# Definition of tensor product

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1. May 27, 2015

### "Don't panic!"

I'm relatively new to differential geometry and would like to check that this is the correct definition for the tensor product of (for simplicity) two one-forms $\alpha,\;\beta\;\;\in V^{\ast}$: $$(\alpha\otimes\beta)(\mathbf{v},\mathbf{w})=\alpha (\mathbf{v})\beta (\mathbf{w})$$ where $\alpha\otimes\beta\;\;\in V^{\ast}\otimes V^{\ast}$ and $\mathbf{v},\;\mathbf{w}\;\;\in V$.
Given this, is it correct to write, $$(dx^{\mu}\otimes dx^{\nu})(\mathbf{v},\mathbf{w})=dx^{\mu}(\mathbf{v})dx^{\nu}(\mathbf{w})=V^{\mu}W^{\nu}$$
where we have expressed $\mathbf{v}=V^{\mu}\partial_{\mu}$ and $\mathbf{w}=W^{\nu}\partial_{\nu}$ in terms of a coordinate basis $\lbrace\partial_{\mu}\rbrace$ for $V$, and $\lbrace dx^{\mu}\otimes dx^{\nu}\rbrace$ is a coordinate basis for $V^{\ast}\otimes V^{\ast}$ (with $\lbrace dx^{\mu}\rbrace$ a basis for $V^{\ast}$). As such, if we express $\alpha$ and $\beta$ in terms of the coordinate basis $\lbrace dx^{\mu}\rbrace$ as $\alpha = \alpha_{\mu}dx^{\mu}$ and $\beta = \beta_{\nu}dx^{\nu}$, respectively, we have $$(\alpha\otimes\beta)(\mathbf{v},\mathbf{w})=\alpha (\mathbf{v})\beta (\mathbf{w})=\alpha_{\mu}\beta_{\nu}dx^{\mu}\otimes dx^{\nu}(\mathbf{v},\mathbf{w})=\alpha_{\mu}\beta_{\nu}V^{\mu}W^{\nu}$$.

Would this be correct at all?

2. May 27, 2015

### Orodruin

Staff Emeritus
Yes.

3. May 27, 2015

### "Don't panic!"

Excellent, thanks.