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Bruno Martelli
Bruno Martelli
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You need geometrisation to prove this fact. See Corollary 12.9.5 here for a reference.

You can't prove this without Perelman, at least with our present knowledge. For instance, if the orientable cover is $S^3$, then you must ensure that $M$ be elliptic, and that's precisely the space form conjecture, which is "one third" of geometrisation. But even when the finite cover is some other Seifert space, I don't see an easy argument to conclude without using geometrisation.

Edit. I overlooked the "with boundary" hypothesis. In that case Thurston's proof of geometrisation suffices.

You need geometrisation to prove this fact. See Corollary 12.9.5 here for a reference.

You can't prove this without Perelman, at least with our present knowledge. For instance, if the orientable cover is $S^3$, then you must ensure that $M$ be elliptic, and that's precisely the space form conjecture, which is "one third" of geometrisation. But even when the finite cover is some other Seifert space, I don't see an easy argument to conclude without using geometrisation.

You need geometrisation to prove this fact. See Corollary 12.9.5 here for a reference.

You can't prove this without Perelman, at least with our present knowledge. For instance, if the orientable cover is $S^3$, then you must ensure that $M$ be elliptic, and that's precisely the space form conjecture, which is "one third" of geometrisation. But even when the finite cover is some other Seifert space, I don't see an easy argument to conclude without using geometrisation.

Edit. I overlooked the "with boundary" hypothesis. In that case Thurston's proof of geometrisation suffices.

Source Link
Bruno Martelli
Bruno Martelli
  • 10.5k
  • 2
  • 39
  • 70

You need geometrisation to prove this fact. See Corollary 12.9.5 here for a reference.

You can't prove this without Perelman, at least with our present knowledge. For instance, if the orientable cover is $S^3$, then you must ensure that $M$ be elliptic, and that's precisely the space form conjecture, which is "one third" of geometrisation. But even when the finite cover is some other Seifert space, I don't see an easy argument to conclude without using geometrisation.