Demystifying the tensor product

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

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

defined

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

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

problem

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!

• 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. – Matematleta Jun 25 '16 at 22:11
• Could you expand? Perhaps with an example – RhythmInk Jun 25 '16 at 22:12
• check math.stackexchange.com/questions/1750015/… for humble approach – 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.