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Suppose that $A$ is a $d^2$ dimensional algebra over $\mathbb{C}$ and we know the multiplication tensor $c_{ij}^k$ and the unit $u^k$ in some basis. If $A$ is semi-simple and has a single simple representation, then computing an explicit isomorphism $A \cong M_d(\mathbb{C})$ is often called the explicit isomorphism problem. This is equivalent to constructing an explicit $d$ dimensional representation of $A$.

Most of the literature I have been able to find about the explicit isomorphism problem assumes you are working over some number field. I want to know the state of the art for working over $\mathbb{C}$. If $d=2$, Grobner bases work fine, but I don't know what I am doing and this approach takes too long when $d>2$.

Question: What is known about the explicit isomorphism problem over $\mathbb{C}$?

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When specifying an algorithmic problem "over $\mathbb{C}$", you should say precisely what the input and output should be.

The isomorphism problem is less studied over $\mathbb{C}$ because it is much easier than over a number field.

Here is the basic idea: if you can find an element $a\in A$ such that right multiplication by $a$ on $A$ has rank $d$ (i.e. the corresponding matrix, which you don't know yet, has rank $1$), then the left ideal $Aa$ is isomorphic to $\mathbb{C}^d$ as a vector space, and left multiplication on $Aa$ induces an isomorphism $A\cong M_d(\mathbb{C})$. If you are able to draw "random" elements in your algebra, then with probablility $1$ a random element $b\in A$ will correspond to a matrix with $d$ distinct eigenvalues; then the characteristic polynomial of right multiplication of $b$ on $A$ is $\prod_{i=1}^d(X-\lambda_i)^d$ and from this factorisation you can find a polynomial $P$ such that $a=P(b)$ corresponds to a matrix of rank $1$.

Some references :

  • Eberly, Decomposition of algebras over R or C
  • de Graaf, Ivanyos, Finding maximal tori and splitting elements in matrix algebras.
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  • $\begingroup$ That makes sooooo much sense. Thank you very much :D $\endgroup$ Commented Feb 21, 2019 at 8:08
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    $\begingroup$ A problem here seems to be that in practice one won't be able to factor the characteristic polynomial. $\endgroup$ Commented Feb 21, 2019 at 8:36
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    $\begingroup$ It still works really nicely. I have been using the same trick for commutative algebras for ages, it just never occurred to me that it generalized $\endgroup$ Commented Feb 21, 2019 at 8:44
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    $\begingroup$ yeah it doesn't work once d is too big, but i just did a d=4 $\endgroup$ Commented Feb 21, 2019 at 8:58
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    $\begingroup$ i am generating random elements like this: (x1/100,x2/100,...) where x1,x2,... are indipendently sampled from between 1 and 100 $\endgroup$ Commented Feb 21, 2019 at 8:59

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