The same index showing up twice in the charge conservation law $\nabla_a j^a = 0$, as stated using abstract index notation, highly confuses me.

If we chose a coordinate basis $\{\partial_\rho\}_{\rho}$, expressed $j^b = j^\nu \left(\partial_\nu\right)^b$ in that basis and picked $e^a = \left(\partial_\mu\right)^a$, we would compute

$$\nabla_a j^b \overset{(Notation)}{=} \nabla_{e^a}j^b = j^\nu_{,\mu}\left(\partial_\nu\right)^b + j^\nu\Gamma^\rho_{\nu\mu}\left(\partial_\rho\right)^b $$

Now, how should the result be adapted, when $b$ is replaced by $a$? A repeated index stands for contraction in abstract index notation, usually. However, if that was the case here, what would be contracted?


1 Answer 1


As Qmechanic pointed out in the comments, you're mixing Einstein and abstract index notation a bit. To make things absolutely clear, we will use early Latin indices for abstract indices $(abc)$ and Greek indices for component indices $(\mu\nu\rho)$ and will always indicate Einstein summation explicitly.

First and foremost, an abstract index is nothing more than a label for what type of tensor an object is. For instance, $v^a$ is literally the whole vector, not the $a$-th component of $\mathbf{v}$.

To see why this is important, we consider the partial derivative operator in come chart. We can show that the set of partial derivatives are linearly independent and are also derivations on the smooth functions, so they're a natural choice for our basis vectors for the tangent space $V$. The partial derivative vector is $(\partial_\mu)^a$. The $\mu$ tells us what coordinate the vector belongs to and the $a$ tells us this is a vector. The expansion of a vector $v^a$ in terms of this basis is $$v^a=\sum_\mu v^\mu(\partial_\mu)^a.$$ The coordinate components $\{v^\mu\}$ are determined as follows: Let $\{(\theta^\mu)_a\}$ be a basis for $V^*$, the algebraic dual of $V$, such that $$\tag{1}(\theta^\mu)_a(\partial_\nu)^a:=\theta^\mu(\partial_\nu)=\delta^\mu{}_\nu.$$ Here $\theta^\mu(\partial_\nu)$ is just the application of $\theta^\mu\in V^*$ viewed as a linear map $V\to \mathbb{R}$. Then $$v^\mu:=(\theta^\mu)_av^a:=\theta^\mu(v)=v(\theta^\mu).$$ The last equality comes from the natural identification $V^{**}=V$.

Note that $\partial_a$ can also have meaning. For instance, for $f$ a function, $\partial_a f$ is the covector defined by $$\partial_af:=\sum_\mu\partial_\mu f(\theta^\mu)_a$$ where $\partial_\mu f=\partial f/\partial x^\mu$ are just "numbers" now.

Ok, let us unpack $\nabla_aj^b$ a bit: $$\tag{2}\nabla_aj^b=\sum_{\mu\nu}(\partial_\mu j^\nu+\sum_\rho\Gamma^\nu{}_{\mu\rho}j^\rho)\,(\theta^\mu)_a(\partial_\nu)^b$$ Now, when we contract the indices, i.e. set $a=b$, then we end up with $$\nabla_aj^a=\sum_\mu(\partial_\mu j^\mu+\sum_\rho\Gamma^\mu{}_{\mu\rho}j^\rho)$$ which just follows from (1).

Comment/rant: An object of the form $\partial_a v_b$ is not valid in abstract index notation. It is obvious that the components should be of the form $\partial_\mu v_\nu$, but since they don't transform appropriately for other coordinate systems, it's not a well defined object. The same is true for $\partial_a g_{bc}$ or $\Gamma^a{}_{bc}$. Thus (2) is the correct formula for the covariant derivative. Some books (notably Wald) write $$\nabla_a v^b=\partial_a v^b+\Gamma^b{}_{ac}v^c$$
which is inconsistent with the spirit of abstract index notation at best.

  • $\begingroup$ Great answer, thank you very much! The missing piece was the definition of $\partial_a f$. This makes the result a $\binom{1}{1}$-tensor field that can be contracted. Before, I interpreted $\nabla_a$ as $\nabla_{(e_\mu)^a}$, so I expected the result to be a vector field (since $\nabla_XY$ is a vector field, if $X$, $Y$ are so). $\endgroup$
    – A. Keller
    Commented Jan 12, 2016 at 22:05
  • $\begingroup$ @bpl Yes, indeed, in abstract-index formalism one usually regards $\nabla_X Y$ as a shorthand for $X^a \nabla_a Y^b.$ $\endgroup$
    – CR Drost
    Commented Jan 12, 2016 at 22:14
  • 1
    $\begingroup$ @bpl Yeah, $\nabla_a$ by itself is just $\nabla$ for a mathematician. So $\nabla_a:\mathcal{T}^r{}_s\to\mathcal{T}^r{}_{s+1}$ where $\mathcal{T}^r{}_s$ is the type $(r,s)$ tensor bundle. $\endgroup$
    – Ryan Unger
    Commented Jan 12, 2016 at 22:15
  • $\begingroup$ @0celo7 The information you provided is amazing! I guess the reason for which $\nabla$ is indexed by $a$ is that $\nabla_a Y^b \in \mathcal{T}^1_1$, while $\nabla: \mathcal{T}^r_{s} \to \mathcal{T}^r_{s+1}$ itself is not a covariant tensor (in particular cannot be expressed in the cotangent basis). Moreover, $X^c \otimes \nabla_a Y^b$ results in the covariant derivative $X^a\nabla_a Y^b$, when the indices $a$, $c$ are contracted. $\endgroup$
    – A. Keller
    Commented Jan 13, 2016 at 12:19
  • $\begingroup$ @bpl Note: one omits the tensor product when writing expressions in abstract index notation. For example, if $X\in\mathcal{T}^1{}_0$ and $\omega\in\mathcal{T}^0{}_1$, then $X\otimes \omega$ becomes $X^a\omega_b$. Note further that while $X\otimes\omega\ne\omega\otimes X$ in general, $X^a\omega_b=\omega_bX^a$. $\endgroup$
    – Ryan Unger
    Commented Jan 13, 2016 at 13:12

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