Graduate How to switch from tensor products to wedge product

[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:

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

 

[fresh_42]

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

 

[victorvmotti]

Is this computation below for the case of $n=2$ correct?$$\epsilon= \frac{1}{2} ( \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\wedge dx^{\mu_{2}})$$
$$= \frac{1}{2} (\epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} - \epsilon_{\mu_{2}\mu_{1}}dx^{\mu_{2}}\otimes dx^{\mu_{1}})$$
$$= \frac{1}{2} ( \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} - (-1) \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}} )$$
$$= \frac{1}{2}(2\epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}})$$
$$= \epsilon_{\mu_{1}\mu_{2}}dx^{\mu_{1}}\otimes dx^{\mu_{2}}$$

 

[dextercioby]

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

 

[fresh_42]

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.

 

[Ayush_Hazarika]

Point 2. how we are getting 1/2 in front of the tensor product? it should have been A^B=AxB-BxA