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In the lecture notes A brief introduction to Dirac manifolds, Henrique Bursztyn recall the notion of a Courant bracket on section of the generalised tangent bundle $TM\oplus T^*M$.

For $X+\xi, Y+\eta\in \Gamma(M,TM\oplus T^*M)$, the Courant bracket is given by

$$[X+\xi, Y+\eta]=[X,Y]+L_X\eta-L_Y\xi+\frac{1}{2}d(\xi(Y)-\eta(X)).$$

It also mentions that,

One may alternatively use, instead of (3.2), the non-skew-symmetric bracket $$((X,\alpha),(Y,\beta))\mapsto ([X,Y],L_X\beta − i_Y d\alpha)$$ for condition (ii); (3.2) is the skew-symmetrization of this bracket, and a simple computation shows that both brackets agree on sections of subbundles satisfying (i).

In the lecture Introduction to Poisson geometry and Lie algebroids Eckhard Meinrenken defines (7 mins to 10 mins) Courant bracket as $((X,\alpha),(Y,\beta))\mapsto ([X,Y],L_X\beta − i_Y d\alpha)$ instead of $[X+\xi, Y+\eta]\mapsto ([X,Y], L_X\eta-L_Y\xi+\frac{1}{2}d(\xi(Y)-\eta(X))$.

He said the following:

"a weird formula which takes some time to get used to".

"I prefer to work with the bracket that is not skew-symmetric."

"There are some ways of motivating why this is the right formula, but we do not have time to do that"

So, I would like to understand the motivation in considering this non skew-symmetric bracket.

Please suggest some references that gives motivation for this.

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  • $\begingroup$ Another approach is to regard Leibniz algebroids as acyclic dg Lie algebroids concentrated in degrees 0, 1 and 2. There's an equivalence of categories sending a Leibniz algebra $S$ to a complex with degree 0 equal to liefication of $S$ (quotient by an ideal spanned by $[x, x]$), degree 1 being $S$ and degree 2 the kernel of natural projection. So, you can rewrite that non-skew symmetric bracket as a honest dg Lie bracket. $\endgroup$
    – Denis T
    Commented Aug 21, 2021 at 11:48
  • $\begingroup$ @DenisT. Can you please give reference for seeing Leibniz algebroids as dg Lie algebroids. Thank you :) $\endgroup$ Commented Aug 22, 2021 at 7:35
  • $\begingroup$ Differential graded Lie algebras and Lebniz algebra cohomology, Jacob Mostovoy arxiv.org/abs/1910.03754v1 The paper doesn't mention algebroids, but usual Lie-Reinhart definition of an algebroid translates directly to this setting. $\endgroup$
    – Denis T
    Commented Aug 22, 2021 at 11:18

1 Answer 1

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The point is that the non-skew-symmetric bracket (Dorfman bracket, or "generalised Lie derivative" in physics) satisfies the Leibniz rule in the following sense.

Write your $(X,\alpha)=\varphi$, $(Y,\beta)=\psi$ and denote your quoted formula for this bracket as $L_\varphi \psi$. Then if you introduce $(Z,\gamma)=\chi,$ $$ L_\varphi L_\psi \chi= L_{L_\varphi \psi} \chi + L_{\psi} L_{\varphi}\chi\,. $$ This Leibniz property means that the Dorfman bracket endows the generalised tangent bundle with the structure of a Leibniz algebroid.

Side remark: If this were also antisymmetric, it would define a Lie bracket. Therefore you can either work with the Courant bracket (that is antisymmetric but has no Leibniz rule) or with the Dorfman bracket (that is the exact opposite).

To put this in context: Leibniz algebroid structures arise quite frequently. In this case you get a Leibniz algebroid because the generalised tangent bundle has a symplectic $L_2$-algebroid structure (or a QP manifold as a physicist would call it) due to a result of Roytenberg and those canonically produce Leibniz algebroids via the derived bracket construction of Kosmann-Schwarzbach. See this nice review by her and references therein.

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  • $\begingroup$ Thanks for your answer.. It is interesting to learn one more algebroid.. I am aware of Lie algebroid, Courant algebroid.. Now, I know there is something called Leibniz algebroid.. I do not know anything about symplectic L_2-algebroid.. I will read the reference you suggested. $\endgroup$ Commented Aug 22, 2021 at 7:28

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