Consider the real Grassmannian as the symmetric space $\operatorname{Gr}(n,k) \cong \operatorname{O}(n)/(\operatorname{O}(k) \times \operatorname{O}(n-k))$ for $n \geq 3$, $k \geq 2$, where the metric is that induced from the bi-invariant metric on $\operatorname{O}(n)$, $\langle X,Y\rangle =\frac{1}{2}\operatorname{tr}(X^\intercal Y)$. The sectional curvature on $\operatorname{O}(n)$ with this bi-invariant metric is given by

$$ \operatorname{sec}_{\operatorname{O}(n)}(X, Y) = \frac{1}{4}\lVert[X,Y]\rVert^2. $$ where the norm is that induced by the scalar product.

Writing $\mathfrak{h} = \mathfrak{o}(n-k) \oplus \mathfrak{o}(k)$ and $\mathfrak{m} = \mathfrak{h}^\perp \subset \mathfrak{o}(n)$, by the O'Neill's formula and identifying a tangent space of the Grassmannian with a subspace of the Lie algebra of $\operatorname{O}(n)$, we have that the sectional curvature of $\operatorname{Gr}(n,k)$ for a pair of orthonormal vectors $X, Y \in \mathfrak{m}$

$$ \operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) = \frac{1}{4}\lVert[X,Y]\rVert^2 + \frac{3}{4}\lVert [X,Y]_{\mathfrak{h}}\rVert^2 = \lVert[X,Y]\rVert^2 $$ since $[\mathfrak{m}, \mathfrak{m}] \subset \mathfrak{h}$. Using now the bounds for the Lie bracket in $\operatorname{O}(n)$ (see this other MO answer) $$ \operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) \leq 4. $$

This bound is not tight, as it can be seen by looking at the equality cases for the inequality used, as per Lemma 2.5 in this article.

On the other hand, in this paper the author announces (without proof) in Theorem 3a that $$ \operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) \leq 2. $$ and even gives examples in Theorem 5a of a submanifold where this bound is achieved.

Is there any reference in which the tighter bound of $2$ is computed? Is there a reference where the tightness of the bound $2$ is also derived?


1 Answer 1


A proof can be found in this article by Hildebrandt, Jost, and Widman. I reproduce here the proof for completeness.

Consider the usual representation of $\mathfrak{m}$ as matrices of the form

$$ \mathfrak{m} = \left\{ \begin{pmatrix} 0 & A \\ -A^\intercal & 0 \end{pmatrix} \bigm\vert A \in \mathbb{R}^{(n-k)\times k} \right\} $$

We can write the sectional curvature of the Grassmannian at $A, B \in \mathbb{R}^{(n-k)\times k}$ for two matrices such that $\operatorname{tr}(A^\intercal B)$ in terms of their Frobenius norm as

$$ \operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) = \frac{ \lVert AB^\intercal - BA^\intercal \rVert_F^2 + \lVert A^\intercal B - B^\intercal A \rVert_F^2 }{ 2\lVert A \rVert_F^2 \lVert B \rVert_F^2 } $$

In the rest, we will be deliberately imprecise with the limits of the indices to not clutter innecessarily the proof.

Considering the SVD of A, we may assume that $A$ just has non-zero elements in its main diagonal. We can then write the numerator of the sectional curvature as

$$ N = \sum_{i\neq j} (a_{ii}b_{ji}-a_{jj}b_{ij})^2 + \sum_{i\neq j} (a_{ii}b_{ij}-a_{jj}b_{ji})^2 $$ $$ D = 2(\sum_i a_{ii}^2)(\sum_{j,k}b_{jk}^2) $$

We can bound $N$ using $(a+b)^2 \leq 2(a^2 + b^2)$ on the summation terms so that

$$ N \leq 4\sum_{i\neq j} a_{ii}^2b_{ji}^2 + 4\sum_{i\neq j} a_{ii}^2b_{ij}^2 \leq 4\sum_{i} a_{ii}^2(\sum_{j \neq i}b_{ji}^2 + \sum_{j \neq i}b_{ij}^2) $$

and we can bound the denominator as

$$ D = 2(\sum_i a_{ii}^2)(\sum_j b^2_{ji} + \sum_k\sum_{j \neq i}b_{kj}^2) \geq \frac{1}{2}N. $$

In the paper they also show the tightness of this bound considering $A = \mathrm{Id}$ and $B =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.


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