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This is Lemma 1.4.4 from Scharlau's book Quadratic and Hermitian Forms.

Let $GL(V)=Aut(V)$ denote the group of invertible endomorphisms of $V$. Further one can check that $$\mathbb H \colon GL(V) \to O(\mathbb H(V)), \qquad \alpha\to \alpha\oplus(\alpha^*)^{-1}$$ is an injective group homomorphism.

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$H$ is hyperbolic space on $V$ and one can define bilinear form on this like below:

Let $V$ be a vector space and $V^*$ its dual space. On the vector space $V\oplus V^*$ we consider the following symmetric bilinear form: \begin{gather*} h=h_V\colon (V\oplus V^*)\times(V\oplus V^*) \to K\\ h((x,f),(y,g)):= fy+gx \end{gather*}

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Also, I know from this lemma that

4.4. Lemma. Let $\alpha\colon V\to W$ be a vector space isomorphism and $\alpha^*\colon W^*\to V^*$ be the dual isomorphism. Then $$\mathbb H(\alpha):= \alpha\oplus(\alpha^*)^{-1} \colon \mathbb H(V) \to \mathbb H(V)$$ is a bijective isometry.

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Question is how to show it is an injective group homomorphism. First of all how to show $O(H(V))$ is a group.

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Why is $O(\mathbb H(V))$ a group? If you already know that $\mathbb H(V)=(V\oplus V^*, \mathbb h_V)$ is a bilinear space, then this follows from the fact that $O(V,b)$ is a group for any bilinear space $(V,b)$.

$\mathbb H(\alpha)$ belongs to $O(\mathbb H(V))$. The fact that $\mathbb H(\alpha)\in O(\mathbb H(V))$ for $\alpha\in GL(V)$ is exactly the claim in Proposition 4.4.

$\mathbb H$ is a group homomorphism. We just want to show that $\mathbb H(\beta\circ\alpha)=\mathbb H(\beta)\circ \mathbb H(\alpha)$ for any $\alpha,\beta\in GL(V)$. We just compute \begin{align*} (\beta\circ\alpha)\oplus ((\beta\circ\alpha)^*)^{-1} &= (\beta\circ\alpha)\oplus (\alpha^*\circ\beta^*)^{-1} \\ &= (\beta\circ\alpha)\oplus ((\beta^*)^{-1}\circ(\alpha^*)^{-1}) \\ &= (\beta,(\beta^*)^{-1})\circ(\alpha,(\alpha^*)^{-1}) \end{align*}

$\mathbb H$ is injective. It suffices to check that $\mathbb H$ has trivial kernel. Clearly, $$\mathbb H(V)=\alpha\oplus(\alpha^*)^{-1}=id_V\oplus id_{V^*}$$ implies $\alpha=id_V$.

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  • $\begingroup$ How the third step in homomorphism is true? $\endgroup$ – maths student Jul 12 '18 at 7:25
  • $\begingroup$ By "the third step" do you mean $(\beta\circ\alpha)\oplus ((\beta^*)^{-1}\circ(\alpha^*)^{-1}) = (\beta,(\beta^*)^{-1})\circ(\alpha,(\alpha^*)^{-1})$? $\endgroup$ – Martin Sleziak Jul 12 '18 at 7:28
  • $\begingroup$ yes that step only. $\endgroup$ – maths student Jul 12 '18 at 7:29
  • $\begingroup$ Is it just component wise breaking of direct sum? $\endgroup$ – maths student Jul 12 '18 at 7:31
  • $\begingroup$ Perhaps (depending on your notation and on how you interpret $\oplus$) you might prefer the notation $\beta\oplus(\beta^*)^{-1}$ instead of $(\beta,(\beta^*)^{-1})$. $\endgroup$ – Martin Sleziak Jul 12 '18 at 7:31

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