Clifford algebra isomorphic to tensor algebra or exterior algebra?

[precondition]

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~~

 

[Hurkyl]

One of the names for \Lambda(V) 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]

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 Cl(V) and \Lambda(V) is exploited by defining a second product on (vector space) \Lambda(V) that makes (vector space) \Lambda(V) with new product isomorphic to Cl(V).