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I'm teaching myself the algebraic side of tensors and I was wondering if you would be able to clarify a few things for me.

I prefer to think of a tensor in the set theoretic manner as a multi-linear mapping taking several copies of a vector space and its dual space to the base field.

I'm trying to convince myself about all of the commonly used tensor index laws from this. If we consider the basis vectors of each space as tensors themselves, then it is clear that a basis for the (k,l) tensor space can be constructed by suitable tensor products of the basis vectors and their duals.

Consider any basis vector [itex]\hat{\theta}^{(\alpha)}[/itex] of the dual space. This basis vector forms a (0,1) tensor

[itex]\hat{\theta}^{(\alpha)} : T_p \to \mathbb{R}[/itex] such that [itex]\hat{\theta}^{(\alpha)}: \hat{e}_{(\beta)} \mapsto 1[/itex] if [itex]\alpha = \beta[/itex] and zero otherwise.

If we take the tensor product [itex]\hat{\theta}^{(\alpha)} \otimes \hat{e}_{(\beta)}[/itex] we get a (1,1) tensor [itex]T_p^\ast \times T_p \to \mathbb{R}; (\omega,v) : \mapsto \hat{\theta}^{(\alpha)}(v)\hat{e}_{(\beta)}(\omega) = v_\alpha \hat{e}_{(\beta)}(\omega) [/itex].

Now I think we can say that [itex]\hat{e}_{(\beta)}(\omega) = \omega^\beta[/itex] if we treat [itex]T_p[/itex] as the dual space of [itex]T_p^\ast[/itex] which gives [itex]\hat{\theta}^{(\alpha)}(v)\hat{e}_{(\beta)}(\omega) = v_\alpha \omega^\beta[/itex].

I'm not eactly sure how to show that this the Kronecker delta. Is this just the way Kronecker delta is defined?

Thanks

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# Understanding tensor contraction

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