Timeline for Multizeta function values
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 27, 2016 at 7:51 | vote | accept | Nguyen lan Lee | ||
| Dec 27, 2016 at 5:27 | comment | added | Vesselin Dimitrov | (continued.) Grothendieck's conjecture contains many of the classical theorems and conjectures in transcendence. It is completely open of course. On MZV, it states that all the relations lift to relations among the $\zeta^{\mathbb{m}}(\boldsymbol{s})$. Determining the latter is easier as there is additional structure available that is not seen on the level of numbers (the periods). This may be compared to proving the algebraic independence of functions vs. proving algebraic independence of their special values, as in Schanuel's conjecture. | |
| Dec 27, 2016 at 5:19 | comment | added | Vesselin Dimitrov | @johnmangual: The construction looks complicated, but the idea is to start not with the series expression of the MZV, but with the alternative iterated integral representation (which is elementary, and discovered by Kontsevich). Ultimately this expresses an MZV as a period of a mixed Tate motive. Now, with periods of motives, there is a set of algebraic relations that are said to 'come from geometry' (or have 'a motivic origin'). Grothendieck's period conjecture states that all the algebraic relations have a motivic origin. | |
| Dec 26, 2016 at 19:19 | comment | added | john mangual | i try not to comment too much on here. what does it mean motivic? $\zeta(\mathbf{s})^\mathfrak{m}$ the definition is too difficult yet Brown was successful there rather with classical zeta values $\zeta(\mathbf{s})$ with $\mathbf{s} \in \{2,3 \}^\times$ | |
| Dec 26, 2016 at 17:39 | comment | added | T. Amdeberhan | Here is one place to start people.mpim-bonn.mpg.de/zagier/files/scanned/… | |
| Dec 26, 2016 at 17:15 | comment | added | Nguyen lan Lee | thank you for your answer. How it is related to the modular forms of weight n ? | |
| Dec 26, 2016 at 14:38 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Dec 26, 2016 at 14:07 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Dec 26, 2016 at 14:00 | history | answered | Vesselin Dimitrov | CC BY-SA 3.0 |