## Why only three classical matrix ensembles in RMT? (Newbie question)

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

I haven't been able to find a good expository reference for this, so any thoughts along those lines are also welcome.

-

## 4 Answers

The short answer is that there are three kinds of positive-definite elementary inner products:

1. symmetric on $\mathbb{R}^n$, giving rise to the orthogonal ensemble;
2. hermitian on $\mathbb{C}^n$, giving rise to the unitary ensemble; and
3. hermitian on $\mathbb{H}^n$, giving rise to the symplectic ensemble.

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.

-

1
Incidentally, there was a recent question concerning classical Lie groups: mathoverflow.net/questions/50610/… – José Figueroa-O'Farrill Dec 29 2010 at 22:38
That helped a lot, thanks! But I am still unclear as to whether there could be more than three different cases. Could there be, for example, a fourth analogous ensemble arising from an Hermitian inner product over the octonions? – Jiahao Chen Jan 2 2011 at 1:56

2
@Jiahao, there is work out there on what are known as $\beta$ ensembles. I've seen a few talks but not participated in it myself. Where $\beta = 1$ is the orthogonal case, $\beta = 2$ is the unitary case, and $\beta = 4$ the symplectic case. However $\beta$ is usually allowed the run through all positive reals. As I recall there is no special behavior at $\beta=8$ which would indicate something something special for octonians. – BSteinhurst May 1 2011 at 15:53
@Jiahao, @BSteinhurst: I seem to recall a recent discussion on how the five exceptional Lie groups can be thought of as belonging an "infinite" series like ${\rm O}(n)$, ${\rm U}(n)$ and ${\rm Sp}(n)$ that fails and this is related to the failure of associativity for the octonians. Also, a reference that starts with the definition of inner product and builds all the groups that can rise from the definition is chapter 3 of Rossmann, "Lie Groups: An Introduction through Linear Groups". – WetSavannaAnimal aka Rod Vance Jul 26 2011 at 0:50

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Here is the answer: http://arxiv.org/abs/cond-mat/9602137

-

3
Would you care to explain how this paper answers the question? – S. Carnahan♦ Jan 6 2011 at 18:13

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: http://arxiv.org/abs/1001.0722

-

Thanks for the reference! It really brought to the fore the relationship between the relevant vector spaces and the resulting matrix ensembles. – Jiahao Chen May 8 2011 at 0:15

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 ...)

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9217249751091003, "perplexity_flag": "head"}