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I am reading about the orthogonal group $O(V)$ of a real finite dimensional quadratic vector space $(V,Q)$ with $Q$ nondegenerate. By definition $$O(V)=\{f:V\mapsto V |\quad Q(f(v))=Q(v) \quad \forall v\in V\}.$$ I don't know if this definition is enough to derive that $O(V)\subset GL(V)$. Also can we imply from definition that $|\det(f)|=1$ for every $f\in V$ ? (For the case $V$ is positive definite, i.e. the bilinear form associated to $Q$ is an inner product, it's true, but I don't know in general case.)

Thanks for any help!

Seirios
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mapping
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1 Answers1

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A quadratic form $Q$ can be given by a symmetric matrix $A$ as $$Q(x)=x^TAx\,.$$ This being nondegenerate means that $A$ is invertible, i.e. $\det A\ne 0$.

As you commented, $Q$ determines a nondegenerate (but not necessarily positive definite) bilinear form $B$: $$B(x,y):=\frac12(Q(x+y)-Q(x)-Q(y))\,.$$ Thus, $B(x,y)=x^TAy$.

If $f:V\to V$ is linear and preserves $Q$ (i.e. $f\in O(V,Q)$), then it also preserves $B$, so, also writing $f$ for its matrix (w.r.t. a fixed basis of $V$), $$x^Tf^TAfy=(fx)^TA(fy)=x^TAy$$ for all $x,y$. Applying it to the basis vectors, they imply that $f^TAf=A$. So, we also have $$(\det f)^2\cdot\det A=\det A$$ and hence $|\det f|=1$ as $\det A\ne 0$.

Berci
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  • By definition of quadratic form, we don't have $Q(x)>= 0$, so how can we define the norm on $V$ as you did? – mapping Mar 19 '13 at 11:04
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    Anyway $Q$ defines a bilinear form (but not positive definite) $$B(x,y)=\frac{1}{2}(Q(x+y)-Q(x)-Q(y).$$ So if $f\in V$ then we have $B(f(x),f(y))=B(x,y)$. But we don't know if $B$ is positive definite, I don't know if we can derive $f$ is linear, and injective or not. – mapping Mar 19 '13 at 11:14
  • Oh, we don't have $Q(x)\ge 0$? I had in mind so.. I will edit my answer accordingly.. Also, you mean $f\in O(V)$ instead of $f\in V$, don't you? – Berci Mar 19 '13 at 11:15
  • Excuse me that's what I mean. Thanks for your correction. Additionally, can we prove that $f$ is linear with only condition of preserving the quadratic form? – mapping Mar 19 '13 at 13:15
  • It's a good question. I think, it doesn't follow. Probably, if continuous, but I would also doubt this, too. – Berci Mar 19 '13 at 13:36