# Intuition for a certain tensor product.

Tensor products occur in lots of places and until recently I thought I understood them at least reasonably well. During the past few weeks, however, I've attended several talks where the tensor product $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}$ played some role and I noticed that I don't actually quite know how to think of this. I therefore wonder if there's some particular way of thinking about this object (this specific tensor product) that makes it easier to understand. How can I get some intuition for it? Might there be some nice geometric interpretation, perhaps?

• "I've attended several talks where the tensor product $\mathbb{R} \otimes_{\mathbb{Q}} \mathbb{R}$ played some role" - Really? I thought that this kind of tensor product is pathological and useless. For example, it is not noetherian and has lots of unexpected prime ideals. Could you give some background? Of course this would also make it easier to answer the actual question. – Martin Brandenburg Jul 2 '14 at 7:30
• By the way, $\mathbb{R} \otimes_{\mathbb{Q}} \mathbb{R} = \mathbb{R} \otimes_{\mathbb{Z}} \mathbb{R}$ is just the coproduct of two copies of $\mathbb{R}$ in $\mathsf{CRing}$. So this is the canonical choice of a commutative ring which contains two copies of the real numbers. Does this (trivial) remark help? – Martin Brandenburg Jul 2 '14 at 7:36
• Unless I'm much mistaken and am confusing talks, the last time it occured was in relation to the Dehn invariant (in the context of scissor congruences and Hilbert's third problem). Before that, it must've been during a talk about rational homotopy theory. I think the remark helps a bit, indeed, but isn't quite what I'm looking for. – HSN Jul 2 '14 at 9:20
• Interesting. If you want to, we can discuss this in person. I've just read that you are also a PhD student in Münster. – Martin Brandenburg Jul 2 '14 at 12:07

Linear algebra over ℚ actually represents an ancient idea of commensurability. Although tensor products are not linear algebra, this way of thinking helps enough for understanding “⊗”, with some application of order theory.

First of all, how should we visualize elementary tensors? Consider for clarity only positive ones, i.e. representable as ℓ ⊗ ℎ where ℓ, ℎ ∈ (0, +∞). It is natural to put it onto “Cartesian plane” as a rectangle of width ℓ and height ℎ:

ℎ↕■■
←→
ℓ


Which geometry should our “Cartesian plane” have for a good ℝ ⊗  ℝ? It is convenient to assume translations, but it definitely may not permit to rotate shapes, since ℓ ⊗ ℎ ≠ ℎ ⊗ ℓ, in general (from now on, “⊗” refers only to “⊗”).

We see, our shapes are equivalent up to dissecting to equal pieces and assembling in different order. For example, the rectangle from the drawing above is equivalent to

 ■↑2ℎ
■↓
↔
ℓ ∕ 2


When is the sum ℓ1 ⊗ ℎ1 + ℓ2 ⊗ ℎ2 equivalent to an elementary tensor? Purely algebraically, if and only if (ℓ1 and ℓ2 are commensurable or1 and ℎ2 are commensurable). In the former case we can reassemble both rectangles to narrow and tall ones, of the same width, and then connect vertically. In the latter case we can reassemble both rectangles to long and low ones, of the same height, and then connect horizontally, such as

[ ] + [ ] = [ ][ ]

Obviously, we can not only add shapes, but also subtract them in a similar way:

[ ][ ] − [ ] = [ ]

What about sums and differences that are not elementary tensors, i.e. rectangles? There is no problem with addition of two arbitrary rectangles: just place them without overlap and consider resulting shape bounded by no more than 4 horizontal strokes and no more than 4 vertical strokes. It can be made either connected or disjoint by translation of the parts, as one wishes. It is easy to realize that any shape bounded by a finite number of horizontal and vertical strokes is a sum of finite number of rectangles.

Subtraction of arbitrary rectangles is a bit trickier. From the Archimedean postulate follows that if ℓ ℎ > ℓ ℎ, ℓ, ℎ, ℓ, ℎ ∈ (0, +∞), then ∃ℓ′, ℎ′ ∈ (0, +∞) such that ℓ′ ⊗ ℎ′ = ℓ ⊗ ℎ and ℓ′ ≥ ℓ and ℎ′ ≥ ℎ, that implies we can subtract ℓ ⊗ ℎ of ℓ′ ⊗ ℎ′ in a geometrically evident way, by carving it out. It also gives us a hint that ℝ ⊗  ℝ has a natural partial order with an epimorphism m: ℝ ⊗  ℝ →  ℝ of ordered abelian groups having the

b ≠ a ∧ b ≥ a ⇔ m(b) > m(a), ∀b, a ∈ ℝ ⊗  ℝ

property, but this reasoning does not constitute a rigorous proof, of course. “m” is, obviously, the area of a shape.

Now we know enough to claim that positive elements of ℝ ⊗  ℝ are the same as shapes on the Cartesian plane bounded by a finite number of horizontal and vertical strokes, up to reassembly (that means finite dissection, translation of parts, and joining again). The full proof remains as an exercise, but it is worth noting that, as a corollary, we have the fact that any positive element of ℝ ⊗  ℝ can be represented as a finite sum of only positive elementary tensors (i.e. rectangles).

A sketch of the proof follows, as an algorithm to add and subtract any finite number of rectangles, providing that total area of added rectangles surpasses the total area of subtracted ones.

1. Add all rectangles with “+” signs.

2. If there are no rectangles to subtract, then all done.

3. Choose some rectangle ℓ ⊗ ℎ (with “−” sign).

4. Reassemble (if needed) our shape to cover some ℓ × ℎ rectangle completely.

5. Carve ℓ ⊗ ℎ from the shape.

6. Go to Step 2.

The only remaining question is: what to do with (non-zero) elements of ℝ ⊗  ℝ that are neither positive nor negative? Due to the m(·) = 0 thing (see the partial order stuff above) the positive-area-shape-based interpretation fails for this interesting case.

Update: to handle both signs, one may consider a real vector space of all piecewise constant functions of a real variable with finite number of pieces and bounded support, and then make a quotient space by all one-dimensional translations (by the span of images of all Tx − I, x ∈ ℝ operators). In this setting ℓ ⊗ ℎ, ℓ > 0 is represented by a function that gives ℎ between 0 and ℓ (endpoints are not important) and is zero elsewhere. Vector space structure captures the right-hand “ℝ ” in the tensor product, and “m” became Riemann integral.

• Thanks for a great answer! A few more questions: Is a negative tensor one of the form $l\otimes h$, $l,h\in(-\infty,0)$? Do I understand carving out, if I think that $l_1\otimes h_1 - l_2\otimes h_2 = l_1\otimes (h_1-h_2) + (l_1-l_2)\otimes h_2$ after ensuring $l_1>l_2, h_1>h_2$? More generally: doesn't this subtraction depend on the coice of $l', h'$? For instance, consider $2\otimes 2-3\otimes 1$. The obvious choice would be to replace $2\otimes 2$ by $4\otimes 1$, resulting in $1\otimes 1$. But what if $l'=\pi$, $h'=\frac{4}{\pi}$. Why is the resulting 'L-shape' equivalent to $1\otimes 1$? – HSN Oct 31 '14 at 15:35
• (There are a few more questions later on, but I prefer taking this a few steps at a time. Hopefully you're willing to aid in this.) – HSN Oct 31 '14 at 15:37
• Both ℓ and ℎ from (−∞, 0) yield a positive ℓ ⊗ ℎ because of ℓ ⊗ ℎ = (−ℓ) ⊗ (−ℎ) identity. No, a negative elementary tensor has one positive factor and one negative (no difference in which order). Yes, your formula for carving out is correct. The shape of the difference, of course, depends on choice of ℓ′, ℎ′, but elements of ℝ ⊗  ℝ are equivalence classes, not shapes themselves. – Incnis Mrsi Oct 31 '14 at 15:57
• “ℓ′ = π, ℎ′ = 4/π” is an invalid choice for ℓ = 4, ℎ = 1. You may not permute multiplication over ℝ with outer product over ℚ – ℓ ⊗ ℎ and ℓ′ ⊗ ℎ′ would be different. Of course, ℓ  and ℓ′ must be commensurable (I deemed it obvious). – Incnis Mrsi Oct 31 '14 at 16:08
• Thanks, this clarifies a lot. I don't know why I didn't see this last remark myself, but anyway. In that case, the remaining question is if the $>$ on the right-hand side of the definition of the order relation should be $\geq$, since there's a $\geq$ on the left-hand side as well. – HSN Oct 31 '14 at 16:43