text:
\section{Introduction}

A framework of Finite Element Systems (FES) has been developed, in \cite{Chr08M3AS,Chr09AWM,ChrMunOwr11,ChrRap15} to construct mixed finite elements, generalizing those of \cite{RavTho77,Ned80,BreDouMar85}. Cast in the language of differential forms, following \cite{Hip99,Chr07NM,ArnFalWin06,ArnFalWin10}, it allows for polyhedral meshes and non-polynomial differential forms.

A minimal compatible FES (mcFES) has three key properties: it is compatible, it contains certain prescribed functions, and it has the smallest dimension among all possible finite element systems with these properties.
Each property relates to a practical computational purpose: compatibility is used for the design of provably stable mixed methods, function containment is used for estimation of approximation error, and dimension minimality is used to maximize computational efficiency.

In this paper, we do the following:
\begin{itemize}
\item In Section \ref{sec:back}, we recall the main concepts of FES. We also illustrate them with some comments on the mixed Virtual Element Method \cite{BeiEtAl14}.

\item In Section \ref{sec:con}, we show how the dimension of a mcFES can be computed in terms of certain cohomology groups and how a mcFES that contains a given set of functions can be constructed, within a larger compatible FES. These results were announced, mostly in French and without proofs, in \cite{Chr10CRAS}.

\item In Section \ref{sec:ex}, we apply this analysis and construction process to show that:
\begin{enumerate}[label=(\roman*)]
\item the trimmed polynomial spaces $\mathcal{P}_r^-\Lambda^k$, defined in \cite{Hip99},  form a mcFES containing $\mathcal{P}_{r-1}\Lambda^k$ on simplices;
\item the serendipity spaces $\mathcal{S}_r\Lambda^k$, defined in \cite{ArnAwa14}, form a mcFES containing $\mathcal{P}_{r-k}\Lambda^k$ on hypercubes;
\item the TNT elements, defined in \cite{CocQiu14}, form a mcFES containing $\mathcal{Q}_r\Lambda^k$ on hypercubes;
\item the dimension of a mcFES on hypercubes that contains $\mathcal{P}_r\Lambda^k$ can be given in closed form.
\end{enumerate}
\end{itemize}

\section{Background on finite element systems\label{sec:back}}

The notion of finite element systems is presented in detail in~\cite{Chr08M3AS,Chr09AWM,ChrMunOwr11,ChrRap15} so we will only recall the main definitions and key results that are most relevant to this paper.

A \emph{cellular complex} is a collection $\mathcal{T}$ of cells $T$, where each cell is either a singleton or homeomorphic to the unit ball of some Euclidean space. The cells are subject to some gluing conditions that make them into a regular CW complex, but where topologists use continuous maps for these conditions, we require the maps to be at least Lipschitz continuous. We stress that in the collection $\mathcal{T}$, cells of all dimensions are included. Typically, they are taken to be flat-faced polytopes, all sitting in some given $\mathbb{R}^n$, but this is not necessary for the theory to hold.
See \cite[Definition 2.1]{ChrRap15} for details.

Given a cellular complex $\mathcal{T}$, a \emph{finite element system} is defined as in \cite[Definition 2.2]{ChrRap15}.
If $T$ is a cell in a cellular complex $\mathcal{T}$,  we denote by $\mathfrak{E}^k(T)$ the set of $k$-forms on $T$ with the following property:  for any $T' \in \mathcal{T}$ included in $T$ (including $T$ itself),  the pullback of the form to $T'$ is in $\mathrm{L}^2(T')$ and has its exterior derivative in $\mathrm{L}^2(T')$.

An \emph{element system} on $\mathcal{T}$, is a family of  closed subspaces $E^k(T)  \subseteq \mathfrak{E}^k(T)$, one for each $k\in \mathbb{N}$ and each $T \in \mathcal{T}$, subject to the following requirements:
\begin{itemize}
\item The exterior derivative should induce maps:
\begin{equation}
\mathrm{d}: E^k(T) \to E^{k+1}(T).
\end{equation}
\item If $T' \subseteq T$ are two cells in $\mathcal{T}$ and  $i_{TT'}: T' \to T$ denotes the canonical injection, then pullback by $i_{TT'}$ should induce a map:
\begin{equation}
i_{TT'}^\star: E^k(T) \to E^k(T').
\end{equation}
\end{itemize}
For instance the spaces $\mathfrak{E}^k(T)$ constitute an element system. A finite element system (FES) is one in which all the spaces are finite dimensional.

We define $E^k(\mathcal{T})$ as follows :
\begin{equation}
E^k(\mathcal{T})= \{u \in \bigoplus_{T \in \mathcal{T}} E^k(T) \ : \  \forall T, T' \in \mathcal{T} \quad  T' \subseteq T \Rightarrow u_{T}|_{T'} = u_{T'}\}.
\end{equation}
In this definition, which can be interpreted as encoding a continuity property of differential forms,  $u_{T}|_{T'}$ denotes the pullback of $u_T$ to $T'$ by the inclusion map.

Not all finite element systems yield good spaces $E^k(\mathcal{T})$. As in \cite[Definition 2.3]{ChrRap15}, we consider the following two conditions on an element system $E$ on a cellular complex $\mathcal{T}$:
\begin{itemize}
\item \emph{Extensions.} For each $T\in \mathcal{T}$ and $k\in \mathbb{N}$, the restriction operator (pullback to the boundary) $E^k(T) \to E^k(\partial T)$ is onto. The kernel of this map is denoted $E^k_0(T)$.
\item \emph{Local exactness.} The following sequence is exact for each $T\in \mathcal{T}$:
\begin{equation}\label{eq:coh}
\xymatrix{
0 \ar[r] & \mathbb{R} \ar[r] & E^0(T) \ar[r]^-\mathrm{d} & E^1(T) \ar[r]^-\mathrm{d} & \cdots \ar[r]^-\mathrm{d} &  E^{\dim T}(T) \ar[r] & 0.
}
\end{equation}
The second arrow sends an element of $ \mathbb{R}$ to the constant function on $T$ taking this value.
\end{itemize}
We will say that an element system \emph{admits extensions} if the first condition  holds, is \emph{locally exact} if the second condition holds and is \emph{compatible} if both hold.

In \cite[Proposition 2.6]{ChrRap15}, it is shown that for finite element systems admitting extensions, local exactness is equivalent to the combination of the following properties:
\begin{itemize}
\item For each $T \in \mathcal{T}$,  $E^0(T)$ contains the constant functions.
\item For each $T\in \mathcal{T}$, the following sequence (with boundary condition) is exact:
\begin{equation} \label{eq:coh0}
\xymatrix{
0 \ar[r] & E^0_0(T) \ar[r] & E^1_0(T) \ar[r] & \cdots \ar[r] & E^{\dim T}(T) \ar[r] & \mathbb{R} \ar[r] & 0.
}
\end{equation}
The second to last arrow is integration.
\end{itemize}

We take it for granted that a good finite element method consists of defining a compatible finite element system that contains certain prescribed functions and with which one can somehow compute.
That we want a \emph{compatible} finite element system is justified for instance in \cite[Section 2.4]{ChrRap15},  where the underlying conditions are related to the existence of degrees of freedom and commuting interpolators, the main tools of analysis of mixed methods. Typically, one would also want the finite element system to contain polynomials of a certain degree, to ensure corresponding best approximation properties in Sobolev spaces. We also remark that in some cases it is more desirable to contain certain exponentials \cite{Chr13FoCM,ChrHalSor14}.

As was already the case in \cite{Chr08M3AS}, one of the basic tools of our constructions is harmonic extension:
\begin{proposition}\label{prop:intext} Suppose $E$ is a FES where each $E^k(T)$ is equipped with a scalar product, denoted $a$. We suppose that $T$ is a cell such that  (\ref{eq:coh0}) is exact.
For each  $\alpha \in \mathbb{R}$ there is a unique $u$ in $E^{\dim T}(T)$ such that:
\begin{equation}
\int_T u = \alpha  \quad \textrm{ and } \quad  \forall v \in E^{\dim T -1}_0(T)\quad  a(u, \mathrm{d} v ) = 0.
\end{equation}
Fix $k<\dim T$. Any $u \in E^k(\partial T)$ that has an extension in $E^k(T)$, has a unique extension in $ E^k(T)$ such that:
\begin{equation}\label{eq:harm}
\forall v \in E^k_0(T) \quad a(\mathrm{d} u, \mathrm{d} v)=0
 \quad \textrm{ and } \quad
\forall v \in E^{k-1}_0(T) \quad a(u, \mathrm{d} v)=0.
\end{equation}
\end{proposition}
An element $u$ of $E^k(T)$ such that (\ref{eq:harm}) holds will be called \textit{$E$-harmonic}.  The proposition above asserts that elements having an extension have a unique $E$-harmonic extension. Orthogonality with respect to $a$ will be denoted $\perp$, so that (\ref{eq:harm}) can also be written:
\begin{equation}
\mathrm{d} u \perp \mathrm{d}  E^k_0(T)  \quad \textrm{ and } \quad  u \perp \mathrm{d}  E^{k-1}_0(T).
\end{equation}

We are also interested in dimension counts, so we recall \cite[Proposition 2.1]{ChrRap15}:
\newpage
\begin{proposition}\label{prop:extdim}
Let $E$ be a FES on a cellular complex $\mathcal{T}$. Then:
\begin{itemize}
\item We have:
\begin{equation}\label{eq:bounddim}
\dim E^k (\mathcal{T}) \leq \sum_{ T\in \mathcal{T}}\dim E^k_0(T).
\end{equation}
\item Equality holds in (\ref{eq:bounddim})  if and only if $E$ admits extensions for $k$-forms on each $T \in \mathcal{T}$.
\end{itemize}
\end{proposition}

Finally we recall some results on tensor products, following \cite{Chr09AWM,ChrMunOwr11}.

Suppose $\mathcal{U}$ and $\mathcal{V}$ are cellular complexes, equipped with FES systems $B$ and $C$, respectively. Consider the product cellular complex:
\begin{equation}
\mathcal{T} = \{ U \times V \ : \ U \in \mathcal{U} \textrm{ and } V \in \mathcal{V} \}.
\end{equation}
For $U \in \mathcal{U}$ and $V\in \mathcal{V}$, define spaces $A^k(U\times V)$ by:
\begin{equation}\label{eq:tensordef}
A^k(U\times V) = \{ p_U^\star u \wedge p_V^\star v \ : \ u \in B^l(U), \quad v \in C^{k-l}(V), \quad 0 \leq l \leq k \},
\end{equation}
where $p_U: U\times V \to U$ and $p_V : U\times V \to V$ are the canonical projections.

One checks that  $A$ is a FES on $\mathcal{T}$, and we call it the tensor product of $B$ and $C$. This name is motivated by the fact that the formula:
\begin{equation}
u \otimes v = p_U^\star u \wedge p_V^\star v
\end{equation}
indeed defines a tensor product.  Then (\ref{eq:tensordef}) can be written in compact form:
\begin{equation}
A^{\scriptscriptstyle \bullet}(U\times V) = B^{\scriptscriptstyle \bullet}(U) \otimes C^{\scriptscriptstyle \bullet}(V).
\end{equation}
One recognizes a tensor product of graded spaces.

The important result is:
\begin{proposition}\label{prop:tensorfes}

The tensor product construction satisfies:
\begin{itemize}
\item If $B$ and $C$ admit extensions, then so does $A$.
\item If $B$ and $C$ are locally exact, then so is $A$.
\item If $B$ and $C$ are compatible, then so is $A$.
\end{itemize}
\end{proposition}
These facts were proved in \cite{Chr09AWM} and with some simplifications in \cite{ChrMunOwr11}. It is possible to simplify the proof of the first bullet point further by using the only if part of Proposition \ref{prop:extdim}.

\subsection{Comments on virtual element methods}

The Virtual Element Method is presented for scalar function spaces in \cite{BeiEtAl13}. It has recently been extended to the mixed setting, producing $\mathrm{H}(\curl)$ and $\mathrm{H}(\div)$ conforming spaces \cite{BeiEtAl14}. As we see it, the mixed virtual element method does essentially two things: it defines a finite element system and it provides a way of computing with it, that avoids reconstructing all the basis functions from the degrees of freedom. As an illustration of the methods of FES, we provide some details on the first point. The second point however, we leave open.

The constructions of \cite{Chr08M3AS} were based on a notion of harmonic extension for differential forms (both continuous and discrete). Canonical degrees of freedom for compatible FES can be defined, as in projection based interpolation, by non-homogeneous harmonic extensions -- see \cite[Proposition 3.23]{Chr09AWM} and \cite[Proposition 5.44]{ChrMunOwr11}. As we shall see, the mixed VEM spaces are defined by a similar technique.

\begin{lemma}\label{lem:zdef}
Let $T$ be a cell of dimension $n$.
\begin{itemize}
\item Fix $k <n$. Choose $f\in \mathrm{L}^2 \Lambda^k(T)$ and $g\in \mathrm{L}^2 \Lambda^{k-1} (T)$. The system, with unknown $u \in \mathfrak{E}^k_0(T)$:
\begin{align}
\mathrm{d}^\star \mathrm{d} u & = f,\\
\mathrm{d}^\star  u & = g.
\end{align}
has a solution if and only if $\mathrm{d}^\star f = 0$ and $\mathrm{d}^\star g = 0$. In this case the solution is unique.
\item The case $k = n$. Fix $f \in \mathbb{R}$ and $g\in \mathrm{L}^2 \Lambda^{k-1} (T)$. The system, with unknown $u \in \mathfrak{E}^k(T)$:
\begin{align}
\int u & = f,\\
\mathrm{d}^\star  u & = g,
\end{align}
has a solution if and only if $\mathrm{d}^\star g = 0$. In this case the solution is unique.
\end{itemize}
\end{lemma}
\begin{proof}
Left to the reader.
\end{proof}

Consider now a cellular complex $\mathcal{T}$. We suppose we have, for each integer $k$ and cell $T \in \mathcal{T}$, with $k < \dim T$, a finite dimensional subspace $Z^k(T)$ of $\mathrm{L}^2 \Lambda^k(T)$, whose elements $g$ satisfy $\mathrm{d}^\star g = 0$. For $k = \dim T$, we put $Z^k(T) = 0$ and note that for $u \in \mathrm{L}^2\Lambda^k(T)$ we have $ \mathrm{d}^\star \mathrm{d} u = 0$. We define:
\begin{equation}
A^k(T) = \{ u \in \mathfrak{E}^k(T) \ : \ \forall~T' \lhd} % {\vartriangleleft T  \quad \mathrm{d}^\star \mathrm{d} u|_{T'} \in Z^k(T') \textrm{ and } \mathrm{d}^\star u|_{T'} \in Z^{k-1}(T') \},
\end{equation}
where $T'\lhd} % {\vartriangleleft T$ indicates that $T'$ is a face of $T$.
\begin{proposition}
The above defined spaces $A^k(T)$ constitute a compatible finite element system and
for $k < \dim T$ :
\begin{equation}
\dim A^k_0(T) = \dim Z^k(T) + \dim Z^{k-1}(T),
\end{equation}
whereas for $k = \dim T$:
\begin{equation}
\dim A^k_0(T) = 1 + \dim Z^{k-1}(T).
\end{equation}
\end{proposition}
\begin{proof}
The definition of $A^k(T)$ ensures that it is an element system.
The technique of harmonic extension shows that $A$ has the extension property.
The local exactness in the form (\ref{eq:coh0}) is also trivial to check.
Thus $A$ defines a compatible element system.
The dimension counts follow from Lemma \ref{lem:zdef}.
\end{proof}

There are several variants of mixed VEM, but the main one seems to correspond to the following choice of $Z$. We suppose that each cell is flat, so that polynomials are well defined objects. Fix $r \geq 1$ and put, for $k < \dim T$.
\begin{equation}
Z^k(T) = \{f \in \calP_{r-1} \Lambda^k(T) \ : \ \mathrm{d}^\star f = 0 \}.
\end{equation}
Then $A$, defined as above, is a compatible FES by the preceding proposition, and $A^k(T)$ contains polynomial $k$-forms of degree $r$.
There is a notion of degrees of freedom for FES. Since it is rather intuitive we omit the definition; details can be found in \cite[Section 2.4]{ChrRap15}.

\begin{proposition}
For any integer $k$ and cell $T$ of dimension $n$, consider the linear forms, defined on $k$-forms by:
\begin{equation}
u \mapsto \int u \wedge v,
\end{equation}
for some $v \in \calP^-_r \Lambda^{n-k}(T)$.

These linear forms constitute unisolvent degrees of freedom on the FES $A$.
\end{proposition}

\begin{proof}
Let $ {\scriptscriptstyle \bigstar}$ denote the Hodge star operator.
We notice that $u \in \calP_{r-1} \Lambda^k(T)$ iff $ {\scriptscriptstyle \bigstar} u \in \calP_{r-1} \Lambda^{n-k}(T)$. Moreover, for such $u$ we have:
\begin{equation}
\mathrm{d}^\star u = 0 \iff \mathrm{d}  {\scriptscriptstyle \bigstar} u = 0.
\end{equation}

Therefore, for $k < \dim T = n$ (recalling \cite[Theorem 5.4]{ArnFalWin10}):
\begin{align}
\dim Z^k(T) &= \dim \{ u \in \calP_{r-1} \Lambda^{n-k}(T) \ : \ \mathrm{d} u = 0\},\\
& = \dim \{ u \in \calP^-_{r} \Lambda^{n-k}(T) \ : \ \mathrm{d} u = 0\}.
\end{align}
We also have:
\begin{align}
\dim Z^{k-1}(T) & = \dim \{ u \in \calP_{r-1} \Lambda^{n-k+1}(T) \ : \ \mathrm{d} u = 0\},\\
& = \dim \{ \mathrm{d} v : v \in \calP^-_{r} \Lambda^{n-k}(T)\}.
\end{align}
It follows that:
\begin{equation}
\dim Z^k(T) + \dim Z^{k-1}(T) = \dim \calP^-_{r} \Lambda^{n-k}(T).
\end{equation}
In the case $ k= \dim T$ we may also check:
\begin{equation}
1 + \dim Z^{k-1}(T) = \dim \calP^-_{r} \Lambda^{n-k}(T).
\end{equation}
Therefore:
\begin{equation}
\dim A^k_0(T) = \dim \calP^-_{r} \Lambda^{n-k}(T).
\end{equation}
Suppose now that $u \in A^k_0(T)$ satisfies:
\begin{equation}
\forall v \in \calP^-_{r} \Lambda^{n-k}(T) \quad \int u \wedge v = 0.
\end{equation}
We have $\mathrm{d}^\star \mathrm{d} u \in \calP_{r-1} \Lambda^k(T)$, hence $\mathrm{d}  {\scriptscriptstyle \bigstar} \mathrm{d} u \in \calP_{r-1} \Lambda^{n-k}(T) \subseteq \calP^-_{r} \Lambda^{n-k}(T)$ . Hence:
\begin{equation}
0 = \int u \wedge \mathrm{d}  {\scriptscriptstyle \bigstar} \mathrm{d} u = \pm \int \mathrm{d} u \wedge  {\scriptscriptstyle \bigstar} \mathrm{d} u = \pm \int |\mathrm{d} u|^2.
\end{equation}
Therefore $\mathrm{d} u = 0$.

Let $\mathfrak{h}$ denote the standard homotopy operator used to prove the Poincar\'e lemma, as in e.g.\ \cite{Tay96I}. Recall that $\mathfrak{h}$, which is proportional to $\kappa$ on homogeneous polynomial differential forms, maps $\calP^-_r \Lambda^{n-k+1}(T)$ to $\calP^-_r \Lambda^{n-k}(T)$.
We have $\mathrm{d}^\star u \in \calP_{r-1} \Lambda^{k-1}(T)$, so that $ \mathrm{d}  {\scriptscriptstyle \bigstar} u \in \calP_{r-1} \Lambda^{n-k+1}(T)$ and hence $ \mathfrak{h} \mathrm{d}  {\scriptscriptstyle \bigstar} u \in \calP^-_{r} \Lambda^{n-k}(T)$. We write:
\begin{align}
0 = \int u \wedge  \mathfrak{h} \mathrm{d}  {\scriptscriptstyle \bigstar} u & =  - \int u \wedge \mathrm{d} \mathfrak{h}  {\scriptscriptstyle \bigstar} u  + \int u \wedge  {\scriptscriptstyle \bigstar} u ,\\
 & =  \pm \int \mathrm{d} u \wedge \mathfrak{h}  {\scriptscriptstyle \bigstar} u  + \int |u|^2=   \int |u|^2.
\end{align}
Hence $u = 0$.

It follows that the integrated wedge product is an invertible bilinear form on the product $A^k_0(T) \times \calP^-_r \Lambda^{n-k}(T)$. This concludes the proof.
\end{proof}

\section{Construction of minimal finite element systems\label{sec:con}}

The material presented in this section is an expanded version in English of material that appeared previously, in French and without proofs, in~\cite{Chr10CRAS}.
Given any sequence
\[
\xymatrix{
\cdots \ar[r]^-\mathrm{d} & X^{k-1}(T) \ar[r]^-\mathrm{d} & X^k(T) \ar[r]^-\mathrm{d} & X^{k+1}(T) \ar[r]^-\mathrm{d} & \cdots
}
\]
satisfying $\mathrm{d} \circ \mathrm{d}=0$, we use the notation $\ker \mathrm{d} | X^k$ to denote the kernel of the $\mathrm{d}$ map whose domain is $X^k$ and $\mathrm{d} X^k$ to denote the image of that map.
The $k$-th cohomology group associated to this sequence is the quotient:
\[
\mathrm{H}^k(X^{\scriptscriptstyle \bullet}) =(\ker \mathrm{d} | X^k) /\mathrm{d} X^{k-1}.
\]
\subsection{Sufficient conditions for minimality}
\begin{proposition}
Suppose that $A$ is a finite element system, and that $B$ is a compatible finite element system containing $A$. Then we have:
\begin{equation}\label{eq:mindim}
\dim B^k_0(T) \geq \dim A^k_0(T) + \dim \mathrm{H}^{k+1}(A^{\scriptscriptstyle \bullet}_0(T)).
\end{equation}
\end{proposition}
\begin{proof}
By the rank-nullity theorem and the definition of cohomology groups, we have that:
\begin{align}
\dim A^k_0(T) & = \dim \mathrm{d} A^k_0(T) + \dim \ker \mathrm{d} | A^k_0(T),\\
\dim \mathrm{H}^{k+1}(A^{\scriptscriptstyle \bullet}_0(T)) & = \dim \ker \mathrm{d} | A^{k+1}_0(T) - \dim \mathrm{d} A^k_0(T) .
\end{align}
Hence:
\begin{equation}\label{eq:comp}
\dim A^k_0(T) + \dim \mathrm{H}^{k+1}(A^{\scriptscriptstyle \bullet}_0(T)) = \dim \ker \mathrm{d} | A^{k+1}_0(T) + \dim \ker \mathrm{d} | A^k_0(T).
\end{equation}
Now, since $B^{\scriptscriptstyle \bullet}_0(T)$ is exact we may choose a subspace $V$ of $B^k_0(T)$ such that the map:
\begin{equation}
\mathrm{d} : V \to \ker \mathrm{d} | A^{k+1}_0(T),
\end{equation}
is an isomorphism. Then it is clear that:
\begin{equation}
V \cap \ker \mathrm{d} | A^k_0(T) = 0.
\end{equation}
It follows that:
\begin{equation}
\dim B^k_0(T) \geq \dim V + \dim \ker \mathrm{d} | A^k_0(T).
\end{equation}
Since $\dim V=\dim \ker \mathrm{d} | A^{k+1}_0(T)$, we apply (\ref{eq:comp}) to complete the proof.
\end{proof}

We will use this as follows:

\begin{corollary}\label{cor:minnec}
Suppose that $A$ is a finite element system, and that $B$ is a compatible finite element system containing $A$. Suppose that:
\begin{equation}\label{eq:mindimeq}
\dim B^k_0(T) = \dim A^k_0(T) + \dim \mathrm{H}^{k+1}(A^{\scriptscriptstyle \bullet}_0(T)).
\end{equation}
Then $B$ is minimal among compatible finite element systems containing $A$.
\end{corollary}

We will show in the next two sections that, given a finite element system $A$, there exists a compatible finite element system $B$ containing $A$ such that (\ref{eq:mindimeq}) holds. Thus the condition is not only sufficient, it is necessary for minimality. Moreover, in the case that $A$ is included in a compatible finite element system $B$, it is always possible to find a minimal one containing $A$, inside $B$. Finally, in this context, we will actually \emph{construct} a particular minimal FES. We will show how other authors have proposed different definitions of finite element spaces, which turn out to be minimal among those containing certain differential forms.

\subsection{First step \label{sec:consminfir}}

Given a finite element system $A$, the first step is to construct a finite element system $\tilde A$ containing $A$ such that $\tilde A_0$ is exact.
In the case that $A_0$ is exact already, this step can be skipped.

For $k < \dim T$ choose a subspace $H^{k+1}(T)$ such that:
\begin{equation}
\mathrm{d} A^k_0(T) \oplus H^{k+1}(T) =   \ker \mathrm{d} | A^{k+1}_0(T).
\end{equation}
Then choose a subspace $E^k(T)$ of $B^k_0(T)$ such that $\mathrm{d}$ restricts to an isomorphism:
\begin{equation}
\mathrm{d} : E^k(T) \to H^{k+1}(T).
\end{equation}
For $k=\dim T$, if $A^k(T)$ contains an element with integral $1$, set $E^k(T)= 0$, if not choose an element in $B^k(T)$ with integral $1$ and let $E^k(T)$ be its linear span.

We remark that $A^k(T) \cap E^k(T) = 0$ and set:
\begin{equation}
\tilde A^k(T) := A^k(T) \oplus E^k(T).
\end{equation}
We remark that we have an induced isomorphism:
\begin{equation}\label{eq:disoe}
\mathrm{d} : E^k(T) \to \mathrm{H}^{k+1}(A^{\scriptscriptstyle \bullet}_0(T)),
\end{equation}
where if $k=\dim T$, the exterior derivative operation is replaced by integration. In fact we might take this, together with $E^k(T) \subseteq B^k_0(T)$,  as the defining properties of $E^k(T)$, and the space $\mathrm{H}^k(T)$ introduced above merely as a device to make it clear that such an $E^k(T)$ exists.

We could also have specified $E^k(T)$ from a choice of scalar products as follows:
If $k < \dim T$, set:
\begin{equation}\label{eq:defe}
E^k(T) := \{ u \in B^k_0(T) \ : \ \mathrm{d} u \in A^{k+1}_0(T), \ \mathrm{d} u \perp \mathrm{d} A^k_0(T) \textrm{ and } u \perp \mathrm{d} B^{k-1}_0(T) \}.
\end{equation}
If $k=\dim T$ and if $A^k(T)$ contains an element with integral $1$, set $E^k(T):= 0$, if not set:
\begin{equation}\label{eq:defebis}
E^k(T):= \{ u \in B^k(T) \ : \ u \perp \mathrm{d} B^{k-1}_0(T) \}.
\end{equation}
Then one checks immediately that $E^k(T)$ is a subspace of $B^k_0(T)$  such that (\ref{eq:disoe}) is an isomorphism.

\begin{proposition}
Under the above hypothesis, $\tilde A$ is a FES containing $A$ such that the sequences:
\begin{equation}
0 \to \tilde A^0_0(T) \to \tilde A^1_0(T) \to \cdots \to \tilde A^{\dim T}(T) \to \mathbb{R} \to 0,
\end{equation}
are exact.
\end{proposition}

\begin{proof}
That $\tilde A$ is stable under restrictions and the exterior derivative is immediate, so we only need to prove the exactness property.

Consider first $k < \dim T$.
Suppose $u \in \tilde A^{k}_0(T) $ is such that $\mathrm{d} u = 0$. We have $\tilde A^k_0(T) = A^k_0(T) \oplus E^k(T)$, so we can write $u = v + w$ with $v \in A^k_0(T)$ and $w\in E^k(T)$. Then $\mathrm{d} w = - \mathrm{d} v \in \mathrm{d} A^k_0(T)$ so $\mathrm{d} w = 0$, so $w= 0$.

Now we have $\mathrm{d} v = 0$.  If $k = 0$ this gives $v = 0$. If $k >0$,  we have  already remarked that $\mathrm{d}  A^{k-1}_0(T) \oplus \mathrm{d} E^{k-1} (T) =   \ker \mathrm{d} | A^{k}_0(T)$, so we can write $v = \mathrm{d} t$ with $t \in \tilde A^{k-1}_0(T)$. This concludes the case $k < \dim T$.

If $k = \dim T$, the same proof goes through, replacing the exterior derivative by the integral for the involved $k$-forms.
\end{proof}

\subsection{Second step \label{sec:consminsec}}
We now consider the situation of a finite element system $A$ included in a compatible finite element system $B$, such that the following sequences are exact:
\begin{equation}
0 \to A^0_0(T) \to A^1_0(T) \to \cdots \to  A^{\dim T}(T) \to \mathbb{R} \to 0.
\end{equation}

For each cell $T$ we will augment the spaces $A^{\scriptscriptstyle \bullet} (T)$ in such a way that the extension property holds, while $A^{\scriptscriptstyle \bullet}_0(T)$ is preserved (i.e. no trace-free elements will be added). We start by carrying out this process on 1D cells, then on 2D cells, and so forth. We will use the following result, where $\tr$ denotes the trace (pull-back) operator $B^\bullet(T) \to B^\bullet(\partial T)$.

\begin{proposition}\label{prop:mixext}
Let $ l\geq 0$ and suppose $T$ is a cell of dimension $l+1$. Fix $k\leq l$. Any $u \in B^k(\partial T)$ such that  $\mathrm{d} u \in \tr A^{k+1}(T)$ has a unique extension $\tilde u\in B^k(T)$ satisfying:
\begin{equation}\label{eq:mixharm}
 \mathrm{d} \tilde u \in A^{k+1}(T), \ \mathrm{d} \tilde u \perp \mathrm{d} A^k_0(T)  \quad \textrm{ and } \quad  \tilde u \perp \mathrm{d} B^{k-1}_0(T).
\end{equation}
\end{proposition}
\begin{proof}
\emph{Existence.} Suppose first $k<l$. Let $v$ be the $A$-harmonic extension of $\mathrm{d} u$. Since $\mathrm{d} v$ is $A$-harmonic and zero on the boundary we have $\mathrm{d} v = 0$. Choose $u'\in B^k(T)$ such that $\mathrm{d} u' = v$. We have $\mathrm{d}(\tr u' -u)= 0$.
If $k=0$, $\tr u' - u$ is a constant $c$ and  $u'-c$ is an extension of $u$ with exterior derivative $v$. If on the other hand $k>0$ choose $u''\in B^{k-1}(T)$ such that  $\mathrm{d} \tr u'' = \tr u' -u$. Then $u' -\mathrm{d} u''$ is an extension of  $u$ with exterior derivative $v$. Adding an element of $\mathrm{d} B^{k-1}_0(T)$ we can ensure that it becomes orthogonal to $\mathrm{d} B^{k-1}_0(T)$.

Suppose now $k=l$. Pick $v \in A^{k+1}(T)$ such that $\int v = \int u$ and $v \perp \mathrm{d} A^k_0(T)$. Pick $u'\in B^k(T)$ such that $\mathrm{d} u' = v$. By Stokes we have $\int(\tr u' - u) = 0$, so we can choose $u''\in B^{k-1}(T)$ such that $\mathrm{d} \tr u'' = \tr u' -u$. Then $u' -\mathrm{d} u''$ is, as before, an extension of $u$ with exterior derivative $v$ and by adding an element of $\mathrm{d} B^{k-1}_0(T)$ we can ensure orthogonality to $\mathrm{d} B^{k-1}_0(T)$.

\emph{Uniqueness.} If $\tilde u\in B^k_0(T)$ satisfies (\ref{eq:mixharm}) we have $\mathrm{d} \tilde u = 0$. The condition $\tilde u \perp \mathrm{d} B^{k-1}_0(T)$ then gives $\tilde u=0$.

\end{proof}

Suppose we have realized our plan for cells of dimension at most $l$ ($l\geq 0$), the augmented FES being denoted $\tilde A$. We consider a cell $T$ of dimension $l+1$.

We equip $\tilde A^k(\partial T)$ with a scalar product, for instance the sum of the local ones, defined for $T' \lhd} % {\vartriangleleft \partial T$. For $k \leq l$ we put:
\begin{equation}
F^k= \{u \in \tilde A^k(\partial T) \ : \ u \perp \tr A^k(T)  \quad \textrm{ and } \quad  \mathrm{d} u \in \tr A^{k+1}(T) \},
\end{equation}
and denote by $\tilde F^k$ the extensions of elements of $F^k$ defined by Proposition \ref{prop:mixext} above. We also denote:
\begin{equation}
G^k= \{u \in \tilde A^k(\partial T) \ : \ \mathrm{d} u \perp \tr A^{k+1}(T)  \quad \textrm{ and } \quad  u \perp \{ v\in  \tilde A^k(\partial T) \ : \ \mathrm{d} v = 0 \} \},
\end{equation}
and denote by $\tilde G^k$ the $B$-harmonic extensions of elements of $G^k$. Finally we put:
\begin{equation}\label{eq:tilde}
\tilde A^k(T) = A^k(T) + \tilde F^k + \tilde G^k.
\end{equation}
For $k= l+1$ we take simply $\tilde A^k(T) = A^k(T)$.

\begin{proposition}
On the $(l+1)$-skeleton of $\mathcal{T}$, $\tilde A$ is a compatible finite element system intermediate between $A$ and $B$ such that  $\tilde A^k_0(T) = A^k_0(T)$ for all $k$ and $T$.
\end{proposition}

\begin{proof}
We remark that $\mathrm{d} \tilde F^k \subseteq A^{k+1}(T)$. If $k<l$, $\mathrm{d} \tilde G^k \subseteq \tilde F^{k+1}(T)$ and if $k=l$, $\tilde G^k =0$. Therefore $\mathrm{d} \tilde A^k(T) \subseteq \tilde A^{k+1}(T)$. We have a direct sum decomposition:
\begin{equation}
\tilde A^k(\partial T) = \tr A^k(T) \oplus F^k \oplus G^k.
\end{equation}
It follows in particular that $\tr : \tilde A^k(T) \to \tilde A^k(\partial T)$ is surjective. Moreover if $ u \in A^k(T)$, $ v\in \tilde F^k $ and $w \in \tilde G^k$ satisfy $\tr (u + v + w) = 0$, then $\tr v =0$ and $\tr w =0$, which gives $v=0$ and $w=0$. If in fact  $u + v + w = 0$, we will also have $u=0$. Therefore the sum (\ref{eq:tilde}) is direct and:
\begin{equation}
\tilde A^k_0(T) =  A^k_0(T).
\end{equation}
This completes the proof.
\end{proof}

The construction is therefore complete, and provides a minimal compatible finite element system containing $A$, by Corollary \ref{cor:minnec}.

\section{Minimality of some finite element systems \label{sec:ex}}

We now consider more concrete examples, based on polynomial differential forms on simplices or hypercubes.
The two steps of our general construction enter differently in them:
\begin{itemize}
\item In Section \ref{sec:mintrim}, on trimmed differential forms, the local cohomology groups $ \mathrm{H}^k(A^{\scriptscriptstyle \bullet}_0(T))$ are non-trivial, but once this problem is fixed, the obtained spaces turn out to have the extension property.
\item In Section \ref{sec:minser}, on serendipity elements,  the local cohomology groups are trivial, so one only needs to add spaces of forms that ensure the extension property.
\item In Section \ref{sec:mintnt}, on TNT elements, the extension property holds for the spaces one starts with, but local sequence exactness fails.  By adding tiny bubble functions $E^k(T)$ to fix the cohomology requirement, one also needs to add forms that extend them to higher-dimensional cells.
\item In Section \ref{sec:minsp}, in which we introduce ``small pleasures'' elements, the sequence one starts with has nontrivial cohomology and does not satisfy the extension property.
\end{itemize}

\subsection{Minimality of trimmed polynomial differential forms\label{sec:mintrim}}

We now adopt notations from finite element exterior calculus~\cite{ArnFalWin06}.
We first consider finite element systems on a simplicial complex $\mathcal{T}$.
Let $\calP_r \Lambda^k(T)$ denote the space of $k$-forms on $T\in\mathcal{T}$ whose coefficients are polynomials of degree at most $r$. The space $\calP_r^-\Lambda^k(T)$ is defined with the help of the operator $\kappa$ on forms, which is contraction by the vector field $x \mapsto x$ (different choices of origin yield the same space). Then:
\begin{equation}
\calP_r^-\Lambda^k(T) = \{ u \in \calP_r \Lambda^k(T) \ : \  \kappa u \in  \calP_r \Lambda^{k-1}(T) \}.
\end{equation}

In keeping with our previous notations, $\calP_r^-\Lambda^k_0(T)$ is the subspace of $\calP_r^-\Lambda^k(T)$ consisting of differential forms whose pullback to the boundary $\partial T$ of $T$ is $0$. We also use freely that these spaces form exact sequences under the exterior derivative, which follows from the fact that they constitute a \emph{compatible} FES. The underlying facts are proved in standard finite element language in \cite{ArnFalWin06}, but it is also checked with the general tools of finite element systems in \cite{ChrRap15}.

Similar statements hold for the $\calP_r\Lambda^k(T)$ spaces, but here sequence exactness requires decreasing $r$ through the complex. That is, under the exterior derivative, the sequence:
\begin{equation}
\calP_{r+1} \Lambda^{k-1}_0(T) \to \calP_r \Lambda^k_0(T) \to \calP_{r-1} \Lambda^{k+1}_0(T),
\end{equation}
is exact, whereas the sequence:
\begin{equation}
\calP_{r} \Lambda^{k-1}_0(T) \to \calP_r \Lambda^k_0(T) \to \calP_{r} \Lambda^{k+1}_0(T),
\end{equation}
in general is not. The cohomology group of the latter sequence is denoted:
\begin{equation}
\mathrm{H}^{k} \calP_{r}\Lambda^{\scriptscriptstyle \bullet}_0(T).
\end{equation}

\begin{lemma} We have :
\begin{equation}
\mathrm{d} \calP_r^-\Lambda^k_0(T) = \mathrm{d} \calP_r\Lambda^k_0(T) .
\end{equation}
\end{lemma}
\begin{proof}
Pick $u \in \calP_r\Lambda^k_0(T)$ and set $v:= \mathrm{d} u$.
Then $v \in \calP_{r-1}\Lambda^{k+1}_0(T)\subseteq \calP_{r}^-\Lambda^{k+1}_0(T)$ and $\mathrm{d} v = 0$. Hence there is $w \in \calP_{r}^-\Lambda^{k}_0(T)$ such that $\mathrm{d} w = v$.
\end{proof}

The following identity shows that  the spaces $\calP_r^-\Lambda^k(T) $ constitute a \emph{minimal} compatible FES containing the spaces $\calP_{r-1}\Lambda^k(T)$.
\begin{proposition} We have:
\begin{equation}
\dim  \calP_r^-\Lambda^k_0(T) = \dim  \calP_{r-1}\Lambda^k_0(T) + \dim \mathrm{H}^{k+1} \calP_{r-1}\Lambda^{\scriptscriptstyle \bullet}_0(T).
\end{equation}
\end{proposition}
\begin{proof}
We use the vertical line notation $|$ to denote restriction of an operator to a subspace.
We have:
\begin{align}
\dim  \calP_r^-\Lambda^k_0(T) &= \dim \mathrm{d} \calP_r^-\Lambda^k_0(T) + \dim \ker \mathrm{d} | \calP_r^-\Lambda^k_0(T),\\
& = \dim \mathrm{d} \calP_r\Lambda^k_0(T) + \dim \mathrm{d} \calP_r^-\Lambda^{k-1}_0(T),\\
&  =  \dim \mathrm{d} \calP_r\Lambda^k_0(T)  + \dim \mathrm{d} \calP_r\Lambda^{k-1}_0(T),
\end{align}
and:
\begin{align}
\dim  \calP_{r-1}\Lambda^k_0(T) &= \dim \mathrm{d} \calP_{r-1}\Lambda^k_0(T) + \dim \ker \mathrm{d} | \calP_{r-1} \Lambda^{k}_0(T),\\
&= \dim \mathrm{d} \calP_{r-1}\Lambda^k_0(T) + \dim \mathrm{d} \calP_r\Lambda^{k-1}_0(T).
\end{align}
Moreover:
\begin{align}
\dim \mathrm{H}^{k+1} \calP_{r-1}\Lambda^{\scriptscriptstyle \bullet}_0(T) & = \dim \ker \mathrm{d} | \calP_{r-1}\Lambda^{k+1}_0(T)  - \dim \mathrm{d} \calP_{r-1}\Lambda^k_0(T),\\
&= \dim \mathrm{d} \calP_r\Lambda^k_0(T)  - \dim  \mathrm{d} \calP_{r-1}\Lambda^k_0(T).
\end{align}
The result follows by addition of the above equalities.
\end{proof}

\subsection{Minimality of serendipity elements\label{sec:minser}}

In \cite{ArnAwa14}, Arnold and Awanou define spaces of \textit{serendipity} finite element differential forms on $n$-dimensional cubes, denoted $\mathcal{S}_r\Lambda^k(I^n)$.
They prove a subcomplex property (Theorem 3.3) and trace property (Theorem 3.5), which is equivalent, in our notation, to showing that the spaces $\mathcal{S}_{r-k} \Lambda^k(I^n)$
constitute a FES, for any fixed $r\geq n$. Arnold and Awanou also define these spaces for $r < n$, but then one does not get a full complex, but one that stops at $k = r$. In many applications this is useful, since one does not require the full complex but rather two consecutive spaces, possibly three. But the FES framework has not been designed for this.

They also prove that the spaces $\calP_{r -k - 2(n-k)} \Lambda^{n-k} (I^n)$ provide unisolvent degrees of freedom (Theorem 3.6).  By \cite[Proposition 2.5]{ChrRap15}, this guarantees that the extension property holds. They also check the sequence exactness property (page 1566). Thus, in our language, they have shown that the serendipity finite element spaces constitute a compatible finite element system. The inclusion $\calP_{r-k}\Lambda^k(I^n) \subseteq A^k(I^n)$ is obvious from their construction and it is implicit from their Proposition 3.7 that $A^k_0(I^n) = \calP_{r-k}\Lambda^k_0(I^n)$.

Thus one can interpret many of their results, by the statement that the serendipity elements define a minimal compatible finite element system containing $ \calP_{r-k}\Lambda^k(I^n)$. We now explain this, proving the main statements above, in our framework.

It should be noted that the construction in \cite{ArnAwa14} is very different from the one we propose in Section \ref{sec:consminsec} (the first step in our construction is not necessary here, for reasons provided below). Our construction depends on a choice of compatible finite element system $B$ containing $\calP_{r-k}\Lambda^k(I^n)$. For that purpose we could take the spaces:
\begin{equation}
B^{\scriptscriptstyle \bullet}(I^n) = \calP^-_{r} \Lambda^{\scriptscriptstyle \bullet}(I)  \otimes \ldots \otimes \calP^-_{r} \Lambda^{\scriptscriptstyle \bullet}(I).
\end{equation}
We would also need a scalar product. One can use $\mathrm{L}^2$ scalar products, or more algebraic ones, for instance one that makes some preferred basis in $B$ orthonormal.

\begin{remark}
We will take $I$ to be the interval $[0, 1]$ while Arnold and Awanou take it to be $[-1,1]$.
Also, we will only address the cases $r\geq n$, although the arguments can be adapted to treat lower order cases as well.
\end{remark}

\begin{lemma}\label{lem:zerotens}
Let $B_\alpha^k$ be a space of differential $k$-forms on $I^{n-1}$, for each $0 \leq \alpha \leq r$ and for each $k$.
Define the sum of graded tensor product spaces of forms on $I^n$:
\begin{equation}
A^{\scriptscriptstyle \bullet} = \sum_{\alpha = 0}^r B_\alpha^{\scriptscriptstyle \bullet} \otimes \calP_{\alpha} \Lambda^{{\scriptscriptstyle \bullet}}(I).
\end{equation}
Consider $\delta_n^0$ and $\delta_n^1$, the $n$-th trace operators, defined respectively as pullbacks by the injections $ I^{n-1} \times \{0\} \to I^n$   and $I^{n-1} \times  \{1\} \to I^n$. Then the intersections of the kernels of $\delta_n^0$ and $\delta_n^1$ on $A^{\scriptscriptstyle \bullet}$ is :
\begin{equation}
(\ker \delta_n^0 | A^{\scriptscriptstyle \bullet} ) \cap (\ker \delta_n^1 | A^{\scriptscriptstyle \bullet})  = \sum_{\alpha = 0}^r B_\alpha^{\scriptscriptstyle \bullet} \otimes \calP_{\alpha} \Lambda_0^{{\scriptscriptstyle \bullet}}(I).
\end{equation}
\end{lemma}
\begin{proof}
We let $\lambda_0$ (resp. $\lambda_1$) denote the element of $\calP_{1} \Lambda^{0}(I)$ taking the values $1$ (resp. $0$) at $0$ and $0$ (resp. $1$) at $1$. Observe that for $\alpha \geq 1$ we have a direct sum decomposition:
\begin{equation}
\calP_{\alpha} \Lambda^{{\scriptscriptstyle \bullet}}(I) = \mathbb{R} \lambda_0 \oplus \mathbb{R} \lambda_1 \oplus \calP_{\alpha} \Lambda^{{\scriptscriptstyle \bullet}}_0(I).
\end{equation}
We also have:
\begin{equation}
\calP_{0} \Lambda^{{\scriptscriptstyle \bullet}}(I) = \mathbb{R} (\lambda_0  + \lambda_1) +  \mathbb{R} \mathrm{d} \lambda_0.
\end{equation}

Now take an element $u$ of $A^k$ and write it as a sum:
\begin{equation}
u =     v_0 \otimes  (\lambda_0  + \lambda_1) + v'_0 \otimes \mathrm{d} \lambda_0 + \sum_{\alpha = 1}^r (v_{\alpha_0} \otimes \lambda_0 + v_{\alpha_1} \otimes \lambda_1 + v_\alpha),
\end{equation}
with:
\begin{equation}
v_0  \in B_0^k,\ v'_0  \in B_1^{k-1}, \textrm{ and } v_\alpha  \in \sum_l B_\alpha^{k-l} \otimes \calP_{\alpha} \Lambda^{l}_0(I).
\end{equation}
Observe that if $u$ is in the intersection of the kernels, we have:
\begin{align}
0 = \delta_n^0 u & = v_0 + \sum_{\alpha = 1}^r v_{\alpha_0},  \\
0 = \delta_n^1 u & = v_0 + \sum_{\alpha = 1}^r v_{\alpha_1}.
\end{align}
Then we are left with:
\begin{equation}
u = v'_0 \otimes \mathrm{d} \lambda_0  + \sum_{\alpha = 0}^r  v_\alpha,
\end{equation}
and this completes the proof.
\end{proof}

\begin{proposition}\label{prop:polyzc}
On $I^n$ let $x_j$ denote the $j$-th coordinate function. We have:
\begin{equation}\label{eq:polyrz}
\calP_r \Lambda^k_0(I^n) =  \mathrm{span} \left\{ u \prod_{j \not \in J} x_j (1- x_j) \mathrm{d} x_J \ :  \begin{array}{l} u \in \calP_{r - 2(n-k)}(I^n), \\  J \subseteq \{1, \ldots, n\}, \\  \card J = k. \end{array}\right\},
\end{equation}
where, for $J =\{ j_1, \ldots, j_k \}$ we define $\mathrm{d} x_J = \mathrm{d} x_{j_1} \wedge \ldots \wedge \mathrm{d} x_{j_k}$.
\end{proposition}
\begin{proof}
We have:
\begin{equation}
\calP_r \Lambda^{\scriptscriptstyle \bullet}(I^n) = \sum_{\alpha_1 + \ldots + \alpha_n = r} \calP_{\alpha_1}\Lambda^{\scriptscriptstyle \bullet} (I) \otimes \ldots \otimes \calP_{\alpha_n}\Lambda^{\scriptscriptstyle \bullet} (I),
\end{equation}
where we sum over integer $\alpha_i$ in the range $[0, r]$.

In order to determine $\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I^n)$ we apply Lemma \ref{lem:zerotens}, or its equivalent, in each coordinate direction. We get:
\begin{equation}
\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I^n) = \sum_{\alpha_1 + \ldots + \alpha_n = r} \calP_{\alpha_1}\Lambda^{\scriptscriptstyle \bullet}_0(I) \otimes \ldots \otimes \calP_{\alpha_n}\Lambda^{\scriptscriptstyle \bullet}_0(I),
\end{equation}
Now, concerning each factor, we have with $x$ the coordinate function on $I$:
\begin{align}
\calP_{q}\Lambda^0_0(I) & = \{ u(x) x(1-x) \ : \ u \in \calP_{q-2}(I)  \},\\
\calP_{q}\Lambda^1_0(I)& = \calP_{q}\Lambda^1(I).
\end{align}

It follows that the left hand side of (\ref{eq:polyrz}) is included in the right hand side. The inclusion in the other direction is trivial, so the proof is complete.
\end{proof}

\begin{remark}
The result just shown seems to be taken as self-evident in \cite[page 1569]{ArnAwa14}.
\end{remark}

\begin{proposition}\label{prop:zerodual}
The integrated wedge product defines an invertible bilinear form:
\begin{equation}
(\int \cdot \wedge \cdot) \ : \  \calP_r \Lambda^k_0(I^n) \times \calP_{r - 2(n-k)} \Lambda^{n-k} (I^n) \to \mathbb{R}.
\end{equation}
\end{proposition}
\begin{proof}
It follows from the preceding proposition that the two spaces have the same dimension.

Fix $u \in \calP_r \Lambda^k_0(I^n)$. We want to show that the map sending $v\mapsto \displaystyle \int u\wedge v\in\mathbb{R}$ is non-zero only if $u$ is zero. Write $u$ in the form:
\begin{equation}
u = \sum_J \left(u_J \prod_{j \not \in J} x_j (1- x_j) \mathrm{d} x_J\right),
\end{equation}
with:
\begin{equation}
u_J \in \calP_{r - 2(n-k)}(I^n),\ J \subseteq \{1, \ldots, n\} \textrm{ and } \card J = k.
\end{equation}
Letting $J'$ denote the complement of $J$ in $\{1, \ldots, n\}$ we define:
\begin{equation}
v= \sum_J \epsilon_J u_J \mathrm{d} x_{J'} \in \calP_{r - 2(n-k)} \Lambda^{n-k} (I^n).
\end{equation}
Then we have:
\begin{equation}
u \wedge v = \sum_J \epsilon_J u_J^2 \prod_{j \not \in J} x_j (1- x_j)\mathrm{d} x_J \wedge \mathrm{d} x_{J'} .
\end{equation}
We then choose the signs $\epsilon_J$ such that $\epsilon_J \mathrm{d} x_J \wedge \mathrm{d} x_{J'} $ is the volume form on $I^n$. Then $\int u\wedge v$ is zero iff $u$ is zero.
\end{proof}

\begin{proposition}\label{prop:zeroce} Fix $r \geq 0$.

For $0 \leq k < n$, the following sequence, for the exterior derivative, is exact:
\begin{equation}
\calP_{r+1} \Lambda^{k-1}_0(I^n) \to \calP_r \Lambda^k_0(I^n) \to \calP_{r-1} \Lambda^{k+1}_0(I^n).
\end{equation}

The sequence:
\begin{equation}
\calP_{r+1} \Lambda^{n-1}_0(I^n) \to \calP_r \Lambda^n_0(I^n) \to \mathbb{R},
\end{equation}
is also exact (where the last arrow is integration).

\end{proposition}
\begin{proof}
Consider the first statement. By Proposition \ref{prop:zerodual}, the sequence is paired with the following one:
\begin{equation}
\calP_{r-1- 2(n-k)} \Lambda^{n-k+1}(I^n) \leftarrow \calP_{r - 2(n-k)} \Lambda^{n-k}(I^n) \leftarrow \calP_{r+1- 2(n-k)} \Lambda^{n-k-1}(I^n).
\end{equation}
The differential in this sequence is again the exterior derivative. Recall also that:
\begin{equation}
\int_T \mathrm{d} u \wedge v = \pm \int_T u \wedge \mathrm{d} v,
\end{equation}
when (for instance) $u$ is zero on $\partial T$.
Since the latter sequence is exact, the former one must be too.

Consider now the second statement. The sequence is paired with:
\begin{equation}
\calP_{r-1} \Lambda^{n-1}(I^n) \leftarrow \calP_{r} \Lambda^{0}(I^n) \leftarrow \mathbb{R}.
\end{equation}
On the right we have inclusion of the constants. This sequence is exact.
\end{proof}

We may summarize our results as follows:
\begin{proposition}
Fix $r \geq n$. Let $A^k(I^m)$ be a compatible finite element system on hypercubes of dimension at most $n$, such that:
\begin{equation}\label{eq:sercp}
\calP_{r-k}\Lambda^k(I^m) \subseteq A^k(I^m).
\end{equation}
Then $A^k(I^m)$ is a minimal compatible finite element system with this property if and only if:
\begin{equation}\label{eq:minpz}
A^k_0(I^m) = \calP_{r-k}\Lambda^k_0(I^m).
\end{equation}
In this case, the spaces $\calP_{r +k - 2m} \Lambda^{m-k} (I^m)$ provide unisolvent degrees of freedom on $A$.
\end{proposition}

\begin{proof}
The first statement follows from the dimensional characterization of minimality provided by Corollary~\ref{cor:minnec}, given the exactness property proved in Proposition \ref{prop:zeroce}. The second statement recalls Proposition~\ref{prop:zerodual}.
\end{proof}

Again we point out that \cite{ArnAwa14} and Section \ref{sec:con} use quite different means to define a compatible FES $A$ such that (\ref{eq:sercp}) and (\ref{eq:minpz}) hold.

\subsection{Minimality of TNT elements\label{sec:mintnt}}

We now consider the work of Cockburn and Qiu regarding the `TiNiest spaces containing Tensor product spaces of polynomials' on cubes or \emph{TNT} elements for short~\cite{CocQiu14}.
We will show how to derive their minimality claims from the framework of FES. Moreover we cast their constructions in the language of differential forms, which is uniform in the treatment of dimensions and degree of forms. We will also make explicit use of other facts from homological algebra such as the Kunneth theorem.

Fix $r\geq 1$.  Define the spaces, on hypercubes:
\begin{equation}
\label{eq:a-tnt-def}
A^{\scriptscriptstyle \bullet}(I^n) = \mathcal{Q}_r\Lambda^{\scriptscriptstyle \bullet}(I^n) = \calP_r \Lambda^{\scriptscriptstyle \bullet}(I)  \otimes \ldots \otimes \calP_r \Lambda^{\scriptscriptstyle \bullet}(I).
\end{equation}
We remark that on an interval $\calP_r \Lambda^{\scriptscriptstyle \bullet}(I)$ has the extension property. Therefore, by Proposition \ref{prop:tensorfes}, $A$ also has the extension property. However the sequence exactness fails. We want to define a minimal compatible finite element system containing $A$. This is accomplished for dimension $n=3$ in \cite{CocQiu14}.

By \cite[Lemma 3.10]{Chr09AWM} we have:
\begin{equation}
A^{\scriptscriptstyle \bullet}_0(I^n) = \calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I)  \otimes \ldots \otimes \calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I).
\end{equation}

Looking at the sequence:
\begin{equation}\label{eq:zerocohr}
0 \to \calP_r \Lambda^0_0(I) \to \calP_r \Lambda^1_0(I) \to 0,
\end{equation}
we notice that there is a non-zero cohomology group only in the 1-forms, and that it has dimension $2$. We write out the Kunneth theorem:
\begin{equation}
\mathrm{H}^{\scriptscriptstyle \bullet} A^{\scriptscriptstyle \bullet}_0(I^n) \approx \mathrm{H}^{\scriptscriptstyle \bullet} (\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I)) \otimes \ldots \otimes \mathrm{H}^{\scriptscriptstyle \bullet} (\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I)).
\end{equation}
Since, as we remarked above, $\mathrm{H}^0 (\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I)) = 0$, most of the terms cancel. There is only one non-zero cohomology group in $A^{\scriptscriptstyle \bullet}(I^n)$, namely the one in $n$-forms. Explicitly:
\begin{align}
\mathrm{H}^{k} A^{\scriptscriptstyle \bullet}_0(I^n) & = 0,\ \textrm{ for } k < n, \label{eq:aexn}\\
\mathrm{H}^n A^{\scriptscriptstyle \bullet}_0(I^n) & \approx \mathrm{H}^1 (\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I)) \otimes \ldots \otimes \mathrm{H}^1 (\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I)), \label{eq:hnaz}
\end{align}
and the latter space has dimension $2^n$.

Now, the sequence that we want to make exact is really:
\begin{equation}
0 \to A^0_0(I^n) \to A^1_0(I^n) \to \ldots \to A^n(I^n) \to \mathbb{R} \to 0,
\end{equation}
where the second to last arrow is integration. Thus the notation in (\ref{eq:hnaz}) is slightly different from the one used in the beginning of the article, e.g in (\ref{eq:mindimeq}), the difference being whether we consider that there is an $\mathbb{R}$ in the end  of $A^{\scriptscriptstyle \bullet}_0(I^n)$ or not. We notice that $A^n(I^n)$ contains elements with non-zero integral. We thus have:

\begin{proposition}\label{prop:cocmincond}
Let $B$ denote a compatible finite element system containing $A$, as defined in (\ref{eq:a-tnt-def}). Then $B$ is a minimal one containing $A$ if and only if, for all $k$ and $n$:
\begin{align}
B^k_0(I^n) & = A^k_0(I^n), \ \textrm{ for } k \neq  n-1 ,\\
\dim B^{n-1}_0 (I^n) & = \dim A^{n-1}_0(I^n)  + 2^n -1.
\end{align}
\end{proposition}
This provides the dimension count for any minimal finite element system containing $A$, using Proposition \ref{prop:extdim}.

Any construction of a minimal compatible finite element system $B$ containing $A$ would need to add spaces $E^{n-1}(I^n)$ to $A^{n-1}(I^n)$ such that:
\begin{align}
\dim E^{n-1}(I^n) & = 2^n -1,\\
B^{n-1}_0(I^n) & = A^{n-1}_0(I^n) \oplus E^{n-1}(I^n),
\end{align}
as in the `first step' outlined in Section \ref{sec:consminfir}. The second step, outlined in Section \ref{sec:consminsec},  would have to ensure that the extension property holds, after these additions, since they also take place on the faces of $I^n$.

To be more specific, we notice that if $x$ denotes the coordinate function on the interval $I$, we can choose two generators of $\mathrm{H}^1 (\calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I))$, to be $\mathrm{d} x $ and $P(x)\mathrm{d} x$, where $P$ is a polynomial of degree $r$ with integral $0$ on $I$. We also impose $P(1-x) = P(x)$ if $r$ is even and $ P(1-x) = - P(x)$ if $r$ is odd (to ensure that our construction becomes invariant under reflections in the coordinate directions). Then we let $Q$ be the polynomial of degree $r+1$ such that $Q(0)= Q(1) = 0$ and $Q' = P$.

Denote by $x_i$ the $i$-th coordinate map on $I^n$. Let $J$ denote a non-empty subset of $\{1, \ldots, n\}$. Define the $n$-form $f_J$ on $I^n$:
\begin{equation}
f_J = \bigotimes_{j \in J} P(x_j) \mathrm{d} x_j \bigotimes_{j \not \in J} \mathrm{d} x_j,
\end{equation}
where the notation is taken to mean that we have the form $P(x) \mathrm{d} x$ in the directions of $J$ and the form $\mathrm{d} x$ in the other ones (thus there is a permutation in the indices going on). Since $J$ is non-empty, $f_J$ has integral $0$. Notice that we have thus defined $2^n-1$ linearly independent forms in $A^n(I^n)$ with zero integral.

We now observe that:
\begin{equation}
f_J = \frac{1}{\card J} \mathrm{d} g_J, \textrm{ with } g_J = \sum_{i \in J}  Q(x_i)\bigotimes_{j \in J\setminus \{i\}} P(x_j) \mathrm{d} x_j \bigotimes_{j \not \in J} \mathrm{d} x_j.
\end{equation}
This notation is taken to mean that we can obtain $g_J$ from $f_J$ by replacing, for each index $i$ in $J$, the term $P(x_i)\mathrm{d} x_i$ by $Q(x_i)$, and summing over $i \in J$.

We define:
\begin{equation}
E^{n-1}(I^n) = \mathrm{span} \{ g_J \ : \  J \subseteq \{1, \ldots, n\}, \ J \neq \emptyset \}.
\end{equation}
We remark that:
\begin{lemma}\label{lem:egood}
The elements of $E^{n-1}(I^n)$ are zero on $\partial I^n$ and the map defined in  (\ref{eq:disoe}) is an isomorphism.
\end{lemma}

Thus they are adequate for the first step of our general construction. Notice also that  these spaces are invariant under permutation of indices and reflections $x_i \mapsto 1-x_i$.

For the second step, one should keep in mind that we have added spaces $E^{d-1}(T)$ for all faces $T$ of dimension $d$ in $I^n$, and that the extension property is therefore no longer guaranteed. We could proceed by our general method. But inspired by \cite{CocQiu14}, we prefer the following explicit approach:

We determine a face $T$ of $I^n$ by a choice of indices $J_T\subseteq \{1, \ldots , n \}$ and for the indices $i \in \{1, \ldots , n \}\setminus J_T$ a number $N_T(i) \in \{0, 1 \}$. Then the associated face $T$ is:
\begin{equation}
T = \{ (x_1, \ldots , x_n ) \  : \ \textrm{ for }  i \not \in J_T \quad x_i = N_{T}(i) \}.
\end{equation}
Let $T$ be a $d$ dimensional face of $I^n$. An element of $E^{d-1} (T)$ is extended to $I^n$ as follows. Let $u$ be the element in question. Its extension is chosen to be:
\begin{equation}\label{eq:tildeu}
\tilde u = u \bigotimes_{i \not \in J_T} x_i^{N_T(i)}(1-x_i)^{1-N_T(i)}.
\end{equation}
We then define:
\begin{equation}\label{eq:defcoc}
B^k(I^n) = A^k(I^n) + \sum_{\dim T = k+1} \tilde E^k(T),
\end{equation}
where the sum is over $k+1$ dimensional faces of $I^n$ and $\tilde E^k(T)$ denotes the space of extensions of elements of $E^k(T)$ defined by $(\ref{eq:tildeu})$.

\begin{proposition}
The definition (\ref{eq:defcoc}) provides a minimal compatible FES containing $A$.
\end{proposition}
\begin{proof}
Suppose that (\ref{eq:tildeu}) holds for some $u \in E^k(T)$.  Then:
\begin{align}
\mathrm{d} \tilde u   & =~ u \bigotimes_{i \not \in J_T} x_i^{N_T(i)}(1-x_i)^{1-N_T(i)}~ +\notag \\
 & \qquad\quad \sum_{j \not \in J_T} u \bigotimes_{i \not \in J_T}  x_i^{N_T(i)}(1-x_i)^{1-N_T(i)} (\pm \mathrm{d} x_i)_{i= j},\\
& \in ~~A^{k+1}(I^n)  + \sum_{\dim T = k+2} \tilde E^{k+1}(T).
\end{align}
Thus $B$ satisfies the requirements to be an element system.
Let $T$ be a $k+1$ dimensional face of $I^n$ and let $T'$ be another one. Let's take an element $u$ of $ E^k(T)$ and restrict $\tilde u$ to $T'$. Choose $i \in J_{T} \setminus J_{T'}$.  The restrictions of $u$ to the faces $x_i = 0$ and $x_i = 1$ inside $T$ are $0$. Therefore the restriction of $\tilde u$ to $T'$ is $0$.
From this we deduce that the sum in (\ref{eq:defcoc}) is direct, and that:
\begin{align}
B^k_0(I^n) & = A^k_0(I^n), \ \textrm{ for } k \neq  n-1 ,\\
B^{n-1}_0 (I^n) & = A^{n-1}_0(I^n)  \oplus E^{n-1}(T) .
\end{align}
Recalling that $A$ has the extension property, we see in particular that:
\begin{align}
\dim B^k(I^n) & = \dim A^k(I^n) +  \sum_{\dim T = k+1} \dim E^k(T),\\
& = \sum_{\dim T \neq k+1} A^k_0(T) + \sum_{\dim T = k+1} \left(\dim A^k_0(T) + \dim E^k(T)\right) ,\\
& = \sum_{T} \dim B^k_0(T),
\end{align}
where $T$ runs through the set of faces of $I^n$. This proves that $B$ has the extension property by Proposition \ref{prop:extdim}.
The sequence exactness in the form (\ref{eq:coh0}) has already been proved, see (\ref{eq:aexn}) and Lemma \ref{lem:egood}.
Thus we have proved that $B$ is a compatible finite element system.
It is minimal containing $A$ by application of Proposition \ref{prop:cocmincond}.

\end{proof}

If we want degrees of freedom for this compatible FES that yield commuting interpolators, we can take the ones defined in \cite[Proposition 2.8]{ChrRap15}.

\subsection{Dimension count for ``small pleasures'' elements\label{sec:minsp}}

Neither the construction of Arnold and Awanou nor the one of Cockburn and Qiu resolves the question that seems most natural to us, namely, defining the minimal compatible FES on hypercubes containing the sequences
\begin{equation}
\label{eq:minseqPrLkIn}
\calP_r\Lambda^0(I^n) \to \calP_r\Lambda^1(I^n) \to \ldots \to \calP_r\Lambda^n(I^n).
 \end{equation}
The preceding general construction provides such a mcFES. Being neither the serendipity elements, nor the tiniest elements in the Cockburn-Qiu sense, we refer to these as the ``Small Pleasures'' elements.
We provide a dimension count of the Small Pleasures elements with the following result.

\begin{lemma}
\begin{equation}
\label{eq:dimHkPrL0In}
\dim \mathrm{H}^k \calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I^n)  = { r + 2k-n -1 \choose k-1}{r +k -n -1 \choose n - k}.
\end{equation}
\end{lemma}
\begin{proof}

We have an exact sequence:
\begin{align}
0 \to \calP_{r+k-1} \Lambda^0_0(I^n) & \to  \calP_{r+k-2}\Lambda^1_0(I^n) \to \ldots \notag \\
& \ldots \to \calP_{r+1} \Lambda^{k-2}_0(I^n)   \to \calP_r\Lambda^{k-1}_0(I^n) \to \mathrm{d} \calP_r\Lambda^{k-1}_0(I^n) \to 0.
 \end{align}
This gives:
\begin{equation}
\dim \mathrm{d} \calP_r\Lambda^{k-1}_0(I^n) = \sum_{i= 0}^{k-1} (-1)^i \dim \calP_{r+i}\Lambda^{k-1 -i }_0(I^n).
\end{equation}

We also have:
\begin{equation}
\ker \mathrm{d}  | \calP_r \Lambda^k_0(I^n)  =  \mathrm{d}  \calP_{r+1} \Lambda^{k-1}_0(I^n).
\end{equation}
Hence, by the same formula as previously:
\begin{equation}
\dim \ker \mathrm{d} | \calP_r\Lambda^{k}_0(I^n) = \sum_{i= 0}^{k-1} (-1)^i \dim \calP_{r+1+i}\Lambda^{k-1 -i }_0(I^n).
\end{equation}

Therefore:
\begin{align}
\dim \mathrm{H}^k \calP_r \Lambda^{\scriptscriptstyle \bullet}_0(I^n) & = \sum_{i= 0}^{k-1} (-1)^i (\dim \calP_{r+1+i}\Lambda^{k-1 -i }_0(I^n) - \dim \calP_{r+i}\Lambda^{k-1 -i }_0(I^n)),\\
& = \sum_{i= 0}^{k-1} (-1)^i (\dim \calP_{r+1+i  - 2(n-k+1+i)}  \Lambda^{n -k+ 1 + i }(I^n) -\\
& \qquad \dim \calP_{r+i - 2(n-k+1+i)}\Lambda^{n-k+1 +i }(I^n)),\\
& = \sum_{i=0}^{k-1} (-1)^i \dim \mathcal{H}_{r-1-i  - 2(n-k)}  \Lambda^{n -k+ 1 + i }(I^n).
\end{align}

Now, following \cite[Section 3.2]{ArnFalWin06}, the Koszul sequence on $I^n$:
\begin{equation}
0 \to \mathcal{H}_t \Lambda^n \to \mathcal{H}_{t+1} \Lambda^{n-1} \to \ldots \to \mathcal{H}_{t+ k-1} \Lambda^{n-k+1} \to \kappa \mathcal{H}_{t+ k-1} \Lambda^{n-k+1} \to 0,
\end{equation}
is exact. We take $t= r+ k -2n$, and deduce:
\begin{align}
 \sum_{i = 0}^{k-1} (-1)^i \dim\mathcal{H}_{t+k-1-i}\Lambda^{n-k+1+i} & =  \dim \kappa \mathcal{H}_{t+ k-1} \Lambda^{n-k+1},\\
& = { n + t + k -1 \choose k - 1} {t +n -1 \choose n-k},\\
& = { r + 2k-n -1 \choose k-1}{r +k -n -1 \choose n - k}.
\end{align}

This concludes the proof.
\end{proof}

Using (\ref{eq:dimHkPrL0In}) and (\ref{eq:mindimeq}), we can compute the dimension of $B^k_0(I^n)$ where $B$ is a minimal compatible finite element system containing (\ref{eq:minseqPrLkIn}).
We give some some sample values in Table~\ref{tab:dimcount}.
These spaces will be investigated further in future work.

\begin{table}
\[
\begin{tabular}{l|ccccccc}
 & $r=4$ & $r=5$ & $r=6$ & $r=7$ & $r=8$ & $r=9$ & $r=10$ \\
 \hline
$k=0$ & 0 & 1 & 4 & 10 & 20 & 35 & 56 \\
$k=1$ & 11 & 27 & 54 & 95 & 153 & 231 & 332 \\
$k=2$ & 45 & 81 & 133 & 204 & 297 & 415 & 561 \\
$k=3$ & 35 & 56 & 84 & 120 & 165 & 220 & 286
\end{tabular}
\]
\caption{Some computed values for $\dim B^k_0(I^3)$ where $B$ is a minimal compatible finite element system containing (\ref{eq:minseqPrLkIn}).}
\label{tab:dimcount}
\end{table}

\section*{Acknowledgements}
SHC was supported by the European Research Council through the FP7-IDEAS-ERC Starting Grant scheme, project 278011 STUCCOFIELDS.
AG was supported in part by NSF Award 1522289.

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 Snorre Christiansen, \today\\
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redpajama_set_name: RedPajamaArXiv


Well, my first day at work was actually kind of boring, but that's a good thing, and it was fully expected. After leaving a job where they had to replace me with two people that are doing about half the work I was expected to do, it's a nice change of pace to be a little bored during the day.
I arrived at work to find that the IT department had not yet set up my username or my computer, so I spent the first few hours pouring through two large manilla folders full of paperwork they had been filing for the "future webmaster" for over a year. The folders were full of great articles on various design tips, etc., as well as correspondences, notes, etc.
Once the IT guys got my new computer set up, I had a few meetings, so I didn't get to do much at that point. Then, when I finished with all of my meetings, I hopped onto the computer, looked through all of the emails that had been sent to the "webmaster" email account, and started fixing things on the site.
Developing a "plan of action", detailing all the steps that need to be taken in order to develop the new site, all the tools and training I will need in order to get it done, how much each thing might cost, and how long it should take me to get the stuff done.
Migrating the site, which is currently made up of over 35,000 separate HTML and CSS files, into a database-driven application.
Implementing the new design, which will be prepared by our graphic designer.
I really think I'm going to enjoy this job. It will be so much easier to prioritize my tasks in this environment. Of course, I'll still have quite a few people all clamoring for their projects to get done first, and I will have to balance quite a bit, but it will be so much easier. Rather than switching between 20 or 30 different tasks, all requiring different parts of my brain, I'll simply be switching between various tasks that all require the same skillset. I won't have to "switch gears" nearly as much, which will make me a great deal more efficient.
Yesterday morning, they set up a brand new computer for me. I have the entire Adobe Creative Suite installed (v2), as well as the entire Macromedia development suite (Fireworks, Flash, Dreamweaver, etc.). With the current state of the web site, it looks as though DreamWeaver will be an invaluable tool for me to use. While I love writing my own code by hand, it can be a real pain to try to pick through someone else's code in a text editor, especially when the only thing you want to do is fix a typo.
Obviously, I'll probably use my own code editing tools, along with whatever's included in the new CMS, once I start working on the new site, but, for now, Dreamweaver is great.
It's kind of strange, though, as I'm sitting at home right now with nowhere to go, since the inclement weather has caused the college to close for the day. My second day of work, and I've already got a snow day! I'm so used to being in situations where, no matter what the weather was, we were required to be at work. Heck in one of my previous jobs, they would encourage you to either spend the night at work, or walk to work if there was inclement weather.
Great overview Curtiss – Really detailed. The first few days can always be tough because everyone will see you as superman and that you can fix everything in 10 minutes or less.
it will be interesting to watch your journey as you grow in this new role and learn and so forth.


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Pampa 's first resident doctor was Dr. Vittorio E. von Brunow, who was born at Charleston , South Carolina, on October 27, 1862. He was the son of Count Phillip von Brunow and Countess von Brunow. Count von Brunow was prominent in diplomatic circles in the service of Russia for many years. Count von Brunow's family returned to their old home in East Russia in 1864.
Dr. von Brunow received his elementary education in Wittenberg , Saxony, and later studied in Austria and Russia . He was a graduate of the University of Vienna and the University of Warsaw , where he took a degree in 1887. This was followed by study in the medical clinics in Berlin , and at Kaln and Hamburg .
In 1892 he returned to America and took a very important part in medical research and the scientific development of therapeutics. His medical attainments were the result of exceptional training, ripened by a broad experience and constant study.
He practiced at New Orleans for a time, and later at St. Louis and in Chicago . He then went into Indian Territory, and after a short period of practice there, moved to Gainesville in 1900 and three years later came on to Pampa.
For nearly forty years he served the people in and around Pampa with ceaseless effort and intelligent application. His practice was along general lines since he felt that he could better serve his community this way.
Dr. Brunow (as he was known in Pampa ) liked hunting and fishing but his real hobby was in research work. He was also a philanthropist but only his closest friends knew of his many gifts and kindnesses.
He organized the Republican party in Gray County and he was its only chairman until the time of his death. He was a member of the Elks Club and the Kiwanis Club.
Dr. Brunow took a keen interest in current events and subscribed to numerous magazines and newspapers. He was one of the best informed men in the Panhandle and had a remarkable memory.
Dr. Brunow was married on February 25, 1937, to Mrs. Lonna D. Lan Franco at Pampa . He had four children by a previous marriage: Julia D. Wilson of Pampa , Edward of Freeport, Fred of Benton, Louisiana, and J.H. of Port Arthur . Mrs. Lonna D. Brunow had a daughter, Vera Lee (Mrs. Bob Andis), who was adopted by Dr. Brunow.
Dr. Brunow died on May 7, 1941, and was buried in Fairview Cemetery . Mrs. Brunow, age 95, is living in Amarillo .
Dr. Brunow came from Gainesville to the Texas Panhandle in a surrey. Being German, he had probably heard of the Thut family who were German speaking Swiss, and he stayed at first in the Thut Hotel near Lefors.
Soon after arriving in Pampa (1903), Dr. Brunow had a white frame building constructed at 101 South Cuyler. The Brunow's lived upstairs, and the lower floor was used as a doctor's office, drug store, post office and telephone office.
The building at 101 South Cuyler was the third location of the Pampa post office which was moved from the Johnson Mercantile Building at 105 North Cuyler. Records in the National Archives show that Vittorio E. von Brunow was appointed the fourth postmaster of Pampa on October 23, 1903. However, people have always referred to Mrs. Brunow (Josye Lucille Barnes) as the postmistress.
Whenever a bag of mail was thrown from a train, some one brought it to the post office where people stood around and waited while the postmistress sorted the letters and other items and placed them in the 24 pigeon holes reserved for mail.
Mrs. Brunow also had the first telephone service in Pampa . There were 24 plugs which connected Pampa with Miami and Panhandle. Some ranchers in Roberts County ran a telephone line into Pampa where the "central" part of the service was in Dr. Brunow's drugstore. The line was run on the fences, with wires over the gates and places where it could not be used on the fences. This caused some confusion as the cowboys would come by and staple the wire to the post, not knowing it was the telephone line.
In 1913, the post office was moved to 115 North Cuyler. Congressman John Stephens suggested that an election would be the best way for Democrats to select their postmasters, and this method was followed all over Texas that year. Claudine Barnes, sister of Mrs. Brunow, was elected postmistress by popular vote of the patrons. Her tenure began on June 27, 1914, and she served until January 1, 1922, as the fifth postmaster of Pampa .
In 1926, when Pampa was growing rapidly because of the oil boom, the white frame Brunow building was moved from 101 South Cuyler to 825 West Kingsmill. Dr. Brunow gave the house to his daughter, Mrs. Roy J. (Julia D.) "Dollie" Wilson . The Wilsons lived in the house until about 1942.
The house was bought by James W. Roden in 1950, and some of that family has owned it until recently. In the summer of 1991, Joey Roden, owner at that time, had begun to have the house razed when he was told of its historical significance. Recently he gave the building to June Thomas, who plans to have the house moved and restored.
The ground floor of the red brick Brunow Building built in 1926 was occupied by L.T. Hill department store until 1932. Levine's department store occupied the lower floor and basement from 1934 until 1966.
The second story had apartments or offices. Dr. Brunow's address was 108 East Foster, apartment 7 .
The red brick building burned on Christmas night in 1981. An electrical short was blamed for the blaze which was estimated to have caused close to $1 million in damage.
At the time of the fire, tenants of the Brunow Building on the Foster Street side were Pampa Travel Center , Joe Cree Agency, Billie's Originals, Betty's Botique and King's Row Barber Shop. The Carousel Shoe Store was the only business with a Cuyler Street entrance.


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Upgrade to a Kuranda cot.
We can trim your dog's nails for you.
Send your dog home clean and smelling great! Included: wash, choice of shampoo, ear wipes, eye wipes, and fragrant spritz.
Includes: bath (with choice of shampoo, ear wipes, eye wipes, and fragrant spritz), nail trim, and dessert.
We exclusively use earthbath products.
Yes, please add an after wash deoderizing spritz.
Yes, please use eye wipes to clean and remove stains around the eyes.
Yes, please use ear wipes to clean inside the ears.
This is a 15 minute individualized walk with a member of our team for added socialization and enjoyment outdoors. This service is especially beneficial for dogs that are overly stimulated in the dog daycare environment, dogs that are shy and could benefit from added socialization, and dogs that are boarding with us for an extended period of time.
This is a 15 minute individualized session with your dog where a member of our daycare team will practice basic manners to reinforce already established behaviors such as sit, down, come when called/attention, walking nice on leash, polite greetings/no jumping. This is not to be confused with our day training services which are more comprehensive and performed by a member of our dog training team. Treats are provided but if your dog has a sensitive stomach you may want to supply the treats.
What would you like us to practice?
This is a 10 minute individualized session where your dog will be given additional mental stimulation with an opportunity to interact with various puzzle toys, activity mat, or novel equipment. Treats are provided but if your dog has a sensitive stomach you may want to supply the treats.
Any additional information related to Add On Services?


text:
Letter from Australia: The problem with women
by Julie Hare
Julie-Hare-Wonkhe
Julie Hare
Associate Editor (Australia)
Julie is Wonkhe's Associate Editor in Australia.
Women and Wonkery
International Women's Day has been and gone for another year. It seemed to have an added vigour this year, underscored by the #metoo movement and the fact domestic violence is now, finally, getting some political traction.
A really interesting report was released last week. It came from Zoe Whitton, Citi Bank's head of environmental, social and governance research.
It looked at the retention and promotion of women in 137 of the top 200 companies listed on the Australian Securities Exchange.
What it found was shocking. Male-dominated sectors such as mining, construction and energy were surprisingly good at getting junior women into more senior positions. However, sectors such as healthcare, with huge numbers of junior female employees, fail miserably. And not just as a proportion of total employees, but in absolute numbers. Worse than that, over the four years of data Whitton looked at, the numbers were actually in decline.
"Healthcare [has] heaps of women, but then there's a huge drop-off," Whitton told the Australian Financial Review. "Even though utilities starts at 30% (females in junior roles), it ends up at 15% (in senior roles). Healthcare starts in the vicinity of 70% women but only 10% representation in senior roles. That trend is getting worse."
Searching for equality
What's doing my head in about all of this is not just the unfairness of it, but the fact that females make up 60% of all students in universities. If we look at commencements, about 50% of first-year women each year at in three areas: healthcare, education and humanities (excluding law and economics).
So while education is a good space for women – they see it as a strategy for economic security – what happens once they reach the workforce is fiendishly cruel. Due to a range of reasons, there is a disconnect between educational outcomes and career progression.
Why? Self-segregation in the careers women choose. Structural sexism in the workplace. The fact that men's roles are more highly valued than traditionally female roles. Caring responsibilities, go without saying.
In fact, average full-time salaries for women are lower than for men in every occupation and industry in Australia, according to the Workplace Gender Equality Agency. That is a horrible sentence to write. The exception to the rule appears to be at Google, which reported last week its male employees were earning less than women – although that has been disputed.
The one bright light in all of this is that the gender pay gap is narrowing. But very slowly.
Last year, the Grattan Institute found that female university graduates are now expected to earn 27 per cent less than men – or $750,000 less – over their career. Ten years earlier, the gap was 30 per cent.
This IWD, Prime Minister Scott Morrison bought into the debate. The Liberal Party has been under intense pressure in recent times over the lack of female representation generally (23%) and a historical lack of senior women in the ministry.
It went something like this: "We want to see women rise [but] we don't want to see women rise only on the basis of others doing worse."
Somehow he couldn't summon the "a rising tide lifts all boats" metaphor but stumbled awkwardly over the very conservative notion that middle-aged, middle-class white men are the new victims of leftist, feminist, multicultural, pro-Marxist thinking. Or something like that.
The brick wall
Unsurprisingly, Morrison's comments were condemned around the country and also reached an international audience. He was forced to clarify his thoughts the next day – only to make it worse.
He said that as a husband and a father of two daughters he had a vested interest in the advancement of women. Which is interesting given he first met his wife when she was 12, started dating at 16, and married at 21. She is/was a registered nurse, but is currently a stay-at-home mum. This is not a criticism of Jen Morrison. But perhaps her PM husband doesn't really have his head around the economic necessities that drive women into education and the workforce.
Anyway, I offer this up not just merely as an example of the banality of federal politics in Australia and as momentary relief from the rolling catastrophe that is Brexit, but as evidence of the fact that some blokes, including those who run the country, just don't seem to get it.
There is so much structural sexism at play in this country that women can't even get ahead in the fields in which they dominate by sheer numbers.
As the wonderful Gloria Steinem once said: "I've yet to be on a campus where most women aren't worrying about some aspect of combining marriage, children and a career. I've yet to find one where many men were worrying about the same thing."
I've been wondering if nursing/teaching/humanities degrees should contain components that address sexism in the workplace and also some fundamentals in management. Or whether this is the responsibility of employers to ensure there is a female pipeline to move into more senior positions. Whitton said she thought complacency in female dominated sectors might the the problem.
After all, if you have a workforce of 70% women, surely some are going to float to the top. Or maybe not.
One response to "Letter from Australia: The problem with women"
Gavin Moodie says:
Mar 11 2019 at 12:34 pm
See also this study which found that more than half of the gender earnings gap in Canada is due to the under-representation of women in top 10% of earners, not to women being paid less for the same lower level job.
Bonikowska, Aneta, Drolet, Marie, and Fortin, Nicole M (2019) Earnings inequality and the gender pay gap in Canada: the role of women's under-representation among top earners. Statistics Canada catalogue number 11-626-X No. 088,
https://www150.statcan.gc.ca/n1/pub/11-626-x/11-626-x2019002-eng.htm
Silver tortoises will help us to narrow the gender pay gap
by Helen Scott
Comment, Long read
Tackling sexual violence and harassment in higher education
by Debbie McVitty
Comment, People
No more steps. It's time for a leap on gender equality
by Minto Felix
A poor deal for women in higher education
Image: Joy Elliot-Bowman
#Wonklife – Joy Elliot-Bowman
by Catherine Boyd
Look again at International Women's Day
by Jess Moody
Getting the gang together – reflections on women in HE leadership
by Linda Naughton
Women wonks: why it's time to get writing

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Как запушить один коммит, чтоб не заливалась вся история Такая ситуация. Проект перенесен на другой удаленный репозиторий. И при выполнении задачи и попытке ее залить, у меня заливается вся история гита. Подскажите пожалуйста, как правильно поступить в данной ситуации.

A: Данная проблема решилась благодаря удаление папки .git и клонирования ее с нового удаленного репозитория.
Спасибо всем, кто принимал участие в обсуждении.

A: Как вариант можно установить SourceTree - Визуализация и управление репозиториями через простой интерфейс SourceTree. В нем можно выборочно запушить отдельные файлы.

redpajama_set_name: RedPajamaStackExchange


text:
using Foundation.Build.Activities;
using Microsoft.VisualStudio.TestTools.UnitTesting;

namespace Foundation.Tests.Build.Activities
{
    [TestClass]
    public class BuildQueryProviderTests
    {
        [TestMethod]
        public void Regex()
        {
            var provider = new BuildQueryProvider();

            var number = "Blah blah 1.0.11012.4 blah";

            var result = provider.BuildNumberRegex.Match(number);

            Assert.IsTrue(result.Success);
            Assert.AreEqual("1", result.Groups["Major"].Value);
            Assert.AreEqual("0", result.Groups["Minor"].Value);
            Assert.AreEqual("11012", result.Groups["Build"].Value);
            Assert.AreEqual("4", result.Groups["Revision"].Value);
        }
    }
}

redpajama_set_name: RedPajamaGithub


Original Furla Candy Bag from Australia.. slightly used, good as new. Good price.
Rush! Still in good condition.
Bought at Furla Hong Kong. Used a couple of times. Good condition. No return no exchange. Sf- c/o buyer.
Original. Excellent condition. Only used ONCE. Comes with dustbag. Adjustable chain strap. Bought in 2015. Price when bought was around ₱24,000.
100% Original and authentic furla handbag Was my mom's! Patent/shiny leather Serial number indicated inside Minor flaws, needs cleaning but overall still usable and in good condition! Pm for more info.
Pre-loved Authentic Furla for sale and negotiable!!
Pre owned authentic Furla two way bag Color: beige Dimension: 10"X 8"X 4" Very good used condition Guaranteed authentic or your money back. Almost new condition except for the slight line ink stain at the back of the bag Price posted is slightly negotiable.


One of the biggest weekends of the year is upon us and we're all gearing up for the big game in our own way, especially here in St. Louis. And by gearing up, I mean doing doing other things if you're a bitter Rams fan. A couple of guys who WILL be watching the Super Bowl on Sunday are country stars Luke Bryan and Jordan Davis. Jordan is still a bit upset that his Saints won't be making their Super Bowl appearance this year, but still plans on watching, while Luke plans around hovering around the snack table.
What are your plans for Sunday? Are you having people over for the game or boycotting it altogether this year? Let us know at the New Country 92.3 Facebook page and give us your Super Bowl predictions as well!


Embrace and accept yourself with this healing three-stone turquoise bracelet. Help understand and encourage the daily challenges we face in life by believing there is something greater. Considered a stone of life and beauty, continue to believe, continue to hope. The gold-filled chain is oval shaped and is supported by hematite as well as a single glass piece in teal.


text:
Palm Paper will build a new power station near King's Lynn
The Palm Paper Factory near King's Lynn. - Credit: Matthew Usher
Palm Paper can build a new power station at its site near King's Lynn.
The company was yesterday given development consent by the Department of Energy and Climate Change to build the 162MW gas-powered plant next to its paper mill on the banks of the tidal Ouse at Saddlebow.
Due to its size, the application was decided by the planning inspectorate, and it was the 50th nationally significant infrastructure project to be decided following examination by the inspectorate. The company wants to reduce its carbon emissions and its reliance on imported electricity, by generating its own.
A report produced by West Norfolk Council concluded the combined cycle gas turbine plant would help secure Palm Paper's operations in Lynn and create up to 50 jobs during its 18-month construction phase.
Palm Paper King's Lynn mill began production of newsprint paper in August 2009 and produced the one millionth tonne three years later.


text:
WDFB can refer to:

 WDFB (AM), a radio station at 1170 AM licensed to Junction City, Kentucky
 WDFB-FM, a radio station at 88.1 FM licensed to Danville, Kentucky

redpajama_set_name: RedPajamaWikipedia


The wonderful lake has an length of 16.5 km and the nicest way to explore it is with a motor boat.
The round trip is gonna be with our 409 Ps motor boat. For 9 people maximum.
Click here for our different destinations.
You have booked a Candle light dinner at one of the restaurants along the lake? We will drive you there.
We also offer Shuttle service by motor boat.
The motorboat is not for rent.


text:
Novodnistrovsk (,  – Novodněstrovsk) je město v Černovické oblasti na Ukrajině. Žije v něm  obyvatel.

Poloha a doprava
Novodnistrovsk leží na pravém, západním břehu Dněstru, který zde tvoří hranici mezi Černovickou a Vinnyckou oblastí a který je zde přehrazen hrází Dněsterské přečerpávací elektrárny, po které zároveň vede silnice spojující oba břehy. Od Černovic je Novodnistrovsk vzdálen přibližně 116 kilometrů severovýchodně.

Dějiny
Sídlo zde bylo založeno 3. dubna 1973 z rozhodnutí ministerstva energetiky Sovětského svazu v souvislosti s výstavbou Dněsterské přečerpávací elektrárny. V roce 1975 získalo status sídlo městského typu a od roku 1982 je městem. Dříve bylo součástí Sokyrjanského rajónu a od roku 2000 bylo pod přímou oblastní správou. Po administrativně-teritoriální reformě v červenci 2020 patří do nově vzniklého Dněsterského rajónu.

Odkazy

Reference

Externí odkazy

Města v Černovické oblasti


Every month, the Chronicle's film critics select a theme and offer movie recommendations. In honor of Star Wars Day, we look back on our favorite performances by Star Wars actors in lightsaber-free films.
I lived for a year in the shadow of Stirling Castle - a former royal residence for the Scottish crown, later regimental headquarters for the Argyll and Sutherland Highlanders, and (grim Stirlingshire weather permitting) occasional concert venue. It was also the shooting location for one of the future Obi-Wan Kenobi's greatest performances.
For more Chronicle Recommends picks, visit our Movie Picks page.


text:
Eidg. Forschungsanst Created with Sketch. Swiss Federal Institute for Forest, Snow and Landscape Research WSL Eidg. Forschungsanst Created with Sketch. Swiss Federal Institute for Forest, Snow and Landscape Research WSL Eidg. Forschungsanst Created with Sketch. Swiss Federal Institute for Forest, Snow and Landscape Research WSL
About WSL
1200 years of water balance data challenge climate models
Water availability in the Northern Hemisphere has seen much larger changes during the past twelve centuries than during twentieth century global warming, a new study in Nature reports.
A new study has reconstructed dry and wet periods over the past 1200 years. It concludes that water availability has fluctuated more in previous centuries than in the twentieth century. This challenges current climate models which show greater fluctuations for the twentieth century. (photo: Markus Bolliger)
The researchers from Sweden, Germany, and Switzerland have for the first time reconstructed the variations in water availability across the Northern Hemisphere seamless for the past twelve centuries. This allows for comparisons between various parts of Europe, Asia, and North America.
The study shows that hydroclimate extremes have been stronger and covered larger areas in some earlier centuries than during the twentieth century, explains lead author Fredrik Charpentier Ljungqvist from Stockholm University.
Water availability through the centuries
The researchers reconstructed changes in water availability – the hydroclimate – by statistically analysing evidence for changes in precipitation and drought, such as varying lake levels, soil humidity or river runoff. To do this the researchers compiled hundreds of records from various climate archives across the Northern Hemisphere: tree-rings, speleothems, lake sediments, and written historical records.
The scientists compared their reconstructed hydroclimate variations with a new temperature reconstruction they also developed, to understand links between the two. It turned out that only a few regions showed clear correlations between changes in temperature and hydroclimate. For instance, drought was most widespread during both the relatively warm twelfth century and the relatively cold fifteenth century.
Test for climate models
According to Ljungqvist, the study shows the importance of a millennium-long perspective on hydroclimate changes. Meteorological measurements of precipitation and drought date back at most a few centuries – too short to tell if today's changes fall outside the range of natural variability. "They are also too short to test if climate models are correct when predicting that drier regions get drier and wetter regions get wetter with global warming", says Ljungqvist.
Both the new temperature reconstruction and the climate model simulations agree that the twentieth century likely was the warmest in at least the past millennium. "But unlike the climate models our reconstruction does not show any dramatic increase in hydroclimate extremes", says Ljungqvist.
"The current models seem to simulate precipitation characteristics well over the pre-industrial period, but do not appear to capture as well the more recent anthropogenically driven changes in the climate system," explaines co-author David Frank from WSL. More research is needed to find out why, he says.
This does not necessarily mean the mechanisms driving hydroclimate changes in climate models are wrong, points out Ljungqvist. "It could be that the global warming is not yet strong enough to trigger the changes in precipitation patterns that climate models simulate". The authors conclude that while the reconstructions based on climate archive data also have their uncertainties, they are still a very valuable tool to test models used to predict future changes in water availability.
Prof. Dr. David Frank
Forest Dynamics
Dendrosciences
Dr. phil. Fredrik C. Ljungqvist
University of Stockholm
fredrik.c.l@historia.su.se
Mobile phone: +4670-662 07 28
Fredrik Charpentier Ljungqvist, Paul J. Krusic, Hanna S. Sundqvist, Eduardo Zorita, Gudrun Brattström & David Frank: "Northern Hemisphere hydroclimate variability over the past twelve centuries", Nature, vol. 532, pp. 94–98, doi:10.1038/nature17418
WSL and SLF provide the artwork for imaging of press articles relating to this media release for free. Transferring and saving the images in image databases and saving of images by third parties is not allowed.
Swiss Federal Institute for Forest, Snow and Landscape Research WSL
WSL websites
Knowhow for forestry experts and all others interested in the forest
Avalanche knowledge and prevention, e-learning an tour planning
Swiss National Forest Inventory NFI - a forest information service
natural-hazards.ch
Warnings on natural hazards by Swiss federal authorities
Access to environmental monitoring and research data
WSL Institute for Snow and Avalanche Research SLF in Davos
Research for People and the Environment: The Swiss Federal Institute for Forest Snow and Landscape Research WSL monitors and studies forest, landscape, biodiversity, natural hazards and snow and ice. WSL is a research institute of the Swiss Confederation and part of the ETH Domain. The WSL Institute for Snow and Avalanche Research SLF is part of the WSL since 1989.


The Leadership Team (formerly Ad Board) meets on a monthly basis. This is the governing body of the church. They also function as the SPRC (Staff Parish Relations Committee) The focus of the meetings are assessing the church's mission and ministry and how well we are living into our vision. They also vote on important church matters. All are welcome to attend this meeting. All meeting are listed in the bulletin the week prior.


Published 04/26/2019 02:59:41 am at 04/26/2019 02:59:41 am in Medicine Cabinet Locking.
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Fabio, real name Fitzroy Heslop, has been at the forefront of the rave scene in the nineties. He strated his career as a pirate radio DJ in the mid 80s and joining together with long time DJ partner Grooverider, the pair happened to be in the right place at the right time when acid house exploded across London and the UK. Fabio and Grooverider are sometimes referred to as the "gruesome twosome". He started with early house music then moved towards jungle and drum and bass, Fabio is considered as one of the pioneers and is credited with pioneering the subgenre known as liquid funk as well as coining the name.


text:
Q: GetAncestor: Undeclared identifier I'm using Visual Studio 6 for backward compatibility reasons and I need
to use GetAncestor() like this to create a dll :
HWND ancestorhandle = GetAncestor(myvar, 2);

I've included these header files:
#include <windows.h>
#include "stdafx.h"
#include "offlinelib.h"
#include <stdio.h>

As per the documentation:
GetAncestor function
But while compiling, I get an error:
GetAncestor: Undeclared identifier

What am I doing wrong?

A: Try moving the #include "stdafx.h" (precompiled header) up, making it the first include line.

A: Solved by adding
#ifdef WINVER
#undef WINVER
#endif
#define WINVER 0x500

on top of stdafx.h


Caps Doorman Maxima looks great in Spain!
The Caps Mairy's are going back on the American brood cow Jo-Wal Cleitus Macey. Goldwyn dtr Caps Mairy 4 belongs to the greatest Genomic Giants in the breed.
Caps Mairy 14 VG-87-NL 6yr.
The #1 GTPI Jeeves in the World!
Welcome to the site of Caps Farm Caps Farm is owned by the Family Captein and established in the middle part of the Netherlands in Biddinghuizen. Caps Farm is the home for the best cow families worldwide with members of the families of Markwell Durham Daisy, Jo-Wal Cleitus Macey, Lynncrest BW Marshl Horizon, D-R-A August, Snow N Denises Dellia, Sunnylodge Prelude Spottie and many more! We always have live stock, bulls and embryos available, do not hesitate to contact us for more information.


Deb-Sorry Amelia and I can't make it tonight, but have fun and I think you will probably pack the house!
Lovely! I went to Joe Coffee Bar today, on 13th street. A coffee drinking friend of mine says they make the best latte around. I stupidly had tea but I will go again and scope out the scene. There was a line in there at 10:30 a.m.!
Rad - Joe's in NYC is a fantastic coffee and espresso bar. Never had a latte there but like tazza d'oro they brew their daily coffee in a French Press -very, very yummy. We promise, the next time Amelia and I are visiting you in Jersey City we will all go into the city for a coffee tour. No tea allowed though! Love to Izzy and A.
Just wanted to say that I've had great coffee and espresso drinks at several Joe the Art of Coffee locations (if that's what we're talking about). Impressively consistent.
Yep, Joe the Art of Coffee is what we are discussing and it is a great place in NYC. I've been to the location on East 13th St.