# Why $H$ is a (1,2) tensor field?

1. Apr 6, 2015

### CAF123

I have a conceptual question associated with one of the worked examples in my notes. The question is:
'Let $\nabla$ and $\nabla^*$ be connections on a manifold $M$. Show that $H(X,Y) = \nabla_X Y - \nabla_X^* Y$ where $X,Y$ are vector fields defines a (1,2) tensor on M.

To show it is a tensor, one needs to check linearity of the arguments X and Y. This is clear, but I don't understand why these arguments show that H is necessarily of rank (1,2)? The notation H(X,Y) seems to suggest it is a (1,1) tensor since the map H acts on one covector and one vector argument. The solution says we may define a (1,2) tensor by $(\lambda, X,Y) \mapsto \langle \lambda, H(X,Y)\rangle$, but I am not sure why it must be (1,2) and why this means H is of rank (1,2)?

Thanks!

2. Apr 6, 2015

### MarcusAgrippa

The inputs are two vector fields, and the output is a vector field. In index notation, that means that two upper index quantities are converted into an upper index quantity. The components of the tensor that does this must thus have two lower indices and one upper.

Another way to view this, because of the antisymmetry on exchange of X and Y, is to say that H is a vector-valued 2-form.

If you define a tensor as a mapping from $T : \underbrace{V \times \cdots \times V}_{\mbox{r factors}} \times \underbrace{ V^* \times \cdots \times V^*}_{\mbox{s factors}} \rightarrow R$, you get equivalent definitions of the same tensor by removing factors of $V \mbox{ and } V^*$ and taking their duals when you place them on the RHS of the arrow. Thus $H: V^* \times V \times V \rightarrow R$ is equivalent to $H:V \times V \rightarrow V$

Last edited: Apr 6, 2015
3. Apr 6, 2015

### CAF123

Thanks! So can I write the equation $H(X,Y) \rightarrow \nabla_XY - \nabla_X^* Y$ in components as $H^a_{\,\,bc}X^b Y^c = Z^a - Z^{*\,a}$?

Also, why do we define the mapping to be $(\lambda, X,Y) \rightarrow \langle \lambda, H(X,Y)\rangle$? As far as I understand the notation $\langle \lambda, H(X,Y) \rangle$ means $\lambda ( H(X,Y))$. Is it because $H(X,Y) \in V$ and therefore since $\lambda \in V^*$ we have a covector acting on a vector which is in R (a co-vector is a mapping $\lambda: V \rightarrow \mathbb{R}$, so $\lambda(H(X,Y)) \in \mathbb{R}$) and hence the mapping $H: (\lambda, X,Y) \mapsto \langle \lambda, H(X,Y)\rangle$ is a mapping $V^* \times V \times V \rightarrow \mathbb{R}$ as you wrote?

4. Apr 6, 2015

Exactly so.