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](https://mathoverflow.net/a/343406/145929))
$$
\operatorname{sec}\_{\operatorname{Gr}(n,k)}(X, Y) \leq 4.
$$


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](https://www.sciencedirect.com/science/article/pii/S000187081300385X).


On the other hand, in [this paper](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC539132/pdf/pnas00119-0088.pdf) 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?



Upper bounds on the sectional curvature of the real Grassmannian
A proof can be found in [this article by Hildebrandt, Jost, and Widman](https://link.springer.com/content/pdf/10.1007/BF01389161.pdf). 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 $B =\begin{pmatrix}
 0 & -1 \\
 1 & 0
\end{pmatrix}$.


