Re: What's the defenition of Wedge product ?


Thomas Mautsch wrote:

Thanks, but why is that so? - Doesn't the "usual" naive definition
of the projection operator from the p-fold tensor product
to the p-fold exterior product of a vector space
contain a factor

1/p!

in front of a certain sum of terms?

Sorry I misunderstood the question. Obviously, if p divides n, then in
characteristic p you can't divide by n! and hence you can't chop
something up and spread it evenly over all n! permutations. You can't,
likewise, average something over all n! permutations.

.

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