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 $\endgroup$
    – KP99
    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.

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