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Does $VP \neq VNP$ imply $P \neq NP$?

In his GCT papers Mulmuley first starts with the #P/NC question. And then gets into setting up a roadmap for how to show obstructions ( some sort of Weyl modules I gathered, if they can be constructed) could potentially rule out embedding of a class variety into another.

Is it clear, and almost obvious how that chain of argument ( if and when fleshed out) will automatically extend for P/NP separation? I am not a TCS expert - so I wonder if I am missing something obvious.

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Arnab
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    See this question: http://cstheory.stackexchange.com/questions/529/does-vp-neq-vnp-imply-p-neq-np – Aaron Sterling Jul 10 '11 at 20:52
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    FYI, @Arnab asked about how the GCT chain of argument extends to $P$ vs $NP$, and there is actually a reasonable answer, beyond the answers to the question of whether $VP \neq VNP$ implies $P \neq NP$. In brief, one can associate certain functions to $P$ and $NP$ and use those functions in place of $det$ and $perm$, and then the whole GCT chain of reasoning goes through for those functions. There are a few important technical issues to be discussed about this, but that is the idea, and the technical issues do not fit in a comment. – Joshua Grochow Aug 07 '11 at 03:23

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