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I just started reading Wald's "General Relativity" and I am on his section regarding tensors. He defines the outer product as an operation on tensors of type of $(k,l)$ and $(k', l')$ which gives a tensor of type $(k + k', l + l')$. The definition is straightforward. But then in the next paragraph he goes on to say that one way of constructing tensors is to take outer products of vectors and dual vectors and that if $V$ is a finite-dimensional real vector space and $\{v_\mu\}$ is a basis for $V$ and $\{v^{{\nu}^*}\}$ is its dual basis, then the simple tensors $\{v_{\mu_1} \otimes \cdots \otimes v_{\mu_k} \otimes v^{{\nu_1}^*} \otimes \cdots \otimes v^{{\nu_l}^*}\}$ yields a basis for $\mathcal{T}(k,l)$.

How is he taking the outer product of vectors? The outer product on the dual vectors makes sense, as they are just covariant tensors, but what sense does it make to take the outer product on vectors?

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For finite dimensional vector spaces, we have an isomorphism $V \cong V^{**}$, so you are really using a basis of $V^{**}$

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  • $\begingroup$ Ah, so when taking the tensor product of two vectors, we are really taking the tensor product of two dual dual-vectors (linear functionals on $V^*$), which is of course a tensor. Correct? $\endgroup$
    – Jonathan Gafar
    May 4, 2016 at 18:11
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    $\begingroup$ @JonathanGafar Correct. You can have tensors between non function vectors, but in the context of $\mathcal{T}(k,l)$, where you want to make sense of $\nu(f),\, f\in V^*$, you need to take $V^{**}$, where $\nu(f)$ makes sense. $\endgroup$
    – Luis Vera
    May 4, 2016 at 18:18

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