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There are many $n$-fold coverings (For instance $\Bbb R/n\Bbb Z$ as $S^1$ cover) but I can't find one with trivial automorphism group.

Thank you for your help and comments.

Conjecture
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2 Answers2

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By automorphism group do you mean the deck transformation group? In general, the group of deck transformations of the covering space corresponding to a subgroup $H \leq \Pi_1(X)$ is equal to the quotient of $\Pi_1(X)$ by the normalizer of $H$ (Edit: see comments). An observation you can make knowing some group theory is that if $[\Pi_1(X):H]$ is a prime of smallest order dividing $|\Pi_1(X)]$, then the automorphism group of its cover cannot be trivial (since the normalizer is bigger than $H$). So a two fold cover never has trivial automorphism group.

Connor Malin
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    I think the deck transformations of $\tilde{X}/H$ should be $N(H)/H$, not $\pi_1(X) /N(H)$. In fact the normalizer of a subgroup isn't always a normal subgroup of the supergroup. The key property is that $H$ is a normal subgroup of $N(H)$. – William Feb 25 '19 at 14:24
  • Yeah you are correct I just wrote it incorrectly. What I said after that is based off your correct interpretation. – Connor Malin Feb 25 '19 at 15:59
  • @ConnorMalin: By $\Pi_1$ you meant fundamental group $\pi_1$? – C.F.G Mar 24 '22 at 19:02
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In your other question I constructed a family of self-coverings over the Klein bottle $K \to K$ which should have trivial automorphism groups if $n$ is odd. ($\pi_1(K)$ is generated by elements $a, b$ were $bab^{-1} = a^{-1}$, take the covering corresponding to the non-normal subgroup generated by $a^n$ and $b$ for $n>2$.)

William
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