text:
To get more accurate information about the caustics of
iterations, we have to rely on lengthy calculations.  Our
investigation is indeed partly theoretical and partly
experimental.  We will first describe some interesting
properties with illustrations.  The technical details of
justification are left to the interested reader in \S\ref{Sec-details}.
\subsection{Observations}
\label{Sec-observations}
We first put forward a conjecture
about the exact picture of the caustic when $R_r^n$ is still a
diffeomorphism.
Then we look at the bifurcation process of the structure of the
caustics when $r$ varies.
Finally, we compare the caustics for different $n$.
We will soon see that the 2-cycles of $R_r$ play a special role
(propositions~\ref{PropTgtReven} and~\ref{PropQuadRinfty}).  The
2-cycles are $\Set{0,\pi}$ and $\Set{\pm\phi_c}$ where $\phi_c \in
(0,\pi)$ and $\cos\phi_c = \dfrac{1-\sqrt{1+8r^2}}{4r}$.  We will
often refer to this notation.

\begin{conjecture}\label{ConjAllRr}
For $0 < r \leq 1/3$, the caustic of $R_r^{2m+1}$ is a C$^\infty$
curve with exactly four cusp singularities, with two of them
occurring at $\phi=0, \pi$.
On the other hand, the caustic of $R_r^{2m}$ is a differentiable curve;
C$^\infty$ everywhere except at exactly the four 2-periodic
points of $R_r$, where the caustic is tangent to the unit circle.
\end{conjecture}
The conjecture is demonstrated by the following pictures,
which are produced by programming in Mathematica.
The purple (thin) curve is the unit circle and the
blue (thick) curve is the caustic with the light source
at the red dot, $(r,0)$.
Besides these pictures, we also have
partial results that support our conjecture.

\medskip
{\parindent=0mm
\parbox{52mm}{\centering
\mbox{\epsfxsize=50mm \epsfbox{caust02.ps}}\\
{\tiny caustic for $n=2$, $r=1/3$.}
} \hfill
\parbox{52mm}{\centering
\mbox{\epsfxsize=50mm \epsfbox{caust03.ps}}\\
{\tiny caustic for $n=3$, $r=1/3$.}
} \hfill
\parbox{52mm}{\centering
\mbox{\epsfxsize=50mm \epsfbox{caust04.ps}}\\
{\tiny caustic for $n=4$, $r=1/3$.}
}}
\medskip

\begin{proposition}\label{PropTgtReven}
Any caustic curve of $R_r^{2m}$ is tangential to the unit
circle at
$(\cos\phi,\sin\phi)$ at any point $\phi$ satisfying
$R_r^{2m}(\phi) = \phi$. In particular, this includes
the points $0$, $\pi$, and $\pm\phi_c$.
Moreover, if $r\leq 1/3$, these are the only four points
that the caustic
meets the circle.
\end{proposition}
\begin{proof}
By substitution of such $\phi$ into equations
(\ref{Eqncausticf}) with
$f=R_r^n$,
we have $x(\phi) = \cos\phi$ and
$y(\phi) = \sin\phi$.  Applying lemma~\ref{LemCuspAtFixes} to
such $\phi$'s, the assertion about the tangential property
of the caustics of even
iterations of $R_r$ follows easily.
Secondly, since
$x^2 + y^2 = 1$, we have
\begin{align*}
(R_r^n)'(\phi)\cos\phi - \cos R_r^n(\phi) &= 0 \\
(R_r^n)'(\phi)\sin\phi - \sin R_r^n(\phi) &= 0.
\end{align*}
This leads to $R_r^n(\phi) = \phi$.  If $r\leq 1/3$, by
lemma~\ref{LemCalRn}, $n$ must be even and $\phi$ is one
of the four 2-periodic points.
\end{proof}
We have already seen from symplectic topology that caustics
of odd
iterations, $R_r^{2m+1}$, always have at least four cusps for
sufficiently small $r>0$.  Moreover,
for all $R_r$, two of the cusps occur at
$\phi=0,\pi$.
Now, we may extend this result to odd iterations of $R_r$
with isolated exceptional values of $r$.
It turns out these exceptional values only occur in
$r>1/3$.
\begin{proposition}\label{PropCusp0piRodd}
For $0<r< 1/3$ and for generic $1/3 < r < 1$, the caustic of
$R_r^{2m+1}$ always has cusps at $0, \pi$.
\end{proposition}
\begin{proof}
Note that for both $\phi_a = 0, \pi$, one has
$R_r(\phi_a) = \phi_a + \pi \mod 2\pi$, thus we may apply
lemma \ref{LemCuspAtFixes} to check whether there are cusps.
Furthermore, $R_r^2(\phi_a)=\phi_a$ and $R_r''(\phi_a)=0$,
we can simplify the chain rules (lemma~\ref{LemChainRuleRn}) and
obtain $(R_r^{2m+1})''(\phi_a)=0$ and
\begin{align*}
(R_r^{n+2})^{(3)}(\phi_a) &=
R_r'(R_r(\phi_a)) (R_r^n)'(\phi_a) R_r^{(3)}(\phi_a) + \\
&\qquad {} +
   {{R_r'(\phi_a)}^3} \left[ (R_r^n)'(\phi_a)
   R_r^{(3)}(R_r(\phi_a)) +
	 {{R_r'(R_r(\phi_a))}^3} (R_r^n)^{(3)}(\phi_a)
	 \right].
\end{align*}
Let us temporarily define, for positive integers $n$,
$$
A_{n} = -(R_r^{n})' + {(R_r^{n})'}^3 + 2 (R_r^{n})^{(3)}.
$$
So it suffices to check that $A_{2m+1}(\phi_a) \ne 0$.
We will proceed by induction on $m$.  First, by
direct computation,
$$
A_1(0) = \frac{24r^2}{(1-r)^2} > 0, \qquad\qquad
A_1(\pi) = \frac{24r^2}{(1+r)^2} > 0.
$$
Then, it can be shown that
\begin{align*}
A_{n+2}(\phi_a)
&= \left(\frac{1-9r^2}{1-r^2}\right)^3 A_n(\phi_a) +
\left[2R_r^{(3)}(\phi_a+\pi)R_r'(\phi_a)^3 +
2R_r'(\phi_a+\pi)R_r^{(3)}(\phi_a) \right. \\
&\qquad \qquad \left. {} - R_r'(\phi_a)R_r'(\phi_a+\pi) +
R_r'(\phi_a)^3R_r'(\phi_a+\pi)^3 \right]
(R_r^n)'(\phi_a) \\
&= \left(\frac{1-9r^2}{1-r^2}\right)^3 A_n(\phi_a) +
A_2(\phi_a)\cdot(R_r^n)'(\phi_a). \\
\intertext{In particular,}
A_{2m+1}(0) &= \left(\frac{1-9r^2}{1-r^2}\right)^3
A_{2m-1}(0) + A_2(0) \left(\frac{1-3r}{1-r}\right)^{m}
\left(\frac{1+3r}{1+r}\right)^{m-1}; \\
A_{2m+1}(\pi) &= \left(\frac{1-9r^2}{1-r^2}\right)^3
A_{2m-1}(\pi) + A_2(\pi) \left(\frac{1-3r}{1-r}\right)^{m-1}
\left(\frac{1+3r}{1+r}\right)^{m}; \\
\intertext{where}
A_2(0) &= \frac{48r^2(1+r)}{(1-r^2)^3}(1-3r+13r^2-15r^3), \\
A_2(\pi) &= \frac{48r^2(1+r)}{(1-r^2)^3}(1+3r+13r^2+15r^3).
\end{align*}
It can be easily computed that $A_2(0) > 0$ for $0<r<1/3$ and
$A_2(\pi)>0$ for all $r>0$.  Thus,
$A_{2m+1}(\phi_a) \geq \left(\dfrac{1-9r^2}{1-r^2}\right)^3
 A_{2m-1}(\phi_a) > 0$.

For $r > 1/3$, $A_{2m+1}$ may have zeros.  We can only
conclude from above that $A_{2m+1}$ is a rational function
in $r$ with denominator being a power of $(1-r^2)$.
Therefore, it has only isolated zeros and cusps at $0, \pi$ occur
for $r > 1/3$ generically.
\end{proof}
We observe that at cusp points, (\ref{Eqncausticftgt})
gives a set of ``homogeneous'' equations which has zero determinant.
Thus, except at a couple of $\phi$'s, it is sufficient to solve
only one equation, say, $x'(\phi)=0$.  It is likely that this equation
has exactly four solutions for $0<r<1/3$.  However, it is still hard
to solve explicitly especially for high iterations of $R_r$.
Here are two pictures of the functions
$(R_r^3)''/((R_r^3)'(1+(R_r^3)'))$ and $(\sin\phi + \sin
R_r^3(\phi))/(\cos\phi - \cos R_r^3(\phi))$ for $r = 0.1$ and
$r=0.33$.  Exactly four cusp solutions are demonstrated in each of
them.  We
do not have a proof of this graphical fact.  Perhaps it
may be proved by detailed
curve sketching argument and comparison of $R_r^{2m-1}$ and
$R_r^{2m+1}$ using the known properties of $R_r^2$ given in
lemma~\ref{LemCalRn}.

{\parindent=0mm
\parbox{72mm}{\centering
\mbox{\epsfxsize=70mm \epsfbox{exact4a.ps}}
} \hfill
\parbox{72mm}{\centering
\mbox{\epsfxsize=70mm \epsfbox{exact4b.ps}}
}
}

\noindent
It is probably worthwhile to compare the situation here with the
so-called Jacobi conjecture promoted by Arnold \cite{A1,A5}. As mentioned
in the beginning of \S\ref{Sec-caustics}, the caustics of
$R_r$ agrees with the locus
of conjugate points of $(r,0)$ on a flat flying disk. Although
the loci of higher order conjugate points are not the same as the
caustics of odd iterations of $R_r$, their common contact geometric
nature
indicates that two problems of whether there are exactly 4
cusps on the loci of higher order conjugate points and whether
there are exactly 4 cusps of the caustics of odd iteration of $R_r$
might
be related. In both cases, since we do not have exact nice formulae
for the
loci of higher order conjugate points and the caustics of odd
iterations
of $R_r$, it would be very difficult to have an exact count of
cusps. Of course, Conjecture~\ref{ConjAllRr} deals with a
very special situation. From graphical
evidence, it is tempting to think that a proof should not be out of
reach
with brute force calculation. Nevertheless, after spending much
effort on
this temptation, we think some more conceptual understanding of the
caustics of iterations of $R_r$ is needed in order to get an exact
count of cusps.

Furthermore, in the proof above, we see that for
$n\geq 3$, except $A_3(\pi)$, the quantity
$A_{n}(0) = 0 = A_n(\pi)$ always at $r=1/3$.  There is a
possible structural change on the caustic of $R_r^{n}$
occurring at $r=1/3$ for $n\geq 3$.
We observe from experiment that, for the caustic of odd
iterations, once $r > 1/3$, bifurcation
of cusp may occur.  Interestingly, from the computed pictures (only
that of $R_r^3$ is shown), bifurcation only occur at the cusp
corresponding to $\phi=0$ but not others.  Would
the different properties between $A_2(0)$ and
$A_2(\pi)$ be part of the reasons?

\medskip
{\parindent=0mm
\parbox{72mm}{\centering
\mbox{\epsfysize=50mm \epsfbox{caust03b.ps}} \\
{\tiny $n = 3$, $r=0.36$.}
} \hfill
\parbox{80mm}{\centering
\mbox{\epsfysize=50mm \epsfbox{bifur3a.ps}} \\
{\tiny its enlargement, $\phi=0$.}
}}
\medskip

On the other hand, bifurcation into cusps also occurs
for even iterations of $R_r$ at $r=1/3$.
We will discuss this by beginning with some Taylor expansions.
Since $R_r$ and its iterations are $2\pi$-periodic odd functions, they
have particular nice expansions at $\phi_a=0,\pi$.  This enables us
to see the local properties of the caustics more clearly.

Let $f$ be any even iteration of $R_r$, then $f(\phi_a) = \phi_a$.
We write $\vartheta = \phi-\phi_a$ and $g(\vartheta) = f(\phi) -
\phi_a$ and suppose it has an expansion
$$
g(\vartheta) = \sum_{k=0} a_{2k+1}\vartheta^{2k+1}.
$$
One can inductively work out the coefficients of the expansions of
$1+f'(\phi) = 1+g'(\vartheta)$, $\cos f(\phi) = \pm\cos
g(\vartheta)$, etc.  If $P_k, Q_k$ denote
polynomials with $P_k(0,\ldots,0)=0=Q_k(0,\ldots,0)$, one has
\begin{align*}
x(\phi) &= \pm 1 \mp \frac{a_1}{2}\vartheta^2 + \sum_{k=1}
\frac{P_k(a_1,\ldots,a_{2k+1})}{(1+a_1)^{2k+1}} \vartheta^{2k}; \\
y(\phi) &= \pm \frac{2a_1}{1+a_1}\vartheta + \sum_{k=1}
\frac{Q_k(a_1,\ldots,a_{2k+1})}{(1+a_1)^{2k+1}}\vartheta^{2k+1}.
\end{align*}
These expansions are helpful to understand the caustics of
$R_r^{2m}$ at $\phi_a=0,\pi$.  It would be convenient to look at the
pictures before we go on.

\medskip
\noindent{\parindent=0mm
\parbox{52mm}{\centering
\mbox{\epsfysize=50mm \epsfbox{caust02b.ps}} \\
{\tiny $n = 2$, $r=0.36$.}
} \hfill
\parbox{73mm}{\centering
\mbox{\epsfysize=50mm \epsfbox{bifur2a.ps}} \\
{\tiny its enlargement, $\phi=0$.}
}\hfill\\
\medskip
\noindent\parbox{52mm}{\centering
\mbox{\epsfysize=50mm \epsfbox{caust04b.ps}} \\
{\tiny $n = 4$, $r=0.36$.}
} \hfill
\parbox{73mm}{\centering
\mbox{\epsfysize=50mm \epsfbox{bifur4a.ps}} \\
{\tiny its enlargement, $\phi=0$.}
}}
\medskip

\noindent In the above pictures, cusps are born near
$\phi=0$ and $\pi$.  From the enlargement, the
caustic bifurcates into $2m$~cusps when $r$ increases
across $1/3$, where $R_r$ changes from a diffeomorphism to a
degree~1 map.  In the expansion of $R_r^{2m}$, $a_1 = \left(
\frac{1-9r^2}{1-r^2} \right)^m$.  By this, we will see that
$r=1/3$ is the value that the
caustic of $R_r^{2m}$ changes at $\phi=0,\pi$.
In fact, the caustics of
$R_{1/3}^2$ has the following Taylor expansions.  At $\phi_a = 0$,
\begin{align*}
x(\phi) - x(0) &=
\frac{-27}{4}\vartheta^4 + \text{O}(\vartheta^6) \\
y(\phi) - y(0) &=
18\vartheta^3 + \text{O}(\vartheta^5); \\
\intertext{and at $\phi_a=\pi$,}
x(\phi) - x(\pi) &=
\frac{243}{16}\vartheta^4 + \text{O}(\vartheta^6) \\
y(\phi) - y(\pi) &=
\frac{-81}{2}\vartheta^3 + \text{O}(\vartheta^5).
\end{align*}
This shows that the caustic of $R_r^2$ undergoes a
swallowtail bifurcation
at $0,\pi$ when $r=1/3$.
We may further work out the
expansion of $R_{1/3}^{2m}$ as the
$m$-iteration of $R_{1/3}^2$. Using $a_1=0$ and $a_3\ne 0$
for $R_r^2$, we have
$$
R_{1/3}^{2m}(\phi) = \phi_a + \vartheta^{3^m} U(\vartheta)
$$
for some function $U$ with $U(0)\ne 0$.  The bifurcation of
the caustics of $R_r^{2m}$ at $\phi=0,\pi$
should be of the type $(\vartheta^{3^m+1},\vartheta^{3^m})$ when $r$
passes $1/3$.

In the above, we consider the behavior of the caustics $R_r^n$
with parameter $r$ and $n$ fixed.
What happens to the caustics if $r$ is
fixed and $n$ is allowed to vary?  There is an interesting
phenomenon for $r \leq 1/3$.  Although
there is fundamental difference when $n$ is odd and even,
this difference
disappears as $n$ goes to infinity.
Here are the pictures showing how the caustics
of $R_r^{2m+1}$ and $R_r^{2m}$ change--they tend
to the same quadrilateral.

\medskip
{\parindent=0mm
\parbox{52mm}{\centering
\mbox{\epsfxsize=50mm \epsfbox{caust81.ps}} \\
{\tiny caustics of $R_{1/3}^{\text{odd}}$}
}\hfill
\parbox{52mm}{\centering
\mbox{\epsfxsize=50mm \epsfbox{caust82.ps}} \\
{\tiny caustics of $R_{1/3}^{\text{even}}$}
}}
\medskip

\noindent The green (dashed) vertical line from $\phi_c$ to
$-\phi_c$ in the picture of even or
odd caustics is determined by $r$ but not the number of
iterations.  When $n$ is even, it is where the caustic is
tangent to the circle.  When $n$ is odd, every caustic is
tangent to this vertical line because it is the line joining
$\phi_c$ and $R_r^{2m+1}(\phi_c) = -\phi_c$.  The point of tangency
occurs exactly at $\phi_c$ by definition.
\begin{proposition}\label{PropQuadRinfty}
For $0<r\leq 1/3$, as $m\to\infty$, both the caustics of
$R_r^{2m+1}$ and $R_r^{2m}$ approach the same quadrilateral
defined by the four points $0$, $\pi$, and $\pm\phi_c$,
which are the only 2-periodic points of $R_r$.
\end{proposition}
\begin{proof}
Let us first see that the caustics of $R_r^{2m+1}$ at the four
points tend to the circle as $m\to\infty$.
These four $\phi$'s are the solution to $R_r^2(\phi) =
\phi$.  Thus,
we have
$(R_r^{2m+1})'(\phi) = R_r'(\phi)^{m+1}\cdot
R_r'(R_r(\phi))^m$.

At $\phi=0, \pi$, the coordinates of the caustic are
$$
x(\phi) =
\frac{\pm\left((R_r^n)'(\phi)-1\right)}
{1+(R_r^n)'(\phi)}, \qquad\qquad
y(\phi) = 0.
$$
It is clear that $(R_r^{2m+1})'(0) =
R_r'(0)^{m+1}R_r'(\pi)^m
= \left(\frac{1-3r}{1-r}\right)^{m+1}\cdot
\left(\frac{1+3r}{1+r}\right)^m$.  Thus, $x(0) \to -1$ as
$m\to\infty$.  The situation at $\pi$ is similar.

At the point $\phi_c$ with
$R_r^n(\phi_c) = -
\phi_c$, the coordinates of the caustic at this point are
$$
x(\phi_c) = \cos \phi_c, \qquad\qquad
y(\phi_c) = \frac{
\left(-1+(R_r^n)'(\phi_c)\right)\sin\phi_c}
{1+(R_r^n)'(\phi_c)}.
$$
Since $R_r$ is odd and $R_r^2(\phi_c) = \phi_c$, it follows
that
$(R_r^n)'(\phi_c) = R_r'(\phi_c)^n$.  Moreover, by
$4r\cos\phi_c =
1-\sqrt{1+8r^2}$,  one may show that
$$
\frac{\left(-1+(R_r^n)'(\phi_c)\right)}{1+(R_r^n)'(\phi_c)}
\to 1
\qquad\quad \text{as $n\to\infty$.}
$$
Again, these two cusps approach to the unit circle.

Secondly, from the lemmas~\ref{LemCalR} and~\ref{LemCalRn}, $0$ and $\pi$
are the attracting fixed points of $R_r^2$ while $\pm\phi_c$
are repelling.  Moreover, the attracting basins for $0$ and $\pi$ are
$(-\phi_c,\phi_c)$ and $(\phi_c,2\pi-\phi_c)$
respectively.  Thus, for any given neighborhood of $0$,
for sufficiently large $m$, for any neighboring $\phi_1, \phi_2 \in
(-\phi_c,\phi_c)$, $R_r^{2m}(\phi_1)$ and
$R_r^{2m}(\phi_2)$ lie in that neighborhood of $0$.
Hence, the intersection of the lines from $\phi_j$ to
$R_r^{2m}(\phi_j)$ lies in a neighborhood of the
quadrilateral.  The proof for the cases at $\pi$ and of odd
iterations are similar.
\end{proof}
\endinput


TITLE: Solving ODE - Space curve Frame
QUESTION [5 upvotes]: I'm trying to calculate the parallel frame $\{T, U, V\}$ of a space curve $\alpha : I \mapsto \mathbb{R}^3$. It's similar to Frenet frame, except we have instead the projection of torsion $\tau$ on $U$: $\tau_U = 0$
It's described by the three equations,
$$T' = \kappa_U U + \kappa_V V \quad \quad \quad \quad \ \ (1)$$
$$U' = -\kappa_U T = -\langle \kappa,U \rangle \ T \quad \quad (2)$$
$$V' = -\kappa_V T = -\langle \kappa,V \rangle \ T \quad \quad (3)$$
where $\kappa = T'$ is the curvature vector and $\kappa_U, \kappa_V$ are its components.
I know $T$, $\kappa$ and the initial conditions $\{T(0), U(0), V(0)\}$ at $\alpha(0)$ but I don't know how to solve the ODE $(2)$ and $(3)$ in order to get $U$ and $V$.
Any suggestion or a note on what to search for would be great!

REPLY [0 votes]: To find $U$ and $V$ first consider the Frenet frame $\{T,N,B\}$ along a unit speed curve (or where it exists). We know this frame explicitly since $N=T'/\|T'\|$ and $B=T\times N$.  Also note that there exists locally a smooth function $\theta\colon I \to \mathbb{R}$ such that
$$U = \cos \theta N + \sin \theta B.$$
Deriving $U$ gives
$$
  U' = -\kappa \cos\theta T + (\theta'+\tau)(-\sin\theta N + \cos\theta B).
$$
We see that $U$ satisfies ODE $(2)$ iff $\theta' = -\tau$. Also the initial value $\theta(0)$ is determined by $U(0)$, so we find $\theta$ from this:
$$
  \theta = -\int \tau\,dt + \mathrm{const.}
$$
The vector field $V$ is determined similarly. Note that the inner product $U\cdot V$ is constant along the curve. So if $U(0)$ and $V(0)$ are orthogonal, we can just take $V=\pm T \times U$.
Remark: You can do the same calculation starting with an arbitrary adapted frame $\{T,N_1,N_2\}$. You will find the condition $\theta' = -N_1' \cdot N_2$.
You will probably find this article interesting:
There is more than one way to frame a curve. by R.L. Bishop (The American Mathematical Monthly, 1975).


TITLE: Is $a^2b^2-a^2+b^2$ factorable or simplifiable?
QUESTION [3 upvotes]: I'm just wondering if the expression $a^2b^2-a^2+b^2$ is simplifiable or factorable. It just looked like it might have been factorable but I wasn't sure.

REPLY [5 votes]: Maybe the following.
$$a^2b^2-a^2+b^2=a^2b^2-a^2+b^2-1+1=1+(b^2-1)(a^2+1).$$
It's not factoring of course.


text:
{\bf Problem.} Write the following expression as a polynomial: $$(2x^2+3x+7)(x+1)-(x+1)(x^2+4x-63)+(3x-14)(x+1)(x+5).$$
{\bf Level.} Level 5
{\bf Type.} Algebra
{\bf Solution.} Factoring out $(x+1)$, we have:

\begin{align*}
&(x+1)((2x^2+3x+7)-(x^2+4x-63)+(3x-14)(x+5))\\
=\text{ }&(x+1)(2x^2+3x+7-x^2-4x+63+3x^2+x-70) \\
=\text{ }&(x+1)(2x^2-x^2+3x^2+3x-4x+x+7+63-70) \\
=\text{ }&(x+1)(4x^2+0x+0) \\
=\text{ }&4x^2(x+1) \\
=\text{ }&\boxed{4x^3+4x^2}.
\end{align*}

set_name: MATH


TITLE: Is any given triplet spin state an eigenstate of some $j^z$ in the suitable basis?
QUESTION [3 upvotes]: Imagine you have a triplet spin state, which, in general, can be written as
$$|\psi \rangle = \alpha | \uparrow \uparrow \rangle +
 \beta ( | \downarrow \uparrow \rangle+ | \uparrow \downarrow \rangle) +
 \gamma | \downarrow \downarrow \rangle,$$
with some complex numbers $\alpha, \beta, \gamma$.
Here, I've written the given state as a superposition of the $m_z=0,-1$ and $+1$ triplet states in the $z$-basis; so, for example, $|\uparrow \uparrow \rangle$ is an eigenstate of $S^{z} =\tfrac{1}{2}( \sigma_1^{z} + \sigma_2^{z}) $ with eigenvalue $m_z = 1$.
My question is: Can you find a total spin operator $S^{\hat{n}} =\tfrac{1}{2}( \sigma_1^{\hat{n}} + \sigma_2^{\hat{n}}) $, defined by some quantization axis $\hat{n}$, such that the above state $|\psi \rangle$ is an eigenstate of $S^{\hat{n}}$?

REPLY [3 votes]: No, this isn't possible for generic states.
To see the underlying structure in more detail, let's start off by using a correct normalization for your state,
$$|\psi \rangle = \alpha | {\uparrow \uparrow} \rangle +
 \beta \frac{ | {\downarrow \uparrow} \rangle+ | {\uparrow \downarrow} \rangle}{\sqrt{2}} +
 \gamma | {\downarrow \downarrow} \rangle,$$
and then, as Cosmas Zachos noted (in a now-deleted answer), noting that there's nothing special about the tensor-product structure that underlies this state, and that it can be equally well thought of as a general state
$$
|\psi \rangle = \alpha | 1 \rangle + \beta |0 \rangle + \gamma | {-1} \rangle
$$
within an $\ell=1$ dipolar representation of $\rm SO(3)$.
Moreover, since all of the (complex) irreducible representations of $\rm SO(3)$ with equal total angular momentum quantum number $\ell$ are equivalent, your question can equally well be phrased in the language of spherical harmonics, which carry one such representation, i.e. anything that you can say about the state $|\psi\rangle$ as you've phrased it can also be made to apply to the linear combination
$$
f(\theta,\phi) = \alpha \, Y_{11}(\theta,\phi) + \beta \, Y_{10}(\theta,\phi) + \gamma \, Y_{1,-1}(\theta,\phi)
$$
of dipolar spherical harmonics. And in yet another transformation, I will multiply each of these $Y_{\ell m}(\theta, \phi)$ with $r$ to get solid harmonics, which are homogeneous linear combinations of $x$, $y$ and $z$: in other words, anything you can say about the state $|\psi\rangle$ as you've phrased it can also be made to apply to the linear combination
\begin{align}
g(x,y,z)
& = -\alpha \frac{x+iy}{\sqrt{2}} + \beta z + \gamma \frac{x-iy}{\sqrt{2}}
\\ & = \frac{-\alpha + \gamma}{\sqrt{2}} x + \frac{\alpha + \gamma}{\sqrt{2}\, i} x + \beta z
\\ & = \left( \frac{-\alpha + \gamma}{\sqrt{2}}, \frac{\alpha + \gamma}{\sqrt{2}\, i}, \beta \right) \cdot (x,y,z)
\\ & = \mathbf n \cdot \mathbf r,
\end{align}
i.e. to a complex-valued linear functional $g:\mathbb R^3 \to \mathbb C$ which is fully characterized by the (complex) gradient vector $\mathbf n$.
In these terms, then your question boils down to the following:

Does $\mathbf n$ look like $\mathbf n = (0,0,1)$? (which would imply that $|\psi\rangle$ is an eigenstate of some $S_z$ with eigenvalue $0$)
Or does $\mathbf n$ look like $\mathbf n = \frac{1}{\sqrt{2}}(1,i,0)$? (which would imply that $|\psi\rangle$ is an eigenstate of some $S_z$ with eigenvalue $1$)
Or does $\mathbf n$ look like neither of those? (which would imply that $|\psi\rangle$ is not an eigenstate of any $S_z$)

A key component of these questions is the words "looks like", which here means "equal up to a rigid rotation in $\mathbb C^3$ and up to a global phase", with that rotation drawn from $\rm SO(3)$.
And here the answers are relatively easy to provide:

If $\mathbf n$ looks like $\mathbf n = (0,0,1)$, then $\mathrm{Re}(\mathbf n)$ and $\mathrm{Im}(\mathbf n)$ are parallel, in which case $|\psi\rangle$ is an eigenstate of $\mathbf n \cdot S_z$ with eigenvalue $0$.
If $\mathbf n$ looks like $\mathbf n = \frac{1}{\sqrt{2}}(1,i,0)$, then $\mathrm{Re}(\mathbf n)$ and $\mathrm{Im}(\mathbf n)$ are exactly orthogonal, in which case $|\psi\rangle$ is an eigenstate of $(\mathrm{Re}(\mathbf n)\times \mathrm{Im}(\mathbf n))\cdot S_z = \frac{1}{2}(\mathbf n \times \mathbf n^*)\cdot S_z$ with eigenvalue $1$.
And, conversely, if $\mathrm{Re}(\mathbf n)$ and $\mathrm{Im}(\mathbf n)$ are neither parallel nor orthogonal, then neither of those conclusions is possible.

Examples from point 3 are trivial to construct, and they form almost all of the states in the space (by which I mean: the set of complex numbers $(\alpha,\beta,\gamma)$ which yield an eigenstate of some $S_\hat{n}$ forms a set of measure zero under the regular Lebesgue measure of the space $\mathbb C^3$ that encases them).
I should also note that this conclusion changes if you force $\beta\in \mathbb R$, and $\alpha = -\gamma^*$ (which is a not-unreasonable restriction), in which case $\mathbf n$ is also real-valued and everything collapses so that all states are zero-$S_z$ eigenstates along some direction. The fact that this is possible on the (real-valued) $\ell = 1$ forms the premise of my question How many truly different multipolar charge distributions are there?, which goes on to ask what happens at $\ell = 2$ and higher ─ with the conclusion that at higher angular momentum, even if you restrict to real-valued representations, there are nontrivial states that cannot be reduced to the canonical form of a rotated spin eigenvector.


TITLE: $x \in \mathbb Q , 0 \ne y \in \mathbb R$ , $\lim_{n \to \infty} (yx^n-m_n)=0$ , $\{m_n\}$ a sequence of positive integers , then is $x$ an integer?
QUESTION [3 upvotes]: Let $x$ be a rational number and $y$ be a non-zero real number and $\{m_n\}$ be a sequence of positive integers such that $\lim_{n \to \infty} (yx^n-m_n)=0$ ; then is it true that $x$ is an integer ?

REPLY [1 votes]: First, $y$ cannot be irrational: To see this, note that either $m_n=\lfloor yx^n\rfloor$ or $m_n=\lceil yx^n\rceil$ minimizes $|yx^n-m_n|$; By the irrationality of $y$, however, the remainder of this subtraction will not converge to zero with increasing $n$. Hence, $y\in\mathbb{Q}$.
That is just the first part of the proof, there is still something left to do. However, we now know that $y=q/p$. If $x=p$, then $yx^n=qp^{n-1}$ is integer and by choosing $m_n=qp^{n-1}$ the limit is zero. It remains to show that also the converse holds.


TITLE: Convergence in $L^p(\Omega)$ Norm
QUESTION [1 upvotes]: Let $\Omega\subset\mathbb{R}^n$ be a bounded domain. Let $u,v,v_n\in L^p(\Omega)$ and suppose that $$\|u+v_n\|_p\rightarrow\|u+v\|_p$$
Is true that $$\|v_n\|_p\rightarrow\|v\|_p$$
Thanks

REPLY [2 votes]: This is false. Consider the constant functions $u=-1,v_n=1+(-1)^n,v=0$.


TITLE: Detect Wrong Proof by strong induction $a^{n-1}=1$ for all $n$
QUESTION [3 upvotes]: Obviously there is some error in the steps, but i can't figure it out, appreciate some hint.
We show using strong induction that
$$ a^{n-1}=1 \hspace{1cm} \forall n\in \mathbb{N}$$
$$ n=1 \hspace{.5cm} \Longrightarrow \hspace{.5cm} a^{1-1}=1$$
We assume that is true for k with $1\leq k< n-1$, then
$a^{n-1}=\frac{a^{n-2}a^{n-2}}{a^{n-3}}=\frac{1.1}{1}=1$
Therefore by strong induction principle. $\hspace{1cm}a^{n-1}=1\hspace{1cm} \forall n \in \mathbb{N}$
thanks

REPLY [1 votes]: You have assumed that $n-3\ge 1$ or $n\ge4$. So you need to check for $n=2$ and $n=3$. Clearly, $a^{2-1}=a\neq 1$.


TITLE: $A$ is bounded if $\displaystyle\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{H}\|=0$
QUESTION [0 upvotes]: I am looking for the solution to the problem:
Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$. Show that $A$ is bounded if and only if $$\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{H}\|=0.$$ Here $h \in\mathbb R$ and $i^2=-1$.
The if part is easy by Taylor expansion of $e^{ihA}$ and some estimation. But I do not know how to prove the only if part. I tried to prove the equivalent: $A$ is unbounded then $$\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{H}\|\ne 0.$$
I stared with definition of unbounded operator $A$, i.e. there exists a sequence $x_n\in D(A)$ such that $$\dfrac{\|Ax_n\|}{\|x_n\|}\to \infty.$$ But how to derive from this that there exists a sequence $y_n\in D(A)$ such that $$\lim_{h\to 0}\frac{\|e^{ihA}y_n-y_n\|}{\|y_n\|}\to \infty$$

REPLY [1 votes]: By the Spectral Theorem we can assume that $\mu$ is a measure on (some $\sigma$-algebra of subsets of ) $X$, $D\subset L^2(\mu)$, and $$A:D\to L^2(\mu)$$is given by $$Af(x)=m(x)f(x)$$for some measurable function $m:X\to\Bbb R$.
Now we have to show that $m\in L^\infty$ if and only if $$||e^{ihm}-1||_\infty\to0\quad(h\to0),$$which is not hard.


TITLE: Complement of set $\{ z \mid (b+z) \in A, b \in B \}$
QUESTION [0 upvotes]: I'm trying to express the complement $X^C$ of the following set:
$$X = \{ z \mid (b+z) \in A, b \in B \}$$
From the definition of the complement $A^C = \{ a \mid a \notin A \}$, I am not sure which of the following is the complement $X^C$:

$\{ z \mid (b+z) \notin A, b \notin B \}$
$\{ z \mid (b+z) \notin A, b \in B \}$
$\{ z \mid (b+z) \in A, b \notin B \}$

Could someone make this clear to me?

REPLY [0 votes]: Your $x$ contains all $z$ for which $b+z \in A$ and for which $b \in B$. So, the complement would be all $z$ for which that is not true, i.e. for which $b+z \not \in A$ or for which $b \not \in B$. Which translates to:
$$X^C = \{ z | (b+z) \not \in A \lor b \not \in B \}$$
... which is neither one of your options, unless you take the union of all the ones you have:
$$X^C = \{ z | (b+z) \not \in A, b \not \in B \} \cup \{ z | (b+z) \not \in A, b \in B \} \cup \{ z | (b+z) \in A, b \not \in B \}$$


TITLE: integer $m$ has primitive root if and only if the only solutions of the congruence $x^{2} \equiv 1 \pmod m$ are $x \equiv \pm 1\pmod m$.
QUESTION [4 upvotes]: Show that the integer $m$ has primitive root if and only if the only solutions of the congruence $x^{2} \equiv 1 \pmod m$ are $x \equiv \pm 1\pmod m$.
I don't quite understand what this question is asking for, I thought primitive roots had order $\phi(m)$.

REPLY [2 votes]: Let $m\geq 3$ and $g$ is a primitive root modulo $m$. Then $\mathbb Z^*_m$ can be fully represented by $\{g_0,\dots,g^{\varphi(m)-1}\}$. Hence, any $x\in \mathbb Z^*_m$ can be written as  $x=g^k$ for some $0\leq k<\varphi (m)$ and working in $\mathbb Z_m$:
$$x^2=1 \Leftrightarrow (g^k)^2= 1 \Leftrightarrow g^{2k}= 1 \Leftrightarrow \varphi(m) \mid 2k$$
which is only possible for at most two values of $k$, namely $0$ and $\frac{\varphi(m)}{2}$. As $\pm 1$ are always solutions and distinct, those are exactly the two solutions.

The other direction is most likely more difficult, I don't have a  proof yet, but you can of course deduce this from Euler's theorem:
If there is no primitive root modulo $m$, then by Euler's theorem $m$ can not be of the form
$$2, 4, p^k, 2p^k$$
where $p$ is an odd prime and $k$ is a positive integer. Check carefully, that this implies that $m$ is either a power of two or can be written as product of two coprime numbers which are both at least $3$. In both cases we find more than two solutions:

$m=2^k$. Define $x=2^{k-1}-1$. Then $x\not\equiv \pm 1\pmod{2^k}$, but
$$x^2=(2^{k-1}-1)^2=2^k-2\cdot 2^{k-1} +1 \equiv 1\pmod{2^k}$$
$m=ab, (a,b)=1, a,b\geq 3$: Then $x^2\equiv 1\pmod a$ and $x^2\equiv 1\pmod b$ have both at least two distinct solutions ($\pm 1$) and the Chinese remainder theorem tells that $x^2\equiv 1\pmod{ab}$ has at least four solutions.


TITLE: Square Numbers general methods
QUESTION [0 upvotes]: Let this equation be given:
$$a^x+a^y=b^2$$
With which method can one determine if there are infinitely many or no solutions for a random number $a$ so that $b^2$ is a square number.
$a, b, x$ and $y$ have to be positive whole numbers.

REPLY [2 votes]: A fuller exposition of solutions to $a^x+a^y=b^2$
WLOG, let $x\le y$ $$a^x+a^y=b^2 \Rightarrow a^x(a^{y-x}+1)=b^2$$
Now $\gcd(a^x,a^{y-x}+1)=1$, and since $b^2$ is a perfect square, its relatively prime factors $a^x$ and $a^{y-x}+1$ must each be perfect squares. So the exponent  $x=2n$. This further means for some integer $c$, $c^2-(a^{y-x})=1$. However, Catalan's conjecture establishes that if $y-x>1$, the only positive integer solution to that equation is $c=3,a=2,(y-x)=3$. This affords the first family of solutions: $$2^{2n}+2^{2n+3}=2^{2n}(8+1)=b^2 \\ (3\cdot 2^n)^2=b^2 \\ a=2,b=(3\cdot 2^n),x=2n,y=2n+3$$
Examples in this family include $4+32=36,\ 16+128=144$
Next, we look at the case $x=y,\ (y-x)=0$. This corresponds to $2a^x=b^2$ implying first $2\mid b$ and next that $a^x$ must have a factor of $2$ to an odd exponent. That further requires that $x$ be odd, and that $a$ must have a factor of $2$ to an odd exponent. Finally, any odd prime factor of $a$ must be present an even number of times. To recap: $x=y=(2k-1); \ a=2^{2n-1}c^{2m}$ where $c$ is some odd integer. This affords a second family of solution: $$2\cdot (2^{2n-1}c^{2m})^{2k-1}=b^2$$
Examples in this family include $8+8=16, \ 18^3+18^3=108^2,\ 800^5+800^5=25600000^2$
Finally we look at the remaining case $y-x=1$. This requires $a^{2n}(a+1)=b^2$. Plainly, this holds whenever $a=d^2-1$, yielding $b=d(d^2-1)^n$
Examples in this family include $48^5+48^4=16128^2,\ 168^3+168^2=2184^2$
Choices of $a$ other than $2,\ 2^{2n-1}c^{2m},\ d^2-1$ will fail to afford solutions.


TITLE: Prime ideal $P$ of $\mathbb{Z}[x]$ such that $P \cap \mathbb{Z}=\{0\}$ is principal
QUESTION [5 upvotes]: The problem stated more precisely is this:

Let $P$ be a prime ideal of $\mathbb{Z}[x]$ such that $P \cap \mathbb{Z} =\{0\}$. Show that $P$ is a principal ideal.

I think there is a problem with my proof.
So I said that if $m(x)$ is a nonzero polynomial in $P$ of minimum degree then $P=(m(x))$. If $f(x) \in P$ is any polynomial then in $\mathbb{Q}[x]$ there is $q(x),r(x) \in \mathbb{Q}[x]$ such that $f(x)=q(x)m(x)+r(x) \implies r(x)=f(x)-q(x)m(x)$ with $r=0$ or the degree of r less than the degree of $m$. Letting $L$ be the LCM of the denominators of the coefficients we have $Lr(x)=Lf(x)-Lq(x)m(x) \implies Lr(x) \in P$.

This is where the problem is. I want to claim that $Lr(x) \in P \implies L\in P$ or $r(x) \in P$. Obviously this then forces $r(x) \in P$ by the intersection assumption which then forces $r=0$. The problem is that $r(x)$ is not necessarily in $\mathbb{Z}[x]$ so the prime condition on $P$ does not necessarily apply here. Is there a way to strengthen the condition on $m(x)$ to ensure that $r(x) \in \mathbb{Z}[x]$?

REPLY [1 votes]: You know that $r(x)\in\mathbb Q[x]$ and $Lr(x)\in P$ for some $L\in\mathbb Z$, $L\ne0$. Now multiply $Lr(x)$ by a non-zero integer $a$ such that $ar(x)\in\mathbb Z[x]$, and find $Lar(X)\in P$. Then $ar(x)\in P$ and $\deg ar(x)=\deg r(x)<\deg m(x)$.
You have to be careful when start with $m\in P$, $\deg m\ge1$ of minimal degree to suppose that $m$ is primitive if want $P=(m(x))$, that is, the $\gcd$ of coefficients of $m$ is $1$. (You also need this assumption when $r=0$.) If this is not the case, then $m(x)=kn(x)\in P$ with $k\ge 2$ and $n(x)$ primitive, and then $n(x)\in P$ is a generator.


TITLE: Show that each of the following are homomorphisms
QUESTION [1 upvotes]: Show that the following maps are group homomorphisms and find their kernels:
1) $\theta: \Bbb Z \rightarrow GL_2$
$\theta(n) = $$
        \begin{pmatrix}
        1 & n \\
        0 & 1 \\
        \end{pmatrix}
$
My attempt:
Let $y\in\Bbb Z$ such that
$\theta(y) = $$
        \begin{pmatrix}
        1 & y \\
        0 & 1 \\
        \end{pmatrix}
$
Then $\theta(n) \theta(y) = $$
        \begin{pmatrix}
        1 & y \\
        0 & 1 \\
        \end{pmatrix}
 $$
        \begin{pmatrix}
        1 & n \\
        0 & 1 \\
        \end{pmatrix}
$$
=  \begin{pmatrix}
        1 & y+n \\
        0 & 1 \\
        \end{pmatrix}
$ = $\theta(n+y)$
So $\theta: \Bbb Z \rightarrow GL_2$ is a homomorphism. And I think ker$\theta = \theta(0)$ as  \begin{pmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{pmatrix} is the identity of the $GL_2$
(Is that an efficient enough proof?)
2) $\theta:\Bbb Q$ \ {0} $\rightarrow GL_2(\Bbb Q)$ given by
$\theta(a) =  \begin{pmatrix}
        a & 0 \\
        0 & 1 \\
        \end{pmatrix}
$
My attempt:
Let there exist $b \in \Bbb Q$ \ {0} such that $\theta(b) =  \begin{pmatrix}
        b & 0 \\
        0 & 1 \\
        \end{pmatrix}
$
Then we have:
$\theta(a)\theta(b)= \begin{pmatrix}
        a & 0 \\
        0 & 1 \\
        \end{pmatrix}
\begin{pmatrix}
        b & 0 \\
        0 & 1 \\
        \end{pmatrix}
$ = $ \begin{pmatrix}
        ab & 0 \\
        0 & 1 \\
        \end{pmatrix}
$ = $\theta(ab)$
ker$\theta= \theta(1)$
etc
Is this correct way to answer this question?
This isn't homework, by the way. I'm revising for an exam I have on monday and these questions were in our practice sheets. If you have more tips for me on my first ever Abstract Algebra exam please let me know!

REPLY [1 votes]: You’re fine, if a bit clumsy, with verifying that the maps are homomorphisms, but there are some problems with the kernels.
Let’s look at the first problem. To show that $\theta$ is a homomorphism, you really should start with two arbitrary elements of $\Bbb Z$ and verify that $\theta$ has the homomorphism property with respect to these two elements.

Let $m,n\in\Bbb Z$. Then $$\theta(m+n)=\pmatrix{1&m+n\\0&1}=\pmatrix{1&m\\0&1}\pmatrix{1&n\\0&1}=\theta(m)\theta(n)\;,$$ so $\theta$ is a homomorphism.

Your statement that $\ker\theta=\theta(0)$ doesn’t make sense: $\ker\theta$ is by definition a subset of $\Bbb Z$, while $\theta(0)$ is an element of $GL_2$, so they can’t possibly be equal. Go back to the definition:
$$\ker\theta=\left\{n\in\Bbb Z:\theta(n)=\pmatrix{1&0\\0&1}\right\}\;,$$
since $\pmatrix{1&0\\0&1}$ is the identity element of $GL_2$. Now you can argue as follows.

Suppose that $n\in\ker\theta$; then $\theta(n)=\pmatrix{1&0\\0&1}$ by the definition of kernel. On the other hand, $\theta(n)=\pmatrix{1&n\\0&1}$ by the definition of $\theta$, so $n=0$. Thus, $\ker\theta=\{0\}$.

You can also conclude that $\theta$ is injective (one-to-one), since its kernel is trivial, but that’s not part of the problem.
The other problem is dealt with similarly.


TITLE: If a mix of elements in a plasma are frozen rapidly, does the crystalline structure of the frozen material reflect the plasma phase?
QUESTION [0 upvotes]: If you have a mix of elements in a plasma, with ions flying around freely, and you freeze it, would you see reflections of that plasma phase in the crystalline structure of the frozen material? Plasma tends to organize into plasmoids and helices, would that be reflected after freezing?

REPLY [3 votes]: No. In the process of cooling the plasma down, the positive and negative ions it is composed of would recombine and revert to just being a cloud of very hot gas. Cool that further and it would condense into a very cold liquid; cool it still more and (depending on what it is made from) it would freeze into a solid. At each stage of this freezing process, whatever structure there might have existed in the previous phase gets erased as the new phase forms.