## 1 Introduction

Let $$\mathcal Q\subset \mathbb {R}^3$$ denote the three-dimensional second-order cone (also known as the “ice-cream” cone or the Lorentz cone):

\begin{aligned} \mathcal Q= \{ (x,t) \in \mathbb {R}^2 \times \mathbb {R}: \Vert x\Vert \le t \}. \end{aligned}

It is known that $$\mathcal Q$$ is linearly isomorphic to the cone of $$2\times 2$$ real symmetric positive semidefinite matrices. Indeed we have:

\begin{aligned} (x_1,x_2,t) \in \mathcal Q\quad \Longleftrightarrow \quad \begin{bmatrix} t-x_1&\quad x_2\\ x_2&\quad t+x_1\end{bmatrix} \succeq 0. \end{aligned}
(1)

Despite its apparent simplicity the second-order cone $$\mathcal Q$$ has strong expressive abilities and allows us to represent various convex constraints that go beyond “simple quadratic constraints”. For example it can be used to express geometric means ($$x\mapsto \prod _{i=1}^n x_i^{p_i}$$ where $$p_i \ge 0$$, rational, and $$\sum _{i=1}^n p_i = 1$$), $$\ell _p$$-norm constraints, multifocal ellipses (see e.g., [11, Equation (3.5)]), robust counterparts of linear programs, etc. We refer the reader to [4, Section 3.3] for more details.

Given this strong expressive ability one may wonder whether the general positive semidefinite cone can be represented using $$\mathcal Q$$. This question was posed in particular by Adrian Lewis (personal communication) and Glineur, Parrilo and Saunderson [7]. In this paper we show that this is not possible, even for the $$3\times 3$$ positive semidefinite cone. To make things precise we use the language of lifts (or extended formulations), see [8]. We denote by $$\mathcal Q^k$$ the Cartesian product of k copies of $$\mathcal Q$$:

\begin{aligned} \mathcal Q^k = \mathcal Q\times \dots \times \mathcal Q\quad \text {({ k} copies)}. \end{aligned}

A linear slice of $$\mathcal Q^k$$ is an intersection of $$\mathcal Q^k$$ with a linear subspace. We say that a convex cone $$K \subset \mathbb {R}^m$$ has a second-order cone lift of size k (or simply $$\mathcal Q^k$$-lift) if it can be written as the projection of a slice of $$\mathcal Q^k$$, i.e.:

\begin{aligned} K = \pi \left( \mathcal Q^k \cap L\right) \end{aligned}
(2)

where $$\pi :\mathbb {R}^{3k} \rightarrow \mathbb {R}^m$$ is a linear map and L is a linear subspace of $$\mathbb {R}^{3k}$$. Let $$\mathbf {S}^n_+$$ be the cone of $$n\times n$$ real symmetric positive semidefinite matrices. In this paper we prove:

### Theorem 1

The cone $$\mathbf {S}^3_+$$ does not admit any $$\mathcal Q^k$$-lift for any finite k.

Actually our proof allows us to show that the slice of $$\mathbf {S}^3_+$$ consisting of Hankel matrices does not admit any second-order representation (see Sect. 4 for details). Note that higher-dimensional second order cones of the form

\begin{aligned} \{(x,t) \in \mathbb {R}^{n} \times t : \Vert x\Vert \le t\} \end{aligned}

where $$n \ge 3$$ can be represented using the three-dimensional cone $$\mathcal Q$$, see e.g., [5, Section 2]. Thus Theorem 1 also rules out any representation of $$\mathbf {S}^3_+$$ using the higher-dimensional second-order cones. Moreover since $$\mathbf {S}^3_+$$ appears as a slice of higher-order positive semidefinite cones Theorem 1 also shows that one cannot represent $$\mathbf {S}^n_+$$, for $$n \ge 3$$ using second-order cones.

## 2 Preliminaries

The paper [8] introduced a general methodology to prove existence or nonexistence of lifts in terms of the slack matrix of a cone. In this section we review some of the definitions and results from this paper, and introduce the notion of a second-order cone factorization and the second-order cone rank.

Let E be a Euclidean space with inner product $$\langle \cdot , \cdot \rangle$$ and let $$K \subseteq E$$ be a cone. The dual cone $$K^*$$ is defined as:

\begin{aligned} K^* = \{x \in E: \langle x, y \rangle \ge 0 \quad \forall y \in K \}. \end{aligned}

We also denote by $${{\mathrm{ext}}}(K)$$ the extreme rays of a cone K. The notion of slack matrix plays a fundamental role in the study of lifts.

### Definition 1

(Slack matrix) The slack matrix of a cone K, denoted $$S_K$$, is a (potentially infinite) matrix where columns are indexed by extreme rays of K, and rows are indexed by extreme rays of $$K^*$$ (the dual of K) and where the (xy) entry is given by:

\begin{aligned} S_K[x,y] = \langle x,y \rangle \quad \forall (x,y) \in {{\mathrm{ext}}}(K^*) \times {{\mathrm{ext}}}(K). \end{aligned}
(3)

Note that, by definition of dual cone, all the entries of $$S_K$$ are nonnegative. Also note that an element $$x \in {{\mathrm{ext}}}(K^*)$$ (and similarly $$y \in {{\mathrm{ext}}}(K)$$) is only defined up to a positive multiple. Any choice of scaling gives a valid slack matrix of K and the properties of $$S_K$$ that we are interested in will be independent of the scaling chosen.

The existence/nonexistence of a second-order cone lift for a convex cone K will depend on whether $$S_K$$ admits a certain second-order cone factorization which we now define.

### Definition 2

($$\mathcal Q^k$$-factorization and second-order cone rank) Let $$S \in \mathbb {R}^{|I|\times |J|}$$ be a matrix with nonnegative entries. We say that S has a $$\mathcal Q^k$$-factorization if there exist vectors $$a_i \in \mathcal Q^k$$ for $$i\in I$$ and $$b_j \in \mathcal Q^k$$ for $$j \in J$$ such that $$S[i,j] = \langle a_i, b_j \rangle$$ for all $$i \in I$$ and $$j \in J$$. The smallest k for which such a factorization exists will be denoted $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(S)$$.

### Remark 1

Recall that for any $$a,b \in \mathcal Q$$ we have $$\langle a, b \rangle \ge 0$$. This means that any matrix with a second-order cone factorization is elementwise nonnegative.

### Remark 2

It is important to note that the second-order cone rank of any matrix S can be equivalently expressed as the smallest k such that S admits a decomposition

\begin{aligned} S = M_1 + \dots + M_k \end{aligned}
(4)

where $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(M_l) = 1$$ for each $$l=1,\dots ,k$$ (i.e., each $$M_l$$ has a factorization $$M_l[i,j] = \langle a_i, b_j \rangle$$ where $$a_i,b_j \in \mathcal Q$$). This simply follows from the fact that $$\mathcal Q^k$$ is the Cartesian product of k copies of $$\mathcal Q$$.

We now state the result from [8] that we will need.

### Theorem 2

(Existence of a lift, special case of [8]) Let K be a convex cone. If K has a $$\mathcal Q^k$$-lift then its slack matrix $$S_K$$ has a $$\mathcal Q^k$$-factorization.

This theorem can actually be turned into an if and only if condition under mild conditions on K (e.g., K is proper), see [8], but we have only stated here the direction that we will need.

The cone $$\mathbf {S}^3_+$$ In this paper we are interested in the cone $$K = \mathbf {S}^3_+$$ of real symmetric $$3\times 3$$ positive semidefinite matrices. The extreme rays of $$\mathbf {S}^3_+$$ are rank-one matrices of the form $$xx^T$$ where $$x \in \mathbb {R}^3$$. Also $$\mathbf {S}^3_+$$ is self-dual, i.e., $$(\mathbf {S}^3_+)^* = \mathbf {S}^3_+$$. The slack matrix of $$\mathbf {S}^3_+$$ thus has its rows and columns indexed by three-dimensional vectors and

\begin{aligned} S_{\mathbf {S}^3_+}[x,y] = \langle xx^T, yy^T \rangle = \left( x^T y\right) ^2 \quad \forall (x,y) \in \mathbb {R}^3 \times \mathbb {R}^3. \end{aligned}
(5)

In order to prove that $$\mathbf {S}^3_+$$ does not admit a second-order representation, we will show that its slack matrix does not admit any $$\mathcal Q^k$$-factorization for any finite k. In fact we will exhibit a sequence $$(A_n)$$ of submatrices of $$S_{\mathbf {S}^3_+}$$ where $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(A_n)$$ grows to $$+\infty$$ as $$n\rightarrow +\infty$$.

Before introducing this sequence of matrices we record the following simple (known) proposition concerning orthogonal vectors in the cone $$\mathcal Q$$ which will be useful later.

### Proposition 1

Let $$a,b_1,b_2 \in \mathcal Q$$ nonzero and assume that $$\langle a,b_1 \rangle = \langle a,b_2 \rangle = 0$$. Then $$b_1$$ and $$b_2$$ are collinear.

### Proof

This is easy to see geometrically by visualizing the “ice cream” cone. We include a proof for completeness: let $$a = (a',t) \in \mathbb {R}^2 \times \mathbb {R}$$ and $$b_i = (b'_i,s_i) \in \mathbb {R}^2 \times \mathbb {R}$$ where $$\Vert a'\Vert \le t$$ and $$\Vert b'_i\Vert \le s_i$$. Note that for $$i=1,2$$ we have $$0 = \langle a, b_i \rangle = \langle a', b'_i \rangle + t s_i \ge - \Vert a'\Vert \Vert b'_i\Vert + t s_i \ge 0$$ where in the first inequality we used Cauchy-Schwarz and in the second inequality we used the definition of the second-order cone. It thus follows that all the inequalities must be equalities: by the equality case in Cauchy-Schwarz we must have that $$b'_i = \alpha _i a'$$ for some constant $$\alpha _i < 0$$ and we must also have $$t = \Vert a'\Vert$$ and $$s_i = \Vert b'_i\Vert$$. Thus we get that $$b_i = (\alpha _i a', |\alpha _i| \Vert a'\Vert ) = |\alpha _i| (-a',\Vert a'\Vert )$$. This shows that $$b_1$$ and $$b_2$$ are both collinear to the same vector $$(-a',\Vert a'\Vert )$$ and thus completes the proof. $$\square$$

## 3 Proof of Theorem 1

A sequence of matrices We now define our sequence $$A_n$$ of submatrices of the slack matrix of $$\mathbf {S}^3_+$$. For any integer i define the vector

\begin{aligned} v_i = (1,i,i^2) \in \mathbb {R}^3. \end{aligned}
(6)

Note that this sequence of vectors satisfies the following:

\begin{aligned} \text {For all distinct integers } i_1,i_2,i_3 \; \det (v_{i_1},v_{i_2},v_{i_3}) \ne 0. \end{aligned}
(7)

Our matrix $$A_n$$ has size $$\left( {\begin{array}{c}n\\ 2\end{array}}\right) \times n$$ and is defined as follows (rows are indexed by 2-subsets of [n] and columns are indexed by [n]):

\begin{aligned} A_n[\{i_1,i_2\},j] := & {} \left( ( v_{i_1} \times v_{i_2})^T v_j \right) ^2 \nonumber \\= & {} \det (v_{i_1},v_{i_2},v_j)^2 \quad \forall \{i_1,i_2\} \in \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) , \; \forall j \in [n] \end{aligned}
(8)

where $$\times$$ denotes the cross-product of three-dimensional vectors. It is clear from the definition of $$A_n$$ that it is a submatrix of the slack matrix of $$\mathbf {S}^3_+$$. Note that the sparsity pattern of $$A_n$$ satisfies the following:

\begin{aligned} \begin{aligned}&A_n[e,j] = 0\quad \text {if } j \in e\\&A_n[e,j] > 0\quad \text {otherwise} \end{aligned} \qquad e \in \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) , \; j \in [n]. \end{aligned}
(9)

Also note that $$A_n$$ satisfies the following important recursive property: for any subset C of [n] of size $$n_0$$ the submatrix $$A_n[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C]$$ has the same sparsity pattern as $$A_{n_0}$$ (up to relabeling of rows and columns). In our main theorem we will show that the second-order cone rank of $$A_n$$ grows to infinity with n.

### Remark 3

(Geometric interpretation of (9)) The property (9) of the matrices $$A_n$$ will be the key to prove a lower bound on their second-order cone rank. Geometrically, the property (9) reflects a certain 2-neighborliness property of the extreme raysFootnote 1$${{\mathrm{ext}}}(\mathbf {S}^3_+)$$ of $$\mathbf {S}^3_+$$ : for any two distinct extreme rays $$xx^T$$ and $$yy^T$$ of $$\mathbf {S}^3_+$$, there is a supporting hyperplane H to $$\mathbf {S}^3_+$$ that touches $${{\mathrm{ext}}}(\mathbf {S}^3_+)$$ precisely at $$xx^T$$ and $$yy^T$$. This 2-neighborliness property turns out to be the key geometric obstruction for the existence of second-order cone lifts for $$\mathbf {S}^3_+$$.

Covering numbers Our analysis of the matrix $$A_n$$ will only rely on its sparsity pattern. Given two matrices A and B of the same size we write $$A \overset{supp}{=}B$$ if A and B have the same support (i.e., $$A_{ij} = 0$$ if and only if $$B_{ij} = 0$$ for all ij). We now define a combinatorial analogue of the second-order cone rank:

### Definition 3

Given a nonnegative matrix A, we define the $$soc$$-covering number of A, denoted $${{\mathrm{cov_{soc}}}}(A)$$ to be the smallest number k of matrices $$M_1,\dots ,M_k$$ with $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(M_l) = 1$$ for $$l=1,\dots ,k$$ that are needed to cover the nonzero entries of A, i.e., such that

\begin{aligned} A \overset{supp}{=}M_1+\dots +M_k. \end{aligned}
(10)

### Proposition 2

For any nonnegative matrix A we have $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(A) \ge {{\mathrm{cov_{soc}}}}(A)$$.

### Proof

This follows immediately from Remark 2 concerning $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}$$ and the definition of $${{\mathrm{cov_{soc}}}}$$. $$\square$$

A simple but crucial fact concerning $$soc$$-coverings that we will use is the following: in any $$soc$$-covering of A of the form (10), each matrix $$M_l$$ must satisfy $$M_l[i,j] = 0$$ whenever $$A[i,j] = 0$$. This is because the matrices $$M_1,\dots ,M_k$$ are all entrywise nonnegative.

We are now ready to state our main result.

### Theorem 3

Consider a sequence $$(A_n)$$ of matrices of sparsity pattern given in (9). Then for any $$n_0 \ge 2$$ we have $${{\mathrm{cov_{soc}}}}(A_{3n_0^2}) \ge {{\mathrm{cov_{soc}}}}(A_{n_0})+1$$. As a consequence $${{\mathrm{cov_{soc}}}}(A_{n})\rightarrow +\infty$$ when $$n\rightarrow +\infty$$.

The proof of our theorem rests on a key lemma concerning the sparsity pattern of any term in a $$soc$$-covering of $$A_n$$.

### Lemma 1

(Main) Let n be such that $$n \ge 3n_0^2$$ for some $$n_0 \ge 2$$. Assume $$M \in \mathbb {R}^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) \times n}$$ satisfies $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(M) = 1$$ and $$M[e,j] = 0$$ for all $$e\in \left( {\begin{array}{c}n\\ 2\end{array}}\right)$$ and $$j \in [n]$$ such that $$j \in e$$. Then there is a subset C of [n] of size at least $$n_0$$ such that the submatrix $$M[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C]$$ is identically zero.

Before proving this lemma, we show how this lemma can be used to easily prove Theorem 3.

### Proof of Theorem 3

Let $$n = 3n_0^2$$ and consider a $$soc$$-covering of $$A_n \overset{supp}{=}M_1 + \dots + M_r$$ of size $$r = {{\mathrm{cov_{soc}}}}(A_n)$$ (note that we have of course $$r \ge 1$$ since $$A_n$$ is not identically zero). By Lemma 1 there is a subset C of [n] of size $$n_0$$ such that $$M_1[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] = 0$$. It thus follows that we have $$A_n[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] \overset{supp}{=}M_2[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] + \dots + M_r[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C]$$. Also note that $$A_n[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] \overset{supp}{=}A_{n_0}$$. It thus follows that $$A_{n_0}$$ has a $$soc$$-covering of size $$r-1$$ and thus $${{\mathrm{cov_{soc}}}}(A_{n_0}) \le {{\mathrm{cov_{soc}}}}(A_{3n_0^2})-1$$. This completes the proof. $$\square$$

For completeness we show how Theorem 1 follows directly from Theorem 3.

### Proof of Theorem 1

Since for any $$n \ge 1$$, $$A_n$$ is a submatrix of the slack matrix of $$\mathbf {S}^3_+$$, Theorem 3 shows that the slack matrix of $$\mathbf {S}^3_+$$ does not admit any $$\mathcal Q^k$$-factorization for finite k. This shows, via Theorem 2, that $$\mathbf {S}^3_+$$ does not have a $$\mathcal Q^k$$-lift for any finite k. $$\square$$

The rest of the section is devoted to the proof of Lemma 1.

### Proof of Lemma 1

Let $$M \in \mathbb {R}^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) \times n}$$ and assume that M has a factorization $$M_{e,j} = \langle a_e, b_j \rangle$$ where $$a_e, b_j \in \mathcal Q$$ for all $$e \in \left( {\begin{array}{c}[n]\\ 2\end{array}}\right)$$ and $$j \in [n]$$, and that $$M_{e,j} = 0$$ whenever $$j \in e$$.

Let $$E_0 := \{e \in \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) : a_e = 0\}$$ be the set of rows of M that are identically zero and let $$E_1 = \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) \setminus E_0$$. Similarly for the columns we let $$S_0 := \{j \in [n] : b_j = 0\}$$ and $$S_1 = [n] \setminus S_0$$.

In the next lemma we use the sparsity pattern of $$A_n$$ together with Proposition 1 to infer additional properties on the sparsity pattern of M.

### Lemma 2

Let C be a connected component of the graph with vertex set $$S_1$$ and edge set $$E_1(S_1)$$ (where $$E_1(S_1)$$ consists of elements in $$E_1$$ that connect only elements of $$S_1$$). Then necessarily $$M[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] = 0$$.

### Proof

We first show using Proposition 1 that all the vectors $$\{b_j\}_{j \in C}$$ are necessarily collinear. Let $$j_1,j_2 \in S_1$$ such that $$e = \{j_1,j_2\} \in E_1$$. Note that since $$M_{e,j_1} = M_{e,j_2} = 0$$ then we have, by Proposition 1 that $$b_{j_1}$$ and $$b_{j_2}$$ are collinear. It is easy to see thus now that if $$j_1$$ and $$j_2$$ are connected by a path in the graph $$(S_1,E_1(S_1))$$ then $$b_{j_1}$$ and $$b_{j_2}$$ must be collinear.

We thus get that all the columns of M indexed by C must be proportional to each other, and so they must have the same sparsity pattern. Now let $$e \in \left( {\begin{array}{c}C\\ 2\end{array}}\right)$$. If $$a_e = 0$$ then $$M[e,C] = 0$$ since the entire row indexed by e is zero. Otherwise if $$a_e \ne 0$$ let $$e=\{j_1,j_2\}$$ with $$j_1,j_2 \in C$$. Since, by assumption, $$M_{e,j_1} = 0$$ it follows that for any $$j \in C$$ we must have $$M_{e,j} = 0$$, i.e., $$M[e,C] = 0$$. This is true for any $$e \in \left( {\begin{array}{c}C\\ 2\end{array}}\right)$$ thus we get that $$M[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] = 0$$. $$\square$$

To complete the proof of Lemma 1 assume that $$n \ge 3n_0^2$$ for some $$n_0 \ge 2$$. We need to show that there is a subset C of [n] of size at least $$n_0$$ such that $$M[\left( {\begin{array}{c}C\\ 2\end{array}}\right) ,C] = 0$$.

First note that if the graph $$(S_1,E_1(S_1))$$ has a connected component of size at least $$n_0$$ then we are done by Lemma 2. Also note that if $$S_0$$ has size at least $$n_0$$ we are also done because all the columns indexed by $$S_0$$ are identically zero by definition.

In the rest of the proof we will thus assume that $$|S_0| < n_0$$ and that the connected components of $$(S_1,E_1(S_1))$$ all have size $$< n_0$$. We will show in this case that $$E_0$$ necessarily contains a clique of size at least $$n_0$$ (i.e., a subset of the form $$\left( {\begin{array}{c}C\\ 2\end{array}}\right)$$ where $$|C| \ge n_0$$) and this will prove our claim since all the rows in $$E_0$$ are identically zero by definition. The intuition is as follows: the assumption that $$|S_0| < n_0$$ and that the connected components of $$(S_1,E_1(S_1))$$ have size $$< n_0$$ mean that the graph $$(S_1,E_1(S_1))$$ is very sparse. In particular this means that $$E_1$$ has to be small which means that $$E_0 = E_1^c$$ must be large and thus it must contain a large clique.

More precisely, to show that $$E_1$$ is small note that it consists of those edges that are either in $$E_1(S_1)$$ or, otherwise, they must have at least one node in $$S_1^c = S_0$$. Thus we get that

\begin{aligned} |E_1| \le |E_1(S_1)| + |S_0|(n-1) \le |E_1(S_1)| + (n_0-1) (n-1). \end{aligned}

where in the second inequality we used the fact that $$|S_0| < n_0$$. Also since the connected components of $$(S_1,E_1(S_1))$$ all have size $$<n_0$$ it is not difficult to show that $$|E_1(S_1)| < n_0 n/2$$ (indeed if we let $$x_1,\dots ,x_k$$ be the size of each connected component we have $$|E_1(S_1)| \le \frac{1}{2} \sum _{i=1}^k x_i^2 < \frac{1}{2} \sum _{i=1}^k n_0 x_i \le \frac{1}{2} n_0 n$$). Thus we get that

\begin{aligned} |E_1| \le \frac{n_0 n}{2} + (n_0 - 1)(n-1) \le \left( \frac{3}{2} n_0 - 1\right) n \end{aligned}

Thus this means, since $$E_0 = \left( {\begin{array}{c}[n]\\ 2\end{array}}\right) \setminus E_1$$:

\begin{aligned} |E_0| \ge \left( {\begin{array}{c}n\\ 2\end{array}}\right) - \left( \frac{3}{2} n_0 - 1\right) n > \frac{n^2}{2} - \frac{3}{2} n_0 n \end{aligned}

We now invoke a result of Turán to show that $$E_0$$ must contain a clique of size at least $$n_0$$:

### Theorem 4

(Turán, see e.g., [2]) Any graph on n vertices with more than $$\left( 1-\frac{1}{k}\right) \frac{n^2}{2}$$ edges contains a clique of size $$k+1$$.

By taking $$k = n_0-1$$ we see that $$E_0$$ contains a clique of size $$n_0$$ if

\begin{aligned} \frac{n^2}{2} - \frac{3}{2} n_0 n > \left( 1-\frac{1}{n_0-1}\right) \frac{n^2}{2} \end{aligned}

This simplifies into

\begin{aligned} n > 3n_0(n_0-1) \end{aligned}

which is true for $$n \ge 3n_0^2$$. $$\square$$

## 4 Slices of the $$3\times 3$$ positive semidefinite cone

Hankel slice The proof given in the previous section actually shows the following more general statement.

### Theorem 5

Let $$\mathcal { H}$$ denote the cone of $$3\times 3$$ positive semidefinite Hankel matrices:

\begin{aligned} \mathcal { H}= \{X \in \mathbf {S}^3_+ : X_{13} = X_{22}\}. \end{aligned}

Assume K is a convex cone that is “sandwiched” between $$\mathcal { H}$$ and $$\mathbf {S}^3_+$$, i.e., $$\mathcal { H}\subseteq K \subseteq \mathbf {S}^3_+$$. Then K does not have a second-order cone representation.

### Proof

The proof follows from the observation that the matrices $$A_n$$ considered in Sect. 3 (see Eq. (8)) are actually submatrices of the generalized slack matrix of the pair of nested cones $$(\mathcal { H},\mathbf {S}^3_+)$$, the definition of which we now recall (see e.g., [9, Definition 6]): The generalized slack matrix of a pair of convex cones $$(K_1,K_2)$$ with $$K_1 \subseteq K_2$$ is a matrix whose rows are indexed by $${{\mathrm{ext}}}(K_2^*)$$ (the valid linear inequalities of $$K_2$$) and its columns indexed by $${{\mathrm{ext}}}(K_1)$$ and is defined by

\begin{aligned} S_{K_1,K_2}[x,y] = \langle x, y \rangle \quad x \in {{\mathrm{ext}}}(K_2^*), y \in {{\mathrm{ext}}}(K_1). \end{aligned}

When $$K_1 = K_2$$ this is precisely the slack matrix of $$K_1=K_2$$. The following theorem is a generalization of Theorem 2 and can be proved using very similar arguments (see e.g., [9, Proposition 7]).

### Theorem 6

(Generalization of Theorem 2 to nested cones) Let $$K_1,K_2$$ be two convex cones with $$K_1 \subseteq K_2$$, and assume there exists a convex cone K with a $$\mathcal Q^k$$-lift such that $$K_1 \subseteq K \subseteq K_2$$. Then $$S_{K_1,K_2}$$ has a $$\mathcal Q^k$$-factorization.

The main observation is to see that the vector $$v_i$$ defined in (6) satisfies $$v_i v_i^T \in \mathcal { H}$$, and so this shows that $$A_n$$ defined in Equation (8) is a submatrix of the generalized slack matrix of the pair $$(\mathcal { H},\mathbf {S}^3_+)$$. Since $${{\mathrm{{{\mathrm{rank}}}_{soc}}}}(A_n)$$ grows to infinity with n, this gives the desired result. $$\square$$

The dual cone of $$\mathcal { H}$$ is the cone of nonnegative quartic polynomials on the real line (see e.g., [3, Section 3.5]). It thus follows that the latter is also not second-order cone representable using the second-order cone. More generally we can prove:

### Corollary 1

Let $$\Sigma _{n,2d}$$ be the cone of polynomials in n variables of degree at most 2d that are sums of squares. Then $$\Sigma _{n,2d}$$ is not second-order cone representable except in the case $$(n,2d)=(1,2)$$.

For the proof we recall that in the cases $$n=1$$ (univariate polynomials) and $$2d=2$$ (quadratic polynomials), nonnegative polynomials are sums of squares.

### Proof

For $$2d=2$$, $$\Sigma _{n,2d}$$ is the cone of nonnegative quadratic polynomials in n variables. By homogenization, this cone is linearly isomorphic to $$\mathbf {S}^{n+1}_+$$, the cone of nonnegative quadratic forms in $$n+1$$ variables. By Theorem 1, this shows that $$\Sigma _{n,2}$$ is not second-order cone representable for $$n \ge 2$$. The case $$(n,2d)=(1,2)$$ is clearly second-order cone representable because $$\mathbf {S}^2_+$$ is linearly isomorphic to the second-order cone.

If $$2d \ge 4$$ then the cone of nonnegative quartic polynomials on the real line can be obtained as a section of $$\Sigma _{n,2d}$$ by setting to zero the coefficients of some appropriate monomials. This shows that $$\Sigma _{n,2d}$$ is not second-order cone representable when $$2d \ge 4$$. $$\square$$

Other slices of$$\mathbf {S}^3_+$$that are second-order cone representable There are certain slices of $$\mathbf {S}^3_+$$ of codimension 1 that are, on the other hand, known to admit a second-order cone representation. For example the following second-order cone representation of the slice $$\{X \in \mathbf {S}^3_+ : X_{11} = X_{22}\}$$ appears in [7]:

\begin{aligned} \begin{bmatrix} t&\quad a&\quad b\\ a&\quad t&\quad c\\ b&\quad c&\quad s \end{bmatrix} \succeq 0 \quad&\Longleftrightarrow \quad \exists u, v \in \mathbb {R}\text { s.t. } \begin{bmatrix} t+a&\quad b+c\\ b+c&\quad u\end{bmatrix} \succeq 0, \nonumber \\&\quad \quad \quad \quad \begin{bmatrix} t-a&\quad b-c\\ b-c&\quad v\end{bmatrix} \succeq 0, \quad u+v = 2s. \end{aligned}
(11)

(The $$2\times 2$$ positive semidefinite constraints can be converted to second-order cone constraints using (1)). To see why (11) holds note that by applying a congruence transformation by $$\frac{1}{\sqrt{2}}\left[ {\begin{matrix} 1 &{} 1 &{} 0\\ 1 &{} -1 &{} 0\\ 0 &{} 0 &{} 2\end{matrix}}\right]$$ on the $$3\times 3$$ matrix on the left-hand side of (11) we get that

\begin{aligned} \begin{bmatrix} t&\quad a&\quad b\\ a&\quad t&\quad c\\ b&\quad c&\quad s \end{bmatrix} \succeq 0 \Longleftrightarrow \begin{bmatrix} t+a&\quad 0&\quad b+c\\ 0&\quad t-a&\quad b-c\\ b+c&\quad b-c&\quad 2s \end{bmatrix} \succeq 0. \end{aligned}

The latter matrix has an arrow structure and thus using results on the decomposition of matrices with chordal sparsity pattern [1, 6, 10] we get the decomposition (11).