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Definition of the wedge product on the exterior algebra of a vector space

  1. Jan 2, 2013 #1
    Hi,

    I am currently reading about differential forms in "Introduction to Smooth Manifolds" by J. M. Lee, and I was wondering exactly how you define the wedge product on the exterior algebra [itex]\Lambda^*(V) = \oplus_{k=0}^n\Lambda^k(V)[/itex] of a vector space [itex]V[/itex]. I understand how the wedge product is defined as a map
    [itex]\Lambda^k(V)\times \Lambda^l(V) \rightarrow \Lambda^{k+l}(V)[/itex].
    Is the idea simply to extend this definition bilinearly to [itex]\Lambda^*(V)[/itex]? I.e. given elements [itex](T_0, ..., T_n)[/itex] and [itex](S_0, ..., S_n)[/itex] in [itex]\Lambda^*(V)[/itex], do we simply calculate all the combinations [itex]T_j \wedge S_k[/itex], add the obtained forms that have the same degree, and finally let these sums make up the new element of [itex]\Lambda^*(V)[/itex]? This seems to me like an intuitive definition, but Lee wasn't very specific about this, so I thought I should make sure this is correct.

    Please let me know if there is something in my post that seems unclear.
     
  2. jcsd
  3. Jan 2, 2013 #2

    dextercioby

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    Actually, the wedge product is placed as an internal operation on the exterior algebra right from the definition. That's the whole purpose of defining the direct sum of vector spaces, to make an external operation (from the pov of the vector spaces) internal. The wedge product takes an element of V* in another (typically different) element of V*, hence it's an internal operation on the exterior algebra. The direct sum vector space together with the wedge product is a Z_2 graded algebra.
     
  4. Jan 12, 2013 #3
    Thanks for your reply! I'm not sure if I see what you mean, though. I was primarily interested in whether my "rule" for calculating the wedge product between two elements of the exterior algebra was correct or not.
     
  5. Jan 12, 2013 #4

    quasar987

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    Yes, that is it!
     
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