Is Hom(Lambda^k(V),Lambda^(k+1)(V)) and element of End(Lambda(V))?

  1. Sorry I don't know how to write in symbols so I'm using Latex codes, anyway,
    the question is as in title:

    Is an element of Hom(Lambda^k(V),Lambda^(k+1)(V)) and element of End(Lambda(V))?

    In words, is an element of the homomorphism between k th grade exterior algebra over V and k+1 th exterior algebra over V also an element of endomorphism of exterior algebra over vector space V?

    I appreciate your help
     
  2. jcsd
  3. Hurkyl

    Hurkyl 16,090
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    What is the domain of an element of

    Hom(Lambda^k(V),Lambda^(k+1)(V))?

    What is the domain of an element of

    End(Lambda(V))?

    Can a function have two different domains?
     
  4. matt grime

    matt grime 9,396
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    I'm afraid I'm going to disagree. Let n be the dimension of V. Then canonically

    [tex] \Hom(\Lambda(V),\Lambda(V))= \oplus\Hom(\Lambda^r(V),\Lambda^s(V)[/tex]

    0<=r,s<=n.

    Thus any element of the vector space Hom(\Lambda^k(V),\Lambda^{k+1}(V)) is canonically an element of End(\Lambda(V)).

    The question boils down to: if x is in X, is x also an element of X\oplus Y, and it is. I think the function notion is very misleading at times.
     
  5. I agree with matt grime

    Yes I agree with what matt grime said,
    When I carefully read the text again it said,
    "Let V-->Hom(Lambda^kV, Lambda^(k+1)V) be the action of V on LambdaV"
    and since action of V on LambdaV is precisely the homomorphism between V and End(LambdaV) I can see that an element of Hom(Lambda^kV, Lambda^(k+1)V) is precisely an element of End(Lambda V)

    Thank you for your help anyways
     
  6. Hurkyl

    Hurkyl 16,090
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    But being canonically an element isn't the same as actually being an element... which is what I had asumed the OP meant.

    And there's a problem with this argument: it would also prove, for example, that every 1x1 real matrix is canonically an element of any group of mxn real matrices.

    I had also assumed End(\Lambda V) was the ring of algebra endomorphisms, not vector space endomorphisms. The canonical embedding you describe doesn't live in the subset of algebra endomorphisms.

    I'm pretty sure that the text doesn't mean

    For any particular k, V-->Hom(Lambda^kV, Lambda^(k+1)V) is the action of V on Lambda V​

    but instead means

    The collection of all V-->Hom(Lambda^kV, Lambda^(k+1)V) constitutes the action of V on Lambda V​

    which makes more sense, because to have an endomorphism of V, you have to know what happens to every Lambda^j V. (The canonical embedding matt describes for a particular k assumes the zero map on every j unequal to k)
     
  7. matt grime

    matt grime 9,396
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    It doesn't prove that at all. There is nothing canonical about the description you just gave. Anything that invokes matrices is by definition not canonical. Although I agree it is a wooly question: if 1 is a real number is 1 an element of C?
     
    Last edited: Dec 30, 2006
  8. Hurkyl

    Hurkyl 16,090
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    Canonically,

    [tex]
    (\mathbb{R}^m, \mathbb{R}^n)
    \cong \bigoplus_{\substack{1 \leq i \leq m \\ 1 \leq j \leq n}} ( \mathbb{R}, \mathbb{R} )
    [/tex]

    I guess my real point, though, is that it's an abuse of language to say "An element of X is an element of [itex]X \oplus Y[/itex]", since what we really mean is "we have an embedding [itex]X \mapsto X \oplus Y[/itex] which we will invoke implicitly as needed". Using an example with a repeated summand was just an indirect way to get at this.
     
    Last edited: Dec 30, 2006
  9. matt grime

    matt grime 9,396
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    I never have liked canonical. How are you picking out a summand, canonically? What if I change bases? I then change that decomposition. If you start writing down copies of R like that it stops being canonical in any meaningful sense. The fact is that a 2 dimensional real vector space is not canonically isomorphic to R\oplus R.


    The decomposition of Lambda(V) does not depend on any choice of basis in V.
     
    Last edited: Dec 30, 2006
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