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Let $G$ be a Lie group and $f = f(t, s) : \mathbb{R}^2 \to G$ smooth. Consider $\theta \in \Omega^1(G, \mathfrak{g})$ the Maurer-Cartan form of $G$. I'm trying to understand why

$\displaystyle\frac{d}{dt} \Bigr|_{t = 0} f^{*}\theta \left(\displaystyle\frac{d}{ds} \Bigr|_{s = 0}\right) = \displaystyle\frac{d}{ds} \Bigr|_{s = 0} f^{*}\theta \left(\displaystyle\frac{d}{dt} \Bigr|_{t = 0}\right) + \left[ \theta \left( \frac{df}{ds}(0, 0) \right), \theta \left( \frac{df}{dt}(0, 0) \right) \right]$

which is supposed to be a consequence of the Maurer-Cartan structure equation. The left hand side can be thought of as $\frac{d}{dt} \Bigr|_{t=0}$ acting on the differential of the function $f^{*}\theta \left( \frac{d}{ds} \Bigr|_{s = 0} \right)$, which is a function $\mathbb{R} \to \mathfrak{g}$. Here I suppose we could identify this differential as $f^{*}(d\theta)\left(\frac{d}{ds} \Bigr|_{s = 0}\right)$. Then the left hand side would be $f^{*}(d\theta)\left(\frac{d}{ds} \Bigr|_{s = 0}, \frac{d}{dt} \Bigr|_{t = 0}\right) = d\theta\left(\frac{df}{ds}(0, 0), \frac{df}{dt}(0, 0)\right) = -\left[ \theta \left( \frac{df}{ds}(0, 0) \right), \theta \left( \frac{df}{dt}(0, 0) \right) \right]$, but this result doesn't agree with the one stated above. I'm sure the error is in understanding the differential of $f^{*}\theta \left( \frac{d}{ds} \Bigr|_{s = 0} \right)$ but I've been confused on how to do it.

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    $\begingroup$ The Maurer-Cartan structure equation is, indeed, $d\theta = [\theta,\theta]$. But who are $s$ and $t$ here? Presumably you're doing some variant of evaluating this expression on vector fields $X$ and $Y$. Are you wanting to apply the formula $d\omega(X,Y) = X(\omega(Y))-Y(\omega(X))-\omega([X,Y])$, assuming that your vector fields commute? $\endgroup$ – Ted Shifrin Jul 1 '16 at 0:14
  • $\begingroup$ @TedShifrin, $s$ and $t$ are the (standard) coordinates in $\mathbb{R}^2$. Doesn't the structure equation come from that formula? $\endgroup$ – Pedro Jul 1 '16 at 15:13
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    $\begingroup$ Note that they've changed the order. They're doing $Y(\theta(X))-X(\theta(Y))$, which takes care of the minus sign. (Here $X=f_*(\partial/\partial s)$ and $Y=f_*(\partial/\partial t)$.) $\endgroup$ – Ted Shifrin Jul 1 '16 at 15:31

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