Timeline for Do convolution and multiplication satisfy any nontrivial algebraic identities?
Current License: CC BY-SA 3.0
8 events
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| Jan 14, 2018 at 7:35 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
corrected minor typos (the question has been bumped anyway)
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| Jan 14, 2018 at 7:30 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
added MathJax (the question has been bumped anyway)
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| Nov 3, 2009 at 2:25 | comment | added | Theo Johnson-Freyd | With just the two ring structure, I don't think you can extract the rest of the structure. Indeed, I have a very hard time imagining identities given your restrictions. For example, imagine scrambling the elements of the group. This does not change the multiplication algebra (which depends only on the set structure), but picks out a different convolution product. | |
| Nov 1, 2009 at 9:06 | comment | added | Darsh Ranjan | I still don't see how to extract that extra structure (Hopf algebra, bialgebra, ...) if all you have are the two ring structures. I guess right now I'm more interested in what you can say about the operations themselves if you have nothing else to work with; on the other hand, the more general question of how the two ring (or k-algebra) structures interact is interesting as well. | |
| Nov 1, 2009 at 7:11 | comment | added | Theo Johnson-Freyd | More generally, I don't think you've quite posed the question well. At least, if the question is: "Let R and S be commutative rings of the same dimension. Is it possible for R and S to be the multiplication and convolution algebras of the same abelian group?" then I'm sure the answer is "no." The reason is that for any abelian group, the two algebra structures along with the canonical pairing are coherent in that they form a Hopf algebra. And I don't believe that any two algebras (even if they individually come from groups) can be made to satisfy the bialgebra identity. | |
| Nov 1, 2009 at 7:06 | comment | added | Theo Johnson-Freyd | So, I agree that I'm using slightly more structure than you: I'm using the canonical pairing &\int;. But I am not using any distinguished elements of the algebra. Here's a better way to explain the construction. Let V be any finite-dimensional inner-product space. The inner product identifies V = V* (the dual space), and so V ⊗ V = V &otimes V* = Hom(V,V). The identity map in Hom(V,V) is absolutely canonical. Pull it back to V ⊗ V, and you get the element I'm calling "the inverse to the pairing". To define it explicitly, I wrote it in a basis. But any basis will do. | |
| Nov 1, 2009 at 4:17 | comment | added | Darsh Ranjan | Theo, thanks for writing this up. I hadn't thought about it this way before. Your examples of identities don't work, though, since you're making use of distinguished elements of the algebra (i. e., nullary operators), namely δ_x for x in G. Actually, since, as you say, convolution and multiplication usually happen in different vector spaces, we probably shouldn't expect any nontrivial identities (and I don't), but it seems like something somebody ought to have proved... | |
| Oct 31, 2009 at 23:33 | history | answered | Theo Johnson-Freyd | CC BY-SA 2.5 |