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What's the functor F and what's the morphism from F(A) to A (and what's A)?

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closed as unclear what you're asking by David White, Andres Caicedo, Willie Wong, Daniel Moskovich, Noah Stein Aug 5 '13 at 13:52

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This question needs to be tightened considerably. The answer to the title question is "yes, of course": any object of any category is an algebra over the identity endofunctor; typically objects can be realized in many ways as algebras over endofunctors $F$ for various $F$. But if you provided more context, there's a good chance you'll get more satisfying answers. –  Todd Trimble Jun 1 '13 at 2:06
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Presumably the question is something like: On the category of vector spaces, is there an endofunctor $F$ whose algebras are precisely the Grassmann (=exterior) algebras? I say "something like" because of course one could replace "vector spaces" by some other category, or ask that $F$ be a monad, or so on. (The latter does not seem to be the common usage en.wikipedia.org/wiki/F-algebra.) Bartosz, please read mathoverflow.net/howtoask, and also look over other MathOverflow questions, and revise your question. –  Theo Johnson-Freyd Jun 1 '13 at 5:57
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Sorry, I'm new to mathoverflow and I'm not a mathematician. I know about Grassman algebras from physics (supersymmetry) and about F-algebras from programming (Haskell), so I'm asking this question from a somewhat narrow point of view. Indeed, I was thinking of vector spaces and the wedge product. Is there a triple (endofunctor F, object A, and morphism F(A)->A) that can be used to define an exterior algebra? Or would Grassman algebra be the initial algebra for such a triple? I'm looking for some intuitions. –  Bartosz Milewski Jun 1 '13 at 19:33
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1 Answer

I want to convince you that this is the wrong question to ask. First, a little background. There is a relationship between algebras in the mathematical sense and $F$-algebras for $F$ an endofunctor, but in mathematics the endofunctor is usually a monad and we usually want algebras over this monad in the monadic sense, where there are more requirements. The basic relevant example here is that on $\text{Vect}$, the category of vector spaces over a field $k$ (suppressed in the notation), there is a monad whose algebras are the associative $k$-algebras. It is a sort of $k$-linear version of the list monad.

One reason we care about monads so much more than arbitrary endofunctors is because we get them from adjunctions, which are ubiquitous. For example, the monad which gives us associative $k$-algebras comes from an adjunction between $\text{Vect}$ and the category $k\text{-Alg}$ of $k$-algebras. There is a similar monad giving us commutative $k$-algebras coming from an adjunction between $\text{Vect}$ and the category of commutative $k$-algebras. In both cases here the second category of the adjunction is naturally equivalent to the category of algebras over the corresponding monad; in this situation the adjunction is said to be a monadic adjunction.

The monad coming from commutative algebras sends a vector space $V$ to the symmetric algebra $S(V)$ over $V$, which is very close to the exterior algebra functor. Here I want to convince you that the right question to ask, in terms of smashing together the concept of exterior algebras with the concept of algebras over an endofunctor, is actually "is the exterior algebra functor a monad, and if so, does it come from a monadic adjunction?"

The answer is yes in the following sense. We need to replace $\text{Vect}$ with the category of super vector spaces (the same super as in supersymmetry). The objects of this category are direct sums $V \cong V_0 \oplus V_1$ where $V_0$ is the even or bosonic part and $V_1$ is the odd or fermionic part, and the morphisms are linear maps preserving parity. There is a monadic adjunction between this category and the category of supercommutative algebras (I'm suppressing the underlying field here again), and the corresponding monad is a super version of the symmetric algebra monad which, on purely odd super vector spaces, reproduces the usual exterior algebra.

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