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My question is very simple: A type $(m,n)$ tensor is an element of $V^{\otimes m}\otimes (V^*)^{\otimes n}$. Is there a reason/motivation, beyond more general definitions, to consider the dual space of $V$ in this definition? Or is it just convention?

I don't expect some mindblowing answer, so to speak, but maybe a clarification of the why the tensor product of $V$ with it's dual would be so interesting.

Pauli
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  • You accidentaly posted the same question twice. I will suggest to delete one of the questions – Marco Feb 18 '24 at 11:38
  • @Marco Thank you for warning – Pauli Feb 18 '24 at 11:40
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    Well, given two vector spaces $V,W$, it is interesting to look at $V\otimes W$ (do you agree?). Next, given a single vector space $V$, it is interesting to look at $V^$ (do you agree?). So by the logic of “interesting +interesting = interesting”, it would be interesting to look at all $V^{\otimes m}\otimes (V^)^{\otimes n}$ :) In some vague sense, these are all spaces for which “the only information you need” is the single vector space $V$, and then everything else just falls right out/created automatically. – peek-a-boo Feb 18 '24 at 11:55
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    Another thing you might do with two vector spaces is look at $\text{Hom}(V,W)$; but this thing is isomorphic to $W^*\otimes V$, assuming finite-dimensionality, so we’re back to tensor products of a space with the dual of another space again :) – peek-a-boo Feb 18 '24 at 11:58
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    You might find some of the answers here relevant: https://math.stackexchange.com/questions/10282/an-introduction-to-tensors. – Hans Lundmark Feb 18 '24 at 12:41

2 Answers2

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Let $V$ a finite dimensional vector space and let $e:=(e_1, \cdots e_n)$ one of its bases. Let $\tilde{e}$ another bases for $V$. The two bases are related to each other by a linear transformation. i.e. there is a $n \times n$ matrix $A$ such that $$ \tilde{e}=Ae $$ or $\tilde{e_i}=\sum_j A^j_i e_j$ Consider now the dual cobases of $V^*$ i.e. $e^*:=({e^*}^1, \cdots {e^*}^n)$ such that $$ {e^*}^i(e_j)= \delta_j^i $$ Let $B$ the matrix of the change of bases of the dual bases, we have $$ \operatorname{Id}_n= \tilde{e^*}(\tilde{e})= B e^*(A e)=BA e^*(e)=BA \operatorname{Id}=BA $$ so $B=A^{-1}$

A bases for the space of $(1,1)$ tensors is given by $$ e_i \otimes {e^*}^j $$ for $i,j \in \{1, \cdots n\}$. So it change bases as $$ \tilde{e_i} \otimes \tilde{{e^*}^j}=\sum_k \sum_h A^k_i (A^{-1})^j_h e_k \otimes {e^*}^h $$ while e.g a $(2,0)$ tensor change bases as $$ \tilde{e_i} \otimes \tilde{e_j}=\sum_k \sum_h A^k_i A^h_i e_k \otimes e_h $$ This is quite used in physics

Marco
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As you probably know, for finite-dimensional spaces, $V$ is isomorphic to $V^\ast$ but such isomorphisms are not canonical. It is often important to work with canonical isomorphisms, for instance when doing vector bundles over manifolds. On the other hand, there are certain canonical isomorphisms such as that between $Hom(V,V)$ and $V^\ast\otimes V$, a fact important for example in Riemannian geometry. That's why one needs to work with the more general tensor products that you mentioned.

Mikhail Katz
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  • Could you provide an usage of this canonical isomorphism between $V^*\otimes V$ and $\textbf{Hom}(V, V)$? I'm starting the study of tensors, so, to me, it's a cool theorem for where I can not seen a concrete aplication – Pauli Feb 18 '24 at 12:30
  • @PauloEstêvão, suppose for example that you want to understand the bundle $Hom(T,T)$ where $T$ is the tangent bundle of the 2-sphere. Then natural isomorphisms tell you that this bundle is the tensor product of the tangent bundle and the cotangent bundle. If you view these as real bundles, it so happens that the tangent bundle and the cotangent bundle are isomorphic. However, viewed as complex line bundles, they are not isomorphic. – Mikhail Katz Feb 18 '24 at 15:42