Let $N$ be a Riemannian manifold, denote $R$ its purely covariant Riemann curvature tensor with sign convention so that the sectional curvature is $K(X,Y) = R(X,Y,X,Y)$ for an orthonormal pair.

Consider the complexified tangent space $TM \otimes \mathbb{C}$ and the complex-linear extension of $R$, which we still denote $R$. By definition, $N$ has nonpositive Hermitian sectional curvature if $R(X, Y, \bar{X}, \bar{Y}) \leqslant 0$ for all $X, Y \in TM \otimes \mathbb{C}$.

Obviously, nonpositive Hermitian sectional curvature is stronger than nonpositive sectional curvature.

QUESTION 1. Is nonpositive Hermitian curvature strictly stronger than nonpositive curvature?

Yes, according to e.g. [1], but I couldn't find an example proving it in the literature, e.g. [1][2][3][4][5][6][7]. Is there such a known example?

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FOLLOW UP QUESTIONS

Following (almost) the terminology of Siu [6], a Riemannian manifold with nonpositive Hermitian sectional curvature has "strongly nonpositive curvature". He also introduces other notions of curvature such as "very strongly nonpositive" as follows. Consider the curvature operator $$ \begin{aligned} Q \colon \Lambda^2 TM \times \Lambda^2 TM \to \mathbb{R} \end{aligned} $$ such that $Q$ is defined for decomposable tensors by $Q(X\wedge Y, Z \wedge W) = R(X , Y, Z , W)$. We still denote $Q$ its complex-linear extension to complexified vectors. By definition, $N$ has very strongly nonpositive curvature if $Q(\sigma, \bar{\sigma}) \leqslant 0$ for all tensors $\sigma \in \Lambda^2 TM \otimes \mathbb{C}$ (not just decomposable ones).

Question 2. Is there an example showing that very strongly nonpositive curvature is strictly stronger than strongly nonpositive curvature?

Question 3. What about the condition that $Q(\sigma, \sigma) \leqslant 0$ for all $\sigma \in \Lambda^2 TM$ (no complexification)? Is it stronger than nonpositive curvature?

Finally, just to be thorough, there is a notion of (very) strongly negative curvature, but it's not simply something like $Q(\sigma, \bar{\sigma}) < 0$ for all nonzero sigma, because that is too much to ask. Assume now that $N$ is a Kähler manifold. Then $Q(\sigma, \bar{\sigma}) = 0$ for any $\sigma$ of type $(2,0)$ or $(0,2)$, e.g. $X \wedge Y$ with $X, Y \in T^{1,0} M$, so $Q \equiv Q^{1,1}$. By definition, $N$ has (strongly/) very strongly negative curvature if $Q(\sigma, \bar{\sigma}) < 0$ for all (decomposable) $\sigma$ of type $(1,1)$.

It is not hard to see that $$\text{very strongly} <0 ~\Rightarrow~ \text{strongly} <0 ~\Rightarrow~ <0 \text{ sectional curvature}$$

Question 4. Are there examples proving that the converses implications are false?

Remark: Of course, there are similar notions of (very) strong nonnegative / positive curvature and one could ask the same questions.

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[1] J. Amorós, M. Burger, K. Corlette, D. Kotschick, and D. Toledo. Fundamental groups of compact Kähler manifolds. 1996.

[2] Eells and Lemaire. Two reports on harmonic maps. 1995

[3] Jost and Yau. Harmonic mappings and Kähler manifolds. 1983.

[4] Mostow and Siu. A compact Kähler surface of negative curvature not coveredby the ball. 1980.

[5] Ohnita and Udagawa. Stability, complex-analyticity and constancy of pluriharmonic maps from compact Kaehler manifolds. 1990.

[6] Siu. The complex-analyticity of harmonic maps and the strong rigidity of compact Kähler manifolds. 1980.

[7] Xin. Geometry of harmonic maps. 1996