Abstract
This paper reformulates Li-Bland’s definition for LA-Courant algebroids, or Poisson Lie 2-algebroids, in terms of split Lie 2-algebroids and self-dual 2-representations. This definition generalises in a precise sense the characterisation of (decomposed) double Lie algebroids via matched pairs of 2-representations. We use the known geometric examples of LA-Courant algebroids in order to provide new examples of Poisson Lie 2-algebroids, and we explain in this general context Roytenberg’s equivalence of Courant algebroids with symplectic Lie 2-algebroids. We study further the core of an LA-Courant algebroid and we prove that it carries an induced degenerate Courant algebroid structure. In the nondegenerate case, this gives a new construction of a Courant algebroid from the corresponding symplectic Lie 2-algebroid. Finally we completely characterise VB-Dirac and LA-Dirac structures via simpler objects, that we compare to Li-Bland’s pseudo-Dirac structures.
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Appendix A: Proof of Theorem 5.15
Appendix A: Proof of Theorem 5.15
Note that in the following computations, we will make use of the identity \(\partial _{Q}=\partial _{Q}^{*}\) without always mentioning it. We begin by proving the following two lemmas.
Lemma A.1
Consider an LA-Courant algebroid \((\mathbb E, Q, B,M)\). The bracket \(\llbracket \cdot ,\cdot \rrbracket _{Q^{*}}\) on sections of the core Q∗ satisfies the following equation:
for all q ∈ Γ(Q) and τ1, τ2 ∈ Γ(Q∗).
Proof
The proof is just a computation using (M1) and (22). We have
Replacing \({\Delta }_{q}{\Delta }_{\partial _{Q}\tau _{1}}\tau _{2}-{\Delta }_{\partial _{Q}\tau _{1}}{\Delta }_{q}\tau _{2}\) by \(R_{\Delta }(q,\partial _{Q}\tau _{1})\tau _{2}+{\Delta }_{\llbracket q,\partial _{Q}\tau _{1}\rrbracket }\tau _{2}\) and reordering the terms yields
Since \(R_{\Delta }(q,\partial _{Q}\tau _{1})\tau _{2}=\langle {\mathbf {i}}_{\partial _{Q}\tau _{1}}{\mathbf {i}}_{q}\omega ,\partial _{B}\tau _{2}\rangle \) by (13), we can now use (22) and ∇q ∘ ∂B = ∂B ∘Δq to replace
by \(-\langle \nabla _{\nabla _{\cdot }\partial _{B}\tau _{2} }q, \tau _{1}\rangle +R(\partial _{B}\tau _{2},\partial _{B}\tau _{1})q\). We use (M1) to replace \({\Delta }_{\llbracket q, \partial _{Q}\tau _{1}\rrbracket -\partial _{Q}({\Delta }_{q}\tau _{1})+\nabla _{\partial _{B}\tau _{1}}q}\tau _{2}\) by \(-{\Delta }_{\partial _{B}^{*}\langle \tau _{1}, \nabla _{\cdot } q\rangle }\tau _{2}\). These two steps yield that the right hand side of our equation is
To conclude, let us show that
vanishes. On \(q^{\prime }\in {\Gamma }(Q)\), this is
We have used (11) and (2) in the first line, as well as for the first equality. To conclude, we have used \(\partial _{B}\circ {\Delta }_{q^{\prime }}=\nabla _{q^{\prime }}\circ \partial _{B}\) by (12. □
Lemma A.2
The bracket on Q∗ satisfies
for all \(f\in C^{\infty }(M)\) and τ ∈ Γ(Q∗).
Proof
By (21), we have \(\llbracket \rho _{Q}^{*}\mathbf {d} f,\tau \rrbracket _{Q^{*}}=\nabla _{\partial _{B}\rho _{Q}^{*}\mathbf {d} f}\tau -{\Delta }_{\partial _{Q}\tau }(\rho _{Q}^{*}\mathbf {d} f)+\rho _{Q}^{*}\mathbf {d}((\rho _{Q}\partial _{Q}\tau )f)\). But \(\partial _{B}\rho _{Q}^{*}=0\) by (11) and \({\Delta }_{\partial _{Q}\tau }(\rho _{Q}^{*}\mathbf {d} f)=\rho _{Q}^{*}\mathbf {d}(\rho _{Q}(\partial _{Q}\tau )(f))\) by (3). □
Now we check that the bracket \(\llbracket \cdot ,\cdot \rrbracket _{\mathbb B}\) in Theorem 5.15 is well-defined. We have for all υ ∈ Γ(U°), τ ∈ Γ(Q∗) and u ∈ Γ(U):
By (M1), the properties of 2-representations and (21), this is
Since υ ∈ Γ(U°) and ∇b preserves Γ(U) for all b ∈ Γ(B), the section 〈υ,∇⋅u〉 of B∗ vanishes and we get
Because Δu preserves as well Γ(U°), the sum \(\nabla _{\partial _{B}\tau }^{*}\upsilon +{\Delta }_{u}\upsilon \) is a section of U°, and so ⟦u ⊕ τ, (−∂Qυ) ⊕ υ⟧ is zero in \(\mathbb B\).
We now check the Courant algebroid axioms (CA1), (CA2) and (CA3). The last one, (CA3), is immediate:
Next we prove (CA2). We have, using 21) to replace \(\llbracket \tau _{1}, \tau _{2}\rrbracket _{Q^{*}}\) by \(-{\Delta }_{\partial _{Q}\tau _{2}}\tau _{1}+\nabla _{\partial _{B}\tau _{1}}\tau _{2}+\rho _{Q}^{*}\mathbf {d}\langle \tau _{1},\partial _{Q}\tau _{2}\rangle \):
We sum \(\langle \llbracket u_{1}\oplus \tau _{1}, u_{2}\oplus \tau _{2}\rrbracket _{\mathbb B}, u_{3}\oplus \tau _{3}\rangle _{\mathbb B}\) with \(\langle u_{2}\oplus \tau _{2}, \llbracket u_{1}\oplus \tau _{1}, u_{3}\oplus \tau _{3}\rrbracket _{\mathbb B} \rangle _{\mathbb B}\), and replace only in the first summand the term \(\langle {\Delta }_{u_{1}}\tau _{2},\partial _{Q}\tau _{3}\rangle \) by ρQ(u1)〈τ2, ∂Qτ3〉−〈τ2,⟦u1, ∂Qτ3⟧〉. This yields
We reorder the remaining terms and replace eight times sums like \(\rho _{Q}(\partial _{Q}\tau _{2})\langle \tau _{1},u_{3}\rangle -\langle {\Delta }_{\partial _{Q}\tau _{2}}\tau _{1}, u_{3}\rangle \) by 〈⟦∂Qτ2, u3⟧, τ1〉 (2), and three times sums like \(\langle \nabla _{\partial _{B}\tau _{1}}u_{2},\tau _{3}\rangle +\langle u_{2},\nabla ^{*}_{\partial _{B}\tau _{1}}\tau _{3}\rangle \) by ρB(∂Bτ1)〈u2, τ3〉. This leads to
The four last terms cancel each other by (M1) and \(\partial _{Q}=\partial _{Q}^{*}\). This yields
Finally we check the Jacobi identity in Leibniz form (CA1). We will check that
with \(\upsilon =(R(\partial _{B}\tau _{1},\partial _{B}\tau _{2})u_{3}-\langle {\mathbf {i}}_{u_{2}}{\mathbf {i}}_{u_{1}}\omega ,\partial _{B}\tau _{3}\rangle )+\text {cyclic permutations}\). Since by Proposition 5.9 \({\mathbf {i}}_{u_{2}}{\mathbf {i}}_{u_{1}}\omega \) has image in U° for all u1, u2 ∈ Γ(U) and R(b1, b2) restricts to a morphism U → U° for all b1, b2 ∈ Γ(B) by Proposition 5.5, υ is a section of U° and so Jac ⟦⋅,⋅⟧(u1 ⊕ τ1, u2 ⊕ τ2, u3 ⊕ τ3) will be zero in \(\mathbb B\).
Using (29), (3), (39) and (11), we find
In the same manner, we compute
Since \(\operatorname {Jac}_{\llbracket \cdot ,\cdot \rrbracket _{U}}(u_{1},u_{2},u_{3})=0\) and ∂B ∘Δq = ∇q ∘ ∂B for all q ∈ Γ(Q), the U-term of Jac ⟦⋅,⋅⟧(u1 ⊕ τ1, u2 ⊕ τ2, u3 ⊕ τ3) equals
Note that since for any b1, b2 ∈ Γ(B), R(b1, b2) restricts to a section of Hom(U, U°) (see Section 5.2), the last summand on the right hand side of (M4) vanishes on sections of U. By sorting out the terms and using (M4) on sections of U, we get
Since R∇ = ∂Q ∘ R, this is − ∂Qυ. We conclude by computing the Q∗-part of Jac ⟦⋅,⋅⟧(u1 ⊕ τ1, u2 ⊕ τ2, u3 ⊕ τ3). Again, because \(\operatorname {Jac}_{\llbracket \cdot ,\cdot \rrbracket _{Q^{*}}}(\tau _{1},\tau _{2},\tau _{3})=0\), we get
The six cancelling terms cancel by (2). Reordering the terms, we get using Lemma A.1:
For the second use of Lemma A.1, we had to replace \(-\llbracket \tau _{2},{\Delta }_{u_{3}}\tau _{1}\rrbracket _{Q^{*}}\) by \(\llbracket {\Delta }_{u_{3}}\tau _{1},\tau _{2}\rrbracket _{Q^{*}}-\rho _{Q}^{*}\mathbf {d}\langle \tau _{2}, \partial _{Q}{\Delta }_{u_{3}}\tau _{1}\rangle \). This is why we get the first term on the third line. Using (13), we get \(\upsilon +\rho _{Q}^{*}\mathbf {d} f\) with \(f\in C^{\infty }(M)\) defined by
By (M1), the first pairing equals \(-\langle \tau _{1},\nabla _{\partial _{B}\tau _{2}}u_{3}\rangle \). Hence, we find f = 0 using ρB ∘ ∂B = ρQ ∘ ∂Q. We have proved Jac ⟦⋅,⋅⟧(u1 ⊕ τ1, u2 ⊕ τ2, u3 ⊕ τ3) = (−∂Qυ) ⊕ υ.
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Jotz Lean, M. On LA-Courant Algebroids and Poisson Lie 2-Algebroids. Math Phys Anal Geom 23, 31 (2020). https://doi.org/10.1007/s11040-020-09355-1
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DOI: https://doi.org/10.1007/s11040-020-09355-1
Keywords
- Positively graded manifolds
- Courant algebroids
- Lie algebroids
- Poisson and symplectic structures
- Double vector bundles
- Representations up to homotopy