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I'm not completely sure if this bunch of questions is the appropriate Level of MO. However at the same time I think that it is at least slightly above the level of stackex. ...


The tensor algebra functor $T: \mathbf{Vec} \to \mathbf{Alg}$ that assigns the Tensor Algebra to any vector space is the left adjoint to the forgetful functor $V:\mathbf{Alg} \to \mathbf{Vec}$ that forgets the algebraic struture and just keeps the linear structure. This is reflected in the universal property of the tensor algebra. That is:

Any linear map $f$ from a vector space $V$ to an algebra $A$ can be uniquely extended to an algebra homomorphism $\bar{f}$ from $T(V)$ to $A$ such that $\bar{f}\circ i = f$ holds for the inclurion $i : V \to T(V)$.

Moreover since the right adjoint is the forgetful functor the tensor algebra is the free algebra over the vector space.

So here is the first question:

1.) The dual situation: On the underlying vector space of the tensor algebra there is more than one coalgebra structure. For example there is the shuffle coproduct and the deconcatenation coproduct. Is one of them (co)universal and moreover cofree? I.e. is the functor $\bar{T}: \mathbf{Vec} \to \mathbf{CAlg}$, that assigns to a vector space the vector space underlying the tensor algebra together with one of those coproducts, a right adjoint to the forgetful functor $V: \mathbf{CAlg}\to \mathbf{Vec}$ that forgets the coalgebraic structure?


Now the exterior algebra as well as the symmetric algebra are defined as quotients of the tensor algebra and as the tensor algebra they have a similar universal property.

2.) What are the right adjoints to the exterior/symmetric algebra? From the universal property they should be left adjoints to something.

3.) Is it a general property of universal free algebras that their quotients are universal algebras?

4.) If 3.) is true what is the dual picture for (co)universal cofree coalgebras? Do they have the property that their sub coalgebras are still (co)universal coalgebras?

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1 Answer 1

1) The answer appears to be no; this was discussed on MO previously.

2) The symmetric algebra functor is left adjoint to the forgetful functor from commutative algebras to vector spaces (over a field of characteristic zero for simplicity). The exterior algebra functor is left adjoint to the forgetful functor from supercommutative algebras to super vector spaces when restricted to odd vector spaces.

3) I'm not sure what you mean by this. Surely you don't just mean quotients but universal quotients, e.g. an appropriate natural transformation...?

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Exterior and Symmetric Algebras are Quotient of the tensor algebra and are universal. That is the way I mean it –  Mark.Neuhaus Jun 1 '12 at 5:08
    
The answer to question one should be 'yes' from your mentioned discussion, since cofree means right adoint –  Mark.Neuhaus Jun 1 '12 at 5:10
2  
@Mark: as far as I can tell, the discussion in that thread shows that the cofree coalgebra on a vector space does not have the same underlying vector space as the free algebra. Also, the exterior and symmetric algebras are more than just quotients of any particular tensor algebra: they're a compatible family of quotients of every tensor algebra (in other words, an appropriate natural transformation from the tensor algebra functor to some other functor). A specific algebra which is a quotient of a specific free algebra isn't even a functor, so it's a type error to ask whether it's universal. –  Qiaochu Yuan Jun 1 '12 at 14:30

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