What does a direct sum of tensor products look like?

Question

I understand that tensor product is distributive over direct sums, but what happens when we have $(v_{1}\otimes w_{1})\oplus (v_{2}\otimes w_{2})$?

Answer 1

Since you mention direct sums, I assume that $v_1,v_2,w_1,w_2$ are vector spaces (even though spaces are usually denoted by upper case letters)

There is no general formula for such case (see the last proposition). This is because there no distribution of addition over multiplication for numbers (real or integers).

On the level of dimensions we know that $$\dim(V\otimes W)=\dim(V)\dim(W)$$ and $$\dim(V\oplus W)=\dim(V)+\dim(W).$$

Thus $$\begin{split}\dim((V\oplus W)\otimes Z)&=(\dim(V)+\dim(W))\dim(Z)\\ &=\dim(V)\dim(Z)+\dim(W)\dim(Z)\\ &=\dim(V\otimes Z)+\dim(W\otimes Z)\\ &=\dim((V\otimes Z)\oplus(W\otimes Z)).\end{split}$$

So in this case, by looking at dimensions, we can guess what $(V\oplus W)\otimes Z$ should be isomorphic with.

On contrary, we do not have a law that says $ab+c=ac+bc$. So we cannot make a transformation on the level on dimensions, cause we are stuck at the first step.

$$\dim((V\otimes W)\oplus Z))=\dim(V)\dim(W)+\dim(Z)$$

More concretely, let $\dim(V)=2$, $\dim(W)=3$ and $\dim(Z)=5$. So

$$\dim((V\otimes W)\oplus Z))=11$$

and it is a prime number. So it cannot be of the form $X\otimes Y$, where $X,Y$ are some combination of $V,W,Z,\oplus,\otimes$ and brackets.

On the other hand, for $\dim(V)=\dim(W)=\dim(Z)=1$ we have that $$\dim((V\otimes W)\oplus Z))=2.$$ Thus if $(V\otimes W)\oplus Z$ is presented as direct sum, it need to have at most $2$ components. This means that the only possibility is that $(V\otimes W)\oplus Z\cong X\oplus Y$ with $X$ and $Y$ being combination of $V,W,Z$ and $\otimes$.

Thus, we conclude

Proposition. If there is a way of presenting $(V\otimes W)\oplus Z$ as a combination of $V,W,Z,\oplus,\otimes$ and brackets that works for $V,W$ and $Z$, then it is of form $X\oplus Y$ where $X$ and $Y$ is a combination of $V,W,Z$ and $\otimes$.

Going back to the case where $\dim(V)=2$, $\dim(W)=3$ and $\dim(Z)=5$ we look at all possible summands of $11$ and we try to fit $X$ and $Y$ according to this decomposition.

$1+10=11$. This case will not work, because all $V,W,Z$ have dimensions greater then $1$. Thus IMPOSSIBLE!

$2+9=11$. In this case $X$ has to be $V$. However there is no way to write $9$ dimensional vector space $Y$ in terms of $V,W,Z,\otimes$ such that $X\oplus Y$ would posses all spaces $V,W,Z$. Thus IMPOSSIBLE!

$3+8=11$. In this case $X$ has to be $W$. However there is no way to write $8$ dimensional vector space $Y$ in terms of $V,W,Z,\otimes$ such that $X\oplus Y$ would posses all spaces $V,W,Z$. Thus IMPOSSIBLE!

$4+7=11$. In this case $X$ has to be $V\otimes V$. However there is no way to write $7$ dimensional vector space $Y$ in terms of $V,W,Z,\otimes$. Thus IMPOSSIBLE!

So we are left with $5+6=11$. However, by looking at the dimensions we see that the only way to write it, is by putting $X=Z$ and $Y=V\otimes W$. And this is precisely our starting point (but with just position changed).

Thus, we finally conclude

Proposition. There is no other representation of $(V\otimes W)\oplus Z$ in terms of $V,W,Z,\oplus,\otimes$ and brackets that works for all vector spaces $V,W,Z$.