


\section{Introduction}\label{intro}
		\input{intro}
  
  \section{Preliminaries}\label{prelim}
    \input{prelim}
    
	\section{Proof of the Main Results}
		\input{proof}
  
 	\section{Corollaries}
   	\input{corollaries}

  \section{Acknowledgements}
   	\input{acknowledgements}
  



\subsection{The $1$-dimensional Case}

This subsection provides a description of valuations which are defined on one dimensional polytopes containing the origin in their interiors. We start with the two special cases where the valuation in question is assumed to be even and odd, respectively. 
\begin{lemma}\th\label{class 1-dim even}
	If $\mu \colon \cP^1_{o} \to \R$ is an even valuation, then
		$$ \mu[-a,b] = \tfrac 1 2\, \mu[-a,a] + \tfrac 1 2\, \mu[-b,b] $$
	for all $a,b > 0$.
\end{lemma}
\begin{proof}
  Let $a,b>0$. The valuation property of $\mu$ implies
		\begin{align*}
	     \mu[-a,a] + \mu[-b,b] = \mu[-a,b] + \mu[-b,a] &\quad\textnormal{for}\quad  a \geq b,\\
			 \mu[-b,b] + \mu[-a,a] = \mu[-b,a] + \mu[-a,b] &\quad\textnormal{for}\quad  a < b.		
    \end{align*}
	But since $\mu$ is even, we obtain for all $a,b > 0$ that
		$$ \mu[-a,a] + \mu[-b,b] = 2 \mu[-a,b] .$$
	The assertion of the lemma follows immediately.
\end{proof}

\begin{lemma}\th\label{class 1-dim odd}
	If $\mu \colon \cP^1_{o} \to \R$ is an odd valuation, then
		$$ \mu[-a,b] = \mu[-1,b] - \mu[-1,a]$$
	for all $a,b > 0$.
\end{lemma}
\begin{proof}
	The valuation property of $\mu$ implies
  \begin{align*}
	  \mu[-a,b] + \mu[-1,1] = \mu[-a,1] + \mu[-1,b] &\quad\textnormal{for}\quad  a,b \geq 1,\\
    \mu[-a,1] + \mu[-1,b] = \mu[-a,b] + \mu[-1,1] &\quad\textnormal{for}\quad  a \geq 1\textnormal{ and } 1 > b > 0,\\
    \mu[-1,b] + \mu[-a,1] = \mu[-1,1] + \mu[-a,b] &\quad\textnormal{for}\quad  b \geq 1\textnormal{ and } 1 > a > 0,\\
    \mu[-1,1] + \mu[-a,b] = \mu[-1,b] + \mu[-a,1] &\quad\textnormal{for}\quad  1 > a,b > 0.
  \end{align*}
	But since $\mu$ is odd, we get in all of the above cases
		$$ \mu[-a,b] = \mu[-1,b] - \mu[-1,a].$$
\end{proof}
Let $\mu:\cP_o^1\to\R$ be a valuation. As was shown in the last section, $\mu$ can be written as the sum of an even and an odd valuation. From \th\ref{class 1-dim even} and \th\ref{class 1-dim odd} we therefore obtain the following theorem.
\begin{theorem}\th\label{class 1-dim}
	If $\mu \colon \cP^1_{o} \to \R$ is a valuation, then
		$$ \mu[-a,b] = \tfrac 1 2 \mu[-a,a] + \tfrac 1 2 \mu[-b,b]
		+ \tfrac 1 2 \left( \mu[-1,b] - \mu[-b,1] \right) - \tfrac 1 2 \left( \mu[-1,a] - \mu[-a,1] \right) $$
	for all $a,b > 0$.
\end{theorem}
An immediate consequence of the last theorem is the following classification result.
\begin{corollary}
	If $\mu \colon \cP^1_{o} \to \R$ is a valuation,
	then there exist functions $F,G \colon \R_+ \to \R$ such that
		$$ \mu[-a,b] = F(a) + G(b) $$
	for all $a,b > 0$.
\end{corollary}

\subsection{The $2$-dimensional Case}

The following definition will be crucial for our derivation. Let $F \colon \R_+ \to \R$ and $\mu: \cQ^2 \to \R$ be given. We say that $F$ describes $\mu$ on $\cQ^2$ if
\begin{equation}\label{eq: class cQ2 even - statement}
		\mu[-ae_1,be_1,-ce_2,de_2] = F(ac) + F(bc) + F(ad) + F(bd)
	\end{equation}
for all $a,b,c,d > 0$. The next result shows that even valuations which are $\SLn[2]$-invariant can always be described by a function.
	
\begin{lemma}\th\label{class cQ2 even}
	Suppose that $\mu \colon \cQ^2 \to \R$ is an $\SLn[2]$-invariant valuation
	which is even with respect to the reflection at a coordinate hyperplane.
	Then there exists a function $F \colon \R_+ \to \R$ which describes $\mu$ on $\cQ^2$.
	
\end{lemma}
\begin{proof}
	Fix $a,b \in \R_+$. Recall that $\mu$ is actually even with respect to both reflections at the
	coordinate hyperplanes by the $\SLn[2]$-invariance. Then $ \nu_1[-c,d] := \mu[-ae_1,be_1,-ce_2,de_2]$ is an even valuation on $\cP^1_{o}$.
	From \th\ref{class 1-dim even} we infer that
	\begin{align*}
		\mu[-ae_1,be_1,-ce_2,de_2]
		&= \nu_1[-c,d] = \tfrac 1 2\, \nu_1[-c,c] + \tfrac 1 2\, \nu_1[-d,d] \\
		&= \tfrac 1 2\, \mu[-ae_1,be_1,-ce_2,ce_2] + \tfrac 1 2\, \mu[-ae_1,be_1,-de_2,de_2].
	\end{align*}
	If we apply a similar argument to $\nu_2[-a,b]:=\mu[-ae_1,be_1,-ce_2,ce_2]$ as well as to    			
	$\nu_3[-a,b]:=\mu[-ae_1,be_1,-de_2,de_2]$, then we arrive at
	\begin{eqnarray*}
		\mu[-ae_1,be_1,-ce_2,de_2] &=& \tfrac 1 4\, \mu[-ae_1,ae_1,-ce_2,ce_2] + \tfrac 1 4\, \mu[-be_1,be_1,-ce_2,ce_2] \\
			&&{}+ \tfrac 1 4\, \mu[-ae_1,ae_1,-de_2,de_2] + \tfrac 1 4\, \mu[-be_1,be_1,-de_2,de_2] .
	\end{eqnarray*}
	The $\SLn[2]$-invariance of $\mu$ now implies that
	\begin{eqnarray*}
		\mu[-ae_1,be_1,-ce_2,de_2] &=& \tfrac 1 4\, \mu[-ace_1,ace_1,-e_2,e_2] + \tfrac 1 4\, \mu[-bce_1,bce_1,-e_2,e_2] \\
			&&{}+ \tfrac 1 4\, \mu[-ade_1,ade_1,-e_2,e_2] + \tfrac 1 4\, \mu[-bde_1,bde_1,-e_2,e_2] .
	\end{eqnarray*}
	Finally, set $F(s) = \frac 1 4 \mu[-se_1,se_1,-e_2,e_2]$ for $ s \in \R_+$.
\end{proof}
For odd, upper semicontinuous, and $\SLn[2]$-invariant valuations more can be said.
\begin{lemma}\th\label{vanish cQ2 odd}
	If $\mu \colon \cQ^2 \to \R$ is an upper semicontinuous and $\SLn[2]$-invariant valuation
	which is odd with respect to the reflection at a coordinate hyperplane,
	then $\mu = 0$. In particular, the function $F = 0$ describes $\mu$ on $\cQ^2$.
\end{lemma}
\begin{proof}
	Fix $a,b\in\R_+$. Recall that $\mu$ is actually odd with respect to both reflections at the
	coordinate hyperplanes by the $\SLn[2]$-invariance. Therefore, the map $[c,d] \mapsto \mu[-ae_1,be_1,-ce_2,de_2]$  
	is an odd valuation on $\cP_o^1$. By \th\ref{class 1-dim odd} we have
		$$\mu[-ae_1,be_1,-ce_2,de_2]	= \mu[-ae_1,be_1,-e_2,de_2] - \mu[-ae_1,be_1,-e_2,ce_2]. $$
	A similar argument applied to the terms on the right hand side of the last equation shows that
	\begin{eqnarray*}
		\mu[-ae_1,be_1,-ce_2,de_2]	&=& \mu[-e_1,be_1,-e_2,de_2] - \mu[-e_1,ae_1,-e_2,de_2] \\
						& & - \mu[-e_1,be_1,-e_2,ce_2] + \mu[-e_1,ae_1,-e_2,ce_2]
	\end{eqnarray*}
	for all $a,b,c,d \in \R_+$.
	For $x,y\in\R_+$ set $F(x,y) = \mu[-e_1,xe_1,-e_2,ye_2]$. Then
		$$ \mu[-ae_1,be_1,-ce_2,de_2] = F(b,d) - F(a,d) - F(b,c) + F(a,c) .$$
	From the $\SLn[2]$-invariance of $\mu$ and the fact that $\mu$ is odd with respect to reflections at
	coordinate hyperplanes we deduce the following. $F$ is antisymmetric, i.e. $F(x,y) = - F(y,x)$ for all $x,y\in\R_+$, $F(\cdot\,,1) = 0$, $F(1,\cdot) = 0$, and $F(x,1/x) = 0$ for all $x \in \R_+$. Moreover, for $a,b,c,d \in \R_+$ the quantity
		$$F(rb,d/r) - F(ra,d/r) - F(rb,c/r) + F(ra,c/r) $$
	is independent of $r\in\R_+$.
	For simplicity we will consider the function $G(x,y) = F(\exp(x),\exp(y))$ for $x,y \in \R$.
	Clearly, $G$ inherits the properties just established for $F$. Thus $G$ is antisymmetric and
		\begin{equation}\label{eq: properties G}
			G(\cdot\,,0) = 0,\qquad G(0,\cdot) = 0,\quad\textnormal{ and }\quad G(x,-x) = 0 \,\,\textnormal{ for all }x\in\R.
		\end{equation}
	Moreover, for $a,b,c,d \in \R$ the quantity
	\begin{equation}\label{eq: vanish cQ2 odd - functional equation}
	 	G(b+r, d-r) - G(a+r, d-r) - G(b+r, c-r) + G(a+r, c-r)
	 \end{equation}
	is independent of $r \in \R$.
	We have to show that $G = 0$. In order to do so,
	let $k,l \in \mathbb{N}_0$ and $x \in \R$.
	Since the function in \eqref{eq: vanish cQ2 odd - functional equation} is independent of $r$, 
	the choice $a = 0$, $b = kx$, $c = 0$, $d = lx$ for $r = 0$ and $r = -x$, respectively, shows 
	with the aid of \eqref{eq: properties G} that
	\begin{equation}\label{eq: vanish cQ2 odd - induction first equation}
		G(kx,lx) = G \left( (k-1)x, (l+1)x \right) - G \left( -x, (l+1)x \right) - G \left( (k-1)x, x \right).
	\end{equation}
	Similarly, set $a = 0$, $b = kx$, $c = 0$, $d = -lx$ for $r = 0$ and $r = -x$, respectively, to get
	\begin{equation}\label{eq: vanish cQ2 odd - induction second equation}
		G(kx,-lx) = G \left( (k-1)x, -(l-1)x \right) - G \left( -x, -(l-1)x \right) - G \left( (k-1)x, x \right).
	\end{equation}
	We will simultaneously prove $G(kx,lx) = 0$ and $G(kx,-lx) = 0$ for all $k,l \in \mathbb{N}_0$ and each $x \in \R$ by induction over $m = \max(k,l)$.
	For $m=0,1$ this is a direct consequence of \eqref{eq: properties G} and the antisymmetry of $G$.
	Let $m \geq 2$.
	Since $G$ is antisymmetric we can assume without loss of generality that $l \leq k$.
	We have to treat several different cases for $l$.
	If $l = 0$ or $l = k$, then the induction statement again follows from \eqref{eq: properties G} and the antisymmetry of $G$.
	For $l \in \{1,\ldots,k-1\}$ we first take a glance at \eqref{eq: vanish cQ2 odd - induction second equation}.
	Then we can use the induction hypothesis to see that $G(kx,-lx) = 0$.
	Next, we have a look at \eqref{eq: vanish cQ2 odd - induction first equation}.
	For $l = k-1$ we use the induction hypothesis and what we already have shown to get
		$$ G(kx,(k-1)x) = G \left( (k-1)x, kx \right) .$$
	By the antisymmetry of $G$ again, we obtain $G(kx,(k-1)x) = 0$.
	For $l \in \{1,\ldots,k-2\}$, the induction hypothesis and \eqref{eq: vanish cQ2 odd - induction first equation} 
	directly yield $G(kx,lx) = 0$. This finishes the induction. Hence, we conclude that 
		\begin{equation}\label{eq: G almost zero}
			G(kx,lx) = 0 \quad\textnormal{ for all }\,\, k,l\in\mathbb{Z} \,\,\textnormal{ and }\,\, x\in\R.
		\end{equation}	
	Let $y,z\in\mathbb{Q}\backslash\{0\}$. Then there exist $p,q,r,s \in \mathbb{Z}\backslash\{0\}$ with $y = p/q$ and $z = r/s$. Set $k = ps$,
	$l = qr$, and $x=1/qs$. From \eqref{eq: G almost zero} we deduce that $G(y,z)=0$. Thus $G$ vanishes on $\mathbb{Q}^2$.
	In addition, $G$ is upper semicontinuous because $\mu$ is upper semicontinuous. 
	Thus $G$ must be nonnegative.
	But $G$ is also antisymmetric, so it has to vanish on $\R^2$.
\end{proof}

We need an extension of the concept that a function $F$ describes a map $\mu$ on $\cQ^2$. Let $F : \R_+ \to \R$, $k \in \R$, and $\mu \colon \cR^2 \to \R$. We say that $F$ and $k$ describe $\mu$ on $\cR^2$, if
\begin{equation}\label{eq: class cR2 - statement}
		\mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right]
		= F(ac) + F(bc) + F(ad) + F(bd) + k \left( b^{-2} - a^{-2} \right) (x+y)
	\end{equation}
for all $a,b,c,d \in \R_+$ and $x,y \in \R$ with
		\begin{equation}\label{admiss R}
			\left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right] \cap (\R \times \{0\}) = [-ae_1, be_1] .
	  \end{equation}
\begin{lemma}\th\label{class cR2}
	Suppose that $\mu \colon \cR^2 \to \R$ is a measurable $\SLn[2]$-invariant valuation.
	If $F$ describes $\mu$ on $\cQ^2$, then there 
	exists a constant $k \in \R$ such that $F$ and $k$ describe $\mu$ on $\cR^2$.
	
\end{lemma}
\begin{proof} We denote by $A$ the set of all points $(a,b,c,d,x,y) \in \R_+^4 \times \R^2$
	for which (\ref{admiss R}) holds. Let $(a,b,c,d,x,y)\in A$.
	From the valuation property of $\mu$ we infer
	\begin{align*}
		&  \mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right]
		+  \mu \left[ -ae_1, be_1, -t \ptwo y 1, t' \ptwo y 1 \right] \\
		=& \mu \left[ -ae_1, be_1, c \ptwo x {-1}, t' \ptwo y 1 \right]
		+  \mu \left[ -ae_1, be_1, -t \ptwo y 1, d \ptwo y 1 \right]
	\end{align*}
	for sufficiently small $t,t' > 0$. By assumption there exists 
	a function $F \colon \R_+ \to \R$ which satisfies \eqref{eq: class cQ2 even - statement}.
	For this $F$ the $\SLn[2]$-invariance of $\mu$ gives
	\begin{align*}
		&  \mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right]
		+  F(at') + F(bt') \\
		=& \mu \left[ -ae_1, be_1, c \ptwo x {-1}, t' \ptwo y 1 \right]
		+  F(ad) + F(bd) .
	\end{align*}
	This equality shows that the function
		$$ \mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right] - F(ad) - F(bd) $$
	does not depend on $d$.
	Similarly it follows that
		$$ \mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right] - F(ac) - F(bc) $$
	does not depend on $c$.
	Combining the last two statements we obtain that the expression
	\begin{equation}\label{eq: class cR2 - not depend}
		\mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right] - F(ac) - F(bc) - F(ad) - F(bd) 
	\end{equation}
	is independent of $c$ and $d$. Fix $a,b \in \R_+$. Given $x,y \in \R$, choose sufficiently small $c$ and $d$ such that $(a,b,c,d,x,y) \in A$ and set
			$$ f(x,y) = \mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right] - F(ac) - F(bc) - F(ad) - F(bd). $$
	By what we have already shown, the function $f: \R^2 \to \R$ is well defined. Let $\phi \in \SLn[2]$ be the     map given by $\phi e_1 = e_1$ and $\phi e_2 = -y e_1 + e_2$. Note that $\phi (x,-1) = (x+y, -1)$ 
	and $\phi (y,1) = (0,1)$. Moreover, it follows immediately from the definition of the set $A$ that $(a,b,c,d,x,y) \in   A$ if and only if $(a,b,c,d,x+y,0) \in A$.
	By the $\SLn[2]$-invariance of $\mu$ we therefore have 
	\begin{equation}\label{eq: f sln variant}
		f(x,y) = f(x+y,0) \qquad \textnormal{for all } \quad x,y \in \R.
	\end{equation}
	Moreover, if $(a,b,c,d,x,y)\in A$ then
	\begin{align*}
		&  \mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right]
		+  \mu \left[ -ae_1, be_1, -re_2, re_2 \right] \\
		=& \mu \left[ -ae_1, be_1, c \ptwo x {-1}, re_2 \right]
		+  \mu \left[ -ae_1, be_1, -re_2, d \ptwo y 1 \right]
	\end{align*}
	for sufficiently small $r > 0$.
	Relation \eqref{eq: class cQ2 even - statement} and the definition of $f$ yield
		$$ f(x,y) = f(x,0) + f(0,y) $$
	for all $x , y \in \R $. Set $g(x) = f(x,0)$. By \eqref{eq: f sln variant} we get
		$$ g(x+y) = g(x) + g(y) $$
	for all $x , y \in \R $. This is Cauchy's functional equation. Recall that the only measurable solutions of Cauchy's 		functional equation are the linear ones.
	Since the measurability of $\mu$ implies the measurability of $g$, it follows that $g$ has to be linear.
	Thus there exists a $\nu \colon \cP^1_{o} \to \R$ such that for $(a,b,c,d,x,y)\in A$
		\begin{equation}\label{eq: representation mu} 
			\mu \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right]
			= F(ac) + F(bc) + F(ad) + F(bd) + \nu[-a,b] (x+y) .
		\end{equation}
	Since $\mu$ is a valuation, it is easy to verify that $\nu$ is also a valuation.
	For $r > 0$ and suitably small $t,t' > 0$, the $\SLn[2]$-invariance of $\mu$ implies that
		\begin{eqnarray*}
			\mu \left[ -rae_1, rbe_1, t \ptwo x {-1}, t' \ptwo y 1 \right]
			&=&  \mu \left[ -ae_1, be_1, t \ptwo {\frac x r} {-r}, t' \ptwo {\frac y r} r \right]\\
			&=&  \mu \left[ -ae_1, be_1, rt \ptwo {\frac x {r^2}} {-1}, rt' \ptwo {\frac y {r^2}} 1 \right].
		\end{eqnarray*}	
	This and \eqref{eq: representation mu} show that $\nu$ is positively homogeneous of degree $-2$.
	The $\SLn[2]$-invariance of $\mu$ also implies that $\nu$ is odd.
	By \th\ref{class 1-dim odd} there exists a function $G : \R_+ \to \R$ with
		\begin{equation}\label{eq: nu g} \nu[-a,b] = G(b) - G(a) .\end{equation}
	and $G(1) = 0$. The homogeneity of $\nu$ implies that
	\begin{align*}
		G(rb) - G(ra) &= \nu[-ra,rb] = r^{-2} \nu[-a,b] = r^{-2} G(b) - r^{-2} G(a)
	\end{align*}
	for all $a,b,r \in \R_+$.	Now take $a = 1$ in order to arrive at
		$$ G(rb) = r^{-2} G(b) + G(r) .$$
	By symmetry we also have
		$$ G(rb) = b^{-2} G(r) + G(b) $$
	for all $b,r \in \R_+$. Combining these two equations leads to
	
		$$r^{-2} G(b) + G(r) = b^{-2} G(r) + G(b) . $$
	for all $b,r \in \R_+$. Choose $b=\sqrt{2}$. An elementary calculation shows that
		$$ G(r) = 2(1- r^{-2})G(\sqrt{2})$$
	for all $r \in \R_+$.
	It follows from \eqref{eq: nu g} that there exists a constant $k \in \R$ such that
		$$ \nu[-a,b] = k \left( b^{-2} - a^{-2} \right) .$$
\end{proof}
The next lemma provides an explicit description of the function $F$ from \th\ref{class cR2}.
\begin{lemma}\th\label{class cR2 improved}
	Suppose that $\mu \colon \cR^2 \to \R$ is a measurable $\SLn[2]$-invariant valuation 
	and let $F \colon \R_+ \to \R$ and $k\in\R$ describe $\mu$ on $\cR^2$.
	Then there exist constants $c_1, c_2, c_3 \in \R$ such that
		$$ F(r) = \frac {c_1} r + c_2 + c_3 r $$
	for all $r > 0$, where $c_1 = -2k$.
\end{lemma}
\begin{proof}
	For $a,b,r,s\in\R_+$ consider the convex polytope
		$$ S = \left[ -r e_1, -s e_2, \ptwo a b \right] .$$
	Observe that $S$ can be rewritten as
		$$ S = \left[ -r e_1, \frac {sa} {s+b} e_1, s \ptwo 0 {-1}, b \ptwo {\frac a b} 1\right]\quad\textnormal{and}\quad S = \left[ -s e_2, \frac {rb} {r+a} e_2, a         \ptwo 1 {\frac b a}, r \ptwo {-1} 0 \right] .$$
	Using the $\SLn[2]$-invariance of $\mu$ and \eqref{eq: class cR2 - statement}
	we can therefore calculate $\mu(S)$ in two different ways, namely
		$$ \mu(S) = F(rs) + F \left( \frac {s^2a} {s+b} \right) + F(rb) + F \left( \frac {sab} {s+b} \right)
		+ k \left( \left( \frac {sa} {s+b} \right)^{-2} - r^{-2} \right) \frac a b $$
	and
		$$ \mu(S) = F(sa) + F \left( \frac {rab} {r+a} \right) + F(sr) + F \left( \frac {r^2b} {r+a} \right)
		+ k \left( \left( \frac {rb} {r+a} \right)^{-2} - s^{-2} \right) \frac b a .$$
	This yields the following functional equation
	\begin{align*}
		&  F \left( \frac {s^2a} {s+b} \right) + F \left( \frac {sab} {s+b} \right) - F(sa)
		+  k \left( \left( \frac {sa} {s+b} \right)^{-2} \frac a b + s^{-2} \frac b a \right) \\
		=& F \left( \frac {r^2b} {r+a} \right) + F \left( \frac {rab} {r+a} \right) - F(rb)
		+ k \left( \left( \frac {rb} {r+a} \right)^{-2} \frac b a + r^{-2} \frac a b \right) 
	\end{align*}
	for all $a,b,r,s\in\R_+$. An elementary calculation shows that $r \mapsto \frac {c_1} r$ with $c_1 := -2k$ solves the 	above equation. Moreover, this functional equation is an inhomogeneous linear one. Thus, the function $G(r)=F(r)-\frac{c_1}{r}$ 		  solves 
	\begin{equation*}
		  G \left( \frac {s^2a} {s+b} \right) + G \left( \frac {sab} {s+b} \right) - G(sa)
		= G \left( \frac {r^2b} {r+a} \right) + G \left( \frac {rab} {r+a} \right) - G(rb).
	\end{equation*}
	Note that the left hand side of this equation only depends on $s$, while the right hand side only depends on $r$,
	hence they must be a function depending only on $a,b$. So
	\begin{equation}\label{eq: G}
		G \left( \frac {s^2a} {s+b} \right) + G \left( \frac {sab} {s+b} \right) - G(sa) = g(a,b)
	\end{equation}
	for some $g:\R_+^2\to\R$ and all $a,b,s\in\R_+$. Now choose $s = b$ in order to arrive at
		$$ G \left( \frac {ab} 2 \right) + G \left( \frac {ab} 2 \right) - G(ab) = g(a,b) .$$
	We see that $g(a,b) = h(ab)$ for some $h \colon \R_+ \to \R$.
	Let $x,y\in\R_+$. Take $s = 1$, $a = x+y$ and $b = \frac x y$ in \eqref{eq: G} to get
		\begin{equation}\label{eq: G cauchy}
			G(y) + G(x) - G(x+y) = h \left( \frac {x(x+y)} y \right). 
		\end{equation}	
	Choosing $s = 1$, $a = x+y$ and $b = \frac y x$ in \eqref{eq: G} yields
		$$ G(x) + G(y) - G(x+y) = h \left( \frac {y(x+y)} x \right) .$$
	Combining the last two equations gives
		$$ h \left( \frac {x(x+y)} y \right) = h \left( \frac {y(x+y)} x \right) $$
	for all $x,y\in\R_+$. In particular, for $x\in(0,1)$ and $y = \frac {x^2} {1-x}$ we therefore have
		$$ h(1) = h \left( \frac {x^2} {(1-x)^2} \right) .$$
	Since $\lim_{x \to 0^+} \frac {x^2} {(1-x)^2} = 0$ and $\lim_{x \to 1^-} \frac {x^2} {(1-x)^2} = +\infty$,
	we infer that $h$ is constant.
	Set $c_2 = h(1)$. Then by \eqref{eq: G cauchy} the function $r \mapsto G(r) - c_2$ solves Cauchy's functional equation.
	Since $\mu$ is measurable, $G - c_2$ is measurable, too. Thus there exists a constant $c_3$ such that $G(r)=c_2+rc_3$. Finally, recall that $G(r)=F(r)-\frac{c_1}{r}$. Thus $F(r) = \frac {c_1} r + c_2 + c_3 r$.
\end{proof}
Now, we are in a position to establish \th\ref{main2} in dimension two. First, we will settle the even case. Afterwards, the odd case will be treated and finally these two results will be combined to obtain the assertion in general.
\begin{theorem}\th\label{class cP2 even}
	Suppose that $\mu \colon \cP^2_{o} \to \R$ is a measurable and $\SLn[2]$-invariant valuation
	which is even with respect to the reflection at a coordinate hyperplane.
	Then there exist constants $c_1, c_2, c_3 \in \R$ such that
		$$ \mu(P) = c_1 V_2(P^*) + c_2 + c_3 V_2(P) $$
	for all $P \in \cP^2_{o}$.
\end{theorem}
\begin{proof} Let $P\in \cR^2$ be given by
		$$ P = \left[ -ae_1, be_1, c \ptwo x {-1}, d \ptwo y 1 \right] $$
	such that $P \cap (\R\times\{0\}) = [-ae_1, be_1]$.	In this case, it readily follows that
		$$ V_2(P) = \frac12 \left( ac + bc + ad + bd\right)$$ 
	and
		$$ V_2(P^*) = \frac{1}{ac} + \frac{1}{bc} + \frac{1}{ad} + \frac{1}{bd} 
						- \frac 12 (b^{-2} - a^{-2})(x+y).$$
	From \th\ref{class cQ2 even}, \th\ref{class cR2} and \th\ref{class cR2 improved} we therefore obtain the existence of constants $c_1,c_2,c_3\in\R$ such that
		$$\mu(P) = c_1 V_2(P^*) + c_2 + c_3 V_2(P). $$
	Since every $P\in\cR^2$ has, up to a transformation in $\SLn[2]$, a representation of the form considered above, the last equality actually holds for all $P\in\cR^2$. 
	An application of \th\ref{Ludwig Extension} concludes the proof.
\end{proof}
\begin{theorem}\th\label{class cP2 odd}
	If $\mu \colon \cP^2_{o} \to \R$ is an upper semicontinuous $\SLn[2]$-invariant valuation
	which is odd with respect to the reflection at a coordinate hyperplane,
	then $\mu\equiv 0$.
\end{theorem}
\begin{proof}
	It remains to remove the invariance property with respect to reflections at coordinate hyperplanes from the last theorem. 
  By \th\ref{vanish cQ2 odd} and \th\ref{class cR2} there exists a constant $k\in\R$ such that $F \equiv 0$
	and $k$ describe $\mu$ on $\cR^2$. From \th\ref{class cR2 improved} we deduce the existence of constants $c_2$ and
	$c_3$ with $-2k/r+c_2+c_3r=0$ for every $r \in \R_+$. Thus $k=c_2=c_3=0$ and hence $\mu(P) = 0$ for every 	   
	$P\in\cR^2$. \th\ref{Ludwig Extension} implies that $\mu(P) = 0$ for every $P\in\cP_o^2$. 
\end{proof}	

\begin{theorem}\th\label{class cP2}
	If $\mu \colon \cP^2_{o} \to \R$ is an upper semicontinuous and $\SLn[2]$-invariant valuation,
	then there exist constants $c_1, c_2, c_3 \in \R$ such that
		$$ \mu(P) = c_1 + c_2 V_2(P) + c_3 V_2(P^*) $$
	for all $P \in \cP^2_{o}$.
\end{theorem}
\begin{proof}
	Write $\mu = \mu_+ + \mu_-$, where $\mu_+$ and $\mu_-$ are even and odd with respect to
	the reflection at a coordinate hyperplane, respectively. It follows directly from the definition of 
	$\mu_+$ and $\mu_-$ that both are $\SLn[2]$-invariant valuations. Moreover, since $\mu$ is upper semicontinuous, 
	it is measurable and hence $\mu_+$ and $\mu_-$ are also measurable.
	\th\ref{class cP2 even} implies in particular that $\mu_+$ is continuous.
	Being the difference of an upper semicontinuous and a continuous map, $\mu_-$ is upper semicontinuous.
	The desired result is now a direct consequence of \thref{class cP2 even} and \thref{class cP2 odd}.
\end{proof}
\subsection{The $n$-dimensional Case}
The next result reveals that an $\SLn$-invariant valuation which vanishes on doublepyramids, actually vanishes on $\cR^n$.
\begin{lemma}\th\label{vanish cRn}
	Let $n \geq 3$. If $\mu \colon \cR^n \to \R$ is a measurable $\SLn$-invariant valuation such that
		\begin{equation}\label{eq: mu vanish}
			\mu[P,-ae_n,be_n] = 0 \qquad \textnormal{ for all } P \in \cP_o^{n-1} \textnormal{ and } a,b \in \R_+,
		\end{equation}	
	i.e. it vanishes on $\cQ^n(x_n)$, then $\mu$ vanishes on $\cR^n$.
\end{lemma}
\begin{proof}
	We denote by $A$ the set of all $(P,c,d,x,y) \in
	\cP_o^{n-1}\times\R_+^2 \times \R^{2(n-1)}$ 
	for which 
		\begin{equation}\label{admiss R ndim}
			\left[ P, c \ptwo x {-1}, d \ptwo y 1 \right] \cap e_n^{\bot} = P \times \{0\} 
		\end{equation}
	holds. Let $(P,c,d,x,y) \in A$.
	From the valuation property of $\mu$ we infer
	\begin{align*}
		&  \mu \left[ P, c \ptwo x {-1}, d \ptwo y 1 \right] + \mu \left[ P, -t \ptwo y 1, t' \ptwo y 1 \right] \\
		={}& \mu \left[ P, c \ptwo x {-1}, t' \ptwo y 1 \right] + \mu \left[ P, -t \ptwo y 1, d \ptwo y 1 \right]
	\end{align*}
	for suitably small $t,t' > 0$.
	Because $\mu$ is $\SLn$-invariant and vanishes on $\cQ^n(x_n)$, we have
		$$ \mu \left[ P, c \ptwo x {-1}, d \ptwo y 1 \right] = \mu \left[ P, c \ptwo x {-1}, t' \ptwo y 1 \right]. $$
	This shows that the quantity
	\begin{equation*}
		\mu \left[ P, c \ptwo x {-1}, d \ptwo y 1 \right]
	\end{equation*}
	does not depend on $d \in \R_+$.
	Similarly we see that it does not depend on $c \in \R_+$ either.
	Fix $P\in\cP_o^{n-1}$. Given $x,y \in \R^{n-1}$, choose sufficiently small $c$ and $d$ such that $(P,c,d,x,y) \in A$ and set
		$$ f(x,y) = \mu \left[ P, c \ptwo x {-1}, d \ptwo y 1 \right] .$$
	By what we have shown before, the function $f: \R^{2(n-1)} \to \R$ is well defined. Let $\phi \in \SLn$ be the 
	map given by 
		$$\phi e_n = -y_1e_1 - \cdots  -y_{n-1}e_{n-1}+e_n, \quad \phi e_i = e_i \textnormal{ for } 1\leq i\leq n-1 .$$
	Note that $\phi (x,-1) = (x+y, -1)$ 
	and $\phi (y,1) = (0,1)$. Moreover, it follows immediately from the definition of the set $A$ that $(P,c,d,x,y) \in   A$
	if and only if $(P,c,d,x+y,0) \in A$.
	By the $\SLn$-invariance of $\mu$ we therefore have 
	\begin{equation}\label{eq: f sln variant n-dim}
		f(x,y) = f(x+y,0) \qquad \textnormal{for all } \quad x,y \in \R^{n-1}.
	\end{equation}
	If $(P,c,d,x,y)\in A$, then the valuation property of $\mu$ yields
	  $$\mu \left[ P, c\ptwo x {-1}, d\ptwo y 1 \right]	+  \mu \left[ P, -r e_n, r e_n \right] 
			= \mu \left[ P, c\ptwo x {-1}, r e_n \right] +  \mu \left[ P, -r e_n, d \ptwo y 1 \right] 
	  $$
	for suitably small $r \in \R_+$.
	By the definition of $f$, \eqref{eq: f sln variant n-dim}, and the assumption that $\mu$ vanishes on $\cQ^n(x_n)$ we obtain
		$$ f(x+y,0) = f(x,0) + f(0,y) $$
	for all $x , y \in \R^{n-1} $. Set $g(x) = f(x,0)$. Since $f(0,y) = f(y,0)$ by \eqref{eq: f sln variant n-dim}, we have
		$$ g(x+y) = g(x) + g(y) $$
	for all $x , y \in \R^{n-1} $. This is Cauchy's functional equation. Recall that the only measurable solutions of Cauchy's 		functional equation are the linear 	 ones.
	Since the measurability of $\mu$ implies the measurability of $g$, it follows that $g$ has to be linear.
	Thus there exists a $\nu \colon \cP^{n-1}_{o} \to \R^{n-1}$ such that for $(P,c,d,x,y)\in A$
		\begin{equation}\label{eq: def nu n-dim}
		   \mu \left[ P, c\ptwo x {-1}, d\ptwo y 1 \right] = \s{\nu(P)}{x+y} .
		\end{equation}   
	Since $\mu$ is a measurable valuation, it is easy to verify that $\nu$ is also a measurable valuation.
	The $\SLn$-invariance of $\mu$ implies that
		\begin{eqnarray*}
			\mu \left[ r P, c\ptwo x {-1}, d\ptwo y 1 \right]
		  &=&  \mu \left[ P, c\ptwo {r^{-1} x} {-r^{n-1}}, d \ptwo {r^{-1} y} {r^{n-1}} \right]\\
		  &=&  \mu \left[ P, r^{n-1} c \ptwo {r^{-n} x} {-1}, r^{n-1} d \ptwo {r^{-n} y} 1 \right] 
		\end{eqnarray*}  
	for all $r \in \R_+$, each $x,y \in \R^{n-1}$ and sufficiently small $c,d \in \R_+$.
	From \eqref{eq: def nu n-dim} we deduce that $\nu$ is $(-n)$-homogeneous.
	Moreover, the $\SLn$-invariance of $\mu$ also implies that $\nu(\phi P) = \phi^{-t} \nu(P)$ for all 
	$\phi \in \SLn[n-1]$. Consequently,
	\begin{equation}\label{eq: nu sln contravariant}
		\nu(\phi P) = |\det \phi^{-t}| \phi^{-t} \nu (P)
	\end{equation}	
	for all $P \in \cP_o^{n-1}$ and each $\phi \in \GLn[n-1]$ with positive determinant. For a map $\phi \in \GLn[n-1]$ denote by $\hat{\phi} \in \GLn[n]$ the extension given by $\hat{\phi}|_{\R^{n-1}}=\phi$ and $\hat{\phi}e_n= -e_n$. Let $\psi \in \GLn[n-1]$
	be defined by $\psi e_1 = -e_1$ and $\psi e_i = e_i$ for $2 \leq i \leq n-1$. Choose $x=o$ and $y=e_1$. 
	Then \eqref{eq: def nu n-dim} and the $\SLn$-invariance of $\mu$ with respect to $\hat{\psi}$ imply for the first component of $\nu$
	that $\nu_1 (\psi P)=- \nu_1 (P)$. 
	Suppose that $\phi \in \GLn[n-1]$ has negative determinant. Then $\psi \phi $ has positive determinant and by 
	\eqref{eq: nu sln contravariant} we obtain
		$$ \nu_1 (\phi P) = - \nu_1(\psi \phi P) = - |\det \phi^{-t}| \left( \psi \phi^{-t} \nu (P) \right)_1 = 
		   |\det \phi^{-t}| \left( \phi^{-t}\nu (P) \right)_1. $$
	Similarly, the last relation holds for the other components of $\nu$. Thus \eqref{eq: nu sln contravariant} holds for each 
	$\phi \in \GLn[n-1]$. \th\ref{Ludwig moment} implies that $\nu = c' m^*$ for some constant $c' \in \R$.
	Let $P = [P', -e_{n-1}, e_{n-1}]$ for some $P' \in \cP^{n-2}_o$ and choose suitably small vectors $x,y \in\R^{n-1}$ with
	$x_{n-1}, y_{n-1} = 0$. On the one hand, by the $\SLn$-invariance of $\mu$ and \eqref{eq: def nu n-dim} we get for suitable 
	$\tilde{x},\tilde{y} \in \R^{n-1}$
	  $$ \mu \left[ P, \ptwo x {-1}, \ptwo y 1 \right]
		   = \mu \left[ P', -e_n, e_n, \ptwo {\tilde x} 0, \ptwo {\tilde y} 0 \right] 
		   = \s {\nu \left[ P', \ptwo {\tilde x} 0, \ptwo {\tilde y} 0 \right]} {o} 
		   = 0 . $$
	On the other hand, we have
		$$ \mu \left[ P, \ptwo x {-1}, \ptwo y 1 \right] = 
		   \mu \left[ P', -e_{n-1}, e_{n-1}, \ptwo x {-1}, \ptwo y 1 \right]
		   =  \s {\nu \left[ P', -e_{n-1}, e_{n-1} \right]} {x+y} .$$
	Therefore we arrive at
		\begin{equation}\label{eq: cm zero}
		   c' \s {m^* \left[ P', -e_{n-1}, e_{n-1} \right]} {x+y} = 0 
		\end{equation}
	for all $P' \in \cP^{n-2}_{o}$ and suitably small vectors $x,y \in \R^{n-1}$ with $x_{n-1}, y_{n-1} = 0$. Now
	take 
		$$P' = [-e_1, 2e_1, -e_2, e_2, \ldots, -e_{n-2}, e_{n-2}].$$
	Using \eqref{eq: polar product} repeatedly, we see that $(P')^* = [-1,1/2]\times[-1,1]\times\ldots\times[-1,1]$. Again by
	\eqref{eq: polar product} we arrive at
		$$m^* \left[ P', -e_{n-1}, e_{n-1} \right] = -\tfrac38 e_1.$$
	Relation \eqref{eq: cm zero} proves $c'=0$.
	Hence $\nu \equiv 0$ and a glance at \eqref{eq: def nu n-dim}, together with the $\SLn$-invariance of $\mu$, concludes the proof.
\end{proof}
Inductively, we will now prove \th\ref{main2} in dimensions greater or equal than three. We will thereby proceed as in the two dimensional case.
\begin{theorem}\th\label{class cPn even}
	Let $n \geq 2$.
	If $\mu \colon \cP^n_{o} \to \R$ is a measurable $\SLn$-invariant valuation
	which is even with respect to the reflection at a coordinate hyperplane,
	then there exist constants $c_1, c_2, c_3 \in \R$ such that
		$$ \mu(P) = c_1 V_n(P^*) + c_2 + c_3 V_n(P) $$
	for all $P \in \cP^n_{o}$.
\end{theorem}
\begin{proof}
	By the $\SLn$-invariance, $\mu$ is actually even with respect to each coordinate hyperplane. 
	We proceed by induction on the dimension $n$. For $n = 2$, 	 the assertion is an immediate consequence 
	of \th\ref{class cP2 even}. Assume that the theorem holds in dimension $n-1$. For $P\in\cP_o^{n-1}$, define 
	$\nu (P)=\mu[P,-e_n,e_n]$. Clearly, $\nu$ is a measurable $\SLn[n-1]$-invariant valuation which is even with 
	respect to the reflection at a coordinate hyperplane. By the induction assumption there exist constants
	$\tilde{c}_1,\tilde{c}_2,\tilde{c}_3\in\R$ such that 
		$$ \nu(P) = \tilde{c}_1 V_{n-1}(P^*) + \tilde{c}_2 + \tilde{c}_3 V_{n-1}(P) $$
	for all $P \in \cP^{n-1}_{o}$. From \eqref{eq: polar product} and Fubini's theorem we obtain that
		$$ \mu[P,-e_n,e_n] = \frac{\tilde{c}_1}{2} V_{n}([P,-e_n,e_n]^*) + \tilde{c}_2 + \frac{n \tilde{c}_3}{2} V_{n}[P,-e_n,e_n]. $$
	Set $c_1 = \tilde{c}_1/2$, $c_2=\tilde{c}_2$, and $c_3=n\tilde{c}_3/2$. The map
		$$ \rho(P) = \mu(P) - c_1 V_n(P^*) - c_2 -c_3 V_n(P),\qquad P \in \cP_o^n, $$
	is a measurable $\SLn$-invariant valuation which is even with respect to the reflection at each coordinate
	hyperplane. Moreover, we know that $\rho[P,-e_n,e_n] = 0$ for all $P \in \cP_o^{n-1}$. So by the $\SLn$-invariance
	of $\rho$ we have
		\begin{equation}\label{eq: rho vanish}
			\rho[P,-ae_n,ae_n] = 0 \qquad \textnormal{ for all } P \in \cP_o^{n-1} \textnormal{ and } a \in \R_+.
		\end{equation}
	Note that, for each $P \in \cP_o^{n-1}$, the map $[-a,b] \mapsto \rho[P,-ae_n,be_n]$ is an even valuation on $\cP_o^1$. From 
	\th\ref{class 1-dim even} we obtain
		$$ \rho[P,-ae_n,be_n] = \tfrac12 \rho[P,-ae_n,ae_n] + \tfrac12 \rho[P,-be_n,be_n] .$$
	By \eqref{eq: rho vanish} we therefore have $\rho[P,-ae_n,be_n] = 0$ for each $P \in \cP_o^{n-1}$ and all $a,b \in \R_+$.
	Thus \thref{vanish cRn} implies that $\rho(P) = 0$ for all $P\in\cR^n$. By \thref{Ludwig Extension} we finally get 
	$\rho \equiv 0$, which immediately gives the desired result.
\end{proof}
\begin{theorem}\th\label{class cPn odd}
	Let $n \geq 2$.
	If $\mu \colon \cP^n_{o} \to \R$ is an upper semicontinuous $\SLn$-invariant valuation
	which is odd with respect to the reflection at a coordinate hyperplane,
	then $\mu\equiv 0$.
\end{theorem}
\begin{proof}
	By the $\SLn$-invariance, $\mu$ is actually odd with respect to each coordinate hyperplane. 
	We proceed by induction on the dimension $n$. For $n = 2$, the assertion is an immediate consequence 
	of \th\ref{class cP2 odd}. Assume that the theorem holds in dimension $n-1$. Fix $a,b \in \R_+$. 
	For $P\in\cP_o^{n-1}$, define $\nu (P)=\mu[P,-ae_n,be_n]$. Clearly, $\nu$ is an upper
	semicontinuous $\SLn[n-1]$-invariant valuation which is odd with 
	respect to the reflection at a coordinate hyperplane. By the induction assumption we therefore have 
	$\nu \equiv 0$. Thus $\mu[P,-ae_n,be_n] = 0$ for each $P \in \cP_o^{n-1}$ and all $a,b \in \R_+$.
	\thref{vanish cRn} implies that $\mu(P) = 0$ for all $P\in\cR^n$. By \thref{Ludwig Extension} we finally get 
	$\mu \equiv 0$.
\end{proof}	
As in the two dimensional case, one can combine the last two theorems in order to obtain the desired result.
\begin{theorem}\th\label{class cPn}
	Let $n \geq 2$.
	If $\mu \colon \cP^n_{o} \to \R$ is an upper semicontinuous $\SLn$-invariant valuation,
	then there exist constants $c_1, c_2, c_3 \in \R$ such that
		$$ \mu(P) = c_1 +c_2V_n(P)+ c_3 V_n(P^*) $$
	for all $P \in \cP^n_{o}$.
\end{theorem}
Thus, we proved \th\ref{main2} in all dimensions greater or equal than two. Finally, let us show how this result implies \th\ref{main1}. 

\vspace{0.4cm}
\noindent\emph{Proof of \th\ref{main1}.} Suppose that $\mu:\cK_o^n \to \R$
is an upper semicontinuous and $\SLn$-invariant valuation. By \th\ref{main2}, there exist constants
$c_0,c_1,c_2\in\R$ such that 
 $$\nu(K):= \mu(K) - c_0 V_0(K) - c_1 V_n(K) - c_2 V_n(K^*)$$
vanishes on $\cP_o^n$. Moreover, $\nu$ is an upper semicontinuous and $\SLn$-invariant valuation. From 
\th\ref{Ludwig Reitzner Orlicz} we infer that there exists a function $\varphi\in\textnormal{Conc}(\R_+)$ such that
$\nu (K)=\Omega_{\varphi} (K)$ for all $K\in\cK_o^n$. The definition of $\nu$ concludes the proof of \th\ref{main1}. \qed
