lecture 9

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< An introduction to Lie algebra cohomology
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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam
Author: Dr. Sara Lombardo, Vrije Universiteit Amsterdam

Contents

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the symplectic form

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definition

If a^2\in Z^2(\mathfrak{g},\mathbb{C}) is nondegenerate, then one says that a^2 is a proto-symplectic form. If, moreover, a^2\in Z_\wedge^2(\mathfrak{g},\mathbb{C}), one says that a^2 is a symplectic form

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definition

Let the symplectic operator \mathcal{S}:\mathfrak{g}\rightarrow \mathfrak{g}^\star=C^1(\mathfrak{g},\mathbb{C}) be defined by

\mathcal{S}(x)=\iota^2(x)a^2

If c^1\in C^1(\mathfrak{g},\mathbb{C}) can be written as

c^1=S(x)

one defines the cosymplectic operator \mathcal{C} by

C(c^1)=x

One has, by definition, that \mathcal{C}(\mathcal{S}(x))=x, or \mathcal{C}\mathcal{S} equals the identity on \mathfrak{g}.

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definition

One defines a Leibniz algebra structure on the domain of \mathcal{C} in \mathfrak{g}^\star as follows. Let

\{a_1^1,a_2^1\}=\mathcal{S}([\mathcal{C}(a_1^1),\mathcal{C}(a_2^1)])
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proof

\{\{a_1^1,a_2^1\},a_3^1\}=\mathcal{S}([\mathcal{C}(\{a_1^1,a_2^1\}),\mathcal{C}(a_3^1)])
=\mathcal{S}([\mathcal{C}(\mathcal{S}([\mathcal{C}(a_1^1),\mathcal{C}(a_2^1)])),\mathcal{C}(a_3^1)])
=\mathcal{S}([[\mathcal{C}(a_1^1),\mathcal{C}(a_2^1)],\mathcal{C}(a_3^1)])
=\mathcal{S}([\mathcal{C}(a_1^1),[\mathcal{C}(a_2^1)],\mathcal{C}(a_3^1)])-\mathcal{S}([\mathcal{C}(a_2^1),[\mathcal{C}(a_1^1)],\mathcal{C}(a_3^1)])
=\mathcal{S}([\mathcal{C}(a_1^1),\mathcal{C}(\mathcal{S}([\mathcal{C}(a_2^1)],\mathcal{C}(a_3^1)])))-\mathcal{S}([\mathcal{C}(a_2^1),\mathcal{C}(\mathcal{S}([\mathcal{C}(a_1^1)],\mathcal{C}(a_3^1)])))
=\mathcal{S}([\mathcal{C}(a_1^1),\mathcal{C}(\mathcal{S}([\mathcal{C}(a_2^1)],\mathcal{C}(a_3^1)])))-\mathcal{S}([\mathcal{C}(a_2^1),\mathcal{C}(\mathcal{S}([\mathcal{C}(a_1^1)],\mathcal{C}(a_3^1)])))
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definition

If a^2(x,y)=K(x,\omega^1 y), one says that \sharp \mathcal{S}=\omega^1\in C^1(\mathfrak{g},\mathfrak{g}) is a (proto-)symplectic map.

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definition

When K is nondegenerate, and \alpha\in End(\mathfrak{g}), then \alpha^\star (the adjoint of \alpha) is defined by

K(\alpha^\star x,y )=K(x,\alpha y)
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proposition

When a^2 is a symplectic form, \omega^1 is an antisymmetric map, that is to say,

\omega^{1\star}=-\omega^1
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proof

K(x,\omega^1 y)= a^2(x,y)= - a^2(y,x)= -K(y,\omega^1 x)=-K(\omega^{1\star} y ,x )= -K(x,\omega^{1\star}y)

It follows that \omega^1 y = - \omega^{1\star}y.

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lemma

When K is nondegenerate,

\omega^1([y,z]) = [y,\omega^1 (z)]-[\omega^{1\star}( y),z]
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proof

0 = d^2 a^2 (x,y,z)=-a^2([x,y],z)-a^2(y,[x,z])+a^2(x,[y,z])
=-K([x,y],\omega^1 z)-K(y,\mathcal{O}[x,z])+K(x,\omega^1 [y,z])
=-K(x,[y,\omega^1z])-K(\omega^{1\star} y,[x,z])+K(x,\omega^1[y,z])
=-K(x,[y,\omega^1 z])-K([x,z],\omega^{1\star} y)+K(x,\omega^1[y,z])
=-K(x,[y,\omega^1z])-K(x,[z,\omega^{1\star} y])+K(x,\omega^1[y,z])
=-K(x,[y,\omega^1z])+K(x,[\omega^{1\star} y,z])+K(x,\omega^1[y,z])
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lemma

d^1\omega^1(y,z)=-d_-^{(0)}(z)(\omega^1(y)+\omega^{1\star}( y))
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proof

d^1\omega^1(y,z) = [y,\omega^1 (z)]+[\omega^{1}( z),y]-\omega^1([y,z])
= [\omega^1(y)+\omega^{1\star}( y),z]
= -d_-^{(0)}(z)(\omega^1(y)+\omega^{1\star}( y))
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corollary

When a^2 is symplectic, \omega^1\in Z_{\wedge}^1(\mathfrak{g},\mathfrak{g}).

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