2
$\begingroup$

Suppose $TM\to M$ is the tengent bundle over the closed Riemannian manifold $M$. Let $\nabla$ be the Levi-Civita connection, $S$ and $T$ are two $(1,1)$-tensor, i.e. at each point $x\in M$, we can view $S_x$ and $T_x$ as linear homomorphisms between $T_xM$ to itself.

For any vector $v\in T_xM$ one can compose this two tensors, i.e. $(S\circ T)(v)= S(T(v))$. Now my question is the composition $S\circ T$ should be $(1,1)$ tensor, and what is the covariant derivative of it?

My understand is $S \circ T$ is different with $S\otimes T$, right? (for the second one, I know the answer)

$\endgroup$
  • $\begingroup$ Yes, $S\circ T$ is a (1,1) tensor, and since $S\otimes T$ is (2,2) they cannot possibly be the same. In fact $S\circ T$ is a contraction of $S\otimes T$. Which format do you expect the answer to the covariant derivative of $S\circ T$ to have? $\endgroup$ – Henning Makholm Oct 21 '11 at 21:48
  • $\begingroup$ I am wondering whether the 'product rule' holds? i.e. $\nabla_X(S\circ T)=\nabla_XS\circ T+ S\circ \nabla_XT$ $\endgroup$ – user17150 Oct 21 '11 at 21:51
  • $\begingroup$ My concern is: since this is a composition, the 'chain rule' should holds instead of 'product rule'. However, it seems the product is correct, which is given above. I just try to understand why. $\endgroup$ – user17150 Oct 21 '11 at 21:56
  • 3
    $\begingroup$ No, you'd need a chain rule if you were trying to differentiate $(S\circ T)(v)$ with respect to a variable tangent vector $v$ at each point. However that's quite different from the spatial variation of the map $S\circ T$ itself. Locally $S\circ T$ is just a matrix product, or $(S\circ T)^i_k=S^i_j T^j_k$ in Einstein notation, so differentiating that needs the product rule. $\endgroup$ – Henning Makholm Oct 21 '11 at 22:03
  • $\begingroup$ I see, thank you Henning! $\endgroup$ – user17150 Oct 21 '11 at 23:11
0
$\begingroup$

Just compute: $$\begin{align}\require{cancel} (\nabla_X(S\circ T))(Y) &= \nabla_X((S\circ T)(Y)) - (S\circ T)(\nabla_XY) \\ &= \nabla_X(S(T(Y))) - S(T(\nabla_XY)) \\ &= (\nabla_XS)(T(Y)) + \cancel{S(\nabla_X(T(Y)))} - S(\cancel{\nabla_X(T(Y))} - (\nabla_XT)(Y)) \\ &= (\nabla_XS)(T(Y)) + S((\nabla_XT)(Y)). \end{align}$$This means that $\nabla_X(S\circ T) = (\nabla_XS) \circ T + S\circ (\nabla_XT)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.