text:
\section{Introduction and background}

Let $\Delta$ be a connected finite simplicial complex whose geometric realization $|\Delta|$ is a topological disk in $\mathbb R^2$. Let $r\geq0$ be an integer. The space of splines of smoothness $r$ and degree $d$ is the $\mathbb R-$ vector space $$C_d^r(\Delta)=\{F:|\Delta|\longrightarrow\mathbb R: F|_{\sigma}=\mbox{ polynomial of degree }\leq d,$$ $$\forall \sigma\in\Delta_2,\mbox{ and }F\in C^r\}.$$ A very nice accessible introduction to Polynomial Splines is \cite[Chapter 8]{clo}.

One of the major questions in spline approximation is to find the dimension of this vector space; even when $d=3$ and $r=1$, this dimension is not known for arbitrary triangulations. If $d\geq 3r+1$, for almost all triangulations, Alfeld and Schumaker (\cite{as}) give a beautiful, yet complicated formula for this dimension in terms of combinatorial and local geometric data (data depending only on local geometry at the interior vertices of $\Delta$):

$$\dim C_d^r(\Delta)={{d+2}\choose{2}}+{{d-r+1}\choose{2}}f_1^0-\left({{d+2}\choose{2}}-{{r+2}\choose{2}}\right)f_0^0 +\sigma,$$ where $f_1^0$ is the number of interior edges of $\Delta$, $f_0^0$ is the number of interior vertices of $\Delta$, and $\sigma=\sum\sigma_i$, where $\sigma_i=\sum_{j\geq 1}\max\{(r+1+j(1-n(v_i))),0\}$, and $n(v_i)$ is the number of distinct slopes at the interior vertex $v_i$. For further reference this formula will be denoted by $L(\Delta,r,d)$.

In \cite{ss1}, by showing that a certain zeroth local cohomology is zero, Schenck and Stillman prove that if $\Delta$ has only pseudoedges (such a triangulation is called {\em quasi-cross-cut}), the Alfeld-Schumaker formula is true for any $d$. With different methods, in \cite{swy} the same is true, but for the more general case when instead of a triangulation, one has a partition. Lemma 2.5 in \cite{ss1} also says that if $\Delta$ has at least one non-pseudoedge, the local cohomology module considered is not zero. In fact, Schenck and Stiller conjectured that for any $\Delta$, this local cohomology module vanishes in degree $d\geq 2r+1$.

\cite{ss} considered a simplicial complex $\Delta_S$ with exactly one non-pseudoedge, and in \cite{to} it was shown that for this particular example, the above conjecture is tight: for any $r\geq 1$, $\dim C_{2r}^r(\Delta_S)\neq L(\Delta_S,r,2r)$. The present notes are a followup of \cite{to}. We show that for the same simplicial complex $\Delta_S$, $$\dim C_{d}^r(\Delta_S)= L(\Delta_{S},r,d),\mbox{ for any }r \mbox{ and }d\geq 2r+1,$$ and therefore the Schenck-Stiller conjecture is true for this first non-trivial triangulation $\Delta_S$\footnote{See Subsection 1.2.}.

The confirmation of the Schenck-Stiller conjecture in this case proved to be surprisingly challenging, and in fact, it took several years to establish the main result of this paper. The reason for this difficulty seems to lie in rather deep connections with representation theory, matrix theory and commutative algebra.

The proof is subtle in several places, but we put considerable effort into making our exposition clear and readable also for the non-specialists. In the first part we provide the solution of the main problem that relies on classical concepts in commutative algebra (e.g., regular sequences, monomial order, etc.). In the second part of the paper we investigate the connections with Schur functors, Roth's equation in matrix theory and lower-upper triangular matrix decompositions.

\subsection{Homological approach to spline approximation}

Following \cite{br91}, consider $\mathbb R^2$ embedded in $\mathbb R^3$, and let $\hat{\Delta}$ be the cone of $\Delta$ with its origin in $\mathbb R^3$ and let $R=\mathbb R[x,y,z]$ be the ring of polynomials in variables $x,y,z$ with coefficients in $\mathbb R$. Since we consider the cone of $\Delta$, from now on if $e$ is an edge of $\Delta$ we will think of $\ell_e$ to be the homogenized equation of the equation of the line in $\mathbb R^2$ where $e$ is placed; also, abusing the terminology a bit, we are going to say that the linear form $\ell_e$ defines the edge $e$.

Consider the finitely generated graded $R-$module: $$C^r(\hat{\Delta})=\{F:|\hat{\Delta}|\longrightarrow\mathbb R: F|_{\hat{\sigma}}\in R, \forall \sigma\in\Delta_2, \mbox{ and }F\in C^r\}.$$ Then $$\dim_{\mathbb R}C_d^r(\Delta)=\dim_{\mathbb R}C^r(\hat{\Delta})_d,$$ the dimension of the degree $d$ piece of graded module $C^r(\hat{\Delta})$. So, by taking $\hat{\Delta}$ the cone of $\Delta$, we homogenized our polynomials and therefore the problem is translated into a homological algebra problem: to find the Hilbert function of a graded module \footnote{ If $M=\oplus_{d\in\mathbb Z}M_d$ is a finitely generated graded $R-$module, the Hilbert function of $M$ in degree $d$ is by definition $HF(M,d)=\dim_{\mathbb R}M_d$. See, for example, \cite{s} for more background on Hilbert functions.}.

Let $F\in C^r(\hat{\Delta})_d$. Piecewise, on each triangle $T_i$ of $\Delta$, $F$ is defined by a homogeneous polynomial of degree $d$: $F_i\in R_d$. For $F$ to be a $C^r-$function, since polynomials are $C^{\infty}-$functions, whenever we have two triangles $T_i$ and $T_j$, with a common (interior) edge of equation $\ell_{ij}=0$, then $$F_i-F_j\in\langle \ell_{ij}^{r+1}\rangle.$$ For example, if $r=0$ (i.e., $F$ is continuous), one must have $F_i(P)=F_j(P),$ for all $P\in V(\ell_{ij})$. But this means exactly that $F_i-F_j\in \langle \ell_{ij}\rangle.$

With the above idea in mind, Billera and Rose (\cite{br2}) place $C^r(\hat{\Delta})$ in the following exact complex of graded $R-$modules: $$0\rightarrow C^r(\hat{\Delta})\rightarrow R^{f_2}\oplus R^{f_1^0}(-r-1)\stackrel{\phi}\rightarrow R^{f_1^0}\rightarrow N\rightarrow 0,$$ where $N=coker(\phi)$, and $$\phi=\left(\begin{array}{ccccc}
&&\ell_{e_1}^{r+1}&&\\
\partial_2&|&&\ddots&\\
&&&&\ell_{e_{f_1^0}}^{r+1}
\end{array}\right).$$ $\partial_2$ is the simplicial reduced boundary map $R^{f_2}\rightarrow R^{f_1^0}$, where $f_2$ is the number of triangles of $\Delta$ and $\ell_{e_i}$ is the linear form that defines the interior edge $e_i$.

Using the properties of the Hilbert function, we obtain $$\dim_{\mathbb R}C^r(\hat{\Delta})_d={{d+2}\choose{2}}f_2+{{d-r+1}\choose{2}}f_1^0-{{d+2}\choose{2}}f_1^0+HF(N,d).$$

In \cite{ss2}, Schenck and Stillman place the graded $R-$module $N$ in the following short exact sequence of graded $R-$modules: $$0\rightarrow H_{\langle x,y,z\rangle}^0(N)\rightarrow N\rightarrow \oplus_{v\in\Delta_0^0}R/J(v)\rightarrow 0,$$ where $H_{\langle x,y,z\rangle}^0(N)$ is the zeroth local cohomology module of $N$ at the maximal ideal $\langle x,y,z\rangle$, $\Delta_0^0$ denotes the set of interior vertices of $\Delta$, and $$J(v)=\langle \ell_{v,1}^{r+1},\ldots,\ell_{v,n(v)}^{r+1}\rangle,$$ with $\ell_{v,i}$ being the linear forms defining the interior edges of different slopes with one of the vertices being $v$.

Theorem 3.1 in \cite{ss1} describes the minimal graded free resolution of $R/J(v)$, so $HF(R/J(v)),d)$ is known. In Corollary 4.5 in \cite{ss1} all of this information is combined to obtain $$\dim_{\mathbb R}C^r(\hat{\Delta})_d=L(\Delta,r,d)+HF(H_{\langle x,y,z\rangle}^0(N),d).$$ Since $H_{\langle x,y,z\rangle}^0(N)$ is a module of finite length, we have for $d$ sufficiently large $HF(H_{\langle x,y,z\rangle}^0(N),d)=0$. In fact, Alfeld and Schumaker show that it is enough to take $d\geq 3r+1$. The Schenck-Stiller conjecture claims that one can take $d\geq 2r+1$.

\subsection{The Schenck-Stiller example}

In this section we reduce our problem to calculating the Hilbert function of some ideal in degree $r$. We will provide a detailed picture of our approach which is also accessible to a non-specialist in this area.

Let $R=\mathbb R[x,y,z]$ be the ring of polynomials with real coefficients and let $\Delta_S:=\Delta$ be the following simplicial complex that triangulates a bounded connected region in the real plane:
\begin{center}
\epsfig{figure=fig2.eps,height=1.3in,width=2in}
\end{center}

The goal is to compute $\dim_{\mathbb R}C^r(\hat{\Delta})_d, d\geq 2r+1$ by showing that $HF(H_{\langle x,y,z\rangle}^0(N),d)=0, d\geq 2r+1$.

For $\Delta_S$, because it has exactly one totally interior edge, $H_{\langle x,y,z\rangle}^0(N)$ is isomorphic to $R/I$, where $I$ is a certain ideal (see the assertions before Note \ref{note:note}), and therefore it is enough to prove that $HF(H_{\langle x,y,z\rangle}^0(N),2r+1)=0$. Since our triangulation is specific, our analysis will be more direct, building on basic definitions which, via careful considerations, will lead us to our results.

Let $d=2r+1$, and let us label the triangles of $\Delta$ with $T_1,\ldots,T_8$, clockwise: for example the first triangle has vertices $(0,2),(2,2)$ and $(1,1)$. Let $\ell_{ij}$ be the linear form defining the common edge between the triangles $T_i$ and $T_j$. We have 9 interior edges, one of them being a non-pseudoedge (i.e., the edge common to the triangles $T_2$ and $T_6$; this edge is placed on the line in $\mathbb R^2$ of equation $y=1$, hence, after homogenization, $\ell_{26}=y-z$):
\begin{eqnarray} \ell_{12}&=&\ell_{78}=x-y\nonumber\\
\ell_{23}&=&\ell_{45}=x+y-4z\nonumber\\
\ell_{34}&=&\ell_{56}=x-y-2z\nonumber\\
\ell_{67}&=&\ell_{18}=x+y-2z\nonumber\\
\ell_{26}&=&y-z.\nonumber
\end{eqnarray}

An element $F\in C^r(\hat{\Delta})_d$ is an $8-$tuple: $$F=(F_1,\ldots,F_8),$$ where $F_i\in R_d$ are subject to the conditions:

\begin{eqnarray}
F_1-F_2&=&\ell_{12}^{r+1}G_{12}\nonumber\\
F_2-F_3&=&\ell_{23}^{r+1}G_{23}\nonumber\\
F_3-F_4&=&\ell_{34}^{r+1}G_{34}\nonumber\\
F_4-F_5&=&\ell_{45}^{r+1}G_{45}\nonumber\\
F_5-F_6&=&\ell_{56}^{r+1}G_{56}\nonumber\\
F_6-F_7&=&\ell_{67}^{r+1}G_{67}\nonumber\\
F_7-F_8&=&\ell_{78}^{r+1}G_{78}\nonumber\\
F_8-F_1&=&\ell_{18}^{r+1}G_{18}\nonumber\\
F_2-F_6&=&\ell_{26}^{r+1}G_{26},\nonumber
\end{eqnarray}where $G_{ij}$ are homogeneous polynomials in $R$ of degree $d-(r+1)=r$.

One should observe that once we know $F_1$ and the $G_{ij}$'s, $F$ is completely determined. The $G_{ij}$'s are subject to the following relations:
\begin{eqnarray}
(G_{12}+G_{78})\ell_{12}^{r+1}+(G_{67}+G_{18})\ell_{67}^{r+1}+G_{26}\ell_{26}^{r+1}=0,\nonumber\\
(-G_{23}-G_{45})\ell_{23}^{r+1}+(-G_{34}-G_{56})\ell_{34}^{r+1}+G_{26}\ell_{26}^{r+1}=0.\nonumber
\end{eqnarray}

From the discussions above, considering first the possibilities for $F_1$, then the possibilities for $G_{ij}$'s with $G_{26}$ fixed, and finally the possibilities for $G_{26}$ we see that the dimension we are looking for is $$\dim C^r(\hat{\Delta})_d={{d+2}\choose{2}}+4{{r+2}\choose{2}}+\epsilon,$$ where $\epsilon=\dim I_r$ with $$I=\langle \ell_{12}^{r+1},\ell_{67}^{r+1}\rangle :\ell_{26}^{r+1}\cap \langle \ell_{23}^{r+1},\ell_{34}^{r+1}\rangle :\ell_{26}^{r+1}.$$

The Alfeld-Schumaker formula says that $$L(\Delta,r,d)={{d+2}\choose{2}}+9{{d-r+1}\choose{2}}-2\left({{d+2}\choose{2}}-{{r+2}\choose{2}}\right)+\sigma,$$ and from the proof of Corollary 4.5 in \cite{ss1}, $$\sigma=2r\alpha-2\alpha^2,\mbox{ where }\alpha=\lfloor\frac{r+1}{2}\rfloor,$$ the smallest integer larger than $\frac{r+1}{2}$.

We have $d=2r+1$.

If $r+1=2n$, then $\sigma=2n^2-2n$. For this case $$\dim C^r(\hat{\Delta})_d-L(\Delta,r,d)=\epsilon-n.$$

If $r+1=2n+1$, then $\sigma=2n^2$. For this case  $$\dim C^r(\hat{\Delta})_d-L(\Delta,r,d)=\epsilon-(n+1).$$

So the Schenck-Stiller conjecture is true for this $\Delta$, if one can show that $\epsilon=n$ if $r+1=2n$, and $\epsilon=n+1$ if $r+1=2n+1$. Our goal is to prove these equalities.

\vskip .3in

In the language of the previous subsection, one needs to show that $$HF(H_{\langle x,y,z\rangle}^0(N),2r+1)=0.$$ Indeed our goal matches this new goal as we can see below.

First, by making the change of variables suggested in \cite{to}: $y-z=\frac{1}{2}\bar{y}$, $x-y=\bar{x}$, $x-y-2z=\bar{z}$, we can assume that $\ell_{26}=y$, $\ell_{12}=x$, $\ell_{67}=x+y$ and $\ell_{34}=z$, $\ell_{23}=x+z$. So the ideals $J(v_i)$ at the two interior vertices become: $$J(v_1)=\langle x^{r+1},(x+y)^{r+1},y^{r+1}\rangle,$$ and $$J(v_2)=\langle z^{r+1},(z+y)^{r+1},y^{r+1}\rangle.$$

Second, by \cite{ss1}, Theorem 3.1, the graded minimal free resolution for $R/J(v_i), i=1,2$ is $$0\rightarrow R(-a)\oplus R(-b)\stackrel{\psi_i}\rightarrow R^3(-r-1)\rightarrow R\rightarrow R/J(v_i)\rightarrow 0,$$ where $a$ and $b$ are some shifts (see Note \ref{note:note} below) and $\psi_i=\left(\begin{array}{cc} A_i&D_i\\ B_i&E_i\\C_i&F_i\end{array}\right).$ For example, if $i=1$ this means that $A_1x^{r+1}+B_1(x+y)^{r+1}+C_1y^{r+1}=0 \mbox{ and }D_1x^{r+1}+E_1(x+y)^{r+1}+F_1y^{r+1}=0.$ We should note that $$\langle C_1,F_1\rangle=\langle x^{r+1},(x+y)^{r+1}\rangle : y^{r+1},$$ and $$\langle C_2,F_2\rangle=\langle z^{r+1},(z+y)^{r+1}\rangle : y^{r+1}.$$

Third, by \cite{ss2}, $$H_{\langle x,y,z\rangle}^0(N)\approx R(-r-1)/\langle C_1,F_1,C_2,F_2\rangle.$$ With this, the Schenck-Stiller conjecture for this $\Delta$ reduces to showing that $$HF(R/\langle C_1,F_1,C_2,F_2\rangle,r)=0.$$

\vskip .2in

\begin{note}\label{note:note} By \cite{ss1}, Theorem 3.1, if $r+1=2n$, then $\deg(C_i)=\deg(F_i)=n,i=1,2$, and if $r+1=2n+1$, then $\deg(C_i)=n,\deg(F_i)=n+1, i=1,2$. Furthermore, the ideals $\langle C_i,F_i\rangle$ are complete intersections (see \cite{to}).
\end{note}

\vskip .2in

We have an exact sequence of $R-$ modules: $$0\rightarrow \frac{R}{\langle C_1,F_1\rangle\cap\langle C_2,F_2\rangle} \rightarrow \frac{R}{\langle C_1,F_1\rangle}\oplus \frac{R}{\langle C_2,F_2\rangle}\rightarrow \frac{R}{\langle C_1,F_1,C_2,F_2\rangle}\rightarrow 0.$$ Since $\langle C_i,F_i\rangle, i=1,2$ is a complete intersection, we have the graded minimal free resolution $$0\rightarrow R(-(\deg(C_i)+\deg(F_i))\rightarrow R(-\deg(C_i))\oplus R(-\deg(F_i))\rightarrow R\rightarrow R/\langle C_i,F_i\rangle\rightarrow 0,$$ and the Hilbert function of $\frac{R}{\langle C_1,F_1\rangle}\oplus \frac{R}{\langle C_2,F_2\rangle}$ can be computed using these resolutions.

Therefore, to prove the claim it will be enough to show \begin{eqnarray} HF(\frac{R}{\langle C_1,F_1\rangle\cap\langle C_2,F_2\rangle},r)&=&
2HF(\frac{R}{\langle C_1,F_1\rangle},r)\nonumber\\
&=& \left\{
                                          \begin{array}{ll}
                                           2n^2, & \hbox{if } r=2n-1; \\
                                           2n(n+1), & \hbox{if } r=2n.
                                          \end{array}
                                          \right.\nonumber
\end{eqnarray}

Equivalently, our main goal is to prove

\begin{thm}\label{thm:main} $$HF(\langle C_1,F_1\rangle\cap\langle C_2,F_2\rangle,r)=\epsilon= \left\{
                                          \begin{array}{ll}
                                           n, & \hbox{if } r=2n-1; \\
                                           n+1, & \hbox{if } r=2n.
                                          \end{array}
                                          \right.$$
\end{thm}

\section{Proof of the main result}

Let us denote $$K(r)=((\langle x^{r+1},(x+y)^{r+1}\rangle\cap\langle z^{r+1},(z+y)^{r+1}\rangle):y^{r+1})_r.$$ We want to show that $$\dim K(2n-1)=n\mbox{ and } \dim K(2n)=n+1.$$

We have the following sequence of useful lemmas.

\begin{lem}\label{lem:important0} For any $n\geq 1$ we have $$\dim K(2n-1)\geq n\mbox{ and } \dim K(2n)\geq n+1.$$
\end{lem}
\begin{proof} Since $HF(\frac{R}{\langle C_1,F_1,C_2,F_2\rangle},r)\geq 0$, then from the exact sequence above we have $HF(R,r)-\dim K(r)\leq 2HF(\frac{R}{\langle C_1,F_1\rangle},r)$ and hence the result.
\end{proof}

We have $$\langle x^{r+1},(x+y)^{r+1}\rangle :y^{r+1}=\langle C_1,F_1\rangle$$ and $$\langle z^{r+1},(z+y)^{r+1}\rangle :y^{r+1}=\langle C_2,F_2\rangle.$$

Observe that $C_1,F_1$ are polynomials in variables $x$ and $y$ so they are elements in $A=\mathbb R[x,y]$. $A$ is a subring of $R=\mathbb R[x,y,z]$. We will denote by $\langle C_1,F_1\rangle A$ the ideal in $A$ generated by $C_1,F_1$, and we will denote by $\langle C_1,F_1\rangle R$ the ideal in $R$ generated also by $C_1,F_1$. We also have $\langle C_1,F_1\rangle R\cap A=\langle C_1,F_1\rangle A$.

Similarly, $C_2,F_2$ are polynomials in variables $y$ and $z$, and so they belong to $B=\mathbb R[y,z]\subset R$. Again, $\langle C_2,F_2\rangle B$ will denote the ideal in $B$ generated by $C_2,F_2$, and  $\langle C_2,F_2\rangle R$ will denote the ideal in $R$ generated by $C_2,F_2$.

\begin{lem}\label{lem:important1} Let $F\in K(r)$. Then
\begin{enumerate}
\item If we write $F=z^rf_0+z^{r-1}f_1+\cdots+ zf_{r-1}+f_r$ with $f_i\in A=\mathbb R[x,y]$, then $f_i\in \langle C_1,F_1\rangle A$.
\item If we write $F=x^rg_0+x^{r-1}g_1+\cdots+ xg_{r-1}+g_r$ with $g_j\in B=\mathbb R[y,z]$, then $g_j\in \langle C_2,F_2\rangle B$.
\end{enumerate}
\end{lem}
\begin{proof} It is enough to prove the first part, since we can switch the roles of $x$ and $z$ afterwards. Let $$F=z^rf_0+z^{r-1}f_1+\cdots+ zf_{r-1}+f_r\in K(r).$$

Since $F\in \langle C_1,F_1\rangle R$, substituting $z=0$ we obtain $f_r\in\langle C_1,F_1\rangle A$. Therefore $$z(z^{r-1}f_0+\cdots+f_{r-1})\in \langle C_1,F_1\rangle R.$$ Since $\langle C_1,F_1\rangle A\subset A$ is a complete intersection, then $\{z,C_1,F_1\}$ is an $R-$regular sequence, and therefore  $z^{r-1}f_0+\cdots+f_{r-1}\in \langle C_1,F_1\rangle R$. Again, by making $z=0$, and with the same argument as above, we obtain inductively that for all $i=0,\ldots,r$, one has $f_i\in\langle C_1,F_1\rangle A$. \end{proof}

A useful consequence of the proof of the above lemma is that if $F\in K(r)$ and if $x^ay^bz^c$ is a nonzero monomial in the expression of $F$, since the ideals $\langle C_i,F_i\rangle, i=1,2$ are generated in degree $\geq n$, then $a\leq r-n$ and $c\leq r-n$.

The key result is the following lemma. For convenience we assume that $0$ has any degree. In this section most of the arguments use induction on $r$ (the base cases are simple calculations, and we left them out of the notes), and the third part of this next result will help resolve the inductive step.

\begin{lem}\label{lem:important} The following statements are true:
\begin{enumerate}
  \item $y^r\in K(r)$.
  \item For any $j\leq r-1$, $y^j\notin \langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1}$ and $y^j\notin \langle z^{r+1},(z+y)^{r+1}\rangle:y^{r+1}$.
  \item If $F\in K(r)$, then $y\frac{\partial^2F}{\partial x\partial z}=yF_{xz}\in K(r-1)$.
\end{enumerate}
\end{lem}
\begin{proof} Let $I=\langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1}=\langle C_1,F_1\rangle\subset A=\mathbb R[x,y]$.
\vskip .2in
(1). From the exact sequence $$0\rightarrow A(-(\deg(C_1)+\deg(F_1))\rightarrow A(-\deg(C_1))\oplus A(-\deg(F_1))\rightarrow A\rightarrow A/\langle C_1,F_1\rangle\rightarrow 0,$$ we have $$HF(A/I,r)=r+1-(r+1-\deg(C_1) + r+1-\deg(F_1))+(r+1-\deg(C_1)-\deg(F_1))=0,$$ so $y^r\in I$. Similarly, replacing $x$ with $z$, we have $y^r\in K(r)$.
\vskip .2in
(2). Let $j\leq r-1$. If $y^j\in \langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1}$, then there exist $P,Q\in A$ of degree $j$ such that $$y^{j+r+1}=Px^{r+1}+Q(x+y)^{r+1}.$$ Taking the partial derivative with respect to $x$ we obtain $$0=((r+1)P+xP_x)x^r+((r+1)Q+(x+y)Q_x)(x+y)^r.$$ Therefore $(r+1)P+xP_x$, which has degree $j$, is a multiple of $(x+y)^r$. But $j\leq r-1$, and therefore $$(r+1)P+xP_x=0.$$ If $P=\sum a_ix^iy^{j-i}$, we have $$a_0=0\mbox{ and }(r+1+i)a_i=0, i\geq 1.$$ Hence $P=0$, and therefore we have a contradiction.
\vskip .2in
(3). Let $F\in K(r)$. So there exist $P,Q\in R$ such that $$y^{r+1}F=Px^{r+1}+Q(x+y)^{r+1}.$$ Taking the partial derivative with respect to $z$ we have $$y^{r+1}F_z=P_zx^{r+1}+Q_z(x+y)^{r+1}.$$ Taking the partial derivative with respect to $x$ we have $$y^r(yF_{xz})\in\langle x^r,(x+y)^r\rangle.$$ Similarly, we have $$y^r(yF_{xz})\in\langle z^r,(z+y)^r\rangle.$$ Since $\deg(yF_{xz})=1+r-2=r-1$, we indeed obtain that $yF_{xz}\in K(r-1)$.
\end{proof}

\vskip .2in

\begin{prop}\label{prop:symmetry} We have $$K(r)=W:=\{f\in (\langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1})_r|f(x,y,z)=f(z,y,x)\}.$$
\end{prop}
\begin{proof} For convenience denote with $\bar{f}(x,y,z)=f(z,y,x)$.

The inclusion ``$\supseteq$'' is immediate, by switching $x$ and $z$, so $\dim K(r)\geq \dim W$.

For the other inclusion, consider $\{f_1,\ldots,f_m\}$ to be a basis for $K(r)$. Since $f_i\in K(r)$, then $\bar{f}_i\in K(r)$, and therefore $g_i:=f_i+\bar{f}_i\in K(r)$. Since $g_i=\bar{g}_i$, and since $$g_i\in K(r)\subset (\langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1})_r,$$ we have that $g_i\in W, i=1,\ldots,m$.

Suppose that these $g_i$ are linearly dependent. Then there exist $c_1,\ldots,c_m\in\mathbb R$ not all zero, such that $$c_1(f_1+\bar{f}_1)+\cdots+c_m(f_m+\bar{f}_m)=0.$$

We have $F:=c_1f_1+\cdots+c_mf_m\in K(r)$ and $F=-\bar{F}$. We show by induction on $r$ that in these conditions $F=0$. From Lemma \ref{lem:important} (3), we have $yF_{xz}\in K(r-1)$. Also $yF_{xz}=-y(\bar{F})_{xz}$, and by induction $$yF_{xz}=0,$$ which means $$F=P(x,y)+Q(y,z)=a_n(x^{r-n}-z^{r-n})y^n+a_{n+1}(x^{r-n-1}-z^{r-n-1})y^{n+1}+\cdots+a_{r-1}(x-z)y^{r-1}.$$

From Lemma \ref{lem:important1} we have that for $j=n,\ldots,r-1$, $$a_jy^j\in \langle z^{r+1},(z+y)^{r+1}\rangle:y^{r+1},$$ and therefore from Lemma \ref{lem:important} (2), $a_j=0$.

This means $F=0$, and since $f_1,\ldots,f_m$ are linearly independent, all the $c_i$ must vanish. But this implies that $\dim W\geq m=\dim K(r)$, and with the previous inclusion we obtain the desired equality. \end{proof}

\begin{prop}\label{prop:mindegree} The ideal $$\mathcal I(r):=(\langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1})\cap (\langle z^{r+1},(z+y)^{r+1}\rangle:y^{r+1})$$ is minimally generated in degree $r$.
\end{prop}
\begin{proof} We show by induction on $r\geq 2$, that $\dim_{\mathbb R} \mathcal I(r)_{r-1}=0$.

Let $F\in \mathcal I(r)_{r-1}$. The same proof as for Lemma \ref{lem:important} (3), gives us that $$yF_{xz}\in \mathcal I(r-1)_{r-2}.$$ By induction this must vanish and therefore $$F_{xz}=0.$$

So $$F=a_{r-1}x^{r-1}+a_{r-2}x^{r-2}y+\cdots+a_1xy^{r-2}+a_0y^{r-1}+b_1zy^{r-2}+\cdots+b_{r-1}z^{r-1}\in \mathcal I(r).$$ A similar proof as for Lemma \ref{lem:important1} will yield $$a_{r-1},a_{r-2}y,\ldots,a_1y^{r-2}\in \langle z^{r+1},(z+y)^{r+1}\rangle:y^{r+1}.$$ These must vanish because of Lemma \ref{lem:important} (2). Similarly, $b_1=\cdots=b_{r-1}=0$. So $F=a_0y^{r-1}\in \langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1}$, and this must vanish as well, from the same lemma. \end{proof}

Now we can prove our desired equalities.

\begin{thm}\label{thm:odd} For all $n\geq 2$, $$\dim K(2n-1)=n.$$
\end{thm}
\begin{proof} From Lemma \ref{lem:important0}, it will be enough to show that $\dim K(2n-1)\leq n$.

As we denoted before, let $$\mathcal I(2n-1)=(\langle x^{2n},(x+y)^{2n}\rangle:y^{2n})\cap (\langle z^{2n},(z+y)^{2n}\rangle:y^{2n}).$$ Then we need to show that $$HF(\mathcal I(2n-1),2n-1)\leq n.$$

Let $\succ$ be the Graded Reverse Lexicographic order on the monomials of $R=\mathbb R[x,y,z]$, with $x\succ y\succ z$. Let $$in_{\succ}(\mathcal I(2n-1))$$ be the initial ideal of $\mathcal I(2n-1)$ with respect to $\succ$.

From Proposition \ref{prop:mindegree}, $\mathcal I(2n-1)$ is minimally generated in degree $2n-1$. Also, since by Proposition \ref{prop:symmetry}, the polynomials of degree $2n-1$ in $\mathcal I(2n-1)$ are symmetric in $x$ and $z$, we have that $$\{y^{2n-1},xy^{2n-2},\ldots,x^{n-1}y^n, xzM_1,\ldots,xzM_p\}$$ includes the monomials that generate $in_{\succ}(\mathcal I(2n-1))$. Here we also used the remark after Lemma \ref{lem:important1} that the power of $x$ in a nonzero monomial of an element of degree $2n-1$ in $\mathcal I(2n-1)$ is $\leq r-n=2n-1-n=n-1$.

Next we show that if $xzM$ is the leading monomial of an element $F\in \mathcal I(2n-1)$, then $\deg(F)\geq 2n$. Let $F\in \mathcal I(2n-1)_{2n-1}=K(2n-1)$ be such that $in_{\succ}(F)=xzM$. By Proposition \ref{prop:symmetry}, $F$ is symmetric in $x$ and $z$, so $$F=xzG,$$ for some $G\in R$.

We have $$y^{2n}(xzG)=Ax^{2n}+B(x+y)^{2n}\mbox{ and } y^{2n}(xzG)=Cz^{2n}+D(z+y)^{2n}.$$ We have $\{z,x^{2n},(x+y)^{2n}\}$ and $\{x,z^{2n},(z+y)^{2n}\}$ are $R-$regular sequences so $$y^{2n}(xG)=A'x^{2n}+B'(x+y)^{2n}\mbox{ and } y^{2n}(zG)=C'z^{2n}+D'(z+y)^{2n}.$$ From this we have that $x|B'$ and $z|D'$, and therefore $$y^{2n}G=A'x^{2n-1}+B''(x+y)^{2n}\mbox{ and } y^{2n}G=C'z^{2n-1}+D''(z+y)^{2n}.$$

Taking the partial derivative with respect to $y$ of the two equations above, we obtain

$$y^{2n-1}(2nG+yG_y)\in\langle x^{2n-1},(x+y)^{2n-1}\rangle\mbox{ and } y^{2n-1}(2nG+yG_y)\in\langle z^{2n-1},(z+y)^{2n-1}\rangle.$$

We obtained that $2nG+yG_y$ is an element of degree $2n-3$ in $\mathcal I(2n-2)$, or is equal to zero. So from Proposition \ref{prop:mindegree}, $$2nG+yG_y=0.$$ If $G=\sum a_{i,j,k}x^iy^jz^k$, then $(2n+j)a_{i,j,k}=0$, and so $G=0$.

We obtained that the leading monomials of elements of degree $2n-1$ in $\mathcal I(2n-1)$, belong to the set $\{y^{2n-1},xy^{2n-2},\ldots,x^{n-1}y^n\}$, and therefore $$HF(\mathcal I(2n-1),2n-1)\leq n.$$ \end{proof}

\begin{thm}\label{thm:even} For all $n\geq 2$, $$\dim K(2n)=n+1.$$
\end{thm}
\begin{proof} From Lemma \ref{lem:important0}, it will be enough to show that $\dim K(2n)\leq n+1$.

Since $y^{2n}\in K(2n)$, from Lemma \ref{lem:important} (1), we can find a basis for $K(2n)$: $$\{y^{2n},H_1,\ldots,H_m\}.$$

Suppose $m\geq n+1$. From Lemma \ref{lem:important} (3), $y(H_1)_{xz},\ldots,y(H_m)_{xz}\in K(2n-1)$. From Theorem \ref{thm:odd}, we have $\dim K(2n-1)=n$ and therefore, these elements must be linearly dependent. So there exist constants $c_1,\ldots,c_m\in\mathbb R$, not all zero, such that $$c_1y(H_1)_{xz}+\cdots+c_my(H_m)_{xz}=0.$$ This implies that $$(c_1H_1+\cdots+c_mH_m)_{xz}=0.$$

So $H:=c_1H_1+\cdots+c_mH_m\in K(2n)$, which is symmetric in $x$ and $z$ from Proposition \ref{prop:symmetry}, is of the form $$H=a_{2n}(x^{2n}+z^{2n})+\cdots+a_2(x^2+z^2)y^{2n-2}+a_1(x+z)y^{2n-1}+a_0y^{2n}.$$

Using Lemma \ref{lem:important1} and Lemma \ref{lem:important} (2), we obtain $$a_1=\cdots=a_{2n}=0,$$ which leads to the linear dependency $$-a_0y^{2n}+c_1H_1+\cdots+c_mH_m=0.$$ This is a contradiction since $\{y^{2n},H_1,\ldots,H_m\}$ is a basis for $K(2n)$. We obtain that $m\leq n$ which proves the theorem: $$\dim K(2n)\leq n+1.$$ \end{proof}

\section{Connections with Schur functors and Roth's equation}

Our initial approach to prove Theorem \ref{thm:main} was more direct: consider an arbitrary element in $\langle C_1,F_1\rangle_r$ and we require it to belong to $\langle C_2,F_2\rangle_r$; this will lead to a comparison of polynomial coefficients.

\vskip .2in

Let $$F=\sum_{i+j+k=r}a_{i,j,k}x^iy^jz^k$$ be an element in $\langle C_1,F_1\rangle\cap\langle C_2,F_2\rangle\subset R=\mathbb R[x,y,z]$ of degree $r$.

We can write $$(*)\mbox{ }F=z^rf_0+z^{r-1}f_1+\cdots+ zf_{r-1}+f_r,$$ where for $k=0,\ldots,r,$ $$f_k=a_{0,k,r-k}y^k+a_{1,k-1,r-k}xy^{k-1}+\cdots+a_{k,0,r-k}x^k$$ is a homogeneous polynomial of degree $k$ in $A=\mathbb R[x,y]$.

From Lemma \ref{lem:important1}, we obtain that for all $k=0,\ldots,r$, $$f_k\in\langle C_1,F_1\rangle A.$$

Since $\langle C_1,F_1\rangle A$ is minimally generated in degree $\geq n$ (see Note \ref{note:note}), then we obtain $$f_0=\cdots=f_{n-1}=0.$$ We are going to use these equations later when we switch the roles of $x$ and $z$. For now let us consider $k=n,\ldots,r$. We have

$$f_k\in\langle C_1,F_1\rangle=\langle x^{r+1},(x+y)^{r+1}\rangle:y^{r+1},$$ and therefore $$(**)\mbox{ }y^{r+1}f_k=P_kx^{r+1}+Q_k(x+y)^{r+1},$$ where $P_k=\sum_{i=0}^kp_{i,k-i}x^iy^{k-i}$ and $Q_k=\sum_{i=0}^kq_{i,k-i}x^iy^{k-i}$ are some homogeneous polynomials of degree $k$ in $A$.

Writing $(x+y)^{r+1}=\sum_{i=0}^{r+1}{{r+1}\choose{i}}x^iy^{r+1-i}$ and identifying the coefficients of $x^uy^v$ in both the left and right-hand sides of $(**)$, for each $k=n,\ldots,r$, we obtain:

\vskip .1in

\noindent (1) For the monomial $x^uy^v$ with $0\leq u\leq k$ we can determine the coefficients of $f_k$ from the coefficients of $Q_k$, as follows:

$$\left(\begin{array}{c}a_{0,k,r-k}\\a_{1,k-1,r-k}\\\vdots\\a_{k,0,r-k}\end{array}\right)= \left(
\begin{array}{ccccc}
1&0&\cdots&0&0\\
{{r+1}\choose{1}}&1&\cdots&0&0\\
{{r+1}\choose{2}}&{{r+1}\choose{1}}&\cdots&0&0\\
\vdots&\vdots& &\vdots&0\\
{{r+1}\choose{k}}&{{r+1}\choose{k-1}}&\cdots&{{r+1}\choose{1}}&1
\end{array}
\right)\left(\begin{array}{c}q_{0,k}\\q_{1,k-1}\\\vdots\\q_{k,0}\end{array}\right).$$

\vskip .1in

\noindent (2) For the monomial $x^uy^v$ with $k+1\leq u\leq r$ we obtain the following conditions on the coefficients of $Q_k$:

$$\left(
\begin{array}{cccc}
{{r+1}\choose{k+1}}&{{r+1}\choose{k}}&\cdots&{{r+1}\choose{1}}\\
{{r+1}\choose{k+2}}&{{r+1}\choose{k+1}}&\cdots&{{r+1}\choose{2}}\\
\vdots&\vdots& &\vdots\\
{{r+1}\choose{r}}&{{r+1}\choose{r-1}}&\cdots&{{r+1}\choose{r-k}}
\end{array}
\right)\left(\begin{array}{c}q_{0,k}\\q_{1,k-1}\\\vdots\\q_{k,0}\end{array}\right)= \left(\begin{array}{c}0\\0\\\vdots\\0\end{array}\right).$$

\vskip .1in

\noindent (3) For the monomial $x^uy^v$ with $u\geq r+1$ the coefficient identifications will express the coefficients of $P_k$ in terms of the coefficients of $Q_k$: $$\left(\begin{array}{c}p_{0,k}\\p_{1,k-1}\\\vdots\\p_{k,0}\end{array}\right)= -\left(
\begin{array}{ccccc}
1&{{r+1}\choose{r}}&{{r+1}\choose{r-1}}&\cdots&{{r+1}\choose{r+1-k}}\\
0&1&{{r+1}\choose{r}}&\cdots&{{r+1}\choose{r+1-(k-1)}}\\
\vdots&\vdots&\vdots& &\vdots\\
0&0&0&\cdots&1
\end{array}
\right)\left(\begin{array}{c}q_{0,k}\\q_{1,k-1}\\\vdots\\q_{k,0}\end{array}\right).$$ These equations will not be useful for our computations. Basically they show how to create $P_k$ from $Q_k$ to have $(**)$ be valid.

\vskip .2in

Let us consider our initial polynomial $F$ expanded by the powers of $x$: $$(***)\mbox{ } F=x^rg_0+x^{r-1}g_1+\cdots+xg_{r-1}+g_r,$$ where $$g_i=a_{r-i,i,0}y^i+a_{r-i,i-1,1}y^{i-1}z+\cdots+a_{r-i,0,i}z^i\in B=\mathbb R[z,y]$$ are homogeneous of degree $i$. Using the fact that $F\in\langle C_2,F_2\rangle R$, Lemma \ref{lem:important1} yields $g_i\in\langle C_2,F_2\rangle B$. Since this ideal is minimally generated in degree $n$ (see Note \ref{note:note}) we have $g_0=\cdots=g_{n-1}=0$. Equivalently,
\begin{eqnarray}
a_{r,0,0}&=&0\nonumber\\
a_{r-1,1,0}&=&a_{r-1,0,1}=0\nonumber\\
&\vdots&\nonumber\\
a_{r-(n-1),n-1,0}&=&a_{r-(n-1),n-2,1}=\cdots=a_{r-(n-1),0,n-1}=0.\nonumber
\end{eqnarray}

These mean that for each $k=n,\ldots,r$, the last $n+k-r$ of the coefficients $a_{i,j,k}$ in (1) above must vanish. So for each $k=n,\ldots,r$, we obtain $n+k-r$ more linear relations among the parameters $q_{a,b}$, that combined with the relations already obtained in (2), yield that for each $k=n,\ldots,r$, the vector $\left(\begin{array}{c}q_{0,k}\\q_{1,k-1}\\\vdots\\q_{k,0}\end{array}\right)$ is in the kernel of a $n\times (k+1)$ matrix $\mathcal M(k)$. We will see this matrix in more detail later on when we study the cases $r=2n-1$ and $r=2n$.

What remains from (1) are the first $(k+1)-(n+k-r)=r-n+1$ of the coefficients $a_{i,j,k}$. So for all $k=n,\ldots,r$,

$$\left(\begin{array}{c}a_{0,k,r-k}\\a_{1,k-1,r-k}\\\vdots\\a_{r-n,k-(r-n),r-k}\end{array}\right)= \left(
\begin{array}{ccccc}
1&0&\cdots&0&0\\
{{r+1}\choose{1}}&1&\cdots&0&0\\
{{r+1}\choose{2}}&{{r+1}\choose{1}}&\cdots&0&0\\
\vdots&\vdots& &\vdots&0\\
{{r+1}\choose{r-n}}&{{r+1}\choose{r-n-1}}&\cdots&{{r+1}\choose{1}}&1
\end{array}
\right)\left(\begin{array}{c}q_{0,k}\\q_{1,k-1}\\\vdots\\q_{r-n,k-(r-n)}\end{array}\right).$$

Combining the above for each $k=n,\ldots,r$ we obtain

$$\left(
\begin{array}{ccc}
a_{0,n,r-n}&\cdots&a_{0,r,0}\\
a_{1,n-1,r-n}&\cdots&a_{1,r-1,0}\\
\vdots& &\vdots\\
a_{r-n,2n-r,r-n}&\cdots&a_{r-n,n,0}
\end{array}
\right)=\mathcal D\cdot \left(
\begin{array}{ccc}
q_{0,n}&\cdots&q_{0,r}\\
q_{1,n-1}&\cdots&q_{1,r-1}\\
\vdots& &\vdots\\
q_{r-n,2n-r}&\cdots&q_{r-n,n}
\end{array}
\right),$$ where $\mathcal D$ is the lower-triangular matrix\footnote{A square matrix is called {\em lower (upper) triangular} if the entries above (below) the main diagonal are all zero.} in the previous vector equation. Denote with $\mathcal A$ the matrix to the left of the above equation and with $\mathcal Q$ the matrix of $q_{a,b}$'s.

$$\mathcal A=\mathcal D\cdot\mathcal Q.$$

\vskip .2in

At this moment we interchange the roles of $x$ and $z$, and we start with $F$ expanded by the powers of $x$, as we've seen in $(***)$. At this point we mentioned that $g_i\in\langle C_2,F_2\rangle B$, and therefore for $i=n,\ldots,r,$ $$g_i\in\langle z^{r+1},(z+y)^{r+1}\rangle:y^{r+1}.$$ Similarly as before $$y^{r+1}g_i=R_ix^{r+1}+S_i(x+y)^{r+1},$$ where $R_i=\sum_{k=0}^ir_{i-k,k}y^{i-k}z^k$ and $S_i=\sum_{k=0}^is_{i-k,k}y^{i-k}z^k$ are some homogeneous polynomials of degree $i$ in $B$.

With the same arguments as before (here we use the fact that $f_0=\cdots=f_{n-1}=0$), we obtain that for $i=n,\ldots,r$, the vector $\left(\begin{array}{c}s_{i,0}\\s_{i-1,1}\\\vdots\\s_{0,i}\end{array}\right)$ is in the kernel of the same $n\times (i+1)$ matrix $\mathcal M(i)$. Furthermore, we have for all $i=n,\ldots,r$,

$$\left(\begin{array}{c}a_{r-i,i,0}\\a_{r-i,i-1,1}\\\vdots\\a_{r-i,i-(r-n),r-n}\end{array}\right)= \left(
\begin{array}{ccccc}
1&0&\cdots&0&0\\
{{r+1}\choose{1}}&1&\cdots&0&0\\
{{r+1}\choose{2}}&{{r+1}\choose{1}}&\cdots&0&0\\
\vdots&\vdots& &\vdots&0\\
{{r+1}\choose{r-n}}&{{r+1}\choose{r-n-1}}&\cdots&{{r+1}\choose{1}}&1
\end{array}
\right)\left(\begin{array}{c}s_{i,0}\\s_{i-1,1}\\\vdots\\s_{i-(r-n),r-n}\end{array}\right).$$

Combining these vector equations for all $i=n,\ldots,r$ we obtain

$$\left(
\begin{array}{ccc}
a_{r-n,n,0}&\cdots&a_{0,r,0}\\
a_{r-n,n-1,1}&\cdots&a_{0,r-1,1}\\
\vdots& &\vdots\\
a_{r-n,2n-r,r-n}&\cdots&a_{0,n,r-n}
\end{array}
\right)=\mathcal D\cdot \left(
\begin{array}{ccc}
s_{n,0}&\cdots&s_{r,0}\\
s_{n-1,1}&\cdots&s_{r-1,1}\\
\vdots& &\vdots\\
s_{2n-r,r-n}&\cdots&s_{n,r-n}
\end{array}
\right),$$ where $\mathcal D$ is the same invertible matrix as before. Denote with $\mathcal B$ the matrix to the left of this matrix equation and with $\mathcal S$ the matrix of $s_{u,v}$'s.

$$\mathcal B=\mathcal D\cdot\mathcal S.$$

Observe that the entries of $\mathcal B$ are the same as the entries of $\mathcal A$, but in different positions. For example, the last column of $\mathcal B$ is the first row of $\mathcal A$ written backwards. This pattern is true for all the columns of $\mathcal B$. In matrix form this relation can be expressed as $$\mathcal J\cdot\mathcal A^{T}\cdot\mathcal J=\mathcal B,$$ where $\mathcal J$ is the exchange matrix $\left(
\begin{array}{cccc}
0&\cdots&0&1\\
0&\cdots&1&0\\
\vdots& & &\vdots\\
1&\cdots&0&0
\end{array}
\right)$.

Everything put together gives $$\mathcal D\cdot\mathcal S=\mathcal J\cdot\mathcal Q^{T}\cdot\mathcal D^{T}\cdot\mathcal J.$$

\vskip .1in

To summarize, we parameterized the elements in $\langle C_1,F_1\rangle_r\cap\langle C_2,F_2\rangle_r$ by two sets of parameters $q_{a,b}$ and $s_{u,v}$, both in the kernel $K$ of the same matrix, and with $(r-n+1)^2$ relations among them given by the above matrix equation. As we will see in the next subsections, the real challenge is not to find $\dim K$ (we will use powerful results from representation theory to do this), but it is to answer the following question: given any two matrices $\mathcal S$ and $\mathcal Q$ satisfying the above matrix equation, can these two matrices be extended to two sets of parameters $q_{a,b}$ and $s_{u,v}$ that are in the kernel $K$?

\vskip .3in

To find the dimension of the kernel $K$ of the matrix with diagonal blocks $\mathcal M(n),\ldots,\mathcal M(r)$ one has to use relevant facts about Schur functors. We briefly recall some of these in the next subsection. We follow the nice exposition in \cite{e}, A.2.5, but we also refer the reader to \cite{m}, Chapter 1.

\subsection{Schur functors} Let $V$ be a vector space over a field of characteristic 0. Let $\dim V=t$. The finite dimensional representations of $SL(V)$ decompose into a direct sum of irreducible representations. These summands are called {\em Schur modules} and can be viewed as functors (called {\em Schur functors}). To a sequence of numbers $t>d_1\geq\cdots\geq d_s>0$ one can associate the Schur module $S^{\{d_1,\ldots,d_s\}}V$, which is a nontrivial, irreducible finite-dimensional representation of $SL(V)$.

Let $Y$ be a Young diagram containing $s$ rows of boxes, row $i$ having $d_i$ boxes. Label the boxes in matrix notation; thus row $i$ has the boxes labeled $i1,i2,\ldots,id_i$. Set $A_i=\{i1,i2,\ldots,id_i\}$ and $A=A_1\cup A_2\cup\cdots\cup A_s$.

Similarly, we label the columns of $Y$ as $B_1,\ldots,B_{d_1}$. For example $B_1=\{11,21,\ldots,s1\}$. Thus $A=B_1\cup\cdots\cup B_{d_1}$.

For each index $ij\in A$ we consider a copy $V_{ij}\cong V$, viewed as a representation of $G=SL(V)$. For each set $A_i$ (or $B_j$) as above, we can consider $S_{A_i}V,\Lambda^{A_i}V,T_{A_i}V$ symmetric, exterior or tensor products of copies of $V$ labeled by indices in $A_i$ (or $B_j$).

For the sake of simplicity, assume $C=\{1,2,\ldots,m\}$ and let us recall that we have an embedding $$\Lambda^cV\rightarrow T_cV,$$ given via $$v_1\wedge\cdots\wedge v_c\mapsto \sum_{\sigma\in S_c}sgn(\sigma)v_{\sigma(1)}\otimes\cdots\otimes v_{\sigma(c)},$$ where $S_c$ is the symmetric group on $c$ letters.

Thus considering the induced natural maps, let $\Phi$ be the composition $$\Lambda^{A_1}V\otimes\cdots\otimes\Lambda^{A_s}V\rightarrow T_{A_1}V\otimes\cdots\otimes T_{A_s}V\cong T_{B_1}V\otimes\cdots\otimes T_{B_{d_1}}V\rightarrow S_{B_1}V\otimes\cdots\otimes S_{B_{d_1}}V.$$

We set $S^{\{d_1,\ldots,d_s\}}$ to be the image of $\Phi$. The classical convention to create $S^{\{d_1,\ldots,d_s\}}V$ from the Young diagram $Y$ is to take antisymmetric products on columns and symmetric product on rows. Observe that in \cite{e}, the construction is reversed.

The action of $G=SL(V)$ on $V$ extends naturally to the action of $G$ on $T_cV$ for each $c\in\mathbb N$, by acting $\sigma\in G$ on each factor, and this action induces an action of $G$ on $S^{\{d_1,\ldots,d_s\}}$.

\begin{exm} $S^{\{2,1\}}V$. The Young tableau is

\begin{center}\epsfig{figure=youngtab.eps,height=0.5in,width=0.5in}\end{center}

We have $A_1=\{11,12\},A_2=\{21\},B_1=\{11,21\},B_2=\{12\}$. So $A=A_1\cup A_2=B_1\cup B_2$.

$\Phi$ is the composition $$\Lambda^{A_1}V\otimes \Lambda^{A_2}V\stackrel{\alpha}\rightarrow T_{A_1}V\otimes T_{A_2}V \cong T_{B_1}V\otimes T_{B_2}V\stackrel{\beta}\rightarrow S_{B_1}V\otimes S_{B_2}V,$$ where $$\alpha((v_{11}\wedge v_{12})\otimes v_{21})=v_{11}\otimes v_{12}\otimes v_{21}-v_{12}\otimes v_{11}\otimes v_{21},$$ and therefore

$$\Phi(v_{11}\wedge v_{12}\otimes v_{21})= v_{11}\cdot v_{21}\otimes v_{12}-v_{12}\cdot v_{21}\otimes v_{11}.$$ $v_{11}\cdot v_{21}$ and $v_{12}\cdot v_{21}$ are products in $S_2V$. We have the fact that $Im(\Phi)=S^{\{2,1\}}V$.

Skipping details that can be found in \cite{fh}, Chapter 1, Lecture Six, one can show that $$\Lambda^2V\otimes V\simeq \Lambda^3V\oplus S^{\{2,1\}}V.$$ One idea of showing this is to embed everything in $T_3V$, and use the conventions and the approach in \cite{fh}. This is based on showing that $S^{\{2,1\}}V$ is the kernel of the map: $$\rho:\Lambda^2V\otimes V\rightarrow\Lambda^3V,$$ given by $\rho(v_1\wedge v_3\otimes v_2)=v_1\wedge v_2\wedge v_3$.

Hence we obtain $$\dim S^{\{2,1\}}V=\frac{(t-1)t(t+1)}{2}.$$
\end{exm}

In fact, in general the formula for $\dim S^{\{d_1,\ldots,d_s\}}V$ is the determinant of the matrix $M$, where $M$ is the $s\times s$ matrix with $$M_{i,j}={{t}\choose{d_j+i-j}}.$$ (Observe that in \cite{e}, Theorem A 2.9, there is a misprint in the entry $M_{s,s-1}$ where it should read ${{t}\choose{d_{s-1}+1}}$.)

\vskip .2in

\subsection{Finding $\dim K$} We have to study the two cases: $r=2n-1$ and $r=2n$.

\subsubsection{The case $r=2n-1$.} If $r=2n-1$, for $k=n,\ldots, 2n-1$ we have
$$\mathcal M(k)=\left(\begin{array}{cccccccc}
m_n&m_{n-1}&\cdots&m_1&m_0&0&\cdots&0\\
m_{n+1}&m_{n}&\cdots&m_2&m_1&m_0&\cdots&0\\
\vdots&\vdots& &\vdots&\vdots&\vdots& &\vdots\\
m_{2n-1}&m_{2n-2}&\cdots&m_n&m_{n-1}&m_{n-2}&\cdots&m_{2n-1-k}
\end{array}\right)$$ is an $n\times(k+1)$ matrix. We denoted $m_{\ell}={{2n}\choose{\ell}},\ell=0,\ldots,2n-1$.

We also have $$K=\oplus_{k=n}^{2n-1}Ker(\mathcal M(k)).$$

\begin{lem} We have $$\dim K=\frac{n(n+1)}{2}.$$
\end{lem}
\begin{proof} First we have $$\dim K=\sum_{k=n}^{2n-1}\dim Ker(\mathcal M(k))=\sum_{k=n}^{2n-1}(k+1-rank(\mathcal M(k))).$$

One should notice the $n\times n$ block in the left part of $\mathcal M(k)$:
$$\mathcal N=\left(\begin{array}{cccc}
{{2n}\choose{n}}&{{2n}\choose{n-1}}&\cdots&{{2n}\choose{1}}\\
{{2n}\choose{n+1}}&{{2n}\choose{n}}&\cdots&{{2n}\choose{2}}\\
\vdots&\vdots&\ddots&\vdots\\
{{2n}\choose{2n-1}}&{{2n}\choose{2n-2}}&\cdots&{{2n}\choose{n}}
\end{array}\right).$$

We have $m_{\ell}=m_{2n-\ell}$. From the above considerations, the determinant of this block is the dimension of the nontrivial, irreducible representation $S^{\{\lambda_1,\ldots,\lambda_n\}}V, \lambda_i=n,$ of $SL(V)$, where $V$ is a vector space of dimension $2n$.

With this fact, since these representations are nontrivial, these determinants are never zero, and therefore, for all $k=n,\ldots,2n-1$, $rank(\mathcal M(k))=n$. So $$\dim K=\sum_{k=1}^n k=\frac{n(n+1)}{2}.$$
\end{proof}

\vskip .1in

\subsubsection{The case $r=2n$.} If $r=2n$, for $k=n,\ldots, 2n$ we have
$$\mathcal M(k)=\left(\begin{array}{cccccccc}
m_{n+1}&m_n&\cdots&m_1&m_0&0&\cdots&0\\
m_{n+2}&m_{n+1}&\cdots&m_2&m_1&m_0&\cdots&0\\
\vdots&\vdots& &\vdots&\vdots&\vdots& &\vdots\\
m_{2n}&m_{2n-1}&\cdots&m_n&m_{n-1}&m_{n-2}&\cdots&m_{2n-k}
\end{array}\right),$$ is an $n\times(k+1)$ matrix. Now, $m_{\ell}={{2n+1}\choose{\ell}},\ell=0,\ldots,2n$.

We have $$K=\oplus_{k=n}^{2n}Ker(\mathcal M(k)).$$

\begin{lem} We have $$\dim K=\frac{(n+1)(n+2)}{2}.$$
\end{lem}
\begin{proof} Consider the left-most $n\times n$ matrix block of the matrix $\mathcal M(k)$: $$\mathcal N=\left(\begin{array}{cccc}
{{2n+1}\choose{n+1}}&{{2n+1}\choose{n}}&\cdots&{{2n+1}\choose{2}}\\
{{2n+1}\choose{n+2}}&{{2n+1}\choose{n+1}}&\cdots&{{2n+1}\choose{3}}\\
\vdots&\vdots&\ddots&\vdots\\
{{2n+1}\choose{2n}}&{{2n+1}\choose{2n-1}}&\cdots&{{2n+1}\choose{n+1}}
\end{array}\right).$$

As in the previous case, the determinant of this block is the dimension of the nontrivial representation $S^{\{\lambda_1,\ldots,\lambda_n\}}V, \lambda_i=n+1,$ of $SL(V)$, where $V$ is a vector space of dimension $2n+1$. Therefore, for all $k=n,\ldots,2n$, $rank(\mathcal M(k))=n$.

We have $$\dim K=\sum_{k=n}^{2n}\dim Ker(\mathcal M(k))=\sum_{k=n}^{2n}(k+1-rank(\mathcal M(k)))=\frac{(n+1)(n+2)}{2}.$$
\end{proof}

\vskip .1in

\subsection{Roth's equation and LU-decompositions} In this subsection we will discover how our problem relates to LU-decompositions of matrices and to solving certain types of matrix equations. Again, we divide the analysis into the two cases: $r=2n-1$ and $r=2n$, and we use the results in the previous two subsections: we have the $n\times n$ block denoted with $\mathcal N$ that is invertible.

\subsubsection{The case $r=2n-1$.} For each $k=n,\ldots,2n-1$ we have $$\mathcal M(k)\cdot \left(\begin{array}{c}q_{0,k}\\\vdots\\q_{n-1,k-n+1}\\ \hline q_{n,k-n}\\\vdots\\q_{k,0}\end{array}\right)=\left(\begin{array}{c}0\\\vdots\\0\end{array}\right).$$

Multiplying to the left the above equation by $\mathcal N^{-1}$, putting everything together, we obtain $$\mathcal Q = -\mathcal N^{-1}\cdot \left(\begin{array}{cccc}
m_0&0&\cdots&0\\
m_1&m_0&\cdots&0\\
\vdots&\vdots& &\vdots\\
m_{n-1}&m_{n-2}&\cdots&m_{0}
\end{array}\right)\cdot \left(\begin{array}{cccc}
q_{n,0}&q_{n,1}&\cdots&q_{n,n-1}\\
0&q_{n+1,0}&\cdots&q_{n+1,n-2}\\
\vdots&\vdots& &\vdots\\
0&0&\cdots&q_{2n-1,0}
\end{array}\right).$$

Observe that the lower triangular matrix above is exactly the matrix $\mathcal D$. Also denote with $\tilde{\mathcal Q}$, the upper triangular matrix we see above. The entries of this matrix consist of all the $q_{a,b}$ not occurring in $\mathcal Q$.

We obtain $$\mathcal Q = -\mathcal N^{-1}\mathcal D\tilde{\mathcal Q}.$$

We have the same result for the parameters $s_{u,v}$: $$\mathcal S = -\mathcal N^{-1}\mathcal D\tilde{\mathcal S}.$$

We had that $\mathcal D\mathcal S=\mathcal J\mathcal Q^{T}\mathcal D^{T}\mathcal J$. Denoting with $\mathcal E=\mathcal D\mathcal N^{-1}\mathcal D$, we have $$\mathcal E\tilde{\mathcal S}=\mathcal J(\tilde{\mathcal Q})^T\mathcal E^T\mathcal J,$$ and since $\mathcal J=\mathcal J^{-1}=\mathcal J^T$, denoting with $\mathcal U=\mathcal J\mathcal E$, we obtain $$\mathcal U\tilde{\mathcal S}=(\mathcal U\tilde{\mathcal Q})^T.$$

Our original problem translates into the following question regarding the solution of a certain type of matrix equation:

\begin{prop} We have $\dim K(2n-1)=n$ if and only if the matrix equation $$\mathcal UX-Y^T\mathcal U^T=C,$$ has a solution consisting of two upper-triangular matrices for any $n\times n$ matrix $C$.
\end{prop}
\begin{proof} The matrix equation has the desired solution if and only if the $\mathbb R-$linear map $$\phi: \mathbb R^{\frac{n(n+1)}{2}}\oplus \mathbb R^{\frac{n(n+1)}{2}}\rightarrow \mathbb R^{n^2},$$ given by $\phi(\tilde{\mathcal S},\tilde{\mathcal Q})= \mathcal U\tilde{\mathcal S}-\tilde{\mathcal Q}^T\mathcal U^T$, is surjective. \footnote{Here we have used the fact that an upper-triangular $n\times n$ matrix is described by $\frac{n(n+1)}{2}$ parameters.}

But this is equivalent to the dimension of $ker(\phi)\simeq K(2n-1)$ being equal to $n(n+1)-n^2=n$.
\end{proof}

\vskip .1in

\subsubsection{The case $r=2n$.} For this case the same things occur. For each $k=n,\ldots,2n$ we have $$\mathcal M(k)\cdot \left(\begin{array}{c}q_{0,k}\\\vdots\\q_{n-1,k-n+1}\\ \hline q_{n,k-n}\\\vdots\\q_{k,0}\end{array}\right)=\left(\begin{array}{c}0\\\vdots\\0\end{array}\right).$$

Multiplying to the left the above equation by $\mathcal N^{-1}$ and putting everything together, we obtain $$\left(\begin{array}{ccc}
q_{0,n}&\cdots&q_{0,2n}\\
q_{1,n-1}&\cdots&q_{1,2n-1}\\
\vdots& &\vdots\\
q_{n-1,1}&\cdots&q_{n-1,n+1}
\end{array}\right) = -\mathcal N^{-1}\left(\begin{array}{ccccc}
m_1&m_0&0&\cdots&0\\
m_2&m_1&m_0&\cdots&0\\
\vdots&\vdots&\vdots& &\vdots\\
m_n&m_{n-1}&m_{n-2}&\cdots&m_{0}
\end{array}\right)\left(\begin{array}{ccc}
q_{n,0}&\cdots&q_{n,n}\\
0&\cdots&q_{n+1,n-1}\\
\vdots&\ddots&\vdots\\
0&\cdots&q_{2n,0}
\end{array}\right).$$

The matrix to the left of the equality is an $n\times (n+1)$ matrix consisting of the first $n$ rows of the matrix $\mathcal Q$ we saw before. The missing row occurs as the first row in the upper-triangular matrix we see on the right.

To correct this inconvenience, let $\mathcal N'$ be the $(n+1)\times (n+1)$ matrix $$\mathcal N'=\left(\begin{array}{ccc}
0&|&\mathcal N^{-1}\\ \hline
-1&|&0
\end{array}\right),$$ and since $m_0=1$ observe that $$\mathcal Q=-\mathcal N'\left(\begin{array}{ccccc}
m_0&0&0&\cdots&0\\
m_1&m_0&0&\cdots&0\\
m_2&m_1&m_0&\cdots&0\\
\vdots&\vdots&\vdots& &\vdots\\
m_n&m_{n-1}&m_{n-2}&\cdots&m_{0}
\end{array}\right)\left(\begin{array}{ccc}
q_{n,0}&\cdots&q_{n,n}\\
0&\cdots&q_{n+1,n-1}\\
\vdots&\ddots&\vdots\\
0&\cdots&q_{2n,0}
\end{array}\right).$$

Observe that the lower triangular matrix above is exactly the matrix $\mathcal D$. Also denote with $\tilde{\mathcal Q}$, the upper triangular matrix we see on the right side of the equality.

We obtained $$\mathcal Q = -\mathcal N'\mathcal D\tilde{\mathcal Q}.$$

We have the same taking place for the parameters $s_{u,v}$: $$\mathcal S = -\mathcal N'\mathcal D\tilde{\mathcal S}.$$

We had that $\mathcal D\mathcal S=\mathcal J\mathcal Q^{T}\mathcal D^{T}\mathcal J$. Denoting with $\mathcal E=\mathcal D\mathcal N'\mathcal D$, and with $\mathcal U= \mathcal J\mathcal E$ similarly to the case $r=2n-1$, we have $$\mathcal U\tilde{\mathcal S}=(\mathcal U\tilde{\mathcal Q})^T.$$

As for the other case, with a similar proof, we have:

\begin{prop} We have $\dim K(2n)=n+1$ if and only if the matrix equation $$\mathcal UX-Y^T\mathcal U^T=C,$$ has a solution consisting of two upper-triangular matrices for any $(n+1)\times (n+1)$ matrix $C$.
\end{prop}

\vskip .3in

The equations in the two propositions above are a particular case of {\em Roth's equation} (see \cite{r}): $$AX-YB=C.$$ This equation has solutions in $X$ and $Y$ if and only if the matrices $\left(\begin{array}{ccc}
A&|&0\\ \hline
0&|&B
\end{array}\right)$ and $\left(\begin{array}{ccc}
A&|&C\\ \hline
0&|&B
\end{array}\right)$ have the same rank (see \cite{p2}, Theorem 44.3, page 198).
\vskip .1in

Though in our case $A=\mathcal U$ and $B=\mathcal U^T$, and they are invertible matrices, our goal, and challenge, is to find a special type of solution: we need $X$ to be upper-triangular and $Y$ to be lower-triangular. The next lemma presents one instance when this goal is achieved.

First, we say that an invertible matrix $W$ admits an {\em LU-decomposition} if $W$ has a decomposition: $W=VU$, with $V$ a lower-triangular matrix and $U$ an upper-triangular matrix. It is known that for any invertible matrix $W$ there exists a permutation matrix $P$ such that $PW$ has an LU-decomposition.

\begin{lem}\label{lem:triang} Let $C$ be a $p\times p$ matrix, and $W$ be an invertible $p\times p$ matrix that admits a LU-decomposition. Then there exist two upper triangular matrices $X$ and $Y$, such that $$C=WX-Y^TW^T.$$
\end{lem}
\begin{proof} We have the classical known properties of triangular matrices: (1) the transpose of an upper (lower) triangular matrix is a lower (upper) triangular matrix; (2) the inverse of an upper (lower) triangular matrix is an upper (lower) triangular matrix; (3) the product of two upper (lower) triangular matrices is an upper (lower) triangular matrix.

We have $W=VU$. Let $C'=V^{-1}C(V^T)^{-1}$ and write $C'=C'_u-C'_l$, where $C'_u$ is upper triangular and $C'_l$ is lower triangular (this decomposition is not unique).

Let $X=U^{-1}C'_uV^T$ and $Y=U^{-1}(C'_l)^TV^T$. Both $X$ and $Y$ are upper triangular and we have $$WX-Y^TW^T = VUU^{-1}C'_uV^T-VC'_l(U^T)^{-1}U^TV^T=VC'V^T=C.$$ \end{proof}

Based on possibly not enough experimentation, we ask the following question:

\begin{question} Let $W$ be an invertible matrix. If for any square matrix $C$ the equation $C=WX-Y^TW^T$ has a solution consisting of two upper-triangular matrices, is it true that $W$ has an LU-decomposition?
\end{question}

Regardless if the above question has an affirmative answer or not, one has the following classical criterion: an invertible matrix has LU-decomposition if and only if its leading principal minors are nonzero (see \cite{g}, page 35).

\vskip .3in

Now we go back to our problem. First, let us denote with $\bar{\mathcal N}$ the matrix $\mathcal N^{-1}$ (when $r=2n-1$), and also with $\bar{\mathcal N}$ the matrix $\mathcal N'$ (when $r=2n$). Then our invertible matrix that we would like to show has LU-decomposition is $$\mathcal U=\mathcal J\mathcal D\bar{\mathcal N}\mathcal D.$$

\begin{prop}\label{prop:propertiesU} \begin{enumerate}
                                       \item $\mathcal U$ is symmetric.
                                       \item $\mathcal J\mathcal U\mathcal J$ has LU-decomposition.
                                     \end{enumerate}
\end{prop}
\begin{proof} To prove (1), observe that in both cases for $r$, the matrices $\mathcal J\mathcal N$ and $\mathcal J\mathcal D$ are symmetric; this is true from the fact that multiplying a matrix to the left by the exchange matrix $\mathcal J$, we reverse the rows in the matrix. If we multiply to the right, we reverse the columns.

$\mathcal J^2$ is the identity matrix, so $\mathcal N^{-1}\mathcal J$ is also symmetric. Now, writing $$\mathcal U=(\mathcal J\mathcal D)(\bar{\mathcal N}\mathcal J)(\mathcal J\mathcal D),$$ we obtain the result.

\vskip .2in

To prove (2), observe that $\mathcal D\mathcal J=\mathcal J\mathcal D^T$. With this we have $$\mathcal J\mathcal U\mathcal J=\mathcal D(\bar{\mathcal N}\mathcal J)\mathcal D^T.$$ It becomes enough to show that $\bar{\mathcal N}\mathcal J$ has LU-decomposition, since $\mathcal D$ is lower-triangular and hence $\mathcal D^T$ is upper-triangular.

By the way we denoted what $\bar{\mathcal N}$ is, it will be enough to show that $(\mathcal J\mathcal N)^{-1}$ has LU-decomposition, or equivalently, that $\mathcal J\mathcal N$ has upper-lower decomposition (i.e., UL-decomposition).

In terms of minors, we have to show that the minors of $\mathcal N$ starting from the north-east corner and moving down along the anti-diagonal to the south-west corner, are nonzero: $$det(\mathcal N_{\{1\}\times\{n\}})\neq 0, det(\mathcal N_{\{1,2\}\times\{n-1,n\}})\neq 0,\ldots,det(\mathcal N_{\{1,\ldots,n\}\times\{1,\ldots,n\}})\neq 0.$$

Our matrix $\mathcal N$ is a submatrix of the Toeplitz matrix $$A_T=(a_{j-i})_{i,j}$$ associated to the polynomial $$T(x)=(1+x)^{r+1}=a_0+a_1x+\cdots+a_{r+1}x^{r+1}.$$

The polynomial $T(x)$ has all roots real numbers, equal to $-1<0$, and has coefficients $a_i>0$. From a theorem of Aissen-Schoenberg-Whitney from 1952 (see \cite{p}, Theorem 4.5, page 105), this will imply that all the minors of $A_T$, and therefore of $\mathcal N$, are strictly larger than zero.\footnote{This also gives an alternative proof that $det(\mathcal N)\neq 0$, and so $\mathcal N$ is invertible.}

In particular, the minors of our interest are different than zero, and hence the claim.
\end{proof}

Though Proposition \ref{prop:propertiesU} (2) does not resolve our problem, it gives lower-triangular solutions for our equation $$\mathcal UX-Y^T\mathcal U=C.$$ Part (1) of Proposition \ref{prop:propertiesU} allows us to replace $\mathcal U^T$ by $\mathcal U$.

\begin{cor} For any square matrix $C$, the equation $$\mathcal UX-Y^T\mathcal U=C$$ has a solution consisting of two lower-triangular matrices $X$ and $Y$.
\end{cor}
\begin{proof} From Proposition \ref{prop:propertiesU}, for any square matrix $C$, there exist two upper-triangular matrices $X'$ and $Y'$ such that $$(\mathcal J\mathcal U\mathcal J)X'-(Y')^T(\mathcal J\mathcal U\mathcal J)=\mathcal JC\mathcal J.$$

Then $X=\mathcal J X'\mathcal J$ and $Y=\mathcal J Y'\mathcal J$ are lower-triangular solutions of the equation $$\mathcal UX-Y^T\mathcal U=C.$$
\end{proof}

Is the corollary above enough to show that $\dim K(2n-1)=n$ and $\dim K(2n)=n+1$?

\vskip .1in

\textbf{Acknowledgements:} We would like to thank the anonymous referee for the useful corrections, suggestions and comments.

\renewcommand{\baselinestretch}{1.0}
\small\normalsize

\bibliographystyle{amsalpha}


text:
David McNamee: Hype-free and ready to deliver in St George
David McNamee will start the IRONMAN World Championship in St George, Utah on Saturday in a very select group. The Girona-based Scot is one of only five men who will be racing with a previous IRONMAN World Championship podium finish on their C.V.
In the case of McNamee, he has two – third place in Kona in both 2017 and 2018. Five years on, but still just 34 years of age, can he call upon that experience again to add another?
I spoke to David this past weekend via Zoom, and he gave me his thoughts on why he feels this race will be 'very different from Kona'.
With a full breakdown of the course, his dark horse to watch – and why overlooking Alistair Brownlee would be a big mistake – it's a great insight into what to look out for during the race.
Nice confirmed as dual host of IRONMAN World Championship to 2026
Welcome back World Champs
McNamee has never hidden the fact that preparing for the biggest race of the year – typically Kona – is what excites him. While a lot is different about this build-up, I started by asking if he was finding a similar sort of motivation and excitement ahead of St George.
"Of course. It's not Kona, and I think that everybody knows that I love the conditions of Kona, but it's great to finally have a World Championship after two-and-a-half years. I'm excited about the course.
"I'm a big fan of humidity – of which there is absolutely none here! – but I think that everyone is just happy and relieved to be racing a World Championships after such a long, long time."
Photo Credit: Activ' Images
Third time lucky?
David has raced twice already this year, with a DNF at IRONMAN 70.3 Lanzarote and a 12th-place finish at IRONMAN 70.3 Oceanside to show for his efforts. A modest return on the surface, but the typically relaxed McNamee is not too concerned.
"In Lanzarote, I just got sick two days before the race. I came down with a really bad sort of virus. I probably shouldn't have even tried to start the race, but it's always tough when you are already there at the race event… you just sort of hope that you wake up on race morning and everything will be ok. I knew very quickly into the race that it wasn't happening.
"Oceanside – I don't know if it was a mixture of bad luck or my own stupidity – but I ended up hitting a massive pothole on the course and I think because my weight was so far forward on the bike, the next thing I knew my saddle was pointing downwards instead of horizontal and that made riding near on impossible. I had to end up waiting for the mechanic, so it's not been a great start to the year which seems to be the story of the last couple of years for me, to be honest. I'm hoping of course next week changes that."
While those bike woes put him out of competitive contention for the top places, I noted that he did continue and finish the race. A 1:11 half marathon when you are eight-plus minutes back is pretty solid – was that an attempt to still make it a strong training day?
"Yeah, I think 1:11 is a good time. I was happy that even though I was out of the race, I still committed to giving it my best performance. At one point if you ran a 1:11 you'd probably have the quickest run split, but now guys are running like a 1:08. It's not crazy… but the sport has taken another step forward."
Hype? No thanks!
One thing that is different this year is the degree of spotlight on McNamee. Ahead of both the 2018 and 2019 Kona races, he was a top-three finisher 12 months previously. More than two-and-a-half years on, new names have earned the headlines and McNamee hasn't been featuring in too many predictions articles.
Given David could well be the most chilled athlete in the sport, it probably won't come as a surprise that he's not too concerned…
"For sure, I just sort of try and concentrate on my own things. I don't need other people to be hyping me up, I just always try and have trust and confidence in the training process.
"I know I love the big championship races and I'm hoping that I have an incident-free race next week. I know my fitness is there to deliver a good result. That's all that matters to me – I'm not one for the big hype!"
Photo by Nils Nilsen/Getty Images for IRONMAN
A very different race, a different sport
A new venue, a new calendar slot and new athletes racing – there are a lot of questions to be answered on Saturday. David believes that we won't be simply taking the Kona playbook and doing a copy/paste in Utah.
"I think the race will actually be very different from Kona. I think also it's been two-and-a-half years, which in sport is almost like a lifetime.
"It's the first time in years that we've had all the best guys – the healthy best guys anyway – on a start line together. I've never looked at the start line before in an IRONMAN World Championship and seen such a mixture of ages. I think the youngest guy is 22, Sam (Laidlow), and then you have people in their early to mid-40's too, like Andy Potts.
"I remember doing my first Kona, I'm pretty sure I was the youngest guy in the field at 27. That adds a different dimension and dynamic to the race. I think the sport is changing. It's always been seen as like an older person's sport, and if you come from an ITU background you'd never consider combining ITU racing and Ironman. It used to be you typically retire from ITU racing and move to an Ironman event, maybe in your early to mid 30's.
"I know when I switched from ITU at 27, people thought I was crazy, that I was too young. Now you have people just skipping that phase entirely, or the Norwegians who decide to do everything they want!"
Swim: absent Amberger and the dark horse
We talked at length through each discipline, and potentially how the course and the entries (and absent names too), could impact things:
"I think the swim will be interesting, in that when you take Josh out of the race [Ed. Amberger], it's very different. His take-out speed is incredible, so much so that even the very good swimmers struggle to stay with him. That means that the guys who are sometimes front pack, sometimes second pack, have no chance of being with him.
"With him out, it could then be a bigger pack than usual. The people that are usually borderline… with Josh not there, they could make the front pack. Depending how big that group is, it could shape the dynamics of the first 80-90km on the bike.
"I think if someone like Kristian Hogenhaug [Ed. World Triathlon Long Distance Champion], if he makes the front pack, he's such a strong cyclist. Having raced against him at IRONMAN Frankfurt last year and seen how explosive he can be, if he can be at the front from the start of the race, he's a guy that could make the bike very traumatic for everyone else.
"If you asked me who is the dark horse that nobody is really speaking about, it's probably him. He could easily go top-five, if not better. I think him and Leon Chevalier are very similar in that if they do end up in the front pack of the swim, they could both on that type of bike course do a lot of damage to everybody else."
What's interesting to note about Hogenhaug, is that Rudy Von Berg also marked him out as his dark horse to watch when we spoke in January. He also won, with a "best ever power output on the bike and a half marathon PB" recently at Challenge Salou.
Kristian Hogenhaug – Photo Credit: Jose Luis Hourcade (www.joseluishourcade.com)
Bike: different course, different dynamic
"I think the course profile, especially on the bike, makes it a very different event. I think in Kona the bike course rewards someone who can hold a stable power output. Somebody who can sit at that 280-300 watts and push that consistently for four-and-a-bit hours.
"I think here there will be a lot more dynamics to the racing. There's a lot of short, sharp, steep climbs where you have to put down big power to get up and over them without losing all momentum. The guys that can do the big efforts and then recover from them – like the guys who have come across from ITU recently – they are used to that. This course suits them more than Hawaii would.
"There will be these three to five-minute efforts on climbs, but then the ability to recover from them. I think a few guys will sort of ruin their race by following the more explosive characters and not be able to recover from the efforts. I've ridden the course a few times – it's sort of like IRONMAN UK, in that it doesn't matter if you are at the front of the race or the back, to get up the climbs is still a bloody big effort.
"Hawaii is not like that, it's more gradual and suits the diesel engines."
David winning IRONMAN UK 2017: Photo by Stephen Pond/Getty Images
Run: expect explosions
While the run course for the IRONMAN World Championship removes some of the very steep descents we witnessed for the 70.3 Championship last year, it will still provide a stern test.
"Basically the whole marathon is on some sort of gradient and will be challenging. I think a lot of how fast it will be, will depend upon how hot it is, whether there is cloud cover or not. I've been here a few days and some days it's been 30 degrees and no clouds, then the next day 23-24 and clouds. It varies massively how hard it is to run.
"If it's a hot day, especially starting the marathon, there's going to be a lot of explosions in the last 10-15km.
"The descents here are a lot more gradual than those on the 70.3 course, however in the last 10/15km in any Ironman, any gradient feels horrific. There's been a few times in Hawaii that I've been running down Palani Drive and thought that I might not be able to make it. While people are training on the run course now, I think they will get a big shock to how it feels on race day."
Don't underestimate Brownlee
We finished with a final word on those that seem to have dismissed the prospects of a double Olympic champion from contending on Saturday.
"I think Alistair Brownlee will be a lot closer than people think on this course. It's a course for him, it's a course for the Norwegians. I think coming off the bike that Kristian, Gustav and Alistair will be the front three."
Will David be right with them?
"I'm a realist – I know that we have to go up Snow Canyon near the end, and that's going to cause a lot of pain. I'm a good climber, but I know how good Alistair and Gustav are. You saw it in 70.3 Worlds last year, Gustav was incredible going up the climb.
"We'll see, I'm ready to give it my best shot."

redpajama_set_name: RedPajamaCommonCrawl


Good luck, my name is Axel Helbig. I am a pioneer, strategist and lateral thinker – my passions have been multichannel business and my customers for the last 26 years. As a member of the management board I am responsible for the location Essen of the foryouandyourcustomers group and I am already looking forward to the discussion and the exchange with you as our customer. Besides my profession I enjoy spending quality time with my wife and our children, going on extensive mountain bike tours and am interested in arts and culture as well.


text:
UNICEF Raises World Hunger Awareness
Senior Tenzin Yontin and Nam Vo enjoying their "first world" meal. Picture by Rose Nguyen '20.
Sophie Sharp, Social Media Editor
Food is such a basic necessity that it's often easy for MacDuffie students to take it for granted, despite millions of people around the world who go hungry each day. In an effort to raise awareness of world hunger, members of the school's UNICEF (United Nations International Children's Emergency Fund) Club hosted a "Hunger Banquet" this past Tuesday and invited all students and faculty to attend.
At 5:30 on Tuesday afternoon, twenty students made their way to the cafeteria where they were assigned a country categorized as either first, second, or third world. Students assigned to first world countries were escorted to a table with a white tablecloth, folded cloth napkins, and tall water glasses where they waited to be served a variety of food by students posing as waiters.
Those appointed second world countries gathered at a nearby table with a tablecloth before getting up to serve themselves food. Those who were designated third world were asked to sit on the floor or on two straw bags and were given four bowls of rice, soy sauce, and four spoons to split between the group of them. During the meal, Senior Taylor Pluta, a member of the UNICEF club, read a script primarily written by the club's President Cindy Nguyen, elaborating on the consequences of world hunger.
Junior Carrie Lu, another member of the UNICEF club, summed up the objective of the event, saying, "The main goal was basically to…place people in the real life scenario and have them feel how third world kids are feeling."
Senior Mojolaoluwa Oke, who was also a UNICEF club member at the event, was pleased that participants had such strong reactions to the events that played out. "When people who were set up in third world countries had to go get water and then died from cholera…there were some people that reacted like 'Wow…this really happens?' And that's really what I was looking for," Oke said.
If there had been more time, Lu would have liked to hold a discussion after the Banquet was over. "I think it was super good except for the time limit, it was too short. The purpose is to make this interactive and to have people share their comments, but since we have limited time, we didn't do that."
Despite the time limit, Oke categorized the event as a success. "I felt that we affected most of the people who attended."
Seniors Tenzin Yonten and Nam Vo enjoying their "first world" meal. Picture by Rose Nguyen '20.
New Librarian Joins the MacDuffie Faculty
MacDuffie Math Department Adds a Familiar Face
Ms. Tuleja Departs From MacDuffie to Pursue Other Opportunities in Higher Education
Ms. Panico Leaves MacDuffie After Five Years as Teacher, RA, and Magnet Advisor
"Yeah, I'll give that a try!" Tew Reflects on her Time at MacDuffie and Future Endeavors
MacDuffie Plans For Modified In-Person Graduation Ceremony
COVID-19 Testing in Effect at MacDuffie
Tech Improvements Benefit Online Learners
Ms. Amanda Putnam joins the MacDuffie Arts Department
Is MacDuffie Nearing a Remote Schooling Future?


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Perfect finish and seamless surface: It makes the surface incredibly smooth ad seamless. It is easy to clean and wipe. Hence, it is perfect for food industries, hotels, and restaurants. It looks quite attractive and gives the premises a stylish appearance. There are innumerable patterns, colors and décor are available in the market. One can pick up the best fit one. It is possible to create patterns using different shades of flooring.
High resistance to chemicals and anti-skid properties: Epoxy flooring is ideal for industries, warehouses, and manufacturing plant because it is greatly resistant to chemicals. Even if it has a glossy finish, you get good skid resistance.
Stylish look: People in Chennai are typically crazy for trendy and mod floorings. Jaichitra satisfies their needs with a mind-boggling collection of epoxy flooring in Chennai.
Epoxy flooring is an easy to install, but a stylish alternative for household and industrial environment. However, there are some cases when people go for epoxy flooring but it comes out a nightmare. Experts say that it happens because people go for it without preparation. A few probable reasons for the fiasco are as follows.
Epoxy flooring needs a thorough preparation before installation. The surface should be perfectly plain and smooth. Even if the ups and downs are minor and negligible, you won't get satisfactory results. Make the surface clean using a short blast or grind. It is equally important to clean the surface thoroughly after making it plane. Dust and dirt should be removed before flooring.
Humidity results in improper adhesion. Epoxy coating doesn't bond with the surface if the surface is not fully dry. Make sure there are no taps dipping or no leakage in the water pipes. In case of water seepage from the ground, waterproofing should be done before Epoxy flooring.
When the substrate is weaker than adhesive, the layer of Epoxy gets detached after some time. Tensile strength and adhesion of the substrate should be compatible with Epoxy. For example, wood is not an appropriate substance because their rate of expansion is different which causes cracking.
Do not forget to apply a high-quality primer on the floor.
There is a great contamination on old floors, factories, hotels and, restaurants, etc. Don't expect great results if the contamination is not removed properly.
Sometimes, mismatched resin and hardeners cause disappointing results. It is important that they are thoroughly mixed before application.
Call Jaichitra today if you want a makeover of the flooring. You will get the perfect advice and top-quality epoxy flooring in Chennai.


Rarely does everything always go to plan as we discovered in Munich at the third and final World Cup regatta. Here we were beaten by Australia, our main and closest competition, ending a 15-race winning streak for me in the coxless four. We were beaten in a very tight race by a very fast, efficient crew who look good together and row very well. These Australians have a very effective rhythm that allows them to use the power they produce in their stroke to maximum benefit and lets the boat travel a great distance with each stroke. In contrast, we have a powerful stroke and put in a lot of effort but so far we haven't been able to transfer that into maximum efficiency and boat speed. We need to be more patient with each stroke and allow the boat to move for us, flow from stroke to stroke and resist the temptation to rush the hull. There is a balance in rowing that is nearly impossible to see from the bank. It is a feeling of relaxed movement and flow, reducing the negative forces on the hull while putting every ounce of energy into the stroke. It sounds like a contradiction but to a rower makes perfect sense. So far this season we have got away with our way of rowing but coming up against a crew with such pedigree and experience as the Australians, we need to make improvements.
This result was not actually a shock and as much as we were disappointed as a crew not to maintain our winning ways, we came off the water feeling positive about the result. It was nothing but a good thing for us to get beaten, better here than in the Olympic final, and it forces us to make much-needed technical changes. It is so easy to win a race and be happy with what you have and with what you are doing but you may not be reaching your potential. I have learnt that most improvements come from the mistakes we make and so this has come at a perfect time. We have new fire in our bellies and desire to significantly improve. We are not far off where we need to be, no doubt the Australians will gain more speed in the coming weeks but I am confident that we will too. There is one thing for sure, an Olympic gold is not an easy thing to come by, and come the 4th August on Dorney Lake, there will be an almighty battle to the finish line. The important thing for me and my crew is to enjoy every second of these coming weeks, we are facing a huge challenge but it is something we relish. We now head off to Austria for a gruelling altitude training camp where the last big push in strength and fitness will be made.


Something happens when you step into a volcanic mud pool or a thermal mud pool. It's a mystery but after you spend some time soaking in one of these pools, you feel rejuvenated! Nobody knows why, it just happens.
Hot springs are nestled at the base of the Arenal Volcano. You can pick any of the pools that you want, depending on the temperature you feel comfortable sitting in. All you have to do to take advantage of the pools' healing power is just sit back and relax. Make sure you get a spot under the cascading water. Watch as the waters from the Arenal River flow by through the beautiful gardens. The view is second to none. There's beauty wherever you look. And the pulsating, heated water is sure to make you feel refreshed and invigorated.
Costa Rica, however, is much more famous for its thrilling canopy tours and whitewater tubing adventures. Its reputation for being a haven for relaxation is starting to grow as well. Still, most people come for the eco-tourism and opportunities for some heart-thumping, adrenaline-pumping activities, not to mention observing one of the most unique environments in the world.
This country is rich with the most breathtaking landscapes on the planet. It abounds with waterfalls, rivers, mountains and beaches. And the combination of all of these features has created an ecosystem in which all kinds of animals and plant life thrive. In fact, there is more than 500,000 types of species that live in Costa Rica.
Surfers flock to this country because of its opportunities to catch the perfect wave, whether they like them tall or short. There's a beach for whatever your skill level happens to be. That's because there are two oceans that surround Costa Rica.
And when you're finished exploring the countryside or soaking in a thermal pool, it's time to relax at your private accommodations at Ollie's Point Surf Camp. Get some sleep or catch up on your favorite TV shows or a movie with the 40-inch TV or DVD player. You can also surf the Web with the wireless Internet access offered by the resort or take a late-night dip at the resort's gorgeous outdoor pool.
Vacation packages also include a chef to prepare your dinners and transportation to and from the airport. Relaxation is the name of the game at this resort. If relaxing is the best thing for your body and soul, there's no better place than Ollie's Point Surf camp. Book your vacation today!


income tax return(s), annex(es), statement(s), etc. and a brief introduction of these forms for the tax annex an application for claim of refund in... Individuals who hire professional tax return preparers are Tax return preparation fees vary by price for a corporate and personal return with that income?


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Home > News > Walla Walla Valley winery at home with Oregon Tempranillo
Walla Walla Valley winery at home with Oregon Tempranillo
By Eric Degerman on January 26, 2016
Chris Castillo, winemaker for Castillo de Feliciana Vineyard and Winery in Milton-Freewater, Ore., made the eight-hour, cross-state drive to attend the inaugural Oregon Tempranillo Alliance Celebration on Jan. 22, 2016, in Ashland. (Photo by Eric Degerman/Great Northwest Wine)
ASHLAND, Ore. — Chris Castillo made sure the Walla Walla Valley was represented at the inaugural Oregon Tempranillo Celebration.
The winemaker of Castillo de Felicana Vineyard and Winery traveled the farthest to reach Ashland near the California border, but he and his parents felt at home as members of the new Oregon Tempranillo Alliance.
"We really wanted to be a part of this because there's the opportunity in Oregon to showcase high-end domestic Tempranillo," Castillo told Great Northwest Wine.
Friday's conference was the brainchild of Earl Jones, founding winemaker of Abacela in Oregon's Umpqua Valley, who planted the state's first commercial Tempranillo vineyard in 1995.
"This came together beautifully, thanks to all of the Tempranillo producers," said Jones, who noted the grape now has a $9 million presence in Oregon. "It's remarkable growth in such a short period of time."
Les Martin, co-owner of Red Lily Vineyards in Jacksonville, said, "Obviously, we are always grateful to Earl. He's probably tired of me calling him 'The Godfather of Tempranillo,' but he really is. He started it all here. We've piggybacked on what he started, and now some people are piggybacking off of us."
Jones recruited Martin, Scott Steingraber of Kriselle Cellars and Eric Weisinger of Weisinger Family Winery in January 2015 to help develop and coordinate the program for the alliance's Oregon Tempranillo Celebration.
"We were super happy with the way things turned out," Martin said. "We now have 42 members starting from scratch, and that's out of 57 Tempranillo producers in the state, so we've gotten the majority of the wineries to sign on with us. And we had 30 of those wineries sign up and participate for the Celebration."
Tempranillo symposium features 3 blind tastings
Rachael Martin, winemaker for Red Lily Vineyards in Jacksonville, Ore., earned a double gold at the 2016 San Francisco Chronicle Wine Competition for her 2012 Red Lily Tempranillo. (Photo by Eric Degerman/Great Northwest Wine)
The single-day conference included a series of blind Tempranillo tastings to study the influence of soil types and aging as well as analyze wines made in several American Viticultural Areas throughout Oregon — including the Walla Walla Valley.
"The focus was Southern Oregon rather than the Willamette Valley, and as Milton-Freewater we were a bit of an outside example," Castillo said. "From a wine standpoint, I thought the Southern Oregon wines were expressive and more fruit-driven and balanced than we what have to work with in the Walla Walla Valley at this point."
Earlier this month, Martin's winemaking wife, Rachael, earned a double gold for her Red Lily 2012 Tempranillo at the 2016 San Francisco Chronicle Wine Competition. That wine was among those inserted into one of the tasting seminars, which featured brief informal presentations by those who made the individual wines. There also were a number of Tempranillo bottles poured by wineries in the North Willamette Valley — Pinot Noir country.
"We want to come back, and when we come back we hope there will be more producers," Castillo said. "We can get Tempranillo the spotlight it deserves and get consumers to try more than their standard Pinot Noirs, Cabs and Syrahs. Try some of these other varietals that deserve the attention — like today. All the wines we tasted were phenomenal."
A year ago, one of the Northwest's more fascinating examples of Tempranillo was produced by Walla Walla Community College. It released the 2014 Estate Tempranillo within months of production and earned the equivalent of a gold medal in a Wine Press Northwest magazine peer-judging. It matched the traditional young, low-oak joven style wines of Spain.
"The people who come into our tasting room are looking for varieties other than Cabernet, Syrah and Merlot, and wineries are telling them to go see us on the other side of the stateline — this family that is growing Tempranillo and Albariño," Castillo said. "We get a lot of recommendations from other wineries because of that. We're glad to be doing something different because it helps, and I'm trying to be the best winemaker I can."
Castillo, a 2012 graduate of WWCC's College Cellars, said he's enjoyed that 2014 joven-style Temp from his alma mater and has held onto one bottle of it. However, he expects a much different style to emerge from his family's 1 acre planting in The Rocks District of Milton-Freewater, a 10-minute drive from the winery.
"We're looking at producing a single-vineyard Rocks District Tempranillo that will be out next winter," Castillo said. "I was telling my dad that we need to capitalize on The Rocks District. It's Tempranillo. It's Oregon, and we can use The Rocks on our label. It really has a unique flavor because of the soil."
Abacela offers 4 examples of Tempranillo
Earl Jones, founding winemaker and co-owner of Abacela in Roseburg, Ore., offers four expressions of Tempranillo at his Umpqua Valley tasting room. (Photo by Eric Degerman/Great Northwest Wine)
The Castillo family also has a 2-acre parcel Tom Waliser manages for it near their stateline winery — "It's the first to break bud in the spring," Castillo said. It also produces a vineyard-designate Tempranillo from famed Rosebud Vineyard on the warm Wahluke Slope. All combined, Castillo de Feliciana makes about 2,000 cases of Tempranillo, accounting for 40 percent of its 5,000-case production.
"As Tempranillo producers, there are challenges in getting our wines in a sizable production on the market beyond our tasting room and wine club," Castillo said. "You have the Spanish imports at the lower prices."
Perhaps the most competitive domestic example is the Abacela 2013 Fiesta Tempranillo, a $23 bottling with a production of 1,675 cases. The 2012 vintage merited a spot on Great Northwest Wine's Top 100 list of 2015 at No. 63.
"Where he sells that wine is different from where we sell our wines," Castillo said. "We have our challenges because people want Tempranillo, but they want it as such a low price. That's how it is."
In 2017, expect the Oregon Tempranillo Celebration to move near Portland and include a consumer tasting. There seems to be little room to grow at conference center of the Ashland Hills Hotel & Suites, as the venue was maxed out at 75 people.
"I think that's the next logical step," Martin said. "Everybody loved the conference, and there's certainly a lot of enthusiasm about where the future might lead us.
"I think the big takeaway for us after talking to the winery people is that there's a great opportunity here for Oregon to be identified with Tempranillo — to quote, 'Own it' to some extent — because arguably we can produce Tempranillo here better than anywhere else in the United States."
Weather data proprietary in much of Europe
Les Martin of Red Lily Vineyards in Jacksonville, Earl Jones of Abacela in Roseburg, Scott Steingraber of Kriselle Cellars in White City and Eric Weisinger of Weisinger Family Winery in Ashland founded the Oregon Tempranillo Alliance in 2015. (Photo by Eric Degerman/Great Northwest Wine)
Greg Jones, a Southern Oregon University professor and arguably the international wine community's leading expert on climate research, pointed out stark ways in which weather data is handled throughout Europe vs. the United States.
"I often tell people it took me four years to do my dissertation, and three of those were all about getting the data," said Jones, who helped his father at Abacela. "Weather and climate data is a valued resource for Europe — and they sell it for large sums of money. The data you get immediately when you touch the button on the National Weather Service site or another site, that is free to you — we pay for it through taxes, of course. In Portugal, they have over 300 weather stations. They only make nine of them public. The rest of them are for sale."
MEMBERS: The Oregon Tempranillo Alliance's inaugural Celebration listed its members as Abacela Winery, Aurora Vines/Aguila Vineyard, Belle Fiore Winery, Castillo de Feliciana, Coventina Vineyards, Dana Campbell Vineyard, Delfino Vineyards, EdenVale Winery, Folin Cellars, Girardet Wine Cellars, Greg Jones, Grizzly Peak Winery, Holloran Vineyard Wines, Jaxon Vineyards, Kriselle Cellars, Ledger David Cellars, Naked Wines, Naumes Crush and Fermentation, Paradox Vineyard, Paul O'Brien Winery, Pebblestone Cellars, Plaisance Ranch, Platt Anderson Cellars, Purple Cow Vineyards, Quady North, Red Lily Vineyards, Reustle Prayer Rock Vineyards, RoxyAnn Winery, Schmidt Family Vineyards, Schultz Wines, Serra Vineyards, Silvan Ridge, South Stage Cellars, Southern Oregon Wine Scene, Stone River Vineyards, TeSóAria Vineyard & Winery, 2Hawk Vineyard and Winery, Upper Five Vineyard, Valley View Winery, Weisinger Family Winery and Zerba Cellars.
NOTES: Earl Jones pointed out the meaning behind the naming of Tempranillo in the Rioja. "Tempra means 'timely' or 'early', and nillo is a term of endearment," he said. "These guys in an area where they couldn't grow a lot of grapes were really pleased to have a grape that ripened within their time period." … The grape accounts for 20 percent of Spain's acreage, is grown in 80 percent of the country's wine growing regions, and is known by more than 60 synonyms. "And the longer the growing season, the poorer the quality of the wine," Jones said. "Very interesting." … There were multiple references at the conference to University of California-David professors Maynard Amerine and A. J. Winkler, who discouraged the planting of Tempranillo in California. "Lest you think they were picking on Tempranillo," Jones said. "They said the same thing about Pinot Noir, Cabernet Franc and several other grapes that needed a cooler climate." … The first commercial vintage of Tempranillo at Abacela was 1997, which generated 243 cases of wine from 4 acres of vines. "I didn't know how to make wine," Jones said. "This wine won gold medals, and I think the point was that the climate made the wine. We just harvested what we planted."
About Eric Degerman
Eric Degerman is the President and CEO of Great Northwest Wine. He is a journalist with more than 30 years of daily newspaper experience and has been writing about wine since 1998. He co-founded Wine Press Northwest magazine with Andy Perdue and served as its managing editor for a decade. He is a frequent wine judge at international wine competitions throughout North America and orchestrates 10 Northwest competitions each year.
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Q: PROGNAME(x) references in manpages -- What does the digit in ()s mean?
Possible Duplicate:
Why do programs in Unix-like environments have numbers after their name?

I have seen several programs, such as GREP(3) and PING(8), listed in manpages.  What is the significance of the digit in ()s?

A: The number indicates which section the manpage is in. For your examples:
grep(3)

To get the documentation, type
man 3 grep

More commonly, if there is no grep(2) or grep(1), you can get away with
man grep

However, I should note that grep is in section 1. Section 3 is generally reserved for C functions. An example is getopt: getopt(1) refers to the command-line utility getopt, but getopt(3) refers to the C function getopt. Likewise, read(1) is a program that reads from standard input, but read(2) is a POSIX system call for use in programs - it is one of the lowest-level forms of input you can get on most Linux (and other Unix) systems.

A: If you run man man you will see the following information in the man page:
1   Executable programs or shell commands
2   System calls (functions provided by the kernel)
3   Library calls (functions within program libraries)
4   Special files (usually found in /dev)
5   File formats and conventions eg /etc/passwd
6   Games
7   Miscellaneous (including macro  packages  and  conven‐
    tions), e.g. man(7), groff(7)
8   System administration commands (usually only for root)
9   Kernel routines [Non standard]

Some names are associated with multiple entries, eg on my system 'sleep' has an entry in section 1 and an entry in section 3.  You can specify the one you want with e.g.
man 3 sleep

Sometimes I just guess with
man -a sleep

which displays each entry associated with sleep in turn.  I just go through them until I find the one I want.  You can also try
man -k sleep

to get a slightly bigger listing of pages involving the term 'sleep'

A: It's to tell you what man page section help is in... 8 is typically the location of Administrative related utilities (/sbin, /usr/sbin, etc.)
So help for GREP(3) is in man page section 3, and you could type man 3 grep to get the help for grep(3) directly.

redpajama_set_name: RedPajamaStackExchange


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Published on Jamaica Gleaner (https://jamaica-gleaner.com)
Home > 'Facehunter Was Here' - Top fashion photographer films travel show in JA
Published:Thursday | October 5, 2017 | 12:00 AMKimberley Small [1]
"Cultural exploration around the globe" is currently the motto associated with Yvan Rodic's Facebook page.
Fashion and style aficionados would be more familiar with Rodic's brand and moniker Facehunter. Beginning in 2006, Rodic has been at the vanguard of street-style blogging, one of the decade's most remarkable trends. When he began in 2006, the practice was still in its infancy, but Rodic managed to transform his blog into a repository of haute couture, high-fashion imagery, to become one of the most sought-after image-takers in the fashion world.
This year, Rodic made his way to Jamaica for the first time to feature the island destination as part of his new project.
"I came to Jamaica to shoot an episode of my YouTube travel show called Facehunter Was Here," he told The Sunday Gleaner. Rodic came by invitation of local visual artist Taj Francis, who became acquainted with him at a design conference that took place three years ago in Canc?n, Mexico.
"We were both invited at a design conference and then we connected. And I've always wanted to go to Jamaica - and I told him maybe I'll go one day," Rodic told The Sunday Gleaner.
Sign up for The Gleaner's morning and evening newsletters. [2]
"We got along. It's always been something I had in mind because Jamaica has such a unique identity and I thought to come and explore the culture. I've always been fascinated by Jamaica and Jamaican culture. I have Jamaican friends in London, and I've seen videos of dancehall parties, and I think, 'wow! this is incredible. I want to experience that!'" Rodic said he always had superficial knowledge of what Jamaica could be, "and I always thought, 'I need to go and see myself'."
Rodic started blogging while living in Paris.
"I went to a lot of art openings and thought it would be interesting to document how people look like. Maybe people around the world want to see what's happening in Paris," he said. Now widely considered one of the most influential movers and shakers in fashion, Rodic's original blog and more recent Visual Diary receive approximately one million hits per month. From blogging success, travel invitations poured in, and Rodic found himself hitting the road to London, New York, Berlin, Stockholm, Geneva, Zurich, Copenhagen, Reykjavik, Oslo, Gdansk and countless other destinations.
Fashion journey
In early 2010, he was commissioned by Vogue Italy to cover Milan and Paris fashion weeks for the new website vogue.it. More recently stepping outside the fashion bubble and into the cinematic world, Rodic was asked to cover festivities and happenings at the world-famous Sundance Film Festival for its eponymous Sundance Film Channel's Full Frontal Fashion magazine.
In the first episode of Facehunter Was Here, posted last month, Rodic takes viewers around Albania. With the intent of posting one episode per month, Jamaica is early in the pipeline.
From this frequent flyer's observations, Jamaican style is one of a kind.
"On one hand, style is more and more connected between different places around the world, thanks to the Internet. But at the same time, you can still define some specificities that are unique to places, and I think Jamaica is one of a kind. I think it's so heavily musically influenced," Rodic deduced. "Music makes people look the way they look in Jamaica in many ways. But I think it's hard to define one style in Jamaica. It's quite eclectic, but you can see a whole section of the population going for bold and colourful, loud and experimental and daring. And then you see more laid-back, casual, easy-going."
This summer, Spice, Capleton, and Protoje were highlighted in international media, adding heft to the observations shared by the Facehunter.
In July, a few days after her performance at Reggae Sumfest 2017, Spice was showcased by the world's leading authority on all things fashion, Vogue. For the magazine, Deidre Dyer wrote: "While there was a veritable runway of festival style on the ground at Red Stripe's Reggae Sumfest last weekend, the most show-stopping look appeared when Spice hit centre stage for her late-night set. She quickly ditched her pink boudoir robe for a chain-encrusted gold bustier then kicked off her heels. It was a perfect example of the artiste's fearless, ever-changing wardrobe - one that is redefining dancehall style. Her electrifying stage shows are always kicked up a notch by elaborate bodysuits and a rotation of colourful wigs that recall an early Nicki Minaj. All it takes is a scroll through her Instagram feed and YouTube channel to see that the future of dancehall looks bright and bold."
In a street-style round-up of Reggae Sumfest, Vogue writers surmised that the weekend's style standouts included Beenie Man, who came dressed in head-to-toe Gucci, and Capleton, "who stepped out in a custom three-piece look that could have easily come down the runway at Comme des GarÁons." Reggae revivalist was also caught in this summer's style round-up by GQStyle at the annual Afropunk Music Festival.
entertainment@gleanerjm.com [3]
Source URL: https://jamaica-gleaner.com/article/entertainment/20171008/facehunter-was-here-top-fashion-photographer-films-travel-show-ja
[1] https://jamaica-gleaner.com/authors/kimberley-small
[2] http://member.jamaica-gleaner.com/subscription/
[3] mailto:entertainment@gleanerjm.com


» Basketball Plays » Free Basketball Plays, Tips & Drills » How do I teach young kids basketball plays while keeping all players interested?
How do I teach young kids basketball plays while keeping all players interested?
I have a team of 10 players, how do I teach them a basketball play without the other 5 getting bored? Should I have them doing something else while i work with the 1st group and rotate when they are done?
Very good question.. This is one of the challenges of teach youth basketball. Keeping the kids interested while coach is instructing. I've had the most success with splitting them up into two groups. The first group is learning the plays while the other is working on individual skills (drills) or defense. I usually have one of my other coaches take the group to the other basket and work on defensive as a team. We run the plays for a while then switch the teams to provide the same training to both. Depending upon the complexity of the play and the length of time needed to instruct, we may combine the players at the beginning by positions, so they can hear the instruction of the plays together, before splitting to practice defense/offense. Hope this helps. Good Luck.
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