# Covariant derivative of composition of two tensors

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)

• 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? – Henning Makholm Oct 21 '11 at 21:48
• I am wondering whether the 'product rule' holds? i.e. $\nabla_X(S\circ T)=\nabla_XS\circ T+ S\circ \nabla_XT$ – user17150 Oct 21 '11 at 21:51
• 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. – user17150 Oct 21 '11 at 21:56
• 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. – Henning Makholm Oct 21 '11 at 22:03
• I see, thank you Henning! – user17150 Oct 21 '11 at 23:11

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