Exercise 2.9.5

Question

Prove that the map that sends a pair $(r,f)$ consisting of a real number and a form $f$ to their product $rf$ defines an action of the set $\mathbb{R}_{>0}$ of all positive real numbers on the set of positive definite forms. Also show that this map defines an action of $\mathbb{R}$ on the set of all indefinite forms. Show in both cases that the number of orbits is infinite.

Richard Ganaye · Accepted Answer

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ewcommand{\U}{\mathbb{U}}

\begin{proof}
Note that if $f$ is a positive definite form, and $r >0$, then $rf = (ra,rb,rc)$ is a positive definite form, since $\Delta(rf) = r^2 \Delta(f)<0$, and $ra>0$.

Since $1\cdot f = f$, and $s\cdot (r \cdot f) = s(rf) = (sr)\cdot f$, the group $\R_+^* = \R_{>0}$ acts on the set of positive definite forms.

If $f$ is a indefinite form, then $\Delta(f) >0$. If $r 
e 0$, then $\Delta(rf) = r^2 \Delta(f) >0$, thus $rf$ is indefinite by Proposition 1.2.10, part 5. Thus the group $(\R^*,	imes)$ acts on the set of indefinite forms.

Take the positive definite forms $f_n = (1,0,n)$, where $n \geq 1$. If $n 
e m$,  $f_n$ and $f_m$ are not in the same orbit, otherwhise $(1,0,m ) = r(1,0,n)$ for some $r>0$. This gives $r= 1$ and $m=n$, which is a contradiction. Thus the number of orbits for the first action is infinite.

Take the indefinite forms $g_n = (1,0,-n)$, where $n \geq 1$. For the same reason, $g_n$ and $g_m$ are not in the same orbit if $n 
e m$. The number of orbits for the second action is infinite.
\end{proof}

Exercise 2.9.5

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Exercise 2.9.5

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