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LSpice
LSpice
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Per the request to post it as an answer.

Notice that the Ad representation is a polynomial representation into $GL(Lie(G))$$\operatorname{GL}(\operatorname{Lie}(G))$. We do know that $Ad(G)$$\operatorname{Ad}(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem. Hence $Ad\mid_{\Gamma}$$\operatorname{Ad}\rvert_{\Gamma}$ is also irreducible.

Per the request to post it as an answer.

Notice that the Ad representation is a polynomial representation into $GL(Lie(G))$. We do know that $Ad(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem. Hence $Ad\mid_{\Gamma}$ is also irreducible.

Per the request to post it as an answer.

Notice that the Ad representation is a polynomial representation into $\operatorname{GL}(\operatorname{Lie}(G))$. We do know that $\operatorname{Ad}(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem. Hence $\operatorname{Ad}\rvert_{\Gamma}$ is also irreducible.

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Asaf
Asaf
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Per the request to post it as an answer.

Notice that the Ad representation is a polynomial representation into $GL(Lie(G))$. We do know that $Ad(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem. Hence $Ad\mid_{\Gamma}$ is also irreducible.