# How do fractional tensor products work?

[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, 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].

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?

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.

• A couple very minor comments: (a) I find your term "positive line" very confusing, because to me that implies that every element is positive. I'd call it a "signed line" instead. (b) $L_{>0}$ isn't a subspace of a signed line, but rather a convex cone. – tparker Oct 29 '18 at 1:26
• This answer only considers lines, i.e. 1D vector spaces over the real numbers. Your universal property $(x \ell)^{\otimes r} = x^r \ell^{\otimes r}$ "reuses" all the structure of the reals, e.g. the (rather complicated) definition of exponentiation of an arbitrary positive real base to an arbitrary real exponent. Do you think that Tao made a mistake in assuming an arbitrary ordered field rather than just the reals, or can your construction be extended to that case? – tparker Oct 29 '18 at 1:35
• @tparker (a) The only reason I used the word "positive" was for the analogy with real numbers: positive things are the things you can take arbitrary powers of. (b) It is not a vector subspace. It is a subspace for any sufficiently general meaning of the word "space", e.g. "set". (c) As you point out in your question, $\mathbb Q$ is ordered, but does not allow fractional, let alone irrational, powers, so you cannot take fractional powers of lines over $\mathbb Q$. I assume Tao meant that the line itself should be ordered. – Theo Johnson-Freyd Oct 29 '18 at 3:01
• (d) Vector spaces of (finite) dimension not equal to $1$ certainly do not admit fractional/negative/etc powers, because tensor powers act on dimensions by powers, and there are not vector spaces of irrational dimension. I think, but might be misremembering, that there are certain types of infinite-dimensional vector spaces that show up in the theory of von Neumann algebras that have a version of "dimensions" which are arbitrary real numbers; these might allow fractional powers. – Theo Johnson-Freyd Oct 29 '18 at 3:05

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.

• Sorry, where'd you use completeness? – tparker Oct 29 '18 at 1:08