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  What Exactly are Supersymmetric Cycles in String Theory?

+ 5 like - 0 dislike

In a lot of physics papers I hear talk of "wrapping a stack of D-branes on a supersymmetric cycle in a Calabi-Yau threefold $X$" and I'm wondering if I understand this correctly from a mathematical perspective.  

Let's assume we want to use D-branes to engineer BPS particles in four dimensions.  Therefore, our D$p$-branes will have $p$ real spatial dimensions in $X$, and only the worldline supported in four dimensions.  Okay, in Type IIA we therefore have D6, D4, D2, and D0 branes.  The mass of the particle should be proportional to the volume of the brane.  Therefore, if we want BPS particles, we want minimal mass and hence, minimal volume.  We therefore want our branes wrapping calibrated submanifolds (these minimize volume within their homology class).  Since $X$ is Calabi-Yau, the calibrated submanifolds are either holomorphic submanifolds or special Lagrangian submanifolds.  

It follows that using Type IIA, we can get BPS particles with D-branes wrapping holomorphic cycles in $X$ and using Type IIB we get BPS particles with D3-branes wrapping special Lagrangians.  (Please correct me if any of this is mistaken!) 

(In the Type IIA case) are supersymmetric cycles simply holomorphic cycles?  Meaning we necessarily get BPS particles?

The reason I'm confused is because on page 31 of (https://arxiv.org/abs/math/0412328) they write "...in order to give a supersymmetric configuration the mass $M$ of a D-brane and its central charge have to satisfy the inequality $M \geq |Z(Q)|$."  

So are supersymmetric cycles when that bound is saturated or when that bound is satisfied?  Is the idea that supersymmetric cycles are those homologous to a holomorphic submanifold?  So wrapping a brane on the holomorphic submanifold itself gives you a BPS state whereas wrapping it on a cycle homologous will just give you a supersymmetric state?  

asked Aug 14, 2018 in Theoretical Physics by Benighted (310 points) [ no revision ]

1 Answer

+ 6 like - 0 dislike

A supersymmetric cycle is such that, wrapping a brane on it, the resulting theory on the non-compact directions of the brane is supersymmetric. If you restrict yourself to particles (one non-compact direction: the time one), then you are right: supersymmetric cycles define BPS particles and they are holomorphic cycles in IIA and Lagrangian submanifolds in IIB.

But if one does not restrict to particles, other combinations are possible. For example, wrapping a D4 brane on a Lagrangian submanifold in IIA will produce a BPS string in 4d non-compact spacetime.

You are right to be confused by the paper you linked to because I think this phrase is just incorrect. Any configuration satisfies the BPS inequality $M \geqslant |Z(Q)|$ (this follows from the supersymmetry of the theory without branes, i.e. of the Calabi-Yau condition). The geometric translation is that the volume of any submanifold in an homology class is bounded below by the volume of an holomorphic or Lagrangian submanifold in the same homology class. A supersymmetric/BPS configuration is one saturating this BPS bound.

A final remark: this geometric description of supersymmetric cycles in terms of calibrated geometry is in general only valid in the large volume limit.

answered Aug 16, 2018 by 40227 (5,140 points) [ no revision ]

Can you elaborate on your closing comment about the large volume limit?

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