Tensor Product : Meaning of Element $ u \overline v $ in U\ overline V

Question

Just wanted to get a better feel for the Tensor Product (basic here, Vector Spaces over the Reals). I ( think I ) understand that given f.d ( Finite Dimensional ) V.Spaces $U,V$ over the same field $F$, the tensor product $ U \otimes_F V $ is a Vector Space maps every bilinear map $B: U \times V \rightarrow Z $; $Z$ a f.d Vector Space over F into a linear map $L : V \otimes W \rightarrow Z $ through $L( v \otimes w):=B(v,w) $ (right?)

Is it fair to say that an element $ u \otimes v = 1 u \otimes 1 v$ from the Tensor Product $ U \otimes V $ of Vector Spaces over F is a representative B of the class of bilinear maps from $ U \times V $ that send $(u,v)$ to (1,1)? This seems to fit with the quotient relations in the tensor products :

$(\alpha u \otimes v) = \alpha (v \otimes w)= B(v \otimes \alpha v) $ : Any bilinear map $B(v,w)$ that sends (v,w) to (1,1) will map $(\alpha u ,v) $ to $( \alpha,1) $ , etc.

Is this an accurate description of a singleton $ u \otimes v $?

Thanks.

Answer 1

Is it fair to say that an element $ u \otimes v = 1 u \otimes 1 v$ from the Tensor Product $ U \otimes V $ of Vector Spaces over F is a representative B of the class of bilinear maps from $ U \times V $ that send $(u,v)$ to (1,1)?

No, that doesn't sound right.

First, elements of the tensor product space are not themselves bilinear maps, so they're definitely not representatives of classes of such bilinear baps.

Second $(1,1)$ is not generally something that can be the output of a bilinear map $U\otimes V\to Z$, unless $Z$ is in particular the vector space $\mathbb F^2$.

Any bilinear map $B(v,w)$ that sends $(v,w)$ to $(1,1)$ will map $(\alpha u ,v) $ to $( \alpha,1) $ , etc.

No, that can't be. If $B$ is bilinear, that requires that $B(\alpha u,v)=\alpha\cdot B(u,v)$. So if $B$ sends $(u,v)$ to $(1,1)\in\mathbb F^2$, then it must send $(\alpha u,v)$ to $(\alpha,\alpha)$ rather than to $(\alpha,1)$.

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A better understanding of the tensor elements would be that $u\otimes v$ represents a recipe for applying some bilinear map you don't know what is yet. The recipe says, "once you find out what the bilinear map is, apply it to the arguments $u$ and $v$".

In this view, far from being something that represents a bilinear map, an element of $U\otimes V$ is something that will eat a bilinear map and gives you something in that map's codomain.

The reasons tensors are different from just expressions with a variable ranging over bilinear maps is that equality between tensors knows about the bilinearity. So, for example $2u\otimes v = u\otimes 2v$ because when you apply a bilinear map to either $(2u,v)$ or $(u,2v)$ you're guaranteed you get the same result.