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Clifford algebra isomorphic to tensor algebra or exterior algebra?

  1. Dec 22, 2006 #1
    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: Dec 22, 2006
  2. jcsd
  3. Dec 22, 2006 #2

    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.
     
  4. Dec 23, 2006 #3

    George Jones

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    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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