Let A be a Jordan algebra (with identity). If x is in A let A[x] be the subalgebra generated by x and the identity. An element x is regular if the dimension of A[x] is maximal.

For x regular, denote by tr(x) the trace of the operator "multiplication by x" restricted to A[x].

The function tr can be extended to all of A. This is explained for instance in the book by Faraut and Koranyi, "Analysis on symmetric cones".

My question is: how does one prove that this extended function tr is linear? (the point is to prove that it is additive, the homogeneity being easy)

In the book by Faraut and Koranyi, this seems to be assumed, there is no explanation. May be this is easy and I am missing something.

Ultimately, for instance if A is Euclidean and simple, one prove that tr is proportional to the function that maps x to the trace of the operator L(x) of multiplication by x (on all of A) which is clearly linear. But the proof of this last fact uses that tr is linear.


  • $\begingroup$ Is it implicit that $A$ is finite-dimensional over a field? Some specific field? what do you mean by "can extended to all of $A$"? as a linear map, it is clear. You seem to refer to some particular extension, but it's hard to answer your question if you don't say how it's extended (except for the few readers who have the book). $\endgroup$
    – YCor
    May 20, 2016 at 22:22
  • $\begingroup$ Let's say that A is finite dimensional over R. By "can be extended" I mean "can be extended continuously". Linearity is not clear a priori (with the definition I wrote) because you take the trace of the restriction of the multiplication by x on some subspace which depends on x. But actually I think I can answer my question myself now.... I think that the answer is contained in the book, I just have to understand what is written there... $\endgroup$
    – Pierre
    May 20, 2016 at 23:36
  • $\begingroup$ Is the set of regular elements dense? indeed in this case there's at most one continuous extension. $\endgroup$
    – YCor
    May 20, 2016 at 23:51
  • $\begingroup$ Yes it's dense. Actually what the authors of the book prove is the following. Call m the maximal dimension of A[x]. So on the set of regular elements you have functions $a_i$ ($1\le i \le m$) where a_i is equal to the coefficient of degree $m-i$ of the minimal polynomial (up to $(-1)^i$ ). What the authors prove is that a_i is the restriction to the set of regular elements of a homogeneous polynomial on A of degree i. $\endgroup$
    – Pierre
    May 21, 2016 at 0:10
  • $\begingroup$ so $a_1=tr$ is the restriction to the set of regular elements of a homogeneous polynomial of degree 1... so this answers my question. Somehow I had missed the statement that the extension of $a_i$ was polynomial.... so my question was stupid. $\endgroup$
    – Pierre
    May 21, 2016 at 0:12


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