Clifford algebra isomorphic to tensor algebra or exterior algebra?

Unfortunately there seems to be a misprint in the paper I'm reading which is an introduction to clifford algebra, it says:(I highlighted in red possible misprint, either one of them has to be true misprint if you know what I mean)

The Clifford algebra C(V) is isomorphic to the tensor algebra Lambda(V) and is therefore a 2^{dim(V)} dimensional vector space with generators blah blah blah...

Now, I know C(V) is defined as T(V)/I with you know what "I" so I'm wondering how can there be isomorphism between C(V) and T(V) but on the other hand dimension 2^{dim(V)} is indeed dimension of tensor algebra right? Also the author said tensor algebra but then wrote Lambda...-_-

I'm confused~~
 
Last edited:

Hurkyl

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One of the names for [itex]\Lambda(V)[/itex] is the "antisymmetric tensor algebra (over V)".


Incidentally, while they are always isomorphic as vector spaces, I think they are only isomorphic as algebras when the Clifford algebra is built from the zero quadratic form.
 

George Jones

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Incidentally, while they are always isomorphic as vector spaces, I think they are only isomorphic as algebras when the Clifford algebra is built from the zero quadratic form.
Yes.

Somtimes the vector space isomorphism between [itex]Cl(V)[/itex] and [itex]\Lambda(V)[/itex] is exploited by defining a second product on (vector space) [itex]\Lambda(V)[/itex] that makes (vector space) [itex]\Lambda(V)[/itex] with new product isomorphic to [itex]Cl(V).[/itex]
 

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