text:
\begin{document}

\maketitle

    \begin{abstract}
We provide further evidence to favor the fact that rank-one convexity does not imply quasiconvexity for two-component maps in dimension two. We provide an explicit family of maps parametrized by $\tau$, and argue that, for small $\tau$, they cannot be achievable by lamination.
In this way, Morrey's conjecture might turn out to be correct in all cases.
    \end{abstract}

\section{Introduction}

One of the main ingredients of the direct method of the Calculus of Variations (\cite{DacorognaH}) to show existence of minimizers for an integral functional of the kind
$$
I(\bu)=\int_\Omega\psi(\nabla \bu(\bx))\,d\bx
$$
is its weak lower semicontinuity. Here $\Omega\subset\R^N$ is a regular, bounded domain, and feasible mappings $\bu:\Omega\to\R^m$ are smooth or Lipschitz, so that $\nabla \bu$ is a $m\times N$-matrix at each point $\bx\in\Omega$. The weak lower semicontinuity property is in turn equivalent to suitable convexity properties of the continuous integrand $\psi:\cM\to\R$. Morrey (\cite{Morrey}, \cite{MorreyB}) proved that this weak lower semicontinuity (in $W^{1, \infty}(\Omega; \R^m)$) is equivalent to the quasiconvexity of the integrand $\psi$, namely,
$$
\psi(\bF)\le\frac1{|D|}\int_D\psi(\bF+\nabla \bv(\bx))\,d\bx
$$
for every $\bF\in\cM$, and every test map $\bv$ in $D$. This concept does not depend on the domain $D$, and can, equivalently, be formulated in terms of periodic mappings so that such a density $\psi$ is quasiconvex if
$$
\psi(\bF)\le\int_D\psi(\bF+\nabla \bv(\by))\,d\by
$$
for all $\bF\in\cM$, and every periodic mapping $\bv:D\to\R^m$. Now $D\subset\R^N$ is the unit cube.

Unfortunately, the issue is far from settled by simply saying this, since even Morrey realized that it is not at all easy to decide when a given density $\psi$ enjoys this property. For the scalar case, when either of the two dimensions $N$ or $m$ is unity, quasiconvexity reduces to usual convexity. But for genuine vector situations, it is not so. As a matter of fact, necessary and sufficient conditions for quasiconvexity in the vector case ($N, m>1$) were immediately sought, and important new convexity conditions were introduced:
\begin{itemize}
\item Rank-one convexity. A continuous integrand $\psi:\cM\to\R$ is said to be rank-one convex if
$$
\psi(t_1\bF_1+t_2\bF_2)\le t_1\psi(\bF_1)+t_2\psi(\bF_2),\quad t_1+t_2=1, t_1, t_2\ge0,
$$
whenever the difference $\bF_1-\bF_2$ is a rank-one matrix.
\item Polyconvexity. Such an integrand $\psi$ is polyconvex,  if it can be rewritten in the form $\psi(\bF)=g(\bM(\bF))$ where $\bM(\bF)$ is the vector of all minors of $\bF$, and $g$ is a convex (in the usual sense) function of all its arguments.
\end{itemize}

One of the main fields where vector variational problems are relevant is non-linear elasticity (\cite{Ball}). In particular, polyconvexity has played a major role in existence results. See also \cite{Ciarlet}. A main hypothesis to assume in this area is the rotationally invariance, as well as the behavior for large deformations. See \cite{DacorognaKoshigoe} for a discussion on all these notions of convexity under this invariance. Higher-order theories have also been explored, at least from an abstract point of view (\cite{Dal Maso et al.}, \cite{Meyers}). More general concepts of quasiconvexity have been introduced in \cite{Fonseca-Muller}. Recent interesting results about approximation by polynomials are worth mentioning \cite{Heinz}.
Explicit examples of rank-one convex functions can be found in various works: \cite{BandeiraPedregal}, \cite{DacorognaDouchetGangboRappaz}, \cite{SverakB}, among others. See also \cite{VanHove1}, \cite{VanHove2}.

It was very soon recognized that quasiconvexity implies rank-one convexity (by using a special class of test fields), and that polyconvexity is a sufficient condition for quasiconvexity. The task suggested itself as trying to prove or disprove the equivalence of these various kinds of convexity. In the scalar case, all three coincide with usual convexity, so that we are facing a purely vector phenomenon. It turns out that these three notions of convexity are different, and counterexamples of various sorts have been found over the years. See \cite{AlibertDacorogna}, \cite{DacorognaDouchetGangboRappaz}, \cite{SerreA}, \cite{Terpstra}.

If we focus on the equivalence of rank-one convexity and quasiconvexity, Morrey conjectured that they are not equivalent (\cite{Morrey}), though later he simply stated it as an unsolved problem (\cite{MorreyB}). The issue remained undecided until the surprising counterexamble by V. Sverak (\cite{Sverak}) after some other additional and very interesting results (\cite{SverakB}, \cite{SverakC}, \cite{SverakE}). What is quite remarkable is that the original counterexample is only valid when $m\ge3$, and later attempts to extend it for $m=2$ failed (\cite{bandeiraornelas}, \cite{PedregalH}, \cite{PedregalSverakB}). Other counterexamples have not been found. References \cite{faracolaszlo}, and \cite{KristensenB} are also relevant here.

The situation for two-component maps has stayed unsolved, though some evidence in favor of the equivalence has been gathered throughout the years. See \cite{ChaudMuller}, \cite{Muller}, \cite{MullerB}, \cite{ParryA}. It is also interesting to point out that for quadratic densities, rank-one convexity and quasiconvexity are equivalent regardless of dimensions. This has been known for a long time (\cite{Ball}, \cite{MorreyB}), and it is not difficult to prove it by using Plancherel's formula. A different point of view is taken in \cite{BandeiraPedregal}.  Another field where the resolution of this equivalence for two components maps would have an important impact is the theory of quasiconformal maps in the plane. There is a large number of references for this topic. See \cite{astalaiwaniecmartin} for a recent account. In particular, if the equivalence between rank-one convexity and quasiconvexity for two component maps turns out to be true, then the norm  of the corresponding Beurling-Ahlfors transform equals $p^*-1$ \cite{iwaniec}.

In this note, we support that Morrey was right in all cases: for dimensions $m$ and $N$ greater than unity, rank-one convexity does not imply quasiconvexity.
We provide an explicit family of maps for $m=N=2$, parametrized by $\tau$, and argue that, for $\tau$ small, they cannot be achieved by lamination. As it is well-known (\cite{PedregalI}), this is equivalent to showing that there are rank-one convex functions which are not quasiconvex. The genesis of such family of maps goes back to many efforts of the author in trying to show the equivalence of rank-one convexity and quasiconvexity, up to a point that it was clear the structure of a potential counterexample. A principal issue is how one could rigorously prove that a certain probability measure $\nu\equiv\nu_\tau$, supported in matrices, with a vanishing first moment, and generated by gradients, cannot be achieved by lamination, without finding or constructing a suitable rank-one convex function $\Psi$ for which
\begin{equation}\label{desigualdad}
\langle\Psi, \nu\rangle<\Psi(\cero).
\end{equation}
For our candidate $\nu\equiv\nu_\tau$, it may be possible to find one such explicit function $\Psi$ (even in the form of a four-degree polynomial  in $\dM$ as in \cite{Sverak} ) for which (\ref{desigualdad}) holds, and so such $\Psi$, though rank-one convex, cannot be quasiconvex. The author has not succeeded in doing so, and so the emphasis is placed in arguing that, for small $\tau$, $\nu_\tau$ is not, in fact, reachable by lamination.

The structure of the paper is as follows. We immediately introduce that family of gradients parametrized by a real parameter $\tau$, which is bound to be small. Since our strategy focuses in showing that, for sufficiently small $\tau$, those gradients cannot be achieved by lamination, we next recall the definition, and the basic facts about the class of laminates. In Section \ref{seccuatro}, we revise the case of $3\times2$-matrices so as to stress how one can conclude that a gradient is not a laminate without finding a rank-one convex which is not quasiconvex. Our discussion for the $3\times2$ case has been designed to follow as close as possible our strategy for the $2\times2$ case. For this important $2\times2$ situation, the proof has to be harder as a specific clear-cut example after the $3\times2$ counterpart is not possible, but rather one has to rely on a asymptotic situation as a certain parameter $\tau$ becomes small.
We then concentrate in the main goal of convincing that $\nu_\tau$ cannot be reached by lamination for small $\tau$.

\section{The counterexample}\label{secdos}

We will consider the following family of piecewise affine maps $\bu_\tau:D\subset\R^2\to\R^2$ parametrized by a real parameter $\tau$.
$D$ is the unit cube in $\R^2$. We in fact give their gradients $\nabla \bu_\tau:D\to\dM$ in which  matrices
\begin{gather}
\bX_1=\begin{pmatrix} 1-\tau&\tau\\ \tau&1-\tau\end{pmatrix}, \quad
\bX_2=\begin{pmatrix} 1+\tau&3\tau\\ 3\tau&1+\tau\end{pmatrix}, \nonumber\\
\bX_3=\begin{pmatrix} -1+\tau&3\tau\\ -\tau&1+\tau\end{pmatrix}, \quad
\bX_4=\begin{pmatrix} -1-\tau&\tau\\ -3\tau&1-\tau\end{pmatrix},\nonumber\\
\bX_5=\begin{pmatrix} 1+\tau&-\tau\\ 3\tau&-1+\tau\end{pmatrix}, \quad
\bX_6=\begin{pmatrix} 1-\tau&-3\tau\\ \tau&-1-\tau\end{pmatrix}, \nonumber\\
\bX_7=\begin{pmatrix} -1-\tau&-3\tau\\ -3\tau&-1-\tau\end{pmatrix}, \quad
\bX_8=\begin{pmatrix} -1+\tau&-\tau\\ -\tau&-1+\tau\end{pmatrix}.\nonumber
\end{gather}
participate according to the geometry provided in Figure 1. Subindices are only shown in this figure. Notice that the underlying barycenter is the zero matrix for all $\tau$.

The appropriate rank-one connections with the appropriate normals occur among them. This is elementary, but worth checking. We need to compute the differences
\begin{gather}
\bX_1-\bX_2=\begin{pmatrix}-2\tau&-2\tau\\-2\tau&-2\tau
\end{pmatrix},\quad \bX_1-\bX_4=\begin{pmatrix}2&0\\4\tau&0
\end{pmatrix}, \nonumber\\
\bX_1-\bX_6=\begin{pmatrix}0&4\tau\\0&2
\end{pmatrix},\quad \bX_2-\bX_3=\begin{pmatrix}2&0\\4\tau&0
\end{pmatrix},\nonumber\\
\bX_2-\bX_5=\begin{pmatrix}0&4\tau\\0&2
\end{pmatrix},\quad \bX_3-\bX_4=\begin{pmatrix}2\tau&2\tau\\2\tau&2\tau
\end{pmatrix},\nonumber\\
\bX_3-\bX_8=\begin{pmatrix}0&4\tau\\0&2
\end{pmatrix},\quad \bX_4-\bX_7=\begin{pmatrix}0&4\tau\\0&2
\end{pmatrix},\nonumber\\
\bX_5-\bX_6=\begin{pmatrix}2\tau&2\tau\\2\tau&2\tau
\end{pmatrix},\quad \bX_5-\bX_8=\begin{pmatrix}2&0\\4\tau&0
\end{pmatrix}, \nonumber\\
\bX_6-\bX_7=\begin{pmatrix}2&0\\4\tau&0
\end{pmatrix},\quad \bX_7-\bX_8=\begin{pmatrix}-2\tau&-2\tau\\-2\tau&-2\tau
\end{pmatrix}.\nonumber
\end{gather}
We therefore see that these differences are rank-one matrices with the appropriate normals according to Figure 1. In addition, we realize that, because only three rank-one matrices occur in those differences, the subspace spanned by those eight matrices is three-dimensional, with basis
\begin{equation}\label{base}
\begin{pmatrix}1&0\\2\tau&0
\end{pmatrix},\quad \begin{pmatrix}0&2\tau\\0&1
\end{pmatrix},\quad \begin{pmatrix}\tau&\tau\\ \tau&\tau
\end{pmatrix}.
\end{equation}
As a matter of fact, the coordinates of the $\bX_i$ matrices are easy to compute
\begin{gather}
\bX_1\mapsto (1, 1, -1),\quad \bX_2\mapsto (1, 1, 1),\quad
\bX_3\mapsto (-1, 1, 1),\quad \bX_4\mapsto (-1, 1, -1),\nonumber\\
\bX_5\mapsto (1, -1, 1),\quad \bX_6\mapsto (1, -1, -1),\quad
\bX_7\mapsto (-1, -1, -1),\quad \bX_8\mapsto (-1, -1, 1).\nonumber
\end{gather}
In this way, if $\nu_\tau$ designates the underlying (gradient Young) probability measure, then it is supported in the vertices of the unit cube of the subspace $\bL_\tau$ generated by (\ref{base}). More specifically
$$
\nu_\tau=\frac3{16}\left(\delta_{\bX_2}+\delta_{\bX_4}+\delta_{\bX_6}+\delta_{\bX_8}\right)
\frac1{16}\left(\delta_{\bX_1}+\delta_{\bX_3}+\delta_{\bX_5}+\delta_{\bX_7}\right),
$$
or  through the above identification
\begin{align}
\nu_\tau=&\frac3{16}\left(\delta_{(1, 1, 1)}+\delta_{(-1, 1, -1)}+\delta_{(1, -1, -1)}+\delta_{(-1, -1, 1)}\right)\label{tau}\\
&+\frac1{16}\left(\delta_{(1, 1, -1)}+\delta_{(-1, 1, 1)}+\delta_{(1, -1, 1)}+\delta_{(-1, -1, -1)}\right)\nonumber.
\end{align}

\bigskip
\begin{center}
\includegraphics[width=6cm]{def}

Figure 1. The family of mappings $\bu_\tau$.
\vspace*{0.5cm}
\end{center}

Note that the representation of $\nu_\tau$ in (\ref{tau}) does not depend on $\tau$, but rank-one directions in $\bL_\tau$ do, namely, rank-one directions $(x, y, z)$ must satisfy the equation
\begin{equation}\label{rangouno}
(1+2\tau)xy+\tau xz+\tau yz=0
\end{equation}
if $(x, y, z)$ are coordinates in $\bL_\tau$ with respect to the basis in (\ref{base}).

We would like to argue that this gradient Young measure $\nu_\tau$ cannot be achieved by lamination for $\tau$ sufficiently small. See \cite{PedregalI}) for a full discussion about gradient Young measures.

\section{Laminates}

\begin{definition}\label{laminado}
A probability measure $\nu$, supported in $\cM$  with barycenter $\cero$, is called a laminate if
there are matrices $\bX_{i, j}$, and weights $\lambda_{i, j}\ge0$, for $i\ge0$, $1\le j\le2^i$ with:
\begin{enumerate}
\item $\bX_{0, 1}=\cero$.
\item Main decomposition step:
$$
(\lambda_{i+1, 2j}+\lambda_{i+1, 2j-1})\bX_{i, j}=\lambda_{i+1, 2j}\bX_{i+1, 2j}+\lambda_{i+1, 2j-1}\bX_{i+1, 2j-1},
$$
and every difference
$$
\bY_{i, j}\equiv\bX_{i+1, 2j}-\bX_{i+1, 2j-1}
$$
is a rank-one matrix.
\item $\lambda_{i, j}=\lambda_{i+1, 2j}+\lambda_{i+1, 2j-1}$.
\item Limit behavior:
$$
\sum_{j=1}^{2^i}\lambda_{i, j}\delta_{\bX_{i, j}}\rightharpoonup \nu
$$
as $i\to\infty$.
\end{enumerate}
\end{definition}

The rank-one condition for each of the differences $\bY_{i, j}$ is the main structural condition. This condition implies that
for each such matrix $\bY_{i, j}$ there are vectors $\bv_{i, j}\in\R^m$, and $\bn_{i, j}\in\R^N$ so that $\bY_{i, j}=\bv_{i, j}\otimes\bn_{i, j}$. In particular, if we were to go through this decomposition process for each projection of $\nu$ onto each of the $k$-th row of matrices, $k=1, 2, \dots, m$, decomposition directions for all components  would simultaneously correspond to $\bn_{i, j}$. This identical decomposition directions for all rows is what ties together the (vector) measure $\nu$. The common direction $\bn_{i, j}$ also has a very clear geometrical interpretation as the ``normal to the layers" when a laminate is understood as a gradient Young measure (see \cite{PedregalI}).

\begin{definition}
A real function $\Psi:\cM\to\R$ is rank-one convex if
$$
\Psi(t\bZ_1+(1-t)\bZ_0)\le t\Psi(\bZ_1)+(1-t)\Psi(\bZ_0)
$$
whenever the difference matrix $\bZ_1-\bZ_0$ is of rank one.
\end{definition}

With these two basic definitions, it is elementary to show Jensen's inequality.
\begin{lemma}
Let $\nu$ be a laminate as above, and $\Psi$, a rank-one convex function. Then
\begin{equation}\label{jensen}
\langle\Psi, \nu\rangle\ge\Psi(\cero).
\end{equation}
\end{lemma}
What is interesting is that this inequality is a characterization of laminates (\cite{Pedregalam}, \cite{PedregalI}).
\begin{theorem}\label{caracrangouno}
Let $\nu$ be a probability measure as before. Then $\nu$ is a laminate in the sense of Definition (\ref{laminado}) if and only if for every rank-one convex function $\Psi$, (\ref{jensen}) is valid.
\end{theorem}
This result suggests a potential strategy to prove that there are rank-one convex functions which are not quasiconvex, if we can show that a certain probability measure $\nu$, underlying a gradient $\nabla\bu$ as in Section \ref{secdos}, is not a laminate, for then, according to Theorem \ref{caracrangouno}, there must be at least one rank-one convex function $\overline\Psi$ for which (\ref{jensen}) cannot be correct. This function $\overline\Psi$ cannot be quasiconvex because
$$
\int_Q\overline\Psi(\nabla\bu(\bx))\,d\bx=\langle\overline\Psi, \nu\rangle<\overline\Psi(\cero).
$$
This pretends to be our strategy.

Since we will be working with the full family of gradients parametrized by $\tau$ introduced in Section \ref{secdos}, it is important to write down explicitly such situation.
Assume $\nu_\tau$ is a laminate for all positive $\tau$. Then there are matrices $\bX^{(\tau)}_{i, j}$, and weights $\lambda^{(\tau)}_{i, j}\ge0$, for $i\ge0$, $1\le j\le2^i$ with:
\begin{enumerate}
\item $\bX^{(\tau)}_{0, 1}=\cero$, for all $\tau$.
\item Main decomposition step:
$$
(\lambda^{(\tau)}_{i+1, 2j}+\lambda^{(\tau)}_{i+1, 2j-1})\bX^{(\tau)}_{i, j}=\lambda^{(\tau)}_{i+1, 2j}\bX^{(\tau)}_{i+1, 2j}+\lambda^{(\tau)}_{i+1, 2j-1}\bX^{(\tau)}_{i+1, 2j-1},
$$
and every difference
$$
\bY^{(\tau)}_{i, j}\equiv\bX^{(\tau)}_{i+1, 2j}-\bX^{(\tau)}_{i+1, 2j-1}
$$
is a rank-one matrix.
\item $\lambda^{(\tau)}_{i, j}=\lambda^{(\tau)}_{i+1, 2j}+\lambda^{(\tau)}_{i+1, 2j-1}$.
\item Limit behavior:
$$
\sum_{j=1}^{2^i}\lambda^{(\tau)}_{i, j}\delta_{\bX^{(\tau)}_{i, j}}\rightharpoonup \nu_\tau
$$
as $i\to\infty$, for all $\tau>0$, or else
\begin{equation}\label{descomposicion}
\nu_\tau=\sum_{j=1}^{2^i}\lambda^{(\tau)}_{i, j}\delta_{\bX^{(\tau)}_{i, j}}
\end{equation}
for some finite $i$, that may depend on $\tau$.
\end{enumerate}
We also know that all matrices $\bX^{(\tau)}_{i, j}$ participating in these decompositions should belong to the convex hull of the eight matrices in the support of $\nu_\tau$.

Since our main argument will be asymptotic as $\tau\searrow0$, the following concept will play a relevant role. It identifies the rank-one directions we cannot be dispensed with in the limit.

\begin{definition}\label{limite}
With the notation just introduced, we say that a certain unit, rank-one matrix $\bY$ is a proper decomposition direction for the family of laminates $\{\nu_\tau\}$ as $\tau\searrow0$, if there is $i\ge0$, $1\le j\le 2^i$, and a subsequence $\tau_k\searrow0$ such that
$$
\frac{\bY^{(\tau_k)}_{i, j}}{|\bY^{(\tau_k)}_{i, j}|}\to\bY,\quad\frac{\lambda_{i+1, 2j}^{(\tau_k)}}{\lambda_{i+1, 2j}^{(\tau_k)}+\lambda_{i+1, 2j-1}^{(\tau_k)}}\to\overline\lambda_{i, j}\in(0, 1),\quad
\hbox{ as }k\to\infty.
$$
\end{definition}
The emphasis is placed in the fact that $0<\overline\lambda_{i, j}<1$.

\section{The $3\times2$ situation}\label{seccuatro}

As a preliminary step to better understand the main argument behind a proof to check that a certain probability measure is not reachable by lamination, we are going to treat, from this viewpoint, the case of $3\times2$ matrices. This material is taken from \cite{PedregalH}.

Consider the piecewise-affine map $\bu:D\subset\R^2\to\R^3$ determined by its gradient $\nabla \bu:D\to\bM^{3\times2}$ in which matrices
\begin{gather}
\bX_1=\begin{pmatrix} 1&0\\ 0&1\\ -1&-1\end{pmatrix}, \quad
\bX_2=\begin{pmatrix} 1&0\\ 0&1\\ 1&1\end{pmatrix}, \nonumber\\
\bX_3=\begin{pmatrix} -1&0\\0&1\\1&1\end{pmatrix}, \quad
\bX_4=\begin{pmatrix} -1&0\\ 0&1\\ -1&-1\end{pmatrix},\nonumber\\
\bX_5=\begin{pmatrix} 1&0\\ 0&-1\\ 1&1\end{pmatrix}, \quad
\bX_6=\begin{pmatrix} 1&0\\ 0&-1\\ -1&-1\end{pmatrix}, \nonumber\\
\bX_7=\begin{pmatrix} -1&0\\ 0&-1\\ -1&-1\end{pmatrix}, \quad
\bX_8=\begin{pmatrix} -1&0\\ 0&-1\\ 1&1\end{pmatrix}.\nonumber
\end{gather}
participate according again to the geometry in Figure 1. The underlying barycenter is the zero matrix $\cero$. Once again, it is elementary to check the various rank-one relations among those matrices with the corresponding normals, so that the underlying (homogeneous, gradient Young) probability measure is
$$
\nu=\frac3{16}\left(\delta_{\bX_2}+\delta_{\bX_4}+\delta_{\bX_6}+\delta_{\bX_8}\right)
\frac1{16}\left(\delta_{\bX_1}+\delta_{\bX_3}+\delta_{\bX_5}+\delta_{\bX_7}\right).
$$
Specifically, those differences are
\begin{gather}
\bX_1-\bX_2=\begin{pmatrix} 0&0\\ 0&0\\ -2&-2\end{pmatrix}=\begin{pmatrix}0\\0\\-2\end{pmatrix}\otimes\begin{pmatrix}1& 1\end{pmatrix},\nonumber\\ \bX_1-\bX_4=\begin{pmatrix} 2&0\\ 0&0\\ 0&0\end{pmatrix}=\begin{pmatrix}2\\0\\0\end{pmatrix}\otimes\begin{pmatrix}1& 0\end{pmatrix}, \nonumber\\
\bX_1-\bX_6=\begin{pmatrix} 0&0\\ 0&2\\ 0&0\end{pmatrix}=\begin{pmatrix}0\\2\\0\end{pmatrix}\otimes\begin{pmatrix}0& 1\end{pmatrix},\nonumber
\end{gather}
and then
\begin{gather}
\bX_2-\bX_3=\bX_1-\bX_4,\quad
\bX_2-\bX_5=\bX_1-\bX_6,\quad
\bX_3-\bX_4=\bX_2-\bX_1,\nonumber\\
\bX_3-\bX_8=\bX_1-\bX_6,\quad
\bX_4-\bX_7=\bX_1-\bX_6,\quad
\bX_5-\bX_6=\bX_1-\bX_2,\nonumber\\
\bX_6-\bX_7=\bX_1-\bX_4,\quad
\bX_7-\bX_8=\bX_1-\bX_2.\nonumber
\end{gather}
It is again easy to check that all those eight matrices belong to the subspace $\bL$ spanned by the basis
\begin{equation}\label{basee}
\left\{\begin{pmatrix}1&0\\0&0\\0&0\end{pmatrix},
\begin{pmatrix}0&0\\0&1\\0&0\end{pmatrix},
\begin{pmatrix}0&0\\0&0\\1&1\end{pmatrix}\right\},
\end{equation}
and that $\nu$ can be identified, using coordinates with respect to this basis, with
\begin{align}
\nu=&\frac3{16}\left(\delta_{(1, 1, 1)}+\delta_{(-1, 1, -1)}+\delta_{(1, -1, -1)}+\delta_{(-1, -1, 1)}\right)\label{contra}\\
&+\frac1{16}\left(\delta_{(1, 1, -1)}+\delta_{(-1, 1, 1)}+\delta_{(1, -1, 1)}+\delta_{(-1, -1, -1)}\right)\nonumber.
\end{align}
Matrices in $\bL$ are represented in the form
$$
\begin{pmatrix}x&0\\0&y\\ z&z\end{pmatrix},
$$
and so rank-one directions in $\bL$ correspond exactly to the three basis matrices with coordinates $(1, 0, 0)$, $(0, 1, 0)$, $(0, 0, 1)$. This means that in building laminates supported in $\bL$, we are only entitled to use these three decomposition directions. It is then clear that the asymmetry of weights in (\ref{contra}) cannot be achieved by using just these three directions in $\bL$ in the process determining a laminate (Definition \ref{laminado}). In fact, the only laminate supported in those eight matrices with barycenter $\cero$ is the one with symmetric weights
\begin{align}
\nu=&\frac1{8}\left(\delta_{(1, 1, 1)}+\delta_{(-1, 1, -1)}+\delta_{(1, -1, -1)}+\delta_{(-1, -1, 1)}\right)\nonumber\\
&+\frac1{8}\left(\delta_{(1, 1, -1)}+\delta_{(-1, 1, 1)}+\delta_{(1, -1, 1)}+\delta_{(-1, -1, -1)}\right)\nonumber.
\end{align}
So, $\nu$ in (\ref{contra}) cannot be a laminate. See Figure 2.

\bigskip
\begin{center}
\includegraphics[width=7cm]{tresxdos}

Figure 2. The $3\times2$ situation.
\vspace*{0.5cm}
\end{center}

\section{The $2\times2$ case}

We go back to the problem which is the main concern of this contribution. Notice how in this case, a subspace $\bL$ like the one in the preceding section cannot be found. This is essentially what makes the search for a counterexample in the $2\times2$ situation much more involved.
\subsection{The limit measure}
Because the probability measure $\nu_\tau$ in (\ref{tau}), associated with $\nabla \bu_\tau$, is independent of $\tau$, we can take advantage of this feature to examine the (singular) limit as $\tau\searrow0$. Indeed, the limit measure will be the same
\begin{align}
\nu_0=&\frac1{16}(3\delta_{(1, 1, 1)}+3\delta_{(1, -1, -1)}+3\delta_{(-1, 1, -1)}+3\delta_{(-1, -1, 1)}\nonumber\\
&+\delta_{(-1, 1, 1)}+\delta_{(1, -1, 1)}+\delta_{(1, 1, -1)}+\delta_{(-1, -1, -1)}),\nonumber
\end{align}
though the cone $\Lambda_0$ of rank-one directions will degenerate to the condition $xy=0$ which is the limit of the equation (\ref{rangouno}) as $\tau\searrow0$. $\nu_0$ should be a laminate with respect to this cone of directions. It is elementary to check that this is indeed so:
\begin{gather}
(0, 0, 0)=\frac12(0, -1, 0)+\frac12(0, 1, 0),\quad (0, 1, 0)-(0, -1, 0)=(0, 2, 0)\in\Lambda_0,\nonumber\\
(0, -1, 0)=\frac12(1, -1, -1/2)+\frac12(-1, -1, 1/2),\quad (1, -1, -1/2)-(-1, -1, 1/2)=(2, 0, -1)\in\Lambda_0,\nonumber\\
(0, 1, 0)=\frac12(1, 1, 1/2)+\frac12(-1, 1, -1/2),\quad (1, 1, 1/2)-(-1, 1, -1/2)=(2, 0, 1)\in\Lambda_0, \nonumber\\
(1, -1, -1/2)=\frac14(1, -1, 1)+\frac34(1, -1, -1),\quad (1, -1, 1)-(1, -1, -1)=(0, 0, 2)\in\Lambda_0,\nonumber\\
(-1, -1, 1/2)=\frac14(-1, -1, -1)+\frac34(-1, -1, 1),\quad (-1, -1, -1)-(-1, -1, 1)=(0, 0, -2)\in\Lambda_0,\nonumber\\
(1, 1, 1/2)=\frac14(1, 1, -1)+\frac34(1, 1, 1),\quad (1, 1, -1)-(1, 1, 1)=(0, 0, -2)\in\Lambda_0,\nonumber\\
(-1, 1, -1/2)=\frac14(-1, 1, 1)+\frac34(-1, 1, -1),\quad (-1, 1, 1)-(-1, 1, -1)=(0, 0, 2)\in\Lambda_0.\nonumber
\end{gather}
There is another possibility that can easily found by replacing the decomposition directions $(2, 0, -1)$ and $(2, 0, 1)$, by $(1, 0, -2)$ and $(1, 0, 2)$, respectively, and then using the direction $(1, 0, 0)$ instead of $(0, 0, 1)$. Figure 3 can help in visualizing the situation. But there are some other possibilities.

\bigskip
\begin{center}
\includegraphics[width=8cm]{lim}

Figure 3. The limit situation.
\vspace*{0.5cm}
\end{center}

\begin{lemma}
There is a constant $c_0>0$, such that proper decomposition directions to achieve $\nu_0$ through lamination can only be of the form $(c, 0, 1)$ or $(0, c, 1)$ with either $c=0$, $c=\infty$, or else $c_0\le |c| \le 1/c_0$.
\end{lemma}
\begin{proof}

Recall that $\Lambda_0$ is the cone of directions $(x, y, z)$ with $xy=0$.
The statement is elementary once we realize that only decomposition directions $(x, y, z)$ with $x=0$ or $y=0$ can be used, and that the partial barycenter for the four mass points corresponding to either $y=-1$, $y=1$, $x=-1$, or $x=1$ is in all cases the center of the corresponding square. The asymetric distribution of weights on each of these four faces of the cube takes place along the directions $(1, 0, 1)$, $(-1, 0, 1)$, $(0, -1, 1)$, and $(0, 1, 1)$, respectively. This asymetry forbids decomposition directions to be arbitrarily close to $(0, 0, 1)$, to $(1, 0, 1)$, or to $(0,1, 0)$, which are the directions corresponding to $c=0$, and to $c=\infty$.
\end{proof}

Let $\Lambda_\tau$ be the set of rank-one directions in $\bL_\tau$ complying with equation (\ref{rangouno}).
The feasible directions in the lemma
do not belong to $\Lambda_\tau$, though, by continuity, they are close to belonging to it. The directions
\begin{equation}\label{direcciones}
\left(c, \frac{-\tau c}{c+\tau(1+2c)}, 1\right),\quad
\left(\frac{-\tau c}{c+\tau(1+2c)}, c, 1\right),
\end{equation}
do belong to $\Lambda_\tau$. As already pointed out above, for $c=0$, we have $(0, 0, 1)$, while for $c=\infty$, the directions are $(1, 0, 0)$, and $(0, 1, 0)$.

\subsection{The contradiction}

Suppose the probability measure $\nu_\tau$ in (\ref{tau}) can be reached by lamination for all small $\tau$. There might be many decomposition directions involved in the  process for the laminate, some of which may even depend upon $\tau$, but, by continuity, only those which are close to the ones used for the limit case $\tau=0$ in the preceding subsection, can be proper according to Definition \ref{limite}.
By this we mean that corresponding relative weights stay away from zero and one, as $\tau\searrow0$. Those proper decomposition directions are
$$
(1, 0, 0),\quad (0, 1, 0),\quad (0, 0, 1),
$$
and those in (\ref{direcciones}) for $c_0\le|c|\le 1/c_0$. The true proper, decomposition directions for small $\tau$ might not be exactly these  ones, but their differences ought to converge to zero with $\tau$. Let us examine those decomposition directions, and find the real rank-one matrices underlying those directions through the identification associated with the basis in (\ref{base})
$$
(x, y, z)\mapsto \begin{pmatrix}x+\tau z&2\tau y+\tau z\\
2\tau x+\tau z&y+\tau z\end{pmatrix}.
$$
Indeed, we have
$$
(1, 0, 0)\mapsto\begin{pmatrix}1&0\\2\tau&0\end{pmatrix}\mapsto (1, 0),\quad
(0, 1, 0)\mapsto\begin{pmatrix}0&2\tau\\0&1\end{pmatrix}\mapsto (0, 1),\quad
(0, 0, 1)\mapsto\begin{pmatrix}\tau&\tau\\ \tau&\tau\end{pmatrix}\mapsto (1, 1),
$$
for either $c=0$, or $c=\infty$, while for finite $c$, vectors and matrices are
\begin{gather}
\left(c, \frac{-\tau c}{c+\tau(1+2c)}, 1\right)\mapsto\begin{pmatrix}c+\tau&\tau(c+\tau)/(c+\tau(1+2c))\\ \tau(1+2c)&\tau^2(1+2c)/(c+\tau(1+2c))\end{pmatrix}\mapsto (1, \tau/(c+\tau(1+2c))),\nonumber\\
\left(\frac{-\tau c}{c+\tau(1+2c)},  c, 1\right)\mapsto\begin{pmatrix}\tau(c+\tau)/(c+\tau(1+2c))&c+\tau\\ \tau^2(1+2c)/(c+\tau(1+2c))&\tau(1+2c)\end{pmatrix}\mapsto (\tau/(c+\tau(1+2c)), 1).\nonumber
\end{gather}
The final vectors written for each case correspond to the true lamination directions, i.e. the normal to the parallel layers.
Hence, we conclude that the proper normals used in the lamination process for $\nu_\tau$ have to be close to one of the three directions $(1, 0)$, $(0, 1)$, or $(1, 1)$.
The contradiction arises because it is impossible to get the asymmetry in the weights for $\nu_\tau$ by using these three decomposition directions simultaneously in both components, much in the same way as with the $3\times2$ situation.

To argue further on this issue, let us focus on the probability measures $\nu_\tau^{(l)}$, $l=1, 2$, corresponding to the two components in (\ref{tau}). Namely,
\begin{equation}\label{componentes}
\nu_\tau^{(l)}=\frac3{16}\left(\delta_{\bX_2}^{(l)}+\delta_{\bX_4}^{(l)}+\delta_{\bX_6}^{(l)}+\delta_{\bX_8}^{(l)}\right)
\frac1{16}\left(\delta_{\bX_1}^{(l)}+\delta_{\bX_3}^{(l)}+\delta_{\bX_5}^{(l)}+\delta_{\bX_7}^{(l)}\right),\hbox{ for }l=1, 2.
\end{equation}
Here $\bX_s^{(l)}$ is the $l$-th row of $\bX_s$:
\begin{gather}
\bX_1^{(1)}=\begin{pmatrix} 1-\tau&\tau\end{pmatrix},
\bX_1^{(2)}=\begin{pmatrix}\tau&1-\tau\end{pmatrix}, \quad
\bX_2^{(1)}=\begin{pmatrix} 1+\tau&3\tau\end{pmatrix},
\bX_2^{(2)}=\begin{pmatrix}3\tau&1+\tau\end{pmatrix}, \nonumber\\
\bX_3^{(1)}=\begin{pmatrix} -1+\tau&3\tau\end{pmatrix},
\bX_3^{(2)}=\begin{pmatrix}-\tau&1+\tau\end{pmatrix}, \quad
\bX_4^{(1)}=\begin{pmatrix} -1-\tau&\tau\end{pmatrix},
\bX_4^{(2)}=\begin{pmatrix}-3\tau&1-\tau\end{pmatrix},\nonumber\\
\bX_5^{(1)}=\begin{pmatrix} 1+\tau&-\tau\end{pmatrix},
\bX_5^{(2)}=\begin{pmatrix}3\tau&-1+\tau\end{pmatrix}, \quad
\bX_6^{(1)}=\begin{pmatrix} 1-\tau&-3\tau\end{pmatrix},
\bX_6^{(2)}=\begin{pmatrix}\tau&-1-\tau\end{pmatrix}, \nonumber\\
\bX_7^{(1)}=\begin{pmatrix} -1-\tau&-3\tau\end{pmatrix},
\bX_7^{(2)}=\begin{pmatrix}-3\tau&-1-\tau\end{pmatrix}, \quad
\bX_8^{(1)}=\begin{pmatrix} -1+\tau&-\tau\end{pmatrix},
\bX_8^{(2)}=\begin{pmatrix}-\tau&-1+\tau\end{pmatrix}.\nonumber
\end{gather}

\bigskip
\begin{center}
\includegraphics[width=12cm]{geom1}

Figure 4. The first component of  $\bu_\tau$.

\end{center}

\bigskip
\begin{center}
\includegraphics[width=4cm]{geom2}

Figure 5. The second component of $\bu_\tau$.
\vspace*{0.5cm}
\end{center}

The eight vectors participating in $\nu_\tau^{(1)}$ are
\begin{gather}
\bX_1^{(1)}=\begin{pmatrix} 1-\tau&\tau\end{pmatrix},
\quad
\bX_2^{(1)}=\begin{pmatrix} 1+\tau&3\tau\end{pmatrix},
\nonumber\\
\bX_3^{(1)}=\begin{pmatrix} -1+\tau&3\tau\end{pmatrix},
\quad
\bX_4^{(1)}=\begin{pmatrix} -1-\tau&\tau\end{pmatrix},
\nonumber\\
\bX_5^{(1)}=\begin{pmatrix} 1+\tau&-\tau\end{pmatrix},
\quad
\bX_6^{(1)}=\begin{pmatrix} 1-\tau&-3\tau\end{pmatrix},
\nonumber\\
\bX_7^{(1)}=\begin{pmatrix} -1-\tau&-3\tau\end{pmatrix},
\quad
\bX_8^{(1)}=\begin{pmatrix} -1+\tau&-\tau\end{pmatrix},
\nonumber
\end{gather}
while those occurring in $\nu_\tau^{(2)}$ are
\begin{gather}
\bX_1^{(2)}=\begin{pmatrix}\tau&1-\tau\end{pmatrix}, \quad
\bX_2^{(2)}=\begin{pmatrix}3\tau&1+\tau\end{pmatrix}, \nonumber\\
\bX_3^{(2)}=\begin{pmatrix}-\tau&1+\tau\end{pmatrix}, \quad
\bX_4^{(2)}=\begin{pmatrix}-3\tau&1-\tau\end{pmatrix},\nonumber\\
\bX_5^{(2)}=\begin{pmatrix}3\tau&-1+\tau\end{pmatrix}, \quad
\bX_6^{(2)}=\begin{pmatrix}\tau&-1-\tau\end{pmatrix}, \nonumber\\
\bX_7^{(2)}=\begin{pmatrix}-3\tau&-1-\tau\end{pmatrix}, \quad
\bX_8^{(2)}=\begin{pmatrix}-\tau&-1+\tau\end{pmatrix}.\nonumber
\end{gather}
It is easy to represent both probability measures in the plane. See Figures  4 and 5.

Since $\tau$ tends to zero, the proper decomposition directions are, as indicated above, $(1, 0)$, $(0, 1)$, $(1, 1)$.
When we use a decomposition rank-one matrix in $\nu_\tau$, associated with a direction close to $(1, 0)$ in both components, that direction might help in creating the asymmetry of weights for $\nu_\tau^{(1)}$ in (\ref{componentes}) (see Figure 6), but not for $\nu_\tau^{(2)}$, precisely because the support of $\nu_\tau^{(1)}$ is stretched out  in that direction, but the support of $\nu_\tau^{(2)}$ is arbitrarily thin in that same direction. Since the asymmetry of weights has to be created simultaneously in both components, it is impossible for a direction close to $(1, 0)$ to do so. Exactly the same argument, used symmetrically in both components, shows that a decomposition direction close to $(0, 1)$ cannot produce the asymmetry of weights in $\nu_\tau$. But under these circumstances, neither a direction close to $(1, 1)$ by itself can produce that asymmetry. Only can the probability measure
$$
\overline\nu_\tau=\frac18\left(\delta_{\bX_2}+\delta_{\bX_4}+\delta_{\bX_6}+\delta_{\bX_8}
+\delta_{\bX_1}+\delta_{\bX_3}+\delta_{\bX_5}+\delta_{\bX_7}\right),
$$
be a laminate for small $\tau$.

\bigskip
\begin{center}
\includegraphics[width=12cm]{geom11}

Figure 6. Assymetry of weights with the $(1, 0)$ direction.
\vspace*{0.5cm}
\end{center}

\section{A final remark}

It is interesting to note that if we are allowed to perturbed a bit some of the matrices $\bX'$s occurring in $\nu_\tau$, then we would have a laminate. Indeed, taking advantage of the decompositions for the limit rank-one cone $\Lambda_0$ written earlier, and using the corresponding approximating directions from the cone $\Lambda_\tau$, one can
redo essentially the same decompositions replacing the directions $(2, 0, 1)$ and $(2, 0, -1)$ by the two first ones in (\ref{direcciones}), respectively, and adjusting a little bit the weights.
See Figure 7.

Specifically, one arrives at
\begin{gather}
(0, 0, 0)=a_1(\tau)(0, -1-\frac\tau{2+3\tau}, 0)+a_2(\tau)(0, 1+\frac\tau{2+5\tau}, 0),\nonumber\\
(0, 1+\frac\tau{2+5\tau}, 0)-(0, -1-\frac\tau{2+3\tau}, 0)=(0, 2+4\tau\frac{1+2\tau}{(2+3\tau)(2+5\tau)}, 0)\in\Lambda_\tau,\nonumber\\
a_1(\tau)=\frac{(2+6\tau)(2+3\tau)}{8+36\tau+38\tau^2},\quad a_2(\tau)=\frac{(2+4\tau)(2+5\tau)}{8+36\tau+38\tau^2},\nonumber\\
(0, -1-\frac\tau{2+3\tau}, 0)=\frac12(1, -1, -1/2)+\frac12(-1, -1-\frac{2\tau}{2+3\tau}, 1/2),\nonumber\\
(1, -1, -1/2)-(-1, -1-\frac{2\tau}{2+3\tau}, 1/2)=(2, \frac{2\tau}{2+3\tau}, -1)\in\Lambda_\tau,\nonumber\\
(0, 1+\frac\tau{2+5\tau}, 0)=\frac12(1, 1, 1/2)+\frac12(-1, 1+\frac{2\tau}{2+5\tau}, -1/2),\nonumber \\
(1, 1, 1/2)-(-1, 1+\frac{2\tau}{2+5\tau}, -1/2)=(2, \frac{-2\tau}{2+5\tau}, 1) \in\Lambda_\tau, \nonumber\\
(1, -1, -1/2)=\frac14(1, -1, 1)+\frac34(1, -1, -1),\quad (1, -1, 1)-(1, -1, -1)=(0, 0, 2)\in\Lambda_\tau,\nonumber\\
(-1, -1-\frac{2\tau}{2+3\tau}, 1/2)=\frac14(-1, -1-\frac{2\tau}{2+3\tau}, -1)+\frac34(-1, -1-\frac{2\tau}{2+3\tau}, 1),\nonumber\\
(-1, -1-\frac{2\tau}{2+3\tau}, -1)-(-1, -1-\frac{2\tau}{2+3\tau}, 1)=(0, 0, -2)\in\Lambda_\tau,\nonumber\\
(1, 1, 1/2)=\frac14(1, 1, -1)+\frac34(1, 1, 1),\quad (1, 1, -1)-(1, 1, 1)=(0, 0, -2)\in\Lambda_\tau,\nonumber\\
(-1, 1+\frac{2\tau}{2+5\tau}, -1/2)=\frac14(-1, 1+\frac{2\tau}{2+5\tau}, 1)+\frac34(-1, 1+\frac{2\tau}{2+5\tau}, -1),\nonumber\\
(-1, 1+\frac{2\tau}{2+5\tau}, 1)-(-1, 1+\frac{2\tau}{2+5\tau}, -1)=(0, 0, 2)\in\Lambda_\tau.\nonumber
\end{gather}

The laminate that stems from this decomposition process is
\begin{align}
\mu_\tau=&\frac{3a_1(\tau)}{8}\delta_{(1, -1, -1)}+
\frac{a_1(\tau)}{8}\delta_{(1, -1, 1)}+\frac{3a_2(\tau)}{8}\delta_{(1, 1, 1)}+\frac{a_2(\tau)}{8}\delta_{(1, 1, -1)}\nonumber\\
&+\frac{3a_1(\tau)}{8}\delta_{(-1, -1-\frac{2\tau}{2+3\tau}, 1)}
+\frac{a_1(\tau)}{8}\delta_{(-1, -1-\frac{2\tau}{2+3\tau}, -1)}\nonumber\\
&+\frac{3a_2(\tau)}{8}\delta_{(-1, 1+\frac{2\tau}{2+5\tau}, -1)}
+\frac{a_2(\tau)}{8}\delta_{(-1, 1+\frac{2\tau}{2+5\tau}, 1)}.\nonumber
\end{align}
Notice that
$$
a_1(\tau), a_2(\tau)\to1/2,\quad a'_2(\tau), a'_2(\tau)\to0
$$
as $\tau\to0$. It is the fact that mass points in $\nu_\tau$ are fixed, independent of $\tau$, and symmetrically located that prevents $\nu_\tau$ from being a laminate.

\bigskip
\begin{center}
\includegraphics[width=8cm]{limtorcido}

Figure 7. The $\tau$-laminate.
\vspace*{0.5cm}
\end{center}

config: arxiv


TITLE: Question about Intermediate Value Theorem
QUESTION [4 upvotes]: In the solution, it says that $f(a)\ge a$ and $f(b)\le b$  but it do not seem obvious for me. If I am just given that a $f:[a,b]\to[a,b]$, how do I know is this function increasing, decreasing or just a horizontal line?

REPLY [0 votes]: Go by contradiction, let $f:[a,b]\rightarrow [a,b]$be continuous with no fixed point, so $\forall x\in[a,b]:f(x)\ne x$.

If you consider $a\in[a,b]$, then it has to follow $f(a)\ne a \Rightarrow f(a) > a$.
If there exists $x\in[a,b]$ such that $f(x) < x$, then continuity would imply $\exists x'\in[a,b]: f(x') = x'$, so $\forall x\in[a,b]:f(x) > x$.
Since $f(b)\in[a,b]$, either $f(b) = b$ or $f(b) < b$. First case is an instant contradiction, second case would follow (2) and give a contradiction.

So $f$ has to have a fixed point.


TITLE: the convergence of $\int_0^\infty t^{2k}e^{-xt^2}dt $ being uniform on $[\delta,\infty)$ for $\delta\gt 0$
QUESTION [0 upvotes]: Can somebody explain to me the convergence of $\int_0^\infty t^{2k}e^{-xt^2}dt $ being uniform on $[\delta,\infty)$ for $\delta\gt 0$ ?
Also according to the theorem above shouldn't be $t^{2k}e^{-xt^2}\leq t^{2k}e^{-\delta t^2}$ for $t\geq \delta$ instead of $x\geq \delta$  ? And how do we show $\int_\delta^{\infty} t^{2k}e^{-\delta t^2}dt \lt \infty$?

REPLY [1 votes]: The theorem you quote is sometimes called 'dominated convergence,' and the idea is that if the family of functions lives within a region cut out by an integrable function, then the family of functions cannot escape and be wild.
The observation is that as a function in $x$, we have $f(x, \cdot) = t^{2k}e^{-xt^2}$ which is monotonically decreasing as $x$ increases and $t$ is held constant, but not integrable at $0$, as now you're just integrating an increasing polynomial over a ray to $\infty$.  All one has to do then is check that this function is integrable, and then the claim follows on any ray $[a, \infty)$ which does not contain $0$, since the function is not defined there. Thus, taking $\delta > 0$, no matter how small, suffices.
The value of that last integral is not easy to find by hand, but if you know about the Gamma function, then you can show that it can be expressed in those terms. There's a calculation available here, if you enter the integral.


text:
\begin{document}
\title{A note on the Baker--Campbell--Hausdorff series in terms of right-nested commutators

}

\author{Ana Arnal\footnote{ Email: \texttt{ana.arnal@uji.es}. ORCD: 0000-0002-3283-3379}
 \and
 Fernando Casas\footnote{Email: \texttt{Fernando.Casas@uji.es}. ORCID: 0000-0002-6445-279X}
 \and Cristina Chiralt\footnote{Email: \texttt{chiralt@uji.es}. ORCID: 0000-0003-0925-4034}

}

\date{}

\maketitle

\begin{center}
Institut de Matem\`atiques i Aplicacions de Castell\'o (IMAC) and  De\-par\-ta\-ment de
Ma\-te\-m\`a\-ti\-ques, Universitat Jaume I,
  E-12071 Cas\-te\-ll\'on, Spain.
\end{center}

\vspace*{0.5cm}

\begin{abstract}

We get compact expressions for the Baker--Campbell--Hausdorff series $Z = \log(\e^X \, \e^Y)$ in terms of right-nested commutators. The reduction in the number of terms
originates from two facts: (i) we use as a starting point an explicit expression directly involving independent commutators and (ii) we derive
a complete set of identities arising among right-nested commutators.  The procedure allows us to obtain the series with fewer terms than when expressed in the classical
Hall basis at least up to terms of grade 10.

\

\noindent
\textit{Keywords}: Baker--Campbell--Hausdorff formula, exponentials, commutators

\noindent
\textit{MSC (2010)}: 22Exx, 17Bxx

\end{abstract}\bigskip

\section{Introduction}

Exponentials of non-commuting operators appear in many areas of physics and mathematics, ranging from quantum mechanics to
the theory of Lie groups and Lie algebras to the numerical analysis of differential equations. It is then natural to consider products
of such exponentials and how to express such products as the exponential of a new operator. This of course is closely related with the celebrated Baker--Campbell--Hausdorff
theorem \cite{bonfiglioli12tin}.

In the most basic algebraic setting,
one considers the associative algebra $\mathbb{K}\langle X, Y \rangle$ of formal power series in  the non-commuting variables
$X$ and $Y$ over a field $\mathbb{K}$ of characteristic zero. Then  $\e^X \, \e^Y = \e^{\Phi(X,Y)}$, with
\begin{equation}  \label{zwc}
\aligned
   \Phi(X,Y)  & = \log( \e^X \, \e^Y) = \sum_{k \ge 1} \frac{(-1)^{k-1}}{k} (\e^{X} \e^{Y} -1)^k = \sum_{k \ge 1} \frac{(-1)^{k-1}}{k}
   \left( \sum_{p+q > 0} \frac{X^p Y^q}{p! q!} \right)^k \\
   & =  \sum_{k \ge1} \frac{(-1)^{k-1}}{k}
   \sum \frac{X^{p_1} Y^{q_1} \ldots X^{p_k} Y^{q_k}}{p_1! \,
      q_1! \ldots p_k! \, q_k!},
\endaligned
\end{equation}
where, in the last expression, the inner summation extends over all non-negative integers
$p_1$, $q_1$, \ldots, $p_k$, $q_k$ for which $p_i + q_i > 0$
($i=1,2, \ldots, k$). The first terms read explicitly
 \[
 \aligned
  & \Phi =  (X + Y + X Y + \frac{1}{2} X^2 + \frac{1}{2} Y^2 + \cdots )
 - \frac{1}{2} ( XY + YX + X^2 + Y^2 + \cdots) + \cdots \\
  & =  X + Y + \frac{1}{2} (X Y - Y X) + \cdots = X + Y + \frac{1}{2} [X,Y] + \cdots
\endaligned
 \]
The Baker--Campbell--Hausdorff (BCH) theorem states that $\Phi(X,Y)$ in (\ref{zwc}) can be expressed
as
 \begin{equation}  \label{eq.1.3}
   \Phi(X,Y)  = X + Y +
    \sum_{m \ge 2} \, \Phi_m(X,Y),
\end{equation}
where $\Phi_m(X,Y)$ is a \emph{homogeneous Lie polynomial} in $X$ and $Y$
of degree $m$, i.e., a linear combination
of commutators of the form $[V_1,[V_2, \ldots,[V_{m-1},V_m]
\ldots]]$ with $V_i \in \{X,Y\}$ for $1 \le i \le m$, the
coefficients being rational constants. The formal power series (\ref{eq.1.3}) is called the Baker--Campbell--Hausdorff series, and
plays a fundamental role not only in the theory of Lie groups and Lie algebras \cite{bonfiglioli12tin}, but also in linear differential equations, control theory,
quantum and statistical mechanics, and numerical analysis (see e.g. \cite{bonfiglioli10aov,sornborger99hom,strichartz87tcb,varadarajan84lgl,wilcox67eoa}).

An explicit expression for $\Phi_n$ in the BCH series was provided
 by Dynkin \cite{dynkin47eot,dynkin00spo} in the form
\begin{equation}  \label{eq.1.2}
  \Phi_m(X,Y) =  \sum_{p_i, q_i}
    \frac{(-1)^{m-1}}{m} \frac{[X^{p_1} Y^{q_1} \ldots
      X^{p_m} Y^{q_m} ]}{(\sum_{i=1}^m (p_i + q_i)) \,
      p_1! \, q_1! \ldots p_m! \, q_m!},
\end{equation}
where the summation is taken over all non-negative
integers $p_1, q_1, \ldots$, $p_m$, $q_m$ such that
$p_1+q_1>0, \ldots, p_m + q_m > 0$ and
$[ X^{p_1} Y^{q_1} \ldots X^{p_m} Y^{q_m}]$ denotes the right-nested commutator based on the \emph{word}
$ X^{p_1} Y^{q_1} \ldots X^{p_m} Y^{q_m}$, i.e.,
\[
   [XY^2X^2 Y] \equiv [X Y Y X X Y] \equiv [X,[Y,[Y,[X,[X,Y]]]]].
\]
Expression (\ref{eq.1.2}) can be used in principle to compute $\Phi_n$ in the BCH series  up to
any desired order. One should notice, however, is that not all the terms are independent, due to
the many existing redundancies. Thus, for instance,
$[X^3 Y^1] = [X^1 Y^0 X^2 Y^1] = [X,[X,[X,Y]]]$. An additional source of redundancies arises from the Jacobi identity \cite{varadarajan84lgl}:
\begin{equation}  \label{jacobi}
    [X_1,[X_2,X_3]] + [X_2,[X_3,X_1]] + [X_3,[X_1,X_2]] = 0,
\end{equation}
for any three variables $X_1, X_2, X_3$, and other identities obtained from it. From this perspective, a procedure allowing to remove at once all the superfluous
terms in (\ref{eq.1.2}) would be of great value for practical applications.

Although different procedures exist in the literature to construct the BCH series up to an arbitrary degree in terms of commutators,
all of them have a basic limitation, as is the case with the Dynkin presentation (\ref{eq.1.2}): not all the commutators are
independent, and so a rewriting process has to be
carried out to express the results in terms of a basis of the free Lie algebra $\mathcal{L}(X,Y)$ generated by $X$ and $Y$. This process, of course, although can be
carried out by computer algebra systems,
requires a good deal of
computational time and memory resources.

One of the most efficient
algorithms was proposed in \cite{casas09aea}, where explicit expressions of $\Phi_{m}$ up to $m=20$
in terms of the classical Hall and Lyndon basis of $\mathcal{L}(X,Y)$ were obtained with relatively modest computer requirements.
In any event, the fact that no basis in the free Lie algebra
exists that eases the calculation of the BCH series is one major problem when dealing with problems where this series
plays a role \cite{burgunder08eia}.

Expressing the BCH series in terms of right-nested commutators presents several advantages, especially when the series is
considered in some particular physical settings. There are problems whose structure leads in a natural way to consider the
BCH series of two operators $X$, $Y$ satisfying $[Y,[Y,[X,Y]]] \equiv 0$. This happens, in particular, when designing
splitting methods for the numerical integration of classical Hamiltonian systems and also for the time-dependent Schr\"odinger
equation. Although in this case it is still possible to construct a generalized Hall basis \cite{mclachlan19tla}, it is much simpler
to identify the non-vanishing terms when using right-nested commutators. On the other hand, and contrary to Hall--Viennot
bases, there is no a straightforward procedure to construct a set of independent right-nested commutators generating each
homogeneous subspace of $\mathcal{L}(X,Y)$.

Several attempts have been made to directly remove in (\ref{eq.1.2}) redundant terms and therefore to express $\Phi_m$ only as
a linear combination of independent right-nested commutators. Thus, we can mention in particular references \cite{oteo91tbc}
and \cite{kolsrud93mri}, where compact expressions up to $m=8$ and $m=9$, respectively, have been reported, after identifying highly
non-trivial commutators identities arising when $m \ge 4$.

In this work we show that it is indeed possible to get directly rather compact expressions for $\Phi_m$ in terms of right-nested
commutators without much computational effort, sometimes with fewer terms than when expressed in the classical Hall basis.
This reduction is still more remarkable if the existing commutator identities are introduced at each degree. In addition, the procedure
can be easily extended to the BCH series involving any number of variables,
\begin{equation}  \label{bch-n}
  \exp(X_1) \, \exp(X_2) \cdots \, \exp(X_n) = \exp \big( \Phi(X_1, X_2, \ldots, X_n) \big).
\end{equation}
This can be achieved by considering, instead of the Dynkin presentation (\ref{eq.1.2}), another explicit expression of $\Phi_m$ as a linear
combination of products of $m$ operators $X$ and $Y$ ordered according with the group of permutations. It turns out that such a formula
was also originally obtained by Dynkin and published in his somehow unnoticed
paper \cite{dynkin49otr}\footnote{We are grateful to Prof. M. M\"uger for bringing  this
reference  to our attention in his ``Notes on the theorem of Baker--Campbell--Hausdorff--Dynkin", available at \newline
\url{https://www.math.ru.nl/~mueger/PDF/BCHD.pdf}}.

One could also consider, of course, left-nested commutators instead and the same results would be still valid with a factor $(-1)^k$, if $k$
is the number of nested commutators.

\section{The BCH series in terms of permutations}

We consider the general case (\ref{bch-n}), i.e.,
\begin{equation}  \label{bchg1}
  \Phi(X_1, X_2, \ldots, X_n) = \sum_{m \ge 1} \Phi_m(X_1, \ldots, X_n),
\end{equation}
where $\Phi_m(X_1, \ldots, X_n)$ is a homogeneous polynomial of degree $m$ in the non-com\-mu\-ta\-ti\-ve
variables $X_1, \ldots, X_n$.

Let us denote  by $\varphi_n(X_1, X_2, \ldots, X_n)$ the multilinear part of $\Phi_n(X_1, \ldots, X_n)$, i.e.,
the part obtained by replacing $X_i^2$ by $0$ for all $i$ in
$\Phi_n(X_1, \ldots, X_n)$. Then one has the following remarkable result (see e.g. \cite{loday94sdh}):

\begin{proposition}  \label{propo.1}
  It holds that
\begin{equation}   \label{ei.4}
   \Phi_m(X_1, \ldots, X_n) = \sum_{i_1 + \cdots + i_n = m \atop i_j \ge 0}
   \frac{1}{i_1! \cdots i_n!} \, \varphi_m(\underbrace{X_1, \ldots, X_1}_{i_1}, \ldots,
   \underbrace{X_n, \ldots, X_n}_{i_n})
\end{equation}
\end{proposition}
As far as we know, the first proof of Proposition \ref{propo.1} is due to Dynkin \cite{dynkin49otr}. Later on, it was shown that the notion of
Eulerian idempotent leads to a shorter proof \cite{loday94sdh,burgunder08eia}.
The explicit expression of $\varphi_n(X_1, X_2, \ldots, X_n)$
 can be obtained as follows \cite{loday94sdh}.
Since we are only interested in the
multilinear part of $\Phi_n$, we have to replace
\[
   \exp(X_1) \, \exp(X_2) \cdots \, \exp(X_n)
\]
by
\[
    (1+X_1) (1+ X_2)  \cdots (1+ X_n)
\]
and analyze $\log\big((1+X_1) (1+ X_2)  \cdots (1+ X_n)\big)$, or more specifically, its multilinear part. In other words, we have
to deal with $\log(1+Z)$, where
\[
  Z = \sum_i X_i + \sum_{i < j} X_i X_j + \sum_{i < j < k} X_i X_j X_k + \cdots X_1 X_2 \cdots X_n.
\]

It is then clear that $\varphi_n$ is of the form
\begin{equation}  \label{egs}
   \varphi_n(X_1, X_2, \ldots, X_n) = \sum_{\sigma \in S_n} c_{\sigma} \, X_{\sigma(1)} X_{\sigma(2)} \cdots X_{\sigma(n)}
\end{equation}
where the sum is extended over all permutations $\sigma$ of $\{1, 2, \ldots, n \}$. The coefficients $c_{\sigma}$ in (\ref{egs}) can be obtained by
analyzing the contribution coming from each power $Z^k$ in the expansion $\log(1+Z) = Z - \frac{Z^2}{2} + \cdots + \frac{(-1)^{k-1}}{k}
Z^k + \cdots$. The situation is similar to the computation of the explicit expression for the Magnus expansion as carried out, in
particular, in \cite{strichartz87tcb}: it turns out that
\[
   c_{\sigma} =  \sum_{i_1 + \cdots + i_m = n  \atop \sigma \in S(i_1, \ldots, i_m)} \frac{(-1)^{m-1}}{m},
\]
where the sum is extended over all ordered partitions $i_1 + \cdots + i_m = n$ of $n$ such that $\sigma \in S(i_1, \ldots, i_m)$,
with
\[
   S(i_1, \ldots, i_m) = \{ \sigma \in S_n \, | \, \sigma(j) < \sigma(j+1) \, \mbox{ for all } \, j \ne i_1 + \cdots + i_{\ell}, \;\; \ell = 1, \ldots, m-1 \}
\]
and $S_n$ denotes the permutation group.

A straightforward computation shows that this number is  $\binom{n - d_{\sigma} - 1}{m-1-d_{\sigma}}$, where $d_{\sigma}$ is the number of
\emph{descents} in $\sigma$. We recall that $\sigma \in S_n$ has an ascent in $i$ if $\sigma(i) < \sigma(i+1)$, $i=1,\ldots, n-1$
and it has a descent in $i$ if $\sigma(i) > \sigma(i+1)$.

In this way one arrives at \cite{strichartz87tcb,loday94sdh}
\[
  c_{\sigma} = \sum_{m = d_{\sigma} + 1}^n \frac{(-1)^{m-1}}{m} \binom{n - d_{\sigma} - 1}{m-1-d_{\sigma}} =
  \frac{(-1)^{d_{\sigma}}}{n}  \, \frac{1}{ \binom{n-1}{d_{\sigma}}}
\]
and finally
\begin{equation}   \label{ei.1}
  \varphi_n(X_1, X_2, \ldots X_n) = \frac{1}{n}    \sum_{\sigma \in S_n} \, (-1)^{d_{\sigma}} \, \frac{1}{ \binom{n-1}{d_{\sigma}}} \,
    X_{\sigma(1)} X_{\sigma(2)} \cdots X_{\sigma(n)}.
\end{equation}

At this point some remarks are in order:
\begin{itemize}
 \item The existing relationship between the multilinear part $\varphi_n$ with the Eulerian idempotent can be both ways: either to compute the coefficients $c_{\sigma}$
 in (\ref{egs}) by applying
 different descriptions of this object  \cite{solomon68otp,loday89osl} or by providing an explicit combinatorial expression
 for this Eulerian idempotent with (\ref{ei.1}) that allows in particular to characterize its symmetries \cite{burgunder08eia}.
 \item Goldberg \cite{goldberg56tfp} analyzed the formal power series (\ref{bch-n}) when $n=2$ characterizing  the coefficient of the
 general term $X_1^{s_1} X_2^{s_2} \cdots$ in terms of certain polynomials. This result was generalized to an arbitrary $n$ in
 \cite{kobayashi98gta} (see also \cite{helmstetter89sdh}). It turns out that the coefficients $c_{\sigma}$ and the
 explicit expression (\ref{ei.1}) reproduce these previous results.
\end{itemize}

If we restrict ourselves to the case of two variables $X_1 \equiv X$, $X_2 \equiv Y$, then eq. (\ref{ei.4}) reads
\begin{equation}  \label{ei.5}
  \Phi_m(X,Y) = \sum_{i+j = m \atop i,j \ge 1} \, \frac{1}{i!} \frac{1}{j!} \, \varphi_m(\underbrace{X, \ldots, X}_{i}, \underbrace{Y, \ldots, Y}_{j}),
\end{equation}
and, according with eq. (\ref{ei.1}), one gets for the first terms
\begin{equation}  \label{fterms}
\aligned
 & \Phi_2(X,Y) = \varphi_2(X, Y) =  \frac{1}{2} X Y - \frac{1}{2} Y X\\
  & \Phi_3(X,Y) = \frac{1}{2} \, \varphi_3(X, X, Y) +  \frac{1}{2} \, \varphi_3(X, Y, Y) \\
  &  \qquad \qquad = \frac{1}{12} XXY - \frac{1}{6} XYX + \frac{1}{12} XYY + \frac{1}{12} YXX - \frac{1}{6} YXY + \frac{1}{12} YYX \\
  & \Phi_4(X,Y) = \frac{1}{6} \, \varphi_4(X,X,X,Y)  + \frac{1}{4} \, \varphi_4(X,X,Y,Y) + \frac{1}{6} \, \varphi_4(X,Y,Y,Y) \\
  &  \qquad \qquad = \frac{1}{24} XXYY - \frac{1}{12} XYXY + \frac{1}{12} YXYX - \frac{1}{24} YYXX
\endaligned
\end{equation}

\section{The BCH series in terms of commutators}

Application of Dynkin--Specht--Wever (DSW) theorem to (\ref{ei.1}) allows one to express $\varphi_n(X_1,\ldots, X_n)$ in terms
of commutators and get an alternative expression for the homogeneous Lie polynomial $\Phi_n$. If we define the
Lie bracketing from right to left by the unique linear map $r$ such that for any word $w = a_1 a_2 \ldots a_{n-1} a_n$ of length $n$
one has $r(w) = [a_1,[a_2,\ldots, [a_{n-1}, a_n] \cdots]]$, the DSW theorem states that
for each homogeneous Lie polynomial $P$ of degree $n$, it is true that $r(P) = n P$ \cite{reutenauer93fla}. In consequence,
\begin{equation}   \label{ei.1b}
  \varphi_n(X_1, X_2, \ldots, X_n) = \frac{1}{n^2}    \sum_{\sigma \in S_n} \, (-1)^{d_{\sigma}} \, \frac{1}{ \binom{n-1}{d_{\sigma}}} \,
    [X_{\sigma(1)}, [X_{\sigma(2)} \cdots [X_{\sigma(n-1)},X_{\sigma(n)}] \cdots ]].
\end{equation}

Notice that, as is the case with (\ref{eq.1.2}), not all the commutators (Lie brackets) appearing in
(\ref{ei.1b}) are linearly independent
among each other, due to antisimmetry and the Jacobi identity. Thus, if one aims to get
a expression in terms of independent commutators, then a particular basis of the vector subspace
spanned by those commutators in which each generator appears exactly once has to be considered. If we denote this subspace by
$L_n(X_1,\ldots, X_n)$, then $\dim L_n = (n-1)!$.

Among the possible bases of $L_n$,
the class considered by Dragt \& Forest \cite{dragt83con} is particularly appropriate.
In forming such a basis,
one uses only those right-nested brackets ending with a particular but otherwise arbitrary variable selected from the
 collection $X_1, X_2, \ldots, X_n$.  If this
variable is chosen as $X_n$, then the basis is formed by the right-nested brackets of the form
  \[
      [X_k, [X_{j},\ldots [X_i, X_n] \ldots ]],
  \]
  where the indices $k, j, \ldots i$ are all possible permutations of $\{1, 2, \ldots n-1 \}$. Of course, there are $n$ different such bases, depending on the particular
ending operator one selects. What makes this class of bases specially compelling is the following property. Suppose we have an expression in terms of
products of $n$ distinct operators $X_1, \ldots, X_n$ which is
known to be written as a linear combination of right-nested commutators in $L_n(X_1,\ldots, X_n)$. This is the case, in particular, of $\varphi_n$. Suppose
all the right-nested commutators ending with, say, $X_n$ are used as a basis. Then, in this linear combination, the
coefficient of the right-nested commutator
  $ [X_k, [X_{j},\ldots [X_i, X_n] \ldots ]]$ is \emph{precisely} the coefficient of the permutation $\alpha = (k j \ldots i n)$
in the original expression.

 In consequence, if apply this observation to eq. (\ref{ei.1}), we end up with
\begin{equation}  \label{ei.2}
  \varphi_n(X_1, X_2, \ldots, X_n) = \frac{1}{n}    \sum_{\sigma \in S_{n-1}} \, (-1)^{d_{\sigma}} \, \frac{1}{ \binom{n-1}{d_{\sigma}}} \,
    [X_{\sigma(1)}, [X_{\sigma(2)}, \cdots , [X_{\sigma(n-1)}, X_n] \cdots ]],
\end{equation}
just involving the $(n-1)!$ permutations of $S_{n-1}$. Of course,
similar formulas can be obtained for $\varphi_n$ if instead of choosing $X_n$ as the last element in the right-nested commutator of the
basis one takes any other element $X_j$, $j=1,\ldots, n-1$. In any case, a more compact expression than (\ref{ei.1b}) for $\varphi_n$
is obtained in this way, since all the commutators are now independent.

\section{Reducing the number of commutators}

Applying formula (\ref{ei.2}) and taking into account eq. (\ref{ei.5}) to the case of two variables we get
\begin{equation}  \label{bchg1}
  \Phi_m(X,Y)  = \sum_{i+j = m \atop i,j \ge 1} \, \frac{1}{i!} \frac{1}{j!} \, \frac{1}{m}
    \sum_{\sigma \in S_{m-1}} \,  \frac{(-1)^{d_{\sigma}} }{ \binom{m-1}{d_{\sigma}}} \,
    [X_{\sigma(1)}, [X_{\sigma(2)}, \cdots , [X_{\sigma(m-1)}, Y] \cdots ]],  \nonumber
\end{equation}
where $X_{\sigma(i)}$, $i=1,\ldots,m-1$, can be either $X$ or $Y$ according with the particular permutation $\sigma$ considered. In particular, for the first
terms we have
 \begin{align}  \label{ft1}
  & \Phi_2(X,Y) = \frac{1}{2} [X,Y] \nonumber \\
  & \Phi_3(X,Y) = \frac{1}{12}  [X,[X,Y]] + \frac{1}{6} [X,[Y,Y]] - \frac{1}{12} [Y,[X,Y]] \\
  & \Phi_4(X,Y) = \frac{1}{24} [X,[X,[Y,Y]]] - \frac{1}{24} [X,[Y,[X,Y]]] + \frac{1}{36} [X,[Y,[Y,Y]]] - \nonumber  \\
   &  \qquad \qquad \;\; - \frac{1}{36} [Y,[X,[Y,Y]]]  \nonumber
\end{align}
If we take into account, however, the obvious property $[z,z] =0$,  then  we get the correct formula for $\Phi_m(x,y)$ up to $m=4$ and a much reduced number of terms in
$\Phi_m$ for $m \ge 5$ than formula (\ref{eq.1.3}) with (\ref{eq.1.2}).
To substantiate this claim, we have elaborated the code presented in Appendix \ref{apenA}  for the computation of $\varphi_m$
and the functions $\Phi_m(X,Y)$. The number of terms in $\Phi_m$ produced by this code
 is collected in the fifth line of Table
\ref{tab1} up to $m=10$. It is labelled as ``no identities'' to emphasize the fact that no existing identities among commutators have been yet implemented. We include for
comparison the corresponding number of terms of $\Phi_m(X,Y)$ in the classical Hall (third line) and Lyndon bases (fourth line) as obtained by applying the procedure
of \cite{casas09aea}. For completeness, we also write the dimension of each homogeneous subspace $\mathcal{L}_m(X,Y)$ of the free Lie algebra  $\mathcal{L}(X,Y)$.

\begin{table}
\begin{center}
\begin{tabular}{|c|ccccccccc|}  \hline
$m$  &    2  &  3  &  4  &  5  &  6  &  7  &  8  &  9  &  10   \\ \hline\hline
$\dim \mathcal{L}_m(X,Y)$  &    1  &  2  &  3  &  6  &  9  &  18  &  30  &  56  &  99   \\  \hline
\# terms Hall basis  &   1  &  2  &  1  &  6  &  6  &  18  &  24  &  56  &  86  \\
\# terms Lyndon  basis &    1  &  2  &  1  &  6  &  5  &  18  &  17  &  55  &  55   \\  \hline
no identities  &   1  &  2  &  1  &  8  &  7  &  32  &  31  & 96   &  97  \\
Grade 4 &   1  &  2  &  1  &  6  &  5  &  24  &  23  &  78  &  78  \\
Grade 6  &  1  &  2  &  1  &  6  &  4  &  18  &  17  &  67  &  65 \\
Compact  &  1  &  2  &  1  &  6  &  4  &  18  &  13  &  38 &   52 \\  \hline
Symmetric compact  & 0  &  2  & 0  &  6  & 0  &  18  & 0  & 42 &   0 \\  \hline

\end{tabular}
\end{center}
\caption{Number of terms in the homogeneous Lie polynomial $\Phi_m(X,Y)$ in the BCH series for the first values of $m$, together with the dimension of $\mathcal{L}_m(X,Y)$.
The last line refers to the symmetric BCH series, eq. (\ref{sy.4}).}
 \label{tab1}
\end{table}

If we incorporate the identity
\[
     [Y,[X,[X,Y]]] = [X,[Y,[X,Y]]]
\]
appearing at $m=4$ into the procedure, we get the numbers collected in the line labelled ``Grade 4''. It is remarkable that this number agrees with the corresponding to
the Lyndon basis up to $m=6$, whereas $\Phi_8$ and even $\Phi_{10}$ contain a smaller number of terms than in the classical Hall basis.

Whereas no further identities exist at $m=5$, the following three appear at $m=6$, namely \cite{oteo91tbc}
\begin{eqnarray*}
 \mathrm{(i)} &  & [X,X,X,Y,X,Y] - 2 \, [X,Y,X,X,X,Y] + [Y,X,X,X,X,Y]  = 0, \\
 \mathrm{(ii)} & & [X,X,Y,Y,X,Y] + 3 \, [Y,X,X,Y,X,Y]   - 3 \, [X,Y,X,Y,X,Y]  \\
  & & \;  \qquad\qquad -  [Y,Y,X,X,X,Y] = 0, \\
 \mathrm{(iii)} & & [Y,Y,X,Y,X,Y]  - 2 \, [Y,X,Y,Y,X,Y]  + [X,Y,Y,Y,X,Y] = 0,
\end{eqnarray*}
where we have denoted  $ [X,X,X,Y,X,Y]  :=  [X,[X,[X,[Y,[X,Y]]]]]$, etc., for simplicity.
By incorporating them into the algorithm we get a further reduction, as the line labelled ``Grade 6'' in Table \ref{tab1} clearly shows.

As a matter of fact, a systematic procedure to generate all the existing identities  at a given $m$ can be designed by
using tools of liner algebra, and in particular Gaussian elimination, as is explained in Appendix \ref{apenB} (see also \cite{rencangli95rto}).
The algorithm can also be used to construct bases of the homogeneous subspace
$\mathcal{L}_m(X,Y)$ for any $m \ge 1$ formed by right-nested commutators in a quite straightforward manner. For completeness, we have collected all such
existing identities up to $m=10$ in the reference \cite{explicitidentities}.

Once the identities have been obtained, to get compact expressions a particular basis has to be identified at each $m$ so that the number of
vanishing coefficients of $\Phi_m(X,Y)$ is as large as possible. This can be done either by inspection (for small $m$) or applying the technique proposed in
\cite{kolsrud93mri}. By proceeding in this way, we have been able to get rather compact expressions for
$\Phi_m$, as shown in Table \ref{tab1} (line labelled ``Compact"). The corresponding explicit expressions can be found at \cite{explicitidentities}.
In the reduction process a relevant role is played by the existing symmetries, namely
\[
\begin{aligned}
  &  \Phi_m(-X,-Y) = (-1)^m \, \Phi_m(X,Y) \\
  & \Phi_m(X,Y) = (-1)^{m+1} \, \Phi_m(Y,X)
\end{aligned}
\]

\section{Further considerations}

The above procedure can also be easily generalized to any number of variables.
In particular, from eqs. (\ref{ei.4}) and (\ref{ei.2})  we get for the case of three variables
\begin{equation}   \label{sy.3}
\begin{aligned}
  & \Phi_2(x_1,x_2,x_3) = \frac{1}{2} [x_1,x_2] + \frac{1}{2} [x_1,x_3] + \frac{1}{2} [x_2,x_3] \\
  & \Phi_3(x_1,x_2,x_3) = \frac{1}{12} [x_1,[x_1,x_2]] + \frac{1}{12} [x_1,[x_1,x_3]] + \frac{1}{3} [x_1,[x_2,x_3]] - \frac{1}{12} [x_2,[x_1,x_2]]  \\
  & \quad - \frac{1}{6} [x_2,[x_1,x_3]] + \frac{1}{12} [x_2,[x_2,x_3]] - \frac{1}{12} [x_3,[x_1,x_3]] - \frac{1}{12} [x_3,[x_2,x_3]],
\end{aligned}
\end{equation}
etc. This can be applied to get directly the so-called symmetric BCH formula,
\begin{equation}  \label{sy.4}
  \exp(\frac{1}{2} X) \, \exp(Y) \,  \exp(\frac{1}{2} X) = \exp(\Psi(X,Y)) = \sum_{m=1}^{\infty} \Psi_m(X,Y)
\end{equation}
of great relevance in the design of time-symmetric splitting and composition methods (see e.g. \cite{blanes16aci}, \cite{hairer06gni} and references therein).
In this case it is easy to show that in general $\Psi_{m}(X,Y) = 0$ when $m$ is even.

Of course, to get compact expressions we have to apply the same procedure as before to the corresponding formulas (\ref{sy.3}) with the obvious replacements
$x_1 \rightarrow X/2$, $x_2 \rightarrow Y$, $x_3 \rightarrow X/2$. It is more advantageous, however, to start with a different expression for (\ref{sy.4}) involving less terms
before applying the reduction procedure. This can be achieved by connecting $\Phi(X,Y)$ and $\exp(\Psi(X,Y))$ as follows:
\[
  \e^{\Psi(X,Y)} = \e^{-\frac{X}{2}} \, \e^X \, \e^{Y} \, \e^{\frac{X}{2}} = \e^{-\frac{X}{2}} \, \e^{\Phi(X,Y)}  \, \e^{\frac{X}{2}},
\]
so that
\begin{equation} \label{sy.5}
   \Psi(X,Y) = \e^{-\mathrm{ad}_{\frac{X}{2}}} \, \Phi(X,Y) = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^k k!} \, \mathrm{ad}_X^k \Phi(X,Y)
\end{equation}
where
\[
    \mathrm{ad}_A B = [A,B], \qquad \mathrm{ad}_A^j B = [A, \mathrm{ad}_A^{j-1} B], \qquad    \mathrm{ad}_A^0 B = B.
\]
If we use the compact expressions for $\Phi_m(X,Y)$ obtained in the previous section, then the number of terms in the corresponding $\Psi_m$ determined according with
(\ref{sy.5}) diminishes considerably. For instance, when $m=9$ we get  52 terms instead of 121. By applying the existing identities, this number is further reduced to 42. For
comparison, $\Psi_9$ in the Hall basis contains 56 terms \cite{casas09aea}

The last line in Table \ref{tab1} contains the number of terms of $\Psi_m(X,Y)$  up to $m=9$, whereas the explicit expressions can also be found at \cite{explicitidentities}.

On the other hand, formula (\ref{bchg1}) for $\Phi_m(X,Y)$ can be in fact obtained when the Magnus expansion is used to construct the formal solution of the
differential equation
\begin{equation}  \label{eqdif}
Y^{\prime}(t)=A(t)Y(t),\qquad\qquad Y(0)=I,
\end{equation}
when $A(t)$ is defined as
\[
   A(t) = \theta(t-1) X + \big( \theta(t) + \theta(t-1) \big) Y,
\]
$\theta(t)$ being the step function. As is well known, the solution of (\ref{eqdif}) can be written as
\[
   Y(t,0) = \exp \Omega(t,0),
\]
where $\Omega$ is an infinite series
   \begin{equation}
\Omega(t,0)=\sum_{m=1}^{\infty}\Omega_{m}(t,0), \qquad \mbox{ with } \qquad \Omega_m(0,0) = 0,  \label{ME}
\end{equation}
whose terms are increasingly complex expressions involving time-ordered integrals  of nested commutators of $A$ evaluated at different times.
An explicit expression for $\Omega_m(t,0)$, $m \ge 1$, in terms of iterated integrals of linear combinations of independent commutators has been obtained
in \cite{arnal18agf}, namely
\begin{equation}   \label{me.com3}
\begin{aligned}
  & \Omega_m(t,0) = \frac{1}{m} \sum_{\sigma \in S_{m-1}} \, (-1)^{d_b} \frac{1}{\binom{m-1}{d_b}} \,
    \int_0^t dt_1 \int_0^{t_1} dt_2 \cdots \int_0^{t_{m-1}} dt_m \,  \\
    &  \qquad\qquad\qquad  [A(t_{\sigma(1)}), [A(t_{\sigma(2)}) \cdots
    [A(t_{\sigma(m-1)}), A(t_m)] \cdots ]]
\end{aligned}
\end{equation}
where $\sigma$ and $d_{\sigma}$ have the same meaning as in (\ref{ei.2}). Notice that, since $Y(t=2,0) = \e^X \e^Y$, then $\log ( \e^X \e^Y ) = \Omega(2,0)$,
and $\Omega_m(2,0)$ as given by (\ref{me.com3}) reproduces exactly the expression of $\Phi_m(X,Y)$ given by (\ref{bchg1}). This can be checked order by order.

\subsection*{Acknowledgements}

FC would like to thank the Isaac Newton Institute for Mathematical Sciences for support and hospitality during the programme ``Geometry,
compatibility and structure preservation in computational differential equations", when work
on this paper was undertaken. This work was supported by EPSRC Grant Number EP/R014604/1 and by
Ministerio de Econom{\'i}a y Com\-pe\-ti\-ti\-vi\-dad (Spain) through project MTM2016-77660-P (AEI/FEDER, UE).

\appendix
\section{Appendix}  \label{apenA}

The following code in \emph{Mathematica} implements (perhaps not in the most efficient way) formula (\ref{ei.4}):
{\small
\begin{alltt}
fact[n_,m_]:=Product[k[j]!,\{j,1,n-1\}](m-Sum[k[j],\{j,1,n-1\}])!;
table[\{h_\,_\,_\}]:= Table[h];
intable[n_,m_]:= Join[Table[\{X[j], \{r[j],1, If[j == n,
   m-Sum[k[j], \{j,1,n-1\}], k[j]]\}\}, \{j,1,n\}]];
count[n_,m_]:= Table[\{k[j],0,If[j == 1, m,
   m-Sum[k[i],\{i,1,j-1\}]]\}, \{j,1,n-1\}];
Phinm[f_][n_,m_][\{counter_\,_\,_\}]:= Sum[1/fact[n,m] f[Flatten[
   Join[Table[table[intable[n,m][[j]]],\{j,1,n\}]],2]], counter];
Phi[f_][m_][\{h_\,_\,_\}]:= Phinm[f][Length[\{h\}],m][count[Length[\{h\}],
     m]]/. Table[X[j]-> \{h\}[[j]], \{j,1,Length[\{h\}]\}];
\end{alltt}
}
Next we compute $\Phi_m(X_1, \ldots, X_n)$ with the explicit expression (\ref{ei.1}) for $\varphi_m $ in terms of non-commutative products \texttt{Mm}. This is contained in
\texttt{PhiMm[m][\{$X_1, \ldots, X_n$\}]}:

{\small
\begin{alltt}
s[n_]:= Permutations[Range[1, n]];
coef[p_]:= coef[p] = Module[\{b, n\}, b = 0; n = Length[p];
   Do[If[p[[i]] > p[[i + 1]], b = b + 1], \{i, 1, n - 1\}];
   (-1)^b/(n Binomial[n - 1, b])];
ei[\{h_\,_\,_\}]:= Sum[coef[s[Length[\{h\}]][[j]]] Apply[Mm,
   Permutations[\{h\}][[j]]],   \{j,1,Length[\{h\}]!\}];
Ei[\{h_\,_\,_\}]:= ei[Table[x[i], \{i,1,Length[\{h\}]\}]]/.
     Table[x[j]-> \{h\}[[j]], \{j,1,Length[\{h\}]\}];
PhiMm[m_][\{h_\,_\,_\}]:= PhiMm[m][\{h\}]= Phi[Ei][m][\{h\}];
\end{alltt}
}
Then we express $\varphi_n(X_1, \ldots, X_n)$ in terms of commutators, equation (\ref{ei.2}), and finally $\Phi_m$ by computing \texttt{PhiCmt[m][\{$X_1, \ldots, X_n$\}]}:
{\small
\begin{alltt}
Cmt[x_,x_]:= 0;
Cmt[a_\,_\,_,0,b_\,_\,_]:= 0;
Cmt[a_\,_\,_,x_+y_,b_\,_\,_]:= Cmt[a,x,b] + Cmt[a,x,b];
Cmt[a_\,_\,_,n_ x_Cmt,b_\,_\,_]:= n Cmt[a,x,b];
Cmt[a_\,_\,_,t^n_ x_,b_\,_\,_] := t^n Cmt[a,x,b];
Cmt[a_\,_\,_,n_(x_+y_),b_\,_\,_] := Cmt[a,n x,b] + Cmt[a,n y,b];
(* Cmt /: Format[Cmt[x_, y_]]:= SequenceForm["[",x,",",y,"]"]; *)

cmtt[\{b_,a_\}]:=Cmt[b,a];
cmtt[\{a_,h_\,_\,_\}]:=Cmt[a,cmtt[\{h\}]];
basisn[n_]:= basisn[n]= Partition[Flatten[Tuples[
   \{Permutations[Range[1, n-1]], \{n\}\}]], n];
Varphi[\{h_\,_\,_\}]:= Sum[coef[basisn[Length[\{h\}]][[j]]]
   cmtt[Map[x, basisn[Length[\{h\}]][[j]]]]/. Table[x[j]->
   \{h\}[[j]], \{j, 1, Length[\{h\}]\}], \{j,1,(Length[\{h\}]-1)!\}];
PhiCmt[m_][\{h_\,_\,_\}]:= PhiCmt[m][\{h\}]= Phi[Varphi][m][\{h\}];
\end{alltt}
}
The first block defines the commutator (just the linearity property and the antisymmetry) with the correct format for output if necessary. Linearity properties in an analogous way should be implemented for the non commutative product \texttt{Mm}. The second block defines the basis and the generic term
$\varphi_n(X_1,\ldots, X_n)$.
Finally, let us remark that since the number of variables is free, the same code allows one to compute both the BCH  \texttt{PhiCmt[m][\{$X,Y$\}]} and the
symmetric BCH series, \texttt{PhiCmt[m][\{$\frac{1}{2} X,Y, \frac{1}{2} X$\}]}.

\section{Appendix}  \label{apenB}
The algorithm we have applied to generate the identities among commutators and a basis of the homogeneous subspace $\mathcal{L}_m(X,Y)$ formed by
right-nested commutators is a generalization of a procedure proposed in \cite{rencangli95rto}, and
 can be summarized as follows:

\

\noindent
For each $j=2, \ldots, m$, do
\begin{enumerate}
  \item Generate all possible right-nested commutators $C_i$ involving $j$ operators $X$ and $m-j$ operators $Y$.
 For example, with $m=4$,
 $$\mathcal{B}_ 4=\{ [X,[X,[X,Y]]], [Y,[X,[X,Y]]], [X,[Y,[X,Y]]], [Y,[Y,[X,Y]]]\}$$

  \item Generate the corresponding element $\langle C_i \rangle$ in the homogeneous subspace $\mathcal{U}_m(X,Y)$ of the universal enveloping
  algebra associated with $\mathcal{L}(X,Y)$. This is done by expanding each commutator $[A,B] = A B - B A$. {For example, on the previous list, for  $C_1=[X,[X,[X,Y]]]$,
  $$<C_1>=XXXY-3XXYX+3XYXX-YXXX$$}
  \item The element $\langle C_i \rangle$ is then a linear combination of words $X_{i_1} X_{i_2} \cdots X_{i_{m}}$, where $X_{i_j}$ is either $X$ of $Y$.
  The total number of words is $\binom{m}{j}$. Once all these words are arranged in a prescribed order, the element commutator $C_i$ can be
  identified with the vector $(a_1, a_2, \ldots, a_p)$ formed by the linear combination. {Then $C_1$ would be $C_1\equiv (1,-3,0,3,0,0,-1,0,0,0,0,0)$.}

  \item { Define a matrix $A$ whose rows are formed by these coefficient vectors and augment it to the right with the $m \times m$ identity matrix, forming an block matrix $\left(A | I\right)$. Apply Gauss-Jordan elimination and get the block matrix $\left(M | P\right)$. For $m=4$ we have,

  	\footnotesize{
$$\left(A | I\right)=\left(
\begin{array}{cccccccccccc|cccc}
1 & -3 & 0 & 3 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & -1 & 0 & 2 & 0 & 0 & -2 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 & 2 & 0 & 0 & -2 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -3 & 0 & 3 & -1 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
$$}
\normalsize{and}
\footnotesize{$$\left(M | P\right)=\left(
\begin{array}{cccccccccccc|cccc}
	1 & -3 & 0 & 3 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 1 & 0 & -2 & 0 & 0 & 2 & 0 & -1 & 0 & 0 & 0 & 0 & -1 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -3 & 0 & 3 & -1 & 0 & 0 & 0 & 1 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\
\end{array}
\right)
$$
}
}
  \item Now the non-vanishing rows on $M$ give the commutators of the basis. The identities we want to find out are obtained after making equal to zero the linear combinations on $P\cdot\mathcal{B}_ n$ corresponding to the vanishing rows.  In our example, since there is one null row on $M$ we get one Grade 4 identity, we equal to zero the last element on the product $P\cdot\mathcal{B}_ 4$, that is,

$$0=[Y,[X,[X,Y]]]-[X,[Y,[X,Y]]].
$$

\end{enumerate}

\bibliographystyle{siam}


text:
\subsection{Plate with a hole}
As the first example, we consider the well-known benchmark of an infinite plate with a hole, e.g.~\cite[\S58]{muskhelishvili:77}. Due to symmetry, only a quarter of the plate is considered, and the infinite geometry is cut with the exact traction being applied as a boundary condition. The exact setting is illustrated in Figure~\ref{fig:plateSetup}.

We consider the equations of linear elasticity, where the Cauchy stress $\stress(\mathbf u)$  depends linearly on the strain $\strain(\mathbf u) =\sym{ \nabla \mathbf{u}}$ via Hooke's law as $\stress(\mathbf u) =2\lameM \strain(\mathbf u) + \lameL \mattrace \strain(\mathbf u) \idmat$, with the trace operator $\mattrace \strain = \sum_i \strain_{ii}$ and the Lam\'e parameters $\lameL, \lameM$.
The Lam\'e parameters can be computed by
\[
\lambda= \frac{\nu E}{(1+\nu)(1-2\nu)},\qquad \mu = \frac{E}{2(1+\nu)}.
\]
We measure convergence in the energy norm
\[
\left\| \mathbf{u} - \mathbf{u}_h \right\|_{\rm E}^2  = \sum_k \int_{\Omega_k} \stress(\mathbf{u} - \mathbf{u}_h ) : \strain(\mathbf{u} - \mathbf{u}_h ) \dx,
\]
where the optimal convergence order is $\mathcal{O}(h^{p})$, see~\cite[Theorem 6]{brivadis:15}.

\begin{figure}[h!]
\begin{center}
\includegraphics[width=.5\textwidth]{figures/plate/InfPlateSetup.eps}
\caption{Plate with a hole: Geometry, boundary conditions and material parameters.}
\label{fig:plateSetup}
\end{center}
\end{figure}

Different geometry parametrizations are considered, as shown in Figure~\ref{fig:plateGeometries}.
The first case (Figure~\ref{plateGeometriesA})  is a two-patch setting with a straight interface, where the parametrization of the interface is the same on both subdomains.
In the second case (Figure~\ref{plateGeometriesB}) the same subdomains are considered, but with a different parametrization, such that the parametrizations along the interface do no longer match.
We note, that this is a situation, where the construction of~\cite{zou:17} is not exact, but requires additional steps of refinement.
In the third case (Figure~\ref{plateGeometriesC}), the subdomains are coupled across a curved interface.

\begin{figure}[h!]
\subfloat[straight interface, matching parametrization]
{
\hbox{
\setlength{\unitlength}{0.04300bp}
 \begin{picture}(2300,2200)(0,0)
  \put(750,1800){\makebox(0,0){$\xi=1/2$}}
  \put(1050,1650){\vector(1,-1){300}}
  \put(1350,770){\makebox(0,0){$\xi=1/2$}}
  \put(1220,930){\vector(1,3){130}}
   \put(0,0){\includegraphics[width = 2300\unitlength]{figures/plate/patches_straight_2_3_Matching-crop}}
\end{picture}
}
\label{plateGeometriesA}
}\hfill
\subfloat[straight interface, non-matching parametrization]
{
\hbox{
\setlength{\unitlength}{0.04300bp}
 \begin{picture}(2300,2200)(0,0)
  \put(750,1800){\makebox(0,0){$\xi=1/2$}}
  \put(1050,1650){\vector(1,-1){300}}
  \put(1060,470){\makebox(0,0){$\xi=1/2$}}
  \put(930,630){\vector(1,3){130}}
   \put(0,0){\includegraphics[width = 2300\unitlength]{figures/plate/patches_straight_2_3_nonMatching-crop}}
\end{picture}
}
\label{plateGeometriesB}
}\hfill
\subfloat[curved interface, matching parametrization]
{
\hbox{
\setlength{\unitlength}{0.04300bp}
 \begin{picture}(2300,2200)(0,0)
  \put(550,1950){\makebox(0,0){$\xi=1/2$}}
  \put(850,1800){\vector(1,-1){300}}
  \put(1195,920){\makebox(0,0){$\xi=1/2$}}
  \put(1065,1080){\vector(1,3){130}}
   \put(0,0){\includegraphics[width = 2300\unitlength]{figures/plate/patches_curved_2_3_Matching-crop}}
\end{picture}
}
\label{plateGeometriesC}
}
\caption{Different geometric setups of the plate with hole, shown for the mesh ratio $2:3$. }
\label{fig:plateGeometries}
\end{figure}

For all three setups, convergence for quadratic splines and different Lagrange multiplier bases is presented in Figure~\ref{fig:convergence_p_2}.
We observe an optimal order convergence for the newly constructed biorthogonal basis functions ('optimal'), with similar error values as with a standard Lagrange multiplier ('std'). In comparison, the naive, element-wise biorthogonal basis functions ('ele dual') as considered in~\cite{seitz:16} show suboptimal convergence, especially when  the slave mesh is coarser than the master mesh.

\begin{figure}
\scalebox{.5}{
\input{figures/plate/straight_MatchingParam_2nd/convergence.tex}
\input{figures/plate/straight_MatchingParam_2nd_2.9/convergence.tex}
}
\\
\scalebox{.5}{
\input{figures/plate/straight_nonMatchingParam_2nd/convergence.tex}
\input{figures/plate/straight_nonMatchingParam_2nd_2.9/convergence.tex}
}\\
\scalebox{.5}{
\input{figures/plate/curved_MatchingParam_2nd/convergence.tex}
\input{figures/plate/curved_MatchingParam_2nd_2.9/convergence.tex}
}
\caption{Convergence of the energy error for $p=2$. Comparison of the new biorthogonal basis ('optimal'), standard dual basis ('ele dual') and standard Lagrange multipliers ('std'). Left: mesh ratio $2:3$. Right: mesh ratio $2:9$. From top to bottom the three mesh cases (a) to (c) from Figure \ref{fig:plateGeometries} are considered. }
\label{fig:convergence_p_2}
\end{figure}

The same comparisons for cubic splines are  shown in Figure~\ref{fig:convergence_p_3}. Again, we see an optimal order convergence of the optimal biorthogonal basis functions as expected theoretically, while the suboptimality of the naive biorthogonal basis functions from~\cite{seitz:16} becomes even more apparent.
However, when the slave mesh is coarser than the master mesh, the values of the error are larger than for the standard Lagrange multiplier case. Since the gap gets smaller with further refinements, this seems to be a pre-asymptotical effect. We note that for a fine slave mesh, no significant suboptimality can be observed in the pre-asymptotics.

\begin{figure}
\scalebox{.5}{
\input{figures/plate/straight_MatchingParam_3rd/convergence.tex}
\input{figures/plate/straight_MatchingParam_3rd_2.9/convergence.tex}
}
\\
\scalebox{.5}{
\input{figures/plate/straight_nonMatchingParam_3rd/convergence.tex}
\input{figures/plate/straight_nonMatchingParam_3rd_2.9/convergence.tex}
}\\
\scalebox{.5}{
\input{figures/plate/curved_MatchingParam_3rd/convergence.tex}
\input{figures/plate/curved_MatchingParam_3rd_2.9/convergence.tex}
}
\caption{Convergence of the energy error for $p=3$. Comparison of the new biorthogonal basis ('optimal'), standard dual basis ('ele dual') and standard Lagrange multipliers ('std'). Left: mesh ratio $2:3$. Right: mesh ratio $2:9$. From top to bottom the three mesh cases (a) to (c) from Figure \ref{fig:plateGeometries} are considered. }
\label{fig:convergence_p_3}
\end{figure}

In summary,  we have observed optimal convergence rates in all cases for the newly constructed biorthogonal basis functions, while suboptimal rates were seen for the naive biorthogonal basis from \cite{seitz:16}. When the finer side is chosen as the slave side, the error values were the same as for standard Lagrange multipliers. Only when the slave side is coarser, a suboptimal preasymptotic evolves. Hence, for the optimal biorthogonal basis, it is especially important to choose the finer side of the interface as the slave side, whenever this is possible.


TITLE: Does there exist a commutative monoid without non-trivial idempotents but with a trivial Grothendieck group?
QUESTION [3 upvotes]: Does there exist a non-trivial commutative monoid that does not have non-trival idempotents yet has a trivial Grothendieck group?

To partially explain my motivation, a cute little monoid! Consider a commutative monoid $M = \{0, k, a\},\ a + a = a + k = k + k = k$. It is very trivial, but at the same time interesting, because its Grothendieck group is trivial, yet $a$ is not an idempotent (which illustrates how idempotents can 'drag down' other elements to the kernel of the canonical map $x \mapsto x - 0$), and $a$ is also not a sum of an invertible and an idempotent element.

REPLY [4 votes]: Yes.  Let $\mathbb{Z}_+ = \{1,2,3,\ldots\}$, let $M = \{(0,0)\}\cup (\mathbb{Z}_+\times\mathbb{Z}_+)$, and consider the monoid structure on $M$ defined by
$$
(x_1,y_1) + (x_2,y_2) \;=\; \begin{cases}(x_2,y_2) & \text{if }x_1<x_2 \\ (x_1,y_1+y_2) & \text{if }x_1 = x_2 \\(x_1,y_1) & \text{if }x_1  > x_2\end{cases}
$$
Then the Grothendieck group for this monoid is trivial, since $(x,y) + (x+1,y) = (x+1,y)$ for all $(x,y)\in\mathbb{N}\times\mathbb{N}$.  However, the monoid itself is non-trivial and commutative, and has no nontrivial idempotent elements.
Incidentally, this monoid can also be described in the following way: let $\mathbb{N}[x]$ be the monoid of all polynomials with natural number coefficients under addition, and let $\sim$ be the congruence relation on $\mathbb{N}[x]$ defined by $p(x)\sim q(x)$ if and only if $p(x)$ and $q(x)$ have the same leading term.  Then the quotient $\mathbb{N}[x]/{\sim}$ is isomorphic to the monoid $M$ defined above.
Though I do not have a proof, I would guess that any example must be infinite.  The example above isn't even finitely-generated.


TITLE: Why is this statement about uniform convergence to $\cos x$ false?
QUESTION [1 upvotes]: There is a sequence of polynomials $P_n(x)$ which converge uniformly
  to $f(x)=\cos x$ on $\mathbb{R}$.

Given $\cos x\in C[a,\ b]$ and $\epsilon>0$, by the Weierstrass Approximation theorem, there is a polynomial $p$ such that $\|\cos x-p\|_\infty<\epsilon$. So, there is a sequence of polynomials $P_n(x)$ which converge uniformly to $f(x)=\cos x$ on $[a,\ b]$. I don't know why $[a,\ b]$ cannot be enlarged into $\mathbb{R}$.

REPLY [2 votes]: It can't be extended to all of $\mathbb R$ because $\cos(x) $ is bounded whereas nonconstant polynomials are never bounded. As you increase the size of the interval, the same polynomials will no longer work for the larger interval because eventually the difference will exceed $\epsilon$.
In general no sequence of polynomials  on an unbounded set of reals can converge uniformly to a nonconstant  bounded function. The theorem only applies on an interval of finite length.


TITLE: Find identities for $ζ^\star(2,1,2)$ and $ζ^\star(\bar{2},\bar{1},2)$.
QUESTION [0 upvotes]: Could someone help find identities for these two?
I started with
$ζ^\star(2,1,2)$ = $\sum_{k=1}^{\infty}(\frac{1}{k^2})\sum_{l=1}^{k}(\frac{1}{l})\sum_{m=1}^{l}(\frac{1}{m^2})$
and
$ζ^\star(\bar{2},\bar{1},2)$ = $\sum_{k=1}^{\infty}(\frac{((-1)^k}{k^2})\sum_{l=1}^{k}(\frac{(-1)^l}{l})\sum_{m=1}^{l}(\frac{1}{m^2})$
but I've only worked on zeta functions with depth 2, so I'm not really sure how to proceed.
Thanks!

REPLY [1 votes]: With a bit of patience, summation by parts reduces
$$ \zeta^\star(2,1,2) = \sum_{k\geq 1}\frac{1}{k^2}\sum_{l=1}^{k}\frac{H_l^{(2)}}{l} $$
 to $\sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2}$, which is computable through the Maclaurin series of $\arcsin^4(x)$, and
$$ \sum_{k\geq 1}\frac{1}{k^2}\sum_{l=1}^{k}\frac{H_l}{l^2}= 2\zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_{k+1}H_k^{(2)}}{(k+1)^2}, $$
so the determination of $ \zeta^\star(2,1,2)$ can be considered as already solved by nospoon.


TITLE: Effect of elementary row operations on determinant?
QUESTION [60 upvotes]: 1) Switching two rows or columns causes the determinant to switch sign
2) Adding a multiple of one row to another causes the determinant to remain the same
3) Multiplying a row as a constant results in the determinant scaling by that constant.
Using the geometric definition of the determinant as the area spanned by the columns of the matrix, could someone give me a geometric interpretation of the above theorems? Thanks.

REPLY [48 votes]: Row operations can be thought of as acting on the ENTIRE space by reflection, shearing, or dilation.
For 2, have you heard of Cavalieri's principle? It says that shearing things while holding each cross section steady maintains volume.
For 3, you are just stretching or shrinking along each axis.
For 1, you are just reflecting in the plane $x_i=x_j$.
Edit: Every row operation is the effect of multiplying on the left by an elementary matrix. We can think of this elementary matrix as a map from $R^n$ to $R^n$, which changes every vector in $R^n$ including the column vectors. Thus, each row operation corresponds to a way of changing the whole space.
The row operation in 1 interchanges two rows. This corresponds to interchanging two coordinates in the space. It is not obvious, but it has been shown that interchanging two coordinates is the same thing as reflecting the entire space around the subset where the two coordinates are equal. This does not change volume.
The row operation in 3 corresponds to stretching one coordinate by the multiple given, which multiplies volume by the same amount.
The operation in 2 can be thought of as follows: say that you add a multiple of the second row to the first. Imagine the space sliced into 'pancakes', one for each value of the second coordinate. The map doesn't interchange pancakes, it just slides each pancake 'horizontally'. This doesn't change the volume of anything.


TITLE: How do I calculate the greatest guarantee period through normal distribution?
QUESTION [0 upvotes]: The full question says:
"If the life of a car engine, calculated in miles, is normally distributed, with a mean of 240,000 miles and a standard deviation of 18,500 miles, what is the greatest the guarantee period should be if the company wants less than 2% of the engines to fail while under warranty?"
I've been learning standard distribution for the past couple weeks but have really struggled when I actually get questions. I understand the z score equation to solve it:
Z score equation
I also know how to use the table but I need to have a better understanding of questions like this. Any help would be very much appreciated :)

REPLY [0 votes]: Assuming that engine life is normally distributed (which is questionable in practice), you have $X \sim \mathsf{Norm}(\mu = 240, \sigma =18.5),$ in thousands of miles. You want $k$ such
that $P(X <k) = 0.02.$
Using software: In R, where qnorm is a normal quantile function (inverse CDF):
qnorm(.02, 240, 18.5)
[1] 202.0056
pnorm(202, 240, 18.5)
[1] 0.01998523
pnorm(200, 240, 18.5)
[1] 0.01530284

So 202,000 miles is the largest limit. Maybe easier to
say it's a 200,000 mile guarantee.
Using printed normal tables: Using standardization and printed normal CDF tables you would have the following.
$$P(X < k) = P\left(\frac{X-\mu}{\sigma} < \frac{k - 240}{18.5}\right) = 0.02.$$
Then from a printed table find the z-score that cuts probability $0.02$ from the lower tail of the standard normal distribution. Look in the body of the table to find a value (nearly) equivalent to probability  $0.0200$ (of being smaller), and then read the z-score from the margins.
Maybe $z = -2.05.$ Then solve $\frac{k-240}{18.5} = -2.05$
to get a reasonable approximation of $k.$


TITLE: How to find algebraic simplification for recurrence relation with closed-form solution, specifically for the Lucas-Lehmer primality test
QUESTION [3 upvotes]: I have a question based on the section Proof of correctness in the article Lucas-Lehmer primality test, see following link.
https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test#Proof_of_correctness
"We begin by noting that $\langle s_i\rangle$ is a recurrence relation with a closed-form solution. Define $\omega = 2 + \sqrt{3}$ and $\bar{\omega} = 2 - \sqrt{3}; ...$"
How does $\omega$ and $\bar{\omega}$ come in to the picture? I have no problem following the prof ones $\omega$ and $\bar{\omega}$ is defined but I don't understand how they where defined in the first place.

REPLY [3 votes]: The whole setup began with Lucas. For given integers $P,Q$ we get Lucas sequences $U_n, V_n,$ where $V_n = a^n + b^n.$
In your case, $P=14, \; Q=1.$
Among the relations we find $$ V_{2n} = V_n^2 - 2 Q^n, $$ the fact that $Q=1$ takes us to
$$  V_{2n} = V_n^2 - 2;  $$ furthermore $V_0 = 2, V_1 = P.$
For your application, define
$$ s_j = V_{2^{j-1}},  $$ so that $$  s_1 = V_1 = P, \; s_2 = V_2 = P^2 - 2, \; s_3 = V_4 = P^4 - 4P^2 +2.   $$
$$  s_{j+1} = s_j^2 - 2. $$
In case $P+2$ is a square, we may also define $s_0 = \sqrt{P+2}$ as another integer. With $P=14$ this works.
So, with $Q=1,$ we get the roots $$ a = \frac{P + \sqrt{P^2 - 4}}{2}, \;  b = \frac{P - \sqrt{P^2 - 4}}{2}  $$
$$ s_j = a^{\left( 2^{(j-1)} \right)} + b^{\left( 2^{(j-1)} \right)}.  $$
Go Figure.
EDIT: Well, I like this, and want to remember it in a simple format.
Given any real number $t_0 > 2,$ define the sequence $t_1 = t_0^2 - 2$ and, generally, $t_{n+1} = t_n^2 - 2.$ Then,
taking
$$  \color{magenta}{ A = \frac{t_0 + \sqrt {t_0^2 - 4}}{2}, \; \; \; B = \frac{t_0 - \sqrt {t_0^2 - 4}}{2} } $$
One may check quickly that $$ AB = 1.  $$
Indeed, both are positive, $A$ is the larger, and
$$ A + B = t_0.   $$
So, think about it, using $AB= 1,$ also $A^m B^m = 1$ for any positive integer $m,$
$$ t_1 = t_0^2 - 2 = A^2 + 2 A B + B^2 - 2 =  A^2 + 2 A B + B^2 - 2 AB = A^2 + B^2.  $$
Then $$ t_2 = t_1^2 - 2 = A^4 + 2 A^2 B^2 + B^4 - 2 =  A^4 + 2 A^2 B^2 + B^4 - 2 A^2 B^2 = A^4 + B^4.  $$
Then $$ t_3 = t_2^2 - 2 = A^8 + 2 A^4 B^4 + B^8 - 2 =  A^8 + 2 A^4 B^4 + B^8 - 2 A^4 B^4 = A^8 + B^8.  $$
Pattern holds true forever, by induction: given
$$ \color{magenta}{ t_n = A^{\left( 2^n \right)} +  B^{\left( 2^n \right)},}  $$
we get
$$  t_{n+1} = t_n^2 - 2 =  A^{\left( 2^{n+1} \right)} + 2 A^{\left( 2^n \right)}   B^{\left( 2^n \right)}+ B^{\left( 2^{n+1} \right)} -2 =  A^{\left( 2^{n+1} \right)} + 2 A^{\left( 2^n \right)}   B^{\left( 2^n \right)}+ B^{\left( 2^{n+1} \right)} -2 A^{\left( 2^n \right)}   B^{\left( 2^n \right)},   $$
$$ t_{n+1} =   A^{\left( 2^{n+1} \right)} + B^{\left( 2^{n+1} \right)}.  $$


TITLE: Thomason’s homotopy colimit theorem for pseudo-functor
QUESTION [2 upvotes]: Thomason’s homotopy colimit theorem

R W Thomason, Homotopy colimits in the category of small categories,
  Math. Proc. Camb. Phil. Soc. (1) 85 (1979)91-109

says that for a functor $F:I^{op}\to Cat$, the homotopy colimit and pseudo-colimit are the same. More precisely, there exists a natural homotopy equivalence
$$
hocolimt(N F)\to N(\int F),
$$
where $\int F$ is the Grothendieck construction (computes pseudo-colimit).
My question is, what if we consider a pseudo-functor $F: I^{op}\to Cat$.
Now the right hand side is still defined. For the left side, he says it is not defined. But if we compose $F$ with nerve, we obtain a pseudo-functor $F: I^{op}\to SSet$, which induces a functor $hoNF: I^{op}\to hoSSet$, then take colimit in $hoSSet$. Are the two sides homotopical equivalent?

REPLY [2 votes]: The most general result of this form that I know is from 0907.0930: "Classifying spaces for braided monoidal categories and lax diagrams of bicategories" by Carrasco-Cegarra-Garzón.  They deal with lax diagrams of bicategories and include (at least) two useful things:

A rectification functor from lax diagrams to strict diagrams.
A Thomason theorem of the form you want.


text:
\section{The \(S_{n+1}\) action on \(Y_{\Bc(n-1)}\) and the minimality property of  \(\Bc(n-1)\)}
\label{sec:chiusuraazione}

As explained in Section \ref{sec:buildingcombinatorial} the strata of \(Y_{\Bc(n-1)}\) are in correspondence with the elements of \({\mathcal N}'(\Bc(n-1) )\).  The \(S_{n+1}\) action on the open part of \(Y_{\Bc(n-1)}\) can be extended to the boundary.
In fact one  can immediately check that the   geometric    \(S_{n+1}\) action on \(\Bc(n-1) \) can be extended to \({\mathcal N}'(\Bc(n-1) )\): let \(\{\{V\}, \Tc_1,..., \Tc_r\}\) be and element of \({\mathcal N}'(\Bc(n-1) )\) and let \(\sigma \in S_{n+1}\), then \(\sigma\) sends \(\{\{V\}, \Tc_1,..., \Tc_r\}\) to \(\{\{V\}, \Tc_1',..., \Tc_r'\}\) where,
for every \(i\), $\Tc_i'=\sigma \Tc_i$  according to the action on \(\Bc(n-1) \) illustrated in the end of Section \ref{sec:action1}. From the inclusions $\{V\}\subsetneq  \Tc_1\subsetneq \cdots \subsetneq \Tc_r$
it immediately follows that $\{V\}\subsetneq  \Tc_1'\subsetneq \cdots \subsetneq \Tc_r'$, therefore \(\{\{V\}, \Tc_1',..., \Tc_r'\}\) belongs to \({\mathcal N}'(\Bc(n-1) )\).

Now we address the following combinatorial problem:  what is the minimal building set in   \(\Bc(n-1) \) that is closed under the \(S_{n+1}\) action and `contains' \(\Cc_{A_{n-1}}\) ?
We start by  expressing in a precise way what we mean with `contains' \(\Cc_{A_{n-1}}\).

Recall that we write   \({\mathcal N}'(\Cc_{A_{n-1}})\) for the poset given
by the nested sets in \(\Cc_{A_{n-1}}\) that contain $\{V\}$, i.e. the poset that indicizes the strata of
\(Y_{maxA_{n-1}}\).

\begin{proposition}
There is a graded  poset   embedding \(\varphi\) of
\({\mathcal N}'(\Cc_{A_{n-1}})\)
into \({\mathcal N}'(\Bc(n-1) )\).
\end{proposition}
\begin{proof}
Let \(\Tc\) be an element in

\({\mathcal N}'(\Cc_{A_{n-1}})\). Then \(\Tc=\{ B_0=V, B_1,B_2,...,B_r\}\) is a nested set  of the building set \(\Cc_{A_{n-1}}\) containing \(V\). This means that its elements are linearly ordered by inclusion: \(V\supset B_1\supset \cdots \supset B_r\).
Now  we can express  every \(B_i\) as the direct sum of some irreducible subspaces \(A_{ij}\), i.e. elements of  \(\Fc_{A_{n-1}}\) (\(j=1,...,k_i\)).  We notice that, for every \(i=1,...,r\),  the sets  \(\Tc'_i=\{A_{sj}\} \cup \{V\}\) (with  \(s> r-i\) and, for every \(s\), \(j=1,...,k_s\))  is nested in \(\Fc_{{A_{n-1}}}\). The map \(\varphi\) defined by
\[\varphi( \Tc)= \{\{V\}, \Tc'_1,...,\Tc'_r\}\]
if \(r\geq 1\), otherwise \[\varphi( \Tc)=\{\{V\}\}\]
is easily seen to be a poset embedding.
\end{proof}

Given a  complex of nested sets   \(P\), we will  denote by  \(F^k(P)\) the subset made by the nested sets of cardinality \(k+1\).

 The restriction of \(\varphi\)  to \(F^1({\mathcal N}'(\Cc_{A_{n-1}}))\) is an embedding of \(F^1({\mathcal N}'(\Cc_{A_{n-1}}))\) into \(F^1({\mathcal N}'(\Bc(n-1) ))\). Now \(F^1({\mathcal N}'(\Bc(n-1) ))\) can be identified with \(\Bc(n-1)\) (the identification maps \( \{\{V\}, \MS\}\), with \(\MS\) a nested set of \(\Fc_{{A_{n-1}}}\) that strictly contains  \(V\),  to \( \MS\)) and we still call \(\varphi\) the embedding  from

$F^1( {\mathcal N}'(\Cc_{A_{n-1}}))$
 to \(\Bc(n-1)\).
 More explicitly, if $B_1 \in \Cc_{{A_{n-1}}}$ is a subspace which is the direct sum of the irreducible subspaces $A_{11}, \ldots, A_{1k_1} $ then $$\varphi (\{V,B_1\})= \{V, A_{11}, \ldots, A_{1k_1}\}.$$

\begin{theorem}
\label{teo:chiusoazionebuilding}
The minimal building subset of  \(\Bc(n-1) \) which contains the image

$\varphi(F^1({\mathcal N}'(\Cc_{A_{n-1}})))$
and is closed under the \(S_{n+1}\) action  is \(\Bc(n-1) \) itself.

\end{theorem}
\begin{proof}

Let us consider a building subset  \(\Gamma\) of  \(\Bc(n-1)\) that contains $\varphi(F^1({\mathcal N}'(\Cc_{A_{n-1}})))$  and is closed under the  \(S_{n+1}\) action. We will prove the claim by showing  that \(\Gamma=\Bc(n-1)\).

This can be done by induction on the depth of an element of \(\Bc(n-1)\),

which is defined in the following way:

let \(\Tc\) be a \(\Fc_{A_{n-1}}\)-nested set that contains \(V\) and  consider the  levelled graph   associated to \(\Tc\). This graps is  an oriented tree: it coincides with the Hasse diagram of the poset induced by the  inclusion relation,  where  the leaves are the minimal subspaces of \(\MS\) and   the root

is  \(V\) and the orientation goes from the root to the leaves.
A vertex $v$ is in level $k$ if the maximal length of a path that connects $v$ to a leaf is
$k$.
We say that  \(\Tc\) has depth \(k\) if
\(k\) is the  highest level of this tree.
\footnote{We notice that this representation of a nested set by a tree is coherent with the one  introduced  in the Section \ref{secexamplean}: the only difference is that there we added  \(n\)  leaves.}

Now we prove by induction on \(k\) that every element in  \(\Bc(n-1)\) with depth \(k\) belongs to \(\Gamma\).

When \(k=0,1\) this is immediate: given \(B\neq V\in \Cc_{A_{{n-1}}} \), then \(\varphi(\{V, B\})\) is the nested set of depth 1 whose elements are \(V\) and the maximal elements of  \(\Fc_{{A_{n-1}}}\) contained in \(B\). In this way one can show that  all the elements of  \(\Bc(n-1)\)  with depth 1 that contain \(V\) belong to \(\Gamma\).

Let us check the case  \(k=2\).
One first observes that, in view of the definition of the  \(S_{n+1}\) action, every nested set of depth $2$ of the form \(\{V, B, B_1\}\), where \(B_1\subset B\), belongs to \(\Gamma\) since it
can be obtained as \(\sigma \MS\) for a suitable choice of \(\sigma\in S_{n+1}\) and of a the nested set of depth $1$ \(\MS\in \Bc(n-1)\).
Now we show that also all  the nested sets of depth $2$ of the form \(\{V, B, B_1,...,B_j\}\), with $j \geq 2$ and \(B_i\subset B\) for every \(i\), belong to \(\Gamma\). In fact we can obtain \(\{V, B, B_1,...,B_j\}\) as a union, for every $i$, of the nested
sets \(\{V, B, B_i\}\) that belong to $\Gamma$ as remarked above.
Since all these sets have a nontrivial intersection $\{V,B\}$ and \(\Gamma \) is  building in the Feichtner-Kozlov sense (see Section \ref{sec:buildingcombinatorial}), this shows that \(\{V, B, B_1,...,B_j\}\) belongs to
$\Gamma$.

Then let us consider a nested set of depth  $2$ \(\{V, B, B_1,...,B_j\}\) (\(j\geq 2 \) ), where
\begin{itemize}
\item[i)] $B$ is in level $1$;

\item[ii)] \(B_j\)  is  not  included in \(B\);
\item[iii)]$\,$ all the $B_i$'s are in level $0$.
\end{itemize}

This nested set is in \(\Gamma\)  since it can be obtained as a union  of the nested sets (with depth $1$) \(\MS_1=\{V, B_1,...,B_j\}\)  and \(\MS_2\), where \(\MS_2\) is any  nested subset of \(\{V, B, B_1,...,B_j\}\) with   depth $2$. We notice that \(\MS_1\) and \(\MS_2\)  are in \(\Gamma\), and have nonempty intersection, therefore their union belongs to  \(\Gamma \).

Now  we can show that in \(\Gamma\) there are all the  nested sets of depth $2$: if the set \(\{V, C_1,...C_s, B_1,...,B_j\}\) has depth $2$, where \(s\geq 2\) and the subspaces      \(C_1,...,C_s\)   are in level $1$, while the \(B_i\)'s are in level $0$,

we can obtain \(\{V, C_1,...C_s, B_1,...,B_j\}\) as the union of the nested sets \(\{V, C_i, B_1,...,B_j\}\) (for every \(i=1,..,s\)) that   have pairwise nonempty intersection and belong to \(\Gamma\), as we have already shown.

Let us now consider   \(k\geq 2\) and  suppose that every nested set in \(\Bc(n-1)\) with depth  \(\leq k\) belongs to \(\Gamma\).  Let \(\Tc\) be a nested set of depth \(k+1\).
 Let us denote by \(\Tc_k\) the nested set obtained removing from \(\Tc\) the
subspaces in level $0$:

it belongs to \(\Gamma\) by the inductive hypothesis. Then we consider the nested set \(\Tc'\) obtained removing from \(\Tc\) the levels \(2,..., k\): since \(\Tc'\)  has depth 2  it belongs to \(\Gamma\) again by the inductive hypothesis.
We observe that  \(\Tc_k\) and \(\Tc'\) have nonempty intersection, therefore their union \(\Tc\) belongs to \(\Gamma\).
 \end{proof}