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How we can define quaternionic projective space and metric on it using Jordan algebra?

Thank you in advance!

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2 Answers

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I am not much educated in this subject so please take my answer with a grain of salt.

The appropriate Jordan algebra is $J_n(\mathbb{H}) := \{ A \in M(n,\mathbb{H})\\,|\\, \overline{A}^T=A \}$ with multiplication defined as $A\circ B = (AB+BA)/{2}$. The metric is defined by $\mathrm{Tr}(A^2)$. The quaternionic projective space can be then defined as the space of elements of $J_n(\mathbb{H})$ of rank one and trace one (think of these matrices as projectors to one-dimensional subspaces) or alternatively as the (real) projectivization of rank one matrices.

I am little bit more familiar with the case of octonionic projective plane, so let me explain that case. There the relevant algebra is $J_3(\mathbb{O})$ and the plane $\mathbb{OP}^2$ and its metric is defined in the same way[1] as in the quaternionic case. Now there is a well defined cubic form on $J_3(\mathbb{O})$ which is basically the determinant. The group that preserves this cubic form is $E_6$. In fact one defines some sort of cross product (I think it is called Jordan cross product.) out of the cubic form which then gives the incidence relation of the projective geometry of $\mathbb{OP}^2$. If I remember correctly, the product $A\times B$ is defined by the relation $(A\times B,C) = (A,B,C)$, where on the left hand side the brackets denote the polarization of the quadratic form while on the right hand side the brackets denote the polarization of the cubic form. The group $E_6$ is then the group of collineations of $\mathbb{OP}^2$

The group that preserves the determinant and the quadratic form (the norm) is the group $F_4$. This sheds some light on the fact that $F_4$ is in fact the isometry group of $\mathbb{OP}^2$. It can be proven that $F_4$ is in fact the automorphism group of the Jordan algebra $J_3(\mathbb{O})$!

In the quaternionic case one finds that $E_6$ is replaced by $Sl(n,\mathbb{H})$ and $F_4$ by $Sp(n)$.

The projective plane $\mathbb{OP}^2$ was much studied by Freudenthal and others but I do not know of any reference where the quaternionic case is treated via Jordan algebras.

[1] Of course one needs to be a little careful with the definition of rank because of the nonassociativity of $\mathbb{O}$. Either one uses the fact that any element of $J_3(\mathbb{O})$ can be mapped by the action of $F_4$ to a diagonal matrix and then one defines the rank by the number of nonzero elements in this diagonalization. Or one can experiment a little bit and discover that it is possible to define determinants of 2x2 minors in such a way that their vanishing is equivalent to matrix of being of rank one according to first definition.

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How does the structure of the Jordan algebra comes into play ? Is the space of rank 1 closed with respect to Jordan's product ? I would be very thankful for further comments... For the long time I keep some papers on Jordan algs nearby me but always does not see key how to open them :) –  Alexander Chervov May 16 '12 at 10:28
    
@robot Still I am not clear about the role of Jordan algebra structure. If you will omit this word and just write matrices A^t=A , norm Tr(A^2)... nothing changes.. Any way thank you for yours answer, it useful (at least for me). –  Alexander Chervov May 16 '12 at 12:00
    
I do not know of a nice direct way to connect Jordan algebra structure and the geometry of the projective space. In the octonionic case, the incidence relation is given by the cross product which can be expressed in terms of Jordan product and traces. (See arxiv.org/abs/0902.0431) –  Vít Tuček May 16 '12 at 12:05
    
You are right, the conditions $\overline{A}^t=A$ and being of rank one and trace one are sufficient to define the projective space and its metric. The Jordan multiplication lurks in the background, because - at least in the octonionic case - the group of isometries is the group of automorphisms of the Jordan algebra. –  Vít Tuček May 16 '12 at 12:08
    
Is the condition $\overline{A}^{T}=A$ here because these matrices preserve $H$-inner product? Is there any connection between $Sp(n)$ and and $J_{n}(H)$? –  Mirjana May 17 '12 at 9:33
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(Perhaps my answer will be too general for your purposes.)

The description of the projective space that you mention, is only one special case of a very general construction (due to O. Loos) of a projective space from a Jordan pair, which is a pair of vector spaces $(V^+, V^-)$ equipped with certain quadratic operators. The notion of a Jordan pair generalizes the notion of a Jordan algebra, in the sense that any Jordan algebra $J$ can be made into a Jordan pair (but the converse is not true).

If $V = (V^+, V^-)$ is such a Jordan pair, then the projective space $X(V)$ of $V$ is the quotient of $V^+ \times V^-$ by projective equivalence, defined as $$ (x, y) \sim (x', y') \iff (x, y - y') \text{ is quasi-invertible, and } x' = x^{y-y'}, $$ where $x^{y-y'}$ denotes the quasi-inverse of the pair $(x, y-y')$.

In the case where $V$ is the Jordan pair arising from the Jordan algebra of hermitian matrices over any skew field $D$ (of which $D = \mathbb{H}$ is just a particular case), the projective space $X(V)$ is the classical projective space coordinatized by the skew field $D$.

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what is "projective space" ? Some nice manifold ? Symmetric space ? homogeneous space ? –  Alexander Chervov May 16 '12 at 12:29
    
For general fields, it's just an algebraic object, but for instance if $k = \mathbb{C}$, then one obtains the compact hermitian symmetric spaces. –  Tom De Medts May 16 '12 at 12:33
    
@Tom thank you ! Do all compact hermitian spaces can be obtained in this way ? –  Alexander Chervov May 16 '12 at 13:00
    
I guess so (but I'm not very familiar with symmetric spaces though). You might be interested in the lecture notes "Jordan pairs and bounded symmetric domains" by O. Loos, available on the website homepage.uibk.ac.at/~c70202/jordan/archive/irvine/index.html. –  Tom De Medts May 16 '12 at 13:10
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