Exercise on left invariant 2-forms I'm trying to solve a problem about the Lie group of transformation over $\mathbb{R}$:
\begin{equation}
x\mapsto ax+b,
\end{equation}
where $a,b\in\mathbb{R}, a\neq 0.$
I'm asked to find the space of left invariant 2-forms.
Can anybody explain me how to get the solution?
 A: The element, call it $g(a,b)$, of your Lie group can be viewed as the matrix
$$
g(a,b)=\left(\begin{array}{rr}a&b\\0&1\end{array}\right)
$$
acting on column vectors $(x,1)^T$. We see that the matrix of $1$-forms
$$
g(a,b)^{-1}dg(a,b)=\frac1a\left(\begin{array}{rr}1&-b\\0&a\end{array}\right)\left(\begin{array}{rr}da&db\\0&0\end{array}\right)=
\frac1a\left(\begin{array}{rr}da&db\\0&0\end{array}\right)
$$
is left-invariant. Namely, if $h$ is a fixed element of the group, then
$$
(hg)^{-1}d(hg)=g^{-1}h^{-1}hdg=g^{-1}dg.
$$
As $\omega_1=\dfrac1a\,da$ and $\omega_2=\dfrac1a\,db$ are invariant 1-forms, their wedge product
$$
\omega_1\wedge\omega_2=\frac1{a^2}da\wedge db
$$
is an invariant 2-form.
On a locally compact 2-dimensional group the left-invariant 2-forms form a 1-dimensional space (IIRC) as integrating them gives a left Haar measure.
A: Another approach, ultimately equivalent to Jyrki's but from a different point of view, proceeds by first choosing arbitrarily a non-zero 2-form at the identity element (i.e., a vector in $\bigwedge^2T_e^*$) and then translating that vector to all other points by means of left-translations, so as to enforce left-invariance.  (A different initial choice would just multiply the result by a non-zero scalar.)
In more detail: Let me write $u_1$ and $u_2$ for the obvious two coordinate functions on your group $G$, i.e., $u_1(a,b)=a$ and $u_2(a,b)=b$, and let me choose, as my 2-form at the identity, the simplest one I can think of, namely $du_1\land du_2$.  Then the desired value of the 2-form at any other element $(a,b)$ in $G$ is obtained by considering the left-translation $L$ that sends $(a,b)$ to the identity $(1,0)$ and applying $L^*$ to $du_1\land du_2$. Of course, $L$ is left-translation by the inverse of $(a,b)$, namely $(\frac 1a,-\frac ba)$.  Using the multiplication rule of $G$, I compute that $L^*(u_1)=\frac1au_1$ and $L^*(u_2)=\frac1au_2-\frac ba$. Since $L^*$ commutes with $d$ and with $\land$, it follows that the value of the desired form at $(a,b)$ is $L^*(du_1\land du_2)=\frac1{a^2}du_1\land du_2$. Finally, to get a formula for the global form, use the fact that the $a$ in the preceding formula is the value of $u_1$ at the point $(a,b)$ under consideration. So our form is $\frac1{{u_1}^2}du_1\land du_2$.
