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Algebra Problems

  1. Feb 6, 2009 #1
    ok. today we discussed the following:

    The wedge product defines a bilinear multiplication in [itex]\Lambda(V^{*})[/itex] which is associateive i.e. [itex](\alpha \wedge \beta) \wedge \gamma = \alpha \wedge (\beta \wedge \gamma) \forall \alpha,\beta,\gamma \in \Lambda(V^{*})[/itex] and which is supercommutative i.e. if [itex]\alpha^{k} \in \Lambda^{k}(V^{*}),\beta^{l} \in \Lambda^{l}(V^{*})[/itex] then [itex]\alpha^{k} \wedge \beta^{l} = (-1)^{kl} \beta^{l} \wedge \alpha^{k}[/itex]

    Question 1: What is a bilinear multiplication?
    Question 2: Is my definition of supercommutative ok? I wasn't sure if it was [itex](-1)^{kl}[/itex] or [itex](-1)^{k+l}[/itex]? What does supercommutative mean?

    Then as an example of this, we considered the following:

    [itex]\alpha=e_1^* \wedge e_2^* - 3 e_2^* \wedge e_4^* = e_{12}^* -3 e_{24}^* \in \Lambda^{2}(V^{*}), \beta=3 e_1^* + 4e_2^*[/itex]

    Then,

    [itex]\alpha \wedge \beta=6 e_{12}^* \wedge e_1^* +8 e_{12}^* \wedge e_2^* - 9 e_{24}^* \wedge e_1^* - 12 e_{24}^* \wedge e_2^*[/itex]

    but we have[itex]e_{ij}^* \wedge e_j^* = e_i^* \wedge (e_j^* \wedge e_j^*)=0[/itex] so 3 terms drop out and we're left with

    [itex]\alpha \wedge \beta =- 9 e_{24}^* \wedge e_1^* = -9 e_2^* \wedge e_4^* \wedge e_1^8=\mathbf{- (-1)^{2} e_1^* \wedge e_2^* \wedge e_4^*=-e_{124}^*}[/itex]

    Question 3: Is [itex]\beta \in \Lambda^{1}(V^{*})[/itex]?
    Question 4: I don't understand how he got the part in bold. I have a feeling he just forgot to write in the 9 but the rearranging of that step uses the supercommutative algebra described above and I don't follow the logic there???

    Any help would be greatly appreciated

    cheers
     
  2. jcsd
  3. Feb 7, 2009 #2

    CompuChip

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    I don't really know much about this topic, but since nobody has replied yet I will try to help you with my general knowledge :smile:

    If you consider for example a vector space V, then a bilinear function on V is a function f on V x V which is linear in both its arguments. So if v, w, u are elements of V and x is a number, then
    f(a v + w, u) = a f(v, u) + f(w, u)
    f(v, a w + u) = a f(v, w) + f(v, u)

    You can see the wedge product as a function of two arguments:
    [tex]f(v, w) = v \wedge w[/tex]
    then bilinearity simply means that for example
    [tex]f(a v + w, u) = (a v + w) \wedge u = a (v \wedge u) + w \wedge u [/tex]

    I think so. Supercommutative probably either refers to the fact that you are commuting a k- and an l-form, or to the fact that it isn't precisely a commutation but some generalization where an extra sign may occur. Since I recall something like
    [tex]v \wedge w = - w \wedge v [/tex]
    when v and w are one-forms, I think it should be (-1)^(k l). That is: if both arguments are even or odd in grade you can switch them, if one is even and the other odd you get a minus sign.

    Yes. You can see it by the fact that it is expressed in the basis [itex]\{ e_1^*, e_2^* \}[/itex]. For a two-form you would need a basis like [itex]\{ e_{ij} = e_i \wedge e_j \mid i < j \}[/itex] (I am dropping the asterisks, too much work to put them in), for an n-form in general [itex]\{ e_{i_1 \cdots i_n} = e_{i_1} \wedge \cdots \wedge e_{i_n} \mid i_1 < \cdots < i_n \}[/itex]. (Note that this implies that if V is k-dimensional, there are no n-forms for [itex]n > k[/itex].

    Yes, it looks like he just forgot the 9. He commutes the e1 to the front, to have all the elements in a logical order.
     
  4. Feb 7, 2009 #3
    thanks a lot. quick questions though.

    (i)if you agree with me that [itex]\alpha \in \Lambda^2(V^*),\beta \in \Lambda^1(V^*)[/itex], then why don't we write them as [itex]\alpha^2[/itex] and [itex]\beta^1[/itex] since at the start of the definition we were considering an [itex]\alpha^k \in \Lambda^k(V^*)[/itex] etc?

    (ii)If we are calling them [itex]\alpha^k[/itex] just because it's in that set, how do we distinguish it from being [itex]\alpha \cdot ... \cdot \alpha[/itex] k times?

    (iii) What is [itex]\Lambda^{n}(V^*)[/itex] - is it a set?

    thanks for your help.
     
  5. Feb 7, 2009 #4
    I have a tiny bit of experience here as well, from a very basic study of exterior calculus in diffy geo. The only reason we used [itex]\alpha^k[/itex] in the previous scenario was to illustrate supercommutativity. It does not represent the wedge product of the same vector with itself k times. From the definition of the wedge product, [itex]\alpha^k = \alpha\wedge\dots\wedge\alpha[/itex] for any k > 1 is the 0 vector.

    The nth exterior power of a vector space is the vector space of all n-vectors, rigorously, elements of Vk = [itex]V\times\dots\times V[/itex], quotiented by a subspace defined by the wedge product. That is to say, each vector in [itex]\Lambda^{2}V[/itex] is an ordered pair of vectors in V, quotiented by the subspace made up of vectors of the form {(u+v, w) - [(u,w) + (v,w)], (u, v+w) - [(u,v) + (u,w)], a(u, v) - (au, v), (u, bv) - b(u, v), (u, u)}. As an example, if you regard R (the set of real numbers) as a vector space, you can regard the vector space [itex]\Lambda^{3}R[/itex] as a certain quotient space of R3, albeit with a strange wedge product structure and having all vectors with any identical components identified with the 0 vector. However, this construction is more useful in the development of tensors and tensor calculus, where we do operations on many vectors per tensor.
     
    Last edited: Feb 7, 2009
  6. Feb 8, 2009 #5
    ok cheers. why is that vector space quotiented by that subspace though?
     
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