TITLE: How good is this approximation of $n$ factorial?
QUESTION [10 upvotes]: For fixed $n\geq 1$, consider the sum
$$S_n:=\sum_{k=1}^\infty \frac{k^n}{e^k}.$$
If you compute this sum for some small values of $n$, you will see that it is remarkably close to $n!$. The fact that it is somehat close to $n!$ is not too surprising because $S_n$ "should" be similar to the following integral
$$n! = \int_0^\infty \frac{x^n}{e^x}dx.$$
However, the precision of the approximation seems to be surprisingly good (check some small values on Wolfram Alpha and see for yourself!).
Question: Is there a good way to show that $S_n$ is bounded above by $(1+\varepsilon) n!$ for every $n\geq1$ where $\varepsilon$ is some explicit positive constant which is "close" to $0$?
After checking some small values of $n$, it appears that $n=3$ might be the worst value for an upper bound of this type. Note that, for many values of $n$ (perhaps for most $n$), it is actually the case that $S_n<n!$.
If you think of $S_n$ as a Riemann sum approximating the integral, then the $k$th summand is approximating the area under the curve between $x=k-1$ and $x=k$ by the value at the right endpoint of this interval (namely, by $\frac{k^n}{e^k}$). The function $\frac{x^n}{e^x}$ is increasing for $0\leq x<n$ and decreasing for $x>n$. Therefore, for $k\leq n$, the $k$th summand is overestimating the area and for $k\geq n+1$ it is underestimating it. Magically, it seems that the overestimating terms and the underestimating terms cancel out beautifully and give us a result which is really close to $n!$. Does anyone know whether this phenomenon occurs for all $n$ and whether there is a good reason why this would be occur?
By the way, there is a connection with Eulerian numbers here, if anyone is interested. It turns out that
$$S_n = \frac{e\cdot \sum_{m=0}^{n-1}A(n,m)e^m}{(e-1)^{n+1}}$$
where $A(n,m)$ is the number of permutations of $1,\dots,n$ with exactly $m$ "ascents." Interestingly, $A(n,0)+A(n,1)+\cdots +A(n,n-1)=n!$ (since every permutation has some number of ascents) but this doesn't seem to be terribly helpful since there are these factors of $e$ and $e-1$ floating around everywhere.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\sum_{k = 1}^{\infty}{k^{n} \over \expo{k}}\right\vert_{\ n\ \geq\ 0} & =
-\delta_{n0} + \sum_{k = 0}^{\infty}{k^{n} \over \expo{k}} =
-\delta_{n0} + \int_{0}^{\infty}{k^{n} \over \expo{k}}\,\dd k +
\left.{1 \over 2}{k^{n} \over \expo{k}}\right\vert_{\ k\ =\ 0} -2
\int_{0}^{\infty}{\Im\pars{\bracks{\ic x}^{n}/\expo{\ic x}} \over
\expo{2\pi x} - 1}\,\dd x
\\[5mm] & = -\,{1 \over 2}\,\delta_{n0} + n! -
2\left\{\begin{array}{lcl}
\ds{-\pars{-1}^{n/2}\int_{0}^{\infty}{x^{n}\sin\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ even
\\[3mm]
\ds{\phantom{-\,}\pars{-1}^{\pars{n - 1}/2}\int_{0}^{\infty}{x^{n}\cos\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ odd
\end{array}\right.
\end{align}

$$
\left.\vphantom{\large A}n!\,\right\vert_{\ n\ \geq\ 0} =
\sum_{k = 1}^{\infty}{k^{n} \over \expo{k}} +
{1 \over 2}\,\delta_{n0} +
2\left\{\begin{array}{lcl}
\ds{-\pars{-1}^{n/2}\int_{0}^{\infty}{x^{n}\sin\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ even
\\[3mm]
\ds{\phantom{-\,}\pars{-1}^{\pars{n - 1}/2}\int_{0}^{\infty}{x^{n}\cos\pars{x}
\over \expo{2\pi x} - 1}\,\dd x} & \mbox{if} & \ds{n}\ \mbox{is}\ odd
\end{array}\right.
$$

Relative Error plot:

Absolute Error plot: It's worsening as $\ds{n > 16}$. Such behaviour is fully discussed in
  @Jack D'Aurizio answer.


TITLE: Is the image of a projective variety projective?
QUESTION [5 upvotes]: Suppose $X$ is a projective variety and $f:X\to Y$ a morphism, is the image $f(X)$ projective?(the schematic image is well-defined in this case, or the induced reduced structure on the closed subset $f(X)$) Hartshorne says this property holds for properness. Usually how can we claim a variety(scheme) is projective?

REPLY [9 votes]: No, the image of a projective variety is not always projective.
Indeed Chow's lemma   states that given any complete irreducible variety $X$, there exists a projective variety $X'$ and a birational surjective morphism $f:X'\to X$.
Choosing for $X$ a complete irreducible variety that is not projective thus gives an example of a projective variety $X'$ whose image $X$ is not projective.
To be complete (!), let me mention that an  example due to Hironaka of a complete but not projective smooth $3$-dimensional variety  is given in Shafarevich's Basic Algebraic Geometry, volume 2, page 75.


TITLE: Are there two normal forces here?
QUESTION [0 upvotes]: In this question the normal force direction on the collar was determined to the left, while there is a possibility to be to the right direction also since the collar contacts the rod in both directions, how can someone explain this?

REPLY [1 votes]: The normal forces are not needed in this analysis. If you consider the spring, collar, and earth as the system, then you can consider the system as being isolated. That means total mechanical energy is conserved, or
$$KE_{initial}+GPE_{initial}+EPE_{initial}=KE_{final}+GPE_{final}+EPE_{final}$$
Where $KE$ is kinetic energy of the collar, $GPE$ is gravitational potential energy of the collar, and $EPE$ is elastic potential energy of the spring.
Since this is essentially a homework and exercise type problem, that's as far as I can go. But I think you can proceed from here.
Hope this helps.


TITLE: Find minimum of simple expression over integers
QUESTION [1 upvotes]: I'm interested in minimizing an expression
$F(i) = |1 - \frac{a}{i}|$
where $a > 1.0$ is a known positive real number, and $i$ is a positive integer.
EDIT: In particular, I am interested in how the minimum scales with $a$
EDIT 2: Ok, I found my answer. By noticing that reals and integers can only differ by 0.5, and doing a tiny bit of math, one arrives at
$F_{min} \in [0, \frac{1}{2a}]$
and thus the minimum scales inverse-linearly with $a$

REPLY [1 votes]: Denote with $\lfloor a \rfloor$ the biggest integer no greather than $a$ and with $\lceil a \rceil$ the smallest integer no less than $a$. In other words $\lfloor 2.0 \rfloor = \lceil 2.0 \rceil = 2$, $\lfloor 2.2 \rfloor = 2$, $\lceil 2.2 \rceil = 3$
If $i\ge a$:
$$f(i) = \frac{|i-a|}{a}=\frac{i-a}{a} =1-\frac{a}{i}$$
This epxression reaches minimal value for x=$\lceil a \rceil$ and the minimal value is:
$$f_1 = 1 - \frac{a}{\lceil a \rceil}$$
If $i\lt a$:
$$f(i) = \frac{|i-a|}{a}=\frac{a-i}{a} =1-\frac{i}{a}$$
This epxression reaches minimal value for x=$\lfloor a \rfloor$ and the minimal value is:
$$f_2 = 1 - \frac{\lfloor a \rfloor}{a}$$
Minimum value of $f(i)$ for a given $a$ is:
$$f_{min}=min \left(1 - \frac{a}{\lceil a \rceil},1 - \frac{\lfloor a \rfloor}{a}\right)$$
or:
$$f_{min}=1 - max \left(\frac{a}{\lceil a \rceil},\frac{\lfloor a \rfloor}{a}\right)$$
Here is the graph of $f_{min}$ as a function of $a$:

Red line represents asymptotic boundary of $f_{min}$: $\frac1{2x}$


text:
\section{Tube algebra for gapped boundary excitations in (3+1)d\label{sec:tube3D}}
\emph{In this section, we apply the tube algebra approach to study excitations in the presence of gapped boundaries in (3+1)d. Although the excitation content of the model is rich in (3+1)d, we focus on a special configuration, which turns out to be related to that considered in the previous section via a dimensional reduction argument.}

\subsection{Definition}
The strategy we presented in sec.~\ref{sec:tube2D} applies identically in three dimensions. Given a pattern of two-dimensional gapped boundaries, excitations can be classified by considering boundary conditions of the manifold obtained by removing local neighbourhoods of these excitations. Given that the input data for the bulk theory is a pair $(G,\pi)$, where is $\pi$ a normalized representative of a cohomology class in $H^4(G,\rU(1))$, we are interested in the situation where two two-dimensional gapped boundaries characterized by the boundary conditions $A_\lambda \equiv (A, \lambda)$ and $B_\mu \equiv (B,\mu)$ meet at a one-dimensional interface. The boundary conditions are such that $A,B \subset G$, $d^{(3)}\lambda = \pi^{-1}{\sss |}_{A}$ and $d^{(3)}\mu = \pi^{-1}{\sss |}_{B}$. We denote the Hamiltonian defined according to \eqref{eq:Ham} for these boundary conditions as $\mathbb H^{G,A, B}_{\pi, \lambda,\mu}[\Sigma]$.

Given this situation, several types of excitations could be studied. For instance, we could investigate point-like boundary excitations at the one-dimensional interface. Instead, we consider a bulk string-like excitation that terminates at two (possibly different) gapped boundaries. This situation can be depicted as follows:
\begin{equation}
	\bdryInterfaceThreeExc{1}{1} \;\; \to \;\;
	\bdryInterfaceThreeExc{1}{2} \; ,
\end{equation}
where the dark volume represents a local neighbourhood of the string-like excitation, and thus the region whose energy density is higher than that of the ground state. Removing this local neighbourhood leaves an excitation boundary $\exBdry$ that has the topology of cylinder. Classifiying boundary conditions on such cylinder corresponds to classifying the string-like excitations.

Let us consider the manifold $\exBdry \times \mathbb I$. This manifold has the topology of a hollow cylinder, which has two kinds of boundary components, namely a pair of physical boundary components and a pair of excitation boundary components. Given the 3-ball endowed with two gapped boundaries, the same manifold can be obtained by removing local neighbourhoods of the interface and of a string terminating at the two gapped boundaries:
\begin{equation*}
	\threeDiskExc{1}{2} \;\; \simeq \;\; \tubeThreeDBeauty{1}{1} \! .
\end{equation*}
By construction, this manifold can be glued to the original system along the excitation boundary $\exBdry$ without affecting its topology. It follows from the discussion in sec.~\ref{sec:tube2D} that there is a tube algebra associated with the gluing of two copies of this tube-like manifold, whose irreducible representations classify this special type of string-like excitations.

\subsection{Computation of the tube algebra}

Let us derive the tube algebra for the special configuration described above. As before, we first need to specify  the ground state subspace on $\exBdry \times \mathbb I$ by choosing a discretisation. We choose to discretise $\exBdry \times \mathbb I$ as a triangulated cube with two opposite faces identified. The resulting triangulated manifold is denoted by $\mathfrak{T}[\mathbb S^1 \times \mathbb I]$ and the corresponding ground state subspace explicitly reads
\begin{align}
	\label{eq:basisTube3D}
	\gsThree{\mathfrak{T}[\mathbb S^1 \times \mathbb I]}
	& :=  {\rm Span}_{\mathbb C}\Bigg\{\Bigg| \! \tubeThreeD{0.9}{0}{1}{g}{a}{b}{1} \Bigg\ra  \Bigg\}_{\substack{\hspace{0em}\forall \, g \in G \, | \, g= a_2^{-1}gb_2 \\\hspace{-2.6em} \forall \, a_1,a_2 \in A \\ \hspace{-2.7em} \forall \, b_1,b_2 \in B }}
	\\
	&\equiv {\rm Span}_{\mathbb C}\big\{ \biGr{\big| (g,a_2,b_2)}{a_1}{b_1} \big\ra \big\}_{\substack{\hspace{0em}\forall \, g \in G \, | \, g= a_2^{-1}gb_2 \\\hspace{-2.6em} \forall \, a_1,a_2 \in A \\ \hspace{-2.7em} \forall \, b_1,b_2 \in B }} \; ,
\end{align}
where we make the identifications $\snum{(0)} \equiv \snum{(\tilde 0)}$, $\snum{(0')} \equiv \snum{(\tilde 0')}$, $\snum{(1)} \equiv \snum{(\tilde 1)}$, $\snum{(1')} \equiv \snum{(\tilde 1')}$, $\snum{(00')} \equiv \snum{(\tilde 0 \tilde 0')}$, $\snum{(01)} \equiv \snum{(\tilde 0 \tilde 1)}$, $\snum{(0'1')} \equiv \snum{(\tilde 0' \tilde 1')}$ and $\snum{(11')} \equiv \snum{(\tilde 1 \tilde 1')}$. As before, some labellings are left implicit since they can be deduced from the flatness constraints. Let us now compute the $\star$-product for two such states adapting in the obvious way the definition of the previous section. Firstly, colourings along the gluing interface are identified via the map $\textsf{\small GLU}$, i.e.
\begin{align*}
	&\textsf{\small GLU}
	\Bigg( \Bigg| \! \tubeThreeD{0.9}{0}{1}{g}{a}{b}{1} \! \Bigg\ra
	\otimes
	\Bigg| \! \tubeThreeD{0.9}{1}{2}{g'}{a'}{b'}{1} \!\! \Bigg\ra \! \Bigg)
	\\[-2em]
	& \hspace{11em} =
	\delta_{g',a_1^{-1}gb_1}
	\, \delta_{a_2',a_2^{a_1}} \, \delta_{b_2',b_2^{b_1}}
	 \Bigg| \! \tubeThreeDDouble{0.9}{0}{1}{2}{g}{a}{b}{1} \! \Bigg\ra \; ,
\end{align*}
where we introduced the notation $x^y := y^{-1}xy$.
Secondly, we apply the Hamiltonian projector $\mathbb P_{\mathfrak{T}[\mathbb S^1 \times \mathbb I] \cup_{\mathbb S^1 \times \mathbb I}\mathfrak{T}[\mathbb S^1 \times \mathbb I]}$ in order to enforce the twisted gauge invariance at the physical boundary vertices along the gluing interface. This operator can be expressed by evaluating the partition function \eqref{eq:relativepartitionfunction} on the relevant pinched cobordism. The result reads
\begin{align}
	&\mathbb P_{\mathfrak{T}[\mathbb S^1 \times \mathbb I] \cup_{\mathbb S^2 \times \mathbb I}\mathfrak{T}[\mathbb S^1 \times \mathbb I]}
	\Bigg( \Bigg| \! \tubeThreeDDouble{0.9}{0}{1}{2}{g}{a}{b}{1} \! \Bigg\ra \Bigg)
	\\[-2em]
	\nn
	& \q =
	\frac{1}{|A||B|}\sum_{\substack{(\tilde a ,\tilde b) \in A \times B}}
	\frac{\varrho^{AB}_{g,a_2,b_2}(a_1,\tilde a|b_1,\tilde b)}{\varrho^{AB}_{a_1^{-1}gb_1,a_2^{a_1},b_2^{b_1}}(\tilde a,\tilde a^{-1}a_1'|\tilde b,\tilde b^{-1}b_1')} \,
	\Bigg| \!  \tubeThreeDDouble{0.9}{0}{1}{2}{g}{a}{b}{2} \! \Bigg\ra \; ,
\end{align}
where we introduced the cocycle data
\begin{equation}
	\label{eq:defCoc2Gr3D}
	\varrho^{AB}_{g,a_2,b_2}(a_1,a_1'|b_1,b_1') := 	\frac{{\sfT}_{b_2}(\mu)(b_1,b_1')}{{\sfT}_{a_2}(\lambda)(a_1,a_1')}\frac{{\sfT}_{a_2}(\pi)(a_1,a_1',a_1'^{-1}a_1^{-1}gb_1b_1') \, {\sfT}_{a_2}(\pi)(g,b_1,b_1')}{{\sfT}_{a_2}(\pi)(a_1,a_1^{-1}gb_1,b_1')}
\end{equation}
in terms of the cocycle data ${\sfT}(\lambda)$, ${\sfT}(\mu)$ and ${\sfT}(\pi)$ that are itself defined according to
\begin{align*}
	{\sfT}_{x}(\alpha)(y_1,y_2) &:= \frac{\alpha(x,y_1,y_2) \, \alpha(y_1,y_2,x^{y_1y_2})}{\alpha(y_1,x^{y_1},y_2)} \; ,
	\\
	{\sfT}_x(\pi)(y_1,y_2,y_3) &:=
	\frac{\pi(y_1,x^{y_1},y_2,y_3) \, \pi(y_1,y_2,y_3,x^{y_1y_2y_3})}{\pi(x,y_1,y_2,y_3) \, \pi(y_1,y_2,x^{y_1y_2},y_3)} \; ,
\end{align*}
for any group elements $x,y_1,y_2,y_3 \in H$ in a finite group $H$ and group cochains $\alpha \in C^3(H, \rU(1))$, $\pi \in C^4(H, \rU(1))$. Defining
\begin{align}
	d^{(2)}{\sfT}_{x}(\alpha)(y_1,y_2,y_3)&:=
	\frac{{\sfT}_{x^{y_1}}(\alpha)(y_2,y_3) \, {\sfT}_x(\alpha)(y_1,y_2y_3)}{{\sfT}_x(\alpha)(y_1y_2,y_3) \, {\sfT}_x(\alpha)(y_1,y_2)} \; ,
	\\
	d^{(3)}{\sfT}_{x}(\pi)(y_1,y_2,y_3,y_4)&:=
	\frac{{\sfT}_{x^{y_1}}(\pi)(y_2,y_3,y_4) \, {\sfT}_x(\alpha)(y_1,y_2y_3,y_4) \, {\sfT}_x(\pi)(y_1,y_2,y_3)}{{\sfT}_x(\pi)(y_1y_2,y_3,y_4) \, {\sfT}_x(\alpha)(y_1,y_2,y_3y_4)} \; ,
\end{align}
it follows from the cocycle conditions $d^{(4)}\pi = 1$, $d^{(3)}\lambda = \pi^{-1}{\sss |}_{A}$ and $d^{(3)}\mu = \pi^{-1}{\sss |}_{B}$ that $d^{(3)}{\sfT}(\pi) = 1$, $d^{(2)} {\sfT}(\lambda) = {\sfT}(\pi)^{-1}{\sss |}_{A} $ and $ d^{(2)}{\sfT}(\mu) =  {\sfT}(\pi)^{-1}{\sss |}_{B}$. Utilising the cocycle conditions
\begin{align*}
	d^{(3)}{\sfT}_{a_2}(\pi)(a_1,a_1',a_1'',a_1''^{-1}a_1'^{-1}a_1^{-1}gb_1b_1'b_1'') &= 1
	&
	d^{(3)}{\sfT}_{a_2}(\pi)(a_1,a_1^{-1}gb_1,b_1',b_1'') &= 1
	\\
	d^{(3)}{\sfT}_{a_2}(\pi)(a_1,a_1',a_1'^{-1}a_1^{-1}gb_1b_1',b_1'') &= 1
	&
	d^{(3)}{\sfT}_{a_2}(\pi)(g,b_1,b_1',b_1'') &= 1 \; ,
\end{align*}
we finally obtain that  $\varrho^{AB}$ satisfies
\begin{align}
	d^{(2)}\varrho^{AB}_{g,a_2,b_2}(a_1,a_1',a_1''|b_1,b_1',b_1''):=
	\frac{	\varrho^{AB}_{a_1^{-1}gb_1,a_2^{a_1},b_2^{b_1}}(a_1',a_1''|b_1',b_1'') \, \varrho^{AB}_{g,a_2,b_2}(a_1,a_1'a_1''|b_1,b_1'b_1'') }{\varrho^{AB}_{g,a_2,b_2}(a_1a_1',a_1''|b_1b_1',b_1'') \, \varrho^{AB}_{g,a_2,b_2}(a_1,a_1'|b_1,b_1')}
	= 1
	\; .
\end{align}
Going back to the tube algebra, it remains to apply a triangulation changing isomorphism in order to recover the initial triangulation, and thus a state in $\gsThree{\mathfrak{T}[\mathbb S^1 \times \mathbb I]}$. This is done by evaluating the partition function for the pinched interval cobordism $(012)^+ \times \mathbb S^1 \times \mathbb I$ endowed with the triangulation defined as
\begin{alignat}{5}
	&\snum{(012)}^{+} \times \mathbb S^1 \times \mathbb I
	& \,=  \,& \,
	\snum{(0122'\tilde{2}')}^{+} &\, \cup \,& \snum{(012\tilde{2}\tilde{2}')}^{-} &\, \cup \,& \snum{(01\tilde{1}\tilde{2}\tilde{2}')}^{+} &\, \cup \,& \snum{(0\tilde{0}\tilde{1}\tilde{2}\tilde{2}')}^{-}
	\nn\\
	&{} &\, \cup \,& \,
	\snum{(011'2'\tilde{2}')}^{-} &\, \cup \,& \snum{(011'\tilde{1}'\tilde{2}')}^{+}&\, \cup \,& \snum{(01\tilde{1}\tilde{1}'\tilde{2}')}^{-}&\, \cup \,& \snum{(0\tilde{0}\tilde{1}\tilde{1}'\tilde{2}')}^{+}
	\nn\\
	& {} &\, \cup \,& \,
	\snum{(00'1'2'\tilde{2}')}^{+}&\, \cup \,& \snum{(00'1'\tilde{1}'\tilde{2}')}^{-}&\, \cup \,& \snum{(00'\tilde{0}'\tilde{1}'\tilde{2}')}^{+}&\, \cup \,& \snum{(0\tilde{0}\tilde{0}'\tilde{1}'\tilde{2}')}^{-} \; .
\end{alignat}
The corresponding operator implements the isomorphism
\begin{equation*}
	\Bigg| \!  \tubeThreeDDouble{0.9}{0}{1}{2}{g}{a}{b}{2} \! \Bigg\ra  \simeq
	\frac{1}{|A|^\frac{1}{2}|B|^\frac{1}{2}} \varrho^{AB}_{g,a_2,b_2}(a_1 \tilde a , \tilde a^{-1}a_1' | b_1 \tilde b , \tilde b^{-1} b_1')
	\Bigg| \! \tubeThreeD{0.9}{0}{2}{g}{a}{b}{2} \! \Bigg\ra .
\end{equation*}
Putting everything together, we obtain
\begin{align}
	&\Bigg| \! \tubeThreeD{0.9}{0}{1}{g}{a}{b}{1} \! \Bigg\ra
	\star
	\Bigg| \! \tubeThreeD{0.9}{1}{2}{g'}{a'}{b'}{1} \!\! \Bigg\ra
	\\[-2em]
	\nn
	& \q =
	\frac{\delta_{g',a_1^{-1}gb_1}
	\, \delta_{a_2',a_2^{a_1}} \, \delta_{b_2',b_2^{b_1}}}{|A|^\frac{1}{2}|B|^\frac{1}{2}} \,
	\varrho^{AB}_{g,a_2,b_2}(a_1,a_1'|b_1,b_1') \,
	\Bigg| \! \tubeThreeD{0.9}{0}{2}{g}{a}{b}{2} \! \Bigg\ra ,
\end{align}
where we used the cocycle relation $d^{(2)}\varrho^{AB}_{g,a_2,b_2}(a_1,\tilde a, \tilde a^{-1}a_1'|b_1, \tilde b, \tilde b^{-1}b_1')$.
Using the more symbolic notation introduced in \eqref{eq:basisTube3D}, we obtained
\begin{align*}
	\big| \biGr{(g,a_2,b_2)}{a_1}{b_1} \big\ra
	\star
	\big| \biGr{(g',a_2',b_2')}{a_1'}{b_1'} \big\ra
	=
	\frac{\delta_{g',a_1^{-1}gb_1}
	\, \delta_{a_2',a_2^{a_1}} \, \delta_{b_2',b_2^{b_1}}}{|A|^\frac{1}{2}|B|^\frac{1}{2}} \,
	\varrho^{AB}_{g,a_2,b_2}(a_1,a_1'|b_1,b_1') \,
	\, \big| \biGr{(g,a_2,b_2)}{a_1a_1'}{b_1b_1'} \big\ra \; .
\end{align*}

\subsection{Relative groupoid algebra\label{sec:relatGr}}

Similarly to its (2+1)d analogue, the tube algebra found above can be recast as a twisted groupoid algebra.  Interestingly, due to the topology of the problem, we shall notice how in this language the (3+1)d tube algebra can be recast in terms of the (2+1)d one, unifying both computations. This is reminiscent of the notion of lifted models and lifted tube algebras developed in \cite{Bullivant:2019fmk} in the context of bulk excitations.

An important ingredient of our construction is the notion of \emph{loop groupoid}:
\begin{definition}[Loop groupoid]
	Given a finite groupoid $\mc{G}$, the loop groupoid $\Lambda \mc{G}$ is the groupoid with object set $\{\fr g \in {\rm End}_{\mc{G}}(X) \, | \, \forall \, X \in {\rm Ob}(\mc{G})\}$ and morphisms of the form $\fr h : \fr g \to \fr h^{-1}\fr g \fr h$, for every $\fr g \in {\rm End}_{\mc{G}}(X)$ and $\fr h \in {\rm Hom}_{\mc{G}}(X,Y)$. Composition in $\Lambda \mc{G}$ is inherited from the one in $\mc{G}$.
\end{definition}
\noindent
Specialising to the case where the finite groupoid is taken to be the delooping of a finite group $G$, we obtain that $\Lambda  G$ is the groupoid with object set ${\rm Ob}(\Lambda  G)=G$ and morphism set ${\rm Hom}(\Lambda  G)=\{ g\xrightarrow{a}a^{-1}ga \, | \, \forall \,  g,a\in G\}$. Composition is given by multiplication in $G$ such that
\begin{equation*}
	\biGr{g}{a}{} \biGr{a^{-1}ga}{a'}{} (aa')^{-1}gaa'=g\xrightarrow{aa'} (aa')^{-1}gaa' \; ,
\end{equation*}
for all $g,a,a'\in G$. Using this terminology, we can check that the cocycle data ${\sfT}(\pi)$, ${\sfT}(\lambda)$ and ${\sfT}(\mu)$ defined in \eqref{eq:defCoc2Gr3D} actually correspond to loop groupoid cocycles in $Z^3(\Lambda  G, \rU(1))$, $Z^2(\Lambda  A, \rU(1))$ and $Z^2(\Lambda  B, \rU(1))$, respectively. More generally, for any group $G$, we have a map ${\sfT}: Z^{\bullet}(G,\rU(1)) \to Z^{\bullet -1}(\Lambda  G, \rU(1))$ referred to as the $\mathbb S^1$-transgression map. More details regarding this map can be found in \cite{willerton2008twisted,bartlett2009unitary, Bullivant:2019fmk}. We further require the notion of \emph{relative groupoid}:

\begin{definition}[Relative groupoid]\label{def:relativegroupoid}
	Given a groupoid $\mc{G}$, and a pair of subgroupoids $\mc{A}, \mc{B} \subseteq \mc{G}$, the \emph{relative groupoid} $\mc{G}_{\mc{A} \mc{B}}$ is the groupoid with object set ${\rm Ob}(\mc{G}_{\mc{A} \mc{B}}) := \{ \fr g \in {\rm Hom}(\mc{G})\,|\, {\rm s}(\fr g) \in {\rm Ob}(\mc{A}) ,\, {\rm t}(\fr g )\in{\rm Ob}(\mc{B})\}$ and morphism set provided by
	\begin{align}
		\label{eq:relativeGrNot}
		\biGr{\fr g}{\fr a}{\fr  b}\fr a^{-1}\fr g\fr b \equiv \biGr{\fr g}{\fr a}{\fr b} \;\; ,
	\end{align}
	for all $\fr g\in {\rm Ob}(\mc{G}_{\mc{A}\mc{B}})$, $\fr a\in {\rm Hom}_{\mc A}({\rm s}(\fr g),-)$ and $\fr b\in {\rm Hom}_{\mc B}({\rm t}(\fr g),-)$. Composition is defined by
	\begin{align}
		\biGr{\fr g}{\fr a}{\fr b} \biGr{\fr a^{-1} \fr g \fr b}{\fr a'}{\fr b'} \fr a^{'-1} \fr a^{-1}\fr g \fr b \fr  b'
		=
		\biGr{\fr g}{\fr a \fr a'}{\fr b \fr b'} \fr a^{'-1}\fr a^{-1}\fr g \fr b \fr b' \; ,
	\end{align}
	for all composable pairs $(\fr a, \fr a')\in {\mc A}^{2}_{\rm comp}$ and $(\fr b,\fr b')\in {\mc B}^{2}_{\rm comp}$.
\end{definition}
\noindent
It follows immediately from the definition above that the groupoid $G_{AB}$, whose twisted groupoid algebra is isomorphic to the (2+1)d tube algebra, actually corresponds to the relative groupoid defined for the delooping of the groups. We are almost ready to define the (3+1)d tube algebra in this language. The last item we require is a notion of normalised cocycle for relative groupoid.
To this end we introduce $(\mc{G},\alpha)$-subgroupoids:
\begin{definition}\label{def:twistedsubgroupoid}
	Given a finite groupoid $\mc{G}$ and a normalised 3-cocycle $\alpha\in Z^{3}(\mc{G},\rU(1))$, we call a pair $(\mc A,\phi)$ a $(\mc{G},\alpha)$-subgroupoid when $\mc{A}\subseteq \mc{G}$ is a subgroupoid of $\mc{G}$ and $\phi\in C^{2}(\mc A,\rU(1))$ is a 2-cochain satisfying the condition
	$d^{(2)}\phi(\fr a, \fr a', \fr a'')=\alpha^{-1}(\fr a, \fr a' , \fr a''){\sss |}_{\mc A}$
	for all composable $(\fr a,\fr a',\fr a'')\in \mc{A}^{3}_{\rm comp}$.
\end{definition}
\noindent
For any pair of $(\mc{G},\alpha)$-subgroupoids $(\mc A,\phi)$ and $(\mc B,\psi)$, we construct a normalised 2-cocycle $\vartheta^{\mc A \mc B}\in Z^{2}(\mc{G}_{\mc A \mc B},\rU(1))$ for the relative groupoid $\mc{G}_{\mc A \mc B}$ via:
\begin{align}
	\label{eq:defvartheta}
	\vartheta^{\mc A \mc B}(
	\biGr{\fr g}{\fr a}{\fr b} \, ,
	\biGr{\fr a^{-1}\fr g \fr b}{\fr a'}{\fr b'} \, )
	&:=
	\frac{\psi(\fr b, \fr b')}{\phi(\fr a,\fr a')}
	\frac{\alpha(\fr a,\fr a',\fr a'^{-1}\fr a^{-1}\fr g \fr b\fr b')\, \alpha(\fr g,\fr b,\fr b')}{\alpha(\fr a,\fr a^{-1}\fr g \fr b,\fr b')}
	\\
	&\equiv
	\vartheta^{\mc A \mc B}_{\fr g}(\fr a,\fr a'|\fr b,\fr b')
\end{align}
for all composable morphisms
\begin{equation}
	\biGr{\fr g}{\fr a}{\fr b}\, , \, \biGr{\fr a^{-1}\fr g \fr b}{\fr a'}{\fr b'} \;\, \in \mc{G}_{\mc A \mc B} \; ,
\end{equation}
where we are using the shorthand notation introduced in \eqref{eq:relativeGrNot}.
It follows from $\alpha^{-1}{\sss |}_{\mc A} = d^{(2)} \psi$ and $\alpha^{-1}{\sss |}_{\mc B} = d^{(2)}\phi$, as well as the cocycle conditions
\begin{align*}
	d^{(3)}\alpha(\fr a,\fr a',\fr a'',\fr a''^{-1}\fr a'^{-1}\fr a^{-1}\fr g \fr b \fr b' \fr b'') &= 1
	&
	d^{(3)}\alpha(\fr a,\fr a^{-1}\fr g \fr b,\fr b',\fr b'') &= 1
	\\
	d^{(3)}\alpha(\fr a,\fr a',\fr a'^{-1}\fr a^{-1}\fr g \fr b \fr b',\fr b'') &= 1
	&
	d^{(3)}\alpha(\fr g,\fr b,\fr b',\fr b'') &= 1
\end{align*}
that $\vartheta^{\mc A \mc B}$ satisfies the 2-cocycle relation
\begin{align}
	d^{(2)}\vartheta^{\mc A \mc B}_{\fr g}(\fr a,\fr a',\fr a''|\fr b,\fr b',\fr b''):=
	\frac{	\vartheta^{\mc A\mc B}_{\fr a^{-1}\fr g\fr b}(\fr a',\fr a''|\fr b',\fr b'') \, \vartheta^{\mc A \mc B}_{\fr g}(\fr a,\fr a'\fr a''|\fr b,\fr b'\fr b'') }{\vartheta^{\mc A \mc B}_{\fr g}(\fr a \fr a',\fr a''|\fr b \fr b',\fr b'') \, \vartheta^{\mc A \mc B}_{\fr g}(\fr a ,\fr a'|\fr b,\fr b')}
	= 1
	\; .
\end{align}
Unsurprisingly, this equation mimics \eqref{eq:coc2Gr}.
Furthermore, given that $\alpha$ is a normalized cocycle, we have the normalisation conditions:
\begin{align*}
	\vartheta^{\mc A \mc B}_{\fr g}({\rm id}_{{\rm s}(\fr a')},\fr a'|{\rm id}_{{\rm s}(\fr b')},\fr b') = \vartheta^{\mc A \mc B}_{\fr g}(\fr a,{\rm id}_{{\rm t}(\fr a)}|\fr b , {\rm id}_{{\rm t}( \fr b)} ) & = 1
	\\
	\vartheta^{\mc A \mc B}_{\fr g}({\rm id}_{{\rm s}(\fr a')},\fr a'|\fr b, {\rm id}_{{\rm t}(\fr b)}) = \vartheta^{\mc A \mc B}_{\fr g}(\fr a, {\rm id}_{{\rm t}(\fr a)}|{\rm id}_{{\rm s}(\fr b')},\fr b') & =1
	\; ,
\end{align*}
which further imply
\begin{equation}
	\vartheta^{\mc A \mc B}_{\fr a^{-1}\fr g \fr b}(\fr a^{-1},\fr a|\fr b^{-1},\fr b) = \vartheta^{\mc A \mc B}_{\fr g}(\fr a,\fr a^{-1}|\fr b,\fr b^{-1}) \; .
\end{equation}

\bigskip \noindent
Let $G$ be a finite group and $\pi \in Z^4(G,\rU(1))$. We consider two subgroups $A,B \subset G$ and $\lambda \in C^3(A,\rU(1))$, $\mu \in C^3(B,\rU(1))$ such that $d^{(3)}\lambda = \pi^{-1}{\sss |}_A$ and $d^{(3)}\mu = \pi^{-1}{\sss |}_B$. It follows from the computations in sec.~\ref{sec:tube3D} that $(\Lambda A, \sfT(\lambda))$ and $(\Lambda B, \sfT(\mu))$ are $(\Lambda G, \sfT(\pi))$-subgroupoids. We define $\vartheta^{\Lambda A \Lambda B}$ by applying the formula \eqref{eq:defvartheta} for $\alpha \equiv \sfT(\pi)$, $\phi \equiv \sfT(\lambda)$ and $\psi \equiv \sfT(\mu)$. Putting everything together, we obtain the \emph{twisted relative groupoid algebra} $\mathbb C[\Lambda  G_{\Lambda  A \Lambda  B}]^{\vartheta^{\Lambda  A \Lambda  B}}$. We can show that this twisted relative groupoid algebra is isomorphic to the (3+1)d tube algebra by identifying
\begin{align}
	\label{eq:identGrTube}
	\biGr{(g,a_2,b_2)}{a_1}{b_1} \;\,
	 \equiv \biGr{\fr g}{\fr a_1}{\fr b_1} \; ,
\end{align}
such that $a_{2}\xrightarrow{g}b_{2} \equiv \fr g \in {\rm Ob}(\Lambda  G_{\Lambda A \Lambda B})$, $a_{2}\xrightarrow{a_{1}}a_{2}^{a_{1}} \equiv \fr a_1\in {\rm Hom}_{\Lambda  A}({\rm s}(\fr g),-)$ and $b_{2}\xrightarrow{b_{1}}b_{2}^{b_{1}} \equiv \fr b_1 \in {\rm Hom}_{\Lambda  B}({\rm t}(\fr g),-)$, as well as
$\vartheta^{\Lambda  A \Lambda  B} \equiv \varrho^{AB}$, which was defined in \eqref{eq:defCoc2Gr3D}.

Thereafter, we make use of the shorthand notations $\Lambda(G_{AB}) \equiv \Lambda G_{\Lambda A \Lambda B}$ and $\mathbb C[\Lambda(G_{AB})]^{\vartheta ^{\Lambda(AB)}} \equiv \grAlgL \equiv  \mathbb C[\Lambda  G_{\Lambda  A \Lambda  B}]^{\vartheta^{\Lambda  A \Lambda  B}}  $ to refer to this relative groupoid algebra. We purposefully choose a notation very similar to describe the (2+1)d and (3+1)d tube algebras in order to emphasize the fact that the framework presented in this section unifies both. As a matter of fact, we can obtain the (2+1)d algebra from  the (3+1)d one by restricting the loop groupoid $\Lambda  G$ to morphisms whose source and target objects are the identity in $G$ and by replacing the loop groupoid 3-cocycle $\alpha \equiv \sfT(\pi) \in Z^3(\Lambda  G, \rU(1))$, where $\pi \in Z^4(G,\rU(1))$, by a group 3-cocycle $\alpha \in Z^3(G,\rU(1))$. In virtue of this last remark, we may now focus on the algebra relevant to the (3+1)d scenario, namely $\grAlgL$, and deduce the results for the (2+1)d gapped boundary excitations as a limiting case.

We conclude this section with a remark regarding the notation. Since the morphisms $\fr a_1 \in {\rm Hom}_{\Lambda  A}({\rm s}(\fr g),-)$ and $\fr b_1 \in {\rm Hom}_{\Lambda  B}({\rm t}(\fr g),-)$ in \eqref{eq:identGrTube} are specified by a choice of group variables in the finite groups $A$ and $B$, respectively, we shall often loosely identify both in the following for notational convenience.

config: arxiv


TITLE: How to tell two ideals belong to the same ideal class group
QUESTION [12 upvotes]: Potentially dumb question here: how can you tell if two ideals belong to the same ideal class group?
Let's say we're looking at $\textbf{Z}[\sqrt{10}]$. It has infinitely many ideals, but by the Minkowski bound we need only concern ourselves with ideals with a norm of $2$ or $3$.
My gut tells me $\langle 3, 1 - \sqrt{10} \rangle$ and $\langle 3, 1 + \sqrt{10} \rangle$ are in the same ideal class, but that could be altogether wrong.

REPLY [0 votes]: Let's show that the ideal $(3, 1+ \sqrt{10})\cdot (3, 1-\sqrt{10})^{-1}$ is principal. We have
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = \frac{(3, 1+ \sqrt{10})\cdot(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})\cdot (3, 1+\sqrt{10})}$$
Now, for fractional ideals of number fields we have
$$(\alpha_1, \alpha_2)\cdot( \beta_1, \beta_2) = (\alpha_1 \beta_1, \alpha_1 \beta_2 + \alpha_2 \cdot \beta_1, \alpha_2 \cdot \beta_2)$$
( Gauss lemma for Dedekind rings, see Herman Weyl- Algebraic Number Theory)
Therefore, the denominator in the fraction $(*)$ is
$$(9, 6, 1-\sqrt{10}^2) = (9, 6, -9)= (3)$$
while the numerator is
$$(9, 6(1+ \sqrt{10}), (1+\sqrt{10})^2)=(9, 6(1+ \sqrt{10}), 11 + 2 \sqrt{10})=\\ =(9, 6(1+ \sqrt{10}), 2 + 2\sqrt{10}) =((1-\sqrt{10})(1+\sqrt{10}), 2(1+\sqrt{10})) \\=(1+\sqrt{10})\cdot (1-\sqrt{10}, 2)$$
Now $(1-\sqrt{10}, 2) = (1)$, since $((1-\sqrt{10})(1+\sqrt{10}), 2) = (9,2) = (1)$
We conclude:
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = (\frac{1+\sqrt{10}}{3})$$
or  $\frac{1+\sqrt{10}}{3}\cdot (3, 1-\sqrt{10})= (1+\sqrt{10}, 3)$.
This is obvious, once we know the factor...
$\bf{Added:}$ If $\alpha \cdot \beta =  k^2$ then
$$\frac{\alpha}{k} = \frac{k}{\beta} \ \textrm{so}\ \frac{\alpha}{k} (k, \beta)= (\alpha, k)$$