Let $G$ be a finite group. A map $\rho:G\rightarrow U\_n$ is called an $\epsilon$-homomorphism if and only if for any $g,h\in G$, we have $||\rho(g)\rho(h)-\rho(gh)||\leq \epsilon$ where the $||$ norm is the usual operator norm. Kazhdan proves in his [paper](http://link.springer.com/article/10.1007/BF02761236) that for $\epsilon < 1/1000$ and any $\epsilon$-homomorphism $\rho$ there is a homomorphism $\pi:G\rightarrow U\_n$ such that $||\pi-\rho|| \leq O(\epsilon)$. Can you prove a similar statement if we replace operator norm by Hilbert-Schmidt norm? 


This is what I mean. Suppose $\epsilon\_0$ is a sufficiently small number which doesn't depend on n or $G$. Let $\epsilon<\epsilon\_0$. Assume that $\rho:G\rightarrow U\_n$ is an $n \epsilon$-homomorphism in the Hilbert-Schmidt sense i.e. for any $g,h\in G$ arithmetic mean of singular values of $\rho(g)\rho(h)-\rho(gh)$ is less than $\epsilon$. Can we prove that $\rho$ has distance at most $n O(\epsilon)$ from an exact homomorphism in the Hilbert-Schmidt sense?



Approximating eps-homomorphisms
I just wanted to add that the answer to above question is yes:
<http://arxiv.org/abs/1510.04085>


