Okay, here’s another little puzzle that builds on the ones we’ve done so far.

We started with a dimensional theory $T$ called ‘the theory of two particles, each with a mass’. The corresponding graded commutative algebra $A$ is $\mathbb{Z}$-graded, and elements of degree $d$ correspond to quantities of dimension $mass^d$. The graded commutative algebra $A$ is free on two generators in degree 1, namely the particles’ masses:

$A = k[m_1, m_2]$

Because the algebra $A$ is $\mathbb{Z}$-graded and generated by elements of degree 1, the moduli space of models of the theory $T$ is naturally a projective variety — at least after we discard the ‘irrelevant model’ where both particles have mass 0. And as we’ve seen, this projective variety is just the projective line $k\mathbb{P}^1$.

Then we looked at a theory we could call $T + T$. We actually called it ‘the theory of two particles on a line, each with a mass and velocity’. The corresponding graded commutative algebra is $A \otimes A$, which we think of as $\mathbb{Z} \oplus \mathbb{Z}$-graded. Elements of degree $(d,e)$ correspond to quantities of dimension $mass^d velocity^e$. $A \otimes A$ is freely generated by two quantities of degree $(1,0)$, namely the particles’ masses, and two quantities of degree $(0,1)$, namely their velocities:

$A \otimes A = k[m_1, m_2] \otimes k[v_1, v_2]$

Because the algebra $A \otimes A$ is $\mathbb{Z} \oplus \mathbb{Z}$-graded and generated by elements of degree $(1,0)$ and $(0,1)$, the moduli space of models of the theory $T + T$ is a ‘biprojective variety’ — at least after we discard some ‘irrelevant models’. And as we’ve seen, this biprojective variety is just $k\mathbb{P}^1 \times k\mathbb{P}^1$.

I should admit: I’ve never actually seen the phrase ‘biprojective variety’. What people usually talk about are ‘multiprojective varieties’. These are varieties that live in a product of projective spaces. Just as you can get projective varieties from (sufficiently nice) graded commutative algebras where the grading group is $\mathbb{Z}$, you can get multiprojective varieties from (sufficiently nice) graded commutative algebras where the grading group is $\mathbb{Z}^n$. And right now we’re doing $n = 2$.

Okay, now for the puzzle. Suppose we take our theory $T + T$ and look at the ‘subtheory’ $T'$ consisting only of quantities that have dimensions that are powers of momentum:

$momentum^d = mass^d velocity^d$

for some $d \in \mathbb{Z}$.

This subtheory $T'$ gives a subalgebra $A' \subset A \otimes A$. And we can think of this subalgebra as $\mathbb{Z}$-graded instead of $\mathbb{Z} \oplus \mathbb{Z}$-graded, since everything in this subalgebra has dimension $momentum^d$ for some $d \in \mathbb{Z}$.

And the puzzle is: what’s the moduli space of models of the subtheory $T'$, and how is it related to the moduli space of models of $T + T$?

This is a pretty open-ended puzzle. If algebraic geometry is your cup of tea, you may know some buzzwords that describe what’s going on here. If category theory is your cup of tea, you may want to formalize what I’m doing a bit more.

And here’s a hint: you might guess that we don’t lose much by going from $T + T$ to the subtheory $T'$, because there are four obvious quantities with dimensions of momentum:

$m_1 v_1, m_2 v_1, m_1 v_2 , m_2 v_2$

and we started out with four quantities in the first place: $m_1, m_2, v_1, v_2$.

## Re: Algebraic Geometry for Category Theorists

I’m very curious about the algebraic geometry, but it seems need a private password…