Note that π extends to an involutive antiautomorphism of the universal enveloping algebra π :π(π€)π¨π© βπ(π€) so that
π(π₯π¦)=π(π¦)π(π₯)for π₯,π¦ βπ(π€).
First we prove uniqueness of π.
Clearly ^F1 extends to
π(π₯βπ£,π€)=π(π£,π(π₯)π€)for π₯ βπ(π€) and π£,π€ βπ(π).
Note by the PoincarΓ©-Birkhoff-Witt theorem
π(π€)=π΅πΎπΌπππ(π«β)βπ(π₯)βπ(π«+)=π΅πΎπΌππ(πβπ«βπ(π«β))βπ(π₯)β(πβπ(π«+)π«+)=π΅πΎπΌπππ«βπ(π«β)βπ(π₯)βπ(π«+)π«+βπ«βπ(π«β)βπ(π₯)βπ(π₯)βπ(π«+)π«+βπ(π₯)=π΅πΎπΌππ(π«βπ(π«β)π(π₯)π(π«+)+π«βπ(π«β)π(π₯)+π(π₯)π(π«+)π«+)βπ(π₯)=π΅πΎπΌππ(π«βπ(π€)+π(π€)π«+)βπ(π₯)so we may define the projection operator
π:π(π€)βπ(π₯)=π΄π ππ π
πππβ(π₯)Given any π£,π€ βπ(π) by irreducibility
π£=π₯βπ£ππ€=π¦βπ£πfor some π₯,π¦ βπ(π€) so
π(π£,π€)=π(π₯βπ£π,π¦βπ£π)=π(π£π,π(π₯)π¦βπ£π)Since π£π is a vacuum vector and
π(π£π,π«βπ(π€)βπ£π)=π(π«+βπ£π,π(π€)βπ£π)=0it follows
π(π£,π€)=π(π£π,π(π₯)π¦βπ£π)=π(π£π,π(π(π₯)π¦)βπ£π)=π(π£π,π(π(π(π₯)π¦))π£π)=π(π(π(π₯)π¦))Therefore the behaviour of π is completely determined by properties ^F1 and ^F2,
so if π exists it is unique.
To prove existence, consider the annihilator of π£π
π=π(π€)(π«++βββπ₯π(ββπ(β)1))which is a left-ideal π(π€).
Thus
π:π(π€)/πβπ(π)π₯+ββ¦π₯βπ£πis a π€-module isomorphism.
Now
π(π₯π¦)=π(π₯)π¦π(π¦π₯)=π¦π(π₯)for π₯ βπ(π€) and π¦ βπ(π₯),
whence
π(π(π))=π(π(π(π)))=0Thus the above formula
π(π₯βπ£π,π¦βπ£π)=π(π(π(π₯)π¦))for π₯,π¦ βπ(π€) is well-defined, since
π(π₯π¦)π§=π(π¦)π(π₯)π§π(π(π(1)1))=1for π₯,π¦,π§ βπ(π€).
Therewithal since
πβπ=πβπit follows
π(π(π(π¦)π₯))=π(π(π(π(π₯)π¦)))=π(π(π(π(π₯)π¦)))=π(π(π(π₯)π¦))for π₯,π¦ βπ(π€), so π is symmetric.