most recent 30 from http://mathoverflow.net 2013-06-20T03:00:55Z http://mathoverflow.net/feeds/question/50684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50684/why-only-three-classical-matrix-ensembles-in-rmt-newbie-question Jiahao Chen 2010-12-29T22:09:11Z 2011-07-28T19:05:06Z

<p>I am just starting out on understanding random matrix theory from a background in applied mathematics. I have a very basic question about the Gaussian ensembles: why are there only three classical Gaussian ensembles? This seems very mysterious to me. Is it historically motivated from applications, or is there a deeper reason? I thought it might arise from exhausting all possible classes of diagonalizable matrices of a certain symmetry, but I have no idea if this is true or not.</p> <p>I haven't been able to find a good expository reference for this, so any thoughts along those lines are also welcome.</p>

http://mathoverflow.net/questions/50684/why-only-three-classical-matrix-ensembles-in-rmt-newbie-question/50686#50686 José Figueroa-O'Farrill 2010-12-29T22:37:31Z 2010-12-30T21:43:03Z

<p>The short answer is that there are three kinds of positive-definite elementary inner products: </p> <ol> <li>symmetric on $\mathbb{R}^n$, giving rise to the orthogonal ensemble;</li> <li>hermitian on $\mathbb{C}^n$, giving rise to the unitary ensemble; and</li> <li>hermitian on $\mathbb{H}^n$, giving rise to the symplectic ensemble.</li> </ol> <p>Each one gives rise to a compact classical Lie group: $\mathrm{O}(n)$, $\mathrm{U}(n)$ and $\mathrm{Sp}(n)$, respectively. Compactness makes the integrals defining the matrix model convergent.</p>

http://mathoverflow.net/questions/50684/why-only-three-classical-matrix-ensembles-in-rmt-newbie-question/51288#51288 Eugene 2011-01-06T07:47:21Z 2011-01-06T07:47:21Z

<p>Here is the answer: <a href="http://arxiv.org/abs/cond-mat/9602137" rel="nofollow">http://arxiv.org/abs/cond-mat/9602137</a></p>

http://mathoverflow.net/questions/50684/why-only-three-classical-matrix-ensembles-in-rmt-newbie-question/63566#63566 Carlo Beenakker 2011-04-30T22:29:41Z 2011-04-30T22:29:41Z

<p>Every Riemannian symmetric space can be associated with a random matrix ensemble, producing a total of 10 ensembles, as is nicely explained by Martin Zirnbauer: <a href="http://arxiv.org/abs/1001.0722" rel="nofollow">http://arxiv.org/abs/1001.0722</a></p>

http://mathoverflow.net/questions/50684/why-only-three-classical-matrix-ensembles-in-rmt-newbie-question/71278#71278 Adrien Hardy 2011-07-26T00:31:15Z 2011-07-28T19:05:06Z

<p>These three ensembles are hermitian matrices over a (finite dimensional real) field of numbers, and it is known that the only finite dimensional real fields are the real numbers, the complex numbers ($2$-dimensional) and the quaternionic numbers ($4$-dimensional). Octonions are not a field of number since you do not have associativity. The motivation in physics comes from the fact that an hermitian matrix represents a finite dimentional Hamiltonian (an Hermitian operator) in quantum mechanics (then you add randomness, in order to take in account the lack of information about your system, and you let the size of the matrix, that is the dimension of your state space where your Hamiltionian is acting on, going to infinity). In this setting, $N\times N$ quaternionic matrices have to be seen as subclasses of complex hermitian matrices (but of size $2N\times 2N$) and both real symmetric and quaternionic hermitian matrices are a subclass of complex hermitian matrices, with extra symmetries. Anyway, you may imagine many different matrix models relevant for studying (look for Wigner matrices, the answer of Beenakker about other symmetries in physics, the generalized $\beta$-ensemble of Edelman, etc ...)</p>