Tensors, tensor products

[wrobel]

Just found a really good lecture on tensors. It starts with the absolute basics and goes all the way up to universal mapping:

Spoiler: an additional note

I just want to recall the standard realization of the tensor product.

Let $X, Y$ be vector spaces, say over $\mathbb{R}$. These spaces need not be finite-dimensional.
Let $T$ denote the space of bilinear maps $b:X\times Y\to \mathbb{R}$, and let $T^*$ stand for the corresponding algebraic dual space. By definition, $x\otimes y$ is the element of $T^*$ defined by $(x\otimes y)(b):=b(x,y).$
Thus, we have a mapping $\psi:X\times Y\to T^*$ given by $\psi(x,y)=x\otimes y.$
We define the tensor product of the spaces as $X\otimes Y:=\text{span}\,\psi(X\times Y)$

 

[mathwonk]

nice spoiler. took me a while to grasp, the key for me being that the map psi is itself bilinear, indeed it is the universal bilinear map.

 

[jedishrfu]

Brings back memories of reading Wheeler’s Gravitation tome preprints as a junior physics major.

 

[Matterwave]

My first gut reaction was, wouldn't you want $x\otimes y$ to be in $T$ not $T^*$?

Then I realized I flipped your $T$ and $T^*$ in my head as I'm used to attaching a star to the space of maps of vectors.

 

[jedishrfu]

There's a more modern field of math, Geometric Calculus, that encompasses Vector and Tensor Analysis, which I find interesting. One thing they address is the vector cross product, which they subsume under the geometric product.

 

[WWGD]

My take is that the Tensor product of Vector Spaces $$V, W$$ over the same field F, gives rise to a new Tensor Vector space , where every bilinear map on $$V \times W$$ into a base field(ring?) factors into a linear map from $$V \otimes W$$ into the same base ring/field.