Problem 6, UMD August 2018 Topology/Geometry exam
Let $\xi = \frac{d}{dx}, \eta = x\frac{d}{dx}$ (on the real line). Find local flows for these vector fields.
a) Prove or disprove: These integrate to global flows.
b) Compute the Lie brackets $[\xi ,\xi ], [\eta ,\eta], [\xi ,\eta]$.
c) Find a nonzero external differential form of positive degree that is invariant under the local flow of $\xi$. Is it possible to choose it so that it is also invariant under $\eta$? Hint: consider $L_\xi (\omega )$.
Primarily, I believe I did this problem correctly, but am unsure and would like to verify that I did.
a) The flows should be, for $\xi$: $\frac{d\Phi _t}{dt}=1$, ie $\Phi _t(x)=x+t$ and, for $\eta$: $\frac{d\Psi _t}{dt}=x(t)$, ie $\Psi _t(x)=e^tx$. These both exist for all time t and any starting point x.
b) $[\xi ,\xi]$ and $[\eta ,\eta]$ are zero, by definition. $[\xi ,\eta] = \xi (\eta )-\eta (\xi )$, which should be $\xi (x)\frac{d}{dx}-\eta (1)\frac{d}{dx} = 1\frac{d}{dx}-0\frac{d}{dx} = \frac{d}{dx} = \xi $, since all of the second-order terms cancel.
c) Since the manifold is 1-d, a nonzero differential form, $\omega$, of positive degree must have degree 1. I believe (and this is the part I am most unsure of) the invariance under local flow corresponds to $L_\xi (\omega)=0$, hence $L_\xi (f dx) = i_\xi (d(fdx))+d(i_\xi (fdx)) = i_\xi (0) + d(f) = \frac{df}{dx}dx = 0$, hence $f$ is a constant, eg $\omega =1dx$. For the sake of completeness, $d(\omega )=0$ because it is a 2-form on a 1-d manifold, hence zero.
Similarly, to get a 1-form invariant under $\eta$, it should be $0 = L_\eta (gdx) = i_\eta (d(gdx))+d(i_\eta (gdx)) = 0+d(x*g) = (g+x\frac{dg}{dx})dx$, hence $g = \frac{C}{x}$, any constant $C$. This is only compatible with $g$ constant for $g=0$, so it should not be possible to pick such a nonzero differential form of positive degree.