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\title{Analiticka geometrija \textbf{ispitni zadatak}}
\author{Lazar Savic(dobio-ukrao savete-pomoc oko uvoznih listi) }


\begin{document}
\pagestyle{empty}
the coset space. For any measurable subset $E\subset G/H$ , we may easily choose
a measurable function $\delta_E:G\to\mathbb{C}$ so that
%formula 1, viticasta    x[zavrseno]
\begin{equation}
\nonumber
\delta_E(g)=\delta_E^H(gH) = \left\{
  \begin{array}{l l}
    1 & \quad \text{if $gH\in E$,}\\
    0 & \quad \text{if $gH\notin E$.}
  \end{array} \right.
\end{equation}
We may then define an $H$–invariant quotient measure $\tilde{\mu}$ satisfying:
%formula2, integral      x[zavrseno]
\begin{equation}
\nonumber
\tilde{\mu}=\int_G \delta_E(g)d\mu(g) = \int_{G/H} \delta_E^H(gH)d\tilde{\mu}(gH),
\end{equation}
and
%formula3, integral      x[zavrseno]
\begin{equation}
\nonumber
\int_Gf(g)d\mu(g)=\int_{G/H}f^H(gH)d\tilde{\mu}(gH),
\end{equation}
for all integrable functions $f:G\to\mathbb{C}$. \hfill $\square$

\textbf{Remarks} There is an analogous version of Theorem 1.5.1 for left coset spaces $H\backslash G$. Note that we are not assuming that $H$ is a normal subgroup of $G$. Thus $G/H$ (respectively $H\backslash G$) may not be a group.

\textbf{Example 1.5.2} (\textbf{Left invariant measure on} $GL(n,\mathbb{R})/(O(n,\mathbb{R})\cdot \mathbb{R}^x)$)

For $n\geq 2$, we now explicitly construct a left invariant measure on the generalized upper half-plane $\mathfrak{h}^n=GL(n,\mathbb{R})/(O(n,\mathbb{R})\cdot \mathbb{R}^x)$.Returning to the Iwasawa decomposition (Proposition 1.2.6), every $z\in\mathfrak{h}^n$ has a representation in the form $z = xy$ with
%formula4, teska, pmatrix             x[zavrseno]
\begin{equation}
\nonumber
z = \begin{pmatrix}
1 & x_{1,2} & x_{1,3} & \cdots & & x_{1,n}\\
& 1 & x_{2,3}& \cdots & & x_{2,n}\\
& & & \ddots && \vdots\\
&&&&1&x_{n-1,n}\\
&&&&&1
\end{pmatrix},
y=\begin{pmatrix}
y_1y_2\cdots y_{n-1}&&&&\\
&y_2y_3\cdots y_{n-2}&&&\\
&&\ddots&&\\
&&& n &\\
&&&&1
\end{pmatrix},
\end{equation}
with $x_{i,j}\in\mathbb{R}$ for $1\leq i\leq j\leq n$ and $y_i>0$ $1\leq i\leq n$. Let $d^*z$ denote the left invariant measure on $\mathfrak{h}^n$. Then $d^*z$ has the property that
%formula 5, kratka                     x[zavrseno]
\begin{equation}
\nonumber
d^*(gz)=d^*z
\end{equation}
for all $g\in GL(n,\mathbb{R}).$

\textbf{Proposition 1.5.3} \emph{The left invariant $GL(n,\mathbb{r})-$measure $d^*z$ on $\mathfrak{h}^n$ can be given explicitly by the formula \[d^z=d^*xd^*y\] where
%formula 6, Perioda nad, ispod         x[zavrseno]
\[d^*x = \prod_{1\leq i\leq j\leq n}dx_{i,j}, \>\>\> d^*y = \prod_{k=1}^{n-1}y_k^{-k(n-k)-1}dy_k. \tag{1.5.4}\]}

For example, for $n=2$, with $z = \begin{pmatrix}
y & x\\
0 & 1
\end{pmatrix}$,we have $d^*z = \frac{dxdy}{y^2}$ while for $n=3$ with
%formula 7, 3*3 pmatrica               x[zavrseno]
\begin{equation}
\nonumber
z= \begin{pmatrix}
y_1y_2 & x_{1,2} y_1 & x_{1,3}\\
0 & y_1 & x_{2,3}\\
0&0&1
\end{pmatrix},
\end{equation}
we have
%formula 8, d*z                        x[zavrseno]
\begin{equation}
\nonumber
d^*z = dx_{1,2}dx_{1,3}dx_{2,3}\frac{dy_1dy_2}{(y_1y_2)^3}.
\end{equation}
\emph{Proof} We sketch the proof. The group $GL(n,\mathbb{R})$ is generated by diagonal matrices, upper triangular matrices with 1s on the diagonal, and the Weyl group $W_n$ which consists of all $n\times n$ matrices with exactly one 1 in each row and column and zeros everywhere else. For example,
%poravnaj W2 W3;                       x[zavrseno]
\begin{align*}
W_2 &= \left\lbrace \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}, \begin{pmatrix}0&1\\1&0\end{pmatrix} \right\rbrace,\\
W_3 & = \left\lbrace \begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{pmatrix}, \begin{pmatrix}
1&0&0\\
0&0&1\\
0&0&1
\end{pmatrix}, \begin{pmatrix}
0&1&0\\
1&0&0\\
0&0&1
\end{pmatrix}, \right.\\
&\>\>\>\>\>\>\>\>\>\>\left. \begin{pmatrix}
0&1&0\\
0&0&1\\
1&0&0
\end{pmatrix},\begin{pmatrix}
0&0&1\\
1&0&0\\
0&1&0
\end{pmatrix},\begin{pmatrix}
0&0&1\\
0&1&0\\
1&0&0
\end{pmatrix}\right\rbrace
\end{align*}
%poslednji tekst ubaciti              x[zavrseno]
Note that the Weyl group $W_n$ has order $n!$ and is simply the symmetric group on $n$ symbols. It is clear that $d^*(gz)=d^*z$ if $g$ is an upper triangular matrix with 1s on the diagonal. This is because the measures $dx_{i,j}$ (with $1\leq i<j\leq n$)are all invariant under translation. It is clear that the differential $d^*z$ is $Z_n-$ invariant where $Z_n\cong \mathbb{R}^x$ denotes the center of $GL(n,\mathbb{R})$.So, without loss of generality,
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