TITLE: Scalar curvature and the degree of symmetry
QUESTION [10 upvotes]: Let $M$ be a closed connected smooth manifold. 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 $N(T^n)\neq 0$.
Is there a closed connected manifold $M$ with $N(M)=0$ such that $M$  admits a metric with positive scalar curvature?  That is, does admitting a metric with positive scalar curvature imply its degree of symmetry is nonzero?

REPLY [17 votes]: 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 Wiemeler  https://arxiv.org/pdf/1108.5327.pdf.