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I am reading "Analysis on Manifolds" by James R. Munkres.

Definition:
Let $f$ be a $k$-tensor on $V$ and let $g$ be an $l$-tensor on $V$. We define a $k+l$ tensor $f \otimes g$ on $V$ by the equation $$(f \otimes g)(v_1, \cdots, v_{k+l}) = f((v_1, \cdots, v_{k}) \cdot g(v_{k+1}, \cdots, v_{k+l}).$$

It is easy to check that the function $f \otimes g$ is multilinear; it is called the tensor product of $f$ and $g$.

In the above definition, the operands for $\otimes$ are tensors on $V$.

When I checked "Linear Algebra" by Ichiro Satake, I found the operands for $\otimes$ are vector spaces.

What is the relation between these two $\otimes$s?

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If $V$ and $W$ are vector spaces, their tensor product is a vector space $V \otimes W$ with the following (informally stated) property:

$(\star)$ Linear maps $V \otimes W \to U$ are in bijective correspondence with bilinear functions $V \times W \to U$.

The motivation for this is to encode multilinear maps as linear maps, which we understand thoroughly.

Moreover, we have

$$ (V \otimes W)^\ast \simeq \hom(V \otimes W, \Bbbk) \simeq \operatorname{Bil}(V \times W,\Bbbk). $$

If $f_1, \dots, f_n$ and $g_1, \dots, g_m$ are dual basis of basis $\{v_i\}$ and $\{w_i\}$ in $V$ and $W$, in particular they define linear functions $V \times W \to \Bbbk$ which I will denote with the same letters. For example $f_1$ induces the map $(v,w) \mapsto f_1(v)$.

Given $f_i$ and $g_j$, we can also define the bilinear function

$$ f_i \otimes g_j(v,w) = f_i(v)g_j(w). $$

We will call this the tensor product of $f_i$ and $g_j$ because of their relation with respect to the roles of $V$ and $W$ in their tensor product,

Proposition: the set $\{f_i \otimes g_j\}_{i,j}$ is a basis for $(V \otimes W)^\ast$.

Proof. Linear independence comes from the fact that

$$ f_i \otimes g_j(v_k,w_l) = \delta_{ik}\delta_{jl}. $$

If $d : V \times W \to \Bbbk$ is bilinear and $x = \sum a_iv_i, y = \sum b_iw_i$, then

$$ d(x,y) = \sum_{i,j}a_ib_jd(v_i,w_j) $$

and thus $d$ only depends on its values in each pair $(v_i,w_j)$. Noting $d_{ij} = d(v_i,w_j)$ and writing $\varphi = \sum_{i,j}d_{ij}f_i \otimes g_j$, we thus obtain that $d = \varphi$ since $d(v_i,w_j) = \varphi(v_i,w_j)$ for each $i,j \ \square$.

In the same fashion, fixing a vector space $V$ we can consider the $n$ (tensor) powers of $V$,

$$ V^{\otimes n} := V \otimes \cdots \otimes V, $$

whose dual space turns out to be the space of $n$-multilinear maps $V^n \to \Bbbk$. Now, in general, if $v \in V^{\otimes k \ast}$ is a $k$-multiniear function and $w \in V^{\otimes l \ast}$ is an $l$-multilinear one, we can form their tensor product

$$ v \otimes w(v_1, \dots, v_k,{v'}_1,\dots,{v'}_l) = v(v_1,\dots,v_k)w({v'}_1,\dots,{v'}_l)$ $$

which is an element of ${V^{\otimes (k+l)}}^\ast$.

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  • $\begingroup$ Thank you very much for your detailed answer. $\endgroup$ – tchappy ha Mar 7 '20 at 7:58
  • $\begingroup$ I have to say I truly don't think it is a good idea to identify a vector space and its dual. Saying that $v\in V^{\otimes k}$ is a $k$-multilinear form on $V$ is in my opinion bound to lead to errors. $\endgroup$ – Captain Lama Mar 7 '20 at 23:09
  • $\begingroup$ Would you care to elaborate your point? I'm only using that definition for the answer to be self contained. $\endgroup$ – guidoar Mar 9 '20 at 4:00

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