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Vitali Kapovitch
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Without a lower volume bound this is false in dimensions 4 and up (and true in dimensions 2, 3). First counterexamples were constructed by Anderson. It's Theorem 0.4 in "Hausdorff perturbations of Ricci-flat manifolds and the splitting theorem".

You can take a sequence of flat tori of the form $T^3\times S^1(\epsilon)$ with $\epsilon\to 0$ and do several surgeries along various $p\times S^1(\epsilon)$ by cutting out $D^3\times S^1$ and gluing back in $S^2\times D^2$. One can put almost Ricci flat metrics on the resulting manifolds by gluing in Schwarzschild Ricci flat metrics on $R^2\times S^2$ because a Schwarzschild metric is asymptotic to $R^3\times S^1$. As $\epsilon\to 0$ you can glue in arbitrary many of these. With a lower volume bound the statement is true in dimension 4 by Andrerson-Cheeger-Naber in dimension 4"Regularity of Einstein Manifolds and the Codimension 4 Conjecture". In fact, there is even diffeomorphism finiteness in this case. In dimensions above 4 as far as I know this is an open question but I would guess it's most likely false.

Without a lower volume bound this is false in dimensions 4 and up (and true in dimensions 2, 3). First counterexamples were constructed by Anderson. It's Theorem 0.4 in "Hausdorff perturbations of Ricci-flat manifolds and the splitting theorem".

You can take a sequence of flat tori of the form $T^3\times S^1(\epsilon)$ with $\epsilon\to 0$ and do several surgeries along various $p\times S^1(\epsilon)$ by cutting out $D^3\times S^1$ and gluing back in $S^2\times D^2$. One can put almost Ricci flat metrics on the resulting manifolds by gluing in Schwarzschild Ricci flat metrics on $R^2\times S^2$ because a Schwarzschild metric is asymptotic to $R^3\times S^1$. As $\epsilon\to 0$ you can glue in arbitrary many of these. With a lower volume bound the statement is true by Andrerson-Cheeger-Naber in dimension 4. In fact, there is even diffeomorphism finiteness in this case. In dimensions above 4 as far as I know this is an open question but I would guess it's most likely false.

Without a lower volume bound this is false in dimensions 4 and up (and true in dimensions 2, 3). First counterexamples were constructed by Anderson. It's Theorem 0.4 in "Hausdorff perturbations of Ricci-flat manifolds and the splitting theorem".

You can take a sequence of flat tori of the form $T^3\times S^1(\epsilon)$ with $\epsilon\to 0$ and do several surgeries along various $p\times S^1(\epsilon)$ by cutting out $D^3\times S^1$ and gluing back in $S^2\times D^2$. One can put almost Ricci flat metrics on the resulting manifolds by gluing in Schwarzschild Ricci flat metrics on $R^2\times S^2$ because a Schwarzschild metric is asymptotic to $R^3\times S^1$. As $\epsilon\to 0$ you can glue in arbitrary many of these. With a lower volume bound the statement is true in dimension 4 by Andrerson-Cheeger-Naber "Regularity of Einstein Manifolds and the Codimension 4 Conjecture". In fact, there is even diffeomorphism finiteness in this case. In dimensions above 4 as far as I know this is an open question but I would guess it's most likely false.

Source Link
Vitali Kapovitch
  • 7.8k
  • 2
  • 34
  • 47

Without a lower volume bound this is false in dimensions 4 and up (and true in dimensions 2, 3). First counterexamples were constructed by Anderson. It's Theorem 0.4 in "Hausdorff perturbations of Ricci-flat manifolds and the splitting theorem".

You can take a sequence of flat tori of the form $T^3\times S^1(\epsilon)$ with $\epsilon\to 0$ and do several surgeries along various $p\times S^1(\epsilon)$ by cutting out $D^3\times S^1$ and gluing back in $S^2\times D^2$. One can put almost Ricci flat metrics on the resulting manifolds by gluing in Schwarzschild Ricci flat metrics on $R^2\times S^2$ because a Schwarzschild metric is asymptotic to $R^3\times S^1$. As $\epsilon\to 0$ you can glue in arbitrary many of these. With a lower volume bound the statement is true by Andrerson-Cheeger-Naber in dimension 4. In fact, there is even diffeomorphism finiteness in this case. In dimensions above 4 as far as I know this is an open question but I would guess it's most likely false.