From Killing equation $$\nabla_\nu \xi_\mu + \nabla_\mu \xi_\nu = 0$$ it can be shown that $\nabla_\nu \xi_\mu$ is antisymmetric.

From it we can construct an antisymmetric tensor $\mathcal{A}_{\mu\nu}$ such that

$$\mathcal{A}_{\mu\nu} = \nabla_\mu \xi_\nu - \nabla_\nu \xi_\mu = \partial_\mu \xi_\nu - \partial_\nu \xi_\mu$$ is independent of the connection. Where $\nabla_\mu$ is a covariant derivative in the $\mu$-direction and $\partial_\mu$ is a normal derivative in the $\mu$-direction.

Considering that the Killing vectors are the directions along which the metric is invariant, I'm trying to find out if $\mathcal{A}_{\mu\nu}$ can be given any geometrical meaning.

Can somebody shed light on its meaning?

Addendum: I've found that in Geometric Algebra language, $\mathcal{A}_{\mu\nu}$ can be expressed as a bivector originating from the covariant curl of Killing vectors

$$\mathcal{A} = \mathcal{A}_{\mu\nu}g^\mu \wedge g^\nu = \nabla \wedge \xi.$$

Thus, I'm thinking about something along the lines of some kind of rotation (or boost) or rotational property of Killing vector fields. But any help is welcomed.

  • $\begingroup$ $A_{\mu\nu}$ can represent angular momentum for system of particles in a curved space-time (See eq. 6.3.16 from Spinors and Space-time Vol-II). I will expand on this later $\endgroup$
    – paul230_x
    Oct 20, 2023 at 8:14
  • $\begingroup$ The closest physical analogue is the electromagnetic potential and the corresponding tensor. $\mathcal{A}$ can be written as an exterior derivative, which is something that can be integrated, so there is some similarity with the volume element and the Stokes theorem. $\endgroup$
    – auxsvr
    Nov 1, 2023 at 20:07

2 Answers 2


Let's consider the simple case of Minkowski space time in $d=4$, $\mathcal{M}_4$. We know that $\mathcal{M}_4$ has six non-trivial Killing vectors:

  1. Three rotations: $\xi_{\mu}^R$

    $\xi_{\mu}^{R_x}=(0,0,-y,z)$, $\xi_{\mu}^{R_y}=(0,z,0,-x)$, $\xi_{\mu}^{R_z}=(0,-y,x)$

  2. Three boosts: $\xi_{\mu}^B$

    $\xi_{\mu}^{B_x}=(-x,t,0,0)$, $\xi_{\mu}^{B_y}=(-y,0,t,0)$, $\xi_{\mu}^{B_z}=(-z,0,0,t)$

Note that we have taken $g_{\mu\nu}=(-1,1,1,1)$.

It is easy too check that

$\nabla_{\mu}\xi_{\nu}^{R_x}=\begin{pmatrix} 0&0&0&0\\ 0&0&1&0\\ 0&-1&0&0\\ 0&0&0&0\\ \end{pmatrix}\, $, $\nabla_{\mu}\xi_{\nu}^{R_y}=\begin{pmatrix} 0&0&0&0\\ 0&0&0&-1\\ 0&0&0&0\\ 0&1&0&0\\ \end{pmatrix}\, $, $\nabla_{\mu}\xi_{\nu}^{R_z}=\begin{pmatrix} 0&0&0&0\\ 0&0&0&1\\ 0&0&-1&0\\ 0&0&0&0\\ \end{pmatrix}\, $ and $\nabla_{\mu}\xi_{\nu}^{B_x}=\begin{pmatrix} 0&1&0&0\\ -1&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix}\, $, $\nabla_{\mu}\xi_{\nu}^{B_y}=\begin{pmatrix} 0&0&1&0\\ 0&0&0&0\\ -1&0&0&0\\ 0&0&0&0\\ \end{pmatrix}\, $, $\nabla_{\mu}\xi_{\nu}^{B_z}=\begin{pmatrix} 0&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&0\\ \end{pmatrix}\, $

As you mentioned, $\nabla_{\mu}\xi_{\nu}$ is anti-symmetric. If instead of $\nabla_{\mu}\xi_{\nu}$, we calculate $\nabla_{\mu}\xi^{\nu}$, it turns out that for the rotation, the result remains unchanged while for the boost, the minus signs become plus; And clearly, these are the generators of rotation and boost in $\mathcal{M}_4$: $$ "\text{$\mathcal{A}_{\mu\nu}=2\nabla_{\mu}\xi_{\nu}$ corresponds to generators of the symmetries associated with $\xi_{\mu}$}"$$


I am not sure if this answer's OP's question, but if we set $A_{\mu\nu}=\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$, then $A_{\mu\nu}=2\nabla_\mu\xi_\nu$, since $A_{\mu\nu}=\nabla_\mu\xi_\nu-\nabla_\nu\xi_\mu=\nabla_\mu\xi_\nu+\nabla_\mu\xi_\nu$.

Now suppose that $\phi_t$ is the flow corresponding to the Killing vector field $\xi$, and let's assume that $x\in M$ is a fixed point of the flow, so that $\phi_t(x)=x$ for all $t$. This implies that $\xi^\mu(x)=0$, i.e. the Killing vector vanishes at $x$. Then the tangent map $T_x\phi_t:T_xM\rightarrow T_xM$ at that point is an endomorphism, i.e. it maps the tangent space into itself (rather than into the tangent space at a different point).

Since $\phi_t$ is an isometry of the Riemannian structure $(M,g)$ (for each fixed $t$), it then follows that $T_x\phi_t$ is a linear isometry of the tangent space $(T_xM,g_x)$, so the $t$-derivative at $0$ gives an infinitesimal isometry of $(T_xM,g_x)$ i.e. an element of the Lie algebra $\mathfrak o(g_x)$ (which is isomorphic, although not canonically so, to $\mathfrak o(m)$ if $g$ is positive definite, or $\mathfrak o(s,m-s)$ if $g$ has index $s$, here $m=\dim M$).

The $t$-derivative at $0$ can be expressed in coordinates as $$ \frac{\mathrm d}{\mathrm dt}(T_x\phi_t)^{\mu}_{\ \nu}|_{t=0}=\partial_\nu\xi^\mu(x)=\nabla_\nu\xi^\mu(x)=\frac{1}{2}A_{\nu}{}^\mu(x), $$ where switching to covariant derivatives was possible because $\xi^\mu(x)=0$, so$$ \nabla_\nu\xi^\mu(x)=\partial_\nu\xi^\mu(x)+\Gamma^\mu_{\nu\rho}\xi^\rho(x)=\partial_\nu\xi^\mu(x). $$

We then understand that $\nabla_\mu\xi_\nu$ is antisymmetric, because an infinitesimal rotation (linear isometry) is always given by an antisymmetric matrix (after raising/lowering all indices to the same level).

Unfortunately, this particular interpretation requires to have $\phi_t(x)=x$, i.e. for the isometry flow to have a fixed point, because otherwise we have $T_x\phi_t:T_xM\rightarrow T_{\phi_t(x)}M$, i.e. the tangent map is between two different vector spaces. The interpretation can be recovered partially, in the sense that $T_x\phi_t$ is still a linear isometry from $(T_xM,g_x)$ to $(T_{\phi_t(x)}M,g_{\phi_t(x)})$, but because we have two different vector spaces, it is subtle how to take the "infinitesimal" part of this action.

For brevity, let $\Phi_t:=T_x\phi_t$, and let $u\in T_xM$ be a totally arbitrary vector. What we can do to quantify the infinitesimal action of $\Phi_t$ is to first compute $\Phi_tu$, and then parallel transport this vector backwards from $\phi_t(x)$ to $x$ through the curve $t\mapsto\phi_t(x)$. Let $P_t:T_{\phi_t(x)}M\rightarrow T_xM$ is this backwards transport. Then we have now a self-map $\Psi_t:T_xM\rightarrow T_xM$ given by $\Psi_tu=P_t\Phi_tu$. This is again a linear isometry of $(T_xM,g_x)$, so its derivative at $t=0$ is again a transformation whose matrix is skew-symmetric.

Although not trivial, it can be computed that in coordinates, to first order in $t$, we have $$ (\Psi_t)^\mu{}_\nu=\delta^\mu_\nu+\nabla_\nu\xi^\mu(x)t+O(t^2), $$ so the infinitesimal transformation corresponding to $\Psi_t$ is$$ \frac{\mathrm d}{\mathrm dt}(\Psi_t)^\mu{}_\nu=\nabla_\nu\xi^\mu(x)=\frac{1}{2}A_\nu{}^\mu(x).$$

To reiterate:

  1. The Killing condition means that $\nabla_\mu\xi_\nu$ is a "pure curl", i.e. its symmetric part vanishes, so the interpretation of $\nabla_\mu\xi_\nu$ coincides with the interpretation of its curl.
  2. For each point $x\in M$, one can obtain a one-parameter self-map $\Psi_t$ of the tangent space $T_xM$ by letting a vector in $T_xM$ flow along the integral curves of $\xi$, and then parallel transporting backwards. This is a composition of linear isometries, so this map is a linear isometry. The infinitesimal generator of this map is just $\nabla\xi=(1/2)A$. $\nabla\xi$ is antisymmetric precisely because this linear transformation is an infinitesimal isometry.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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