text:
\begin{document}

\title{Resummed Wentzel-Kramers-Brillouin Series: Quantization and Physical Interpretation}

\author{B.~Tripathi}
\affiliation{Department of Physics, University of Wisconsin-Madison, Madison, Wisconsin 53706, USA}
\email{btripathi@wisc.edu}

\date{\today}

\begin{abstract}
The Wentzel-Kramers-Brillouin (WKB) perturbative series, a widely used technique for solving linear waves, is typically divergent and at best, asymptotic, thus impeding predictions beyond the first few leading-order effects. Here, we report a closed-form formula that exactly resums the perturbative WKB series to all-orders for two turning point problem. The formula is elegantly interpreted as the action evaluated using the product of spatially-varying wavenumber and a coefficient related to the wave transmissivity; unit transmissivity yields the Bohr-Sommerfeld quantization.

\end{abstract}

\maketitle

\section{Introduction}

Linear waves are ubiquitous in the world of physics with applications ranging from quantum mechanics \cite{berry1972} to electromagnetism, fluid dynamics, and astrophysics \cite{iyer1987}. The properties of these waves are encoded in their dispersion relations, which reveal the nature of the medium they traverse, as well as their generating source \cite{iyer1987,fuller2015} (e.g., gravitational waves). This problem of obtaining the dispersion relation is traditionally addressed with the Wentzel-Kramers-Brillouin (WKB) perturbative series \cite{benderorszag}, which however is typically divergent and at best, asymptotic \cite{berry1972, robnik1997}. It thus presents challenges to predict phenomena beyond the leading-order effects. Obtaining an expression for this series to all-orders in perturbation theory, preferably in a closed-form, is therefore desirable. Despite its usefulness in finding previously unknown physical interpretations of fully quantized wave, it has remained elusive. Here, we will be accomplish both the tasks of finding a closed-form formula and assigning a physical meaning to it.

In the quest for developing a closed-form quantization condition (dispersion relation) for linear waves, several insightful but hitherto-unsuccessful attempts have been undertaken through various ways like investigating structures of higher order expressions in the WKB series  \cite{bender1977, barclay1994}, utilization of supersymmetric WKB method \cite{susy1985,susy1995}, complex WKB method \cite{voros1983}, and phase integral method \cite{froman1965}. We present here, a simple and insightful method to achieve this aim successfully.

The unexpected simplicity (both in mathematical structure and geometric-optical interpretation) of the closed-form formula reported here, arising out of the unwieldy WKB series, is what we believe to be the most striking about this work.

Our principal result is that the one-dimensional wave equation\footnote{
The book-keeping parameter $\epsilon$ can be set equal to $1$ at the outset or at the end of the perturbative calculations.} (or Schr\"{o}dinger equation)
\begin{equation} \label{eq:WKB_diff}
\epsilon^2 \frac{d^2 \psi(x)}{dx^2} = Q\left(x\right) \psi(x),\hspace{1.3cm} \psi\left(\pm \infty \right)=0,
\end{equation}
 with $Q\left(x\right) = - k^2(x) = 2m[V(z)-E]/\hslash^2$ where $k(z)$ is the local wavenumber, $V(z)$ is the potential, $E$ is the energy eigenvalue, $m$ and $\hslash$ are the mass and reduced Planck's constant respectively, for the case of two turning points [locations where $Q(z)=0$ with $z$ being a complex variable] has an exact closed-form quantization condition

\begin{equation} \label{eq:quantpot}
\oint_{\Gamma} k(z) \cdot \tau(z) dz = \left(K +\frac{1}{2}\right) 2\pi, \hspace{0.8cm} (K = 0,1,2,...),
\end{equation}
where $T(z)=\tau^2(z)$ is the wave-traversing medium's transmissivity of a layer of width $1/k(z)$ given as \cite{bremmer1951}
\begin{equation} \label{eq:transmissivitydefn}
\tau(z) = \sqrt{1 -\left(\frac{1}{2} \frac{d (k^{-1})}{dz}\right)^{2}} = \sqrt{1+ \left(\frac{S_{1}'}{S_{0}'}\right)^{2}}.
\end{equation}
The contour $\Gamma$ encircles the two turning points in anticlockwise direction. ($S_{0}'$ and $S_{1}'$ are explained immediately below, but presented here due to their elegant appearances.) Unit transmissivity reproduces the commonly known leading-order WKB approximation.

\section{Conventional WKB}
To begin with, consider the traditional transformation that is applied to the Schr\"{o}dinger equation \eqref{eq:WKB_diff},
\begin{subequations}
\begin{align} \label{eq:sprime}
    \psi(z,\epsilon)&=\exp\left[ \frac{1}{\epsilon} S(z,\epsilon) \right],\\
    \mathrm{or,\ } S'(z,\epsilon)&=\frac{\epsilon}{\psi(z,\epsilon)} \frac{d\psi(z,\epsilon)}{dz},
\end{align}
\end{subequations}
to obtain the Riccati equation: $\left(S'\right)^2 + \epsilon S^{''}=Q(z)$.

The quantization condition for energy eigenvalue $E$ in Eq.~\eqref{eq:WKB_diff} is given in terms of the WKB eigenfunction's exponent in Eq.~\eqref{eq:sprime} as \footnote{
One way to derive this is by integrating $dS/dz = S'(z,\epsilon)$ in Eq.~\eqref{eq:sprime} along a contour $\Gamma$ such that it encloses, for the $K^{\mathrm{th}}$ energy level, all $K$ zeros of the eigenfunction on the real axis, between the classical turning points. This leads to
$\oint_\Gamma S'(z,\epsilon) dz = \oint_\Gamma \epsilon \frac{d\mathrm{\ ln}\psi(z)}{dz} dz= \epsilon \mathrm{\ ln}\psi(z) \Big|_{\mathrm{evaluated\ once\ around\ }\Gamma}= K \cdot 2 \pi i \epsilon$.}:
\begin{equation} \label{eq:logintegrated}
    \frac{1}{\epsilon} \oint_\Gamma S'(z,\epsilon) dz = K \cdot 2 \pi i.
\end{equation}

This equation, although exact, is not useful unless we know what $S'(z,\epsilon)$ is [or be able to solve the above Riccati equation or Eq.~\eqref{eq:WKB_diff} exactly]. We, therefore, proceed with the perturbative method to compute $S'(z,\epsilon)$ as
\begin{equation} \label{eq:pertWKB}
S'(z,\epsilon) =  \sum_{n=0}^{\infty} \epsilon^n S_{n}'(z).
\end{equation}
Substituting this ansatz in the aforementioned Riccati equation and equating like-powers of $\epsilon$, one finds the $S_{n}'(z)$ to obey the recurrence relation,
\begin{align} \label{eq:recur}
 &S_{0}'(z) =  \sqrt{ Q(z)} = i k(z) ,\\ \label{eq:recurS1}
 &S_{1}'(z) =  -\frac{1}{2} \frac{d}{dz} \mathrm{ln} [S_0^{'}(z)],\\  \label{eq:recur2}
&2 S_{0}' S_{n}' + \sum_{j = 1}^{n-1} S_{j}' S_{n-j}' + S_{n-1}'' = 0, \hspace{0.8cm} (n \geq 2 ).
\end{align}
Eq.~\eqref{eq:logintegrated} thus, to all-orders in perturbation theory, is (it was first written in this form by Dunham \cite{dunham1932}):
\begin{equation} \label{eq:Dunham1}
\frac{1}{2 i \epsilon} \oint_{\Gamma} \sum_{n=0}^{\infty} \epsilon^n S_n'\left( z \right) dz = K \pi, \hspace{1cm} (K = 0,1,2,...).
\end{equation}

This series on the left hand side (LHS) is now to be summed up. However, it is typically a divergent asymptotic series \cite{brezin1977, stone1978}. One way to circumvent this challenge is to employ the Borel summation technique and assign a physical meaning to such a series. When employed, the analytic continuation of the Borel transform, however, presents another difficulty -- singularities on the integration contour (see, e.g., \cite{zj1981, zj1984, unsal2012, aniceto2015}). Avoiding such with contour deformation yields ambiguous imaginary terms that plague the energy eigenvalues $E$. Significant progresses have been made to address such problems and more recently via exact WKB and uniform WKB methods, following advances of resurgence theory, developed by Ecalle and others in the 1980s \cite{ecalle1981, voros1983, sueishi2020, dunnetransseries2014}. In such works, the ambiguous imaginary terms arising from the Borel summation are made to cancel each other systematically to all-orders by considering a ``resurgent trans-series" for the energy eigenvalues \cite{dunnetransseries2014}, as opposed to a perturbative series for it as we have done here. Such a path although very insightful and useful will not be pursued here as we wish to present an alternative simpler way to resum the diverging series of Eq.~\eqref{eq:Dunham1}, for a class of potentials, and assign a physical meaning to the resummed series.

Note that, in Eq.~\eqref{eq:Dunham1}, the term of first order in $\epsilon$ [i.e., $S'_1(z)$] can be integrated exactly \cite{benderorszag}:
\begin{equation} \label{eq:piover2}
\begin{aligned}
&\frac{1}{2 i} \oint_{\Gamma}  dz S_1'\left( z \right) = -\frac{1}{8 i} \oint_{\Gamma}  dz \frac{d}{dz} \mathrm{ln}[Q(z)] \\
&= -\left. \frac{1}{8 i} \mathrm{ln}\mathrm{\ }Q(z)\right\rvert_{\substack{\mathrm{evaluated\ once} \\ {\mathrm{around\ contour\ } \Gamma }}} = - \frac{1}{8 i} \left(2 \cdot 2\pi i \right) = -\frac{\pi}{2},
\end{aligned}
\end{equation}
where evaluating the logarithmic function around the contour $\Gamma$, enclosing the two turning points of $Q(z)$ yields $4 \pi i$.

This total contribution of $-\pi/2$ on the LHS of Eq.~\eqref{eq:Dunham1} correctly accounts for the zero-point energy of the simple harmonic oscillator. The series in Eq.~\eqref{eq:Dunham1}, truncated at the first order, is the Bohr-Sommerfeld quantization relation \cite{berry1972}. It has been considered as an exceptional case that all other higher order terms for the simple harmonic oscillator turn out to be zeros. However, in general, this is not the case. Fr\"{o}man and Fr\"{o}man \cite{froman1965} have shown that all \textit{other} higher \textit{odd}-order terms in the WKB series, Eq.~\eqref{eq:Dunham1}, can be written as exact derivatives, regardless of the type of potential, which upon contour integrating yield zeros. Setting $\epsilon=1$, we can, therefore, rewrite Eq.~\eqref{eq:Dunham1} as
\begin{equation} \label{eq:Dunham2}
\frac{1}{2 i} \oint_{\Gamma} \sum_{n=0}^{\infty} S_{2n}'\left( z \right) dz = \left(K +\frac{1}{2}\right) \pi, \qquad (K = 0,1,2,...).
\end{equation}

Several attempts have been undertaken in the past \cite{bender1977, romanovski2000} to infer the general expression for $S'_{2n}$ with expectations of summing up the series afterwards.  It, however, has turned out to be, heretofore, insurmountable. It is the objective of this work to present such a summation in an exact manner for an arbitrary potential with two turning points. (Note that such a route has been possible only for a very few special kinds of potentials, for e.g., the Eckart and the Morse potentials \cite{bender1977, romanovski2000}). Next, we outline our method of summing up the WKB series up to all-orders and interpret its physical meaning thereafter.

Let us recast Eq.~\eqref{eq:Dunham2} as
\begin{equation} \label{eq:Bohrcorrection2}
\frac{1}{2 i}  \oint_{\Gamma} S_0'\left( z \right) \cdot \sum_{n=0}^{\infty} T_{2n}\left( z \right)  dz = \left(K +\frac{1}{2}\right) \pi,
\end{equation}
where $T_{2n}(z) = S_{2n}'\left( z \right)/S_0'\left( z \right)$ and the summation over $T_{2n}$ will be achieved below. Introduce an economical notation  $L(z) \equiv 1/S_0'(z)$. ($L(z)$ can be regarded as having the dimension of length, found using Eqs.~\eqref{eq:recur} and \eqref{eq:WKB_diff}, and so does $D^{-1}$ where $D \equiv d/dz$.) Dividing both sides of Eq.~\eqref{eq:recur2} by $\left(S_0'\right)^2$ to rewrite it in these new notations of $T, D,$ and $L$, we find,
\begin{align}
&2 T_{n} + \sum_{j = 1}^{n-1} T_{j} T_{n-j} + L^2 \frac{d}{dz} \left[ \frac{S_{n-1}'}{S_0'}\cdot \frac{1}{L} \right]= 0,\\  \label{eq:recur3}
&2 T_{n} = - \sum_{j = 1}^{n-1} T_{j} T_{n-j} - LDT_{n-1} + T_{n-1} DL.
\end{align}

\section{Pattern Searching Campaign}
Note $T_0=1$ and $T_1=DL/2$. This allows to cast Eq.~\eqref{eq:recur3} finally in a neat way as
\begin{align}  \label{eq:recur4}
& T_{n} = T_{n-1} T_1 - \frac{1}{2}\sum_{j = 1}^{n-1} T_{j} T_{n-j} - \frac{1}{2}LDT_{n-1}, \qquad \hspace{0.3cm} (n \geq 2) .
\end{align}

We provide below the expressions for $T_n$.\footnote{It is worth highlighting that $T_{n}$ is a dimensionless function as it has exactly the same number of $D$'s and $L$'s; see Ref.~\cite{appendix:equalnoofLandD}.} Notice the appearance of $L$ below,
\begin{subequations}
\begin{alignat}{3} \label{eq:T2}
T_2 &= &&\textrm{\ \ \ }\frac{T_1^2}{2} \quad &&- L \times \left[\frac{DT_1}{2}\right], \\
T_3 &=  && \quad &&-L\times \left[\frac{DT_2}{2}\right],\\
T_4 &=  &&-\frac{T_1^4}{8} \quad &&- L \times \left[\frac{DT_3}{2} - \frac{T_1^2\cdot DT_1}{4} + \frac{LDT_1 \cdot DT_1}{8} \right],\\
T_5 &=  && \quad &&-L \times \left[ \frac{DT_4}{2} - \frac{T_2 \cdot DT_2}{2} \right],\\
T_6 &= &&\textrm{\ \ \ }\frac{T_1^6}{16} \quad &&- L\times \left[...\right],
\end{alignat}
\end{subequations}
where ellipsis with a square bracket, $\left[...\right]$, represents a collection of functions of lower order $T_n$. Note that all terms of odd order (in $n$) of $T_n$ necessarily begin with $L$ because these expressions when substituted in Eq.~\eqref{eq:Bohrcorrection2} yield cancellation of such $L$ with $L$ in the denominator of $S_0'=1/L$. The remaining part of the integrand can be shown to be (the sum of) the product of exact derivatives or expressions that can be changed into them \cite{appendixB}. Such integrand with the product of exact derivatives is trivially zero upon contour integrating \cite{appendixC} because they are single-valued functions for the defined contour path (no logarithmic derivatives are involved here as $L$ in the denominator of $S_0'=1/L$ has been cancelled out). This is tantamount to stating that all odd-order terms of $T_n$ contribute to the wavefunction's amplitude (i.e., they do not play role in the quantization condition \cite{bender1977}) and $T_n$'s even-order terms modulate the phase of the wavefunction -- thus the quantization condition involves thereof. To reiterate, for $n \geq 1$, $T_{\mathrm{odd\ order}} = T_{2n+1}= -L \times \left[...\right]$ and $T_{\mathrm{even\ order}} = T_{2n} = ...T_1^{2n} - L \times \left[...\right]$. After explicitly computing higher order terms (e.g., $T_8 = 5T_1^8/128 - L  \times \left[...\right]$), we recognize a completely unexpected but instructive pattern, based upon which we propose the following hypothesis and prove it subsequently.\footnote{We are very grateful to Michael V. Berry for questions that prompted us to propose this hypothesis.}

\section{Inductive Hypothesis}
Proposition:
\textit{For any $n \in \mathbb {N}$,}

\begin{equation} \label{eq:proposition}
\mathrm{P}(n)\mathrm{:\ } T_{2n} = \binom{n-\frac{3}{2}}{-\frac{3}{2}}  \left(\frac{T_1}{i}\right)^{2n} - L \times \left[...\right],
\end{equation}
\textit{where $ \mathrm{\binom{n-\frac{3}{2}}{-\frac{3}{2}}} $ is the binomial coefficient and $i^2=-1$.}

This statement is not challenging to prove using the principle of mathematical induction (see section A in the Appendix for the proof).

Although not immediately obvious, the inductive hypothesis, presented in the precise form as above, has paramount consequence. In the proposition, in Eq.~\eqref{eq:proposition}, the first term of $T_{2n}$ on the right hand side lacks $L$ in front of it unlike the second term and hence, when substituted in Eq.~\eqref{eq:Bohrcorrection2}, it yields a function that does not vanish upon doing the contour integration (as terms like $\oint_\Gamma dz S_0' \cdot T_1^{2n} \sim \oint_\Gamma dz (DL)^{2n}/L$ contribute to a logarithmic derivative of $L$ and thus the contour encloses poles of a logarithmic function, which are the zeros of $Q(z)$). In contrast thereof, the second term of $T_{2n}$ that begins with $L$, written as $L \times \left[...\right]$, contributes exactly zero upon contour integrating as the cancellation of this $L$ with $S_0'=1/L$ in Eq.~\eqref{eq:Bohrcorrection2} modifies the WKB integrand into (a sum of) the product of exact derivatives (importantly, without any logarithmic derivative) \cite{appendixB}. Resulting product of exact derivatives, lacking logarithmic term, amount to zero upon contour integrating; the reasoning here can follow the same as was aforementioned for the odd-order terms in the WKB series \cite{appendixC, appendixD}. Thus this campaign of searching terms which begin with $L$ and others which don't is unexpectedly helpful. We shall, therefore, deal with only the power series in $T_1$ in Eq.~\eqref{eq:Bohrcorrection2}. By straightforward summation of this special series up to all-orders in perturbation theory, we obtain a closed-form expression for Eq.~\eqref{eq:Bohrcorrection2} as presented below in Eq.~\eqref{eq:summedupWKB}.

\begin{widetext}
\begin{equation}  \label{eq:summedupWKB}
\begin{aligned}
\hspace{-0.2cm}\left(K +\frac{1}{2}\right) \pi
= \frac{1}{2 i} \oint_{\Gamma} \frac{dz}{L} \left[1+ \sum_{n=1,2,..}^{\infty} \binom{n-\frac{3}{2}}{-\frac{3}{2}} \left(\frac{\dl}{2 i}\right)^{2n} \right]
=\frac{1}{2 i}\oint_{\Gamma}   \frac{dz}{L\left( z \right)} \sqrt{1+\left(\frac{DL}{2}\right)^2} &= \frac{1}{2} \oint_{\Gamma} k(z)  \cdot \sqrt{1+ \left(\frac{S_{1}'}{S_{0}'}\right)^{2}}  dz\\
&\hspace{-0.5cm}=\frac{1}{2} \oint_{\Gamma} k(z) \cdot \sqrt{1 -\left(\frac{1}{2} \frac{d (k^{-1})}{dz}\right)^{2}} dz.
\end{aligned}
\end{equation}
\end{widetext}

We emphasize that our summation in Eq.~\eqref{eq:summedupWKB} involves a power series, which is also the case in all the special problems for which the WKB series has been summed up exactly \cite{romanovski2000, salasnich1997}. We believe this to be the reason why the first few terms of the WKB expansion often approximate the correct eigenvalues despite the full series being divergent (this is, in part, also an answer to why the WKB expansion is asymptotic).

We are also able to physically interpret the expressions involved on the last line of Eq.~\eqref{eq:summedupWKB} using the notion of geometric optics; however, the equation in its entirety is non-trivial to elucidate, possibly owing to the quantum effects embodied in all-orders (in perturbation theory).

\section{Geometric-Optical Meaning}
We borrow here the illustration by Bremmer \cite{bremmer1951} where the author demonstrates, \textit{mutatis mutandis}, the similarity of each order of WKB series (consider its $n^{\mathrm{th}}$ order) to the transmitted waves in an infinitesimally-discretized inhomogeneous medium that undergo $n$-number of reflections. At each reflection, the waves change their direction by $180^{{\circ}}$. Thus a wave that begins from a point far to the left gets continually transmitted to the right while suffering reflection at each discretized boundary (and consider for now only the directly transmitted waves -- not doubly or quadruply or even number of multiply reflected ones that also can eventually transmit to the right). Such resulting wavefunction at the rightmost end yields the $1^{\mathrm{st}}$ order WKB approximation \cite{bremmer1951}. Each of the above-mentioned reflected waves can undergo further reflection(s) and keep continually transmitting to the right. The more the number of reflections they suffer before they arrive to the rightmost end, the higher they belong to in the order of WKB expansion. Referring readers to the original paper \cite{bremmer1951} for additional interesting details, we present now heuristically how Bremmer arrives at the reflection coefficient of a layer of width $1/k(z)$. The well-known reflection coefficient in one-dimension is (Bremmer \cite{bremmer1951} uses $R$ notation, which we shall reserve for the reflectivity to avoid potential confusions),
\begin{equation}
r(z) = \frac{k_s - k_{s+1}}{k_s + k_{s+1}} \approx \frac{k_s - k_{s+1}}{2 k_s}  \approx-\frac{dk/d\xi}{2k},
\end{equation}
where $s$ is the layer-number in the infinitesimally discretized inhomogeneous medium (within which $k_s$ remains constant) and $\xi$ is proportional to the number of wavelengths over which the wavelength (or $k$) changes appreciably (from $k_s$ to $k_{s+1}$). Following Bremmer \cite{bremmer1951}, $d\xi = k(z) dz$.
Therefore, the transmissivity of a layer of width $1/k(z)$ is
\begin{equation}
    T(z) = 1-r^2(z) = 1-\left(-\frac{dk/d\xi}{2k}\right)^2=1 -\left(\frac{1}{2} \frac{d (k^{-1})}{dz}\right)^{2},
\end{equation}
which is exactly what appears on the last line of Eq.~\eqref{eq:summedupWKB}.

Note that the wavenumber, obtained via resummation of WKB series to all-orders (i.e., integrand in Eq.~\eqref{eq:summedupWKB}), vanishes even before we reach the classical turning points. Interestingly, for \textit{all} potentials, it vanishes exactly at the locations where $p_{\mathrm{cl}}(x) \cdot x = \hslash/2$ where $p_{\mathrm{cl}} = \sqrt{2m[E-V(x)]}$ (cf. with the Heisenberg uncertainty principle).

\section{Applications}
Our novel closed-form quantization condition yields exact energy eigenvalues to potentials with two turning points. We demonstrate the efficacy of our formula in an example below and present other cases like simple harmonic oscillator, $3$-dimensional harmonic oscillator, Coulomb potential, Eckart potential, and Morse potential in section D of the Appendix. We also find in cases of $3$-dimensional spherically symmetric potentials, the Langer-correction factor \cite{langer1937, koike2009} appears naturally upon performing the contour integration of our formula that resums all-orders perturbative effects. This corroborates the previous claim that the Langer-modification comes from the higher order corrections in the WKB series \cite{salasnich1997}.

Consider an asymmetric Rosen-Morse potential, $V(x)$, with $\hslash^2=2m$:
$V(x)=-U_0 \textrm{sech}^2\left( x/a \right) +U_1\tanh\left( x/a \right)$. Let $z$ be a complex variable such that $z=\tanh\left( x/a\right)$. Then, $z_a$ and $z_b$ are the two classical turning points, satisfying $z_a+z_b = -U_1/U_0$; $z_a z_b = -\left(E+U_0\right)/U_0$. Using Eq.~\eqref{eq:quantpot},

\begin{equation}  \label{eq:example1}
\frac{1}{2 i}\oint_{\Gamma}  \frac{\sqrt{16 \left[ V(z)-E \right]^{3}+\left[V'(z)\right]^{2} }}{4 \left[ V(z)-E \right]} dz = \left(K +\frac{1}{2}\right)\pi,
\end{equation}
where $K=0,1,2,...$ represent different energy levels.
The poles of the integrand are at $z=z_a,z_b,1,-1,\textrm{ and } \infty$. So, calculating residue at each pole with a proper principal value yields
\begin{align}
\begin{split}
-\frac{1}{4}+\frac{1}{4}  &{-} \frac{a\sqrt{(z_a-1)(z_b-1)U_0}}{2} {-} \frac{a\sqrt{(z_a+1)(z_b+1)U_0}}{2}\\
&\hspace{2cm}+ \frac{\sqrt{1+4a^2 U_0}}{2} = K+\frac{1}{2},
\end{split}
\end{align}
\begin{equation}
\begin{split}

\therefore \frac{1}{2}\sqrt{\frac{-E-U_1}{U_0}} {+} \frac{1}{2}\sqrt{\frac{-E+U_1}{U_0}} = -\frac{1}{a\sqrt{U_0}} \left( K+\frac{1}{2} \right) \\
+ \frac{\sqrt{1+4a^2U_0 }}{2a\sqrt{U_0}},
\end{split}
\end{equation}
which agrees with Ma \& Xu  \cite{maxu2005} and the individual terms directly manifest from the residues at poles whose locations are precisely predicted by Eq.~\eqref{eq:example1}.

At last, we remark that, for reasons not fully understood, the proposed quantization relation works only for two turning point problems. Nevertheless,
understanding the nature of the exactly quantized action in two turning point problems, considered here, is likely to benefit the extension of our geometric-optically interpretable equation to problems with multiple turning points. Such interesting extensions will be investigated in a future study. Our proposed quantization condition might also engender rethinking of quantization in higher dimensions in terms of geometry \cite{stone2005}.

\section{Conclusion}
This article presents an exact closed-form quantization relation by summing up the WKB series to all-orders in perturbation theory for arbitrary one-dimensional potentials having two turning points. The new resummation procedure utilized herein reveals an unexpectedly simple pattern in the general term of the WKB series, leading to an inductive hypothesis, which we are able to prove by the principle of mathematical induction. The presented formula is then physically interpreted as the action of a wave with wavenumber corrected by a factor related to the wave transmissivity. Unit transmissivity recovers the Bohr-Sommerfeld quantization relation. This closed-form expression for the quantization might also be useful in problems with more than two turning points where non-perturbative effects that give rise to tunneling phenomena come into play, i.e., spectral curves with non-zero genus \cite{basar2017}. For such problems, some of the neglected terms in the series resummation appear to be necessary. In the light of resurgent perturbative/nonperturbative relations \cite{dunne2014, Gahramanov2016, basar2017}, collecting such terms seem interesting and further investigation is merited. This will, however, be left for the future as it is beyond the objective of the present article. Arguably the most important advancement through this work is the discovery of an elegant and physically-interpretable equation emerging from a myriad of complicated terms in the WKB series that become increasingly unmanageable at each higher order of the series. It is gratifying to find that the spectral problem with genus-0 spectral curve (i.e., with two turning points and no tunneling phenomenon) can be reduced to an exact equation as simple and economical as Eq.~\eqref{eq:quantpot}. Analyzing this equation might lead to a deeper understanding of quantum geometry \cite{basar2017}.

\begin{acknowledgments}
I am pleased to thank Dhrubaditya Mitra for enkindling interest in this work and for offering discussions and guidance; it is difficult to imagine NORDITA (Sweden) without his friendly teaching style and advocacy for perturbation methods. I am also indebted to Michael V. Berry and Carl M. Bender for their appraisals of this work that greatly improved the manuscript. Thanks are also due to Paul W. Terry, MJ Pueschel, and Dibyendu Nandi for their valuable feedback and suggestions. Correspondences with Mithat \"{U}nsal and Luca Salasnich helped to find connections of this work with others. I also acknowledge support from the Van Vleck Fellowship in Physics at UW-Madison and the conducive environment herein.
\end{acknowledgments}

\begin{widetext}
\vspace{50cm}
\section{Appendix}

\subsection{Proof by the principle of mathematical induction}

Below, we present the proof for the proposition put forth for $T_{2n}$ in Eq.~\eqref{eq:proposition}. The statement, P($n$), is trivially satisfied for $n=1$ [cf. Eq.~\eqref{eq:proposition} with Eq.~\eqref{eq:T2}]. Now, we assume it to be valid for an arbitrary $n$ and prove that it implies the proposition, in Eq.~\eqref{eq:proposition}, is true for $n+1$ as well [i.e., P$(n+1)$ is true].  We begin with the WKB recurrence relation, Eq.~\eqref{eq:recur4},
\begin{subequations}
\begin{align}
 T_{2(n+1)}
&= - \frac{1}{2}\sum_{j = 2}^{2(n+1)-2} T_{j} T_{2(n+1)-j} - \frac{LDT_{2(n+1)-1}}{2}\\
&= -\sum_{\substack{j = 2,4,... \\ \mathrm{\textbf{even}}}}^{2n} \frac{T_{j} T_{2n+2-j}}{2}  - \sum_{\substack{j = 3,5,...\\ \mathrm{\textbf{odd}}}}^{2n-1} \frac{T_{j} T_{2n+2-j}}{2} - \frac{LDT_{2n+1}}{2}\\
&= \left\{-\sum_{j/2 = 1,2,...}^{n} \binom{\frac{j}{2}-\frac{3}{2}}{-\frac{3}{2}} \binom{n+1-\frac{j}{2}}{-\frac{3}{2}} \left(\frac{T_{1}}{i}\right)^{2n+2} - L \times \left[...\right] \right\} - \sum_{\substack{j = 3,5,...\\ \mathrm{\textbf{odd}}}}^{2n-1} \frac{T_{j} T_{2n+2-j}}{2} - \frac{LDT_{2n+1}}{2}\\
&= -\sum_{j/2 = 1,2,...}^{n} \binom{\frac{j}{2}-\frac{3}{2}}{-\frac{3}{2}} \binom{n+1-\frac{j}{2}}{-\frac{3}{2}} \left(\frac{T_{1}}{i}\right)^{2n+2} - L \times \left[...\right] -L \times \left[...\right] - \frac{LDT_{2n+1}}{2} \hspace{0.5cm} \left[ \because T_{\mathrm{odd}} \mathrm{\ begins\ with\ } L \right] \\
&= \binom{n+1-\frac{3}{2}}{-\frac{3}{2}}  \left(\frac{T_1}{i}\right)^{2(n+1)} - L \times \left[...\right] \QEDA
\end{align}
\end{subequations}
This proves the proposition.

\subsection{$T_n$ beginning with $L$ convertible to a product of exact derivatives}
Here, we detail the recipe of casting any expression in $T_n$ that begins with $L$ into a product of exact derivatives (without a logarithmic derivative) \cite{appendixB}. Consider the term in $T_n$ that begins with $L$:
\begin{equation} \label{eq:appB1}
    \begin{aligned}
    \oint_\Gamma dz S_0' T_n  = \oint_\Gamma dz S_0' L \times \left[...\right] &= \oint_\Gamma dz \times \frac{1}{L} \times L \times \left[...\right] \\
    &= \oint_\Gamma dz \times \frac{1}{L} \times L \times \left[\left(D^{p_1}L^{q_1}D^{p_2}L^{q_2}...\right) \times ... \times \left(D^{p_{j-1}}L^{q_{j-1}}D^{p_{j}}L^{q_{j}}...\right)\right]\\
    &= \oint_\Gamma dz  \left[\left(D^{p_1}L^{q_1}D^{p_2}L^{q_2}...\right) \times ... \times \left(D^{p_{j-1}}L^{q_{j-1}}D^{p_{j}}L^{q_{j}}...\right)\right],
    \end{aligned}
\end{equation}
where $D^{p_1}L^{q_1}D^{p_2}L^{q_2}...$ represents $D^{p_1}\left(L^{q_1}D^{p_2}\left(L^{q_2}...\right)\right)$; $p_1, q_1, p_2, q_2, ..., p_j, q_j, ...$ are all integers in between (and including) $0$ and $n$, satisfying the constraints: $p_1+p_2+...+p_j+... = n$ (for $n$ number of $D$'s) and $q_1+q_2+...+q_j+... = n-1$ (for $n-1$ number of $L$'s); and one extra $L$ lies in the very beginning of $T_n$, which has gotten cancelled with $S_0'=1/L$. This makes $T_n$ to have $n$ number of $D$'s and $L$'s, as argued in Ref.~\cite{appendix:equalnoofLandD}. If all $p_1, p_2, ..., p_j, ...$ are equal to or greater than $1$, this integrand is already a product of exact derivatives (without a logarithmic derivative as $S_0'=1/L$ has already been cancelled with $L$ of $T_n$). If any of $p_1,p_2, ...,p_j, ..$ is zero, the integrand in Eq.~\eqref{eq:appB1}, using chain rule, becomes (say $p_1=0$): \begin{equation}
    \begin{aligned}
    \mathrm{integrand\ } &= \left(D^{0}L^{q_1}D^{p_2}L^{q_2}...\right) \times ... \times \left(D^{p_{j-1}}L^{q_{j-1}}D^{p_{j}}L^{q_{j}}...\right) \\
    &= \hspace{0.5cm}\left(L^{q_1}D^{p_2}L^{q_2}...\right) \times ... \times \left(D^{p_{j-1}}L^{q_{j-1}}D^{p_{j}}L^{q_{j}}...\right)\\
    &= D\left[\left(L^{q_1}D^{p_2}L^{q_2}...\right) \times ... \times \left(D^{p_{j-1}}L^{q_{j-1}}D^{p_{j}}L^{q_{j}}...\right)\right] \bm{-} D\left[\left(L^{q_1}D^{p_2}L^{q_2}...\right) \times ...  \right] \times \left(D^{p_{j-1}-1}L^{q_{j-1}}D^{p_{j}}L^{q_{j}}...\right),
    \end{aligned}
\end{equation}
thus turning the integrand into (a product of) exact derivatives. This process can be repeated if $p_{j-1}-1$ also happens to be $0$ (and if several $L$ multiplies each other, its exponent can be raised to abridge this procedure). It should be emphasized that it is guaranteed to find such a transformation as there are exactly the same number of $D$'s and $L$'s in any $T_n$ \cite{appendix:equalnoofLandD}. In this regard, searching a general expression for $T_{2n}$ as stated in the proposition in Eq.~\eqref{eq:proposition} is surprisingly helpful.

\subsection{Product of exact derivatives yields zero}
We demonstrate here that the product of exact derivatives (without a logarithmic function) amounts to zero on contour integrating, expounding on Ref.\cite{appendixC}. Let us consider two functions $f$ and $g$ (without a logarithm). Then, $\oint_\Gamma dz ~ df/dz  = 0$.

Now, using the Cauchy integral formula to represent the exact derivatives and thereafter employing the partial fraction decomposition,
\begin{subequations}
\begin{align}
\oint_\Gamma dz \frac{df}{dz} \frac{dg}{dz}  &= \oint_\Gamma dz \frac{df}{dz} \frac{dg}{dz} \\
&= \oint_\Gamma dz \left\{\frac{1}{2\pi i} \oint_{\gamma_1} \frac{f(u)}{\left(u-z\right)^{1+1}} du \right\} \left\{\frac{1}{2\pi i} \oint_{\gamma_2} \frac{g(v)}{\left(v-z\right)^{1+1}} dv \right\} \hspace{1cm} [\gamma_1\ \& \ \gamma_2\ \mathrm{enclose\ the\ pole\ at\ } z]\\
&= -\frac{1}{4 \pi^2} \oint_{\gamma_1} \oint_{\gamma_2} du dv f(u) g(v)  \oint_\Gamma dz {\left[\frac{1}{(u-z)(v-z)}\right]}^2\\
&= -\frac{1}{4 \pi^2} \oint_{\gamma_1} \oint_{\gamma_2} du dv f(u) g(v)  \oint_\Gamma dz {\left[\frac{1}{v-u} \left(\frac{1}{u-z}-\frac{1}{v-z} \right)\right]}^2\\
&= -\frac{1}{4 \pi^2} \oint_{\gamma_1} \oint_{\gamma_2} du dv \frac{f(u) g(v)}{\left(v-u\right)^2}  \oint_\Gamma dz { \left(\frac{1}{u-z}-\frac{1}{v-z} \right)}^2\\
&= -\frac{1}{4 \pi^2} \oint_{\gamma_1} \oint_{\gamma_2} du dv \frac{f(u) g(v)}{\left(v-u\right)^2} \left[ \oint_\Gamma dz \frac{1}{\left(u-z\right)^2} -2\oint_\Gamma dz \frac{1}{\left(u-z\right)\left(v-z\right)} + \oint_\Gamma dz \frac{1}{\left(v-z\right)^2} \right]\\
&= -\frac{1}{4 \pi^2} \oint_{\gamma_1} \oint_{\gamma_2} du dv \frac{f(u) g(v)}{\left(v-u\right)^2} \left[ 0 -2\oint_\Gamma dz \frac{1}{\left(u-z\right)\left(v-z\right)} + 0 \right]\\
&= \frac{1}{2 \pi^2} \oint_{\gamma_1} \oint_{\gamma_2} du dv \frac{f(u) g(v)}{\left(v-u\right)^2}  \oint_\Gamma dz  \frac{1}{v-u} \left(\frac{1}{u-z}- \frac{1}{v-z} \right)\\
&= \frac{1}{2 \pi^2} \oint_{\gamma_1}  \oint_{\gamma_2} du dv \frac{f(u) g(v)}{\left(v-u\right)^3} \left[ \oint_\Gamma dz \frac{1}{u-z}-  \oint_\Gamma dz \frac{1}{v-z} \right]\\
&= \frac{1}{2 \pi^2} \oint_{\gamma_1}  \oint_{\gamma_2} du dv \frac{f(u) g(v)}{\left(v-u\right)^3} \left[- 2 \pi i+ 2 \pi i\right]\\
&=0.
\end{align}
\end{subequations}
It is straightforward to prove, in the similar manner, that the product of three (or more) exact derivatives ((without a logarithmic function) under contour integration is also zero. We are thus left with a power series in $\left(T_1\right)^{2n}$, which we sum up in Eq.~\eqref{eq:summedupWKB}.

\end{widetext}
\subsection{More Examples}
Let us now test the validity of our novel formula. We find that this formula gives the exact energy eigenvalues $E$ for all the following potentials and many more which are not listed here (but all having exactly two turning points). We choose, without the loss of generality, $\hslash^2 = 2m$ in all of the following calculations.

\subsection*{I. Simple Harmonic Oscillator}
Even though we already know that the leading-order WKB is exact for simple harmonic oscillator, it is desirable to test if the extra factor in the integrand of Eq.~\eqref{eq:example1} would cause any deviation from the correct eigenvalues. Consider,
\begin{subequations}
\begin{align}
V(z)&=z^2,\\
V-E &=z^2-E,\\
V'(z) &=2 z .
\end{align}
\end{subequations}

From Eq.~\eqref{eq:example1},
\begin{equation}
\frac{1}{2 i}\oint  \frac{\sqrt{16 \left( V-E \right)^{3}+\left(V'\right)^{2} }}{4 \left( V-E \right)} dz = \left(K +\frac{1}{2}\right)\pi,
\end{equation}

which leads to
\begin{equation}
\frac{1}{2 i}\oint  \frac{\sqrt{16 \left( z^2-E \right)^{3}+\left( 2z \right)^{2} }}{4 \left( z^2-E \right)} dz = \left(K +\frac{1}{2}\right)\pi. \hspace{0.6cm}
\end{equation}
The poles of the integrand are at $z=\sqrt{E},-\sqrt{E},\textrm{ and } \infty$. So, calculating residues at each of the poles respectively,
\begin{align}
 \frac{1}{4}-\frac{1}{4} + \frac{E}{2} &= K+\frac{1}{2},\\
\therefore E &=2 \left( K+\frac{1}{2}\right).
\end{align}

\subsection*{II. $3$-D harmonic oscillator  }
Consider the potential,
\begin{subequations}
\begin{align}
V(r)&=r^2+\frac{b}{r^2}+\frac{l(l+1)}{r^2},\\
V-E &= r^2+\frac{b}{r^2}+\frac{l(l+1)}{r^2}-E.
\end{align}
\end{subequations}

Let $u=r^2$; $u' = 2r= 2\sqrt{u}$.
\begin{subequations}
\begin{align}
\therefore V-E &= \frac{1}{u} \left[ u^2-E u + \left\{b+l\left(l+1\right)  \right\}  \right]\\
&= \frac{1}{u} \left(u-u_a  \right)\left(u-u_b  \right),
\end{align}
\end{subequations}
with
\begin{subequations}
\begin{align}
u_{a,b} &= \frac{E}{2}\pm \sqrt{\left(\frac{E}{2}\right)^2 -\left\{b+l\left(l+1\right) \right\} },\\
u_a+u_b &= E,\\
u_a u_b &= b+l\left(l+1\right).
\end{align}
\end{subequations}
Now,
\begin{subequations}
\begin{align}
V' &= \frac{dV}{du}u'\\
&= \left[ \frac{2u-u_a-u_b}{u} -\frac{(u-u_a)(u-u_b)}{u^2}\right] 2\sqrt{u}.
\end{align}
\end{subequations}

Using Eq.~\eqref{eq:example1},
\begin{equation}
\frac{1}{2 i}\oint  \frac{\sqrt{16 \left( V-E \right)^{3}+\left(V'\right)^{2} }}{4 \left( V-E \right)} dz = \left(K +\frac{1}{2}\right)\pi.
\end{equation}
The poles of the integrand are at $u=u_a,u_b,0,\textrm{ and } \infty$. So, calculating residues at each of the poles respectively,

\begin{align}
 \frac{1}{4}-\frac{1}{4} - \frac{1}{4}&\sqrt{1+4u_a u_b}+ \frac{u_a+u_b}{4} = K+\frac{1}{2},\\
\therefore E &=2 \left[ 2K+1+\sqrt{{\color{red}\bm{\left(l+\frac{1}{2}\right)}^2}+b}\right].
\end{align}

This solution agrees with Rosenzweig \& Krieger \cite{Rosenzweig1968}. \\

Note that the correct Langer correction factor, shown in the bold typeset, emerges naturally from our all-orders resummed WKB series, which Langer proposed to replace $l(l+1)$ by $(l+1/2)^2$ to obtain the correct eigenvalues.

\subsection*{III. Coulomb potential}
Now, assume the Coulomb potential,
\begin{subequations}
\begin{align}
V(r)&=-\frac{V_0}{r}+\frac{b}{r^2}+\frac{l(l+1)}{r^2},\\
V-E &= \frac{1}{r^2}\left[-V_0 r + b+ l(l+1)-Er^2 \right]\\
&= \frac{-E}{r^2}\left[r^2+\frac{V_0}{E}r-\frac{b+l(l+1)}{E}  \right]\\
&= \frac{-E}{r^2}\left(r-r_a\right)\left(r-r_b\right),
\end{align}
\end{subequations}
with
\begin{subequations}
\begin{align}
r_{a,b} &= -\frac{V_0}{2E}\pm \sqrt{\left(\frac{V_0}{2E}\right)^2 +\frac{b+l\left(l+1\right) }{E} },\\
r_a+r_b &= \frac{-V_0}{E},\\
r_a r_b &= -\frac{b+l\left(l+1\right)}{E}.
\end{align}
\end{subequations}
Now,
\begin{align}
V' &= \frac{2E(r-r_a)(r-r_b)}{r^3} - \frac{E}{r^2}\left(2r-r_a-r_b\right).
\end{align}

We use Eq.~\eqref{eq:example1} below,
\begin{equation}
\frac{1}{2 i}\oint  \frac{\sqrt{16 \left( V-E \right)^{3}+\left(V'\right)^{2} }}{4 \left( V-E \right)} dz = \left(K +\frac{1}{2}\right)\pi.
\end{equation}
The poles of the integrand are at $r=r_a,r_b,0,\textrm{ and } \infty$. So, calculating residues at each of the poles with proper principal value,

\begin{align}
- \frac{1}{4}+\frac{1}{4} {-} \frac{1}{2}&\sqrt{1-4E r_a r_b}+ \frac{\sqrt{-E}}{2}\left(r_a+r_b\right) = K+\frac{1}{2},\\
\therefore E &= \frac{-V_0^2}{4 \left[ K+1/2 {+}\sqrt{b+{\color{red}\bm{\left(  l+\frac{1}{2}\right)}^2}}  \right]^2}.
\end{align}

This solution agrees with Rosenzweig \& Krieger \cite{Rosenzweig1968}.\\

Notice again that the correct Langer correction factor, shown in the bold typeset, emerges naturally from our all-orders resummed WKB series.

\subsection*{IV. Eckart potential}
Let us consider the Eckart potential,
\begin{align}
V(x)&=\frac{-\lambda e^{-\alpha x}}{1-e^{-\alpha x}}+\frac{b e^{-\alpha x}}{\left(1-e^{-\alpha x}\right)^2}.
\end{align}

Using a transformation, $u=e^{\alpha x}-1$; $u' = \alpha(u+1)$; we write,
\begin{subequations}
\begin{align}
V-E &= \frac{-\lambda}{u} +\frac{b(u+1)}{u^2}-E\\
&= \frac{-E}{u^2}\left[u^2+\frac{\lambda-b}{E}u -\frac{b}{E}  \right]\\
&= \frac{-E}{u^2}(u-u_a)(u-u_b),
\end{align}
\end{subequations}
with
\begin{subequations}
\begin{align}
u_{a,b} &= \frac{b-\lambda}{2E}\pm \sqrt{\left(\frac{b-\lambda}{2E}\right)^2 +\frac{b}{E} },\\
u_a+u_b &= \frac{b-\lambda}{E},\\
u_a u_b &= -\frac{b}{E}.
\end{align}
\end{subequations}
Now,
\begin{subequations}
\begin{align}
V' &= \frac{dV}{du}u'\\
&= \left[\frac{2E}{u^3}(u-u_a)(u-u_b) -  \frac{E}{u^2}(2u-u_a-u_b) \right] \alpha (u+1).
\end{align}
\end{subequations}

Using Eq.~\eqref{eq:example1},
\begin{equation}
\frac{1}{2 i}\oint  \frac{\sqrt{16 \left( V-E \right)^{3}+\left(V'\right)^{2} }}{4 \left( V-E \right)} dz = \left(K +\frac{1}{2}\right)\pi.
\end{equation}
The poles of the integrand are at $u=u_a,u_b,0,-1,\textrm{ and } \infty$. So, calculating residues at each of the poles with proper principal value,

\begin{eqnarray}
\begin{split}
 -\frac{1}{4}+\frac{1}{4} {-} \frac{1}{2\alpha}\sqrt{\alpha^2-4E u_a u_b}+& \frac{1}{\alpha}\sqrt{-E(1+u_a)(1+u_b)}\\
&{-} \frac{\sqrt{-E}}{\alpha} = K+\frac{1}{2},
 \end{split}\\
\therefore {-} \frac{1}{2}\sqrt{1+\frac{4b}{\alpha^2}}+ \frac{\sqrt{\lambda-E}}{\alpha}{-} \frac{\sqrt{-E}}{\alpha} = K+\frac{1}{2}\hspace{1cm}
\end{eqnarray}

This solution agrees with Romanovski \& Robnik \cite{romanovski2000}.

\subsection*{V. Morse potential}
Finally, consider the potential of the form,
\begin{subequations}
\begin{align}
V(x)&=Ae^{-2\alpha x}-Be^{-\alpha x},\\
V-E &= Ae^{-2\alpha x}-Be^{-\alpha x}-E.
\end{align}
\end{subequations}

Let $u=e^{\alpha x}$, $u' = \alpha u$, which leads us to,
\begin{subequations}
\begin{align}
\therefore V-E &= \frac{1}{u^2} \left[ A-Bu-Eu^2 \right]\\
&= \frac{-E}{u^2} \left[ u^2+\frac{B}{E}u -\frac{A}{E}\right]\\
&= \frac{-E}{u^2} (u-u_a)(u-u_b),
\end{align}
\end{subequations}
with
\begin{subequations}
\begin{align}
u_{a,b} &= -\frac{B}{2E}\pm \sqrt{\left(\frac{B}{2E}\right)^2 +\frac{A}{E} },\\
u_a+u_b &= \frac{-B}{E},\\
u_a u_b &= \frac{-A}{E}.
\end{align}
\end{subequations}
Now,
\begin{subequations}
\begin{align}
V' &= \frac{dV}{du}u'\\
&= \left[\frac{2E}{u^3}(u-u_a)(u-u_b) -  \frac{E}{u^2}(2u-u_a-u_b) \right] \alpha u.
\end{align}
\end{subequations}

Employing Eq.~\eqref{eq:example1},
\begin{equation}
\frac{1}{2 i}\oint  \frac{\sqrt{16 \left( V-E \right)^{3}+\left(V'\right)^{2} }}{4 \left( V-E \right)} dz = \left(K +\frac{1}{2}\right)\pi.
\end{equation}
The poles of the integrand are at $u=u_a,u_b,0,\textrm{ and } \infty$. So, calculating residues at each of the poles with proper principal value,

\begin{align}
 \frac{-1}{4}+\frac{1}{4} + \frac{1}{2 \alpha}\sqrt{\frac{-E}{u_a u_b}}(u_a+u_b) {-}\frac{\sqrt{-E}}{\alpha} &= K+\frac{1}{2},\\
\therefore\frac{B}{2\alpha \sqrt{A}} {-}\frac{\sqrt{-E}}{\alpha} = K+\frac{1}{2}.
\end{align}

This solution agrees with Romanovski \& Robnik \cite{romanovski2000}.


text:
{\bf Problem.} Let the first term of a geometric sequence be $\frac{3}{4}$, and let the second term be $15$. What is the smallest $n$ for which the $n$th term of the sequence is divisible by one million?
{\bf Level.} Level 5
{\bf Type.} Algebra
{\bf Solution.} The common ratio is $$\frac{15}{\frac{3}{4}} = 20$$Therefore, the $n$th term is $(20^{n-1}) \left(\frac{3}{4}\right)$.

If one million (aka $10^6$) divides the $n$th term, then it must be divisible by $5^6$. This can only happen if $n-1$ is at least $6$, or $n \ge 7$.

The $7$th term is $$\left(20^6\right) \left(\frac{3}{4}\right) = \left(4\right)^6\left(5\right)^6\left(\frac{3}{4}\right) = (2)^{10}(5)^6(3),$$which is divisible by $(2)^6(5)^6=10^6$, so the answer is indeed $\boxed{7}$.

set_name: MATH


TITLE: Expressing the integral $\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{\left(1-x^3\right)\left(1-a^6x^3\right)}}$ in terms of elliptic integrals
QUESTION [28 upvotes]: Define the function $\mathcal{I}:\left(0,1\right)\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(p\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{4p}{\sqrt{\left(1-x^{3}\right)\left(1-p^{6}x^{3}\right)}}.\tag{1}$$

Question: Given $p\in\left(0,1\right)$, is there a (relatively) simple way to express $\mathcal{I}{\left(p\right)}$ in terms of elliptic integrals in Legendre normal form? Can $\mathcal{I}{\left(p\right)}$ possibly be written as a simple scalar multiple of a complete elliptic integral of the first kind?

Recall that for amplitude $\theta\in\left(0,\frac{\pi}{2}\right)$ and elliptic modulus $\kappa\in\left(0,1\right)$, the incomplete elliptic integral of the first kind $F{\left(\theta,\kappa\right)}$ is defined as
$$F{\left(\theta,\kappa\right)}:=\int_{0}^{\theta}\mathrm{d}\varphi\,\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}}.\tag{2}$$

Motivation:
I have noticed that of the questions on this site related to proving special closed form values of the Gauss hypergeometric function, ${_2F_1}{\left(a,b;c;z\right)}$, an unusually large number of the more difficult ones can be shown equivalent to proving the particular hypergeometric function with parameters $a=\frac12$, $b=\frac13$, and $c=\frac56$, i.e., ${_2F_1}{\left(\frac12,\frac13;\frac56;z\right)}$, takes some special (usually algebraic) value for some particular (usually rational) value of $z$, where $0<z<1$. So in the interest of generalization, I wanted to see how close one could get to obtaining a closed form for $f(z):={_2F_1}{\left(\frac12,\frac13;\frac56;z\right)}$ for arbitrary values of $z\in(0,1)$.
Using Euler's integral representation for the Gauss hypergeometric function with the assumption that $0<z<1$, we have:
$$\begin{align}
f(z)
&={_2F_1}{\left(\frac12,\frac13;\frac56;z\right)}\\
&=\frac{1}{\operatorname{B}{\left(\frac13,\frac12\right)}}\int_{0}^{1}t^{-\frac23}(1-t)^{-\frac12}(1-zt)^{-\frac12}\mathrm{d}t\\
&=\frac{3}{\operatorname{B}{\left(\frac13,\frac12\right)}}\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{(1-x^3)(1-zx^3)}},\tag{3}\\
\end{align}$$
where the beta function factor can alternatively be written in terms of the gamma function as,
$$\operatorname{B}{\left(\frac13,\frac12\right)}=\frac{\sqrt{3}}{2^{4/3}\pi}\left[\Gamma{\left(\frac13\right)}\right]^3.$$
The integral factor in the last line of $(3)$ is already superficially quite similar to the following integral representation of the complete elliptic integral of the first kind:
$$K(k)=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}.$$
Moreover, the last line of $(3)$ implies that in order for $f(z)$ to take an algebraic value, the integral factor will be the product of an algebraic number $A$ with an elliptic integral singular value:
$$\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{(1-x^3)(1-zx^3)}}=A\,K{\left(k_3\right)}=A\frac{3^{1/4}}{2^{7/3}\pi}\left[\Gamma{\left(\frac13\right)}\right]^3.$$
This made me suspect that the integral $\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{(1-x^3)(1-zx^3)}}=\mathcal{I}{(z^{1/6})}$ might be capable of being written as a linear combination of complete elliptic integrals in general.

REPLY [17 votes]: General Procedure:

I will give an outline of a possible procedure (that I learned from a calculus book some time ago) to solve integrals of the form
$$\mathcal{J}=\int\frac{{\rm d}x}{\sqrt{(x^2+ax+b)(x^2+cx+d)}}$$
First, let $x=t+\lambda$.
\begin{align}
\mathcal{J}=\int\frac{{\rm d}t}{\sqrt{(t^2+(2\lambda+a)t+\lambda^2+\lambda a+b)(t^2+(2\lambda+c)t+\lambda^2+\lambda c+d)}}
\end{align}
Next, let $\lambda=\dfrac{d-b}{a-c}$, then make the substitution $\displaystyle t=\mu v$ where $\displaystyle \mu=\sqrt{\lambda^2+\frac{ad-bc}{a-c}}$.
$$\mathcal{J}=\frac{1}{\mu}\int\frac{{\rm d}v}{\sqrt{\left(v^2+Av+1\right)\left(v^2+Cv+1\right)}}$$
where $\displaystyle A=\frac{2\lambda+a}{\mu}$ and $\displaystyle C=\frac{2\lambda+c}{\mu}$. Follow this up with the substitution $\displaystyle v=\frac{1+w}{1-w}$.
$$\mathcal{J}=\frac{2}{\mu}\int\frac{{\rm d}w}{\sqrt{((2-A)w^2+(2+A))((2-C)w^2+(2+C))}}$$
By applying the appropriate trigonometric substitution, it can be seen that $\mathcal{J}$ can be reduced to incomplete elliptic integrals.

Parameters

I will define the following parameters to save space and typing.
\begin{align}
\alpha_1=&\ \frac{1}{\sqrt{\left(a+1\right)^2\left(3a^2+2\sqrt{3}\sqrt{a^4+a^2+1}+3\right)}}\\
\alpha_2=&\ \frac{1}{\sqrt{\left(a-1\right)^2\left(3a^2+2\sqrt{3}\sqrt{a^4+a^2+1}+3\right)}}\\
\beta_1=&\ \frac{\sqrt{3}a^2-\sqrt{a^4+a^2+1}+\sqrt{3}a+\sqrt{3}}{\sqrt{3}a^2+\sqrt{a^4+a^2+1}+\sqrt{3}a+\sqrt{3}}\\
\beta_2=&\ \frac{\sqrt{3}a^2-\sqrt{a^4+a^2+1}-\sqrt{3}a+\sqrt{3}}{\sqrt{3}a^2+\sqrt{a^4+a^2+1}-\sqrt{3}a+\sqrt{3}}\\
\xi_1=&\ \frac{\sqrt{a^4+a^2+1}-\sqrt{3}a}{\sqrt{a^4+a^2+1}+\sqrt{3}a}\\
\xi_2=&\ \frac{\sqrt{a^4+a^2+1}+\sqrt{3}a}{\sqrt{a^4+a^2+1}-\sqrt{3}a}
\end{align}

Evaluation of $\ \mathcal{I}(a)$

It was derived by the OP that
\begin{align}
\mathcal{I}(a)
=&\underbrace{\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a-1)^2}{2a}x-\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}}_{\mathcal{I}_1}\\
&+\underbrace{\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a+1)^2}{2a}x+\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}}_{\mathcal{I}_2}
\end{align}
Applying the aforementioned procedure and using Wolfram Alpha to do all the algebra, we get
\begin{align}
\mathcal{I}_1
=&\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a-1)^2}{2a}x-\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}\\
=&\frac{1}{\sqrt{3}\sqrt{a^4+a^2+1}}\int^\infty_\frac{\sqrt{3}(a^2+a+1)}{\sqrt{a^4+a^2+1}}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{2(2a^2+a+2)}{\sqrt{3}\sqrt{a^4+a^2+1}}x+1\right)\left(x^2-\frac{2\sqrt{3}\ a}{\sqrt{a^4+a^2+1}}+1\right)}}\\
=&\alpha_1\int^1_{\beta_1}\frac{{\rm d}x}{\sqrt{\left(x^2-\beta_1^2\right)\left(x^2+\xi_1\right)}}
=\alpha_1\int^{\arccos{\beta_1}}_{0}\frac{\sec{\varphi}\tan{\varphi}}{\sqrt{(\sec^2{\varphi}-1)(\beta_1^2\sec^2{\varphi}+\xi_1)}}{\rm d}\varphi\\
=&\alpha_1\int^{\arccos{\beta_1}}_{0}\frac{{\rm d}\varphi}{\sqrt{\beta_1^2+\xi_1\cos^2{\varphi}}}
=\alpha_1\int^{\arccos{\beta_1}}_{0}\frac{{\rm d}\varphi}{\sqrt{(\beta_1^2+\xi_1)-\xi_1\sin^2{\varphi}}}\\
=&\frac{\alpha_1}{\sqrt{\beta_1^2+\xi_1}}\int^{\arccos{\beta_1}}_{0}\frac{{\rm d}\varphi}{\sqrt{1-\frac{\xi_1}{\beta_1^2+\xi_1}\sin^2{\varphi}}}
=\color{red}{\frac{\alpha_1}{\sqrt{\beta_1^2+\xi_1}}\mathbf{F}\left(\arccos{\beta_1},\sqrt{\frac{\xi_1}{\beta_1^2+\xi_1}}\right)}
\end{align}
for $\mathcal{I}_1$, and we obtain (I omit the steps of trigonometric manipulations since they are the same)
\begin{align}
\mathcal{I}_2
=&\frac{1}{4a}\int^\infty_\frac{a^2+1}{2a}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{(a+1)^2}{2a}x+\frac{a^2+1}{2a}\right)\left(x^2+\frac{a^2+1}{2a}x+\frac{a^4-a^2+1}{4a^2}\right)}}\\
=&\frac{1}{\sqrt{3}\sqrt{a^4+a^2+1}}\int^\infty_{\frac{\sqrt{3}(a^2-a+1)}{\sqrt{a^4+a^2+1}}}\frac{{\rm d}x}{\sqrt{\left(x^2-\frac{2(2a^2-a+2)}{\sqrt{3}\sqrt{a^4+a^2+1}}x+1\right)\left(x^2+\frac{2\sqrt{3}\ a}{\sqrt{a^4+a^2+1}}x+1\right)}}\\
=&\alpha_2\int^1_{\beta_2}\frac{{\rm d}x}{\sqrt{\left(x^2-\beta_2^2\right)\left(x^2+\xi_2\right)}}=\color{blue}{\frac{\alpha_2}{\sqrt{\beta_2^2+\xi_2}}\mathbf{F}\left(\arccos{\beta_2},\sqrt{\frac{\xi_2}{\beta_2^2+\xi_2}}\right)}
\end{align}
Therefore
$$\mathcal{I}(a)=\frac{\alpha_1}{\sqrt{\beta_1^2+\xi_1}}\mathbf{F}\left(\arccos{\beta_1},\sqrt{\frac{\xi_1}{\beta_1^2+\xi_1}}\right)+\frac{\alpha_2}{\sqrt{\beta_2^2+\xi_2}}\mathbf{F}\left(\arccos{\beta_2},\sqrt{\frac{\xi_2}{\beta_2^2+\xi_2}}\right)$$
I have substituted in some random values of $a$ into $\mathcal{I}_1$ and it seems to match numerically. However, I have not verified $\mathcal{I}_2$ yet so there might be a mistake.


TITLE: Compute $\operatorname{Hom}_{\mathbb{Z}} ( \mathbb{Q}, \mathbb{Z})$
QUESTION [12 upvotes]: I think this question could be a little idiot, however I could not solve this after some hours. I need to find all homomorphism between the additive group $\mathbb{Q}$ and the additive group $\mathbb{Z}$, i.e., $\operatorname{Hom}_{\mathbb{Z}}  ( \mathbb{Q}, \mathbb{Z})$. I tried to find a good set of generators for $\mathbb{Q}$, however none of them  satisfied a general homomorphism.

REPLY [9 votes]: $\,\Bbb Q\,$ is a divisible group , so is any of its homomorphic images. But $\,\Bbb Z\,$ is not a divisible group (why?) , so it must be that the image of any group homomorphism  $\,\Bbb Q\to\Bbb Z\,$ is the trivial subgroup (which, BTW, is the only finite group which is divisible and the only subgroup of $\,\Bbb Z\,$ that is divisible), and thus $\,\operatorname{Hom}_{\Bbb Z}(\Bbb Q,\Bbb Z)=0\,$


text:
\begin{document}

\title{Knoto-ID: a tool to study the entanglement of open protein chains using the concept of knotoids}

\author{Julien Dorier}
\affiliation{Vital-IT, SIB Swiss Institute of Bioinformatics, 1015 Lausanne, Switzerland}
\author{Dimos Goundaroulis}
\affiliation{Center for Integrative Genomics, University of Lausanne, 1015 Lausanne, Switzerland}
\affiliation{Swiss Institute of Bioinformatics, 1015 Lausanne, Switzerland}
\author{Fabrizio Benedetti}
\affiliation{Center for Integrative Genomics, University of Lausanne, 1015 Lausanne, Switzerland}
\affiliation{Vital-IT, SIB Swiss Institute of Bioinformatics, 1015 Lausanne, Switzerland}
\author{Andrzej Stasiak}
\affiliation{Center for Integrative Genomics, University of Lausanne, 1015 Lausanne, Switzerland}
\affiliation{Swiss Institute of Bioinformatics, 1015 Lausanne, Switzerland}

  \begin{abstract}
 \noindent {\footnotesize The authors wish it to be known that, in their opinion, the first two authors should be regarded as joint First Authors.} \\

  \noindent \textbf{Abstract:}  The backbone of most proteins forms an open curve.
    To study their entanglement, a common strategy consists in searching for the presence of knots in their backbones using topological invariants.
    However, this approach requires to close the curve into a loop, which alters the geometry of curve.
    Knoto-ID allows evaluating the entanglement of open curves without the need to close them, using the recent concept of knotoids which is a generalization of the classical knot theory to open curves.
    Knoto-ID can analyse the global topology of the full chain as well as the local topology by exhaustively studying all subchains or only determining the knotted core.\\
    \textbf{Availability and Implementation:}  Knoto-ID is written in C++ and includes R (\href{www.R-project.org}{www.R-project.org}) scripts to generate plots of projections maps, fingerprint matrices and disk matrices.
    Knoto-ID is distributed under the GNU General Public License (GPL), version 2 or any later version and is available at \href{https://github.com/sib-swiss/Knoto-ID}{https://github.com/sib-swiss/Knoto-ID}. A binary distribution for Mac OS X and Linux with detailed user guide and examples can be obtained from \href{https://www.vital-it.ch/software/Knoto-ID}{http://www.vital-it.ch/software/Knoto-ID}.\\
  \textbf{Contact:} \href{julien.dorier@sib.swiss}{julien.dorier@sib.swiss}

\end{abstract}

\maketitle

\begin{figure*}[!tbp]
\centerline{\includegraphics{figure1.eps}}
\caption{ (a) Projection map on a plane using spherical coordinates or (b) on a globe using Cartesian coordinates. (c) Fingerprint matrix and (d) a disc matrix. This figure was generated with Knoto-ID using the protein 3KZN (\cite{shi2006}) as input.}\label{fig:01}
\end{figure*}

\section{Introduction}

The observation that protein backbones can form knots(\cite{mans}) initiated numerous studies of their nature and potential advantages or disadvantages that they may provide (e.g. \cite{virnau, dabrstas}). In this context, it was important to classify protein knots in terms of their topology.

A knot is a closed curve in 3-dimensional space that doesn't intersect itself and it can be freely deformed as long as it does not pass through itself (\cite{adams}). However, the backbone of many biomolecules and specifically of many proteins correspond to open spatial curves and so, in strict topological sense, such curves are classified as unknotted. Until recently, the only way to study the topology of an open protein chain was to first close them and then proceed with the study of the entanglement. Of course, closing the chain alters its  geometry. In 2012, V. Turaev introduced the concept of knotoids as a generalization of the classical knot theory to open knots (\cite{turaev}). Knotoids were studied further by N. G\"ug\"umc\"u and L. H. Kauffman (\cite{guka}).  As a consequence, a number of studies emerged that implemented this new mathematical tool in the analysis of a protein backbone (\cite{dennis, dgound, dgound2}).

 In this note we introduce Knoto-ID, a command line tool that is able to analyse and classify  open spatial curves using this new mathematical concept. Moreover, we provide the possibility of closing an open 3D curve, if a knot analysis is required, with either a direct closure (e.g. \cite{taylor}) or using the uniform closure technique (e.g. \cite{sulk}). This note focuses on individual open protein chains, however Knoto-ID can be used to analyse any open linear conformation in 3-space such as chromosomes (\cite{siebert2017}), synthetic polymers, random walks.

\section{Implementation}
To analyse a protein, the coordinates of the $C\alpha$ atoms of the protein backbone have to be extracted from a .pdb file downloaded from the PDB (\href{www.rcsb.org}{www.rcsb.org}, \cite{pdb}) and then stored to a text file, which is the input of Knoto-ID. The backbone is then placed inside a large enough enclosing sphere. Each point of the sphere defines a direction of projection. For a given direction of projection, two infinite lines are introduced that pass from each of the termini of the curve and are parallel to the chosen direction. A triangle elimination method based on the KMT algorithm (\cite{koniaris, taylor}) is then applied to simplify the curve while preserving the its underlying topology with respect to the two parallel lines.
A knotoid diagram is obtained by projecting the curve on an oriented surface (a plane or a sphere).
Finally, a topological invariant is evaluated on the knotoid diagram. For curves projected on a sphere the topological invariant is the Jones polynomial for knotoids (\cite{turaev}), while for curves projected on a plane it is the Turaev loop bracket polynomial (\cite{turaev}) (see the Knoto-ID user guide for a brief description of the theory). The knotoid type corresponding to the resulting polynomial can be optionally obtained using a list knotoid types distributed with Knoto-ID.
Different choices of projection directions may yield different  diagrams and so one has to sample an adequate number of projection directions in order to approximate the spectrum of knotoid types that corresponds to the spatial curve. The spectrum of knotoid types can be visualized using projection maps generated by Knoto-ID (Figure~\ref{fig:01}a and \ref{fig:01}b).
Knoto-ID is also able to handle closed chains as input, or to create a closed chain from an open one using either direct and uniform closure.

Knoto-ID is also able to analyse all subchains of a given curve to produce a fingerprint matrix (\cite{yeates}), for the case of open chains or a disc matrix (\cite{rawdon}), for the case of closed chain (see Figure~\ref{fig:01}c and \ref{fig:01}d). In addition, Knoto-ID can also find the knotted core of the chain, which is the shortest subchain obtained by progressively altering the length of the input chain by 1 point without changing the dominant knot or knotoid type in the process.

\section{Conclusion}
Knoto-ID is the first tool that is able to handle, analyse as well as classify open linear conformations in 3-space such as proteins in terms of their topology  without requiring them to be closed into a loop, using the concept of knotoids.\vspace*{-17pt}

\section*{Acknowledgements}
This is the Author's Original Version of the  article has been accepted for publication in Bioinformatics Published by Oxford University Press.

We thank Louis H. Kauffman for fruitful discussions, Frédéric Sch\"utz for his advice on Knoto-ID packaging. We also thank Eric Rawdon, Elizabeth Annoni and Nicole Lopez for kindly providing the list of projections distributed with Knoto-ID.\vspace*{-2pt}

\section*{Funding}
The work was funded in part by Leverhulme Trust (RP2013-K-017) and by the Swiss National Science Foundation (31003A-138267), both credited to Andrzej Stasiak. \vspace*{-14pt}

config: arxiv


TITLE: value of $abc$ in algebraic expression
QUESTION [1 upvotes]: If $a+b+c=0$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 3$. where $a,b,c$ are non zero real numbers
then $ab(a+b)+bc(b+c)+ca(c+a)$ is
Attempt: from $a+b+c=0$ we have to find value of $-3abc$
and from $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 3$ we have $\displaystyle \frac{ab+bc+ca}{abc} = 3$
could some help how to get value of $abc,$ thanks

REPLY [3 votes]: We have $$(a+b+c)^2 =a^2+b^2+c^2+2 (ab+bc+ca) =0$$ $$\Rightarrow ab+bc+ca = -0.5 (a^2+b^2+c^2) $$
We already know $ab + bc + ca =3abc $, so hope you can take it from here.
On a side note, observe that $$ (a+b+c)^3 =a^3+b^3+c^3+ (a+b+c)(ab+bc+ca) \Rightarrow a^3+b^3+c^3=0$$


TITLE: Help me understand about the calculation of the Dominium in a function (?)
QUESTION [0 upvotes]: I really don't understand why the argument of the natural logarithm is greater than or equal to one. I don't know if this question sounds stupid, and I apologize in that case, but I am very new to funcions, so I am wondering if someone can explain me this passage here. Thanks a lot!

REPLY [1 votes]: Crostul explains it well (in Italian), but for anyone not being able to decipher it, it is roughly:
If we have $\ln(t) \geq 0$, it follows (from exponentiating both sides) that $$e^{\ln(t)}=t\geq e^0=1$$


TITLE: Bounding a convergent sum
QUESTION [2 upvotes]: Let $\lg = \log_2$ and $n$ be a positive integer. I am trying to find an upper bound on the growth rate of the following sum:
$$S_n=\sum_{i=1}^{n} \frac{1}{2^ii}e^{-n/2^i}$$
Clearly the sum converges, and furthermore it seems like $\lim_{n\to \infty}S_n = 0$. It seems like the terms achieve a maximum around $i\approx \lg(n)$ and most other terms are much smaller (i.e., don't contribute much to the sum). Given this, it seems like we have $$S_n \leq c\frac{1}{n\lg n}$$ for some $c>0$.
If we were to omit the factor of $i$ in the denominator, then the sum could be easily bounded by integrating $e^{-nx}$ over a suitable interval. Similarly, if we were to drop the factor of $e^{-n/2^i}$, then we could bound the sum by the closed form for $\sum\limits_{i=1}^\infty \frac{x^i}{i} \bigg|^{x=\frac{1}{2}}$. But with both of them together, I am not sure how to proceed.

REPLY [4 votes]: For each $n$, let $f_n(x) = \frac{\exp(-n/2^x)}{2^x x}$. Then we can show that(1),(2)
$$ \left| \sum_{i=1}^{n} f_n(i) - \int_{1}^{n} f_n(x) \, \mathrm{d}x \right| \leq \max_{1\leq x \leq n} f_n(x). $$
(This is essentially becaues $f_n$ is unimodal.) To estimate the maximum, we note that
\begin{align*}
f_n'(x) = 0
&\quad\iff\quad (\log f_n(x))' = 0 \\
&\quad\iff\quad \frac{n}{2^x} = 1+ \frac{1}{x \log 2} \\
&\quad\iff\quad \log n - x \log 2 = \log\left(1 + \frac{1}{x \log 2}\right).
\end{align*}
Inspecting the last equation, it is not hard to show that the equation has a unique zero $x_*$ in $[1, \infty)$. Moreover, for $n$ large enough,
\begin{align*}
x = \log_2(n/2) &\quad\implies\quad \frac{n}{2^x} = 2 > 1+ \frac{1}{\log(n/2)} = 1+ \frac{1}{x \log 2}, \\
x = \log_2 n &\quad\implies\quad \frac{n}{2^x} = 1 < 1+ \frac{1}{\log n} = 1+ \frac{1}{x \log 2},
\end{align*}
So $\log_2(n/2) < x_* < \log_2 n$. Using this, we obtain
$$ \max_{1\leq x \leq n} f_n(x) \leq \frac{e^{-1}}{(n/2) \log_2(n/2)} $$
Moreover, substituting $u = n/2^x$, or equivalently, $x = \log_2(n/u)$, we get
\begin{align*}
\int_{1}^{n} f_n(x) \, \mathrm{d}x
&= \int_{n/2^n}^{n/2} \frac{e^{-u}}{n \log(n/u)} \, \mathrm{d}u
= \frac{1}{n\log n} \int_{n/2^n}^{n/2} \frac{e^{-u}}{1 - \frac{\log u}{\log n}} \, \mathrm{d}u.
\end{align*}
It is not hard to show that this integral is asymptoticall $\frac{1}{n\log n}$ as $n \to \infty$. Combining altogether and letting $A_n = \left( n \log n \right) S_n$, we get
$$ 1 - \frac{2\log 2}{e} \leq \liminf_{n\to\infty} A_n \leq \limsup_{n\to\infty} A_n \leq 1 + \frac{2\log 2}{e}. $$
Since $\frac{2 \log 2}{e} \approx 0.509989$, this shows that $S_n = \Theta\left(\frac{1}{n\log n} \right) $.
(Simulation seems suggesting that $A_n$ actually converges as $n\to\infty$, but I will leave this question to others.)


TITLE: Uniformly continuous in the compact variable
QUESTION [1 upvotes]: What is this theorem/where can I find a proof of the following?

Let $f: K \times X \to Y$ be continuous, with $K$ compact. Using uniform continuity ideas, for $x \in X$ and $\epsilon > 0$, there exists $\delta = \delta_{x, \epsilon} > 0$ such that $\|x - x'\| < \delta$ forces $\|f(t, x) - f(t, x') \| < \epsilon$ for all $t \in K$. That is, we can pick $\delta$ in a manner which is uniform across $K$ for fixed $x \in X$ and $\epsilon > 0$. In other words, $f$ is "uniformly continuous in the compact variable."
Using this, we can conclude that for a continuous map $f: [a, b] \times X \to \mathbb{R}$, the function $X \to \mathbb{R}$ defined by$$x \mapsto \int_a^b f(t, x)\,dt$$is continuous.

REPLY [0 votes]: We first justify the application to integrals. For $f: [a, b] \times X \to \mathbb{R}$ a continuous map, the function $f_x: [a, b] \to V$ given by $t \mapsto f(t, x)$ is continuous, so the integral $I(x) = \int_a^b f_x = \int_a^b f(t, x)\,dt$ makes sense. Using the estimate$$|I(x') - I(x)| \le \int_a^b |f(t, x) - f(t, x')|\,dt \le \epsilon(b-a)$$of $\|x' - x\| < \delta = \delta_{x, \epsilon}$ with $\delta$ as in the main part of the problem, we deduce the desired continuity of $I$.
Now we turn to the construction of $\delta_{x, \epsilon}$ for continuous maps $f: K \times X \to Y$ with $K$ compact. We fix $x$ and $\epsilon$, and we use $\|\cdot\|$ to denote the norm arising from choices of inner products on the ambient vector spaces containing each of $K$, $X$, and $Y$. For each $t \in K$, we get some $\delta_t > 0$ such that$$\max(\|t - t'\|, \|x - x'\|) < \delta_t \implies \|f(t, x) - f(t', x') \| < \epsilon.$$The opens $U_t = B_{\delta_t/2}(t)$ form an open covering of the compact $K$, so there is some finite sub covering $U_{t_1}, \dots, U_{t_n}$. Let $\delta = \min \delta_{t_j} > 0$. Now pick any $x' \in X$ with $\|x - x'\| < \delta$, and we wish to prove$$\|f(t, x) - f(t, x') \| < \epsilon$$for all $t \in K$. Fix a choices of $t$, so $t \in U_{t_j} = B_{\delta_j}(t_j)$ for some $j$. We have the estimate$$\|f(t, x) - f(t, x')\| \le \|f(t, x) - f(t_j, x)\| + \|f(t_j, x) - f(t, x')\|$$with the second term on the right at most $\epsilon$ since $\max(\|t_j - t\|, \|x - x'\|) < \max(\delta_{t_j}, \delta) \le \delta_{t_j}$. Meanwhile, we have$$\max(\|t - t_j\|, \|x - x\|) = \|t - t_j\| < \delta_{t_j},$$so $\|f(t, x) - f(t_j, x)\| \epsilon$. Putting everything together, when $\|x - x'\| < \delta$, we get$$\|f(t, x) - f(t, x')\| \le 2\epsilon$$for all $t \in K$. This completes the proof.


TITLE: Deformation theory of representations of an algebraic group
QUESTION [29 upvotes]: For an algebraic group G and a representation V, I think it's a standard result (but I don't have a reference) that

the obstruction to deforming V as a representation of G is an element of H2(G,V⊗V*)
if the obstruction is zero, isomorphism classes of deformations are parameterized by H1(G,V⊗V*)
automorphisms of a given deformation (as a deformation of V; i.e. restricting to the identity modulo your square-zero ideal) are parameterized by H0(G,V⊗V*)

where the Hi refer to standard group cohomology (derived functors of invariants). The analogous statement, where the algebraic group G is replaced by a Lie algebra g and group cohomology is replaced by Lie algebra cohomology, is true, but the only proof I know is a big calculation. I started running the calculation for the case of an algebraic group, and it looks like it works, but it's a mess. Surely there's a long exact sequence out there, or some homological algebra cleverness, that proves this result cleanly. Does anybody know how to do this, or have a reference for these results? This feels like an application of cotangent complex ninjitsu, but I guess that's true about all deformation problems.
While I'm at it, I'd also like to prove that the obstruction, isoclass, and automorphism spaces of deformations of G as a group are H3(G,Ad), H2(G,Ad), and H1(G,Ad), respectively. Again, I can prove the Lie algebra analogues of these results by an unenlightening calculation.
Background: What's a deformation? Why do I care?
I may as well explain exactly what I mean by "a deformation" and why I care about them. Last things first, why do I care? The idea is to study the moduli space of representations, which essentially means understanding how representations of a group behave in families. That is, given a representation V of G, what possible representations could appear "nearby" in a family of representations parameterized by, say, a curve? The appropriate formalization of "nearby" is to consider families over a local ring. If you're thinking of a representation as a matrix for every element of the group, you should imagine that I want to replace every matrix entry (which is a number) by a power series whose constant term is the original entry, in such a way that the matrices still compose correctly. It's useful to look "even more locally" by considering families over complete local rings (think: now I just take formal power series, ignoring convergence issues). This is a limit of families over Artin rings (think: truncated power series, where I set xn=0 for large enough n).
So here's what I mean precisely. Suppose A and A' are Artin rings, where A' is a square-zero extension of A (i.e. we're given a surjection f:A'→A such that I:=ker(f) is a square-zero ideal in A'). A representation of G over A is a free module V over A together with an action of G. A deformation of V to A' is a free module V' over A' with an action of G so that when I reduce V' modulo I (tensor with A over A'), I get V (with the action I had before). An automorphism of a deformation V' of V as a deformation is an automorphism V'→V' whose reduction modulo I is the identity map on V. The "obstruction to deforming" V is something somewhere which is zero if and only if a deformation exists.
I should add that the obstruction, isoclass, and automorphism spaces will of course depend on the ideal I. They should really be cohomology groups with coefficients in V⊗V*⊗I, but I think it's normal to omit the I in casual conversation.

REPLY [5 votes]: About what Anton said at the end about deformations of a group.    Let $m_0$ be the standard multiplication.  Then I want to consider a deformation of the form $m:(G \times \epsilon \mathfrak{g}) \times (G \times \epsilon \mathfrak{g}) \to G \times \epsilon \mathfrak{g}$ where $m(g_1, g_2) = m_{0}(g_1,g_2) + \epsilon m_1 (g_1,g_2)$.  When you write out the associativity condition $m\circ (m \times 1) = m \circ (1 \times m)$ it seems that you find that
$(g_1,g_2) \mapsto (m_{1}(g_{1},g_{2}))(g_{1}g_{2})^{-1}$ is a group cohomology cocycle for G acting on $\mathfrak{g}$ by the adjoint representation.  Now one has to identify $H^{2}(G,Ad)$ with $H^{2}(BG,Ad)$ (taking care of the topology somehow).


text:
\begin{document}

\begin{abstract}
  Is every locally compact abelian group which admits a Heisenberg central extension isomorphic to the product of a locally compact abelian group and its Pontryagin dual?
  An affirmative answer is obtained for all the commonly occurring types of abelian groups having Heisenberg central extensions, including Lie groups and certain finite Cartesian products of local fields and ad\`eles.
  Furthermore, for these types of groups, it is found that the isomorphism class of the abelian group determines the Heisenberg group up to isomorphism, thereby providing a classification of such Heisenberg groups.
\end{abstract}

\maketitle

\section{Introduction}\label{sec:introduction}
In his mathematical formulation of quantum kinematics \cite[Chap.~IV, Sec.~14]{Weyl50}, Hermann Weyl introduced the concept of Heisenberg group as a special kind of central extension of an abelian group (which is thought of as phase space) by the circle group.
His formulation allowed the phase space to be quite general, and besides real vector spaces, he considered finite abelian groups.
A natural class of Heisenberg groups are the standard Heisenberg groups defined in Section~\ref{sec:defin-heis-groups}, where the phase space is the product of a locally compact abelian group (thought of as position space) with its Pontryagin dual (thought of as momentum space).
However, the theory of Heisenberg groups can be formulated independently of such a decomposition (see, for example \cite{MR1116553}).
In some important applications, the phase space associated to a Heisenberg group does not come with any natural decomposition into position and momentum spaces.
In fact, it is not known to the authors whether such a decomposition exists in general.

In this article it is shown that, in a large number of cases including many of the standard examples, such a decomposition of the phase space exists and that the Heisenberg group is isomorphic to a standard Heisenberg group.
In fact, the isomorphism class of the phase space (as a locally compact abelian group) determines the Heisenberg group up to isomorphism.
We rely on known structure theorems for certain families of locally compact abelian groups.
Since no structure theorem is available for locally compact abelian groups in general (even self-dual ones), new ideas may be needed to tackle the general case.

The organization of this paper is as follows:
In Section~\ref{sec:defin-heis-groups}, we recall the definition of Heisenberg group following \cite{MR1116553} and introduce standard Heisenberg groups.
In Section~\ref{sec:heis-lie-groups}, we treat the case where the phase space is a Lie group.
Isomorphism classes of Lie Heisenberg groups are found to be in canonical bijective correspondence with isomorphism classes of abelian Lie groups with finitely many connected components.
In Section~\ref{sec:divis-tors-free}, we treat the case where the phase space is a divisible and torsion-free locally compact abelian group.
This includes Heisenberg groups whose phase space is a finite Cartesian product of local fields of characteristic zero and ad\`eles of algebraic number fields.
Isomorphism classes of Heisenberg groups with divisible torsion-free phase space are found to be in canonical bijective correspondence with divisible torsion free groups whose identity component is a Lie group.
In Section~\ref{sec:groups-prime-expon}, we treat the case where the phase space is of prime exponent (every non-trivial element has order $p$ for a fixed prime number $p$).
This includes Heisenberg groups whose phase space is a finite Cartesian product of local fields and ad\`eles of global function fields of characteristic $p$.
Isomorphism classes of Heisenberg groups with phase space of exponent $p$ are found to be in canonical bijective correspondence with direct products of cyclic groups of order $p$.

Our results provide a simple minimal framework within which the study of many questions involving a large class of Heisenberg groups can be carried out.
The classification of finite Heisenberg groups can be interpreted as the fact that the number of irreducible $n$\dash dimensional abelian groups of unitary rotations in ray space (see \cite[p.~276]{Weyl50}) is the same as the number of isomorphism classes of abelian groups of order $n$ (see \cite{cow}).

\section{Definition of Heisenberg Group}
\label{sec:defin-heis-groups}
Let $K$ be a locally compact abelian group, and
$U(1) =\{ z \in \C \: :\: \abs{z} = 1 \}$.
Consider a locally compact group $G$ which is a central extension:
\begin{equation}\label{hg}
\xymatrix{
1 \ar[r] & U(1) \ar[r]^i & G \ar[r]^\pi & K \ar[r] & 0
}
\end{equation}
The commutator map $G\times G\to U(1)$ defined by $(g,h)\mapsto ghg^{-1}h^{-1}$ descends to an alternating bicharacter $e:K\times K\to U(1)$ known as the commutator form.
Thus, $e$ is a homomorphism in each variable separately, and $e(k,k)=1$ for all $k\in K$.
Let $\widehat{K}$ denote
the Pontryagin dual of $K$.  Then a homomorphism $e^\flat: K \to \widehat{K}$
is obtained by setting
$$
(e^\flat(l))(k) = e(k,l).
$$
\begin{defn*}
  If $e^\flat$ is an isomorphism, then $G$ is called a {\em Heisenberg group}.
\end{defn*}
In this case, $e$ is said to be a non-degenerate alternating bicharacter.
\begin{eg*}[Standard Heisenberg group]
  Let $A$ be a locally compact abelian group.
  Denote its Pontryagin dual by $\hat A$.
  For each $x\in A$, let $T_x$ denote the translation operator $T_xf(y)=f(y-x)$ on $L^2(A)$.
  For each $\chi \in \hat A$, let $M_\chi$ denote the modulation operator $M_\chi f(y)=\chi(y)f(y)$ on $L^2(A)$.
  Consider
  \begin{equation*}
    G=\{zT_xM_\chi\;|\;z\in U(1),\; x\in A\text{ and } \chi \in \hat A\}
  \end{equation*}
  as a subgroup of the group of unitary operators on $L^2(A)$.
  With $i(z)=zI$ for each $z\in U(1)$, $K=A\times \hat A$, and $\pi(zT_xM_\chi)=(x,\chi)$, $G$ is a Heisenberg group as in (\ref{hg}) with commutator form
  \begin{equation*}
    e((x,\chi),(x',\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A \text{ and } \chi,\chi'\in \hat A.
  \end{equation*}
  We call $G$ the standard Heisenberg group associated to $A$.
\end{eg*}
In general, if $K$ is a locally compact abelian group and $c:K\times K\to U(1)$ is a $2$\dash cocycle (often called a multiplier) so that $G$ can be realized as $U(1)\times K$ with the group structure given by
\begin{equation*}
  (z,k)(w,l)=(zwc(k,l), k+l),
\end{equation*}
then $e(k,l)=c(k,l)c(l,k)^{-1}$.
It is not difficult to see that $c\mapsto e$ gives rise to an embedding of $H^2(K,U(1))$ into the group of alternating bicharacters of $K$.
Thus the data $(K,e)$, where $K$ is a locally compact abelian group and $e$ is a non-degenerate alternating bicharacter, determine the Heisenberg group $G$ up to isomorphism of central extensions.
We refer to $K$ as the phase space associated to $G$.
\begin{rmk*}
  If $x\mapsto 2x$ is an automorphism of $K$, then given an alternating bicharacter $e:K\times K\to U(1)$, we may define $c(k,l)=e(k/2,l/2)^2$ to construct a central extension with alternating bicharacter $e$.
  But if $x\mapsto 2x$ is not an automorphism of $K$, it is not clear that every non-degenerate alternating bicharacter is the commutator form of a Heisenberg group.
  When $K$ is Lie or of exponent $2$, this is a consequence of Theorems~A and C.
\end{rmk*}

\section{Lie Heisenberg groups}
\label{sec:heis-lie-groups}
A Lie group will be assumed to have countably many connected components.
\begin{thma}
  Let $K$ be an abelian Lie group and let $e:K\times K\to U(1)$ be a non-degenerate alternating bicharacter.
  Then there exists an abelian Lie group $A$ and an isomorphism $\phi:A\times \hat A\to K$ such that
  \begin{equation*}
    e(\phi(x,\chi),\phi(x',\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A\text{ and }\chi,\chi'\in \hat A.
  \end{equation*}
\end{thma}
\begin{rmka}
  In the above theorem, the isomorphism class of $A$ is not uniquely determined by the isomorphism class of $(K,e)$.
  For example, $A=\Z$ and $A=U(1)$ give rise to isomorphic Heisenberg groups.
  However, $A$ is uniquely determined up to isomorphism if we stipulate that $A$ have only finitely many connected components.
\end{rmka}
\begin{proof}
  $K$ can be decomposed as
  \begin{equation*}
    K=K_f\times K_\R\times K_t \times K_\Z,
  \end{equation*}
  where $K_f$ is a finite abelian subgroup, $K_\R$ is a subgroup isomorphic to the additive group of a finite dimensional real vector space, $K_t$ is a subgroup isomorphic to a compact real torus, and $K_\Z$ is a finitely generated free abelian subgroup.
  Thus $\hat K$ inherits a decomposition
  \begin{equation*}
    \hat K = \hat K_f \times \hat K_\R\times \hat K_t\times \hat K_\Z,
  \end{equation*}
  where $\hat K_f$, $\hat K_\R$, $\hat K_t$ and $\hat K_\Z$ are subgroups of $\hat K$ isomorphic to the Pontryagin duals of $K_f$, $K_\R$, $K_\Z$ and $K_t$ respectively.
  The isomorphism $e^\flat$ can be written as a matrix $(e_{ij})$, where $e_{ij}:K_j\to \hat K_i$ for $i,j\in \{f, \R, t, \Z\}$.
  Many of these entries are forced to be trivial, as there are no non-trivial homomorphisms from some types of groups to others.
  We see immediately that $e^\flat$ has the form:
  \begin{equation*}
    e^\flat=
    \begin{pmatrix}
      e_{ff} & 0 & 0 & e_{f\Z}\\
      0 & e_{\R\R} & 0 & e_{\R\Z}\\
      e_{tf} & e_{t\R} & e_{tt} & e_{t\Z}\\
      0 & 0 & 0 & e_{\Z\Z}
    \end{pmatrix}
    .
  \end{equation*}
  Noting that $(e^\flat)^{-1}$ must also have a similar matrix form, we conclude that $e_{ii}$ must be an isomorphism for each $i\in \{f,\R,t,\Z\}$.
  Given an automorphism $\alpha$ of $K$, let $e^\alpha (k,l)=e(\alpha(k),\alpha(l))$.
  The automorphism $\alpha$ itself can be represented by a matrix $(\alpha_{ij})$ of a similar form, with $\alpha_{ii}$ an automorphism of $K_i$.
  We have
  \begin{equation}
    \label{eq:4}
    (e^\alpha)^\flat = \alpha^* e^\flat \alpha.
  \end{equation}
  Here $\alpha^*: \hat K\to \hat K$ is the transpose of $\alpha$, defined by $\alpha^*(\xi)(k)=\xi(\alpha(k))$ for $k\in K$, $\xi\in \hat K$.
  For the purpose of proving Theorem~A, it is always permissible to replace $e$ by $e^\alpha$.

  Since $e_{ii}$ is an isomorphism for each $i\in \{f,\R,t\}$, we may take $\alpha_{i\Z}:K_\Z\to K_i$ such that $e_{i\Z}=e_{ii}\alpha_{i\Z}$ for each $i\in \{f,\R\}$,
  and $\alpha_{t\Z}:K_\Z\to K_t$ such that
  \begin{equation*}
    e_{tt}\alpha_{t\Z}=-\alpha^*_{f\Z}e_{f\Z}-\alpha^*_{\R\Z}e_{\R\Z}+e_{t\Z}-\alpha^*_{t\Z}e_{\Z\Z}.
  \end{equation*}
  Consider
  \begin{equation}
    \label{eq:3}
    \alpha=
    \begin{pmatrix}
      1 & 0 & 0 & -\alpha_{f\Z}\\
      0 & 1 & 0 & -\alpha_{\R\Z}\\
      0 & 0 & 1 & -\alpha_{t\Z}\\
      0 & 0 & 0 & 1
    \end{pmatrix}
    .
  \end{equation}
  Its transpose, in matrix form, is
  \begin{equation}
    \label{eq:2}
    \alpha^*=
    \begin{pmatrix}
      1 & 0 & 0 & 0\\
      0 & 1 & 0 & 0\\
      -\alpha^*_{f\Z} & -\alpha^*_{\R\Z} & 1 & -\alpha^*_{t\Z}\\
      0 & 0 & 0 & 1
    \end{pmatrix}
  \end{equation}
  Furthermore, the fact that $e$ is alternating means that $(e^\flat)^*(l)(k)=e^\flat(k)(l)=e(l,k)=e(k,l)^{-1}=e(-k,l)=e^\flat(l)(-k)$.
  Consequently,
  \begin{equation}
    \label{eq:1}
    \begin{pmatrix}
      e^*_{ff} & 0 & 0 & e^*_{tf}\\
      0 & e^*_{\R\R} & 0 & e^*_{t\R}\\
      e^*_{f\Z} & e^*_{\R\Z} & e^*_{\Z\Z} & e^*_{t\Z}\\
      0 & 0 & 0 & e^*_{tt}
    \end{pmatrix}
    =
    \begin{pmatrix}
      -e_{ff} & 0 & 0 & -e_{f\Z}\\
      0 & -e_{\R\R} & 0 & -e_{\R\Z}\\
      -e_{tf} & -e_{t\R} & -e_{tt} & -e_{t\Z}\\
      0 & 0 & 0 & -e_{\Z\Z}
    \end{pmatrix}
  \end{equation}
  Using (\ref{eq:4}), (\ref{eq:2}) and (\ref{eq:1}), and by our choice of $\alpha$ in (\ref{eq:3}), we see that
  \begin{equation}
    \label{eq:5}
    (e^\alpha)^\flat=
    \begin{pmatrix}
      e_{ff} & 0 & 0 & 0\\
      0 & e_{\R\R} & 0 & 0\\
      0 & 0 & e_{tt} & 0\\
      0 & 0 & 0 & e_{\Z\Z}
    \end{pmatrix}
    .
  \end{equation}
  Therefore, it suffices to prove Theorem~A in the cases where $K=K_f$, $K=K_\R$ and $K=K_t\times K_\Z$ separately.
  \subsection*{Analysis of the finite part}
  The argument from \cite{iafhg} is repeated here for the sake of completeness.
  Choose a basis $x_1,\ldots,x_n$ of $K_f$ such that the order of $x_i$ is $d_i$ with $d_n|d_{n-1}|\cdots|d_1$.
  Let $\zeta$ denote a fixed primitive $d_1$\dash th root of unity.
  For each $i,j\in \{1,\ldots,n\}$, there exists a non-negative integer $q_{ij}$ such that
  \begin{equation*}
    e(x_i,x_j)=\zeta^{q_{ij}}.
  \end{equation*}
  Observe that $q_{ij}$ is determined modulo $d_1$ and is constrained to be divisible by $d_1/(d_i,d_j)$.
  Let $Q(e)$ denote the matrix $(q_{ij})$ associated to $e$ as above.

  Any automorphism $\alpha_f$ of $K_f$ is of the form
  \begin{equation*}
    \textstyle \alpha_f\big(\sum_{j=1}^n a_jx_j\big) = \sum_{i=1}^n \sum_{j=1}^n \alpha_{ij} a_j x_i
  \end{equation*}
  for a suitable matrix $(\alpha_{ij})$ of integers,
  where $\alpha_{ij}$ is determined modulo $d_i$ and is constrained to be divisible by $d_i/(d_i,d_j)$.
  Furthermore,
  \begin{equation*}
    Q(e^\alpha)=\alpha^* Q(e) \alpha.
  \end{equation*}
  In the above expression, $\alpha$ is identified with its matrix $(\alpha_{ij})$, so that $\alpha^*$ is identified with the transpose matrix.
  In particular, $Q(e^\alpha)$ can be obtained from $Q(e)$ by a sequence of column operations followed by the corresponding sequence of row operations.
  The following operations are permitted:
  \begin{enumerate}
  \item [$\beta_{i\sigma}$: ] multiplication of the $i$th row and $i$th column by an integer $\sigma$ such that $(\sigma,d_i)=1$.
  \item [$\pi_{ij}$: ] interchange of $i$th and $j$th rows and columns when $d_i=d_j$.
  \item [$\alpha_{ij\sigma}$: ] addition of $\sigma$ times $i$th row to the $j$th row, and $\sigma$ times the $i$th column to the $j$th column when $d_i/(d_i,d_j)$ divides $\sigma$.
  \end{enumerate}

  Since $e$ is alternating, $Q(e)$ can be taken to be skew-symmetric with zeroes along the diagonal.
  Since $e$ is non-degenerate, there must be an integer in the first row of $Q(e)$ that is coprime to $d_1$.
  First, $\beta_{i\sigma}$ for appropriate $\sigma$ can be used to make this entry equal to $1$.
  If this entry is in the $j$th column, then the constraints on $q_{ij}$ force $d_j=d_1$.
  Now, $\pi_{2j}$ transforms the top-left corner of the matrix into $\left(\begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix}\right)$.
  In particular $d_1=d_2$.
  Finally, $\alpha_{1j\sigma}$ and $\alpha_{2j\sigma}$ can used to make the remaining entries of the first two rows and columns vanish.

  The form $e$ identifies the Pontryagin dual of the subgroup generated by the first (transformed) generator with the subgroup generated by the second one.
  After splitting off the subgroup spanned by these two generators, we are left with a finite abelian group with $n-2$ generators.
  By induction on $n$, we conclude that there exists a finite abelian group $A_f$, together with an isomorphism $\phi_f: A\times \hat A \to K_f$ such that
  \begin{equation*}
    e(\phi_f(x,\chi),\phi_f(x'\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A\:\chi,\chi'\in \hat A.
  \end{equation*}
  \subsection*{Analysis of the real part} Since $K_\R$ is the additive group of a finite dimensional vector space, $e(k,l)=e^{iQ(k,l)}$ for all $k,l\in K_\R$ with $Q:K_\R\times K_\R\to \R$ a symplectic form.
  It is well known in this case that $K_\R$ must be an even dimensional vector space, and that $Q$ can be transformed into the standard symplectic form by an automorphism of $K_\R$.
  Hence there exists a vector space $A_\R$ and an isomorphism $\phi_\R:A_\R\times \hat A_\R\to K_\R$ such that
  \begin{equation*}
    e(\phi(x,\chi),\phi(x',\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A_\R\; \chi,\chi'\in \hat A_\R.
  \end{equation*}
  \subsection*{Analysis of the toric and free parts}
  By (\ref{eq:5}), we may assume that $e_{t\Z}=0$.
  Note that $e_{\Z\Z}$ is an isomorphism of $K_\Z$ onto the Pontryagin dual of $K_t$.
  When $A_t=K_t$, and $\phi_t:A_t\times \hat A_t\to K_t\times K_\Z$ is $\mathrm{id}\times e_{\Z\Z}^{-1}$,
  we have
  \begin{equation*}
    e(\phi_t(x,\chi),\phi_t(x',\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A,\:\chi,\chi'\in \hat A.
  \end{equation*}
\end{proof}
\section{Divisible torsion-free phase space}
\label{sec:divis-tors-free}
\begin{thmb}
  Let $K$ be locally compact, divisible and torsion-free, and let $e:K\times K\to U(1)$ be a non-degenerate alternating bicharacter.
  Then there exists a locally compact divisible torsion-free abelian group $A$ and an isomorphism $\phi:A\times \hat A\to K$ such that
  \begin{equation*}
    e(\phi(x,\chi),\phi(x',\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A\text{ and }\chi,\chi'\in \hat A.
  \end{equation*}
\end{thmb}
\begin{rmkb}
  As in the case of Lie Heisenberg groups, $K$ does not determine $A$ up to isomorphism in Theorem~B. For example, $A=\Q$ and $A=\hat \Q$ give rise to isomorphic Heisenberg groups.
However, $A$ is uniquely determined if we stipulate that every connected component of $A$ is a Lie group.
\end{rmkb}
\begin{proof}
  According to \cite[25.33]{MR551496}, $K$ can be decomposed as
  \begin{equation*}
    K=K_\R \times K_\A \times K_\Q\times K_s,
  \end{equation*}
  where $K_\R$ is a subgroup isomorphic to the additive group of a finite dimensional real vector space,  $K_\A$ is a restricted direct product (over all primes $p$) of finite dimensional vector spaces over $p$\dash adic fields, $K_\Q$ is a subgroup isomorphic to the additive group of a finite dimensional vector space over the rational numbers with the discrete topology, and $K_s$ is a subgroup isomorphic to a finite Cartesian power of $\hat\Q$.
  The finite dimensionality of the vector spaces over $\Q_p$ is not part of the classification of divisible torsion-free groups in \cite{MR551496}, where the minimal divisible extension of an arbitrary Cartesian power of $\Z_p$ is allowed, but is a consequence of the fact that $K$ is isomorphic to its own Pontryagin dual.

  Likewise, $\hat K$ can be decomposed as
  \begin{equation*}
    \hat K = \hat K_\R\times \hat K_\A\times \hat K_\Q \times \hat K_s,
  \end{equation*}
  where $\hat K_\R$, $\hat K_\A$, $\hat K_\Q$ and $\hat K_s$ are subgroups of $\hat K$ isomorphic to the Pontryagin duals of $K_\R$, $K_\A$, $K_s$ and $K_\Q$ respectively.
  As in Section~\ref{sec:heis-lie-groups}, the isomorphism $e^\flat$ can be written as a matrix some of whose entries are forced to be $0$.
  It is not difficult to see that $e^\flat$ must have the form:
  \begin{equation*}
    e^\flat =
    \begin{pmatrix}
      e_{\R\R} & 0 & 0 & e_{\R s}\\
      0 & e_{\A\A} & 0 & e_{\A s}\\
      e_{\Q\R} & e_{\Q\A} & e_{\Q\Q} & e_{\Q s}\\
      0 & 0 & 0 & e_{ss}
    \end{pmatrix}
    ,
  \end{equation*}
  where $e_{ij}$ is a homomorphism $K_j\to \hat K_i$.
  Proceeding as in Section~\ref{sec:heis-lie-groups}, we reduce Theorem~B to the cases where $K=K_\R$, $K=K_\A$ and $K=K_\Q\times K_s$.
  The first case, the analysis of the real part, has already been dealt with in Section~\ref{sec:heis-lie-groups}, and the third case is similar to the analysis of the toric and free parts treated there.
  It remains to address the case where $K=K_\A$.
  We know that $K_\A$ is a restricted direct product:
  \begin{equation*}
    K_\A = {\prod_p}' K_p,
  \end{equation*}
  where, for each prime $p$, $K_p$ is a finite dimensional vector space over $\Q_p$.
  For $p\neq q$, there are no non-trivial homomorphisms $K_p\to \hat K_q$.
  Therefore $e_{\A\A}$, when restricted to $K_p$, gives rise to an isomorphism $e_p:K_p\to \hat K_p$, and we are reduced the case where $K=K_p$ is a finite dimensional vector space over $\Q_p$.
  Let $\psi$ be a non-trivial character $\psi\in \hat \Q_p$.
  Then we may write $e(k,l)=\psi(Q(k,l))$ for all $k,l\in K_p$, where $Q:K_p\times K_p\to \Q_p$ is a non-degenerate alternating bilinear form.
  As in the real case, this form can be transformed into the standard symplectic from by an automorphism of $K_p$, and the result follows.
\end{proof}
\section{Phase space of prime exponent}
\label{sec:groups-prime-expon}
\begin{thmc}
  Let $p$ be a prime number and suppose that $K$ is a locally compact abelian group of exponent $p$.
  Let $e:K\times K\to U(1)$ be a non-degenerate alternating bicharacter.
  Then there exists a locally compact abelian group $A$ of exponent $p$ and an isomorphism $\phi:A\times \hat A\to K$ such that
  \begin{equation*}
    e(\phi(x,\chi),(x',\chi'))=\chi'(x)\chi(x')^{-1} \text{ for all } x,x'\in A\text{ and } \chi,\chi'\in \hat A.
  \end{equation*}
\end{thmc}
\begin{rmkc}
  Once again, $K$ does not determine $A$ up to isomorphism, unless we stipulate that $A$ is a Cartesian power of a cyclic group of order $p$ with the product topology.
  This will be evident from the proof of Theorem~C.
\end{rmkc}
\begin{proof}
  Using a result of Braconnier (see \cite[25.29]{MR551496}) and the fact that $K$ is self-dual, we see that there exists a unique cardinal number $m$ such that $K$ can be decomposed as
  \begin{equation*}
    K=K_c\times K_d,
  \end{equation*}
  where $K_c$ is a direct product of $m$ copies of $\Z/p\Z$, a compact group under the product topology, and $K_d$ is the direct sum (in the category of abelian groups) of $m$ copies of $\Z/p\Z$, and is endowed with the discrete topology.
  Therefore, $\hat K$ has a decomposition
  \begin{equation*}
    \hat K = \hat K_c \times \hat K_d,
  \end{equation*}
  where $\hat K_c$ and $\hat K_d$ are subgroups of $\hat K$ isomorphic to the Pontryagin duals of $K_d$ and $K_c$ respectively.
  Write $e^\flat$ as a matrix
  \begin{equation}
    \label{eq:7}
    e^\flat =
    \begin{pmatrix}
      e_{cc} & e_{cd}\\
      e_{dc} & e_{dd}
    \end{pmatrix}
    ,
  \end{equation}
  where $e_{ij}$ is a homomorphism $K_j\to \hat K_i$.
  The image of $e_{dc}$, being the continuous image of a compact group in a discrete one, is finite and may be thought of as a finite dimensional vector space over $\Z/p\Z$.
  Any lift of a basis extends to a homomorphism $s:\mathrm{Image}(e_{dc})\to K_j$.
  Let $F=s\circ e_{dc}(K_j)$.
  Then $K_c=\ker e_{dc}\times F$.
  Now let $K_c'=\ker e_{dc}$ and let $K_d'=F\times K_d$.
  Suppose that, with respect to the decompositions $K=K_c'\times K_d'$ and $\hat K=\hat K_c'\times \hat K_d'$ (where $\hat K_c'$ and $\hat K_d'$ are isomorphic to the Pontryagin duals of $K_d'$ and $K_c'$ respectively), $e^\flat$ has the matrix
  \begin{equation*}
    e^\flat =
    \begin{pmatrix}
      e_{cc}' & e_{cd}'\\
      e_{dc}' & e_{dd}'
    \end{pmatrix}
    ,
  \end{equation*}
  then $e_{dc}'=0$.
  Now Theorem~C can be obtained as in the analysis of the real and toric parts in Section~\ref{sec:heis-lie-groups}.
\end{proof}
\subsection*{Acknowledgments}
The authors thank P.~P.~Divakaran and George Willis for their help and encouragement.

\bibliographystyle{amsplain}
\bibliography{refs}

\end{document}


TITLE: Why does $b^{\log_bx} = x$?
QUESTION [3 upvotes]: Why does $b^{\log_bx} = x$?
Can someone break this down by showing me the steps as to why this is true?

REPLY [1 votes]: Another way to think of this is to define a family of functions for any $b > 0$:  $$f_b(x) = b^x.$$  Then there is an inverse function $$f_b^{-1}(x)$$ such that $$f_b^{-1}(f_b(x)) = x$$ for all real $x$, and $$f_b(f_b^{-1}(x)) = x$$ on the domain of the inverse function, which happens to be for $x > 0$.  The inverse function is the logarithm:  $$f_b^{-1}(x) = \log_b x.$$  Thus, $\log_b x^b = x$ and $b^{\log_b x} = x$ are consequences of the above and merely use different notation to express the same concept.
Another way to think about the base-$b$ logarithm is that logarithms are exponents.  That is to say, $\log_b x$ is the exponent or power to which $b$ must be raised in order to obtain $x$.