text:
\begin{document}

\title{A short note on the multiplier ideals of monomial space curves}
\date{\today}
\author{Howard M Thompson}
\address{Mathematics Department\\ 402 Murchie Science Building \\ 303 East Kearsley Street \\ Flint, MI 48502-1950}
\email{\href{mailto:hmthomps@umflint.edu}{hmthomps@umflint.edu}}
\urladdr{\url{http://homepages.umflint.edu/~hmthomps/}}
\thanks{The author would like to thank Karen E. Smith and Roc\'{i}o Blanco as well as too many others to name, for putting up with his near endless and largely fruitless discussion of this topic.}
\subjclass[2010]{Primary: 14F18; Secondary: 14H50, 14M25}
\keywords{Multiplier ideals, toroidal varieties}
\begin{abstract}
  Thompson~(2014) exhibits a formula for the multiplier ideal with multiplier $\lambda$ of a monomial curve $C$ with ideal $I$ as an intersection of a term coming from the $I$-adic valuation, the multiplier ideal of the term ideal of $I$, and terms coming from certain specified auxiliary valuations. This short note shows it suffices to consider at most two auxiliary valuations. This improvement is achieved through a more intrinsic approach, reduction to the toric case.
\end{abstract}
\maketitle

\section{Introduction}

Let $(Y, \Delta)$ be a pair, consisting of a normal variety $Y$ over an algebraically closed field of characteristic zero and a $\QQ$-divisor $\Delta$ such that $K_Y+\Delta$ is $\QQ$-Cartier. Let $\pi:X\rightarrow Y$ be a log resolution of the ideal sheaf $\sheaf{I}\subseteq\sheaf{O}_Y$ that is also a log resolution of the pair $(Y, \Delta)$. That is, $\pi$ is a proper birational morphism such that $X$ is smooth, the union of the exceptional set of $\pi$ and $\pi^{-1}(\Delta)$ is a divisor with simple normal crossing support, and $\sheaf{I}\cdot\sheaf{O}_X=\sheaf{O}_X(-F)$ is also a divisor with simple normal crossing support. In this setting, we define the multiplier ideal of $\sheaf{I}^\lambda$ on the pair $(Y,\Delta)$ to be
\[
    \multideal\left((Y,\Delta),\sheaf{I}^\lambda\right)=\pi_*\sheaf{O}_X(K_X-\lfloor\pi^*(K_Y+\Delta)+\lambda F\rfloor).
\]
This ideal sheaf on $Y$ does not depend upon the choice of log resolution.

In recent years, researchers have begun to study which divisors on a log resolution contribute jumping numbers. See Alberich-Carrami\~{n}ana, \`{A}lvarez Montaner and Dachs-Cadefau~\cite{Alberich-Carraminana2014}, Galindo and Monserrat~\cite{MR2671187}, Hyry and J{\"a}rvilehto~\cite{MR2844818}, Naie~\cite{MR2470185}, Naie~\cite{MR3035120}, Smith and Thompson~\cite{MR2389246}, and Tucker~\cite{MR2592954}. This paper refines the result of Thompson~\cite{MR3168880} by finding a smaller set of divisors that contains all the divisors that contribute jumping numbers for a monomial space curve.

\vref{S:factorizing} of this paper recalls a strengthening of the notion of an embedded resolution of singularities known as a factorizing resolution and uses it to provide a proposition (\vref{P:b-ideal}) about the structure of multiplier ideals.

\vref{S:toric_case} of this paper recalls the Howald-Blickle Theorem (\vref{P:Blickle}) that provides a formula for the multiplier ideals of a monomial ideal on a normal affine toric variety, provides a reinterpretation (\vref{P:toroidal}) of that theorem, and provides a formula (\vref{P:principal}) for the multiplier ideals of a principal binomial ideal.

\vref{S:space_curves} applies the ideas of the previous sections to refine the result of Thompson~\cite{MR3168880}.

\section{Using factorizing resolutions to compute multiplier ideals} \label{S:factorizing}

\begin{defn}
  Let $Z$ be a generically smooth subscheme of any variety $Y$. A \emph{factorizing resolution} of $Z$ is an embedded resolution $\pi:X\rightarrow Y$ of $Z$ such that
  \[
    \sheaf{I}_Z\cdot\sheaf{O}_X=\sheaf{I}_{\widetilde{Z}}\cdot\sheaf{L}
  \]
  where $\widetilde{Z}$ is the strict transform of $Z$, $\sheaf{L}$ is an invertible sheaf, and the support of $\sheaf{I}_Z\cdot\sheaf{O}_X$ is a simple normal crossings variety.
\end{defn}

Recall that $\pi$ is an embedded resolution of $Z$ if it is proper birational morphism $\pi:X\rightarrow Y$ such that: $X$ is smooth and $\pi$ is an isomorphism over the generic points of the components of $Z$, the exceptional locus $\exc(\pi)$ of $\pi$ is a divisor with simple normal crossing support, and the strict transform $\widetilde{Z}$ is smooth and transverse to $\exc(\pi)$. For an embedded resolution, we always have $\sheaf{I}_Z\cdot\sheaf{O}_X=\sheaf{I}_{\widetilde{Z}}\cap\sheaf{L}$ for some invertible sheaf $\sheaf{L}$. Here we require the intersection to be a product. Typically, this is achieved by blowing up embedded components of $\sheaf{I}_Z\cdot\sheaf{O}_X$. Here is a theorem on the existence of factorizing resolutions.

\begin{prop}(Theorem~1.2 of Bravo~\cite{MR3031261}, Section~3 of Eisenstein~\cite{Eisenstein2010})
  Let $Z$ be a generically smooth subscheme of any variety $Y$ over an algebraically closed field of characteristic zero such that there exists a birational morphism $\mu_1:Y'\rightarrow Y$ from a smooth variety $Y'$ that is an isomorphism over the generic points of the components of $Z$.  If $D$ is a divisor on $Y'$ with simple simple normal crossing support such that no component of the strict transform of $Z$ is contained in $D$, then there exists a factorizing resolution $\pi:X\rightarrow Y$ of $Z$ that factors through $\mu_1$, $\pi=\mu_2\circ\mu_1$, such that $\widetilde{Z}\cup\exc(\pi)\cup\mu_2^{-1}(D)$ has simple normal crossing support.
\end{prop}

Notice that if $\pi:X\rightarrow Y$ is such a factoring resolution of $Z$, then the blowup of $\widetilde{Z}$ is a log resolution of $Z$ and that the exceptional locus of this blowup consists of a collection of prime divisors in one-to-one correspondence with the components of $Z$ with codimension at least two.

\begin{prop} \label{P:b-ideal}
  Let $Z_1,\ldots,Z_r$ be the components of $Z$ and suppose $e_i$ is the codimension of $Z_i$ for all $i$. Fix a factorizing resolution $\pi:X\rightarrow Y$ of $Z$ that is also a log resolution of the pair $(Y,\Delta)$ and let $\mathfrak{b}=\pi_*(\sheaf{L})$ where $\sheaf{I}_Z\cdot\sheaf{O}_X=\sheaf{I}_{\widetilde{Z}}\cdot\sheaf{L}$ as above. Then,
  \[
    \multideal\left((Y,\Delta),\sheaf{I}_Z^\lambda\right)=\multideal\left((Y,\Delta),\mathfrak{b}^\lambda\right)\cap\bigcap_{i=1}^r\sheaf{I}_{Z_i}^{(\lfloor\lambda+1-e_i\rfloor)}
  \]
\end{prop}

\begin{proof}
  Since $\sheaf{I}_Z\subseteq\mathfrak{b}$, it is clear that $\multideal\left((Y,\Delta),\sheaf{I}_Z^\lambda\right)\subseteq\multideal\left((Y,\Delta),\mathfrak{b}^\lambda\right)$. Let us now show $\multideal\left((Y,\Delta),\sheaf{I}_Z^\lambda\right)\subseteq\sheaf{I}_{Z_i}^{(\lfloor\lambda+1-e_i\rfloor)}$ for each $i$. Since $Z$ is generically smooth, the $\sheaf{I}_{Z_i}$ are prime and the $\sheaf{I}_{Z_i}^{(\lfloor\lambda+1-e_i\rfloor)}$ are primary. Since $\sheaf{I}_{Z_i}^{(\lfloor\lambda+1-e_i\rfloor)}$ is primary, it suffices to check the inclusion generically along the corresponding component (that is, after localizing at $\sheaf{I}_{Z_i}$). Because $Z$ is generically reduced, $\sheaf{I}_Z\cdot\sheaf{O}_{Z_i}=\sheaf{I}_{Z_i}\cdot\sheaf{O}_{Z_i}$. It is simple calculation based on the fact that one can resolve $\sheaf{O}_{Z_i}$ generically by blowing up $\sheaf{O}_{Z_i}$, and the relative canonical divisor for this blowup is $(e_i-1)E_i$, where $E_i$ is the resulting exceptional divisor.

  On the other hand, the extension of the contraction of an ideal is contained in the ideal. So, $\mathfrak{b}\cdot\sheaf{O}_X\subseteq\sheaf{L}$. Thus,
  \[
    \mathfrak{b}\cdot\sheaf{O}_X\cap\bigcap_{i=1}^r\sheaf{I}_{\widetilde{Z}_i}\subseteq\sheaf{L}\cap\bigcap_{i=1}^r\sheaf{I}_{\widetilde{Z}_i}=\sheaf{I}_Z\cdot\sheaf{O}_X.
  \]
  Therefore, we see
  \[
    \multideal\left((Y,\Delta),\mathfrak{b}^\lambda\right)\cap\bigcap_{i=1}^r\sheaf{I}_{Z_i}^{(\lfloor\lambda+1-e_i\rfloor)}\subseteq\multideal\left((Y,\Delta),\sheaf{I}_Z^\lambda\right)
  \]
  using any log resolution of $\mathfrak{b}$, $Z$ and $(Y,\Delta)$ that factors through $\pi$.
\end{proof}

\section{Exploiting the toric case} \label{S:toric_case}

This paragraph is, essentially, a direct quote of Blickle~\cite{MR2092724}. Let $(Y,\Delta)$ be a pair such that $Y$ is a normal (affine) toric variety (say $Y=\spec R$ for some normal semigroup ring $R\subseteq \Bbbk[x_1^{\pm1},\ldots,x_n^{\pm1}]$) and $\Delta$ is a torus invariant $\QQ$-divisor. Since $K_Y+\Delta$ is $\QQ$-Cartier and torus invariant, there is a monomial $\mathbf{x}^\mathbf{u}$ such that $\divisor\mathbf{x}^\mathbf{u}=r(K_Y+\Delta)$ for some integer $r$. Set $\mathbf{w}=\mathbf{u}/r$. Blickle's version of Howald's~\cite{MR1828466} formula is the following.

\begin{prop}(Theorem~1 of Blickle~\cite{MR2092724}) \label{P:Blickle}
  Let $\mathfrak{a}$ be a monomial ideal on $Y$. Then, if $\newt(\mathfrak{a})$ is the Newton Polyhedron of $\mathfrak{a}$,
  \[
    \multideal\left((Y,\Delta),\mathfrak{a}^\lambda\right)=\langle\mathbf{x}^\mathbf{v}\in R\mid\mathbf{v}+\mathbf{w}\in\text{ interior of }\lambda\newt(\mathfrak{a})\rangle
  \]
  for all $\lambda > 0$.
\end{prop}

This means that the multiplier ideals of a monomial ideal on a toric variety are contributed by (divisors supported on unions of) the Rees divisors of the ideal. (See Thompson~\cite{Thompson2003} for  a quick overview of the relationship between toric blowups and Newton polyhedra.) Other divisors that may appear on a log resolution do not contribute.

\begin{defn}
  We will say a sheaf $\sheaf{F}$ (respectively a Weil divisor $D$) on $X$ is \emph{locally monomial} if $X$ can be covered by open subschemes $U$ such that each $U$ is isomorphic to an open subscheme of a normal toric variety in such a way that $\sheaf{F}(U)$ is identified with a torus invariant sheaf (resp. a torus invariant divisor).
\end{defn}

\begin{prop} \label{P:toroidal}
  Let $(Y, \Delta)$ be a pair, consisting of a normal variety $Y$ over an algebraically closed field of characteristic zero and a $\QQ$-divisor $\Delta$ such that $K_Y+\Delta$ is $\QQ$-Cartier. If $\pi:X\rightarrow Y$ is a proper birational morphism such that $\pi^{-1}(\Delta)$ and $\sheaf{I}\cdot\sheaf{O}_X$ are locally monomial, then the multiplier ideals of $\sheaf{I}$ are contributed by the Rees divisors of $\sheaf{I}\cdot\sheaf{O}_X$.
\end{prop}

\begin{proof}
  Since the question is local on $X$, it suffices to consider the case where $X$ is a normal affine toric variety and $\mathfrak{a}=\sheaf{I}\cdot\sheaf{O}_X$ is a monomial ideal. Let $\mu:X'\rightarrow X$ be a toric log resolution of the ideal $\mathfrak{a}$ that is also a log resolution of the pair $(X, \pi^{-1}(\Delta))$. Evidently, $\mu$ factors through the blowup of $\mathfrak{a}$ and, as in the toric case, orders of vanishing on any exceptional divisor of $\mu\circ\pi$ are determined by those on the blowup of $\mathfrak{a}$. This is just the fact that when one represents a polyhedron as an intersection of half-spaces it suffices to consider only the  facet-defining half-spaces.
\end{proof}

Consider the case of any principal binomial ideal $I=\langle\mathbf{x}^{\mathbf{v}_1}-\mathbf{x}^{\mathbf{v}_2}\rangle\subseteq\Bbbk[x_1,\dots,x_n]$.

\begin{prop} \label{P:principal}
  If $I=\langle\mathbf{x}^{\mathbf{v}_1}-\mathbf{x}^{\mathbf{v}_2}\rangle$ and $\termideal=\langle\mathbf{x}^{\mathbf{v}_1},\mathbf{x}^{\mathbf{v}_2}\rangle$ is the term ideal of $I$, then
  \[
    \multideal(\affspace^n,I^\lambda)=\multideal(\affspace^n,\termideal^\lambda)\cap I^{(\lfloor\lambda\rfloor)}.
  \]
\end{prop}

\begin{proof}
  Let $\mathbf{v}$ be a primitive lattice vector such that $r\mathbf{v}=\mathbf{v}_1-\mathbf{v}_2$ for some positive integer $r$. Cover the normalized blowup of $\termideal$ with affine open toric varieties $U_1,\ldots,U_s$. Now, consider the covering consisting of the open sets of the form $\spec\Bbbk[x_1,\dots,x_n,\mathbf{x}^{\pm\mathbf{v}}] \setminus V(f)$ where $f=\frac{\mathbf{x}^{r\mathbf{v}}-1}{\mathbf{x}^\mathbf{v}-\zeta}$ for an $r$th root of unity $\zeta$ and the open subsets obtained by removing the closure of $V(\mathbf{x}^{r\mathbf{v}}-1)\subseteq\spec\Bbbk[x_1^{\pm1},\dots,x_n^{\pm1}]$ from each $U_i$.

  Note that since $\mathbf{v}$ is primitive, $\ZZ\mathbf{v}$ splits from $\ZZ^n$. So, $\NN^n+\ZZ\mathbf{v}\cong S\times\ZZ$ where $S$ is the image of $\NN^n$ in $\ZZ^n/\ZZ\mathbf{v}$. Each open set of the form $\spec\Bbbk[x_1,\dots,x_n,\mathbf{x}^{\pm\mathbf{v}}] \setminus V(f)$ is of the form $\spec\Bbbk[S][t]_{t+1}$ where $t=\mathbf{x}^{\mathbf{v}}-1$.

  And, on each open set of the form $U_i\setminus V(\mathbf{x}^{r\mathbf{v}}-1)$, $I\cdot\sheaf{O}_X=\termideal\cdot\sheaf{O}_X$ is monomial already. Thus, \vref{P:toroidal} applies. Moreover, it is clear that the components of the closure of $V(\mathbf{x}^{r\mathbf{v}}-1)\subseteq\spec\Bbbk[x_1^{\pm1},\dots,x_n^{\pm1}]$ are smooth and meet the boundary transversely.

  So, a toric desingularization of the blowup of $\termideal\cdot\sheaf{O}_X$ is a factorizing resolution of $I$. Now, apply \vref{P:b-ideal}.
\end{proof}

This result is not new. Principal binomial ideals are nondegenerate. For an alternate proof, see Howald~\cite{Howald2003}.

\section{Application to the monomial space curve case} \label{S:space_curves}

Using the previous ideas, one can refine the result of Thompson~\cite{MR3168880}. The case where the the monomial space curve is contained in a smooth toric surface follows from the principal toric case by using adjunction and inversion of adjunction.

Let $\Bbbk$ be a field of characteristic zero, let $C=\{(t^{n_1},t^{n_2},t^{n_3})\}\subset\affspace_\Bbbk^3$ be a monomial space curve not contained in a smooth toric surface. Assume $\mathbf{n}=\begin{bmatrix}n_1 & n_2 & n_3\end{bmatrix}\in\ZZ_{>0}^3$ is a primitive vector, let $\ord_\mathbf{n}$ be the monomial valuation given by the standard pairing, $\mathbf{x}^\mathbf{m}\mapsto\langle\mathbf{n},\mathbf{m}\rangle$, and let $I\subset\Bbbk[x_1,x_2,x_3]$ be the ideal of $C$. We may assume there exist irreducible binomials $f_1=\mathbf{x}^{\mathbf{m}_1^+}-\mathbf{x}^{\mathbf{m}_1^-}$, $f_2=\mathbf{x}^{\mathbf{m}_2^+}-\mathbf{x}^{\mathbf{m}_2^-}$, and  $f_3=\mathbf{x}^{\mathbf{m}_3^+}-\mathbf{x}^{\mathbf{m}_3^-}$ such that $\{f_1,f_2,f_3\}$ or $\{f_1,f_2\}$ is a minimal generating set for $I$. Let $\termideal=\left(\mathbf{x}^{\mathbf{m}_1^+},\mathbf{x}^{\mathbf{m}_1^-},\mathbf{x}^{\mathbf{m}_2^+},\mathbf{x}^{\mathbf{m}_2^-},\mathbf{x}^{\mathbf{m}_3^+},\mathbf{x}^{\mathbf{m}_3^-}\right)$ be the term ideal of $I$. Let $d_i=\ord_\mathbf{n}(f_i)$ for $i=1,2,3$. Order the generators so that $d_1< d_2< d_3$ and order the $n_i$ so that $n_i|d_i$ for $i=1,2$ (and $n_3|d_3$ when $f_3$ is a minimal generator). See Section~3 of Shibuta and Takagi~\cite{MR2533766} for a more detailed setup. Let $\mathbf{m}_1=\mathbf{m}_1^+-\mathbf{m}_1^-$ and let $\mathbf{q}=\begin{bmatrix}q_1 & q_2 & 0\end{bmatrix}\in\NN^3$ be the primitive vector such that $\langle\mathbf{q},\mathbf{m}_1\rangle=0$. And, let $e_i=\ord_\mathbf{q}(f_i)$ for $i=1,2,3$.

\begin{prop} \label{P:curvefix}
  Let $\mathfrak{a}_1=\left(\mathbf{x}^{\mathbf{m}_1^+},\mathbf{x}^{\mathbf{m}_1^-}\right)$, let $\mathfrak{a}_2=(x_1^{n_2n_3},x_2^{n_1n_3},x_3^{n_1n_2})$, and let the toric variety $X=X_\Sigma$ be the normalized blowup of $\mathfrak{a}_1\mathfrak{a}_2$. Then the ideal sheaf $I\cdot\sheaf{O}_X$ is locally monomial.
\end{prop}

\begin{proof}
  The blowup of $\mathfrak{a}_1$ is the partial desingularization of the toric surface $V(f_1)$ identified in Gonz{\'a}lez P{\'e}rez and Teissier~\cite{MR1892938}, and the normalized blowup of $\mathfrak{a}_2$ is  the partial desingularization of $C$. The fan $\Sigma_1$ of the blowup of $\mathfrak{a}_1$ has two maximal cones $\left\{\mathbf{v}\in\RR_{\geq0}^3\mid\langle\mathbf{v},\mathbf{m}_1\rangle\leq0\right\}$ and $\left\{\mathbf{v}\in\RR_{\geq0}^3\mid\langle\mathbf{v},\mathbf{m}_1\rangle\geq0\right\}$. The normalized blowup of $\mathfrak{a}_2$ is stellar subdivision along the ray $\rho=\RR_{\geq0}\mathbf{n}$. Note that $\mathbf{n}$ is in the intersection of the two maximal cones of $\Sigma_1$. So, the two operations on fans, stellar subdivision along $\mathbf{n}$ and cutting with the plane $\left\{v\in\RR_{\geq0}^3\mid\langle\mathbf{v},\mathbf{m}_1\rangle=0\right\}$ commute. And, $\Sigma$ is the stellar subdivision along $\rho$ of $\Sigma_1$. (Any toric desingularization of $X$ provides a common embedded desingularization of $C$ and the surface $V(f_1)$.)

  First, consider the affine open $U_\rho$ of $X$ and fix an element of the affine semigroup $\mathbf{m}_\rho\in\mathsf{S}_\rho$ such that $\langle\mathbf{m}_\rho,\mathbf{n}\rangle=1$. I claim, $\{\mathbf{m}_1,\mathbf{m}_2\}$ is a basis of the kernel of the matrix $\begin{bmatrix}n_1 & n_2 & n_3\end{bmatrix}$, $\mathsf{S}_\sigma=\NN^3+\ZZ\mathbf{m}_1+\ZZ\mathbf{m}_2$, $f_i=(\mathbf{x}^{\mathbf{m}_\rho})^{d_i}(\mathbf{x}^{\mathbf{m}_i}-1)$ for each $i=1,2,3$, and $f_3\in(f_1,f_2)\Bbbk[\mathsf{S}_\sigma]$. So,
  \begin{multline*}
    I\cdot\sheaf{O}_{U_\rho}=\left((\mathbf{x}^{\mathbf{m}_\rho})^{d_1}(\mathbf{x}^{\mathbf{m}_1}-1),(\mathbf{x}^{\mathbf{m}_\rho})^{d_2}(\mathbf{x}^{\mathbf{m}_2}-1)\right) \\
     =\left(\mathbf{x}^{d_1\mathbf{m}_\rho}\right)\cap\left(\mathbf{x}^{\mathbf{m}_1}-1,\mathbf{x}^{d_2\mathbf{m}_\rho}\right)\cap(\mathbf{x}^{\mathbf{m}_1}-1,\mathbf{x}^{\mathbf{m}_2}-1)
  \end{multline*}
  is monomial in $\mathbf{x}^{\mathbf{m}_\rho}$, $\mathbf{x}^{\mathbf{m}_1}-1$, and $\mathbf{x}^{\mathbf{m}_2}-1$. Since $d_1<d_2$, there is an embedded component supported on the intersection of the strict transform of the surface $\widetilde{V(f_1)}$ and the divisor $D_\rho$. Away from this embedded component, $I\cdot\sheaf{O}_X=\termideal\cdot\sheaf{O}_X$. Thus, it suffices to check the closed points where the curve $\overline{V(\mathbf{x}^{\mathbf{m}_1}-1,\mathbf{x}^{\mathbf{m}_\rho})}$ meets $X\setminus U_\rho$.

  Let $p\in X$ be one of these two points, and let $\sigma$ be the smallest cone of $\Sigma$ such that $p\in U_\sigma$. Evidently, $\rho\subsetneq\sigma$ since $p\notin U_\rho$. After possibly replacing $\mathbf{m}_1$ with $-\mathbf{m}_1$, we may assume $\mathbf{x}^{\mathbf{m}_1}-1\in\mathfrak{m}_{X,p}$. Since $\mathbf{x}^{\mathbf{m}_1}-1\in\mathfrak{m}_{X,p}$, $p$ is not a torus-fixed point and $\sigma$ is two-dimensional. Let $\sigma=\RR_{\geq0}^2\begin{bmatrix}n_1 & n_2 & n_3 \\ r_1 & r_2 & r_3\end{bmatrix}$. Note that $\mathbf{m}_1$ is a basis for the kernel of $\begin{bmatrix}n_1 & n_2 & n_3 \\ r_1 & r_2 & r_3\end{bmatrix}$, and $\langle\mathbf{n},\mathbf{m}_i\rangle=0$ for $i=1,2,3$. After possibly replacing $\mathbf{m}_2$ with $-\mathbf{m}_2$ and $\mathbf{m}_3$ with $-\mathbf{m}_3$, we may assume $\mathbf{m}_2,\mathbf{m}_3\in\mathsf{S}_\sigma=\NN^3+\ZZ \mathbf{m}_1$. As in \vref{P:principal}, the affine semigroup is a product $\mathsf{S}_\sigma\cong\mathsf{S}\times\ZZ\mathbf{m}_1$ where $\mathsf{S}$ is the image of $\NN^3$ in the quotient $\ZZ^3/\ZZ\mathbf{m}_1$. Thus, $\Bbbk[\mathsf{S}_\sigma]\cong\Bbbk[\mathsf{S}][t_3]_{t_3+1}$ where $t_3=\mathbf{x}^{\mathbf{m}_1}-1$.
\end{proof}

Recall $\mathbf{q}=\begin{bmatrix}q_1 & q_2 & 0\end{bmatrix}\in\NN^3$ is the primitive vector such that $\langle\mathbf{q},\mathbf{m}_1\rangle=0$ and $e_i=\ord_\mathbf{q}(f_i)$ for $i=1,2,3$.  Here is the improvement to the main theorem of Thompson~\cite{MR3168880}.

\begin{prop} \label{P:main}
  \begin{enumerate}[(i)]
    \item If $I$ is a complete intersection or if $e_2(d_3-d_1)\leq e_1(d_3-d_2)$, then
      \[
        \multideal(I^\lambda)=I^{\left(\lfloor\lambda-1\rfloor\right)}\cap\multideal(\termideal^\lambda)\cap\left(f\mid\nu_1(f)\geq\lfloor a_1\lambda-k_1\rfloor\right)
      \]
      where $\nu_1$ is the valuation given by the generating sequence $x_i\mapsto n_i$ for $i=1,2,3$, $f_1\mapsto d_2$. Thus, $a_1=\nu_1(I)=d_2$ and $k_1=\nu_1(J_{R_{\nu_1}/\Bbbk[x]})=n_1+n_2+n_3+d_2-d_1$ where $J_{R_{\nu_1}/\Bbbk[x]}$ is the Jacobian of the discrete valuation ring $R_{\nu_1}$ of $\nu_1$ over $\Bbbk[x]$.
    \item Otherwise,
      \[
        \multideal(I^\lambda)=I^{\left(\lfloor\lambda-1\rfloor\right)}\cap\multideal(\termideal^\lambda)\bigcap_{i=1,2}\left(f\mid\nu_i(f)\geq\lfloor a_i\lambda-k_i\rfloor\right)
      \]
      where $\nu_1$ is as before and $\nu_2$ is given by the generating sequence $x_1\mapsto e_2n_1+(d_3-d_2)q_1$, $x_2\mapsto e_2n_2+(d_3-d_2)q_2$, $x_3\mapsto e_2n_3$, $f_1\mapsto e_2d_3$. Thus, $a_2=\nu_2(I)=e_2d_3$, and $k_2=\nu_2(J_{R_{\nu_2}/\Bbbk[x]})=e_2(n_1+n_2+n_3)+(d_3-d_1)(q_1+q_2)+e_2(d_3-d_1)-e_1(d_3-d_2)$ where $J_{R_{\nu_2}/\Bbbk[x]}$ is the Jacobian of the discrete valuation ring $R_{\nu_2}$ of $\nu_2$ over $\Bbbk[x]$.
  \end{enumerate}
\end{prop}

\begin{proof}
  Apply \vref{P:curvefix} and the convex geometry computation in the appendix.
\end{proof}

\begin{cor}
  \begin{enumerate}[(i)]
    \item If $I$ is a complete intersection or if $e_2(d_3-d_1)\leq e_1(d_3-d_2)$, then the log canonical threshold of $I$ (at the origin) is
      \[
        \lct_0(I)=\min\left(\lct_0(\termideal),\frac{k_1+1}{a_1}\right).
      \]
    \item Otherwise,
      \[
        \lct_0(I)=\min\left(\lct_0(\termideal),\frac{k_1+1}{a_1},\frac{k_2+1}{a_2}\right).
      \]
  \end{enumerate}
\end{cor}

Note that when $e_2(d_3-d_1)=e_1(d_3-d_2)$, $\nu_2$ is monomial in the $\mathbf{x}$-variables and both formulas apply. In Example~5.3 of Blanco and Encinas~\cite{Blanco2014}, $e_2(d_3-d_1)=e_1(d_3-d_2)$, $\nu_2$. I do not know an example where $e_2(d_3-d_1)>e_1(d_3-d_2)$. A Macaulay2 package that implements this calculation, as presented in Thompson~\cite{MR3168880}, is described in Teitler~\cite{Teitler2013}.

\section*{Appendix}

Recall that $d_i=\ord_\mathbf{n}(f_i)$ and let $\mathbf{u}_i=\begin{bmatrix}d_i \\ e_i\end{bmatrix}=\begin{bmatrix}n_1 & n_2 & n_3 \\ r_1 & r_2 & r_3\end{bmatrix}\mathbf{m}_i^-$ for $i=1,2,3$. In the local monomial coordinates, we find the Rees valuations from the facets of the Newton polyhedron. It suffices to consider the ideal $\left(\mathbf{t}^{u_1}t_3,\mathbf{t}^{u_2},\mathbf{t}^{u_3}\right)\subset\Bbbk[\mathsf{S}]ty_3]$ (see the conclusion of the proof of \vref{P:main}). We know $e_1>e_2$ by examining Section~3 of Shibuta and Takagi~\cite{MR2533766}. And, $r_2=0$ or $r_3=0$.

If  $r_2=0$, then $e_2=0$, $\mathbf{t}^{\mathbf{u}_3}\in(\mathbf{t}^{\mathbf{u}_2})$, and the facets of the Newton polyhedron $\newt\left(\mathbf{t}^{\mathbf{u}_1}t_3,\mathbf{t}^{\mathbf{u}_2}\right)$ are orthogonal to the rows of the matrix
\[
  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    e_1 & d_2-d_1 & 0 \\
    1 & 0 & d_2-d_1
  \end{bmatrix}
\]
This includes the complete intersection case. Note that the only two rows of our matrix that have a nonzero last entry are $\begin{bmatrix}0 & 0 & 1\end{bmatrix}$ and $\begin{bmatrix}1 & 0 & d_2-d_1\end{bmatrix}$. The other vectors correspond to valuations that are monomial in the original $\mathbf{x}$-variables. Our ideal $\left(\mathbf{t}^{\mathbf{u}_1}t_3,\mathbf{t}^{\mathbf{u}_2},\mathbf{t}^{\mathbf{u}_3}\right)$ has order zero on the valuation corresponding to $\begin{bmatrix}0 & 0 & 1\end{bmatrix}$. And, the row $\begin{bmatrix}1 & 0 & d_2-d_1\end{bmatrix}$ corresponds to $\nu_1$.

If $\mathbf{t}^{\mathbf{u}_2}\notin\overline{(\mathbf{t}^{\mathbf{u}_1},\mathbf{t}^{\mathbf{u}_3})}$ and $e_2\neq0$, then the facets of the Newton polyhedron $\newt\left(\mathbf{t}^{\mathbf{u}_1}t_3,\mathbf{t}^{\mathbf{u}_2},\mathbf{t}^{\mathbf{u}_3}\right)$ are orthogonal to the rows of the matrix
\[
  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    e_1-e_2 & d_2-d_1 & 0 \\
    1 & 0 & d_2-d_1 \\
    e_2 & d_3-d_2 & 0
  \end{bmatrix}
\]
and these rows all have nonnegative integer entries. In terms of the parameters introduced in Section~3 of Shibuta and Takagi~\cite{MR2533766}, $\alpha\leq\gamma$ in this case.

If $\mathbf{t}^{\mathbf{u}_2}\in\overline{(\mathbf{t}^{\mathbf{u}_1},\mathbf{t}^{\mathbf{u}_3})}$ and $e_2\neq0$, then $r_3=0$ and the facets of the Newton polyhedron $\newt\left(\mathbf{t}^{\mathbf{u}_1}t_3,\mathbf{t}^{\mathbf{u}_2},\mathbf{t}^{\mathbf{u}_3}\right)$ are orthogonal to the rows of the matrix
\[
  \begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    e_1-e_3 & d_3-d_1 & 0 \\
    1 & 0 & d_2-d_1 \\
    e_2 & d_3-d_2 & e_2(d_3-d_1)-e_1(d_3-d_2)
  \end{bmatrix}
\]
and these rows all have nonnegative integer entries. Note that the only three rows that have a nonzero last entry are $\begin{bmatrix}0 & 0 & 1\end{bmatrix}$, $\begin{bmatrix}1 & 0 & d_2-d_1\end{bmatrix}$, and
\[
  \begin{bmatrix}e_2 & d_3-d_2 & e_2(d_3-d_1)-e_1(d_3-d_2)\end{bmatrix}
\]
corresponding to the only bounded facet of  $\newt\left(\mathbf{t}^{\mathbf{u}_1}t_3,\mathbf{t}^{\mathbf{u}_2},\mathbf{t}^{\mathbf{u}_3}\right)$. The other vectors correspond to valuations that are monomial in the original $\mathbf{x}$-variables. And, the bounded facet corresponds to $\nu_2$. For $\nu_2$, the orders of vanishing of the $x$-variables are given by the entries of
\begin{multline*}
  \begin{bmatrix}
    e_2 & d_3-d_2 & e_2(d_3-d_1)-e_1(d_3-d_2)
  \end{bmatrix}
  \begin{bmatrix}
    n_1 & n_2 & n_3 \\
    q_1 & q_2 & 0   \\
    0   & 0   & 0
  \end{bmatrix} \\
  =
  \begin{bmatrix}
    e_2n_1+(d_3-d_2)q_1 & e_2n_2+(d_3-d_2)q_2 & e_2n_3
  \end{bmatrix}
\end{multline*}
and $\nu_2(f_i)=e_2d_3$ for all $i=1,2,3$
\begin{multline*}
  \begin{bmatrix}
    e_2 & d_3-d_2 & e_2(d_3-d_1)-e_1(d_3-d_2)
  \end{bmatrix}
  \begin{bmatrix}
    d_1 & d_2 & d_3 \\
    e_1 & e_2 & 0   \\
    1   & 0   & 0
  \end{bmatrix} \\
  =
  \begin{bmatrix}
    e_2d_3 & e_2d_3 & e_2d_3
  \end{bmatrix}
\end{multline*}

\end{document}

config: arxiv


TITLE: Could the Fourier transform of a one-sided non-negative function be compactly supported?
QUESTION [2 upvotes]: I know the Fourier transform of a compactly supported function cannot be compacted supported. But what if the function is supported on $(0,+\infty)$ or $(-\infty,0)$? Is there any chance for its Fourier transform be compactly supported?

REPLY [3 votes]: The inverse Fourier transform of an $L^1$ function of compact support is an entire function.  Therefore it can't be zero on any set with a finite limit point, in particular $(0,\infty)$ or $(-\infty,0)$.


TITLE: Integration help - find $f(x)$ (2nd derivative is $f''(x)=6x+6$, stationary point at $(3,19))$
QUESTION [1 upvotes]: I've done part a, it's a local minimum as substituting 3 into $f''(x)$ gives a value > 0.
However, for the 2nd part, here's my working so far.
$f'(x)=\int 6x+6 \ dx=3x^2+6x+c$
$f(x)=\int3x^2+6x+c \ dx = x^3+3x^2+cx+c$
I substituted 3 and 9 in, so:
$19=(3)^3+3(3)^2+(3)c+c$
$-35=4c$ which doesn't look right.
The answer in the back says:
$f(x)=x^3+3x^2-45x+100$
I am going wrong somewhere, can someone help please>

REPLY [1 votes]: Hint:
$$\int\left(\int(6x+6)dx\right)dx=\int(3x^2+6x+c_1)dx=x^3+3x^2+c_1x+c_2$$


TITLE: How to show two random variables have the same distribution?
QUESTION [3 upvotes]: Let $\{X_n\}$ and $\{Y_n\}$ be sequences of random variables such that the pairs $(X_i,X_j)$ and $(Y_i,Y_j)$ have the same distributions for all $i,j$. If $X_n \rightarrow X$ in probability, show that $Y_n$ converges in probability to some limit $Y$ having the same distribution
I am trying to prove it in the following steps:
Step1: Show that $P(|Y_n-Y_m|> \epsilon) = P(|X_n-X_m|> \epsilon)$ for arbitrary $\epsilon > 0$. This is true since $(X_i,X_j)$ and $(Y_i,Y_j)$ have the same distributions for all $i,j$.
Step2: $X_n \rightarrow X$ in probability implies $X_n$ is Cauchy convergent in probability(meaning that for all $\epsilon >0 $, $P(|X_n-X_m|> \epsilon)$ goes to zero as m,n go to $\infty$). So $Y_n$ is Cauchy convergent in probability as well.
Step3: Show that there exists a random variable $Y$ such that $Y_n \rightarrow Y$ in probability. Actually, $Y(\omega) = \lim_{k\rightarrow \infty} Y_{n_k}(\omega)$ for a subsequence such that $P(|Y_{n_k} - Y_{n_{k+1}} | > \epsilon) \leq 2^{-k}$. And we can show $Y_n \rightarrow Y$ in probability using Borel Cantelli Lemma.
Step4: Show $Y$ have the same distribution as $X$.
I am having trouble about Step4. So do $(X_i,X_j)$ and $(Y_i,Y_j)$ have the same distributions imply $X_n$ and $Y_n$ are identically distributed?

REPLY [1 votes]: $(X_n,X_{n+1}) \to (X,X)$ in probability and $(Y_n,Y_{n+1}) \to (Y,Y)$ in probability. Hence $(X,X)$ has same distribution as $(Y,Y)$ which implies that $X$ has the same distribution a $Y$.


TITLE: Limit Help: $\lim_{x\to\infty} xe^{-a\frac{x}{\ln x}}$
QUESTION [0 upvotes]: I feel dumb for asking this, but I couldn't quite show that this limit is 0 (which I think is correct) whenever $a>0$:
$$\lim_{x\to\infty} xe^{-a\frac{x}{\ln x}}.$$
I tried using L'Hospital's rule which was mildly annoying given the amount of terms.  I also couldn't figure out a good way to use comparison either.

REPLY [2 votes]: Hint: try to use the fact that $\log x < \sqrt{x}$ for $x \gg 1$


TITLE: Chance letter a next to b in circle with whole alphabet such that no vowels next to each other
QUESTION [7 upvotes]: Here's a question from a book on probability I'm working through:

If the $26$ letters of the alphabet are written down in a ring so that no two vowels come together, what is the chance that a is next to b?

Here's what I did. Let's fix a. Since b can be immediately to the left or right of a, there's $2$ choices for b. Without loss of generality let's say we have ab, and so we have $4$ vowels remaining and $20$ consonants remaining. With the condition that no $2$ vowels come together: We want to find the number of possible places where we can place a vowel, which is in between consonants. With the $20$ consonants, there's $19$ possible "gaps" between, plus the $2$ on the end, for a total of $21$. However, because our b already occupies one of them, we have to subtract $1$, getting us $20$. So out of these $20$ places we're choosing $4$, so there's $\binom{20}{4}$ ways to place our $4$ vowels among the $20$ consonants subject to the condition no $2$ vowels come together. There's $4!$ ways to order the remaining vowels and $20!$ ways to order the remaining consonants. So our numerator is$${{2 \binom{20}{4} 20!4!}}$$Now let's calculate the denominator. Fix a again. This time we have $21$ consonants and $4$ vowels remaining, but only the $20$ possible "gaps" between the consonants to places our $4$ vowels (there's no $2$ at the end this time around), so our numerator is$$\binom{20}{4}21!4!$$Therefore the chance that a is next to b is$${2\over{21}}$$However, the answer in the back of my book (which is known to be wrong in many places in the answers in the back section) is ${1\over{10}}$. So who's correct? And if I'm wrong, where did I specifically go wrong?

REPLY [6 votes]: A simpler reasoning: We have $26-5=21$ consonants. We know that, in the allowed configurations, each vowel (in particular, $a$) will have two consonants on each side. By symmetry, any pair of consonants is equiprobable. There are $\binom{21}{ 2}=210$ such (unordered) pairs. Among them, there are $20$ pairs that include $b$. Then the probability of our event (letter $a$ is next to $b$) is
$$ p=\frac{20}{210}=\frac{2}{21}=0.95238\cdots$$
which agrees with your answer.


text:
\section{Representation theory of \(\SymplTwogQ\)}
\label{Representation theory of SymplTwogQ}

The following is a classical theorem in representation theory
(see e.g. \cite{Fulton1991}).

\begin{theorem}
    There is a bijection
    \begin{equation*}

        \includegraphics{reptheory_defmap_bijection-crop.pdf}
    \end{equation*}
    where \(V_\lambda\) is a certain submodule of
    \((\wedge^{\lambda^\prime_1} H ) \tensor \cdots \tensor (\wedge^{\lambda^\prime_d} H )\).
    \(V_\lambda\) is generated by the element
    \begin{equation*}
        a_\lambda \defeq (a_1 \wedge \cdots \wedge a_{\lambda^\prime_1}) \tensor \cdots
        \tensor (a_1 \wedge \cdots \wedge a_{\lambda^\prime_d})
        \ \in (\wedge^{\lambda^\prime_1} H ) \tensor \cdots \tensor (\wedge^{\lambda^\prime_d} H )
    \end{equation*}
    as an \(\SymplTwogQ\)-module.
    Here
    \(\transpose{\lambda} = [\lambda^\prime_1 \cdots \lambda^\prime_d] \quad
         (g \geq \lambda^\prime_1 \geq \cdots \geq \lambda^\prime_d \geq 1)\)
    is the transpose of \(\lambda\),
    and \(\left[ V_\lambda \right]\) is the isomorphism class of \(V_\lambda\).
    This \(a_\lambda\) is called the \textit{highest weight vector}
    with respect to the ordered symplectic basis
    \(a_1, \dots, a_g, b_1, \dots, b_g\).
    For example,
    \begin{equation}
        a_{[32]} = (a_1 \wedge a_2) \tensor (a_1 \wedge a_2) \tensor a_1
        ,\quad
        a_{[221]} = (a_1 \wedge a_2 \wedge a_3) \tensor (a_1 \wedge a_2).
    \end{equation}
\end{theorem}

\par Now we describe
each \(\Symp\)-irreducible component of \(\wedge^\bullet \cgplus\)
by a Young diagram.
We simply write an \(\Symp\)-module isomorphic to \(V_\lambda\)
as \(\lambda\) when we focus mainly on its isomorphism class.

\begin{lemma}
    \begin{itemize}
        \item[(i)] If \(k > l \geq 1\),
            \begin{equation*}
                \cg(k) \tensor \cg(l) \isom
                    \bigoplus_{0 \leq \lambda_2 \leq l+2} \
                    \bigoplus_{0 \leq \rho \leq l+2-\lambda_2}
                    \lbrack (k+l+4-\lambda_2-2\rho) \quad \lambda_2 \rbrack
            \end{equation*}
            In particular, if \(V \subset \cg(k) \tensor \cg(l)\) is
            an \(\Symp\)-irreducible submodule corresponding to a Young diagram
            \(\lambda = \lbrack \lambda_1 \lambda_2\rbrack\), then the following conditions hold:
            \begin{align}
                l + 2 \geq \rho + \lambda_2, & &
                k + 2 \leq \rho + \lambda_1 & &
                \text{where } \rho \defeq \frac{1}{2} (k+l+4-\lambda_1-\lambda_2) \in \ZInteger_{\geq 0} \label{possible young diagrams in cgk cgl}
            \end{align}
        \item[(ii)] If \(k \geq 1\),
        \begin{equation*}
            \cg(k) \wedge \cg(k) \isom
                \bigoplus_{0 \leq \lambda_2 \leq k+2} \
                \bigoplus_{\substack{0 \leq \rho \leq k+2-\lambda_2 \\ \rho + \lambda_2 \text{ is odd}}}
                \lbrack (2k+4-\lambda_2-2\rho) \quad \lambda_2 \rbrack
        \end{equation*}
        In particular, if \(V \subset \cg(k) \tensor \cg(l)\) is
        an \(\Symp\)-irreducible submodule corresponding to a Young diagram
        \(\lambda = \lbrack \lambda_1 \lambda_2\rbrack\), then the following conditions hold:
        \begin{align}
            \rho + \lambda_2 \text{ is odd}, & &
            \rho + \lambda_2 \leq k + 2 \leq \rho + \lambda_1 & &
            \text{where } \rho \defeq \frac{1}{2} (2k+4-\lambda_1-\lambda_2) \in \ZInteger_{\geq 0} \label{possible young diagrams in cgk cgk}
        \end{align}
    \end{itemize}
\end{lemma}
\begin{proof}
(i) follows from the Littlewood-Richardson rule and branching rules.
\par (ii) follows from the same rules and the plethysm of symmetric polynomials
\begin{equation*}
    e_2 \circ h_{k+2} = \sum_{1 \leq j \leq 2k+4-j, \ j \text{ is odd}}
        s_{\lbrack (2k+4-j) \ j\rbrack}
\end{equation*}
where \(e_2\) is the second elementary symmetric function,
\(h_{k+2}\) is the \((k+2)\)-nd complete symmetric function and
\(s_{\lambda}\) is the Schur function corresponding to a Young diagram \(\lambda\)
(see \cite{MacDonald1995}).
\end{proof}
This lemma shows, for fixed \(k\) and \(l\),
each Young diagram \(\lambda\) satisfying
\eqref{possible young diagrams in cgk cgl} or \eqref{possible young diagrams in cgk cgk}
corresponds to \textit{exactly one} \(\Symp\)-irreducible component in \(\cg(k) \wedge \cg(l)\),
so that we identify such Young diagrams with them.
From now, we denote simply by \(\lambda\) an \(\Symp\)-irreducible component
\(V \subset \cg(k) \wedge \cg(l)\) isomorphic to \(V_\lambda\), like
\(\lambda \subset \cg(k) \wedge \cg(l)\).


TITLE: How can I prove this equality?
QUESTION [1 upvotes]: I'm not sure how I can prove this. Any help would be appreciated

REPLY [2 votes]: $Let\; x\in (V+W)^\perp \Rightarrow (x,v+w)=0$ where $v\in V \; and\; w\in W$
$(x,v+w)=0 $
$\Rightarrow (x,v)+(x,w)=0$
$\Rightarrow (x,v)=0\;and\;(x,w)=0$
$\Rightarrow x\in V^\perp\;and\;x\in W^\perp$
$\Rightarrow x\in V^\perp \cap W^\perp$
and all these steps holds backwards too.
This solves your problem.


TITLE: Pressure $p$ in the Bernoulli equation
QUESTION [1 upvotes]: I know the Bernoulli equation that states that the energy along a streamline is constant:
$$ \frac{v^{2}}{2}+gh+\frac{p}{\rho}=C $$
However I can't seem to figure out what exactly $p$ is in this equation. In every book I look and also on Wikipedia it only says "$p$ is the (static) pressure". I always considered $p$ to be the internal pressure of the liquid, i.e. the pressure exerted by the fluid on its surroundings (that includes covering fluid). But this raises a problem that I wasn't sure of how to solve.

Lets assume that the flow in the picture is an incompressible flow, like water. When the water flows from the wide to the narrow part of the tube, the speed $v$ increases, lowering the pressure $p$ (as $gh$ and $\rho$ remain constant). As this is an incompressible fluid, the volume cannot change. But if the pressure is lowered and the volume does not change, doesn't that mean that the temperature is lowered too? This just seems weird to me.
What goes wrong in this thought process?

REPLY [1 votes]: If we think of the forces $(F=PA)$ acting on the fluid as it flows along the pressure gradient the problem becomes clearer.
An incompressible fluid flowing from a region of higher pressure to a region of lower pressure feels a net force in the direction of the pressure gradient. This net force causes the fluid to accelerate (bulk motion).
Thus there is an increase in kinetic energy. On the assumption that the water is incompressible there will not be a change in temperature.


TITLE: Prove that the volume of a cylinder is pi*a^2*h using triple integration and spherical coordinates
QUESTION [1 upvotes]: We have a cylinder of radius a and height h. We need to prove its volume to be equal to pi*a^2*h using triple integral and spherical coordinates. The best way to solve this problem is to divide the cylinder to two volume regions , the first region is the one defined by phi to range from 0 to arctan(a/h) and the second region is the one defined by phi to range from arctan(a/h) till pi/2. Surley , for both regions theta will vary from 0 till 2*pi. However , i tried many times to find the proper limits for the radius for each region but i failed. So how do i  properly define the radius for each region ?
Edit : I know the limits of each region for the radius r= 0 till asecθ and r= 0 till bcscθ but my question exactly is how to drive them ?

REPLY [1 votes]: I would say that to find the volume of a cylinder use cylindrical coordinates.
But you want to use spherical.
Your boundaries
$x^2 + y^2 = a^2\\
z = 0\\
z = h$
$x = \rho\cos\theta\sin\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho\cos \phi$
$\rho^2\cos^2 \theta\sin^2\phi + \rho^2\sin^2 \theta\sin^2\phi = a^2\\
\rho^2\sin^2\phi = a^2\\
\rho = a\csc\phi$
$\rho\cos \phi = h\\
\rho = h\sec \phi$
$a\csc\phi = h\sec \phi\\
\tan\phi = \frac ah$
$\int_\limits0^{2\pi}\int_\limits0^{\arctan \frac {a}{h}}\int_\limits0^{h\sec \phi} \rho^2\sin \phi \ d\rho\ d\phi\ d\theta\\ + \int_\limits0^{2\pi}\int_\limits{\arctan \frac {a}{h}}^\frac {\pi}{2}\int_\limits0^{a\csc \phi} \rho^2\sin \phi \ d\rho\ d\phi\ d\theta$


TITLE: Uniformly continuous function approximated by Lipschitz
QUESTION [2 upvotes]: Let $f~:~\mathbb R \to \mathbb R$ a uniformly continuous function.
How could I construct a sequence $(f_n)_{n\ge 0}$ of Lipschitz functions such that $(f_n)$ uniformly converges towards $f$?

REPLY [5 votes]: f uniformly continuous $\Rightarrow$ for all $\epsilon > 0$ there exists a $\delta = \delta(\epsilon) > 0$ such that $|x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon$.
The idea is to build a piecewise linear approximation to $f$ (which will clearly be Lipschitz). For each $n$, we can build $f_n$ in the following way:

Choose $\epsilon_n = \frac{1}{n}$. Then by the above defn. of uniform continuity, there exists a corresponding $\delta = \delta_n$ such that $|x-y| < \delta_n \Rightarrow |f(x) - f(y)| < \frac{1}{n}$
Divide $\mathbb{R}$ into intervals of size $\frac{\delta_n}{2}$. Build $f_n$ by just evaluating $f$ at the endpoints of these intervals and linearly interpolating within each interval.
We know that $f$ can't grow by more than $\frac{1}{n}$, within one of these subintervals. so all of the linear pieces have a bounded slope. To fully check whether $f_n$ is Lipschitz, you have to choose arbitrary $x,y \in \mathbb{R}$ and check the Lipschitz condition, not just $x,y$ within one of these subintervals, so I'll leave it to you to check that.
Within each interval, $f$ can't differ from the value of $f$ at the left endpoint by more than $\frac{1}{n}$ by uniform continuity, and similarly, the value of $f_n$ can't differ by the value of $f$ at the left endpoint by linearity within the interval, so by the triangle inequality, $||f_n - f||_{L^\infty} \le \frac{2}{n}$.
Taking the limit as $n \rightarrow \infty$ shows uniform convergence.


TITLE: How to find the linear application given the function on the basis vectors.
QUESTION [0 upvotes]: Say I am given a linear application $f$ from $R^2$ to $R^2$, and I am told it maps $e_1$ to $(1,3)$ and $e_2$ to $(-2,7)$
In this case I know how to find how the linear application acts on a generic vector $(x,y)$:
$$f((x,y)) = f(x e_1 + ye_2) = (x -2y, 3x + 7y)$$
But say I swap $e_1, e_2$ with $(5,3),(-2, 7)$ then how could I do to find how the linear application acts on a generic vector $(x,y)$?

REPLY [1 votes]: in this case it must finding the passage matrix from the bases
$B_2=\{(1,3),(-2,7)\}$ to  the $B_1=\{(1,0),(0,1)\}$;  as the
passage matrix  from $B_2$ to base $B_1$ is the matrix inverse of
the passage matrix from $B_1$ to $B_2$  and you  know explicitly
the
 matrix  passage $P$ form $B_1$ to $B_2$ that is the matrix   whose colons are vectors $(1,3)$ and $(-2,7)$,  then the wanted  matrix
is  $P^{-1}=\left(
\begin{array}{cc}
{\frac
7{13}} & {\frac 2{13}} \\
{-\frac 3{13}}& {\frac 1{13}}
\end{array}
\right)$ and so $ f (x, y) = \left(
\begin{array}{cc}
{\frac
7{13}} & {\frac 2{13}} \\
{-\frac 3{13}}& {\frac 1{13}}
\end{array}
\right) (x, y) =({\frac 7{13}x+\frac 2{13}y},-\frac 3{13}x+\frac
{1}{13}y)$.


text:
\subsection{Shifted cotangent stacks}

\subsubsection{Linear stacks}\label{linst}

Given a quasi-coherent sheaf $E$ on a derived stack $X$, one can define a derived stack $\mathbb{A}_E$ over $X$ as follows:
\[
\mathbb{A}_E\big(\scat{Spec}(A)\overset{u}{\to}X\big):=|u^*E|\,.
\]
The construction is functorial in $E$, and we write $\pi_X:\mathbb{A}_E\to X$ for the map corresponding to $E\to 0$.
The construction is also compatible with pull-backs: for every $f:Y\to X$, $\mathbb{A}_{f^*E}\simeq f^*\mathbb{A}_E$.
\begin{remark}
If $X$ is Artin and if $E$ is perfect, then $\mathbb{A}_E$ is Artin too, and it coincides with the derived stack
$\mathbb{V}(E^\vee)$ from \cite[\S3.3]{ToEMS}.
\end{remark}
If $X$ is an Artin derived $k$-stack and $f:E\to F$ is a morphism between perfect quasi-coherent sheaves on $X$, then
the relative tangent complex of $\mathbb{A}_f:\mathbb{A}_E\to\mathbb{A}_F$ is
\[
\mathbb{T}_{\mathbb{A}_f}\simeq \pi_X^*\cat{fib}(f)\,.
\]
In particular, if $F=0$, then $\mathbb{T}_{\pi_X}\simeq\pi_X^*E$, so that we get equivalences
\[
\mathbb{A}_{\mathbb{T}_{\pi_X}}
\simeq \mathbb{A}_{\pi_X^*E}
\simeq\mathbb{A}_E\underset{X}{\times}\mathbb{A}_E
\simeq\mathbb{A}_{E\oplus E}
\quad\textrm{and}\quad \mathbb{A}_{\mathbb{L}_{\pi_X}}\simeq \mathbb{A}_{E\oplus E^\vee}\,.
\]
\begin{example}\label{exGmod}
Let $G$ be an algebraic group, and let $V$ be $k$-module with a linear $G$-action; this defines a quasi-coherent
sheaf $\mathbf{V}$ on $BG$.
One has $\mathbb{A}_{\mathbf{V}}\simeq \big[\mathbb{A}_V/G\big]$, which is sometimes just denoted $[V/G]$.
Then $\mathbb{L}_{[V/G]}$ is the pull-back along $p_{V,G}\,:\,[V/G]\to BG$ of the quasi-coherent sheaf on $BG$ corresponding
to the dual representation $V^*$. Hence
\[
\mathbb{A}_{\mathbb{L}_{p_{V,G}}}\simeq [V\oplus V^*/G]\,.
\]
\end{example}
The above situation specializes to the following
\begin{example}\label{exQuiv}
Let $Q$ be a quiver, with vertex set $V(Q)$ and edge set $E(Q)$. The edges are oriented, we denote by $s(e)$ and $t(e)$ the
source and target of an edge $e$. Let $\vec{n}=(n_v)_{v\in V(Q)}$ be a dimension vector.
We define
\[
Rep_k(Q,\vec{n}):=\prod_{e\in E(Q)}\cat{Hom}_{\cat{Mod}_k}(k^{n_{s(e)}},k^{n_{t(e)}})\,,
\]
which is a $GL_{\vec{n}}(k)$-module. The relative cotangent complex of the morphism
\[
p_{k,Q,\vec{n}}\,:\,\big[Rep_k(Q,\vec{n})/GL_{\vec{n}}(k)\big]\longrightarrow BGL_{\vec{n}}(k)
\]
is therefore the pull-back (along $p_{k,Q,\vec{n}}$) of the $GL_{\vec{n}}(k)$-module $Rep_k(Q^*,\vec{n})$, where $Q^*$
stands for the opposite quiver (built from $Q$ by reversing all edges), and where we view a $GL_{\vec{n}}(k)$-module as
a quasi-coherent sheaf on $BGL_{\vec{n}}(k)$.
As a consequence, we have
\[
\mathbb{A}_{\mathbb{L}_{p_{k,Q,\vec{n}}}}\simeq \big[Rep_k(\overline{Q},\vec{n})/GL_{\vec{n}}(k)\big]\,,
\]
where $\overline{Q}$ stands for the doubled quiver (built from $Q$ by adjoining to every edge $e$ a new edge $e^*$ going
the reverse way).
\end{example}
The following example is a variation on the previous one.
\begin{example}\label{exQuiv2}
Let us consider the derived $k$-stack $\scat{Perf}_k$ of perfect complexes from \cite{HAG-II}, that is defined by
\[
\scat{Perf}_k(A):=(\scat{Mod}_A^{perf})^\simeq\,,
\]
where $\scat{Mod}_A^{perf}$ is the full subcategory spanned by perfect $A$-modules, and the superscript $(-)^\simeq$
means that we only keep the $1$-morphisms that are equivalences. A morphism $X\to\scat{Perf}_k$ is by definition the
data of a perfect sheaf on $X$; therefore the identity morphism $\scat{Perf}_k\to\scat{Perf}_k$ classifies a tautological
perfect sheaf $E$ on $\scat{Perf}_k$.

For a finite quiver $Q$, with vertex set $V(Q)$ and edge set $E(Q)$, we consider the perfect sheaf
\[
E^Q:=\prod_{e\in E(Q)}\textsc{Map}\big(p_{s(e)}^*E,p_{t(e)}^*E\big)
\]
on $\scat{Perf}_k^{V(Q)}$, where $p_v:\scat{Perf}_k^{V(Q)}\to \scat{Perf}_k$ is the $v$-th projection ($v\in V(Q)$).
The relative cotangent complex of the morphism
\[
p_Q\,:\,\mathbb{A}_{E^Q}\longrightarrow \scat{Perf}_k^{V(Q)}
\]
is therefore $(E^Q)^\vee\simeq E^{Q^*}$, where $(-)^\vee=\textsc{Map}(-,\mathcal O)$ is the dualization functor.
As a consequence,
\[
\mathbb{A}_{\mathbb{L}_{p_Q}}\simeq \mathbb{A}_{E^{\overline{Q}}}\,.
\]
\end{example}
\begin{remark}\label{labelresBG}
Later on, we will identify $\mathbb{A}_{E^Q}$ with the moduli of objects $\scat{Perf}_{kQ}$. For the moment,
it is sufficient to observe that the restriction of $E$ to the open substack $BGL_n(k)\subset\scat{Perf}_k$ is $k^n$,
so that the restriction of $E^Q$ to the open substack $BGL_{\vec{n}}(k)\subset \scat{Perf}_k^{V(Q)}$ is $Rep_k(Q,\vec{n})$.
\end{remark}

\subsubsection{Shifted cotangent stacks are shifted symplectic}

The \textit{$n$-shifted cotangent stack} of a derived Artin $k$-stack $X$ is
\[
\mathbf{T}^*[n]X:=\mathbb{A}_{\mathbb{L}_X[n]}\,.
\]
In other words, a morphism $Y\to\mathbf{T}^*[n]X$ is determined by a morphism $u:Y\to X$ and a global
section of $u^*\mathbb{L}_X[n]$.
The identity morphism $\mathbf{T}^*[n]X\to\mathbf{T}^*[n]X$ is thus determined by a global
section $\ell_X$ of $\pi_X^*\mathbb{L}_X[n]$. Let us denote $\lambda_X$ the image of $\ell_X$
through $\pi_X^*\mathbb{L}_X[n]\longrightarrow \mathbb{L}_{\mathbf{T}^*[n]X}[n]$:
\[
\lambda_X\in \mathcal A^1(\mathbf{T}^*[n]X,n)\,.
\]
We know (see \cite[Theorem 2.2(1)]{CalCot}) that $\omega_X:=d_{dR}\lambda_X\in\mathcal A^{2,\cl}(\mathbf{T}^*[n]X,n)$
defines an $n$-shifted symplectic structure on $\mathbf{T}^*[n]X$.
\begin{remark}\label{remark: 1-form}
The $1$-form $\lambda_X$ can also be described as follows using the functor of points.
Recall that a $B$-point of $\mathbf{T}^*[n]X$ is determined by a point $x\in X(B)$ and a section $s\in|x^*\mathbb{L}_X[n]|$.
Then
\[
\lambda_X(x,s)=x^*s\in|\mathbb{L}_B[n]|=\mathcal{A}^1(B,n)\,.
\]
\end{remark}
\begin{example}\label{BGex}
For a reductive affine algebraic group $G$, $\mathbb{L}_{BG}$ is the quasi-coherent sheaf associated with the shifted
coadjoint $G$-module $\mathfrak{g}^*[-1]$. Therefore
\[
\mathbf{T}^*[n](BG)\simeq \big[\mathfrak{g}^*[n-1]/G\big]\,.
\]
Moreover, the section $\ell_{BG}$ has a simple expression
\[
\ell_{BG}=\mathrm{id}_{\mathfrak{g}[1-n]}\in\big|(\mathfrak{g}^*[n-1]\otimes\mathfrak{g}[1-n])^G\big|\subset
\big|\big(\mathfrak{g}^*[-1]\otimes\cat{Sym}(\mathfrak{g}[1-n])\big)^G[n]\big|\simeq|\Gamma(\pi_{BG}^*\mathbb{L}_{BG})[n]|\,.
\]
Actually, the above still makes sense without assuming $G$ reductive; we still have a map
\[
\big|\big(\mathfrak{g}^*[-1]\otimes\cat{Sym}(\mathfrak{g}[1-n])\big)^G[n]\big|\longrightarrow
|\Gamma(\pi_{BG}^*\mathbb{L}_{BG})[n]|\,,
\]
which is not necessarily an equivalence (as there might be non-trivial higher $G$-cohomology).
Notice that in the reductive case, one can choose a non-degenerate invariant quadratic form $c\in\mathrm{Sym}^2(\mathfrak{g}^*)^G$
and thus get an isomorphism $c^\flat:\mathfrak{g}\to\mathfrak{g}^*$ of $G$-modules, so that
\[
\mathbf{T}^*[n](BG)\simeq \big[\mathfrak{g}[n-1]/G\big]\,.
\]
Through this identification, we have
\[
\ell_{BG}=c\in (\mathfrak{g}^{\otimes2})^G\subset\big|\big(\mathfrak{g}^*[-1]\otimes\cat{Sym}(\mathfrak{g}^*[1-n])\big)^G[n]\big|
\simeq|\Gamma(\pi_{BG}^*\mathbb{L}_{BG})[n]|\,.
\]
\end{example}
\begin{example}\label{linPerfk}
It is known (see \cite[Corollary 3.29]{ToVa}) that the derived $k$-stack $\scat{Perf}_k$ is locally geometric
and locally of finite presentation, and that
\[
\mathbb{T}_{\scat{Perf}_k}\simeq\textsc{Map}(E,E)[1]\,.
\]
Therefore
\[
\mathbf{T}^*[n]\scat{Perf}_k\simeq\mathbb{A}_{\textsc{Map}(E,E)^\vee[n-1]}\simeq \mathbb{A}_{\textsc{Map}(E,E)[n-1]}
\]
as there is an isomorphism of perfect sheaves
\[
c^\flat\,:\,\textsc{Map}(E,E)\tilde\longrightarrow \textsc{Map}(E,E)^\vee\,;\, f\longmapsto \mathrm{tr}(f\circ-)\,.
\]
Through the above identification, on $\mathbb{A}_{\textsc{Map}(E,E)[n-1]}$, $\ell_{\scat{Perf}_k}$ becomes a section of
\[
\pi_{\scat{Perf}_k}^*\textsc{Map}(E,E)^\vee[n-1]\simeq\pi^*_{\scat{Perf}_k}\mathbb{L}_{\scat{Perf}_k}\,,
\]
which is the section classifying $c^\flat[n-1]$.
\end{example}

\subsubsection{Lagrangian correspondences for shifted cotangent stacks}\label{lagshicotst}

Let $f:X\to Y$ be a morphism of derived Artin $k$-stacks.
We have a morphism $f^*\mathbb{L}_Y\to \mathbb{L}_X$ leading, for every $n\in\mathbb{Z}$, to a correspondence
\begin{equation}\label{eq:corrcot}
\bT^*[n]X\overset{q}\longleftarrow f^*\bT^*[n]Y\overset{p}{\longrightarrow} \bT^*[n]Y
\end{equation}
of derived Artin $k$-stacks.
Observe that, by definition, $p^*\ell_Y$ is the section of $p^*\pi_Y^*\mathbb{L}_Y[n]\simeq q^* \pi_X^*f^*\mathbb{L}_Y[n]$
classifed by $p$, and it coincides with the one classified by $\mathrm{id}:f^*\mathbf{T}^*[n]Y\to f^*\mathbf{T}^*[n]Y$.
Hence the image of $p^*\ell_Y$ through the morphism $q^* \pi_X^*f^*\mathbb{L}_Y[n]\to q^* \pi_X^*\mathbb{L}_X[n]$ is classified by
$q$, and thus it coincides with $q^*\ell_X$. As a consequence, we get that $p^*\lambda_Y$ coincides with $q^*\lambda_X$
(as sections of $\mathbb{L}_{f^*\mathbf{T}^*[n]Y}$), and that the correspondence \eqref{eq:corrcot} lifts to an exact
isotropic one:
\[
\xymatrix{
& f^*\bT^*[n]Y \ar[rd]^-{p} \ar[dl]_-{q} & \\
\bT^*[n]X \ar[rd]_-{\lambda_X} && \bT^*[n] Y  \ar[ld]^-{\lambda_Y} \\
&\mathcal A^1[n] &}
\]
\begin{remark}
For a $B$-point given by $x\in X(B)$ and $s\in|x^*f^*\mathbb{L}_Y[n]|$, this boils down to the commutativity
of the diagram of $B$-modules
\[
\xymatrix{
& x^*f^*\mathbb{L}_Y[n] \ar@{=}[rd] \ar[dl] & \\
x^*\mathbb{L}_X[n] \ar[rd] && (f\circ x)^*\mathbb{L}_Y[n]  \ar[ld] \\
&\mathbb{L}_B[n] &}
\]
\end{remark}
It can be shown (along the lines of \cite[Theorem 2.8]{CalCot}) that this isotropic correspondence is in fact lagrangian.
\begin{example}
When $Y=\mathrm{pt}$, we get back the exact lagrangian structure on the zero section $X\to\bT^*[n]X$ from
\cite[Theorem 2.2(2)]{CalCot}.
\end{example}
\begin{proposition}\label{prop:cotangentfunctor}
The above construction defines a functor
\[
\bT^*[n]\,:\,Ho(\scat{dSt}_k^{Art})\longrightarrow Ho(\scat{ExLag}_{[n]})\,,
\]
where $\scat{dSt}_{k}^{Art}$ is the full subcategory of $\scat{dSt}_{k}$ spanned by Artin stacks.
\end{proposition}
\begin{proof}
First of all, the identity map $X\to X$ is easily seen to be sent to the identity correspondence
\[
\xymatrix{
& \bT^*[n]X \ar[rd]^-{\mathrm{id}} \ar[dl]_-{\mathrm{id}} & \\
\bT^*[n]X \ar[rd]_-{\lambda_X} && \bT^*[n] X  \ar[ld]^-{\lambda_X} \\
&\mathcal A^1[n] &}
\]
Then, given a sequence $X\overset{f}{\longrightarrow} Y\overset{g}{\longrightarrow} Z$, we do have the following
commuting diagram
\[
\xymatrix{
&& (g\circ f)^*\bT^*[n]Z \ar[rd]\ar[ld] && \\
& f^*\bT^*[n]Y \ar[rd]\ar[ld] && g^*\bT^*[n]Y \ar[rd]\ar[ld] & \\
\bT^*[n]X \ar[rrd]^-{\lambda_X} && \bT^*[n]Y \ar[d]^-{\lambda_Y} && \bT^*[n]Z \ar[lld]_-{\lambda_Z} \\
&& \mathcal A^{1}[n] &&
}
\]
The fact that it commutes follows from the commutativity of the diagram of $B$-modules
\[
\xymatrix{
&& x^*(g\circ f)^*\mathbb{L}_Z[n] \ar@{=}[rd] \ar[ld] && \\
& x^*f^*\mathbb{L}_Y[n] \ar@{=}[rd]\ar[ld] && (f\circ x)^*g^*\mathbb{L}_Z[n] \ar@{=}[rd]\ar[ld] & \\
x^*\mathbb{L}_X[n] \ar[rrd] && (f\circ x)^*\mathbb{L}_Y[n]\ar[d] && (g\circ f\circ x)^*\mathbb{L}_Z[n] \ar[lld] \\
&& \mathbb{L}_B[n] &&
}
\]
where $x\in X(B)$.
\end{proof}
\begin{remark}
We neither prove nor use use it, but one can actually show that $\bT^*[n]$ lifts to a functor
\[
\scat{dSt}_{k}^{Art}\longrightarrow\scat{ExLag}_{[n]}\,,
\]
at the level of $\infty$-categories.
\end{remark}
\begin{definition}
The conormal exact $n$-shifted lagrangian $\bT^*_X[n]Y\to\bT^*[n]Y$ is the composition of $f^*\bT^*[n]B$ with the zero
section $X\to\bT^*[n]X$ in $\scat{ExLag}_{[n]}$:
\[
\xymatrix{
&& \bT^*_X[n]Y \ar[rd]\ar[ld] && \\
& X \ar[rd]\ar[ld] && f^*\bT^*[n]Y \ar[rd]\ar[ld] & \\
{\mathrm{pt}} \ar[rrd]^-{0} && \bT^*[n]X \ar[d]^-{\lambda_X} && \bT^*[n]Y \ar[lld]_-{\lambda_Y} \\
&& \mathcal A^{1}[n] &&
}
\]
\end{definition}
Then one proves that
\begin{align}\label{cotformula}
\bT^*_X[n]Y\underset{{\bT^*[n]Y}}{\times}Y\simeq\bT^*[n-1]X\,,
\end{align}
as exact $(n-1)$-shifted symplectic stacks. This is a special case of \cite[Remark 2.7]{CalCot}, but we can also
prove it thanks to the following diagram of exact lagrangian correspondences
\begin{align}\label{grodiag}
\xymatrix{
&&& \bT^*[n-1]X \ar[rd]\ar[ld] &&& \\
&& \bT^*_X[n]Y \ar[rd]\ar[ld] && X \ar[rd]\ar[ld] && \\
&X \ar[rd]\ar[dl] && f^*\bT^*[n]Y \ar[rd]^-{p} \ar[dl]_-{q} && Y \ar[ld]\ar[dr] &\\
\mathrm{pt} \ar@/_/[rrrd]_-0 && \bT^*[n]X \ar[rd]^-{\lambda_X} && \bT^*[n] Y  \ar[ld]_-{\lambda_Y} && \mathrm{pt} \ar@/^/[llld]^-0 \\
&&& \mathcal A^{1}[n] &&&
}
\end{align}
Indeed $\bT^*[n-1]X\simeq X\underset{\bT^*[n]X}{\times}X\simeq\bT^*_X[n]Y\underset{\bT^*[n]Y}{\times}Y$.
\begin{example}\label{exRedux}
Our aim is to describe the cotangent to the quotient stack $[V/G]$, where $G$ is a reductive affine algebraic group
and $V$ is a finite dimensional $G$-module. According to \eqref{cotformula}, and Examples~\ref{exGmod} and~\ref{BGex},
it is given by the derived exact lagrangian intersection
\[
\xymatrix{
\bT^*[V/G] \ar[r] \ar[d]	& BG 		\ar[d]										\\
[V\oplus V^*/G] \ar[r]	& [\mathfrak{g}^*/G]
}\]
The bottom morphism is given by a $G$-equivariant map $V\times V^*\to \mathfrak{g}^*$.
We let the reader check that this map is the usual moment map, sending $(v,v^*)$ to the linear form $x\mapsto v^*(x\cdot v)$,
where $x\cdot$ is the $\mathfrak{g}$-action.
The exact lagrangian structure of this morphism can be described as follows. Recall that
\[
\Gamma(\mathbb{L}_{[V\oplus V^*/G]})\simeq
\mathrm{hofib}\Big(\Omega^1(V\times V^*)\to \mathcal O(V\times V^*)\otimes\mathfrak{g}^*\Big)^G\,,
\]
where the map is the transpose of the infinitesimal action.
Therefore, by definition, the pull-back of the canonical element $\mathrm{id}_{\mathfrak{g}^*}$ is the tensor of the
$\mathfrak{g}$-action, lying in $(V\otimes V^*\otimes\mathfrak{g}^*)^G$, which is the image of the tautological $1$-form
on $V\times V^*=T^*V$ under the transpose of the infinitesimal action. Hence the exact lagrangian structure on
$[V\oplus V^*/G] \to [\mathfrak{g}^*/G]$ is given by the tautological $1$-form on $V\times V^*$, which is $G$-equivariant.
\end{example}

\subsubsection{Symplectic structure on derived quiver representation schemes}\label{quivsch}

First for a given quiver $Q$, let us introduce the \emph{path algebra} $kQ$ generated by paths, with concatenation as product.
For each $v\in V(Q)$ we denote by $e_v$ the idempotent path of length $0$. They generate a subalgebra of $kQ$ isomorphic to
$k^{V(Q)}$ as a vector space, that will be denoted by $R$ when there is no ambiguity. We will make use of the Jordan quiver $L$
with one vertex and one loop, which satisfy $kL=k[x]$ for some degree $0$ variable $x$. Note that $kL^{V(Q)}$ is an $R$-algebra.

For a fixed dimension vector $\vec{n}=(n_v)_{v\in V(Q)}$, we want to describe the cotangent to
the non-derived moduli stack $\big[Rep_k(Q,\vec{n})/GL_{\vec{n}}(k)\big]$ of $\vec{n}$-dimensional representations
of $Q$, following the notation of Example \ref{exQuiv}. According to Example~\ref{exRedux}, it is obtained as the derived
lagrangian intersection
\[
\xymatrix{
\bT^*\big[Rep_k(Q,\vec{n})/GL_{\vec{n}}(k)\big] \ar[r] \ar[d]	& BGL_{\vec{n}}(k) 		\ar[d]										\\
\big[Rep_k(\overline{Q},\vec{n})/GL_{\vec{n}}(k)\big] \ar[r]	& \big[\mathfrak{gl}_{\vec{n}}(k)/GL_{\vec{n}}(k)\big]
}\]
where the bottom map is given by the moment map
\[
Rep_k(\overline{Q},\vec{n})\longrightarrow \mathfrak{gl}_{\vec{n}}(k)=Rep_k(L^{V(Q)},\vec{n})\,,
\]
which is just pulling-back representations along the $R$-algebra morphism $kL^{V(Q)}\to k\overline Q$ given by
\[
\sum_{v\in V(Q)}x_v\,\longmapsto\,\sum_{e\in E(Q)}(ee^*-e^*e)\,.
\]
One can compute the above derived fiber product by first taking the derived fiber at $0$ of the moment map,
and then taking the quotient by $GL_{\vec{n}}(k)$.

\medskip

Let us now explain shortly that the derived fiber at $0$ of the moment map can be described as the derived representation
scheme of a certain dg-algebra. Let us consider the model category $\cat{dgAlg}_{R}$ of dg-$R$-algebras, which contains
$kQ$, $kL^{V(Q)}$ or $\mathfrak{gl}_{\vec{n}}(k)$.
Inspired by \cite{BKR},
we define, for a dg-$R$-algebra $A$, the derived representation stacks
\[
\scat{DRep}(A,\vec{n})\,:\,B\mapsto \cat{Map}_{\scat{dgAlg}_{R}}\big(A,\mathfrak{gl}_{\vec{n}}(B)\big)\,.
\]
One can show that
\begin{itemize}
\item If $A$ is sitting in non-positive degree, these derived stacks are in fact derived schemes;
\item If $A$ is free and sitting in degree $0$, i.e.~$A$ is the path algebra of a quiver $Q$, then
\[
\scat{DRep}(A,\vec{n})\simeq Rep_k(Q,\vec{n})\,;
\]
\item The construction defines a limit preserving functor
\[
\scat{DRep}(-,\vec{n})\,:\,(\scat{dgAlg}_{R})^{\op}\longrightarrow \scat{dSt}_k
\]
\end{itemize}
As a consequence we get that the homotopy fiber at $0$ of the moment map is the derived representation scheme
(of dimension $\vec{n}$) of the derived push-out
\[
\mathcal G_2(kQ)	:=		{k}\overline{Q}\overset{\mathbb{L}}{\underset{{k}[x]^{V(Q)}}{\coprod}} k^{V(Q)}
							\simeq	{k}\overline{Q}\overset{\mathbb{L}}{\underset{{k}[x]}{\coprod}} k\,,
\]
where the second tensor product takes place in $\scat{dgAlg}_k$, and the morphism ${k}[x]\to k\overline{Q}$
sends $x$ to
\[
\sum_{e\in E(Q)}(ee^*-e^*e)\,.
\]
We thus obtain that
\[
\bT^*\big[Rep_k(Q,\vec{n})/GL_{\vec{n}}(k)\big]\simeq \big[\scat{DRep}\big(\mathcal{G}_{2}(kQ),\vec{n}\big)/GL_{\vec{n}}(k)\big]\,.
\]

\begin{remark}\label{remG2}
The dg-algebra $\mathcal G_2(kQ)$ is the $2$-Calabi--Yau Ginzburg's dg-algebras (see~\cite{Gi}) associated to $Q$ with no
potential. It is generated by some variable $x$ in degree $-1$ and $k\overline Q$ in degree $0$, with
$\mathfrak d(x)=\sum_{e\in E(Q)}ee^*-e^*e$. A precise study of this push-out will be done in~\S\ref{subsubginz}.
Its $0$-th cohomology is the so-called \emph{preprojective algebra} $\Pi_k(Q)=k\overline Q/ (\sum_{e\in E(Q)}[e,e^*])$.
\end{remark}

\begin{remark}\label{remlinquiv}
We now describe a variation on the above, allowing complexes of representations, in the spirit of Examples \ref{exQuiv2}
and \ref{linPerfk}. More precisely, we compute the cotangent stack of $\mathbb A_{E^Q}$.
According to~\eqref{cotformula}, it is given by derived (exact) lagrangian intersection
\[
\xymatrix{
\bT^* \mathbb A_{E^Q} \ar[r]	\ar[d] 	& \bPerf_{ k^{ V(Q)}} \ar[d] \\
\mathbb A_{E^{\overline{Q}}}													\ar[r]		&  \mathbb{A}_{\textsc{Map}(E,E)^{V(Q)}}}
\]
since $\bT^*_{\mathbb A_{E^Q}}[1] \bPerf_{ k^{ V(Q)}}\simeq\mathbb A_{E^{\overline{Q}}}$ thanks to Example~\ref{exQuiv2},
and $\bT^*[1] \bPerf_{ k^{ V(Q)}}\simeq\mathbb A_{\textsc{Map}(E,E)^{V(Q)}}$ thanks to Example~\ref{linPerfk}.
The bottom line $\mathbb A_{E_{\overline Q}} \to  \mathbb A_{   \textsc{Map}(E,E)^{V(Q)}}$ is induced by
$E_{\overline Q}\to\textsc{End}(E)^{V(Q)}$, $(x_e,x_{e^*})\mapsto\sum_e[x_e,x_{e^*}]$.
We will see in Corollary~\ref{theorem: compareGinzburg} that $\bT^* \mathbb A_{E^Q}$ can be identified with the moduli of objects
of $\mathcal{G}_2(kQ)$.

\end{remark}

\subsection{Relative derived critical loci}\label{ss:critrel}

\subsubsection{Definition}\label{critrel}

Inspired by~\cite{RS}, as explained in the introductory part of section~\ref{sec:motiv}, we want to define a relative version
of critical loci. We do so by replacing the zero section in~\eqref{grodiag} by the graph $X\to \bT^*X$ of a closed $1$-form.
We thus start with the following datum:
\begin{itemize}
\item a closed $1$-form $\alpha$ on $X$;
\item a morphism $\pi:X\to B$ of derived $k$-stacks.
\end{itemize}
For simplicity and because it will always be the case in our applications, we assume that $\alpha=d\phi$ for some function $\phi:X\to\mathbb{A}^1$. We know from section~\ref{lagshicotst} that we have an exact isotropic correpondence
$T^*X\leftarrow \phi^*T^*\mathbb A^1=X\times \mathbb{A}^1 \to T^*\mathbb{A}^1=\mathbb{A}^1\times\mathbb{A}^1$.
When we compose it with the lagrangian correspondence $T^*\mathbb{A}^1\leftarrow \{(q,1)|q\in\mathbb{A}^1\}\to *$
we get the lagrangian graph of $d\phi:X\to \mathbf{T}^*X$.
 The composition
 \begin{align}\label{critreldiag}
  \xymatrix{& & \bcrit_\pi(\phi) \ar[rd] \ar[dl] && \\
&X \ar[rd]^-{d\phi}\ar[ld] && \pi^* \bT^* B  \ar[ld]_-q\ar[rd]^-p& \\
~~~\mathrm{pt}~~~&&\bT^* X &&~\bT^*B}\end{align}
 defines the \emph{relative derived critical locus} of $\phi$ with respect to the constraint $\pi$.
 We get the usual derived critical locus when the base $B$ is a point.

\subsubsection{The quiver case}\label{pot3CY}

Assume that $\pi$ is given by the restriction morphism $\iota^*:Rep_k( Q,\vec n)\to Rep_k(D,\vec m)$ induced by an inclusion
of finite, arbitrary quivers $D\subset Q$ (not necessarily a full inclusion), with $m_v=n_v$ for every $v\in V(D)$.

We then consider a \emph{potential} $W$ of the quiver $Q$, that is a linear combination of cycles in $Q$.
It determines a $GL_{\vec{n}}$-invariant function $\phi=\mathrm{tr}(W)$ on $Rep_k( Q,\vec n)$. For any $e\in E(Q)$ we denote
by $\partial_eW$ the linear combination of paths obtained by removing $e$ from the cycles of $W$.

If we do not perform any quotient, $X$, $\bT^*X$ and $\pi^*\bT^*B$ are genuine vector spaces and
\[
\bcrit_\pi(\mathrm{tr}(W))=\scat{DRep}_k\left(k Q\overset{\bbL}{\underset{k\overline Q}{\coprod}} k(Q\sqcup D^*),\vec n\right)\,,
\]
where $\delta W:k\overline Q\to kQ$ is generated by
\[
E(Q)\ni e\mapsto e\quad,\quad E(Q^*)\ni e^*\mapsto \partial_eW
\]
and $k\overline Q\to k(Q\sqcup D^*)$ is the obvious projection.
We resolve $kQ$ by the free $k\overline Q$-dg-algebra generated by $e'\in E(Q^*)$ in degree $-1$ with differential\[
e'\mapsto e^*-\partial_eW\]
 and get that the derived push-out\[
 k Q\underset{k\overline Q}{\overset{\bbL}{\coprod}} k(Q\sqcup D^*)\]
  is generated by $e'\in E(Q^*)$ in degree $-1$ and $e\in E(Q), e^*\in E(D^*)$ in degree $0$ with differential given by:\[
 E(D^*)\ni e'\mapsto e^*-\partial_eW\quad,\quad E(Q^*\setminus D^*)\ni e'\mapsto-\partial_eW.\]

Hence, the $0$-th truncation of $\bcrit_\pi(\mathrm{tr}(W))$ is\[
\tau_0\bcrit_\pi(\mathrm{tr}(W))=\left\{(x,x^*)\in Rep_k(Q\sqcup D^*,\vec n)\middle|\begin{aligned}x_{\partial_eW}&=x^*_{e}\text{ if }e\in E(D)\\x_{\partial_eW}&=0\text{ otherwise}\end{aligned}\right\}.\]

\begin{example}\label{linex}
Let us consider some specific linear examples.\begin{enumerate}[label=(\roman*)]
\item $D=A_2=$\raisebox{-0.10\height}{
\begin{tikzpicture}\hspace{0cm}
\node(A)at(0,0){$\circ$};
\node(B)at(2,1){$\circ$};
\draw[->](A)to[bend left]node[midway,below]{$a$}(B);
\end{tikzpicture}}
$\subset Q=\tilde A_2=$\raisebox{-0.48\height}{
\begin{tikzpicture}\hspace{0cm}
\node(A)at(0,0){$\circ$};
\node(B)at(2,1){$\circ$};
\draw[->](A)to[bend left]node[midway,below]{$a$}(B);
\node(D)at(2,-1){$\circ$};
\draw[->](B)to[bend left]node[midway,left]{$b$}(D);
\draw[->](D)to[bend left]node[midway,above]{$c$}(A);
\end{tikzpicture}}, $W=abc$. We have
\[
\tau_0\bcrit_\pi(\mathrm{tr}(W))=\left\{(x_a,x_{a^*},x_b,x_{c})\middle|\begin{aligned}x_bx_c&=x_{a^*}\\x_cx_a&=0\\x_ax_b&=0\end{aligned}\right\}\,.
\]
One can show that the image of $\bcrit_{\pi}(\mathrm{tr}(W))$ in $T^*Rep_k(A_2,\vec m)$ is actually the classical Lusztig
 lagrangian subvariety~\cite[§12]{Lusztig1} for finite type quiver
\[
\Lambda_{\vec m}(A_2):=\{(x_a,x_{a^*})\mid x_ax_{a^*}=0,~x_{a^*}x_a=0\}\,.
\]
\item The case of~\S\ref{lagcomm}: $S_1=\!\!$\raisebox{-0.46\height}{
\begin{adjustbox}{max totalsize=4cm}
\begin{tikzpicture}
\node(1)at(0,0){$\circ$};
\draw[-latex](1)edge[out=-40,in=40,looseness=12,right]node{$a$}(1);
\end{tikzpicture}
\end{adjustbox}}
$\subset
S_3=\!\!\!\!\!\!\!\!$\raisebox{-0.46\height}{
\begin{adjustbox}{max totalsize=4cm}
\begin{tikzpicture}
\node(1)at(0,0){$\circ$};
\draw[-latex](1)edge[out=-40,in=40,looseness=12,right]node{$a$}(1);
\draw[-latex](1)edge[out=80,in=160,looseness=12,above ]node{$b$}(1);
\draw[-latex](1)edge[out=200,in=280,looseness=12,below]node{$c$}(1);
\end{tikzpicture}
\end{adjustbox}}, $W=[a,b]c$. This time\[
\tau_0\bcrit_\pi(\mathrm{tr}(W))=\left\{(x_a,x_{a^*},x_b,x_{c})\middle|\begin{aligned} [x_b,x_c] &=x_{a^*}\\ [x_c,x_a]&=0\\ [x_a,x_b]&=0\end{aligned}\right\}\]
and observe that its image in $T^*Rep_\mathbb C(S_1,n)$ is the subvariety $\Lambda_n$ defined in~\S\ref{lagcomm}.
\end{enumerate}
\end{example}

Now consider \[\pi:[Rep_k( Q,\vec n)/GL_{\vec n}(k)]\to[Rep_k( D,\vec n)/GL_{\vec m}(k)]\]
 where $[Rep_k( Q,\vec n)/GL_{\vec n}(k)]$ is still endowed with the function $\phi=\mathrm{tr}(W)$.
We want to describe the underlying dg-algebra hidden behind $\bcrit_\pi(f)$, as announced by Remark~\ref{annder}(ii).
To that end, we first recall the definition of Ginzburg dg-algebras~\cite{Gi}.

\begin{definition}\label{defW3CY}
Consider $(Q,W)$ a quiver with potential.
We denote by $\mathcal G_3(kQ,\delta W)$ the dg-$R$-algebra generated by $e\in E(Q)$ in degree $0$, $e'\in E(Q^*)$ in
degree $-1$ and $x'$ in degree $-2$,

with concatenation as product and the following differential
\[
\mathfrak d: \quad x'\mapsto \sum_{e\in E(Q)}(ee'-e'e)\quad,\quad e'\mapsto\partial_eW\,.
\]

\end{definition}

Our construction recovers these dg-algebras thanks to the following.

\begin{proposition}\label{critasginz}
If $D=\emptyset$, then
\[
 \bcrit_\pi(\mathrm{tr}(W))=\big[\scat{DRep}(\mathcal{G}_3(kQ,\delta W),\vec{n})/GL_{\vec{n}}(k)\big]\,.
\]
\end{proposition}

\begin{proof}
From~\S\ref{quivsch}, we know that
\[
 \bcrit_\pi(\mathrm{tr}(W))=\big[\scat{DRep}(\mathcal A,\vec{n})/GL_{\vec{n}}(k)\big]\,,
\]
where $\mathcal A$ is defined as the following homotopy push-out
\[
\xymatrix{
\mathcal G_2(kQ)\ar[r]^{\delta1}\ar[d]_{\delta W}&kQ\ar[d]\\kQ\ar[r]&\mathcal A.}
\]

As pointed out in~\cite[Remark 6.4]{Keller}, this homotopy push-out defines $\mathcal G_3(kQ,\delta W)$.
It can be seen by resolving $kQ$ with the quasi-free $\mathcal G_2(kQ)$-dg-algebra $\mathcal B(kQ)$ generated by

$e'\in E(Q^*)$ in degree $-1$ and $x'\in V(Q)$ in degree $-2$
with differential
\[
\mathfrak d:\quad x'\mapsto x-\sum_{e\in E(Q)}(ee'-e'e),\quad e'\mapsto e^*-\partial_eW\,.
\]
One can then compute the desired push-out using $\mathcal B(kQ)$ and check that we get $\mathcal G_3(kQ,\delta W)$.
\end{proof}

When $D$ is non trivial, we define $\mathcal G_2(D\subseteq Q)$ as the homotopy push-out
\[
\xymatrix{
kD\ar[r]\ar[d]&kQ\ar[d]\\
\mathcal G_2(kD)\ar[r]&\mathcal G_2(D\subseteq Q)}
\]
so that
\[
\pi^*\bT^*B=\big[\scat{DRep}(\mathcal G_2(D\subseteq Q),\vec{n})/GL_{\vec{n}}(k)\big]\,.
\]

As a consequence, keeping~\eqref{critreldiag} in mind, we get the following.

\begin{proposition}\label{defG3QDW}
 The dg-algebra $\mathcal G_3(kQ|kD,\delta W)$ defined by the homotopy push-out
\[
\xymatrix{
\mathcal G_2(kQ)\ar[d]_{\delta W}\ar[r]&\mathcal G_2(D\subseteq Q)\ar[d]\\kQ\ar[r]&\mathcal G_3(kQ|k D,\delta W)}
\]
satisfies
\[
\bcrit_\pi(\mathrm{tr}(W))=\big[\scat{DRep}(\mathcal G_3(kQ|k D,\delta W),\vec{n})/GL_{\vec{n}}(k)\big]\,.
\]
\end{proposition}

Using the same resolution as in the proof of Proposition~\ref{critasginz}, one can check that $\mathcal G_3(kQ|k D,\delta W)$
is generated by

\begin{itemize}
\item $e\in E(Q)$ and $e^*\in E(D^*)$ in degree $0$;
\item $e'\in E(Q^*)$ and $x_v$ a loop in degree $-1$ to every vertex $v\in V(D)$;
\item $x'_v$ a loop in degree $-2$ to every vertex $v\in V(Q)$
\end{itemize}
with differential
\[
x_v'\mapsto \left\{\begin{aligned} &x_v-\sum_{e\in E(Q)}e_v(ee'-e'e)e_v\text{ if }v\in V(D)\\
&-\sum_{e\in E(Q)}e_v(ee'-e'e)e_v\text{ otherwise,}\end{aligned}\right.\]
and\[
x_v\mapsto\sum_{e\in E(D)}e_v(ee^*-e^*e)e_v\,,\quad
e'\mapsto \left\{\begin{aligned} &e^*-\partial_eW\text{ if }e\in E(D)\\
&-\partial_eW\text{ otherwise.}\end{aligned}\right.
\]

\begin{remark} \label{remark: section56}
\begin{enumerate}
\item We retrieve our linear examples when taking the representation spaces of the $0$-th cohomology of
$\mathcal G_3(kQ|k D,\delta W)$.
\item We generalize these constructions in future sections. In section~\ref{sec:CY} we define dg-categories
$\cG_{n+1}(\mathcal B|\cA,c)$ associated with any dg-functor $\cA\to\cB$. In section~\ref{sec:comparison} we investigate
their moduli stacks of objects $\bPerf_{\cG_{n+1}(\cB|\cA,c)}$, and in the case of a quiver embedding $kD\subset kQ$,
\[
\Big[\scat{DRep}(\mathcal G_3(kQ|k D,\delta W),\vec{n})/GL_{\vec{n}}(k)\Big]\subset\bPerf_{\cG_{3}(kQ|kD,\delta W)}
\]
will be recovered as an open substack for the same reason as in Remark~\ref{labelresBG}.
\end{enumerate}
\end{remark}

\subsubsection{Lagrangian subvarieties from lagrangian morphisms}\label{lag2lagsec}

We want a direct proof of the lagrangian nature of the families of varieties studied in section~\ref{sec:motiv}, relying on
the lagrangian structure of $\bcrit_\pi(\phi)\to\bT^*B$. By lagrangian subvariety, as in section~\ref{sec:motiv}, we mean
isotropic of pure dimension half the ambient one.
Thanks to the following proposition and Example~\ref{linex}(ii) we get that $\Lambda_n$ is lagrangian in $T^*Rep_\mathbb C(S_1,n)$
as wished.

\begin{proposition}\label{lag2lag}
Consider a lagrangian morphism $g:L\to Y$, with $Y$ a genuine smooth and symplectic algebraic variety. Denote by $\tau_0(L)\to L$
the $0$-th truncation of $L$, and $Z\subset Y$ the smallest closed subscheme factorizing the following diagram
\[
\xymatrix{ & \tau_0(L) \ar[rd] \ar[dl] && \\
L \ar[rd]&&Z  \ar[ld] \\
&Y&}
\]
\textit{i.e.}\ $Z$ is the schematic image of $\tau_0(L)\to Y$. Then $Z\subset Y$ is a lagrangian subvariety.
\end{proposition}

\begin{proof}
The lagrangian structure yields the following fibre sequence\[
\bbT_L\to g^*T_Y\simeq g^*\Omega^1_Y\to\bbL_L\]
which in turn gives, since $T_Y$ is a complex concentrated in $0$, the following five terms sequence\[
0\to H^{-1}(\bbL_L)\to H^0(\bbT_L)\to g^*T_Y\to H^0(\bbL_L)\to H^1(\bbT_L)\to0\]
and then\begin{align*}
H^{-1}(\bbL_L)&\simeq\ker\big(H^0(\bbT_L)\to g^*T_Y\big)\\
H^1(\bbT_L)&\simeq\operatorname{coker}\big(g^*\Omega^1_Y\to H^0(\bbL_L)\big).\end{align*}
Now denote by $p$ the dominant morphism $\tau_0(L)\to Z$. Consider $\mathcal Z$ and $\mathcal L$ irreducible components of $Z$ and $\tau_0(L)$ respectively such that $p:\mathcal L\to\mathcal Z$ is still dominant. The two following commutative diagrams\[
\begin{tikzcd}T_\mathcal L\arrow[r]\arrow{d}[pos=0.1,rotate=-90,yshift=-0.8ex]{\sim}&p^*T_\mathcal Z\arrow[d,hook']\\H^0(\bbT_L)\arrow[r]&g^*T_Y\end{tikzcd}~~\text{ and }~~
\begin{tikzcd}g^*\Omega^1_Y\arrow[r]\arrow[d,two heads]&H^0(\bbL_L)\arrow{d}[pos=0.1,rotate=-90,yshift=0.8ex]{\sim}\\p^*\Omega^1_\mathcal Z\arrow[r]&\Omega^1_\mathcal L\end{tikzcd}\]
 imply\begin{align*}
H^{-1}(\bbL_L)&\simeq\ker\big(T_\mathcal L\to p^*T_\mathcal Z\big)\\
H^1(\bbT_L)&\simeq\operatorname{coker}\big(p^*\Omega^1_\mathcal Z\to \Omega^1_\mathcal L\big).\end{align*}
Now, by~\cite[Lemma 10.5]{Hartshorne}, $p$ is smooth on some non-empty open subscheme $\mathcal U\subseteq\mathcal L$, which make $T_\mathcal U\to p^*T_\mathcal Z$ and $p^*\Omega^1_\mathcal Z\to\Omega^1_\mathcal U$ respectively surjective and injective. Thus\begin{align*}
p^*T_{\mathcal Z}&\simeq\dfrac{T_\mathcal U}{H^{-1}(\bbL_L)}\simeq\dfrac{H^0(\bbT_L)}{H^{-1}(\bbL_L)}\\
p^*\Omega_{\mathcal Z}^1&\simeq\ker\big(\Omega^1_\mathcal U\to H^1(\bbT_L)\big)\simeq\ker\big(H^0(\bbL_L)\to H^1(\bbT_L)\big)\end{align*}
and\[
0\to p^*T_\mathcal Z\to g^*T_Y \to p^*\Omega^1_\mathcal Z\to0
\]is exact as expected.
\end{proof}

Unfortunately it does not directly apply to the example of~\S\ref{laghilb}.
With the same notations as therein, we consider $S_1^+\subset S_3^+$, and
\[
\pi :X=[Rep_{\mathbb C}(S_3^+,(n,1))/PGL_{n,1}(\mathbb C)]\rightarrow B=[Rep_{\mathbb C}(S_1^+, (n,1))/PGL_{n,1}(\mathbb C)]
\]
endowed with $\phi=\mathrm{tr}(W)$ where $W$ is the potential induced in Example~\ref{linex}(ii) through
$S_3\subset S_3^+$. The issue is that
\[
\bT^*B=\big[\scat{DRep}\big(\mathcal{G}_{2}(\mathbb CS^+_1),{(n,1)}\big)/PGL_{n,1}(\mathbb C)\big]
\]
is no longer concentrated in degree $0$.

Using the stability defined in~\S\ref{laghilb}, we consider the stable locus $Y=(\tau_0\bT^*B)^\text{st}\simeq(\mathbb C^2)^{[n]}$
of the $0$-th truncation of $\bT^*B$. We know that $Y$ is this time a smooth and symplectic variety.
Set
\[
L:=\bcrit_\pi(\phi)\underset{\bT^*B}{\times}Y
\]
and note that $L\to Y$ is lagrangian as the composition of $\bcrit_\pi(\phi)\to\bT^*B$ with the graph
$Y\leftarrow {gr}(s) \rightarrow\bT^* B$ of the symplectic embedding $s:Y\hookrightarrow \bT^*B$,
which is a lagrangian correspondence. Applying Proposition~\ref{lag2lag} to $L\to Y$, we get that the schematic
image $\Lambda_{n,1}$ of $\tau_0L$ in $Y$ is a lagrangian subvariety of $Y=(\mathbb C^2)^{[n]}$, as obtained
in~\S\ref{laghilb}.


TITLE: Decomposition of Lie subspaces
QUESTION [0 upvotes]: If $M=G/H$ is a reductive homogeneous space then we can write $\frak{g}=\frak{m}+\frak{h}$
where $[\frak{h}, \frak{m}]\subset \frak{m}$. Here $\frak{g}$ and $\frak{h}$ are the Lie algebras of $G$ and $H$ respectively. Let $T_o$ be the torsion of the canonical connection (of the second kind) w.r.t. this decomposition at $o=\pi(e)$.
My questions is: when can we decompose $\frak{m}$ as
$$\frak{m}=\frak{m}_0+\frak{m}_1$$
where  $T_o(X_1,X_2)=0$ for $X_1,X_2\in \frak{m}_0$, $T_o(Y_1,Y_2)\ne 0$ for $Y_1,Y_2\in \frak{m}_1$, and $T_o(X,Y)=0$ for $X\in \frak{m}_0$ and $Y\in \frak{m}_1$? For example, any compact connected Lie group $G$ is a semidirect product of its commutator subgroup $G^\prime$ and an Abelian subgroup $G_0$. We can therefore take $\frak{m}_0$ to be the Lie algebra of $G_0$ and $\frak{m}_1=[\frak{g},\frak{g}]$, the Lie algebra of $G^\prime$. It's well known that the torsion for $G$ is given by $T_e(X,Y)=-[X,Y]$.

REPLY [1 votes]: I am starting to think that you can always do this, but perhaps I am missing something at this later hour.
The torsion of the canonical connection at the origin defines a skewsymmetric bilinear map $T_0: \Lambda^2 \mathfrak{m} \to \mathfrak{m}$ by
$$
T_0(X,Y) = -[X,Y]_{\mathfrak{m}}
$$
where the subscript indicates the natural projection $\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{m} \to \mathfrak{m}$.  In turn this defines a linear map $\tau : \mathfrak{m} \to \operatorname{End}(\mathfrak{m})$ by
$$
\tau(X) = - [X, -]_{\mathfrak{m}}
$$
We define $\mathfrak{m}_0 = \ker \tau$.  I claim that $\mathfrak{m}_0$ is an $\operatorname{Ad}(H)$-module.  I assume $H$ is connected, so it is enough to show that if $\tau(X) = 0$, then for all $Z \in \mathfrak{h}$, $\tau([Z,X])=0$.  This follows from the following calculation (using the Jacobi identity of $\mathfrak{g}$):
$$
\tau([Z,X]) = [[Z,X],-]_{\mathfrak{m}} = [Z,[X,-]]_{\mathfrak{m}} - [X,[Z,-]]_{\mathfrak{m}}
$$
The first term in the RHS vanishes because it can be rewritten as
$$[Z,[X,-]_{\mathfrak{m}}]_{\mathfrak{m}}$$
and we use that $\tau(X) = 0$.  The second term in the RHS vanishes because of reductivity and the fact that $\tau(X)=0$, since it can be rewritten as
$$-[X,[Z,-]_{\mathfrak{m}}]_{\mathfrak{m}}$$
By reductivity again, $\mathfrak{m}_0$ has an $\operatorname{Ad}(H)$-module complement $\mathfrak{m}_1$ in $\mathfrak{m}$.


TITLE: Explicit Formula for a Recurrence Relation for {2, 5, 9, 14, ...} (Chartrand Ex 6.46[b])
QUESTION [1 upvotes]: Consider the sequence $a_1 = 2, a_2 = 5, a_3 = 9, a_4 = 14,$ etc...
  (a) The recurrence relation is: $a_1 = 2$ and $a_n = a_{n - 1} + (n + 1) \; \forall \;n \in [\mathbb{Z \geq 2}]$.
  (b) Conjecture an explicit formula for $a_n$. (Proof for conjecture pretermitted here)

I wrote out some $a_n$ to compass to cotton on to an idea or pattern. It seems bootless.
$\begin{array}{cc}
 a_2 = 5 & a_3 = 9 & a_4 = 14 & a_5 = 20 & a_6 = 27\\
\hline \\
5 = 2 + (2 + 1) & 9 = 5 + (3 + 1) & 14 = 9 + (4 + 1) & 20 = 14 + (5 + 1) & 27 = 20 + (7 + 1) \\
\end{array}$
The snippy answer only says $a_n = (n^2 + 3n)/2$. Thus, could someone please explicate the (missing) motivation or steps towards this conjecture? How and why would one envision this?

REPLY [1 votes]: Let $2\leq n$, then you can write
$$a_{2}-a_{1}=2+1$$
$$a_{3}-a_{2}=3+1$$
$$...$$
$$a_{n}-a_{n-1}=n+1.$$
Now addition of all this equality implies that
$$a_{n}-a_{1}=(2+3+..+n)+(n-1),$$
$$a_{n}=(2+3+..+n)+(n+1)-1,$$
we know that $1+2+..+n=\frac{n^{2}+n}{2}$, so
$$a_{n}=\frac{(n+1)(n+2)}{2}-1$$