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A How to switch from tensor products to wedge product

  1. Sep 22, 2016 #1
    Suppose we are given this definition of the wedge product for two one-forms in the component notation:

    $$(A \wedge B)_{\mu\nu}=2A_{[\mu}B_{\nu]}=A_{\mu}B_{\nu}-A_{\nu}B_{\mu}$$

    Now how can we show the switch from tensor products to wedge product below:

    $$\epsilon=\epsilon_{\mu_{1}...\mu_{n}}dx^{\mu_{1}}\otimes...\otimes dx^{\mu_{n}}$$
    $$=\frac{1}{n!}\epsilon_{\mu_{1}...\mu_{n}}dx^{\mu_{1}}\wedge...\wedge dx^{\mu_{n}}$$
     
  2. jcsd
  3. Sep 22, 2016 #2

    fresh_42

    Staff: Mentor

    What happens to your equation, if ##A=B##?
     
  4. Sep 22, 2016 #3
    Is this computation below for the case of ##n=2## correct?


    $$\epsilon= \frac{1}{2} ( \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\wedge dx^{\mu_{2}})$$
    $$= \frac{1}{2} (\epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} - \epsilon_{\mu_{2}\mu_{1}}dx^{\mu_{2}}\otimes dx^{\mu_{1}})$$
    $$= \frac{1}{2} ( \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} - (-1) \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} )$$
    $$= \frac{1}{2}(2\epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}})$$
    $$= \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}}$$
     
  5. Sep 22, 2016 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes, the computation is correct. The wedge product is a particular type of tensor product.
     
  6. Sep 22, 2016 #5

    fresh_42

    Staff: Mentor

    In this literally universal view every product is a tensor product.

    Edit: ##A \wedge A = 0## whereas ##A \otimes A## is not. This is a crucial difference.
     
    Last edited: Sep 22, 2016
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