These two questions are from my exam practice question sets , which are quite similar. I got some problem understanding and solving both of them .

For (a) , I can only substite $dx\wedge dy\wedge dz$ with $d(-x)\wedge d(e^{-x}(xz-y))\wedge d(-e^{-x}z)$ and it comes to $e^{-2x}dx\wedge dy \wedge dz$ by direct computation. I wonder if there is a more general method to solve the question. e.g compute a general form using left-variance property and then use the value of $\Omega_0$ to determine the exact form. But I coudnt figure it out. Thanks for any help !

(a)Let G denote the Lie group with underlying smooth manifold $R^3$ and group operation $\star$ defined by $(x,y,z)\star(x',y',z'):=(x+x',y+e^xy'+ xe^xz',z+e^xz').$

Compute the unique n-form $\Omega \in \Gamma(\wedge^3 T^\ast M)$ such that $\Omega_0 = dx\wedge dy\wedge dz$ and $L^\ast_g\Omega=\Omega$.(Elementary algebra, which you need not reproduce here, gives that the inverse in G is given by $(x,y,z)^{-1}=(-x,e^{-x}(xz-y),-e^{-x}z$)).

(b)Let $H_3$ denote the smooth manifold $R^3$ endowed with the group operation $(a,b,c)\ast(p,q,r):=(a+p,b+q,c+r+aq)$.Checking shows that $H_3$ is a Lie group with the identiy (0,0,0), which we just denote 0.

Compute the left-invariant metric g on $H_3$ whose value at the identity is $g|_0 = (dx\otimes dx+dy\otimes dy+dz\otimes dz)$


In general, a left-invariant n-form $\Omega$ in a n-dimensional Lie group will have to be given by $$\Omega_g(v_1,\dots,v_n)= \Omega_e (d_g\mathcal{L}_{g^{-1}}(v_1),\dots, d_g\mathcal{L}_{g^{-1}}(v_n)),$$ so if you are given $\Omega_e$ you can compute $\Omega$ everywhere. Similarly for the left-invariant metric $g$.

This applies perfectly to this situation, where you can explicitly compute the differential of the left translation.


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