TITLE: filtered algebra vs graded algebra
QUESTION [1 upvotes]: BACKGROUND
When reading Okounkov-Olshanski's paper about shifted symmetric functions, they define $\Lambda^*$ to be the algebra of shifted symmetric functions.
They say that $\Lambda_n^*$ is a filtered by degree of polynomials and $$\phi: \Lambda^*_{n+1}\to \Lambda^*_n : x_{n+1}\mapsto 0$$ is a morphism of filtered algebras.
They define $$\Lambda^* = \varprojlim\Lambda^*_n$$
taken in the category of filtered algebras, with respect to the morphisms defined above.
MY QUESTIONS

I have seen that when we work with the algebra of symmetric functions, we need to take the inverse limit in the category of graded rings in order to have the necessary elements. Otherwise there are elements that are not symmetric functions on the limit. Why do we take the inverse limit in the category of filtered algebras and not in the category of graded algebras (when we are working with shifted symmetric)? 
I have seen that the filtered algebras are a generalization of the graded algebras. Could anyone explain it to me in a clearer way?

REPLY [4 votes]: I will answer my own question after some time of investigation.

We take the inverse limit in the category of filtered algebras since
it makes no sense to take it in the category of graded algebras. The morphisms
$\rho_n:(\Lambda^*(n))^k \to (\Lambda^*(n-1))^k$ are not compatible
with the specialization $x_n\mapsto 0$. In the new variables, we
have constants that do not go to zero. Hence, we may obtain things
that are not homogeneous of degree $k$. (Taking the inverse limit in
the category of algebras would give us more than what we want.
Infinite products would be allowed and we could have things that are
not finite sum of the elements of the basis).
For a short explanation of what filtered algebras are and how to
define the corresponding graded algebra, I will give the two
definitions. A filtered algebra is a generalization of the notion of
a graded algebra. 
A filtered algebra over the field $k$ is an algebra
$A$ over $k$ which has an increasing sequence
$\{0\}\subset F_0\subset F_1\subset\cdots \subset F_i\subset\cdots\subset A$ of substructures of $A$ such that: $$A=\bigcup_{i\in
\mathbb{N}}F_i$$
and that is compatible with the multiplication in the following
sense:
$$\forall m,n\in \mathbb{N},\ F_n \cdot F_m\subset F_{n+m}.$$ 
Now, let $A$ be a filtered algebra. Then, its corresponding graded algebra $\mathrm{gr}(A)$ is defined as $$\mathrm{gr}(A)= \bigoplus_{n\in\mathbb N}
\frac{F_n}{F_{n-1}}$$ where $\frac{F_n}{F_{n-1}}=G_n$ and
$G_0=F_0$. It is also compatible with the multiplication in the
following sense:
$$\forall m,n\in \mathbb{N},\ G_n \cdot G_m\subset G_{n+m}.$$
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