Orthogonal group isomorphisam

Question

First we take identity element from set which is Identity matrix so S=I for which b(σ(x),σ(y))=b(x,y) which is identity transformation in O(V,b) so kernal becomes identity and so it is surjective since it is finite dimensional. Is it correct or not please guide regarding this.

Answer 1

$O(V,b)$ is the set of linear maps $\sigma$ that verify $b(\sigma(x),\sigma(y))=b(x,y)$ for every $x,y\in V$ and $B$ is the matrix associated to $b$ in the canonical base (for example) Now you can prove that when you fix a base on $V$

$b(x,y)=x^tBy$

And so if you consider $A_\sigma$ the matrix associated to $\sigma \in O(V,b)$ in that base, than

$b(\sigma(x),\sigma(y))=x^t(A_\sigma^tBA_\sigma) y=b(x,y)=x^tBy$

So $A_\sigma$ verify the property $ x^t(A_\sigma^tBA_\sigma)y=x^tBy$ for every $x,y\in V$

and in this case it is simple prove that you get that $A_\sigma$ verify the property

$A_\sigma^t BA_\sigma=B$

Now if $b$ is a definite bilinear form than you have the property $y^tBy\neq 0$ for every $y\in V/ \{0\}$ and so if $det(A_\sigma)=0$ than there exist an $x\in V/ \{0\}$ such that $A_\sigma x=0$ but $x^tBx =x^t(A_\sigma^tBA_\sigma)x= x^t(A_\sigma^tB)(A_\sigma x)=0$ and it is impossibile.

So you have also that $det(A_\sigma)\neq 0$ .

Now if you define $O(V,B)=\{A\in M_n(\mathbb{R}) : det(A)\neq 0, A^tBA=B\}$

there exists a map $\Psi: O(V,b) \to O(V,B)$ that maps every $\sigma\in O(V,b)$ to $A_\sigma$

This map is oviously bijective and your sets are also a Groups and so it is important to ask oneself if it is also a morphism of group.

Answer 2

We want prove the surjectivity of the map $\Psi$.

We had fix a base $\beta:=\{v_1,...,v_n\}$ on $V$ and the function $\Psi$ maps every $\sigma\in O(V,b)$ to matrix $A_\sigma$ associate with itself in the base $\beta$.

If $A=(a_{ij})\in O(V,B)$ we can define a function

$\sigma’: \beta\to V$

that maps every $v_i\in \beta$ to $ \sigma’(v_i):=\sum_{k=1}^{n}a_{ki}v_k$

In this way there exists a unique linear map $\sigma: V\to V$ that extends $\sigma’$.

Now we want prove that $\sigma\in O(V,b)$.

We know that the matrix associate to $\sigma$ in the base $\beta$ is the matrix that have on the $j$-th column the coefficients of $\sigma(v_j)$ written in the base $\beta$ and so, by construction, the matrix associate to $\sigma$ is $A$. But now for all $x,y\in V $

$b(\sigma(x),\sigma(y))=(Ax)^tB(Ay)=$

$=x^t(A^tBA)y=x^tBy=b(x,y)$

So $\sigma \in O(V,b)$ and oviously $\Psi(\sigma)=A$.

In other words the map $\Psi$ is surjective.