On representing the positive semidefinite cone using the second-order cone

Abstract

The second-order cone plays an important role in convex optimization and has strong expressive abilities despite its apparent simplicity. Second-order cone formulations can also be solved more efficiently than semidefinite programming problems in general. We consider the following question, posed by Lewis and Glineur, Parrilo, Saunderson: is it possible to express the general positive semidefinite cone using second-order cones? We provide a negative answer to this question and show that the $3\times 3$ positive semidefinite cone does not admit any second-order cone representation. In fact we show that the slice consisting of $3\times 3$ positive semidefinite Hankel matrices does not admit a second-order cone representation. Our proof relies on exhibiting a sequence of submatrices of the slack matrix of the $3\times 3$ positive semidefinite cone whose “second-order cone rank” grows to infinity.

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 (x, y) 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 rays¹${{\mathrm{ext}}}(\mathbf {S}^3_+)$ 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 i, j). 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).