2
$\begingroup$

Given a connection $\nabla$ on the tangent bundle, the divergence of a vector field $X\in C^{\infty}(TM)$ is defined as $div(X) = C_1^1(\nabla X)$ where $C$ is the contraction.

If $\{e_i\}$ is a local reference frame and $\{\eta^i\}$ is the induced coframe, then locally $$(1)\quad \quad div(X) = \nabla X(e_i,\eta^i)=\eta^i(\nabla_i X) = e_i(X^i) + X^r\Gamma_{i r}^i $$ where $\Gamma_{i r}^k$ are the Christoffel's symbols.

Given a pseudo-Riemannian metric, if instead of a vector field we have a 1-form $\omega\in C^{\infty}(T^*M)$, we can define $$(2) \quad \quad div(\omega) := div(\omega^\#).$$

Sometimes I have encountered this notation for the divergence of a 1-form : $$ (3)\quad \quad \text{div} \ \omega = \nabla^i \omega_i$$ What does the symbol $\nabla^i$ mean? Is $\omega_i$ defined by $\omega = \omega_i dx^i$? Or it is $\text{div} \ \omega = (\nabla^i \omega)_i$ the i-component of the covariant tensor $\nabla^i \omega$?
I also need to explain why , in the same paper, the author states that $\Delta = -\nabla^i\nabla_i$ as if $\Delta f = -\nabla^i\nabla_if$ since $\nabla_i f $ is a function it seems that $\nabla^i$ act on functions.

Locally $(2)$ is written as $div(\omega) = e_i( g^{ij}\omega_j)+ g^{r j}\omega_j\Gamma_{i r}^i $ I can't see clearly how is defined $\nabla^i$.

$\endgroup$
3
$\begingroup$

This is index notation as traditionally used in the Ricci calculus, these days typically interpreted by mathematicians as abstract index notation.

When we write $\mathrm{div}\,X = \nabla_i X^i$, what is meant is that we first take the covariant derivative of $X$ and then perform the contraction indicated by the repeated indices; i.e. $\nabla_i X^i = (\nabla X)^i_i$, where the RHS can be interpreted either as abstract index notation denoting the contraction $C(\nabla X)$ or as a coordinate formula using Einstein summation notation. Importantly, this is not the same thing as first taking the coordinate components $X^i$ (which are scalars) and then differentiating, which would yield instead $\partial_i X^i$, a coordinate-dependent quantity. Thus we usually don't think of the expression $\nabla_i X^i$ as being an operator $\nabla_i$ acting on the component functions $X^i$ - instead, it's the operator $\nabla$ acting on the vector field $X$, with the indices just telling us how to contract. (Hopefully this answers your last question regarding $\nabla^i \nabla_i f.$)

In the case of a one-form $\omega$, the covariant derivative is a $(0,2)$-tensor $\nabla\omega = (\nabla_j \omega_i) \eta^j \otimes \eta^i$. (Note again that in this notation, $\nabla_j \omega_i$ is not the same as $\partial_j \omega_i$ - we are taking the covariant derivative before plugging in the reference frame.) To take the trace of this to obtain a divergence, we first raise an index with the metric as you described, yielding $\mathrm{div}(\omega^\sharp) = \nabla_i \omega^i := \nabla_i (\omega^\sharp)^i$ where of course $\omega^i = (\omega^\sharp)^i = g^{ij} \omega_j.$ Since the covariant derivative is $g$-compatible and the metric is symmetric, we can think of this as $\mathrm{tr}_g \nabla \omega = g^{ij} (\nabla \omega)_{ij};$ so we could just as fairly raise the first index instead of the second, yielding $\nabla^i \omega_i.$

$\endgroup$
  • $\begingroup$ If I get it right $\nabla_j\omega_i$ is just a notation for $(\nabla \omega)_{ij}$ the coefficient relative to $\eta^{i}\otimes \eta^j$ in $\nabla \omega$. And then raising and lowering indices works as usual. But when you wrote $\nabla_i\omega^i:=\nabla_i g^{ij}\omega_j$ did you mean $\nabla_i\omega^i:=g^{ij} \nabla_i \omega_j$? $\endgroup$ – Warlock of Firetop Mountain Dec 21 '17 at 14:54
  • $\begingroup$ I guess I meant $\nabla_i(g^{ij}\omega_j)$ - I was trying to avoid writing this because this parenthesized notation is sometimes used to mean $\partial_i(g^{ij}\omega_j)$. To be perfectly clear, I mean $\nabla_i X^i$ where $X^i = g^{ij} \omega_j$. $\endgroup$ – Anthony Carapetis Dec 21 '17 at 14:58
  • $\begingroup$ Thank you Anthony, can you recommend me some good books where I can find this kind of notations or the contracted Bianchi identity or Pohozaev identity? $\endgroup$ – Warlock of Firetop Mountain Dec 22 '17 at 9:35
-5
$\begingroup$

For a function, $\omega$, of n variables, $x_1$, $x_2$, ..., $x_n$, "grad $\omega$", or $\nabla \omega$, is defined as $\frac{\partial \omega}{\partial x_1}+ \frac{\partial \omega}{\partial x_2}+ \cdot\cdot\cdot+ \frac{\partial \omega}{\partial x_n}$. That can be thought of as the "scalar product" of the "vector operator" $\nabla= \frac{\partial}{\partial x_1}+ \frac{\partial}{\partial x_2}+ \cdot\cdot\cdot+ \frac{\partial}{\partial x_n}$ with $\omega$ which is denoted, in Einstein summation notation, as "$\nabla_i \omega$".

Similarly, divergence of a vector valued function (one-form) $\vec{\omega}= f_1(x_1,x_2,...,x_n)\vec{i_1}+ f_2(x_1,x_2,...,x_n)\vec{i_2}+ \cdot\cdot\cdot+ f_n(x_1,x_2,...,x_n)\vec{i_n}$, where "$i_m$" is the mth basis vector, is the scalar vector $\frac{\partial f_1}{\partial x_1}+ \frac{\partial f_2}{\partial x_2}+ \cdot\cdot\cdot+ \frac{\partial f_n}{\partial x_n}$ which can be thought of as the dot product of the vector operator $\nabla$ and the vector $\vec{\omega}$. That is denoted, in Einstein summation notation, as $\nabla_i\omega^i$.

$\endgroup$
  • 1
    $\begingroup$ We are speaking of operators defined over a Riemannian manifold. The definition you gave of the gradient is valid only in $\mathbb{R}^n$ with the standard metric. Please read the question again. $\endgroup$ – Warlock of Firetop Mountain Dec 21 '17 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.