I'm trying to learn by working through an example but I got stuck with the line of reasoning provided in the text.

Let $G$ be the group of matrices

$$g= \begin{bmatrix} 1 &0 &\log{x} \\ y &x &z \\ 0 & 0 &1 \end{bmatrix} , x>0$$

Let $H$ be the closed subgroup containing matrices of the form

$$h= \begin{bmatrix} 1 &0 &0 \\ y &1 &0 \\ 0 & 0 &1 \end{bmatrix} $$

The goal is to find left-invariant forms on the space $G/H$. Here is the approach taken by the text.

First, we obtain left-invariant 1-forms on $G$ by computing $g^{-1} dg$. We obtain

$$\omega_1 = dx/x, \omega_2 = dy/x, \omega_3 = -yx^{-2} dx + x^{-1} dz$$

$H$ is defined by $x = 1, z=0$, and is an embedded submanifold (by closed subgroup theorem). Letting $i:H \to G$ be the inclusion map, we have $i^*\omega_1 =0$ and $i^*\omega_3 = 0$ ($i^*$ here refers to pullbacks).

I understand everything thus far. Here comes the next two steps that I do not understand:

1) Thus, the invariant form on $G/H$, if it exists, must be $\omega_1 \wedge \omega_3 = x^{-2} dx \wedge dz$. (Why did we just wedge the two zero 1-forms together?)

2) This candidate works because $d(\omega_1 \wedge \omega_3) = 0$. (I have no problem with this computation, but why does this computation verify that we have an invariant form on $G/H$?)

May I obtain some explanation on the above two steps?

  • $\begingroup$ what's the text where this example is taken from? Thanx $\endgroup$ – Dac0 Aug 4 '16 at 6:30

In the chart $x,y,z$, the cosets $G/H$ are the lines $x=\mathrm{const.}, z=\mathrm{const.}$ Hence a well-defined two-form on the coset space cannot depend on $y$ or $dy$ and must be on the form $f(x,z)dx\wedge dz$. The two steps taken is to ensure that this is indeed the case.

1) $i^*\omega=0$ says that $\omega$ does not depend on $dy$.

2) $d(\omega_1\wedge\omega_3)=0$ says that $\omega_1\wedge\omega_3$ does not depend on $y$, since if $\omega_1\wedge\omega_3=f(x,y,z)dx\wedge dz$, then $d(\omega_1\wedge\omega_3)=f_y\,dy\wedge dx\wedge dz$.


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