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Wedge product

  1. Oct 11, 2014 #1
    If I want to take the wedge product of $$\alpha = a_i\theta^i $$ and $$\beta = b_j\theta^j$$ I get after applying antisymmetrization,$$ \alpha \Lambda \beta = \frac{1}{2}(a_ib_j - a_jb_i)\theta^i\theta^j$$

    My question is it seems to me that antisymmetrization technique doesn't apply to the basis $$\theta^i , \theta^j$$ right? Is it that the wedge product antisymmetrization jumps over those basis only affecting the components? Or is there something I am missing?

    Thanks!


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    Last edited by a moderator: May 7, 2017
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  3. Oct 11, 2014 #2

    WannabeNewton

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    ##\alpha \wedge \beta = \alpha_i \beta_j \theta^i \wedge \theta^j = \frac{1}{2}\alpha_i \beta_j (\theta^i \theta^j - \theta^j \theta^i) = \frac{1}{2}(\alpha_i \beta_j - \alpha_j \beta_i)\theta^i \theta^j ## is how the wedge product works on the basis 1-forms.
     
  4. Oct 11, 2014 #3
    Thank you for the reply, I guess you meant $$a_i , b_j $$ instead of $$\alpha_i , \beta_j$$?

    And so now, what I said in my question was totally wrong, it means according to this that it works on the basis rather than on the components. And then we flip indices accordingly, right?
     
  5. Oct 11, 2014 #4

    vanhees71

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    Shouldn't it be (concerning the factors 1/2):
    [tex]\alpha \wedge \beta=\alpha_i \beta_j \Theta^{i} \wedge \Theta^{j}=\frac{1}{2} (\alpha_i \beta_j - \alpha_j \beta_i) \Theta^{i} \wedge \Theta^{j} = (\alpha_i \beta_j-\alpha_j \beta_i) \Theta^{i} \otimes \Theta^{j}.[/tex]
     
  6. Oct 11, 2014 #5

    pervect

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    I'm not sure what the question is. I'd interpret what you wrote as the wedge product of a vector with a multiple of itself, which would be zero.
     
  7. Oct 12, 2014 #6

    vanhees71

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    No, [itex]\alpha[/itex] and [itex]\beta[/itex] are different vectors. Note that the Einstein summation convention is used here, and [itex]\Theta^j[/itex] is a basis of the dual space (space of linear forms).
     
  8. Oct 12, 2014 #7
    @vanhees71 thanks for your reply. But where are the components$$ a_i , b_j $$ And is there a general rule to this wedge product?
     
  9. Oct 12, 2014 #8

    vanhees71

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    I called them [itex]\alpha_i[/itex] and [itex]\beta_j[/itex]. Why should I introduce more symbols?

    The wedge product is defined as follows. If you have two antisymmetric tensors [itex]\omega_1[/itex] and [itex]\omega_2[/itex], i.e., antisymmetric linear mappings [itex]\omega_1:V^p \rightarrow \mathbb{R}[/itex] and [itex]\omega_2:V^q \rightarrow \mathbb{R}[/itex] (also called [itex]p[/itex]- and [itex]q[/itex]-forms respectively), the wedge product is a [itex](p+q)[/itex] form which is defined by
    [tex](\omega_1 \wedge \omega_2)(\vec{v}_1,\ldots,\vec{v}_p,\vec{v}_{p+1},\ldots,\vec{v}_{p+q})=\sum_{\tau \in S_{p+q}} \frac{\text{sign}(\tau)}{p! q!} \omega_1((\vec{v}_1,\ldots,\vec{v}_p) \omega_2(\vec{v}_{p+1},\ldots,\vec{v}_{p+q}).[/tex]
    Here [itex]S_{p+q}[/itex] denotes the permutation of [itex](p+q)[/itex] elements.
     
  10. Oct 12, 2014 #9

    Matterwave

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    There's a mismatched parenthesis in the last expression here.
     
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