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## Main Question or Discussion Point

What is meant by taking the tensor product of vectors? Taking the tensor product of two tensors is straightforward, but I am currently reading a book where the author is talking about tensor product on tensors then in the next paragraph declares that tensors can then be constructed by taking tensor products on vectors and dual vectors. Taking tensor products on dual vectors makes sense to me, but what sense does it make to take tensor products of vectors?

EDIT: Okay, I think it is because a vector space $V$ is isomorphic to $V^{**}$. So when I am taking the tensor product of vectors, I am really taking the tensor product of linear functionals on $V^*$, correct?

I also have another question now, which I may as well ask in this thread. The book I am reading is "General Relativity" by Wald. At first he defines a tensor of type [itex](k,l)[/itex] to be a multilinear map [itex]T: V^* \times \cdots \times V^* \times V \times \cdots \times V \rightarrow \mathbb{R}[/itex], but then he literally says this on the next page: "Thus, one way of constructing tensors is to take outer products of vectors and dual vectors. A tensor which can be expressed as such an outer product is called simple. If [itex]\{v_{\mu}\}[/itex] is a basis of [itex]V[/itex] and [itex]\{v^{\nu^*}\}[/itex] is its dual basis, it is easy to show that the [itex]n^{k+l}[/itex] simple tensors [itex]\{v_{\mu_1} \otimes \cdots \otimes v_{\mu_k} \otimes v^{{\nu_1}^*} \otimes \cdots \otimes v^{{\nu_k}^*} \}[/itex] yield a basis for [itex]\mathcal{T}(k,l)[/itex]."

My question is, when he first defined a tensor, he defined it as a multilinear map on [itex](V^*)^k \times V^l[/itex]. But then if you look at the quote above, in his basis for the simple tensors, he starts the tensor product with the vectors first and the dual vectors last. Is this not incorrect, because the tensor product does not, in general, commute? Shouldn't he have put the dual vectors first and the vectors last in that tensor product?

EDIT: Okay, I think it is because a vector space $V$ is isomorphic to $V^{**}$. So when I am taking the tensor product of vectors, I am really taking the tensor product of linear functionals on $V^*$, correct?

I also have another question now, which I may as well ask in this thread. The book I am reading is "General Relativity" by Wald. At first he defines a tensor of type [itex](k,l)[/itex] to be a multilinear map [itex]T: V^* \times \cdots \times V^* \times V \times \cdots \times V \rightarrow \mathbb{R}[/itex], but then he literally says this on the next page: "Thus, one way of constructing tensors is to take outer products of vectors and dual vectors. A tensor which can be expressed as such an outer product is called simple. If [itex]\{v_{\mu}\}[/itex] is a basis of [itex]V[/itex] and [itex]\{v^{\nu^*}\}[/itex] is its dual basis, it is easy to show that the [itex]n^{k+l}[/itex] simple tensors [itex]\{v_{\mu_1} \otimes \cdots \otimes v_{\mu_k} \otimes v^{{\nu_1}^*} \otimes \cdots \otimes v^{{\nu_k}^*} \}[/itex] yield a basis for [itex]\mathcal{T}(k,l)[/itex]."

My question is, when he first defined a tensor, he defined it as a multilinear map on [itex](V^*)^k \times V^l[/itex]. But then if you look at the quote above, in his basis for the simple tensors, he starts the tensor product with the vectors first and the dual vectors last. Is this not incorrect, because the tensor product does not, in general, commute? Shouldn't he have put the dual vectors first and the vectors last in that tensor product?

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