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TH`ESE A NEW INVARIANT OF QUADRATIC LIE ALGEBRAS AND ...

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3.4. Quadratic Lie superalgebras with 2-dimensional even part<br />

(1) g s 4,1 is the double extension of the 2-dimensional symplectic vector space q = C2 by the<br />

map having matrix:<br />

C =<br />

<br />

0 1<br />

0 0<br />

in a canonical basis {E1,E2} of q where B(E1,E2) = 1.<br />

(2) g s 4,2 the double extension of the 2-dimensional symplectic vector space q = C2 by the<br />

map having matrix:<br />

C =<br />

<br />

1 0<br />

0 −1<br />

in a canonical basis {E1,E2} of q where B(E1,E2) = 1.<br />

(3) g s 6 is the double extension of the 4-dimensional symplectic vector space q = C4 by the<br />

map having matrix:<br />

⎛<br />

0<br />

⎜<br />

C = ⎜0<br />

⎝0<br />

1<br />

0<br />

0<br />

0<br />

0<br />

0<br />

⎞<br />

0<br />

0 ⎟<br />

0⎠<br />

0 0 −1 0<br />

in a canonical basis {E1,E2,E3,E4} of q where B(E1,E3) = B(E2,E4) = 1, the other are<br />

zero.<br />

Let (q,B) be a symplectic vector space. We recall that Sp(q) is the isometry group of the<br />

symplectic form B and sp(q) is its Lie algebra, i.e. the Lie algebra of skew-symmetric maps<br />

with respects to B. The adjoint action is the action of Sp(q) on sp(q) by conjugation (see<br />

Chapter 1).<br />

Proposition 3.4.10. Let (q,B) be a symplectic vector space. Let g = (CX0 ⊕ CY0 ) ⊥<br />

⊕ q and<br />

g ′ = (CX ′<br />

⊥<br />

⊕CY ′ ) ⊕ q be double extensions of q, by skew-symmetric maps C and C<br />

0 0 ′ respectively.<br />

Then:<br />

(1) there exists a Lie superalgebra isomorphism between g and g ′ if and only if there exist an<br />

invertible map P ∈ L (q) and a non-zero λ ∈ C such that C ′ = λ PCP −1 and P ∗ PC = C<br />

where P ∗ is the adjoint map of P with respect to B.<br />

(2) there exists an i-isomorphism between g and g ′ if and only if C ′ is in the Sp(q)-adjoint<br />

orbit through λC for some non-zero λ ∈ C.<br />

Proof. The assertions are obvious if C = 0. We assume C = 0.<br />

(1) Let A : g → g ′ be a Lie superalgebra isomorphism then A(CX0 ⊕ CY0) = CX ′ 0<br />

A(q) = q. It is obvious that C ′ = 0. It is easy to see that CX0 = Z(g) ∩ g0 and CX ′ 0 =<br />

Z(g ′ ) ∩ g ′ 0 then one has A(CX0) = CX ′ 0 . It means A(X0) = µX ′ for some non-zero µ ∈ C.<br />

0<br />

Let A|q = Q and assume A(Y0) = βY ′<br />

0 + γX ′ . For all X, Y ∈ q, we have A([X,Y ]) =<br />

0<br />

. Also,<br />

µB(C(X),Y )X ′<br />

0<br />

A([X,Y ]) = [Q(X),Q(Y )] ′ = B(C ′ Q(X),Q(Y ))X ′<br />

0 .<br />

80<br />

⊕ CY ′<br />

0 and

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