How to switch from tensor products to wedge product

  • #1
victorvmotti
153
5
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:

$$\epsilon=\epsilon_{\mu_{1}...\mu_{n}}dx^{\mu_{1}}\otimes...\otimes dx^{\mu_{n}}$$
$$=\frac{1}{n!}\epsilon_{\mu_{1}...\mu_{n}}dx^{\mu_{1}}\wedge...\wedge dx^{\mu_{n}}$$
 

Answers and Replies

  • #3
victorvmotti
153
5
Is this computation below for the case of ##n=2## correct?


$$\epsilon= \frac{1}{2} ( \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\wedge dx^{\mu_{2}})$$
$$= \frac{1}{2} (\epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} - \epsilon_{\mu_{2}\mu_{1}}dx^{\mu_{2}}\otimes dx^{\mu_{1}})$$
$$= \frac{1}{2} ( \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} - (-1) \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} )$$
$$= \frac{1}{2}(2\epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}})$$
$$= \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}}$$
 
  • #4
dextercioby
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Yes, the computation is correct. The wedge product is a particular type of tensor product.
 
  • #5
fresh_42
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Yes, the computation is correct. The wedge product is a particular type of tensor product.
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.
 
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