Isomorphism of orthogonal groups

Question

1.5 Lemma. If $(V,b) \cong (V', b')$, then $\operatorname{O}(V,b) \cong \operatorname{O}(V',b')$.

Proof: If $\sigma \colon V \to V'$ is a bijective isometry, then $\tau \mapsto \sigma \tau \sigma^{-1}$ is an isomorphism of the orthogonal groups.

(Original image here.)

Since vector space are isomorphic there exist bijective isometry between those two vector space. I am confused whether $\tau$ is map from $V$ to $V$ or something else and how it become isomorphic.

Answer 1

The text provides the explicit map of the isomorphism between the orthogonal groups namely

$$\tau \mapsto \sigma \tau \sigma^{-1}$$

There are 3 things you'd need to show to flesh out the details:

1. If $\tau$ in the orthogonal group of $V$ then $\sigma \tau \sigma^{-1}$ is in the orthogonal group $V'$

2. $\tau \mapsto \sigma \tau \sigma^{-1}$ is a homomorphism

3. $\tau \mapsto \sigma \tau \sigma^{-1}$ is bijective

You seemed most concerned about 3 in the comments so lets start with that.

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To SHOW 1)

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$\tau \mapsto \sigma \tau \sigma^{-1}$ is injective

Suppose that $\sigma \tau \sigma^{-1}=I$ where $I$ is the identity transformation of $V'$ then $\tau=\sigma^{-1}I\sigma$. Pick an arbitrary element $v \in V$ then

$$\begin{split} \tau(v)&=(\sigma^{-1}I\sigma)(v)\\ &=\sigma^{-1}I(\sigma(v))\\ &=\sigma^{-1}(\sigma(v))\\ &=v\\ \end{split}$$ So $\tau$ is the identity transformation on $V$. Therefore the matp is injective.

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$\tau \mapsto \sigma \tau \sigma^{-1}$ is surjective

Let $\alpha:V'\to V'$ be an invertible orthogonal transformation. Then consider $\tau:V \to V$ defined by $\tau=\sigma^{-1}\alpha\sigma$. It's clear that $\tau$ is invertible but we need to show it preserves the bilinear form $b$.

Note that $\sigma:V \to V'$ being a linear isometry means that for all $a,b \in V$

$$ b(a,b)=b'(\sigma(a),\sigma(b)) $$

and for all $x,y \in V'$

$$ b'(x,y)=b(\sigma^{-1}(a),\sigma^{-1}(b)) $$

Also note since $\alpha$ is in the orthogonal group of $V'$ $$ b'(\alpha (x),\alpha(y))=b'(x,y) $$

So let $v,w \in V$ and consider $$\begin{split} b(\tau(v),\tau(w))&=b(\sigma^{-1}\alpha\sigma(v),\sigma^{-1}\alpha\sigma(w)) \\ &=b(\sigma^{-1}(\alpha\sigma(v)),\sigma^{-1}(\alpha\sigma(w))) \\ &=b'(\alpha\sigma(v),\alpha\sigma(w))\\ &=b'(\sigma(v),\sigma(w))\\ &=b(v,w)\\ \end{split} $$

So $\tau$ preserves the bilinear form $b$. Note that usually $\sigma$ is assumed to be linear. Often it is called a linear isometry. I don't think this is true or even 1) is true if $\sigma$ is not linear.

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To SHOW 2) $\alpha,\beta$ in the orthogonal group of $V$. Then $$ \sigma \alpha \beta \sigma^{-1}=(\sigma \alpha\sigma^{-1}) (\sigma\beta \sigma^{-1} ) $$

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To SHOW 1)

Let $v,w \in V'$, tau in the orthogonal group of $V$

$$\begin{split} b'(\sigma\tau\sigma^{-1}(v),\sigma\tau\sigma^{-1}(w)) &=b'(\sigma(\tau\sigma^{-1}(v)),\sigma(\tau\sigma^{-1}(w))) \\ &=b(\tau\sigma^{-1}(v),\tau\sigma^{-1}(w)) \\ &=b(\sigma^{-1}(v),\sigma^{-1}(w)) \\ &=b'(v,w)\\ \end{split} $$

The linearity and invertibility of $\sigma\tau\sigma^{-1}$ follow from $\sigma,\tau,\sigma^{-1}$ each being linear and invertible.