Timeline for Do 1-additive maps admit tensor products?
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19 events
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| Dec 23, 2019 at 16:54 | comment | added | Yemon Choi | Tomek: It's been too long since I read the paper, although MCW did once tell me that he thought a factor of 2 was missing somewhere, and I couldn't remember whether this meant the value stated by K+R was too small or too large :) | |
| Dec 23, 2019 at 16:48 | comment | added | Tomasz Kania | @YemonChoi, $K=44.5$ is perfectly fine. ;-) | |
| Dec 23, 2019 at 16:46 | comment | added | Yemon Choi | @PietroMajer I have not thought about Tomek's question or your modification, but the paper of Kalton and Roberts which is alluded to in the question shows that every $\delta$-additive measure has total variation distance at most $K\delta$ from a $0$-additive measure, where $K$ is some universal constant independent of $\mathcal F$; I think maybe $K=100$ is good enough? (One of the ideas suggested to me by my PhD supervisor was to find a "better" or "more cohomological" proof, but 15 years later I still haven't succeeded) | |
| Dec 23, 2019 at 13:04 | vote | accept | Tomasz Kania | ||
| Dec 23, 2019 at 12:50 | comment | added | Pietro Majer | Maybe a reasonable question is: for what $\epsilon$ does there exist the product of two given $\delta$-additive measures? | |
| Dec 23, 2019 at 12:33 | answer | added | Andrea Marino | timeline score: 3 | |
| Dec 23, 2019 at 10:43 | comment | added | Andrea Marino | Well I mean, I was just pointing that this allows to search for a unique tensor product and not "a" tensor product. The approach I was thinking is inductive: cover your big rectangle with the all the maximal proper subrectangles (which are of the form $X \setimus \{x\} \times Y$ and the same for Y). If you take a minimal tensor product on each rectangle in the sense above, they coincide on sets which are contained in more than one subrectangle. In general this is not true if you take an arbitrary tensor on each subrectangle! Now we are left with trying to extend this to the rest. | |
| Dec 23, 2019 at 9:08 | comment | added | Tomasz Kania | @AndreaMarino, yes, but the point is to show that such $h$ exists :-) Now, I think it should not be possible in general. | |
| Dec 23, 2019 at 9:07 | history | edited | Tomasz Kania | CC BY-SA 4.0 |
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| Dec 23, 2019 at 9:00 | comment | added | Andrea Marino | No I mean for the h on the product, so that f and g on X and Y are fixed. In this case you have $h(S) \le \sum_{s \in S} h(s) + n$ and similarly for the other inequality. Isn't it? I am not saying that 1-additive functions are compact but that the space of "possible tensor products" is. | |
| Dec 23, 2019 at 8:46 | comment | added | Andrea Marino | I found something that can be useful. The space of such functions is a subset of $\mathbb{R}^{N}$ for some $N$, and I claim that it is compact and convex. Compactness come from repeated use of the inequality with singleton, that have assigned measure because they are rectangles. Furthermore, a convex combination preserve both the conditions. Thus, you can add the further condition on h having minimum modulo; such function will exist and will be unique provided that there exist a 'tensor' product. This allows to 'glue' solutions by uniqueness on intersection. | |
| Dec 23, 2019 at 0:37 | comment | added | Andrea Marino | Have you tried with some simple cases like X,Y having two elements? Or maybe with arbitrary number of elements but every subset having measure 1? | |
| Dec 22, 2019 at 20:41 | comment | added | Sylvain JULIEN | Hence the requirement is $\sigma(h(A\times B)=h(\sigma(A\times B))=h(\sigma(A)\times\sigma(B))=\sigma(f(A).g(B))=f(\sigma(A)).g(\sigma(B))$. | |
| Dec 22, 2019 at 20:28 | comment | added | Sylvain JULIEN | I don't know the subject, but from an algebraic point of view, the number of "canonical admissible decompositions" of a given set should depend on the symmetries thereof. I think one should study the group of permutations $\sigma$ of $1$-additive "building blocks" that commute to $f$, $g$ and $h$. | |
| Dec 22, 2019 at 19:23 | history | edited | Tomasz Kania |
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| Dec 22, 2019 at 19:17 | history | edited | Tomasz Kania | CC BY-SA 4.0 |
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| Dec 22, 2019 at 11:03 | history | edited | Tomasz Kania | CC BY-SA 4.0 |
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| Dec 22, 2019 at 5:57 | history | edited | Tomasz Kania | CC BY-SA 4.0 |
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| Dec 21, 2019 at 20:33 | history | asked | Tomasz Kania | CC BY-SA 4.0 |