Isometries of a hyperbolic quadratic form I am reading an article that says "The group of isometries (of a hyperbolic space) of a hyperbolic quadratic form in two variables is isomorphic to the semi-direct product $\mathbb{R} \rtimes \mathbb{Z}/2\mathbb{Z}$".
Could someone help me to understand that fact? First, it seems that the semi-direct product $\mathbb{R} \rtimes \mathbb{Z}/2\mathbb{Z}$ is equal to the direct product $\mathbb{R} \times \mathbb{Z}/2\mathbb{Z}$. And, what means a hyperbolic quadratic form? Is it simply a quadratic form defined in a hyperbolic space?
Thanks a lot.
 A: A hyperbolic quadratic form is a quadratic form such that the matrix, in a suitable basis, is 
$$
A =
 \left(  \begin{array}{rr}
  0  &  1       \\
  1  &  0       
\end{array}
  \right).
  $$
A two dimensional vector space over $\mathbb R$ with this bilinear form is often called a hyperbolic plane. Given two column vectors $x,y$ the value of the bilinear form is just
$B(x,y) = y^t A x.$ The associated quadratic form is $q(x) = x^t A x = B(x,x).$ We are using this way of writing the quadratic form as in page 13 of GERSTEIN and page 31 of GROVE. The result, for fields not of characteristic 2, is $$ B(x,y) = \frac{q(x+y) - q(x) - q(y)}{2} $$  
Let's see, you did not type the result correctly. The orthogonal group, or group of isometries, or automorphism group, or group of automorphs of $q$ are those matrices $T$ of deteminant $\pm 1$ such that
$$ T^t A T = A,   $$ page 26 of Gerstein. 
You can work this part out yourself: the isometries with positive determinant, also called the special orthogonal group or the rotation group, are 
 $$
T =
 \left(  \begin{array}{rr}
  a  &  0       \\
  0  &  \frac{1}{a}       
\end{array}
  \right)
  $$
for nonzero $a.$
The isometries with negative determinant are 
 $$
T =
 \left(  \begin{array}{rr}
  0  &  b       \\
   \frac{1}{b} & 0       
\end{array}
  \right)
  $$
for nonzero $b.$
The thing you missed in the statement is that it is not $\mathbb R$ being used, it is the multiplicative group $\mathbb R^\ast$ of nonzero reals. Otherwise you have it. This is Theorem 6.1 on page 45 of Grove, or example 2.31 on page 27 of Gerstein. 
The two other books I would recommend here are LAM and my favorite (because of number theoretic information) CASSELS 
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page 45 from Grove. If anyone knows how to crop the image, please feel free.

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