A How to switch from tensor products to wedge product

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1. Sep 22, 2016

victorvmotti

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. Sep 22, 2016

Staff: Mentor

What happens to your equation, if $A=B$?

3. Sep 22, 2016

victorvmotti

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. Sep 22, 2016

dextercioby

Yes, the computation is correct. The wedge product is a particular type of tensor product.

5. Sep 22, 2016

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