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arXiv 1208.3582math.FA
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拟凸自由多项式

Quasi-Convex Free Polynomials

Sriram Balasubramanian, Scott McCullough

摘要:

令$\Rx$表示$g$中自由非交换变量$x=(x_1,...,x_g)$的多项式环。存在由$x_j^*=x_j$和$(pq)^*=q^* p^*$确定的$\Rx$上的自然对合 *,并且自由多项式$p\in\Rx$是对称的,如果它在这个对合下是不变的。如果$X=(X_1,...,X_g)$是对称$n\times n$矩阵的$g$元组,则评估$p(X)$自然被定义,并且进一步被定义为$p^*(X)=p(X)^*$。特别地,如果$p$是对称的,则$p(X)^*=p(X)$。本文的主要结果是:如果$p$是对称的,$p(0)=0$和对于每个$n$和每个对称正定的$n\times n$矩阵$A$,集合${X:A-p(X)\succ 0}$是凸的,那么$p$的度至多为2,并且它本身是凸的,或者$-p$是埃尔米特平方和。

英文摘要:

Let $\Rx$ denote the ring of polynomials in $g$ freely non-commuting variables $x=(x_1,...,x_g)$. There is a natural involution * on $\Rx$ determined by $x_j^*=x_j$ and $(pq)^*=q^* p^*$ and a free polynomial $p\in\Rx$ is symmetric if it is invariant under this involution. If $X=(X_1,...,X_g)$ is a $g$ tuple of symmetric $n\times n$ matrices, then the evaluation $p(X)$ is naturally defined and further $p^*(X)=p(X)^*$. In particular, if $p$ is symmetric, then $p(X)^*=p(X)$. The main result of this article says if $p$ is symmetric, $p(0)=0$ and for each $n$ and each symmetric positive definite $n\times n$ matrix $A$ the set ${X:A-p(X)\succ 0}$ is convex, then $p$ has degree at most two and is itself convex, or $-p$ is a hermitian sum of squares.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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