4
$\begingroup$

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 $g|_0 = (dx\otimes dx+dy\otimes dy+dz\otimes dz)$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.