Good morning, I am looking for a particular product between matrices and vectors defined in the following way:

$ (A\star v)_i=\prod_{j}v_j^{a_{ij}} $

I can give you also and example: consider

$ A=\left(\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right) \qquad v=\left(\begin{array}{c} a \\ b \end{array}\right) $

The star product gives

$ A\star v=\left(\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right)\star \left(\begin{array}{c} a \\ b \end{array}\right)=\left(\begin{array}{c} ab^2 \\ a^3b^4 \end{array}\right) $

I met this kind of definition when I was dealing with change of variables for some kind parametrization of the characters of a representation. In particular they used the $\mathfrak{SU}(3)_{\mathbb{C}}$ Cartan matrix as the previous $A$ matrix in such a way they can change the variables for a given representation in variables in the representation of the Cartan matrix. Is this something that I can find in any book? I tried on google but I think that it is not called star product, or anyway it is something so specific that I could not be able to find any proper definition.

Last question: If this kind of change of variables makes sense, is it possible to make it in such a way it is not invertible? I can explain: if I have some variables in a representation, for instance $x_1,x_2,x_3$, it is possible to find a not square matrix in such a way the they can be expressed as one variable $z$? I think that I can reformulate: there are not square Cartan matrices?

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    $\begingroup$ You can take logarithm of the $v$, do the normal product and then take element-wise exponentiation. $\endgroup$ – jeckerya Mar 13 '17 at 8:51
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    $\begingroup$ I don't know what you mean by the "not invertible" question. But I definitely have seen this "star product" and used it for variable substitutions in Laurent polynomial rings. I have even used a star for it in my notations! See Definition 0.1 in dropbox.com/s/ow1tk0xb85d562r/exe3.pdf?dl=0 (link is not permanent; this is not a proper preprint). $\endgroup$ – darij grinberg Mar 13 '17 at 8:53
  • $\begingroup$ @darijgrinberg I think that the star product described in your .pdf is exactly what I was looking for. Thank you very much. I will read all the document so that I can understand better in what consist. $\endgroup$ – Alessandro Mininno Mar 13 '17 at 9:43
  • $\begingroup$ FYI you tagged this question as Cartan geometry but it is unrelated. Cartan geometry studies spaces that are locally like Klein geometries with their Maurer-Cartan forms. $\endgroup$ – ಠ_ಠ Mar 13 '17 at 21:33
  • $\begingroup$ @darijgrinberg I would like to cite you (or your references) in my thesis regarding this star product you gave to me the notes. Could you give to me some information about the documents, who you are, when they have been written, or whatever you surely know is necessary in order to add you in my bibliography? Anything, books, complete notes, documents, names ecc. Also in private, tell me if there is a way to contact you and I can write you an email, you have been very useful to me. Thank you in advance $\endgroup$ – Alessandro Mininno Jun 24 '17 at 7:55

If $M$ is a monoid and $k$ is a commutative ring, then we can consider the monoid algebra $k[M]$. When $M = \mathbb{N}^n$ this recovers the polynomial ring over $k$ in $n$ variables, and when $M = \mathbb{Z}^n$ this recovers the Laurent polynomial ring over $k$ in $n$ variables.

If $f : M \to N$ is a homomorphism of monoids, we get an induced homomorphism of monoid algebras $k[f] : k[M] \to k[N]$.

Finally, if $M = \mathbb{N}^m, N = \mathbb{N}^n$, then homomorphisms $M \to N$ correspond exactly to $n \times m$ matrices over $\mathbb{N}$, which compose as matrices do, and similarly for $\mathbb{Z}^m$ and $\mathbb{Z}^n$. These induce homomorphisms on polynomial resp. Laurent polynomial algebras which correspond to your operation.


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