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Let $Q$ be the smooth quadric threefold in $\mathbb{P}^4_{\mathbb{C}}$ defined by the equation $x_0x_4+x_1x_3+x_2^2=0$.

Is it true that the automorphism group of $Q$ is $SO(Q;\mathbb{C})$ which is isomorphic to $SO(5;\mathbb{C})$?

How can I prove $Aut(Q)=SO(Q)$? And can you give a concrete isomorphism between $SO(Q)$ and $SO(5)$?

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    $\begingroup$ Yes: The automorphism group of the quadric is $PO(Q)$. The isomorphism $PO(Q) \cong PO(5)$ can be obtained by diagonalising $Q$. The result follows the fact that the natural map $SO(2k+1) \to PO(2k+1)$ is an isomorphism, for any natural number $k$. $\endgroup$ – Daniel Loughran Aug 20 '14 at 11:12
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I will expand my comment into an answer.

First, any automorphism of $Q$ is induced by an automorphism of the ambient projective space. This is because any automorphism of $Q$ must fix the canonical bundle, which here is $\mathcal{O}_Q(-3)$. In particular any automorphism must fix $\mathcal{O}_Q(1)$, hence is an automorphism of the ambient projective space.

The collection of projective automorphisms which preserve the quadric is exactly $PO(Q)$. The result then follows the fact that $PO(Q) \cong PO(5)$, together with the fact that the natural map $SO(2k+1)\to PO(2k+1)$ is an isomorphism, for any natural number $k$ (see for instance wikipedia.)

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