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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)

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $\nabla_X(S\circ T) = (\nabla_XS) \circ T + S\circ (\nabla_XT)$.

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