# Note on commutation coefficients in two ways


Suppose somebody hands you a collection of n linearly-independent vector fields $\{v_a\}_{a=1}^n$ on an n-dimensional manifold, which you can use as a frame field (not necessarily an orthonormal frame field, because I haven’t said anything about a metric yet!). A natural thing to compute are the commutation coefficients of these vector fields,

\begin{align} \label{eq:vec-c} [v_a, v_b] = c_{ab}{}^d v_d \,, \end{align}

where we decompose the commutators back into the basis of the vector fields themselves. The collection of scalar fields $c_{ab}{}^d$ are called the commutation coefficients. Because of the antisymmetry of the Lie bracket, the commutation coefficients are automatically antisymmetric in the lower two indices.

On the other hand, let’s say somebody hands you a collection of n linearly-independent one-forms $\{\theta^a\}_{a=1}^n$, which you can use as a coframe field (again not necessarily orthonormal, because no metric yet; and this coframe field might not be dual to the frame field). A natural thing to compute is the exterior derivative of each form, $d\theta^a$, which you could then expand in the basis of two-forms made by wedging together the $\theta^a$’s. So you could define another set of coefficients $\tilde{c}$ from

\begin{align} \label{eq:covec-c} d\theta^a = \tilde{c}^a{}_{bd} \frac{1}{2} \theta^b \wedge \theta^d \end{align}

where we have included a factor of 1/2 for future convenience. (This is not to be confused with the connection 1-form $\omega^a{}_b$.)

The wedge product of two one-forms is automatically antisymmetric, so again we have this property that the collection of scalar fields $\tilde{c}^a{}_{bd}$ is automatically antisymmetric in the lower indices.

This should probably lead you to suspect that the two sets of coefficients are related when the vector and covector bases are related. So, let’s now say that the two bases are dual to each other,

\begin{align} \theta^a(v_b) = \langle \theta^a, v_b \rangle = i_{v_b} \theta^a = \delta^a_b \,. \end{align}

Notice that we still haven’t needed a metric: finding a dual basis is possible without metric (roughly, you only need to be able to do matrix inversion).

Now we can extract components of each equation, Eqs. \eqref{eq:vec-c} and \eqref{eq:covec-c}, by contracting with the right type of object. If we contract Eq. \eqref{eq:vec-c} with $\theta^e$, we’ll find

\begin{align} c_{ab}{}^e = \langle \theta^e, [v_a, v_b] \rangle = i_{[v_a, v_b]} \theta^e \,. \end{align}

Similarly, if we insert two vectors into the two slots of Eq. \eqref{eq:covec-c}, we find

\begin{align} \tilde{c}^a{}_{ef} = (d\theta^a)(v_e, v_f) = i_{v_f} i_{v_e} d\theta^a \,. \end{align}

So now the question is: what is the relationship, if any, between

\begin{align} i_{[v,w]} \omega && \textrm{and} && i_v i_w d\omega \ ? \end{align}

In fact, with the use of some differential identities (or a one-liner in my package xTerior, using the function SortDerivations[]) you can show that for any vectors $v, w$ and form $\omega$, we have

\begin{align} \label{eq:fancy} i_v i_w d\omega = i_{[v,w]} \omega + d i_v i_w \omega - \Lie_v i_w \omega + \Lie_w i_v \omega \end{align}

(there are a bunch of equivalent ways to rewrite this). Now let $v=v_a, w = v_b, \omega=\theta^d$. Since $i_{v_a}\theta^d = \delta^d_a$, the last three terms on the right hand side of Eq. \eqref{eq:fancy} will vanish. In this case we’ll find

\begin{align} i_{v_a} i_{v_b} d\theta^d = i_{[v_a, v_b]} \theta^d \end{align}

which immediately tells us that

\begin{align} c_{ab}{}^d = -\tilde{c}^d{}_{ab} \,. \end{align}

So indeed, the same information is encoded in the commutation coefficients of vectors, $c_{ab}{}^d$, and the decomposition of $d\theta^d$ into basis two-forms, $\tilde{c}^d{}_{ab}$.

Note again that everything has been independent of a metric.

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