Let $M$ be a smooth manifold.


  1. $\mathcal T(M)^{(k,l)}$ is the $C^{\infty}(M)$-module of all tensor fields of type $(k,l)$ on $M$ ($k$ indicates the covariant part).
  2. $\mathcal T(M):=\mathcal T(M)^{(1,0)}$ is the $C^{\infty}(M)$-module of all vector fields on $M$.
  3. All tensor fields are smooth.

Now let $\nabla:\mathcal T(M)\times\mathcal T(M)\to \mathcal T(M)$ be a linear connection on $M$; $\nabla$ should be extended in a unique way to a (Koszul) connection, indicated with the same name, $$\nabla:\mathcal T(M)\times\mathcal T(M)^{(k,l)}\to\mathcal T(M)^{(k,l)}$$ $$(X,Y)\mapsto\nabla_XY$$ respecting some properties. I don't understand in which way the tensor field $\nabla_XY$ is defined. By the characterization lemma of tensor fields it is enough to give a multilinear function: $$\varphi:\underbrace{\mathcal T(M)^{(0,1)}\times \mathcal T(M)^{(0,1)}}_{k}\times\underbrace{\mathcal T(M)^{(1,0)}\times\ldots\times \mathcal T(M)^{(1,0)}}_{l}\to C^{\infty}(M)$$

but how can I define such $\varphi$?

  • $\begingroup$ Which book are you reading? I know for sure that it is in do Carmo. I am nearly sure it is in Kobayashi-Nomizu. $\endgroup$ Commented Oct 16, 2013 at 16:01
  • $\begingroup$ I'm reading "J.M. Lee - Riemannian Manifolds: an introduction to curvature". The question is left as an exercise but I have problems to do it. $\endgroup$
    – Dubious
    Commented Oct 16, 2013 at 16:07
  • 1
    $\begingroup$ Then check out do Carmo "Riemannian Geometry". I do not remember the exact formula but it is there. $\endgroup$ Commented Oct 16, 2013 at 16:09
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    $\begingroup$ Isn't there an inconsistency in the notation of the modules? $\mathcal{T}(M)^{(1,0)}$ is the module of 1-times contravariant tensor fields. $\endgroup$
    – gofvonx
    Commented Oct 16, 2013 at 17:51

1 Answer 1


If $\nabla$ is a connection on $\mathcal T(M) = \mathcal T(M)^{(1,0)}$, you can induce a connection on $\mathcal T(M)^{(0,1)}$ by

$$ (\nabla _X \alpha) (Y)= X(\alpha(Y)) - \alpha(\nabla _X Y)$$

where $\alpha$ is a one form and $Y$ is a vector field. You can check that $\nabla _X \alpha$ so defined is $C^\infty(M)$-linear, thus $\nabla_X\alpha \in \mathcal T(M)^{(0,1)}$.

In general if you have two vector bundles $E, F$ with connection $\nabla^E$ and $\nabla^F$, you can define a connection $\nabla$ on $E\otimes F$ by the formula

$$\nabla_X (e\otimes f) = \nabla^E_X e \otimes f + e \otimes \nabla^F_X f\ .$$

Using this you can extend $\nabla$ to all $\mathcal T(M)^{(k,l)}$ .


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