Notions:
- A quadratic form over a field $F$ with characteristic $≠2$ is a vector space $V$ with an application $q:V\to F$ such that:
a) $q(\lambda x)=\lambda ^2q(x)$ for $(\lambda,x)\in F\times V$
b) $(x,y)\mapsto q(x+y)-q(x)-q(y)$ is bilinear
- The bilinear form $\tilde{q}(x,y)=\frac12(q(x+y)-q(x)-q(y))$ is called polar form of $q$: we have $q(x)=\tilde{q}(x,x)$
- A morphism $(V,q)\to(V',q')$ is a linear application $f:V\to V'$ s.t. $q'\circ f=q$. The non-trivial morphisms between quadratic forms are essentially the isomorphisms, also called isometries.
- For any quadratic form $q$, we write $O(q)$ the group of isometries of $q$.
I don't unerstand the following example (example 1.1.8. page 14/305 of this French book https://webusers.imj-prg.fr/~bruno.kahn/preprints/fqbook3.pdf):
Example: Let $x\in V$ s.t. $q(x)\not=0$. Consider $u_x(y)=y-2\frac{\tilde{q}(x,y)}{q(x)}x$. Then $u_x\in O(q)$. In fact, for every $y$, $q(u_x(y))=q(y)-4\frac{\tilde{q}(x,y)}{q(x)}{\tilde{q}(x,y)}+4\frac{\tilde{q}(x,y)^2}{q(x)}q(x)=q(y)$. Let $W$ be the line generated by $x$. On a basis of $V$ adapted to $(W,W^{\perp})$, the matrix of $u_x$ is \begin{bmatrix}-Id&0 \\ 0& Id \end{bmatrix}in particular $u_x^2=Id$
I don't understand this computation: $q(u_x(y))=q(y)-4\frac{\tilde{q}(x,y)}{q(x)}{\tilde{q}(x,y)}+4\frac{\tilde{q}(x,y)^2}{q(x)}q(x)=q(y)$...
How can a general quadratic form be expanded like this?
In particular from last equality I would expect that $q(x)=1$, how can one say that? I thought there is an error in the book, but still I don't know how to compute $q(u_x(y))$ from scratch, can someone help me please?
First of all, there is a typo in the book you cite.
Note that we have that
$$q(x+y)=2\tilde{q}(x,y)+q(x)+q(y).$$
Let $\alpha=2\tilde{q}(x,y)/q(x)$. Then, we have that
$$q(y-\alpha x)=2\tilde{q}(-\alpha x,y)+q(-\alpha x)+q(y).$$
We have that $\tilde{q}(-\alpha x,y)=-\alpha \tilde{q}(x,y)$ and $q(-\alpha x)=\alpha^2 q(x)$.
Expanding this expression yields:
$$q(y-\alpha x)=-4\dfrac{\tilde{q}(x,y)}{q(x)}\tilde{q}(x,y)+4\dfrac{\tilde{q}(x,y)^2}{q(x)^2}q(x)+q(y)=q(y).$$
In the book, there is a square missing at the denominator of the second fraction.