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  Which supersymmetry principles make the axio-dilaton field in F-theory holomorphic?

+ 5 like - 0 dislike

I am following the lecture notes on F-theory by T. Weigand . On page 7 he states that when considering F-theory as the strong coupling limit of type IIB orientifolds with D-branes, the axio-dilaton field defined on the coordinates perpendicular to the brane has to depend holomorphically on these coordinates.

My question is: what are these conditions? A reference would be greatly appreciated as well.

This post imported from StackExchange Physics at 2016-06-17 12:19 (UTC), posted by SE-user saesori

asked Apr 24, 2016 in Theoretical Physics by saesori (25 points) [ revision history ]
edited Jun 17, 2016 by Dilaton

1 Answer

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To make holomorphicity apparent in string theory one uses light-cone coordinates. Then we usually have fields like $\phi(z)$ and their conjugates $\bar{\phi}(\bar{z})$. The same happens with axion and dilaton. When you combine them into one field you get something like $\tau(z) = \rho(z) + i g_s(z)$ where $g_{s}(z) = e^{\phi(z)}$. Then it is very easy to confirm that $\tau(z) \in \mathfrak{H} = \{ (x,y) \in \mathbb{R}^2 | y > 0 \text{ and } z=x+iy \} $ and that $\tau$ transforms under $SL_2(\mathbb{Z})$. Actually, there is another way to see it in F-theory. Since F-theory uses elliptically fibered CY's you can see $\tau$ as the complex parameter of the fiber which has to be holomorphic. This arises from Type IIB and a D7-brane where the tranverse space is $S^1$. Then if we walk around a path on a fixed point on the D7-brane and find that the integral of the RR scalar $C_0=\rho$ is not zero we realize that $\rho$ is not single valued. We can translate this into a torus action on that point - i.e. you see a torus action over each point of the D7-brane where $\rho$ is not single valued. Enriching the fiber over this point with a CY is what actually F-theory is all about. This enrichment also can be done with any Torus or K3. 

If I find some time I will give a more complete answer (I have to run). 

answered Jun 20, 2016 by conformal_gk (3,625 points) [ no revision ]

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