[I asked and bountied this question on Math SE, where it got several upvotes and a comment suggesting it was research-level, but no answers. So I'm reposting here with slight edits, but please feel free to close it if it's inappropriate. Also, I'm a physicist rather than a mathematician, so fancy answers might go over my head.]


In [this blog post](https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/), Terry Tao discusses the $n$-fold tensor product of a one-dimensional vector space $V^L$ ($L$ is just a non-numeric label, not an exponent). He claims that



> 
> With a bit of additional effort (and taking full advantage of the one-dimensionality of the vector spaces), one can also define spaces with fractional exponents; for instance, one can define $V^{L^{1/2}}$ as the space of formal signed square roots $\pm l^{1/2}$ of non-negative elements $l$ in $V^L$, with a rather complicated but explicitly definable rule for addition and scalar multiplication. ... However, when working with vector-valued quantities in two and higher dimensions, there are representation-theoretic obstructions to taking arbitrary fractional powers of [vectors].
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1. What is the "rather complicated but explicitly definable rule for addition and scalar multiplication"?
2. Is it easy to see why this construction doesn't work in higher than one dimension? (Not necessarily a rigorous proof, just intuition for what goes wrong.)
3. Could one extend the construction to include irrational exponents?
4. (a) Tao claims earlier in the blog post that the vector space needs to be totally ordered. Considering that the vector space is 1D, is this equivalent to the requirement that the underlying field be totally ordered? (b) What properties does the vector space (or underlying field) need to satisfy in order for this construction to work? Presumably it doesn't work for arbitrary ordered fields, because you certainly can't define a square root function $\mathbb{Q} \to \mathbb{Q}$. Does it only work for real vector spaces?



How do fractional tensor products work?
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.


