# Covariant Derivative of Kronecker Delta

I am reading Carroll's book on GR right now, and I ran into a little trouble in his chapter 3 on curvature. He is establishing the properties of the covariant derivative, and claims that the fact that the covariant derivative commutes with contraction, i.e $$\nabla_\mu T^\lambda {}_{\lambda \rho} = (\nabla T)_\mu {}^\lambda {}_{\lambda \rho},$$ implies that the covariant derivative of the Kronecker delta is zero: $$\nabla_\mu \delta^\lambda_\sigma = 0.$$ I do not see why this is the case. I would greatly appreciate it if someone could point out what I am missing. Thanks!

• The Kronecker delta is a constant tensor field, so it differentiates to 0... – zzz Jul 20 '17 at 16:23

Note that $\delta^i{}_j=\delta^i{}_k\delta^k{}_j$. Then $$\nabla_l\delta^i{}_j=\nabla_l(\delta^i{}_k\delta^k{}_j)=C(k,m)[\nabla_l(\delta^i{}_k\delta^m{}_j)]=C(k,m)[\delta^i{}_k\nabla_l\delta^m{}_j+\delta^m{}_j\nabla_l\delta^i{}_k]=2\nabla_l\delta^i{}_j$$ where $C(k,m)$[...] is defined as the contraction operator which contracts the $k$ and $m$ indices. Thus $\nabla_l\delta^i{}_j=0$.

• Thinking about this again, this seems to be the only answer that actually uses the fact that the operation commutes with contraction, so I think it should be taken as the answer. – nickodel May 31 '16 at 10:12

The partial derivatives $\partial_\lambda \delta^\mu_\nu$ are clearly zero because the components of the Kronecker delta are constant functions of spacetime coordinates (one or zero).

One may always go to a locally Minkowski frame where the Christoffel symbols vanish and there, the covariant derivative is equal to the partial one and vanishes, too. Because the Kronecker delta is a tensor, its covariant derivative must be the same in all other frames, too.

Alternatively, one may explicitly write the Christoffel symbol terms, too: $$\nabla_\lambda \delta^\mu_\nu = \partial_\lambda\delta^\mu_\nu - \Gamma_{\lambda\kappa}^\mu \delta^\kappa_\nu - \Gamma_{\lambda\nu}^{\kappa} \delta_{\kappa}^\mu = 0 + \Gamma^\mu_{\lambda\nu}-\Gamma^\mu_{\lambda\nu} = 0.$$ The cancellation between the two connection terms is the same cancellation that also allows to ignore contracted pairs of indices in the tensor. I don't think it's helpful to "derive" the second statement from the first one, however, because the first one doesn't explicitly use the Kronecker delta at all so the claim that the first fact "implies" the other must be taken with a grain of salt.

Well, if I had to say what was meant, it was that $$T^\lambda{}_{\lambda\rho} = T^\lambda{}_{\kappa\rho} \delta^\kappa_\lambda$$ and the $\nabla_\mu$ derivative of this sum of products may be written using the Leibniz rule as $$\nabla_\mu T^\lambda{}_{\lambda\rho} = (\nabla T)_{\mu}{}^\lambda{}_{\kappa\rho} \delta^\kappa_\lambda + T^\lambda{}_{\kappa\rho} \nabla_\mu \delta^\kappa_\lambda$$ The assumption is the equation without the last $\nabla\delta$ term, so this last term has to vanish for any tensor $T^{\lambda}_{\kappa\rho}$, which implies that the whole $\nabla_\mu \delta^\kappa_\lambda$ object is zero, too.

• While this is correct, it does not answer OP's question about using contractions to prove this. – Ryan Unger May 31 '16 at 9:29
• I've added an explicit calculation. – Luboš Motl May 31 '16 at 9:31
• I reiterate my first comment. (Unless one uses the contraction rule + Leibnitz rule to prove the covariant derivative rule for $(1,1)$ tensors, as is usual, of course. But that's not stated in this answer.) – Ryan Unger May 31 '16 at 9:31
• I don't need to use all the assumptions. I've still proven the implication. ;-) – Luboš Motl May 31 '16 at 9:33
• yes, I agree with you that this is probably what the author had in mind. Thanks. – nickodel May 31 '16 at 10:23

You can write the Kronecker delta tensor as a product of the metric tensor

$$\nabla_a(\delta^a_b) = \nabla_a (g_{bc} g^{ac}) = \nabla_a (g_{bc} g^{ac}) = g_{bc} \nabla_a g^{ac} + g^{ac}\nabla_a g_{bc}$$

As you may recall, the covariant derivative of the metric tensor is $0$ in general relativity.

Version without using the metricity of the connection :

$\nabla_a (\delta_b^a T^b) = \nabla_a T^a$, but also $\nabla_a (\delta_b^a T^b) = \delta_b^a \nabla_a T^b + T^b \nabla_a \delta_b^a$, so contracting the delta with the connection, this gives us

$$\nabla_a T^a = \nabla_a T^a + T^b \nabla_a \delta_b^a$$

Which will only be true for all $T$ if $\nabla_a \delta_b^a = 0$.

• Why is $\nabla_ag^{ac}=0$? The defining characteristic of the Levi-Civita connection is that $\nabla_a g_{bc}=0$ (lower indices). – Ryan Unger May 31 '16 at 9:30
• Yea this makes sense, however, in Carroll's book, he claims that the covariant derivative of the kronecker delta is zero BEFORE introducing the property of metric compatibility. He seems to be claiming this holds true even if the covariant derivative is not metric compatible. – nickodel May 31 '16 at 9:35
• Try it this way then maybe : $\nabla_a (\delta_b^a T^b) = \nabla_a ( T^a)$ but also $\nabla_a (\delta_b^a T^b) = \delta_b^a \nabla_a (T^b) + T^b \nabla_a (\delta_b^a )$, meaning that $\nabla_a ( T^a) = \nabla_a (T^a) + T^b \nabla_a (\delta_b^a )$, hence $\nabla_a (\delta_b^a ) = 0$ – Slereah May 31 '16 at 9:39
• I just realized I kind of blew this however, because the "a" indices do not need to be contracted. It holds for each component in the kronecker delta individually. I think your proof still holds however if you just turn one of the a's into a c or something. I edited the original question as well. – nickodel May 31 '16 at 9:55
• Also, looking closer, where do you use the "commutation with contraction property" in this proof? This seems only to use the Leibniz rule. – nickodel May 31 '16 at 10:01

Kronecker delta is the (1,1) tensor satisfying $$\delta^{a}\vphantom{\delta}_{b}v^{b}=v^{a}$$ for an arbitrary vector $$v$$. By Leibniz rule of affine connection, we get $$\delta^{a}\vphantom{\delta}_{b}\nabla_{c}v^{b}+v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=\nabla_{c}v^{a}$$. Giving arguments vector $$u$$ and linear form $$\omega$$ to the above, $$\omega_{a}u^{c}\left(\delta^{a}\vphantom{\delta}_{b}\nabla_{c}v^{b}+v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}\right)=\omega_{b}u^{c}\nabla_{c}v^{b}+\omega_{a}u^{c}v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=\omega_{a}u^{c}\nabla_{c}v^{a}$$, we get $$\omega_{a}u^{c}v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=0$$. Because the above holds for arbitrary $$u$$, $$v$$, and $$\omega$$, we conclude that $$\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=0$$.