# Weird Notation for Trace of an Endomorphism

I am having some difficulty understanding a piece of notation from Riemannian Geometry: and Introduction to Curvature by John M. Lee.

In Section 2 just under equation 2.3 Lee defines the trace operator which lowers the rank of a tensor by 2.

He defines the map: $$\mathrm{tr}:T_{l+1}^{k+1}(V)\longrightarrow T_l^k(V)$$ By letting: $$\mathrm{tr}\; F(\omega^1,\dots\omega^l,V_1,\dots,V_k)$$ be the trace of the endomorphism: $$F(\omega^1,\dots,\omega^l,\bullet,V_1,\dots,V_k,\bullet)$$

But how is it that $F(\omega^1,\dots,\omega^l,\bullet,V_1,\dots,V_k,\bullet) \in\mathrm{End}(V)$, it looks like it should belong to $T_{l+1}^{k+1}$, I think my confusion lies with the $\bullet$ in the above expression. Unforturnately I cannot find any explanation of this notation in the textbook. Is this notation common for something that I am not aware of?

For fixed $\omega^1, \ldots, \omega^l \in V^*$ and $V_1, \ldots, V_k \in V$ the notation $F(\omega^1,\dots,\omega^l,\bullet,V_1,\dots,V_k,\bullet)$ signifies an element $G \in T_1^1(V)$ such that $$G(\omega^{l+1}, V_{k+1}) = F(\omega^1, \ldots, \omega^l, \omega^{l+1}, V_1, \ldots, V_k, V_{k+1}).$$
Then $\operatorname{tr} F \in T_l^k(V)$ is defined by $$(\operatorname{tr} F)(\omega^1, \ldots, \omega^l, V_1, \ldots, V_k) = \operatorname{tr}G.$$