}lez - D\'{\i}az}

\address{Laboratorio de Din\'amica no Lineal y Sistemas Complejos. Centro de F\'{\i}sica, Instituto Venezolano de Investigaciones Cient\'{\i}ficas, Caracas 1020 - A, Venezuela.}
\ead{lgonzale@ivic.gob.ve}

\author{Alberto A. D\'{\i}az}

\address{Centro de Estudios Avanzados,
Instituto Venezolano de Investigaciones Cient\'{\i}ficas, Caracas 1020 - A, Venezuela.}
\ead{aldiaz@ivic.gob.ve}

\author{S. D\'{\i}az - Sol\'orzano}

\address{Departamento de F\'{\i}sica, Universidad Sim\'on Bol\'{\i}var,
Sartenejas, Edo. Miranda 89000, Venezuela.}
\ead{sttiwuer@yahoo.es}

\author{J. R. Darias}

\address{Laboratorio de \'{O}ptica y Fluidos, Universidad Sim\'on Bol\'{\i}var,
Apartado Postal 89000, Caracas 1080 - A, Venezuela.}
\ead{jrdarias@usb.ve}

\begin{abstract}
Physical self-adjoint extensions and their spectra of the one-dimensional Dirac type Hamiltonian operator in which both the mass and velocity are constant except for a finite jump at one point of the real axis are correctly found. Different boundary conditions on envelope wave functions are studied, and the limiting case of equal masses (with no mass jump) is reviewed. Transport across one-dimensional heterostructures described by the Dirac equation is considered.
\end{abstract}

\noindent{\it Keywords}: Self-adjoint extensions, Boundary conditions, Mass jump, Heterostructures, Graphene.

\section{Introduction}\label{intro}

The discovery of graphene ended the belief that the Dirac equation useless in condensed matter physics \cite{novoselov1,novoselov2}.

The scientific and technological potential for exploiting charge carriers and quasiparticles with relativistic behavior in tunable condensed matter and atomic physics systems is attracting much attention \cite{zhu,ruostekoski,gomes,zhang}. In this regard, an important question, as yet only partly explored, remains whether quasi - one dimensional graphene systems support exclusively Dirac - Weyl massless or constant - mass Dirac fermions, or they can induce relativistic quantum field behaviors that require the consideration of position - dependent mass term \cite{yannouleas}. The mass jump case was firstly consider in \cite{tamarchenko} for the spherical defects in the $III-V$ semiconductors and then in \cite{firsova1,firsova2,ktitorov} for cylindrically symmetric defects in graphene, where the presence of the mass jump was traced back to the graphene sublattice symmetry violation, which is natural but for a short range defect. Using the Dirac - Weyl equation, the break - down of the sublattice symmetry (the equivalence of the two triangular lattices in graphene) can be described in terms of the effective mass that can be position - dependent. The existence of localized Dirac fermions in graphene with inhomogeneous effective mass was considered in \cite{jakubsky}, where the conditions under which the Dirac fermions are confined are found.

Models with an abrupt discontinuity of the mass and velocity at one point can be used for describing the behavior of a quantum particle moving between two different materials, i.e. an electron moving in a media formed up by two different materials. In each material the particle behaves as if it had a different mass and velocity. The discontinuity point represents the junction between these two materials.

We propose a simple model that describes an electron moving in a media formed up by two different materials given by a one - dimensional system in which the mass and velocity are constant except for a finite jump at one point of the real axis, which is chosen to be the origin for simplicity,
\begin{equation}\label{piecemass}
m(x)=\begin{cases}
m_{l} &\textrm{if}\hspace{.2cm} x<0\\
m_{r} &\textrm{if}\hspace{.2cm} x>0,
\end{cases}
\end{equation}
where $m_{l}$ and $m_{r}$ ($m_{l}\neq m_{r}$) are the masses at rest on the left and right, respectively. In this case, the Hamiltonian operator has the functional form
\begin{equation}\label{piecehamiltonian}
H=\begin{cases}
-i v_{l}\sigma_{x}\frac{d}{dx} + m_{l}v^{2}_{l}\sigma_{z} &\textrm{if}\hspace{.2cm} x<0\\*[.2cm]
-i v_{r}\sigma_{x}\frac{d}{dx} + m_{r}v^{2}_{r}\sigma_{z} &\textrm{if}\hspace{.2cm} x>0,
\end{cases}
\end{equation}
where $v_{l},\,v_{r}$ are the Fermi velocities in each medium and
\begin{equation}\label{pauli}
\sigma_{x}=\begin{pmatrix}0&1 \\ 1&0\end{pmatrix},\;\;\sigma_{z}=\begin{pmatrix}1&0 \\ 0&-1\end{pmatrix}
\end{equation}
are the Pauli matrices.

The finding appropriate of the boundary conditions in this kind of models is very important to describe the correct physics. The study of the role of the boundary conditions of quantum systems has became a recent focus of activity in different branches of physics \cite{asorey,asorey2}. Some examples of quantum physical phenomena which are intimately related to boundary conditions are the Casimir effect \cite{casimir}, the role of edge states \cite{john} and the quantization of conductivity in the Hall effect \cite{thouless}.

In this paper, we show that the operator \eqref{piecehamiltonian}, in a suitable domain, has infinite self-adjoint extensions. All the self-adjoint extensions have real discrete spectrum. Thus, all self-adjoint extensions describe bound states only, but not all the extensions are physically acceptable. We examine which extensions could to play an interesting role according to physical arguments.

The paper is organised as follows: In section \ref{extensions}, we find the set of all possible self-adjoint
extensions of $H$. In section \ref{spectra}, we calculate the reflection and transmission coefficients for all self-adjoint extensions, and we use constraints from physical arguments to reduce the set of all possible self-adjoint extensions. From the equation of the poles of the scattering coefficients, we obtain the spectrum that characterizes each self-adjoint extensions.

\section{Self - adjoint extensions of H}\label{extensions}
We will follow to Reed \cite{simon} and  Naimark \cite{naimark} to construct the self-adjoint extensions of $H$. To construct
the self-adjoint extensions of the operator $H$ we must begin by defining the smaller domain where the action operator makes
sense. In this section we will assume that the operator $H$ is densely defined. The domain of the operator $H$, $\mathcal{D}(H)$, is
\begin{equation}\label{domh}
\mathcal{D}(H)=\left\{\psi\in W^{2,1}(\mathbb{R})\otimes\mathbb{C}^2\, ,\,\psi(0^-)=\psi(0^+)=0\right\},
\end{equation}
where $W^{2,1}(\mathbb{R})$ is the corresponding Sobolev space, and $\psi$ is a two - component spinor wave function
\begin{equation}\label{spinor}
\psi(x)=\begin{pmatrix}\psi_{a}(x) \\ \psi_{b}(x)\end{pmatrix}
\end{equation}
The operator $H$ is symmetric and closed. Let $H^{\dag}$ the adjoint of $H$, with domain
\begin{equation}\label{domhdag}
\mathcal{D}(H^\dag)=\left\{\psi\in W^{2,1}(\mathbb{R} \backslash \left\{0\right\})\otimes\mathbb{C}^2\right\}.
\end{equation}
Note that $\mathcal{D}(H)\subset\mathcal{D}(H^{\dag})$. The deficiency subspaces of $H$ are given by
\begin{equation}\label{subdefecto}
\mathcal{N}_{\pm}=
\left\{\psi_\pm\in\mathcal{D}(H^\dag)\,,\, H^\dag\psi_\pm=\pm i\psi_\pm\right\},
\end{equation}
with the respective dimensions $n_{+}$ and $n_{-}$. These are called the deficiency indices of the operator $H$
and will be denoted by the ordered pair $(n_{+},n_{-})$.

The normalized solutions of
$H^\dag\psi_\pm=\pm i\psi_\pm$ are
\begin{subequations}\label{soldefecto}
\begin{eqnarray}
\label{soldefmas-mas}
\psi_+^{(+)}(x)&=&\left[\tfrac{(1+m_r^2 v_r^4)}{v_r^2}\right]^{1/4}\begin{pmatrix}1 \\ i\tfrac{\sqrt{1+m_r^2 v_r^4}}{i+m_r v_r^2}\end{pmatrix}\theta(x)\,e^{-\tfrac{\sqrt{1+m_r^2 v_r^4}}{v_r}x},\\
\label{soldefmas-men}
\psi_+^{(-)}(x)&=&\left[\tfrac{(1+m_l^2 v_l^4)}{v_l^2}\right]^{1/4}\begin{pmatrix}1 \\ -i\tfrac{\sqrt{1+m_l^2 v_l^4}}{i+m_l v_l^2}\end{pmatrix}\theta(-x)\,e^{\tfrac{\sqrt{1+m_l^2 v_l^4}}{v_l}x},\\
\label{soldefmen-mas}
\psi_-^{(+)}(x)&=&\left[\tfrac{(1+m_r^2 v_r^4)}{v_r^2}\right]^{1/4}\begin{pmatrix}1 \\ i\tfrac{\sqrt{1+m_r^2 v_r^4}}{m_r v_r^2-i}\end{pmatrix}\theta(x)\,e^{-\tfrac{\sqrt{1+m_r^2 v_r^4}}{v_r}x},\\
\label{solskmenmen}
\psi_-^{(-)}(x)&=&\left[\tfrac{(1+m_l^2 v_l^4)}{v_l^2}\right]^{1/4}\begin{pmatrix}1 \\ -i\tfrac{\sqrt{1+m_l^2 v_l^4}}{m_l v_l^2-i}\end{pmatrix}\theta(-x)\,e^{\tfrac{\sqrt{1+m_l^2 v_l^4}}{v_l}x},
\end{eqnarray}
\end{subequations}
where $\theta(x)$ represents the Heaviside step function. Since all the solutions of equations $H^\dag\psi_\pm=\pm i\psi_\pm$ belong to $L^2(\mathbb{R})\otimes\mathbb{C}^2$, the deficiency indices are $(2,2)$ and, the according to Naimark \cite{naimark}, every self-adjoint extensions are parametrized by a $U(2)$ matrix. This matrix defines a unique self-adjoint extension, $H_{U}$, of $H$ with domain characterized by means the set of all functions $\phi\in\mathcal{D}(H^\dag)$ which satisfy the conditions
\begin{equation}\label{conditions}
\begin{pmatrix}
\bar{\psi}_{a2}(0^-) & \bar{\psi}_{a1}(0^-)\\
\bar{\psi}_{b2}(0^-) & \bar{\psi}_{b1}(0^-)
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}=\frac{v_r}{v_l}\begin{pmatrix}
\bar{\psi}_{a2}(0^+) & \bar{\psi}_{a1}(0^+)\\
\bar{\psi}_{b2}(0^+) & \bar{\psi}_{b1}(0^+)
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
\end{equation}
where $\psi(0^\pm)\equiv\lim\limits_{x\rightarrow 0^{\pm}}\psi(x)$ and $\phi(0^\pm)\equiv\lim\limits_{x\rightarrow 0^{\pm}}\phi(x)$, and
\begin{align}\label{basemasmen}
\psi_1(x)&= \psi_+^{(+)}(x)+U_{11}\psi_-^{(+)}(x)+U_{21}\psi_-^{(-)}(x)\\
\psi_2(x)&= \psi_+^{(-)}(x)+U_{12}\psi_-^{(+)}(x)+U_{22}\psi_-^{(-)}(x)
\end{align}
where $\psi_1(x),\psi_2(x)\in\mathcal{D}(H_U)$, and $U_{11},U_{12},U_{21},U_{22}$ are complex numbers that determine $U$. The expression
\eqref{conditions} can be written in the form
\begin{equation}\label{boundc1}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\mathbb{T}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix},
\end{equation}
where the $n_{+}\times n_{-}\;$ matrix $\mathbb{T}$ is given by
\begin{equation}\label{matrixT}
\mathbb{T}=
\frac{v_l}{v_r}
\begin{pmatrix}
\bar{\psi}_{a2}(0^+) & \bar{\psi}_{a1}(0^+)\\
\bar{\psi}_{b2}(0^+) & \bar{\psi}_{b1}(0^+)
\end{pmatrix}^{-1}\begin{pmatrix}
\bar{\psi}_{a2}(0^-) & \bar{\psi}_{a1}(0^-)\\
\bar{\psi}_{b2}(0^-) & \bar{\psi}_{b1}(0^-)
\end{pmatrix},
\end{equation}
whose determinat is given by
\begin{equation}\label{detmatrixT}
\left|\det\mathbb{T}\right|=\frac{v_l}{v_r}.
\end{equation}
The matrix $\mathbb{T}$ gives the matching conditions at the origin. From \eqref{basemasmen}, we can rewrite
the matrix $\mathbb{T}$ in the form
\begin{equation}\label{matrixTu}
\mathbb{T}=\frac{\sqrt{v_l}}{2\bar{U}_{12}\sqrt{v_r}}
\begin{pmatrix}
T_{11}&T_{12}\\
T_{21}&T_{22}
\end{pmatrix}
\end{equation}
with
\begin{subequations}\label{elementosTu}
\begin{align}
T_{11}=&\frac{(1+m_r^2 v_r^4)^\frac{1}{4}((1-i m_l v_l^2)(1+\bar{U}_{11})-(1+i m_l v_l^2)(\bar{U}_{22}+\det(\bar{U})))}{(1+m_l^2 v_l^4)^\frac{1}{4}}\\
T_{12}=&-(1+\det(\bar{U})+\bar{U}_{11}+\bar{U}_{22})((1+m_l^2 v_l^4)(1+m_r^2 v_r^4))^{\frac{1}{4}}\\
T_{21}=&\frac{(i+m_lv_l^2+\bar{U}_{22}(m_l v_l^2-i))(\bar{U}_{11}(m_r v_r^2-i)+m_r v_r^2+i)}{((1+m_l^2 v_l^4)(1+m_r^2 v_r^4))^\frac{1}{4}}\\\nonumber
&-\frac{\bar{U}_{12}\bar{U}_{21}(m_l v_l^2-i)(m_r v_r^2-i)}{((1+m_l^2 v_l^4)(1+m_r^2 v_r^4))^\frac{1}{4}}\\
T_{22}=&\frac{(1+m_l^2 v_l^4)^{\frac{1}{4}}((1-im_r v_r^2)(1+\bar{U}_{22})-(1+im_r v_r^2)(\bar{U}_{11}+\det(\bar{U})))}{(1+m_r^2 v_r^4)^{\frac{1}{4}}}
\end{align}
\end{subequations}

The determinant of \eqref{matrixTu} is given by
\begin{equation}\label{detmatrixTu}
\det\mathbb{T}=\frac{v_l}{v_r}\frac{\bar{U}_{21}}{\bar{U}_{12}}.
\end{equation}
By Comparing \eqref{detmatrixTu} with \eqref{detmatrixT}, we have that $\left|U_{12}\right|=\left|U_{21}\right|$.

\section{Scattering coefficients and the spectra of H}\label{spectra}

In this section we will derive the spectra for the self-adjoint extensions $H_{U}$ from poles of scattering
amplitudes. For this, let us parametrize the unitary matrix $\mathbb{U}$ as
\begin{equation}\label{Uparametrizada}
\mathbb{U}=e^{i\alpha}\mathbb{A},\;
\det(\mathbb{A})=1,
\end{equation}
where
\begin{equation}\label{matrixA}
\mathbb{A}=\begin{pmatrix}
  a_0-ia_3 & -a_2-ia_1 \\
  a_2-ia_1 & a_0+ia_3 \\
\end{pmatrix},
\end{equation}
with $a_0,a_1,a_2,a_3\in\mathbb{R}$, $a_0^2+a_1^2+a_2^2+a_3^2=1$, and $\alpha\in\left[0,\pi\right]$. Notice that the points $\alpha=0$ and
$\alpha=\pi$ have to be identified. Substituting \eqref{Uparametrizada} and \eqref{matrixA} in \eqref{elementosTu}, we obtain the components of matrix $\mathbb{T}$:
\begin{subequations}\label{matrixTa}
\begin{align}

T_{11}=&2ie^{-i\alpha}\frac{(1+m_r^2 v_r^4)^\frac{1}{4}(a_3+\sin\alpha-m_l v_l^2(a_0+\cos\alpha))}{(1+m_l^2 v_l^4)^\frac{1}{4}}\\
T_{12}=&2e^{-i\alpha}(a_0+\cos\alpha)((1+m_l^2 v_l^4)(1+m_r^2 v_r^4))^{\frac{1}{4}}\\
T_{21}=&2e^{-i\alpha}\Big(\frac{a_0-\cos\alpha+m_l m_r  v_l^2 v_r^2(a_0+\cos\alpha)+m_l v_l^2(a_3-\sin\alpha)-m_r v_r^2(a_3+\sin\alpha)}{((1+m_l^2 v_l^4)(1+m_r^2 v_r^4))^\frac{1}{4}}\Big)\\
T_{22}=& -2ie^{-i\alpha}\frac{(1+m_l^2 v_l^4)^{\frac{1}{4}}(a_3-\sin\alpha+m_r v_r^2(a_0+\cos\alpha))}{(1+m_r^2 v_r^4)^{\frac{1}{4}}}
\end{align}
\end{subequations}

In terms of \eqref{matrixTa}, the matching conditions \eqref{boundc1} are
\begin{equation}\label{bounconda}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\begin{pmatrix}
T_{11}&T_{12}\\
T_{21}&T_{22}
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}.
\end{equation}
\indent
Let us assume that an incoming monochromatic wave $\begin{pmatrix}1\\ \sqrt{\frac{E-m_lv_l^2}{E+m_lv_l^2}}\end{pmatrix}e^{ik_{l}x}$, $k_{l}=\frac{\sqrt{E^2-m_l^2v_l^2}}{v_l}$, $E>max (m_l v_l^2,m_r v_r^2)$,
comes from the left, so that the wave function for $x<0$ is $\begin{pmatrix}1\\ \sqrt{\frac{E-m_lv_l^2}{E+m_lv_l^2}}\end{pmatrix}e^{ik_{l}x}+r_l \begin{pmatrix}1\\ -\sqrt{\frac{E-m_lv_l^2}{E+m_lv_l^2}}\end{pmatrix}e^{-ik_{l}x}$, and the wave function
for $x>0$ is $t_l\begin{pmatrix}1\\ \sqrt{\frac{E-m_rv_r^2}{E+m_rv_r^2}}\end{pmatrix}e^{ik_{r}x}$, $k_{r}=\frac{\sqrt{E^2-m_r^2v_r^2}}{v_r}$, $E>max (m_l v_l^2,m_r v_r^2)$, where $r_l$ and $t_l$ are the
reflection and transmission amplitudes, respectively, for an incoming wave come from the left. Then, the matching
conditions \eqref{bounconda} at the origin give
\begin{equation}\label{condRT}
\begin{pmatrix}1+r_l \\*[.4cm] \sqrt{\frac{E-m_lv_l^2}{E+m_lv_l^2}}(1-r_l)\end{pmatrix}=
\mathbb{T}
\begin{pmatrix}t_l \\*[.5cm] \sqrt{\frac{E-m_rv_r^2}{E+m_rv_r^2}}\,t_l\end{pmatrix}
\end{equation}
and then one finally obtains the expressions of $r_l$ and $t_l$ as
\begin{eqnarray}
\label{rl}
&r_l=\frac{\mathfrak{N}}{\mathfrak{D}}\\
\label{tl}
t_l=&2\sqrt{\frac{v_l}{v_r}}\frac{\sqrt{a_1^2+a_2^2}\;e^{-i\tan ^{-1}\left(\frac{a_2}{a_1}\right)}}{\mathfrak{D}}\mathfrak{T}
\end{eqnarray}
with
\begin{multline}\label{n}
\mathfrak{N}=-\sqrt{m_r^2 v_r^4+1} \sqrt{E -m_r v_r^2}\Big(\sqrt{E +m_l v_l^2} (a_3+\sin\alpha-m_l v_l^2(a_0+\cos\alpha))\\
+i(a_0+\cos\alpha)\sqrt{(m_l^2 v_l^4+1) (E-m_l v_l^2)}\Big)
-\sqrt{m_l^2v_l^4+1}\sqrt{E -m_l v_l^2} \sqrt{E +m_r v_r^2}(a_3\\
-\sin\alpha+ m_rv_r^2(a_0+\cos\alpha))-i \sqrt{E +m_l v_l^2}\sqrt{E +m_r v_r^2}(m_l m_r v_l^2 v_r^2(a_0+\cos\alpha)+a_0\\
-\cos\alpha+(m_l v_l^2(a_3-\sin\alpha)-m_r v_r^2 (a_3+\sin\alpha)),
\end{multline}

\begin{multline}\label{d}
\mathfrak{D}=\sqrt{m_r^2 v_r^4+1}\sqrt{E-m_r v_r^2}\Big(\sqrt{E+m_l v_l^2}(a_3+\sin\alpha-m_l v_l^2(a_0-\cos\alpha))\\
-i(a_0+\cos\alpha)\sqrt{(m_l^2 v_l^4+1)(E-m_l v_l^2)}\Big)
-\sqrt{m_l^2v_l^4+1} \sqrt{E -m_l v_l^2} \sqrt{E +m_r v_r^2}(a_3\\
-\sin\alpha+m_rv_r^2(a_0+\cos\alpha))
+i \sqrt{E +m_l v_l^2}\sqrt{E +m_r v_r^2}(m_l m_r v_l^2 v_r^2(a_0+\cos\alpha)+a_0\\
-\cos\alpha)+(m_l v_l^2(a_3-\sin\alpha)-m_r v_r^2 (a_3+\sin\alpha)),
\end{multline}
\begin{equation}\label{te}
\mathfrak{T}=\sqrt[4]{\left(m_l^2 v_l^4+1\right) \left(m_r^2 v_r^4+1\right)} \sqrt{\left(E -m_l
   v_l^2\right) \left(E +m_r v_r^2\right)}
\end{equation}

Since the matrix $\mathbb{T}$ is not real, the transmission amplitudes are different and the self-adjoint extensions are not explicitly time reversal invariant \cite{nogami,coutinho1}.
\indent

Physically, the term $e^{-i\arctan\left(\frac{a_2}{a_1}\right)}$ in \eqref{tl} does not add new information to the phase shift, since that $a_0$, $a_1$, $a_2$ and $a_3$
are independent of the energy, so we can put $a_2 = 0$ without any loss of information. The matrix $\mathbb{T}$ coincides with the corresponding one in the nonrelativistic case \cite{gonzalez-diaz}  precisely when $a_2=0$. In this situation, we have that the matching conditions \eqref{bounconda} are
\begin{equation}\label{bouncondaa20}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\mathbb{T}\Big|_{\scriptscriptstyle a2=0}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}.
\end{equation}
whose determinant is
\begin{equation}\label{detmatrixTa20}
\det\mathbb{T}\Big|_{\scriptscriptstyle a2=0}=\frac{v_l}{v_r}
\end{equation}

Making use of $a_0^2+a_1^2+a_3^2=1$, we have $\left|r_l\right|^2+\left|t_l\right|^2\frac{\sqrt{E+m_lv_l^2}\sqrt{E-m_rv_r^2}}{\sqrt{E-m_lv_l^2}\sqrt{E+m_rv_r^2}}\frac{v_r}{v_l}=1$.
The poles of $r_l$ and $t_l$ satisfy the following equation
\begin{equation}\label{poles}
\mathfrak{D}=0.
\end{equation}

The poles of $r_r$ and $t_r$ ($r_r$ and $t_r$ are the reflection and transmission amplitudes, respectively, for incoming wave come from the right) also satisfy \eqref{poles}.The zero values of \eqref{rl} correspond to transmission resonances \cite{newton,taylor}. The zero values of \eqref{tl} are called zero momentum resonances \cite{dombey}, and they occur at $E=\pm m_l v_F^2$ and $E=\pm m_r v_F^2$ ($v_F$ is the Fermi velocity). The anti-particle is described by the hole wave function corresponding to the absence of the state with $E=-m_{l,r}v_F^2$ \cite{dombey}.

In the next subsections, we discuss the spectrum of some self-adjoint extension of \eqref{piecehamiltonian} corresponding to one - dimensional spatial Dirac Hamiltonian: (a) with a equally mixed point interaction potential (PIP) at the origin plus mass jump at the same point, (b) with an inverted mixed PIP at the origin plus mass jump at the same point, (c) with a vector PIP at the origin plus mass jump at the same point, and (d) with a scalar PIP at the origin plus mass jump at the same point. The one - dimensional Dirac Hamiltonian with PIPs without mass jump is analyzed in \cite{dominguez-adame}. In this paper, the selfadjoint extensions of the one - dimensional Dirac operator with point interactions can also be obtained (heuristically) starting from the operator $H=-i\sigma_{x}\frac{d}{dx}+m\sigma_{z}+\mathcal{U}(x)$, with $\mathcal{U}(x)=\left(g_v+\sigma_{z}g_s\right)u(x)$, where $u(x)$ is any peaked function at $x=0$ satisfying $\int_{-\infty}^{\infty}u(x)\,dx=1$. $g_v$ and $g_s$ are the strengths of the vector and scalar components of the potential, respectively. When $g_v=0$, $g_s=0$, $g_v=g_s$ $g_v=-g_s$, we say that we have a scalar point interaction potential, vector point interaction potential, equally mixed point interaction potential and inverted mixed point interaction potential, respectively. For simplicity and comparison, the following sections we will impose that $v_l = v_r = v_F$. Thus, \eqref{detmatrixTa20} equals one, similarly to the case of equal masses.

\subsection{One - dimensional spatial Dirac Hamiltonian  with a equally mixed PIP at the origin plus mass jump at the same point}\label{equallypip}
The boundary conditions corresponding to one - dimensional spatial Dirac Hamiltonian with a vector PIP at the origin plus mass jump at the same point are obtained by the following ids:
\begin{subequations}\label{aes-vectorpip}
\begin{align}
a_0&=-\cos\alpha,\\
a_1&=\sin\alpha,\\
a_3&=0,\\
\cot\alpha&=-\frac{\delta+\left(m_l+m_r\right)v_F^3}{2v_F},\,\delta<0,
\end{align}
\end{subequations}
where $\delta$ is the strength of PIP.
By inserting \eqref{aes-vectorpip} in \eqref{bouncondaa20}, we obtain the matching conditions for this self -adjoint extension:
\begin{equation}\label{bcsequallypip}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\begin{pmatrix}
\frac{\sqrt[4]{m_r^2v_F^4+1}}{\sqrt[4]{m_l^2v_F^4+1}}&0\\
-\frac{i \delta }{v_F \sqrt[4]{m_l^2 v_F^4+1}\sqrt[4]{m_r^2v_F^4+1}}&\frac{\sqrt[4]{m_l^2v_F^4+1}}{\sqrt[4]{m_r^2v_F^4+1}}
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}
\end{equation}
\begin{figure}[!ht]
\centerline{\includegraphics[scale=0.35,clip]{esdelta.eps}}
\caption{Solution curves of \eqref{spdelta} (solid line) and \eqref{spdeltaaproxsolre} (dashed line) as a function of $\delta$, respectively.}\label{fspdelta}
\end{figure}
By inserting \eqref{aes-vectorpip} in \eqref{poles}, we obtain the spectral equation (bound states energy equation)
\begin{multline}\label{spdelta}
\frac{\delta}{v_F} \sqrt{\left(E +v_F^2 m_l\right)\left(E +v_F^2 m_r\right)}+\sqrt{\left(v_F^4 m_l^2+1\right)\left(v_F^2 m_l-E \right) \left(E +v_F^2 m_r\right)}\\
+\sqrt{\left(v_F^4 m_r^2+1\right) \left(E +v_F^2m_l\right) \left(v_F^2 m_r-E \right)}=0
\end{multline}
The energy of bound states lies between $-min(m_l v_F^2,m_r v_F^2)$ and $min(m_l v_F^2,m_r v_F^2)$. Note that equation \eqref{spdelta} is invariant under the change of $m_l$ by $m_r$.

For $m_l=1,\,m_r=2$ and $v_F=1$, the solution curve of \eqref{spdelta} as a function of $\delta$ is represented in the Figure \ref{fspdelta}. The solution curves intersect at a point, which means that they have the same energy for a given value of $\delta$. The value of the energy is insensitive to ratio of the masses.

As stated in \cite{dominguez-adame}, the boundary of the lower continuum is never reached for finite values of $\delta$ due to the presence of the scalar potential term, but the energy level crosses zero because of the vector potential term (see Figure \ref{fspdelta}).

For $m_l\approx m_r\equiv m$, \eqref{spdelta} becomes
\begin{equation}\label{spdeltaaprox}
2 v_F \sqrt{\left(v_F^4 m^2+1\right) \left(v_F^4 m^2-E^2\right)}+\delta\sqrt{\left(E +v_F^2m\right)^2}=0,
\end{equation}
which gives the value of the energy for bound state
\begin{equation}\label{spdeltaaproxsol}
E=mv_F^2\;\frac{4v_F^2+4m^2v_F^6-\delta^2}{4v_F^2+4m^2v_F^6+\delta^2}
\end{equation}
Defining $\tilde{\delta}=\frac{\delta}{v_F\sqrt{1+m^2v_F^4}}$, the above expression can be rewritten as
\begin{equation}\label{spdeltaaproxsolre}
E=mv_F^2\frac{4-\tilde{\delta}^2}{4+\tilde{\delta}^2}
\end{equation}
The energy \eqref{spdeltaaproxsolre} coincides with the one found in \cite{dominguez-adame} and \cite{albeverio} for the self-adjoint extension called equally mixed potential.

At high energies, we have
\begin{equation}\label{Tlasympdelta}
\left|t_l\right|^2\sim\frac{4 v_F^2 \sqrt{\left(v_F^4 m_l^2+1\right) \left(v_F^4 m_r^2+1\right)}}{v_F^2 \left(\sqrt{v_F^4 m_l^2+1}+\sqrt{v_F^4
   m_r^2+1}\right)^2+\delta ^2}
\end{equation}
so that the transmission does not occur as the potential becomes sufficiently strong. Therefore, the interaction equally mixed PIP at the origin plus mass jump at the same point does confine particles. The same conclusion is reported in \cite{dominguez-adame}.

\subsection{One - dimensional spatial Dirac Hamiltonian with a inverted mixed PIP at the origin plus mass jump at the same point}\label{invertedpip}

The matching conditions for this self -adjoint extension are
\begin{equation}\label{bcsinvertedpip}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\begin{pmatrix}
\frac{(1+m_r^2v_F^4)^\frac{1}{4}}{(1+m_l^2v_F^4)^\frac{1}{4}}&\; -i\,\lambda\, v_F\,((1+m_l^2v_F^4)(1+m_r^2v_F^4))^{\frac{1}{4}}\\
0&\;\frac{(1+m_l^2v_F^4)^\frac{1}{4}}{(1+m_r^2v_F^4)^\frac{1}{4}}
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}
\end{equation}
where $\lambda$ is the strength of PIP, $\lambda>0$.
The spectral equation is
\begin{multline}\label{spdeltap}
\sqrt{\left(v_F^4 m_l^2+1\right)\left(v_F^2 m_l-E \right)\left(E +v_F^2
m_r\right)}+\sqrt{\left(v_F^4 m_r^2+1\right)\left(E +v_F^2 m_l\right) \left(v_F^2 m_r-E\right)}\\
-v_F\,\lambda\sqrt{v_F^2m_r-E }\sqrt{v_F^2m_l-E}\sqrt{\left(v_F^4m_l^2+1\right)\left(v_F^4 m_r^2+1\right)}=0,
\end{multline}
where $-min\left(m_l v_F^2,m_r v_F^2\right)<E<min\left(m_l v_F^2,m_r v_F^2\right)$. This equation is invariant under the change of $m_l$ by $m_r$.

For $m_l\approx m_r\equiv m$, \eqref{spdeltap} becomes
\begin{equation}\label{spdeltaaprox}
2\sqrt{\left(v_F^4m^2-E^2\right)}-v_F\,\lambda\sqrt{\left(v_F^4 m^2+1\right)}\left(v_F^2m-E\right)=0,
\end{equation}
which gives the value of the energy for bound state
\begin{equation}\label{spdeltapaproxsolre}
E=-mv_F^2\;\frac{4-\tilde{\lambda}^2}{4+\tilde{\lambda}^2},
\end{equation}
with $\tilde{\lambda}\equiv v_F\,\sqrt{1+m^2v_F^4}\,\lambda$.
The energy \eqref{spdeltapaproxsolre} coincides with the one found in \cite{dominguez-adame} and \cite{albeverio} for the self-adjoint extension called inverted mixed potential.
\begin{figure}[!h]
\centerline{\includegraphics[scale=0.35,clip]{esdeltap.eps}}
\caption{Solution curves of \eqref{spdeltap} (solid line) and \eqref{spdeltapaproxsolre} (dashed line) as a function of $\lambda$, respectively.}\label{fspdeltap}
\end{figure}

For $m_l=1,\,m_r=2$ and $v_F=1$, the solution curve of \eqref{spdeltap} as a function of $\lambda$ is represented in the Figure \ref{fspdeltap}. The solution curves intersect at a point, which means that they have the same energy for $\lambda=0.6324$. For weak coupling, the bound-states energy reaches the lower negative continuum, as distinct from that shown in \cite{dominguez-adame}.

At high energies, the transmission coefficient becomes
\begin{equation}\label{Tlasympdeltap}
\left|t_l\right|^2\sim\frac{4\sqrt{v_F^4 m_l^2+1}\sqrt{v_F^4 m_r^2+1}}{\left(\sqrt{v_F^4 m_l^2+1}+\sqrt{v_F^4
   m_r^2+1}\right)^2+v_F^2\left(v_F^4 m_l^2+1\right)\left(v_F^4
   m_r^2+1\right)\lambda ^2}
\end{equation}
so that the transmission does not occur as the potential becomes sufficiently strong. As in the previous subsection, the interaction inverted mixed PIP at the origin plus mass jump at the same point does confine particles. This same conclusion is reported in \cite{dominguez-adame} for this self-adjoint extension.

\subsection{One - dimensional spatial Dirac Hamiltonian with a pure scalar PIP at the origin plus mass jump at the same point}\label{purescalarpip}
The boundary conditions corresponding to one - dimensional spatial Dirac Hamiltonian with a pure scalar PIP at the
origin plus a mass jump at the same point are
\begin{equation}\label{bcspurescpip}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\begin{pmatrix}
\frac{(1+m_r^2v_F^4)^\frac{1}{4}}{(1+m_l^2v_F^4)^\frac{1}{4}}\cosh{\left(\frac{a}{v_F}\right)}& i\sinh{\left(\frac{a}{v_F}\right)}\\
-i\sinh{\left(\frac{a}{v_F}\right)}&\frac{(1+m_l^2v_F^4)^\frac{1}{4}}{(1+m_r^2v_F^4)^\frac{1}{4}}\cosh{\left(\frac{a}{v_F}\right)}
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}
\end{equation}
where $a$ is the strength of PIP, $a<0$.
The spectral equation is
\begin{multline}\label{sppuresc}
\sqrt[4]{v_F^4 m_l^2+1}\sqrt[4]{v_F^4 m_r^2+1}\Big(\sqrt{v_F^2 m_l-E}
\sqrt{v_F^2 m_r-E }\\
+\sqrt{E +v_F^2 m_l}\sqrt{E +v_F^2 m_r}\Big)\sinh\left(\frac{a}{v_F}\right)+\Big(\sqrt{v_F^4 m_l^2+1}\sqrt{(v_F^2 m_l-E)(E +v_F^2 m_r)}\\
+\sqrt{v_F^4m_r^2+1}
\sqrt{(E +v_F^2 m_l)(v_F^2 m_r-E)}\Big)\cosh\left(\frac{a}{v_F}\right)=0,

\end{multline}
where $-min\left(m_l v_F^2,m_r v_F^2\right)<E<min\left(m_l v_F^2,m_r v_F^2\right)$. This equation is invariant under the change of $m_l$ by $m_r$.

Pairs of allowed energy values appear, which is a common feature of the other scalar-type potentials \cite{coutinho2}. As seen in Figure \ref{fsppuresc}, the same strength $a$ of the scalar potential can bind particles and antiparticles alike. As stated in \cite{dominguez-adame}, the energy level never reaches zero. The positive and negative energy states remain well separated even if the potential becomes strong.
\begin{figure}[!h]
\centerline{\includegraphics[scale=0.35,clip]{espures.eps}}
\caption{Solution curves of \eqref{sppuresc} (solid line) and \eqref{sppurescpaproxsolre} (dashed line) as a function of $a$, respectively, for $m_l=1$, $m_r=2$ and $v_F=1$.}\label{fsppuresc}
\end{figure}

For $m_l\approx m_r\equiv m$, \eqref{sppuresc} becomes
\begin{equation}\label{sppuresccaprox}
\sqrt{v_F^4m^2-E^2}\cosh\left(\frac{a}{v_F}\right)+v_F^2 m\sinh \left(\frac{a}{v_F}\right)=0,
\end{equation}
which gives the value of the energies for bound states
\begin{equation}\label{sppurescpaproxsolre}
E=\pm\,mv_F^2\,\text{sech}\left(\frac{a}{v_F}\right).
\end{equation}

The energy \eqref{sppurescpaproxsolre} coincides with the one found in \cite{dominguez-adame} for the self-adjoint extension called pure scalar potential. For this potential, at hight energy, the transmission coefficient becomes
\begin{equation}\label{Tlasympuresc}
\left|t_l\right|^2\sim 4\,\frac{\sqrt{(1 + v_F^4 m_l^2)(1 + v_F^4 m_r^2)}\,
  }{\left(\sqrt{1 + v_F^4 m_l^2} + \sqrt{1 +v_F^4 m_r^2}\right)^2}\,\text{sech}^2\left(\frac{a}{v_F}\right)
\end{equation}
Thus, the pure scalar potential leads to particle confinement when $\left|a\right|\rightarrow\infty$ .

\subsection{One - dimensional spatial Dirac Hamiltonian with a pure vector PIP at the origin plus mass jump at the same point}\label{purevectorpip}
The boundary conditions corresponding to one - dimensional spatial Dirac Hamiltonian with a pure vector PIP at the
origin plus a mass jump at the same point are
\begin{equation}\label{bcspurevecpip}
\begin{pmatrix}\phi_{a}(0^{+}) \\ \phi_{b}(0^{+})\end{pmatrix}
=\begin{pmatrix}
\frac{m_r}{m_l}\cos{\left(\frac{a}{v_F}\right)}& -i\sin{\left(\frac{a}{v_F}\right)}\\
-i\sin{\left(\frac{a}{v_F}\right)}&\frac{m_l}{m_r}\cos{\left(\frac{a}{v_F}\right)}
\end{pmatrix}\begin{pmatrix}\phi_{a}(0^{-}) \\ \phi_{b}(0^{-})\end{pmatrix}
\end{equation}
with $a>0$, contrary to the assertion in \cite{dominguez-adame}, where the sign of the strength $a$ is immaterial as far as the existence of bound states.

The spectral equation is
\begin{multline}\label{sppurevec}
\left(m_l^2 \sqrt{v_F^2 m_r + E} \sqrt{v_F^2 m_l - E} + m_r^2 \sqrt{v_F^2 m_l + E} \sqrt{v_F^2 m_r - E}\right) \cos\left(\frac{a}{v_F}\right)\\
+m_r m_l \Big( \sqrt{v_F^2 m_r + E} \sqrt{v_F^2 m_l + E} -\sqrt{v_F^2 m_l - E} \sqrt{v_F^2 m_r - E}\Big) \sin\left(\frac{a}{v_F}\right)=0,
\end{multline}
where $-min\left(m_l v_F^2,m_r v_F^2\right)<E<min\left(m_l v_F^2,m_r v_F^2\right)$. Unlike the previous cases, \eqref{sppurevec} is not invariant under the change of $m_l$ by $m_r$. For $m_l=1,\,m_r=2$ and $v_F=1$, the solution curve of \eqref{sppurevec} as a function of $a$ is represented in the Figure \ref{fsppurevec}.
\begin{figure}[!h]
\centerline{\includegraphics[scale=0.35,clip]{espurev.eps}}
\caption{Solution curves of \eqref{sppurevec} (solid line) and \eqref{sppurevecpaproxsolre} (dashed line) as a function of $a$, respectively, for $m_l=1$, $m_r=2$ and $v_F=1$.}\label{fsppurevec}
\end{figure}

For $m_l\approx m_r\equiv m$, \eqref{sppurevec} becomes
\begin{equation}\label{sppureveccaprox}
\sqrt{v_F^4 m^2 - E^2} \cos\left(\frac{a}{v_F}\right)+E\sin\left(\frac{a}{v_F}\right)=0,
\end{equation}
which gives the value of the energies for bound states
\begin{equation}\label{sppurevecpaproxsolre}
E=\pm\,mv_F^2\,\cos\left(\frac{a}{v_F}\right).
\end{equation}
\begin{figure}[!h]
\centerline{\includegraphics[scale=0.35,clip]{resonanpurev1.eps}}
\caption{Transmission resonances for $a=\pi$, $m_l=1$, $m_r=2$ and $v_F=1$.}\label{transres1}
\end{figure}

The transmission coefficient is bounded from below,
\begin{equation}\label{Tlpurevec}
\left|t_l\right|^2\geq\frac{8 m_l^2 m_r^2 \sqrt{E -v_F^2 m_l} \sqrt{E +v_F^2 m_l} \sqrt{E -v_F^2 m_r} \sqrt{E
   +v_F^2 m_r}}{4 m_l^2 m_r^2 \sqrt{E -v_F^2 m_l} \sqrt{E +v_F^2 m_l} \sqrt{E -v_F^2
   m_r} \sqrt{E +v_F^2 m_r}+E ^2 \left(m_l^2+m_r^2\right)^2}
\end{equation}
so that the transmission always occurs. There are transmission resonances or virtual bound states \cite{galindo} for values $\frac{a}{v_F}=n\pi$, $n\in\mathbb{Z}$ (see Figures \ref{transres1} and \ref{transres2}).
\begin{figure}[!h]
\centerline{\includegraphics[scale=0.35,clip]{resonanpurev2.eps}}
\caption{Transmission resonances for $a=\pi$, $m_l=1$, $m_r=2$ and $v_F=1$.}\label{transres2}
\end{figure}

\section{Concluding remarks}\label{remark}
Using the Von Neumann's theory of self-adjoint extensions and eliminating spurious phases of the transmission amplitude, we found the general matching conditions \eqref{bouncondaa20} that describe each one of the different domains of the various self-adjoint extensions of \eqref{piecehamiltonian}. Using the scattering theory, we obtained the spectrum of each one of the extensions, where each corresponds to a different Hamiltonian operator with interaction.

Finally, we found that of the four different self-adjoint extensions of \eqref{piecehamiltonian}, the first three are confining self-adjoint extensions, while the latter is not. For three of the self-adjoint extensions, there is a value of the strength of the interaction point which the energy of the particle is the same for the mass jump case and without it.

\section{Acknowledgments}
This work was supported by IVIC under Project No. 1089.

\section{References}\label{refs}


TITLE: Understaning Euclidean Green's function
QUESTION [2 upvotes]: Consider a scalar field coupled to a source
$$(\Box - m^2)\phi(x) = -J(x)\tag{1}.$$
Then, the response of the source is determined by the Green's function $G(x-y)$, which satisfies
$$(\Box - m^2)G(x-y)=-\delta(x-y) \tag{2}.$$ In Euclidean signature the Green's function which is the solution of the previous equation, can be written as the Fourier transform
$$G(x-y)= \int \frac{d^dk}{(2\pi)^d} \frac {e^{ik\cdot(x-y)}}{k^2+m^2} \tag{3}.$$
I cannot understand how given (3) the solution of (1) can be expressed as the integral
$$\phi(x)=\int d^d y G(x-y)J(y)\tag{4}.$$
My guess is that one has to take (1) and act somehow (4) but right now I cannot see how to arrive to (4). I would appreciate some help.
P.S. My main problem is the fact that in the following semi-proof
\begin{align}
(\Box-m^2)\phi &= (\Box - m^2)\int dy \, J(y) \phi_i(x-y) \\
               &= \int dy \, J(y) {\color{red}{(\Box - m^2)}}\phi_i(x-y) \\
               &= \int dy \, J(y) {\color{red}{\delta(x-y)}} \\
               &= J(x)
\end{align}
Not only I do not get the minus sign of (1) but also I do not understand why we use the homogeneous Klein-Gordon equation to get the red delta function in a problem where we begun with the inhomogeneous one!

REPLY [2 votes]: First, the obvious explanation for the sign is that if $J$ has a minus sign in (1), then there should be a minus sign in (4).
For some reason your $G$ turned into $\phi_i$. Assuming that they are the same thing, then I'm not sure I understand your problem. We didn't use the homogeneous KG equation to get the delta function; we used the inhomogeneous one, with $J(x) = \delta(x-y)$. If $\phi_i$ was a solution to the homogeneous KG equation, $(\Box-m^2)\phi_i$ would be zero, not $\delta(x-y)$.


TITLE: Prove that $2\sqrt 5$ is irrational
QUESTION [1 upvotes]: Prove that $2\sqrt 5$ is irrational

My attempt:
Suppose $$2\sqrt 5=\frac p q\quad\bigg/()^2$$
$$\Longrightarrow 4\cdot 5=\frac{p^2}{q^2}$$
$$\Longrightarrow 20\cdot q^2=p^2$$
$$\Longrightarrow q\mid p^2$$
$$\text{gcd}(p,q)=1$$
$$\Longrightarrow \text{gcd}(p^2,q)=1$$
How can I procced?

REPLY [2 votes]: You have $20|p^2$, thus $2|p^2\wedge 5|p^2$, hence by Euclid's lemma $2|p\wedge 5|p$, hence $10|p$, so we can write $p=10k$ for $k\in\mathbb{Z}$ and then $20q^2=100k^2\Rightarrow q^2=5k^2$, hence $5|q^2$, and by the same lemma $5|q$, thus $\text{gcd}(p,q)=5$, so we got a contradiction.


TITLE: How to evaluate $\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{2013}x}$
QUESTION [2 upvotes]: How to evaluate the integral
$$\int_0^{\frac{\pi}{2}}\frac{\mathrm dx}{1+\tan^{2013}x}$$
It is almost impossible to calculate the antiderivative of this function by hand. Are there any tricks?

REPLY [6 votes]: $$\text{Use }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
Generalization :
$$\text{If }J=\int_a^b\frac{g(x)}{g(x)+g(a+b-x)}dx, J=\int_a^b\frac{g(a+b-x)}{g(x)+g(a+b-x)}dx$$
$$\implies J+J=\int_a^b dx$$ provided $g(x)+g(a+b-x)\ne0$
Here $\displaystyle J=\int_0^{\frac\pi2}\frac{dx}{1+\tan^{2013}x}=\int_0^{\frac\pi2}\frac{\cos^{2013}x}{\sin^{2013}x+\cos^{2013}x}dx$ so $g(x)=\cos^{2013}x$
and in $\displaystyle\left[0,\frac\pi2\right]\sin x,\cos x\ge0\implies \cos^{2013}x+\sin^{2013}x>0$ i.e., $\ne0$

REPLY [3 votes]: Use the identity $$\int_{0}^{\pi/2}f(x)\,\,\mathrm{d}x = \int_{0}^{\pi/2}f\left(\frac{\pi}{2} - x\right)\,\,\mathrm{d}x $$
Now define $$g(x) = f(\pi/2 - x) $$ and note that $f(x) + g(x) = 1$. Thus the integral must equal $\pi/4$, check this yourself!
Note that the exponent 2013 is a red herring - this technique works owing to the symmetry of the tangent, and 2013 could be replaced by any real number and the integral would still equal $\pi/4$.