Consider the orthogonal group $O(n)$ as a Riemannian manifold endowed with the usual (bi-invariant) metric $\langle P, Q \rangle_A = \textrm{Tr}\ P^\top Q$ for tangent vectors $P, Q$, with $$T_A O(n) = \{ P \in \mathbb{R}^{n \times n} : A^\top P\ \text{is skew-symmetric} \}.$$

I'm interested in its sectional curvatures. For $P, Q$ orthonormal, they are given by (see, e.g., [1, Corollary 3.19]) $$K(P, Q) = \frac{\Big\lVert [P, Q] \Big\rVert_F^2}{4}.$$

From this, it's clear that $K(P, Q) \ge 0$, but can we derive an upper bound too? I think it should be less than $1/4$, but I'm not sure how to prove it. Sorry if it's really obvious!

[1]: Jeff Cheeger and David G Ebin. Comparison theorems in Riemannian geometry, volume 365. American Mathematical Soc., 2008

Consider the orthogonal group $O(n)$ as a Riemannian manifold endowed with the usual (bi-invariant) metric $\langle P, Q \rangle_A = \textrm{Tr}\ P^\top Q$ for tangent vectors $P, Q$, with $$T_A O(n) = \{ P \in \mathbb{R}^{n \times n} : A^\top P\ \text{is skew-symmetric} \}.$$

I'm interested in its sectional curvatures. For $P, Q$ orthonormal, they are given by (see, e.g., [1, Corollary 3.19]) $$K(P, Q) = \frac{\Big\lVert [P, Q] \Big\rVert_F^2}{4},$$ where $[\cdot,\cdot]$ is the commutator bracket. From this, it's clear that $K(P, Q) \ge 0$, but can we derive an upper bound too? I think it should be less than $1/4$, but I'm not sure how to prove it. Sorry if it's really obvious!

[1]: Jeff Cheeger and David G Ebin. Comparison theorems in Riemannian geometry, volume 365. American Mathematical Soc., 2008