How does the covariant derivative satisfy the Leibniz rule?

In Carroll's "Spacetime and Geometry", he states on page 95 (section 3.2) that the covariant derivative, $$\nabla$$, is a map from $$\left(k, l\right)$$ tensor fields to $$\left(k, l+1\right)$$ tensor fields which satisfies various properties, one of which is the following Leibniz rule: $$\nabla \left(T \otimes S\right) = \left(\nabla T\right) \otimes S + T \otimes \left( \nabla S\right)\text{.}\tag{p.95}$$ If we express this in terms of the tensors' components, it seems to me that we should have $$\nabla \left(T \otimes S\right)_{\mu}^{\ \ \nu \rho} = \nabla\left(T\right)_{\mu}^{\ \ \nu} S^{\rho} + T_{\mu}\nabla\left(S\right)^{\nu \rho}\text{.}\tag{2}$$

However based on the definition given for $$\nabla$$ on page 97 in terms of the connection coefficients, it seems that, for vectors $$T$$ and $$S$$, we should have $$\nabla \left(T \otimes S\right)_{\mu}^{\ \ \nu \rho} = \partial_{\mu}\left(T^{\nu}S^{\rho}\right) + \Gamma_{\mu \lambda}^{\nu}T^{\lambda}S^{\rho} + \Gamma_{\mu \lambda}^{\rho}T^{\nu}S^{\lambda} = \nabla\left(T\right)_{\mu}^{\ \ \nu} S^{\rho} + T^{\nu}\nabla\left(S\right)_{\mu}^{\ \ \rho}\text{.}\tag{3}$$

Therefore, the definition given doesn't seem to have the required Leibniz rule property. Could someone please point out where I have gone wrong?

Assuming that $$T$$ and $$S$$ are supposed to be $$(1,0)$$ tensor fields we can see that Eq. (2) is wrong immediately, because the expression $$\nabla (S)^{ \nu \rho}$$ has the wrong indices (it should have 1 co- and 1 contra-variant by definition of $$S$$ and $$\nabla$$).

If we write the covariant derivative in abstract index notation, then the Leibniz rule can be stated as $$\nabla_a (T^bS^c) = ( \nabla_a T^b )S^c + T^b ( \nabla_a S^c),$$ which gives us the correct formula (Eq. 3) immediatly.

The expression $$\nabla \left(T \otimes S\right) = \left(\nabla T\right) \otimes S + T \otimes \left( \nabla S\right) \tag{1 }$$ has a subtle problem and should be avoided.

To explain the problem:

$$\nabla \left(T \otimes S\right)$$ is a section of $$T^* (M) \otimes T(M) \otimes T(M)$$, where $$T(M)$$ is the tangent bundle of the base manifold $$M$$ and $$T^* (M)$$ the cotangent bundle of $$M$$. The $$T$$ in $$T(M)$$ and $$T^*(M)$$ has nothing to do with the tensor $$T$$, bad notation here sorry.

The expression $$\left(\nabla T\right) \otimes S$$ is also a section of $$T^* (M) \otimes T(M) \otimes T(M)$$ (after expanding using associativity).

But the expression $$T \otimes \left( \nabla S\right)$$ is not a section of $$T^* (M) \otimes T(M) \otimes T(M)$$, but rather a section of $$T(M) \otimes T^* (M) \otimes T (M)$$.

But this means that we can not do the addition: $$\left(\nabla T\right) \otimes S + T \otimes \left( \nabla S\right)$$, since the two expression are not of the same tensor field type.

Therefore it must be understood that we first change the type of $$T \otimes \left( \nabla S\right)$$ (via permutation of the first and second factor) to a section of $$T^* (M) \otimes T(M) \otimes T(M)$$ and then add them.

To illustrate this consider your example: Let $$C$$ be the type changed version of $$T \otimes \left( \nabla S\right)$$. Then we have $$C_{\mu}{}^{\nu \rho} = (T \otimes ( \nabla S) )^{\nu}{}_{\mu}{}^{\rho}$$. And therefore from formula 1 $$(\nabla T \otimes S)_\mu {}^{\nu \rho} = (\nabla T)_\mu{}^\nu S^\rho +C_{\mu}{}^{\nu \rho} = (\nabla T)_\mu{}^\nu S^\rho +T^\nu (\nabla S)_{\mu}{}^\rho$$ which is the same as your equation 3.

This type changing procedure is explicitly done when writing the Leibniz formula in the abstract index notation as above (this is why $$a,b$$ are inverted in the second summand).

Edit:

As pointed out in the comments by @jawheele we can also formulate the Leibniz rule without ambiguity in the usual notation as follows:

Denote by $$\Gamma(-)$$ spaces of smooth sections.

If $$E$$ is a smooth vector bundle over $$M$$, then we can view a smooth section $$s \in \Gamma(T^*(M) \otimes E )$$ as a $$C^\infty$$-linear map $$\tilde{s}: \Gamma( T(M)) \to \Gamma( E)$$ that is defined by letting $$\tilde{s} (X)$$ be the contraction of the first and second slot of $$X \otimes s$$ for every vector field $$X \in \Gamma (T(M))$$.

Therefore we can view the covariant derivative applied to an arbitrary (smooth) tensor field $$F$$ on $$M$$ as a linear map.

We denote the resulting linear map applied to a vector field $$X \in \Gamma ( T(M))$$ by $$\nabla_X F$$. So to be explicit: $$\nabla_X F$$ is obtained by contracting the first and second slot in $$X \otimes (\nabla F)$$.

Then the Leibniz rule can be restated as:

$$\nabla_X( T \otimes S) = (\nabla_X T) \otimes S + T \otimes (\nabla_X S)$$ for every vector field $$X \in \Gamma( T(M))$$.

• I think this answer could be improved by observing that the index-free notation is perfectly capable of expressing this property in an unambiguous way by writing $$\nabla_X(T \otimes S) = (\nabla_X T) \otimes S + T \otimes (\nabla_X S).$$ As it is, the answer reads as insinuating that abstract index notation is inherently less ambiguous, which I don't think is the case. Commented Sep 10, 2023 at 15:12
• @jawheele I never suggested that the index free notation can not express the Leibniz property. I only wrote that the "type changing" map has to be built into the expression. So if $C$ is the type changing map the Leibniz rule could also be stated as $\nabla (T \otimes S) = ( \nabla T) \otimes S + C ( T \otimes (\nabla S))$. All i want to suggest is, that the abstract index notation is very usefull to express such type changes (and also for contractions, factor permutations, symmetrization and metric type changing).
– jd27
Commented Sep 10, 2023 at 17:55

Carroll on p. 95 means that $$\nabla_X \left(T \otimes S\right) ~=~ \left(\nabla_X T\right) \otimes S + T \otimes \left(\nabla_X S\right),\qquad X\in\Gamma(TM),$$ and hence $$\nabla_{\mu} \left(T \otimes S\right) ~=~ \left(\nabla_{\mu} T\right) \otimes S + T \otimes \left(\nabla_{\mu} S\right).$$ So OP's eq. (3) is right, while OP's eq. (2) is wrong.