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In recent papers the circular Wilson loop in $\mathcal{N}=4$ SYM is always called a 1/2 BPS operator. So, my initial idea was that a 1/2-BPS operator was an operator that preserves half of the supersymmetries, in the $\mathcal{N}=4$ SYM it would be $32\times 1/2=16$ supersymmetries. There are also 1/4-BPS Wilson loops, or 1/8-BPS Wilson loops, for example. Then I suposed that a 1/4-BPS would preserve 8 supersymmetries and a 1/8-BPS, 4 supersymmetries.

Then I was looking for some of the original works. I found

https://arxiv.org/pdf/hep-th/0205160.pdf

He says (page 2) "The circular Wilson loop does not preserve any supersymmetry, but commutes with eight linear combinations of supersymmetry and superconformal generators"

So that confused me, is not it a supersymmetric Wilson loop?

What does it mean that the circular Wilson loop is 1/2-BPS?

Does it mean that it preserves 16 supersymmetries?

Is it right to say that the circular Wilson loop does not preserve any supersymmetry?

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  • $\begingroup$ Minor comment to the post (v2): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Jul 18 '19 at 17:30
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The supersymmetry variations of the circular Wilson loop is not sufficient to establish this but we also need to include special superconformal (note that \mathcal{N} = 4 SYM is a supersymmetric CFT) variations to find linear combinations which leave the operator invariant. See for example https://arxiv.org/abs/0704.2237 where there is sufficient discussion of this point, around Eq.(9) and Eq.(14). We construct linear combinations of $\overline{Q}$ and $\overline{S}$ which leaves the circular loop invariant and hence we get 1/2 BPS object.

For the case of Wilson loops, which are straight lines, it is 1/2 BPS already even without doing this linear combination step. In fact, the expectation value is always 1 i.e. $ \langle W(-)\rangle = 1$ whereas $\langle W(O)\rangle \propto f(\sqrt{\lambda})$ where I have used - to denote straight line and O to denote circular 1/2 BPS loop.

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