Skip to main content
Misha Verbitsky's user avatar
Misha Verbitsky's user avatar
Misha Verbitsky's user avatar
Misha Verbitsky
  • Member for 14 years, 11 months
  • Last seen this week
comment
2-dimensional sublattices with all vectors having very big square (in absolute value)
strike the previous comment, my proof was based on false assumption that any lattice is commeasurable with one which is proportional to unimodular
comment
2-dimensional sublattices with all vectors having very big square (in absolute value)
Yes, and this seems to give an answer indeed. I will post it in a couple of days, once I am entirely sure there are no errors
comment
2-dimensional sublattices with all vectors having very big square (in absolute value)
thanks! anyway, rank $\geq 6$ or $\geq 7$ is a usual assumption in these kind of applications
Loading…
awarded
revised
Loading…
comment
2-dimensional sublattices with all vectors having very big square (in absolute value)
definitely! thanks for pointing this out, I amended the question
revised
Loading…
comment
2-dimensional sublattices with all vectors having very big square (in absolute value)
Sorry! I misstated the question: I should add an assumption that this 2-dimensional lattice is not definite!
Loading…
awarded
awarded
awarded
accepted
answered
Loading…
awarded
awarded
awarded
comment
Complex structure on $S^6$ gets published in Journ. Math. Phys
I apologize for being overly blunt (and thanks to those who pointed this out). The question is edited. For the record, I said "I have read it when it first appeared in arxiv 10 years ago, there were not very subtle errors then. Is it still wrong? "
revised
Loading…
1
10 11
12
13 14
22