How do fractional tensor products work?

You can define the square root of a vector space $V$ simply as a vector space $U$ with a distinguished linear isomorphism $L: U\otimes U \rightarrow V$.

You then want the following universal property to hold: Given any linear map $A: W\otimes W \rightarrow V$, there exists a linear map $B: W \rightarrow U$ such that $(B\otimes B)\circ L: W\otimes W \rightarrow V$ is equal to $A$. If the field is $\mathbb{R}$, then apparently there is no $U$ and $L$ satisfying this property. Even when $V$ is a $1$-dimensional real vector vector space, you have to modify the universal property by restricting it to positive linear maps $A$, where $A$ is positive if, for any $w \in W$, $A(w\otimes w) = aL(u\otimes u)$ for some $u \in U$ and $a > 0$.

This can be extended in the obvious way to all positive fractional powers of $V$. It also can be extended to negative powers by setting $V^{-1} = V^*$, the dual of $V$.

I don't see how to extend this to irrational powers. I don't know how to prove that an ordered field is necessary, but you can see how the ordering and completeness of $\mathbb{R}$ are used above.

I'm not going to try to match Tao's notation.

I will use the word line to mean a one-dimensional real vector space.

Suppose that $L$ is a line. Consider the line $L^{\otimes 2}$. It has the following property: it has a well-defined notion of "positive" element. Indeed, for each $\ell \in L$, we declare that $\ell^{\otimes 2} \in L^{\otimes 2}$ is positive, and $-\ell^{\otimes 2}$ is negative. It is not too hard to show that every nonzero element $L^{\otimes 2}$ is either positive or negative and not both, and that if you multiply a nonzero element of $L^{\otimes 2}$ by a positive real number, then you do not change its sign, whereas multiplication by a negative real number changes the sign.

A line equipped with a notion of "positive" satisfying the above properties will henceforth be called a positive line. Given a positive line $L$, the subspace of nonzero positive elements will be denoted $L_{>0}$.

Suppose now that $L$ is a positive line. For each $r\in \mathbb R$, I will define a positive line $L^{\otimes r}$. The definition is the following. Given $\ell \in L_{>0}$, we declare that there is an element $\ell^{\otimes r} \in L^{\otimes r}_{>0}$. Given $x \in \mathbb R_{>0}$, we declare that $(x\ell)^{\otimes r} = x^r \ell^{\otimes r}$. Note that this makes sense because $x \in \mathbb R_{>0}$ and so $x^r$ is defined.

These declarations are enough to define the line $L^{\otimes r}$ up to isomorphism. Indeed, to define line, it suffices to declare one nonzero element in it — then all other elements are the real multiples of the declared element. So the only other thing needed is to declare how different elements are related, which is what we did.

Example. $L^{\otimes 0}$ is canonically isomorphic to $\mathbb R$, because it has a distinguished element in it. Namely, the element $\ell^{\otimes 0} \in L^{\otimes 0}$ doesn't depend on the choice of element $\ell \in L_{>0}$.

Example. If $L$ is an arbitrary line, then we can define its absolute value to be $|L| := (L^{\otimes 2})^{\otimes 1/2} = \sqrt{L^{\otimes 2}}$. Then we can define things like $|L|^{\otimes r}$ for $r \in \mathbb R$.

Example. When $r \in \mathbb Z$, this definition of $L^{\otimes r}$ agrees with the usual definition (meaning there is a canonical isomorphism). Note that the usual definition of integral powers of a line does not require that the line be positive. This is because, when $r \in \mathbb Z$, the function $x \mapsto x^r$ makes sense on all of $\mathbb R_{\neq 0}$ and not just on $\mathbb R_{>0}$.

There is a representation-theoretic way to say all this. The automorphism group of a positive line (meaning the linear automorphisms that take positive elements to positive elements) is the group $\mathbb R_{>0}$ of positive real numbers, acting by multiplication. It is isomorphic, via the logarithm map, to the group $\mathbb R$ of real numbers under addition. Its irreducible representations are indexed by the Pontryagin dual group, which is also isomorphic to $\mathbb R$. The positive line $L^{\otimes r}$ is "$r$ times the line $L$" in the space of representations of $\mathrm{Aut}(L) \cong \mathbb R_{>0}$.

The operation $L \mapsto |L|$ is important in calculus and manifold theory: it is necessary in order to define volume forms and integration theory on unoriented manifolds. For example, if $M$ is an $n$-dimensional manifold, then a "volume form" on $M$ is not a section of the line bundle $\bigwedge^n T^*_M$ (the top exterior power of the cotangent bundle), but is rather a section of the absolute value of that line bundle. Compare the formulas appearing in "$u$-substitution" in multivariable calculus.