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Why is this form used here?

+ 2 like - 0 dislike
117 views

I was reading this paper, and while I did, I have one question:

Why is that

In the supergravity literature, one often formulates local special geometry in terms of a section of a different bundle. This section is denoted as $V$ and is related to $v$ by

$$V \equiv e^{K/2}v$$, where $K$ is Kaehler potential? That is, why is that it is no longer written as $V$ (aka the symplectic section), but now there is a factor of $e^{K/2}$ hanging around in front of $v$? Why this form?

Please note that

the geometries appearing in $N = 2$ supergravity theories are denoted as local special geometries. In mathemetics literature the local special geometry is called ’projective special geometry’.

Extra notes in order to clarify any vague definitions above:

On p.15 it says

There exists a holomorphic $Sp(2n+ 2;\mathbf{R})$-vector bundle $\mathcal{H}$ over the manifold and a holomorphic section $v(z)$ of $\mathcal{L} \otimes \mathcal{H}$. Note that $\mathcal{L}$ denotes the holomorphic line bundle over the manifold, of which the first Chern class equals the cohomology class of the Kahler form.

This post imported from StackExchange Physics at 2015-12-24 12:30 (UTC), posted by SE-user PhilosophicalPhysics
asked Dec 23, 2015 in Theoretical Physics by PhilosophicalPhysics (20 points) [ no revision ]
Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot.

This post imported from StackExchange Physics at 2015-12-24 12:30 (UTC), posted by SE-user Qmechanic

Hi PhilosophicalPhysics, welcome to PhysicsOverflow!

I saw your comments on the original SE question, if you need access to your imported account and content you can follow the instructions here.

1 Answer

+ 4 like - 0 dislike

On the complex line bundle $\mathcal{L}$, we have two natural geometric structures: an holomorphic structure and an hermitian metric. For a section of $\mathcal{L}$ we thus have to natural constraints, in general incompatible, that we can impose on our section: holomorphy or unitarity (i.e. constant norm with respect to the hermitian metric). Up to some i factor, $e^{-K}$ is the norm squared of the holomorphic section $v$. To construct a unitary section $V$, it is enough to rescale $v$ by the inverse of the square root of its norm squared, i.e. by the (non holomorphic) factor $e^{K/2}$. Depending on what one is trying to do it can be more convenient to work with the holomorphic section $v$ or with the unitary section $V$.

answered Dec 24, 2015 by 40227 (4,660 points) [ revision history ]

Thanks @40227 a lot for your answer. I asked this on another forum and maybe someone moved it here , so it was just today that I saw your answer. I have two questions, first you said "Up to some i factor, $e^{−K}$ is the norm squared of the holomorphic section $v$. " How do we know that? Second, concerning this: "it is enough to rescale $v$ by the inverse of the square root of its norm squared, i.e. by the (non holomorphic) factor $e^{K/2}$", I did not understand here the mathematics, why inverse of the square root of its norm squared?

On the first question: I guess it depends on what is your starting point and indeed the paper linked in the question does not exactly use this language. Let me say the following: if $\omega$ is a Kähler form and $\mathcal{L}$ an holomorphic line bundle of first Chern class equals to the class of $\omega$ then a choice of Hermitian metric on $\mathcal{L}$, such that the curvature of the associated Chern connection is $\omega$, is a global version of a choice of Kähler potential for $\omega$.

On the second question: my formulation was maybe a bit strange but I was simply saying that if I have a vector $v$, then to construct a vector of norm 1, I take $v/||v||$ where $||v||=\sqrt{||v||^2}=\sqrt{<v,v>}$.

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