# Generalised geometry #4: The Courant bracket and the Jacobiator

### The Courant bracket

In addition to the generalised tangent bundle, the next fundamental structure of generalised geometry is the bilinear, skew-symmetric bracket called the Courant bracket.

The Courant bracket is defined on the sections of ${E = TM \oplus T^{\star}M}$ such that it is the generalised analogue of a standard Lie bracket for vector-fields on the tangent-space ${TM}$.

To begin, we may first start with a more general bracket: namely, the Dorfman bracket. Recall the following relations and standard definition for the Lie bracket, $\displaystyle \mathcal{L}_X = d\iota_X + \iota_X d, \ \iota_{[X,Y]} = [\mathcal{L}_X, \iota_Y]. \ \ \ \ \ (1)$

Combining these two equations we find $\displaystyle \iota_{[X,Y]} = d\iota_X \iota_Y + \iota_X d\iota_Y - \iota_Y d\iota_X - \iota_Y \iota_X d, \ \ \ \ \ (2)$

which uniquely defines the Lie bracket. But in the same way vector fields have natural action of contraction on forms, there exists an extension of this action to generalised tangent vectors. This extension is given by Clifford multiplication. By replacing the contractions in (2) by the Clifford action of sections on ${E}$ we can then define a unique bracket operation on those sections of ${E}$.

To determine this bracket we first look to the exterior derivative $\displaystyle \mathcal{L}_{X} d\omega = \mathcal{L}_{x + \xi} d\omega = (x + \xi)d\omega + d((x + \xi)\omega)$ $\displaystyle = \iota_x d\omega + \xi \wedge d\omega + d(\iota_x \omega + \xi \wedge \omega)$ $\displaystyle = \mathcal{L}_x \omega + d\xi \wedge \omega. \ \ \ \ \ (3)$

Using this result then implies that, for the extension to the generalised bundle ${E}$, $\displaystyle [x + \xi, y + \varepsilon] \cdot \omega = [\mathcal{L}_{x + \xi}, (y + \varepsilon)] \cdot \omega$ $\displaystyle = \mathcal{L}_x (\iota_y \omega + \varepsilon \wedge \omega) + d\xi \wedge (\iota_y \omega + \varepsilon \wedge \omega) - (y + \varepsilon)(\mathcal{L}_x \omega + d\xi \wedge \omega)$ $\displaystyle = \iota_{[x,y]} \omega + \mathcal{L}_x (\varepsilon \wedge \omega) - \varepsilon \wedge \mathcal{L}_x \omega + d\xi \wedge \iota_y \omega - \iota_y (d\xi \wedge \omega)$ $\displaystyle = \iota_{[x,y]}\omega + \mathcal{L}_x (\varepsilon) \wedge \omega - \iota_y d\xi \wedge \omega. \ \ \ \ \ (4)$

This defines the Dorfman bracket.

Definition 1 (Dorfman bracket) Formally, the Dorfman bracket is defined on the sections ${\Gamma(E)}$. Its origin comes from the study of Dirac structures, taking the general form $\displaystyle X \circ Y = (x + \xi) \circ (y + \varepsilon) = [x,y] + \mathcal{L}_x \varepsilon - \iota_y d\xi, \ \ \ \ \ (5)$

which satisfies the Leibniz rule $\displaystyle X \circ (Y \circ Z) = (X \circ Y) \circ Z + Y \circ (X \circ Z) \ \ \ \ \ (6)$

for generalised vectors ${X,Y \in \Gamma(E)}$. The Dorfman bracket is closely related to the D-bracket in double field theory, and it can be used to define the generalised Lie derivative such that $\displaystyle \mathcal{L}_X Y = X \circ Y. \ \ \ \ \ (7)$

Remark 2 (Lack of skew-symmetry and the Courant bracket) The Dorfman bracket is not skew-symmetric (this can be easily seen when calculating the generalised Lie derivative for some generalised tensor). In fact, as we’re about to discuss, its skew-symmetrisation ${[X,Y] = \frac{1}{2}[X,Y]_D - \frac{1}{2}[Y,X]_D}$ is the Courant bracket.

Definition 3 (Courant bracket) The Courant bracket is a skew-symmetric, bilinear form defined for smooth sections ${X= x + \xi, Y = y + \varepsilon}$ on ${E = TM \oplus T^{\star}M}$. It is given by $\displaystyle [X,Y]_C = [x + \xi, y + \varepsilon]_C = [x,y]_L + \mathcal{L}_x \varepsilon - \mathcal{L}_y \xi - \frac{1}{2} d(\iota_x \varepsilon - \iota_y \xi), \ \ \ \ \ (8)$

where ${[x,y]_L}$ denotes the usual Lie bracket of vector fields and ${\iota_x}$ denotes contraction with the vector field ${x}$.

Note that the Courant bracket for vector fields reduces to the Lie bracket. The Lie derivative and the contraction ${\iota}$ satisfy the following relations: $\displaystyle \mathcal{L}_x = d\iota_x + \iota_x d, \ \mathcal{L}_{[x,y]_L} = [\mathcal{L}_x, \mathcal{L}_y], \ \iota_{[x,y]_L} = [\mathcal{L}_x, \iota_y]. \ \ \ \ \ (9)$

### Some properties of the Courant bracket

To study some properties of the Courant bracket, note firstly that from the Dorfman bracket we have the relation $\displaystyle [X,Y] = [X,Y]_D - d(Y,X), \ \ (10)$

which follows from the relation $\displaystyle [X,Y]_D + [Y,X]_D = \mathcal{L}_x \varepsilon + \mathcal{L}_y \xi -\iota_x d\varepsilon - \iota_y d\xi = d(\iota_x d\varepsilon + \iota_y \xi) = 2d(X,Y). \ \ (11)$

What this shows is that if ${X,Y}$ are orthogonal with respect to ${( \ , \ )}$ then ${[X,Y]_D = [X,Y]}$. If we define the projection ${\pi : E \rightarrow TM}$ onto the first factor it follows $\displaystyle \pi ([X,Y]_D) = [\pi(X),\pi(Y)], \ \ \ \ \ (12)$

which, for the case of the Courant brackets, works out to be $\displaystyle \pi ([X,Y]) = [\pi X, \pi Y]. \ \ \ \ \ (13)$

Proposition 4 More generally, for sections ${X,Y,Z}$ of ${E}$ we have $\displaystyle \pi (X) (Y,Z) = ([X,Y]_D, Z) + (X, [Y,Z]_D). \ \ \ \ \ (14)$

Proof: Let ${X = x + \xi}$, ${Y = y + \varepsilon}$, and ${Z = z + \zeta}$. Start with the right-hand side $\displaystyle ([x,y] + \mathcal{L}_x \varepsilon - \iota_y d\xi, z + \zeta) + (y + \varepsilon, [x,z] + \mathcal{L}_x \zeta - \iota_z d\xi)$ $\displaystyle = \frac{1}{2}(\iota_{[x,y]} \zeta + \iota_z (\mathcal{L}_x \varepsilon - \iota_y d\xi) + \iota_{[x,z]} \varepsilon + \iota_y (\mathcal{L}_x \zeta - \iota_z d\xi))$ $\displaystyle = \frac{1}{2} ([\mathcal{L}_{\xi}, \iota_y]\zeta + \iota_z \mathcal{L}_x \varepsilon + [\mathcal{L}_x, \iota_z] \varepsilon + \iota_y \mathcal{L}_x \zeta)$ $\displaystyle = \frac{1}{2} (\mathcal{L}_x \iota_y \zeta + \mathcal{L}_x \iota_z \varepsilon)$ $\displaystyle = x(\iota_y \zeta + \iota_z \varepsilon) = \pi (X)(Y,Z). \ \ \ \ \ (15)$ $\Box$

Corollary 5 For sections ${X,Y,Z}$ of ${E}$ we have $\displaystyle \pi(X)(Y,Z) = ([X,Y] + d[X,Y], Z) + (Y, [X,Z] + d(X,Z)), \ \ \ \ \ (16)$

such that, given the Dorfman bracket, if ${X,Y}$ are sections and ${f}$ some arbitrary function we find $\displaystyle [X, fY]_D = f[X,Y]_D + (\pi (X)f)Y \ \ \ \ \ (17)$

and, reversely, $\displaystyle [fY, X]_D = f[Y,X]_D + (\pi (X)f)Y - 2 (X,Y)df. \ \ \ \ \ (18)$

It then follows more generally $\displaystyle [X, fY]_C = f[X,Y] + (\pi (X)f)Y - (X,Y)df. \ \ \ \ \ (19)$

## Jacobi identity and the Jacobiator

Although the Courant bracket satisfies bilinearity and skewness; it does not actually define a Lie bracket on ${E}$. This is because it fails to satisfy the Jacobi identity. To understand this failure let us first define the Jacobiator, which tells us in what way the Courant bracket fails to satisfy the Jacobi identity.

Definition 6 (Jacobiator) Define ${X,Y,Z \in \Gamma (TM \oplus T^{\star}M)}$, then the Jacobiator can be written as $\displaystyle Jac(X,Y,Z) = [[X,Y],Z] + [[Y,Z],X] + [[Z,X],Y]. \ \ \ \ \ (20)$

The Jacobiator is, in fact, the differential of the what is called the Nijenhuis operator. We won’t discuss this in detail here, but the main implication is that the Courant bracket generally satisfies the Jacobi identity up to some exact term.

Proposition 7 To better understand the Jacobiator, we can first look to the following identity for the Dorfman bracket (following lengthy derivation) $\displaystyle [X, [Y,Z]_D]_D = [[X,Y]_D , Z]_D + [Y, [X,Z]_D ]_D, \ \ \ \ \ (21)$

which says that ${[X, \ ]_D}$ acts as a derivative of the Dorfman bracket. If the Dorfman bracket was skew-symmetric, then this would be equivalent to the Jacobi identity.

Proof: Let ${X = x + \xi}$, ${Y = y + \varepsilon}$, and ${Z = z + \zeta}$, then focusing on the right-hand side of (21) $\displaystyle [[X,Y]_D , Z]_D + [Y, [X,Z]_D ]_D$ $\displaystyle = [[x,y] + \mathcal{L}_x \varepsilon - \iota_y d\xi, z + \zeta] + [y + \varepsilon, [x,z] + \mathcal{L}x \zeta - \iota_z d\xi$ $\displaystyle = [[x,y], z] + [y, [x,z]] + \mathcal{L}_{[x,y]} \zeta - \iota_z d (\mathcal{L}_x \varepsilon - \iota_y d\xi) + \mathcal{L}_y (\mathcal{L}_x \zeta - \iota_z d\xi ) - \iota_{[x,z]} d\varepsilon$ $\displaystyle = [x, [y,z]] + \mathcal{L}_x \mathcal{L}_y \zeta - \mathcal{L}_y \mathcal{L}_x \zeta - \iota_z \mathcal{L}_x d\varepsilon - \iota_z \mathcal{L}_y d\xi + \mathcal{L}_y \mathcal{L}_x \zeta - \mathcal{L}_y \iota_z d\xi - \iota_{[x,z]} d\varepsilon$ $\displaystyle = [x,[y,z]] + \mathcal{L}_x \mathcal{L}_y \zeta - \iota_{[y,z]}d\xi - \mathcal{L}_x \iota_z d \varepsilon$ $\displaystyle = [x,[y,z]] + \mathcal{L}_x (\mathcal{L}_y \zeta - \iota_z d\varepsilon) - \iota_{[y,z]}d\xi$ $\displaystyle = [X, [Y,Z]_D]_D. \ \ \ \ \ (22)$ $\Box$

Proposition 8 We can now also relate ${[[X,Y]_D , Z]_D}$ to ${[[X,Y],Z]}$ such that $\displaystyle [[X,Y]_D , Z]_D = [[X,Y] + d(X,Y), Z]_D$ $\displaystyle = [[X,Y], Z]_D$ $\displaystyle = [[X,Y], Z] + d([X,Y],Z), \ \ \ \ \ (23)$

using the fact that ${[a,b]_D = 0}$ when ${a}$ is a closed one-form.

With these relations, we can now calculate the Jacobiator.

Proposition 9 The Jacobiator defined above can be obtained as the differential of the Nijenhuis operator such that $\displaystyle Jac (X,Y,Z) = d(Nij(X,Y,Z)), \ \ \ \ \ (24)$

where $\displaystyle Nij(X,Y,Z) = \frac{1}{3} (\langle [X,Y],Z \rangle + \langle [Y,Z], X \rangle + \langle [Z,X], Y \rangle, \ \ \ \ \ (25)$

which it should be noted is not tensorial in nature.

Proof: $\displaystyle Jac(X,Y,Z) = [[X,Y],Z] + \text{cyclic permutations}$ $\displaystyle = \frac{1}{4} ([[X,Y]_D, Z]_D - [[Y,X]_D, Z]_D - [Z, [X,Y]_D]_D + [Z, [Y,Z]_D]_D + \text{cyclic permutations}$ $\displaystyle = \frac{1}{4} ([[X, [Y,Z]_D]_D - [Y, [X,Z]_D]_D] + (- [Y, [X,Z]_D]_D + [X, [Y,Z]_D]_D) - [Z, [X,Y]_D]_D + [Z, [Y,X]_D]_D + \text{cyclic permutations}$ $\displaystyle = \frac{1}{4} (- [Y, [X,Z]_D]_D + [X,[Y,Z]_D]_D + \text{cyclic permutations}$ $\displaystyle = \frac{1}{4} ([[X,Y]_D, Z]_D + \text{cyclic permutations}$ $\displaystyle = \frac{1}{4} ([[X,Y],Z] + d[[X,Y],Z] + \text{cyclic permutations}$ $\displaystyle = \frac{1}{4} Jac (X,Y,Z) + \frac{1}{4} d([[X,Y],Z] + [[Y,Z], X] + [[Z,X],Y]). \ \ \ \ \ (26)$ $\Box$

And so, to end the proof, we see that we therefore have the structure of (24).

In the next entry, we’ll discuss gerbes and twisting before looking at the twisted Courant bracket.