This question is related to this other question.

When understanding how it is defined the ring of symmetric functions, I can not see why is so much important to take the inverse limit in the category of graded rings.


Consider $\Lambda$ to be the ring of symmetric functions.

$\Lambda_n$ to be the symmetric polynomials in $n$ independent variables.

Moreover, I know that in the category of rings, the objects are rings and the arrows are ring homomorphisms.

I also know that in the category of graded rings, the objects are rings and the arrows are graded rings homomorphisms. I.e. if $f:R\to S$ is a ring homomorphisms. A graded ring homomorphisms is $f$ such that $f(R)\subseteq S$.

Then, in the category of graded rings,

$$\Lambda = \varprojlim\Lambda_n = \left\{a \in \prod_{i\in I}\Lambda_i \mathrel{\Bigg|} \forall i \leq j: f_{i,j}(a_j)=a_i \right\}$$

In the category of rings,

$$\Lambda ^* = \varprojlim\Lambda_n = \left\{a \in \prod_{i\in I}\Lambda_i \mathrel{\Bigg|} \forall i \leq j: f_{i,j}(a_j)=a_i \right\}$$

And $\Lambda \subset \Lambda^*$ (my teacher told me).

But I can not see what makes the difference in considering the inverse limit in these two different categories. I can not see how it affects the arrows in the categories to these sets.

Any help?


4 Answers 4


Taking an inverse limit in the category of graded rings corresponds to taking the inverse limit at each degree and taking the direct sum.

So if $\Lambda_n^k$ is the group of degree $k$ symmetric polynomials in $n$ variables then we let

$$ \Lambda^k = \lim_{\gets} \Lambda_n^k \quad \text{and} \quad \Lambda = \bigoplus_k \Lambda^k.$$

And the effect of this is that we still have a direct sum at the end of the day. I.e. elements of $\Lambda$ consist of symmetric functions with bounded top-degree.

On the other hand, consider the sequence

\begin{align} &1 + x_1, \\ &1+(x_1+x_2)+x_1x_2, \\ &1 + (x_1 + x_2 + x_3) + (x_1x_2 + x_1x_3 + x_2x_3) + (x_1x_2x_3), \\ &\dots \end{align}

In $\Lambda^*$ this is an element, which you can see because if you set $x_n = 0$ in the $n$-th term of the sequence, you get the previous term.

This sequence converges to

$$ 1 + \sum_i x_i + \sum_{i < j} x_ix_j + \sum_{i < j < k} x_ix_jx_k + \cdots $$

but this is not an element of $\Lambda$.

  • $\begingroup$ I see what you mean. But in terms of the definition I have given above, how these arrows affect? I am sorry, I am having a bad time trying to understand this. $\endgroup$ Commented Feb 10, 2019 at 17:55
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    $\begingroup$ @idriskameni You state that $ \{a \in \prod_{i\in I} | f_{i,j}(a_j)=a_i, \forall i \leq j \}$ is the inverse limit in the category of graded rings, but I don't think that's true. $\endgroup$
    – Sera Gunn
    Commented Feb 10, 2019 at 17:56
  • $\begingroup$ That is the general definition of inverse limit. The result should change in each category (or not). Where $f_{i,j}: \Lambda_j \to \Lambda_i$ for all $i \leq j$. $\endgroup$ Commented Feb 10, 2019 at 17:57
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    $\begingroup$ @idriskameni The proper definition of an inverse limit is via the universal property. It is a theorem for rings that this construction satisfies that universal property, but I don't think that construction satisfies the universal property in the category of graded rings. In particular, I don't see why $\{a \in \prod_{i\in I} | f_{i,j}(a_j)=a_i, \forall i \leq j \}$ is even graded. The correct construction of an inverse limit in the category of graded rings has you taking an inverse limit degree by degree and taking the direct sum. $\endgroup$
    – Sera Gunn
    Commented Feb 10, 2019 at 18:00
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    $\begingroup$ @idriskameni "We start with the definition of an inverse system (or projective system) of groups" is what Wikipedia says. $\endgroup$
    – Sera Gunn
    Commented Feb 10, 2019 at 18:02

One should not forget that fundamentally limits are defined using category theory. The expressions given to realise them may be easier to understand for mortals who think in terms of elements, but they can only be used if they can be proved to satisfy the categorical specification.

It turns out that your expression for limits of graded rings is wrong for a rather basic reason: an infinite product (in the set theoretic sense) of graded rings is not (in any natural way) a graded ring to begin with. In a graded ring every element has to be a (by definition finite) sum of homogeneous elements, just like polynomials need to be (finite) linear combinations of monomials. In an infinite product of graded rings (viewed as ring, with subsets of homogeneous elements in each degree defined in the obvious way) one easily finds elements that are not finite sums of homogeneous elements, in the same way that most formal power series are not finite linear combinations of monomials, simply by combining components of infinitely many different degrees from the different factors. Instead, one could form a restricted product, the subring of the product generated by homogeneous elements; that subring obviously can be made into a graded ring. Using this restricted product instead of $\prod_i\Lambda_i$ in the expression for the inverse limit will give you a correct model.

The difference with the inverse limit of rings $\Lambda_n$ (which does use the unrestricted product) is that the latter has far more elements. For instance taking $a_n\in\Lambda_n$ to be the element $\sum_{i=0}^ne_i[X_1,\ldots,X_n]$ for each$~n$, one clearly has $f_{i,j}(a_j)=a_i$ whenever $i\leq j$, so this defines an element of the inverse limit of rings. However this element of the inverse limit (which can also be described more concisely as $\prod_{i\geq 1}(1+X_i)$, even though it requires some additional effort to even make sense of that expression) is not a finite sum of homogeneous elements, and therefore has no corresponding element in the inverse limit of graded rings. And we do not wish to deal with such elements in the ring $\Lambda$ of symmetric functions, which is why it is important to use the graded ring inverse limit construction for$~\Lambda$ (if one wants to use an inverse limit construction at all).

One might wonder how the category theory definition works out for both constructions, and how each of these different rings manages to satisfy the requirements for the inverse limit in one category but not in the other. One part is easy: the unrestricted product construction is not a graded ring in any way that makes the morphisms graded, so it plays no rôle in the graded ring category at all. However the restricted product does define an ordinary ring that has the required morphisms and makes everything commute, so why is it not the limit in the category of rings? Because it fails the universal property: any other ring with such family of morphisms to each $\Lambda_n$ should factor through the inverse limit ring, but the one constructed from the unrestricted product does not factor through the one constructed from the restricted product, as the "unbounded" elements have nowhere to go. In the opposite direction there is no problem: the restricted product construction maps (injectively) to the unrestricted product construction, in a (unique) way that makes everything commute.

  • $\begingroup$ I think I do understand everthing you say. But something is still not clear in my mind. Symmetric polynomials are the elements in $\Lambda_n$. I am ok with that. Now, symmetric functions are $f=(f_n)_{n\geq 0}$ where each $f_n$ is a symmetric polynomial. But this $n$ has to be finite? I do not understand what you said in the other post: one could define them as symmetric formal power series in a countable set of indeterminates with bounded degree terms. I am sorry. Could you try to give some example or something to make it more clear? I know it is difficult. Really appreciate your effort. $\endgroup$ Commented Feb 11, 2019 at 11:57
  • $\begingroup$ When doing the construction in the category of graded rings, should I consider: $$\Lambda^k = \varprojlim\Lambda_n^k = \left\{a \in \prod_{i\in I}\Lambda_i^k \mathrel{\Bigg|} \forall i \leq j: f_{i,j}(a_j)=a_i \right\}$$ with $i\in I$ and $I$ being finite? Then I would obtain an inverse limit which is a graded ring and then I define $\Lambda= \oplus_k \Lambda^k$? $\endgroup$ Commented Feb 11, 2019 at 12:13
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    $\begingroup$ @idriskameni first comment: in $f=(f_n)_{n\geq 0}$ the (bound) variable $n$ runs over all natural numbers, so a symmetric function must correspond to a symmetric polynomial in any number of indeterminates. The idea is that upwards you just increase the symmetric orbits being summed over (so $e_2=\sum_{i<j}X_iX_j$ lets $(i,j)$ range over more and more pairs), and downwards you kill some indeterminates by making them $0$ (which again reduces orbits). When $n\to\infty$ the orbit sums describe formal power series (like $\sum_{i,j\in\Bbb N, i<j}X_iX_j$) but whose terms remain in bounded degree. $\endgroup$ Commented Feb 11, 2019 at 12:57
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    $\begingroup$ @idriskameni 2nd comment: No the product is over $i\in\Bbb N$ here, so it is infinite. But in fixed degree $k$ the maps $f_{i,j}$ become isomorphisms when $i,j\geq k$, so once $a_k$ is known, the remaining components of the product are determined and add no freedom. By the way, this way of doing inverse limits (as Abelian groups) in each degree separately and then taking the direct sum is just one way to get a model for the inverse limit for graded rings, the "restricted product" method of my answer is another. Category theory does not say which one to use, it just specifies up to isomorphism. $\endgroup$ Commented Feb 11, 2019 at 13:07
  • $\begingroup$ Thank you very much. Really appreciate your work. $\endgroup$ Commented Feb 11, 2019 at 19:14

I personally find that it is cleaner to observe the difference between the inverse systems to attain $\Lambda^{k}$ and $\hat{\Lambda}$ rather than directly stating their inverse limits. For the case which will lead to the completion of the ring of symmetric functions $\hat{\Lambda}$ the inverse system is

\begin{equation}\{(\Lambda_{n})_{n \geq 0}, (\rho)_{m \geq n \geq 0}\}.\end{equation}

Whereas the inverse system which leads to $\Lambda^{k}$ is of the form

\begin{equation}\{(\Lambda_{n}^{k})_{n \geq k \geq 0},(\rho_{m,n}^{k})_{m \geq n \geq k \geq 0}\}.\end{equation}

Here $\rho_{m,n}: \Lambda_{m} \rightarrow \Lambda_{n}$ is based upon the stability property of the monomial symmetric functions. So it is done by

\begin{equation}\rho_{m,n}(m_{\lambda}(x_{1},\dots,x_{m})) := m_{\lambda}(x_{1},\dots,x_{n},\underbrace{0,\dots,0}_{m-n}) = m_{\lambda}(x_{1},\dots,x_{n}).\end{equation}

Then the $\rho_{m,n}^{k}$ are defined by restriction with

\begin{equation}\rho_{m,n}^{k} := \rho_{m,n}|_{\Lambda_{m}^{k}}.\end{equation}

So although the $\Lambda_{n}$ and $\Lambda_{n}^{k}$ are fundamentally related (and so are $\rho_{m,n}$ and $\rho_{m,n}^{k}$) they are not equivalent. That each element of $\Lambda_{n}^{k}$ is homogenous and of degree $k$ will ensure bounded degree for the limit. Whereas elements of $\Lambda_{n}$ have no such criteria.

Of course, to complete the construction of the ring of symmetric functions

\begin{equation}\Lambda = \bigoplus_{k \geq 0} \Lambda^{k}.\end{equation}

Note: I am new to posting so I apologise if it is improper to comment on a post that is already three years since its last update. But hopefully someone finds this useful one day.

  • $\begingroup$ So long as you have answered the question, and have provided new ideas that have not been provided in the older answers, answering an old question is perfectly fine (and it looks like you've done that here). Welcome to MSE. : ) $\endgroup$ Commented Nov 2, 2022 at 0:29

The inverse limit in the graded category is the ring of polynomials in the elementary symmetric functions $E_i$, that is, $\Bbb Q[E_1,E_2,E_3,\ldots]$.

The inverse limit in the ungraded category is larger. It contains things like the formal infinite sum $E_1+E_2+\cdots$.


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