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If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.

But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain the Seifert-Weber manifold, which is a rational homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. It has $H_1\cong(\mathbb Z/5)^3$, see Neil Hoffman's comment. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.

If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.

But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain the Seifert-Weber manifold, which is a rational homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. It has $H_1\cong(\mathbb Z/5)^3$, see Neil Hoffman's comment. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.

If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.

But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain another homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.

If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.

But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain the Seifert-Weber manifold, which is a rational homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. It has $H_1\cong(\mathbb Z/5)^3$, see Neil Hoffman's comment. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.

If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is even an integral homology sphere, and to best of my knowledge not a Seifert manifold (I am not sure, however).

But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain another homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.

If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.

But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain another homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.