Exchange coefficient in tensor product

Question

In the highlighted equation I don't understand why we can exchange the position $\langle a_{(2)}\cdot v^{i}, v_{j} \rangle v^{j}$ to the other element in the tensor product? Is this a property of the tensor product that I don't get?

Answer 1

There are several steps going on: first, since $\langle -, - \rangle$ is in $k$ and the tensor product is over $k$, you move that term across the tensor product: $$ (a_{(1)} \cdot v_i) \otimes \langle a_{(2)} \cdot v^i, v_j \rangle v^j = (a_{(1)} \cdot v_i) \langle a_{(2)} \cdot v^i, v_j \rangle \otimes v^j. $$ And then by definition of the action, you get $$ = (a_{(1)} \cdot v_i) \langle v^i, S(a_{(2)}) \cdot v_j \rangle \otimes v^j. $$ Again since $\langle -,- \rangle$ is in $k$, you can move it around: $$ = (a_{(1)} \cdot \langle v^i, S(a_{(2)}) \cdot v_j \rangle v_i) \otimes v^j. $$