TITLE: Schur-Weyl duality and q-symmetric functions
QUESTION [15 upvotes]: Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question will still be clear despite them.
The (integral) representation rings of the symmetric groups can be packed together into a hopf algebra $H_1 = \oplus_n Rep(\Sigma_n)$ where the multiplication (resp. comultiplication) comes from induction (resp. restriction) along $\Sigma_n \times \Sigma_k \to \Sigma_{n+k}$. In fact there's a further structure one can put on $H$ corresponding to the inner product of characters and a notion of positivity (all together its sometimes called a "positive self adjoint hopf algebra"), but for simplicity I will disregard this structure in what follows (of course if its not important for the answer that would be great to know).
Its well known that sending the irreducible specht modules to their corresponding schur functions induces an isomorphism of hopf algebras to the (integral) hopf algebra of symmetric functions. 
Following the "$\mathbb{F}_1$-philosophy" it is tempting to define a ring of "q-symmetric functions" as the hopf algebra $H_q = \oplus_n Rep(GL_n(\mathbb{F}_q))$ equipped with the same structures as above.

Question 1: Is there a hopf algebra over $\mathbb{Z}[q]$ which
  specializes at a prime power $q=p^n$ to $H_{p^n}$ and at $q=1$ to
  $H_1$ the classical ring of symmetric functions?

By schur weyl duality we also know that $H_1 \cong Rep(GL_{\infty}(\mathbb{C})):= colim_n Rep(GL_n(\mathbb{C}))$ (at least as rings). It seems natural to ask if there's any form of schur duality going in the other direction.

Question 2: Is there any kind of relationship between the rings $Rep(\Sigma_{\infty}) := colim_n Rep(\Sigma_n)$ and $\oplus_n Rep(GL_n(\mathbb{C}))$?
Question 3: Is there a $\mathbb{Z}[q]$-algebra which specializes to $Rep(GL_{\infty}(\mathbb{F}_q))$ at a prime power $q = p^n$ and to $Rep(\Sigma_{\infty})$ at $q=1$?

REPLY [14 votes]: As Sam Hopkins says, the category of all representations of $GL_n(\mathbb F_q)$ is too large to give what you want.  Instead, let's consider the category of unipotent representations,  i.e.  those appearing in the irreducible decomposition of $\mathbb Q [GL_n(\mathbb F_q)/B_n(\mathbb F_q)]$.  
Unipotent representations are not closed under the naive induction product,  but they are closed under parabolic induction $V*W = {\rm Ind}_{P(n,m)}^{GL_{n+m}} V \otimes W$.  This gives $\oplus_n  {\rm Rep}^{un}(GL_n(\mathbb F_q))$ the structure of a monoidal category.  Instead of being symmetric monoidal, it is now braided monoidal!  The Grothendieck ring is a $q$ deformation of the ring of symmetric functions.
Finally, by Morita theory,  unipotent representations are equivalent to representations of $\mathcal H_n(q) = {\rm End}_{GL_n}(\mathbb Q GL_n/B_n )$,  here $\mathcal H_n(q)$ is the  Iwahori-Hecke algebra which $q$-deforms the group ring of $S_n$.  It is Schur-Weyl  dual  to representations of the quantum group $U_q(GL_\infty)$.