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

  • $\begingroup$ Have you tried doing the calculation explicitly when $n = 2, 3$? $\endgroup$
    – Deane Yang
    Oct 7, 2019 at 15:05

1 Answer 1


The $n = 3$ case is a straightforward computation using the identification of the cross product on $\mathbb{R}^{3}$ with the Lie bracket on $\mathfrak{so}(3)$. Namely, defined $\hat{x}:\mathbb{R}^{3} \to \mathbb{R}^{3}$ by $\hat{x}y = x \times y$. Then $|\hat{x}|^{2} = 2|x|^{2}$, and the Jacobi identity implies $[\hat{x}, \hat{y}] = \widehat{x \times y}$, so $2|[\hat{x}, \hat{y}]|^{2} = 2|\widehat{x \times y}|^{2} = 4|x \times y|^{2} \leq 2|x|^{2}|y|^{2} = |\hat{x}|^{2}|\hat{y}|^{2}$.

For $n > 3$, the optimal result is not as straightforward. (For $n = 4$ it can be deduced from the $n = 3$ case using the special isomorphism $\mathfrak{so}(4) \simeq \mathfrak{so}(3) \times \mathfrak{so}(3)$; one has only to be careful about factors of $2$.) In general, for arbitrary $n \times n$ matrices $X$ and $Y$ a naive application of Cauchy-Schwarz gives $|[X, Y]|^{2} \leq 4|X|^{2}|Y|^{2}$. That the inequality is true with $2$ in place of $4$ goes back to a paper of Chern, do Carmo, and Kobayashi about minimal immersions into spheres, where they proved it for symmetric matrices (it is straightforward to adapt the proof for skew-symmetric matrices). It was shown with $2$ for general matrices by Böttcher and Wenzel (and also by Lu and probably some others too) to hold for any complex matrices, and the inequality in this form is often called the Böttcher-Wenzel inequality.

For skew-symmetric matrices, the $2$ can be improved to $1$, namely for skew-symmetric $n \times n$ matrices $A, B$, there holds $|[A, B]|^{2} \leq c|A|^{2}|B|^{2}$ with $c = 1/2$ if $n = 3$, and $c = 1$ if $n > 3$. This was shown by Bloch-Iserles in this article (one has to read carefully as they work with the norm on endomorphisms induced from the vector norm rather than the Frobenius norm, and these differ by a factor of $\sqrt{2}$). It was also proved as Lemma 2.5 of this article of J. Ge, which also gives the characterization of the equality case.

  • $\begingroup$ Thanks a lot, Dan! Just one follow-up question, to make sure I understand. The last property that you mentioned shows that the sectional curvatures are bounded by $1/4$ at the identity matrix, right? That's because the tangent vectors are like $P = A X$ with $X$ skew-symmetric, so they are generally not skew-symmetric themselves, but if $A = I_n$, they are. And then, to conclude that the inequality holds everywhere, we could invoke the homogeneity of the space. Is that correct? $\endgroup$
    – Călin
    Oct 8, 2019 at 13:45
  • $\begingroup$ Nevermind, the Lie bracket is already defined for the skew-symmetric part of the tangent vectors, at all points. I'm a novice! $\endgroup$
    – Călin
    Oct 8, 2019 at 14:03

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