6
$\begingroup$

In these days I was trying to minimize a Riemannian functional on the space of Riemannian metrics with non negative scalar curvature over a manifold, and I suddenly realize that I don't know nothing about how this space looks like. In particular I don't even know if it is a vector space, if it is convex, star shaped, path connected or connected. This led me to ask the following questions.

Consider $\mathcal{M}_{\geq 0}(\mathbb{R}^n)$ the topological space of the Riemannian metrics on $\mathbb{R}^n$ that have non negative scalar curvature. We give to it the topology induced by the inclusion $\mathcal{M}_{\geq 0}(\mathbb{R}^n)\subset C^{\infty}(T^0_2 \mathbb{R}^n)$ in the space of sections of (0,2)-tensors. $C^{\infty}(T^0_2 \mathbb{R}^n)$ has the norm given by the usual inner product between (0,2) tensors integrated over the manifold.

  1. Is $\mathcal{M}_{\geq 0}(\mathbb{R}^n)$ connected? Path connected? $C^1$ path connected? Star shaped? Convex? Vector space?

  2. Is there any interesting dense subset? For example is the set of non negative sectional curvature a dense subset?

  3. What happens if instead of considering the scalar curvature, which is the first elementary symmetric function in the eigenvalues of the (2,2) type Riemann tensor we consider other elementary symmetric functions?

  4. If we consider a generic $n$-manifold, the space $\mathcal{M}_{\geq 0}(\mathbb{M}^n)$ can be empty, if not what remains true?

$\endgroup$
  • 2
    $\begingroup$ For closed manifolds of dimension $\ge 4$ the situation is quite complicated, for instance, ${\mathcal M}_{\ge}$ can be disconnected. In contrast, ${\mathcal M}_{\ge}$ for closed 3-manifolds is contractible (modulo the action of the group of diffeomorphisms). This is, however, is not even completely written AFAIK... Not sure about $R^n$. $\endgroup$ – Moishe Kohan Mar 2 '18 at 18:23
  • 2
    $\begingroup$ The subject is quite technical and difficult. Start by reading this blog post: matheuscmss.wordpress.com/2012/12/02/… $\endgroup$ – Moishe Kohan Mar 2 '18 at 18:41
  • 1
    $\begingroup$ I don't know if this is of interest to you, but in general $\mathcal{M}_{\geq 0}(M)$ is not closed under addition, so it is not convex. $\endgroup$ – Michael Albanese Mar 7 '18 at 15:40
  • $\begingroup$ @MichaelAlbanese Thank you this is absolutely of interest. $\endgroup$ – Warlock of Firetop Mountain Mar 7 '18 at 17:24
  • 1
    $\begingroup$ How could it possibly be a vector space?! Multiplication by -1 is not going to work, no? $\endgroup$ – Mariano Suárez-Álvarez Mar 25 '18 at 4:21
2
$\begingroup$

Here's an example that shows that $\mathcal{M}_{\geq 0}(M)$ is not closed under addition in general. There may be easier examples, but this is the only one I know.


Let $\pi : \widetilde{M} \to M$ be a covering map. Suppose $g$ is a Riemannian metric on $\widetilde{M}$ such that $f^*g = g$ for all deck transformations $f$. Then there is a Riemannian metric $h$ on $M$ such that $\pi^*h = g$. Note that $s_g = s_{\pi^*h} = \pi^*s_h =s_h\circ \pi$. As $\pi$ is surjective, the functions $s_g$ and $s_h$ are determined by one another.

Now, let $X$ be a non-spin complex surface arising as a complete intersection in some complex projective space (e.g. a smooth hypersurface of $\mathbb{CP}^3$ with odd degree $d \geq 5$), and let $N = S^2\times S^2/\mathbb{Z}_2$ where the $\mathbb{Z}_2$ action is generated by $\sigma(x, y) = (-x, -y)$. Set $M = X\# N$. Then the universal cover of $M$ is $\widetilde{M} = X\# X \#(S^2\times S^2)$, and $\pi : \widetilde{M} \to M$ is a double covering.

In Scalar curvature, covering spaces, and Seiberg-Witten theory, LeBrun showed that $Y(M) < 0$ and $Y(\widetilde{M}) > 0$ where $Y$ denotes the Yamabe invariant. In particular, $\widetilde{M}$ admits a positive scalar curvature metric, while $M$ does not admit a metric with non-negative scalar curvature.

Let $f : \widetilde{M} \to \widetilde{M}$ be the non-trivial deck transformation of $\pi$. If $g$ is a positive scalar curvature metric on $\widetilde{M}$, then $f^*g \neq g$, otherwise there would be a positive scalar curvature metric $h$ on $M$ with $\pi^*h = g$. Note that $f^*g$ is another Riemannian metric on $\widetilde{M}$, and as $s_{f^*g} = s_g\circ f$, it also has positive scalar curvature. Now consider the metric $\tilde{g} := g + f^*g$. As $f^*\tilde{g} = \tilde{g}$, there is a Riemannian metric $h$ on $M$ with $\pi^*h = \tilde{g}$. As $M$ does not admit metrics with non-negative scalar curvature and $s_{\tilde{g}} = s_h\circ\pi$, the metric $\tilde{g}$ does not have non-negative scalar curvature.

So $g, f^*g \in \mathcal{M}_{> 0}(\widetilde{M}) \subset \mathcal{M}_{\geq 0}(\widetilde{M})$, but $g + f^*g \not\in \mathcal{M}_{\geq 0}(\widetilde{M})$.


If you only wanted an example to show (the weaker result) that $\mathcal{M}_{> 0}(M)$ is not closed under addition in general, then this follows from earlier work by Bérard Bergery. He pointed out that there are examples of finite coverings $\pi : \widetilde{M} \to M$ such that $\widetilde{M}$ admits positive scalar metrics, but $M$ doesn't.

For example, one could take $M = (S^2\times\mathbb{RP}^7)\#\Sigma$ where $\Sigma$ is an exotic $9$-sphere with $\alpha(\Sigma) \neq 0$; here $\alpha$ denotes the Hitchin-Lichnerowicz obstruction $\alpha : \Omega^{\text{spin}}_n \to KO_n(\text{pt})$. Hitchin showed that if $M$ is a spin manifold which admits metrics of positive scalar curvature, then $\alpha(M) = 0$; in particular, $\alpha(S^2\times\mathbb{RP}^7) = 0$. Now $M$ is spin and $\alpha(M) = \alpha(S^2\times\mathbb{RP}^7) + \alpha(\Sigma) = \alpha(\Sigma) \neq 0$, so $M$ does not admit metrics of positive scalar curvature. On the other hand $\widetilde{M} = (S^2\times S^7)\#\Sigma\#\Sigma$ is diffeomorphic to $S^2\times S^7$ (because $\Theta_9 = \mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_2$) which does admit metrics of positive scalar curvature.

In the same way as we did above, we can use this example to construct two metrics $g, f^*g \in \mathcal{M}_{> 0}(S^2\times S^7)$ with $g + f^*g \not\in \mathcal{M}_{> 0}(S^2\times S^7)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.