I'm taking a course in differential geometry and my professor used some notation that confused me.
Let $X$ be a vector field and let $J$ be a $(1,1)$ tensor (defined as being an element of $V\otimes V^*$). My question is this: what does the notation $J(X)$ mean?
I know that there have been lots of questions on this site relating $(1,1)$ tensors to linear maps and the answers to the questions have only somewhat helped. I understand that there is a map $\langle v,v^* \rangle \to \mathbb{R}$ taking $ v\otimes v^* \mapsto \sum_{i} v_iv^*_i$ and that this can somehow also be viewed as a map taking derivations to derivations via $X|_p \mapsto \langle X|_p \cdot \rangle$ but I'm not entirely sure if I'm understanding this correctly.
Any help would be greatly appreciated.
This is a matter of linear algebra, more so than one of differential geometry, so let's just consider a vector space $V$. There is an isomorphism $V\otimes V^*\cong\text{End}(V)$ (assuming $V$ is of finite dimension). To see this, pick any basis $\{v_i\}$ for $V$ and let $L\in\text{End}(V)$ be arbitrary. We map this to an element of $V\otimes V^*$, namely $\sum_iL(v_i)\otimes v_i^*$ (where $\{v_i^*\}$ is the dual basis). This map is injective and surjective.
The same applies to vector bundles. Given $E\to M$, there is an isomorphism $\text{End}(E)\cong E\otimes E^*$ be applying the above fibrewise, in particular for $E=TM$. So when someone takes a $(1,1)$-tensor $J$, which is a section of $TM\otimes T^*M$, they mean what I wrote above, when they say $J(X)$.
Explicitly, it means that, in local coordinates, one can write $J$ and $X$ as $$J=J^i_j\partial_i\otimes dx^j\quad\quad X = X^k\partial_k$$ and $J(X)$ is $$\sum_{i,j,k}J^i_j\partial_i\otimes dx^j(X^k\partial_k)= \sum_{i,j} X^jJ^i_j\partial_i$$