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One can show that such a map in the question doesn't exist (no need to assume simply-conntednessconnectedness). As abx pointed out, any finite map between smooth projective varieties with trivial canonical bundle must be étale.

Let $\chi(X)$ be the topoloigicaltopological Euler charateristiccharacteristic of $X$ and $d$ be the degree of the map. Since $X$ is rigid, $$\chi(X) = 2 \left( h^{1,1}(X)- h^{1,2}(X) \right) =2 h^{1,1}(X) > 0.$$

Noting $d \cdot \chi(X) = \chi(X)$, one has $d=1$.

One can show that such a map in the question doesn't exist (no need to assume simply-conntedness). As abx pointed out, any finite map between smooth projective varieties with trivial canonical bundle must be étale.

Let $\chi(X)$ be the topoloigical Euler charateristic of $X$ and $d$ be the degree of the map. Since $X$ is rigid, $$\chi(X) = 2 \left( h^{1,1}(X)- h^{1,2}(X) \right) =2 h^{1,1}(X) > 0.$$

Noting $d \cdot \chi(X) = \chi(X)$, one has $d=1$.

One can show that such a map in the question doesn't exist (no need to assume simply-connectedness). As abx pointed out, any finite map between smooth projective varieties with trivial canonical bundle must be étale.

Let $\chi(X)$ be the topological Euler characteristic of $X$ and $d$ be the degree of the map. Since $X$ is rigid, $$\chi(X) = 2 \left( h^{1,1}(X)- h^{1,2}(X) \right) =2 h^{1,1}(X) > 0.$$

Noting $d \cdot \chi(X) = \chi(X)$, one has $d=1$.

Source Link
Basics
  • 1.8k
  • 10
  • 14

One can show that such a map in the question doesn't exist (no need to assume simply-conntedness). As abx pointed out, any finite map between smooth projective varieties with trivial canonical bundle must be étale.

Let $\chi(X)$ be the topoloigical Euler charateristic of $X$ and $d$ be the degree of the map. Since $X$ is rigid, $$\chi(X) = 2 \left( h^{1,1}(X)- h^{1,2}(X) \right) =2 h^{1,1}(X) > 0.$$

Noting $d \cdot \chi(X) = \chi(X)$, one has $d=1$.