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.

  • $\begingroup$ 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? $\endgroup$
    – Kurt G.
    Feb 11 at 18:45
  • $\begingroup$ @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) $\endgroup$
    – nspace
    Feb 11 at 18:53
  • $\begingroup$ 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 $J(\,.\,,X)\,?$ $\endgroup$
    – Kurt G.
    Feb 11 at 18:57
  • $\begingroup$ @KurtG. Thank you for the clarification, I believe you understood correctly. $\endgroup$
    – nspace
    Feb 11 at 19:22

1 Answer 1


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

  • $\begingroup$ +1 -- note that the isomorphism $\mathrm{End}(V) \simeq V\otimes V^*$ only holds if $V$ is finite dimensional (which is the case here) $\endgroup$
    – Didier
    Feb 11 at 18:59
  • $\begingroup$ 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. $\endgroup$
    – nspace
    Feb 11 at 19:20
  • $\begingroup$ @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. $\endgroup$
    – Kurt G.
    Feb 11 at 19:26
  • $\begingroup$ @KurtG. Thank you, I think I understand now. That part was confusing me $\endgroup$
    – nspace
    Feb 11 at 19:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – Deane
    Feb 11 at 20:53

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