Here's a chaser to this question.

Recall the proof that the number of partition of an integer $n$ into at most $k$ addends is the same as the number of partition of an integer $n$ into integers no larger than $k$: you flip the corresponding Young's diagrams (drawn in French notation) over the main quadrant diagonal. Is this flipping operation related to a Fourier transform of something out there?

You can also add a couple of binary operations on Young diagrams: "multiplication" and "convolution". "Multiplication" draw the 1st diagram above the 2nd one and drops the columns to close the gaps: $(a*b)_i=a_i+b_i$, where $a_i,b_i$ are $i^{th}$ addends in the partitions. Similarly, "convolution" draw the 1st diagram on the right of the 2nd one and "pulls" rows to the left until the gaps are closed. Clearly, the above "flipping operation" interchanges multiplication and convolution. Is that property enough to model the "flipping operation" as a Fourier transform on something?

  • $\begingroup$ You could also write this in fermionic fock space terms. $\endgroup$
    – AHusain
    Commented Aug 30, 2016 at 21:26

1 Answer 1


A sort of answer:

There are two ways to realise the Springer correspondence in the language of perverse sheaves: one by restriction (see [BM]), and one by Fourier transform (see [HK]).

[BM] Borho, Walter; MacPherson, Robert Partial resolutions of nilpotent varieties. Astérisque 101-102, 23-74 (1983).

[HK] Hotta, R.; Kashiwara, M. The invariant holonomic system on a semisimple Lie algebra. Invent. Math. 75, No. 2, 327-358 (1984).

It turns out that these two correspondences are related by tensoring with the sign representation which in type A is "flipping the young diagram". Thus it is not too much of an exaggeration to say that under the Springer correspondence for $S_n$, flipping the Young diagram is the Fourier transform.

PS: I haven't thought if this explains your "multiplication" and "composition" operations...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.