Difference between 'free algebra' and 'absolutely free algebra'

Question

What is the difference between a free algebra and an absolutely free algebra? Wikipedia and Encyclopaedia of Maths are not very clear on the subject...

Answer 1

Honestly, I reckon the notion of an absolutely free algebra makes absolutely no sense. For instance:

• The phrase "absolutely free group" makes no sense
• The phrase "absolutely free ring" makes no sense
• The phrase "absolutely free magma" kind of makes sense, but really, these are just free magmas; the qualifier 'absolutely' doesn't convey any actual information.

Anyway, I have a suspicion that the phrase "absolutely free algebra" is really just a vague way of gesturing at the notion of a free monad, or perhaps it is really a way of gesturing at the notion of a "monad freely generated by a polynomial endofunctor." For example, the process that takes free groups is a monad, and the process that takes free magmas is a monad; the latter is free as a monad, but the former is not.

I'd be interested to hear the community's opinion on this matter.

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After thinking about it a bit, I've decided that I don't really agree with what others have written here.

Please allow me to expand on this thought here.

Andreas Blass summarizes the position I object to succinctly. He writes:

"Absolutely free algebra" means "free algebra in a very particular variety V", namely the variety defined by the signature Ω and no identites. Other varieties, like the variety of groups or of rings, also have free algebras, but, since the definitions of these varieties include identities, those free algebras are not absolutely free.

I respectfully object to this viewpoint. Think of it like this. Imagine the phrase "vector space" didn't exist. All we had was a notion of $R$-module. In this hypothetical universe, what if someone tried to coin the term "vector space over $k$" to mean "module over $k$, where $k$ is a field"? My guess is that every last reader would instantly object that there's already terminology for this, they're called modules over $k$, and the distinction that the writer is drawing has precious little value. If I'm right in saying this, then it's probably not a stretch to say that defending the terminology "vector space over $k$" in the modern world is a bit hypocritical, given that the only reason this terminology exists is historical accident, and that if we lived in a world where it didn't exist and someone tried to bring it into existence, the same people who defend the terminology today would, in this hypothetical alternative universe, probably be its foremost critics.

In a similar way, if lived in an alternative universe where all we had was a notion of a free algebra over a theory $T$, and someone tried to coin the phrase "absolutely free algebra over a theory $T$" to mean "free algebra over $T$, where $T$ has no identities," then we'd all oppose the new terminology as redundant. If so, then it's probably not a stretch to consider defending this terminology as a bit hypocritical, at least in those cases where this defense comes from someone who, in that ohter hypothetical universe, would oppose the introduction of this new, redundant terminology.

Just my 2 cents.

Answer 2

Absolutely free $\Omega$-algebra is the same thing as the term-algebra , where $\Omega$ is a given signature (in the sense of universal algebra).

We have the category $\textbf{$\Omega$-Alg}$, where objects are $\Omega$-algebras and arrows are $\Omega$-algebra homomorphisms. There is the canonical forgetful functor $U:\textbf{$\Omega$-Alg} \rightarrow \textbf{Set}$, which maps $\Omega$-algebras to their underlying set. The absolutely-free algebra is simply a functor $T:\textbf{Set} \rightarrow \textbf{$\Omega$-Alg}$ which is left-adjoint to $U$.

A variety $\textbf{V}$ of $\Omega$-algebras is a full-subcategory of $\textbf{$\Omega$-Alg}$ which is closed under products, subalgebras and homomorphic images (in $\textbf{$\Omega$-Alg}$). There is also a canonical forgetful funtor $U_V:\textbf{V} \rightarrow \textbf{Set}$, which is just the restriction of $U$ to $\textbf{V}$. The free-algebra in $\textbf{V}$, is simply a functor $F_V:\textbf{Set} \rightarrow \textbf{V}$, which is left-adjoint to $U_V$.

Various explicit constructions of term-algebras and free algebras exist, however, they all are left adjoints to their respective forgetful functors, and are thus unique up to (unique) isomorphism.

So, to summarize:

• Absolutely-free $\Omega$-algebra = left adjoint to $U:\textbf{$\Omega$-Alg}\rightarrow \textbf{Set}$

and

• Free-algebra in $\mathbf{V}$ = left adjoint to $U_V:\textbf{V} \rightarrow \textbf{Set}$.

Read here for information about category theory.