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Let $Y$ be an oriented closed $3$-manifold, with trivial homology group, i.e. integer homological sphere.

Q: If $Y$ can be embedded into $\mathbb R^4$, is there any example, that such a $Y$ admits a negative scalar curvature?

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    $\begingroup$ If you mean the metric induced from the ambient Euclidean metric on $\mathbb R^4$, the answer is no. Take a point of maximal distance from the origin. Then all sectional curvatures are $> 0$ at this point, because $Y$ touches a round sphere from the interior. $\endgroup$ – Sebastian Goette Mar 29 '17 at 10:25
  • $\begingroup$ @SebastianGoette Yes, you are right. I mean any metric, actually I guess, such a manifold $Y$ admits a positive scalar curvture. $\endgroup$ – DLIN Mar 29 '17 at 10:51
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There are certainly closed hyperbolic 3-manifolds that embed in $\mathbb R^4$, that also have arbitrarily big volume. A construction goes as follows: construct a Mazur manifold with one 1-handle and one 2-handle. More specifically, by attaching a 1-handle to $D^4$ you get $D^3 \times S^1$. Then you attach a 2-handle along a knot $K \subset \partial (D^3 \times S^1) = S^2 \times S^1$ that is homotopic to the knot $\{ pt\} \times S^1$. The result is a contractible 4-manifold $W$ whose double is $S^4$ (see the Wikipedia page). Therefore $W$ embeds in $\mathbb R^4$ and hence also $\partial W$ does (smoothly). Moreover $\partial W$ is a homology sphere because it is the boundary of a contractible 4-manifold.

Now you have a lot of freedom here: you can take $K$ sufficiently complicated so that its complement $M = S^2\times S^1 \setminus K$ is hyperbolic and has arbitrarily big volume. You can attach the 2-handles along integral Dehn filling parameters that are arbitrarily big, so that the resulting manifold $\partial W$, which is in fact obtained by integral surgery from $K$, will still be hyperbolic with volume close to that of $M$ thanks to Thurston's Hyperbolic Dehn filling Theorem.

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  • $\begingroup$ Is there any easy way to recognize that a knot $K \subset S^1 \times S^2$ with winding number 1 is hyperbolic? i.e. is there some version of the Thurston's hyperbolic/torus/satellite trichotomy for these knots in $S^1 \times S^2$? $\endgroup$ – PVAL Mar 30 '17 at 16:50
  • $\begingroup$ You can use SnapPy math.uic.edu/t3m/SnapPy $\endgroup$ – Bruno Martelli Mar 30 '17 at 16:59
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L Z Gao and S T Yau showed in Invent Math 85 (1986) 637-652 that any compact 3 manifold admits a metric of negative Ricci curvature .J Lohkamp proved that any manifold of dimension 3 or higher admits a complete metric with Ricci curvature pinched between two negative constants .See Annals of Math 140 (1994) 653-683

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    $\begingroup$ There is an alternate proof of the Gao-Yau result by Robert Brooks, A construction of metrics of negative Ricci curvature. J. Differential Geom. 29 (1989), no. 1, 85–94. $\endgroup$ – Danny Ruberman Mar 29 '17 at 18:31

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