Timeline for To find a DFT for complex functions on a semigroup
Current License: CC BY-SA 4.0
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| Aug 26, 2022 at 16:52 | history | edited | CommonAnts | CC BY-SA 4.0 |
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| Aug 26, 2022 at 15:27 | comment | added | Benjamin Steinberg | Note that $\mathbb C[G]\cong \mathbb C\times I$ where $I$ is the ideal of $\mathbb C[t]/(t^n)$ generated by $t+(t^n)$. The isomorphism takes $k$ to $(1,t^k)$ where I identify $G$ with the basis for $\mathbb C[G]$. | |
| Aug 26, 2022 at 15:25 | comment | added | Benjamin Steinberg | The above representation will be injective on $\mathbb C[G]$ iff $A^{n-1}\neq 0$. | |
| Aug 26, 2022 at 15:22 | comment | added | Benjamin Steinberg | Your question is a bit confusing. First of all a semigroup should be denoted $S$ and not $G$. In any event, there is no invertible homomorphism $\mathbb C[G]\to \mathbb C^{m\times m}$ because $\mathbb C[G]$ is not a unital algebra and also it is commutative and for many other reasons. Do you mean just an injective homomorphism? If I define representation to mean homomorphism then up to equivalence every representation of $G$ sends $k$ to $diag(A^k,I)$ where $A$ is a square matrix with $A^n=0$ and $I$ is an identity matrix. You can of course assume $A$ as a strictly upper triangular. | |
| Aug 26, 2022 at 14:57 | history | edited | CommonAnts | CC BY-SA 4.0 |
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| S Aug 26, 2022 at 14:29 | review | First questions | |||
| Aug 26, 2022 at 15:43 | |||||
| S Aug 26, 2022 at 14:29 | history | asked | CommonAnts | CC BY-SA 4.0 |