# BPS states : Mathematical definition

First of all, let me congratulate the theoretical physics community for this site. I am a mathematics student with very little background in phyiscs. The question I want to ask is:

What is the proper mathematical definition of BPS states?

In string theory the BPS states correspond either to coherent sheaves or special Lagrangians of Calabi- Yau manifold depending upon the type of string theory considered. but in SUSY quantum field theories in 4d, there are no CYs as far I know (which is very little) and in gravity theories, these corresponds to some Black Holes. So what is the general mathematical definition of BPS states which is independent of the theory in consideration, say a general SUSY Quantum field theories, be it QFT, string theory, Gravity and in any dimension.

I am very sorry, if the question is not framed properly due to my ignorance of the physics involved. Thanks

-
J Verma, welcome on TP.SE. Please expand BPS acronym at least once. –  Piotr Migdal Oct 8 '11 at 19:43
BPS = Bogomol'nyi–Prasad–Sommerfield –  José Figueroa-O'Farrill Oct 8 '11 at 20:09

## 2 Answers

In any supersymmetric theory, a BPS state is a state which preserves some of the supersymmetry.

If we take as a definition of a supersymmetric theory, some theory (classical or quantum) which admits a Lie superalgebra of symmetries, then a BPS state (or configuration) of such a theory is one which is annihilated by some nonzero odd element in the superalgebra.

Of course, the original meaning comes from the study of magnetic monopoles. Solutions of the Bogomol'nyi equation are precisely those which saturate a bound, the so-called Bogomol'nyi-Prasad-Sommerfield (BPS) bound.

The relation between the two notions is related to the fact that monopole configurations can be thought of as configurations in a four-dimensional $N=2$ supersymmetric gauge theory which preserve half of the supersymmetry.

-
Here is another question from a mathematician: some physicists have told me that in the context of 4d gauge theory one can think about BPS states as cohomology of the corresponding moduli space of instantons (but some other physicists deny that). In what generality can one claim such thing? For example, what happens if go from pure gauge theory to gauge thery with matter (the instanton moduli space doesn't change...)? What happens if we look at 5d theory compactified on $S^1$? Is there a way to answer such questions in general? –  Alexander Braverman Oct 8 '11 at 21:43
Alexander: the interpretation of the BPS states depends very strongly on the details of the supersymmetric theory one considers. It is not possible to make such sweeping generalisations. In the context of topologically twisted Yang-Mills theory, the BPS states are indeed annihilated by the BRST operator, which can be interpreted as the de Rham differential in the moduli space of instantons. (More precisely it's the equivariant version of BRST cohomology which can be so interpreted, but this is typical in the gauge theoretic context.) –  José Figueroa-O'Farrill Oct 8 '11 at 22:27
Is it important which topological twist you are taking? I suppose you meant the one which leads to Donaldson theory? Also, the context in which I met this statement did not mention topological twist - for example in the last section of Witten's paper "Geometric Langlands from 6 dimensions". –  Alexander Braverman Oct 9 '11 at 14:08
Alexander: indeed, different twists correspond to different differentials. For 4-d N=2 supersymmetric Yang-Mills, though, all topological twists are equivalent. The story is different in N=4, though, which might be the context of Witten's paper. –  José Figueroa-O'Farrill Oct 9 '11 at 14:23
Alexander: I suggest you ask this question in the site. I'm not sure that I can give you better answers than you might get from others here. (I have not looked at that paper of Witten's in a while, but I would be surprised if this were not in the context of a topologically twisted $N=4$ theory. The geometric Langlands correspondence is basically electromagnetic duality applied to one of the possible $N=4$ twisting, if I recall correctly.) –  José Figueroa-O'Farrill Oct 9 '11 at 21:32

The BPS bound was discovered independently of supersymmetry, but it was then better understood as general feature of the supersymmetry algebra. Look at the original paper by Witten and Olive. BPS states are states which saturate the BPS bound, forming "short" representations an extended supersymmetry algebra. Such representations have special properties, which often can be thought of as consequence of some fraction of the supersymmetry which remains in their presence. The examples you cite are special cases, in all those cases the special property of the objects you mention is a deduced from the requirement that the states involved form a short representation of the supersymmetry algebra.

-