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I have seen written that the space of $m\times m$-complex matrices $M_m(\mathbb C)$ endowed with the usual Jordan product is isomorphic, as a Jordan algebra to the complexification of the space $Herm(m,\mathbb C)$ of hermitian $m\times m$ matrices.
Is it possible (easy) to write down an explicit isomorphism between the two?

For example, for the $2\times 2$ matrices, the elements of $Herm(2,\mathbb C)\otimes_{\mathbb R}\mathbb C$ are of the form $\left(\begin{smallmatrix}\alpha &\lambda + I\mu\\\lambda-I\mu &\beta\end{smallmatrix}\right)$ with $I=i\otimes_{\mathbb R}1$ and $\alpha,\beta,\lambda,\mu\in \mathbb C$ and the elements of $M_2(\mathbb C)$ are of the form $\left(\begin{smallmatrix}\alpha &c\\d &b\end{smallmatrix}\right)$ with $a,b,c,d\in \mathbb C$. The question is: is there a nice expression of $a,b,c,d$ in terms of $\alpha,\beta,\lambda,\mu$?

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    $\begingroup$ Try the canonical map induced by inclusion. $\endgroup$
    – YCor
    Commented Mar 1, 2021 at 6:15
  • $\begingroup$ @Ycor: Thank you for your answer but I cannot see it: (complexified) hermitian matrices are of the form (for example $2\times 2$ $\left(\begin{smallmatrix}\alpha &a+Ib \\a-Ib &\beta\end{smallmatrix}\right)$ where $I$ stands for $i\otimes_{\mathbb R} \mathbb C$ and $\alpha,\beta,a,b\in \mathbb C$ but the $2\times 2$-matrices are just of the form $\left(\begin{smallmatrix}m &n \\o &p\end{smallmatrix}\right)$. What I would like to see is an expression of $m,n,o,p$ in terms of $\alpha,\beta, a,b$. $\endgroup$
    – pi_1
    Commented Mar 1, 2021 at 12:28
  • $\begingroup$ @dodd: They have the same dimension over $\mathbb C$ ($Herm(m,\mathbb C)$ is complexified) $dim_{\mathbb C}(M_m(\mathbb C))=m^2$ and for $dim(Herm(m,\mathbb C)\otimes_{\mathbb R}\mathbb C$, there the $m$ complex numbers on the diagonal and the $\frac{m(m-1)}{2}$ pair of complex numbers outside the daigonal (see for example p.158 of Analysis on symmetric cones by Faraut and Koranyi) $\endgroup$
    – pi_1
    Commented Mar 1, 2021 at 12:33
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    $\begingroup$ View an $n$-dimensional algebra as an $n^3$-tuple of numbers, namely coefficients of the law in a basis [and isomorphism means equivalence up to a suitable rule basis change]. The space $M_m(C)$ has a $C$-basis consisting of Hermitian matrices, in which the structure constants are real (and even integral). This basis is also a basis of the real algebra of Hermitian matrices, and the coefficients are just the same. This is all what you need. $\endgroup$
    – YCor
    Commented Mar 1, 2021 at 13:12
  • $\begingroup$ In concrete terms, take $ c = \lambda + i \mu$ and $d = \lambda- i \mu$. $\endgroup$
    – Will Sawin
    Commented Mar 3, 2021 at 21:53

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