A fully evaluated tensor is a smooth function, so why does using the Leibniz rule give a bunch of Gammas?

Question

According to how the covariant derivative acts on smooth functions:

$ \nabla_V(T(w, Y)) = V^k \partial_k (T^i_j w_i Y^j) = V^k [ \partial_k T^i_j w_i Y^j + T^i_j \partial_k w_i Y^j + T^i_j w_i \partial_k Y^j ]. $

This is because we get the value of the fully evaluated tensor at a point by multiplying and summing the coefficient functions as above. I mean, the sum of products $T^i_j w_i Y^j$ gives us the smooth function. Now, the above formula holds because we are just multiplying smooth functions. Also the covariant derivative of a sum of two tensors is the sum of the covariant derivatives. (Functions are (0,0) tensors.)

Now according to Leibniz rule:

$\nabla_V (T(w,Y)) = (\nabla_V T)(w,Y) + T(\nabla_V w,Y) + T(w, \nabla_V Y)$,

Term by term:

$ (\nabla_V T)(w, Y) = [V^k \partial_k T^i_j + \Gamma^i_{mk} T^m_j V^k - \Gamma^n_{jk} T^i_n V^k] w_i Y^j, (1) $

$ T(\nabla_V w, Y) = T^i_j (V^k \partial_k w_i - \Gamma^l_{ik} w_l V^k) Y^j,(2) $

$ T(w, \nabla_V Y) = T^i_j w_i (V^k \partial_k Y^j + \Gamma^j_{lk} Y^l V^k).(3) $

Adding these together: $V^k [ \partial_k T^i_j w_i Y^j + T^i_j \partial_k w_i Y^j + T^i_j w_i \partial_k Y^j ]+ V^k([\Gamma^i_{mk} T^m_j - \Gamma^n_{jk} T^i_n ] w_i Y^j - T^i_j \Gamma^l_{ik} w_l Y^j+ T^i_j \Gamma^j_{lk} Y^l w_i )). $

The sum of the first terms on the right hand side of equations 1,2,3 is the same as the right hand side of the first equation, but in total the sum is not the same. What's wrong?

Answer 1

Nothing is wrong; you’re just confused by the tons of indices. You need to shuffle some of them around and relabel the dummy indices. Your “extra” term consists of 4 parts (I ungrouped your brackets): \begin{align} V^k\left(\Gamma^i_{mk} T^m_jw_iY^j - \Gamma^n_{jk} T^i_nw_iY^j - \Gamma^l_{ik} T^i_j w_l Y^j+ \Gamma^j_{lk} T^i_j Y^l w_i \right). \end{align} The first and third term cancel, and the second and fourth cancel. To see this,

• in the third term, replace $(i,j,k,l)\to (m,j,k,i)$, i.e simply $(i,l)\to (m,i)$.
• in the fourth term, replace $(i,j,k,l)\to (i,n,k,j)$, i.e simply $(j,l)\to (n,j)$.

Answer 2

It's easy to get confused about this kind of formula that you mentioned in a comment from Wikipedia:

\begin{align} \begin{aligned}(\nabla _{Y}T)\left(\alpha _{1},\alpha _{2},\ldots ,X_{1},X_{2},\ldots \right)=&{}\nabla _{Y}\left(T\left(\alpha _{1},\alpha _{2},\ldots ,X_{1},X_{2},\ldots \right)\right)\\&{}-T\left(\nabla _{Y}\alpha _{1},\alpha _{2},\ldots ,X_{1},X_{2},\ldots \right)-T\left(\alpha _{1},\nabla _{Y}\alpha _{2},\ldots ,X_{1},X_{2},\ldots \right)-\cdots \\&{}-T\left(\alpha _{1},\alpha _{2},\ldots ,\nabla _{Y}X_{1},X_{2},\ldots \right)-T\left(\alpha _{1},\alpha _{2},\ldots ,X_{1},\nabla _{Y}X_{2},\ldots \right)-\cdots \end{aligned} \end{align}

when we don't write the basis elements of the tensor space explicitly.

To take a very simple example, we can evaluate a tensor in a specific basis by letting it act on the basis vectors and co-vectors as arguments:

$$ T(\mathbf{e^j},\mathbf{e_k},\mathbf{e_l}) $$

This will make it easier to see what's happening, and where the confusion is, since our thinking won't get mixed up with additional sums of vectors and co-vectors components.

Now, I think your confusion stems from not distinguishing between two possible ways to treat this object:

1. As its components, where we simply write $T(\mathbf{e^j},\mathbf{e_k},\mathbf{e_l})=T^{j}_{~~kl}$ and then, if taken as a smooth function of some variable, indeed the covariant derivative will equal its partial derivative. But this is not the covariant derivative of a tensor! To see that even more clearly, note that the covariant derivative takes a $(p,q)$ tensor and gives back a $(p,q+1)$ tensor. But in your first equation, you are taking the covariant derivative of an object where all the indices are contracted:

   $$V^k \color{red}{\nabla_k} (T^i_j w_i Y^j)$$

   Where I took the liberty of slightly misquoting you and changing $\partial \rightarrow \nabla$ as indicated in red, because for the case of covariantly differentiating a scalar it doesn't matter.

2. As a tensor, where we write $$ \mathbf{T}=T^j_{~~kl}~\mathbf{e_j}\otimes\mathbf{e^k}\otimes\mathbf{e^l} $$ Taking the covariant derivative of this object is what yields the true covariant derivative of the tensor, which will include the connection coefficient terms that get added to the partial derivative term of the tensor components.

The bottom line is, the covariant derivative of a tensor is only meaningful if you don't only differentiate the tensor components, but also its basis elements. That is also the sense in which the formula I cited from Wikipedia should be read, e.g. every application of the covariant derivative in it, apart from the LHS, is also applied to the basis vector of every argument, and that's what yields the $\Gamma$ terms which are missing in your first equation. To put it perhaps even more clearly, a more explicit way to write the Leibniz rule is:

$$ \nabla_{v}(S\otimes T) = (\nabla_{v}S) \otimes T + S \otimes (\nabla_{v} T) $$

This, together with the knowledge of how the covariant derivative acts on basis vectors (which is by definition what yields the connection coefficients terms!) should make it crisply clear how to apply the Leibniz rule correctly.

(Possibly important!) Edit:

In light of @peek-a-boo's comments and answer, I think that I have been using a different meaning of the Leibniz rule from what you (and probably Wikipedia too) had intended. I was referring to the following way to define the covariant derivative via the Leibniz rule:

\begin{align} \nabla_{\small{V}} (T^{ij}{\bf e}_i\otimes {\bf e}_j) &= (\nabla_{\small{V}} T^{ij}){\bf e}_i\otimes {\bf e}_j+ T^{ij}(\nabla_{\small{V}}{\bf e}_i)\otimes {\bf e}_j+ T^{ij}{\bf e}_i\otimes (\nabla_{\small{V}} {\bf e}_j)\\ \\ &=V^{k} (\partial_k T^{ij}+ {\Gamma^i}_{lk} T^{lj} + {\Gamma^j}_{lk} T^{il}){\bf e}_{i}\otimes {\bf e}_j \end{align}

However, @peek-a-boo's observation is completely correct, since it appears that your entire goal was to show that the covariant derivative commutes with the contraction of its argument, i.e. that it doesn't matter if you first contract the argument and then covariantly differentiate or vice versa, then @peek-a-boo's answer is all you need to look at ;-)