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Let $\omega$ be a 1-form and $X$ and $Y$ the vector fields on $M$, a smooth manifold. I know that there is a well-known identity:

$d\omega(X,Y)= X(\omega(Y))-Y(\omega(X))-\omega[X,Y]$

which illustrates the relationship between exterior derivative and Lie bracket. The proof of this follows from taking co-ordinates of $X$ and $Y$ and running some computations, which is not difficult. However, I don't get the intuitive meaning behind this.

Is it simply meant to be seen as things like Leibniz rule, where the result, although computational, is quite fundamental that we are to take it as theoretical building-blocks? Indeed, this identity is key in showing that given a connection $\nabla$ with connection 1-form matrix $\Omega$ and curvature tensor $R$, $R(X,Y)(Z)=\nabla^2(Z)(X,Y)$(or the curvature matrix of $R$ is indeed $d\Omega-\Omega \wedge \Omega$). Since Lie bracket $[X,Y]$ measures(in some sense) how far the vector fields $X$ and $Y$ are off from forming local co-ordinates, we can think of the term $\omega[X,Y]$ as making 'adjustments', but I wonder whether there are better interpretations of this identity. The cases where $X=\partial/\partial x^i$ and $Y=\partial/\partial x^j$ do return what is expected- that is, $d\omega(x,y)=\partial \omega^j/\partial x^i-\partial \omega^i/\partial x^j$, but it doesn't really contribute to explaining the $\omega[X,Y]$ term.

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The bracket term is needed to make the right-hand side tensorial (i.e., linear over $C^\infty$ functions), inasmuch as the left-hand side is clearly a tensor. (I like thinking of this identity in terms of Stokes's Theorem on a small parallelogram.)

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  • $\begingroup$ Can you explain how to see it in terms of Stokes? $\endgroup$ – Pedro Mar 4 '17 at 22:14
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    $\begingroup$ @Pedro: Intuitively, we apply Stokes's Theorem to a small "parallelogram" $R_t$ in $M$ determined by flowing by $\phi_t$ (the $X$-flow) and $\psi_t$ (the $Y$-flow) for $t$ small. It's well known that $\psi_{-t}\circ\phi_{-t}\circ\psi_t\circ\phi_t(p) \approx [X,Y]_pt^2$. So the little parallelogram needs that bracket term to close up. To second order, $\int_{R_t} d\omega$ then is the integral of $\omega$ over the four sides plus the "close-up" side coming from the Lie bracket. With some care, we see that when we divide by $t^2$ and take the limit, we get the desired equality. $\endgroup$ – Ted Shifrin Mar 5 '17 at 7:40

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