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This question is a follow-up to a question about the theory of polynomials.

It should be quite clear by now that matrix theory and linear algebra are quite different topics. As the various answers to that question clearly state, the difference is largely because of basis choice. But we know how to do a lot of abstract linear algebra, without ever having to choose a basis.

But when one switches to polynomials, the situation changes: even very general definitions of polynomials (like that in Lang's Algebra, for example) still use monomials in the standard basis. Why that basis, and not the rising factorial basis, or falling factorial, or Chebyshev basis, or any other? And the most important question: why, in fact, choose a basis at all?

Which leads to the question in the title: Are plethories a theory of basis-free polynomials?

I think the motivation should be clear: if going 'basis free' was so successful for linear algebra, shouldn't we also expect similar success with polynomials? Unfortunately, I have not been able to find any work which I could readily understand as being about developing a theory of basis-free polynomials.

EDIT: After reading the current comments and answers, I am starting to think that perhaps what I am really seeking is a theory for basis-independent polynomials, or even basis-generic rather than leaping right away to basis-free. There are many applications of polynomials where bases other than the monomial one greatly simplify both reasoning and computations (which is what I am eventually after), but the tools for doing this seem to be hard to find.

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Is "plethory" a real tag? Or "polynomial" for that matter? Too many tags to keep track of. –  Jim Humphreys Oct 27 '11 at 22:21
    
@Jim: there were not real tags, until now... I welcome suggestions on that front. –  Jacques Carette Oct 27 '11 at 22:59
    
There is a tag "polynomials" already in use... –  Gjergji Zaimi Oct 27 '11 at 23:37
    
I'm unclear as to what you would want to be able to do with a basis-free description of polynomials. My inner category theorist wants to know what you would consider to be the morphisms between spaces of polynomials - if there are any: my experience (which isn't much) is that we tend to think of spaces of polynomials one at a time, whereas we work with lots of vector spaces at the same time. Plethories are more like rings: the right way to move between them is via bimodules. If that's the sort of thing you want, then maybe they are the right description for you. –  Loop Space Oct 28 '11 at 7:06
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The Grothendieck ring of finite dimensional representations of $SL(2)$ is isomorphic to the polynomial ring. The monomial basis is given by taking tensor powers of the fundamental representation. A more natural basis are the irreducible representations (which happen to be the symmetric powers of the fundamental representation). The irreducible representations then correspond to the Chebychev polynomials; so this example, which you mention, is studied. –  Bruce Westbury Oct 28 '11 at 8:09

1 Answer 1

Some naive remarks. It seems to me that the simplest reason to choose the standard basis is because it exhibits the universal property of a polynomial ring.

One way to exhibit a partially basis-free definition of a polynomial ring is to define it as the symmetric algebra on a vector space. This retains the grading but is otherwise basis-free. If one wants to ignore the grading then I am not sure what options are available and depending on the application it may not be natural to ignore the grading.

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Let me add that many basic (and some not so basic) homological constructions have basis free definitions - for example one can easily define the Koszul resolution using the exterior algebra. –  Alexander Woo Oct 28 '11 at 3:56
    
+1. The distinction seems to be whether you take the left adjoint of the "underlying set" functor or the "underlying module" functor from commutative rings. –  S. Carnahan Oct 28 '11 at 4:20
    
One may ignore the grading when describing a symmetric algebra through its universal property: a module map from the vector space to a commutative algebra extends uniquely to an algebra map. It is a happy and useful circumstance that this ungraded problem has a graded solution. –  Wilberd van der Kallen Oct 28 '11 at 7:11
    
I do not want to ignore the grading - it is there in all choices of basis, so I don't see a good a priori reason why it would disappear when we go basis-free. –  Jacques Carette Oct 28 '11 at 12:53
    
@Jacques: I don't know what you mean by "it is there in all choices of basis." Certainly all the examples you've written down implicitly give the same filtration on polynomials, but it is possible to write down bases where this is not the case (for example $x, x^3 - x^2, x^3 + x^2, x^4, x^5, ...$). @Wilberd: what I meant by "ignore the grading" is giving an abstract description of a polynomial ring that doesn't distinguish any of its elements (in the symmetric algebra description we distinguish the elements of degree $1$). –  Qiaochu Yuan Oct 28 '11 at 14:51

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