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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?

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2 Answers 2

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.

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It is probably not necessary to view $g(a,b)$ as a matrix to do this. But it is the way of constructing invariant 1-forms I'm familiar with :-/ –  Jyrki Lahtonen Aug 21 '14 at 16:59

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 $\frac1{{u_1}^2}du_1\land du_2$.

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