Talk:Connes embedding problem

CorenSearchBot is indeed in error. Henry Delforn (talk) 08:51, 28 November 2009 (UTC)[reply]

How so? This appears to be a close copy of [1], unless I'm missing something. — The Earwig @ 15:56, 28 November 2009 (UTC)[reply]

i think the bot missed the context in which the wording is used. The reference cited by the bot appears to be an advertisement or flyer for a colloquium, the "non-free" terminology (as mentioned in the tag) is with regards to attending the colloquium not the paper itself which is just a flyer. Doing a general search for key words in the previous version of the article turns up several other exact advertisements. Regardless, article copy was edited substantially to remove any close paraphrasing. Henry Delforn (talk) 20:44, 28 November 2009 (UTC)[reply]

Ambiguity

In noncommutative mathematics, the Connes embedding problem or conjecture asks whether every finite von Neumann algebra (separable II1 factor) can be embedded into the ultrapower of the hyperfinite II1 factor (countably generated finite factors)[...]

I rewrote the sentence to read as above. Formerly it say "any finite von Neumann algebra". In this context, "any" is ambiguous. Is it the case that "any finite von Neumann algebra" can be so embedded? If so, then "every" is true. But "asks whether any finite von Neumann algebra" can be so embedded could be construed as: "asks whether there is any finite von Neumann algebra" that can be so embedded. If there is one, then the answer is "yes", even if "every" is not true. Michael Hardy (talk) 01:15, 30 November 2009 (UTC)[reply]