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Stream: learning: questions

Topic: Constructing a rack from a shelf

John Baez (Jun 23 2026 at 13:12):

Chad Nester said:

Hello,

I've encountered a binary operation that, somewhat unexpectedly, forms a right-shelf. That is, satisfies the equation: $(a \odot b) \odot c = (a \odot c) \odot (b \odot c)$.

The potential connection to knot theory is intriguing, and I'm interested to know if it can be made into a rack in a sensible way.

Is anything known about this?

I've never seen anyone study the free rack on a shelf - and since they're both rather little-studied algebraic structures, there might not have been any study of this.

But both shelves and racks can be described as algebras of Lawvere theories if we include an additional operation in the definition of a shelf, rather than stating the extra clause as a mere property. I made sure the nLab article does this. And I believe that whenever we have two Lawvere theories $T$ and $T'$, with the latter having more operations and/or equations than the former, the map $T \to T'$ gives rise to a functor $T' \mathsf{Alg} \to T \mathsf{Alg}$ that has a left adjoint. So, the forgetful functor $\mathsf{Rack} \to \mathsf{Shelf}$ should have a left adjoint.

Kevin Carlson (Jun 23 2026 at 15:00):

Yeah, that general statement is true: the algebra categories are locally finitely presentable and the forgetful functor preserves limits and filtered colimits because they're created on underlying sets, so it has a left adjoint.

John Baez (Jun 23 2026 at 15:02):

I wish I understood all that stuff. But not enough to actually put energy into learning it! :laughing:

Kevin Carlson (Jun 23 2026 at 15:15):

It's always fun to hang on to a mystical incantation to chant at people.

Todd Trimble (Jun 23 2026 at 15:19):

In fact it's easy to write down the coequalizers that give you the left adjoint to $T'\mathsf{Alg} \to T\mathsf{Alg}$. It looks something like this:

$(A: T\mathsf{Alg}) \mapsto \mathrm{coeq} (T' T A \rightrightarrows T' A)$

(the coequalizer being computed in $T'\mathsf{Alg}$), which I'm going to leave to your imagination what those two arrows are. (But you just follow your canonical nose.)

Todd Trimble (Jun 23 2026 at 15:20):

It's sort of like taking a "tensor product" $T' \circ_T A$.

Chad Nester (Jun 24 2026 at 08:46):

Thanks everyone. If I ever get around to working any of this out I'll report back!