TITLE: Checking an "equivalence of products" in the notes on deformation quantization by Allen C. Hirshfeld and Peter Henselder
QUESTION [1 upvotes]: These notes are a great introduction to deformation quantization but I failed to check the validity of the statement p.9, right before (5.18).
Context: let $(\mathcal{A},+,\mu)$ be an algebra. $\mu:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ standing for multiplication. Deformation consists in considering a family (paramatrized by $\nu$ in a yet to be chosen space) of product on $\mathcal{A}[[\nu]]$ (formal power series with coefficients in $\mathcal{A}$) generically given by
$$ \forall\ f,g\in \mathcal{A},\quad  \mu_{\nu}(f,g) := \mu(f,g) + \sum_{k=1}^{+\infty} \nu^k \mu_k (f,g) \label{1}\tag{1}$$
i.e. by a family of bilinear maps $\mu_k:\mathcal{A}\times \mathcal{A} \to \mathcal{A}$ satisfying some conditions and extended to elements $F, G\in \mathcal{A}[[\nu]]$ of the form $F=\sum_{k=1}^{+\infty} \nu^k f_{k},\ f_k \in \mathcal{A}$ by $\mathbb{K}[[\nu]]$-bilinearity. (the idea behind formal power series, as far as I understand, is to ignore convergence issues but still have a structure where one can compare terms of the same degree in $\nu$").
Two of these star-product are equivalent if there exists an invertible algebra isomorphism (transition map) $T:(\mathcal{A}[[\nu]],+,\mu_{\nu}) \longrightarrow (\mathcal{A}[[\nu]],+,\rho_{\nu})$, i.e. a map such that
$$ \forall\ F,G \in \mathcal{A}[[\nu]], \quad T\big(\mu_{\nu}(F,G)\big)= \rho_{\nu}(T(F),T(G))$$

Question: Let $\mathcal{A}=\mathcal{C}^{\infty}(\mathbb{R}^2)$ and denote $(a, \overline{a})$ or $(b, \overline{b}$) the variables of the functions. I want to check that the normal product ( (5.4) p.8; with the more usual notation for products)
$$f \ast_N g := \sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \frac{\partial^k f}{\partial a^k} \frac{\partial^k g}{\partial \overline{a}^k} = f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}}}\, g \label{2}\tag{2}$$
is equivalent to the Moyal product ((5.15) p.9, one can consider $\hbar$ as the deformation parameter... although there is usually the factor as in (\ref{5}))
$$ f \ast_M g := \sum_{k=0}^{+\infty} \left(\frac{\hbar}{2} \right)^k \frac{1}{k!}  \left. \left( \frac{\partial }{\partial a} \frac{\partial }{\partial \overline{b}} - \frac{\partial }{\partial \overline{a}} \frac{\partial }{\partial b}\right)^k f\big(a, \overline{a}\big) g\big(b, \overline{b}\big)  \right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}} = f\, e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)}\, g \label{3}\tag{3}$$
i.e.
$$ T\big(f \ast_N g \big) = T(f) \ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial a\, \partial \overline{a}} \right)\genfrac{}{}{0pt}{0}{"}{} \label{4}\tag{4}$$
Remarks:

In fact I already checked (\ref{4}) up to second order in $\hbar$ but it did not work at order 3 (although I'm not sure as the calculations were quite tedious...). It was not a priori clear that (\ref{4}) hold, one could have the other way round $ T\big(f \ast_M g \big) = T(f) \ast_N T(g)$ instead but this seems to fail at order 1. I only want to check the first few orders, but I would gladly take a proof for all order. I will soon write what I have done, but as I mentionned, it's tedious.
The Moyal product is first defined in the text (3.5-3.6) p.5 by
$$ f \ast_M g := \sum_{k=0}^{+\infty} \frac{\nu^k}{k!} \underbrace{\left(\frac{\partial }{\partial q_1} \frac{\partial }{\partial p_{2}} - \frac{\partial }{\partial p_{1}} \frac{\partial }{\partial q_2} \right)^k f(q_1,p_1)\, g(q_2,p_2)}_{\mu_k(f,g)}\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{q_1=q_2}{p_{1}=p_{2}}}\quad \text{with}\quad \nu = \frac{i\hbar}{2} \label{5}\tag{5}$$
and it does coincide with (\ref{3}) via (these are the correct $\sqrt{2}$ factors...)
$$ \left\lbrace \begin{aligned}
 a & := \frac{1}{\sqrt{2}} \left(q  + i\hspace{.5pt} p \right) \\
 \overline{a} & := \frac{1}{\sqrt{2}} \left( q - i\hspace{.5pt} p \right)
 \end{aligned} \right. \enspace \Longrightarrow\quad
 \left\lbrace \begin{aligned}
 \frac{\partial}{\partial\hspace{.7pt} q} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} q} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} =\frac{1}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} +  \frac{\partial}{\partial\hspace{.7pt} \overline{a}} \right) \\
 \frac{\partial}{\partial\hspace{.7pt} p} & = \frac{\partial\hspace{.7pt} a}{\partial\hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} a} + \frac{\partial\hspace{.7pt} \overline{a}}{\partial \hspace{.7pt} p} \frac{\partial}{\partial\hspace{.7pt} \overline{a}} = \frac{i}{\sqrt{2}} \left( \frac{\partial}{\partial\hspace{.7pt} a} - i\, \frac{\partial}{\partial\hspace{.7pt} \overline{a}}\right)
 \end{aligned} \right. $$
To make (\ref{3}) (same for (\ref{5})) more explicit, let me write the $k=2$ term: (notation $\displaystyle \partial_a=\frac{\partial}{\partial a},\ \partial_{ab}= \frac{\partial^2}{\partial a \partial b}$ etc.)
$$\begin{split}
 \mu_2(f,g) &= \Big(\partial_{aa\overline{b}\overline{b}} - 2 \partial_{a\overline{a}b\overline{b}} + \partial_{\overline{a}\overline{a}bb} \Big) f(a,\overline{a})g(b,\overline{b})\left.\vphantom{\frac{T}{T}}\right|_{\genfrac{}{}{0pt}{1}{a=b}{\overline{a}=\overline{b}}}\\
 &= (\partial_{aa}f)(\partial_{\overline{a}\overline{a}}g) - 2 (\partial_{a\overline{a}}f)(\partial_{b\overline{b}}g) + (\partial_{\overline{a}\overline{a}}f) (\partial_{aa}g)
\end{split} \label{6}\tag{6}$$
One can also use the $\overleftarrow{\partial}$ or $\overrightarrow{\partial}$ notations or a tensorial notation.

REPLY [0 votes]: It is in fact possible to prove
$$\begin{split} T\big(f \ast_N g \big) = T(f) &\ast_M T(g)\quad \text{with}\quad T = \genfrac{}{}{0pt}{0}{"}{}\!\!\exp\left(-\frac{\hbar}{2} \frac{\partial^2}{\partial_{a}\, \partial_{\overline{a}} } \right)\genfrac{}{}{0pt}{0}{"}{} \\
\Longleftrightarrow \quad e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left(f\, e^{\hbar \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} }\, g \right) & = \left(e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} f\right)  e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)} \left(e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} g\right)
\end{split}\tag{4}$$
by brute force. The idea is to reckognize the simplification that occurs for the exponents of exponential functions suggested by Cosmas but in terms of series:
$$\begin{split}
\text{L.h.s.} &= e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left(f\, e^{\hbar m s }\, g \right) = e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \left( e^{\big(ma + n\overline{a} \big) + \hbar m k  + \big(ka + s\overline{a} \big)} \right)\\
& = e^{-\frac{\hbar}{2} (m+k)(n+s)} \left(f\, e^{\hbar m s }\, g \right)\\
&= \left( e^{-\frac{\hbar}{2} m(n+s)} f\right) e^{\hbar m s } \left(\,e^{-\frac{\hbar}{2} k(s+n)} g \right)
\end{split} \label{4L}\tag{4L}$$
$$ \begin{split}
\text{R.h.s.} &= \left(e^{-\frac{\hbar}{2} mn\ + \big(ma + n\overline{a}\big)} \right)  e^{\frac{\hbar}{2}\big( \overleftarrow{\partial}_a \overrightarrow{\partial}_{\overline{a}} - \overleftarrow{\partial}_{\overline{a}} \overrightarrow{\partial}_{a} \big)} \left(e^{-\frac{\hbar}{2} ks\ + \big(ka + s\overline{a}\big)}\right) \\
&= \left(e^{-\frac{\hbar}{2} mn\ + \big(ma + n\overline{a}\big)} \right)  e^{\frac{\hbar}{2}\big( ms - nk \big)} \left(e^{-\frac{\hbar}{2} ks\ + \big(ka + s\overline{a}\big)}\right)
\end{split} \label{4R}\tag{4R}$$
The simplification for the exponents to go from L.h.s. to R.h.s. $-\frac{\hbar}{2} ms + \hbar m s -\frac{\hbar}{2} kn = \frac{\hbar}{2} (ms-nk) $.
Remark: One could probably exploit this argument by decomposing a certain class of functions in terms of exponentials, probably with a Laplace transform in analogy with the case of complex exponentials: a tempered distribution can be written $u =\mathcal{F}^{-1}(\hat{u})=\genfrac{}{}{0pt}{0}{"}{}\!\!\frac{1}{(2\pi)^2}\int_{\mathbb{R}^2}\hat{u}(k,\overline{k})\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}\, \genfrac{}{}{0pt}{0}{"}{}$.
Notice that Leibniz' rule can be written: $\partial^j_{a}(fg) =f \left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j g$ and similarly with two variables: $\partial^j_{a\overline{a}}(fg)=f \left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j g$. Hence
$$\begin{split}
\text{L.h.s.} &=\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2} \partial_{ a\, \overline{a}}\right)^j \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \partial^k_a f\, \partial^k_{\overline{a}} g\right)\\
&= \sum_{j=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^j \frac{1}{j!} f\left(\overleftarrow{\partial}_a +\overrightarrow{\partial}_a \right)^j \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)  g \\
&= f \left( \sum_{j=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^j \frac{1}{j!} \left(\sum_{0\leq i\leq j} {j\choose i} \overleftarrow{\partial}_a^{j-i} \overrightarrow{\partial}_a^i \right) \left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)  \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^j \right) g \\
&= f \left( \sum_{\genfrac{}{}{0pt}{1}{j,k=0}{i+l=j}}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!}  \overleftarrow{\partial}_a^{i}\ \left(-\frac{\hbar}{2}\right)^l\!\! \frac{1}{l!} \overrightarrow{\partial}_a^l \left( \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)  \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^{i+l} \right) g \\
&=f \underbrace{\left( \sum_{i=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!}  \overleftarrow{\partial}_a^{i} \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^i \right)}_{"e^{-\frac{\hbar}{2} m(n+s)}"}  \underbrace{\left(\sum_{k=0}^{+\infty} \frac{\hbar^k}{k!} \overleftarrow{\partial}^k_a \, \overrightarrow{\partial}_{\overline{a}} ^k \right)}_{"e^{\hbar ms}"} \underbrace{\left( \sum_{l=0}^{+\infty}  \left(-\frac{\hbar}{2}\right)^l\!\! \frac{1}{l!} \overrightarrow{\partial}_a^l  \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^{l} \right)}_{"e^{-\frac{\hbar}{2} k(s+n)}"} g
\end{split}$$
Now it suffices to check the equivalent of $e^{-\frac{\hbar}{2} m(n+s)} = e^{-\frac{\hbar}{2} mn} e^{-\frac{\hbar}{2} ms},\ e^{-\frac{\hbar}{2} ms} e^{\hbar ms}= e^{\frac{\hbar}{2} ms}$ and $e^{\frac{\hbar}{2} ms} e^{-\frac{\hbar}{2} nk}= e^{\frac{\hbar}{2} (ms-nk)}$. Since they are all similar, let us just write the first one:
\begin{equation}
\begin{split}
& f \left( \sum_{i=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^i \!\! \frac{1}{i!}  \overleftarrow{\partial}_a^{i} \left(\overleftarrow{\partial}_{\overline{a}} +\overrightarrow{\partial}_{\overline{a}} \right)^i \right) g\\
&= f \left( \sum_{\genfrac{}{}{0pt}{1}{i=0}{0\leq h\leq i}}^{+\infty} \left(-\frac{\hbar}{2}\right)^{i-h+h} \!  \overleftarrow{\partial}_a^{i-h+h} \, \frac{1}{(i-h)!} \overleftarrow{\partial}^{i-h}_{\overline{a}}\, \frac{1}{h!}  \overrightarrow{\partial}_{\overline{a}}^h  \right) g\\
&= f \left( \sum_{\genfrac{}{}{0pt}{1}{i=0}{h +l =i}}^{+\infty} \left(-\frac{\hbar}{2}\right)^{l} \! \frac{1}{l!} \overleftarrow{\partial}^{l}_{a\overline{a}}\ \left(-\frac{\hbar}{2}\right)^{h} \! \frac{1}{h!}  \overleftrightarrow{\partial}^h_{\overline{a}}  \right) g\\
&= f \left( \sum_{l=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^{l} \! \frac{1}{l!} \overleftarrow{\partial}^{l}_{a\overline{a}}\right) \left(\sum_{h=0}^{+\infty} \left(-\frac{\hbar}{2}\right)^{h} \! \frac{1}{h!} \overleftarrow{\partial}^h_a \overrightarrow{\partial}_{\overline{a}}^h  \right) g
\end{split}
\end{equation}

I also thought about the fact that a linear map from certain classes of functions of $\mathbb{R}^n$ (different versions, e.g. $\mathcal{C}^{\infty}_c(\mathbb{R}^n)$) which commutes with translations is a convolution operator (there is also a continuity condition, so one has to specify the topology on the space of functions, and this will constrain the convolution operator: convolution with what kind of function or distribution). Unfortunately, a quick computation shows that it would be a convolution with the Fourier transform of $e^{\frac{\hbar}{2}k^2}$ which does not make sense...
$$ \begin{split}
T(f) &= e^{-\frac{\hbar}{2} \partial_{ a\, \overline{a}}} \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \hat{f}(k,\overline{k})\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k} = \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \hat{f}(k,\overline{k})\,e^{\frac{\hbar}{2} k\overline{k}}\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}
\\
&= \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} \left(\int_ {\mathbb{R}^2} f(b,\overline{b})\,e^{-ikb -i\overline{k}\overline{b}} db\, d\overline{b}\right) e^{\frac{\hbar}{2} k\overline{k}}\, e^{ika +i\overline{k}\overline{a}}\, dk\, d\overline{k}\\
&=  \frac{1}{(2\pi)^2} \int_ {\mathbb{R}^2} f(b,\overline{b}) \left(\int_ {\mathbb{R}^2}  e^{\frac{\hbar}{2} k\overline{k}}\, e^{ik(a-b) +i\overline{k}(\overline{a}-\overline{b})}\, dk\, d\overline{k}\right) db\, d\overline{b}
\end{split}$$
Writing things differently also leads to the same result:
$$\begin{split}
T(f) &= \mathcal{F}^{-1}\circ\mathcal{F}\left(\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2} \partial_{ a\, \overline{a}}\right)^j f\right) \\
&=\sum_{j=0}^{+\infty} \frac{1}{j!} \left(-\frac{\hbar}{2}\right)^j (ik)^j (i\overline{k})^j f
\end{split}$$
Way out: $T$ seems to be unbounded but one can always imagine that its "inverse" (we didn't define its domain and target space...) could be convolution by a Gaussian. Another possibility would be to take a modified Fourier transform, $ \hat{f}(k,\overline{k}) := \int_ {\mathbb{R}^2} f(a,\overline{a})\, e^{- ika +i\overline{k}(\overline{a}-\overline{b})}\, da\, d\overline{a}$ or as remarked, take a Laplace transform. cf. also Inverse Weierstrass transform


text:
\begin{document}

\begin{abstract}
Let $Z$ be a  germ  of a reduced  analytic space of pure dimension.
We provide an analytic proof of the
uniform Brian\c con-Skoda theorem for the local ring $\Ok_Z$;
a result which was previously proved by Huneke by
algebraic methods.

For ideals with few generators we also get much
sharper  results.
\end{abstract}

\maketitle

\section{Introduction}\label{introduction}
Let $\mathfrak{a}=(a)=(a_1,\ldots, a_m)$ be an ideal in the local ring $\Ok=\Ok_0$ of
holomorphic functions
at $0\in\C^d$ and let $|\mathfrak{a}|=\sum_j|a_j|$. Up to constants, this function is
independent of the choice of generators of $\mathfrak{a}$.
In \cite{BS},  Brian\c{c}on and Skoda proved:

\smallskip
\noindent {\it If $\phi\in\Ok$ and
\begin{equation}\label{bsvillkor}
|\phi|\le C|\mathfrak{a}|^{\min(m,d)+\ell-1},  \quad  \ell=1,2,3,\ldots,
\end{equation}
then $\phi\in \mathfrak{a}^\ell$.}

\smallskip
If $m\le d$, then the statement  follows directly from Skoda's $L^2$-estimate in
\cite{Sko};  if $m>d$ one  uses that there is an ideal $\mathfrak{b}\subset \mathfrak{a}$
such that $|\mathfrak{a}|\sim |\mathfrak{b}|$, a so-called reduction of $\mathfrak{a}$,  with
$n$ generators.
\smallskip

If  $\mathfrak{b}$ is any ideal in $\Ok$ then
$|\phi|\le C|\mathfrak{b}|$ if (and in fact only if)  $\phi$ is in theintegral closure
$\overline{\mathfrak{b}}$.
Therefore, the statement implies (is equivalent to) the inclusion
\begin{equation}\label{smaka}
\overline{\mathfrak{a}^{\min(m,d)+\ell-1}}\subset \mathfrak{a}^{\ell}.
\end{equation}
This  is a notable example of a purely algebraic theorem  that was first proved
by transcendental methods. It took several years before
algebraic proofs appeared,  \cite{LT} and
\cite{LS}.
In \cite{BGVY} there is a proof by integral formulas  and residue theory.

\smallskip

Assume now that  $Z$ is  a  germ of an analytic space of pure dimension $d$
and let $\Ok_Z$ be its  structure ring of germs of
(strongly) holomorphic functions.
It is non-regular if (and only if) $Z$ is non-regular.
It is easy to see that the usual Brian\c con-Skoda theorem  cannot hold in general
in the non-regular case,  not even for $m=1$, see Example~\ref{pucko} below.
However,
Huneke proved in \cite{Hu}  that there is a number $\mu$ only
depending on $Z$ such that for any ideal $\mathfrak{a}\subset \mathcal{O}_Z$, and integer
$\ell\ge 1$,
\begin{equation}\label{hun}
\overline{\mathfrak{a}^{\mu+\ell-1}}\subset \mathfrak{a}^{\ell}.
\end{equation}
Huneke's proof is completely algebraic (and holds for some more general rings as well),
so it is natural to look for an analytic proof.
In this paper we give a proof  by means of  residue calculus,
and the membership can be realized by an  integral formula
on  $Z$.
A problem of general interest, see, e.g., p.\ 657 in \cite{HyVi} and Remark 4.14 in \cite{Hu},
is to estimate the Brian\c{c}on-Skoda
number, $\mu$, in Huneke's theorem in terms of invariants of the ring.
Our proof relates $\mu$ to the complexity of a free resolution of $\mathcal{O}_Z$.
We have also a sharper statement in case
$\mathfrak{a}$ has  ``few'' generators, and the zero set, $Z^{\mathfrak{a}}$, of the ideal does not
overlap the singular set of $Z$ ``too much''.
To formulate this we first have to introduce  certain (germs of) subvarieties,
$Z^r$, associated with $Z$:

To begin with we  choose an embedding of $Z$ and
consider it as a subvariety at, say,
the origin of $\C^n$ for some $n$. If $\I$ is the corresponding radical ideal
in $\Ok=\Ok_{\C^n,0}$, then $\Ok_Z=\Ok/\I$.
Let
\begin{equation}\label{upplosning}
0\to \Ok(E_N)\stackrel{f_N}{\longrightarrow}\ldots
\stackrel{f_3}{\longrightarrow} \Ok(E_2)\stackrel{f_2}{\longrightarrow}
\Ok(E_1)\stackrel{f_1}{\longrightarrow}\Ok(E_0)
\end{equation}
be a free resolution of  $\Ok/\I$.  Here  $E_k$ are trivial vector bundles
and $E_0$ is a  trivial line bundle.
Thus $f_k$ are just   holomorphic matrices in a neighborhood of $0$.

We let
$Z_k$ be the set of  points $x$ such that $f_k(x)$ does not have optimal rank.
These varieties  are,  see, \cite{Eis} Ch.~20,
independent of the choice of resolution, and  we have the inclusions
$$
\cdots\subset Z_{p+2}\subset Z_{p+1}\subset Z_{sing}\subset Z_p= \dots =Z_1=Z,
$$
where $p=n-d$.

Now let
\begin{equation}\label{hoppet}
Z^0=Z_{sing}, \quad  Z^r=Z_{p+r}, \ r>0.
\end{equation}
Since any  two minimal embeddings are equivalent, and any embedding factors
in a simple way over a minimal embedding, one can verify that these subsets
$Z^r$ are intrinsic subvarieties of the analytic space $Z$, that  reflect  the
degree of complexity of $Z$.
To begin with, since $Z$ has pure dimension (Corollary 20.14 in \cite{Eis}),
$$
\codim Z^r\ge r+1,\quad r>0.
$$
Moreover, $Z^r=\emptyset$ for $r > d-\nu$ if and only if the
depth of the ring $\mathcal{O}_Z$ is  at least $\nu$.
In particular,
$Z^r=\emptyset$ for $r>0$ if and only if $Z$ (i.e., $\Ok_Z$)  is
Cohen-Macaulay.

\begin{thm}\label{boll}
Let $Z$ be a germ of an  analytic space  of pure dimension.

\noindent (i)\  There is a natural number $\mu$, only depending on $Z$,
such that for any ideal $\mathfrak{a}=(a_1,\ldots,a_m)$ in $\Ok_Z$ and $\phi\in\mathcal{O}_Z$,
\begin{equation}\label{bolla}
|\phi|\le C|\mathfrak{a}|^{\mu+\ell-1}
\end{equation}
implies that $\phi\in \mathfrak{a}^{\ell}$.

\smallskip
\noindent (ii) \   If  for a given ideal $\mathfrak{a}=(a_1,\ldots,a_m)$
\begin{equation}\label{bolla2}
\quad {\rm codim}\,(Z^r\cap Z^{\mathfrak{a}})\ge m+1+r,\quad   r\ge 0,
\end{equation}
then for any $\phi\in\Ok_Z$,
\begin{equation}\label{bolla3}
|\phi|\le C|\mathfrak{a}|^{m+\ell-1}
\end{equation}
implies that $\phi\in \mathfrak{a}^{\ell}$.
\end{thm}

Huneke's theorem \eqref{hun}  follows immediately from part (i) of
Theorem~\ref{boll}, since  even in the non-regular case
$\phi\in \overline{(b)}$ immediately implies that
$|\phi|\le C|b|$. The less obvious implication
$|\phi|\le C|b| \Rightarrow \phi\in \overline{(b)}$ also holds, see,
e.g., \cite{LeTe}, and so Theorem~\ref{boll} (i) is in fact equivalent
to Huneke's theorem.

\begin{ex}\label{pucko}
If $Z$ is the zero set of $z^p-w^2$ in $\C^2$, where $p>2$ is a prime,
then $|w|\le|z|^{[p/2]}$ on $Z$, but $w$ is not in $(z)$. However,
if $|\phi|\le C|z|^{(p+1)/2}$, then $\phi\in (z)$, i.e., $\phi/z$ is strongly
holomorphic on $Z$.
\end{ex}

\begin{remark}
The important point in Huneke's  theorem  is the uniformity
in $\mathfrak{a}$ and $\ell$. Notice that \eqref{hun} implies the slightly weaker statement
\begin{equation}\label{hun2}
\overline{\mathfrak{a}}^{\mu +\ell -1}\subset \mathfrak{a}^{\ell}.
\end{equation}
It is quite easy to prove such an inclusion for fixed $\mathfrak{a}$ and $\ell$.

In fact, assume that $Z$ is a germ of a subvariety in $\C^n$
and choose a tuple $f$ such that  $Z=\{f=0\}$.  Let $A=(A_1,\ldots,A_m)$
and $\Phi$ denote
fixed representatives in $\Ok_{\C^n}$ of $\mathfrak{a}=(a_1,\ldots,a_m)$ and
$\phi\in \overline{\mathfrak{a}}$. Then

\begin{equation*}
|\Phi(z)|\le C d(z,Z^{\mathfrak{a}}\cap Z)\le C'(|A|+|f|)^{1/M}
\end{equation*}
for some $M$ by  Lojasiewicz' inequality, and hence $\Phi^{Mn}$ is in
the ideal $(A)+(f)$ by the usual Brian\c{c}on-Skoda theorem
in the ambient space.  Thus $\phi^{Mn}\in \mathfrak{a}$
and
therefore $\phi^{Mn\ell}\in\mathfrak{a}^\ell$. Thus
$\overline{\mathfrak{a}}^{Mn\ell}\subset \mathfrak{a}^\ell$.

\end{remark}

From  Theorem~\ref{boll} (ii) we get:

\begin{cor}
If
\begin{equation}\label{bolla4}
{\rm codim}\,Z^r\ge m+1+r,\quad r\ge 0,
\end{equation}
then  \eqref{bolla3} implies that $\phi\in\mathfrak{a}^\ell$
for any $\mathfrak{a}$ with $m$ generators.
\end{cor}

Assume that \eqref{bolla4} holds for $m=1$. The conclusion for $\ell =1$ then is that
each weakly holomorphic function is indeed holomorphic, i.e.,
$Z$ (or equivalently $\Ok_Z$) is normal. In fact,
if $\phi$ is weakly holomorphic, i.e., holomorphic on $Z_{reg}$ and
locally bounded, then it is meromorphic, so $\phi=g/h$ for some
$g,h\in\Ok_Z$. The boundedness means that $|g|\le C|h|$ and by the
corollary thus $\phi$ is in $\Ok_Z$.
One can check that \eqref{bolla4} with $m=1$ is equivalent to
Serre's condition for normality of the local ring $\Ok_Z$ and therefore
both necessary and sufficient.

\smallskip

The basic tool in our proof is the residue calculus
developed in \cite{A0},
\cite{AW1},  and \cite{AW2},  and we recall the necessary material
in  Section~\ref{utter}. Given an ideal sheaf $\J$ one
can associate a current $R$ such that a holomorphic function
$\phi$ is in $\J$ as soon as $\phi R=0$. We use  such
a current $R^{\mathfrak{a},\ell}$ associated with the ideal $\mathfrak{a}^\ell$.
For $\ell=1$ it is the current of Bochner-Martinelli type
from \cite{A0}, whereas for $\ell>1$ we use a
variant  from \cite{A2}. Since we are to prove the membership
on $Z$ rather than on some ambient space, thinking of $Z$ as embedded in some
$\C^n$, we will also use a  current $R^Z$
associated to the  radical ideal $I$ of the embedding.  For the analysis
of this current we rely  on  results from \cite{AS},
described in Section~\ref{zres}.
It turns out that one can form the ``product'' $R^{\mathfrak{a},\ell}\w R^Z$ such that
$\phi R^{\mathfrak{a},\ell}\w R^Z$ only depends on the values of $\phi$ on $Z$;
moreover,  if the hypotheses in Theorem~\ref{boll} are fulfilled
then it vanishes (Proposition~\ref{main}),
which  in turn implies that $\phi$ belongs to the ideal $\mathfrak{a}$ modulo $I$.
In the last section we  present   an integral formula that
provides an explicit representation of the membership.

\section{Currents obtained from locally free complexes}\label{utter}

Let
\begin{equation}\label{vcomplex}
0\to E_N\stackrel{f_N}{\longrightarrow} E_{N-1}\stackrel{f_{N-1}}{\longrightarrow}\cdots
\stackrel{f_2}{\longrightarrow}E_1\stackrel{f_1}{\longrightarrow} E_0\to 0
\end{equation}
be a generically exact holomorphic complex of  Hermitian vector bundles
over a  complex manifold $X$, say a \nbh of the origin in $\C^n$. We
assume that $E_0$ is a trivial line bundle so that
$\Ok(E_0)=\Ok$.  There is an associated complex, like
\eqref{upplosning},  of (locally) free sheaves
of $\Ok$-modules,  and  we let $\J=f_1\Ok(E_1)\subset\Ok$ be
the ideal sheaf generated by (the entries in) $f_1$.
Let $Z$ be the analytic set where  \eqref{vcomplex} is not pointwise
exact. In $X\setminus Z$ we let $\sigma_k$ be the section of
$\Hom(E_{k-1},E_k)$ that vanishes on the orthogonal complement
of the pointwise image of $f_k$ and is  the minimal left inverse
of $f_k$ on the image of $f_k$. If $E=\oplus E_k$,
$f=\oplus f_k$, and $\sigma=\oplus \sigma_k$, then
$
\sigma f+f\sigma=I,
$
where $I$ is the identity on $E$.
Since $E_0$ is trivial  we identify $\Hom(E_0,E)$ with $E$.

Following \cite{AW1}, in $X\setminus Z$ we define the form-valued sections
\begin{equation}\label{startmal}
u=\sum_{k=1}^N u_k, \quad  u_k=(\dbar\sigma_k)\cdots(\dbar\sigma_2)\sigma_1,
\end{equation}

of $E$.
If $\nabla_f=f-\dbar$ we have that $\nabla_f u=1$.
It turns out that $u$ has a current  extension $U$ to
$X$ as a principal value current: If $F$ is a tuple
of holomorphic functions such that $F=0$ on $Z$, then
$|F|^{2\lambda} u$ has a current-valued  analytic continuation
to $\Re \lambda>-\epsilon$ and $U$ is the value at $\lambda=0$.
Alternatively one can take a smooth approximand of the characteristic
function $\chi$ for $[1,\infty)$, and  let $\chi_\delta=\chi(|F|^2/\delta^2)$.
Then $U$ is the weak limit of $\chi_\delta u$ when $\delta\to 0$
(see, e.g., the proofs of Theorems 16 and 21 in \cite{Sam}).
In this paper the latter definition will be more convenient.
Clearly $\nabla_f$ also applies to currents, and
\begin{equation}\label{bulla}
\nabla_f U=1-R,
\end{equation}
where $R$ is a residue current with support on $Z$;  more precisely
$R=\lim_{\delta\to 0} R^\delta$,
where
$$
R^\delta=R^\delta_0+R^\delta_1+R^\delta_2+ \cdots =
(1-\chi_\delta)  +\dbar\chi_\delta\w u_1+\dbar\chi_\delta\w u_2+\cdots;
$$
notice that  $R^\delta_k$ is an $E_k$-valued $(0,k)$-current.

A basic observation is that the annihilator sheaf, $\ann R$, of
$R$ is contained in the sheaf $\J$, i.e.,
\begin{equation}\label{pulla}
\ann R\subset \J.
\end{equation}
In fact, if  $\phi\in\Ok$ and $\phi R=0$, then by \eqref{bulla},
$
\nabla_f(\phi U)=\phi-\phi R=\phi.
$
By  solving a sequence of $\dbar$-equations, which is always possible
locally at least, we get a holomorphic solution $\psi\in\Ok(E_1)$
to $f_1\psi=\phi$, which means that $\phi$ is in the ideal $\J$.
One can also prove \eqref{pulla} by an integral formula
that gives an explicit realization of the membership of $\phi$ in $\J$,
see Section~\ref{intrep}.

In general the converse inclusion is not true. However,
if the associated sheaf complex
is exact, i.e., a resolution of $\Ok/\J$, then
indeed  $\ann R=\J$ (Theorem~1.1 in \cite{AW1}).

\begin{ex}\label{bsex}
Let $a_1,\ldots,a_m$ be holomorphic functions in $X$.
Choose a nonsense basis $\{e_1,\ldots,e_m\}$ and  consider
$E_1=sp\{e_j\}$ as a trivial vector bundle of rank $m$, let $e_j^*$
be the dual basis, and  consider
$a=a_1e_1^*+\cdots +a_me_m^*$ as a section of the dual bundle $E_1^*$.
If $E_k=\Lambda^k E_1$
we then get a complex \eqref{vcomplex},
the Koszul complex,
with the mappings  $f_k$ as interior multiplication $\delta_a$ with $a$.
Following the recipe above (with the trivial metric on the $E_k$) we get, cf.,  \cite{AW1} Example~1,
the corresponding form
\begin{eqnarray}\label{equ}
u^a=\sum_{k=1}^{m}
\frac{(\sum_{j=1}^m \bar{a}_je_j)\wedge (\sum_{j=1}^m \dbar \bar{a}_j\wedge e_j)^{k-1}}{|a|^{2k}}
\end{eqnarray}
outside $\{a=0\}$
and the associated residue current $R^a=\lim_{\delta\to 0} R^{a,\delta}$
where
$R^{a,\delta}= (1-\chi_\delta)+
\dbar\chi_\delta\wedge u^a$ and $\chi_\delta=\chi(|a|^2/\delta^2)
$.
This current of so-called Bochner-Martinelli type
was introduced already in \cite{PTY}, and its  relation to the
Koszul complex and division problems was noticed in  \cite{A0}.
Now \eqref{pulla} means that
\begin{equation}\label{ain}
\ann R^a\subset (a).
\end{equation}
Except for the case when $a$ is a complete intersection,
in which case the Koszul complex provides a resolution of
$\Ok/(a)$, the   inclusion \eqref{ain} is strict, see \cite{W} and \cite{JW}.
Nevertheless, the singularities of  $R^a$
reflect the
characteristic varieties associated to the ideal, see \cite{JW} and \cite{AG},

which are  closely related to the integral closure
of powers of $(a)$,  and therefore
$R^a$   is well suited for the Brian\c{c}on-Skoda theorem.

A slight modification of the Koszul complex, derived  from the so-called
Eagon-Northcott complex, with associated ideal sheaf $\J=(a)^\ell$,
was introduced in \cite{A2}.

The associated form $u^{a,\ell}$ is a sum of terms like
$$
\frac{\bar a_{I_1}\cdots\bar a_{I_\ell}\dbar\bar a_{I_{\ell+1}}\w\ldots\w \dbar\bar a_{I_{k+\ell-1}}}
{|a|^{2(k+\ell-1)}},\quad k\le m,
$$
see the proof of Theorem~1.1 in \cite{A2} for
a  precise description of $u^{a,\ell}$ and the corresponding residue current $R^{a,\ell}$.
It turns out that  $\phi$ annihilates   $R^{a,\ell}$
if \eqref{bsvillkor} holds,   and thus $\phi\in(a)^\ell$, so the
classical Brian\c con-Skoda theorem follows.
The most expedient way to  prove
this annihilation  is to use a  resolution
of singularities where $a$ is principal.

However, it is not really necessary to define
the current $R^{a,\ell}$ in itself;
it is actually enough to make sure that $\phi R^{a,\ell,\delta}\to 0$
when $\delta\to 0$, and this can be proved
essentially  by integration by part in an ingenious way, thus providing
a proof of  the Brian\c{c}on-Skoda theorem by
completely elementary means, see \cite{Sz}.

\end{ex}

In  \cite{AW2} was  introduced the sheaf of  {\it pseudomeromorphic}
currents $\mathcal{PM}$.  For the definition,
see \cite{AW2}.
It is closed under $\dbar$ and  multiplication with
smooth forms.  In particular, the  currents $U$ and $R$ are pseudomeromorphic.
The following fact (Corollary~2.4 in \cite{AW2})  will be used
repeatedly.

\begin{prop}\label{dimer}
If $T\in\mathcal{PM}$ has bidegree $(r,k)$ and the support of $T$ is contained
in a variety of codimension strictly larger than $k$, then $T=0$.
\end{prop}

In particular, this means that if $Z$ (the variety where \eqref{vcomplex} is not pointwise
exact) has codimension $p$ then $R=R_p+R_{p+1}+\cdots$.

\smallskip

As mentioned in the introduction, we need to form products of currents associated to complexes.
Assume therefore that $(\Ok(E^g_{\bullet}), g_{\bullet})$ and $(\Ok(E^h_{\bullet}), h_{\bullet})$
are two complexes as above
and $\I$ and $\J$ are the corresponding ideal sheaves.
We can  define a   complex  \eqref{vcomplex}  with
\begin{equation}\label{sven}
E_k=\bigoplus_{i+j=k} E^g_i\otimes E^h_j,
\end{equation}
and $f=g+h$, or more formally,
$f=g\otimes I_{E^h}+ I_{E^g}\otimes h$,
such that
\begin{equation}\label{hoppsan}
f(\xi\otimes\eta)=g\xi\otimes\eta +(-1)^{\deg\xi}\xi\otimes h\eta.
\end{equation}
Notice that $E_0=E_0^g\otimes E_0^h=\C$ and that
$f_1\Ok(E_1)=\I+\J$.
One can extend \eqref{hoppsan} to form-valued  or current-valued sections $\xi$ and $\eta$
and $\deg\xi$ then means  total degree. It is  natural to write
$\xi\w \eta$ rather than $\xi\otimes\eta$, and
we  define $\eta\w\xi$  as $(-1)^{\deg \xi \deg\eta}\xi\w\eta$.

Notice that
\begin{equation}\label{rregel}
\nabla_f (\xi\otimes\eta)=\nabla_g\xi\otimes\eta
+(-1)^{\deg \xi}\xi\otimes \nabla_h\eta.
\end{equation}

Let  $u^g$ and $u^h$ be the corresponding
$E^g$-valued and $E^h$-valued
forms, cf.\ \eqref{startmal}.
Then $u=u^h\w u^g$ is an  $E$-valued form
outside $Z^g\cup Z^h$.
Following the proof of  Proposition~2.1 in \cite{AW2} we can define
$E$-valued pseudomeromorphic currents
$$
R^h\w  R^g=\lim_{\delta\to 0} R^{h,\delta}\w  R^g,
\quad
U^h\w R^g=\lim_{\delta\to 0} U^{h,\delta}\w R^g,
$$
where
$U^{h,\delta}=\chi_\delta u^{h}$ and $R^{h,\delta}=1-\chi_{\delta}+\dbar\chi_\delta \w  u^{h}$,
and $\chi_\delta=\chi(|H|^2/\delta^2)$ as before.
The ``product''  $R^h\w R^g$ so defined is {\it not} equal to
$R^g\w R^h$ in general. It is also understood here that
$H$ only vanishes where it has to, i.e., on the set where
the complex $(E^h_{\bullet}, h_{\bullet})$ is not pointwise exact. If we use
an $H$ that vanishes on a larger set, the result will be
affected. It is worth to point out that  a certain component
$R^h_k\w R^g$ may be nonzero even if $R^h_k$ itself vanishes.

\begin{prop}\label{bellman}
With the notation above we have that
\begin{equation}\label{armar}
\nabla_f (U^g+ U^h\w R^g)=1-R^h\w R^g.
\end{equation}
Moreover,  $\phi R^h\w R^g=0$ implies that
$\phi\in\I+\J$.
\end{prop}

\begin{proof}
Recall that $\nabla_h U^{h,\delta}=1-R^{h,\delta}$, $\nabla_g U^g=1-R^g$ and
$\nabla_gR^g=0$.
Therefore,
$$
\nabla_f (U^g+ U^{h,\delta}\w R^g)=1-R^g+(1-R^{h,\delta})\w R^g=
1-R^{h,\delta}\w R^g.
$$
Taking limits, we get \eqref{armar}. The second statement now follows
in the same way as \eqref{pulla} above.
\end{proof}

\section{The residue  current associated to the variety $Z$}\label{zres}

Consider a subvariety $Z$ of a \nbh of the origin in $\C^n$ with
radical ideal sheaf $\I$ and let \eqref{upplosning}
be a resolution of $\Ok/\I$.  Let $R^Z$ be the associated
residue current obtained as in the previous section.
We then know that $R^Z$ has support on $Z$ and that
$\ann R^Z=\I$.

Outside the set $Z_k$, cf., Section~\ref{introduction},  the mapping $f_k$ has constant
rank, and hence $\sigma_k$ is smooth there. Outside
$Z_k$ we therefore have that
\begin{equation}\label{strumpa}
R^Z_{k+1}=\alpha_{k+1} R_k^Z
\end{equation}
where $\alpha_{k+1}=\dbar\sigma_{k+1}$
is a smooth $\mbox{Hom}(E_k,E_{k+1})$-valued $(0,1)$-form,
cf., \eqref{startmal}.

Locally on $Z_{reg}$, the current
$R^Z$ is essentially the integration current $[Z]$.
We have the following more precise statement that
gives a Dolbeault-Lelong-type representation, in the sense of \cite{jeb1},
of the current $R^Z$.

Let  $\chi$ be  a  smooth regularization of the characteristic function of
$[1,\infty)$ and $p=\textrm{codim}\, Z$ as before.

\begin{prop}\label{DL}
For each given $x\in Z_{reg}$, there  is a hypersurface $\{h=0\}$ in $Z$, avoiding $x$ but
containing $Z_{sing}$ and intersecting
$Z$ properly,  and  $E_k$-valued $(n-p,k-p)$-forms
$\beta_k$,  smooth outside $\{h=0\}$, such that
\begin{equation*}
R^Z_k .\, (dz\wedge \xi)=\lim_{\epsilon\to 0}
\int_{Z}\chi(|h|/\epsilon)\beta_k \wedge \xi,\,\,\,\,\,\, \xi \in \mathcal{D}_{0,n-k}(X),
\end{equation*}
for $p\leq k \leq n$.
Moreover, in a suitable resolution $\pi\colon \tilde Z\to Z$
the forms $\beta_k$ locally have  the form $\alpha_k/m_k$, where $\alpha_k$ are  smooth
and $m_k$ are   monomials.
\end{prop}

Here, $dz =dz_1\wedge \cdots \wedge dz_n$.

\begin{proof}
Following  Section~5 in \cite{AS} (the proof of Proposition~2.2)
one can find, for each given $x\in Z_{reg}$, a holomorphic function $h$ such that $h(x)\neq 0$
and $h$ does not vanish identically on any component of $Z_{reg}$. Moreover, for $k\ge p$,
$$
R_k^Z=\gamma_k\lrcorner [Z],
$$
where $\gamma_k$ is an  $E_k$-valued and $(0,k-p)$-form-valued
$(p,0)$-vector field that is smooth outside $\{h=0\}$.
Let $\xi$ be a test form of bidegree $(0,n-k)$.
The current $R^Z$  has the so-called standard extension property, SEP, see \cite{AW2}
Section~5, which means that
$$
R^Z_k. (\xi\w dz)=\lim_{\epsilon\to 0}\int\chi(|h|/\epsilon)\gamma_k\lrcorner[Z]\w \xi\w dz=
\pm
\lim_{\epsilon\to 0}\int_Z\chi(|h|/\epsilon)\xi\w \gamma_k\lrcorner dz.
$$

Thus we can take $\beta_k=\pm  \gamma_k\lrcorner dz$.

More precisely,  according to the  last paragraph of Section~5 in \cite{AS},
$\gamma_p$ is a
meromorphic  $(p,0)$-field (with poles where $h=0$)
composed by the  orthogonal projection
of $E_p$ onto the orthogonal complement in $E_p$ of the pointwise image of $f_{p+1}$.
This projection is given by
$$
I_{E_p}-f_{p+1}\sigma_{p+1}.
$$
Furthermore, cf., \eqref{strumpa},
$$
\gamma_k=(\dbar\sigma_k)\cdots(\dbar\sigma_{p+1})\gamma_p
$$
for $k>p$.
Now choose a resolution of singularities
$\tilde Z\to Z$ such that for each $k$ the
the determinant ideal of $f_k$ is principal. On $\tilde Z$, then
each  $\sigma_k$  (locally) is
a smooth form over a monimial, see Section~2 in \cite{AW1}, and thus
$\beta_k=\gamma_k\lrcorner dz$  has this form as well.
\end{proof}

We can choose the resolution of singularities $\tilde{Z} \to Z$
so that also $\tilde h=\pi^*h$ is a monomial.
By a partition of unity  it follows that
$R^Z_k.(dz\w\xi)$ is a finite sum of terms like
\begin{equation}\label{armada}
\lim_{\epsilon\to 0} \int_s\chi(|\tilde h|/\epsilon)
\frac{ds_1\w\ldots\w ds_{\nu}}{s_1^{\alpha_1+1}\cdots s_{\nu}^{\alpha_{\nu}+1}}
\w\tilde\xi\w\psi,
\end{equation}
where $s_1,\ldots, s_{n-p}$ are local holomorphic coordinates and $\nu\le n-p$,
$\tilde\xi=\pi^*\xi$, and $\psi$ is a smooth form with compact support.
It is easily checked that this limit  is the
tensor product of the one-variable principal value currents
$ds_i/s_i^{\alpha_i+1}$,
$1\le j\le\nu$,
acting on $\tilde\xi\w\psi$. Therefore \eqref{armada}  is equal
to (a constant times)
\begin{equation}\label{armada2}
\int
\frac{ds_1\w\ldots\w ds_{\nu}}{s_1\cdots s_{\nu}}\w\partial^\alpha_s(\tilde \xi\w\psi),
\end{equation}
if $\partial^\alpha_s=\partial^{\alpha_1}_{s_1}\cdots \partial^{\alpha_{\nu}}_{s_\nu}$.

\section{Proof of Theorem \ref{boll}}\label{pf}

To prove Theorem~\ref{boll} we are going to apply the idea
in Example~\ref{bsex} but performed on $Z$.

To this end we assume that $Z$ is embedded in $\C^n$ and we let
$R^Z$ be the current introduced in the previous section.
Let $\mathfrak{a}=(a)$ be the ideal in $\Ok_Z$ and suppose for the moment
that $a$ also denotes representatives in $\Ok$ of the generators.
If $R^{a,\ell}=\lim_{\delta\to 0} R^{a,\ell,\delta}$ denotes
the current from  Example~\ref{bsex} we can form,
cf., the end of Section~\ref{utter}, the product
$$
R^{a,\ell}\w R^Z=\lim_{\delta\to 0}R^{a,\ell,\delta}\w R^Z.
$$
Since $R^Z$ annihilates $\I$ it follows that $R^{a,\ell}\w R^Z$ only
depends on $\mathfrak{a}\subset\Ok_Z$.  For the same reason,
$\phi R^{a,\ell}\w R^Z$
is well-defined for $\phi\in\Ok_Z$.  We know from Proposition~\ref{bellman}
that
$\phi$ belongs to $\mathfrak{a}$ if it annihilates this current, and thus
Theorem~\ref{boll} follows from the following proposition.

\begin{prop}\label{main}
If the hypotheses of Theorem~\ref{boll} are fulfilled
i.e., either \eqref{bolla}, or \eqref{bolla3} together with the geometric
conditions \eqref{bolla2}, then
$\phi R^{a,\ell}\w R^Z=0$.
\end{prop}

\begin{remark}
It is natural to try to use the Lelong current $[Z]$ rather than $R^Z$.
There is, see  \cite{AB} Example~1,  a holomorphic $E_p$-valued
form $\xi$ such that $[Z]=\xi\cdot R^Z_p$. Thus the hypotheses
in Theorem~\ref{boll} imply that
$\phi R^a\w[Z]=0$. However, this in turn does not imply that
$\phi$ is in $(a)$.

In fact, if $m=1$ so that $a$ is just one function, then
$$
0=\phi R^a\w [Z]=\phi\dbar\frac{1}{a}\w[Z],
$$
and this means that $\phi/a$ is in  $\omega_Z^0$ introduced by
Barlet, see, e.g., \cite{HP},
and this class  is wider than $\Ok_Z$ in general.
\end{remark}

\begin{proof}[Proof of Proposition~\ref{main}]

We first  assume that \eqref{bolla2} and \eqref{bolla3} hold.
Considering $\phi R^{a,\ell}$ as an intrinsic  current on
the submanifold $Z_{reg}$ (cf.\ the beginning of this section)
it follows from the residue proof of the Brian\c{c}on-Skoda theorem
in the regular case that $\phi R^{a,\ell}$ must vanish on $Z_{reg}$
since \eqref{bolla3} holds. Thus, $\phi R^{a,\ell}\wedge [Z]$ vanishes on $Z_{reg}$
and so,
in view of  Proposition~\ref{DL}, it follows that
the support of $\phi R^{a,\ell}\w R^Z$ is contained in $Z_{sing}$.
On the other hand it is readily verified that
$R^{a,\ell}\w R^Z$ must vanish if $a$ is nonvanishing.
Thus the support of $\phi R^{a,\ell}\w R^Z$ is contained in $Z_{sing}\cap Z^a$.

The current  $R^{a,\ell}$ has (maximal)  bidegree $(0,m)$ and hence
$R^{a,\ell}\w R^Z_p$ has (maximal) bidegree $(0,m+p)$. Since it has support
on  $Z_{sing}\cap Z^a$ that has codimension $\geq p+m+1$ by \eqref{bolla2},
it follows that $\phi R^{a,\ell}\w R^Z_p=0$.
Outside $Z_{p+1}$ we have that $R^Z_{p+1}=\alpha_{p+1} R^Z_p$ for a smooth form $\alpha_{p+1}$,
and hence
$$
\phi R^{a,\ell}\w R^Z_{p+1}=\phi R^{a,\ell}\w \alpha_{p+1} R^Z_p=\alpha_{p+1}\phi R^{a,\ell}\w R^Z_p =0
$$
there. Thus $\phi R^{a,\ell}\w R^Z_{p+1}$ has support on $Z_{p+1}\cap Z^a$, and again for degree
reasons we find  that $\phi R^{a,\ell}\w R^Z_{p+1}=0$. Continuing in this way
we can conclude that $\phi R^{a,\ell}\w R^Z=0$.

\smallskip
We now assume that \eqref{bolla} holds.
We have to prove that $R^Z.(dz\w\xi)\to 0$ when $\delta\to 0$, for
\begin{equation}\label{pelle}
\xi=\phi R^{a,\ell,\delta}\w\eta,
\end{equation}
with  test forms $\eta$ of bidegree $(0,*)$.

In view of the comments after the proof of
Proposition~\ref{DL} it is enough to prove that
each term \eqref{armada2} tends to zero if
\eqref{bolla} holds and $\mu$ is large enough
(independently of $(a)$ and $\ell$).
For this particular term we will see that we need $\mu\ge\mu_0$, where
\begin{equation}\label{val}
\mu_0=|\alpha|+2\min(m,n-p).
\end{equation}
For simplicity we omit all snakes from now on
and write $\phi$ rather than $\tilde \phi$ etc.
Moreover, we assume that $\ell=1$, the general case follows completely analogously.
Since $\tilde Z$ is smooth, by the usual Brian\c{c}on-Skoda theorem we have
that
\begin{equation}\label{pelle1}
 \phi\in( a)^{|\alpha|+\min(m,n-p)+1}.
\end{equation}
Notice that
$$
R^{a,\delta}_{k}=\chi'(|a|^2/\delta^2)\w\frac{\dbar|a|^2}{\delta^2}\w u^a_k, \quad k>0,
$$
and thus $R^{a,\delta}_k$ is a sum of terms like
$$
\chi' \frac{\dbar\bar a_{I_1}\w\ldots\w\dbar\bar a_{I_k}}{\delta^2|a|^{2k}}\bar a a\w \omega
$$
for $|I|=k$, where in what follows
$a^r$ denotes a product of $r$ factors $a_i$, and similarly with $\bar a^r$,
and $\omega$ denotes a smooth form.
For degree reasons $k\le \nu=\min(m, n-p)$.
In view of \eqref{pelle1} therefore
$\phi R^{a,\delta}_{k}$ is a sum of terms like
$$
\chi' \frac{\dbar\bar a_{I_1}\w\ldots\w\dbar\bar a_{I_\nu}}{\delta^2|a|^{2\nu}}
\bar a a^{2+\nu+|\alpha|}\w \omega
$$
plus lower order terms.

A straight forward computation yields that
$\partial_s^\alpha (\phi R^{a,\delta}_{k})$ is a finite sum of terms like
$$
\chi^{(r+1)} \frac{\dbar\bar a_{I_1}\w\ldots\w\dbar\bar a_{I_\nu}}
{\delta^{2(r+1)} |a|^{2(\nu+|\gamma|-r)}} \bar a^{1+|\gamma|} a^{2+\nu+|\gamma|}\w \omega,
$$
where $\gamma\le\alpha$ and $r\le|\gamma|$,
plus lower order terms.

We thus  have to see that each
\begin{equation}\label{stolle}
\int_s
\frac{ds_1\w\ldots\w ds_{\nu}}{s_1\cdots s_{\nu}}
\chi^{(r+1)} \frac{\dbar\bar a_{I_1}\w\ldots\w\dbar\bar a_{I_\nu}}
{\delta^{2(r+1)} |a|^{2(\nu+|\gamma|-r)}} \bar a^{1+|\gamma|} a^{2+\nu+|\gamma|}\w \omega
\end{equation}
tends to $0$ when $\delta\to 0$.
After a suitable further resolution we may assume that locally
$a=a_0a'$ where $a_0$ is holomorphic and $a'$ is a non-vanishing
tuple.
Then
$$
\dbar\bar a_{I_1}\w\ldots\w\dbar\bar a_{I_\nu}=\bar a_0^{\nu-1}\w \omega.
$$
Also notice that  the expression
\begin{equation}\label{stall}
\frac{ds_1\w\ldots\w ds_{\nu}}{s_1\cdots s_{\nu}}
\end{equation}
becomes a sum of similar expressions in this new  resolution.
Altogether we end up with a finite sum of terms like
$$
\int_s
\frac{ds_1\w\ldots\w ds_{\nu}}{s_1\cdots s_{\nu}}
\chi^{(r+1)} (|a|^2/\delta^2)\w \Ok(1),
$$
and each such integral tends to zero by dominated convergence.

The term corresponding to
$R^{a,\delta}_0=1-\chi(|a|^2/\delta^2)$ is handled in a similar but easier way.
\end{proof}

\section{Integral representation of the membership}\label{intrep}

Finally  we describe  how one can obtain an explicit integral
representation of the membership provided that the residue is annihilated.
The starting point is the formalism in \cite{A1} to generate integral
representations for holomorphic functions. Let $\delta_\eta$ denote
interior multiplication with the vector field
$$
2\pi i\sum_1^n (\zeta_j-z_j)\frac{\partial}{\partial\zeta_j}
$$
and let $\nabla_\eta=\delta_\eta-\dbar$.
A smooth form $g=g_0+g_1+\cdots +g_n$, where $g_k$ has bidegree $(k,k)$, is
called a {\it weight}  (with respect to $z$) if $\nabla_\eta g=0$ and $g_0(z,z)=1$.
Notice that the product of two weights is again a weight.

\begin{ex}\label{sopp}
Let $\chi$ be a cutoff function that is identically $1$ in a \nbh of the
closed unit ball, and let
$$
s=\frac{1}{2\pi i}\frac{\partial|\zeta|^2}{|\zeta|^2-\bar\zeta\cdot z}.
$$
Then $\nabla_\eta s=1-\dbar s$ and therefore
$$
g=\chi -\dbar\chi\w [s+s\w\dbar s+\cdots +s\w(\dbar s)^{n-1}]
$$
is a weight with respect to $z$ for each $z$ in the ball, with compact support,
and  it depends holomorphically on $z$.
\end{ex}

If $g$ is a weight with compact support and $z$ is holomorphic on the support,
then
$$
\phi(z)=\int g\phi =\int g_n\phi.
$$

Now consider a complex like \eqref{vcomplex}  in Section~\ref{utter},
 defined in a \nbh
of the closed ball, and let $U^\delta$ and $R^\delta$ be the associated
$E$-valued forms.
One can find, see \cite{A11} Proposition~5.3, holomorphic $E^*_k$-valued
$(k,0)$-forms $H^0_k$ and $\textrm{Hom}(E_k,E_1)$-valued
$(k-1,0)$-forms $H^1_k$ such that $\delta_{\eta} H^0_k=H^0_{k-1} f_k(\zeta)-f_1(z) H^1_k$
and $H^j_j=\textrm{Id}_{E_j}$.
Using that $\nabla_{f}U^{\delta}=1-R^{\delta}$ one verifies that
$$
f_1(z)HU^\delta+HR^\delta = 1-\nabla_{\eta}(\sum H^0_kU_k^{\delta}),
$$
where
$$
H U^{\delta}=\sum H^1_k U_k^\delta, \quad H R^\delta=\sum H^0_k R_k^\delta.
$$
It follows that $g^{\delta}:=f_1(z)HU^\delta+HR^\delta$ is a weight with respect to $z$.

If $g$ is, e.g., the weight from Example~\ref{sopp}  we thus get the
representation
$$
\phi(z)=\int g^{\delta}\wedge g \phi=f_1(z)\int HU^\delta\w g \phi+\int HR^\delta\w g\phi.
$$
Taking limits we obtain  the interpolation-division formula
\begin{equation}\label{pup}
\phi(z)=f_1(z)\int HU\w g \phi+\int HR\w g\phi.
\end{equation}
To be precise, the integrals here are the action of currents on smooth forms.
In particular, \eqref{pup} implies  that $\phi$ belongs to the ideal generated by $f_1$
if $\phi R=0$.

If we now choose as our complex the resolution of the sheaf $I=I_Z$, we get
the formula
$$
\phi(z)=\int g\w H^ZR^Z \phi, \quad z\in Z,
$$
for $\phi\in\Ok_Z$.
We then   replace $g$ by the weight
$g^{a,\ell,\delta}\w g$,
where

\begin{equation*}
g^{a,\ell,\delta}=a(z)^\ell\cdot  H^{a,\ell} U^{a,\ell,\delta}+ H^{a,\ell} R^{a,\ell,\delta};
\end{equation*}
here  $a(z)^\ell$ denotes  the first mapping in the complex
associated with $(a)^\ell$, cf., Example~\ref{bsex},  so that its entries
are elements in the ideal $(a)^\ell$.

We get

\begin{eqnarray*}\label{eq2}
\phi(z) &=& a(z)^\ell\cdot\int_{\zeta} H^a U^{a,\ell,\delta}\w
H^ZR^Z \phi \wedge g  \\
& &
+\int_{\zeta} H^a R^{a,\ell,\delta}\w
H^ZR^Z\wedge g \phi. \nonumber
\end{eqnarray*}
If the hypotheses in Theorem~\ref{boll}
are fulfilled, since $H^Z$, $H^a$ and $g$ are smooth, the second integral tends
to zero when $\delta\to 0$, and the  first integral on the right hand side
converges to an $E_1^{a,\ell}$-valued
holomorphic function.
Thus we get the  explicit representation
$$
\phi(z)=
 a(z)^\ell\cdot\int_{\zeta} H^a U^{a,\ell}\w
H^ZR^Z \phi \wedge g
$$
of the membership.

\def\listing#1#2#3{{\sc #1}:\ {\it #2},\ #3.}

config: arxiv


text:
\begin{document}

\begin{abstract}
P-E. Caprace and N. Monod isolate the class $\ms{X}$ of locally compact groups for which relatively amenable closed subgroups are amenable. It is unknown if $\X$ is closed under group extension. In this note, we exhibit a large, group extension stable subclass of $\X$, which suggests $\X$ is indeed closed under group extension. Along the way, we produce generalizations of the class of elementary groups and obtain information on groups outside $\X$.
\end{abstract}

\maketitle

\section{Introduction}
In \cite{CM14}, P-E. Caprace and N. Monod introduce the notion of relative amenability:

\begin{defn}
For a locally compact group $G$, a closed subgroup $H\sleq G$ is \textbf{relatively amenable} if $H$ fixes a point in every non-empty convex compact $G$-space.
\end{defn}
\noindent A convex compact $G$-space is a convex compact subset of a locally convex topological vector space such that the subset has a continuous affine $G$ action.

They go on to study the relationship between amenability and relative amenability. In particular, they isolate a large and interesting class of locally compact groups.

\begin{defn}
The class $\X$ is the collection of locally compact groups for which every relatively amenable closed subgroup is amenable.
\end{defn}

\begin{thm}[Caprace--Monod, {\cite[Theorem 2]{CM14}}]\label{thm:closure}
\
\begin{enumerate}[(a)]
\item $\X$ contains all discrete groups.
\item $\X$ contains all groups amenable at infinity.
\item $\X$ is closed under taking closed subgroups.
\item $\X$ is closed under taking (finite) direct products.
\item $\X$ is closed under taking adelic products.
\item $\X$ is closed under taking directed unions of open subgroups.
\end{enumerate}
Let $N\trianglelefteq G$ be a closed normal subgroup of a locally compact group $G$.
\begin{enumerate}[resume*]
\item If $N$ is amenable, then $G\in\X \Longleftrightarrow G/N\in\X$.
\item If $N$ is connected, then $G\in\X \Longleftrightarrow G/N\in\X$.
\item If $N$ is open, then $G\in\X \Longleftrightarrow N\in\X$.
\item If $N$ is discrete and $G/N\in\X$, then $G\in\X$.
\item If $N$ is amenable at infinity and $G/N\in\X$, then $G\in\X$.
\end{enumerate}
\end{thm}

Two questions concerning the class $\X$ arise:
\begin{quest}[Caprace--Monod]\label{quest:ext}
\
\begin{enumerate}[(1)]
\item Is $\X$ stable under group extension?
\item Are there locally compact groups outside of $\X$?
\end{enumerate}
\end{quest}

Our note contributes primarily to the study of the former. Indeed, let $\mathscr{Y}$ be the smallest collection of locally compact groups such that
\begin{enumerate}[(1)]
\item $\Y$ contains all compact groups, discrete groups, and connected groups,

\item $\Y$ is closed under group extensions, and

\item $\Y$ is closed under directed unions of open subgroups. That is to say, if $G=\bigcup_{i\in I}O_i$ where $\{O_i\}_{i\in I}$ is a directed system of open subgroups of $G$ such that $O_i\in \Y$ for each $i$, then $G\in \Y$.

\end{enumerate}

\noindent We prove the following:

\begin{thm}\label{thm:Y}
The class $\Y$ is contained in $\X$ and enjoys the following additional permanence properties:
\begin{enumerate}[(a)]
\item $\Y$ is closed under taking closed subgroups.

\item $\Y$ is closed under taking quotients by closed normal subgroups.
\end{enumerate}
\end{thm}

This provides evidence the class $\X$ may be stable under group extensions. Furthermore, it gives information on how groups $G \not \in \X$ look like; see Remark~\rm\ref{rem:counterexample} below.

\begin{ack*}
The author would like to thank Pierre-Emmanuel Caprace for his many helpful suggestions and for supporting a week research stay at the Universit\'{e} catholique de Louvain during which this note was partially developed. This note was further developed during a visit to the Fields Institute; the author thanks the Institute for its hospitality.
\end{ack*}

\section{Generalities on locally compact groups}

All groups are taken to be Hausdorff topological groups. We abbreviate ``locally compact" by ``l.c.", ``totally disconnected" by ``t.d.", and ``second countable" by ``s.c.". We write $H\sleq _oG$ to indicate $H$ is an open subgroup of $G$. We denote by $\U(G)$ the collection of compact open subgroups of $G$.

As its statement is not the obvious generalization, we recall the first isomorphism theorem for locally compact groups (see \cite[(5.33)]{HR79}):

\medskip
\textit{
Let $G$ be a locally group, $A\sleq G$ be closed, and $H\trianglelefteq G$ be closed. If $A$ is $\sigma$-compact  and $AH$ is closed, then $AH/H\simeq A/A\cap H$ as topological groups.}

\medskip
Locally compact $\sigma$-compact groups are close to being second countable.

\begin{thm}[{Kakutani--Kodaira, see \cite[(8.7)]{HR79}}]\label{KK}
If $G$ is a $\sigma$-compact \tdlc group, then there is a compact $K\trianglelefteq G$ such that $G/K$ is metrizable, hence second countable.
\end{thm}

A \tdlc group $G$ is said to be a \textbf{small invariant neighborhood group}, denoted SIN, if $G$ admits a basis at $1$ of compact open normal subgroups. These groups when compactly generated admit a useful characterization.

\begin{thm}[Caprace--Monod, {\cite[Corollary~4.1]{CM11}}]\label{CM_SIN}
A compactly generated \tdlc group is SIN if and only if it is residually discrete.
\end{thm}

The \textbf{discrete residual} of a \tdlc group $G$, denoted $\Res{}(G)$, is the intersection of all open normal subgroups. When $G$ is compactly generated, Theorem~\rm\ref{CM_SIN} implies $G/\Res{}(G)$ is a SIN group.

Groups ``built by hand" from profinite and discrete groups often appear when studying \tdlc groups. The class of elementary groups captures the intuitive idea of such groups.
\begin{defn}
The class of \textbf{elementary groups} is the smallest class $\E$ of \tdlcsc groups such that
\begin{enumerate}[(1)]

\item $\E$ contains all second countable profinite groups and countable discrete groups.

\item $\E$ is closed under taking group extensions of second countable profinite or countable discrete groups. I.e. if $G$ is a \tdlcsc group and $H\trianglelefteq G$ is a closed normal subgroup with $H\in \E$ and $G/H$ profinite or discrete, then $G\in \E$.

\item If $G$ is a \tdlcsc group and $G=\bigcup_{i\in \N}O_i$ where $(O_i)_{i\in \N}$ is an $\subseteq$-increasing sequence of open subgroups of $G$ with $O_i\in\E$ for each $i$, then $G\in\E$. We say $\E$ is \textbf{closed under countable increasing unions}.
\end{enumerate}
\end{defn}

The class $\E$ enjoys robust permanence properties, which supports the thesis $\E$ is exactly the groups ``built by hand."

\begin{thm}[{\cite[Theorem 3.18, Theorem 5.7]{W_1_14}}]\label{thm:closure_main}
$\E$ enjoys the following permanence properties:
\begin{enumerate}[(a)]
\item $\E$ is closed under group extension.

\item If $G\in \E$, $H$ is a \tdlcsc group, and $\psi:H\rightarrow G$ is a continuous, injective homomorphism, then $H\in \E$. In particular, $\E$ is closed under taking closed subgroups.

\item $\E$ is closed under taking quotients by closed normal subgroups.

\item If $G\in \E$, $H$ is a \tdlcsc group, and $\psi:G\rightarrow H$ is a continuous, injective homomorphism with dense normal image, then $H\in \E$.
\end{enumerate}
\end{thm}

The class $\E$ admits a canonical rank function:

\begin{defn}
The \textbf{decomposition rank} $\xi:\E\rightarrow [1,\omega_1)$ is an ordinal-valued function satisfying the following properties:
	\begin{enumerate}[(a)]
		\item $\xi(G)=1$ if and only if $G \simeq \{1\}$;

		\item If $G \in \E$ is non-trivial and $G=\bigcup_{i\in \N}O_i$ with $(O_i)_{i\in \N}$ some $\subseteq$-increasing sequence of compactly generated open subgroups of $G$, then
		\[
		\xi(G)=\sup_{i\in \N}\xi(\Res{}(O_i))+1.
		\]
	\end{enumerate}
\end{defn}

By \cite[Theorem 4.7]{W_1_14} and \cite[Lemma 4.10]{W_1_14}, such a function $\xi$ exists, is unique, and is equivalent to the decomposition rank as it is defined in \cite{W_1_14}.

The decomposition rank has a useful permanence property.

\begin{lem}[{\cite[Lemma 2.9]{RW_LC_16}}]\label{lem:compact_ext}
Suppose $G$ is a \tdlcsc group and $N$ is a closed cocompact normal subgroup of $G$. If $N\in \E$, then $G\in \E$ with $\xi(G)=\xi(N)$.
\end{lem}

\section{The class $\Es$}

We first argue $\E$ is a subclass of $\X$. To this end, we require a lemma due to Caprace and Monod.

\begin{lem}[{\cite[Lemma 13]{CM14}}]\label{lem:grp_ext}
Suppose $G$ is a locally compact group, $H\trianglelefteq G$ is closed, and $O\sleq_o G$ contains $H$. If $O\in \X$ and $G/H\in \X$, then $G\in \X$.
\end{lem}

\begin{thm}\label{key lem}
$\E\subseteq \X$.
\end{thm}
\begin{proof}We argue by induction on $\xi(H)$ for $H\in \X$. As the base case is obvious, suppose $\xi(H)=\beta+1$ and write $H=\bigcup_{i\in \N}O_i$ with $(O_i)_{i\in \N}$ an inclusion increasing sequence of compactly generated open subgroups of $H$.

Fixing $i\in \N$, the quotient $O_i/\Res{}(O_i)$ is a SIN group via Theorem~\ref{CM_SIN}. It is therefore compact-by-discrete, and Theorem~\ref{thm:closure} implies it is in $\X$. On the other hand, the definition of the decomposition rank gives that $\xi(\Res{}(O_i))\leq \beta$, so taking a compact open subgroup $V\leq O_i$, the group $V\Res{}(O_i)$ is elementary with rank at most $\beta $ via Lemma~\ref{lem:compact_ext}. Applying the induction hypothesis, we deduce that $V\Res{}(O_i)\in \X$. Lemma~\ref{lem:grp_ext} now ensures that $O_i\in \X$, and since $\X$ is closed under direct limits, we conclude that $H\in \X$, completing the induction.
\end{proof}

We next relax the second countability assumption on $\mathscr{E}$ by introducing the following class:

\begin{defn}
The class $\mathscr{E}^*$ is the smallest collection of \tdlc groups such that
\begin{enumerate}[(1)]
\item $\Es$ contains all profinite groups and discrete groups,

\item $\Es$ is closed under group extensions of profinite and discrete groups, and

\item $\Es$ is closed under directed unions of open subgroups.

\end{enumerate}
\end{defn}
\noindent There is an ordinal rank on $\Es$. For $G\in \mathscr{E}$, define

\begin{enumerate}[$\bullet$]
\item $G\in \Es_0$ if and only if $G$ is profinite or discrete.

\item Suppose $\Es_{\alpha}$ is defined. Put $G\in (\Es)_{\alpha}^e$ if and only if there exists $N\trianglelefteq G$ such that $N\in \Es_{\alpha}$ and $G/N\in \Es_{0}$. Put $G\in (\Es)_{\alpha}^l$ if and only if $G=\bigcup_{i\in I}H_i$ where $(H_i)_{i\in I}$ is an $\subseteq$-directed set of open subgroups of $G$ and $H_i\in \Es_{\alpha}$ for each $i\in I$. Define $\Es_{\alpha+1}:=(\Es)_{\alpha}^e\cup (\Es)_{\alpha}^l$.

\item For $\lambda$ a limit ordinal, $\Es_{\lambda}:=\bigcup_{\beta<\lambda}\Es_{\beta}$.
\end{enumerate}

Certainly, $\Es=\bigcup_{\alpha\in \ord} \Es_{\alpha}$, so for $G\in \Es$, we define
\[
\rk(G):=\min\{\alpha\in\ord\;|\;G\in \Es_{\alpha}\}.
\]
We call $\rk(G)$ the \textbf{construction rank} of $G$.

The construction rank has a number of nice properties.
\begin{lem} Let $G\in \Es$. Then
\begin{enumerate}[(a)]
\item $\rk(G)$ is a successor ordinal when $\rk(G)$ is non-zero.

\item If $G$ is compactly generated and has non-zero rank, then $\rk(G)$ is given by a group extension. I.e.\ if $\rk(G)=\beta+1$, there is $H\trianglelefteq G$ such that $\rk(H)=\beta$ and $\rk(G/H)=0$.

\item If $O\sleq_oG$, then $O\in \Es$ and $\rk(O)\sleq \rk(G)$.
\end{enumerate}
\end{lem}

\begin{proof}
These follow by transfinite induction on $\rk(G)$.
\end{proof}

\begin{prop}\label{Es E}
Let $G\in \Es$ be  $\sigma$-compact. If $K\trianglelefteq G$ is compact with $G/K$ second countable, then $G/K\in \E$.
\end{prop}
\begin{proof} We induct on the construction rank of $G$. As the proposition is immediate if $\rk(G)=0$, suppose $\rk(G)=\alpha +1$.

Suppose $\rk(G)$ is given by a directed union, so there is $(O_i)_{i\in I}$ a directed system of open subgroups of $G$ such that $G=\bigcup_{i\in I}O_i$ with $\rk(O_i)\sleq \alpha$ for each $i\in I$. Certainly,
\[
G/K=\bigcup_{i\in I}O_iK/K,
\]
and we may find a countable subcover $(O_iK/K)_{i\in \N}$ since $G/K$ is Lindel\"{o}f. One checks this cover may be taken to be an increasing $\subseteq$-chain. Further, $O_iK/K\simeq O_i/(O_i\cap K)$, $\rk(O_i)\sleq \alpha$, and $O_i\cap K$ is a compact normal subgroup of $O_i$ whose quotient is second countable. The induction hypothesis implies $O_i/(O_i\cap K)\in \E$, and as $G/K$ is the countable increasing union of $(O_iK/K)_{i\in \N}$, we conclude $G/K\in \E$.

Suppose the construction rank of $G$ is given by a group extension; say $H\trianglelefteq G$ is such that $\rk(H)=\alpha$ and $\rk(G/H)=0$. Since $H$ is $\sigma$-compact and $K$ compact, we have $HK/K\simeq H/(H\cap K)$, and as $\rk(H)= \alpha$, the induction hypothesis implies $H/(H\cap K)\in \E$. On the other hand, $HK/H\trianglelefteq G/H$ is closed, and the quotient
\[
(G/H)/(HK/H)\simeq G/HK\simeq (G/K)/(HK/K)
\]
is second countable and either discrete or compact. We conclude $G/K$ is a group extension of $G/HK$ by $HK/K$, hence $G/K\in \E$. This completes the induction, and we conclude the proposition.
\end{proof}

\begin{thm}\label{thm:es_x}
$\Es\subseteq \X$.
\end{thm}
\begin{proof}
Take $G \in \Es$. Every locally compact group is the directed union of its open compactly generated subgroups. Since $\X$ is closed under directed unions, we may assume $G$ is compactly generated and therefore $\sigma$-compact. Applying Theorem~\rm\ref{KK}, there is $K\trianglelefteq G$ such that $G/K$ is second countable. Proposition~\rm\ref{Es E} implies $G/K\in \E$, and by Theorem~\rm\ref{key lem}, we have that $G/K\in \X$. Since $K$ is compact, we deduce from Theorem~\rm\rm\ref{thm:closure} that $G\in \X$.
\end{proof}

We conclude this section by noting three permanence properties of $\Es$.

\begin{thm}\label{thm:es_closure}
$\Es$ enjoys the following permanence properties:
\begin{enumerate}[(a)]
\item $\Es$ is closed under group extension.

\item If $G\in \Es$, then every \tdlc group admitting a  continuous, injective homomorphism into $G$ also belongs to $ \Es$. In particular, $\Es$ is closed under taking closed subgroups.

\item $\Es$ is closed under taking quotients by closed normal subgroups.
\end{enumerate}
\end{thm}

\begin{proof}
Claim $(a)$ follows just in the case of $\E$; see the proof of \cite[Proposition 3.5]{W_1_14}.

\medskip \noindent
For $(b)$, let $G\in \Es$, let $H$ be a \tdlc group, and let $\psi \colon H \to G$ be a continuous, injective homomorphism.  Every locally compact group is the directed union of its compactly generated open subgroups. Since $\Es$ is closed under directed unions, we may therefore assume $H$ is compactly generated. Therefore, $P := \overline{\psi(H)}$ is also compactly generated.

Since $H$ is $\sigma$-compact, it has a compact normal subgroup $K$ such that $H/K$ is second countable; likewise, $P$ admits a compact normal subgroup $L\normal P$ so that $P/L$ is second countable. The subgroup $L\psi(K)$ is then a compact normal subgroup of $P$ so that $P/L\psi(K)$ is second countable. The group $H/\psi^{-1}(L)K$ is also second countable, and $\psi$ induces an injective, continuous homomorphism $\tilde{\psi}:H/\psi^{-1}(L)K\rightarrow P/L\psi(K)$.  Applying Proposition~\ref{Es E}, we conclude $P/L\psi(K) \in \E$ and, via Theorem~\rm\ref{thm:closure_main}, $H/\psi^{-1}(L)K \in \E$.

On the other hand, we have an injection $\psi^{-1}(L)K\rightarrow \psi(K)L$ with the latter group residually discrete. The group $\psi^{-1}(L)K$ is then residually discrete, and it follows from Theorem~\rm\ref{CM_SIN} that $\psi^{-1}(L)K\in \Es$. In view of part $(a)$, we deduce $H \in \Es$ as desired.

\medskip \noindent
For $(c)$, suppose $L\trianglelefteq G$ is closed; as above, it suffices to consider the case that $G$ is compactly generated. In view of Proposition~\rm\ref{Es E}, there is $K\trianglelefteq G$ compact such that $G/K\in \E$. The group $LK/K$ is a closed normal subgroup of $G/K$, so Theorem~\rm\ref{thm:closure_main} implies
\[
(G/L)/(LK/L)\simeq G/LK \simeq (G/K)/(LK/K) \in \E
\]
On the other hand, $LK/L\simeq K/(K\cap L)$ with the latter group compact. We conclude $G/L$ is compact-by-$\E$ and, via $(a)$, belongs to $\Es$.
\end{proof}

\section{The class $\Y$}

In order to prove Theorem~\rm\ref{thm:Y}, we introduce the class  $\Ys$  consisting of those locally compact groups $G$ such that $G/G^\circ \in \Es$ where $G^{\circ}$ is the connected component of the identity. By definition, we have that $\Es \subseteq \Y$, hence $\Ys \subseteq \Y$ since $\Y$ is stable under group extensions and contains all connected \lc groups. We will eventually show that $\Ys = \Y$. The proof of this relies on the following.

\begin{prop}\label{prop:Ys}
The class $\Ys$ enjoys the following permanence properties:
\begin{enumerate}[(a)]
\item $\Ys$ is closed under directed unions of open subgroups.

\item $\Ys$ is closed under taking closed subgroups.

\item $\Ys$ is closed under taking group extensions.

\item $\Ys$ is closed under taking quotients by closed normal subgroups.
\end{enumerate}
\end{prop}

The proof of (c)  requires the following subsidiary fact.

\begin{lem}\label{quot lem}
Let $G$ be a $\sigma$-compact \lc~group with closed $L,P\trianglelefteq G$ so that $L \geqslant G^\circ$.  If $P\in \Es$, then $\overline{PL}/L\in \Es$.
\end{lem}

\begin{proof} Set $H:=\overline{PL}/L$ and let $\pi:P\rightarrow H$ be the restriction of the usual projection. The image of $\pi$ is a dense normal subgroup of $H$. Theorem~\ref{thm:es_closure} ensures $\Es$ is closed under taking quotients, so we may assume $\pi$ is injective.

As $G$ is $\sigma$-compact, $P$ is $\sigma$-compact, and additionally, $H$ is $\sigma$ compact since $P$ has dense image in $H$. We may thus find $K\normal P$ and $R\normal H$ compact subgroups so that $P/K$ and $H/R$ are second countable. The subgroup $\pi(K)$ is then a compact normal subgroup of $H$ with $H/\pi(K)R$ second countable.

The induced map $\tilde{\pi}:P/K\pi^{-1}(R)\rightarrow H/\pi(K)R$ is an injective, continuous homomorphism with dense normal image. Appealing to Proposition~\ref{Es E}, we have $P/K\pi^{-1}(R)\in \E$. Theorem~\ref{thm:closure_main} thus implies $H/\pi(K)R\in \E$, and as $\Es$ is closed under group extensions, we conclude that $H\in \Es$, verifying the lemma.
\end{proof}

\begin{proof}[Proof of Proposition~\rm\ref{prop:Ys}]
(a) Let $G$ be the directed union of $(O_i)_{i\in I}$ with $O_i\sleq _oG$ and $O_i \in \Ys$ for each $i\in I$. For each $i$, we have $G^\circ = O_i^\circ\sleq O_i$ since $O_i$ is open. So $G/G^\circ=\bigcup_{i\in I}O_i/O_i^\circ$, and therefore, $G/G^\circ \in \Es$.

\medskip \noindent
(b) Given $H \sleq G$, we have that $H/H\cap G^{\circ}$ embeds continuously into $G/G^{\circ}$. By Theorem~\rm\ref{thm:es_closure}, this implies $H/H\cap G^{\circ} \in \Es$. On the other hand, $ H\cap G^\circ\sleq G^\circ$ with $G^\circ$ a connected locally compact group. By the solution to Hilbert's fifth problem, $G^\circ$ is pro-Lie. Let $K\trianglelefteq G^\circ$ be compact such that $G^\circ/K$ is a Lie group. In view of Cartan's theorem, see \cite[LG 5.42]{Se64}, $(H\cap G^\circ)K/K$ is a Lie group. Putting $J:=K\cap H\cap G^\circ$, we have
\[
H^\circ J/J=\left(H\cap G^\circ /J\right)^\circ \sleq_o(H\cap G^\circ)/J,
\]
so $(H\cap G^\circ)/ H^\circ J$ is discrete. Since we may find such $J$ inside arbitrarily small neighborhoods of $1$, the group $(H\cap G^\circ)/ H^\circ$ is residually discrete. Via Theorem~\rm\ref{CM_SIN}, we infer $(H\cap G^\circ)/H^{\circ}$ is a directed union of SIN groups and so in $\Es$. Since $\Es$ is closed under group extension, $H/H^\circ\in \Es$, whereby $H \in \Ys$ as desired.

\medskip \noindent
(c) Let $H \trianglelefteq G$ be such that $H$ and $G/H$ both belong to $\Ys$. To show $G \in \Ys$, we may reduce to $G$ compactly generated. Indeed, if $O\sleq_oG $ is compactly generated, then
\[
H^\circ=(O\cap H)^\circ\sleq O\cap H\trianglelefteq O
\]
and $(O\cap H)/H^\circ\sleq_o H/H^\circ$. We thus deduce that $(O\cap H)/H^\circ\in \Es$, whereby $O\cap H \in \Ys$. Similarly,
$O/(O\cap H)\in \Ys$. The group $O$ is thus a group extension of groups in $\Ys$. Therefore, if the claim holds for all compactly generated groups, then $O\in \Ys$, and since $\Ys$ is closed under directed unions, $G\in \Ys$.

Suppose $G$ is compactly generated, put $\tilde{G}:=G/(H\cap G^\circ)$, and let $\pi:G\rightarrow \tilde{G}$ be the usual projection. Since $H^\circ\sleq H\cap G^\circ$, $\tilde{H}:=\pi(H)$ is a quotient of $H/H^\circ$ and, via Theorem~\rm\ref{thm:es_closure}, a member of $\Es$. Setting $\tilde{G^\circ}:=\pi(G^\circ)$, we apply Lemma~\rm\ref{quot lem} to conclude $\overline{\tilde{H}\tilde{G^\circ}}/\tilde{G^\circ}\in \Es$. Since
\[
\overline{\tilde{H}\tilde{G^\circ}}/{\tilde G^\circ}\simeq \overline{HG^\circ}/G^\circ,
\]
we indeed have that $\overline{HG^\circ}/G^\circ\in \Es$.

The connected component of any locally compact group coincides with the intersection of all its open subgroups. This implies $\overline{HG^\circ}/H=(G/H)^\circ$, so
\[
(G/G^\circ)/(\overline{HG^\circ}/G^\circ)\simeq G/\overline{HG^\circ} \simeq (G/H)/(\overline{HG^\circ}/H)\in \Es.
\]
We conclude $G/G^\circ$ is an extension of a group in $\Es$ by another group in $\Es$ and thus a member of $\Es$. Therefore, $G\in \Ys$.

\medskip \noindent
(d) Let $G \in \Ys$ and $N \trianglelefteq G$. As noticed above, we have $\overline{NG^\circ}/N = (G/N)^\circ$. Therefore, in order to show that $G/N \in \Ys$, it suffices to show that $G/\overline{G^\circ N} \in \Es$. The latter is isomorphic to a quotient of $G/G^\circ  \in \Es$, so the desired conclusion follows from Theorem~\rm\ref{thm:es_closure}.
\end{proof}

\begin{cor} \label{cor:Y=Ys}
$\Y = \Ys \subseteq \X$.
\end{cor}
\begin{proof}
We have already observed that $\Ys \subseteq \Y$. Since $\Ys$ is stable under group extensions and directed unions of open subgroups by Proposition~\rm\ref{prop:Ys}, the reverse inclusion follows. Taking $G\in \Ys$, we see $G/G^{\circ}\in \Es\subseteq \X$ and, via Theorem~\rm\ref{thm:closure}, $G\in \X$.
\end{proof}

\begin{proof}[Proof of Theorem~\rm\ref{thm:Y}]
We have $\Y \subseteq \X$ by Corollary~\rm\ref{cor:Y=Ys}. The permanence properties of $\Y$ follow from Proposition~\rm\ref{prop:Ys}.
\end{proof}

\begin{rmk}\label{rem:counterexample}
The results of this note give new information concerning locally compact groups $G\notin\X$. As remarked in \cite{CM14}, we may take $G\notin \X$ to be a compactly generated \tdlc group. Applying Theorem~\rm\ref{KK}, we have a compact $K\trianglelefteq G$ such that $G/K$ is second countable. Theorem~\rm\ref{thm:closure} implies $G/K$ must also lie outside of $\X$. We may thus take $G\notin \X$ to be a compactly generated \tdlcsc group.

By \cite[Theorem 7.8]{W_1_14}, a \tdlcsc group $G$ admits a unique maximal closed normal elementary subgroup, denoted $\Rad_{\E}(G)$ and called the \textbf{elementary radical} of $G$. Suppose $G$ is a \tdlcsc group outside of $\X$ and fix $U\in \U(G)$. By Theorem~\rm\ref{key lem}, $\Rad_{\E}(G)\in \X$, and since $U\Rad_{\E}(G)\in \E$, we further have $U\Rad_{\E}(G)\in \X$. In view of Lemma~\rm\ref{lem:grp_ext}, it must be the case $G/\Rad_{\E}(G)\notin \X$. We may thus suppose $G\notin \X$ has trivial elementary radical, and via \cite[Corollary 9.12]{W_1_14}, $G$ is $[A]$-semisimple. The definition and a discussion of $[A]$-semisimple groups may be found in \cite{CRW_1_13}. Here we merely recall that $[A]$-semisimple groups have a canonical action on a lattice and, in many cases, a non-trivial boolean algebra \cite{CRW_1_13}.

\end{rmk}
\bibliography{biblio1}
\bibliographystyle{bibgen}
\end{document}


TITLE: According to the epsilon delta definition, Is a function differentiable at its end points $[a,b]$?
QUESTION [1 upvotes]: $$∀x\in A,   ∀ϵ>0, ∃δ>0, \text{ s.t. } |x−c| <δ ⟹  \left|\frac{f(x) − f(c)}{x-c}  −L\right|  < ϵ $$
My doubt is that given the epsilon delta definition of differentiability, if a function is right differentiable at $x=a$ in the interval $[a,b]$ and it is continuous at $x=a$, then it should be differentiable at $x=a$. I think so because for all points left of $x=a$, the statement is vacuously true, because all points left of $x=a$ are outside the domain $A$ of the function. Thus those points fail to qualify the $∀x\in A$ requirement. Hence, if a function is right differentiable and continuous at $x=a$, the derivative should exist at $x=a$.
Is my logic correct?

REPLY [0 votes]: Yes, you are correct. Actually, you do not need to assume that $f$ is continuous at $a$. If $f$ is right differentiable at $a$, then that will occur automatically.


TITLE: Solving Using Integrating Factors Found By Inspection
QUESTION [0 upvotes]: Can the Differential Equation below be solve using Integrating Factors Found By Inspection? If yes, how?
$$2x^5y'=y(3x^4+y^2)$$

REPLY [0 votes]: OK OK another way!
substitute : $y=(zx)^{2}$
$y'=2xz^{2}+2zz'x^{2}$
$2x^{5}(2xz^{2}+2zz'x^{2})=(zx)^{2}(3x^{4}+(zx)^{4})$
$z(2(2z+2z'x)-z(3+z^{4}))=0$
$z=0$,$y=(zx)^{2}$,$y=0$
$2(2z+2z'x)-z(3+z^{4})=0$
$4z'x=z^{5}-z$
$4{\int (z^{5}-z)^{-1}dz}={\int x^{-1} dx}$
$ln(-z^{4}+1)-4ln(z)=ln(x)+ln(c)$
$z^{-4}-1=cx$
$y=(zx)^{2}$
$x^{4}y^{-2}-1=cx$
$x^{3}y^{-2}-x^{-1}=C$


TITLE: Show there is a continuous isomorphism $l_{\infty}\rightarrow \left(l_1\right)^* $
QUESTION [3 upvotes]: Let $\left(l_1\right)^*$ be the dual space to $l_1$. Each $f \in \left(l_1\right)^*$ is a continuous linear functional over $l_1$. There is constant $C \in \Bbb R$ such that $|f(x)|\le C|x|_1, \forall x\in l_1$, (note that we denote the $L^1$-norm of $x\in l_1 $ by $|x|_1$).
Also, we define the norm for $f \in \left(l_1\right)^*$ by$$||f||:= sup \{|f(x)|:\forall x\in l_1 \text {such that }|x|_{1}=1\}$$
For each $\left(b_n\right)\in l_{\infty}$, define its norm by $|(b_n)|_{\infty}:=sup\{a_n:\forall n\}$, where $l_{\infty}$ is the vector space of all bounded real sequences.
Show that there is a continuous vector space isomorphism $M:l_{\infty}\rightarrow \left(l_1\right)^* $ such that $$||M(x)||\le C|x|_{\infty} $$ for some constant $C$.
My thought is to show that as in the finite-dimensional case, check that a bounded sequence $b:=(b_n)$ defines a linear functional $$f_b:(a_1,a_2,...)\in l_1 \mapsto \sum_{n=1}^{\infty} b_na_n \in \Bbb R$$, but I'm not sure how to show this exactly, and stuck on showing the following steps. Could someone give a complete proof of the question please? Any help is appreciated. Thanks a lot.

REPLY [1 votes]: You have the right idea. Check that the partial sums $\sum^n_{i=1} |b_na_n|$ are bounded above by $\|b\|_\infty \|a\|_1$ to see that the series you used to define $f$ converges and gives a functional of norm at most $\|b\|_\infty$. To see that $\|f_b\| = \|b\|_\infty$ consider what happens when you apply $f_b$ to basis vectors of $l_1$. That'll tell you that $b \mapsto f_b$ is an isometric linear map of $l_\infty$ into $l_1^*$. To see that it's surjective, fix a functional $f$ in $l_1^*$ and find an element $b$ of $\ell_\infty$ such that $f_b$ and $f$ agree on basis vectors for $l_1$.


TITLE: How do I prove that a statistic is a pivot?
QUESTION [1 upvotes]: In this example I have a sample out of a distribution with density $P_{\theta}(x) = 2x^{\theta} e^{−\theta x^2} I_{x \ge 0}$.
I know that for all $i \ge 1$ we have $X_i^2 \sim Exp(\theta)$.
If $\theta = 1$, then $T_1 = 2\sum_{1 \le i \le n}X_i^2 \sim X_{2n}^2$.
And now I have to show that $T_{\theta} = 2\sum_{1 \le i \le n} \theta X_i^2$ is a pivot.
A pivot is a function of which the probability function does not depend on $\theta$.
But how do I show that $T_{\theta}$ does not depend on $\theta$?
Btw: I tried to plug it first in LaTeX, so that it's easier to read. But that doesn't work.

REPLY [0 votes]: Fix $x \ge 0$. As $X_i^2 \sim exp(\theta)$ it follow that
$$P(\theta X_1^2 \le x) = P(X_1^2 \le \frac{x}{\theta }) = P( exp(\theta) \le \frac{x}{\theta })=1 - e^{- \frac{x}{\theta } \theta} = 1 - e^{-x} =P( exp(1) \le x).$$
So, we get that $\theta X_1^2 \sim \exp(1)$. Thus $Y_1 = \theta X_1^2$ is pivot. Hence $2 \sum_i Y_i$ is pivot as sum of independent pivot statistics
.


TITLE: Example of a topology on R except usual topology, with exactly one limit point for each converging sequence
QUESTION [0 upvotes]: A convergent sequence in R has exactly one limit point, if R is under usual topology. Give an example of another topology on R with this property.

REPLY [1 votes]: Given a topological space $(Y,\tau)$ and a bijective function $f:X\to Y$, we can define a topology $\tau_f$ on $X$ such that $f:X\to Y$ is a homeomorphism.
 Namely:
$$\tau_f=\{U\subseteq X\mid f(U)\in\tau\}$$
This is useful, because a lot of topological spaces have the same cardinality as $\mathbb R$.
Now, while it is not true for all topologies, all metric spaces have unique limits. So if we can find a metric space with cardinality equal to $\mathbb R$ which has properties different from $\mathbb R$, we are done.
All of the examples in comments above (other than the half-open intervals) are such metric spaces:

$[0,1]$ is a compact metric space with cardinality $\mathbb R$, so there is a topology on $\mathbb R$ which is compact. Since the usual topology on $\mathbb R$ is not compact, this new topology is not homeomorphic to the usual.
$\mathbb R\setminus \{0\}$ with the usual topology is not path-connected, and has the same cardinality as $\mathbb R$, so there is a topology on $\mathbb R$ which is not path-connected. $\mathbb R$ is path connected, so this new topology is not homeomorphic to the usual topology.
The discrete topology on any set $X$ is metrizable with $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ if $x\neq y$. If $|X|=|\mathbb R|$, then this gives a discrete topology on $\mathbb R$.
If $X=\mathbb R^2$ with the standard topology, then $|X|=|\mathbb R|$ and we can see that $X$ is not homeomorphic to $\mathbb R$ with the standard topology because if we remove a point from $X$, it is still path-connected, but that is not true for $\mathbb R$ with the usual topology.

The half-open topology is interesting because it is a "bigger" topology than the standard topology. It has, as basis, the half-open intervals $[a,b)$, but it includes all standard open sets by seeing that $(a,b)=\bigcup_{n>0}[a+\frac{1}{n}+a,b)$. Limits are unique here from the more general result:

Given a space $(X,\tau)$ with unique limits, and another topology $\tau'\supset \tau$ on $X$, we get that $(X,\tau')$ also has unique limits.

This is easy to prove. Essentially, if $x_n\to x$ in $(X,\tau')$ then $x_n\to x$ in $(X,\tau)$, but not necessarily visa versa. So adding to the open sets gives no additional convergences, but it can remove convergences.
For example, in the half-open topology, $x_n=\frac{-1}{n}$ doesn't converge, because $[0,1)$ is an open set containing $0$ which doesn't contain any $x_n$. But it can't converge to any other value because if $a\neq 0$ then the interval $(a-|a|/2,a+|a|/2)$ contains only finitely many $x_i$. Essentially, for each $a\neq 0$ we already have an open set $U\in\tau$ which is a "counterexample" showing that $x_n$ does not converge to $a$.
(In the half-open topology, a sequence $x_n\to x$ if and only if $x_n\to x$ in the usually topology and all but finitely many $x_i$ are greater than $x$.)
[This general result also proves the case of the discrete topology, again.]