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For a random quantum state on $\cH=\C^d \otimes \C^d$ obtained by partial tracing a random pure state on $\cH \otimes \C^s$, we consider the question whether it is typically separable or typically entangled. For this problem, we show the existence of a sharp threshold $s_0=s_0(d)$ of order roughly $d^3$. More precisely, for any $\e>0$ and for $d$ large enough, such a random state is entangled with very large probability when $s \leq (1-\e)s_0$, and separable with very large probability when $s \geq (1+\e) s_0$. One consequence of this result is as follows: for a system of $N$ identical particles in a random pure state, there is a threshold $k_0 = k_0(N)\sim N/5$ such that two subsystems of $k$ particles each typically share entanglement if $k>k_0$, and typically do not share entanglement if $k<k_0$. Our methods work also for multipartite systems and for ``unbalanced'' systems such as $\C^{d_1} \otimes \C^{d_2} $, ${d_1} \neq {d_2} $. The arguments rely on random matrices, classical convexity, high-dimensional probability and geometry of Banach spaces; some of the auxiliary results may be of reference value.

No abstract uploaded!

No abstract uploaded!

Let $K/k$ be a finite Galois extension and $\pi = \fn{Gal}(K/k)$. An algebraic torus $T$ defined over $k$ is called a $\pi$-torus if $T\times_{\fn{Spec}(k)} \fn{Spec}(K)\simeq \bm{G}_{m,K}^n$ for some integer $n$. The set of all algebraic $\pi$-tori defined over $k$ under the stably birational equivalence forms a semigroup, denoted by $T(\pi)$. We will give a complete proof of the following theorem due to Endo and Miyata \cite{EM4}. Theorem. Let $\pi$ be a finite group. Then $T(\pi)\simeq C(\Omega_{\bm{Z}\pi})$ where $\Omega_{\bm{Z}\pi}$ is a maximal $\bm{Z}$-order in $\bm{Q}\pi$ containing $\bm{Z}\pi$ and $C(\Omega_{\bm{Z}\pi})$ is the locally free class group of $\Omega_{\bm{Z}\pi}$, provided that $\pi$ is isomorphic to one of the following four types of groups : $C_n$ ($n$ is any positive integer), $D_m$ ($m$ is any odd integer $\ge 3$), $C_{q^f}\times D_m$ ($m$ is any odd integer $\ge 3$, $q$ is an odd prime number not dividing $m$, $f\ge 1$, and $(\bm{Z}/q^f\bm{Z})^{\times}=\langle \bar{p}\rangle$ for any prime divisor $p$ of $m$), $Q_{4m}$ ($m$ is any odd integer $\ge 3$, $p\equiv 3 \pmod{4}$ for any prime divisor $p$ of $m$).