text:
\begin{document}
\footnotetext{
{\rm 2020 MSC:}\  14M15, 14N15, 15A75.\\
	{\em Keywords and Phrases:} Hasse-Schmidt Derivation, Exterior Algebra, Universal Decomposition Algebra, Giambelli's formula, Bosonic and Fermionic Representations by Date-Jimbo-Kashiwara-Miwa.\\
}
\maketitle

\begin{abstract} \noindent The goal of this paper is to supply an explicit description  of the universal decomposition algebra of the generic polynomial of degree $n$ into the product of two monic polynomials, one of degree $r$,  as a representation of  Lie algebras of $n\times n$ matrices with polynomial entries. This is related with the bosonic vertex representation of the Lie algebra $gl_\infty$ due to Date, Jimbo, Kashiwara and Miwa.

\end{abstract}

\section*{Introduction}
\claim{\bf The Goal.} Let $X^n(c)$ denote the generic monic polynomial of degree $n\geq 1$
$$
X^n-c_1X^{n-1}+\cdots+(-1)^nc_n\in B_0(c)[X],
$$
where $B_0(c):=\QQ[c_1,\ldots,c_n]$ is the polynomial ring in the $n$
indeterminates $(c_1,\ldots,$ $ c_n)$.
The main purpose of this paper is {\em to provide an explicit description of the
universal decomposition algebra $B_{r,n}(c)$ of $X^n(c)$, $ 1\leq r\leq  n$,
into the product of two monic polynomials, one of degree $r$, as  a
representation of the Lie algebra $gl_n(B_0(c))$ of the $B_0(c)$-valued $n
\times n$ matrices}. Our description is achieved by computing the action on
$B_{r,n}(c)$ of the generating function of the  basis elements of  $gl_n(B_0(c))$
formed by the generic monic polynomials $X^i(c)\in V(c):=B_0(c)[X]$ of
degree  $0\leq i\leq n-1$, combining techniques and results by Laksov and Thorup on one hand and the notion of Schubert derivation on exterior algebras introduced by Gatto in \cite{SCHSD}. Putting all the $c_i=0$ one obtains, as a
particular case,  \cite[Theorem 4.7]{gln} by Gatto and Salehyan,
describing the singular cohomology $B_{r,n}$ of the complex Grassmannian
$G(r,n)$ as a module over the Lie algebra of $\QQ$-valued $n\times n$
matrices. The latter, as pointed out by those authors, can be seen as a finite dimensional counterpart of the celebrated
vertex representation of the Lie algebra $gl_n(\CC)$ due to Date, Jimbo,
Kashiwara and Miwa (DJKM) \cite{DJKM01,jimbomiwa}, which in turn  can be recovered as the limit for $r,n\sra \infty$, as shown e.g. in \cite{SDIWP}.

\claim{\bf Relationship with related literature.} It is well  known that the singular cohomology of the complex Grassmannian $G(r,n)$ is the universal decomposition algebra $B_{r,n}$ of the polynomial $X^n$ as the product of two monic polynomials, one of degree $r$ (see \cite{LakTh02}).  Thus, our main Theorem \ref{thm44} may be understood as  the extension of \cite[Theorem 4.7]{gln}, describing the cohomology of Grassmannians  as a representation of  the Lie algebra of the $n\times n$ square $\QQ$-valued matrices, to the cohomology of Grassmann bundles  $G(r,E)\sra X$ over a smooth variety of dimension at least $n$, once the coefficients of the generic polynomial $X^n(c)$  are interpreted as the Chern classes of $E\sra X$. We have chosen, however, to adopt in our exposition an  entirely algebraic language, to keep at a minimum the  prerequisites required to the  readership.

The  approach we propose, consists in blending and, in a sense, reconciling two alternatives ways to look at Schubert Calculus via exterior powers due, on one hand,  to Laksov and Thorup \cite{LakTh01,LakTh02,LakTh06} (see also \cite{LakEq1,LakEq2} for the equivariant version) and, on the other hand, to Gatto and collaborators who rely on the notion of Hasse-Schmidt derivations on a exterior algebra \cite{SCHSD,SCGA,ESC, HSDGA}. The special case relevant for our applications  is the Schubert derivation,  whose kinship with the Boson-Fermion correspondence and the vertex operators arising in the representation theory  of Heisenberg algebras has been highlighted in \cite{HSDGA,pluckercone,gln,SDIWP}. The rest of this introduction shall be devoted to describe more precisely our main results and methods while the last part will be concerned with the content of the paper, section by section.
\claim{} The universal decomposition algebra $B_{r,n}(c)$ of $X^n(c)$ is a quotient of $B_r(c):=B_0(c)[e_1,\ldots,e_r]$. Let $\pi_{r,n}:B_r(c)\sra B_{r,n}(c)$ be the canonical projection and let
$$E_r(z)=1-e_1z+\cdots+(-1)^re_rz^r.
$$ Let $V(c):=B_0(c)[X]$ and $V_n(c)=V(c)/(X^n(c))$. The first observation is that $B_{r,n}(c)$ is a $gl(V_n(c))$-module. The reason is that by \cite[The main Theorem 0.1]{LakTh01}, there is a $B_0(c)$-module isomorphisms
\be
B_{r,n}(c)\sra \bw^rV_n(c),
\ee
mapping certain deformations of Schur polynomials $\Delta_\blamb(H_{r,n}
(c))$, forming a basis of $B_{r,n}(c)$ parametrised by partitions whose
Young diagram is contained in a $r\times (n-r)$ rectangle, to the basis
element $\wXc^r_\blamb:=X^{r-1+\lambda_1}(c)\w\cdots \w X^{\lambda_r}(c)$
(Cf. \ref{thm:lakth}).
As the exterior $\bw V_n(c)$ represents canonically the $B_0(c)$
endomorphisms of $V_n(c)$ (Cf. Section~\ref{sec2:glnwr}) and  the exterior
powers are irreducible sub-representations, it follows that $\bw^rV_n(c)$ is
itself  a $gl(V_n(c))$-module for all $r\geq 0$. Thus $B_{r,n}(c)$ gets
a $gl(V_n(c))$-module structure by transporting that of $\bw^rV_n(c)$ via
the Laksov-Thorup isomorphism. The core of our paper consists in providing
the explicit description of  the $B_{r,n}(c)$ representation of $gl(V_n(c))$
by exploiting the notion of Schubert derivation introduced in \cite{SCHSD},
extensively treated in \cite{HSDGA} and used in \cite{SDIWP} to supply an
alternative deduction of the generating function accounting for the bosonic
representation of $gl_\infty$ after DJKM.
Let $(\d^i(s))_{0\leq i\leq n-1}$
be the basis of $V_n(c)^*$, dual of $(X^i(c))_{0\leq i\leq n-1}$. Thus
$$
gl(V_n(c))=\bigoplus_{i,j=0}^{n-1}B_0(c)\cdot X^i(c)\otimes \d^j(s),
$$
and we consider the generating function of the basis of $gl(V_n(c))$,
$$
\Ecal_n(z,w^{-1})=\displaystyle{\sum_{i,j=0}^{n-1}}X^i(c)\otimes \d^j(s)z^iw^{-j}.
$$
Our main result is the description of the action of $\Ecal_n(z,w^{-1})$ on any basis element of $B_{r,n}(c)$.

\medskip
\noindent
{\bf Theorem \ref{thm44}}
\begin{eqnarray*}
	&&\Ecal_n(z,w^{-1})(\Delta_\blamb(H_{r,n}(c)))\cr\cr
	&=&\pi_{r,n}{c(z)z^{r-1} \over c(w)E_r(z)}\begin{vmatrix}w^{-r+1-\lambda_1}(c)&w^{-r+2-\lambda_2}(c)&\cdots &w^{-\lambda_r}(c)\cr\cr
	\ovsig_-(z)h_{\lambda_1+1}(c)&\ovsig_-(z)h_{\lambda_2}(c)&	\ldots&\ovsig_-(z)h_{\lambda_r-r+2}(c)\cr\cr
	\vdots&\vdots&\ddots&\vdots\cr
	\ovsig_-(z)h_{\lambda_1+r-1}(c)&\ovsig_-(z)h_{\lambda_1+r-2}(c)&\ldots&\ovsig_-(z)h_{\lambda_r}(c)\end{vmatrix}
\end{eqnarray*}
where $\sum_{j\in\ZZ}h_j(c)z^j=\displaystyle{1\over E_r(z)}\in B_r(c)[[z]]$ and $\ovsig_-(z)$ (which is the differential part of a vertex operator) acts on $h_j(c)$ according to the rule $\ovsig_-(z)h_j(c)=h_j(c)-h_{j-1}(c)z^{-1}$ and $c(z)=1-c_1z+\cdots+(-1)^nc_nz^n$.
For instance if $n=4$ and $r=2$, we show that
\begin{eqnarray*}
 \Ecal_{4}(z,w^{-1})\Delta_{(2,1)}(H_{2,4}(c))&=&{h_2(c)\over w} + h_1(c)h_2(c){z\over w}+ h_2(c)^2 {z^2\over w} -{1\over w^3} \cr\cr &+& (h_1(c)^2-h_2(c)){z^2\over w^3}
+ h_1(c) h_2(c){z^3\over w^3}\label{eq4:eq25}
\cr\cr\cr
   &=&h_2(c)\,{1\over w}+\Delta_{(2,1)}(H_{2,4}(c)){z\over w}+\Delta_{(2,2)}(H_{2,4}(c)){z^2\over w}\cr\cr
   &-&{1\over w^3}+\Delta_{(1,1)}(H_{2,4}(c)){z^2\over w^3}+ \Delta_{(2,1)}(H_{2,4}(c)){z^3\over w^3}.
\end{eqnarray*}

Theorem \ref{thm44} is obtained from \ref{thm.thm210}, computing the $gl(V(c))$-structure of $B_r(c)$ by checking that the result can be harmless projected on $B_{r,n}(c)$.

\claim{\bf Organisation of the paper.} The paper as a whole is organised as
follows. Section \ref{sec1:sec1} is devoted to preliminaries and notation.
This section is quite necessary because in spite there are very precise
reference to Hasse-Schmidt derivations of exterior algebras (see also
\cite{GSCH} in addition to the previously cited ones), most of the terminology is not
standard yet,  although has already been mentioned by other authors like
e.g. in \cite{Nigran,tango} or, more recently,  in \cite[p.~116]{CoDuGu}. In
this section we also recall the theory of Laksov and Thorup together with
their definition of residue of an $r$-tuple of Laurent polynomials, as in
\cite{LakTh01}. Section \ref{sec2:sec2} begins by recalling the definition
of the Schubert derivation through the canonical $\bw V(c)$ representation
of the Lie algebra $gl(V(c))$ and  in Theorem  \ref{thmgmblt} we recall the
Laksov and Thorup version of the Giambelli formula holding in $B_r(c)$. In
Section \ref{sec3:sec3} we describe the $gl(V(c)$-module structure of
$B_r(c)=B_0(c)[e_1,\ldots,e_r]$. The outputs are Theorem \ref{thm.thm210}
and  Corollary \ref{cor3:cor22}, whose proofs, far from being a trivial
repetition of the main theorems of \cite{gln}, need to be suitably adapted
and require the overpassing of some subtleties. One of the main tools is an
easy but powerful formalism to describe the contraction of an exterior
monomial against a linear form, for which we do not know any explicit
reference to the literature, described in details and with examples in
Section~\ref{maintool}. The paper concludes with Section \ref{sec4:sec4}
devoted to our main result of describing through suitable generating
functions the universal decomposition algebra $B_{r,n}(c)$ as a
representation of the Lie algebra $gl(V_n(c))$. Here the main issue is to
check that the generating function $\Ecal_n(z,w^{-1})$ for $n=\infty$
preserves the ideal of relations passing from $B_r(c)$ to $B_{r,n}(c)$. The
other consists in getting rid in a efficient way of the combinatorics needed
to prove Theorem~\ref{thm44}, which we then illustrates with an explicit
Example for special low values of $r$ and $n$. This enables the readers to
check that the  formula  works by computing the image of a basis element of
$B_{r,n}(c)$ through a specific basis element of $gl(V_n(c))$. As we have said in case $c_i=0$ one gets back the main theorems of \cite{gln} and the case $r,n=\infty$ does correspond, as explained in \cite{gln,SDIWP}, with the DJKM representation of the Lie algebra $gl_\infty(\CC)$ on a polynomial ring $B$ (the bosonic Fock space) of polynomials in infinitely many indeterminates (as pedagogically explained  e.g. in \cite[Section 5.4]{KacRaRoz}).

\section{Preliminaries and Notations}\label{sec1:sec1}
\claim{} Let $B_0(c):=\QQ[c_1,\ldots,c_n]$ be the polynomial ring in the $n$
indeterminates $(c_1,\ldots,$ $ c_n)$ and  $V(c):=B_0(c)[X]$ the polynomial
ring with coefficients in $B_0(c)$ in the indeterminate $X$. Note that  $V(c) $ is a free  $B_0(c)$-module with basis either $\bfX:=(X^i)_{i\geq 0}$ or  $
\bfX(c):=(X^j(c))_{j\geq 0}$, where
$
X^j(c)=X^j+\sum_{i=1}^j(-1)^ic_iX^{j-i}.
$
By convention, we put $X^j=0$ if $j<0$ and $c_j=0$ if $j>n$.
Setting
\be
\bfc(z):=1-c_1z+c_2z^2+\cdots+(-1)^rc_rz^r.
\ee
Let $V(c)^*:=\bigoplus B_0(c)\d^j$ be the restricted dual of $V(c)$, where
$$
\d^j:={1\over j!}\left.{d^j\over dX^j}\right|_{X=0},
$$
such that $(\d^j(X^i)=\delta_j^i)$. It follows that
\be
gl(V(c)):=V(c)\otimes V(c)^*=\bigoplus_{i,j\geq 0}B_0(c)\cdot X^i\otimes \d^j.\label{eq1:xidi}
\ee
\claim{} We denote by $\bw V(c)$ the exterior algebra of $V(c)$. Notice that $\bw^0V(c)=B_0(c)$. For $r\geq 1$ we shall use the following two bases $(\wX^r_\blamb)_{\blamb\in\Pcal_r}$ and $(\wXc^r_\blamb)_{\blamb\in\Pcal_r}$, where $\Pcal_r$ denotes the set of all partitions of length at most $r$, and
$$
\wX^r_\blamb:=X^{r-1+\lambda_1}\w X^{r-2+\lambda_2}\w\cdots\w X^{\lambda_r},
$$
and
$$
\wXc^r_\blamb:=X^{r-1+\lambda_1}(c)\w X^{r-2+\lambda_2}(c)\w\cdots\w X^{\lambda_r}(c).
$$
\claim{\bf Hasse-Schmidt Derivations on $\bw V(c)$} Recall from \cite{HSDGA}
that a Hasse-Schmidt derivation n $\bw V(c)$ is a map $\Dcal(z): \bw V(c)
\sra \bw V(c)[[z]]$ such that $\Dcal(z)(u\w v)=\Dcal(z)u\w \Dcal(z)v$ for
all $u,v\in \bw V(c)$. By \cite{HSDGA, GSCH} any endomorphism $f$ of $V(c)$
uniquely determines a HS-derivation, whose restriction to $V(c)=\bw^1V(c)$ is precisely $f$.
\claim{\bf The universal decomposition algebra.} \label{sec1:uda}
Let $(e_1,\ldots,e_r)$ be indeterminates over $B_0(c)$. Let $B_r(c):=B_0(c)[e_1,\ldots,e_r]$. In the ring $B_r(c)[z]$ consider the generic polynomial  $E_r(z):=1-e_1z+\cdots+(-1)^re_rz^r$ and $H_r(z)$ implicitly defined by
\be
H_r(z)=\sum_{j\in\ZZ}h_jz^j={1\over E_r(z)}\in B_r(c)[[z]].\label{eq:Hrz}
\ee
In particular $h_j=0$ if $j<0$ and  $h_0=1$.
Let $H_r(c):=(h_0(c),h_1(c),\ldots)$ be the sequence defined by the equality
\be
\sum_{j\in\ZZ}h_j(c)z^j=\bfc(z)H_r(z)={\bfc(z)\over E_r(z)}.
\ee
In particular $h_j(c)=0$ if $j<0$, $h_0(c)=1$ and $h_1(c)=h_1-c_1$,
$h_2(c)=h_2-c_1h_1+c_2$, \ldots, $h_k(c)=h_k-\sum_{j\geq 0}c_jh_{k-j}$, with
the convention that $c_j=0$ if $j\geq n-1$. We also need the notation
\be
\ttp_r(X)=X^rE_r\left(X^{-1}\right)\label{eq1:ttprX}.
\ee
Let
\be
B_{r,n}(c):={B_r(c)\over (h_{n-r+1}(c),\ldots, h_n(c))},\label{eq2:brnc}
\ee
 and $\pi_{r,n}:B_r(c)\sra B_{r,n}(c)$, where $B_r(c):=B_0(c)[e_1,\ldots,e_r]$.
\bclm{\bf Proposition.} {\em The $B_0(c)$-algebra $B_{r,n}(c)$ is the universal decomposition algebra of the generic monic polynomial $X^n(c)$.}
\eclm
\proof The claim means that $X^n(c)$ decomposes in $B_{r,n}(c)[X]$ as the product of two monic polynomials, one of degree $r$. To see this, firts notice that in $B_{r,n}(c)$ we have the following decomposition
$$
\bfc(z)=E_r(z)(1+h_1(c)z+h_2(c)z^2+\cdots+h_{n-r}(c)z^{n-r}),
$$
using the variable $z=X^{-1}$ we have
$$
{X^n(c)\over \ttp_r(X)}=X^{n-r}(1+h_1(c)X^{-1}+\cdots+h_{n-r}(c)X^{n-r})=X^{n-r}+h_1(c)X^{n-r-1}+\cdots+h_{n-r}(c).
$$
We claim that
\be
\ttp_r(X)(X^{n-r}+h_1(c)X^{n-r-1}+\cdots+h_{n-r}(c)),\label{eq1:undec}
\ee
is the universal decomposition of $X^n(c)$ in $B_{r,n}(c)$.
 Indeed,  if $A$ is any $B_0(c)$--algebra where $X^n(c)=p(X)q(X)$ for some  monic polynomials
$$
p(X)=X^r-e_1(p)X^{r-1}+\cdots+(-1)^re_r(p),\qquad e_i(p)\in A,
$$
and
$$
q(X)=X^{n-r}+h_1(q)X^{n-r-1}+\cdots+h_{n-r}(q),\qquad h_i(q)\in A,
$$
of degree $r$ and $n-r$ respectively, then the unique $B_0(c)$-algebra
homomorphism $\varphi: B_{r,n}(c)\sra A$ defined by $e_i\mapsto e_i(p)$,
maps the universal decomposition \eqref{eq1:undec} of $X^n(c)$ to $p(X)q(X)
$.
As a matter of
$$
q(X)={X^n(c)\over p(X)}={\phi(X^n(c))\over \phi(\ttp_r(X))}=\phi\left(X^n(c)\over \ttp_r(X)\right)=\phi(X^{n-r}+h_1(c)X^{n-r-1}+\cdots+h_{n-r}(c)).
$$
\qed

A crucial result for our analysis is provided by the following:

\bclm{\bf Theorem.} \label{thm:lakth}{\em  There is a  $B_0(c)$-module isomorphism
$$
\bw^rV(c)\lra B_r(c),
$$
which turns $\bw^rV(c)$ into a free $B_r(c)$-module of rank $1$ generated by $\wX^r_0$.
}
\eclm
\proof
It is a consequence of \cite[Theorem 0.1 (The Main Theorem)]{LakTh01}, once
one rename $n$ by $r$ and identifies $B_0(c)$ with the algebra $S$ of
symmetric functions in $r$-indeterminates over the ring $A=B_0(c)$. Notice
that the above identification is possible due to the fundamental theorem of
symmetric functions \cite[(2.4) and Remark following it]{MacDonald}\qed

\bclm{\bf Definition.}\label{def1:def17} {\em Let $g_0(X),g_1(X),
\ldots,g_{r-1}(X)$ elements of $V(c)[[X^{-1}]]$. The {\em residue} of the
ordered $r$-tuple $(g_{r-1}(X),g_{r-2}(X),g_{0}(X))$ is defined as follows.
If $r=1$,
	\begin{center}
		$\Res (g_0(X))=$ coefficient in $B_0(c)$ of the power $X^{-1}$
		occurring in $g_0(X)$.
\end{center}
For $r>1$:
\begin{eqnarray*}
\Res\left(g_{r-1}(X),\ldots, g_{0}(X)\right)=\begin{vmatrix}\Res(g_{r-1}
(X))&\ldots&\Res(g_0(X))\cr\cr
	\Res(Xg_{r-1}(X))&\ldots&\Res(Xg_0(X))\cr
	\vdots&\ddots&\vdots\cr
	\Res(X^{r-1}g_{r-1}(X))&\ldots&\Res(X^{r-1}g_0(X))\end{vmatrix}.
	\end{eqnarray*}
}
\eclm

Let now  $f_0(X),f_1(X),\ldots,f_{r-1}(X)\in V(c)$ and recall notation  \eqref{eq1:ttprX}.
\bclm{\bf Theorem.}\label{thm:resthm}
\be
f_{r-1}(X)\w\cdots\w f_0(X)=\Res\left({f_{r-1}(X)\over\ttp_r(X)},\ldots,{f_{0}(X)\over\ttp_r(X)}\right)\wX^r_0\label{eq:resfor}
\ee
\eclm
\proof \cite[Main Theorem]{LakTh01}.

\section{The $B_r(c)$-module structure of $\bw^rV(c)$}\label{sec2:sec2}

\claim{} \label{sec2:glnwr} The module structure alluded to in Theorem~\ref{thm:lakth} can be described as follows.
Let us consider the standard representation of the Lie algebra
$$
\delta:gl(V(c))\sra \End_{B_0(c)}(\bw V(c)),
$$
mapping $\phi\mapsto \delta(\phi)$, where $\delta(\phi)(u)=\phi(u)$ for all $u\in V(c)$ and
$$
\delta(\phi)(v\w w)=\delta(\phi)v\w w+ v\w \delta(\phi)w,\qquad \forall v,w\in\bw V(c).
$$
In particular, we have:
\bclm{\bf Proposition.} {\em
The exterior power $\bw^rV(c)$ is an $gl(V(c))$-module.
}
\eclm
\proof Indeed, one gets a representation via the map $\phi\mapsto \delta(\phi)_{|\bw^rV(c))}$.\qed

\claim{} Relying on the above representation, we now describe more explicitly the module structure of $\bw^rV(c)$ over $B_r(c)$. Consider the endomorphism of $V(c)$ given by the multiplication by $X$. Define
$$
\sigma_+(z):=\sum_{i\geq 0}\sigma_iz^i=\exp\left(\sum_{i=1}{1\over i}\,\,\delta(X^i)z^i\right):\bw V(c)\lra \bw V(c)[[z]],
$$
which is the unique Hasse-Schmidt (HS) derivation on $\bw V(c)$ \cite{HSDGA} such that
$$
{\sigma_i}_{|V(c)}:V(c)\sra V(c),
$$
 is the multiplication by $X^i$.
We denote by $\ovsig_+(z)$ the inverse HS derivation:
$$
\ovsig_-(z):=\sum_{i\geq 0}(-1)^i\ovsig_iz^i=\exp\left(-\sum_{i=1}{1\over i}\,\,\delta(X^i)z^i\right):\bw V(c)\lra \bw V(c)[[z]].
$$

Then, the $B_r$-module structure of $\bw^rV(c)$ is achieved by declaring that

\be
h_iu=\sigma_iu\qquad \mathrm{or, \,\, equivalently}\qquad e_iu=\ovsig_iu,\label{eq:IImosstr}
\ee
for all $u\in\bw^rV(c)$, where
 $h_i\in B_r$ is defined by \eqref{eq:Hrz}. This is precisely the same considered by Laksov and Thorup as one infer by comparing the references \cite[Theorem 2.4]{SCHSD}  and  \cite[Sections 2.1--2.2]{LakTh01}.

In the following result we prove the Giambelli's formula\cite[Formula 5.43]{HSDGA}
\bclm{\bf Theorem.}\label{thmgmblt} {\em Let
\begin{eqnarray*}
\Delta_\blamb(H_r(c))=\begin{vmatrix}
		h_{\lambda_1}(c) & h_{\lambda_2-1}(c) &\ldots & h_{\lambda_r -r+1}(c) \\
		h_{\lambda_1+1}(c) & h_{\lambda_2}(c) &\ldots & h_{\lambda_r-r+2}(c) \\
		\vdots  & \vdots & \ddots & \vdots \\
		h_{\lambda_1+r-1}(c) & h_{\lambda_2+r-2}(c) &\ldots &  h_{\lambda_r}(c) \\
	\end{vmatrix}.
\end{eqnarray*}
Then $\wXc^r_\blamb=\Delta_\blamb(H_r(c))\wXc^r_0$.}
\eclm
\proof  By Laksov--Thorup residue formula \cite[Theorem 0.1 (2)]{LakTh01}:
\begin{eqnarray*}
	\wXc^r_\blamb&=&X^{r-1+\lambda_1}(c)\w X^{r-2+\lambda_2}(c)\w\cdots\w X^{\lambda_r}(c)\cr\cr
	&=&\Res\left({X^{r-1+\lambda_1}(c)\over \ttp_r(X)},{X^{r-2+\lambda_2}(c)\over \ttp_r(X)}, \ldots,{X^{\lambda_r}(c)\over \ttp_r(X)}\right)\wX^r_0.
\end{eqnarray*}
Now:
\begin{eqnarray*}
	&&\Res \left({X^k \cdot X^{r-j+\lambda_j}(c)\over \ttp_r(X)}\right)\cr\cr&=&
	\Res \left({X^k \cdot X^{r-j+\lambda_j}(c)\over X^r}\left(1+{h_1 \over X}+{h_2 \over X^2} + \cdots \right)\right)\cr\cr
	&=&\Res  X^{k-r}\left( X^{r-j+\lambda_j}-c_1 X^{r-j+\lambda_j-1}+c_2 X^{r-j+\lambda_j-2}-\cdots
	+(-1)^{r-j+\lambda_j}c_{r-j+\lambda_j}\right)\cr\cr
	&\cdot & 	\left(1+{h_1 \over X}+{h_2 \over X^2} + \cdots \right) \cr\cr
	&=&h_{\lambda_j+k-j+1}-c_1 h_{\lambda_j+k-j}+c_2 h_{\lambda_j+k-j-1}- \cdots + (-1)^{r-j+\lambda_j}c_{r-j+\lambda_j}\cr\cr
	&=&h_{\lambda_j+k-j+1}(c).
\end{eqnarray*}

The  definition \ref{def1:def17} of residue  and a simple computation gives that

\begin{eqnarray*}
	&&\Res\left({X^{r-1+\lambda_1}(c)\over \ttp_r(X)},{X^{r-2+\lambda_2}(c)\over \ttp_r(X)}, \ldots,{X^{\lambda_r}(c)\over \ttp_r(X)}\right)\hskip 20pt
	\cr\cr\cr &=&\begin{vmatrix}
		h_{\lambda_1}(c) & h_{\lambda_2-1}(c) &\ldots & h_{\lambda_r -r+1}(c) \\
		h_{\lambda_1+1}(c) & h_{\lambda_2}(c) &\ldots & h_{\lambda_r-r+2}(c) \\
		\vdots  & \vdots & \ddots & \vdots \\
		h_{\lambda_1+r-1}(c) & h_{\lambda_2+r-2}(c) &\ldots &  h_{\lambda_r}(c) \\
	\end{vmatrix}=\Delta_\blamb(H_r(c)).
 \end{eqnarray*}
 \qed

\bclm{\bf Proposition.} {\em  The Schur determinants $\Delta_\blamb(H_r(c))$ are $B_0(c)$-bases of $B_r(c)$. More precisely \be
B_r(c)=\bigoplus_{\blamb\in\Pcal_r}B_0(c)\cdot\Delta_\blamb(H_r(c)).
\ee
}
\eclm
\proof Indeed, a $B_0(c)$-basis of $\bw^rV(c)$ is given either by $\wXc^r_\blamb$ or $\wXc^r_\blamb$. Thus if $P(e_1,\ldots,e_r)\in B_r(c)$, then
$P(e_1,\ldots,e_r)\wXc^r_0\in \bw^rV(c)$ and can be then uniquely expressed as a  $B_0(c)$--linear combination of $\wXc^r_\blamb$ ($\wXc^r_\blamb$ respectively).
$$
P(e_1,\ldots,e_r)\wX^r_0=\sum_{\blamb}a_\blamb\wXc^r_\blamb=\sum_{\blamb}a_\blamb\Delta_\blamb(H_r)\wXc^r_0,
$$
from which
$$
P(e_1,\ldots,e_r)=\sum_{\blamb}a_\blamb\Delta_\blamb(H_r),
$$
for some unique $a_\blamb\in B_0(c)$. \qed

\bclm{\bf Corollary.} {\em The ring $B_r(c)$ is a $gl(V(c))$-module}.
\eclm
\proof It follows from transporting the $gl(V(c))$-module structure via the isomorphism claimed in Theorem~\ref{thm:lakth}.\qed

\section{The $gl(V(c))$-structure of $B_r(c)$}\label{sec3:sec3}

The purpose of this section is to appropriately define the $gl(V(c))$-structure of $B_r(c)$, having in view the description of the structure of $B_{r,n}$ as a module over the Lie algebra $gl(V_n(c))$, where $V_n(c):=V(c)/(X^n(c))$.
The isomorphism $B_r(c)\sra \bw^rV(c)$ implies the following:
\bclm{\bf Corollary.} {\em The ring $B_r(c)$ is a $gl(V(c))$-module.}
\eclm
\proof It follows from transporting the $gl(V(c))$-module structure of $\bw^rV(c)$ via the isomorphism claimed in Theorem~\ref{thm:lakth}.\qed

\bclm{\bf Definition.}\label{def:mndef}  {\em Let $f,g\in B_{0}(c)[t]$. The equality:
	\be
	\big[(f(X)\otimes g(\d))\star\Delta_\blamb(H_r(c))\big]\wXc^r_0=f(X)\w g(\delta)\lrcorner \wXc^r_\blamb,
	\ee
	defines $gl(V_n(c))$-structure of $B_r(c)$, i.e. the left hand side is defined through the right hand side.}

\claim{}\label{maintool} Our main tool to determine the explicit structure within a more general context than the one dealt with by Gatto and  Salehyan is a clever way to express the classical well known contraction of an exterior monomial against (a generating series of) linear forms.
Let $\d\in V(c)^*$ arbitrarily chose. Then an easy application of the definition shows that
\be
\d\lrcorner ( P_1(X)\w \cdots\w P_r(X))=\begin{vmatrix}\d(P_1(X)&\cdots&\d(P_r(X))\cr\cr P_1(X)&\cdots&P_r(X)\end{vmatrix}\label{eq:conset}.
\ee

The right-hand side of \eqref{eq:conset} means  an alternating sum  where the coefficient of $\d(P_i(X))$ is the element of $\bw^{r-1}V(c)$ obtained by removing the $j$-th wedge factor from $P_1(X)\w\cdots\w P_r(X)$. For instance, the coefficient of $\d(P_1(X))$ is $P_2(X)\w\cdots\w P_r(x)$ while that of $\d(P_2(X)$ is $-P_1(X)\w P_3(X)\w\cdots\w P_r(X)$.
\bclm{\bf Example.}

\medskip
\begin{eqnarray*}
	\d \lrcorner (P_1(X)\w P_2(X)\w P_3(X))&=&\begin{vmatrix}\d(P_1(X))&\d(P_2(X))&\d(P_3(X))\cr\cr\cr
		P_1(X)&P_2(X)&P_3(X)\end{vmatrix}\cr\cr\cr&=&\d(P_1(X))\cdot P_2(X)\w P_3(X)-\d(P_2(X))\cdot P_1(X)\w P_3(X)\cr\cr &+&\d(P_3(X)\cdot P_1(X)\w P_2(X).
\end{eqnarray*}
Notice that $\d(P(X))\in B_0(c)$.
\eclm

\bclm{\bf Example.} Using this formalism, let us see how Definition~\ref{def:mndef} works by computing  $(X^3\otimes \d^2)\star\Delta_{(2,1)}(H_2(c))$.
By definition:
\begin{eqnarray*}
	(X^3\otimes \d^2)\star\Delta_{(2,1)}(H_2(c))\wXc^2_0&=&X^3\w (\d^2 \lrcorner (X^{3}(c)\w X^1(c))\cr\cr
	&=&X^3\w \begin{vmatrix}\d^2(X^3(c))&\d^2(X(c))\cr\cr X^3(c)&X(c)
	\end{vmatrix}

	=X^3\w \begin{vmatrix}-c_1&0\cr\cr X^3(c)&X(c)
	\end{vmatrix}
	\cr	\cr\cr
	&=&-c_1X^3\w X(c)

	=-c_1X^3\w X+c_1^2X^3\w 1\cr\cr
	&=&(-c_1\Delta_{(2,1)}(H_2)+c_1^2h_2)\wXc^2_0,
\end{eqnarray*}
from which, finally:
$$
(X^3\otimes \d^2)\star\Delta_{(2,1)}(H_2(c))=-c_1\Delta_{(2,1)}(H_2)+c_1^2h_2=-c_1(h_1h_2-h_3)+c_1^2h_2.
$$
\eclm
\claim{} Define $\Ecal(z,w^{-1}):B_r(c)\sra B_r(c)[[z,w^{-1}]$ through:
\be
(\Ecal(z,w)\star\Delta_\blamb(H_r(c)))\wX^r_0=\sum_{i,j\geq 0}X^iz^i\w\d^jw^{-j}\lrcorner \wXc^r_\blamb.\label{eq1:defac}
\ee
\eclm
Our purpose is to find an expression for $\Ecal(w,c)\Delta_\blamb(H_r(c))$ in such a way that $(X^i\otimes\d^j)\Delta_\blamb(H_r(c))$ is the coefficient of $z^iw^{-j}$ in the product $\Ecal(z,w)\Delta_\blamb(H_r(c))$.

\bclm{\bf Lemma.} {\em  Let $f\in V(c)$. Then
$$
{\bm\partial}(w^{-1})(f(X))=f(w^{-1}).
$$
}
\eclm
\proof
Assume $f$ of degree $d\geq 1$. Then $f(X)=\sum_{i=0}^da_iX^d$, where $a_i\in
B_0(c)$. Thus
\begin{eqnarray*}
	{\bm\d}(w^{-1})(f(X))&=&\sum_{j\geq 0}w^{-j}\d^j(\sum_{i=0}^da_iX^i)\cr\cr
	&=&\sum_{i=0}^da_i\sum_{j\geq 0}w^{-j}\d^j(X^i)\cr\cr
	&=&\sum_{i=0}^da_i\sum_{j\geq 0}w^{-j}\delta^{ij}=\sum_{i=0}^da_iw^{-i}=f(w^{-1}).\hskip50pt \qed
\end{eqnarray*}
As a consequence, we get
\bclm{\bf Corollary.} ${\bm\partial}(w^{-1})(X^j(c))=w^{-j}(c)$.\qed
\eclm
\bclm{\bf Proposition.}\label{lem:lem26}{ \em
	\begin{eqnarray}
		&&\hskip-20pt {\bm \d}(w^{-1})\lrcorner (f_{r-1}(X)\w\cdots\w f_0(X))
		=\begin{vmatrix}f_{r-1}(w^{-1})& f_{r-2}(w^{-1})&\ldots&f_0(w^{-1})\cr\cr\cr
			f_{r-1}(X)& f_{r-2}(X)&\ldots&f_0(X)\cr\end{vmatrix}\cr\cr\cr

		&=&\begin{vmatrix}f_{r-1}(w^{-1})& f_{r-2}(w^{-1})&\ldots&f_0(w^{-1})\cr\cr
			\Res\,\displaystyle{f_{r-1}(X)\over \ttp_{r-1}(X)}&\Res\displaystyle{f_{r-2}(X)\over  \ttp_{r-1}(X)}&\ldots&\Res\displaystyle{f_{0}(X)\over \ttp_{r-1}(X)}\cr\cr
			\vdots&\vdots&\ddots&\vdots\cr\\
			\Res\displaystyle{X^{r-2}\cdot f_{r-1}(X)\over \ttp_{r-1}(X)}&\Res\displaystyle{X^{r-2}\cdot f_{r-2}(X)\over \ttp_{r-1}(X)}&\ldots&\Res\displaystyle{X^{r-2}f_{0}(X)\over \ttp_{r-1}(X)}\end{vmatrix}\wXc^{r-1}_0.\hskip40pt \label{eq:lakgmb}
	\end{eqnarray}
}\eclm

\proof The first equality of \eqref{eq:lakgmb} is obvious by virtue of \eqref{eq:conset}. To prove the second, first observe that $f_{r-1}(w^{{-1}})$ is the coefficient of the determinant
\begin{eqnarray*}
	&&\begin{vmatrix}\Res\,\displaystyle{f_{r-2}(X)\over  \ttp_{r-1}(X)}&\ldots& \Res\,\displaystyle{f_{0}(X)\over  \ttp_{r-1}(X)}\cr\vdots&\ddots&\vdots\cr\cr \Res\,\displaystyle{X^{r-2}f_{r-2}(X)\over  \ttp_{r-1}(X)})&\ldots& \Res\,\displaystyle{X^{r-2}f_{0}(X)\over  X^{r-2}\ttp_{r-1}(X)}
	\end{vmatrix}\wXc^{r-1}_0=\cr\cr\cr
	&=&\Res\left(f_{r-2}(X),\ldots, f_{0}(X)\right)\wXc^{r-1}_0=f_{r-2}(X)\w\cdots\w f_{0}(X),
\end{eqnarray*}
which is precisely the coefficients of $f_{r-1}(w^{-1})$ in the first member of \eqref{eq:lakgmb}. Then the property is proved because the determinant is alternating with respect to the columns.\qed

\bclm{\bf Definition.} {\em
	For all $\blamb\in \Pcal_r$, define
	\be
	\Delta_\blamb(\bfw^{-1}(c),H_{r-1}(c)):={w^{r-1}}\,{\bm\d}(w^{-1})\star \Delta_\blamb(H_r).\label{eq:3411}
	\ee
}
\eclm

\bclm{\bf Corollary.} {\em The explicit expression of the right hand side of \eqref{eq:3411} is:
	\begin{eqnarray*}
		\Delta_\blamb(\bfw^{-1}(c),H_{r-1}(c))
		&=&{w^{r-1}}\begin{vmatrix}w^{-r+1-\lambda_1}(c)&w^{-r+2-\lambda_2}(c)&\cdots& w^{-\lambda_r}(c)\cr\cr
			h_{\lambda_1+1}(c)&h_{\lambda_2}(c)&\cdots&h_{\lambda_r+r-2}(c)\cr\cr\vdots&\vdots&\ddots&\vdots
			\cr\cr
			h_{\lambda_1+r-1}(c)&h_{\lambda_2+r-2}(c)&\cdots&h_{\lambda_r}(c)
		\end{vmatrix}.
	\end{eqnarray*}
}
\eclm

\proof It is a straightforward consequence of Proposition~\ref{lem:lem26}, by choosing $f_{r-j}(X)=X^{r-j+\lambda_j}(c)$, for all $1\leq j\leq r$. \qed

For the next proposition it is useful to remark $\wX^k_0=\wXc^k_0$.

\bclm{\bf Lemma.} \label{lemw} {\em The following equality holds for all $k\geq 1$ and all $\blamb\in\Pcal_r$:
	\begin{eqnarray*}
		&&\wX^{k}_0\w \begin{vmatrix}f_{r-1}(w^{-1})&f_{r-2}(w^{-1})&\cdots& f_0(w^{-1})\cr\cr
			f_{r-1}(X)&f_{r-2}(X)&\ldots&f_0(X)\end{vmatrix}\cr\cr\cr
		&=&(-1)^{r+k-1}\begin{vmatrix}f_{r-1}(w^{-1})&f_{r-2}(w^{-1})&\cdots& f_0(w^{-1})&0&\ldots&0\cr\cr
			f_{r-1}(X)&f_{r-2}(X)&\ldots&f_0(X)&X^{k-1}&\ldots&X^0
		\end{vmatrix}.\\

	\end{eqnarray*}

}
\eclm
\proof By just comparing the coefficients of the expression $f_j(w^{-1})$ ($0\leq j\leq r-1$) on eithe side of the equality. \qed

\bclm{\bf Corollary.}
\begin{eqnarray*}
	&&\wX^{k}_0\w \begin{vmatrix}w^{-r+1-\lambda_1}(c)&w^{-r+2-\lambda_2}(c)&\cdots& w^{-\lambda_r}(c)\cr\cr
		X^{r-1+\lambda_1}(c)&X^{r-2+\lambda_2}(c)&\cdots&X^{\lambda_r}(c)
	\end{vmatrix}\cr\cr\cr
	&=&(-1)^{r+k-1}\begin{vmatrix}w^{-r+1-\lambda_1}(c)&w^{-r+2-\lambda_2}(c)&\cdots& w^{-\lambda_r}(c)&0&\ldots&0\cr\cr
		X^{r-1+\lambda_1}(c)&X^{r-2+\lambda_2}(c)&\cdots&X^{\lambda_r}(c)&X^{k-1}&\ldots&X^0
	\end{vmatrix}\\ &=&\wXc^{r+k}_{\blamb-(k^r)}.
\end{eqnarray*}

\eclm

\proof The formula is a straightforward consequence of Lemma~\ref{lemw}. Once one checks the formula for $k=1$, the other cases follow by induction. Clearly the result is zero if $\blamb-(k^r)$ were not a partition, i.e. if $\lambda_r-k<0$. \qed

\bclm{\bf Proposition.}\label{prop:lem} {
	\begin{eqnarray*}
		&&\sum_{i\geq 0} X^iz^i\w {\bm \d}(w^{-1}(c))\lrcorner \wXc^r_\blamb\cr&=&{z^{r-1}\over w^{r-1}}{1\over E_r(z)}\begin{vmatrix}w^{-\lambda_1}(c)&w^{1-\lambda_2(c)}&\ldots&w^{r-1-\lambda_r}(c)&0\cr\cr
			\ovsig_-(z)\cdot X^{r+\lambda_1}(c)&\ovsig_-(z) X^{r-1+\lambda_2}(c)&\ldots&\ovsig_-(z) X^{\lambda_r+1}(c)&1\end{vmatrix}\cr\cr\cr\cr
		&=&{z^{r-1}\over w^{r-1}}{1\over E_r(z)}\begin{vmatrix}w^{-\lambda_1}&\ldots&w^{r-1-\lambda_r}&0\cr\cr X^{r+\lambda_1}(c)-\displaystyle{X^{r-1+\lambda_1}(c)\over z}&\ldots&  X^{1+\lambda_r}(c)
			-\displaystyle{X^{\lambda_r}(c)\over z}&1\end{vmatrix}.
	\end{eqnarray*}
}
\eclm
\proof
It follows by noticing that

\begin{eqnarray*}
	&&\sum_{i\geq 0} X^iz^i\w {\bm \d}(w^{-1})\lrcorner \wXc^r_\blamb
	\cr\cr
	&=&\sigma_+(z)X^0\w w^{r-1}\begin{vmatrix}w^{-\lambda_1}&w^{1-\lambda_2}&\ldots&w^{r-1-\lambda_r}&0\cr\cr  X^{r-1+\lambda_1}(c)&X^{r-2+\lambda_2}(c)&\ldots&X^{\lambda_r}(c)&1\end{vmatrix}\cr\cr\cr\cr
	&=&w^{r-1}\sigma_+(z)\left(X^0\w \ovsig_+(z)\begin{vmatrix}w^{-\lambda_1}&w^{1-\lambda_2}&\ldots&w^{r-1-\lambda_r}\cr\cr  X^{r-1+\lambda_1}(c)&X^{r-2+\lambda_2}(c)&\ldots&X^{\lambda_r}(c)\end{vmatrix}\right),

\end{eqnarray*}
where in the last equality we have used integration by parts. The fact that $\ovsig_+(z)$ is a HS derivation translate in filtering the $\ovsig_+(z)$ inside the diagram, by letting it acting singularly on each element of the second row:

\begin{eqnarray*}
	&=&w^{r-1}\sigma_+(z)\left(X^0\w\begin{vmatrix}w^{-\lambda_1}&w^{1-\lambda_2}&\ldots&w^{r-1-\lambda_r}\cr\cr   \ovsig_+(z)X^{r-1+\lambda_1}(c)& \ovsig_+(z)X^{r-2+\lambda_2}(c)&\ldots& \ovsig_+(z)X^{\lambda_r}(c)\end{vmatrix}\right).
\end{eqnarray*}

The last equality is equivalent to
\begin{eqnarray}
	&=&w^{-r+1}z^{r-1}\sigma_+(z)\begin{vmatrix}w^{-\lambda_1}&w^{1-
	\lambda_2}&\ldots&w^{r-1-\lambda_r}&0\cr\cr   \ovsig_-(z)X^{r+\lambda_1}
	(c)& \ovsig_-(z)X^{r-1+\lambda_2}(c)&\ldots& \ovsig_-(z)X^{1+\lambda_r}
	(c)&1\end{vmatrix}\cr\cr\cr
	&=&{z^{r-1}\over w^{r-1}}\sigma_+(z)\ovsig_-(z)\begin{vmatrix}w^{-
	\lambda_1}&w^{1-\lambda_2}&\ldots&w^{r-1-\lambda_r}&0\cr\cr   X^{r+
	\lambda_1}(c)& X^{r-1+\lambda_2}(c)&\ldots&X^{1+\lambda_r}
	(c)&1\end{vmatrix}.
	\label{eq:psdet}
\end{eqnarray}

Since the ``determinant'' occurring in \eqref{eq:psdet} is a linear
combination of vectors of $\bw^{r}V$, which is an eigenspace of $\sigma_+(z)
$ having $\displaystyle{1\over E_r(z)}$ as eigenvalue, we eventually have:
$$
{z^{r-1}\over w^{r-1}}{1\over E_r(z)}\ovsig_-(z)\begin{vmatrix}w^{-
\lambda_1}&w^{1-\lambda_2}&\ldots&w^{r-1-\lambda_r}&0\cr\cr   X^{r+
\lambda_1}(c)& X^{r-1+\lambda_2}(c)&\ldots&X^{1+\lambda_r}
(c)&1\end{vmatrix}.\hskip100pt \qed
$$

\bclm{\bf Theorem.}\label{thm.thm210} {\em One has:
	\be
	\Ecal(z,w^{-1})\Delta_\blamb(H_r(c))= {z^{r-1}\over w^{r-1}}{1\over
	E_r(z)}\Delta_\blamb(\bfw^{-1}(c),\ovsig_-(z)H_r(c)).
	\label{eq:finalformula}
	\ee
}\eclm
\proof
By Definition~\ref{def:mndef}, the formal power series $\Ecal(z,w^{-1})
\Delta_\blamb(H_r(c))$ is the unique element of $B_r[[ z,w^{-1}]]$ such that
$$
\Ecal(z,w^{-1})\Delta_\blamb(H_r(c))\wXc^r_0=\sum_{i\geq 0} X^iz^i\w {\bm
\d}(w^{-1})\lrcorner \wXc^r_\blamb.
$$
Now, by Propostion~\ref{prop:lem}, one has
\begin{eqnarray*}
	&&\sum_{i\geq 0}X^iz^i\otimes {\bm\d}(w^{-1})\lrcorner \wXc^r_\blamb\cr\cr\cr
	&=&{z^{r-1}\over w^{r-1}}{1\over E_r(z)}\ovsig_-(z)\begin{vmatrix}w^{-\lambda_1}&w^{1-\lambda_2}&\ldots&w^{r-1-\lambda_r}&0\cr\cr   X^{r+\lambda_1}(c)& X^{r-1+\lambda_2}(c)&\ldots&X^{1+\lambda_r}(c)&1\end{vmatrix},\cr\cr
\end{eqnarray*}
and by invoking Lemma~\ref{lem:lem26}, eventually:
\be
{z^{r-1}\over w^{r-1}}{1\over E_r(z)}\Delta_\blamb(\bfw^{-1}(c),\ovsig_-(z)H_r(c)),
\ee
(Notice that we are now with coefficients in $B_r(c)$). Now by ~\cite[Theorem 5.7]{pluckercone}, $\ovsig_-(z)$ can filter inside the determinant, to give finally the required expression~(\ref{eq:finalformula}).\qed
\claim{\bf Remark.} Notice that for $c_1=\ldots=c_k=0$, we obtain the main formula of \cite{gln}. However our proof is more general and neater and we will see how in a while will allow us to describe the $gl(V_n(c))$ structure of the universal decomposition algebra $B_{k,n}(c)$ of the polynomial $X_n(c)\in B_0(c)[X]$.

\bclm{\bf Example.} Let us check the formula for $r=3$, applied to $1\in B_r$. It gives:
\begin{eqnarray}
	&&\Ecal(z,w^{-1})1={z^{2}\over w^{2}}{1\over E_3(z)}\Delta_{(0)}(\bfw^{-1}(c),H_3(c))\cr\cr\cr
	&=&{z^2\over E_3(z)}\begin{vmatrix}w^{-2}(c)&w^{-1}(c)&1\cr\cr h_1(c)-\displaystyle{1\over z}&1&0\cr\cr h_2(c)-\displaystyle{h_1(c)\over z}&h_1(c)-\displaystyle{1\over z}&1\end{vmatrix}\cr\cr\cr
	&=&{z^2\over E_3(z)}\left\{w^{-2}(c)-w^{-1}(c)\left(h_1(c)-{1\over z}\right)+\left[\left(h_1(c)-{1\over z}\right)^2-h_2(c)+{h_1(c)\over z}\right]\right\}\cr\cr\cr
	&=&{1\over E_3(z)}\left(1-h_1z+(h_1^2-h_2)z^2+{z\over w}(1-h_1z)+{z^2\over w^2}\right)\cr\cr\cr
	&=&{1\over E_3(z)}\left(E_2(z)+{z\over w}E_1(z)+{z^2\over w^2}\right).\label{eq:allbm}
	\end{eqnarray}

\eclm
For example
\begin{eqnarray*}
((X^5\otimes \d^1)1)\wX^3_0&=&X^5\w (\d^1\lrcorner X^2\w X^1\w X^0)\cr\cr
&=&-X^5\w X^2\w X^0=-\Delta_{(3,1)}(H_3)\wX^3_0,
\end{eqnarray*}
 from which
$(X^5\otimes \d^1)1=-\Delta_{(3,1)}(H_3)=h_4-h_1h_3$, which in turns is  precisely  the coefficient of $z^5w^{-1}$ in formula \eqref{eq:allbm}.

\claim{} Let $\bfs(w)=\displaystyle{\sum_{i\geq 0}}s_iw^i:=\displaystyle{1\over \bfc(w)}$
and define  $(\d^j(s))_{j\geq 0}$  through the equality:
\be
\sum_{i\geq 0}\d^j(s)w^{-j}=\bfs(w)\d(w^{-1})={\d(w^{-1})\over \bfc(w)}.\label{eq:eq16}
\ee
Then $(\d^j(s))_{j\geq 0}$ is the dual basis of $\bfX(c)=(X^i(c))_{i\geq 0}$ as it is easy to check.
\bclm{\bf Corollary.} \label{cor3:cor22} {\em  In terms of the adapted bases $\wXc$ and $(\d^j(s))_{j\geq 0}$ one has:
\be
(\sum_{j\geq 0}X^j(c)z^j\otimes \d^j(s)w^{-j})\star\Delta_\blamb(H_r(c))={z^{r-1}\over w^{r-1}}{c(z)\over c(w)}{1\over E_r(z)} \Delta_\blamb(\bfw^{-1},\ovsig_-(z)H_r(c)).\label{eq:eqcrl}
\ee
}
\eclm
\proof
Indeed, $\displaystyle{\sum_{j\geq 0}}X^j(c)z^j=\bfc(z)\sigma_+(z)1$ and by \eqref{eq:eq16}:
\begin{eqnarray*}
	(\sum_{j\geq 0}X^j(c)z^j\otimes \d^j(s)w^{-j})\star\Delta_\blamb(H_r(c))&=&\left(\bfc(z)\sigma_+(z)1\otimes {\d^j(w^{-j})\over \bfc(w)}\right)\Delta_\blamb(H_r(c))\cr\cr
	&=&{\bfc(z)\over \bfc(w)}(\sigma_+(z)\otimes \d^j(w^{-1})\Delta_\blamb(H_r(c)),
\end{eqnarray*}
whence \eqref{eq:eqcrl}, after applying Theorem~\ref{thm.thm210}.\qed

\section{The $gl(V_n(c))$-structure of $B_{r,n}(c)$.}\label{sec4:sec4}
\claim{} For all $k\geq 0$, denote
$$
V_k(c)=\bigoplus_{j=0}^{k-1}B_0(c)\cdot X^j(c)\qquad \mathrm{and}\qquad V^k(c)=\bigoplus_{j\geq k}B_0(c)\cdot X^{j}(c).
$$
Therefore:
$$
V(c)=V_k(c)\oplus V^k(c).
$$

By  \cite[Main Theorem 0.1]{LakTh01}, or our previous remarks,  there is a $B_0(c)$-module isomorphism between the universal decomposition algebra
$$
B_{r,n}(c):={B_r(c)\over I_{r,n}(c)},
$$
of the monic polynomial $X^n(c)$, where
$I_{r,n}(c)$ is the ideal $(h_{n-r+1}(c),\ldots, h_n(c))$, and  $\bw^rV_n(c)$, where
$$
V_n(c):={V(c)\over (X^n(c))}={B_0(c)[X]\over (X^n(c))}=\bigoplus_{i=0}^{n-1}B_0(c)\cdot X^i(c).
$$
Since $\bw^rV_n(c)$ represents $gl(V_n(c))$ via the map $\delta:gl(V_n(c))\sra End_{B_0(c)}\bw V_n(c))$, it turns out that $B_{r,n}(c)$ is itself a $gl(V_n(c))$-module.
Let $\pi_{r,n}:B_{r}(c)\sra B_{r,n}(c)$ the canonical epimorphism.
\bclm{\bf Lemma.} \label{lem:lem352} {\em Suppose $\blamb\in\Pcal_r$ such that $\lambda_1\geq n-r+1$. Then $\Delta_\blamb(H_r(c))\in I_{r,n}(c)$.
}
\eclm

\proof It is  an adaptation of similar proof in the book \cite[p.~113--114]{HSDGA}.
Recall that by its very definition,  $\Delta_\blamb(H_r(c))$ belongs to the ideal $(h_{\lambda_1}(c), h_{\lambda_1+1}(c),\ldots,h_{\lambda_1+r-1}(c))$. Thus, if $\lambda_1\geq n-r+1$, $\Delta_\blamb(H_r(c))\in I_{r,n}(c)$, as claimed.\qed

Let
\be
\Delta_\blamb(H_{r,n}(c))=\Delta_\blamb(H_{r}(c))+I_{r,n}(c)\label{eq:defhrn}.
\ee
Clearly  $\Delta_\blamb(H_{r,n}(c))=0$ in $B_{r,n}(c)$ if $\lambda_1\geq n-r+1$. But if $\blamb\in\Pcal_{r,n}$, i.e. the Young diagram of $\blamb$ is contained in a $r\times (n-r)$ rectangle, then $\Delta_\blamb(H_{r,n}(c)$ are linearly independent, because  they are in $B_r$. So $(\Delta_\blamb(H_{r,n}(c))$ is a $B_0(c)$-basis of $B_{r,n}(c)$.

We consider
the generating function
\be
\Ecal_n(z,w^{-1})=\sum_{i,j= 0}^{n-1}X^i(c)z^i\otimes \partial^j(s)w^{-j}:B_{r,n}\sra B_{r,n}[z,w^{-1}].
\ee Clearly this is a polynomial in $z$ and $w^{-1}$, because $(X^0,X(c),\ldots,X^{n-1}(c))$ is a basis of $V_n(c)$ and $X^{n+i}(c)=0$ for all $i\geq 0$.

\bclm{\bf Proposition.}\label{prop:prop43} {\em For all $0\leq j\leq n-1$
	$$
	X^i(c)\otimes \d^j(s)(I_{r,n}(c))\subseteq I_{r,n}(c).
	$$
}
\eclm
\proof
First recall that $I_{r,n}(c)\bw^rV(c)= V^n(c)\w \bw^{r-1}V(c)$. In fact
for all $\blamb\in\Pcal_r$ one has
$$
I_{r,n}(c)\wXc^r_\blamb=I_{r,n}(c)\Delta_\blamb(H_r(c)\wXc^r_0\subseteq I_{r,n}(c)\wXc^r_0.
$$
So, it suffices to show that $I_{r,n}(c)\wXc^r_0\in V^n(c)\w \bw^{r-1}V(c)$, i.e. that $h_{n-r+j}(c)\wXc^r_0\in V^n(c)\w \bw^{r-1}V(c)$ for all $1\leq j\leq r$.
But
$$
h_{n-r+j}(c)\wXc^r_0=X^{n+j-1}(c)\w X^{r-2}(c)\w\cdots\w X^0(c)\in V^n(c)\w\bw^{r-1}V(c),
$$
 which proves the inclusion $I_{r,n}(c)\bw^rV(c)\subseteq V^n(c)\w \bw^{r-1}V(c)$.  Conversely if $\wXc^r_\blamb\in V^n(c)\w\bw^{r-1}V(c)$, then $\lambda_1\geq n-r+1$. Thus $\wXc^r_\blamb=\Delta_\blamb(H_r(c))\wXc^r_0$, with $\Delta_\blamb(H_r(c))\in I_{r,n}(c)$, because $\lambda_1\geq n-r+1$ and Lemma~\ref{lem:lem352}.
Now if $j\leq n-1$
\begin{eqnarray*}
	(\d^j(s)\star I_{r,n}(c))\wXc^r_0&=&\d^j(s)(V^n(c)\w\bw^{r-1}V(c)\subseteq V^n(c)\w \d^j(s)\lrcorner \bw^{r-1}V(c)\cr\cr
	&\subseteq& V^n(c)\w\bw^{r-1}V(c)=I_{r,n}(c)\wXc^r,
\end{eqnarray*}
whence $\d^j(s)\star I_{r,n}(c)$. It follows that
\begin{eqnarray*}
	X^i(c)\otimes \d^j(s)\star(\Delta_\blamb(H_r(c))+I_{r,n})&=& X^i(c)\otimes \d^j(s)(\Delta_\blamb(H_r(c))+ I_{r,n}\cr\cr
	&=&V^n(c)\w \bw^{r-1}V(c). \hskip 60pt \qed
\end{eqnarray*}

We have then the following:
\bclm{\bf Theorem.}\label{thm44} {\em
	Let $\Ecal_n(z,w^{-1})=\displaystyle{\sum_{i,j=0}^{n-1}}X^i(c)\otimes \partial_j(s)z^iw^{-j}$.
	Then
\be
	\Ecal_n(z,w^{-1})(\Delta_\blamb(H_{r,n}(c)))={z^{r-1}\over w^{r-1}}\left(1+\sum_{i=1}^{n-1}s_iw^i\right){c(z)\over E_r(z)}\pi_{r,n}\Delta_\blamb(\bfw^{-1}(c),\ovsig_-(z)H_r(c)).\label{eq:brn}
\ee
}
\eclm
\proof By definition of $\Delta_\blamb(H_{r,n}(c))$, one has:

\begin{center}
\begin{tabular}{lrll}
&&$\Ecal_n(z,w^{-1})(\Delta_\blamb(H_{r,n}(c)))$&\cr\cr
&$=$&$\Ecal_n(z,w^{-1})(\Delta_\blamb(H_r(c))+I_{r,n}(c))$&(Definition \eqref{eq:defhrn}\cr
&&&of $\Delta_\blamb(H_{r,n})$\cr\cr
&$=$&$\Ecal_n(z,w^{-1})(\Delta_\blamb(H_r(c)))+I_{r,n}(c)$&(Proposition~\ref{prop:prop43})\cr\cr
&$\hskip-5pt =$&$\hskip-10pt \displaystyle{z^{r-1}\over w^{r-1}}\left(\displaystyle{1\over c(w)}\,\mod\, w^{n}\right)\pi_{r,n}\displaystyle{c(z)\over E_r(z)}\Delta_\blamb(\bfw^{-1}(c),\ovsig_-(z)H_r(c)),$&(Corollary 3.12).\cr\cr
&$=$&$\hskip-10pt \displaystyle{z^{r-1}\over w^{r-1}}\left(\displaystyle{1\over c(w)}\,\mod\, w^{n}\right)\pi_{r,n}\displaystyle{c(z)\over E_r(z)}\pi_{r,n}\Delta_\blamb(\bfw^{-1}(c),\ovsig_-(z)H_r(c)),$&
\end{tabular}
\end{center}

where in the last equality we used the fact that $\pi_{r,n}$, the canonical
epimorphism, is a ring homomorphism.\qed
\bclm{\bf Example.} Let us check the formula for $n=4$ and $r=2$ by
computing the generating function of the images of $\Delta_{(2,1)}(H_{2,4}
(c))$.
 Applying Theorem \ref{thm44}, we get
\begin{eqnarray}
	&&\Ecal_4(z,w^{-1})(\Delta_{(2,1)}(H_{2,4}(c)))=
	\Ecal_4(z,w^{-1})\Delta_{(2,1)}(H_2(c))+I_{2,4}(c)\cr\cr\cr
	&=&{z\over w}{c(z)\over c(w)}{1\over E_2(z)}\Delta_{(2,1)}(\bfw^{-1}(c),
	\ovsig_-(z)H_{2,4}(c))\cr\cr\cr
	&=&{z}{1\over c(w)}{c(z)\over E_2(z)}
	\begin{vmatrix}w^{-3}(c)&w^{-1}(c)\cr\cr h_3(c)-\displaystyle{h_2(c)
	\over z}&h_1(c)-\displaystyle{1\over z}\end{vmatrix}+I_{2,4}(c)\cr\cr\cr
	&=&{z}(1+s_1w+s_2w^2+s_3w^3)\left(1+h_1(c)z+ h_2(c)z^2 \right)\cdot\cr
	\cr
	&\cdot& \left(-c_3h_1(c) +{1\over z}(c_3-c_1h_2(c))+{1\over w}c_2h_1(c)\right.\cr\cr&&\left. -{1\over w^2}\,c_1h_1(c)+{1\over zw}(h_2(c)-c_2)+{h_1(c)\over w^3}+{c_1\over w^2 z}-{1\over w^3 z}\right)\cr\cr
	&=&

(1+h_1(c)z+h_2(c)z^2)\left({1\over w^3}(h_1(c)z-1)+{h_2(c)\over w}\right)\cr\cr\cr
&=& {h_2(c)\over w} + h_1(c)h_2(c){z\over w}+ h_2(c)^2 {z^2\over w} -{1\over w^3} + (h_1(c)^2-h_2(c)){z^2\over w^3}\cr\cr
&&
   + h_1(c) h_2(c){z^3\over w^3}\label{eq4:eq25}\\ \cr\cr
   &=&h_2(c)\,{1\over w}+\Delta_{(2,1)}(H_{2,4}(c)){z\over w}+\Delta_{(2,2)}(H_{2,4}(c)){z^2\over w}\cr\cr
   &-&{1\over w^3}+\Delta_{(1,1)}(H_{2,4}(c)){z^2\over w^3}+ \Delta_{(2,1)}(H_{2,4}(c)){z^3\over w^3}.
\label{eq4:eq26}
\end{eqnarray}

\noindent
For instance a direct computation gives:
\begin{eqnarray*}
((X^2(c)\otimes \d^{-1}(s))\Delta_{(2,1)}(H_{2,4}(c))\wXc^2_0&=&X^2(c)\w (\d^{-1}(s)\lrcorner X^{3}(c)\w X^1(c))\cr\cr
&=&X^3(c)\w X^2(c)=X^3\w X^2(c)=h_2(c)^2,
\end{eqnarray*}
which is precisely the coefficient of $z^2w^{-1}$ in either \eqref{eq4:eq26} or \eqref{eq4:eq25}.
\eclm

\bibliographystyle{amsplain}

config: arxiv


TITLE: How to show that finite rank operator on Banach space is compact
QUESTION [0 upvotes]: Let $X,Y$ be Banach space and $T:X\to Y$ bounded . If $T(X)$ has finite dimensional, then $T$ is compact.

Assume $\{y_i\}_{i=1}^N$ be a basis of $T(X)$ , $\{x_n\}$ be a bounded sequence in $X$, we have $$T(x_n)=\sum_1^N a_i(x_n)y_i$$
I want to show $\sup_{1\le i\le N}\sup_n |a_i(x_n)|\le M \lt \infty$. Since $\{z\in C| \,\,|z|\le M\}$ is compact , we can use subsequence trick to extract a convergent subsequence $\{T{x_n}_k\}$ .
Since $T(X)$ is compact in $Y$ , in particular , $T(X)$ is bounded . If for a subsequence ${x_n}_k$ , we have $|a_i({x_n}_k)| \to \infty$ , it seems that this can lead to a contradiction . So let's consider the following question:

Let $Y$ be a Banach space , $A=span\{y_1,...,y_n\}$ be a $n$ dimensional subspace of $Y$ . $B$ is a bounded subspace of $A$. Then for each $y=\sum a_i(y) y_i \in B$ can we show that $$\sup_{1\le i \le n} \sup_{y \in B} |a_i(y)| \lt \infty $$ ?

If $Y$ is a Hilbert space , then we can orthogonormal $\{y_i\}$ and then the result follows . If we only assume $Y$ be a Banach space , can we have the same result?

REPLY [1 votes]: Let $B$ be the closed unit ball of $X$.  Then, $T(B)$ is a bounded subset of a finite-dimensional space and is therefore precompact.  Therefore, $T$ is compact.


TITLE: Eigenvalues of a specific $9\times9$ matrix - a simpler way?
QUESTION [1 upvotes]: Let $ρ$ be the permutation of $\{1, \dots , 9\}$ be given by $$ρ=\bigg(\begin{array}  \\1 2 3 4 5 6 7 8 9 \\2 3 4 1 6 7 5 9 8 \end{array}\bigg)$$ and let $α : C^9 → C^9$ be the linear map defined
  by by $α(e_j) = e_{ρ(j)}$, $j = 1, \dots , 9$. What are the
  eignevalues of $α$?

I know how to find them directly, but I was wondering, if there's a way of doing it without calculating characteristic polynomial?

REPLY [3 votes]: To compute the characteristic polynomial is not that difficult in this case: if you draw the matrix of $\alpha$, you'll see that it's a block-diagonal matrix, with three blocks along the main diagonal. So the characteristic polynomial is the product of the characteristic polynomials of these blocks, which are
$$
t^2-1 \ , \quad -t^3 + 1 \quad \text{and}\quad t^4 -1 \ .
$$


TITLE: Enthalpy vs enthalpy of reaction
QUESTION [0 upvotes]: I have a problem with understanding basic concept about enthalpy.

I know there is an enthalpy (heat) of formation which tells how much energy is going to be released or consumed during chemical reaction. I found information that specific heat of formation for one of the hydrocarbon jet fuel is equal to -200 BTU/lbm. How to interpret that? If all the elemental 'building blocks' of the fuel undergoes reaction of mixing of whatever there will be 200 BTU/lbm heat released?

There must be second 'meaning' of enthalpy because I also found example problem of burning jet fuel with air, where in the given data is information that fuel enthalpy is equal to +189 BTU/lbm (there was no information if that's heat of formation). What does it mean that fuel has enthalpy? It sits in the tank, and before burning process is no chemical reaction

How to interpret both cases? What's the difference here. Generally we talk about enthalpy of fluid. Like compressor inlet flow enthalpy is equal to XXX, compressor exit flow enthalpy = YYY. There is no chemical reaction between inlet/outlet plane so ...
Could you give me a 'dummy' type explanation?
Thank you,
Lukasz from Poland

Let me paste one of my comments here:
This is how book describes combustion in jet engine:
*specific enthalpy of mix air and fuel
hIn = ( mAir * hAir + mFuel * hFuel ) / ( mAir + mFuel)
*specific enthalpy at the end of combustion process
hOut = hIn + (LHV * (mFuel/mAir) * efficiency)
Looks like that I can't understand what hFuel, and hAir mean in that case.

REPLY [2 votes]: The enthalpy of formation of a compound is found relative to some specified reference state (of temperature and pressure) for the elements comprising the compound.  The enthalpy of the elements comprising the compound is taken as zero in this reference state.  So the enthalpy of formation of the compound is the change in enthalpy in going by chemical reaction from an initial state of the pure elements in the reference state to the compound in the reference state; it is equal to the amount of heat that has to be added to the system such that the temperature of the product is the same as that of the elementary reactants.  This heat addition (or removal in the case of a negative heat of formation) is the result of the energy required to make and break chemical bonds.
Once you know the enthalpies of formation of a wide range of elements and compounds, you can determine the heat of reaction (enthalpy of reaction, equal to heat that has to be added to convert reactants into products) of any reactions involving these elements and compounds by applying Hess' law.  One example of such a reaction is combustion.


TITLE: Find $(a_1-\frac{1}{a_1})+(a_2-\frac{1}{a_2})+⋯+(a_5-\frac{1}{a_5})$ if $a_i$'s are eigenvalues of $A$
QUESTION [0 upvotes]: Please help to reach answer for the following problem:

$A∈M_5(R)$ is a matrix and in relation to $A^2-4A-I=0$. if
  $a_1,a_2,a_3,a_4,a_5$ is eigenvalues for A; solve the following
  problem:
$(a_1-\frac{1}{a_1})+(a_2-\frac{1}{a_2})+⋯+(a_5-\frac{1}{a5})$

Thanks a lot.

REPLY [0 votes]: The polynomial $x^2-4x-1$ annihilates the matrix $A$ so $$\sigma(A) \subseteq \{\text{ zeros of } x^2-4x-1\} = \{2-\sqrt5, 2+\sqrt5\}$$
Assume that $2-\sqrt{5}$ is the eigenvalue of $A$ with mulyiplicity $k \ge 0$ and $2+\sqrt5$ is the eigenvalue with multiplicity $5-k$.
Then we have
\begin{align}
\sum_{i=1}^5 \left(a_i - \frac1{a_i}\right) &= k\underbrace{\left(2-\sqrt{5} - \frac1{2-\sqrt{5}}\right)}_{=4} + (5-k)\underbrace{\left(2+\sqrt{5} - \frac1{2+\sqrt{5}}\right)}_{=4} \\
&= 4k + 4(5-k) \\
&= 4\cdot 5 \\
&= 20
\end{align}
We see that the result is independent of $k$.


text:
\subsection{Statement and discussion of main results}

We are now in a position to state and prove our main results for the \ac{GOLD} algorithm under \cref{asm:delay,asm:pay}.
We begin with the algorithm's regret guarantees in the unilateral setting:

\begin{restatable}{theorem}{regret}
\label{thm:regret}
Suppose that an agent is running \eqref{eq:GOLD} with step-size and sampling radius sequences of the form
$\step_{\run} = \step/\run^{c}$ and $\mix_{\run} = \mix/\run^{b}$ for some $\step,\mix > 0$ and
$b = \min\{1/4,1/3-\alpha/3\}$,
$c = \max\{3/4,2/3+\alpha/3\}$.

Then, the agent enjoys the mean regret bound

\begin{equation}
\label{eq:reg-bound}
\overline\reg(\nRuns)
	= \tilde\bigoh\parens[\big]{\nRuns^{3/4} + \nRuns^{2/3 + \alpha/3}}.

\end{equation}

\end{restatable}

\begin{remark*}
In the above, $\tilde\bigoh(\cdot)$ stands for ``$\bigoh(\cdot)$ up to logarithmic factors''.
The actual multiplicative constants that are hidden in the Landau ``big oh'' notation have a complicated dependence on the diameter of $\points$, the dimension of the ambient space, the range of the players' utility functions;
we provide more details on this in the paper's appendix.
\end{remark*}

For the game-theoretic setting, we will focus on games satisfying Rosen's \acl{DSC} condition (\eg as the Kelly auction example described in \cref{sec:model}).
In this general context, we have:

\begin{restatable}{theorem}{Nash}
\label{thm:Nash}
Let $\game$ be a continuous game satisfying \eqref{eq:DSC}, and suppose that each agent follows \eqref{eq:GOLD} with step-size and sampling radius sequences
$\step_{\run} = \step/\run^{c}$ and $\mix_{\run} = \mix/\run^{b}$ for some $\step,\mix > 0$ and
$b,c$ satisfying the conditions:
\begin{subequations}
\label{eq:params}
\begin{align}
2c - b
	&> 1 + \alpha,
	\\
b + c
	&> 1,
	\\
2c - 2b
	&> 1.
\end{align}
\end{subequations}

Then, with probability $1$, the sequence of play $\est\state_{\run}$ induced by \eqref{eq:GOLD} converges to the game's \textpar{necessarily} unique \acl{NE}.
\end{restatable}

The above results are our main guarantees for \eqref{eq:GOLD} so, before discussing their proof, some remarks are in order.
The first concerns the tuning of the algorithm's hyperparameters, \ie the exponents $b$ and $c$.
Even though the conditions stated in \cref{thm:regret} may appear overly precise, we should note that agents have considerably more leeway at their disposal.
Specifically, as part of the proof, we show that any choice of the exponents $b$ and $c$ satisfying \eqref{eq:params} also leads to no regret \textendash\ albeit possibly at a worse rate.
This is particularly important for the interplay between no regret and convergence to \acl{NE} because it shows that the two guarantees are fairly well aligned as long as \eqref{eq:GOLD} is the class of no-regret policies under consideration.

\begin{figure}[t]
\begin{center}
\small
\input{Figures/b_c_a.tex}
\end{center}
\caption{The allowable region (green shaded areas) of possible values of the sampling radius and step-size exponents $b$ and $c$ for various values of the groth exponent $\alpha$ of the encountered delays.
The dashed blue lines corresponding to the last two terms in \eqref{eq:params} indicate hard boundaries leading to logarithmic terms in the regret instead of constants.}
\label{fig:abc}
\end{figure}

We should also note here that

the $\nRuns^{3/4}$ term is the standad regret bound that one obtains in the bandit online convex optimization framework.
On the other hand, the term $\nRuns^{2/3+\alpha/3}$ describes the advent of the delays which, combined with the bias of the \ac{SPSA} gradient estimator, contribute a significant amount of regret over time (recall in particular that $\delay_{\run}$ is a priori unbounded).

This is of particular importance to applications to online advertising where delays can often become arbitrarily large.

For concreteness, we also plot in \cref{fig:abc} the region of allowed step-size and sampling radius exponents.
This plot reveals the interesting property that, if the feedback delays do not grow too large over time \textendash\ specifically, if $\delay_{\run} = o(\run^{1/4})$ \textendash\ then they have no impact on the allowable choices of $b$ and $c$.
This is also reflected in the regret bound \eqref{eq:reg-bound} where, for $\alpha = 1/4$, the regret-specific term becomes $\nRuns^{3/4}$ as well;
in particular, in the \emph{constant regret} case $\delay_{\run} = \bigoh(1)$, the delays are invisible in \eqref{eq:reg-bound}.
These considerations illustrate the impact of each source of feedback scarcity (bandit vs. delays) on the performance of \eqref{eq:GOLD} and provides a clear insight on the different mechanisms affecting the algorithm's regret and convergence guarantees.

\subsection{Analysis and sketch of proof}

The rest of this section is devoted to a high-level sketch of the proof of \cref{thm:regret,thm:Nash}.
We begin by decomposing $\est{\payv}_{\run}$ as a noisy estimate of $ \payv(\state_{\head_{\run}})$ into the following elements:
\begin{equation}
\label{eq:signal}
\est{\payv}_{\run}
	= \payv(\state_{\head_{\run}}) +\noise_{\head_{\run}+1} + \bias_{\head_{\run}}.
\end{equation}
The various terms in \eqref{eq:signal} above are defined as follows:
\begin{enumerate}
\item
First, we set
\begin{equation}
\label{eq:noise}
\noise_{\head_{\run}+1}
	= \est{\payv}_{\run} -  \exof{\est{\payv}_{\run}\given \filter_{\run}}
\end{equation}
where the filtration $\filter_{\run}$ contains all the random variables that have been realized at the beginning of the $\run$-th iteration of the algorithm;
more precisely, we let
\begin{equation}
\label{eq:filter}
\filter_{\run}
	= \sigma(
		\varnothing,
		\state_{\start},

		\dotsc,
		\est\pay_{\head_{\run-1}},\unitvar_{\run-1},\state_{\run})

\end{equation}
with the convention $\est\pay_{\infty} = 0$, $\unitvar_{\infty} = 0$ if $\head_{\run} = \infty$.
We note for posterity that $\noise_{\head_{\run}}$ is a martingale difference sequence relative to $\filter_{\run}$, \ie $\exof{\noise_{\head_{\run}+1} \given \filter_{\run}} = 0$.

\item
Second, we let
\begin{equation}
\bias_{\head_{\run}}
	= \exof{\est\payv_{\run} \given \filter_{\run}} - \payv(\state_{\head_{\run}})
\end{equation}
denote the systematic error of the estimator $\est\payv_{\run}$ relative to the gradient of the dequeued state $\state_{\head_{\run}}$ (\ie the error remaining after any zero-sum component has been averaged out).
In contrast to $\noise$, this term is not zero-mean;
instead, as we discuss in the appendix, the \ac{SPSA} gradient estimation process that we employ induces a bias of order $\norm{\bias_{\head_{\run}}} = \bigoh(\mix_{\head_{\run}})$.
This bias term grows smaller with $\run$ but its variance increases, leading to a bias-variance trade-off in our setting.
\end{enumerate}

With all this in hand, the workhorse of our calculations is the distance of the sequence $\state_{\run}$ to a given ``benchmark'' action $\base\in\points$ (the best fixed action in hindsight, or the game's equilibrium, depending on the context).
Specifically, letting
\begin{equation}
\label{eq:energy}
\breg_{\run}
	= \frac{1}{2} \norm{\state_{\run} - \base}^{2}
\end{equation}
we have the following template inequality:

\begin{restatable}{lemma}{template}
\label{lem:template}
If \eqref{eq:GOLD} is run with assumptions as above, then, for all $\base\in\points$, we have
\begin{subequations}
\label{eq:template}
\begin{align}

\breg_{\run+1}
	\leq \breg_{\run}
	&\label{eq:template-drift}
		+ \step_{\run} \braket{\payv(\state_{\head_{\run}})}{\state_{\run} - \base}
	\\
	&\label{eq:template-noise}
		+ \step_{\run} \braket{\noise_{\head_{\run}+1}}{\state_{\run} - \base}
	\\
	&\label{eq:template-bias}
		+ \step_{\run} \braket{\bias_{\head_{\run}}}{\state_{\run} - \base}
	\\
	&\label{eq:template-error}
		+ \tfrac{1}{2} \step_{\run}^{2} \norm{\hat\payv_{\run}}^{2}
\end{align}
\end{subequations}
\end{restatable}

This lemma follows from the decomposition \eqref{eq:signal}, the nonexpansivity of the projection mapping, and the regularity assumption \eqref{asm:pay} which allows us to control the terms \eqref{eq:template-drift} and \eqref{eq:template-bias} above;
to streamline our discussion, we defer the details to the paper's supplement.
Moving forward, with this estimate at our disposal, the analysis branches for \cref{thm:regret,thm:Nash} as indicated below.

\para{Regret analysis}
To bound the agent's regret, we need to isolate the scalar product in \eqref{eq:template-drift} and telescope through $\run=\running,\nRuns$ after dividing by the step-size $\step_{\run}$.
Deferring the ensuing lengthy calculations to the appendix, we ultimately obtain a bound of the form
\begin{equation}
\label{eq:reg-bound-sum}
\overline\reg(\nRuns)
	= \bigoh\parens*{
		\frac{1}{\step_{\nRuns}}
			\sum_{\run=\start}^{\nRuns}
			\parens*{
				\step_{\run} \sum_{\runalt=\head_{\run}}^{\run-1} \frac{\step_{\runalt}}{\mix_{\runalt}}
				+ \step_{\run} \mix_{\head_{\run}}
				+ \frac{\step_\run^2}{\mix_{\head_{\run}}^2}
			}
		}
\end{equation}
As a result, to proceed, we need to provide a specific bound for each of the above summands.
The difficulty here is the mixing of different quantities at different time-stamps, \eg as in the product term $\step_{\run} \mix_{\head_{\run}}$.
Bounding these terms requires a delicate analysis of the delay terms in order to estimate the maximum distance between $\run$ and $\head_{\run}$.
We will return to this point below;
for now, with some hindsight, we only stress that the terms in \eqref{eq:reg-bound-sum} correspond on a one-to-one basis with the conditions \eqref{eq:params} for the parameters of \eqref{eq:GOLD}.

\para{Game-theoretic analysis}

The game-theoretic analysis is significantly more involved and relies on a two-pronged approach:
\begin{enumerate}

\item
We first employ a version of the Robbins\textendash Siegmund theorem to show that the random variable $\breg_{\run} = (1/2) \norm{\state_{\run} - \eq}^{2}$ converges pointwise as $\run\to\infty$ to a random variable $\breg_{\infty}$ that is bounded in expectation (here $\eq$ denotes the game's unique equilibrium).

\item
Subsequently, we use a series of probabilistic arguments (more precisely, a law of large numbers for martingale difference sequences and Doob's submartingale convergence theorem) to show that \eqref{eq:GOLD} admits a (possibly random) subsequence $\state_{\run_{\runalt}}$ converging to $\eq$.

\end{enumerate}
Once these two distinct elements have been obtained, we can readily deduce that $\state_{\run} \to \eq$ with probability $1$ as $\run\to\infty$.
Hence, given that $\norm{\state_{\run} - \est\state_{\run}} = \bigoh(\mix_{\run})$ and $\lim_{\run} \mix_{\run} = 0$, our claim would follow.

However, applying the probabilistic arguments outlined above requires in turn a series of summability conditions.
Referring to the paper's supplement for the details, these requirements boil down to showing that the sequences
\begin{equation}
\label{eq:3seqs}
A_{\run}
	= \step_{\run} \sum_{\runalt=\head_{\run}}^{\run-1} \frac{\step_{\runalt}}{\mix_{\runalt}},
	\quad
B_{\run}
	= \step_{\run} \mix_{\head_{\run}},
	\quad
\textrm{and}
	\quad
C_{\run}
	= \frac{\step_\run^2}{\mix_{\head_{\run}}^2},
\end{equation}
are all summable.
Importantly, each of these three sums has a clear and concise interpretation in our learning context:
\begin{enumerate}
\item
The first term ($A_{\run}$) is the cumulative error induced by using outdated information.
\item
The second term ($B_{\run}$) is the error propagated from the bias of the \ac{SPSA} estimator.
\item
Finally, the third term ($C_{\run}$) corresponds to the variance (or, rather, the mean square) of the \ac{SPSA} estimator.
\end{enumerate}
As a result, as long as these terms are all summable, their impact on the learning process should be relatively small (if not outright negligible).

Comparing the above term-by-term to \eqref{eq:reg-bound-sum} is where the game-theoretic analysis rejoins the regret analysis.
As we said above, this requires a careful treatment of the delay process, which we outline below.

\para{Delay analysis}

A key difficulty in bounding the sums in \eqref{eq:reg-bound-sum} is that the first term ($A_{\run}$ in \eqref{eq:3seqs} is a sum of $\run-\head_{\run}$ terms, so it can grow quite rapidly in principle.
However, our pooling strategy guarantees that $\run-\head_{\run}$ cannot grow faster than the delay (which is sublinear by assumption).
This observation (detailed in the supplement) guarantees the convergence of the sum.

A further hidden feature of \eqref{eq:template} is in the noise term $\noise_{\run}$:
in the case of batching or reweighted strategies (\eg as in \citealp{ZMBG+17-NIPS}), this term incorporates a sum of terms arriving from different stages of the process, making it very difficult (if not impossible) to control.
By contrast, the pooling strategy that defines the \ac{GOLD} policy allows us to treat this as an additional ``noise'' variable;

we achieve this by carefully choosing the step-size and sampling radius parameters based on the following lemma:

\begin{restatable}{lemma}{threesums}
\label{lem:3sums}
Suppose that \eqref{eq:GOLD} is run with step-size and sampling radius parameters of the form $\step_{\run} \propto \step/\run^{c}$ and $\mix_{\run} \propto \mix/\run^{b}$, with $b,c>0$.
Then:

\begin{enumerate}

\item
If $2c-b \geq 1+\alpha$, then $\sum_{\run=\start}^{\nRuns} A_{t} = \bigoh(\log\nRuns)$;
in addition, if the inequality is strict, $A_{\run}$ is summable.
\item
If $c+b\geq 1$, then $\sum_{\run=\start}^{\nRuns} B_{t} = \bigoh(\log\nRuns)$;
in addition, if the inequality is strict, $B_{\run}$ is summable.
\item
If $2c - 2b \geq 1$, then $\sum_{\run=\start}^{\nRuns} C_{t} = \bigoh(\log\nRuns)$;
in addition, if the inequality is strict, $C_{\run}$ is summable.
\end{enumerate}
\end{restatable}

Proving this lemma requires a series of intermediate results that we defer to the paper's supplement.


TITLE: The space of continuous function with compact support and a norm on it.
QUESTION [0 upvotes]: Denote with $C_0(\Omega)$ the subspace of countinuous functions with compact support, $\Omega\subseteq\mathbb{R}^n$ open. Define an application on $C_0(\Omega)$ as $||\cdot||\colon C_0(\Omega)\to\mathbb{R}_+$ $$||f||_1:=\int_\Omega |f|\;d\lambda_n$$ where $\lambda_n$ is the Lebesgue measure.
I must be prove that $||\cdot||_1$ is a norm on $C_0(\Omega)$. The triangle inequality and positivity is ok. Remain to prove that $$||f||_1=0\iff f=0\;\text{on}\;\Omega.$$
$(\Leftarrow)$ is obvious; I am having difficulty  proving $(\Rightarrow)$.
Suppose that $$||f||_1=0\Rightarrow f=0\;\text{a.e in support}$$
Can I conclude?

REPLY [2 votes]: If $f(x_0)\neq 0$ for some $x_0\in\Omega$ then since $f$ is continuous there must be a neighborhood $U$ of $x_0$ where $f(x)\neq 0$ for all $x\in U$, and now for an $\varepsilon>0$ there is a ball $B_{\varepsilon}(x_0)\subset U$ .Clearly
$$0<\int_{B_{\varepsilon}(x_0
)}|f|d\lambda_n\leq\int_{\Omega}|f|d\lambda_n$$


TITLE: Reference to angular distance metric properties.
QUESTION [3 upvotes]: I know that the Angular Distance is a proper metric but I'm struggling to find a reference that states that and proves all the properties for that distance. In the book Mining of Massive Datasets (Page 95) the authors briefly argue that the properties hold for what they call "Cosine Distance", which is just the angle between two vectors.
However, I've read that the Angular Distance must be normalised in order to be considered a metric.

Does the Angular Distance have to be normalised for it to be a metric?
Can you point me to a reference that proves that the Angular Distance is a proper metric?

REPLY [0 votes]: Metric
Let $X$ be an arbitrary set. A function $d: X \times X \rightarrow \mathbb{R}$ is called a metric on $X$, iff
        the following properties hold true for all $x, y, z \in X$:

identity of indiscernibles: $\forall x, y \in X: d(x, y) = 0 \Leftrightarrow x = y$
symmetry: $\forall x, y \in X: d(x, y) = d(y, x)$
triangle inequality: $\forall x, y, z \in X:d(x, z) \leq d(x,y) + d(y, z)$

Cosine Distance
Cosine distance is defined as:
$$d_{cos}(x, y) := 1 - \cos(\theta) = 1 - {\mathbf{A} \cdot \mathbf{B} \over \|\mathbf{A}\| \|\mathbf{B}\|} = 1 - \frac{ \sum\limits_{i=1}^{n}{A_i  B_i} }{ \sqrt{\sum\limits_{i=1}^{n}{A_i^2}}  \sqrt{\sum\limits_{i=1}^{n}{B_i^2}} }$$
Cosine distance is not a metric in $\mathbb{R}^n$, because the identity of indiscernibles does not hold true:
$$d(0.5, 1) = 1 - \cos(0°) = 1 - \frac{1}{1} = 0\text{, but } 0.5 \neq 1$$
To the question if cosine distance is a metric on $S_n = \{x \in \mathbb{R}^n: ||x|| = 1\}$:
\begin{align}
x &= y\\
\Rightarrow d(x,y) &= d(x,x)\\
&= 1 - \sum_{i=1}^n x_i^2\\
&= 0
\end{align}
So that looks good.
We also know that $\cos(\theta) = 1 \Leftrightarrow \theta \in \{0°, 180°\}$. This means it still does not have the identity of indiscernibles attribute. We have to reduce the space to the non-negative unit sphere $S_n^+ = \{x \in \mathbb{R}^n: ||x|| = 1 \land x^{(i)} \geq 0\}$.


TITLE: Choosing a Cauchy sequence for a real
QUESTION [5 upvotes]: It is easy to form in ZF, for each real $a$, a "canonical" Cauchy sequence that converges to $a$. For example, one can take the sequence of finite initial segments of the decimal expansion of $a$, being careful when $a$ is a base-10 rational to pick one of the two decimal expansions explicitly.
But what if we are given a set of Cauchy sequences of rationals, all converging to the same real. The set might not contain all the Cauchy sequences for that real. How hard is it to pick a "canonical" representative from the sequences that are in the set?
To make this question precise: consider a family of sets $C$ so that each $X \in C$ is a set of Cauchy sequences of rationals that all converge to the same real $a(X)$. Note that $X$ is not required to have the entire set of Cauchy sequences for $a(X)$. Does ZF prove the existence of a choice function for each family $C$ of this kind?
There are two aspects of Cauchy sequences that make this problem interesting. First, every infinite subsequence of a Cauchy sequence is again a Cauchy sequence converging to the same real. So we cannot hope for a "minimal" sequence. Also, we may prepend any finite sequence to a Cauchy sequence to yield a new Cauchy sequence converging to the same real. So we cannot hope for a "maximal" sequence. Cauchy sequences are very slippery in this way.
Dedekind cuts behave differently: once we specify whether cuts for rationals can have a maximum element, we have a unique Dedekind cut for each real, whereas we always have infinitely many Cauchy sequences.
I have a vague memory of encountering something similar to this question in the past, but I cannot remember any details.  It also seems to have a flavor related to Borel equivalence relations, although this question is not written in that way.

REPLY [4 votes]: Such a choice mechanism would translate to a canonical choice for a function $\Bbb N \to \Bbb N_{>0}$ from a given set $S$ of such functions. If I'm not mistaken, this is some nontrivial choice principle (and if I am, please do correct me).
Given $f: \Bbb N \to \Bbb N_{>0}$, define: $$b_f (n) = \begin{cases}2^{-k} &: n = \sum\limits_{i=0}^k f(i) \\0&: \text{otherwise}\end{cases}$$
Now define $s_f(n) = \sum\limits_{i=0}^n b_f(i)$. Then for all $f$: $$\lim_{n\to\infty} s_f(n) = 2$$
and $f \leftrightarrow s_f$ is a bijective correspondence. Hence a choice function for each set $\{s_f: f \in S\}$ amounts to a choice function for each $S \subseteq (\Bbb N_{>0})^{\Bbb N}$.

By defining, for $f: \Bbb N \to 2$: $$b_f(n) = \begin{cases}2^{-k} &: n = k + \sum\limits_{i=0}^k f(i) \\0&: \text{otherwise}\end{cases}$$ the argument seems to carry over to $2^{\Bbb N}$.


TITLE: Does $\lim_{(a, b) \to (0, 0)} \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0$?
QUESTION [3 upvotes]: Does $$\lim_{(a, b) \to (0, 0)} \ \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0 \ ?$$

I'm trying to show that $$\lim_{(a, b) \to (0, 0)} \ \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0 $$
but I am getting stuck. I was thinking that as a starting point I could show that $$\lim_{(a, b) \to (0, 0)} \ \frac{a^2b^2}{a^2 + b^2} = 0 $$
and then conclude that since $$\lim_{x \to 0} \sqrt{x} = 0$$ and $\frac{a^2b^2}{a^2 + b^2} \to 0$ as $(a, b) \to (0, 0)$ we arrive at $$\lim_{(a, b) \to (0, 0)} \ \sqrt{\frac{a^2b^2}{a^2 + b^2}} = 0.$$

Firstly is my approach above a correct one. Secondly how can show that $$\lim_{(a, b) \to (0, 0)} \ \frac{a^2b^2}{a^2 + b^2} = 0. $$
Because I don't see any way to show the above (apart from perhaps proving it from the definition directly, which I would like to avoid if there is an easier way to do it). Also it could be the case that the initial limit doesn't even exist.

REPLY [1 votes]: $a^2+b^2 \ge 2|ab| \ge |ab|.$
$0 \le \sqrt{|ab| \dfrac{|ab|}{a^2+b^2}} \le \sqrt{ |ab| \cdot 1} \le$
$\sqrt{ a^2+b^2}.$
Choose $\delta = \epsilon$.


TITLE: Find value of n if ${(n-1)! \over (n-3)!} = 30$
QUESTION [1 upvotes]: So here's my attempt :
$${(n-1)(n-2)(n-3)! \over (n-3)!} = 30$$
$${(n-1)(n-2)} = 30$$
then $$n-1 = 30$$ or $$n-2 = 30$$
then $$n = 31$$ or $$n = 32$$
but my textbook says
$$(n-1)(n-2) = 6 * 5$$
then $$(n-1) = 6$$ then $$n = 7$$
which does make any sense to me this should have been like any other factorize , can someone please explain the steps of the textbook to me , thanks in advance

REPLY [0 votes]: You've got the first bit spot on, $(n-1)(n-2)=30$
Now, you say that either $(n-1)=30$ or $(n-2)=30$. This would be true only if the other was equal to $1$ at the same time. If $n-1=30$, $n-2=29$ clearly, so this is not met, and likewise if $(n-2)=30$ then $(n-1)=31$ this is not met either.
Instead, expand out $(n-1)(n-2)=n^2-3n+2=30 \to n^2-3n-28=0$
I know $4+(-7)=-3\text{ and } 4\cdot(-7)=-28$ so $(n+4)(n-7)=0\implies n=7, -4$
It can't be negative thus $n=7$


TITLE: Reducible to Separable First Order Differential Equation Word Problem in Analytic Geometry 1.4-29
QUESTION [0 upvotes]: I completed near all problems om a differential equations text chapter on reducing non-separable first order differential equations to separable by using an appropriate substitution for example $u = y/x$ with $y' = u + u'x$ and similar substations in $y$ and $x$ for making other similar problems separable.
I am not asking anyone to do the entire problem for me but I do need a little guidance to begin tackling this problem. I completely understand the procedure for making them separable. What I need help in is setting up the differential equation for the following word problem. Once this is done I can easily finish the problem.
Ch 1.4-29. Show that a straight line through the origin intersects all solution curves of a given differential equation $y = g(y/x)$ at the same angle.

REPLY [1 votes]: Let's say that the line $y = mx$ intersects a solution to $y' = g(\tfrac{y}{x})$ at the point $(x_0,y_0) = (x_0,mx_0)$. The slope of the line is $m$ and the slope of the tangent line to the curve at $(x_0,mx_0)$ is $y' = g(\tfrac{mx_0}{x_0}) = g(m)$. Based on this, can you figure out the angle between the line and the curve? Does this depend on the value of $x_0$? If not, then the line $y = mx$ intersects all solution curves at the same angle.


TITLE: Prove $f $ is identically zero if $f(0)=0$ and $|f'(x)|\le|f(x)|$
QUESTION [13 upvotes]: $f:\Bbb R \to\Bbb R $ is differentiable, $f(0)=0$ and $|f'(x)|\le|f(x)|$ for all $x$ then prove $f$ is identically zero.
I tried to use mean value theorem and end up in $|f(x)|\le |x||f(c)|$ for some constant $c$ which depends on $x$. Can you help me little bit...

REPLY [4 votes]: First, I will show that $f$ is identically zero on $[0,1]$.
Suppose that $|f|$ attains a maximum at a point $m$, which is guaranteed by the Extreme Value Theorem. We can assume that $m < 1$, since
$|f(1)| \le |f(c)|$ for some $c \in (0,1)$.
It follows that
$|f(m)| \le m |f(c_m)|$, where $0<c_m < m$.
If $|f(c_m)|\ne 0$, then we have $|f(m)| < |f(c_m)|$, which contradicts the maximality of $|f(m)|$. Therefore, $f(c_m) = 0$, and $f(m) = 0$, and $f$ must be identically zero on $[0,1]$.
Now, you can apply the mean value theorem to obtain $|f(x)| < |x-1| |f(c)|$ for $c$ in $(1,x)$, and apply the same argument to $[1,2]$, then to $[2,3]$, etc.
Note that same argument can be used to $[-1,0]$, $[-2,-1]$, then to $[-3,-2]$, etc.


TITLE: Confusion between Lie Algebra representation and adjoint representation
QUESTION [1 upvotes]: For the Lie group $SU(2)$ which can be represented as a subset of $GL(\mathbb{C}^2)$, we know that the Lie Algebra $su(2)$ can be represented as a subset of $End(\mathbb{C}^2)$.
I'm confused by what this last bit is referring to, and the difference between the Lie Algebra representation and the adjoint representation.
I've heard that the Lie Algebra of $SU(2)$, $su(2)=span(\sigma_i)$ is the 3D span of the 3 2x2 Pauli matrices. Is the above saying that this is the subset of $End(\mathbb{C}^2)$?
Alternatively the adjoint representation of SU(2) can be represented by 3x3 matrices. Is this what it means by the subset of $End(\mathbb{C}^2)$? (I would assume not?)
I'm also slightly confused of what the point of the adjoint representation is since the Pauli matrices already represent the Lie Bracket correctly with their matrix commutator, so what is the point of having a new 3x3 matrix system of showing the same thing?

REPLY [0 votes]: It is the first case. The span of $2 \times 2$ Pauli matrices is the one that is being referred to by $End( \mathbb{C^2})$


TITLE: How can a bipartite simple graph be non isomorphic
QUESTION [0 upvotes]: From my understanding the bipartite graph is a graph that follows the red blue color scheme. If the graph fails the red blue color, then the graph is not bipartite.
But the question how do you test a bipartite graph that is non isomorphic?
Don't you need two graph to test for non-isomorphism?

REPLY [1 votes]: It doesn't make sense to talk about an individual graph being non-isomorphic.  Informally, two graphs are non-isomorphic if they are structurally different.
If we have two graphs $G_1$ and $G_2$, then if $G_1$ is bipartite and $G_2$ is not bipartite, then they are non-isomorphic.  (I'm guessing this is the link between the two topics you've encountered.)
If they instead both happen to be bipartite, they may or may not be isomorphic.  E.g. $C_8$ and $C_4 \cup C_4$ are two non-isomorphic bipartite graphs.
Graph isomorphism testing is a non-trivial problem, and is often practically solved computationally.


TITLE: Limit of product of matrices with spectral radius less than 1
QUESTION [0 upvotes]: If spectral radius of a square matrix $A$ is less than 1, then we know that $$\lim_{k \to \infty} A^k = 0.$$
Now, I want to know whether we can also conclude the following results?
$$\lim_{k \to \infty} A^k (A^{T})^k = 0$$
where $A^{T}$ is the transpose of matrix $A$.

REPLY [0 votes]: Hint: $A^T$ has the same eigenvalues as $A$. Then use algebraic limit rules.


TITLE: What are the properties of the set of the Real Numbers without the Integers?
QUESTION [1 upvotes]: This question came up in a lunchtime discussion with coworkers.  None of us are professional mathematicians or teachers of math.  I apologize for any incorrect math or sloppy terminology.
We were discussing getting from one number to another along the real number line.  The challenge is that we removed the integers from the number line.
So can an ant get from 0.5 to 1.5 by crawling along this "broken" line?
Searching the web, we discovered this may be a problem in topology, somehow (possibly) related to something called the "Long Line".  This is far outside of any of our knowledge so we would appreciate an explanation at roughly the basic calculus level.

REPLY [2 votes]: The space you describe
$...(-(n+1),-n)...(-2,-1)(-1,0)(0,1)(1,2)...(n,n+1)...$
is homeomorphic to countably infinite ("$\omega$") many copies of $(0,1)$. The Long Line you refer is made of uncountably infinite many copies of $[0,1)$. It can be thought of as a generalization of the nonnegative reals $[0,\infty)$, because we can view this space as  $\omega$-many copies of $[0,1)$:
$[0,1)[1,2)[2,3)...$
Your question about the ant is related to path connectedness (which is equivalent to connectedness here).  There is no continuous path from a point in $(n-1,n)$ to a point in $(n,n+1)$ as the subspace $(n-1,n)(n,n+1)$  is not connected, that is, it is the union of two nonempty open subsets. In the reals this would not be a problem as $(n-1,n+1)$ is connected.


TITLE: Intuition behind grad operator
QUESTION [4 upvotes]: I'm pretty sure this question has a very simple answer and I am just not understanding the geometric picture very well, however I can across a question on the gradient operator which made me a bit unsure as to what it does. The gradient of a scalar field points in the direction of greatest rate of change. Now if I have a surface given by $f(x,y,z)$, then the normal to the sruface can be found by taking the gradient of $f$ at that point. Intuitively, this also makes sense to me. The change is zero along the tangent plane, so the greatest rate of change must be normal to the surface.
However I was doing a question which gave a surface and then asked in what direction a marble would roll if placed at a particular point on the surface. Now I initially thought that the marble would roll in the direction of greatest rate of change- i.e. a ball on a hill will roll down where the potential will decrease most rapidly. I would take the gradient and then take the direction in which f was decreasing as opposed to increasing.
However if the gradient is normal to the surface, then a marble would be ejected directly off of the surface? This is not what happens!
Now I have a hunch as to what might be the problem here although I am not sure. I think the problem is two-fold. First, the mathematical description allows the possibility of the marble passing through the surface or coming off of it. Secondly, I think maybe I am mixing up surfaces and scalar fields which are not surfaces. Perhaps in the physical case I actually have a scalar field (not a surface) describing the potential energy AND a surface on which the marble is constrained to move, and the marble will move in the direction of greatest rate of decrease of potential energy while still being constrained on the surface.
That seems about I think right...

REPLY [0 votes]: Your surface, let's call it $S,$ is not given by $f(x,y,z),$ but by an equation of the form $f(x,y,z) = \text { constant }.$ (Perhaps that's what you meant, but it's good to be clear on this.) If we think of the $z$ variable as representing altitude, we can do this: Let $p\in S.$ What you are looking for is the unit vector $(a,b,c)$ tangent to $S$ at $p$ with the smallest value of $c.$ That unit vector will point in the desired direction.


TITLE: How does a very bright but very small source of light look like?
QUESTION [0 upvotes]: How does a very bright but very small source of light look like?
If its smaller than resolving power of eye but (very very bright and far star) brighter than sun?

REPLY [0 votes]: If the source is a star brighter than the sun but to far away for our eyes to resolve then a telescope will improve the situation. In other words its no different than any source but because of the inverse square law by the time the light reaches us the number of photons are very small.


TITLE: Tension of String Connecting Two Objects
QUESTION [0 upvotes]: I believe the way you find tension in a string connecting 2 objects is by using Newton's Second Law on the object which the tension points to. But I am not completely sure how you would find the tension if the object is not accelerating. Does this mean there is no tension force?

REPLY [1 votes]: If you have two, equal but opposite forces at each end of the string, there will be no acceleration. Think about the game of tug of war. You have persons at each end of the rope pulling with equal but opposite forces so there is no acceleration of the rope.The sum of the external forces on the rope is zero. But tension is an internal force. You know its there because if you cut the rope, the contestants will go flying. So do a free body diagram. Cut the rope at one end eliminating one group of contestants. You need to replace them with the force necessary to maintain equilibrium. That’s the tension in the string.
Hope this helps.


TITLE: Sylow p-subgroup is normal if and only if it is a characteristic subgroup
QUESTION [1 upvotes]: Take finite group $G$ with Sylow p-subgroup $P$. Then, prove that
$P \unlhd G \iff P$ is a characteristic subgroup of $G$.
I know how to show the $\impliedby$ direction (if characteristic, then normal) but can't think of how to prove $\implies$ (if normal, then characteristic).

REPLY [4 votes]: If a Sylow subgroup is normal, then it is the unique subgroup of that order, i.e. if $P\triangleleft G$ then there is only one subgroup of order $|P|$ which implies $P$ is characteristic since automorphisms preserve orders of subgroups.


TITLE: The $n-1$ order principle minor of the adjoint matrix of $A$ has determinant $|A|^{n-2}a_{nn}$.
QUESTION [1 upvotes]: Let $A$ be a $n\times n$ matrix, $A^*$ is its adjoint, that is, $A=\begin{pmatrix}A_{11}&\cdots&A_{n1}\\
\vdots&&\vdots\\
A_{1n}&\cdots&A_{nn}\end{pmatrix}$, where $A_{ij}$ is the cofactor of $a_{ij}$ in $A$. Show that $det\begin{pmatrix}A_{11}&\cdots&A_{n1}\\
\vdots&&\vdots\\
A_{1,n-1}&\cdots&A_{n-1,n-1}\end{pmatrix}=|A|^{n-2}a_{nn}$. How to show? We do not know whether or not $A$ or the $n-1$ matrix is invertible?

REPLY [1 votes]: To prove an identity of this sort, the usual technique is to prove it for a matrix $A$ containing $n^2$ indeterminates, so that $A$ is always invertible. Then, by "substituting" some concrete numbers into the indeterminates, we automatically get an identity for this new $A$ with number entries, even if it is singular.
(There is also a similar but subtly different perspective from Zariski topology, which loosely says that the set of all invertible matrices are "dense" in the set of all matrices, so that in most cases, if a matrix identity holds for all invertible matrices, we can pass the identity to the limit to make it valid for all square matrices. If the field in question is $\mathbb F=\mathbb R$ or $\mathbb C$, you can also ignore Zariski topology and consider the usual topology of $M_n(\mathbb F)$.)
Anyway, suppose $A$ is invertible. The identity then follows from a simple block-matrix argument that was portrayed in Prasolov, Problems and Theorems in Linear Algebra, theorem 2.5.1. Partition $A$ as $\pmatrix{X&y\\ z&a_{nn}}$ and partition $\operatorname{adj}(A)$ in a conforming manner as $\pmatrix{U&v\\ w&m_{nn}}$, where $a_{nn}$ is the bottom right entry of $A$ and $m_{nn}$ is the corresponding $(n-1)$-rowed minor. Since $A\operatorname{adj}(A)=\det(A)I_n$, we have
$$
\pmatrix{X&y\\ z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\det(A)I_n.
$$
Hence
$$
\pmatrix{z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\det(A)\pmatrix{0&1}=\pmatrix{0&\det(A)}
$$
and
$$
\pmatrix{I&0\\ z&a_{nn}} \pmatrix{U&v\\ w&m_{nn}}=\pmatrix{U&v\\ 0&\det(A)}.
$$
Take determinants on both sides, we obtain $a_{nn}\det(\operatorname{adj}(A))=\det(A)\det(U)$. However, from $A\operatorname{adj}(A)=\det(A)I_n$ we also obtain $\det(\operatorname{adj}(A))=\det(A)^{n-1}$. Therefore $a_{nn}\det(A)^{n-2}=\det(U)$.


TITLE: Coefficient of $x^2$ in a polynomial
QUESTION [4 upvotes]: Let $p(x)$ be an polynomial given by
  $$p(x)=(1+x)(1+3x)(1+5x)\cdots(1+19x)$$ Find the coefficient of $x^2$.

I see that this coefficient is the sum of products of roots of $p(x)$ get eight to eight. But, there is ${10 \choose 8}=45$ terms in this sum. Someone know other way more fast?

REPLY [1 votes]: Coefficient of $x^2$ in $(1+x)(1+3x)(1+5x)\cdots (1+(2n-1)x)$ is:
$$\begin{align}
&&1\cdot (3+5+7+9+\cdots +(2n-1))&\\
&&+3\cdot (5+7+9+\cdots +(2n-1))&\\
&&+5\cdot (7+9+\cdots +(2n-1))&\\
&&+7\cdot (9+\cdots +(2n-1))&\\
&&\vdots&\\
&&\cdots +(2n-3)\cdot((2n-1))\\
& && =\sum_{i=1}^{n-1}\sum_{j=i+1}^n(2i-1)(2j-1)\\
& && =\sum_{j=2}^n(2j-1)\;\sum_{i=1}^{j-1}(2i-1)\\
& && =\sum_{j=2}^n(2j-1)\;(j-1)^2\\
& && =\sum_{j=1}^{n-1}(2j+1)\;j^2\\
& && =\sum_{j=1}^{n-1}(2j^3+j^2)\\
& && =2\cdot \left[\frac {n(n-1)}2\right]^2+\frac 16(n-1)n(2n-1)\\
& && =\frac 16n(n-1)(3n^2-n-1)
\end{align}$$
Putting $n=10$ gives
$$[x^2]\; (1+x)(1+3x)(1+5x)\cdots (1+19x)=\color{red}{4335}$$


TITLE: The sum of three all three digit numbers, containing digits : 1,2,3,4,5
QUESTION [0 upvotes]: First of all, there are $5^3 =125$ three-digit numbers (repetition of digits allowed).
We know that there are $555-111+1 = 445$ three-digit numbers between $555$ and $111$
This way, we can determine the sum of the numbers between $111$ and $555$ and then subtract the sum of the three-digit numbers that contain any of $6,7,8,9$.
My question is whether there's more efficient way.

REPLY [3 votes]: Note that there are $125$ possibilities. One fifth of these have $1$ in the hundreds place, one fifth have $2$ in the hundreds place etc. Similarly for the tens and units places. So we have $$\frac {125}5\times (1+2+3+4+5)\times 111 = 41,625$$
For the next problem, there are fewer possibilities $5\times 4 \times 3=60$ but the digits are still equally distributed - one fifth will have $1$ in each place etc. So the calculation becomes $$\frac {60}5\times (1+2+3+4+5)\times 111=19,980$$


TITLE: Involution and Covering space
QUESTION [1 upvotes]: Is there a connected topological space such that admits a free involution, trivial fundamental group and furthermore has the set of real number as it's covering space?

REPLY [1 votes]: A simply connected topological space is its own universal cover, so your question reduces to the following:

Does $\mathbb{R}$ admit a free involution $f$?

I'm going to assume you want the involution $f$ to be continuous. If you don't require continuity, then there are many examples
If $f$ has no fixed points, it is injective and is therefore either increasing or decreasing. If it were increasing, we'd have $x < f(x) < f(f(x)) = x$ which is a contradiction. If it were instead decreasing, we would find that $x = f(f(x)) < f(x) < x$ which is again a contradiction.
Therefore, every continuous involution of $\mathbb{R}$ has a fixed point. In fact, it either has exactly one fixed point, or it is the identity map; see this answer.


TITLE: Proof of a set as a subset of the other set
QUESTION [0 upvotes]: Let $X = \{a\in\mathbb Z:a\equiv2\,\mathrm{mod}\,6\}$ and $Y = \{b\in\mathbb Z:3\mid b−5\}$. Prove that one of these sets is a subset of the other. (State result as a theorem).

So what I have thought so far (I don't know if any of it is correct though) is that $X$ must be the subset of $Y$ since if something divides $6$, then it will also divide $3$. I got to the point where $a-2 = 6k$ and $b-5 = 3j$ for some integers $k,j$. I don't know how to show that $X$ is a subset though from here. Any help appreciated!

REPLY [1 votes]: You can observe that $5\in Y$, but $5\notin X$, so you want to prove that $X\subset Y$.
An element of $X$ is of the form $a=6x+2$ (with integer $x$) and $a-5=(6x+2)-5=\dotsb$
What about a “general theorem”? Observe that $6=2\cdot 3$ and $5=2+3$, so you can consider $mn$ and
$$\textstyle
X=\{a\in\mathbb{Z}: a\equiv m\pmod{mn}\}
\qquad
Y=\{b\in\mathbb{Z}: n\mid \bigl(b-(m+n)\bigr)\}
$$
Does the argument above apply?


TITLE: Different results for same principal argument?
QUESTION [0 upvotes]: I am reading about extraction of roots from complex numbers and here the example that I am having difficulties with:

(the text within the box)
I get that $1-i = \sqrt{2}cis(-\pi/4)$ and thus that the principal argument is $\pi/4$. I also get how the expressions for $z_k$ are derived - using the fact that $z_k = r_0^{1/n}(cis(\frac{\sigma + 2k\pi}{n}))$ for $k = 0, 1, 2, ..., n-1$.
But if the principal argument of $(1-i)$ is $-\pi/4$ - and we use that to derived the expressions after "We therefore have five values:", how does the author end up with the two expressions after "Writing $z_3$ and $z_4$ with principal arguments, we get"? I mean we already used the principal arguments so what is different this time?

REPLY [0 votes]: How do you know all the values are distinct unless you make sure their principal arguments are distinct?
In principle , $z_4$ could have been $\frac{39}{40}\pi$, and that would have been identical with the $z_0$ after conversion to principal arguments.  This was not the case, of course, but I mean that is why you’d compare complex numbers this way.


TITLE: Travel distance of a particle
QUESTION [2 upvotes]: How can we show that a projectile fired at an angle $\theta$ with initial speed $v_0$ travels a total distance $\frac{v_0^2}{g}\sin2\theta$ before hitting the ground?
The way I set it up is: direction is $(\cos\theta,\sin\theta)$. By using Newton's law, $$r(t) = (0,-\frac{1}{2} gt^2)+tv_0(\cos\theta,\sin\theta) = \frac{v_0^2}{g}\sin2\theta$$
I tried to solve the LHS so that it matches the RHS, the total distance. But obviously, there is something wrong with the way I tried to approach it. But cannot see it.
Any hints and pointers in terms of setting it up would be great. Also I do not want the full solution but rather few small hints to be able to set it up correctly.

REPLY [1 votes]: If you have learned differential equation, you can represent the displacement ${\bf r}(t) = (r_1,r_2)$ by an initial value problem:
$${\bf r}''(t) = (0,-g)
\\
{\bf r} (0) = (0,0)
\\
{\bf r}'(0) = {\bf v}(0) = (v_0 \cos(\theta), v_0 \sin(\theta))
$$
solve the y component = zero for a time, then plugging it in the x component. This approach I learned in ODE class is good, for air friction fits perfectly for the model, gravitatiional constant change too :
$$
{\bf r}''(t) = (0,-\frac{GM}{(r_2+R)^2}) - \gamma {\bf r}'(t)
$$
$M$ is the mass of the earth and $R$ is the radius.


TITLE: Growth of the set $\lbrace 2^n+3^m \rbrace$ in the integers
QUESTION [3 upvotes]: Let $A=\lbrace 2^n+3^m \ | \ n,m\in{\mathbb N}\rbrace$, and denote by $a_k$ the $k$-th element of $A$ in order from smallest to greatest ; thus $a_1=2,a_2=3,a_3=4,a_4=5,a_5=7\ldots$. Is it known whether $\frac{a_{k+1}}{a_k}$ has a limit when $k\to\infty$ ?
Weaker question : is it true that $\frac{a_{k+1}}{a_k} \leq \frac{3}{2}$ for every $k$ ? I have checked this for $k\leq 1500$.
My thoughts : if we write $a_k=2^n+3^m$, then $a_k<2^{n+1}+3^m$ so $a_{k+1} \leq 2^{n+1}+3^m \leq 2a_k$, so if the limit exists it is $\leq 2$. Numerical evidence suggests that the $\liminf$ is $1$.

REPLY [1 votes]: See, how many numbers $a_k$ are there below some huge $N$? At least this includes all of those for which $2^n\le{N\over2}$ and $3^m\le{N\over2}$ (and probably more), which is already $O(\log^2N)$, which pretty much forbids the existence of a limit > 1.


TITLE: dividing binomial coefficient by a number
QUESTION [3 upvotes]: so, the question is as follows :
find the remainder when
$${}^{119}C_{33}$$ is divided by $5$
my approach:
well, I don't know about any method except for writing  $${}^{119}C_{33} =  264987608114625679810381761543$$
now it is easy to see that the remainder on dividing this by $5$ should be $3$.
so,is there a better method to get to the solution? kindly help me out.
note: I saw a similar variation of this question at MSE but it had a special case where in
${}^{n}C_{k} , n! = (n-k)!$ hence I hope this qualifies as a new question.

REPLY [2 votes]: I agree with Robert Israel's recommendation in the comments to use Lucas's theorem. Here's the computation:
\begin{align*}
119_{10} &= 434_5,\\
33_{10} &= 113_5.
\end{align*}
Then Lucas tells us that
$$\binom{119}{33}\equiv\binom{4}{1}\binom{3}{1}\binom{4}{3}\equiv 4\cdot 3\cdot 4 \equiv 3\pmod{5}.$$