text:
\section{Introduction}
The relationship between nonnegative polynomials and sums of squares has been studied since Hilbert's seminal paper of 1888. In it Hilbert showed that a nonnegative polynomial in $n$ variables of even degree $2d$ has to be a sum of squares of polynomials only in the following cases: the polynomial is univariate $n=1$, the polynomial is quadratic $2d=2$, or the polynomial is in 2 variables and has degree 4, $n=2$ and $2d=4$. In all other cases he proved existence of nonnegative polynomials that are not sums of squares. It is remarkable that Hilbert's proof was existential and the first explicit nonnegative polynomial that is not a sum of squares was found only 70 years later by Motzkin \cite{Rez1}, \cite{Rez2}.

Hilbert then showed that every nonnegative polynomial in 2 variables is a sum of squares of rational functions and Hilbert's 17th problem asked to show that this is true for any number of variables. This was shown in the 1920's by Artin and Schreier (see \cite{Prest}). However, there is no known algorithm to compute this representation and we may be forced to use denominators and numerators of very large degree compared to the original degree $2d$. While there are bounds on the degree of polynomials used to make rational functions, they are not encouraging from the point of view of efficiently computing such representations (see \cite{Prest}).

A nonnegative polynomial can be homogenized and it will remain nonnegative. Therefore for the remainder of this paper we will restrict our attention to the case of homogeneous polynomials (forms).

Let $P_{n,2d}$ be the vector space of forms in $n$ variables of degree $2d$. For a fixed number of variables $n$ and degree $2d$ nonnegative polynomials and sums of squares form closed convex cones in $P_{n,2d}$. We call these cones $Pos_{n,2d}$ and $Sq_{n,2d}$ respectively:

$$Pos_{n,2d}=\left\{p \in P_{n,2d}\hspace{2mm} | \hspace{2mm} p(x)\geq 0 \hspace{2mm} \text{for all} \hspace{2mm} x \in \mathbb{R}^{n}\right\},$$

$$Sq_{n,2d}=\left\{p \in P_{n,2d} \hspace{2mm} | \hspace{2mm} p(x)=\sum q_i^2  \hspace{2mm} \text{for some} \hspace{2mm} q_i \in P_{n,d} \right\}.$$

These cones are very interesting convex objects but their structure and precise relationship with each other is not very well understood except for the cases of $n=2$, the univariate case for nonhomogeneous polynomials, and the case $2d=2$ (see \cite{Sasha1} Sections II.11 and II.12 for these cases). We studied some convexity properties such as the coefficient of symmetry and maximal volume ellipsoids of these cones in \cite{Me1}.

In \cite{Me2} we have shown that if the degree $2d$ is fixed and the number of variables $n$ grows then asymptotically there are significantly more nonnegative polynomials than sums of squares if the degree $2d$ is at least 4. However the relationship between the cones for small values of $n$ and $d$ is not clear.

The precise relationship between the cones $Pos_{n,2d}$ and $Sq_{n,2d}$ is interesting because of issues of computational complexity and practical testing for nonnegativity. It is known that testing whether a polynomial is nonnegative is NP-hard already when the degree is 4 \cite{NP}. On the other hand testing whether a polynomial is a sum of squares can be reduced to a semidefinite programming problem and it is practically quite fast \cite{pablo}.

In this paper we study the faces of the cone of nonnegative polynomials and the cone of sums of squares. In particular we are interested in the dimensions of the faces. It seems that outside of the simple cases of $n=2$ and $2d=2$ very little is known about the possible dimensions of the faces of these cones.
\subsection{Faces of $Pos_{n,2d}$ and $Sq_{n,2d}$.}

It is easy to describe the faces of $Pos_{n,2d}$. The boundary of the cone $Pos_{n,2d}$ consists of all the forms with at least one zero while the interior consists of strictly positive forms. A facet of $Pos_{n,2d}$ consists of all the forms with a prescribed zero. If we let $S$ be a set of points in $\mathbb{R}^{n}$ (we should think of $S$ projectively as points in $\mathbb{RP}^{n-1}$) then the forms vanishing on all points of $S$ form a face of $Pos_{n,2d}$ which we call $Pos_{n,2d}(S)$:
$$Pos_{n,2d}(S)=\{p \in Pos_{n,2d} \hspace{2mm} \mid \hspace{2mm} p(s)=0 \hspace{2mm} \text{for all} \hspace{2mm} s \in S\}.$$

Moreover any face of $Pos_{n,2d}$ has a description of this form and the set $S$ can be chosen to be finite. We note that despite this simple description the facial structure of $Pos_{n,2d}$ "should" be very difficult to fully describe because the problem of testing for nonnegativity is known to be NP-hard.

The faces of the cone of sums of squares are much harder to describe. We will only look at the faces of $Sq_{n,2d}$ that have description analogous to the faces of $Pos_{n,2d}$. For a (projective) set of points $S$ in $\mathbb{R}^{n}$ we let $Sq_{n,2d}(S)$ be the face of $Sq_{n,2d}$  of forms that vanish on all points of $S$:

$$Sq_{n,2d}(S)=\{p \in Sq_{n,2d} \hspace{2mm} \mid \hspace{2mm} p(s)=0 \hspace{2mm} \text{for all} \hspace{2mm} s \in S\}.$$

We will study the dimensions of the faces $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$ and we will establish large dimensional gaps between the faces for the same set $S$. It is possible to view Hilbert's original proof of existence of nonnegative polynomials that are not sums of squares as establishing a dimensional gap of this type. This dimensional point of view was first made explicit in \cite{Rez2}.

\subsection{Bounds on the Dimensions of Faces.}

For a set of points $S \in \mathbb{R}^{n}$ let $I_{1,d}(S)$ be the vector subspace of $P_{n,d}$ of forms that vanish to order at least 1 on $S$:
$$I_{1,d}(S)=\{p \in P_{n,d} \hspace{2mm} \mid \hspace{2mm} p(s)=0 \hspace{2mm} \text{for all} \hspace{2mm} s \in S\}.$$

If a nonnegative form $p$ vanishes at a point $s$ then $p$ attains its minimum at $s$ and therefore $p$ must be singular at $s$ or in other words $p$ must vanish to order 2 on $s$. We let $I_{2,2d}(S)$ be the vector subspace of $P_{n,2d}$ of forms that vanish to order at least 2 on every point of $S$:
$$I_{2,2d}(S)=\left\{p \in P_{n,2d} \hspace{2mm} \mid \hspace{2mm} \frac{\partial p}{\partial x_i}(s)=0 \hspace{2mm} \text{for all} \hspace{2mm} s\in S \hspace{2mm} \text{and all} \hspace{2mm} i=1\ldots n \right\}.$$

Since every nonnegative form that is zero on $s$ must vanish to order 2 on $s$ it follows that the face $Pos_{n,2d}(S)$ is contained in $I_{2,2d}(S)$. Now $Pos_{n,2d}(S)$ is a face of $Pos_{n,2d}$ and therefore it is a convex cone. We are interested in finding the dimension of $Pos_{n,2d}(S)$ and in particular we will be interested in showing that in many cases $Pos_{n,2d}(S)$ is full-dimensional in $I_{2,2d}(S)$. We will also provide some examples where the full-dimensionality does not hold in Sections \ref{special} and \ref{fewer}.

The dimension of $I_{2,2d}(S)$ has been extensively studied previously. Vanishing to order 2 at a point imposes $n$ conditions on the forms in $P_{n,2d}$. Therefore one would expect that generically $I_{2,2d}(S)$ has codimension $n|S|$ in $P_{n,2d}$. Indeed the Alexander-Hirschowitz Theorem states that generically this is the case, except for a small number of exceptions \cite{Miranda}. By showing full-dimensionality of $Pos_{n,2d}(S)$ in $I_{2,2d}(S)$ we get a handle on the dimension of the face $Pos_{n,2d}(S)$.

We will call a finite set $S \in \mathbb{R}^{n}$ $d$-independent if $S$ satisfies the following two conditions:
\begin{align}
&\label{first cond}\text{The forms in $I_{1,d}(S)$ share no common zeroes outside of $S$. In other words the}\\
&\nonumber \text{conditions of vanishing on $S$ force no additional zeroes on forms of degree $d$.}\\
&\label{second cond}\text{For any $s \in S$ the forms that vanish to order 2 on $s$ and vanish on the rest of $S$}\\
&\nonumber \text{form a vector space of codimension $|S|+n-1$ in $P_{n,d}$.}&
\end{align}

\noindent The second condition simply states that the constraints of vanishing on $S$ and additionally double vanishing at any point $s \in S$ are all linearly independent.

In Section \ref{dimsection} we will show that if a set $S$ is $d$-independent then the face $Pos_{n,2d}(S)$ is full dimensional in $I_{2,2d}(S)$. We will also show that the set of configurations that are $d$-independent is open and therefore in order to show that $d$-independence is a generic condition for sets of fixed size $k$ we simply need to provide a single example.

In Section \ref{dset} we construct a $d$-independent set of size $\binom{n+d-1}{d}-n$. This shows that $d$-independence is a generic condition for finite sets of size at most $\binom{n+d-1}{d}-n$ in $\mathbb{R}^{n}$.

In Section \ref{gapsection} we will establish large gaps between faces $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$. For sums of squares we will use a different dimensional approach. Let $I^{[2]}_{2d}(S)$ be the vector subspace of $P_{n,2d}$ that is spanned by squares of forms from $I_{1,d}(S)$:
$$I_{2d}^{[2]}(S)=\left\{p \in P_{n,2d} \hspace{2mm} \mid \hspace{2mm} p=\sum \alpha_iq_i^2 \hspace{2mm} \text{for some} \hspace{2mm} q_i \in I_{1,d}(S) \hspace{2mm} \text{and}\hspace{2mm} \alpha_i \in \mathbb{R} \right\}.$$

If a form $p$ is a sum of squares, $p=\sum q_i^2$ and $p$ vanishes on $S$ then each $q_i$ must vanish on $S$ and therefore $p$ is a sum of squares of forms from $I_{1,d}(S)$. Thus it follows that $Sq_{n,2d}(S)$ is contained in $I_{2d}^{[2]}(S)$. It is easy to see that $Sq_{n,2d}(S)$ is actually full dimensional in $I_{2d}^{[2]}(S)$ since we can pick a basis of $I_{2d}^{[2]}(S)$ consisting of squares and nonnegative linear combinations of these squares will lie in $Sq_{n,2d}(S)$.

Therefore we can restrict our attention to the dimension of $I_{2d}^{[2]}(S)$. To bound this dimension we observe that the squares from $I_{1,d}(S)$ can span a vector space of dimension at most $\binom{\dim I_{1,d}(S)+1}{2}$. When the set $S$ is fairly small this bound is excessive and it would be interesting to improve on it. However when the set $S$ is large the bound is often optimal.

Using the above bounds we will show that there exist dimensional differences between faces $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$ of size
$$Gap_{n,2d}=\binom{n+2d-1}{2d}-n\binom{n+d-1}{d}+\binom{n}{2}.$$
We note that the numbers $Gap_{n,2d}$ are zero in all the cases where the cones $Pos_{n,2d}$ and $Sq_{n,2d}$ are equal. However $Gap_{n.2d}$ are strictly positive in the cases where there exist nonnegative forms that are not sums of squares. In the smallest cases $n=4$, $2d=4$ and $n=3$, $2d=6$ where $Pos_{n,2d}$ is strictly bigger than $Sq_{n,2d}$ the gap number is 1.

However, as either $n$ or $d$ grow we can see that the dimensional gap $Gap_{n,2d}$ between faces of $Pos_{n,2d}$ and $Sq_{n,2d}$ grows and asymptotically it approaches the full dimension of the vector space $P_{n,2d}$.

In Section \ref{sixpoints} we will completely describe the situation when $S$ is a set of 6 points in $\mathbb{R}^4$. There will be generically a gap of 1 dimension between $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$. We will explicitly describe an extra linear constraint satisfied by sums of squares, that is not satisfied by nonnegative forms and we will also provide explicit examples of forms that double vanish on $S$, but are not spanned by squares. These forms can be used together with Lemma \ref{ext} to construct explicit nonnegative forms that are not sums of squares.

In Section \ref{sectiongenineq} we will use these dimensional differences to derive quadratic inequalities that separate nonnegative forms from sums of squares. These inequalities will hold on the cone of sums of squares $Sq_{n,2d}$ but they will fail for some nonnegative forms. In Section \ref{sectionexplicitineq} we will give an explicit example of such inequality for forms of degree 4 in 4 variables.

We begin by giving some explicit examples of dimensional gaps between nonnegative forms and sums of squares and also examples of dimensional differences between nonnegative and double vanishing forms. Some details of dimension counts will be omitted, but the proofs will be given in more generality in later sections.

\section{Examples}

\subsection{First Gap: Six Points in $\mathbb{R}^4$.}\label{sectionfirstgap}\hspace{1mm}\\
\indent Let $S$ be the set of 6 points in $\mathbb{R}^4$ that have 1 in two coordinates and 0 in the two other coordinates. There are $\binom{4}{2}=6$ such points. We label the points $s_{ij}$ by the two coordinates in which 1 appears.

Let $I_{1,2}(S)$ be the vector space of all forms of degree 2 that vanish on $S$ and let $I_{2,4}(S)$ be the vector space of forms of degree 4 that double vanish on $S$. Also let $I_4^{[2]}(S)$ be the vector space of forms of degree 4 spanned by squares from $I_{1,2}(S)$. This example will be generalized in Section \ref{dset} where our dimension counts will be rigorously proved in more generality.

It is not hard to show that the set $S$ is $2$-independent. In particular, requiring 6 zeroes on the points of $S$ imposes 6 linearly independent conditions on quadratic forms and therefore the dimension of $I_{1,2}(S)$ is $\binom{5}{2}-6=4$. There is a particularly nice basis for $I_{1,2}(S)$ in which every form factors:
\begin{eqnarray*}
Q_1=x_1(x_1-x_2-x_3-x_4), \hspace{4mm}& Q_2=x_2(x_2-x_1-x_3-x_4)\\
Q_3=x_3(x_3-x_1-x_2-x_4), \hspace{4mm}& Q_4=x_4(x_4-x_1-x_2-x_3).
\end{eqnarray*}

Using this basis for $I_{1,2}(S)$ it is easy to check that the dimension of $I_4^{[2]}$ is $\binom{5}{2}=10$, which happens because all pairwise products of $Q_i$'s are linearly independent in $P_{4,4}$. On the other hand the Alexander-Hirschowitz theorem leads us to expect that the dimension of vector space $I_{2,4}(S)$ of double vanishing forms on $S$ is $\binom{7}{4}-4 \cdot 6=11$. It is not hard to verify that this dimension count is correct for our set $S$.

By Corollary \ref{dindep} $2$-independence of $S$ implies that the face $Pos_{4,4}(S)$ is full dimensional in the vector space $I_{2,4}(S)$. Therefore we get a face $Pos_{4,4}(S)$ which has dimension 11, while the corresponding face $Sq_{4,4}(S)$ has dimension 10.

We briefly explain why the face $Pos_{4,4}(S)$ is full dimensional in $I_{2,4}(S)$. Consider the polynomial $$Q=Q_1^2+Q_2^2+Q_3^2+Q_4^2.$$

It is not hard to show that $Q$ has no zeroes outside of $S$ and furthermore at every $s_{ij}\in S$ the Hessian of $Q$ is positive definite on the vector subspace $s_{ij}^{\perp}$ consisting of vectors perpendicular to $s_{ij}$. This suffices to show that $Q$ can be perturbed in any direction by a double vanishing polynomial in $I_{2,4}(S)$ and it will still remain nonnegative (See Lemma \ref{ext}). It follows that the face $Pos_{4,4}(S)$ of the cone of nonnegative polynomials vanishing on $S$ is full dimensional in $I_{2,4}(S)$ and thus has dimension 11, while the face of the cone of sums of squares $Sq_{4,4}(S)$ has dimension 10.

Since there is a gap of one dimension between the faces there exist double vanishing forms in $I_{2,4}(S)$ that are not spanned by sums of squares. Also, it follows that sums of squares must satisfy an extra linear condition that is not satisfied by the double vanishing forms. It is not hard to check that $x_1x_2x_3x_4$ is indeed a doubly vanishing form on $S$, and using our special basis it is easy to show that $x_1x_2x_3x_4$ is not in $I_4^{[2]}(S)$.

The extra linear constraint satisfied by any form $p$ in $I_4^{[2]}(S)$ can be given by:
$$16p(1,0,0,0)=p\left(1,-1,-1,-1\right).$$
On the other hand this constraint is not satisfied by $x_1x_2x_3x_4$.

We show in Section \ref{sixpoints} that the situation in the case of 6 general points in $\mathbb{R}^4$ is very similar. There is a gap of 1 dimension and we also provide explicitly the extra linear condition satisfied by sums of squares and a double vanishing form that is not in the span of sums of squares.

\subsection{A Special Configuration.}\label{special}\hspace{1mm}\\
\indent We now add the point $(1,1,1,1)$ to the six point set $S$ from above to form a 7 point set $S'$ in $\mathbb{R}^4$. We claim that the nonnegative polynomials vanishing on $S'$ do not form a full dimensional convex set in the vector space $I_{2,4}(S')$ of forms of degree 4 double vanishing on $S'$.

The dimension of $\dim I_{2,4}(S')$ has to be at least 7, since we impose at most $7\cdot 4=28$ constraints on a 35 dimensional vector space $P_{4,4}$. It is not hard to show that $\dim I_{2,4}(S')=7$, which is the expected dimension by the Alexander-Hirchowitz Theorem. On the other hand we will see that $\dim Pos_{4,4}(S')=\dim Sq_{4,4}(S')=6$ so both nonnegative polynomials and sums of squares form convex cones of dimension 6 in the 7 dimensional vector space of double vanishing forms. This is a special situation among configurations of 7 points in $\mathbb{R}^4$.

Using the forms $Q_i$ defined above as the basis of $S$ it is clear that $R_1=Q_1-Q_4$, $R_2=Q_2-Q_4$ and $R_3=Q_3-Q_4$ form a basis of $I_{1,2}(S')$, since $Q_i(1,1,1,1)=-2$ for all $i$. After simplification we see that
\begin{align*}
R_1=(x_1-x_4)(x_1&+x_4-x_2-x_3), \hspace{3mm} R_2=(x_2-x_4)(x_2+x_4-x_1-x_3), \\
&R_3=(x_3-x_4)(x_3+x_4-x_2-x_3).
\end{align*}

The set $S'$ is not 2-independent. It is easy to show that the forms $R_i$ have no common zero outside of $S'$. Therefore $S'$ forces no additional zeroes for forms of degree 2 and the first condition \eqref{first cond} for 2-independence is satisfied.

Now lets look at the second condition \eqref{second cond}. Since $|S'|=7$ we would need for any $s \in S'$ the vector space of forms vanishing on $S'$ and double vanishing on $s$ to have codimension 10 in $P_{4,2}$. The dimension of $P_{4,2}$ is 10 and therefore for any $s \in S'$ we would need to have no nonzero forms that are singular on $s$ and vanishing on the rest of $S'$. However the form
$$(x_1-x_2)^2-(x_3-x_4)^2$$
is actually singular at $(1,1,0,0)$ and it vanishes on the rest of $S'$.

It is easy to show that all pairwise products of forms $R_i$ are linearly independent and therefore the dimension of the vector space $I_4^{[2]}(S')$ spanned by squares from $I_{1,2}(S)$  is 6. This implies that the dimension of the face $Sq_{4,4}(S')$ is also 6. We now show that the dimension of the face $Pos_{4,4}(S')$ is 6 as well.

We observe that the set $S'$ is fixed by the action of symmetric group $\mathbb{S}_4$ which acts by permuting the coordinates. Therefore there is a natural action of $\mathbb{S}_4$ on the double vanishing vector space $I_{2,4}(S')$. It is not hard to see that there is a 2 dimensional subspace of $I_{2,4}(S')$ spanned by symmetric forms, i.e. the forms fixed by $\mathbb{S}_4$. A basis of this subspace is given by the forms $F_1$ and $F_2$:

\begin{equation*}
F_1 =\sum_{i <j} (Q_i-Q_j)^2,
\end{equation*}

\begin{equation*}
F_2 =(Q_1+Q_2+Q_3+Q_4)^2-64x_1x_2x_3x_4.
\end{equation*}

The rest of $I_{2,4}(S')$ is made of two irreducible representations of $\mathbb{S}_4$ corresponding to partitions $(2,2)$ and $(3,1)$. This happens since $I_{1,2}(S')$ corresponds to the standard representation of $\mathbb{S}_4$; this is easy to see from the basis of $R_i$. Now $I^{[2]}_4$ is the symmetric square of $I_{1,2}(S')$ and it is known to split into 3 irreducibles: the trivial representation, spanned by $F_1$, and representation corresponding to partitions $(2,2)$ and $(3,1)$. Since $I^{[2]}_4(S')$ has dimension 6 and $F_2$ is not in $I^{[2]}_4(S')$ it follows that $F_2$ and $I^{[2]}_4(S')$ together span $I_{2,4}(S')$.

Thus $I_{2,4}(S')$ splits into two copies of the trivial representation (spanned by $F_1$ and $F_2$) and two more representations of $\mathbb{S}_4$ corresponding to partitions $(2,2)$ and $(3,1)$. For more details on representations of the symmetric group we refer to \cite{Fulton}.

Now lets pick a basis of $I_{2,4}(S')$ consisting of $F_1$, $F_2$ and 5 forms $G_i$ forming a basis of the other two irreducible sub-representations. Let's take a form $p$ in $Pos_{4,4}(S')$. We can write $p=\alpha_1F_1+\alpha_2F_2+\sum \beta_iG_i$. We claim that the coefficient $\alpha_2$ of $F_2$ must be zero, and therefore $Pos_{4,4}(S')$ is not full dimensional in $I_{2,4}(S')$.

Suppose not. Symmetrize $p$ with respect to the action of $\mathbb{S}_4$ to obtain a new polynomial $\bar{p}$:

$$\bar{p}=\frac{1}{24}\sum_{g \in \mathbb{S}_4} p(gx).$$

Since $p$ is nonnegative, it follows that $\bar{p}$ is an average of nonnegative forms and therefore also nonnegative. By elementary representation theory, only the trivial representation components of $p$ survive the averaging process, so

$$\bar{p}=\alpha_1F_1+\alpha_2F_2.$$

Since $F_1$ is a sum of squares it is clearly nonnegative and therefore $F_1$ is in $Pos_{4,4}(S')$. It is easy to show that for $\bar{p}=\alpha_1F_1+\alpha_2F_2$ to be nonnegative the coefficient $\alpha_1$ of $F_1$ must be positive. But then, since both $F_1$ and $\alpha_1F_1+\alpha_2F_2$ are in $Pos_{4,4}(S')$, we see that $F_1 +\epsilon F_2$ must be in $Pos_{4,4}(S')$ for any small enough $\epsilon$ because $Pos_{4,4}(S')$ is a convex cone.

Now lets restrict $F_1$ and $F_2$ to the line $(x,1,1,1)$. We can see that $F_1$ on this line is equal to $3(x-1)^4$ while $F_2$ on this line is equal to $(x-1)^3(x-9)$. Since $F_1$ has a zero of degree 4 at $(1,1,1,1)$ along this line and $F_2$ has zero of degree 3 it follows that $F_1+\epsilon F_2$ cannot be nonnegative for sufficiently small $\epsilon$.

\subsection{No Nonnegative Polynomials among Double Vanishing Forms.} \label{fewer}\hspace{1mm}\\
\indent As we have seen above it does not have to happen that the face $Pos_{n,2d}(S)$ is full dimensional in the vector space of double vanishing forms $I_{2,2d}(S)$. We now provide an example where there are no nonnegative forms with a specified zero set, while dimension count shows that many double vanishing forms with this zero set exist.

In order to exclude nonnegative forms we need to require at least $\binom{n+d-1}{d}$ zeroes, otherwise there will exist squares with the specified zeroes. We restrict ourselves to the case of $2d=4$.

We show below that if take a set $S$ of cardinality $\binom{n+1}{2}$ then generically there will exist no nonnegative polynomials that vanish on $S$ (by generic over the real numbers we mean that it holds on an open set of configurations). This suggests an explanation for one aspect of why it is hard to construct nonnegative polynomials that are not sums of squares: if we require enough zeroes to exclude all squares then generically we will not have any nonnegative polynomials left either.

Let $S_{n,2}$ be the set of all vectors in $\mathbb{R}^{n}$ that are partitions of $2$: $$S_{n,2}=\left\{s=(s_1,\dots,s_n)\in \mathbb{R}^{n} \hspace{2mm} \mid \hspace{2mm} s_i \geq 0, s_i \in \mathbb{Z},\hspace{1mm} \text{and}\hspace{2mm} \sum s_i=2\right\}.$$

$S_{n,2}$ consists of all vectors $s_{ij}$ with 1 in coordinates $i$ and $j$ and 0 in all other coordinates together with vectors $2e_i$, where $e_i$ are the standard basis vectors.

We claim that the vector space $I_{2,4}(S_{n,2})$ of double vanishing forms on $S_{n,2}$ of degree 4 is spanned by forms $x_ix_jx_kx_l$, where $i,j,k,l$ are distinct indices. In particular we need $n \geq 4$ for the vector space to be non-empty.

Let $p$ be a nonzero form in $I_{2,4}(S_{n,2})$. Since $p$ double vanishes at each standard basis vector $e_i$ it follows that $p$ cannot contain monomials $x_i^4$ or $x_i^3x_j$ for any indices $i,j$. Of the remaining monomials of degree 4 only $x_i^2x_j^2$ does not vanish on the point $s_{ij}$. Since $p$ does vanish on $s_{ij}$ it follows that $p$ does not contain any of these monomials either.

The only monomials left that we are allowed to use are of the form $x_i^2x_jx_k$ and $x_ix_jx_kx_l$. We can exclude $x_i^2x_jx_k$ as follows: let $\alpha$ be the coefficient of $x_i^2x_jx_k$, let $\beta$ be the coefficient of $x_ix_j^2x_k$ and let $\gamma$ be the coefficient of $x_ix_jx_k^2$. The monomials $x_i^2x_jx_k$ and $x_ix_j^2x_k$ are the only monomials left whose $k$-th partial derivative does not vanish on $s_{ij}$. Therefore we see that $\alpha=-\beta$, since $p$ double vanishes on $s_{ij}$. Similarly, $\alpha=-\gamma$ and $\beta=-\gamma$. This can only happen if $\alpha=\beta=\gamma=0$. Thus it follows that $p$ can only contain monomials of the form $x_ix_jx_kx_l$. It is easy to see that each of the monomials $x_ix_jx_kx_l$ does indeed double vanish on $S_{n,2}$, which proves our claim.

It is clear that monomials $x_ix_jx_kx_l$ do not span any nonnegative forms so $I_{2,4}(S_{n,2})$ does not contain any nonnegative forms and it is nonempty for $n \geq 4$. To prove that this situation is generic we observe that for a set $S$ the condition that $I_{2,4}(S)$ does not contain any nonnegative forms is clearly "open", i.e. we can perturb the points in $S$ by a small enough $\epsilon$ to obtain a new set $S'$ and $I_{2,4}(S')$ will not contain any nonnegative polynomials either.

We need to be a little bit careful however. The points $S_{n,2}$ are in fact in very special position from the point of view of Alexander-Hirschowitz theorem. The vector space $I_{2,4}(S_{n,2})$ has a larger dimension than is expected generically. Therefore if we perturb the points to a generic configuration we may end up with an empty vector space of double vanishing forms. Indeed, it is easy to check that we need $n \geq 7$ before the generic dimension $\binom{n+3}{4}-n\binom{n+1}{2}$ becomes positive. However for large $n$ this dimension will asymptotically approach the dimension of the whole space of forms of degree 4, which is $\binom{n+3}{4}$.

\section{Dimension of faces of $Pos_{n,2d}$}\label{dimsection}

Let $S$ be a finite set in $\mathbb{R}^{n}$. We will find the dimension of $Pos_{n,2d}(S)$ by establishing that in many cases $Pos_{n,2d}(S)$ is actually full dimensional in the vector space $I_{2,2d}(S)$ of double vanishing forms on $S$ of degree $2d$. The dimension of $I_{2,2d}(S)$ has been well studied and in the case of a generic set $S$ the dimension of $I_{2,2d}(S)$ is provided by Alexander-Hirschowitz theorem \cite{Miranda}.

Generically one expects that every double zero contributes $n$ new linear conditions and Alexander-Hirschowitz theorem states that this is indeed the case, with a small list of exceptions. Therefore generically be expect that:

$$\dim I_{2,2}d(S)=\dim P_{n,2d}-n|S|.$$

\noindent However, for any set $S$ we know that

$$\dim I_{2,2d}(S)\geq \dim P_{n,2d}-n|S|,$$

\noindent since we impose at most $n|S|$ linearly independent conditions.

We will establish full dimensionality of $Pos_{n,2d}(S)$ by finding a form $p \in Pos_{n,2d}(S)$ to which we can add a suitably small multiple of any double vanishing form and it will remain nonnegative:

$$p+\epsilon q \in Pos_{n,2d}(S) \hspace{2mm} \text{for some sufficiently small} \hspace{2mm} \epsilon \hspace{2mm} \text{and any} \hspace{2mm} q \in I_{2,2d}(S).$$

The form $p$ can be viewed as a certificate of full dimensionality of $Pos_{n,2d}(S)$. The important point is that $p$ can be any form, in particular we will focus on finding such $p$ that is a sum of squares. This approach follows that of \cite{Rez2} and indeed it can be traced to the original proof of Hilbert.

For a form $p$ let the Hessian $H_p$ of $p$ be the matrix of second derivatives of $p$:

$$H_p=(h_{ij}), \hspace{5mm} \text{where} \hspace{5mm} h_{ij}=\frac{\partial^2 p}{\partial x_i \partial x_j}.$$

We will need an "extension lemma" which follows from Lemma 3.1 of \cite{Rez2}. We note that if a form $p$ vanishes at a point $s$ then by homogeneity $p$ vanishes on a line through $s$. Therefore the vector $s$ is in the kernel of the Hessian of $p$ at $s$: $H_p(s)s=0$.

If a form $p$ is nonnegative then its Hessian at any zero $s$ is positive semidefinite, since 0 is a minimum for $p$. We call a nonnegative form $p$ \textit{round} at a zero $s$ if the Hessian of $p$ at $s$ is positive definite on the subspace $s^{\perp}$ of vectors perpendicular to $s$:
$$p \hspace{2mm} \text{is round at a zero} \hspace{2mm} s \hspace{2mm} \text{if} \hspace{2mm} H_p(s) \hspace{2mm} \text{is positive definite on} \hspace{2mm} s^{\perp}. $$

\noindent For a form $p$ we will let $Z(p)$ denote the zero set of $p$ in $\mathbb{RP}^{n-1}$.

\begin{lemma}\label{ext}
Let $p$ be a nonnegative form with a finite zero set $Z(p)$, and let $q$ be a form such that $q$ vanishes to order 2 on every point in the zero set of $p$. Furthermore suppose that $p$ is round at every point in $Z(p)$.
Then for a sufficiently small $\epsilon$, the form $p+\epsilon q$ is nonnegative.
\end{lemma}

Now we have an immediate corollary:

\begin{cor} \label{ext2}
Let $S$ be a finite set in $\mathbb{R}^{n}$. Suppose that we can find a nonnegative form $p$ in $Pos_{n,2d}(S)$ such that $S$ is the zero set of $p$ projectively: $Z(p)=S$ and $p$ is round at every point $s \in S$. Then the face $Pos_{n,2d}(S)$ is full dimensional is the vector space $I_{2,2d}(S)$ of double vanishing forms on $S$.
\end{cor}
\begin{proof}
Let $p \in Pos_{n,2d}(S)$ be as above. Then by Lemma \ref{ext} for any $q \in I_{2,2d}(S)$ we have $p+\epsilon q \in Pos_{n,2d}(S)$ for sufficiently small $\epsilon$. Since $Pos_{n,2d}(S)$ is a convex set it follows that is is full dimensional in $I_{2,2d}(S)$.
\end{proof}

\subsection{Sum of Squares Certificate.}Now suppose that for a finite set $S \in \mathbb{R}^{n}$ we can find a sum of squares form $p=q_1^2+\ldots+q_n^2$ such that $Z(p)=S$ and $p$ is round at every $s \in S$. Since $p$ vanishes on $S$ it implies that all $q_i$ vanish on $S$ and the set $S$ has no forced zeroes for forms of degree $d$. Or in other words, $S$ is cut out by the intersection of hypersurfaces $q_i=0$.

The Hessian of $p$ is the sum of the Hessians of $q_i^2$:$$H_p=\sum_i H_{q^2_i}.$$ Since $q_k(s)=0$ for all $k$ and $s \in S$ it follows that

$$\frac{\partial^2 q_k^2}{\partial x_i \partial x_j}(s)=2\frac{\partial q_k}{\partial x_i}(s)\frac{\partial q_k}{\partial x_j}(s).$$

Therefore we see that the Hessian of $q^2_k$ at any $s \in S$ is actually double the tensor of the gradient of $q_k$ at $s$ with itself: $$H_{q^2_k}(s)=2\nabla q_k\otimes \nabla q_k(s).$$ Since the Hessian of $p$ at $s$ is positive definite on $s^{\perp}$, and it is the sum of the gradients of $q_k$ it follows that the gradients of $q_k$ at $s$ span $s^{\perp}$.

The gradients of $q_k$ at any $s \in S$ cannot span more than $s^{\perp}$ since $\ip{\nabla  q_k}{s}=d\cdot q_k(s)=0$. The gradients of $\nabla q_k$ span a vector space of dimension $n-1$ and thus the forms that double vanish at one $s \in S$ and vanish at all other points in $S$ form a vector space of codimension $n-1$ in $I_{1,d}(S)$.

These two conditions are sufficient for $Pos_{n,2d}(S)$ to be full dimensional in $I_{2,2d}(S)$.

\begin{theorem}\label{gapzz}
Let $S$ be a finite set in $\mathbb{R}^{n}$ such that $S$ forces no additional zeroes for forms of degree $d$ and for any $s \in S$ the forms in $I_{1,d}(S)$ that double vanish at $s$ form a vector space of codimension $n-1$ in $I_{1,d}(S)$. Then $Pos_{n,2d}(S)$ is a full dimensional convex cone in $I_{2,2d}(S)$.
\end{theorem}

\begin{proof}
Let $q_1, \ldots, q_k$ be a basis of $I_{1,d}(S)$. We claim that $p=\sum q_i^2$ has the properties of Lemma \ref{ext} and therefore by Corollary \ref{ext2} convex cone $Pos_{n,2d}(S)$ is full dimensional in $I_{2,2d}(S)$.

Since $S$ has no forced zeroes and we picked a basis of $I_{1,d}(S)$ it follows that $q_k$ have no common zeroes outside of $S$ and thus $Z(p)=S$ projectively.

Now choose $s \in S$. Since the forms in $I_{1,d}(S)$ that double vanish at $s$ form a vector space of codimension $n-1$ in $I_{1,d}(S)$ it follows that the gradients of $q_k$ at $s$ must span a vector space of dimension $n-1$. Since $\ip{\nabla q_k}{s}=0$ for all $k$ it follows that they actually span $s^{\perp}$. By the argument about Hessians above it follows that the Hessian of $p$ is positive definite on $s^{\perp}$.
\end{proof}

To avoid working with degenerate configuration instead of requiring that for any $s \in S$ the forms in $I_{1,d}(S)$ that double vanish at $s$ form a vector space of codimension $n-1$ in $I_{1,d}(S)$ we work with configurations where the forms in $I_{1,d}(S)$ that double vanish at $s$ form a vector space of codimension $|S|-n+1$ in $P_{n,d}$. In other words we also require that the conditions of vanishing at all the points of $S$ are linearly independent. This is indeed our definition $d$-independence from before:

\begin{defin}\label{defindindep}
We call a finite set $S$ in $\mathbb{R}^{n}$ \text{d-independent} if $S$ satisfies two following properties:\\
The set $S$ forces no additional zeroes for forms of degree $d$ that vanish on $S$,\\
For any $s \in S$ the forms that vanish to order 2 on $s$ and vanish on the rest of $S$
form a vector space of codimension $|S|+n-1$ in $P_{n,d}$.
\end{defin}

From Theorem \ref{gapzz} we obtain the following immediate corollary:

\begin{cor}\label{dindep}
Suppose that a set $S$ in $\mathbb{R}^{n}$ is $d$-independent. Then the face $Pos_{n,2d}(S)$ is full dimensional in $I_{2,2d}(S)$.
\end{cor}

Using standard methods (vanishing determinants) it is easy to write the set of all configuration of $k$ points in $\mathbb{RP}^{n-1}$ that are $d$-independent as a complement of a closed algebraic set. Therefore we obtain the following proposition:

\begin{prop}\label{open}
The set of configuration of $k$ points in $\mathbb{RP}^{n-1}$ that are $d$-independent is open.
\end{prop}

In the following section we will actually construct an example of a $d$-independent set of cardinality $\binom{n+d-1}{d}-n$. This will show that $d$-independence is a condition that holds on an open set for all $k \leq \binom{n+d-1}{d}-n$, i.e. it is generic. However, we strongly suspect that $d$-independence should be generic in a stronger sense: it should hold on an open set whose closure is all of $\mathbb{RP}^{n-1}$, or in other words "almost any" configuration of $k\leq \binom{n+d-1}{d}-n$ points is $d$-independent.

Indeed we show in Section \ref{sixpoints} that any set of 6 points in $\mathbb{RP}^3$  in general linear position is 2-independent and in Lemma 2.6 of \cite{Rez2} it was shown that any set of 7 points in $\mathbb{RP}^2$ with no 4 on a line and not all on a quadratic is $3$-independent.

\section{A $d$-independent Set.}\label{dset}

Define $\bar{S}_{n,d}$ to be the set of points in $\mathbb{R}^{n} $ that correspond to nonnegative integer partitions of $d$:

$$\bar{S_{n,d}}=\left\{(\alpha_1,\ldots,\alpha_n) \in \mathbb{R}^{n}\hspace{.4mm} | \hspace{.9mm}\alpha_i \in \mathbb{Z}, \alpha_i \geq 0, \sum_{i=1}^n \alpha_i=d\right\}.$$

We can think of the points in $\bar{S}_{n,d}$ as all the possible exponent choices for monomials in $n$ variables of degree $d$. Therefore $\bar{S}$ contains $\binom{n+d-1}{d}$ points.

Also let $S_{n,d}$ be the set of points in $\mathbb{R}^{n}$ that correspond to partitions of $d$ with at least 2 nonzero parts. The points in $S_{n,d}$ again correspond to monomials of degree $d$ but we do not allow monomials of the form $x_i^d$. Therefore $S_{n,d}$ contains $\binom{n+d-1}{d}-n$ points.

The following proposition is taken from \cite{Rez3} p.31 and has been known for at least a hundred years. We reproduce the proof below.

\begin{prop}\label{wtf}
There are no nontrivial forms in $n$ variables of degree $d$ that vanish on $\bar{S}_{n,d}$. In other words $I_{1,d}(\bar{S}_{n,d})=0$.
\end{prop}

\begin{proof}
For every point $s=(s_1,\ldots,s_n) \in \bar{S}_{n,d}$ we will construct a form $p_s \in P_{n,d}$ that vanishes at all points in $\bar{S}_{n,d}$ except for $s$. This shows that the conditions of vanishing at a point in $\bar{S}_{n,d}$ are linearly independent and since $|\bar{S}_{n,d}|=\dim P_{n,d}$ we see that $\dim I_{1,d}(\bar{S}_{n,d})=0$.

Let $M=x_1+\ldots+x_n$. For $i=1 \ldots n$ let $h_i$ be a form defined as follows:
$$h_i(x)=\prod_{k=0}^{s_i-1}(dx_i-kM).$$
\noindent It is clear that the degree of $h_i$ is $s_i$ and $h_i$ vanishes on all partitions in $\bar{S}_{n,d}$ with $i$-th part less than $s_i$. Now let $p_s$ be defined as:

$$p_s=\prod_{i=1}^{n}h_i.$$

\noindent The form $p_s$ has degree $\sum s_i=d$ and it does not vanish on $s$. However for any other partition of $d$ there will be a part that for some $i$ is less than $s_i$. Then $h_i$ will vanish for that $i$ and thus $p_s$ will vanish on any partition of $d$ except for $s$.
\end{proof}

Let $M=x_1+\ldots+x_n$. For $i=1, \ldots ,n$ define a form $Q_i$ as follows:

\begin{equation}\label{qi}
Q_i=\prod_{k=0}^{d-1} (dx_i-{kM}).
\end{equation}

 We observe that each $Q_i$ vanishes on $S_{n,d}$. Let $s=(s_1, \ldots, s_n)\in S_{n,d}$ and consider $Q_i(s)$. We know that $M(s)=d$ because points in $S_{n,d}$ are partitions of $d$, and therefore the term in the defining product of $Q_i$ that corresponds to $k=s_i$ will vanish at $s$, making $Q_i(s)=0$. Therefore $Q_i \in I_{1,d}(S_{n,d})$ for all $i=1 \ldots n$.

We will now show that $Q_i$ actually form a basis of $I_{1,d}(S)$. The fact that we have such a nicely factoring basis is what allows us to prove that $S_{n,d}$ is $d$-independent.

\begin{prop}
The forms $Q_i$ form a basis of $I_{1,d}(S_{n,d})$.
\end{prop}

\begin{proof}
We first show that $Q_i$ are linearly independent. Let $e_1 \ldots e_n$ be the standard basis vectors of $\mathbb{R}^{n}$. Its easy to see that $Q_i(e_j)=0$ when $i \neq j$, since $x_i$ divides $Q_i$. On the other hand, $Q_i(e_i)=d!$. Therefore if a linear combination $\alpha_1Q_1+\ldots+\alpha_nQ_n=0$ for some $\alpha_i \in \mathbb{R}$ then by considering the value of this combination at $e_i$ we see that $\alpha_i=0$ and this works for all $i$. Thus $Q_i$ are linearly independent.

We now show that $Q_i$ span $I_{1,d}(S_{n,d})$. Let $p$ be a form in $I_{1,d}(S_{n,d})$ and let $\beta_i=p(e_i)$. Consider the form $$\bar{p}=p-\sum_{i=1}^{n}\frac{\beta_i}{d!}Q_i.$$ It is clear from the above that $\bar{p}$ vanishes on the standard basis vectors $e_i$. Therefore, $\bar{p}$ vanishes not only on $S_{n,d}$ but also on $\bar{S}_{n,d}$. By Proposition \ref{wtf} it follows that $\bar{p}=0$ and therefore $p$ is in the span of $Q_i$.

\end{proof}

We now show that the set $S_{n,d}$ satisfies the two conditions of $d$-independence from Definition \ref{defindindep}.

\begin{lemma}
The set $S$ forces no additional zeroes for forms of degree $d$.
\end{lemma}

\begin{proof}
Since we know that $Q_i$ form a basis of $I_{1,d}(S_{n,d})$ the statement of the lemma is equivalent to showing that $S_{n,d}$ is projectively equal to $\cap_{i=1}^n Z(Q_i),$
where $Z(Q_i)$ denotes the zero set of $Q_i$.

Let $v=(v_1, \ldots, v_n) \in \cap_{i=1}^n Z(Q_i)$ be a nonzero point and first suppose that $v_1+\ldots+v_n=0$. Then $M(v)=0$ and therefore by equation \eqref{qi} we see that $Q_i(v)=d^dv_i^d$. Since $Q_i(v)=0$ for all $i$ we see that $v=0$ which is a contradiction.

Now suppose that $v_1+\ldots+v_n \neq 0$. By homogeneity we can assume that $v_1+\ldots+v_n=d$. In this case, from equation \eqref{qi} it follows that $Q_i(v)=d^dv_i(v_i-1)\ldots(v_i-d+1).$ Since $Q(v_i)=0$ for all $i$, we see that each $v_i$ is a nonnegative integer between 0 and $d-1$ and $v_1+\ldots+v_n=d$. In other words $v \in S_{n,d}$.
\end{proof}

We know that $|S_{n,d}|=\binom{n+d-1}{d}-n$. For the second condition of $d$-independence we need to show that for any $s \in S_{n,d}$ the vector space of forms double vanishing on $s$ and vanishing on the rest of $S_{n,d}$ has codimension $|S_{n,d}|+n-1=\binom{n+d-1}{d}-1$ in $P_{n,d}$. Since $\dim P_{n,d}=\binom{n+d-1}{d}$ we need to show that the vector space of forms double vanishing at any $s\in S_{n,d}$ and vanishing on the rest of $S_{n,d}$ is one dimensional.

\begin{lemma}
For every point $s \in S_{n,d}$ there is a unique (up to a constant multiple) form in $I_{1,d}(S)$ singular at $s$.
\end{lemma}

\begin{proof}
Let $s=(s_1,\ldots, s_n) \in S_{n,d}$ and let $p \in I_{1,d}(S_{n,d})$ be a form singular at $s$.

Since $Q_i$ form a basis of $I_{1,d}(S_{n,d})$, we may assume that $p=\alpha_1Q_1+\ldots+\alpha_nQ_n$. Now let $A=(a_{ij})$ be a $n \times n$ matrix with entries $$a_{ij}=\frac{\partial Q_i}{\partial x_j}(s).$$

The statement of the lemma is equivalent to showing that the rank of $A$ is $n-1$. Recall from equation \eqref{qi} the definition of $Q_i$:

$$Q_i=\prod_{k=0}^{d-1} dx_i-{kM}.$$

The form $Q_i$ vanishes at $s$ because the term $dx_i-{s_iM}$ corresponding to $k=s_i$ vanishes at $s$. Therefore, the only nonzero term in $\displaystyle \frac{\partial Q_i}{\partial x_j}$ evaluated at $s$ will come from differentiating out $dx_i-{s_iM}$. Now let

$$P_{s_i}=\frac{Q_i}{dx_i-{s_iM}}.$$
We observe that $P_{s_i}(s) \neq 0$ since we removed from $Q_i$ the only factor that vanishes at $s$.

Recall that $M=x_1+\ldots+x_n$ and therefore if we differentiate out $dx_i-{s_iM}$ from $Q_i$ with respect to $x_j$ and evaluate it at $s$ we see that

 \begin{displaymath}
   \frac{\partial Q_i}{\partial x_j}(s) = \left\{
     \begin{array}{lr}
       P_{s_i}(s)\left(d-{s_j}\right) &  \text{if} \hspace{2mm}i=j\\
       -P_{s_i}(s)s_j  & \text{if} \hspace{2mm}i \neq j.
     \end{array}
   \right.
\end{displaymath}

Since $P_{s_i}(s) \neq 0$ we can divide the $i$-th row of $A$ by ${P_{s_i}(s)}$ to obtain matrix $B=(b_{ij})$ where

 \begin{displaymath}
   b_{ij} = \left\{
     \begin{array}{lr}
       d-s_j &  \text{if} \hspace{2mm}i=j\\
       -s_j  & \text{if} \hspace{2mm}i \neq j.
     \end{array}
   \right.
\end{displaymath}

We obtained $B$ from multiplying rows of $A$ by nonzero numbers and therefore the rank of $B$ is equal to the rank of $A$.

Since $s$ is a partition of $d$ it is clear that the vector consisting of all 1's is in the kernel of $B$. Now let $C=(c_{ij})$ be the matrix with $j$-th column having the same entry $s_j$, i.e. $c_{ij}=s_j$. We observe that the rank of $C$ is 1 and $B=dI-C$ where $I$ is the identity matrix. Therefore we know that $\text{rank} \hspace{.5mm} B \geq \text{rank} \hspace{.5mm} I - \text{rank} \hspace{.5mm} C=n-1$. Since we already found a vector in the kernel of $B$ it follows that the rank of $B$ is $n-1$.

\end{proof}

We have now shown that the set $S_{n,d}$ is $d$-independent and together with Proposition \ref{open} this shows that $d$-independence is a generic condition for sets $k$ points in $\mathbb{R}^{n}$ with $k\leq \binom{n+d-1}{d}-n$. We now use this to find large gaps the faces of $Pos_{n,2d}$ and $Sq_{n,2d}$.

\section{Large Dimensional Gaps between $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$}\label{gapsection}
We now establish large gaps between $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$.

\begin{theorem}
Let $S$ be a $d$-independent set of $k$ points in $\mathbb{R}^{n}$. Then the dimension of $Pos_{n,2d}(S)$ is at least $\binom{n+2d-1}{2d}-kn$ while the dimension of $Sq_{n,2d}(S)$ is at most $\binomi{\binom{n+d-1}{d}-k+1}{2}$.
\end{theorem}

\begin{proof}
The dimension of $I_{2,2d}(S)$ is at least $\binom{n+2d-1}{2d}-kn$ since we are imposing at most $kn$ linearly independent conditions by forcing forms to double vanish at all points of $S$. From Corollary \ref{dindep} we know that $Pos_{n,2d}(S)$ is full dimensional in $I_{2,2d}(S)$ and thus the bounds for the dimension of $Pos_{n,2d}(S)$ follows.

Since $S$ is $d$-independent we know that the dimension of $I_{1,d}(S)$ is $\binom{n+d-1}{d}-k$. We can have at most $\binomi{\dim I_{1,d}(S)+1}{2}$ linearly independent  products coming from $I_{1,d}(S)$ and therefore the dimension of $I^{[2]}_{2d}(S)$ is at most $\binomi{\binom{n+d-1}{d}-k+1}{2}$. Since $Sq_{n,2d}(S)$ is contained in $I^{[2]}_{2d}(S)$ the bound for $Sq_{n,2d}(S)$ follows.
\end{proof}

Let $G_{n,2d}(k)$ be the size of the gap that we can show exists between $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$ for a $d$-independent set $S$ of size $k$:

\begin{equation}
G_{n,2d}(k)=\binom{n+2d-1}{2d}-kn-\binomi{\binom{n+d-1}{d}-k+1}{2}.
\end{equation}

From Section \ref{dset} we know that there exist $d$-independent sets of any cardinality $k \leq \binom{n+d-1}{d}-n$. We want to know the smallest $k$ where we can show that the gaps exists and we would also like to know the largest gap size. In other words we want to find the least $k$ for which $G_{n,2d}(k) > 0$ and we want to find the maximum of $G_{n,2d}(k)$.

\begin{prop}\label{uglybounds}
The function $G_{n,2d}(k)$ is maximized at $k=\binom{n+d-1}{n}-n$. Its value and the largest gap are
\begin{equation}\label{lessugly}
\binom{n+2d-1}{2d}-n\binom{n+d-1}{d}+\binom{n}{2}.
\end{equation}
The smallest value of $k$ to make $G_{n,2d}(k)$ positive is the smallest integer strictly greater than:
\begin{equation}\label{fugliness}
\binom{n+d-1}{d}-n+\frac{1}{2}-\sqrt{\left(n-\frac{1}{2}\right)^2+2\binom{n+2d-1}{2d}-2n\binom{n+d-1}{d}}.
\end{equation}

\end{prop}

Before we prove Proposition \ref{uglybounds} we make several remarks. First we observe that the largest gap number
$$\binom{n+2d-1}{2d}-n\binom{n+d-1}{d}+\binom{n}{2}$$
is zero in all the cases where the cones $Pos_{n,2d}$ and $Sq_{n,2d}$ are equal. However it is strictly positive in the cases where exist nonnegative forms that are not sums of squares. In the smallest cases $n=4$, $2d=4$ and $n=3$, $2d=6$ where $Pos_{n,2d}$ is strictly larger than $Sq_{n,2d}$ the gap number is 1.

However, as either $n$ or $d$ we can see that the dimensional gap between faces of $Pos_{n,2d}$ and $Sq_{n,2d}$ grows and asymptotically it approaches the full dimension of the vector space $P_{n,2d}$.

We note that the bound from Equation \ref{fugliness} simplifies remarkably for $n=3$. In this case we get the bound of $\binom{d+2}{2}-d-1$ and we need to take the smallest integer above that which leads to
$$k=\binom{d+2}{2}-d.$$
This is actually the correct bound for the case of $n=3$ and we hope to discuss this elsewhere.

However for $n \geq 4$ the formula does not appear to simplify and the bound given is not going to be optimal. This is due to the overcounting in the bound we use that for the dimension of the vector space of squares $I^{[2]}_{2d}(S)$. It would be interesting to improve the bound which should lead to the optimal value of $k$.

We note that for $k=\binom{n+d-1}{d}-n$, which leads to the largest gap, the bound on the dimension of $I^{[2]}_{2d}(S)$ is also optimal generically. We can see that from the example of $d$-independent set $S_{n,d}$ from Section \ref{dset}, which has this cardinality. Indeed in the case of $S_{n,d}$ it is not hard to show that all pairwise products of $Q_i$, which form the basis of $I_{1,d}(S_{n,d})$ are linearly independent in $P_{n,2d}$. This shows that the dimension of $I_{1,d}(S_{n,d})$ is $\binom{n+1}{2}$, which is exactly equal to the bound we use.

We now prove Proposition \ref{uglybounds}.
\begin{proof}
We observe that $G_{n,d}(k)$ is a quadratic function of $k$ with a negative leading coefficient. It is easy to show that the maximum of $G_{n,d}(k)$ occurs at $\displaystyle k=\binom{n+d-1}{d}-n+\frac{1}{2}$. Therefore the maximum value of $G_{n,2d}(k)$ for an integer $k$ will occur with $k=\binom{n+d-1}{d}-n$ and it is a matter of easy simplification to obtain equation \eqref{lessugly}.

The bound in equation \eqref{fugliness} comes from simply calculating the smallest root of $G_{n,2d}(k)$. We skip the routine application of the quadratic formula.
\end{proof}

We now fully describe the situation with respect to $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$ for 6 points sets $S$ in $\mathbb{R}^4$.

\section{Six points in $\mathbb{R}^4$} \label{sixpoints}

Let $S=\{s_1, \ldots, s_6\}$ be a set of six points in $\mathbb{R}^4$ in general linear position. We will show that $S$ is 2-independent. In particular this implies that the conditions of vanishing at $s_i \in S$ are linearly independent and therefore $\dim I_{1,2}(S)=\binom{5}{2}-6=4$. It follows that the dimension of the vector space $I^{[2]}_4(S)$ spanned by squares from $I_{1,d}(S)$ is at most $\binom{5}{2}=10$. We will show that the dimension of $I^{[2]}_4(S)$ is indeed 10.

On the other hand Alexander-Hirschowitz Theorem tells us that the dimension of $I_{2,4}(S)$ is $\binom{7}{4}-6 \cdot 4=11$ generically. It is not hard to show that for 6 points in $\mathbb{R}^4$ in general linear position this dimension count is actually correct. Therefore we should have a gap of 1 dimension between $Pos_{4,4}(S)$ and $Sq_{4,4}(S)$. In particular, there is a linear constraint that is satisfied by squares and not satisfied by double vanishing polynomials on $S$. There is also a fourth degree form that double vanishes on $S$ but is not in the span of squares. We identify the extra constraint and a double vanishing form not in the span of squares below.

To every $3$ element subset $T=\{t_1,t_2,t_3\}$ of $\{1, \ldots, 6\}$ we can associate the hyperplane $L_T$ spanned by the vectors $s_{t_1}$, $s_{t_2}$ and $s_{t_3}$.

We want to select a double covering of $s_1, \ldots, s_6$ by 4 hyperplanes of the form $L_T$ with some nice combinatorial properties. We select 4 triples $T_i$ such that any two of them intersect at exactly one element of $\{1,\ldots,6\}$ and each element is contained in precisely two triples. Here is an example of such a covering, which is not unique: $$T_1=\{123\}, \hspace{4mm}T_2=\{145\},\hspace{4mm} T_3=\{246\}, \hspace{4mm}T_4=\{356\}.$$ To every such covering we can associate the complementary covering, where we replace the triple $T_i$ with its complement $\overline{T}_i$. So, in our case, $\overline{T}_1=\{456\}$, $\overline{T}_2=\{236\}$, $\overline{T}_3=\{135\}$ and $\overline{T}_4=\{124\}$. We observe that the complementary covering also shares the property any two triples intersect in exactly one point and every point is contained in exactly two triples.

To each triple $T$ we associate the linear functional with kernel $L_T$. We can think of this functional as the inner product with the unit normal vector to $L_T$, which is unique up to a sign. The choice of sign will not make a difference to us. We let $u_i$ be a unit normal to $L_{T_i}$ and $v_i$ be a unit normal to $L_{\overline{T}_i}$:
\begin{align*}
u_i \hspace{2mm} \text{is a unit vector perpendicular to} \hspace{2mm} L_{T_i},\\
v_i \hspace{2mm} \text{is a unit vector perpendicular to} \hspace{2mm} L_{\overline{T}_i}.
\end{align*}

Vectors $u_i$ and $v_i$ form a pair of bases of $\mathbb{R}^4$. The key is to work with the dual configurations. We define $u_i^*$ to be vectors such that
 \begin{displaymath}
   \ip{u_i^*}{u_j} = \left\{
     \begin{array}{lr}
       1 &  \text{if} \hspace{2mm}i=j\\
       0  & \text{if} \hspace{2mm}i \neq j.
     \end{array}
   \right.
\end{displaymath}
One way to think about $u_i^*$ is that if we form matrix $U$ with rows $u_i$ then $u_i^*$ form the columns of $U^{-1}$. We define vectors $v_i^*$ in the same way for $v_i$.

We will show that the four forms
\begin{align*}
Q_1(x)=\ip{x}{u_1}\ip{x}{v_1}, \hspace{4mm} Q_2(x)=\ip{x}{u_2}\ip{x}{v_2}\\
Q_3(x)=\ip{x}{u_3}\ip{x}{v_3}, \hspace{4mm} Q_4(x)=\ip{x}{u_4}\ip{x}{v_4}
\end{align*}
form a basis of $I_{1,2}(S)$. This factoring basis will allow us to prove 2-independence of $S$, and pairwise products $Q_iQ_j$ with $i \leq j$ will form a basis of $I^{[2]}_4(S)$.

Then we will show that the fourth degree form
$$R=\ip{x}{u_1}\ip{x}{u_2}\ip{x}{u_3}\ip{x}{u_4}$$
is singular at each of $s_i$ but it is not in $I^{[2]}_4(S)$.

Finally, let $Q$ be a form in $I^{[2]}_4(S)$. For any $i$ the form $Q$ satisfies:
\begin{equation}\label{equationextraconstraint}
\ip{v_i^*}{u_i}^2Q(u_i^*)=\ip{u_i^*}{v_i}^2Q(v_i^*).
\end{equation}
On the other hand $R$ will not satisfy any of these constraints, which shows that any one of these constraints is independent of being singular at points $s_i$. Of course, by the dimension count there is only 1 "true" extra linear constraint so we can take any one and the rest will cease being "new".

Before we continue with the proofs we will give an explicit example of the extra form and the explicit linear constraint.

\subsection{Explicit Example.}\label{sectionexplicitexample} Let $s_1=(0,0,1,1)$, $s_2=(0,1,0,1)$, $s_3=(0,1,1,0)$, $s_4=(1,0,0,1)$, $s_5=(1,0,1,0)$ and $s_6=(1,1,0,0)$. This is the set from Example \ref{sectionfirstgap}. Our particular numbering of points is chosen to mesh well with our system of covering triples $T_i$ and $\overline{T}_i$.

Vector $u_1$ comes from triple $123$ and therefore is a unit vector perpendicular to $s_1$, $s_2$, and $s_3$ and we may chose choose $u_1=e_1$, the first standard basis vector. Similarly $u_2$ comes from $145$ and is normal to $s_1$, $s_4$ and $s_5$. We may choose $u_2=e_2$. In the same way $u_3=e_3$ and $u_4=e_4$.

For the vectors $v_i$, $v_1$ comes from $456$ and we may choose $v_1=\frac{1}{2}(1,-1,-1,-1)$. In the same way $v_2=\frac{1}{2}(-1,1,-1,-1)$, $v_3=\frac{1}{2}(-1,-1,1,-1)$ and $v_4=\frac{1}{2}(-1,-1,-1,1)$

The extra form $R=\ip{x}{u_1}\ip{x}{u_2}\ip{x}{u_3}\ip{x}{u_4}$ becomes $$R=\ip{x}{e_1}\ip{x}{e_2}\ip{x}{e_3}\ip{x}{e_4}=x_1x_2x_3x_4,$$
as promised in the Example $\ref{sectionfirstgap}$.

We note that the sets $\{u_i\}$ and $\{v_i\}$ form 2 orthogonal bases in $\mathbb{R}^4$ and therefore are self-dual. It follows that $u_i^*=u_i$ and $v_i^*=v_i$. The extra constraint from Equation \eqref{equationextraconstraint} becomes: $\frac{1}{4}Q(u_1)=\frac{1}{4}Q(v_1)$ or after rewriting and using homogeneity of $Q$:
$$16Q(1,0,0,0)=Q(1,-1,-1,-1),$$
again as claimed in Example \ref{sectionfirstgap}.

\subsection{Proofs.} The vectors $u_i$ and $v_i$ are not just two arbitrary sets of bases of $\mathbb{R}^4$. Since they come from a configuration of 6 points in general position they have some structure. The following simple lemma will be crucial to our proofs.

\begin{lemma}\label{inprod}
For all i,j the following hold:
\begin{equation*}
\ip{u_i}{v_j^*}\neq 0 \hspace{2mm} \text{and} \hspace{2mm} \ip{v_i}{u_j^*}\neq 0.
\end{equation*}
\end{lemma}
\begin{proof}
By symmetry it will suffice to prove only one of the two assertions. Also, by symmetry it will suffice to show that $\ip{u_1^*}{v_1}\neq 0$ and $\ip{u_1^*}{v_2}\neq 0$.

Let's suppose that $\ip{u_1^*}{v_1}=0$ then it follows that $v_1$ is in the span of $u_2,u_3,u_4$. Let $$v_1=\alpha_2u_2+\alpha_3u_3+\alpha_4u_4.$$ Now lets consider the inner product $\ip{v_1}{s_4}$. Recall that $v_1$ came from the triple 456, $u_2$ from 145, $u_3$ from 246 and $u_4$ from 356. It follows that $$\ip{v_1}{s_4}=0=\alpha_4\ip{s_4}{u_4}.$$

Since the points $s_i$ are in general position it follows that $\ip{s_4}{u_4}\neq 0$ and therefore $\alpha_4=0$. By considering inner products of $v_1$ with $s_5$ and $s_6$ we can also show that $\alpha_2=\alpha_3=0$ which yields a contradiction.

Similarly, if $\ip{u_1^*}{v_2}=0$ then $v_2$ is in the span of $u_2,u_3,u_4$. Let $$v_2=\alpha_2u_2+\alpha_3u_3+\alpha_4u_4.$$ Recall that $v_2$ came from the triple 236, $u_2$ from 145, $u_3$ from 246 and $u_4$ from 356. By the same argument we can establish that $\alpha_2=0$ by using inner products with $s_6$. Then we use inner product with $s_2$ to show that $\alpha_4=0$ and we will arrive at a contradiction.

\end{proof}

\begin{lemma}
Let $Q_i(x)=\ip{x}{u_i}\ip{x}{v_i}$ for $i=1 \dots 4$. The forms $Q_i$ form a basis of $I_{1,2}(S)$. Furthermore the pairwise products $Q_iQ_j$ with $i \leq j$ form a basis of $I^{[2]}_4(S)$ and the dimension of $I^{[2]}_4(S)$ is 10.
\end{lemma}
\begin{proof}
It is not hard to show that $I_{1,2}(S)$ has dimension 4. Therefore it suffices to show that the polynomials $Q_i$ are linearly independent. Consider the values of $Q_i$ at the points $u_i^*$.

From the definition of the dual points $u_i^*$ and Lemma \ref{inprod} it follows that $Q_i(u_i^*)=\ip{u_i^*}{v_i} \neq 0$ and $Q_i(u_j^*)= 0$ when $i \neq j$. Therefore, if $P=\alpha_1Q_1+\alpha_2Q_2+\alpha_3Q_3+\alpha_4Q_4=0$ then by considering $P(u_i^*)$ we can see that $\alpha_i$ is 0 for each $i$ and therefore $Q_i$ are linearly independent.

Now lets consider pairwise products $Q_iQ_j$ for $i \leq j$. These forms are clearly in $I^{[2]}_4(S)$, and we need to show their linear independence. Of all the pairwise products only $Q^2_i$ does not vanish on $u^*_i$. Therefore the squares $Q_i^2$ are linearly independent from all other pairwise products and we only need to show linear independence of $Q_iQ_j$ for $i < j$.

By Lemma \ref{inprod} only the products $Q_iQ_j$ vanish on $u^*_i$ to order 1. If both indices are distinct from $i$ then the product vanishes to order 2. Therefore if forms $Q_iQ_j$ are linearly dependent it follows that the forms $Q_iQ_j$ for some fixed $i$ are linearly dependent. We can factor out $Q_i$ and it follows that the forms $Q_j$ are linearly dependent. This is a contradiction.

Since pairwise products $Q_iQ_j$ span $I^{[2]}_4(S)$ and are linearly independent it follows $\dim I^{[2]}_4(S)=10$.

\end{proof}

We are now ready to show 2-independence of $S$.

\begin{prop}
Let $S$ be a set of 6 points in $R^4$ in general linear position. Then $S$ is 2-independent.
\end{prop}

\begin{proof}
We first show that $S$ forces no additional zeroes on quadratic forms. Recall that $Q_i=\ip{x}{u_i}\ip{x}{v_i}$ and the forms $Q_i$ form a basis of $I_{1,2}(S)$. It will suffice to show that the forms $Q_i$ have no common zeroes outside of $S$.

Let $z$ be a nonzero point in the intersection $\cap_{i=1}^4 Z(Q_i).$ It follows that for each $i$ we either have $\ip{z}{u_i}=0$ or $\ip{z}{v_i}=0$. Since $u_i$ and $v_i$ form a basis of $\mathbb{R}^4$ the vector $z$ cannot be orthogonal to all four $u_i$ or $v_i$. If $\ip{z}{u_i}=0$ for three indices $i$, which we may assume without loss of generality to be 1,2 and 3, then it follows that $z$ is a multiple of $u_4^*$. But then $\ip{z}{u_4} \neq 0$  and from Lemma \ref{inprod} we know that $\ip{z}{v_4} \neq 0$. Therefore $Q_4(z) \neq 0$, which is a contradiction.

Therefore it must happen that $z$ is orthogonal to two $u_i$ and two $v_i$. Again without loss of generality we may assume that $z$ is orthogonal to $u_1$, $u_2$, $v_3$ and $v_4$. Since $u_1$ comes from the triple $123$, $u_2$ comes from $123$, $v_3$ comes from $135$ and $124$ it follows that $z$ is in the intersection of spans of $\{s_1, s_2, s_3\}$, $\{s_1, s_4, s_5\}$, $\{s_1,s_3,s_5\}$ and $\{s_1,s_2,s_4\}$. Since the points $s_i$ are in general linear position it follows that $s_1$ spans this intersection. The other points $s_i$ arise in the same manner from choosing different pairs of $u_i$'s and $v_i$'s.

For the second condition of $2$-independence we need to show that for any $s_i \in S$ there exists a unique (up to a constant multiple) form in $I_{1,2}(S)$ that is singular at $s_i$. Again by symmetry we only need to prove this for $s_1$. By construction $s_1$ is orthogonal to $u_1$, $u_2$, $v_3$ and $v_4$. Therefore it follows that
$\nabla Q_1(s_1)=\ip{v_1}{s_1}u_1$, $\nabla Q_2(s_1)=\ip{v_2}{s_1}u_2$, $\nabla Q_3(s)=\ip{u_3}{s_1}v_3$ and $\nabla Q_4(s)=\ip{u_4}{s_1}v_4$. The coefficients of vectors $u_1$,$u_2$, $v_3$ and $v_4$ are nonzero and since $s_i$ are in general linear position it follows that $u_1$,$u_2$, $v_3$ and $v_4$ span the vector space $s_1^{\perp}.$ Therefore there is only one (up to a constant multiple) linear combination of gradients of $Q_i$ that vanishes at $s_1$.
\end{proof}

Now we show that the fourth degree form
$$R=\ip{x}{u_1}\ip{x}{u_2}\ip{x}{u_3}\ip{x}{u_4}$$ is not in the span of squares from $I_{1,2}(S)$.

\begin{prop}
Let $R=\ip{x}{u_1}\ip{x}{u_2}\ip{x}{u_3}\ip{x}{u_4}$. The $R$ is not in $I^{[2]}_4(S)$.
\end{prop}

\begin{proof}
We know that products $Q_iQ_j$ with $i \leq j$ form a basis of $I^{[2]}_4(S)$. We observe that $R(u_i^*)=0$ for all $i$, and the only form from the spanning set that doesn't vanish at $u_i^*$ is $Q_i^2$. Therefore, if we assume that $R$ is spanned by $Q_iQ_j$ then $R$ is spanned by products $Q_iQ_j$ with $i$ not equal to $j$.

Now lets look at $R(v_k^*)$. By Lemma \ref{inprod} we know that $R(v_k^*) \neq 0$. However, $Q_iQ_j(v_k^*)=0$ since $\ip{v_i^*}{v_k}=0$ for $i \neq k$. Therefore we arrive at a contradiction.
\end{proof}

Now we focus on an explicit linear constraint on forms in $I^{[2]}_4(S)$ that is independent of vanishing gradients on the points $s_i$.
To establish the constraint it is enough to look at forms in $I^{[2]}_4(S)$ that are squares, since they span $I^{[2]}_4(S)$.

\begin{prop}
Let $Q=\alpha_1Q_1+\alpha_2Q_2+\alpha_3Q_3+\alpha_4Q_4$. Then $Q$ satisfies
$$\ip{v_i^*}{u_i}Q(u_i^*)=\ip{u_i^*}{v_i}Q(v_i^*)$$
for all $i$. It follows that $Q^2$ satisfies:
$$\ip{v_i^*}{u_i}^2Q^2(u_i^*)=\ip{u_i^*}{v_i}^2Q^2(v_i^*)$$
On the other hand the form $R$ will not satisfy any of these constraints.
\end{prop}

\begin{proof} For any $i$,
$$Q(u_i^*)=\alpha_iQ_i(u_i^*)=\alpha_i\ip{u_i^*}{v_i},$$ and $$Q(v_i^*)=\alpha_iQ_i(v_i^*)=\alpha_i\ip{v_i^*}{u_i}.$$ Therefore $$\ip{v_i^*}{u_i}Q(u_i^*)=\ip{u_i^*}{v_i}Q(v_i^*).$$
The constraint for $Q^2$ now follows.

On the other hand by definition of $u_i^*$ and Lemma \ref{inprod} we know that $$R(u^*_i)=0 \hspace{.7cm} \text{while} \hspace{.7cm} R(v_i^*)\neq 0 \hspace{.7cm} \text{and} \hspace{.7cm}\ip{u_i^*}{v_i}\neq 0.$$ Therefore $R$ does not satisfy the relation for any $i$ and any one of these relations is independent of gradient vanishing at points $s_i$.
\end{proof}

\section{Inequalities in the General Setting}\label{sectiongenineq}

Let $S$ be a finite set of points in $\mathbb{R}^{n}$ and suppose that the face $Pos_{n,2d}(S)$ of the cone of nonnegative forms has a higher dimension than the face $Sq_{n,2d}(S)$ of the cone of sums of squares. Then it follows that there must be extra linear constraints that are satisfied by sums of squares in $Sq_{n,2d}(S)$, but are not satisfied by the nonnegative forms in $Pos_{n,2d}(S)$. Since $Sq_{n,2d}(S)$ spans the vector space $I^{[2]}_{2d}(S)$ the extra constraints hold for any polynomial that is spanned by squares.

Let $l_1,\ldots,l_k$ be linear functionals on $P_{n,2d}$ that form a basis of the set of extra constraints. For any $p \in Sq_{n,2d}$ we know that

\begin{equation}\label{forcedcontraints}
\text{if} \hspace{2mm} p(s)=0 \hspace{2mm} \text{for all} \hspace{2mm} s \in S\hspace{2mm} \text{then} \hspace{2mm} l_1(p)=\ldots=l_k(p)=0.
\end{equation}
Let $R_S$ be a quadratic functional on $P_{n,2d}$ given by the sum of squares of linear functionals $l_i$:
$$R_S(p)=\sum_{i=1}^k l_i^2(p).$$

\noindent Let $M_S$ be the linear functional on $P_{n,2d}$ given by summing the values of a form $p$ on the points $s \in S$:

$$M_S(p)=\sum_{s \in S} p(s).$$

\noindent Finally let $T(p)$ be the linear functional given by averaging a form $p$ over the unit sphere:

$$T(p)=\int_{\mathbb{S}^{n-1}}p \hspace{.7mm} d\sigma,$$

\noindent where $\sigma$ is the uniform probability measure on $\mathbb{S}^{n-1}$.

We claim that there exists $\alpha >0$ such that for all $p \in Sq_{n,2d}$
\begin{equation} \label{genineq}
\alpha M_S(p)T(p)-R_S(p) \geq 0.
\end{equation}

We briefly explain why there exists $\alpha$ that makes inequality \eqref{genineq} hold for all $p \in Sq_{n,2d}(S)$. First we observe that we can restrict ourselves to the case of forms of average 1 on the unit sphere, i.e. the case of $T(p)=1$.

The linear functional $M_S$ is clearly nonnegative on $Sq_{n,2d}$
. When $M_S(p)=0$ we know by \eqref{forcedcontraints} all the functionals $l_i(p)$ must vanish and therefore $R_S(p)=0$. Since $M_S(p)>0$ for all $p \in Sq_{n,2d}$ that are not in $Sq_{n,2d}(S)$ it follows that the only obstacle to finding an appropriate $\alpha$ has to come from looking infinitesimally close to the face $Sq_{n,2d}(S)$ where $M_S(p)=0$. Owing to the quadratic nature of the cone of sums of squares $Sq_{n,2d}$ we will be able to argue the existence of $\alpha$ and we will provide an explicit example of such inequality in Section \ref{sectionexplicitineq}.

We observe that regardless of $\alpha$ the inequality \eqref{genineq} will not hold for some $f \in Pos_{n,2d}$. Let $f$ be a form such that $f$ is in $Pos_{n,2d}(S)$ but $f$ is not in the vector space $I^{[2]}_{2d}(S)$ spanned by $Sq_{n,2d}(S)$. We have a dimensional gap between $Pos_{n,2d}(S)$ and $Sq_{n,2d}(S)$  and thus we know that such $f$ exist. Since $f \in Pos_{n,2d}(S)$ it vanishes at every point of $S$ and therefore $M_S(p)=0$. Also, since $f$ is not in $I^{[2]}_{2d}(S)$ it follows that at least one of the functionals $l_i$ is not zero on $f$ and therefore $R_S(f) >0$. We thus see that $M_S(f)T(f)-R_{S}(f) < 0$.

We note that the linear functional $T(p)$ is used to homogenize the inequality and we could have used any other linear functional that is strictly positive on all nonzero forms in $Sq_{n,2d}$ instead.

We first need a preliminary lemma that is very similar in flavor to the Extension Lemma \ref{ext}.

\begin{lemma}\label{prelim}
Let $Q_1$ and $Q_2$ be two quadratic forms on a real vector space $V$, such that $Q_1$ is positive semidefinite and $Q_2$ is positive definite on $V$. Let $R$ be a sum of squares of quadratic forms $p_i$ on $V$: $$R=\sum_i p_i^2,$$
and suppose further that $R$ vanishes whenever $Q_1$ vanishes: $Z(Q_1) \subseteq Z(R)$. Then there exists $\alpha >0$ such that the form
$Q_{\alpha}=\alpha Q_1 Q_2 -R$ is nonnegative on $V$:
\begin{equation}
\label{qu}
Q_{\alpha}(v)=\alpha Q_1(v) Q_2(v) -R(v) \geq 0
\end{equation}
for all $v \in V$.
\end{lemma}

\begin{proof}
The inequality \eqref{qu} is homogeneous and therefore it suffices to prove it for all $v$ on the unit sphere $S_V$ of $V$. Since $Q_1$ is a positive semidefinite quadratic form we know that $Q_1$ vanishes on a subspace $W$ of $V$.  We note that both $Q_1Q_2$ and $R$ vanish to order 2 on $W$, since zero is a global minimum for both forms.

Let's pick a point $w \in W$ that is also on the unit sphere. The Hessian $H_{Q_1Q_2}(w)$ of $Q_1Q_2$ at $w$ has $W$ as the kernel and is positive definite on $W^{\perp}$. This follows from the fact that $Q_1$ is positive semidefinite and $Q_2$ is positive definite. The same is true for the Hessian $H_R(w)$ of $R$ at $w$ since $R$ also vanishes on $W$ and zero is a global minimum of $R$. Therefore by compactness of the unit sphere we can find $\alpha_1$ such that $\alpha_1H_{Q_1Q_2}(w)-H_R(w)$ is positive definite on $W^{\perp}$ for any $w\in W$ on the unit sphere. It follows that the form $Q_{\alpha_1}$ is nonnegative on all points of the unit sphere that are distance at most $\delta$ from $W$, for some $\delta >0$.

Let's consider points $v \in S_v$ that are at least $\delta$ away from the subspace $W$ on which $Q_1$ vanishes. We know that for these points $v$ we will have $Q_1(v)Q_2(v) \geq \epsilon$ for some $\epsilon >0$. Then we can multiply $Q_1Q_2$ by a sufficiently large $\alpha_2$ so that $Q_{\alpha_2}$ will be positive on the points $v$ that are $\delta$ away from $W$. We choose $\alpha=\max (\alpha_1, \alpha_2)$ and $Q_{\alpha}$ is nonnegative on the whole unit sphere $S_V$.

\end{proof}

Now we prove the existence of $\alpha$ that makes inequality \eqref{genineq} true.
\begin{theorem}\label{theoremineqgen}
There exists $\alpha >0$ such that for all $p \in Sq_{n,2d}$
\begin{equation*}
\alpha M_S(p)T(p)-R_S(p) \geq 0.
\end{equation*}
\end{theorem}

\begin{proof}
Let $\overline{Sq}_{n,2d}$ be the section of the cone $Sq_{n,2d}$ with hyperplane of forms of integral 1 on the unit sphere $\mathbb{S}^{n-1}$:
$$\overline{Sq}_{n,2d}=\left\{p \in Sq_{n,2d} \hspace{2mm} \mid \hspace{2mm} \int_{\mathbb{S}^{n-1}}p \hspace{.7mm} d\sigma=1 \right\}$$.

We begin by observing that it suffices to prove \eqref{genineq} for $p \in \overline{Sq}_{n,2d}$, so we restrict our attention to the case $T(p)=1$. We want to find $\alpha >0$ such that for all $p \in \overline{Sq}_{n,2d}$
\begin{equation}
\alpha M_S(p) - R_S(p) \geq 0.
\end{equation}

We note that $R_S$ is a sum of squares of linear functionals and therefore it is a convex functional on $P_{n,2d}$. Then it follows that the functional $M_S-R_S$ is concave. Since $\overline{Sq}_{n,2d}$ is a compact convex set the functional $M_S-R_S$ attains its minimum at an extreme point of $\overline{Sq}_{n,2d}$. We know that an extreme point of $\overline{Sq}_{n,2d}$ must be a square. Therefore it suffices to show \eqref{genineq} for squares.

We observe that $M_S(g^2)=\sum_{s \in S} g^2(S)$ is a positive semidefinite quadratic form on $P_{n,d}$ while $T(g^2)=\int_{\mathbb{S}^{n-1}}g^2 \hspace{.7mm} d\sigma$ is a positive definite quadratic form on $P_{n,d}$. Also, $R_S(g^2)=\sum_i l_i^2(g^2)$ is a sum of squares of quadratic forms and $R_S$ vanishes whenever $M_S$ vanishes. Therefore we can apply Lemma \ref{prelim} and the existence of $\alpha$ follows.
\end{proof}

We note that the problem of finding the value of $\alpha$ for a fixed set $S$ and degree $d$ is a semidefinite programming problem and can be solved fast numerically using semidefinite program solving packages.

\section{An Explicit Inequality}\label{sectionexplicitineq}

We now derive an explicit inequality along the lines described in the previous section. We take $S=(s_{ij})$ to be the set of 6 points in $\mathbb{R}^4$ with $\frac{1}{\sqrt{2}}$ in coordinates $i$ and $j$ and 0 in the other two coordinates. This is exactly the set from Example \ref{sectionfirstgap} but we now make the vectors have unit length.

Let $a=(1,0,0,0)$ and $b=(\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2})$. We recall from Example \ref{sectionfirstgap} and Section \ref{sectionexplicitexample} that the forms $p \in I^{[2]}_{4}(S)$ satisfy the extra linear constraint $p(a)-p(b)=0$.

We now recall from the previous section definitions of functionals $M_S$, $T$ and $R_S$ for $f \in P_{4,2}$:
\begin{align*}
M_S(f)&=\sum_{s_{ij} \in S} f^2(s_{ij}),\\
T(f)&=\int_{S^3}f^2 \hspace{.2mm} d\sigma,\\
R_S(f)&=(f^2(a)-f^2(b))^2.
\end{align*}
The operators are acting on $f^2$, since we know by the proof of Theorem \ref{theoremineqgen} that we only need to establish the inequality for squares.

We will show that
$$15 M_S(f)T(f)-P_S(f) \geq 0$$
for all forms $f \in P_{4,2}$  and 15 is the smallest value of $\alpha$ that makes the inequality hold.

In order to do this we will explicitly calculate the functionals $M_S$, $T$ and $R_S$ using the following inner product on $P_{4,2}$:

$$\ip{f}{g}=\int_{S^3} fg \hspace{.7mm} d\sigma.$$

\noindent For every point $v \in \mathbb{R}^4$ there exists a unique polynomial $P_v \in P_{4,2}$ such that

$$\ip{f}{P_v}=f(v) \hspace{2mm} \text{for all} \hspace{2mm} f \in P_{4,2}.$$

\noindent It is not hard to show that

\begin{equation}\label{innerprod}
P_{v}(x)=12\ip{x}{v}^2-2\|v\|^2\|x\|^2.
\end{equation}

\noindent Also by definition of $P_v$,

\begin{equation}
\ip{P_v}{P_w}=P_v(w)=12\ip{v}{w}^2-2\|v\|^2\|w\|^2.
\end{equation}

\noindent For more information on the integral inner product and the polynomials $P_v$ see \cite{Muller}. For the points $s_{ij}$ we will denote $P_{s_{ij}}$ simply by $P_{ij}$.

Let's first analyze the quadratic form $$M_S(f)=\sum_{s_{ij} \in S} f^2(s_{ij}),$$ for $f \in P_{2,4}$. This is a positive semidefinite quadratic form of rank 6 on $P_{4,2}$. Using our polynomials $P_{ij}$ we can write is as $$M_S(f)=\sum \ip{f}{P_{ij}}^2.$$

We let $V$ be the span of $P_{ij}$ and then $V^{\perp}$ is the kernel of $M_S$. Using equation \eqref{innerprod} it is a matter of routine calculation to show the quadratic form $M_S$ has two eigenspaces $V_1$ and $V_2$. The eigenspace $V_1$ corresponds to eigenvalue 12 and can be chosen to have the following orthogonal basis:
\begin{align*}
&v_1=\sum P_{ij}, \hspace{4mm} v_2=P_{12}+P_{13}+P_{14}-P_{23}-P_{24}-P_{34},\\
&v_3=P_{12}+P_{24}-P_{13}-P_{34}, \hspace{4mm} v_4=P_{12}+P_{13}+2P_{23}-2P_{14}-P_{24}-P_{34}.\\
\end{align*}
The eigenspace $V_2$ corresponds to eigenvalue $6$ and has orthogonal basis:
\begin{equation*}
v_5=P_{12}+P_{34}-P_{23}-P_{14}, \hspace{4mm} v_6=\sum P_{ij}-3(P_{13}+P_{2,4}).
\end{equation*}

If we pick as an orthogonal basis unit vectors in the direction of eigenvectors $v_i$ then the form $M_S$ becomes:
$$M_S(f)=12(x_1^2+x_2^2+x_3^2+x_4^2)+6(x_5^2+x_6^2) \hspace{2mm} \text{where} \hspace{2mm} x_i=\ip{f}{\frac{v_i}{\|v_i\|}}.$$
Now we analyze $R_S(f)=(f^2(a)-f^2(b))^2.$ We can rewrite this as $$P_S(f)=\left(\ip{f}{P_a}^2-\ip{f}{P_b}^2\right)^2.$$
\noindent Then
$$R_S(f)=\left(\ip{f}{P_a}^2-\ip{f}{P_b}^2\right)^2=\ip{f}{P_a+P_b}^2\ip{f}{P_a-P_b}^2.$$
Therefore $R_S(f)$ is a product of two rank 1 quadratic forms.

We know from Section \ref{sixpoints} that for all $f$ with $M_S(f)=0$ we also have $$f(a)=\ip{P_a}{f}=\ip{P_b}{f}=f(p_b).$$ It follows that $P_a-P_b$ is in the span of $P_{ij}$. It is easy to check that $$P_{a}-P_{b}=\frac{v_2}{2} \hspace{2mm} \text{and} \hspace{2mm} \|P_a-P_b\|^2=18.$$
\indent To deal with $P_a+P_b$ we note that its projection onto the span of $P_{ij}$ is equal to $\frac{1}{6}\sum P_{ij}=\frac{v_1}{6}$. It is easy to check using \eqref{innerprod} that $$\|P_a+P_b\|^2=22 \hspace{4mm} \text{while} \hspace{4mm} \|\frac{v_1}{6}\|^2=12.$$
\noindent We can therefore write $\displaystyle P_a+P_b=\frac{v_1}{6}+v_7$ where $v_7$ is a vector is the kernel of $M_S(f)$ and $\|v_7\|^2=22-12=10$. If we extend our basis $\left\{v_i/\|v_i\|\right\}$ of the span of $P_{ij}$ by adding the vector $v_7/\|v_7\|$ and 3 more unit vector to make a basis of $P_{4,2}$ the we can write:

\begin{equation}
R_S(f)=18x_2^2(12x_1^2+10x_7^2) \hspace{2mm} \text{where} \hspace{2mm} x_i=\ip{f}{\frac{v_i}{\|v_i\|}}.
\end{equation}

Since we chose the integral inner product the form $T$ is simply: $$T=\sum_{i=1}^{10}x_i^2,$$
for any choice of orthogonal basis. Therefore $$M_S(f)T(f)=\left(12(x_1^2+x_2^2+x_3^2+x_4^2)+6(x_5^2+x_6^2)\right)\sum_{i=1}^{10}x_i^2.$$

If we want to choose $\alpha$ such that $\alpha M_S(f)T(f)-R_S^2(f)>0$ then we need to choose it so that the coefficient of $x_{2}^2x_7^2$ is nonnegative. If follows that we need $\alpha \geq 18 \cdot 10/12=15,$ and it is easy to see that $\alpha=15$ indeed will suffice.

redpajama_set_name: RedPajamaArXiv


Celebrate Portion Control Side Plates are available in sets of 4.
The side portion control plate has 3 measurements on it.
Love the size (these fit in a picnic basket). Like the black and white. Love and agree on the quotations.
Nice to get a discount at FNCE and to get free shipping!
Thanks for providing a Foundation Basket at FNCE too!
I ordered the side plates because that is the size that I use for my everyday meals. Love the look, the quality and the idea behind them!


text:
The Horror, The Horror!
10 Times American Horror Story Cribbed from Horror History in Season 6
The Juice Is Loose
The Final Twist of American Horror Story Is. . .The People vs. O.J. Simpson
Missing Evan Peters on American Horror Story This Season? This Might Help
How Schitt's Creek Charmed the TV Academy Into Three Major Emmy Nominations
The Horror, The Horror
How American Horror Story Crafted the Perfect Metaphor for the 2016 Election
When troubling reality stars become national heroes.
Left, courtesy of FX; right, by Chip Somodevilla/Getty Images
When creator Ryan Murphy set out to make American Horror Story Season 6, there was no way he could have known, really known, what the 2016 election would become. Fast-forward 10 weeks after its premiere, and we're all living in a surprising new country where someone who was once a reality star is now our president-elect. It's not something Murphy could have seen coming—but even so, it's a phenomenon this season of American Horror Story tackled head-on. The parallels couldn't have been clearer in this week's season finale.
All season long, Murphy has been purporting to use this season of American Horror Story to explore the darker ramifications of reality TV and its flawed stars. It's "a commentary on social reality television, because I do feel like that is the next level where something really bad is—somebody's going to die for the sake of real TV," he explained. And the darker consequence of reality TV is a subject this country has been staring in the face all year.
The high-profile robbery of Kim Kardashian West caused America's most famous TV family to reflect on the price of their own fame. Murphy's other 2016 show, The People v. O.J. Simpson, both chronicled the birth of reality TV and made us reconsider the first victim of sensationalized reality: Marcia Clark. And, of course, there's our next commander in chief, who parlayed his fame hosting The Apprentice into the highest office in the land. Even now, the president-elect is using terminology from his TV-hosting days to describe the business of the White House.
In response to this cultural phenomena, Murphy did what he does best: reflect an American foible back at us through the bleakest mirror possible. His version of reality TV in American Horror Story Season 6 involved multiple murders, ghosts, and every horror trope under the sun. (Or should I say "under the blood moon"?) It's an exaggeration, sure, but there are also rough truths to be found. Addiction to fame is one, obviously. Whether it's millennials desperate for page views or performers frantically seeking the love of their audience, Murphy made it clear that the quest for fame, in this universe, has a fatal price. It's the Kardashian saga taken to its most extreme.
And in presenting his finale through a tapestry of various TV shows, Murphy was able to skewer a number of the ways in which Americans get suckered by what they see on the screen. The trial of Lee Harris (Adina Porter) is told primarily via a show called Crack'd—a tabloid news program in the vein of Hard Copy or, to be more contemporary, TMZ. There, we see footage of the real Lee used interchangeably with the re-enactments from earlier in the season, when the character of Lee was played by Angela Bassett. There's no attempt to distinguish between the two. In retelling Lee's story, the line between fact and fiction vanishes entirely. It should come as no surprise that the Lee character is eventually acquitted for the multiple murders she committed on camera. There's a strong implication that her reality-TV fame—she's described as "equally reviled and celebrated"—helped get her off the hook, the same way O.J. Simpson's celebrity played a large role in his own murder acquittal.
And how stunning, in the wake of the Access Hollywood controversy many thought would derail Trump's presidential chances, to see taped evidence of Lee's wrongdoing being ignored or downplayed by large swaths of the nation. In the end, Lee doesn't get a happily ever after. But, for a time, at least, she enjoys the success of having a "best-selling book" and becoming a "highly paid speaker" in the wake of her criminal scandal. It's an American tale that has become all too common.
But perhaps the most damning of all is the critique Ryan Murphy levels at us, the viewer. Four separate productions—My Roanoke Nightmare, Return to Roanoke, whatever those Blair Witch kids were doing, and Spirit Chasers—were filmed on the property Matt and Shelby bought, with dire and largely fatal consequences. But camera crews kept returning—because the public couldn't get enough. Each in-universe Roanoke show was more wildly popular than the last. The only thing that could stop the endless loop of filmed horrors was for Lee, that reviled and celebrated reality-TV star, to literally burn down the old white house. And that, perhaps, is the greatest election metaphor so far.

redpajama_set_name: RedPajamaCommonCrawl


A bumper podcast for you with Steven Mill and Ray Bradshaw! They have a the latest reaction to the Betfred Cup final and the Scotland Euro 2020 fixtures, plus a full round-up of the SPFL. Another chance to win some Hooch with Guess the Mark Yeardly, and a look ahead to the 2nd big Soccer FM derby of the season.


text:
All Business in Indiana
Eli Eban
Clarinetist
Musical Artists and Musical Artists: Woodwinds
Eli Eban was appointed principal clarinetist of the Jerusalem Symphony Orchestra under Lukas Foss upon graduating from the Curtis Institute of Music. Shortly thereafter he joined the Israel Philharmonic Orchestra at the invitation of Zubin Mehta. During 13 seasons with the Israel Philharmonic, he performed and recorded all the major orchestral repertoire with the world's leading conductors, including Claudio Abbado, Leonard Bernstein, Christoph von Dohnanyi, Charles Dutoit, Christoph Eschenbach, Kurt Masur, Simon Rattle, Georg Solti, Kryzstof Penderecki, and Lorin Maazel. Eban has been soloist with the Israel Philharmonic Orchestra, English Chamber Orchestra, Salzburg Camerata Academica, City of London Sinfonia, Chamber Orchestra of Philadelphia, Louisville Orchestra, and the Israel Camerata/Jerusalem. He has been guest artist with the Alexander, Audubon, Orion, St. Petersburg, Tel Aviv and Ying quartets and often was a participant of the Marlboro Music Festival. While at Marlboro, Eban was selected by legendary violinist Sandor Vegh to be the first wind player to perform at the prestigious IMS Prussia Cove festival in England, drawing acclaim from the London "Guardian" for his "high-powered, electrifying performances." His subsequent recordings for Meridian Records, London, were cited by critics as being "full of life and highly sensitive." He has also recorded for the Saphir, Crystal and Naxos labels. He was a member of "Myriad" (a chamber ensemble formed by members of the Cleveland Orchestra) and has often traveled to Eastern Europe to perform and teach at the invitation of the European Mozart Foundation. Eli Eban was a visiting professor at the Eastman School of Music for two years before joining the faculty of the Jacobs School of Music at Indiana University, where he holds a distinguished ranks' Rudy professorship. His former students are pursuing active solo careers and have won orchestral positions in Israel, Denmark, Korea, Poland, Singapore, and South Africa. In the USA they can be heard in the symphony orchestras of Indianapolis, New Mexico, Toledo, New World Symphony and in the premier service bands in Washington, D.C. He divides his time between teaching at the Jacobs School of Music, touring as a soloist & chamber musician, and serving as principal clarinetist of the Indianapolis Chamber Orchestra. His summers are spent performing and teaching at the Sarasota Music Festival and as principal clarinet in the Chautauqua Symphony Orchestra.


text:
«Highway 61 Revisited» es una canción escrita por Bob Dylan, que da título al álbum publicado en 1965 Highway 61 Revisited. Además, el 21 de diciembre de ese mismo año fue lanzada como cara B del sencillo "Can You Please Crawl Out Your Window?".

En 2004, la revista Rolling Stone incluyó la canción en su Lista de las 500 mejores canciones de todos los tiempos, en el puesto 364.

La Autopista 61 (Highway 61) va desde Duluth, Minnesota, donde creció Dylan, hasta Nueva Orleans. Era una ruta muy transitada por los afroamericanos que viajaban al norte hacia Chicago, Saint Louis y Memphis, siguiendo el valle del río Misisipi por más de 2250 km (1400 millas).

Letra

La canción tiene cinco estrofas. En cada una de ellas, un personaje describe un problema, que se resuelve finalmente en la Autopista 61.

En la primera estrofa, Dios ("God") le pide a Abraham que mate a su hijo en su nombre ("kill me a son"), y pide que lo haga en la Autopista 61. Alude claramente al Génesis 22:1-24, donde Dios pide a Abraham que mate a su único hijo Isaac para probar su fe. Además, Abraham es el nombre del padre de Dylan.

La segunda estrofa describe a un hombre, Georgia Sam, que pide ropa al departamento de asistencia social, pero como no se la dan, otro pobre, Howard, le dice que vaya a la Autopista 61.

La tercera estrofa trata del problema de "Mack the Finger" ("Mack el Dedo"), que tiene cuarenta cordones rojos, blancos y azules, y mil teléfonos que no suenan ("I got forty red white and blue shoe strings / And a thousand telephones that don't ring"). "Louie the King" soluciona el problema diciéndole que se deshaga de todo en la Autopista 61.

La cuarta estrofa trata sobre la "segunda madre" ("second mother") y el "séptimo hijo" ("seventh son"), ambos juntos en la Autopista 61.

La quinta y última estrofa trata de la historia de un jugador, que intenta crear la siguiente guerra mundial. Un promotor le dice que ponga algunas gradas al sol, y que lo haga en la Autopista 61. Detrás de este cuento sin sentido aparente se esconde una referencia política.

Hay una pausa en cada verso, donde Dylan espera que alguno de los sucesos de la historia termine. En la tercera estrofa, por ejemplo, la pausa está detrás de la frase "Y Louie the King dijo espera que piense un momento hijo" ("And Louie the King said let me think for a minute son").
Al principio de la canción, entre cada estrofa, y al final, se puede escuchar el efecto sonoro de Al Kooper imitando a una sirena de policía.

Versiones

 Johnny Winter, en su segundo álbum en 1969, Second Winter, y en The 30th Anniversary Concert Celebration, grabado en 1992 y lanzado al año siguiente.
 PJ Harvey, en su álbum de 1992 Rid of Me
 Billy Joel y Johnny Cash, para la película The Hunted.
 Karen O y The Million Dollar Bashers, para la banda sonora del biopic de Dylan de 2007 I'm Not There

Referencias

Enlaces externos
 "Highway 61 Revisited" Letra en bobdylan.com
  Review por William Ruhlmann en Allmusic

Canciones de Bob Dylan
Canciones de 1965

redpajama_set_name: RedPajamaWikipedia


text:
Cpl. Bonnie Brodie (Mrs. John F. Hassebroek)
Ten Sioux City Women Learn About Army in Waac Centers
Many Hope to See Service Abroad After Training
Studying war maps, military customs and courtesies, protection against air and chemical attacks and other phases of basic training, 10 Sioux City women are learning about army life in Waac training centers.
Auxiliary Bonnie Jean Brodie enlisted the day she became 21, which is the minimum age for entry. She hopes to serve as a nurse's aid overseas, so that she may serve close to her cousin, Lieut. Gordon Brodie of LeMars, Iowa, who is in an infantry division.
Source: The Sioux City Journal, March 9, 1943 (photo included)
Pvt. Bonnie Brodie, of the W.A.C., is spending an 11-day furlough in the home of her sister, Mrs. Harley Hall. Bonnie is stationed in Des Moines.
Source: LeMars Sentinel, July 30, 1943
PVT. BONNIE BRODIE PUT ON CORPORAL'S STRIPES LAST WEEK
Fort Des Moines, Ia. Oct. 4: Pfc. Bonnie Jean Brodie sewed on her Corporal's stripes this week, the reward of her excellent record as a member of the Women's Army Corps. Daughter of Mr. and Mrs. S. M. Brodie of LeMars, Private Brodie went to Administrative School in downtown Des Moines after completing basic training at first WAC Training Center, Fort Des Moines.
Pvt. Brodie is stationed in downtown Des Moines where she is acting supply sergeant for a WAC company at the Savory Hotel.
Her cousin, Lt. Gordon Brodie, is serving with the United States Army overseas.
Source: LeMars Globe-Post, October 4, 1943
Cpl. Bonnie B. Hassebroek, of LeMars, has been awarded the Good Conduct Medal. She is stationed at First WAC Training Center, Fort Des Moines, Iowa. This was one of the first awards of this nature ever presented to a WAC at Fort Des Moines. The Good Conduct medals were awarded to WACS who had shown exemplary behavior, efficiency and fidelity and who had served in the Army for a year or more. Since no medals are presented by the War Department during the war, the WACS received red and white striped ribbons and will be awarded the medals when the war ends. The ribbons are worn over the upper left pocket.
Source: LeMars Semi-Weekly Sentinel, March 21, 1944
An Army headquarters release from England states that Corporal Bonnie (Brodie) Hassebroek of LeMars is among members of the first mobile army unit, composed entirely of WACS. They are the first WACS assigned to a tactical army ground forces unit in England, and will follow hard on the heels of the invasion troops as they chase Hitler's armies in the general direction of Berlin.
Source: LeMars Globe-Post, June 22, 1944
RIPLEY SPOTS 'EM -- M. Sgt John F. Hassebroek of Buffalo Center and his WAC wife recently were subjects of one of Ripley's "Believe it not" marvels. The sergeant received a 3-day pass to marry a WAC corporal at Des Moines where both were stationed. When he went overseas, they lost track of each other until one night in France he made a convoy to the front lines and went to a farm house to spend the night. There he met his wife on the exact day and hour of their first wedding anniversary. His wife's home was at LeMars. ~Note his wife was the former, Bonnie Brodie.
Source: Mason City Globe-Gazette, April 4, 1945 (photo of the couple included)
Transcribed by Linda Ziemann, May 2013; updated Nov 2015; updated Sep 2017 by WWII researcher, Terry Hirsch; updated Jul 2019 by C. Swearingen


One of the most iconic scenes in Scotland. It's simply impossible to drive through Glencoe without stopping to photograph this mountain and the ones beyond! The best view however is from the top!
This guided walk will take you to the summit and bags the two Munros, Stob Dearg and Stob na Bròige.
Total distance: 13km, climb of 1100m, taking approximately 8 hrs.
A chance to bag another two Munros, Creise and Meall a'Bhuiridh. These Munros offer excellent views to the neighbouring hills and down Glen Etive.
Total distance: 11km, height gain of 1090m, taking approximately 7 hours.
Please select each day you wish to walk with us when booking. Cost for each day £60, if you book both days you will be refunded £20! Limited to 6 places per day. Detailed joining instructions with kit advice will be emailed closer to the date.
Contact us for more information. Campsite, hostels and hotels near by too.


text:
Zooanzoo
Interview by Nicki Stein
Issue 66 • August 2018 • Harrisonburg
Experimental and eccentric, this electronic artist brings a freewheeling playfulness to his rich soundscapes—and most other parts of his life.
"Since I can remember I always wanted to be a cartoon character," Zach Williams earnestly explained. "I always connected to them more." As the artist at the helm of ethereal trip-hop project Zooanzoo, Williams—otherwise known as Zoo—creates small worlds out of sound for his listeners to inhabit, offering up melodies that can be worn like a second skin. The pulsating rhythms of Zooanzoo's frenzied, hypnotic refrains allow the swirling thoughts of a buzzing brain to melt away, the contents dripping like a sweaty-sweet popsicle onto hot pavement, the candy-colored liquid sizzling for a moment, then evaporating into the throbbing summer heat.
Zach "Zoo" Williams
Zoo began experimenting as a solo artist around 2012, but his first connection to music extends back to his childhood, standing next to his grandmother in church singing Southern gospel. "Belting out songs with her signature vibrato," he remembered fondly, "that's my first contact high." As a kid, Zoo picked up any instrument-like object he could get his hands on. "The first time I held a guitar was my sister's," he recounted. "I went to experiment with the tuning and snapped a string," he giggled at the memory, but that spirit of experimentation is clearly threaded through Zoo's exploratory style of music-making.
In middle school, Zoo found punk and electronic music. From there, he continued to experiment sonically, sometimes with the help of his partner Josh Hebdon (AKA Sprit) on drums, until the current iteration of Zooanzoo's sound announced itself. "Creating under Zooanzoo came about as a necessity," Zoo said. The name he performs by, although originally bestowed upon him as a teasing nickname by classmates, is now a core part of Zoo's identity. "I identify with Zoo more than my birth name," he said simply, "It seems even closer to my true self." Zoo toys with a sense of escapism by cloaking himself under the blanket of that better-fitting identity and tucking the warmth he's pulled from it into the fabric of the songs he writes.
Josh "Sprit" Hebdon
After years of honing his sound, experimenting with operatic overtures, toy instruments, and recorded voice memos, Zooanzoo released an official full-length album, LOUD MOUTH, in 2017. "That's when it began to feel a bit more real, or intentional," Zoo said, referring to his career as a professional musician. "It was like I wasn't the only person who resonated with the tones I was creating. It reached out, connected me to people."
Building on the success of the first release, Zooanzoo's second full-length NECK OUT was released late last month on July 27 from Beau Travail Records. "A lot of the songs that are created through me have two souls in them," Zoo said of the release. "The album has that dichotomy: a tender, awfully sensitive side that wants to heal and hold, make love, and recognize our mortal vulnerabilities. The other side is a loud mouth growling the twisted gospel, rather spit acid than drink OJ, vulgar, over the top, hound toothed, bat eyed, demon."
Cover art for NECK OUT
The songs on NECK OUT do seem to tug at opposing poles: the soft plunking of acoustic guitar strings that open the album's first song "Dionysus" are a soothing lullaby, blending with the ethereal drone of high melodic voices like a ghostly choir. Then, a few bars in, the tune turns dark, almost cinematic in its chiaroscuro, exaggerating the shadowy side of Zoo's rhythmic vocal patter. Contradictions abound, but that's what makes it interesting.
In addition to writing and performing the music of NECK OUT, Zoo animated a music video to accompany the single "May I?". He used a technique called rotoscoping where the animator traces over video footage frame-by-frame to create a fluid, live-action effect. "Animation frame-by-frame will drive you mad," he said with a smile, "but the payoff is totally worth it." In the video, a cartoon version of Zoo swims across the screen as bright shapes, some recognizable and some not, squiggle along to a smooth beat. Pictures swirl into one another, characters dance in barren landscapes and flail their arms. The imagery is joyus but jittery, and best of all, Zoo is himself a cartoon character at long last.
As for his next moves, Zoo shows no sign of slowing down. With a collaborative album in the works, as well as a record release show to look forward to, Zoo's life as a performer continues to pick up momentum. But even amidst the hubbub of his unfurling career, Zoo thinks of music as a healing force: "This energy that strengthens me, teaches me even, allows me to escape," he explained. On NECK OUT, Zoo has tried to stay true to that sense of emotional vulnerability inherent in making and listening to music. "That moment your spirit may evaporate or explode from pure honesty and vulnerability. To strip away the skin, the sensation of being naked, exposing yourself to the sun—or judgement. Peeking from the shell, taking those risks ..." he mused. "It's what drives humanity along. It's what carries this album."
Zooanzoo will perform an album release show on Friday, August 3 at Restless Moons Brewing in Harrisonburg. NECK OUT is available on vinyl at zooanzoo.bandcamp.com.
Photography by Alex Kent
The Whurk Week
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Whurk Newsletter
More From Issue 66
Southern Grit
What started as a frank conversation about food has shifted to a frank conversation about culture—but are the two really all that different?
Life's Work
Comic by Jamie Douglas
I Hate Long Good-byes
Cover by Nick Candela
More Interview Features
First Time's the Charm RVA
Seasoned musicians and neophytes alike team up to rock out in this annual showcase of new bands, new music, and traditionally marginalized voices.
Charlottesville Playwrights Collective
When these writers couldn't find the right organization to produce their work, they did what any self-respecting creative group would do — team up with a local venue and do it themselves.
Richmond Zine Fest
Celebrating populist self-expression in all its analog glory, this DIY festival holds the zine torch high.
Other Recent Stuff
When he left the world of finance to become a chef, it wasn't because he needed a break—he just wanted to apply his fast-paced approach to life to a more fulfilling pursuit.
VMFA on the Road
To bring art to the most Virginians possible, this museum is reviving a long dormant outreach program. The game plan? Just keep on truckin'.
Samhaim
Comic by Christine Skelly
Whurk Staff
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You will soon receive an updated Health Benefits ID card from Express Scripts listing your new $0 co-payments. If you need to see your healthcare provider before you receive your new card, ask him or her to call the Benefit Fund to verify the co-payment amount.
If you paid any co-payments for these visits or services on or after January 1, we will reimburse you for the amount you paid. Contact our Member Services Department at (646) 473-9200 or MemberServices@1199Funds.org and we will issue you a check for the co-payment amount. Please provide your Health Benefits ID card number, patient name and date of the visit, and, if emailing, indicate "GNY Co-pay Reimbursement" in the subject field.
If you have any questions, or if you don't receive your Health Benefits ID card by March 1, please call our Member Services Department at (646) 473-9200 from 8:00 am to 6:00 pm, Monday through Friday.


Ronda Rousey might not be on Monday Night Raw every week, but she's likely going to be right in the mix for the main event at WrestleMania.
After her debut at the Royal Rumble, Rousey beat up Triple H at the Elimination Chamber and eventually settled for a mixed tag team match with Kurt Angle against the King of Kings and his wife, Stephanie McMahon.
Rousey's transition into WWE from the world of mixed martial arts has yet to be fully seen, but boxing legend and one-time WrestleMania enforcer (and actual WWE Hall of Famer) Mike Tyson thinks she'll do just fine.