# About two definitions of tensor product. (in James R. Munkres's book and Ichiro Satake's book)

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?

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 $${V^{\otimes (k+l)}}^\ast$$.

• Thank you very much for your detailed answer. – tchappy ha Mar 7 '20 at 7:58
• 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. – Captain Lama Mar 7 '20 at 23:09
• Would you care to elaborate your point? I'm only using that definition for the answer to be self contained. – guidoar Mar 9 '20 at 4:00