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Ladder of Algebraic Structures

jwkennington.com

51–59 of 59 posts

Re: Ladder of Algebraic Structures

#51
post #10

Most algebraic structures are best understood by which axioms it satisfies. For example basically every subset of axioms of an abelian group is useful enough to have a name. Wiki has a really nice table: Semigroupoid Small Category Groupoid Magma Quasigroup Unital Magma Loop Semigroup Inverse Semigroup Monoid Commutative monoid Group Abelian group https://en.wikipedia.org/wiki/Abelian_group

I'd like to seen an extension of this table with the negation of these axioms

Here is an beautiful old paper about data structures built using a binary join operator, exploring the 16 possible outcomes for properties: unit, idempotent, associative, commutative.

The resulting grid can be factored around set, bag, list and binary tree, with empty/non-empty variants.

Then there is interaction of the structures with binary operators on the data elements themselves, giving a nice analysis of map, filter, fold (reduce) in functional programming.

A.Bunkenburg, The Boom Hierarchy

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.49....

There is a rich literature if you chase the references back and forth, starting from Bird-Meertens Formalism (Squiggol), Hoogendijk, through Backhouse and Malcolm, to Meijer and McBride.

Re: Ladder of Algebraic Structures

#52
post #19
post #7

Out of clarity this is an "algebra over a field" vs a more general concept of an algebra over a ring. More generally an algebra A, over a ring R, an R-algebra , is a ring A equipped with a map Hom(A,Z(R)). Algebra over a field is a special case. Here's a "fun" object for you to consider: https://en.wikipedia.org/wiki/Field_with_one_element

> More generally an algebra A, over a ring R, an R-algebra, is a ring A equipped with a map Hom(A,Z(R)). I don't think that's the usual definition of an algebra. For example, it would mean that there is no difference between an algebra over a non-commutative ring and over its centre, which seems weird; and it clashes with the usual habit to regard every non-0 commutative ring as a non-trivial ℤ-module, whereas, for e…

At long last, my disastrous and scarring grad-school experience in representation theory can save the day!

First, regarding the OP: Having spent many years studying algebra, I don't find the hierarchy of axioms to be very useful in thinking about these things. Sure, you can think of a field as a "commutative ring with inverses", but rings and fields present themselves so differently that this connection doesn't end up being all that useful. Fields are not rich enough on their own to support much interest. You'll find them mostly as building blocks rather than powerful tools in and of themselves. Ditto for modules and vector spaces. Sure, a module is "like a vector space but over a ring", but vector spaces are so boring by themselves that they show up mostly as scaffolding. The study of modules, on the other hand, is its own branch of mathematics. It's much more useful to think of them in terms of what you actually do with them.

Now, on to definitions. The following few paragraphs are all very small-minded and look far more complicated than they actually are. It all encodes pretty much what you'd expect.

If you want to define algebras over commutative rings, you need to start with left- and right-algebras. A left-algebra is an abelian group A equipped with a map \phi: R -> End(A). The abelian group structure defines the addition in the algebra, and the map defines the left-multiplication: if r \in R, and a \in A, then you define a times r as \phi(r)(a), where \phi(r) is an endomorphism on A.

A right-algebra is the same, only the map is from R to the opposite ring of End(A), where the opposite ring is the one you get by just reversing the multiplication. You need to do this because associativity demands that you compute ((a)r)s, where a \in A, r,s \in R, by first acting on a with r, then by s. But with the usual conventions of composition of functions, \phi(r) \circ \phi(s) means you first "do" s, then r. So you need to flip it. Working with left- and right-algebras is a pain in the butt because you have to carry around a ton of left-right nonsense.

A bialgebra (in the literature I read) is a an abelian group that is both a left- and right-algebra. A central bialgebra is one where the left and right multiplication are the same, which is not a given. Noncentral bialgebras are especially annoying, mostly because you have to figure out how to do pre-subscripts in LaTeX so you can write nonsense like "_R M_S".

Obviously, all of these things collapse if R is commutative. Noncommutative ring theory requires a special kind of patience. And don't even get me started on noncommutative geometry.

Re: Ladder of Algebraic Structures

#53
post #19

Earlier quoted context omitted.

> More generally an algebra A, over a ring R, an R-algebra, is a ring A equipped with a map Hom(A,Z(R)). I don't think that's the usual definition of an algebra. For example, it would mean that there is no difference between an algebra over a non-commutative ring and over its centre, which seems weird; and it clashes with the usual habit to regard every non-0 commutative ring as a non-trivial ℤ-module, whereas, for e…

At long last, my disastrous and scarring grad-school experience in representation theory can save the day! First, regarding the OP: Having spent many years studying algebra, I don't find the hierarchy of axioms to be very useful in thinking about these things. Sure, you can think of a field as a "commutative ring with inverses", but rings and fields present themselves so differently that this connection doesn't end u…

You are speaking with a representation theorist, too. It seems to me that we agree on the definition. (I agree that I was writing as if `R` were automatically commutative, after having made a big fuss about the possibility that it wasn't. Incidentally, if you feel insufficiently scarred, you might like to expand your stable of algebras: there is the notion of a coalgebra, which is dual to that of an algebra; and I believe that the usual notion of bialgebra is of a ring equipped with the structure both of an algebra and a coalgebra: https://en.wikipedia.org/wiki/Bialgebra ; but maybe it's different in the world of algebras over non-commutative rings, which is not my speciality. Then among the bialgebras are the Hopf algebras, etc.)

Re: Ladder of Algebraic Structures

#55
post #35

Earlier quoted context omitted.

What I would love are examples of how they are useful.

For me the first example where I really got why algebraic structures was useful this video on using abstract algebra in analytics[0]. This helped me grasp something that I had read from Alexander Stepanov[1] that I hadn't fully understood before (not being familiar with the algebraic terminology): > I suddenly realized that the ability to add numbers in parallel depends on the fact that addition is associative...In o…

Thank you!

Re: Ladder of Algebraic Structures

#56
post #35

Earlier quoted context omitted.

What I would love are examples of how they are useful.

This is a really cool presentation where the authors "step up the ladder" to design a really elegant API for animations as semirings (where * is used to sequence animations, and + for animations running in parallel), and then go on to implement it in Swift: https://bkase.github.io/slides/algebra-driven-design/

Thank you!

Re: Ladder of Algebraic Structures

#57
post #41
post #35

Earlier quoted context omitted.

What I would love are examples of how they are useful.

see the second-to-last slide for a mapping from Algebraic Structures to Computer Science Concepts. http://comonad.com/reader/wp-content/uploads/2009/08/Introdu...

Thank you!

Re: Ladder of Algebraic Structures

#58

Earlier quoted context omitted.

I'd like to seen an extension of this table with the negation of these axioms

What would you do with that? For example, I can see the use of commutativity (ab = ba) and anticommutativity (ab = -ba), but I'm not sure what I'd do with the negation of commutativity (ab ≠ ba).

Non-commutative also means that if there are units, there may be different left and right units:

    1L * A = A = A * 1R

Re: Ladder of Algebraic Structures

#59
post #57
post #41

Earlier quoted context omitted.

see the second-to-last slide for a mapping from Algebraic Structures to Computer Science Concepts. http://comonad.com/reader/wp-content/uploads/2009/08/Introdu...

Thank you!

Thank Edward Kmett! That slide is single-handedly responsible for me getting into abstract algebra.
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