On Cones of Nonnegative Quartic Forms

Abstract

Historically, much of the theory and practice in nonlinear optimization has revolved around the quadratic models. Though quadratic functions are nonlinear polynomials, they are well structured and many of them are found easy to deal with. Limitations of the quadratics, however, become increasingly binding as higher-degree nonlinearity is imperative in modern applications of optimization. In recent years, one observes a surge of research activities in polynomial optimization, and modeling with quartic or higher-degree polynomial functions has been more commonly accepted. On the theoretical side, there are also major recent progresses on polynomial functions and optimization. For instance, Ahmadi et al. (Math Program Ser A 137:453–476, 2013) proved that checking the convexity of a quartic polynomial is strongly NP-hard in general, which settles a long-standing open question. In this paper, we proceed to study six fundamentally important convex cones of quartic forms in the space of super-symmetric tensors, including the cone of nonnegative quartic forms, the sums of squared forms, the convex quartic forms, and the sums of fourth-power forms. It turns out that these convex cones coagulate into a chain in a decreasing order with varying complexity status. Potential applications of these results to solve highly nonlinear and/or combinatorial optimization problems are discussed.

Appendix: Proof of Theorem 6.1

Here we only prove the equivalent relation for the maximization problems since the proof for their minimization counterparts is exactly the same. That is, we shall prove the equivalence between (Q) and (RQ).

To start with, let us first investigate the feasible regions of these two problems, to be denoted by $\mathrm{feas}\,(Q)$ and $\mathrm{feas}\,(RQ)$, respectively. The relationship between $\mathrm{feas}\,(Q)$ and $\mathrm{feas}\,(RQ)$ is revealed by the following lemma.

Lemma 6.4

It holds that $\mathrm{conv}\left( \mathrm{feas}\,(Q)\right) \subseteq \mathrm{feas}\,(RQ) = \mathrm{conv}\left( \mathrm{feas}\,(Q)\right) +\mathbf {P}$, where

$$\begin{aligned} \mathbf {P}:=\mathrm{cone}\,\left\{ \left( {\begin{array}{c}x\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}x\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}x\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}x\\ 0\end{array}}\right) \bigg | \begin{array}{ll} (a^i)^{\mathrm{T}}x=0&{}\forall \,1\le i\le m,\\ x^{\mathrm{T}}A^jx=0&{}\forall \,1\le j\le l \end{array} \right\} \subseteq {\varvec{\Sigma }}_{n+1,4}^4. \end{aligned}$$

Proof

First, it is obvious that $\mathrm{conv}\left( \mathrm{feas}\,(Q)\right) \subseteq \mathrm{feas}\,(RQ)$ as (RQ) is a relaxation of (Q) and $\mathrm{feas}\,(RQ)$ is convex. Next, we notice that the recession cone of $\mathrm{feas}\,(RQ)$ is equal to

$$\begin{aligned} \left\{ \bar{\mathcal {X}}\in {\varvec{\Sigma }}_{n+1,4}^4\left| \begin{array}{ll} \mathcal {X}_{n+1,n+1,n+1,n+1}=0,&{}\\ (a^i)^{\mathrm{T}}x=0,\,(a^i\otimes a^i)\bullet X=0,\,(a^i\otimes a^i\otimes a^i\otimes a^i)\bullet \mathcal {X}=0&{}\forall \,1\le i\le m,\\ A^j\bullet X=0,\,(A^j\otimes A^j)\bullet \mathcal {X} =0 &{} \forall \, 1\le j\le l \end{array}\right. \right\} . \end{aligned}$$

Observing that $\bar{\mathcal {X}}\in {\varvec{\Sigma }}_{n+1,4}^4$ and $\mathcal {X}_{n+1,n+1,n+1,n+1}=0$, it is easy to see that $x=0$ and $X=0$. Thus, the recession cone of $\mathrm{feas}\,(RQ)$ is reduced to

$$\begin{aligned} \left\{ \bar{\mathcal {X}}\in {\varvec{\Sigma }}_{n+1,4}^4\left| \begin{array}{ll}\mathcal {X}_{n+1,n+1,n+1,n+1}=0,\,x=0,\,X=0, &{}\\ (a^i\otimes a^i\otimes a^i\otimes a^i)\bullet \mathcal {X}=0&{} \forall \, 1\le i\le m,\\ (A^j\otimes A^j)\bullet \mathcal {X} =0 &{} \forall \, 1\le j\le l \end{array}\right. \right\} \supseteq \mathbf {P} , \end{aligned}$$

which proves $\mathrm{feas}\,(RQ) \supseteq \mathrm{conv}\left( \mathrm{feas}\,(Q)\right) +\mathbf {P}$.

Finally, we shall show the inverse inclusion, i.e., $\mathrm{feas}\,(RQ) \subseteq \mathrm{conv}\left( \mathrm{feas}\,(Q)\right) +\mathbf {P}$. Suppose $\bar{\mathcal {X}}\in \mathrm{feas}\,(RQ)$, and it can be decomposed as

$$\begin{aligned} \bar{\mathcal {X}}=\sum _{k \in K}\left( {\begin{array}{c}y^k\\ \alpha _k\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ \alpha _k\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ \alpha _k\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ \alpha _k\end{array}}\right) , \end{aligned}$$

(25)

where $\alpha _k\in {\mathbf {R}}$, $y^k\in {\mathbf {R}}^n$ for all $k\in K$. Immediately we have

$$\begin{aligned} \sum _{k \in K}{\alpha _k}^4=\mathcal {X}_{n+1,n+1,n+1,n+1}=1. \end{aligned}$$

(26)

Now divide the index set K into two parts $K_0:=\{k\in K\mid \alpha _k=0\}$ and $K_1:=\{ k\in K\mid \alpha _k\ne 0 \}$, and let $z^k=y^k/\alpha _k$ for all $k\in K_1$. The decomposition (25) is then equivalent to

$$\begin{aligned} \bar{\mathcal {X}}=\sum _{k \in K_1}{{\alpha }_k}^4\left( {\begin{array}{c}z^k\\ 1\end{array}}\right) \otimes \left( {\begin{array}{c}z^k\\ 1\end{array}}\right) \otimes \left( {\begin{array}{c}z^k\\ 1\end{array}}\right) \otimes \left( {\begin{array}{c}z^k\\ 1\end{array}}\right) + \sum _{k \in K_0}\left( {\begin{array}{c}y^k\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ 0\end{array}}\right) . \end{aligned}$$

If we can prove that

$$\begin{aligned} \left( {\begin{array}{c}z^k\\ 1\end{array}}\right) \otimes \left( {\begin{array}{c}z^k\\ 1\end{array}}\right) \otimes \left( {\begin{array}{c}z^k\\ 1\end{array}}\right) \otimes \left( {\begin{array}{c}z^k\\ 1\end{array}}\right)\in & {} \mathrm{feas}\,(Q) \quad \forall \,k\in K_1 \end{aligned}$$

(27)

$$\begin{aligned} \left( {\begin{array}{c}y^k\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}y^k\\ 0\end{array}}\right)\in & {} \mathbf {P}\qquad \qquad \forall \,k\in K_0 \end{aligned}$$

(28)

then by (26), we shall have $\bar{\mathcal {X}}\in \mathrm{conv}\left( \mathrm{feas}\,(Q)\right) +\mathbf {P}$, proving the inverse inclusion.

In the following, we shall prove (27) and (28). Since $\bar{\mathcal {X}}\in \mathrm{feas}\,(RQ)$, together with $x=\sum _{k\in K}{\alpha _k}^3y^k$, $X=\sum _{k\in K}{\alpha _k}^2y^k\otimes y^k$ and $\mathcal {X}=\sum _{k\in K}y^k\otimes y^k\otimes y^k\otimes y^k$, we obtain the following equalities:

$$\begin{aligned} \begin{aligned}&\sum _{k \in K}{\alpha _k}^3(a^i)^{\mathrm{T}}y^k=b_i,\, \sum _{k \in K}{\alpha _k}^2\left( (a^i)^{\mathrm{T}}y^k\right) ^2={b_i}^2,\, \sum _{k \in K}\left( (a^i)^{\mathrm{T}}y^k\right) ^4={b_i}^4&\forall \, 1\le i \le m\\&\sum _{k \in K}{\alpha _k}^2(y^k)^{\mathrm{T}}A^jy^k=c_j,\, \sum _{k \in K}\left( (y^k)^{\mathrm{T}}A^jy^k\right) ^2={c_j}^2&\forall \, 1\le j \le l. \end{aligned} \end{aligned}$$

As a direct consequence of the above equalities and (26), we have

$$\begin{aligned} \begin{aligned}&\left( \sum _{k \in K}{\alpha _k}^2\cdot \alpha _k(a^i)^{\mathrm{T}}y^k\right) ^2={b_i}^2=\left( \sum _{k \in K}{\alpha _k}^4\right) \left( \sum _{k \in K}{\alpha _k}^2\left( (a^i)^{\mathrm{T}}y^k\right) ^2\right)&\forall \, 1\le i \le m \\&\left( \sum _{k \in K}{\alpha _k}^2\left( (a^i)^{\mathrm{T}}y^k\right) ^2\right) ^2={b_i}^4=\left( \sum _{k \in K}{\alpha _k}^4\right) \left( \sum _{k \in K}\left( (a^i)^{\mathrm{T}}y^k\right) ^4\right)&\forall \, 1\le i \le m \\&\left( \sum _{k \in K}{\alpha _k}^2(y^k)^{\mathrm{T}}A^jy^k\right) ^2= {c_j}^2=\left( \sum _{k \in K}{\alpha _k}^4\right) \left( \sum _{k \in K}\left( (y^k)^{\mathrm{T}}A^jy^k\right) ^2\right)&\forall \, 1\le j \le l. \end{aligned} \end{aligned}$$

Noticing that the equalities hold for the above Cauchy–Schwarz inequalities, it follows that for every $1\le i \le m$ and every $1\le j \le l$, there exist $\delta _i, \epsilon _i, \theta _j\in {\mathbf {R}}$, such that

$$\begin{aligned} \delta _i{\alpha _k}^2=\alpha _k(a^i)^{\mathrm{T}}y^k,\, \epsilon _i{\alpha _k}^2=\left( (a^i)^{\mathrm{T}}y^k\right) ^2 \text{ and } \theta _j{\alpha _k}^2=(y^k)^{\mathrm{T}}A^jy^k\quad \forall \,k \in K.\quad \end{aligned}$$

(29)

If $\alpha _k=0$, then $(a^i)^{\mathrm{T}}y^k=0$ and $(y^k)^{\mathrm{T}}A^jy^k=0$, which implies (28). Moreover, due to (29) and (26),

$$\begin{aligned} \delta _i=\delta _i\left( \sum _{k\in K}{\alpha _k}^4\right) =\sum _{k\in K}\delta _i{\alpha _k}^2\cdot {\alpha _k}^2=\sum _{k\in K}\alpha _k(a^i)^{\mathrm{T}}y^k\cdot {\alpha _k}^2=b_i\quad \forall \,1\le i\le m. \end{aligned}$$

Similarly, we have $\theta _j = c_j$ for all $1\le j \le l$. If $\alpha _k\ne 0$, noticing $z^k=y^k/\alpha _k$, it follows from (29) that

$$\begin{aligned} \begin{aligned}&(a^i)^{\mathrm{T}}z^k=(a^i)^{\mathrm{T}}y^k/\alpha _k=\delta _i=b_i&\forall \, 1\le i\le m\\&(z^k)^{\mathrm{T}}A^jz^k=(y^k)^{\mathrm{T}}A^jy^k/{\alpha _k}^2=\theta _j=c_j&\forall \,1\le j \le l, \end{aligned} \end{aligned}$$

which implies (27). $\square$

To prove Theorem 6.1, we notice that if $A^j$ is positive semidefinite, then

$$\begin{aligned} x^{\mathrm{T}}A^jx=0\Longleftrightarrow A^jx=0. \end{aligned}$$

Therefore, $\left( {\begin{array}{c}x\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}x\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}x\\ 0\end{array}}\right) \otimes \left( {\begin{array}{c}x\\ 0\end{array}}\right) \in \mathbf {P}$ implies that x is a recession direction of the feasible region for (P). Applying this property and using a similar argument of Theorem 2.6 in [8], Theorem 6.1 follows immediately.