$\Lambda = \varprojlim\Lambda_n$ (ring of symmetric functions)

Question

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.

MY WORK

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?

Answer 1

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).

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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.

Answer 2

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 $E_1+E_2+\cdots$.

Answer 3

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 $\Lambda$.

Answer 4

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.