Timeline for What are the reasons for considering rings without identity?
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| Jun 17, 2019 at 4:25 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
corrected a minor typo (the question has been bumped anyway)
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| Jan 26, 2015 at 2:35 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Apr 11, 2012 at 8:35 | comment | added | Marc Palm | @Kennan Kidwell: There is also one issue that $C_c^\infty(G) \ast C_c^\infty(G) = C_c^\infty(G)$, which does not hold in general. Very important, when you want to write induction functors as $C_c^\infty(G) \otimes_H -$ and such thing. | |
| Apr 26, 2010 at 17:59 | comment | added | Kevin Buzzard | @Harry Gindi: there is, but if you do it like that then you can't interpret the elements as locally-constant functions on the group, it's slightly more elaborate. It's all explained in Cartier Corvallis. If you're going to go for locally-constant functions on the group as elements then you clearly need a Haar measure because if K is a compact subgroup with characteristic function f then f*f tells you mu(K). Different choices of Haar measure give you canonically isomorphic algebras though, so another way of doing it would be taking all Haar measures at once and then taking the projective limit. | |
| Apr 26, 2010 at 12:57 | comment | added | Harry Gindi | I was going was going to give the same answer, but you beat me to it! +1! By the way, isn't there a way to construct the Hecke algebra without a choice of a Haar measure by looking at an operation on the measures themselves (the reason, if I remember correctly (though I may be wrong), that we have to fix a Haar measure is because we're applying a duality from integration theory)? | |
| Apr 26, 2010 at 12:24 | comment | added | Keenan Kidwell | That's a good point. I hadn't even thought of that and it's not clear to me that the bigger ring does have a natural action on V. | |
| Apr 26, 2010 at 12:15 | comment | added | Kevin Buzzard | I'm not so sure that this bigger ring would act on the vector space underlying a smooth representation of $G$. Let me try you on a trivial example. Let $G$ be a locally compact $p$-adic group. Fix once and for all a Haar measure on $G$. Let $V$ be the trivial 1-dimensional representation of $G$. Then the Hecke algebra of locally constant functions with compact support also acts on $V$: an element $f$ in this Hecke algebra acts by the constant equal to the integral of $f$ over $G$. Would your bigger ring also act naturally on $V$? More worryingly, what if $V$ is countably infinite-diml? | |
| Apr 26, 2010 at 11:25 | comment | added | Keenan Kidwell | I'm just curious, Kevin, if you're looking at the space of locally constant complex valued functions on a locally compact group G, which is a subspace of L^1(G), can't you just view it as a subspace of the Banach algebra (with identity) M(G) of complex Borel measures on G with convolution (where the "delta function" does exist as a measure)? Perhaps you don't want to leave the space you're working in? I don't really know anything about the applications you're talking about. | |
| Apr 26, 2010 at 10:41 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |