I'm reading the paper Classification of marginally trapped Lorentzian flat surfaces in $\mathbb{E}^4_2$ and its applications to biharmonic surfaces by B. Y. Chen. Then we have
Lemma 2.1. Let $L\colon M \to \mathbb{E}^4_2$ be an isometric immersion of a Lorentzian surface $M$ into $\mathbb{E}^4_2$. Then the immersion is biharmonic if and only if, with respect to a pseudo-orthonormal frame satisfying (2.11), the mean curvature vector $H$ of $L$ satisfies $$\triangle^D H = h(e_1, A_He_2) + h(e_2, A_He_1) \quad\mbox{and}\quad 2{\rm grad}(\langle H,H\rangle) + 4 {\rm trace}(A_{DH}) = 0,$$where $\triangle^D$ is the Laplace operator associated with the normal connection $D$, i.e., $$\triangle^D = D_{e_1}D_{e_2} + D_{e_2}D_{e_2} - D_{\nabla_{e_2}e_1} - D_{\nabla_{e_1}e_2}.$$
I do not understand the equation $$2{\rm grad}(\langle H,H\rangle) + 4 {\rm trace}(A_{DH}) = 0.$$Clearly ${\rm grad}(\langle H,H\rangle)$ is a smooth vector field along the isometric immersion $L$, but as far as I am concerned, ${\rm trace}(A_{DH})$ is a smooth function. So that sum does not even compile in my head. What does the author mean?
I found what I wanted in page 270 of Chen's Total Mean Curvature and Submanifolds of Finite Type (World Scientific, 1978). He regards $A_{DH}$ as a $(1,2)$-tensor, defined by $$A_{DH}(X,Y) \doteq A_{D_XH}(Y).$$Taking the trace gives a $(1,0)$-tensor or, in other words, a vector field. So summing that with ${\rm grad}\langle H,H\rangle$ makes sense.