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Covering of the orthogonal group
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[QUOTE="Jason Bennett, post: 6254073, member: 641911"] [B]Homework Statement:[/B] see title [B]Relevant Equations:[/B] see below Progress:π:π(3)ββ€2π:π(3)βππ(3)π:π(3)/ππ(3)ββ€2 π(π)=det(π) with πβπ(3), that way π(π)β¦{β1,1}β β€2, where 1 is the identity element.Ker(π) = {πβππ(3)|π(π)=1}=ππ(3), since det(π)=1 for πβππ(3).By the multiplicative property of the determinant function, π = homomorphism. ***What is the form of the canonical homomorphism (π) in this case?I'm used to the coset language,i.e. π:πΊβπΊ/Ker(π) with π(π)=ππΎ for πΎ=ker(π) If this were settled, then π is an isomorphism by the isomorphism theorem. [/QUOTE]
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Covering of the orthogonal group
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