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\begin{document}
\pagestyle{myheadings}

\address{Department of Mathematics, Technion - Israel Institute of Technology, 32000, Haifa, Israel.}
\email{max@tx.technion.ac.il}
\title{A Distributional Treatment of Relative Mirabolic Multiplicity One}
\subjclass[2010]{20G25, 22E50}
\keywords{ distinguished representations, p-adic symmetric spaces, mirabolic subgroup, invariant distributions}
\author{Maxim Gurevich}
\date{\today}

\begin{abstract}
We study the role of the mirabolic subgroup $P$ of $G=\mathbf{GL}_n(F)$ ($F$ a $p$-adic field) in smooth irreducible representations of $G$ that possess a non-zero invariant functional relative to a subgroup of the form $H_{k} = \mathbf{GL}_k(F)\times \mathbf{GL}_{n-k}(F)$. We show that if a non-zero $H_1$-invariant functional exists on a representation, then every $P\cap H_1$-invariant functional must equal to a scalar multiple of it. When $k>1$, we give a reduction of the same problem to a question about invariant distributions on the nilpotent cone of the tangent space of the symmetric space $G/H_k$. Some new distributional methods, which are suitable for a setting of non-reductive groups, are developed.

\end{abstract}

\maketitle

\section{Introduction}
We study the nature of smooth irreducible complex representations of $G_n = \mathbf{GL}_n(F)$ over a non-archimedean field $F$ of characteristic $0$, that are distinguished by subgroups of the form $H_{k,n}=\mathbf{GL}_{k}(F)\times\mathbf{GL}_{n-k}(F)$ (the maximal standard Levi subgroups). Given a smooth irreducible representation $\pi$ of $G_n$ on a space $V$, it is said that $\pi$ is distinguished by a subgroup $H<G_n$ if there is a non-zero linear $H$-invariant functional on $V$. Equivalently, $\pi$ appears as a sub-representation of $C^\infty(G_n/H)$. Apart from standalone interest in the representation theory of $p$-adic groups, distinction comes into play as a local accessory when dealing with periods of an automorphic representation.
\par For such $(\pi,V)$, the dual $V^\ast$ can be embedded as a $G_n$-space into the space of distributions $\mathcal{D}$ on $G_n/H$. This observation allows one to push geometric results on the $p$-adic space $\mathcal{D}$ into representation theory. Among the advantages of such an approach, which has long been known for its potency, is the absence of reliance on a classification of the representations of $G_n$.
\par For one instance, the mentioned motivation from automorphic forms raises the question of multiplicity one, that is, can the $H$-invariants of $V^\ast$ be larger than one-dimensional? In the case of $H_{k,n}$ it was answered negatively by Jacquet and Rallis \cite{jac-ral}, using novel distributional methods. Namely, they applied a transfer of distributions from the symmetric space $G_n/H_{k,n}$ to its linear tangent space. The method was rephrased by Aizenbud and Gourevitch in \cite{aiz-gur1} as a part of a more elaborate mechanism called the Harish-Chandra descent. In this account, we would like to explore further these ideas to address another distributional problem with an implication for representation theory.
\par We focus on the mirabolic subgroup $\mathbf{P}_n(F)<G_n$, that is, the stabilizer subgroup in $G_n$ of the vector $(0,\ldots,0,1)$ in its natural action on the row space $F^n$. Our main result regards the collection of $H_{1,n}$-distinguished representations of $G_n$.
\begin{thm}\label{start-thm}
Every $\mathbf{P}_n(F)\cap (\mathbf{GL}_1(F)\times \mathbf{GL}_{n-1}(F))$-invariant linear functional on a $\mathbf{GL}_1(F)\times \mathbf{GL}_{n-1}(F)$-distinguished irreducible smooth representation of $\mathbf{GL}_{n}(F)$, is also $\mathbf{GL}_1(F)\times \mathbf{GL}_{n-1}(F)$-invariant.
\end{thm}

In particular, this implies multiplicity one for $\mathbf{P}_n(F)\cap H_{1,n}$-invariant functionals on $H_{1,n}$-distinguished representations.
\par Let us mention, that from the results of \cite{venke} some explicit constructions of $H_{1,n}$-distinguished representations can be produced, for which Theorem \ref{start-thm} may be applied. For example, if $\sigma$ is a smooth irreducible representation of $G_2$ with trivial central character and $\pi = \sigma \times 1_{n-2}$ (in the sense of the Bernstein-Zelevinsky classification, where $1_{n-2}$ is the trivial representation of $G_{n-2}$) is irreducible, then $\pi$ is $H_{1,n}$-distinguished. Thus, Theorem \ref{start-thm} concerns what can be roughly described as embeddings of the smooth spectrum of $\mathbf{PGL}_2(F)$ into that of $G_n$. On the other hand, it follows from the results of \cite{matr-linear}, that if $n>3$, any $H_{1,n}$-distinguished $\pi$ must be non-generic.
\par Theorem \ref{start-thm} comes as a corollary of our study into the geometric question about the difference between $P_n(F)\cap H_{k,n}$-invariance and $H_{k,n}$-invariance of distributions on $G_n/X_{k,n}$. We show (Theorem \ref{final-form}) that for all $1\leq k<n-2$ the question can be reduced to that of invariant distributions on a certain cone in a linear $F$-space. Thus, we complete the following reduction.

\begin{lem}\label{start-lem}
Let $1\leq k\leq n-2$ be integers. \\
For $m\leq l$, let $\mathcal{N}_{m,l}$ be the space of nilpotent matrices of the form
$$\left(\begin{array}{cc} 0 & B \\ C & 0 \end{array} \right)\;\quad\; B\in \mathbf{M}_{m,l-m}(F)\quad C\in \mathbf{M}_{l-m,m}(F).$$
Suppose that for all $1\leq m\leq k$ and $m\leq l \leq n$,  every $\mathbf{P}_l(F)\cap (\mathbf{GL}_m(F)\times \mathbf{GL}_{l-m}(F))$-conjugation invariant distribution on $\mathcal{N}_{m,l}$ is also $\mathbf{GL}_m(F)\times \mathbf{GL}_{l-m}(F)$-invariant. Then, every $\mathbf{P}_n(F)\cap (\mathbf{GL}_k(F)\times \mathbf{GL}_{n-k}(F))$-invariant linear functional on a $\mathbf{GL}_k(F)\times \mathbf{GL}_{n-k}(F)$-distinguished irreducible smooth representation of $\mathbf{GL}_{n}(F)$, is also $\mathbf{GL}_k(F)\times \mathbf{GL}_{n-k}(F)$-invariant.
\end{lem}
The role of the mirabolic subgroup is known to be ubiquitous in the representation theory of $G_n$. For example, the subgroup is seen in the Gelfand-Kazhdan theory of derivatives, which served as a foundation for the Zelevinsky classification. In \cite{bern}, Bernstein showed a distributional result regarding this subgroup: Every $\mathbf{P}_n(F)$-conjugation-invariant distribution on the matrix space $\mathbf{M}_{n,n}(F)$ is also $G_n$-conjugation-invariant. Using this result, he showed that integration over the mirabolic group in the Whittaker model of generic representations defines a canonical inner product between a representation and its contragredient. Our current study can be seen as a follow up to these findings. We would like to check to what extent the role of the mirabolic group is preserved in the relative setting.
\par In the context of distinction, Bernstein's distributional result was applied in \cite{off} to prove a similar statement to Theorem \ref{start-thm} for $\mathbf{GL}_n(L)$-distinguished representations, where $F/L$ is a quadratic field extension. In our case, i.e.~$H_{k,n}$-distinction, we show that the geometry of the space requires a similar approach to be supplemented with the assumption of Lemma \ref{start-lem}, that is, $\mathcal{D}(\mathcal{N}_{k,n})^{\mathbf{P}_n(F)\cap H_{k,n}} = \mathcal{D}(\mathcal{N}_{k,n})^{H_{k,n}}$.  The relative analogy to Bernstein's result facilitates the conjecture that the assumption indeed holds (for all $1\leq k\leq n-2$). We were successful in proving it for the case $k=1$ by techniques of prolongation of invariant distributions.
\par Let us expound further on this. In our proof we are able to prolong $H_{1,n}$-invariant distributions on $\mathcal{N}_{1,n}$ from one open orbit to its closure. Yet, in \cite[Section 4]{rr}, an example of a similar case was shown in which a so-called Ranga-Rao type theorem does \textit{not} hold. Namely, if $G/H$ is a $p$-adic symmetric space, the $H$-invariant distributions on the nilpotent orbits of $Lie(G)\ominus Lie(H)$ may not possess such prolongation properties. Indeed, the validity of such a property in our case for $k>1$ remains one of the obstacles when dealing with the general linear nilpotent problem (the assumption of Lemma \ref{start-lem}).
\par Section 2 sets the main distributional tools needed for our analysis. We recall the Frobenius descent (Proposition \ref{frob}) which allows for the most direct transfer of a distribution into a smaller space. Since the mirabolic group is not reductive, we are unable to apply ``traditional" distributional techniques directly. For that reason we introduce a new refinement (Proposition \ref{abstract-lem}) of the descent techniques, which eases the treatment of invariance under general $p$-adic groups. We also sketch the treatment of Luna's Slice Theorem in the setting of symmetric spaces which was developed in \cite{aiz-gur1}.
\par In addition, we give a formulation (Proposition \ref{dense-frob}) of what we call the dense Frobenius descent. Building upon ideas from \cite{bern}, it will serve us as a tool for reducing problems of invariant prolongation of distributions into same problems on smaller spaces.
\par Section 3 studies the geometry of closed $H_{k,n}$-orbits on the space $G_n/H_{k,n}$ and their decomposition to $\mathbf{P}_n(F)\cap H_{k,n}$-orbits. Using the mentioned Harish-Chandra descent techniques we are able to reduce a question on the distribution spaces of $G_n/H_{k,n}$ to that of distributions on its tangent space (Theorem \ref{main-reduction}). Furthermore, we show that the ``heart" of the problem lies in distributions on the nilpotent cones $\mathcal{N}_{k,n}$.

\par In Section 4, using the dense Frobenius descent, we construct the full space of $H_{1,n}$-invariant distributions on $\mathcal{N}_{1,n}$. It is shown that these are also invariant under the smaller $\mathbf{P}_n(F)\cap H_{1,n}$ (Theorem \ref{k1-thm}). That will allow the final deduction of Theorem \ref{start-thm} from Lemma \ref{start-lem}.
\par For completeness, we give in Section 5 the details of the embedding of the linear forms on a $H_{k,n}$-distinguished irreducible representation of $G_n$ into the space of distributions on $G_n/H_{k,n}$. That description is rather classical, and completes our transfer of distributional results into representation theory.
\par The results reported in this account are part of my Ph.D. research. I would like to thank Omer Offen for suggesting me this problem and providing guidance. Special thanks to Dima Gourevitch and Rami Aizenbud for useful discussions and suggestions.

\section{Tools and preliminaries}
\subsection{Notation}
Let $F$ be a non-archimedean local field of characteristic $0$. We will write $\mathbf{V}(F)$ for the $F$-points of an algebraic variety $\mathbf{V}$ defined over $F$. Specifically, we fix the notation $G_n = \mathbf{GL}_n(F)$ and also denote the naturally embedded subgroups $H_{k,n} = G_k\times G_{n-k}<G_n$, for $0\leq k< n$.
\par Let $\theta_{k,n}: \mathbf{GL}_n\to \mathbf{GL}_n$ be the involutive automorphism defined by
$$\theta_{k,n}(g) = \epsilon_{k,n} g\epsilon_{k,n}\qquad \epsilon_{k,n}=\left(\begin{array}{ll} I_k & 0 \\ 0 & -I_{n-k}\end{array}\right).$$
Then, the fixed point subgroup of $\theta_{k,n}$ is exactly $\mathbf{GL}_k\times \mathbf{GL}_{n-k}$, which makes the quotient $\mathbf{GL}_n/(\mathbf{GL}_k\times \mathbf{GL}_{n-k})$ a \textit{symmetric space}. For our current needs this will allow the following construction: Consider the left action of $G_n$ on itself by $\theta_{k,n}$-twisted conjugation: $g\cdot x := gx\theta_{k,n}(g)^{-1}$. The stabilizer of the identity element will be the fixed point subgroup of $\theta_{k,n}$ inside $G_{n}$, i.e.~$H_{k,n}$. Thus, the orbit of the identity relative to this action
$$X_{k,n} = \left\{ g\theta_{k,n}(g)^{-1}\;:\; g\in G_n\right\}\subset G_n$$
is identified with the quotient $G_n/H_{k,n}$. We also denote the symmetrization map $\rho_{k,n}(g) = g\theta_{k,n}(g)^{-1} = g\cdot I_n$ from $G_n$ to $X_{k,n}$.
Notice that the action of $H_{k,n}$ on $X_{k,n}$ is given by usual conjugation, and therefore the stabilizer group inside $H_{k,n}$ of a point $x\in X_{k,n}$ is the centralizer of $x$ which we will denote by $H^x_{k,n}$. A similar notation for a centralizer will be adopted for other groups as well.
\par Observe that on the Lie algebra of $\mathbf{GL}_n$ there is a linear involution $d\theta_{k,n} = Ad(\epsilon_{k,n})$. After identifying the algebra with the matrix space $\mathbf{M}_{n,n}$ it becomes useful to define the linear version of the symmetric space
$$\mathbf{L}_{k,n} = \left\{ A\in \mathbf{M}_{n,n}\;:\; Ad(\epsilon_{k,n})(A) = -A\right\} =$$ $$= \left\{\left(\begin{array}{cc} 0 & B \\ C & 0 \end{array} \right)\;:\; B\in \mathbf{M}_{k,n-k}, C\in \mathbf{M}_{n-k,k}\right\}.$$

\par Consider the natural right action of $\mathbf{GL}_n$ on the row vector space $\mathbb{A}^n$. Let $\mathbf{P}_n< \mathbf{GL}_n$ be the stabilizer group of the vector $(0,\ldots,0,1)$. This is the standard mirabolic group which clearly consists of matrices whose bottom row is $(0 \cdots 0 1)$. We also denote the intersection $P_{k,n} = H_{k,n} \cap \mathbf{P}_n(F)$.

\par For a general locally compact totally disconnected (lctd) space $X$ (such as the $F$-points of an algebraic group), we write $\mathcal{S}(X)$ for the space of locally constant compactly supported complex functions on $X$. The space $\mathcal{D}(X)$ of distributions on $X$ is defined to be the dual space of $\mathcal{S}(X)$. Recall that given an open subset $\Omega$ of such a space $X$, $\Omega$ and $X\setminus \Omega$ become lctd spaces themselves, and we have a short exact sequence
$$0\longrightarrow \mathcal{D}(X\setminus \Omega)\longrightarrow \mathcal{D}(X)\longrightarrow \mathcal{D}(\Omega)\longrightarrow 0$$
Thus, for $T\in \mathcal{D}(X)$ we can talk about its restriction $T|_{\Omega}$, or say it is \textit{supported} in $X\setminus \Omega$. The support $supp\, T$ of a distribution $T$ is the complement of the union of all open sets $\Omega$ for which $T|_{\Omega}=0$.
\par A continuous left action of a lctd group $G$ on a lctd space $X$ induces a left linear action of $G$ on $\mathcal{S}(X)$ by $(g\cdot f)(x) = f(g^{-1}\cdot x)$ ($g\in G, f\in \mathcal{S}(X), x\in X$). Hence, $G$ acts on $\mathcal{D}(X)$ by the dual action.
\par In general, for a group $G$ that acts linearly on a complex space $V$ and a group character $\chi:G\to \mathbb{C}$, we will use the notation $V^{G,\chi} = \{v\in V\;:\; g\cdot v = \chi(g)v\;\forall g\in G\}$ for the space of semi-invariants. As usual $V^G = V^{G,1}$. Given an involution $\sigma$ on any vector space $X$, we will denote by $X^{\widehat{\sigma}}= \{x\in X\;:\; \sigma(x)=-x\}$ the space of anti-invariants of $X$. E.g.~$\mathbf{L}_{k,n}(K) = \mathbf{M}_{n,n}(K)^{\widehat{Ad(\epsilon_{k,n})}}$ for any field $K$.
\par We will denote by $\Delta_G$ the modular character attached to a lctd group $G$, and given a closed subgroup $H<G$ we define the relative modular character $\Delta_{G/H} = \Delta_G|_H \Delta_H^{-1}$ of $H$.
\par Recall that the modular character of $\mathbf{P}_n(F)$ is given by $|\det^{-1}|$, where $|\cdot|$ is the standard absolute value of $F$. Note also, that the same formula remains true for the group $\mathbf{P}_n(E)$, where $F<E$ is a finite field extension, if the determinant of a matrix is taken as an $F$-linear operator.

\subsection{Frobenius Descent}
A major focal point of this work is the analysis of certain spaces of invariant distributions. One valuable tool available to deal with these spaces is the so-called Frobenius descent. It is a well-known rephrasing of the Frobenius reciprocity of group representations. After recalling the original formulation of the descent technique, we develop some adaptations of the tool to our needs.
\begin{prop}\label{frob}\cite[Lemma 1.5]{bern}
Suppose $G$ is a lctd group acting continuously on lctd spaces $\mathcal{X}$ and $\mathcal{O}$, with a $G$-equivariant continuous map $p:\mathcal{X}\to \mathcal{O}$ between them. Suppose furthermore, that the action on $\mathcal{O}$ is transitive. For a chosen point $x\in \mathcal{O}$ let $H<G$ be the stabilizer group of $x$. Then, for any character $\chi:G\to \mathbb{C}$ there is an isomorphism of distribution spaces
$$\mathcal{D}(p^{-1}(x))^{H,\chi \Delta_{G/H}^{-1}} \cong \mathcal{D}(\mathcal{X})^{G,\chi}.$$
If a distribution $T'\in\mathcal{D}(p^{-1}(x))^{H,\chi \Delta_{G/H}^{-1}}$ corresponds to $T\in  \mathcal{D}(\mathcal{X})^{G,\chi}$, we have the equality $supp\, T = G\cdot supp\,T'$.
\end{prop}
We further mention that if $\Delta_{G/H}$ is trivial (with the assumptions of the above proposition), then $\mathcal{O}$ has a non-zero $G$-invariant distribution $\mu$ on it. In that case the Frobenius descent map $\Phi_\mu: \mathcal{D}(p^{-1}(x))^{H} \to \mathcal{D}(\mathcal{X})^{G}$ can be given by the formula
$$(\Phi_\mu(T))(f)= \int_\mathcal{O} T((g_z\cdot f)|_{p^{-1}(x)})\,d\mu(z)\qquad f\in\mathcal{S}(\mathcal{X})$$
where $g_z\in G$ are fixed elements such that $g_z\cdot z = x$.

\par Given a group $G$ acting on $X$ and its subgroup $P$, we want to formulate a criterion for the equality of spaces $\mathcal{D}(X)^G$ and $ \mathcal{D}(X)^P$ based on the Frobenius descent.

\begin{prop}\label{abstract-lem}
Suppose $G$ is an lctd group, and $P<G$ is a closed subgroup. Suppose $\mathcal{X}$ and $\mathcal{O}$ are lctd spaces on which $G$ acts, and $p:\mathcal{X}\to \mathcal{O}$ is an $G$-equivariant surjection. Assume: \\
(i) The action on $\mathcal{O}$ is $G$-transitive. \\
(ii) There is a non-zero $G$-invariant distribution on $\mathcal{O}$.\\
(iii) There are finitely many $P$-orbits in $\mathcal{O}$.\\
If there exists an element $x\in \mathcal{O}$ and an open $P$-orbit $\mathcal{U}\subset \mathcal{O}$ containing $x$ such that:\\
(iv) $\mathcal{D}(p^{-1}(x))^{Stab(P,x)} = \mathcal{D}(p^{-1}(x))^{Stab(G,x)}$.\\
(v) For every $y\in \mathcal{O}\setminus \mathcal{U}$, $\mathcal{D}(p^{-1}(y))^{Stab(P,y),\, \Delta_{P/Stab(P,y)}^{-1}}=\{0\}$.\\
then
$$\mathcal{D}(\mathcal{X})^{P} = \mathcal{D}(\mathcal{X})^G.$$
\end{prop}

\begin{proof}
Let $T\in \mathcal{D}(\mathcal{X})^{P}$ be a distribution. If $0\neq \mu$ is an $G$-invariant distribution on $\mathcal{O}$, then $0\neq \mu|_{\mathcal{U}}$ is $P$-invariant, and $\Delta_{P/Stab(P,x)}=1$ follows.
Thus, we can use Proposition \ref{frob} on the mapping $p: p^{-1}(\mathcal{U})\to \mathcal{U}$ to produce $\xi\in \mathcal{D}(p^{-1}(x))^{Stab(P,x)}$ corresponding to $T|_{p^{-1}(\mathcal{U})}$.
From condition (iv) we know that $\xi$ is also $Stab(G,x)$-invariant, hence, corresponds by another Frobenius descent to an $G$-invariant distribution $S$ on $\mathcal{X}$. Yet, looking on both descents through the formula given above we see that $S|_{p^{-1}(\mathcal{U})} = T|_{p^{-1}(\mathcal{U})}$, which means $T-S$ has its support in $p^{-1}(\mathcal{O}\setminus \mathcal{U})$. Therefore, it is enough to prove that $\mathcal{D}(p^{-1}(\mathcal{O}\setminus \mathcal{U}))^P$ is trivial to show that $T=S$ and finish.\\
This is achieved by induction on the number of $P$-orbits in $\mathcal{O}\setminus \mathcal{U}$. Since this is a finite number, there is an open $P$-orbit $\mathcal{U}_1\subset \mathcal{O}\setminus \mathcal{U}$. Choosing $y\in \mathcal{U}_1$ and combining condition (v) with a Frobenius descent, we see that there are no non-zero $P$-invariant distributions on $p^{-1}(\mathcal{U}_1)$. Thus $\mathcal{D}(p^{-1}(\mathcal{O}\setminus \mathcal{U}))^P =\mathcal{D}(p^{-1}(\mathcal{O}\setminus (\mathcal{U}\cup \mathcal{U}_1)))^P$, and so on.
\end{proof}

\par Another problem we can tackle with descent techniques is the prolongation of invariant distributions. Given a distribution on a locally closed subset $Y$ of a larger space we would like to claim that it is the restriction of a distribution on $\overline{Y}$ with the same invariance properties. The following proposition allows us under suitable conditions to reduce such a question to a smaller space. This argument has appeared implicitly in \cite[4.3]{bern}.
\begin{prop}[``Dense Frobenius descent"]\label{dense-frob}
Let $G$ be a lctd group that acts on a lctd space $\mathcal{X}$. Let $\mathcal{K}$ be a compact totally disconnected space on which $G$ acts transitively.
Suppose $\mathcal{Y}\subset \mathcal{X}$ is an $G$-invariant subset equipped with an $G$-equivariant surjective continuous map $i:\mathcal{Y}\to \mathcal{K}$.\\
For a point $y\in \mathcal{K}$, denote $\mathcal{Y}_y= i^{-1}(y)$ and $B= Stab(y)< G$. If there exists a distribution $T\in\mathcal{D}(\mathcal{X})^{B,\Delta_{G/B}^{-1}}$ with $supp\, T = \overline{\mathcal{Y}_y}$, then there also exists a distribution $T'\in \mathcal{D}(\mathcal{X})^G$ with $supp\, T' = \overline{\mathcal{Y}}$.
\end{prop}
\begin{proof}
Consider the product space $\mathcal{X}\times \mathcal{K}$ with its $G$-action. Let $Q\subset \mathcal{X}\times \mathcal{K}$ be the graph of the map $i$. We apply Proposition \ref{frob} relative to the projection $p_2:\mathcal{X}\times \mathcal{K}\to \mathcal{K}$ on the right component. This gives the correspondence $\mathcal{D}(p_2^{-1}(y))^{B,\Delta_{G/B}^{-1}}\cong \mathcal{D}(\mathcal{X}\times \mathcal{K})^G$. Identifying $p_2^{-1}(y)$ with $\mathcal{X}$, we see that the existence of $T$ as described in the assumption supplies the existence of a $G$-invariant distribution $T''$ on $\mathcal{X}\times \mathcal{K}$ whose support is $\overline{G\cdot (\overline{\mathcal{Y}}_y\times \{y\})}= \overline{Q}$.\\
Consider the projection $p_1:\mathcal{X}\times \mathcal{K}\to \mathcal{X}$ on the left component. Since $\mathcal{K}$ is compact, if $f$ is in $\mathcal{S}(\mathcal{X})$, then $f\circ p_1\in\mathcal{S}(\mathcal{X}\times \mathcal{K})$. Hence, we can push distributions with $p_{1\ast}:\mathcal{D}(\mathcal{X}\times \mathcal{K})\to \mathcal{D}(\mathcal{X})$ defined by $p_{1\ast}(T)(f) = T(f\circ p_1)$. \\
We claim that the support of $T':= p_{1\ast}(T'')$ is exactly $\overline{\mathcal{Y}}$. Indeed, for $x\in \mathcal{Y}$, we have $(x,i(x))\in supp\, T''$. Therefore there exists an open compact neighborhood $U\times V$ of $(x,i(x))$ with $T''(\chi_{U\times V})\neq 0$ (with $\chi$ denoting the characteristic function of a set). From continuity of $i$ we can assume that $i(U\cap \mathcal{Y})\subset V$, hence, $U\times( \mathcal{K}\setminus V)$ lies outside the support of $T''$. So, $T'(\chi_U) = T''(\chi_{U\times V})\neq 0$.
\end{proof}

\subsection{Luna's Slice Theorem and applications}
In \cite{aiz-gur1}, the authors developed a mechanism for the transfer of information on spaces of invariant distributions on $F$-varieties to invariant distributions on $F$-linear spaces. We will now review the required tools from the mentioned reference.
\par Suppose a linear reductive $F$-group $\mathbf{G}$ acts on an affine $F$-variety $\mathbf{V}$. We will say that a subset $U\subset \mathbf{V}(F)$ is \textit{saturated} if there is a $\mathbf{G}(F)$-invariant continuous map $\pi:\mathbf{V}(F)\to F^m$ (such map always exists) and an open set $V\subset F^m$ such that $U = \pi^{-1}(V)$. In particular, saturated subsets are open and $\mathbf{G}(F)$-invariant.
\par In the same setting, suppose $x\in \mathbf{V}(F)$ is such that its $\mathbf{G}(F)$-orbit $\mathcal{O}$ is closed (equivalently, its $\mathbf{G}$-orbit is Zariski closed). The tangent spaces $T_x(\mathbf{G}\cdot x)\subset T_x\mathbf{V}$ are well-defined and are equipped with the action of the stabilizer group $\mathbf{G}^x$ of $x$ (by differentiation). So, $\mathbf{G}^x$ acts on their quotient which is the normal space $\mathbf{N}_x(\mathbf{G}\cdot x)$ whose $F$-points may be identified with the normal space $N_x\mathcal{O}$ of $F$-analytic manifolds.
\begin{thm}\label{luna}\textbf{Luna's Slice Theorem }(see e.g.~\cite[2.3.17]{aiz-gur1}) \\
Let $\mathbf{G}$ be a linear reductive $F$-group acting on an affine $F$-variety $\mathbf{V}$, and $x\in \mathbf{V}(F)$ such that its $\mathbf{G}(F)$-orbit $\mathcal{O}$ is closed. Then there exists an open $\mathbf{G}(F)$-invariant neighborhood $\mathcal{O}\subset U\subset \mathbf{V}(F)$, a $\mathbf{G}(F)$-equivariant continuous retract $p: U\to\mathcal{O}$, and a $\mathbf{G}^x(F)$-equivariant $F$-analytic embedding $\psi: p^{-1}(x)\to N_x\mathcal{O}$ whose image is saturated and $\psi(x)=0$.
\end{thm}

Let us remark that the slices of the variety obtained in the above procedure exhaust the whole variety because of the following fact.
\begin{prop}\label{covered}\cite[2.3.7]{aiz-gur1}
Suppose a linear $F$-group $\mathbf{G}$ acts on an affine $F$-variety $\mathbf{V}$. If $U$ is an open subset of $\mathbf{V}(F)$ which contains all closed $\mathbf{G}(F)$-orbits on $\mathbf{V}(F)$, then $U=\mathbf{V}(F)$.
\end{prop}

We will need to apply this linearization technique on the action of $H_{k,n}$ on $X_{k,n}$. This is possible because we may view $X_{k,n}\cong G_n/H_{k,n}$ as embedded inside the $F$-variety $(\mathbf{GL}_n/\mathbf{GL}_k\times \mathbf{GL}_{n-k})(F)$. As we have mentioned, for $x\in X_{k,n}$ the stabilizer group of the $H_{k,n}$-action is $H_{k,n}^x$.
\begin{prop}\label{closed}
If $x\in X_{k,n}$ has a closed $H_{k,n}$-orbit, then $x$ is a semisimple matrix, and the action of $H^x_{k,n}$ on the normal space $N_x\mathcal{O}$ is isomorphic to its action on $\mathbf{L}_{k,n}^{Ad(x)}(F)$ by conjugation.
\end{prop}
\begin{proof}
This is \cite[Proposition 7.2.1]{aiz-gur1}, after noting that the normal space in $\mathbf{GL}_n$ relative to the action of $\mathbf{GL}_k\times \mathbf{GL}_{n-k}$ on both sides is isomorphic to the normal space of $\mathbf{GL}_n/\mathbf{GL}_k\times \mathbf{GL}_{n-k}$ relative to the one sided action. Moreover, the closed orbits of both actions are in correspondence.
\end{proof}

\section{Reduction to a linear problem}
As will be explained in Section 5, the answer to the representation theoretic problems with which we are dealing can be deduced from the conjectural equalities $\mathcal{D}(X_{k,n})^{P_{k,n}}= \mathcal{D}(X_{k,n})^{H_{k,n}}$ of distribution spaces. We would like to reduce the problem of proving such an equality into an equality of distribution spaces on a given cone of a linear space.
\par Let us introduce some terminology. When a group $G$ acts on a space $X$, we would like to call the pair $(G,X)$ an \textit{action space}. Given a set $\{(G^i,X^i)\}_{i=1}^t$ of action spaces, we will say that an action space $(G,X)$ is their \textit{product}, if there exists a group isomorphism $G\cong G^1\times\cdots\times G^t$ and a bijection $X \to X^1\times\cdots\times X^t$ which intertwines the $G$-action relative to the fixed isomorphism.
\par Suppose an action space $(G,X)$ admits a decomposition into a product of $\{(G^i,X^i)\}_{i=1}^t$. We call a subgroup $H< G$ \textit{admissible} to this decomposition if the fixed isomorphism $G\cong G^1\times\cdots\times G^t$ sends $H$ to a product of subgroups $H^i<G^i$. In this case $(H,X)$ becomes the product of $\{(H^i, X^i)\}_{i=1}^t$.

\subsection{Geometry of closed orbits}
\par Suppose $x\in X_{k,n}$ is a semisimple matrix (for some $k<n-1$). Our first mission is to decompose the conjugation actions of $H_{k,n}^x$ and its subgroup $P_{k,n}^x$ on $\mathbf{L}_{k,n}^{Ad(x)}(F)$ into a product of simpler action spaces. Regarding the group $H_{k,n}^x$, such a decomposition can surely be deduced from similar descriptions in \cite{jac-ral} and \cite{aiz-gur1}, but we prefer to reproduce it here directly for completeness.

\begin{prop}\label{structure}
\textbf{1.} The action space $\left(H_{k,n}^x,\mathbf{L}_{k,n}^{Ad(x)}(F)\right)$ decomposes as a product of the action spaces $\left\{(\mathbf{GL}_{l_i}(E_i), \mathbf{M}_{l_i,l_i}(E_i))\right\}_{i=1}^t$ and of $\{(H_{m_i,l_i}, \mathbf{L}_{m_i,l_i}(F))\}_{t<i\leq s}$,
where $0\leq t \leq s\leq t+2$, for every $1\leq i\leq t$, $F<E_i$ is a finite field extension and $l_i$ is a positive integer; and for every $t< i\leq s$, $m_i\leq k, l_i\leq n$ are positive integers. All actions in the decomposition are by conjugation.\\ \\
\textbf{2.} The subgroup $P^x_{k,n}$ of $H^x_{k,n}$ is admissible to the decomposition above. Therefore, the action space $\left(P_{k,n}^x,\mathbf{L}_{k,n}^{Ad(x)}(F)\right)$ is a product of $\left\{(P^i, \mathbf{M}_{l_i,l_i}(E_i)\right\}_{i=1}^t$ and of $\{(P^i, \mathbf{L}_{m_i,l_i}(F)\}_{t<i\leq s}$, for certain subgroups $P_i$. Those are given as follows:
$$P^i = \left\{\begin{array}{ccc} \mathbf{GL}_{l_i}(E_i) & \alpha_i=0 & 1\leq i\leq t \\ \mathbf{P}_{l_i}(E_i) & \alpha_i =1 \\ H_{m_i,l_i} & \alpha_i=0 & t<i\leq s \\ P_{m_i,l_i} & \alpha_i=1   \end{array} \right.$$

where $\alpha(x) = (\alpha_1,\ldots, \alpha_s)\in \{0,1\}^s$ is a fixed tuple.  \\ \\
\textbf{3.} After identifying $P^x_{k,n}$ with $\prod_{i=1}^s P^i$, the relative modular character of $P^x_{k,n}$ inside $P_{k.n}$ is given by
$$\Delta_{P_{k,n}/P^x_{k,n}}(g_1,\ldots,g_s) = \prod_{\{i\;:\; \alpha_i=0\}} |\det \widetilde{g_i}|^{-1}\; ,$$
where
$$\widetilde{g_i} = \left\{ \begin{array}{lll} g_i & 1\leq i\leq t \\ h_i & t<i\leq s & H_{m_i,l_i} =  G_{m_i}\times G_{l_i-m_i},\, g_i= (f_i,h_i) \end{array} \right. .$$

\end{prop}
\begin{proof}
\textbf{1.} Consider the action of $G_n$ by right-multiplication on the row vector space $F^n$. The semisimple operator $x$ gives rise to a decomposition of $F^n= \oplus_j V_j$ to a direct sum of vector spaces, such that $V_j$ has an $E_j = F(\zeta_j)$-vector space structure for a finite field extension $F<E_j$, and $x|_{V_j}$ is acting as multiplication by $\zeta_j$. Grouping isomorphic extensions, we can assume the actions of the $\zeta_{j}$'s are not isomorphic for distinct $j$'s. Then it is easy to see that such a decomposition of $F^n$ gives rise to a decomposition of the conjugation action space $\left(G_n^x,\mathbf{M}_{n,n}(F)^{Ad(x)}\right)$ into action spaces of the form $(\mathbf{GL}_{l_j}(E_j), \mathbf{M}_{l_j,l_j}(E_j))$, with $l_j\leq n$. It remains to see what happens when we add the requirement for commutation relations with $\epsilon_{k,n}$.
\par Since $x \in X_{k,n}$, we have $\epsilon_{k,n}x\epsilon_{k,n} =\theta_{k,n}(x) = x^{-1}$. Thus, the action of $\epsilon_{k,n}$ on $F^n$ must permute the $V_j$'s. Since $\epsilon_{k,n}$ is an involution, we conclude that $\left(H_{k,n}^x,\mathbf{L}_{k,n}^{Ad(x)}(F)\right)$ decomposes as a product of spaces either of the form
$$\left((\mathbf{GL}_{l_{j_1}}(E_{j_1})\times \mathbf{GL}_{l_{j_2}}(E_{j_2}))^{\epsilon_{k,n}}, (\mathbf{M}_{l_{j_1},l_{j_1}}(E_{j_1})\times \mathbf{M}_{l_{j_2},l_{j_2}}(E_{j_2}))^{\widehat{Ad(\epsilon_{k,n})}}\right)$$
where $j_1\neq j_2$ are such that $\epsilon_{k,n}(V_{j_1}) = V_{j_2}$, or of the form
$$\left(\mathbf{GL}_{l_{j}}(E_{j})^{\epsilon_{k,n}}, \mathbf{M}_{l_{j},l_{j}}(E_{j})^{\widehat{Ad(\epsilon_{k,n})}}\right)$$
where $V_j$ is $\epsilon_{k,n}$-invariant.
Let us show that all of these are isomorphic to the action spaces described in the statement.
\par In the former case (\textit{case (1)}), the acting group is clearly given by $\{(g, \epsilon_{k,n}g\epsilon_{k,n}|_{V_{j_2}})\,:\, g\in \mathbf{GL}_{l_{j_1}}(E_{j_1})\}$, while the space is $\{(A, -\epsilon_{k,n}A\epsilon_{k,n}|_{V_{j_2}})\,:\, A\in \mathbf{M}_{l_{j_1},l_{j_1}}(E_{j_1})\}$ . Hence, the situation is isomorphic to $\mathbf{GL}_{l_{j_1}}(E_{j_1})$ acting on $\mathbf{M}_{l_{j_1},l_{j_1}}(E_{j_1})$.
\par The latter case should itself be separated into two cases. First, assume $x^{-1}|_{V_j} = \theta_{k,n}(x)|_{V_j} \neq x|_{V_j}$ (\textit{case (2)}). Then, on $V_j$, $\theta_{k,n}$ serves as a non-trivial $F$-linear involution of $E_j$. Let $F<L_j<E_j$ be its fixed sub-field ($[E_j:L_j]=2$). This means we can write $V_j\cong E_j\otimes_{L_j} W_j$ for an $L_j$-subspace $W_j\subset V_j$, with $\epsilon_{k,n}$ acting as the non-trivial Galois automorphism in $Gal(E_j/L_j)$ on the $L_j$ component. In these terms, we have
$$ \mathbf{M}_{l_{j},l_{j}}(E_{j})^{\widehat{Ad(\epsilon_{k,n})}} \cong (E_j\otimes_{L_j}\mathbf{M}_{l_{j},l_{j}}(L_{j}))^{\widehat{Ad(\epsilon_{k,n})}} = \tau\otimes_{L_j}\mathbf{M}_{l_{j},l_{j}}(L_{j})$$
where $\tau\in L_i$ is an element for which $\epsilon_{k,n}(\tau\otimes w) = -\tau\otimes w$. Similarly, we have $\mathbf{GL}_{l_{j}}(E_{j})^{\epsilon_{k,n}}\cong \mathbf{GL}_{l_{j}}(L_{j})$, and the action space is isomorphic to $(\mathbf{GL}_{l_j}(L_j), \mathbf{M}_{l_j,l_j}(L_j))$.
\par Finally, still in the case of $\left(\mathbf{GL}_{l_{j}}(E_{j})^{\epsilon_{k,n}}, \mathbf{M}_{l_{j},l_{j}}(E_{j})^{\widehat{Ad(\epsilon_{k,n})}}\right)$, if $x|_{V_j}= x^{-1}|_{V_j}$ (\textit{case (3)}), then $x=\pm I_{V_j}$. That means at once that $E_j = F$ and that there are at most two of such factors in the product. Furthermore, we see that under a suitable change of basis, $\epsilon_{k,n}|_{V_j}$ equals $\epsilon_{m_j,l_j}$. Since $m_j$ is the dimension of the eigenspace of $\epsilon_{k,n}|_{V_j}$ corresponding to the eigenvalue $1$, it obviously cannot exceed $k$.\\

\par \textbf{2.} Let $\{v_j\in V_j\}$ be the decomposition of the vector $e=(0,\ldots,0,1)\in F^n = \oplus_j V_j$. Since the subspaces $V_j$ are invariant under the $H^x_{k,n}$-action on $F^n$ and since $P^x_{k,n}$ is the stabilizer of $e$, the subgroup can also be described as the intersection of the stabilizers of the vectors $v_j$ inside $H^x_{k,n}$. This clearly means that $P^x_{k,n}$ is an admissible subgroup for the above product decomposition. It is left to recognize the stabilizer subgroups of $v_j$ in each of the three cases appearing in the product.
\par In case (1), since $\epsilon_{k,n}(e) = -e$, we see that $v_{j_1} = -v_{j_2}$. If $v_{j_1}=0$, the stabilizer is clearly the whole acting group, and we set $\alpha_i= 0$ for the $i$ with which we are dealing. Otherwise, we set $\alpha_i=1$ and the subgroup of $\mathbf{GL}_{l_{j_1}}(E_{j_1})$ which stabilizes $(v_{j_1},-v_{j_2})$ becomes $\mathbf{P}_{l_{j_1}}(E_{j_1})$, up to a change of basis.
\par In case $V_j$ is $\epsilon_{k,n}$-invariant, again we set $\alpha_i=0$ for the corresponding $i$ when $v_j=0$, and the stabilizer is then the whole acting group. Otherwise we set $\alpha_i=1$. In case (2), the condition $\epsilon_{k,n}(e)=-e$ imposes $v_j = \tau\otimes w$, whose stabilizer is again just a stabilizer of one non-zero vector in $\mathbf{GL}_{l_j}(L_{j})$, i.e.~isomorphic to $\mathbf{P}_{l_j}(L_j)$.
\par As for the last case, it is easily seen that given a vector $v_j$ for which $\epsilon_{k,n}|_{V_j}(v_j) = -v_j$, a suitable change of basis exists such that $H_{m_j,l_j}$ remains at place while the stabilizer of $v_j$ becomes $P_{m_j,l_j}$.\\

\par \textbf{3.} Since $\mathbf{GL}_l(E)$ is a unimodular group and the modular character of $\mathbf{P}_l(E)$ is $|\det^{-1}|$, it is easy to see that $\Delta_{P^x_{k,n}}(g)= \prod_{i\;:\;\alpha_i=1} |\det(\widetilde{g_i})|^{-1}$, for all $g=(g_1,\ldots,g_s)\in \prod_{i=1}^s P^i\cong P^x_{k,n}$. Now, $\Delta_{P_{k,n}}^{-1}$ is given by the norm of the determinant of the lower block, that is, the determinant of the restriction of the operator to the $-1$ eigenspace of $\epsilon_{k,n}$. We need to compute it for each $P^i$.
\par In case (1), we are interested in the restriction of $P^i$ to the space $\{(v,-\epsilon_{k,n}(v))\}_{v\in V_{j_1}}$, which is isomorphic to $\mathbf{GL}_{l_{j_1}}(E_{j_1})$. In case (2), we are looking on its action on the space $\tau\otimes W_j$ which gives the same conclusion. Finally, for case (3), the restriction is the projection on the right component of $H_{m_j,l_j}= G_{m_j}\times G_{l_j-m_j}$.
\par Thus, in all cases we have $\Delta_{P_{k,n}}(g_i) = |\det( \widetilde{g_i})|^{-1}$, and the statement easily follows.

\end{proof}

The above decomposition can also be applied for the study of the geometry of the $P_{k,n}$-action on $X_{k,n}$ in the following way.

\begin{prop}\label{fin-orbit}
Let $\mathcal{O}$ be a closed $H_{k,n}$-orbit in $X_{k,n}$. Then, $\mathcal{O}$ splits into a finite number of $P_{k,n}$-orbits, with a unique open (in $\mathcal{O}$) orbit $A\subset \mathcal{O}$. Furthermore, $A$ consists exactly of those $x\in \mathcal{O}$ for which $\alpha(x) = (1,\ldots,1)$.
\end{prop}
\begin{proof}
Since $H^x_{k,n}$ is the stabilizer of $x\in \mathcal{O}$ in $H_{k,n}$, we can fix one $x\in \mathcal{O}$ and identify $\mathcal{O}\cong H_{k,n}/H^x_{k,n}$. Note, that instead of counting orbits of $P_{n,k}$ on $H_{n,k}/H^x_{k,n}$, we can count the orbits of the right action of $H^x_{k,n}$ on the space $P_{n,k}\backslash H_{k,n}$. Consider $F^{n-k}\setminus \{0\}$ as a row space on which $H_{k,n}$ acts transitively by right matrix multiplication with its lower block. Then, $P_{k,n}$ will be the stabilizer of $e=(0,\ldots,0,1)$, and $F^{n-k}\setminus \{0\}\cong P_{n,k}\backslash H_{n,k}$.
\par By Proposition \ref{closed}, $x$ is a semisimple matrix, hence, we can apply the previous proposition. Since $F^{n-k}$ lies inside $F^{n}$ as the eigenspace of $\epsilon_{k,n}$ related to the value $-1$, we can use the same reasoning as in the previous proof to identify it as the set of vectors of the form $((w_i,-\epsilon_{k,n}(w_i)),(\tau\otimes w_i), w_i)$, with the notation that corresponds to the 3 cases classification of the previous proof. We are left to count the non-zero orbits of $H^x_{k,n}$ on this subspace.
\par Clearly, these are determined by the nullity of each of $\{w_i\}_{i=1}^s$, and we see that there are exactly $2^s-1$ such orbits. In particular, there is only one open orbit among them, namely, the set of vectors for which all $w_i\neq 0$. We denote by $A$ the corresponding unique open orbit in $\mathcal{O}$.
\par The $P_{k,n}$-orbit of $x$ inside $\mathcal{O}$ corresponds to the $H^x_{k,n}$-orbit of the identity in $P_{n,k}\backslash H_{n,k}$, or in other words, of the vector $e$ in $F^{n-k}$. Thus, the question of whether $x$ belongs to $A$ is equivalent to asking whether the components of $e$ in the $\{w_i\}$ (or $\oplus_j V_j$) decomposition are all non-zero. This is equivalent to the condition $\alpha(x)=(1,\ldots,1)$ from the way we defined $\alpha(x)$.
\end{proof}

\subsection{Distributions on $X_{k,n}$}
\begin{cor}\label{first-cor}
If $A$ is the open $P_{k,n}$-orbit inside a closed $H_{k,n}$-orbit $\mathcal{O}\subset X_{k,n}$ and $x\not\in A$, then $\mathcal{D}(N_x\mathcal{O})^{P^x_{k,n},\Delta_{P_{k,n}/P^x_{k,n}}^{-1}}$ is trivial.
\end{cor}
\begin{proof}
First by Proposition \ref{closed}, we may prove that $\mathcal{D}(\mathbf{L}_{k,n}^{Ad(x)}(F))^{P^x_{k,n},\Delta_{P_{k,n}/P^x_{k,n}}^{-1}}$ is trivial.  Since $x\not\in A$, it must have some index $\alpha_i=0$. As observed in Proposition \ref{structure}, it follows there must be a subgroup $M$ inside $P^x_{k,n}$ isomorphic to either $\mathbf{GL}_{l_i}(E)$ or $H_{m_i,l_i}$ whose action on $\mathbf{L}_{k,n}^{Ad(x)}(F)$ is isomorphic to a conjugation action on a specified matrix space. Note also, that by Proposition \ref{structure}.3, $\Delta_{P_{k,n}/P_x}(g) = |\det(\tilde{g})|$ for $g\in M$, with $\tilde{g}$ defined as in the mention proposition. Take $g= \omega I\in M$ for some $\omega \in F$ with $|\omega|>1$. Then it is clear that $\Delta_{P_{k,n}/P_x}(g)\not=1$, while $g$, being a scalar operator, must act trivially on $\mathbf{L}_{k,n}^{Ad(x)}(F)$. This shows there cannot be any non-trivial $P^x_{k,n},\Delta_{P_{k,n}/P_x}^{-1}$-invariant distributions on that space.
\end{proof}

We are now ready to claim the first distributional reduction of the main problem.

\begin{lem}\label{cruc}
Let $k,n$ be positive integers such that $k <n-1$. Suppose that for every semisimple matrix $x\in X_{k,n}$ the equality
$$\mathcal{D}\left(\mathbf{L}_{k,n}^{Ad(x)}(F)\right)^{P^x_{k,n}} = \mathcal{D}\left(\mathbf{L}_{k,n}^{Ad(x)}(F)\right)^{H^x_{k,n}}$$
holds, where the group action is by conjugation. Then, we also have
$$\mathcal{D}(X_{k,n})^{P_{k,n}}= \mathcal{D}(X_{k,n})^{H_{k,n}}$$
\end{lem}
\begin{proof}
Let $\mathcal{O}$ be a closed $H_{k,n}$-orbit in $X_{k,n}$. Since $X_{k,n}\cong G_n/H_{k,n}$, we can consider it as an open and closed subset of $Y:=\mathbf{GL}_n/\mathbf{GL}_k\times \mathbf{GL}_{n-k}(F)$. By the Luna's slice theorem (Theorem \ref{luna}), there exists an open $H_{k,n}$-invariant subset $\mathcal{O}\subset U_{\mathcal{O}}\subset Y$, together with an $H_{k,n}$-equivariant retract $p:U_{\mathcal{O}}\to \mathcal{O}$ possessing certain properties. By Proposition \ref{covered}, the union of all such $U_\mathcal{O}$ over all closed orbits $\mathcal{O}$, together with $Y\setminus X_{k,n}$, cover $Y$. Hence, $X_{k,n}\subset \bigcup_{\mathcal{O}} U_\mathcal{O}$.
\par If we show $\mathcal{D}(U_\mathcal{O})^{P_{k,n}} = \mathcal{D}(U_{\mathcal{O}})^{H_{k,n}}$ for every closed $H_{k,n}$-orbit $\mathcal{O}$, we will know that $\mathcal{D}(\bigcup_{\mathcal{O}} U_\mathcal{O})^{P_{k,n}} = \mathcal{D}(\bigcup_{\mathcal{O}} U_{\mathcal{O}})^{H_{k,n}}$, for example, by \cite[Lemma 3.2]{off}. That will prove the statement, since $X_{k,n}$ is closed and open in the union.
\par Let $A\subset \mathcal{O}$ be the open $P_{k,n}$-orbit which was shown to exist in Proposition \ref{fin-orbit}, and choose $x\in A$. As usual $\mathcal{O}\cong H_{k,n}/H^x_{k,n}$, and $x$ is semisimple by Proposition \ref{closed}. The result will be achieved by applying Proposition \ref{abstract-lem} on the retract $p$ and the groups $P_{k,n}<H_{k,n}$. Indeed, condition (ii) holds because $H_{k,n}$ and $H^x_{k,n}$ are unimodular (this is a consequence of Proposition \ref{structure}.1), and (iii) holds because of Proposition \ref{fin-orbit}. We are left to show that conditions (iv) and (v) hold for $x$.
\par Recall that Luna's slice comes with an $H^x_{k,n}$-equivariant embedding $\iota:p^{-1}(x)\to N_x$ whose image is open and saturated. Hence, there is an $H^x_{k,n}$-invariant map $\pi:N_x\mathcal{O}\to F^m$, and an open $V\subset F^m$ such that $i(p^{-1}(x)) = \pi^{-1}(V)$. Consider $\xi\in \mathcal{D}(p^{-1}(x))^{P^x_{k,n}}$, and suppose there is an $h\in H^x_{k,n}$ with $\xi\not=\xi^h$. Pick $t\in Supp(\xi-\xi^h)$ and an open compact neighborhood $B$ of $\pi(\iota(t))$ in $V$. Then, $\pi^{-1}(B)$ is an open and closed subset of $N_x\mathcal{O}$ that contains $\iota(t)$ and contained in $\iota(p^{-1}(x))$. Now, we have a well-defined operator $\alpha_B:\mathcal{D}(p^{-1}(x))\to \mathcal{D}(N_x\mathcal{O})$ given by $\alpha_B(\eta)(f) = \eta((f\circ \iota)|_{\pi^{-1}(B)})$ for $f\in \mathcal{S}(N_x\mathcal{O})$. Hence, $\alpha_B(\xi)\in \mathcal{D}(N_x\mathcal{O})^{P^x_{k,n}}$, but by Proposition \ref{closed} and our assumption this is the same as $\alpha_B(\xi)\in \mathcal{D}(N_x\mathcal{O})^{H^x_{k,n}}$. Therefore, $\alpha_B(\xi^h) = \alpha_B(\xi)^h = \alpha_B(\xi)$, which gives $\alpha_B(\xi-\xi^h)=0$. This contradicts the clear fact that $\iota(t)\in Supp(\alpha_B(\xi-\xi^h))$. Finally, $\mathcal{D}(p^{-1}(x))^{P^x_{k,n}}= \mathcal{D}(p^{-1}(x))^{H^x_{k,n}}$ holds and condition (iv) of Proposition \ref{abstract-lem} is satisfied.
\par With similar arguments, we can deduce condition (v) from Corollary \ref{first-cor} by noting that $p^{-1}(y)$ can be embedded with an open saturated image inside $N_y\mathcal{O}$ for all $y\in \mathcal{O}$, and directly using \cite[Lemma 3.1.3]{aiz-gur1}.
\end{proof}

\begin{thm}\label{main-reduction}
Let $k,n$ be positive integers such that $k <n-1$. Suppose that the equality
$$\mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{P_{m,l}} = \mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{H_{m,l}}$$
holds, for all positive integers $m\leq k$ and $l\leq n$. Then, we also have
$$\mathcal{D}(X_{k,n})^{P_{k,n}}= \mathcal{D}(X_{k,n})^{H_{k,n}}$$
\end{thm}
\begin{proof}
Suppose $\{G^i\}$ are finitely many lctd groups acting on lctd spaces $\{X^i\}$ respectively, and $H^i<G^i$ are fixed subgroups. Let $(H,X)$ be the product of all $\{(H^i,X^i)\}$'s, and $(G,X)$ the product of $\{(G^i,X^i)\}$'s. If $\mathcal{D}(X^i)^{P^i} = \mathcal{D}(X^i)^{H^i}$ for all $i$, then it is easy to check that $\mathcal{D}(X)^H = \mathcal{D}(X)^G$. See, for example \cite[3.1.5]{aiz-gur3}.
\par So, combining that reasoning with Proposition \ref{structure} and Lemma \ref{cruc}, it is enough to show that $\mathcal{D}(\mathbf{M}_{l,l}(E))^{\mathbf{P}_l(E)} = \mathcal{D}(\mathbf{M}_{l,l}(E))^{\mathbf{GL}_l(E)}$ and that $\mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{P_{m,l}} = \mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{H_{m,l}}$, for certain finite field extensions $F<E$ and integers $m\leq k$, $l\leq n$.
\par The former equality was proved by Bernstein in \cite{bern}, while the latter is the assumption, when $m>0$. When $m=0$ there is, of course, nothing to prove.
\end{proof}

\subsection{Reduction to nilpotents}
Let $\mathcal{N}_{k,n}\subset \mathbf{L}_{k,n}(F)$ be the cone of nilpotent matrices.
\par Before moving to tackle the equality of distribution spaces on the linear space, we would like to reduce our problem further to that of distribution spaces on $\mathcal{N}_{k,n}$. For that cause, we apply some of the ideas which appeared in  \cite[Theorem 3.2.1]{aiz-gur1}.

\begin{lem}\label{closedisclosed}
The set $\mathcal{N}_{k,n}$ is closed in $\mathbf{L}_{k,n}(F)$, and every closed $H_{k,n}$-orbit in $\mathbf{L}_{k,n}(F)\setminus\mathcal{N}_{k,n}$ remains closed in $\mathbf{L}_{k,n}(F)$.
\end{lem}
\begin{proof}
Let $t:\mathbf{L}_{k,n}(F)\to F^{n}$ be the map given by the coefficients of the characteristic polynomials of elements of $\mathbf{L}_{k,n}(F)$ (as matrices). Clearly, $t$ is $H_{k,n}$-invariant, and $\mathcal{N}_{k,n}= t^{-1}(1,0,\ldots,0)$. Every $H_{k,n}$-orbit is contained in a single fiber of $t$, hence, its closure remains in that fiber.
\end{proof}

\begin{prop}\label{redtonil}
Let $k,n$ be positive integers such that $k <n-1$. Suppose that the equality
$$\mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{P_{m,l}} = \mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{H_{m,l}}$$
holds, for all positive integers $m\leq k$ and $l< n$. Then,
$$\mathcal{D}(\mathbf{L}_{k,n}(F)\setminus \mathcal{N}_{k,n})^{P_{k,n}}=\mathcal{D}(\mathbf{L}_{k,n}(F)\setminus \mathcal{N}_{k,n})^{H_{k,n}}.$$
\end{prop}

\begin{proof}
Let $\mathcal{O} \subset \mathbf{L}_{k,n}(F)\setminus \mathcal{N}_{k,n}$ be a closed $H_{k,n}$-orbit. As in the proof of Lemma \ref{cruc}, it is enough to find for each such $\mathcal{O}$ an open $H_{k,n}$-invariant neighborhood $\mathcal{U}$, such that $\mathcal{D}(\mathcal{U})^{P_{k,n}}=\mathcal{D}(\mathcal{U})^{H_{k,n}}$.
\par Consider the identity element $I_n=\rho_{k,n}(I_n)\in X_{k,n}$. The action of $H_{k,n}$ on its normal space is equivalent to that of conjugation on $\mathbf{L}_{k,n}(F)^{Ad(I_n)} = \mathbf{L}_{k,n}(F)$. By applying Luna's Slice Theorem on the trivial closed orbit of $I_n$ in $X_{k,n}$, we get an open $H_{k,n}$-invariant neighborhood $I_n\in U\subset X_{k,n}$ with an $H_{k,n}$-equivariant embedding $\iota:U\to \mathbf{L}_{k,n}(F)$, whose image is open and contains the zero vector.
\par Note, that $\lambda \mathcal{O}\subset \iota(U)$ for some $\lambda\in F^\times $, and $\lambda \mathcal{O}$ is also a closed orbit. Since the action on $\mathbf{L}_{k,n}(F)$ is linear, finding a suitable neighborhood $\mathcal{U}$ for it, would give $\lambda^{-1}\mathcal{U}$ as our desired neighborhood for $\mathcal{O}$. Thus, we can safely assume that $\mathcal{O}\subset\iota(U)$.
\par Another application of Luna's Slice Theorem, this time on the space $\mathbf{L}_{k,n}(F)\setminus \mathcal{N}_{k,n}$, would give a prescribed open neighborhood $\mathcal{U}$ of $\mathcal{O}$. Now, since $\iota$ is an equivariant embedding and since by Lemma \ref{closedisclosed} $\mathcal{O}$ is closed in $\iota(U)$ as well, the normal spaces of elements in $\mathcal{O}$ together with the actions of the stabilizers in $H_{k,n}$ are isomorphic to the normal spaces of elements in $\iota^{-1}(\mathcal{O})$. Thus, using the same arguments as the proof of Lemma \ref{cruc} it can be shown that the equality $$\star \quad \mathcal{D}\left(\mathbf{L}_{k,n}^{Ad(x)}(F)\right)^{P^x_{k,n}} = \mathcal{D}\left(\mathbf{L}_{k,n}^{Ad(x)}(F)\right)^{H^x_{k,n}}$$
for a certain $x\in U$ with $\iota(x)\in \mathcal{O}$, would imply $\mathcal{D}(\mathcal{U})^{P_{k,n}}=\mathcal{D}(\mathcal{U})^{H_{k,n}}$.
\par Indeed, since $\iota(x)\not=0$, we clearly have $H^x_{k,n} = H^{\iota(x)}_{k,n}\subsetneq H_{k,n}$. Hence,
$$\dim \mathbf{L}_{k,n}^{Ad(x)}(F) = \dim N_{x}\iota^{-1}(\mathcal{O}) = \dim X_{k,n} - \,\dim H_{k,n}/H^x_{k,n}< \dim \mathbf{L}_{k,n}(F).$$
We see that the factor $\mathbf{L}_{k,n}(F)$ cannot appear in the decomposition of Proposition \ref{structure} for $\mathbf{L}_{k,n}^{Ad(x)}(F)$. Therefore, using the same technique as in the proof of Theorem \ref{main-reduction}, our assumptions are enough to prove the equality $\star$.

\end{proof}

\begin{cor}\label{nil-cor}
Suppose that the equality
$$\mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{P_{m,l}} = \mathcal{D}\left(\mathbf{L}_{m,l}(F)\right)^{H_{m,l}}$$
holds, for all positive integers $m\leq k$ and $l< n$, and suppose further that $\mathcal{D}(\mathcal{N}_{k,n})^{P_{k,n}}= \mathcal{D}(\mathcal{N}_{k,n})^{H_{k,n}}$. Then, $\mathcal{D}\left(\mathbf{L}_{k,n}(F)\right)^{P_{k,n}} = \mathcal{D}\left(\mathbf{L}_{k,n}(F)\right)^{H_{k,n}}$ holds.
\end{cor}
\begin{proof}
We use the localization principle from \cite[1.4]{bern} with the map $t$ from the proof of Lemma \ref{closedisclosed}. By that principle, it is enough to prove the equality of invariant distribution spaces only on the fibers of $t$. Each of fibers, except $t^{-1}(1,0,\ldots,0)=\mathcal{N}_{k,n}$, is closed in $L_{k,n}(F)\setminus\mathcal{N}_{k,n}$. Hence, the equality of distribution spaces on each of those can be deduced from the previous proposition.
\end{proof}

\begin{thm}\label{final-form}
Let $k,n$ be positive integers such that $k <n-1$. Suppose that the equality
$$\mathcal{D}\left(\mathcal{N}_{m,l}(F)\right)^{P_{m,l}} = \mathcal{D}\left(\mathcal{N}_{m,l}(F)\right)^{H_{m,l}}$$
holds, for all positive integers $m\leq k$ and $l\leq n$. Then, we also have
$$\mathcal{D}(X_{k,n})^{P_{k,n}}= \mathcal{D}(X_{k,n})^{H_{k,n}}$$
\end{thm}
\begin{proof}
By induction on $m$ and $l$, it is easy to use Corollary \ref{nil-cor} at each step in such manner that the assumptions of Theorem \ref{main-reduction} are satisfied.
\end{proof}

\section{The case $k=1$.}
In this section we fix $n\geq3$. We would like to characterize the space of $P_{1,n}$-conjugation-invariant distributions on the nilpotent cone $\mathcal{N}_{1,n}$. The eventual result will be that it is a $2$-dimensional space spanned by the obvious point distribution centered on the zero vector, and by a second distribution $\nu$ which we will need to construct, both of which are also $H_{1,n}$-invariant. To show the existence of such $\nu$ we need to be able to prolong an $H_{1,n}$-invariant distribution on an open orbit onto its closure in such way that it remains invariant. This task of singularity resolution is carried out using the "dense Frobenius descent" (Proposition \ref{dense-frob}) technique, which reduces the problem to that of distribution prolongation from $F^\times$ to $F$.
\par The space $\mathbf{L}_{1,n}(F)$ is the set of matrices given by
$$\left\{\left(\begin{array}{ll} 0 & \overline{v} \\ \overline{w} & 0 \end{array}\right): \overline{v}\in \mathbf{M}_{1,n-1} (F),\;\overline{w}\in \mathbf{M}_{n-1,1}(F)\right\}.$$
Thus we can parametrize this space by pairs $(\overline{v},\overline{w})$, where the first is a row a vector while the second is a column vector. It will sometimes be convenient to write the row vector as $\overline{v} = (\overline{v}'\;a)$ and the column vector as $\overline{w} = \left(\begin{array}{ll} \overline{w}' \\ b\end{array}\right)$, where $a,b\in F$ and $\overline{v}',\overline{w}'$ are of length $n-2$. In this notation we get a parametrization of $\mathbf{L}_{1,n}(F)$ by quadruples $(\overline{v}',a, \overline{w}',b)$.
\par The conjugation action of the group $H_{1,n}$, naturally viewed as $F^\times\times G_{n-1}$, is given in pairs notation by the formula $(\alpha, A)\cdot (\overline{v},\overline{w}) = (\alpha \overline{v}A^{-1}, \alpha^{-1}A\overline{w})$. Also, it can be seen that in these terms the nilpotent cone $\mathcal{N}_{1,n}$ consists of matrices for which $\overline{v}\cdot \overline{w}=0$. We denote by $U\subset \mathcal{N}_{1,n}$ the open $H_{1,n}$-orbit defined by the condition $\overline{v},\overline{w}\neq 0$.

\begin{lem}\label{thereis-dist-gen}
On $\mathcal{N}_{1,n}$ there is an $H_{1,n}$-invariant distribution, whose support is the whole space.
\end{lem}
\begin{proof}
Consider the compact projective space $\mathbb{P}^{n-2}(F)$ together with the $H_{1,n}$-action given by $(\alpha, A)\cdot [p] = [Ap]$ for any line $[p]$ represented by $p\in F^{n-1}$ and $(\alpha,A)\in F^\times\times G_{n-1}$. We define an $H_{1,n}$-equivariant map $i:U\to \mathbb{P}^1(F)$ by $i((\overline{v},\overline{w}))=[\overline{w}]$.
\par Note, that the closure of $i^{-1}\left(\left[\left( 0  \cdots 0 \;1\right)^t\right]\right)$ is
$$Y:= \left\{ (\overline{v}',0, \overline{0}, b)\,:\, \overline{v}'\in \mathbf{M}_{1,n-2}(F),\, b\in F \right\}\cong F^{n-2}\times F.$$
Note also, that
$$S:= Stab\left(H_{1,n}, \left[\left( 0  \cdots 0 \;1\right)^t\right]\in\mathbb{P}^{n-2}(F)\right) =$$ $$= F^\times\times \left\{ \left(\begin{array}{ll} B & 0\\ \overline{\gamma} &d \end{array} \right)\in G_{n-1}(F) \,:\, B\in G_{n-2}(F),\,d\in F^\times\right\}.$$
Since $U$ is dense in $\mathcal{N}_{1,n}$, by Proposition \ref{dense-frob}, it is enough to exhibit a non-zero $S,\Delta_{H_{1,n}/S}^{-1}$-invariant distribution supported on $Y$. The action of $S$ on $Y$ is given by
$$\left(\alpha, \left(\begin{array}{ll} B & 0\\ \overline{\gamma} &d \end{array} \right)\right)\cdot (\overline{v}',0, \overline{0}, b)= (\alpha \overline{v}'B^{-1},0, \overline{0}, \alpha^{-1}d b).$$
Also, we have
$$\Delta_{H_{1,n}(F)/S}^{-1}\left(\alpha, \left(\begin{array}{ll} B & 0\\ \overline{\gamma} &d \end{array} \right)\right) = \Delta_{S}\left(\alpha,\left(\begin{array}{ll} B & 0\\ \overline{\gamma} &d \end{array} \right)\right) = |\det(B)d^{2-n}|.$$
The last equality is easily derived after noting that $S$ is a direct sum of the unimodular $F^\times$, and of a parabolic subgroup of $G_{n-1}$.
\par Let $\chi(x) = |x|^{2-n}$ be a character on $F^\times$, and $0\neq\mu\in \mathcal{D}(F^{\times})^{F^\times, \chi}$ the corresponding distribution. It is known that such $\mu$ can be prolonged into a non-zero distribution $\widetilde{\mu}\in \mathcal{D}(F)^{F^\times, \chi}$ supported on all of $F$ (see \cite[Chapter 2, 2.3]{gel-gp}, or \cite{weil66}). After identifying $Y$ with $F^{n-2}\times F$, we can put the distribution $m\otimes \widetilde{\mu}$ on it, where $m$ is the Haar measure on the vector space $F^{n-2}$. It is easy to see that the $S$-action transforms this distribution according to the formula for $\Delta_{H_{1,n}(F)/S}^{-1}$. Thus, we have our desired distribution.
\end{proof}

Let $\nu\in\mathcal{D}(\mathcal{N}_{1,n})^{H_{1,n}}$ be a distribution as provided by the lemma. Denote also by $\delta_0\in\mathcal{D}(\mathcal{N}_{1,n})^{H_{1,n}}$ the distribution given by $\delta_0(f) = f(\overline{0},\overline{0})$, for all $f\in\mathcal{S}(\mathcal{N}_{1,n})$.

\begin{thm}\label{k1-thm}
The space of $P_{1,n}$-invariant distributions on $\mathcal{N}_{1,n}$ is spanned by $\delta_0$ and $\nu$. In particular,
$$\mathcal{D}(\mathcal{N}_{1,n})^{P_{1,n}}=\mathcal{D}(\mathcal{N}_{1,n})^{H_{1,n}}$$

\end{thm}
\begin{proof}
Let $T\in \mathcal{D}(\mathcal{N}_{1,n})^{P_{1,n}}$ be given.
\par Note that $W=\left\{(\overline{v}',a, \overline{w}',b)\in U\;:\; b\neq 0\right\}$ is an open $P_{1,n}$-orbit in $\mathcal{N}_{1,n}$. Since an orbit can have at most one invariant distribution up to a constant and since $\nu|_W\neq 0$, there must be a constant $c$ for which $T-c\nu|_{W} = 0$. Yet, $T-c\nu$ is still $P_{1,n}$-invariant, which means we can simply assume that $T$ has its support inside $\mathcal{N}_{1,n}\setminus W$.
\par In this set, $O_1= \{(\overline{0},0, \overline{w}',b)\,:\, b\neq0\}$ is an open $P_{1,n}$-orbit lacking a non-zero invariant distribution. Indeed, for $s_1= (\overline{0},0, \overline{0},1)\in O_1$, its stabilizer inside $P_{1,n}$ is the unimodular group $$\{1\}\times\left\{\left(\begin{array}{ll} A \\ &1\end{array}\right)\;:\;A\in G_{n-2}\right\}.$$ It can be seen that $\Delta_{P_{1,n}/Stab(P_{1,n},s_1)}\neq 1$. Thus, $supp\, T\subset Y=\left\{(\overline{v}',a, \overline{w}',0)\right\}$.\\
Note, that we naturally have $Y\cong \mathcal{N}_{1,n-1}\times F$ as $P_{1,n}$-spaces, with the action on the right component given by $(\alpha,A)\cdot a = \alpha a$. Under this decomposition we write $T = T_1\otimes T_2$. Since $\left(\alpha,\left(\begin{array}{ll}\alpha I_{n-2}\\ &1 \end{array}\right)\right)\in P_{1,n}$ acts trivially on the left component of the decomposition, $T_2$ is actually an $F^\times$-invariant distribution on $F$, hence, by a well-known fact $T_2$ is supported on $0$ (see, for example, \cite[0.7]{bern}). In other words, $supp\, T \subset \left\{(\overline{v}',0, \overline{w}',0)\right\}$.
\par Now, there are only two $P_{1,n}$-orbits left that are fully contained inside the above set: The zero orbit, and $O_2 = \left\{(\overline{0}',0, \overline{w}',0)\,:\, \overline{w}\neq 0\right\}$. Thus, $T$ can be viewed as a $G_{n-2}$-invariant distribution on the space $F^{n-2}\cong O_2\cup \{0\}$, which means it must be supported on the zero vector, i.e.~a multiple of $\delta_0$.
\end{proof}

A combination of the above with Theorem \ref{final-form} immediately gives a result about distributions on a non-linear space.
\begin{cor}\label{final-cor}
The equality
$$\mathcal{D}(X_{1,n})^{P_{1,n}}= \mathcal{D}(X_{1,n})^{H_{1,n}}$$
holds.
\end{cor}

\section{Application to Representation Theory}
Our focus will now turn to smooth (admissible) complex representations of the group $G_n$. We will say that an irreducible such representation $(\pi,V)$ is \textit{$H_{k,n}$-distinguished}, if there is a non-zero linear functional on $V$, which is invariant under the dual $H_{k,n}$-action. By Frobenius reciprocity, it easily follows that $\pi$ is $H_{k,n}$-distinguished, if and only if, it can be embedded as a $G_n$-sub-representation of the space of locally constant functions on $G_n/H_{k,n}\cong X_{k,n}$ (with the $G_n$-action given by shifting of functions). Therefore, it is much expected that results on the space of distributions on $X_{k,n}$ would have implications on $H_{k,n}$-distinguished representations of $G_n$.
\par We are interested in the space of $P_{k,n}$-invariant functionals on smooth irreducible representations of $G_n$. More precisely, given an $H_{k,n}$-distinguished such $(\pi,V)$, we would like to claim that $(V^\ast)^{P_{k,n}} = (V^\ast)^{H_{k,n}}$.
\par Note, that since the center $Z$ of $G_n$ is contained in $H_{k,n}$, it must act trivially on all $H_{k,n}$-distinguished representations. In case $k=n-1$, we have $H_{k,n}= P_{k,n}Z$, which trivially implies the equality that we seek. The cases $k=0,n$ are also clearly of no interest. Thus, we deal with the condition $1\leq k\leq n-2$.
\par Let us recount a rather standard argument for a deduction of representation-theoretic statements from the results of previous sections.
\par Let $(\pi,V)$ be a smooth irreducible $H_{k,n}$-distinguished representation of $G_n$. Recall the notion of the smooth contragredient representation $(\widetilde{\pi},\widetilde{V})$. In fact, $\widetilde{\pi}$ can be equivalently realized on the space $V$ with the action $\widetilde{\pi}(g) = \pi(^tg^{-1})$ ($^t$ marks matrix transposition). Since $H_{k,n}=H_{k,n}^t$, the above realization shows that $ (\widetilde{\pi},\widetilde{V})$ is $H_{k,n}$-distinguished as well. Let us fix $0\neq \lambda \in
\left(\left(\widetilde{V}\right)^\ast\right)^{H_{k,n}}$.
\par A choice of Haar measure $dm$ on $G_n$ defines an action of functions in $\mathcal{S}(G_n)$ on $V$ by $f\cdot v= \int_{G} f(g)\pi(g)v\,dm(g)$, hence, induces a surjective map $A_\pi:\mathcal{S}(G_n)\to End(V)\cong V\otimes \widetilde{V}$. Note, that $G_n\times G_n$ has a left action on $G_n$  with $(g_1,g_2)\cdot g = g_1gg_2^{-1}$, which induces an action on $\mathcal{S}(G_n)$. The mapping $A_\pi$ intertwines that action with the action of $G_n\times G_n$ on $V\otimes \widetilde{V}$.  Dualizing, we have an equivariant embedding $A_\pi^\ast$ of $V^\ast\otimes (\widetilde{V})^\ast$ into $\mathcal{D}(G_n)$. The image of the embedding consists of the so-called \textit{Bessel} distributions of $\pi$. In particular,
$$\iota: V^\ast \to \mathcal{D}(G_n)^{ 1 \times H_{k,n} }, \qquad \iota(\nu) = A^\ast_{\pi}(\nu\otimes \lambda)$$
defines an embedding which intertwines the $G_n$-action (the one induced from left multiplication on $G_n$ this time).
\par Finally, it is straightforward to show that $\mathcal{D}(G_n)^{ 1 \times H_{k,n} }$ and $\mathcal{D}(G_n/H_{k,n})$ are isomorphic as $G_n$-spaces (see \cite[3.1]{off}). Going through the identification $G_n/H_{k,n}\cong X_{k,n}$, we see that $V^\ast$ is embedded into $\mathcal{D}(X_{k,n})$ as a $G_n$-space.
\begin{cor}
The equality $\mathcal{D}(X_{k,n})^{P_{k,n}}= \mathcal{D}(X_{k,n})^{H_{k,n}}$ would imply $(V^\ast)^{P_{k,n}} = (V^\ast)^{H_{k,n}}$ for every $H_{k,n}$-distinguished irreducible smooth representation of $G_n$.
\end{cor}
Having shown that deduction, we see that Theorem \ref{final-form} proves Lemma \ref{start-lem}, and Corollary \ref{final-cor} proves Theorem \ref{start-thm}.

\bibliographystyle{abbrv}
\bibliography{propo2}{}

\end{document}


TITLE: How to show that $\gcd(a_1,a_2,\cdots,a_k) = 1$ implies that there exist a non-negative solution to $\sum_{i=1}^{n}a_ix_i = n$ for large $n.$
QUESTION [7 upvotes]: I was reading about the Coin-problem and I am unable to fully understand the following argument:

On the other hand, whenever the GCD equals 1, the set of integers that cannot be expressed as a conical combination of $\{ a_1, a_2, …, a_n \}$ is bounded according to Schur's theorem, and therefore the Frobenius number exists.

Here the author is arguing for the existence of a non-negative solution to the linear Diophantine equation (LDE)
$$\sum_{i=1}^{k}a_ix_i =n \text{}$$
for large enough $n.$ Now I tried to understand the proof of the Schur theorem here enter link description here (Page 98), but I am not sure that I understand it fully. In particular, I don't understand why the generating function associated with the sequence $h_n$ that counts the number of solutions to the LDE is
$$H(x)= \prod_{i=1}^{k}\left(\frac{1}{1-x^{a_i}}\right).$$
Once we have $H(x)$ we can deduce that
$$h_n\sim \frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $n\to \infty.$ How does this exactly show that for large enough $n,$ $h_n>0?$ Is it because the $\frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}>0?$

REPLY [4 votes]: A simple approach to the title:
Thanks to Bezout, there is some (not necessarily) integer combination of the $a_i$s that makes $1$. Add a large enough multiple of $a_1$ to each coefficient to make it non-negative and call the sum $k$; we have $k\equiv 1 \pmod{a_1}$.
Now any number $\ge ka_1$ is a non-negative combination of the $a_i$s. Namely, let $n$ be the desired number, and let $m=n\bmod a_1$. Then $mk$ is a positive combination, and we can add some multiple of $a_1$ to make it $n$ instead, since $mk\equiv n\pmod{a_1}$ and $mk<n$ because $m<a_1$.


TITLE: Is every function $f: C \rightarrow \mathbb{R}$ continuous? (C is Cantor set)?
QUESTION [3 upvotes]: Is every function $f: C \rightarrow  \mathbb{R}$ continuous? (C is Cantor set) We assume that $ f [ C ] \neq \mathbb{R} $.
Continuous image of a compact space is compact and Cantor set is compact, because it is closed subset of [0,1] which is compact.
$|C| = \mathfrak{c} =|(0,1)|$ so there exists a function $f$ s.t. $f[C]=(0,1)$. (0,1) is not compact, thus $f$ is not continuous.
Is that correct?
Incorrect solution:
Let $f(0) = f(1) = 0 $ and $f(x) = 1, $ for $ x \notin \{0,1\} $. Then $ f^{-1}[\{1\}]= C \cap (0,1) $ which is an open set, thus f is not continuous.

REPLY [2 votes]: Claim 1: The relative topology on $C$ is not the discrete topology.
Proof of Claim 1: Prove by contradiction. Suppose that the relative
topology on $C$ is the discrete topology. For each $x\in C$, $\{x\}=C\cap U_x$
for some open set $U_x\subseteq\mathbb{R}$. Note that $C$ is compact and $\{U_x\mid x\in C\}$ is an open covering
for $C$, so there exists a finite subcover, denoted by $\{U_{x_{1}},U_{x_{2}},\ldots,U_{x_{n}}\}$.
Now
$$
C=C\cap\bigcup_{k=1}^{n}U_{x_{k}}=\bigcup_{k=1}^{n}\left(C\cap U_{x_{k}}\right)=\{x_{1},x_{2},\ldots,x_{n}\}
$$
 which is a contradiction because $C$ is an infinite set (actually uncountable, but we do not need this fact).
Claim 2: The relative topology on $C$ is the discrete topology iff
each function $f:C\rightarrow\mathbb{R}$ that satisfies $f[C]\neq \mathbb{R}$ is continuous.
Proof of Claim 2: The $\Rightarrow$ direction is trivial and only
the $\Leftarrow$ needs a proof. Suppose that each function $f:C\rightarrow\mathbb{R}$ that satisfies $f[C]\neq \mathbb{R}$
is continuous. Let $a\in C$ be arbitrary. Define $f:C\rightarrow\mathbb{R}$
by
$$
f(x)=\begin{cases}
x, & \mbox{ if }x\neq a\\
100, & \mbox{ if }x=a
\end{cases}.
$$ Clearly, $f[C]\neq \mathbb{R}$ and $(99,101)$ is an open subset of $\mathbb{R}$, hence $\{a\}=f^{-1}\left((99,101)\right)$
is an open subset of $C$. Therefore, the relative topology on $C$ is the discrete topology.
By Claim 1 and Claim 2, it follows that not all function $f:C\rightarrow\mathbb{R}$
are continuous.


TITLE: Calculating the electric potential in cylindrical coordinates from constant E-field
QUESTION [3 upvotes]: I am having so much trouble with this problem. I feel like I shouldn't be, but I am.

A uniform electric field, $\vec{E} = E_0\hat{x}$. What is the potential, expressed using cylindrical coordinates, $V(s,\phi,z)$?

I asked a question related to this yesterday, and got a solid answer, but when I carry things out, it still doesn't make sense physically (though, after a bit of studying, I now understand the Gradient Theorem).
We know that:
$$\vec{E} = E_0\hat{x} = -\nabla V \implies -\int_{C[x_0,x]} E_0\hat{x}\cdot dl = V(x) - V(x_0)$$
Since the electric field only has quantity in the $\hat{x}$ direction, the dot product inside of the integral comes out to $E_0~dx$, giving us:
$$E_0(x_0 - x) = V(x) - V(x_0)$$
Now, this makes sense. Since the electric field is a conservative one, the path we take shouldn't depend on any other variable but $x$. If we move in some $y$ direction and some $z$ direction, there will be zero change in potential.
Let $V(x_0) = 0$ be our point of reference, so: $$V(x) = E_0(x_0 - x)$$
So far (I think) this all seems fine and dandy. The reason I don't set $E_0x_0$ to zero is because there is no $1/x$ term; otherwise I'd be able to make the reference point at $x=\infty$, and things would cancel out nicely.
Now, I try to convert to cylindrical coordinates. We know:
$$r = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2} = x \\
\theta = \tan^{-1}{\frac{y}{x}} = 0 \\
z = 0
$$
So $V(r) = E_0(r_0 - r)$
But... this doesn't make any sense. With this potential function, moving $r$-distance in the $-x$ direction will give you the same potential as moving $r$-distance in the positive $x$ direction (which will give you the same potential as moving $r$-distance in the $y$ direction, even, since $r$ is just a radial distance from, say, the $z$-axis.
I am doing something terribly wrong and I have no idea what it is, haha.

REPLY [1 votes]: You assumed that $y=0$ and $z=0$ when you did the following steps,
$$r = \sqrt{x^2 + y^2 + z^2} = \sqrt{x^2} = x$$
As a result, you are now looking at all the set of points which  are having coordinates of form $(x,0,0)$ that is, the x-axis. So you cannot find the potential at any other point except the ones on the x-axis using your potential formula in cylindrical coordinates.

If you want to find the general potential in cylindrical coordinates, the you have to use the following transformation:
$$x=r\cos(\phi)$$
$$\Longrightarrow V(r,\phi,z) = E_0(x_0 - r\cos(\phi))$$
Which is the correct form of the potential in cylindrical coordinates


TITLE: Concrete Mathematics: Simplification of quick sort recurrence in preparation of finding summation factor
QUESTION [0 upvotes]: In the analysis of the quicksort algorithm a recurrence is presented in 2.12. I feel I understand the simplification steps down to the point where we have
$$
nC_n - (n - 1)C_{n-1} = 2n + 2C_{n-1},\ \ \ \ \ \text{for}\ n > 1.
$$
Then the book says

It turns out that this relation also holds when $n = 1$, because $C_1 = 2$. Therefore the original recurrence for $C_n$ reduces to a much simpler one:
$$
\begin{align}
C_0 &= 0 \\
nC_n &= (n + 1)C_{n-1} + 2n,\ \ \ \ \ \ \text{for}\ n > 0
\end{align}
$$

I don't understand how the fact that the relation holding for $n = 1$ allows for simplification from the first equation (in this question, not the book section) to the second simpler one. How is this transformation done?

REPLY [0 votes]: $$t_n=nC_n-(n+1)C_n=2n \implies \frac{C_n}{n+1}-\frac{C_{n-1}}{n}=\frac{2}{n+1}$$ $$\implies t_n=f_{n}-f_{n-1}=\frac{2}{n+1}$$
By telescopic summation we have
$$f_N-f_0=\sum_{n=1}^{N}\frac{2}{n+1} \implies \frac{C_N}{N+1}=2\left(H_N-1+\frac{1}{(N+1)}\right)$$
$$\implies C_N=2(N+1)(H_N-1)+2, N >0$$
Here, $H_N=1+1/2+1/3+1/4+...+1/N$ are harmonic numbers.


TITLE: Volume of $E=\{5x^8 \le y \le 7x^8; 2y^5 \le z \le 3y^5; z^7 \le x \le 6z^7\}$.
QUESTION [6 upvotes]: I am stuck with the following problem:
Compute volume of $E=\{5x^8 \le y \le 7x^8; 2y^5  \le z  \le 3y^5; z^7 \le x \le 6z^7\}$.
My progress:
It is easy to see that $x,y,z \ge 0$. Therefore I can add all inequalities and after simple algebra get the following: $$5x^8 + 2y^5+z^7 \le x+y+z \le 7x^8 + 3y^5+6x^7$$
Update: as it was pointed out by A.Γ. adding inequalities does not work because different sets can give the same inequalities. So, it looks like I should solve or at least get some kind of information for integration directly from the system somehow.
Any hint?
Thanks a lot for your help!

REPLY [4 votes]: Hint: the set is given by
$$
\begin{cases}
5\le\frac{y}{x^8}\le 7,\\
2\le\frac{z}{y^5}\le 3,\\
1\le\frac{x}{z^7}\le 6
\end{cases}\quad\rightarrow\quad\text{substitution}\quad\rightarrow\quad
\begin{cases}
5\le u\le 7,\\
2\le v\le 3,\\
1\le w\le 6
\end{cases}.
$$


TITLE: Characteristic room-temperature photon energy - is this nomogram wrong?
QUESTION [2 upvotes]: Reading this recent ars technica article on the James Webb telescope, something kept bothering me about the nomogram - shown below. The credit says it is from The Opensource Handbook of Nanoscience and Nanotechnology.
Is the graphic wrong, or am I missing something? I'm comfortable in rooms where kT is about 1/40 eV - this plot looks like it comes from Pluto!
Question: Approximately where on that graphic should the red room temperature line actually fall?

REPLY [3 votes]: Conversion from K to eV
Let's assume room temperature is at ~300 K, just to make the numbers easy.  To convert to electron volts, one simply does:
$$
T \ [K] \times \left( \frac{ 1.38064852 \times 10^{-23} \ J }{ 1 \ K } \right) \times \left( \frac{ 1 \ eV }{ 1.6021766208 \times 10^{-19} \ J } \right) = \left( k_{B} T \right) \ [eV] \\
\left( k_{B} T \right) \ [eV] \sim \left( \frac{ 8.61733 \times 10^{-5} eV }{ 1 \ K } \right) \times T \ [K]
$$
where $k_{B} = 1.3806505 \times 10^{-23}$ J/K is the Boltzmann constant, and $1.6021766208 \times 10^{-19}$ J/eV is an energy conversion factor equivalent in magnitude to the fundamental charge.  The values shown above are the 2014 CODATA/NIST values.

Question: Approximately where on that graphic should the red room temperature line actually fall?

Answer
So using the above we find that 300 K ~ 0.02585 eV or $\sim 3 \times 10^{-2}$ eV.  Therefore, the red line should be between the $10^{-1}$ and $10^{-2}$ tick marks on the eV scale bar, not the $10^{-2}$ and $10^{-3}$.


TITLE: Evaluating Improper Integral With No Closed-Form Antiderivative
QUESTION [0 upvotes]: I am trying to evaluate $$\int_{0}^{\infty} \frac{\cos(x)-e^{-x}}{x} dx$$
I did manage to get the correct value of this integral through the use of Frullani's Integral Theorem, which states that $$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx = (f(\infty)-f(0))\Big(\ln\Big(\frac{a}{b}\Big)\Big)$$
So from here, I substitute $e^{ix}-i\sin(x)$ for $\cos(x)$ to get this into something of the Frullani Integral form. It seems from the response here: Frullani 's theorem in a complex context., that we can extend the Frullani Theorem to a complex setting with a lot of work and on a case-by-case basis. However, I would like to approach this integral in a way that does not involve complex numbers. This integral was originally meant to be evaluated with just "advanced calc"/elementary real analysis methods, so I am looking for such an approach.

REPLY [0 votes]: In fact this integral is a special case of:
$$\int _0^{\infty }\frac{e^{-ax}\cos \left(cx\right)-e^{-bx}\cos \left(dx\right)}{x}\:dx$$
Which can also be evaluated with Feynman's trick:
$$I\left(a\right)=\int _0^{\infty }\frac{e^{-ax}\cos \left(cx\right)-e^{-bx}\cos \left(dx\right)}{x}\:dx$$
$$I'\left(a\right)=-\int _0^{\infty }e^{-ax}\cos \left(cx\right)\:dx=-\frac{a}{c^2+a^2}$$
$$-\int _a^{\infty }I'\left(a\right)\:da=\int _a^{\infty }\frac{a}{c^2+a^2}\:da$$

$$I\left(\infty \right)=I\left(b \right)=-\int _0^{\infty }\frac{e^{-bx}\cos \left(dx\right)}{x}\:dx$$
$$I'\left(b\right)=\int _0^{\infty }e^{-bx}\cos \left(dx\right)\:dx=\frac{b}{d^2+b^2}$$
$$-\int _b^{\infty }I'\left(b\right)\:db=-\int _b^{\infty }\frac{b}{d^2+b^2}\:db$$
$$I\left(\infty \right)=I\left(b\right)=-\int _b^{\infty }\frac{b}{d^2+b^2}\:db$$

$$-\left(I\left(\infty \right)-I\left(a\right)\right)=\int _a^{\infty }\frac{a}{c^2+a^2}\:da$$
$$I\left(a\right)=\int _a^{\infty }\frac{a}{c^2+a^2}\:da-\int _b^{\infty }\frac{a}{d^2+a^2}\:da$$
$$=\frac{1}{2}\ln \left(c^2+a^2\right)\biggr|_a^{\infty }-\frac{1}{2}\ln \left(d^2+a^2\right)\biggr|_b^{\infty }=\frac{1}{2}\ln \left(\frac{d^2+b^2}{c^2+a^2}\right)$$
Thus:
$$\boxed{I\left(a\right)=\int _0^{\infty }\frac{e^{-ax}\cos \left(cx\right)-e^{-bx}\cos \left(dx\right)}{x}\:dx=\frac{1}{2}\ln \left(\frac{d^2+b^2}{c^2+a^2}\right)}$$

We can verify the result obtained in the other answer with this letting $a=d=0,c=b=1$.
$$\int _0^{\infty }\frac{\cos \left(x\right)-e^{-x}}{x}\:dx=\frac{1}{2}\ln \left(\frac{0+1}{1+0}\right)=0$$


TITLE: Am I counting quantifiers correctly?
QUESTION [6 upvotes]: I think this is right but I want to check.  The theory $\mathsf{WKL}^*_0$ is conservative over EFA for $\Pi^0_2$ sentences.  And the first order part of $\mathsf{WKL}^*_0$ is axiomatized by EFA plus the following  formula scheme, known as the $\Sigma^0_1$ bounding principle.
For every $\Sigma^0_1$ formula $\varphi(i,j)$ ( with $n$ not free):
$$(\forall i<m)\, \exists j\, \varphi(i,j) \rightarrow \exists n\,(\forall i<m)(\exists j<n)\varphi(i,j).$$
So this formula scheme is not  $\Pi^0_2$ and I want to check that I see this correctly.
I was puzzled at first because the quantifier on $i$ in the antecedent is bounded,  so I ignored it.  But I believe the correct point is that this quantifier comes before an unbounded quantifier on $j$, so when you prenex this formula it starts with $\exists i\,\forall j$, placed before a $\Sigma^0_1$ formula.
To state the general rule: when counting quantifier changes, you ignore bounded quantifiers that come after all unbounded ones.  But even a bounded quantifier, if it comes before an unbounded one, must be counted.
Is that right?

REPLY [7 votes]: To state the general rule: when counting quantifier changes, you ignore bounded quantifiers that come after all unbounded ones. But even a bounded quantifier, if it comes before an unbounded one, must be counted.

That's correct, at least when working in weak fragments of PA.
The complication is that when you're working in a theory that includes the bounding axiom, you can use it to move bounded quantifiers inside unbounded ones.  Over PA, that formula is (equivalent to) a $\Pi^0_2$ formula.  So when people are used to working over stronger theories, they often just skip bounded quantifiers entirely when counting quantifier changes, and that's correct...as long as you stick to theories with a strong enough bounding axiom.


TITLE: Radial Schrödinger equation of a scattering in two dimentions
QUESTION [0 upvotes]: The scattering, in two dimensions, of a particle of mass $m$ by a central potential $U(r)$. The hamiltonian of the system is $H= p^2/2m + U(r)$.
Then the radial wave function $ϕ(r)$ is obtained as a solution to
$$((d^2/dr^2)+1/r(d/dr)-(v^2/r^2)+K^2)ϕ(r)=0.$$
where $v^2(L)=L^2+2mλ$,  $K^2=2mE$,  if $U(r)=λ/r^2$.
My first question is why the author chose the separation of variables constant to be $L^2$, where we usually take it as $L(L+1)$? and the second term of the equation i think is $2/r$ not $1/r$?

REPLY [0 votes]: You are in two dimensions, so the free
Schroedinger equation has $E=k^2$
solutions
$$
\psi(r,\theta) =e^{il\theta}J_l(kr),\quad  l\in {\mathbb Z}
$$
The  radial equation is Bessel's equation which  has $l^2/r^2$ rather than $l(l+1)/r^2$.


TITLE: Why we never observe superposition of up and down ferromagnetic ground state of Ising model?
QUESTION [5 upvotes]: I thought it is due to spontaneous symmetry breaking. But isn't that because we never observe the superposition states, then we claim that there is spontaneous symmetry breaking. It looks like spontaneous symmetry breaking is an assumption.

REPLY [5 votes]: As the size of the lattice gets large, the superposition
$$\vert \psi \rangle = \frac{1}{\sqrt{2}}\left(\vert \uparrow \uparrow \uparrow \cdots \rangle + \vert \downarrow \downarrow \downarrow \cdots \rangle\right)$$
in the Ising model becomes unstable to environmental perturbation.  Just like Schrodinger's cat (a superposition $\vert \mathrm{live} \rangle + \vert \mathrm{dead} \rangle$), the Ising ground state is formally possible but in practice impossible to create and preserve because of decoherence.  So-called "pure decoherence" happens when the unitary system-environment interaction takes the form
$$U_{\mathcal{SE}} = \sum_i \vert S_i \rangle \langle S_i \vert \otimes U_{\mathcal{E}}^{(i)}$$
where $\vert S_i \rangle$ is some basis of orthogonal states of the system (the "pointer states") and the $U_{\mathcal{E}}^{(i)}$ are conditional unitaries on the environment.  For the Ising model, the pointer states would be $\vert \uparrow \uparrow \uparrow \cdots \rangle$ and $\vert \downarrow \downarrow \downarrow \cdots \rangle$.  In this idealized case, the  superposition $\vert \psi \rangle$ will decay into a mixed state
$$\rho = \frac{1}{2}\left(\vert \uparrow \uparrow \uparrow \cdots \rangle \langle \uparrow \uparrow \uparrow \cdots \vert + \vert \downarrow \downarrow \downarrow \cdots \rangle \langle \downarrow \downarrow \downarrow \cdots \vert \right)$$
exponentially fast, with a rate proportional to the number of lattice sites (under a reasonable noise model).  Pure decoherence is minimal in the sense that it has "no classical effect" on the system, in the narrow sense that pure decoherence doesn't change the probabilities of various outcomes for a hypothetical measurement of the system in the pointer basis.
Now in general, perturbations from the environment can act in much more complicated ways, and often will have classical effects in addition to pure decoherence.  For instance, when you open the box containing Schordinger's cat, the air currents in the room will rustle that cat's hair (meaning the'll be in a different position state). Likewise the environment's effect on $\psi$ will generally cause it to decay into a mixed state that has non-zero over lap with states that are a short Hamming distance away from $\vert \uparrow \uparrow \uparrow \cdots \rangle$ and $ \vert \downarrow \downarrow \downarrow \cdots \rangle$.
The actual dynamics taken is a complicated question that depends sensitively on how the environment couples to the system, and the environment's initial state.  However, it's intuitive that long range entanglement is unusually sensitive for the following reason. If a single particle in the environment interacts with a single spin in the system, then this will usually reduce the entanglement by a factor of order unity.  (The factor is called a decoherence factor, and it's given by the inner product between $U_{\mathcal{SE}} \vert \uparrow \rangle \vert E_0 \rangle$ and $U_{\mathcal{SE}} \vert \downarrow \rangle \vert E_0 \rangle$, even if the decoherence isn't pure, where $\vert E_0 \rangle$ is the initial state of the particle in the environment.)  As multiple particles interact, the decoherence factors combine multiplicatively, and the entanglement is destroyed exponentially quickly in the size of the environment and system.


TITLE: Understanding Probability Spaces
QUESTION [0 upvotes]: I was looking at the following description of probability spaces, and was trying to understand this better:

In the example of flipping a coin, is the following correct?

The sample space : Heads or Tails

The event space: Some combination of outcomes you interested in observing (e.g. Heads, Heads, Tails)

The probability function: In this example, the binomial probability distribution function

Is my understanding correct?
Thanks
Reference: https://en.wikipedia.org/wiki/Probability_space

REPLY [3 votes]: You are correct about the sample space. We can take the sample space to be $\Omega = \{ H, T \}$.
The event space $\mathcal F$, however, should be the collection of all subsets of $\Omega$. In other words: $\mathcal F = \{ \emptyset, \{H\},\{T\},\{H, T \} \}$.
The probability function $P$ takes as input an event and returns as output the probability of that event. So, assuming the coin is fair:
\begin{align}
P(\emptyset) &= 0,\\
P(\{H\}) &= \frac12,\\
P(\{T\}) &= \frac12,\\
P(\{H, T \}) &= 1.
\end{align}


TITLE: Permission to use Online Notes
QUESTION [16 upvotes]: I am a new professor in Mathematics and I am running an independent study on Diophantine equations with a student of mine.  Online I have found a wealth of very helpful expository notes written by other professors, and I would like to use them for guided reading.  I am wondering whether it is customary to ask permission of the author before using his or her online notes for my own reading course.
Also, if anyone has suggestions for good sources Diophantine Equations please feel free to enlighten me.

REPLY [3 votes]: The Stanford Copyright & Fair Use Overview is pretty clear about what constitutes fair use in the classroom and what doesn't:

Rules for Reproducing Text Materials for Use in Class
The guidelines permit a teacher to make one copy of any of the following: a chapter from a book; an article from a periodical or newspaper; a short story, short essay, or short poem; a chart, graph, diagram, drawing, cartoon, or picture from a book, periodical, or newspaper.
Teachers may photocopy articles to hand out in class, but the guidelines impose restrictions. Classroom copying cannot be used to replace texts or workbooks used in the classroom. Pupils cannot be charged more than the actual cost of photocopying. The number of copies cannot exceed more than one copy per pupil. And a notice of copyright must be affixed to each copy.
Examples of what can be copied and distributed in class include:

a complete poem if less than 250 words or an excerpt of not more than 250 words from a longer poem
a complete article, story, or essay if less than 2,500 words or an excerpt from any prose work of not more than 1,000 words or 10% of the work, whichever is less; or
  one chart, graph, diagram, drawing, cartoon, or picture per book or per periodical issue.

Not more than one short poem, article, story, essay, or two excerpts may be copied from the same author, nor more than three from the same collective work or periodical volume (for example, a magazine or newspaper) during one class term. As a general rule, a teacher has more freedom to copy from newspapers or other periodicals if the copying is related to current events.
The idea to make the copies must come from the teacher, not from school administrators or other higher authority. Only nine instances of such copying for one course during one school term are permitted. In addition, the idea to make copies and their actual classroom use must be so close together in time that it would be unreasonable to expect a timely reply to a permission request. For example, the instructor finds a newsweekly article on capital punishment two days before presenting a lecture on the subject.
Teachers may not photocopy workbooks, texts, standardized tests, or other materials that were created for educational use. The guidelines were not intended to allow teachers to usurp the profits of educational publishers. In other words, educational publishers do not consider it a fair use if the copying provides replacements or substitutes for the purchase of books, reprints, periodicals, tests, workbooks, anthologies, compilations, or collective works.

Note that this does not apply if the work is licensed appropriately. For example, the Stanford Copyright & Fair Use Overview quoted above is licensed CC BY-NC and this answer is licensed CC BY-SA. For course notes, it makes sense to use such a license since that is probably the intended use of the work. Contact the author(s) and recommend that they license their work in a manner they find appropriate.

The Stanford Copyright & Fair Use Overview also has a clear notice regarding online material:

If You Want to Use Material on the Internet
Each day, people post vast quantities of creative material on the Internet -- material that is available for downloading by anyone who has the right computer equipment. Because the information is stored somewhere on an Internet server, it is fixed in a tangible medium and potentially qualifies for copyright protection. Whether it does, in fact, qualify depends on other factors that you would have no way of knowing about, such as when the work was first published (which affects the need for a copyright notice), whether the copyright in the work has been renewed (for works published before 1978), whether the work is a work made for hire (which affects the length of the copyright) and whether the copyright owner intends to dedicate the work to the public domain. If you want to download the material for use in your own work, you should be cautious. It's best to track down the author of the material and ask for permission. Generally, you can claim a fair use right for using a very small portion of text for commentary, scholarship or smilar purposes.

The thing to take away from this is that there is no substitute for proper licensing. If you intend for something to be used in some way or another: make that clear!


TITLE: Second order homogeneous ordinary differential equation with variable coefficients.
QUESTION [1 upvotes]: I am trying to solve the following differential equation,
\begin{equation}
y^{''}_{xx} + a_1 x y^{'}_{x} +(a_0x+b_0)y=0
\end{equation}
This equation is not the Cauchy‐Euler Equidimensional. I found a solution to the following differntial equation in Handbook of exact solutions for ordinary differential equations by Andrei Polyanin and Valentin Zaitsev,
 \begin{equation}
 (ax_2+b2) y^{''}_{xx} + (a_1 x +b_1) y^{'}_{x} + (a_0 x + b_0) y=0
 \end{equation}
The equation that I am trying to solve is special case of the prevous equation ($a_2=b_10, b_2=1$). The solution to this equation is
\begin{equation}
y=e^{hx} z(\zeta), where \, \zeta=\dfrac{x-\mu}{\lambda}
\end{equation}
The authors mentioned that $z(\zeta)$ matches $\mathcal{L}(a,b;x)$ arbitrary solution of the degenerate hyperheometric equation $x y^{''}_{xx}+ (b-x) y^{'}_{x} -ay=0$, yet it is not clear how to do that. I have attached four pages of the books that refere to that solution. The problem is number 103.
Thanks

REPLY [1 votes]: According to Maple, your DE has general solution
$$ y \left( x \right) =c_1 {{\rm e}^{-{\frac {a_{{0}}x}{a_{{1}}}}}}  {
{ U}\left({\frac {b_{{0}}{a_{{1}}}^{2}+{a_{{0}}}^{2}}{2\,{a_{{1}}}^
{3}}},\,{\frac{1}{2}},\,-{\frac { \left( x{a_{{1}}}^{2}-2\,a_{{0}}
 \right) ^{2}}{2\,{a_{{1}}}^{3}}}\right)}+ c_2{{\rm e}^{-{\frac {a_{{0}}x}{a_{{1}}}}}}  {{ M}\left({\frac
{b_{{0}}{a_{{1}}}^{2}+{a_{{0}}}^{2}}{2\,{a_{{1}}}^{3}}},\,{\frac{1}{2}
},\,-{\frac { \left( x{a_{{1}}}^{2}-2\,a_{{0}} \right) ^{2}}{2\,{a_{{1
}}}^{3}}}\right)}
$$
where $U$ and $M$ are Kummer functions.


TITLE: How to Calculate Radius of Circle Given Two Points and Tangential Circle
QUESTION [3 upvotes]: In the image below, the two smallest circles and the chord are given to me.

What I would like to calculate is the radius of the largest circle, such that it includes the endpoints of the chord and is tangential to the smallest circle.  Is there a way to calculate this symbolically?
The following is what I know:

The center of medium circle is at the origin $(0,0)$
The radius of the medium circle is $R_m$, and the small circle is $R_s$
Chord endpoints are given at $(x_1, y_1)$ and $(x_2, y_2)$
The center of the smallest circle lies on the medium circle at point $(S_x, S_y)$
Both intersections of the medium and small circles occur within the chord
The center point of the largest circle can be written as $(L_x, L_y)$, where $L_x=\frac12(x_1+x_2)t$, and $L_y=\frac12(y_1+y_2)t$, for some $t > 1$.

I need to compute the center point of the large circle $(L_x, L_y)$ given this information.  I have created the following system of two equations with two unknowns ($R_I$ and $t$) given this information, but can't figure out how to solve them:

$R_I^2=(\frac12(x_1+x_2)t-x_1)^2+(\frac12(y_1+y_2)t-y_1)^2$
$(R_I-R_s)^2=(\frac12(x_1+x_2)t-S_x)^2+(\frac12(y_1+y_2)t-S_y)^2$

Any help is greatly appreciated.

REPLY [0 votes]: After thinking about this some more, I realized that I could dramatically simplify the problem if I rotate the image such that the line through the center of the chord is fixed as the y-axis.  By doing so, the circle I am looking for is centered at $(0,L_y)$.  Given this translation, the system of two equations is simplified to

$(L_y − S_y)^2 + S_x^2 = (R_I−R_s)^2$
$(L_y - y_1)^2 + x_1^2 = R_I^2$

When I expand this out, the X and Y parameters combine to be the known radius of the medium circle, so the equations become:

$L_y^2 - 2L_yS_y + R_m^2 = (R_I−R_s)^2$
$L_y^2 - 2L_yy_1 + R_m^2 = R_I^2$

Substituting the second equation into the first yields $2L_y(y_1-S_y) = R_s^2 - 2R_s\sqrt{L_y^2-2L_yy_1+R_m^2}$.  Solving for $L_y$ yields the closed-form:
$$L_y=\frac{R_s^2(S_y+y_1) \pm R_s\sqrt{R_s^2(R_s^2 + 4S_yy_1) + 4R_m^2[(S_y-y_1)^2 - R_s^2]}}{2[R_s^2-(S_y-y_1)^2]}$$
Once $L_y$ is known, the remaining quantities are trivial to compute using standard geometry.


TITLE: prove $-a = (-1)a$ ? I've done two possible proofs. Which is valid if at all?
QUESTION [0 upvotes]: I'm self studying math. I'm just starting "A Survey of Modern Algebra" by Birkhoff and Mac Lane because I thought the proof section of this book was more clear than Shilov's first chapter in "Introduction to Real and Complex Analysis". I'm pretty new to proofs. I know this is perhaps very simple but I need some feedback since there are no answers in the back and I have no one else to talk to about proofs.
Are either of these acceptable? Both? Thank you for your patience.
page 8 Problem 3 (d) : Prove $-a = (-1)a$
proof one
let $a = 1 $
$-1 = -1(1) $ by substitution
$-1 = -1 $ by unity
proof two
since $a = 1a$
$(-1)a = (-1)a$ by substitution on the left

REPLY [3 votes]: None of your proofs are actually proofs.
Note that $ -a$ is the additive opposite of $ a $ Therefore to show that $(-1)a=-a$, we have to prove that $$(-1)a+a=0$$
Since $a=(1)a$, we have $$ (-1)a + a = (-1)a + (1)a = (-1+1)a = 0a =0$$
Thus $(-1)a$ is indeed the additive inverse of $a$, that is $(-1)a=-a$

REPLY [2 votes]: They are both wrong. The first one is nonsense. In second one, you don't even deduce what you were supposed to deduce.
Note that\begin{align}(-1)a+a&=(-1)a+1a\\&=\bigl((-1)+1\bigr)a\\&=0\times a=0.\end{align} Therefore, $(-1)a=-a$, since each element has one and only one inverse for the addition.


TITLE: Calculate $\int_{-\infty}^{\infty} \frac{x^4}{(x^2 + a^2)^4} dx$ based on residues in complex analysis.
QUESTION [0 upvotes]: I am reading "An Introduction to Complex Analysis" by Tomoki Kawahira.
There is the following problem in this book.

Calculate the following improper integral based on residues in complex analysis:
$$\int_{-\infty}^{\infty} \frac{x^4}{(x^2 + a^2)^4} dx,$$
where $a$ is a positive real number.

For example, we need to calculate the following integral:
$$\int_{C} \frac{z^4}{(z^2 + a^2)^4} dz,$$
where $C = J_R + H_R$, $J_R : z = x (-R \leq x \leq R)$, $H_R : z = R e^{i \theta} (0 \leq \theta \leq \pi), R > a$.
Let $g(z) := \frac{z^4}{(z + a i)^4}$.
$$\int_{C} \frac{z^4}{(z^2 + a^2)^4} dz = \int_{C} \frac{g(z)}{(z - a i)^4} dz = 2 \pi i \operatorname{Res}(\frac{g(z)}{(z - a i)^4}, a i) \\ = 2 \pi i \frac{1}{3!} g^{(3)}(a i).$$
We need to calculate $g^{(3)}(z)$.
First, I represented $g(z)$ as follows.
$$g(z) = 1 - 4 a i \frac{1}{z+a i} - 6 a^2 \frac{1}{(z+a i)^2} + 4 a^3 i \frac{1}{(z+a i)^3} + a^4 \frac{1}{(z+a i)^4}.$$
Then I calculated $g^{(3)}(z)$ , but the calculation was hard for me.
Is there an easier way to calculate $g^{(3)}(z)$?

REPLY [1 votes]: By definition the residue at $z_0$ is the $c_{-1}$ coefficient in the Laurent series of the function. Thus you first calculate the Laurent Series of each piece centered at $ai$
\begin{align}z^4&=(z-ai+ai)^4=\sum_{n=0}^4{4\choose n}(ai)^n(z-ai)^n\\
\frac{1}{(z-ai)^4}&=(z-ai)^{-4}\\
\frac{1}{(z+ai)^4}&=\left(\frac{1}{z-ai+2ai}\right)^4=\left(\frac{1}{\frac{z-ai}{2ai}+1}\right)^4=\left(\sum \frac{(z-ai)^n}{(-2ai)^n}\right)^4
\end{align}
Then multiplying the three together and collecting only the terms whose power is $-1$ gives you the coefficient. Doing the same for $z+ai$ gives the answer.
In this case it's not fast, but sometimes it's a nice technique


TITLE: Applying the Dominated Convergence Theorem
QUESTION [1 upvotes]: In a question I've got, I found that I need to prove the following equality:
$\lim_{h\rightarrow 0}\int_{0}^{\infty}\frac{e^{-(t+h)x}-e^{-tx}}{hx}dx=\int_{0}^{\infty}\lim_{h\rightarrow 0}\frac{e^{-(t+h)x}-e^{-tx}}{hx}dx$
($t>0$ is a constant.)
I know the limit in the right side is $-e^{-tx}$ if that matters.
Now, by Heine, it's sufficient to prove this result with a sequence $h_n\rightarrow 0$ instead  of a continuous variable. So let's define $f_n(x)=\frac{e^{-(t+h_n)x}-e^{-tx}}{h_nx}$ and now we need to show that the conditions of DCT hold, and all that's left is finding a dominating function.
That's where I failed. Any idea?

REPLY [1 votes]: A very efficient way forward circumvents use of the Dominated Convergence Theorem and instead exploits Frullani's Theorem.  Proceeding, we obtain
$$\frac1h \int_0^\infty \frac{e^{-(t+h)x}-e^{-tx}}{hx} \,dx=\frac1h \log\left( \frac{t}{t+h}\right)$$
The limit as $h\to 0$ is straightforward to evaluate with the answer $-\frac1t$.  And we are done!
If one wishes to proceed through use of the Dominated Convergence Theorem, then the following primer will facilitate analysis.

PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$
for $x<1$.

Herein, we use the inequality on the left-hand side of $(1)$ to establish a coveted dominating function.  To that end, we proceed.
Note that we can write the integrand $\frac{e^{-(t+h)x}-e^{-tx}}{hx}$ as
$$\begin{align}
\frac{e^{-(t+h)x}-e^{-tx}}{hx}&=e^{-tx}\left(\frac{e^{-hx}-1}{hx}\right) \tag{2a}\\\\
&=e^{-(t+h)x}\left( \frac{1-e^{hx}}{hx} \right) \tag{2b}
\end{align}$$
Note that the term in parentheses on the right-hand side of $(2a)$ is non-positive for all $x\in [0,\infty$ and all $h$.  Applying the left-hand side inequality in $(1)$ to $(2a)$, we find immediately that for $h>0$
$$\left|\frac{e^{-hx}-1}{hx}\right|\le 1$$
for $x\ge 0$, $h>0$.
For $h<0$, we take $|h|<t/2$.  Then, applying the left-hand side inequality in $(1)$ to $(2b)$, we see analogously that
$$\left| \frac{1-e^{hx}}{hx} \right|\le 1$$
for $x\ge 0$, $h<0$.
Hence, we have
$$\left|\frac{e^{-(t+h)x}-e^{-tx}}{hx}\right|\le e^{-tx/2}$$
for all $x\ge 0$ and all $|h|\le t/2$.
Inasmuch as $\int_0^\infty e^{-tx/2}\,dx=\frac2t <\infty$, the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{h\to 0}\int_0^\infty \frac{e^{-(t+h)x}-e^{-tx}}{hx}\,dx&=\int_0^\infty \lim_{h\to 0}\left(\frac{e^{-(t+h)x}-e^{-tx}}{hx}\right)\,dx\\\\
&=-\int_0^\infty e^{-tx}\,dx\\\\
&=-\frac1t
\end{align}$$
And we are done!