text:
\section{Introduction}
In \cite{Seg-autoeq-spherical-twist}, Segal proved that every autoequivalence of a triangulated category can be realised as a spherical twist around a spherical functor.
The construction of the spherical functor is explicit but, as pointed out in {\it ibidem}, the source category for this functor is not optimal, in the sense that it could be larger than needed.
\begin{comment}
The author gives a way to deform this category in the case of autoequivalences with a section, {\it i.e.} an autoequivalence $F : \ca{C} \rightarrow \ca{C}$ together with a natural transformation $\sigma : \text{id} \rightarrow F$.
\end{comment}

An example of spherical twist is given by the spherical twist around a spherical object as introduced in \cite{Seidel-Thomas01}.
Let $X$ be a smooth projective variety of dimension $d$ over a field $k$, and consider an object $E \in \derived^{\bounded} (\text{Coh}(X))$ such that $\text{RHom}_{X}(E,E) = k \oplus k[-d]$ and $E \otimes \omega_{X} \simeq E$.
Then, one can define an autoequivalence $T_{E} \colon \derived^{\bounded}(\text{Coh}(X)) \rightarrow \derived^{\bounded}(\text{Coh}(X))$ such that, for every $F \in \derived^{\bounded} (\text{Coh}(X))$, the object $T_E(F)$ fits in a distinguished triangle
\[
	\text{RHom}_{X}(E, F ) \otimes E \rightarrow F \rightarrow T_{E}(F).
\]
In the setting of spherical functors, the functor $T_{E}$ can be realised as the spherical twist around the spherical functor $F \colon \derived^{\bounded}(\text{pt}) \rightarrow \derived^{\bounded}(\text{Coh}(X))$ that sends $k$ to $E$.
It is clear that in this case $\derived^{\bounded}(\text{pt})$ is the nicest category we can look for.

Assume now that we have two spherical objects $E_1,E_2 \in \derived^{\bounded}(\text{Coh}(X))$; we know that the autoequivalence $\Phi = T_{E_2} \circ T_{E_1}$ can be realised as a spherical twist around a spherical functor, therefore it is natural to ask ourselves what is this functor and what is the best source category for it.
As we are dealing with two objects, one might expect the morphisms between them to play a role in the description of the functor $\Phi$.
This expectation is indeed correct, and to represent $\Phi$ as a spherical twist around a spherical functor we will use the dg-algebra
\[
	R = k \, \mathrm{id}_{E_2} \oplus \text{RHom}_{X}(E_2, E_1) \oplus k \, \mathrm{id}_{E_1},
\]
where the multiplication comes from the algebra structure of $\mathrm{RHom}_{X}(E_2 \oplus E_1, E_2 \oplus E_1)$.

Let us now consider the general setting of spherical functors as developed in \cite{Anno-Log-17}: let $\ca{A},\ca{B},\ca{C}$ be three small dg-categories over a field $k$, and $M \in \derived(\ca{A}\text{-}\ca{C})$, $N \in \derived(\ca{B}\text{-}\ca{C})$ be two spherical bimodules, {\it i.e.}, bimodules whose associated functors are spherical.

We aim to find a spherical functor whose twist gives the composition of the twists associated to $M$ and $N$.
The following is our main result.

\begin{comment}
In \cite{Anno-Log-17}, the authors formalized the concept of spherical functors concentrating themselves on functors induced by tensor product with a bimodule.
To get started, we will translate the familiar concept of spherical twists around spherical objects in the language of {\it ibidem}, and we will prove

\begin{thm}[\autoref{n-sph-objs}]
	Let $\ca{C}$ be a small DG-category over a field $k$.
	Let $E_1, E_2$ be two d-spherical objects.
	Then, the smooth and proper DG-algebra $R = k \oplus \textup{RHom}_{\ca{C}}(E_2, E_1) \oplus k$ gives a source category for a spherical functor
	\[
	\begin{tikzcd}[column sep= 6em]
		\derived(R) \ar[r, "f"] & \derived( \ca{C})
	\end{tikzcd}
	\]
	whose twist is given by the composition $t_{E_2} \circ t_{E_1}$, and whose cotwist is given by $- \stackrel{L}{\otimes}_{R} R^{\ast}[-1-d]$, which induces Serre duality on $\derived(R)^{c}$ up to shift.
\end{thm}

The previous result generalizes to the composition of $n$ spherical twists around spherical objects, but instead of proving this generalization by hand, we will obtain it as a corollary of a more general theorem we prove in \autoref{section:general-case}.

After dealing with two spherical objects, we will be ready to tackle the general case.

We put ourselves in the same situation as in \cite{Anno-Log-17}, we consider three small DG-categories $\ca{A},\ca{B},\ca{C}$ and two perfect bimodules $M \in \derived(\ca{A}-\ca{C}) = D ({\bf \ca{A} - Mod - \ca{B}})$, $N \in \derived(\ca{B}-\ca{C})$, and we look for a spherical functor whose twist gives the composition of the twists associated to $ - \stackrel{L}\otimes_{\ca{A}} M$ and $ - \stackrel{L}{\otimes}_{\ca{B}} N$.

Recall that, given two DG categories $\ca{A}$, $\ca{B}$, and a bimodule $\varphi \in {\bf \ca{A} - Mod - \ca{B}}$, we can define the upper triangular DG category $\ca{B} \sqcup_{\varphi} \ca{A}$ following \cite{Tabuada-07}.

Proceeding along the lines we drew in the example of two spherical objects, we obtain
\end{comment}

\begin{thm}[\autoref{twist-twist=twist}]
	Let $\ca{A}$, $\ca{B}$ and $\ca{C}$ be three small dg-categories over a field $k$, and $M \in \derived(\ca{A}\text{-}\ca{C})$, $N \in \derived(\ca{B}\text{-}\ca{C})$ be two  spherical bimodules.
	Then, the bimodule $P := \rmodule{M}{N}^t \in \derived((\ca{B} \sqcup_{\varphi} \ca{A})\text{-}\ca{C})$, $\varphi = \textup{RHom}_{\ca{C}}(N,M)$, with structure morphism
	\[
		\mathrm{RHom}_{\ca{C}}(N,M) \oL{B} N \xrightarrow{\mathrm{tr}} M
	\]
	is spherical, the twist around it is given by the composition $t_N \circ t_M$, and the cotwist is described by the matrix
	\[
		C_P = \left(
		\begin{array}{cc}
			C_M & 0 \\
			\mathrm{RHom}_{\ca{C}}(M,N)[-1] & C_N
		\end{array}
		\right)
		\in \derived((\ca{B} \sqcup_{\varphi} \ca{A})\text{-}(\ca{B} \sqcup_{\varphi} \ca{A}))
	\]
	with non-zero structure morphisms
	\[
		\begin{aligned}
			& \, \mathrm{RHom}_{\ca{C}}(M,N)[-1] \oL{A} \mathrm{RHom}_{\ca{C}}(N,M) \xrightarrow{\sigma \circ \mathrm{cmps}} C_N\\
			& \, \mathrm{RHom}_{\ca{C}}(N,M) \oL{B}  \mathrm{RHom}_{\ca{C}}(M,N)[-1] \xrightarrow{\sigma \circ \mathrm{cmps}} C_M,
		\end{aligned}
	\]
	where $\mathrm{cmps}$ denotes the composition of morphisms.
\end{thm}

For an explanation of the notation we refer to \autoref{subsect:gluing}, \autoref{subsect:modules-on-glued} and \autoref{dfn-sigma}.

\begin{rmk}
	In \cite[Section 7.2]{AL-P-n-func}, to provide an example of a non-split $\mb{P}^n$-functor, the authors perfomed the above construction in the special case where $M = 0$.
\end{rmk}

As a corollary to the previous theorem, we get the following

\begin{cor}[\autoref{cor:commutativity-relation}]
	With the same setting as in the above theorem, we have $t_N \circ t_M \simeq t_{t_N(M)} \circ t_N$.
\end{cor}

Finally, we deal with two explicit examples: spherical objects and $\mb{P}$-objects.
In the case of spherical objects we can prove the following

\begin{thm}[\autoref{thm:cotwist-serre-duality}]
	Let $\ca{C}$ be a small dg-category over a field $k$ such that $\derived(\ca{C})^c$ has a Serre functor, and let $E_1, \dots, E_n \in \derived(\ca{C})$ be $d$-spherical objects.
	Then, the cotwist around the spherical functor realising the composition $t_{E_n} \circ \dots \circ t_{E_1}$ gives a Serre functor up to a shift.
\end{thm}

We refer the reader to \autoref{subsect:spherical-objs} for a more detailed statement of the above theorem.

\addtocontents{toc}{\SkipTocEntry}
\subsection*{Acknowledgments}

I would like to thank my advisor Ed Segal for many helpful conversations.
I would also like to thank Rina Anno and Timothy Logvinenko for carefully reading the paper and providing useful feedback.
Furthermore, I would like to thank Daniel Halpern-Leistner for asking how the main theorem of this paper related to \cite[Theorem 4.14]{Halpern-Shipman16}, Tobias Dyckerhoff for an interesting discussion that resulted in the proof of \autoref{cor:commutativity-relation}, and the referees whose comments and suggestions incredibly improved the paper and brought to a description of the cotwist in full generality.

This project has received funding from the European Research Council (ERC) under the European Union Horizon 2020 research and innovation programme (grant agreement No.725010).


TITLE: Why do quantum eraser and original double-slit setups produce different patterns on the detector?
QUESTION [0 upvotes]: A similar question has been asked here, but there was no answer: Why Delayed choice quantum eraser doesn't produce two lines
I learned that in the original double-slit experiment, an interference pattern is observed after firing many photons on the detector.
I also learned that in Delayed-choice quantum eraser experiment, when you look at the detector, what you initially see is just noise. Only after separating out the photons according to which detector detected their entangled pair, you start to see interference patterns on D1 and D2 detectors.

What I'm wondering is what causes the interference pattern seen on the main detector in the original double-slit experiment to turn into noise in the quantum eraser experiment. I am suspecting it's because the beam splitter BSC but I don't understand why.
Let's consider a simpler experiment than quantum eraser, removing the two initial splitters BSa and BSb. The photons travel from both paths and which-way information is deleted when they pass through beam splitter BSC.
What do we see in the detector D0? Interference pattern or noise? If noise, why? In the original experiment we saw interference pattern, what caused the difference in this case? Since we removed the information, I would expect the result to be equivalent to the one in the original experiment.
In other words, I don't see a difference between this experiment and original experiment, other than the fact that we have entangled photons here which doesn't provide any information thanks to BSC. Why is the outcome different?
Thanks!

REPLY [1 votes]: There's no way to affect the observable behavior of the "upper" photons by manipulating the "lower" photons after the initial generation of the entangled pairs in the Glan-Thompson prism (if there were, you could use it to send a signal faster than light).
There's no interference pattern because the lower photons contain which-path information about the upper photons. Even if you never measure that information, its presence in the world prevents visible interference. There's no way to destroy the information except by reversing the process that originally created the entanglement, which is impossible in practice in this setup.
This answer has a more detailed discussion of a different version of this experiment. The key point is the same: if information about the light is preserved elsewhere in any form then there's no interference, and that information can't be and isn't destroyed, despite the name "quantum eraser".


TITLE: Derivative of integral $\int_0^{\infty} e^{-x \cosh t} dt$
QUESTION [1 upvotes]: I am given the integral $y = \int_0^{\infty}e^{-x\cosh t} dt$ and wish to show that this integral solves the modified Bessell equation: $x^2y'' + xy' -x^2y = 0.$
To do this I need to calculate the derivatives $y'$ and $y''$. I suspect integration by parts is necessary, but do not know how to continue. I have tried substituting $e^-x\cosh t$ and $\cosh t$ by their Taylor series and integrating term by term, but this does not seem to simplify things.
My question is: How do I calculating $y'$?

REPLY [1 votes]: You can check the equation as follows:
\begin{align*}
x^2(y'' - y)
&= x \int_{0}^{\infty} x (\cosh^2 t - 1) e^{-x \cosh t} \, dt \\
&= x \int_{0}^{\infty} x \sinh^2 t \, e^{-x \cosh t} \, dt \\
&= x \left[ -\sinh t \, e^{-x \cosh t} \right]_{0}^{\infty} + x \int_{0}^{\infty} \cosh t \, e^{-x\cosh t} \, dt \\
&= -xy'.
\end{align*}

@anthus, I agree that Leibniz integral rule (interchanging of integration and differentiation) is often not obvious, and is simply false in some cleverly designed examples.
In this case, however, we have a simple idea to circumvent such complication: think everything backward!
For example, suppose that you want to know that $y$ is differentiable with
$$ y'(x) = -\int_{0}^{\infty} \cosh t \, e^{-x \cosh t} \, dt. \tag{*} $$
To this end, let us work backward: That is, we will show that $y$ is an antiderivative of the right-hand side.
Denote the RHS of (*) as $y_1 (x)$. Clearly $y_1 (x)$ is a well-defined function. Now for any $0 < a < b$, Tonelli's theorem guarantees that we can interchange the order of integration and we have
\begin{align*}
\int_{a}^{b} y_1(x) \, dx
&= \int_{0}^{\infty} \int_{a}^{b} (-\cosh t) e^{-x \cosh t} \, dt dx \\
&= \int_{0}^{\infty} (e^{-b \cosh t} - e^{-a \cosh t}) \, dt \\
&= y(b) - y(a).
\end{align*}
This means that $y$ is an antiderivative of $y_1$, hence $y$ is differentiable and $y' = y_1$.


TITLE: Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint
QUESTION [0 upvotes]: Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint

I am not quite sure how to do this proof. My first idea was to use the spectral theorem for normal operators to construct a proof similar to to finite dimensional case.

REPLY [2 votes]: This is also true if we drop the separability condition.
If you know the basics about $C^*$-algebras, there is an easy proof by a standard technique.
Consider $B:= C^*(A,1)$, the unital $C^*$-algebra generated by the operator $A$ in $B(H)$. Since $A$ is a normal operator, $B$ is a commutative $C^*$-algebra, hence $B \cong C(X)$ for a compact Hausdorff space $X$. But in $C(X)$, the statement is trivial: indeed if a function $f\in C(X)$ satisfies $f^3 = f^4$, then $f$ can only take the values $0$ or $1$ and in particular $f$ is real-valued, so self-adjoint. Through the isomorphism $B\cong C(X)$, you find that $A$ must also be self-adjoint, as desired.


TITLE: compute one improper integral involving arctangent
QUESTION [6 upvotes]: Compute $$\int_{0}^{\infty} \frac{\arctan(ax)}{x(1+x^{2})}dx$$
The answer is:
$\frac{\pi}{2}\ln(1+a)$ when $a \geq 0$
$-\frac{\pi}{2}\ln(1-a)$ when $a < 0$

Here is my problem, and I can't even dive into a appropriate first step. The most difficult part is the $\arctan$ which I have no idea to eliminate. Thank you for help.

REPLY [3 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}&\color{#c00000}{\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x}
=\half\,a\verts{a}\int_{-\infty}^{\infty}
{\arctan\pars{x} \over x\pars{x^{2} + a^{2}}}\,\dd x
\\[3mm]&=\half\,a\verts{a}\int_{-\infty}^{\infty}\
\overbrace{{\ic \over 2}\,\ln\pars{1 - \ic x \over 1 + \ic x}}
^{\ds{=\ \arctan\pars{x}}}\
{1\over x\pars{x^{2} + a^{2}}}\,\dd x
=-\,\half\,a\verts{a}\,\Im\int_{-\infty}^{\infty}
{\ln\pars{1 - \ic x} \over x\pars{x^{2} + a^{2}}}\,\dd x
\end{align}

Set $\ds{t \equiv 1 - \ic x\quad\imp\quad x = \pars{t - 1}\ic}$:
  \begin{align}&\color{#c00000}{\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x}
=-\,\half\,a\verts{a}\,\Im\int_{1 + \infty\ic}^{1 - \infty\ic}
{\ln\pars{t} \over \pars{t - 1}\ic\bracks{-\pars{t - 1}^{2} + a^{2}}}\,\ic\,\dd t
\\[3mm]&=-\,\half\,a\verts{a}\,\Im\color{#00f}{%
\int_{1 - \infty\ic}^{1 + \infty\ic}{\ln\pars{t}\over
\pars{t - 1}\pars{t - r_{-}}\pars{t - r_{+}}}\,\dd t}
\quad \mbox{where}\quad r_{\pm} = 1 \pm \verts{a}
\end{align}

Now, we'll evaluate the $\ds{\color{#00f}{\mbox{'blue integral'}}}$.
We take the "$\ds{\ln}$-branch cut" along the negative $\ds{t}$-semi-axis and close the contour in a semi-circle "to the right" $\ds{\pars{~t > 1~}}$:
\begin{align}&\color{#c00000}{\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x}
=-\,\half\,a\verts{a}\,\Im\bracks{-2\pi\ic\,
{\ln\pars{r_{+}} + \ic 0 \over \pars{r_{+} - 1}\pars{r_{+} - r_{-}}}}
\\[3mm]&=\pi\,a\verts{a}\,\bracks{%
{\ln\pars{1 + \verts{a}} \over \verts{a}\pars{2\verts{a}}}}
\end{align}

$$
\color{#66f}{\large\int_{0}^{\infty}
{\arctan\pars{ax} \over x\pars{1+x^{2}}}\,\dd x
={\pi \over 2}\,\sgn\pars{a}\ln\pars{1 + \verts{a}}}
$$


TITLE: Extensions of $M$ by $M$ with same derivation
QUESTION [1 upvotes]: If $M$ is a $A$-module where $A$ is a complete noetherian local ring, for any extension of $A$-modules $0 \rightarrow M \rightarrow X \rightarrow M \rightarrow 0$, we get an object $X$ with a derivation $d:X \rightarrow M \rightarrow X$ ($d$ is the composition of the maps in the exact sequence in the opposite order, so $d\circ d=0$ which is the definition of a differential). If two extensions give isomorphic $(X,d)$, must they be equivalent?

REPLY [0 votes]: No.  For instance, let $A=\mathbb{Z}/(4)$ and $M=\mathbb{Z}/(2)\oplus\mathbb{Z}/(2)$.  Let $X=\mathbb{Z}/(2)\oplus\mathbb{Z}/(2)\oplus\mathbb{Z}/(4)$ be considered as an extension of $M$ by itself in two different ways: one where the $\mathbb{Z}/(4)$ is obtained from the first summand in each copy of $M$, and one where the $\mathbb{Z}/(4)$ is obtained from the second suumand in each copy of $M$.  These two extension induce the same differential $X\to X$ (namely multiplication by $2$).  However, they are not isomorphic as extensions of $M$ by $M$.  (For instance, in the first extension the image of the second summand of $M$ under the inclusion $M\to X$ is a direct summand of $X$, but in the second extension it is not.)
More generally, if you start with an extension $0\to M\to X\to M\to 0$ with differential $d$ and apply an automorphism of $M$ to get a new extension $0\to M\to X'\to M\to 0$ with differential $d'$, then obviously $(X,d)\cong (X',d')$, but the extensions may not be isomorphic.  In fact, every example arises in this way: given an isomorphism between $(X,d)$ and $(X',d')$, we get an isomorphism between the short exact sequences $0\to M\to X \to M\to 0$ and $0\to M\to X'\to M\to 0$ since the inclusion $M\to X$ is just a kernel of $d$ and the quotient $X\to M$ is just a cokernel of $d$, and similarly for $d'$.  Moreover, the two automorphisms $M\to M$ induced on the kernel and cokernel are equal, since it must commute with the isomorphisms $\operatorname{coker}(d)\to \ker(d)$ and $\operatorname{coker}(d')\to \ker(d')$ induced by $d$ and $d'$.  This means exactly $0\to M\to X'\to M\to 0$ is isomorphic to the extension $0\to M\to X\to M\to 0$ obtained by modifying our first extension by this automorphism of $M$.


TITLE: $ J^2=(L+S)^2 $ versus $ (P - \frac{q}{c} A)^2 $
QUESTION [4 upvotes]: The total angular momentum J can be written as $$ J^2 = L^2 + S^2 + 2  L \cdot S .$$ Now, I assume this is a simplification of a more general tensor rule that $$ (M + N )^2 = M^2 + N^2 + M \cdot N + N \cdot M ,$$ where $ M = A \otimes B $, for instance, and $ N = C \otimes D .$
I'm questioning my understanding of this rule now. I thought it came from something as simple as $$( M + N ) (M + N ) = A^2 \otimes B^2 + A C \otimes B D + C A \otimes D B + C^2 \otimes D^2 = M^2 + N^2 + M \cdot N + N \cdot M .$$
This "proof" does not extend to 3 dimensions. For example, for the exercise to compute $ (\bar P - \frac{q}{c} A)^2 $ , one may write this as $ P^2 + A'^2 + P \cdot A' + A' \cdot P $ according to Sakurai, where I've allowed the $ A' $ to swallow the constants to its left.
So, what gives? How would you go about showing this identity? If I blindly multiply as I did in the 2D case, you would receive several cross terms. Even just computing $ (\bar P)^2 $ would result in a 9-term sum since there are 3 separate terms in $ \bar P = P_x \otimes I \otimes I + I \otimes P_y ... $ etc.
I'm guessing I'm either getting the "tensor multiplication" rules wrong, or I'm getting the definition of $ ^2 $ wrong for tensor multiplication.
EDIT: after clarification for a different question, I see that $ P_x \ne P_x \otimes I \otimes I$  Idk why but I associated "different directions" to "different tensor spaces". I think this was what Zero was getting at (thanks). But the two answers below still hold so that you don't get ugly cross terms like $ P_x P_y $ or something when doing the square.

REPLY [0 votes]: I've accepted Dr. Zhou's answer, since I think he was getting at the following, and this is what I think would fill in the dots for my remaining challenges if anyone else is curious. I'm referring to Dr. Z's answer in the linked post as well.
When someone is saying $ whatever^2 $, they really mean you have a vector of equations $\{ whatever_x , whatever_y, whatever_z\} $ that are indexed by $ x, y, z$ even though it may not actually have 3 separate Cartesian directions (for example, $ J = L \otimes I  + I \otimes S $ does not have three separate "sub-Hilbert-spaces", it has two. $ L_x, L_y, L_z $ all live in the same space).
Now, you can list the vector equations for $ \bar J $ as follows: $ J_x = L_x \otimes I + I \otimes S_x $ and so on. You'd compute $ J_x^2 $ the way I explain in the question and Dr. Z also explains in his answer, and so forth.
You could in principle (and somewhat more naturally, since it comes equipped with directions) do it for the canonical momentum.
So, for $ \pi_x^2 $ you'd get $ ((P_x - \frac{q}{c} A_x) \otimes I \otimes I )^2 =(P_x - \frac{q}{c} A_x)^2 \otimes I \otimes I  $ and then you'd expand it as necessary.
And as a last note, my rambling about Sakurai is either deeply misinterpreted on my end or a questionable use of notation from him when he writes $ A \cdot P - P \cdot A$ (I'm using the latest edition, equation 2.341).
This is because writing $ A \cdot P $ should imply $ A_x P_x \otimes A_y P_y \otimes A_z P_z$, which is not what we should get. We should get $ \{ A_x , P_x \} \otimes I \otimes I + I \otimes \{ A_y ,P_y \} ... etc $.
One can see this if you try to work out the answer for $ A = Y \hat x $. If you think Sakurai's got the equation right, then you should get something along the lines of $ \bar P^2 + Y^2 \otimes I \otimes I + \{Y , P_x \} \otimes P_y \otimes P_z  $ which actually should look something like $ P_y^2 + P_z^2 + (P_x - \frac{q}{c} Y \hat x )^2 \otimes I \otimes I $ where I've dropped some of the $ \otimes $ for convenience.


TITLE: Quotient of complex manifold by a free and locally proper action (difficulty with reading German)
QUESTION [3 upvotes]: Let $X$ be a complex manifold with an action of $G=GL(n,\mathbb{C})$ which is free and locally proper (each point of $X$ has a $G$-invariant neighborhood on which $G$ acts properly.)
Satz 24 of the paper
H. Holmann, Quotienten komplexer Ra ̈ume, Math. Ann., 142 (1961), pp. 407–440
asserts that $X/G$ is a complex manifold.
Why is it true?
The biggest issue is, since I cannot read German, I don't know if by "free" he means set-theorecial free or scheme-theoretical free (the latter is in the sense of Mumford's GIT book.)

REPLY [2 votes]: Assuming that by $GL(V)$ you mean $GL(\mathbb{C})$,
then this question concerns the action of a non-compact
complex Lie group on a smooth complex manifold, such that
the action is free
the action is locally-proper
and the question is whether the quotient space $X/G$
can be equipped with a complex manifold structure.
This appears to be true,  a roadmap to (though not the details of) a proof
being given in Lemma 3.1 of
Miebach. C., Oeljeklaus, K. On proper $\mathbb{R}$-action on hyperbolic Stein  manifolds. Documenta Mathematica 14 (2009) 673–689
which is freely accessible online.
Incidentally, it might be instructive if someone could provide
an example of a locally-proper non-proper action of a complex Lie group
on a complex manifold. All such examples I know are actions of real Lie groups
on real manifolds.


TITLE: Prove that for all $n \geq 1$, $F_{-n}$ = $(-1)^{n+1}F_n$ where F is the Fibonacci numbers.
QUESTION [2 upvotes]: Prove that for all $n \geq 1$, $F_{-n}$ = $(-1)^{n+1}F_n$ where F is the Fibonacci numbers.
I've already shown that the formula holds for $n = 1$ and $n = 2$. So I supposed the formula holds for $n$ and started with $F_{-(n+1)}$, and I am trying to get it to equal $(-1)^{n+1 + 1}F_{n+1}$.
Here are the steps I've taken so far:
$F_{-(n+1)}$ = $F_{-n+1} - F_{-n}$
Since the formula holds for $n$ I can change this to
$F_{-n+1} - (-1)^{n+1}F_n$
= $F_{-n+1} + (-1)^{n+2}F_n$
= $F_{-n+1} + (-1)^{n+2}(F_{n+1} - F_{n-1})$
= $F_{-n+1} + (-1)^{n+2}F_{n+1} + (-1)^{n+2}F_{n-1}$
So now I have the part I want in the middle, but with a couple extraneous pieces. Am I close to getting the desired result? Or is there another way I'm supposed to do this problem?

REPLY [1 votes]: For even indices, what you are trying to prove -- i.e., $F_{-n}=-F_n$ -- IS true.
For odd indices, $F_{-n}=F_n$
\begin{array}{l|rrrrrrrrrrrrrrr}
n&-7&-6&-5&-4&-3&-2&-1&0&1&2&3&4&5&6&7\\
F_n&13&-8&5&-3&2&-1&1&0&1&1&2&3&5&8&13
\end{array}