Stringy Things

# Double Field Theory: The Courant Bracket

1. Introduction

In this post we are going to briefly and somewhat schematically discuss the appearance of the Courant bracket in Double Field Theory (DFT), following [1]. The point here is mainly to set the stage, so we jump straight into motivating the Courant bracket. In the next post, we will then study the B-transformations from the maths side and the C-bracket, following and expanding from [1] and others, with an emphasis in the end on how all of this relates to T-duality and strings.

What follows is primarily based on a larger collection of study notes, which I will upload in time.

2. Motivating the Courant Bracket

To understand the appearance of the Courant bracket in DFT, one way to start is by considering some general theory with a metric ${g_{ij}(X)}$ and a Kalb-Ramond field (i.e., an antisymmetric tensor field) ${b_{ij}(X)}$, where ${X \in M}$. The symmetries and diffeomorphisms of ${g_{ij}(X)}$ are generated by vector fields ${V^{i}(X)}$, where ${V \in T(M)}$ with ${T(M)}$ being the tangent bundle. As for ${b_{ij}(X)}$, the transformations are generated by one-forms ${\xi_{i}(X)}$, where ${\xi^{i}(X) \subset T^{\star}(M)}$ with ${T^{\star}(M)}$ being the cotangent bundle. We may combine ${V^{i}(X)}$ and ${\xi^{i}(X)}$ as a sum of bundles, such that (dropping indices) ${V + \xi \in T(M) \oplus T^{\star}(M)}$.

With these definitions, the opening question now is to ask, ‘what are the gauge transformations?’ To make sense of this, consider the following gauge parameters,

$\displaystyle \delta_{V + \xi} g = \mathcal{L}_{V} g$

$\displaystyle \delta_{V + \xi} b = \mathcal{L}_{V} b + d\xi \ \ \ (1)$

Here ${\mathcal{L}}$ is the Lie derivative. Furthermore, note given that ${V}$ generates diffeomorphisms, in (1) we get the Lie derivative in the direction of ${V}$. Also notice that ${\xi}$ does not enter the gauge transformation of ${g}$; however, for the gauge transformation of ${b}$, we do have a one-form ${\xi}$ and so we can take the exterior derivative. We should also note the following important properties of ${\mathcal{L}}$. For instance, when acting on forms the Lie derivative is,

$\displaystyle \mathcal{L}_{V} = i_{V}d + di_{V} \ \ \ (2)$

Where ${iV}$ is a contraction with ${V}$. We’re just following the principle of a contraction with a vector times the exterior derivative. It is also worth pointing out that ${\mathcal{L}}$ and the exterior derivative commutate such that,

$\displaystyle \mathcal{L}_{V}d = d\mathcal{L}_{V} \ \ \ (3)$

There are also some other useful identities that we are going to need. For instance, for the Lie algebra,

$\displaystyle [\mathcal{L}_{V_1}, \mathcal{L}_{V_2}] = \mathcal{L}_{[V_1,V_2]} \ \ (4)$

Where ${[V_1,V_2]}$ is just another vector such that ${[V_1,V_2]^{k} = V_{1}^{p}\partial_{p}V_{2}^{k} - (1,2)}$.

And, finally, we have,

$\displaystyle [\mathcal{L}_{X}, i_{Y}] = i_{[X,Y]} \ \ (5)$

Now follows the fun part. Given the transformation laws provided in (1), we want to determine the gauge algebra. To do this, we must compute in reverse order the gauge transformations on the metric ${g}$ and the ${b}$-field. For the metric we evaluate the bracket,

$\displaystyle [\delta_{V_2 + \xi_2}, \delta_{V_1 + \xi_1}]g = \delta_{V_2 + \xi_2} \mathcal{L}_{V_1}g - (1,2)$

$\displaystyle = \mathcal{L}_{V_1}\mathcal{L}_{V_2}g - (1,2)$

Using the identity (4) we find,

$\displaystyle [\delta_{V_2 + \xi_2}, \delta_{V_1 + \xi_1}]g = \mathcal{L}_{[V_1, V_2]} g \ \ \ (6)$

For the ${b}$-field we have,

$\displaystyle [\delta_{V_2 + \xi_2}, \delta_{V_1 + \xi_1}]b = \delta_{V_2 + \xi_2} (\mathcal{L}_{V_1}b + d\xi_{1}) - (1,2)$

$\displaystyle = \mathcal{L}_{V_1}(\mathcal{L}_{V_2}b + d\xi_2) - (1,2)$

$\displaystyle = \mathcal{L}_{[V_1, V_2]} + d(\mathcal{L}_{V_1}\xi_{2} - \mathcal{L}V_{2}\xi_{1} \ \ \ (7)$

It turns out that this bracket satisfies the Jacobi identity, although it is not without its problems because, as we will see, there is a naive assumption present in the above calculations. In the meantime, putting this aside until later, the idea now is to compare the above with (1) and see what ‘pops out’. Notice that we find,

$\displaystyle [\delta_{V_2 + \xi_2}, \delta_{V_1 + \xi_1}] = \delta_{[V_1,V_2]} + \mathcal{L}_{V_1}\xi_{2} - \mathcal{L}_{V_2}\xi_{1} \ \ \ (8)$

In which we have discovered a bracket defined on ${T(M) \oplus T^{\star}(M)}$,

$\displaystyle [V_{1} + \xi_{1}, V_{2} + \xi_{2}] = [V_1, V_2] + \mathcal{L}_{V_1}\xi_{2} - \mathcal{L}_{V_2} \xi_{1} \ \ \ (9)$

On the right-hand side of the equality we see a vector field in the first term and a one-form given by the final two terms. This Lie bracket is reasonable and, on inspection, we seem to have a definite gauge algebra. Here comes the problem allude a moment ago: there is a deep ambiguity in (9) in that we cannot, however much we try, determine unique parameters in our theory. Notice,

$\displaystyle \delta_{V + \xi}b = \mathcal{L}_{V}b + d\xi$

$\displaystyle = \mathcal{L}_{V + (\xi + d \sigma)} \ \ \ (10)$

The point being that the ambiguity of the one-form ${\xi}$ is so up to some exact ${d\sigma}$. To put it another way, if we change ${\xi}$ by ${d\sigma}$, we’re not actually changing anything at all. We would just get ${\mathcal{L}_{V}b + d(\xi + d\sigma)}$ where, when the exterior derivative hits ${d\sigma}$ we simply get nothing. So, given that ${\xi}$ is ambiguous up to some exact ${d\sigma}$, in a sense what we have is a symmetry for a symmetry. In other words, the present construction is not sufficient.

What we can do to correct the situation is analyse the mistake in (7). Let us, for instance, look at the right-hand side of the summation sign in this equation,

$\displaystyle d(\mathcal{L}_{V_1}\xi_{2} - \mathcal{L}V_{2}\xi_{1}) = d(di_{V_1}\xi_{2} + i_{V_1}d\xi_{2}) - (1,2)$

The logic follows that the first term ${di_{V_1}\xi_{2}}$ is killed by ${d}$. It doesn’t make any contribution and its coefficient is just ${1}$. The trick then is to see, without loss of generality, that we may change the implicit coefficient ${1}$ in front of ${di_{V_1}\xi_{2}}$. It turns out, the coefficient that we can use is ${1 - \frac{\beta}{2}}$,

$\displaystyle = d((1-\frac{\beta}{2} di_{V_1}\xi_{2} + i_{V_1}d\xi_{2}) - (1,2)$

$\displaystyle = d(\mathcal{L}_{V_1}\xi_{2} - \frac{1}{2}\beta di_{V_1}\xi_{2}) - (1,2)$

$\displaystyle = d(\mathcal{L}_{V_1}\xi_{2} - \mathcal{L}_{V_2}\xi_{1} - \frac{1}{2}\beta d [i_{V_1}\xi_{2} - i_{V_2}\xi_{1}]) \ \ \ (11)$

What we end up achieving is the construction of a much more general bracket,

$\displaystyle [V_{1} + \xi_{1}, V_{2} + \xi_{2}]_{\beta} = [V_1, V_2] + \mathcal{L}_{V_1}\xi_{2} - \mathcal{L}_{V_2}\xi_{1} - \frac{1}{2}\beta d(iV_{1}\xi_{2} - iV_{2}\xi_{2}) \ \ \ (12)$

What is so lovely about this result might at first seem counterintuitive. It turns out, as one can verify, for ${\beta \neq 0}$, we do not satisfy the Jacobi identity. So at first (12) may not seem lovely at all! But it makes perfect sense to consider cases of non-vanishing ${\beta}$. In mathematics, the case for ${\beta = 1}$ was introduced by Theodore James Courant in his 1990 doctoral dissertation [5], where he studied the bridge between Poisson geometry and pre-symplectic geometry. The idea here is to forget about the Jacobi identity – consider its loss an artefact of field theory with anti-symmetric tensors and gravity – and impose ${\beta = 1}$. When we do this what we obtain is indeed the famous Courant bracket. That is, given ${\beta = 1}$, the case of maximal symmetry is described by,

$\displaystyle [V_{1} + \xi_{1}, V_{2} + \xi_{2}]_{\beta=1} = [V_1, V_2] + \mathcal{L}_{V_1}\xi_{2} - \mathcal{L}_{V_2}\xi_{1} - \frac{1}{2} d(iV_{1}\xi_{2} - iV_{2}\xi_{2} \ \ \ (13)$

Although the Jacobi identity does not hold, one can show that for ${Z_{i} = V_{i} + \xi_{i}, i = 1,2,3}$, the Jacobiator assumes the form,

$\displaystyle [Z_1, [Z_2,Z_3]] + \text{cyclic} = dN(Z_1, Z_2, Z_3)$

Which is an exact one-form. This gives us a first hint that the unsatisfied Jacobi identity does not provide inconsistencies, because exact one-forms do not generate gauge transformations.

But why ${\beta = 1}$? Courant argued that the correct value of ${\beta}$ is in fact ${1}$ because, as he discovered, there is an automorphism of the bracket. This means that if do an operation on the elements, it respects the bracket. This automorphism is, moreover, an extra symmetry known in mathematics as a B-transformation. What follows from this is, I think, actually quite special. Given the Courant bracket is a generalisation of the Lie bracket, particularly in terms of an operation on the tangent bundle ${T(M)}$ to an operation on the direct sum of ${T(M)}$ and the p-forms of the vector bundle, what we will discuss is how the B-transformation in mathematics relates in a deep way to what in physics, especially string theory, we call T-duality (target- space duality). This is actually one of the finer points where mathematics and physics intersect so wonderfully in DFT.

In the next post we’ll carry on with a discussion of the B-transformation and then also the C-bracket, finally showing how everything relates.

References

[1] Zwiebach, B. (2010). ‘Double Field Theory, T-Duality, and Courant Brackets’ [lecture notes]. Available from [arXiv:1109.1782v1 [hep-th]].

[2] Hohm, O., Hull, C., and Zwiebach, B. (2010). ‘Generalized metric formulation of double field theory’. Available from [arXiv:1006.4823v2 [hep-th]].

[3] Hull, C. and Zwiebach, B. (2009). ‘Double Field Theory’. Available from [arXiv:0904.4664v2 [hep-th]].

[4] Hull, C. and Zwiebach, B. (2009). ‘The Gauge Algebra of Double Field Theory and Courant Brackets’. Available from [arXiv:0908.1792v1 [hep-th]].

[5] Courant, T. (1990). ‘Dirac manifolds’. Trans. Amer. Math. Soc. 319: 631–661. Available from [https://www.ams.org/journals/tran/1990-319-02/S0002-9947-1990-0998124-1/].

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