Algebraic operations with tensor products and direct sums.

Question

When first learning about tensor products and direct sums of vector spaces, the analogy of multiplication and addition of vector spaces is sometimes used to help with the intuition. If we look at the bases of the considered vector spaces, this analogy seems most intuitive:

1. the basis of the direct product of $U$ and $V$ is the concatenation of the bases of $U$ and $V$; and
2. the basis of $U\otimes V$ is the set of all products $u\otimes v$, where $u$ is in the basis of $U$ and $v$ is in the basis of $V$.

This made me wonder, how far can this analogy of addition and multiplication be taken?

More precisely: letting $F$ be a field and $\mathbf V_F$ be the set of all vector spaces over $F$, it is clear that $\oplus$ and $\otimes$ define binary operations on $\mathbf V_F$, and that these operations are closed, in that given $U,V\in\mathbf V_F$, both $U\oplus V$ and $U\otimes V$ gives back an element in $\mathbf V_F$.

If we stretch the concept of two elements in $\mathbf V_F$ being equal to mean that they are isomorphic as vector spaces, then we see that $\oplus$ and $\otimes$ have some properties that the usual $+$ and $\cdot$ have, such as commutativity and associativity, and neutral elements (i.e., $\{0\}\oplus V\cong V$, and $F\otimes V\cong V$).

So then my question would be: Do we have all of the usual properties of addition and multiplication, such as inverse elements, or distributivity? In other words, can $(\mathbf V_F,\oplus,\otimes)$ be seen as a ring? And if not, which properties hold and which doesn't?

Given how natural this question feels for me, it seems that this must have already been investigated by someone.

Answer 1

You're right-- this sort of question is studied a lot. As you have defined things, you're looking at a semiring, instead of a ring because there are no additive inverses to the direct sum operation. Of course distributivity goes through, since $L \otimes (K \oplus J) \cong L \otimes K \oplus L \otimes J$ via $l \otimes(j \oplus k) \mapsto l \otimes j \oplus l \otimes k$.

In fact your semiring is isomorphic to the natural numbers $\mathbb N$ via the map $V \mapsto {\rm dim} \ V$.

Say you wanted to get a ring out instead of a semiring-- there is a canonical procedure (or free functor $F$) that turns any semiring into a ring in a universal way. Applying it to your semiring you would get something isomorphic to $\mathbb Z$

Now we see that for vector spaces over a field, this ring is a bit boring since it just captures dimension. But you could do the same thing for modules over a ring, and all of a sudden things become much more interesting. Or you could do it for vector bundles over some sort of space, for instance an algebraic variety. This ring was first introduced by Grothendieck to prove his version of the Riemann Roch Theorem. All these things often go under the name of $K$-theory, which you should totally check out to learn more!

Answer 2

Vector spaces with $\oplus$ and $\otimes$ do not form a semiring, since associativity etc. do not hold - the laws only hold up to isomorphism. These isomorphisms fit together in a certain way, and what we get is called a $2$-semiring or $2$-rig. Just like a semiring is a "fusion" of two monoids (one being commutative), a $2$-semiring is a "fusion" of two monoidal categories (one being symmetric). Every $2$-semiring induces a semiring by considering the "set" of isomorphism classes. This process of taking isomorphism classes is called decategorification. The reverse process is called categorification. Roughly, categorification produces more structure (which can be very useful), and decategorification "destroys" structure (which might simplify calculations). For example the $2$-semiring of vector spaces over $K$ has much more structure than its underlying semiring, which identifies (via $\dim$) with the class of cardinal numbers $\mathsf{Card}$ with the usual cardinal arithmetics. The $2$-semiring "knows" the group $\mathrm{Aut}(K^n)=\mathrm{GL}_n(K)$, but the underlying semiring does not.