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 $$(\operatorname{tr} F)(\omega^1, \ldots, \omega^l, V_1, \ldots, V_k) = \operatorname{tr}G.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.