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?

In other words, are there examples of Riemannian manifolds with nonpositive sectional curvature, but not nonpositive Hermitian sectional curvature? Yes, according to e.g. [1] or [8], but I couldn't find an example in the literature, e.g. [1][2][3][4][5][6][7][8].

NB: According to [8, Theorem 9.26], the answer is no for Kähler surfaces.

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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 $X, Y \in T^{1,0} M$. By definition, $N$ has very strongly negative curvature if $Q(\sigma, \bar{\sigma}) < 0$ for all nonzero tensors $\sigma$ of type $(1,1)$, and $N$ has strongly negative curvature if $Q(\sigma, \bar{\sigma}) < 0$ for all nonzero tensors $\sigma$ of type $(1,1)$ and of length $\leqslant 2$, e.g. $\sigma = X \wedge \bar{Y} + Z \wedge{\bar{W}}$.

It is clear that $$\text{very strongly negative} ~\Rightarrow~ \text{strongly negative} ~\Rightarrow~ \text{negative sectional curvature}$$

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

Again, ccording to [8, Theorem 9.26], the answer is no for Kähler surfaces

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

[8] F. Zheng, Complex differential geometry, 2000.