I'm currently reading the Seiberg-Witten paper on $N=2$ supersymmetric Yang Mills pure gauge theory (i.e. no hypermultiplets). I have the following question:

How does one understand that the metric on the moduli space of the full quantum theory is the same as the metric obtained from the Kahler potential for the scalar field (or in general the $N=1$ chiral superfield) in the low-energy effective theory? On the face of it, the two things seem quite different - while the moduli space is the space of all gauge-inequivalent vacua in the full theory, the Kahler metric is derived from the Kahler potential in the low energy theory.


1 Answer 1


Dear Onkar, they're the same thing. The (massless or at least light) scalar fields are what parameterizes the moduli spaces - in any theory - and the metric on the moduli space (which is a mathematical concept that doesn't "a priori" exist in physics) is defined from the (ultimate low energy) kinetic terms of these scalar fields. In a supersymmetric theory, these kinetic terms $$\frac{1}{2}g_{ij}(\phi_a) \partial_\mu \phi^i \partial^\mu \phi^j$$ are determined from the Kähler potential, $g_{i\bar j}\sim\partial_i \partial_{\bar j} K$ because of the basic supersymmetric calculus. There is a lot of nontrivial maths here but the particular statement you're quoting is a tautology.

  • $\begingroup$ š Alright, now that I think about it, it is almost obvious that the light scalars parametrize the moduli space of the full quantum theory. Thank you very much for your answer. $\endgroup$
    – Onkar
    Commented May 7, 2011 at 14:53
  • 1
    $\begingroup$ It was a pleasure. Moduli spaces always share the fact that they're scalars - but they appear in very many very different physical manifestations. The scalar fields may parameterize moduli spaces of spacetime fields in unrealistic theories; spacetime itself that may be a moduli space on the world sheets or world volumes or boundaries in CFT; or the moduli spaces of instanton-like solutions or shapes of Riemann surfaces that have to be integrated over, and so on. In all cases, there are some scalars and their kinetic terms give the metric. $\endgroup$ Commented May 7, 2011 at 15:43

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