I found this notation $U^\boxplus$ in the paper 'Superpotentials for superconformal Chern-Simons theories from representation theory' by Paul de Medeiros, José Figueroa-O'Farrill and Elena Méndez-Escobar. Archive: https://arxiv.org/abs/0908.2125. They don't explain this notation while making a lot of effort explaining other notations they introduce, so I guess this one is fairly familiar within mathematical physics. However as a representation theorist I have never seen it before.

The context is the decompostion $\Lambda^2S^2U = \Lambda^4U \oplus U^\boxplus$ where $U$ is a (real) vector space, so it stands to reason that $\Lambda$ and $S$ denote exterior and symmetric powers respectively.

The vector space $U$ is a representation of a Lie algebra, but I do not know whether this plays any role in the definition of $U^\boxplus$.

Can anyone tell me what to make of this notation? Thanks in advance!


I believe $\boxplus$ is a Young tableau. The representation $U^\boxplus$ is what you get if you apply the Schur functor associated to $\boxplus$ on $U$. The conventions should be exactly as set out on page 46 and page 76 of Fulton and Harris.

Since you're a rep theorist, I would imagine that you understand this better than me - it must be the physicists' notation that is confusing. I am not familiar with this particular paper, but I have seen this notation used in many other physics papers.

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    $\begingroup$ I would add that this can also be understood through the symmetric group. If $V$ is any vector space (even one without a Lie action), the rank-2 tensors on $V$ decompose into symmetric and alternating tensors: this is the statement $V^{\otimes 2} = \bigwedge^2 V \oplus S^2 V$. There is an analogous statement for fourth-rank tensors, where you get 5 summands, one for each irreducible character of $S_4$. $\endgroup$ – Joppy May 14 '17 at 7:56
  • $\begingroup$ Yes it all fits together really nicely, I wouldn't have thought about it on my own! This is why MSE is so great. Thank you so much (also Joppy)! $\endgroup$ – Vincent May 15 '17 at 22:51
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    $\begingroup$ This answer is correct. The notation, which is perhaps not standard, is precisely as explained in this answer. I would say, though, that it is defined implicitly by the formula quoted in the OP. It is the kernel of the "algebraic Bianchi" map $S^2\Lambda^2 U \to \Lambda^4 U$; that is, it is the space of algebraic curvature tensors with tangent representation $U$. $\endgroup$ – José Figueroa-O'Farrill May 15 '17 at 22:52

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