# Connection between (1,1) tensors and maps of vector fields

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.

• If every element $v\in V$ is a linear map $V^*\to\mathbb R$ and every element of $V^*$ a linear map $V\to\mathbb R$ what is the map $J=v\otimes v^*\,?$ How many arguments does it take? From which spaces? Is it linear perhaps? Feb 11 at 18:45
• @KurtG. From my understanding, the map $v\otimes v^*: V^*\times V \to \mathbb{R}$ is the defined by $(x^*,y) \mapsto v(x^*)\cdot v^*(y)$ and will be linear. However, this definition takes two arguments, so I'm not sure I have this correct. (also, I apologize for the bad notation, I'm new to this subject) Feb 11 at 18:53
• This is totally correct. I think my $J$ should have been $v^*\otimes v\,.$ It should now be easy to see that $J(x)$ is. In short: tensors eat vectors and covectors. When not all slots are filled a lower order tensor remains. Edit: or did you mean in OP: what is $J(\,.\,,X)\,?$ Feb 11 at 18:57
• @KurtG. Thank you for the clarification, I believe you understood correctly. Feb 11 at 19:22

## 1 Answer

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 $$\sum_{i,j,k}J^i_j\partial_i\otimes dx^j(X^k\partial_k)= \sum_{i,j} X^jJ^i_j\partial_i$$

• +1 -- note that the isomorphism $\mathrm{End}(V) \simeq V\otimes V^*$ only holds if $V$ is finite dimensional (which is the case here) Feb 11 at 18:59
• Thank you @QuareVerum the explicit form really helps. My only confusion is how $J^i_j\partial_i\otimes dx^j(X^k\partial_k)$ is defined since I thought that $J^i_j\partial_i\otimes dx^j$ was a two parameter function. Feb 11 at 19:20
• @nspace It should be stressed that the maps $\partial_i$ and $dx^j$ act only on $dx^k$ resp. on $\partial_k$ and that any coefficients such as $X^k$ are pulled out as if they were constants. This property is called $C^\infty$-linearity of the tensor when it acts on vector and covector fields. Feb 11 at 19:26
• @KurtG. Thank you, I think I understand now. That part was confusing me Feb 11 at 19:35
• Interesting. I prefer to define the isomorphism in the opposite direction. There is a bilinear map \begin{align*} V\times V^* &\rightarrow \operatorname{End}(V)\\ (v,\ell) &\mapsto L, \end{align*} where for each $w \in V$, $$L(w) = \ell(w)v.$$ This then extends uniquely to a linear map $V\otimes V^*\rightarrow \operatorname{End}(V)$ that is easily shown to be an isomorphism. Feb 11 at 20:53