# Scalar curvature and the degree of symmetry

Let $$M$$ be a closed connected smooth manifold, then we define the degree of symmetry of $$M$$ by $$N(M):=\sup_\limits{g}\{\mathrm{dim}(\mathrm{Isom}(M,g)\}$$, where $$g$$ is a smooth Riemannian metric on $$M$$ and $$\mathrm{Isom}$$ is the isometry group of the Riemannian manifold $$(M,g)$$.

The torus $$T^n$$ does not admit a Riemannian metric with positive scalar curvature and has $$N(T^n)\neq 0$$.

Whether there exists $$M$$ with $$N(M)=0$$ such that $$M$$ admits a metric with positive scalar curvature? That is, whether admitting a metric with positive scalar curvature implies its degree of symmetry is nonzero?

• You can replace the supremum in the definition of $N(M)$ with a maximum because $\dim\operatorname{Isom}(M, g) \leq \frac{1}{2}n(n+1)$. Jul 29 at 15:44

It seems that there are examples. By a theorem of Gromov and Lawson every simply connected manifold of dimension $$n \geq 5$$ which is not spin admits a metric of positive scalar curvature.
There are many examples of simply connected, non-spin, closed $$6$$-manifolds which cannot admit a smooth circle action, constructed by Puppe. Theorem 7 of https://arxiv.org/pdf/math/0606714.pdf.
Then, since the isomotetry group of a closed manifold is a compact Lie group, if $$N(M)>0$$ then taking a maximal torus gives a non-trivial circle action, which contradicts the above. So every metric has isometry group of dimension $$0$$.
Edit: A specific example would be a quartic $$3$$-fold $$X \subset \mathbb{CP}^4$$. It admits a metric with positive Ricci curvature (since it is Fano), or alternatively since it is not spin we can apply Gromov-Lawson. It does not admit any smooth circle action due to a Theorem of Dessai and Wiemler https://arxiv.org/pdf/1108.5327.pdf.