It seems to me, through my mathematical immaturity, that the tensor product seems to beg for more well-definition. I am working in vector spaces (so we always have a free module) and here is what my professor has shown me thus far.

We can define the tensor product of two maps (multi-linear) as follows. Let $S \in \mathcal{L}(V_1, \dots, V_n; \mathcal{L}(W;,Z))$ and $T \in \mathcal{L}(V_{n+1}, \dots , V_{n+m};W)$, We define $S \otimes T \in \mathcal{L}(V_1, \dots , V_{n+m};Z)$ by setting

$$S \otimes T(v_1, \dots ,v_{n+m})=S(v_1, \dots, v_n)[T(v_{n+1}, \dots , v_{n+m})]$$

Now, we do have $\mathcal{L}(V_1, \dots , V_{n+m};Z) \cong V^*_1 \otimes \dots \otimes V^*_{n+m} \otimes Z$ I believe. So it is, up to isomorphism, a tensor but not, itself, a tensor.

Further, suppose that $V_1, \dots , V_n$ are vector spaces. We define the tensor product

$$V_1 \otimes \dots \otimes V_n = \mathcal{L}(V^*_1, \dots V^*_n; \mathbb{F})$$

Since we regard $V$ and $V^{**}$ to be identified we have

$$v_1 \otimes \dots \otimes v_n \in V_1 \otimes \dots \otimes V_n$$


$$(v_1 \otimes \dots \otimes v_n)(L_1, \dots L_n)=L_1(v_1)\dots L_n(v_n)$$

Finally, we have defined a tensor of type $m,n$ to be a multi-linear map from $\underbrace{V^* \times \dots \times V^*}_{m \text{ times}}\times \underbrace{V \times \dots \times V}_{n \text{ times}} \to \mathbb{F}$.


So it seems to me that tensor products do not always produce tensors? That a tensor product sometimes is and sometimes is not a map to the field? Which makes me wonder how we can consider the idea to be well-defined? I have to be told by some to think about it in terms of the universal property, i.e., it takes multi-linear maps to linear ones but that isn't as illuminating as some may think. How is one to think about this product and these objects? Thanks for your help!

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    $\begingroup$ I think the UMP approach is the way to go. Since for any bilinear $B:V\times W\to Z$ there is a unique linear $\phi: V\otimes W\to Z$ s.t. $\phi \circ \otimes=B$, the "right" way to define tensor multiplication follows readily. $\endgroup$ – Matematleta Jun 25 '16 at 22:11
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    $\begingroup$ Could you expand? Perhaps with an example $\endgroup$ – RhythmInk Jun 25 '16 at 22:12
  • $\begingroup$ check math.stackexchange.com/questions/1750015/… for humble approach $\endgroup$ – janmarqz Jun 26 '16 at 15:20

My confusion, now all cleared up years later, was one of notation. When we are being slightly less precise we can define the tensor product of maps say from $$f \otimes g: V_1 \otimes V_2 \to V'_1 \otimes V'_2$$ which is defined by $f$ acting on the first coordinate and $g$ on the second. For all those wondering, this is not a tensor. This is (slightly lazy) notation demonstrating how we get certain maps once we have taken the tensor product of vector spaces (modules, more generally). $f \otimes g$ is not a tensor but acts on them.

There is a reason we abuse notation like this as there is a correspondence of sorts between this ``tensor product of maps'' and tensor products between spaces of maps.

So no, the tensor product always gives us tensors and we abuse this notation to say what happens with maps. Hope this helps anyway who has a similar confusion.


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