text:
\begin{document}
\maketitle

\begin{abstract}We study normal reflection subgroups of complex reflection groups. Our point of view leads to a refinement of a theorem of Orlik and Solomon to the effect that the generating function for fixed-space dimension over a reflection group is a product of linear factors involving generalized exponents. Our refinement gives a uniform proof and generalization of a recent theorem of the second author.\end{abstract}

\section{Introduction}

Hopf proved that the (singular) cohomology of a real connected compact Lie group $\mathcal{G}$ is an exterior algebra on $\mathrm{rank}(\mathcal{G})$ generators of odd degree~\cite{hopf1964topologie}.  Its Poincar\'e series is therefore given by \[\mathrm{Hilb}(H^*(\mathcal{G});q) = \prod_{i=1}^n (1+q^{2e_i+1}).\]
Chevalley presented these $e_i$ for the exceptional simple Lie algebras in his 1950 address at the International Congress of Mathematicians~\cite{chevalley1950betti}, and Coxeter recognized them from previous work with real reflection groups~\cite{coxeter1951product}.  This observation has led to deep relationships between the cohomology of $G$, and the invariant theory of the corresponding Weyl group $W=N_\mathcal{G}(T)/T$~\cite{reeder1995cohomology,reiner2019invariant}---notably, $H^*(\mathcal{G}) \simeq \left(H^*(\mathcal{G}/T) \times H^*(T)\right)^W \simeq 􏰁\left(S/S_+^W \otimes \bigwedge V^*\right)^W.$
For more information, we refer the reader to the wonderful survey~\cite{barcelo1994combinatorial}.

It turns out that one method to compute the $e_i$ is the generating function for the dimension of the fixed space $\fixed(w)=\dim(\ker(1-g))$ over the Weyl group:
\[\sum_{w \in W} q^{\fixed(w)} = \prod_{i=1}^r (q+e_i).\]

Shephard and Todd~\cite{shephard1954finite} verified case-by-case that the same sum still factors when $W$ is replaced by a complex reflection group $G$. Let $G$ be a finite complex reflection group, acting by reflections on $V$.  The $e_i$ are now determined by the degrees $d_i$ of the fundamental invariants of $G$ on $V$ as $e_i=d_i-1$.  A case-free proof of this result was given by Solomon~\cite{solomon1963invariants}, mirroring Hopf's result: writing $S=\mathrm{Sym}(V^*)$ and $\Lambda=\bigwedge(V^*)$, $(\left(S \otimes \Lambda\right)^G$ is a free exterior algebra over the ring of $G$-invariant polynomials, which gives a factorization of the Poincar\'e series

\begin{equation}
\label{eq:solomon}
\mathrm{Hilb}(\left(S \otimes \Lambda \right)^G;q,u) = \prod_{i=1}^r \frac{1+u q^{e_i}}{1-q^{d_i}}.
\end{equation}
Computing the trace on $S \otimes \Lambda$ of the projection to the $G$-invariants $\frac{1}{G}\sum_{g\in G} g$, specializing to $u=q(1-x)-1$, and taking the limit as $x \to 1$ gives the Shephard-Todd result:

\begin{theorem}[{\cite{shephard1954finite,solomon1963invariants}}]\label{thm:OS-untwisted}
For any complex reflection group $G$,
\begin{equation}
\label{eq:shephard_todd}
\sum_{g \in G} q^{\fixed(g)} = \prod_{i=1}^r (q+e_i).
\end{equation}
\end{theorem}

More generally, define the \defn{fake degree} of an $m$-dimensional (simple) $G$-module $M$ to be the polynomial encoding the degrees in which $M$ occurs in the coinvariant ring $\mathbb{C}[V]_G$: $f_M(q)=\sum_{i} (\mathbb{C}[V]_G^i,M)q^i =\sum_{i=1}^m q^{e_i(M)}.$

\medskip
Let $G \subset \mathrm{GL}(V)$ be a complex reflection group.  We say that $N \triangleleft G$ is a \defn{normal reflection subgroup} of $G$ if it is a normal subgroup of $G$ that is generated by reflections.  For Weyl groups, nontrivial normal reflection subgroups can be constructed using root lengths.   More generally, normal reflection subgroups are constructed by taking the union of conjugacy classes of reflections.    We give the classification of normal reflection subgroups in~\Cref{sec:classification}, and tie our work with previous work on their numerology in~\Cref{sec:reflexponents}.

The following theorem is a special case of results in~\cite{bessis2002quotients} (where the authors consider the more general notion of ``bon sous-groupe distingu\'e'' in lieu of our normal reflection subgroup $N$ of $G$). We emphasize that our proof of this result in \Cref{sec:quotients} follows the main ideas in~\cite{bessis2002quotients}, specialized to our more restricted setting.

\begin{theorem}
\label{thm:reflection_action}
Let $N$ be a normal reflection subgroup of a complex reflection group $G$ acting on $V$.  Then the quotient group $H=G/N$ acts as a reflection group on the vector space $\VN=V/N$.\end{theorem}

In \Cref{sec:quotients} we will build on our proof of \Cref{thm:reflection_action} to prove the following numerological identities.

\begin{theorem}
\label{thm:numbers}
Let $G,N,H$ be as in~\Cref{thm:reflection_action}. For a suitable choice of indexing of degrees and fake degrees, we have the following equalities:
\begin{align*}
e_i^N(V) {+} e_i^G(\VN) &= e_i^G(V) \\
d_i^N \cdot d_i^H & = d_i^G \\
d_i^N \cdot e_i^H(\VN) & = e_i^G(\VN)
\end{align*}
\end{theorem}

\begin{example}
Take $G=W(F_4)=G_{28}$ and $N$ to be the normal subgroup generated by the reflections corresponding to short roots.  Then $N\simeq W(D_4)$, $G/N\simeq W(A_2)=\mathfrak{S}_3$ acting by reflections on $\CC \oplus \CC \oplus \CC^2$ (trivially on $\CC\oplus\CC$), so that \Cref{thm:numbers} gives the equations
\begin{align*}
(1,5,3,3) {+} (0,0,4,8) &= (1,5,7,11) \\
(2,6,4,4) \cdot (1,1,2,3) & = (2,6,8,12) \\
(2,6,4,4) \cdot (0,0,1,2) & = (0,0,4,8).
\end{align*}
\end{example}

The following result simultaneously generalizes the results of~\cite{williams2019reflexponents} and the Shephard-Todd formula~\Cref{eq:shephard_todd} from \Cref{thm:OS-untwisted}. We state a generalized version that incorporates Galois twists in \Cref{sec:galois}.

\begin{theorem}
\label{thm:main_theorem}
Let $N\triangleleft \ G$ be reflection groups acting by reflections on $V$, and let $\VN=V/N$. Then
\[\sum_{g \in G} q^{\fix_V g} t^{\fix_{\VN} g} = \prod_{i=1}^r \left(qt+e_i^N(V) t + e_i^G(\VN)\right).\]
\end{theorem}

\begin{example}
\label{ex:c2}
The dihedral group $G=G(2,1,2)=\left\langle s,t | s^2=t^2=(st)^4=1\right\rangle$ acts as a reflection group on $V=\mathbb{C}^2$ by $s=\left[\begin{smallmatrix} -1 & 0 \\ 0 & 1 \end{smallmatrix} \right] \text{ and } t=\left[\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right].$  Take $N$ to be the normal subgroup generated by the reflections conjugate to $s$.   Then $N \simeq C_2 \times C_2$ is a normal reflection subgroup, isomorphic to the direct product of the cyclic group of order two with itself, with invariants $N_1=x_1^2$ and $N_2=x_2^2$, and $G$ acts on $\VN$ dual to $\VNd=\mathrm{span}_\CC\{N_1,N_2\}$ as the quotient group $G/N\simeq C_2$ by $s=\left[\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \right] \text{ and } t=\left[\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right].$  In this case, \Cref{thm:main_theorem} expresses the equality
\[\sum_{g \in G} q^{\fix_V g} t^{\fix_{\VN} g} = q^2t^2+2qt^2+2qt+2t+t^2 =(qt+t)(qt+t+2).\]
\end{example}

\noindent
{\bf Acknowledgements.} We thank Theo Douvropoulos for many helpful comments and suggestions. The first author was partially supported by NSF grant CCF-1815108. The second author was partially supported by Simons Foundation award number 585380.

\section{Quotients by Normal Reflection Subgroups}\label{sec:quotients}

Let $V$ be a complex vector space of dimension $r$. A \defn{reflection} is an element of $\mathrm{GL}(V)$ that fixes some hyperplane pointwise. A \defn{complex reflection group} $G$ is a finite subgroup of $\mathrm{GL}(V)$ that is generated by reflections. A complex reflection group $G$ is called \defn{irreducible} if $V$ is a simple $G$-module; $V$ is then called the \defn{reflection representation} of $G$. A \defn{(normal) reflection subgroup} of $G$ is a (normal) subgroup of $G$ that is generated by reflections.

Let $S(V^*)$ be the symmetric algebra on the dual vector space $V^*$, and write $S(V^*)^G$ for its $G$-invariant subring.  By a classical theorem of Shephard-Todd~\cite{shephard1954finite} and Chevalley~\cite{chevalley1955invariants}, a subgroup $G$ of $\mathrm{GL}(V)$ is a complex reflection group if and only if $S(V^*)^G$ is a polynomial ring, generated by $r$ algebraically independent homogeneous $G$-invariant polynomials---the \defn{degrees} $d_1\leq \cdots \leq d_r$ of these polynomials are invariants of $G$.

\begin{theorem}[\cite{shephard1954finite,chevalley1955invariants}]
\label{thm:shephard_todd}
  Let $G \leq \mathrm{GL}(V)$ be finite.  Then $G$ is a complex reflection group if and only if there exist $r$ homogeneous algebraically independent polynomials $\g_1,\ldots,\g_r\in S(V^*)^G$ such that $S(V^*)^G=\CC[\g_1,\ldots,\g_r]$.  In this case, $|G|=\prod_{i=1}^r d_i$, where $d_i=\mathrm{deg}(\g_i)$.
\end{theorem}

Although \Cref{thm:reflection_action} is a special case of results in~\cite{bessis2002quotients} (where they consider the more general notion of ``bon sous-groupe distingu\'e''), the proof is more straightforward in our restricted setting, and also leads directly to a proof of \Cref{thm:numbers}.

{
\renewcommand{\thetheorem}{\ref{thm:reflection_action}}
\begin{theorem}
Let $N$ be a normal reflection subgroup of a complex reflection group $G$ acting on $V$.  Then the quotient group $H=G/N$ acts as a reflection group on the vector space $\VN=V/N$.
\end{theorem}
\addtocounter{theorem}{-1}
}

\begin{proof} We claim that there exist homogeneous generators $\n_1,\dots,\n_r$ of $S(V^*)^N$ such that $\VN^*=\mathrm{span}_\mathbb{C}\{\n_1,\ldots,\n_r\}$ is $H$-stable. By Theorem~\ref{thm:shephard_todd}, $S(V^*)^N=\mathbb{C}[\tilde{\n}_1,\dots,\tilde{\n}_r]$ for some homogeneous algebraically independent $\tilde{\n}_i$. Let $I_+\subset S(V^*)^N$ be the ideal generated by homogeneous elements of positive degree. Then both $I_+$ and $I_+^2$ are $H$-stable homogeneous ideals, and therefore the algebraic tangent space $I_+/I_+^2$ to $\VN=V/N$ at $0$ inherits a graded action of $H$ that is compatible with the (graded) quotient map $\pi:I_+\twoheadrightarrow I_+/I_+^2$. Hence there exists a graded $H$-equivariant section $\varphi:I_+/I_+^2\rightarrow I_+$. Letting $\n_i=\varphi\circ\pi(\tilde{\n}_i)$ we see that $\n_1,\dots,\n_r$ are still homogeneous algebraically independent generators for $S(V^*)^N$ with $\mathrm{deg}(\n_i)=\mathrm{deg}(\tilde{\n}_i)$ and such that $\VN^*=\spn_\CC\{\n_1,\dots,\n_r\}$ is $H$-stable, as claimed.

Write $\g_1, \ldots, \g_r$ for the homogeneous generators of $S(V^*)^G$, again as in Theorem~\ref{thm:shephard_todd}. Consider the action of $H$ on $\VN^*$ defined by $(gN)\n_i\coloneqq  g\n_i.$ Since $S(V^*)^G= (S(V^*)^N)^H=S(E^*)^H$, there exist polynomials $\h_1,\ldots,\h_r \in \CC[\mathbf{\n}]$ such that $\h_i(\mathbf{\n})=\g_i(\mathbf{x})$, where $\mathbf{\n}=\{\n_1,\dots,\n_r\}$ and $\mathbf{x}=\{x_1,\dots,x_r\}$ denote dual bases for $E$ and $V$, respectively.

Since any algebraic relation $f(\h_1,\ldots,\h_r)=0$ would result in an algebraic relation $f(\g_1,\ldots,\g_r)=0$, the $\h_i$ must be algebraically independent. By \Cref{thm:reflection_action}, $H$ is a complex reflection group.\end{proof}

\begin{remark}
Note that the quotient group $H=G/N$ does not necessarily lift to a reflection subgroup of $G$ nor even a subgroup of $G$. A counterexample is given by $G(4,2,2)=N \triangleleft G=G_8$, so that $G/N \simeq \mathfrak{S}_3$.
\end{remark}

\begin{remark}
\label{rem:indexing}
In the course of the proof of \Cref{thm:reflection_action} we showed that the vector space $\VN=V/N$ on which $H$ acts by reflections is dual to $\VNd:=\spn_\CC\{\n_1,\dots,\n_r\}$ for a certain choice of fundamental $N$-invariants $\n_1,\dots\n_r\in S(V^*)^N$ such that $\VNd$ is $G$-stable. The resulting action of $H$ on $\VN$ respects the $\mathbf{x}$-grading on the $N$-invariants $\n_i(\mathbf{x})$, and therefore there is a choice of fundamental $H$-invariants $\h_1,\dots,\h_r\in S(\VNd)^H=S(V^*)^G$ such that each $\h_i(\mathbf{N})=\g_i(\mathbf{x})$ is both $\mathbf{x}$-homogeneous of $\mathbf{x}$-degree $=:d_i^G$ and $\mathbf{N}$-homogeneous of $\mathbf{N}$-degree $=:d_i^H$. Since the action of $H$ on $\VNd$ respects the homogeneous decomposition of $\VNd$ according to $\mathbf{x}$-degree, the $\h_i(\mathbf{N})$ may be chosen such that the $N$-invariants $\n_j(\mathbf{x})\in\mathbf{N}$ occurring non-trivially in $\h_i(\mathbf{N})$ are all of the same $\mathbf{x}$-degree $=:d_i^N$. This relationship between the fundamental invariants for $N$, $G$, and $H$ acting on their corresponding reflection representations leads to interesting numerological identities.
\end{remark}

The following result motivates some of the theoretical ingredients in our proofs.

\begin{theorem}[{\cite{solomon1963invariants}}]
\label{thm:solomon} If $G\subset\mathrm{GL}(V)$ is a complex reflection group, then the ring $\left(S(V^*)\otimes \bigwedge V^*\right)^G$ is a free exterior algebra over the ring of $G$-invariant polynomials:
\[\left(S(V^*)\otimes \bigwedge V^*\right)^G \simeq S(V^*)^G \otimes \bigwedge U_G,\] where $U_G=\spn_\CC\left\{d\g_1,\ldots,d\g_s\right\}$ and $d\g_i=\sum_{j=1}^r \frac{\partial \g_i}{\partial x_j}  \otimes x_j$.\end{theorem}

\begin{comment}
\begin{proof}
Let $u_i:=d\n_i$. Consider the multiplication map $\mu:\UN\to \VN^*$ given by \[\mu(u_i)=\sum_{j=1}^rx_j\frac{\partial \n_i}{\partial x_j},\] and note that $\mu$ is $H$-equivariant of degree $+1$ and $\mu\circ d\n_i=d_i^N\n_i$.
\end{proof}
\end{comment}

{
\renewcommand{\thetheorem}{\ref{thm:numbers}}
\begin{theorem}
Let $G,N,H$ be as in~\Cref{thm:reflection_action}. For a suitable choice of indexing of degrees and fake degrees, we have the following equalities:
\begin{align*}
e_i^N(V) {+} e_i^G(\VN) &= e_i^G(V) \\
d_i^N \cdot d_i^H & = d_i^G \\
d_i^N \cdot e_i^H(\VN) & = e_i^G(\VN)
\end{align*}

\end{theorem}
\addtocounter{theorem}{-1}
}

\begin{proof} Having chosen fundamental $N$-invariants $\n_1,\dots,\n_r\in S(V^*)^N$ such that $\VNd=\spn_\CC\{\n_1,\dots,\n_r\}$ is $G$-stable as in the proof of \Cref{thm:reflection_action} and \Cref{rem:indexing}, we have fundamental $G$-invariants $\g_i(\mathbf{x})=\h_i(\mathbf{N})$ that are $\mathbf{x}$-homogeneous of $\mathbf{x}$-degree $d_i^G$ and $\mathbf{N}$-homogeneous of $\mathbf{N}$-degree $d_i^H$, and where the $\n_j\in\mathbf{N}$ occurring non-trivially in $\h_i(\mathbf{N})$ are all of the same $\mathbf{x}$-degree $d_i^N$. The equality $d_i^Nd_i^H=d_i^G$ is immediate.

Let us show that this same choice of indexing of fundamental invariants for $N$, $G$, and $H$ results in the other two equalities. We begin by comparing $\mathbf{x}$-degrees in \[d\g_i=\sum_{j=1}^r\frac{\partial\g_i}{\partial x_j}\otimes x_j=\sum_{k=1}^r\frac{\partial \h_i}{\partial \n_k}\cdot d\n_k=\sum_{k=1}^r\sum_{j=1}^r\frac{\partial \h_i}{\partial \n_k}\cdot\frac{\partial\n_k}{\partial x_j}\otimes x_j.\] Recall that $e_i^G(V)=d_i^G-1=\mathrm{deg}_\mathbf{x}(d\g_i)$ and $e_i^N(V)=d_i^N-1=\mathrm{deg}_\mathbf{x}(d\n_i)$. Similarly, $e_i^H(E)=d_i^H-1=\mathrm{deg}_\mathbf{N}(d\h_i)$, where this time $d\h_i=\sum_{k=1}^r\frac{\partial \h_i}{\partial \n_k}\otimes\n_k\in  (S(\VNd)\otimes \VNd)^H$. Since $\frac{\partial \h_i}{\partial \n_k}=0$ whenever $\mathrm{deg}_\mathbf{x}(\n_k)\neq d_i^N$, it follows that \[e_i^G(V)=e_i^N(V)+d_i^N\cdot (d_i^H-1)=e_i^N(V)+d_i^N\cdot e_i^H(E).\] It remains to show that $d_i^Ne_i^H(E)=e_i^G(\VN)$.

The $e_i^G(\VN)$ are known to coincide with the $\mathbf{x}$-degrees of any set of homogeneous generators for $(S(V^*)\otimes\VNd)^G$ as a free $S(V^*)^G$-module. Since $\VNd$ consists of $N$-invariants, \[(S(V^*)\otimes\VNd)^G=((S(V^*)\otimes\VNd)^N)^H=(S(\VNd)\otimes\VNd)^H\simeq S(\VNd)^H\otimes U_H=S(V^*)^G\otimes U_H,\] where again $U_H:=\spn_\CC\{d\h_1,\dots,d\h_r\}$ and the non-trivial isomorphism comes from \Cref{thm:solomon} applied to the reflection representation $\VN$ of $H$. Hence $(S(V^*)\otimes\VNd)^G$ is generated by $d\h_i$ as a free $S(V^*)^G$-module, whence $e_i^G(\VN)=\mathrm{deg}_\mathbf{x}(d\h_i)=d_i^Ne_i^H(\VN)$.\end{proof}

\begin{remark}\label{rem:bigraded_solomon}
The same argument used in the proof of \Cref{thm:numbers} shows more generally: \[(S(V^*)\otimes\bigwedge \VNd)^G=((S(V^*)\otimes \bigwedge \VNd)^N)^H=(S(\VNd)\otimes\bigwedge \VNd)^H\simeq S(V^*)^G\otimes\bigwedge U_H.\]
\end{remark}

\section{Poincar\'e Series and Specializations}\label{sec:poincare}

Our goal in this section is to prove our main result:

{
\renewcommand{\thetheorem}{\ref{thm:main_theorem}}
\begin{theorem}
Let $N\triangleleft \ G$ be reflection groups acting by reflections on $V$, and let $\VN=V/N$. Then
\[\sum_{g \in G} q^{\fix_V g} t^{\fix_{\VN} g} = \prod_{i=1}^r \left(qt+e_i^N(V) t + e_i^G(\VN)\right).\]
\end{theorem}
\addtocounter{theorem}{-1}
}

We refer to the left-hand side of~\Cref{thm:main_theorem} as the \defn{sum side}, and to the right-hand side as the \defn{product side}.
We prove \Cref{thm:main_theorem} by computing the Poincar\'e series for $(S(V^*)\otimes\bigwedge \VNd)^G$ in two different (and standard) ways, keeping track of the supplemental grading afforded by the $\mathbf{x}$-degrees of $N$-invariants in $\VNd=\spn_\CC\{\n_1,\dots,\n_r\}$: one way corresponds to the product side (\Cref{sec:prod}), and the other to the sum side (\Cref{sec:sum}).  A subtlety arises when trying to compute the term-by-term specialization for the sum side, which is dealt with in~\Cref{sec:coseti,sec:cosetii}.

A more general version of~\Cref{thm:main_theorem} that incorporates  Galois twists is stated in~\Cref{sec:galois}. For technical reasons that arise in that generalization, we will define the shifted homogeneous decomposition $\VN^*_m:=\spn_\CC\{\n_i \ | \ \mathrm{deg}_\mathbf{x}(\n_i)=m+1\}$, and similarly
\[({\textstyle\bigwedge^p}\VNd)_m:=\spn_\CC\{ \n_{i_1}\wedge\dots\wedge\n_{i_p}\in{\textstyle\bigwedge^p}\VNd \ | \ {\textstyle\sum_{j=1}^p}\mathrm{deg}_\mathbf{x}(\n_{i_j})=m+p\}.\]
Writing $S(V^*)_\ell$ for the homogeneous component of $\mathbf{x}$-degree $\ell$, we define the Poincar\'e series
\begin{equation}\label{eq:p-series-defn}
\mathcal{P}(x,y,u):=\sum_{\ell,m,p\geq 0}\mathrm{dim}_\CC((S(V^*)_\ell\otimes ({\textstyle\bigwedge^p}\VNd)_m)^G)x^\ell y^m u^p.
\end{equation}
We write $\ENs=\{e_1^N(\Vs),\dots,e_r^N(\Vs)\}$ for the set of fake degrees of $\Vs$ as an $N$-representation.

\subsection{Product Side}
\label{sec:prod}
We first obtain the following product formula for the Poincar\'e series $\mathcal{P}(x,y,u)$ defined in \Cref{eq:p-series-defn} from \Cref{thm:numbers} and \Cref{rem:bigraded_solomon}, since $d\h_i\in S(V^*)_{e_i^G(E)}\otimes \VN^*_{e_i^N(V)}$.

\begin{lemma}
\label{lem:product_side}
$\hilb(x,y,u)=\displaystyle\prod\limits_{i=1}\limits^r\frac{1+x^{e_i^G(\VN)}y^{e_i^N(V)}u}{1-x_i^{d_i^G}}.$
\end{lemma}

\begin{corollary}
\label{cor:prod_side_specialization}
$\lim\limits_{x\to1}\hilb\left(x,x^t,qt(1-x)-1\right)=\displaystyle\prod\limits_{i=1}\limits^r \left(qt+e_i^N(V) t + e_i^G(\VN)\right).$
\end{corollary}

\subsection{Sum Side}
\label{sec:sum}
By taking traces, we now compute a formula for the Poincar\'e series $\mathcal{P}(x,y,u)$ defined in \Cref{eq:p-series-defn} as a sum over elements of $G$. To simplify notation, we denote by $\VN_m$ the homogeneous component of $\VN$ corresponding to the dual of $\VN^*_m$.

\begin{lemma} \label{thm:5.2}
$
\hilb(x,y,u)=
\frac{1}{|G|} \displaystyle\sum\limits_{g \in G} \frac{\prod_{m\in\ENs}\mathrm{det}(1+uy^{m}g|_{\VN_m})}{\det(1-xg|_{V})}.$
\end{lemma}

\begin{proof}
Since $\UNsd\simeq \bigoplus_{m\in\ENs} \VN^*_m$, we have that
$\bigwedge \UNsd\simeq \bigotimes_{m\in\ENs}\bigwedge \VN^*_m$ as $G$-modules. hence, for each $g\in G$, \[\sum_{m,p\geq 0} \mathrm{tr}(g|({\textstyle\bigwedge^p}\UNsd)_{m})y^mu^p= \prod_{m\in\EN}\left(\sum_{p\geq 0}\mathrm{tr}(g|{\textstyle\bigwedge^p} \VN^*_m )y^{pm}u^p\right).\]

For each $m\in\ENs$ we have $\displaystyle\sum\limits_{p\geq 0}\left(\mathrm{tr}(g|{\textstyle\bigwedge^p}\VN^*_m)y^{pm}u^p\right)=\mathrm{det}(1+y^{m}ug|_{\VN^*_{m}}).$
 Therefore,
\begin{align*}
\hilb(x,y,u)&=\left(\sum_{\ell\geq 0}\mathrm{tr}(g|(S(\Vd)_\ell)x^\ell\right)\left(\sum_{m,p\geq 0}\mathrm{tr}(g|({\textstyle \bigwedge^p}\UNsd)_m)y^mu^p\right)\\
&=\frac{\prod_{m\in\ENs}\mathrm{det}(1+uy^{m}g^{-1}|_{\UNs_m})}{\det(1-xg^{-1}|_{V})}.\end{align*}
The result follows after taking the average over $G$ on each side.\qedhere
\end{proof}

The story is not quite so simple as just setting the sum over $G$ from~\Cref{thm:5.2} equal to the product, and then specializing.  The trouble is that in this specialized sum over $G$ from~\Cref{thm:5.2}, each element of $G$ \emph{does not} necessarily contribute the ``correct amount'' specified by the sum side of~\Cref{thm:main_theorem}---in particular, $(ng)|_{\VN}$ often has larger fixed space than $(ng)|_{\V}$, which causes many terms in the term-by-term limit to be zero.  It turns out, as we will now show, that the contributions \emph{are} correct when taken coset-by-coset.

\subsection{Sum Side, Coset-by-Coset I}
\label{sec:cosetii}
Fix some $g \in G$.  We find a product formula for \Cref{thm:5.2} restricted to the coset $gN$. Define \[\hilb_{gN}(x,y,u):=\frac{1}{|N|}\sum_{n \in N} \frac{\prod_{m\in\ENs} \det\left(1+u y^{m} (ng)|_{\VN^*_{m}}\right) }{\det(1-x(ng)|_{V^*})}.\]
Given $g\in G$, we can choose the fundamental $N$-invariants $\n_1,\dots,\n_r\in S(V^*)^G$ to also form a $g$-eigenbasis for $\VNd=\spn_\CC\{\n_1,\dots,\n_r\}$, since this space is $G$-stable and $g$ has finite order.  For $g \in G$, let $\epsilon_1^g,\dots,\epsilon_r^g$ denote the eigenvalues of $g$ on $\VNd$, so that $g\n_i=\epsilon_i^g\n_i$.

\begin{proposition}
\label{prop:twisted_graded}
$\hilb_{gN}(x,y,u) = \displaystyle\prod\limits_{i=1}^r \frac{1+ \epsilon_i^g u y^{e_i^N(V)}}{1-\epsilon_i^g x^{d_i^N}}.$
\end{proposition}

\begin{proof}
First, $\displaystyle\prod\limits_{m\in\ENs}\mathrm{det}(1+uy^{m}(ng)|_{\VN^*_{m}})=\displaystyle\prod\limits_{i=1}\limits^r(1+\epsilon_i^guy^{e_i^N(\Vs)})$ uniformly for any $n\in N$, since $\VN^*_m$ is $N$-invariant. It remains to show that \[{\frac{1}{|N|}\sum_{n\in N}\frac{1}{\mathrm{det}(1-x(ng)|_{\Vd})}=\prod_{i=1}^r\frac{1}{1-\epsilon_i^gx^{d_i^N}}}.\]

Since $S(\VN^*)\simeq\bigotimes_{m\in\mathcal{E}_N}\mathrm{Sym}(\VN^*_m),$ where $\VN^*_m$ denotes the span of fundamental $N$-invariants having $\mathbf{x}$-degree $m+1$ and $\mathrm{Sym}(\VN^*_m)$ denotes its symmetric algebra, we have
\[\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|S(\VN^*)_\ell\bigr)x^\ell = \prod_{m\in\mathcal{E}_N}\Bigl(\sum_{\ell\geq 0}\bigl(\mathrm{tr}(g|\mathrm{Sym}^\ell(\VN^*_m))(x^{m+1})^\ell\bigr), \] where $S(\VN^*)_\ell:=S(\VN^*)\cap S(V^*)_\ell$ and $S(V^*)_\ell$ as before denotes the homogeneous subspace of polynomials of $\mathbf{x}$-degree $\ell$. On the other hand,
\[\prod_{m\in\mathcal{E}_N}\Bigl(\sum_{\ell\geq 0}\mathrm{tr}\bigl(g|\mathrm{Sym}^\ell(\VN^*_{m})\bigr)(x^{m+1})^\ell\Bigr)=\prod_{m\in\mathcal{E}_N}\frac{1}{\mathrm{det}(1-x^{m+1}(g|_{\VN^*_{m}}))}=\prod_{i=1}^r\frac{1}{1-\epsilon_i^gx_i^{d_i^N}},\]
since $d_i^N=e_i^N(V)$. Therefore, $\displaystyle\sum\limits_{\ell\geq 0}\mathrm{tr}\bigl(g|S(\VN^*)_\ell\bigr)x^\ell=\displaystyle\prod\limits_{i=1}^r\frac{1}{1-\epsilon_i^gx_i^{d_i^N}}.$ Since for each $n\in N$ we have $\displaystyle\sum\limits_{\ell\geq 0}\mathrm{tr}\bigl(ng|S(V^*)_\ell\bigr)x^\ell = \frac{1}{\mathrm{det}(1-x(ng|_{V^*}))},$ it remains to show that, for each $\ell\geq 0$, \[\frac{1}{|N|}\sum_{n\in N}\Bigl(\mathrm{tr}\bigl(ng|S(V^*)_\ell\bigr)\Bigr)=\mathrm{tr}\bigl(g|S(\VN^*)_\ell\bigr).\]
To see this, note that the operator $\frac{1}{|N|}\displaystyle\sum\limits_{n\in N}ng=g\cdot\left(\frac{1}{|N|}\displaystyle\sum\limits_{n\in N}n\right)=g\circ\mathrm{pr}^N_\ell,$ where $\mathrm{pr}^N_\ell=\frac{1}{|N|}\sum_{n\in N} n$ is the projection from $S(V^*)_\ell$ onto its $g$-stable subspace $S(V^*)_\ell^N=S(E^*)_\ell$, whence $\mathrm{tr}\bigl((g\circ\mathrm{pr}^N_\ell)|S(V^*)_\ell\bigr)=\mathrm{tr}\bigl(g|S(E^*)_\ell\bigr)$. \end{proof}

\subsection{Sum Side, Coset-by-Coset II}
\label{sec:coseti}
We next specialize some results of \cite{bonnafe2006twisted} to the case when $N$ is a normal reflection subgroup of a complex reflection group $G$. They consider the more general situation when $N$ is translated by an arbitrary element in the normalizer of $N$ in $\mathrm{GL}(V)$.  Continue to fix some $g \in G$.

\begin{proposition}[{\cite{bonnafe2006twisted}}]
\label{thm:twisted_os}
\[\frac{1}{|N|}\displaystyle\sum\limits_{n \in N} \frac{\det(1+u (ng)|_\Vd)}{\det(1-x (ng)|_\Vd)} = \displaystyle\prod\limits_{i=1}^r \frac{1+ \epsilon_i^g u x^{e_i^N(V)}}{1-\epsilon_i^g x^{d_i^N}} \left( = \hilb_{gN}(x,x,u)\right).\]
\end{proposition}

Specializing both sides of~\Cref{thm:twisted_os} to $u=q(1-x)-1$ and then taking the limit $x \to 1$ yields the following simple formula for sums over cosets.

\begin{corollary}[{\cite{bonnafe2006twisted}}]
\label{cor:twisted_os}
\[\displaystyle\sum\limits_{n \in N} q^{\fix_V(ng)} = \left(\displaystyle\prod\limits_{\epsilon^g_i =1} q+e_i^N(V)\right) \left(\displaystyle\prod\limits_{\epsilon_i^g \neq 1} d_i^N\right).\]
\end{corollary}

Using~\Cref{cor:twisted_os}, we obtain the following crucial specialization of \Cref{prop:twisted_graded}, exploiting the fact that \Cref{prop:twisted_graded} gives the series $\hilb_{gN}(x,y,u)$, while \Cref{thm:twisted_os} gives the series $\hilb_{gN}(x,x,u)$.

\begin{corollary}
\label{cor:twisted_graded}
$\lim_{x\to 1} \hilb_{gN}(x,x^t,qt(1-x)-1) = t^{\fix_E g}\sum_{n \in N} q^{\fix_V (ng)}.$
\end{corollary}

\subsection{Proof of Theorem \ref{thm:main_theorem}}  We now prove our main theorem.
\begin{proof}[Proof of~\Cref{thm:main_theorem}]
  Equating the formulas from~\Cref{lem:product_side,thm:5.2} gives
\begin{align}\label{eq:sum_eq_product}
\hilb(x,y,u)=\sum_{g \in G} \frac{\prod_{m\in\ENs}\mathrm{det}(1+uy^{m}g|_{\VN_m})}{\det(1-xg|_{V})} = |G|\prod_{i=1}^r\frac{1+x^{e_i^G(\UNs)}y^{e_i^N(V)}u}{1-x_i^{d_i^G}}. \end{align}

Let $\{g_j\}_{j=1}^{|H|}$ be a set of coset representatives for $N$ in $G$. By~\Cref{cor:twisted_graded}, splitting the sum side of~\Cref{eq:sum_eq_product} into a sum over the cosets of $N$ and specializing gives
\begin{align*}
\lim_{x\to 1} \hilb(x,x^t,qt(1-x)-1)&= \sum_{j=1}^{|H|}\lim_{x\to 1} \hilb_{gN}(x,x^t,qt(1-x)-1) \\ &= \sum_{j=1}^{|H|} t^{\fix_{E} g_j}\sum_{n \in N} q^{\fix_V ng_j} = \sum_{g \in G} q^{\fix_V g} t^{\fix_{E} g}.
\end{align*}

The result now follows from~\Cref{eq:sum_eq_product} by equating this specialization of the sum side with the same specialization of the product side from~\Cref{cor:prod_side_specialization}.
\end{proof}

\section{Classification of Normal Reflection Subgroups}
\label{sec:classification}
In the interest of space, we restrict our classification of normal reflection subgroups to $\mathrm{rank} \geq 3$.  In rank 2, there are two connected posets of imprimitive complex reflection groups ordered by normality: one has maximal element $G_{11}$ and minimal elements $G(4,2,2)$, $G_4$, and $G_{12}$, while the other has maximal element $G_{19}$ and minimal elements $G_{16}$, $G_{20}$, and $G_{22}$.

\begin{theorem}[{\cite[Corollary 2.18]{lehrer2009unitary}}]
For $r\geq 3$, the normal reflection subgroups of $G(ab,b,r)$ are $(C_d)^r$ and  $G(ab,db,r)$ for $d|a$, giving quotients $G(ab,b,r)/(C_d)^r = G((a/d) b,b,r)$ and $G(ab,b,r)/G(ab,db,r) = C_d$.
\end{theorem}

As $G_{26}$ and $G_{28}$ are the only exceptional reflection groups with more than a single orbit of reflections, there are three nontrivial exceptional (that is, not imprimitive) examples of normal reflection subgroups in rank greater than two: $G(3,3,3) \triangleleft G_{26}$, with quotient $G_4$; $G_{25} \triangleleft G_{26}$, with quotient $C_2$;  and $G(2,2,4) \triangleleft G_{28} \simeq W(F_4)$, with quotient $\mathfrak{S}_3$.

\section{Reflexponents}
\label{sec:reflexponents}
Fix $G$ a complex reflection group of rank $r$ with reflection representation $V$.  Call an $r$-dimensional representation $M$ of $G$ \defn{factorizing} if $M$ has dimension $r$ and
\[\sum_{g \in G} q^{\mathrm{fix}_V(g)} t^{\mathrm{fix}_M(g)} = \prod_{i=1}^r \Big(qt+(e_i^G(V)-m_i)t+ m_i\Big),\] for some nonnegative integers $m_1,\ldots,m_r$.  More generally, call a representation $M$ of $G$ of dimension $\dim M \leq r$ factorizing if it is factorizing in the above sense after adding in $r-\dim M$ copies of the trivial representation.

We can now give a uniform explanation for certain ad-hoc identities from~\cite{williams2019reflexponents}.  Let $\mathcal{H}$ be an orbit of reflecting hyperplanes, write $\mathcal{R}_\mathcal{H}$ for the set of reflections fixing some $L \in \mathcal{H}$, and let $N_\mathcal{H} = \left\langle \mathcal{R}_\mathcal{H} \right\rangle$ be the subgroup generated by reflections around hyperplanes in $\mathcal{H}$.  Since these reflections form a conjugacy class in $G$, $N_\mathcal{H}$ is a normal reflection subgroup of $G$. Furthermore: the quotient $G/N_\mathcal{H}$ acts as a reflection group on the $N_\mathcal{H}$-invariants of $V$; and this action gives a $G$-representation $M_\mathcal{H}$ that is factorizing by~\Cref{thm:main_theorem}.

\section{Galois Twists}\label{sec:galois}

Let $V$ be an $r$-dimensional complex vector space and $G\subset\mathrm{GL}(V)$ be a complex reflection group. It is known that $G$ can be realized over $\mathbb{Q}(\zeta_G)$, where $\zeta_G$ is a primitive $|G|$-th root of unity, in the sense that there is a basis for $V$ with respect to which $G\subset \mathrm{GL}_r(\mathbb{Q}(\zeta_G))$. For $\sigma\in\mathrm{Gal}(\mathbb{Q}(\zeta_G)/\mathbb{Q})$, the \defn{Galois twist} $V^\sigma$ is the representation of $G$ on the same underlying vector space $V$ obtained by applying $\sigma$ to the matrix entries of $g\in\mathrm{GL}(\mathbb{Q}(\zeta_G))$. Orlik and Solomon found a beautiful generalization of \Cref{eq:shephard_todd} that takes into account these Galois twists. Below we write $\lambda_1(g),\dots,\lambda_r(g)$ for the eigenvalues of $g\in G$ on $V$.

\begin{theorem}[{\cite{orlik1980unitary}}]
\label{thm:orlik_solomon}
Fix $G$ a reflection group and $\sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta_G)/\mathbb{Q})$.  Then
\[\sum_{g \in G} \left(\prod_{\lambda_i(g) \neq 1} \frac{1-\lambda_i(g)^\sigma}{1-\lambda_i(g)}\right) q^{\fix_V g} = \prod_{i=1}^r \left(q+e_i(V^\sigma) \right).\]
\end{theorem}

\begin{definition}
Let $I^G_+\subset S(V^*)$ be the ideal generated by homogeneous $G$-invariant polynomials of positive degree, and let $\mathcal{C}_G$ be a $G$-stable homogeneous subspace of $S(V^*)$ such that $S(V^*)\simeq I^G_+\oplus \mathcal{C}_G$ as $G$-modules. For $\sigma\in\mathrm{Gal}(\zeta_G)$, define the \defn{Orlik-Solomon space} $U_G^\sigma:=(\mathcal{C}_G\otimes (\V^\sigma)^*)^G=\spn_\CC\{u_1^G,\dots,u_r^G\}$, where the $u_i^G$ are homogeneous with $\mathrm{deg}(u_i^G)=e_i^G(V^\sigma)$ and such that $\bigl(S(V^*)\otimes (V^\sigma)^*\bigr)^G\simeq S(V^*)^G\otimes U_G^\sigma$.
\label{def:osspace}
\end{definition}

 We can now state the general form of our main theorem.

\renewcommand{\thetheorem}{\ref{thm:main_theorem}}
\begin{theorem}
Let $N\triangleleft \ G$ be reflection groups acting by reflections on $V$, and let $\VN=V/N$. Let $\sigma\in\mathrm{Gal}(\mathbb{Q}(\zeta_G)/\mathbb{Q})$ and define the Orlik-Solomon space $U_N^\sigma$ as in \Cref{def:osspace}. Then
\[\sum_{g \in G}\left( \prod_{\lambda_i(g) \neq 1} \frac{1-\lambda_i(g)^\sigma}{1-\lambda_i(g)}\right) q^{\fix_V g} t^{\fix_{E} g} = \prod_{i=1}^r \left(qt+e_i^N(V^\sigma) t + e_i^G((U_N^\sigma)^*)\right).\]
\end{theorem}
\addtocounter{theorem}{-1}

\begin{remark}
The Orlik-Solomon space $U_N^\sigma$ playes the role of $\VN^*$ in this generalization of~\Cref{thm:main_theorem}. When $\sigma=1$, a straightforward argument yields a graded $G$-module homomorphism (of degree $-1$) $\VNd\simeq U_N$. This is why we shifted the obvious $\mathbf{x}$-grading of $\VN^*$ by $-1$ in \Cref{sec:poincare}. However, it is not true in general that $(\VN^\sigma)^*\simeq U_N^\sigma$ when $\sigma\neq 1$.
\end{remark}

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TITLE: Find $F'(x)$ when $F(x)= \int_a^{x^2}f(t)dt$
QUESTION [0 upvotes]: Find $F'(x)$ when $f: \mathbb{R} \to \mathbb{R}$ is continous and $F(x)= \int_a^{x^2}f(t)dt$.
The hint given is to use the chain rule but I cannot see how chain rule can be applied in this case.
Thanks for helping.

REPLY [0 votes]: you can use the more general $d \left(\int_a^b f(t) \, dt \right) = f(b) \, db - f(a) \, da$ to find $$d\int_a^{x^2} f(t) \, dt = f(x^2) d(x^2) = 2xf(x^2) \, dx$$ if $a$ is constant. this gives you $\frac{d}{dx}\int_a^{x^2} f(t) \, dt  = 2xf(x^2).$


TITLE: Wire grid polarizer dimension
QUESTION [3 upvotes]: Why should the space between the wires be less than the wavelength of radiation in order for the polarizer to work efficiently?

REPLY [1 votes]: I will provide a heuristic interpretation of what is happening.
Taking a classical view of the electromagnetic field. For the wire grid polarizer any incident electromagnetic wave's electric field can be decomposed into a component perpendicular to the wires and parallel to the wires. For a wire grid polarizer you are counting on the grid to transmit the perpendicular polarization and reflect or absorb the parallel polarization.
The polarization perpendicular to the wires should be transmitted regardless of the spacing given that the wires themselves have a small enough diameter. This is because currents cannot be excited orthogonal to the wires so this component of the electric field won't interact strongly with the polarizer.
For the component that is polarized parallel to the wires, currents will be excited and the field will interact with the wires. The forward scattered fields of the current will cancel the incident field and the reverse scattered fields will show up as the reflected field. How strongly the field interacts with the polarizer as a whole depends on the spacing of the wires though.
If you consider the gaps in the wires independently then they appear like apertures. With an aperture larger than a wavelength significant diffraction will occur and the parallel polarization will leak through. Increasing the transmitted power of the parallel polarized field. Noting that the transmitted fields power is proportional $T$ where, $T=1-|\Gamma|^2$. If $T$ increases then $|\Gamma|$ has decreased so less of the parallel polarized field has been reflected. Therefore, if you want to efficiently polarize the field the wire grid should have a spacing less than a wavelength to limit the amount of the parallel polarized field that diffracts through.


text:
\begin{document}

\maketitle

\begin{abstract}
A triangulation of a closed $d$-manifold is said to be {$r$-stacked} if it is the boundary of a triangulation of a (homology) $(d+1)$-manifold having no interior faces of dimension $\leq d-r$.

In this abstract,
we introduce several interesting properties of $r$-stacked triangulated manifolds.
\end{abstract}

\section*{Introduction}
Recently, the $r$-stacked property of triangulated spheres has been of interest in the study of face numbers of simplicial complexes.
A triangulated $d$-sphere is said to be {\em $r$-stacked} if it is the boundary of a triangulated (homology) $(d+1)$-ball
having no interior faces of dimension $\leq d-r$.
This $r$-stacked property can be naturally extended to triangulations of manifolds,
and it was found that $r$-stacked triangulated $d$-manifolds have many interesting properties when $r < \frac d 2$.
In this abstract, we summarize known nice properties of $r$-stacked triangulated manifolds.

We first introduce necessary notations.
Let $\Delta$ be a (finite abstract) simplicial complex.
We say that a simplicial complex $\Delta$ is a \emph{triangulation} of a topological space $X$ if its geometric realization is homeomorphic to $X$.
We are interested in triangulations of manifolds, but we actually consider slightly larger class of simplicial complexes.
Fix a field $\kk$.
For a simplicial complex $\Delta$ and its face $F \in \Delta$,
the {\em link of $F$ in $\Delta$} is the simplicial complex
$$\lk_\Delta(F)=\{ G \in \Delta: F \cup G \in \Delta \mbox{ and } F \cap G = \emptyset\}.$$
A simplicial complex $\Delta$ of dimension $d$ is said to be a {\em homology $d$-sphere}
if, for all faces $F \in \Delta$ (including the empty face $\emptyset$),
one has $\beta_i(\lk_\Delta(F))=0$ for $i \ne d-\#F$ and $\beta_{d-\#F}(\lk_\Delta(F))=1$,
where $\beta_i(\Delta)=\dim_\kk \widetilde H_i(\Delta;\kk)$ is the $i$th {\em Betti number} of $\Delta$ (w.r.t.\ $\kk$).
A simplicial complex is said to be {\em pure} if all its facets have the same dimension.
A {\em homology $d$-manifold (without boundary)} is a pure $d$-dimensional simplicial complex
all of whose vertex links are homology spheres.
A pure $d$-dimensional simplicial complex $\Delta$ is said to be a {\em homology $d$-manifold with boundary} if it satisfies
\begin{itemize}
\item[(i)] for all $\emptyset \ne F \in \Delta$,
$\beta_i(\lk_\Delta(F))$ vanishes for $i \ne d - \#F$ and is equal to $0$ or $1$ for $i = d - \#F$.
\item[(ii)] the {\em boundary} $\partial \Delta=\{F \in \Delta: \beta_{d-\#F}(\lk_\Delta(F))=0\} \cup \{ \emptyset \}$ of $\Delta$
is a homology $(d-1)$-manifold without boundary.
\end{itemize}
A {\em homology $d$-ball} is an acyclic homology $d$-manifold such that $\partial \Delta$ is a homology $(d-1)$-sphere.
A triangulation of a topological manifold is an example of a homology manifold.
Also, a triangulation of a $d$-sphere (resp.\ $d$-ball) is a homology $d$-sphere (resp.\ $d$-ball).

\section{Examples of stacked polytopes and manifolds}

We say that a homology $d$-manifold $\Delta$ with boundary is {\em $r$-stacked} if it has no interior faces (namely, faces which are not in $\partial \Delta$) of dimension $< d-r$.
Also, a homology manifold without boundary is said to be {\em $r$-stacked} if it is the boundary of an $r$-stacked homology manifold with boundary.
Although we defined $r$-stacked manifolds with boundary, we are mainly interested in homology manifolds without boundary.
So we often omit `without boundary' when it is clear.
We also assume that all homology manifolds are connected.
Here we give two examples of $1$-stacked triangulated manifolds.

\begin{example}
A $d$-polytope is said to be {\em stacked} if it can be obtained from a simplex by adding a pyramid over a facet repeatedly. It is not hard to see that the boundary complexes of stacked polytopes are exactly
$1$-stacked spheres.
\medskip

\begin{center}
\includegraphics[width=100mm]{stacked}\\
construction of stacked polytopes
\end{center}

An important property of stacked polytopes is that it gives a lower bound of face numbers of simplicial polytopes for a fixed dimension and a fixed number of vertices.
See \cite{Ba1,Ba2,Ka}.
\end{example}

\begin{example}[K\"uhnel--Lassmann construction \cite{Ku,KL}]
\label{ex1}
Let $d, n$ be integers with $ n \geq 2d-1$ and let
$K_{d,n}$ be the simplicial complex on $[n]=\{1,2,\dots,n\}$ generated by the facets
$$\{ \{i,i+1,\dots,i+d-1\}: i=1,2,\dots,n\},$$
where $i+k$ means $i+k-n$ if $i+k >n$.
Then $K_{d,n}$ is a combinatorial manifold such that $\partial K_{d,n}$
triangulates either $S^{d-3} \times S^1$ or a non-orientable $S^{d-3}$-bundle over $S^1$ (generalized Klein bottle).
The interior faces of $K_{d,n}$ are faces that contain a face of the form $\{i,i+1,\dots,i+d-2\}$ (mod $n$),
so $K_{d,n}$ and $\partial K_{d,n}$ are $1$-stacked triangulations.

An important property of these triangulations is that they give vertex minimal triangulations of a sphere-bundle over a circle when $n=2d-1$.
For example, $\partial K_{4,7}$ is Cs\'asz\'ar's vertex minimal $7$-vertex triangulation of $S^1 \times S^1$.
We will explain later why the stackedness property appears in minimal triangulations.
\end{example}

\section{Properties of stacked triangulations}

In this section, we list known nice properties of $r$-stacked triangulated manifolds.

\subsection{Construction of $r$-stacked triangulated manifolds with boundary}
One of the most interesting properties of stacked triangulations is that
one can construct the unique $r$-stacked triangulated $d$-manifold with boundary from an $r$-stacked triangulated $(d-1)$-manifold without boundary when $r$ is small enough.
For a simplicial complex $\Delta$ on $V$,
define the simplicial complex
$$\Delta(r)=\{ F \subset V: \mbox{ any subset $G \subset F$ with $\# G \leq r+1$ is contained in }\Delta\}.$$
The following result was proved by Bagchi and Datta for simplicial polytopes and proved in \cite{MN1} for homology spheres. See \cite[Theorem 2.3]{MN1}.

\begin{theorem}
Let $1 \leq r \leq \frac {d} 2$ and let $\Delta$ be an $r$-stacked homology $d$-sphere.
If $\Sigma$ is an $r$-stacked homology $(d+1)$-ball with $\partial \Sigma=\Delta$ then $\Sigma=\Delta(r)$.
\end{theorem}

For triangulated manifolds, a slightly weaker statement was proved in \cite[Theorem 2.20]{BD} and in \cite[Theorem 4.2]{MN2} independently.

\begin{theorem}
\label{4.3}
Let $1 \leq r \leq \frac {d-1} 2$ and
$\Delta$ an $r$-stacked homology $d$-manifold without boundary.
If $\Sigma$ is an $r$-stacked homology $(d+1)$-manifold with $\partial \Sigma=\Delta$ then $\Sigma=\Delta(r+1)$.
\end{theorem}

Note that these theorems also say that the $r$-stacked homology $d$-manifold $\Sigma$ with $\partial \Sigma=\Delta$ is unique under these assumptions.
A more interesting aspect of the above theorems is that if one happens to know that $\Delta$ is an $r$-stacked triangulated $d$-manifold, then the operation $\Delta \to \Delta(r+1)$ gives rise to a homology $(d+1)$-manifold.

\subsection{Local criterion}
We say that a homology $d$-manifold without boundary is {\em locally $r$-stacked}
if all its vertex links are $r$-stacked.
This notion was studies by Walkup \cite{Wa} and Kalai \cite{Ka} when $r=1$
and the class of locally $1$-stacked triangulated manifolds are known as Walkup's class.
The locally $r$-stacked property with $r >1$ also appeared in \cite{Ef}.
It is easy to see that $r$-stacked manifolds are locally $r$-stacked.
The next statement proved by Bagchi--Datta \cite[Theorem 2.19]{BD} and in \cite[Theorem 4.6]{MN2} shows that they are equivalent when $r$ is small.

\begin{theorem}
\label{4.4}
Let $1 \leq r< \frac {d-1} 2$.
Then a homology $d$-manifold without boundary is $r$-stacked if and only if it is locally $r$-stacked.
\end{theorem}

Theorem \ref{4.4} fails for $r= \frac {d-1} 2$.
For example, it is known that the boundary complex of a cyclic $4$-polytope is locally $1$-stacked but not $1$-stacked.
This implies that, for $3$-manifolds, we cannot study the $1$-stacked property by just considering vertex links.

\subsection{Stacked property and face numbers}
The $r$-stacked property was considered in the study of face numbers
of simplicial polytopes.
Let $\Delta$ be a simplicial complex of dimension $d-1$.
Let $f_i(\Delta)$ be the number of $i$-dimensional faces of $\Delta$.
The vector $f(\Delta)=(f_{-1}(\Delta),\dots,f_{d-1}(\Delta))$ is called the {\em $f$-vector} of $\Delta$.
Also, the {\em $h$-vector} $h(\Delta)=(h_0(\Delta),h_1(\Delta),\dots,h_d(\Delta))$ of $\Delta$ is
defined by the relation
$$
\sum_{i=0}^d h_i(\Delta) t^i = \sum_{i=0}^d f_{i-1}(\Delta) t^i (1-t)^{d-i}.
$$
We say that a simplicial $d$-polytope is $r$-stacked if it can be (geometrically) triangulated without introducing faces of dimension $d<r$.
Note that this definition coincides with the $r$-stackedness of its boundary complex when $r \leq \frac {d-1} 2$.
For simplicial polytopes, the following characterization of the $r$-stacked property is known.

\begin{theorem}[Generalized lower bound theorem]
\label{glbc}
Let $P$ be a simplicial $d$-polytope.
\begin{itemize}
\item[(i)] (Stanley \cite{St}) $h_r(P) \leq h_{r+1}$ for all $r \leq \frac {d-1} 2$.
\item[(ii)] (McMullen--Walkup \cite{MW}, Murai--Nevo \cite{MN1}) for $r < \frac {d-1} 2$,
one has $h_r(P)= h_{r+1}(P)$ if and only if $P$ is $r$-stacked.
\end{itemize}
\end{theorem}

The special case of the above theorem when $r=1$ is known to be equivalent to Barnette's Lower bound theorem \cite{Ba1,Ba2}.
It is a natural question to ask if Theorem \ref{glbc} can be extended to triangulations of spheres.
Unfortunately, this seems to be a difficult problem since the proof of the theorem depends on the hard Lefschetz theorem for projective toric varieties.
On the other hand, the following conjecture is suggested.

\begin{conjecture}[{Generalized lower bound conjecture for triangulated manifolds}]
\label{5.1}
Let $\Delta$ be a connected triangulated $(d-1)$-manifold without boundary. Then
\begin{itemize}
\item[(i)] (Kalai) $h_{r+1}(\Delta) \geq h_r(\Delta)+ \binom {d+1} {r+1} \sum_{j=1}^{r+1} (-1)^{r+1-j} \beta_{j-1}(\Delta)$ for $r=1,2,\dots,\lfloor \frac d 2 \rfloor$.
\item[(ii)] (Bagchi--Datta) if equality holds for some $r< \frac d 2 -1$ in (i)
then $\Delta$ is locally $r$-stacked.
\end{itemize}
\end{conjecture}

The case $r=1$ was solved by Novik--Swartz \cite{NS1}.
The conjecture is also known to hold when $\Delta$ is orientable and all the vertex links of $\Delta$ are simplicial polytopes \cite{NS2,MN2}.

\subsection{Vertex minimal triangulations}
Stacked triangulations sometimes appear as minimal triangulations.
Here we explain a reason.
Let $\Delta$ be a homology $d$-manifold without boundary.
It was conjectured by K\"uhnel and proved by Novik--Swartz that
the inequality
\begin{eqnarray}
\label{1-1}
{{d+2} \choose 2 }\beta_1(\Delta)
\leq {f_0(\Delta)-d-1 \choose 2}
\end{eqnarray}
holds.
A triangulated $d$-manifold satisfying \eqref{1-1} with equality is called a {\em tight-neighborly} triangulation.
Clearly a tight-neighborly triangulation must be a vertex minimal triangulation.
The following result follows from the works of Novik--Swartz \cite[Theorem 5.3]{NS1}, Bagchi--Datta \cite[Theorem 2.24]{BD} and Burton--Datta--Singh--Spreer \cite[Theorem 1.2]{BDSS}.

\begin{theorem}
Tight neighborly triangulated manifolds are $1$-stacked.
\end{theorem}

An $r$-stacked analogue of the above theorem is expected to be true.
It is known that all the vertex minimal triangulations of $S^2 \times S^3$ listed in \cite{Lu} are $2$-stacked. It would be interesting to find a result that explains this fact.

\subsection{Topological restriction}
It was proved by Kalai \cite{Ka} that if $\Delta$ is a locally $1$-stacked triangulated manifold then it
is obtained from the boundary of a simplex by repeating the following three operations: (i) stellar subdivision of a facet (i) combinatorial handle addition (iii) taking connected sums of objects obtained from the first two operations.
This gives a strong restriction to topological types of $1$-stacked triangulated $d$-manifolds with $d \geq 4$.
A natural question is ``what can be said about the topological type of $r$-stacked triangulated $d$-manifolds with $r < \frac {d-1} 2$?''
At the moment, we only have information on homology groups. See \cite[Remark 2.22]{BD} and \cite[Theorem 4.4]{MN2}.
Also, we are not sure that, if $\Delta$ is an $r$-stacked homology $d$-manifold with $r \leq \frac {d-1} 2$, then the topological type of $\Delta(r+1)$ in Theorem~2.2 can be determined from the topological type of $\Delta$.

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TITLE: $df_p$ not surjective
QUESTION [0 upvotes]: From Differential Geometry of Curves and Surfaces By Docarmo.
DEFINITION 2. Given a differentiable map $F: U \subset R^n \to R^m$ defined in
an open set U of $R^n$ we say that $p \in U$ is a critical point of F if the differential
$dF_p: R^n \to R^m$ is not a surjective (or onto) mapping. The image $F(p) \in R ^m$
of a critical point is called a critical value of $F$. A point of $R^m$ which is not a
critical value is called a regular value of $F$.
The terminology is evidently motivated by the particular case in which
$F: U \subset R  \to R $ is a real-valued function of a real variable. A point $x_o \in U$
is critical if $f'(x_o ) = 0$, that is, if the differential $df_{x_o}$ carries all the vectors in
R to the zero vector . Notice that any point $a \not\in f(U)$ is trivially a
regular value of $f$.
If $f: U \subset R^3  \to R $ is a differentiable function, then $df_p$ applied to the
vector $(1, 0, 0)$ is obtained by calculating the tangent vector at $f(p)$ to the
curve $x \to f(x,y_o,z_o)$
It follows that
$df_p(1,0,0)=\frac{\partial f}{\partial x} (x_o,y_o,z_o)=f_x$
and analogousaly that
$df_p(0,1,0)=f_y$,and $df_p(0,0,1)=f_z$
We conclude that the matrix of $df_p$ in the basis $(1,0,0), (0, 1,0), (0, 0, 1)$ is
given by  $df_p=(f_x,f_y,f_z)$.
Note, in this case, that to say that $df_p$ is not surjective is equivalent to
saying that $f_x = f_y = f_z = 0$ at p. Hence, $a\in f(U)$ is a regular value of
$f: U \subset R^3 \to R$ if and only if $f_x,f_y$, and $f_z$ do not vanish simultaneously at
any point in the inverse image
$f^{-1}(a)$ = {$(x,y, z) \in U: f(x,y, z) = a$}.
In last pharagraph they have written  "$df_p$ is not surjective is equivalent to
saying that $f_x = f_y = f_z = 0$ at p".
My question is Why $df_p$ is not surjective is equivalent to
saying that $f_x = f_y = f_z = 0$ at p.Can someone explain me.Thanks in advance!

REPLY [2 votes]: $df_p$ is represented by the $1 \times 3$-matrix $(f_x,f_y,f_z)$. In order that $df_p$ be surjective the rank of this matrix at $p$ must be $1$. This is equivalent to $(f_x,f_y,f_z) \ne (0,0,0)$ in $p$.