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Suppose that $Y$ is an integral homology $3$-sphere. Is it true that, for any $n \in \mathbb{N}$, there is always an irreducible representation $\pi_{1}(Y) \rightarrow SU(n)$?

I'm also happy to see the answer when the Lie group is replaced by $SL(n, \mathbb{C})$ or $PSL(n, \mathbb{C})$, but keeping irreducible unchanged.

Edit: In this paper, Zentner proved the statement positively for $SL(2, \mathbb{C})$. It might be true that there is a threshold $n$ such that there is no more $SU(N)$ irreducible representations for $N \geq n$. The question was stated in this way because constructing counter-examples to the vanishing statement might be easier.

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    $\begingroup$ The Poincaré homology sphere has fundamental group the binary icosahedral group, which is finite and therefore admits only finitely many irreducible representations. $\endgroup$ Jul 5 '20 at 13:16
  • $\begingroup$ Yes, my motivation of asking this question is to see if this failure holds in much more generality. $\endgroup$ Jul 5 '20 at 16:56
  • $\begingroup$ So you are asking: for all $n \in \mathbb N$, is there an integral homology 3-sphere $Y$ with an irreducible $\pi_1(Y) \to SU(n)$? $\endgroup$ Jul 5 '20 at 17:29
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    $\begingroup$ No, I am asking if $Y$ is fixed, can you find such representations. As you pointed out this is not true if Y is the Poincar\'e homology sphere, the question is about whether this failure holds for all $Y$. $\endgroup$ Jul 5 '20 at 18:11
  • $\begingroup$ It should be true that for a given closed 3-manifold with infinite fundamental group, there is an irreducible $SU(N)$ representation for $N$ sufficiently large. But the case of homology spheres and $n=2$ is a well-known open question, so I think your question as stated is probably difficult. $\endgroup$
    – Ian Agol
    Jul 5 '20 at 20:15

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