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I am doing my thesis related with symmetric functions and representations.

It is for this reason that I am reading MacDonald's book Symmetric Functions and Hall Polynomials.

When reading chapter 1.2. The ring of symmetric functions, he also talks about symmetric polynomials. And he compares both notions and sometimes treats them as equivalent but sometimes not. Does anyone have a clearer explanation about what is the difference (if there are differences)?

So, what is the difference between symmetric functions and symmetric polynomials?

idriskameni
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The ring of symmetric functions is a direct limit of the rings of symmetric polynomials in $n$ indeterminates, $n$ going to infinity (for certain injective morphisms that are not inclusion maps). There is no such thing as a (non constant) symmetric polynomial in infinitely many indeterminates, unless one bends the definitions, because a polynomial can have only finitely many terms. For this reason it is sensible to make the terminological distinction, even though symmetric functions are no more "functions" than polynomials are.

By the way, the ring of symmetric functions can also be defined as an inverse limit of the rings of symmetric polynomials, which is maybe more natural and what Macdonald does, but in that case it is important to take the limit in the category of graded rings, for otherwise the limit would give you more than is desired.

So since the is some questioning about this in the comments, let me detail. In the direct limit construction, the injective morphisms to rings with more indeterminates send each elementary symmetric polynomial to the corresponding elementary symmetric polynomial, which, them being morphisms, determines them. (They also send each orbit sum of monomials of degree no more than the number of variables to the corresponding larger orbit sum; this helps to show that for instance using complete homogeneous symmetric polynomials rather than elementary ones has the same effect. For higher degrees, the images of orbit sums do gather some additional monomials, for instance going from one indeterminate $X$ to $X,Y$, the image of the monomial $X^n=e_1[X]^n$ is not $X^n+Y^n$, but the full binomial expansion $e_1[X,Y]^n=\sum_{k=0}^n\binom nkX^kY^{n-k}$.) The image of this morphism is complementary to the kernel of a surjective morphism in the opposite direction that is restricted from the morphism of full polynomial rings sending new indeterminates to zero and keeping the old ones intact. Such morphisms are the ones used in the inverse limit construction, but as said the inverse limit must be taken in the category of graded rings lest such beasts as $\prod_{i\in\Bbb N}(1+X_i)$ spring into existence, destroying the graded nature of the ring. It is an easy exercise to see that the direct and indirect limit constructions define isomorphic (graded) rings.

Either definition does not actually say what a symmetric function is, just what the ring of all of them looks like, up to isomorphism. If one wants to define the ring of symmetric functions as a substructure of something familiar, that is also possible: one could define them as symmetric formal power series in a countable set of indeterminates with bounded degree terms. This is the point of view taken by for instance Stanley (Enumerative Combinatorics, 2) and Sagan (The symmetric group; Representation, combinatorial algorithms and symmetric functions).

While I'm at it, let me add why I started with the direct limit construction, even though both kinds of limits can be used. The direct limit construction morally corresponds better with the way I think of symmetric functions: every individual computation in the ring can be faithfully represented inside some ring of symmetric polynomials, and as such the ring behaves like a infinite union of subrings, not as an inverse limit construction which usually brings into existence infinite values of which only shadows exist in the finite domain. The inverse limit of graded rings precisely avoids such values, because in every fixed degree the surjective morphisms ultimately become bijections.

Marc van Leeuwen
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  • And could you give me an example of something that is not a symmetric polynomial? Or an example of something that looks like a symmetric function but is forbidden by definition? – idriskameni Feb 07 '19 at 15:57
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    Hmm, I have never seen it defined as a direct limit -- can you make it so that the arrows involved are ring homomorphisms? – darij grinberg Feb 07 '19 at 17:07
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    @idriskameni: Macdonald gives $\prod_i \left(1+x_i\right)$ as an example of a symmetric formal power series that nevertheless does not count as a symmetric function since it has unbounded degree. On the other hand, most polynomials -- such as $x_5 + x_7$ -- are not symmetric functions because they are not, uhm, symmetric. – darij grinberg Feb 07 '19 at 17:09
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    I think it's actually an inverse limit. – Angina Seng Feb 07 '19 at 17:18
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    I'm going to pile on here: It is the inverse limit. If $\Lambda_n=\mathbb{C}[x_1,\ldots,x_n]^{S_n}$, then there is a surjective map $\Lambda_n\to\Lambda_{n-1}$ obtained by setting $x_n=0$. The ring of symmetric functions is $\Lambda=\lim_{\leftarrow}\Lambda_n$. This ring consists of functions that are symmetric in infinitely many variables: $x_1,x_2,\ldots$. The direct limit is different. It consists of polynomials that are symmetric in finitely many variables: $x_1,\ldots,x_N$ for some $N>0$. – David Hill Feb 07 '19 at 17:33
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    @David Hill: No. As the answer says, the inverse limits as rings is too big and contains $\prod_i(1+X_i)$ (but the corresponding inverse limit of graded rings is OK as definition). And a direct limit of inclusion maps is not defined, because there are no inclusion maps ($X_0+X_1$ is a symmetric polynomial in two indeterminates, but not in three; the collection of polynomials in $x_1,x_2,\ldots$ that you describe is not a ring). – Marc van Leeuwen Feb 07 '19 at 17:49
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    @darijgrinberg: see the added piece of my answer. – Marc van Leeuwen Feb 07 '19 at 17:55
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    Okay, you can define injective morphisms sending the elementary symmetric polynomials in the smaller ring to those in the larger ring. But it will not send monomial orbit sums to monomial orbit sums in general. Otherwise the identity $m_{1,1}^2 = m_{2,2}$ holding in two variables would hold in three as well. Probably monomial sums get preserved up to some degree -- I guess, the number of variables in the smaller ring. So I am not sure how natural these injections are. – darij grinberg Feb 07 '19 at 18:20
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    @darijgrinberg Thank you for pointing out my inaccuracy, which I have now tried to correct in the answer. To my defence, I was writing from memory, but on checking some 15 year old notes, they did point that just filling up orbits does not define ring morphisms, and indicated the correction I now made. – Marc van Leeuwen Feb 08 '19 at 08:36
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    It is a highly useful answer either way :) – darij grinberg Feb 08 '19 at 13:19
  • Thank you very much. Your answer was really helpful. Now, I have another question. Why is so important to take this inverse limit in the category of graded rings and not in the category of rings? A graded ring homomorphism is a ring homomorphisms $f:R\to S$ such that $f(R) \subseteq S$ (if I am not wrong). Why is that property important? @MarcvanLeeuwen – idriskameni Feb 10 '19 at 15:32
  • I have posted another question about it. May be you can help me. @MarcvanLeeuwen – idriskameni Feb 10 '19 at 16:23
  • For information, that question is here – Marc van Leeuwen Feb 10 '19 at 16:49
  • Another question I have related to this one is: In MacDonald's book, he says $\prod_i (1+x_i)$ is not in $\Lambda$ since elements in $\Lambda$ are finite sums of monomial symmetric functions $m_{\lambda}$. But then, a little bit further he says, the elements of $\Lambda$ (unlike those of $\Lambda_n$) are no longer polynomials: they are formal infinite sums of monomials. We have therefore reverted to the older terminology of 'symmetric functions'. I am super confused. – idriskameni Feb 10 '19 at 17:21
  • @idriskameni You just deleted the question you had asked, while I was busy answering it. Please don't do that. – Marc van Leeuwen Feb 10 '19 at 17:24
  • Sorry, I am editing it to be much more easy to understand. @MarcvanLeeuwen – idriskameni Feb 10 '19 at 17:25
  • @idriskameni You don't need to delete a question in order to edit it. But I will wait. – Marc van Leeuwen Feb 10 '19 at 17:26
  • You can see it here @MarcvanLeeuwen. https://math.stackexchange.com/questions/3107701/lambda-lim-leftarrow-lambda-n-ring-of-symmetric-functions – idriskameni Feb 10 '19 at 17:29