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But Schanuel's conjecture hasn't been proven. So would you say this reduces to Schanuel's conjecture?

seliopou 13y ago
Not necessarily. The proof presented depends Schanuel's conjecture, but there may be another proof that doesn't depend on that conjecture, directly or indirectly. The question you're asking is one of metalogic. It's like asking, what is the necessary structure of the proof of some proposition? Mathematicians definitely ask questions like that, but those are tough nuts to crack.
thomasbk 13y ago
No; for that to be the case, equivalence would have to be proven (i.e. construct a proof of Shanuel that works when you assume this theory to be true.) Right now, it could be that there exists a proof that is valid and does not depend on Shanuel's conjecture.
ot 13y ago
The Schanuel's conjecture is widely believed to be true. A friend of mine actually did his Ph.D. on it, and the tools used to attack the problem are truly mind-blowing. Basically, Zilber used model-theory to construct a field that "behaves like C" (the field of complex numbers), and where the conjecture is forced to be true. Not only this field exists, but it is unique. It would be no surprise then if this field was exactly C, and this would prove the conjecture for complex numbers. Note that to prove that the two fields are isomorphic it would suffice to prove that the "artificial" field has a continuous automorphism of order two (that would basically be the conjugation in C); then its fixed set would be the real line, and it would be easy to construct the rest of the isomorphism. The existence of such an automorphism is even less surprising, yet, nobody has managed to prove it.
- seliopou 13y ago
  That sounds so baller. But one question. You say that he constructed a field that "behaves like C." Granted, you put that phrase in quotes for a reason, but it seems to me as though if you pinned down the behavior of multiplication in this artificial field, then conjugation wouldn't be too far away. So what does it mean to "behave like C"? Or better yet, do you have a link to you friend's thesis or other papers?
  - ot 13y ago
    Of course the details are very tricky :) "behaves like C" means that you construct a first-order theory that all the properties of C that you want, plus the conjecture. I don't know much about this, but I think that there is no conjugation in the language of exponential fields, and if you try to add it everything becomes intractable (for one, every proof would first require to prove Schanuel's conjecture). The Wikipedia has a few references: https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber.27s_pseudo-exponentiation https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber...

Not necessarily. The proof presented depends Schanuel's conjecture, but there may be another proof that doesn't depend on that conjecture, directly or indirectly. The question you're asking is one of metalogic. It's like asking, what is the necessary structure of the proof of some proposition? Mathematicians definitely ask questions like that, but those are tough nuts to crack.

thomasbk 13y ago

No; for that to be the case, equivalence would have to be proven (i.e. construct a proof of Shanuel that works when you assume this theory to be true.) Right now, it could be that there exists a proof that is valid and does not depend on Shanuel's conjecture.

ot 13y ago

The Schanuel's conjecture is widely believed to be true. A friend of mine actually did his Ph.D. on it, and the tools used to attack the problem are truly mind-blowing. Basically, Zilber used model-theory to construct a field that "behaves like C" (the field of complex numbers), and where the conjecture is forced to be true. Not only this field exists, but it is unique. It would be no surprise then if this field was exactly C, and this would prove the conjecture for complex numbers. Note that to prove that the two fields are isomorphic it would suffice to prove that the "artificial" field has a continuous automorphism of order two (that would basically be the conjugation in C); then its fixed set would be the real line, and it would be easy to construct the rest of the isomorphism. The existence of such an automorphism is even less surprising, yet, nobody has managed to prove it.

seliopou 13y ago
That sounds so baller. But one question. You say that he constructed a field that "behaves like C." Granted, you put that phrase in quotes for a reason, but it seems to me as though if you pinned down the behavior of multiplication in this artificial field, then conjugation wouldn't be too far away. So what does it mean to "behave like C"? Or better yet, do you have a link to you friend's thesis or other papers?
- ot 13y ago
  Of course the details are very tricky :) "behaves like C" means that you construct a first-order theory that all the properties of C that you want, plus the conjecture. I don't know much about this, but I think that there is no conjugation in the language of exponential fields, and if you try to add it everything becomes intractable (for one, every proof would first require to prove Schanuel's conjecture). The Wikipedia has a few references: https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber.27s_pseudo-exponentiation https://en.wikipedia.org/wiki/Schanuel%27s_conjecture#Zilber...