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In the vector space $V$ of $3\times 3$ symmetric real matrices, we can define a nonassociative algebra structure by the multiplication $$A \bullet B = \frac12 (AB +BA).$$ This turns $V$ into a Jordan algebra.

Question

What is the minimum number of generators of this Jordan algebra? And could you give me one set of such generators?

Thanks a lot!

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I have edited the question to make it more readable. I have added the condition that the algebra be real. If this is not intended, then please delete that. –  José Figueroa-O'Farrill Nov 18 '10 at 19:43
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I don't think the base field $k$ really matters here. The answer is that any two randomly-chosen symmetric matrices should generate "with probability 1". (And it may be clearer to work over $\mathbb{Q}$ or some other field with lots of cubic extensions.) Pick a matrix $A$ with minpoly of degree 3, so $A$ generates a degree 3 (associative!) subalgebra. Pick $B$ not in $k[A]$ also with minpoly of degree 3. (You can take $A$ diagonal, but then $B$ cannot be.) Then I think $A$ and $B$ will generate... –  Skip Nov 18 '10 at 20:25
    
@José Figueroa-O'Farrill:Thank you for your edition.The field does not matter. –  TOM Nov 18 '10 at 22:20
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@skip: This esessentially is in Mark's answer below. He does not give a proof. But if you take two matrices, a matrix unit (1,1) and a generic matrix $A$, then they generate the whole Jordan algebra. Take $A$ with variable coordinates. And start producing matrices with more and more zeroes using the Jordan product and linear combinations. It is clearly possible to get all matrix units, but in the process you need to divide by polynomials of coeff. of A. Hence you need that those are not 0. This gives you the conditions that $A$ must satisfy ($A$ must belong to a complement of a variety). –  Steve Richards Nov 19 '10 at 4:30
    
continued: That is true for any field with significantly many elements. For small finite fields this does not work. –  Steve Richards Nov 19 '10 at 4:53
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4 Answers

up vote 8 down vote accepted

The minimum number of generators is 2. First, it is easy to check that one generator is not enough: every symmetric matrix is diagonalizable, so the subalgebra it generates has dimension at most 3.

Next, the claim is that $$A:=S_{11}+2S_{22}+3S_{33}$$ and $$B:=S_{12}+2S_{13}$$ generate the algebra, where $S_{ij}$ is the usual notation for the symmetric matrices. In fact the powers of $A$ generate the subalgebra $D$ of diagonal matrices (by Vandermonde) and the powers $B$ satisfy $B^2=S_{23}$ mod $D$ and $B^3=S_{12}+\frac12S_{13}$ mod $D$.

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The algebra cannot be generated by one element because a Jordan algebra is power-associative, and your algebra is not associative. –  Mariano Suárez-Alvarez Nov 19 '10 at 1:08
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Also, The $\bullet$ multiplication is the same as ordinary multiplication for matrices which commute (such as powers of one matrix) so a single matrix can only generate a dimension 3 subspace. –  Aaron Meyerowitz Nov 19 '10 at 4:23
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Although Andrei Moroianu has already answered the question, I would like to add an answer I just received from Efim Zelmanov. The Jordan algebra is generated by two matrices. The first is the matrix unit $E_{1,1}$, and the second any "generic" symmetric $3\times 3$-matrix. I checked: it works if the second matrix $$\left\[\begin{array}{ccc} 1 & 2 &3 \\\ 2 & 4 & 5 \\\ 3 & 5 & 6\end{array}\right\].$$

Update 1. An interesting question is whether it is true for $4\times 4$-symmetric matrices. The answer seems to be "yes". How about arbitrary dimension $n\ge 3$?

Update 2. This works for any field.

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This non-associative algebra is the typical example of a Jordan algebra, named after Pascual Jordan, who studied them with J. von Neumann. You should consult a dedicated book. See for instance J. Faraut & A. Korányi: Analysis on symmetric cones, Oxford University Press, New York, 1994.

A complete classification of the so-called Euclidian case is available, but it was disappointing in view of von Neumann & Jordan's expectation.

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This is already well answered, many choices of two symmetric matrices will do. I'll add this observation: $\{I,A,B,A \bullet A,B \bullet B,A\bullet B\}$ is an additive basis where $$A=\left[\begin{array}{ccc} 1 & 1 &1 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{array}\right] \qquad B=\left[\begin{array}{ccc} 0 & 0 &1 \\ 0 & 0 & 1 \\ 1 & 1 & 1\end{array}\right]$$ Note that $A\bullet B$ has rank $3$, not that it matters. I'm not sure if two rank 1 matrices could generate the algebra, but I lean against (I couldn't get past dimension 4), I didn't look hard for a generating set with a rank 1 and a rank 2 matrix.

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