Computation of trace

Question

Show that the trace on any pair of indices is a well-defined linear map from $T^{k+1}_{l+1} (V )$ to $T^k_l(V )$.

I don't have any clue..

Answer 1

Let me just clarify a couple of things.

Your question is exactly Exercise $2.2$ of Lee's Riemannian Manifolds: An Introduction to Curvature. In that text, $T^k_l(V)$ denotes the collection of all $\binom{k}{l}$-tensors ($k$-covariant, $l$-contravariant) which are multilinear maps $(V^*)^l\times V^k \to \mathbb{R}$. Lee defines the trace of $F \in T^{k+1}_{l+1}(V)$ (on it's last covariant and contravariant factors) to be the the element $\operatorname{tr} F \in T^k_l(V)$ given by $$\operatorname{tr} F(\omega^1, \dots, \omega^l, X_1, \dots, X_k) := \operatorname{trace}F(\omega^1, \dots, \omega^l, \cdot, X_1, \dots, X_k, \cdot)$$ where $F(\omega^1, \dots, \omega^l, \cdot, X_1, \dots, X_k, \cdot) \in T^1_1(V) \cong \operatorname{End}(V)$ and $\operatorname{trace}$ indicates the trace of this endomorphism. In terms of a basis, $$(\operatorname{tr} F)^{j_1 \dots j_l}_{i_1 \dotsm i_k} = F^{j_1 \dots j_l m}_{i_1 \dots i_k m}$$ where Einstein notation is being used, so the right hand side is actually a sum over the index $m$. There is nothing special about using the last covariant and contravariant factors/indices though, you can use any pair and obtain a map $T^{k+1}_{l+1}(V) \to T^k_l(V)$ given by the trace in an analogous way. The exercise asks you to verify that, for any pair of indices, this map is well-defined and linear.