I'm wondering if anybody knows where one can find an English translation of Emmy Noether's classical paper E. NOETHER, Hyperkomplexe Grössen und Darstellungstheorie, Math. Zeit. 30(1929), 641–692 ?
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$\begingroup$ there is none, according to the bibliography. $\endgroup$– Carlo BeenakkerApr 19, 2021 at 14:04
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$\begingroup$ @CarloBeenakker could someone have informally made one available over the years on the web? It was mostly sections 23/24 I wanted but I'm afraid to plug it into Google translate and my German is virtually null except for ordering coffee. $\endgroup$– Benjamin SteinbergApr 19, 2021 at 14:10
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$\begingroup$ these are not many pages, I will give it a try. $\endgroup$– Carlo BeenakkerApr 19, 2021 at 15:06
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$\begingroup$ @CarloBeenakker, ah please don't take on too much work. It may very well be that 23 already has everything I want. If you stop there I can tell you if that covers what I was looking for. Maybe if you just give a high level view of what she is doing in 23/24 (and possibly 25) in modern terminology without going into detail that might let me know if it is relevant. And Thanks! I got the impression from a paper of Lam she is defining a Frobenius-Dedekind style determinant for any finite dimensional algebra and proves things about it. I'm trying to get an idea on the definition and results. $\endgroup$– Benjamin SteinbergApr 19, 2021 at 15:26
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$\begingroup$ done with sec. 23, if you need sec. 24 you will let me know (without the equations it's not much work) $\endgroup$– Carlo BeenakkerApr 20, 2021 at 12:40
1 Answer
E. NOETHER, Hyperkomplexe Grössen und Darstellungstheorie [Hypercomplex Quantities and the Theory of Representations], Math. Zeit. 30 (1929), 641–692; informal translation of section 23.
Please feel free to edit as you see fit
original
Determinant of a hypercomplex system.
Let $\mathbb{o}=a_1P+\ldots+a_h P$ be a hypercomplex system. I adjoin to $P$ a number $h$ of undetermined $x_1,\ldots x_h$ and construct
[equation]
The $x$ should be interchangeable with the $a_i$; therefore the computational rules of $\mathbb{o}^\ast$ are already determined.
In $\mathbb{o}^\ast$ there is a "general element of $\mathbb{o}$"
[equation]
If in a certain representation $a_i\mapsto A_i$, then $w$ is associated to
[equation]
$W$ is called the system matrix of the representation (or in particular, when the $a_i$ build a group with ring ${\mathbb{o}}$, the group matrix). If the representation is regular, then one has a regular system matrix.
The elements $w_{ik}$ of $W$ are linear forms of $x$. The "system determinant" $|W|$ is therefore of degree $n$, when the representation has degree $n$. In particular, the regular system determinant is of degree $h$.
The system determinant does not change when one changes to an equivalent representation, since
[equation]
In the transition from $(a_1,\ldots,a_h)$ to a new basis $(b_1,\ldots b_h)$ and from $w=\sum a_ix_i$ to $w=\sum b_iy_i$ one finds new elements of $W$ and therefore also a new determinant, when in the old determinant one substitutes
[equation]
with a regular substitution matrix.
If one has a decomposition of the representation module, the matrix $W$ in a suitable coordinate system would look like this
[equation]
The determinants $|W_i|$ of a particular representation are the same as in the regular representation of $\mathbb{o}$ or even of $\mathbb{o}/\mathbb{c}$, with $\mathbb{c}$ a radical.
If $P$ is algebraically closed, then the determinant $|W_i|$ belonging to an irreducible representation is a prime function in $x$, and inequivalent representations belong to distinct prime factors.
Proof: Because all irreducible representations of $\mathbb{o}$ are also representations of $\mathbb{o}/\mathbb{c}$, we can restrict ourselves to the ring without the radical $\mathbb{c}$. In this ring we take as basis the matrix unities $c_{ik}$; the general element then takes the form:
[equation]
The matrices of the irreducible representations are: $W_\nu=(x_{ik}^{(\nu)})$. The functions $|W_\nu|=|x_{ik}^{(\nu)}|$ are known to be irreducible and evidently distinct.
To calculate the $|W_\nu|$, one can decompose the regular system matrix of $\mathbb{o}/\mathbb{c}$ into prime factors. Each prime factor appears as often as the degree of the irreducible representation, because the corresponding ideal $\mathbb{l}_\nu$ appears equally often in the decomposition of the representation module.
One could also start from the regular system matrix of $\mathbb{o}$, but then one would obtain each irreducible factor more often, namely as often as the corresponding $\mathbb{l}_\nu$ appears as composition factor in the left-ideal. With this regular representation we considered $\mathbb{o}$ as the left-ideal; if instead one would consider $\mathbb{o}$ as the right-ideal, then one obtains a second regular system matrix (the "antistrophe matrix" of Frobenius), which has the same irreducible factors (namely, the system determinants of the collected irreducible representations). Possibly, the exponents of the irreducible factors are different (see the example in paragraph 10).
The system determinants of a commutative system decompose into linear factors, because all irreducible representations are of degree one. These linear factors are themselves the irreducible representations, so they are the characters of an Abelian group. This fact was the starting point of Dedekind's study of the group determinants of non-Abelian groups.
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$\begingroup$ Thanks for this. Can you make sense of what P and the a_i are? Is the a_i a basis? $\endgroup$ Apr 19, 2021 at 15:33
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$\begingroup$ Do you know what she means by a regular substitution matrix? I guess in modern terminology she is just choosing a generic element of the algebra and taking the determinant of its matrix under some representation rather than building a matrix out of the structure constants of the algebra as in one possible definition of the group determinant. I really appreciate your time. $\endgroup$ Apr 19, 2021 at 15:36
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$\begingroup$ @BenjaminSteinberg Did you ever get an English translation of the entire paper? $\endgroup$– TriOct 21, 2022 at 18:45
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$\begingroup$ No, but you translated what I needed. Thanks $\endgroup$ Oct 21, 2022 at 22:45