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A root-graded Manin triple gives dual Lie bialgebras

Statement

Let k be a field of characteristic 0, and let (d,d+,d−;B) be a locally finite root-graded Manin triple in the sense of Lie bialgebras, degreewise duality, and root-graded Manin triples. Define δ+:d+→Λ2d+ and δ−:d−→Λ2d− by transposing the opposite brackets under the degreewise perfect pairing induced by B:

⟨δ+(x),y∧z⟩=B(x,[y,z]),⟨u∧v,δ−(y)⟩=B([u,v],y).

Then these maps are well defined and make d+ and d− dual Lie bialgebras.

Facts & Assumptions

Given: A field k of characteristic 0 and a locally finite root-graded Manin triple.

[F1]

The two complementary isotropic subalgebras have finite-dimensional graded pieces, a degreewise perfect cross pairing, and only finitely many degree decompositions within each subalgebra in any fixed degree (Lie bialgebras, degreewise duality, and root-graded Manin triples).

[F2]

The wedge pairing is the determinant pairing on exterior powers, and a Lie bialgebra cobracket is a cocycle satisfying co-Jacobi (The graded exterior algebra ΛV, Lie bialgebras, degreewise duality, and root-graded Manin triples).

Proof

technique · Transpose the bracket and use the Jacobi identity in the double
1.1F1construct

By [F1], for each homogeneous x∈d+ only finitely many opposite-degree pairs of components of d− can bracket to a degree paired with x. The perfect pairings therefore give a unique finite sum δ+(x)∈Λ2d+ satisfying the first transpose identity; the same construction defines δ−, and both maps are linear and preserve total root degree.

1.2F1algebra

Choose dual bases ei and fa in the finitely many homogeneous pieces involved in a fixed calculation, and write [ei,ej]=∑kcijkek and [fa,fb]=∑kdkabfk. Invariance gives B([ei,fa],fk)=B(ei,[fa,fk])=diak and B([ei,fa],ek)=B(fa,[ek,ei])=−cika; since the two subalgebras are isotropic and pair perfectly, these identities determine [ei,fa]=∑k(diakek−cikafk). For each fixed input pair, grading restricts these sums to two fixed finite-dimensional pieces, so they are finite by [F1].

2.1F1F2step 1.1algebra

Pair Alt⁡(δ+⊗id⁡)δ+(x) with y∧z∧w in the opposite subalgebra. By the defining transpose identity, the result is B(x,[y,[z,w]]+[z,[w,y]]+[w,[y,z]])=0 by Jacobi in d−. Degreewise perfectness makes the co-Jacobi expression zero; the identical argument with signs exchanged proves co-Jacobi for δ−.

3.1F1F2step 1.1step 1.2algebra∎

Jacobi in d for ei,ej,fa, paired with fb, gives ∑kcijkdkab=∑k(cikadjkb+cikbdjak−cjkadikb−cjkbdiak). By the transpose definition, this is exactly the coefficient identity for δ+([ei,ej])=[ei,δ+(ej)]−[ej,δ+(ei)]. Interchanging + and − and using Jacobi for ei,fa,fb proves the cocycle identity for δ−. Since the homogeneous bases were arbitrary and the pairings are perfect, both cocycle identities hold for all elements. Together with step 2.1, this proves that the two transposed maps are dual Lie bialgebra structures.

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