# Tensor spaces / outer product

Most of the examples I have ever seen use Cartesian vectors and matrix representation to illustrate this idea. That's fine and good, and easy to follow, but not very helpful for more abstract cases. For instance, the set of complex linear functions with the standard complex addition and scalar multiplication operations forms a vector space, and thus (by definition) the functions which are its elements are vectors.

$$V = \{ ax + b | a,b \in \mathbb C\}$$

Suppose I took two copies $$V_1$$ and $$V_2$$ of this vector space and formed their tensor product $$W =V_1 \otimes V_2$$; the result ought to be a tensor space of order $$(2,0)$$, but what would the tensors in it actually look like?

How does one go about "taking the outer product" of linear functions $$v_1 \otimes v_2$$ for $$v_1 \in V_1$$ and $$v_2 \in V_2$$?

Further, how would one construct a dual space $$V^*$$ for such a vector space if, for instance, one were interested in constructing tensors of rank $$(0, n)$$ or mixed rank $$(n,m)$$? Examples using Kronecker delta or matrix transposes for Cartesian vectors are not very much help here! Can anyone give me a little guidance to put me on the right path, please? Many thanks.

Let $$V_{1},V_{2}$$ be vector spaces over a field $$k$$ (you can think of it being $$\mathbb{R}$$ or $$\mathbb{C}$$ for concreteness' sake), label each vector in $$V_{1}$$, $$v_{i}$$ where $$i$$ runs over some ordered set, and label each vector in $$V_{2}, w_{j}$$ where $$j$$ also runs over some ordered set, so we've given each vector in our two spaces an index. Now consider $$X$$, the vector space whose basis consists of the symbols $$v_{i}\otimes w_{j}$$ as $$i,j$$ run through their respective sets (In other words, $$X$$ is the vector space whose basis set is $$V_{1}\times V_{2}$$.). So each $$x\in X$$ has the shape $$\sum_{i,j}a_{ij}v_{i}\otimes w_{j}$$

Where the $$a_{ij}\in k$$ are scalars and the sum has finitely many terms. In order to obtain the space $$V_{1}\otimes V_{2}$$ we impose the following relations on vectors in $$X$$, for any vectors $$u,x,y,z\in X$$ and scalars $$a\in k$$ we must have:

$$(x+y)\otimes u=x\otimes u+y\otimes u$$

$$x\otimes (u+z)=x\otimes u+x\otimes z$$

$$(ax)\otimes u = a(x\otimes u)$$

$$x\otimes (au)= a(x\otimes u)$$

So with these relations can you see what the basis vectors for the space $$V_{1}\otimes V_{2}$$ ought to look like? Once you know that, you can get a concrete expression for the tensor product of vectors from each space.

• Thanks Ben P; if I understand correctly, you see Dapto b that there is no specific basis-independent way to define the tensor product, and I'm free to define it any way i like subject to the distributivity and commutativity relations you highlighted (which ensure multilinearity of the resulting tensor space)? Commented May 31, 2020 at 2:15
• Sorry for typo, Ben (sending from mobile, edit timed out) If I define a tensor product over these spaces with something as simple as: $v_i \otimes w_j = (a_i x + b_i) \otimes (c_j y + d_j)$ $= (a_i x + b_i) \times (c_j y + d_j)$ $= (a_i x c_j y + a_i x d_j + c_j y b_i + b_i d_j)$ satisfies the requirements you suggested (necessary for multilinearity in the resulting tensor space, yes?) That is ${a_i c_j y + a_i x d_j + c_j y b_i + b_i d_j}$ given some index sets for $i, j$ provide a basis for $V_1 \otimes V_2$. It can't be that easy? Commented May 31, 2020 at 6:38
• For your specific example $V_{1}$ it looks like we have a 2-dimensional space over the complex numbers with basis given by $\{1,x\}$ so the tensor product with itself would have basis $\{1\otimes 1, 1\otimes x, x\otimes 1, x\otimes x\}$ So all of the vectors in that vector space would be of the form $a_{1}(1\otimes 1)+a_{2}(1 \otimes x)+a_{3}(x\otimes 1)+a_{4}(x\otimes x)$ where $a_{1},a_{2},a_{3},a_{4}\in\mathbb{C}$ Commented May 31, 2020 at 8:22