Understanding the Tensor Product of Two One-Forms in Differential Geometry

["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?

 

[Orodruin]

Would this be correct at all?

Yes.

 

["Don't panic!"]

Excellent, thanks.