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Apr 13, 2021 at 21:32 comment added Benjamin Steinberg @Darijgrinberg me neither. But it depends on what the obstacle is. The projective indecomposables are summands in the group algebra so it is conceivable that if the radical is very complicated we might understand the projective indecomposable better than the simple
Apr 13, 2021 at 21:25 comment added darij grinberg @BenjaminSteinberg: Maybe, but I wouldn't bank on it.
Apr 13, 2021 at 19:47 comment added Benjamin Steinberg @Darijgrinberg probably it is enough to show the projective indecomposables are realizable over Fp. Maybe that is easier combinatorially?
Apr 13, 2021 at 19:15 comment added darij grinberg I don't think there is a combinatorial proof of the $\mathbb F_p$-version of this result -- we don't know the irreps over $\mathbb F_p$ well enough. If anyone has seen such a proof, I'd love to know!
Apr 13, 2021 at 19:14 comment added darij grinberg The combinatorial proof of $\mathbb QS_n\cong \prod_{i=1}^{p_n}M_{d_i}(\mathbb Q)$ using the seminormal form is exposed, e.g., in Murray R. Bremner, Sara Madariaga, Luiz A. Peresi, Structure theory for the group algebra of the symmetric group, with applications to polynomial identities for the octonions, 2016 (I have written some sidenotes to this exposition). I think it's also done in the recent Garsia/Egecioglu book, but probably not as quickly and self-containedly as in the paper mentioned.
Apr 13, 2021 at 18:55 history edited Benjamin Steinberg CC BY-SA 4.0
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Apr 12, 2021 at 21:52 history edited Benjamin Steinberg CC BY-SA 4.0
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Apr 12, 2021 at 21:43 comment added John Baez Nice! This is the cleanest approach.
Apr 12, 2021 at 15:10 history edited Benjamin Steinberg CC BY-SA 4.0
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S Apr 12, 2021 at 14:17 history answered Benjamin Steinberg CC BY-SA 4.0
S Apr 12, 2021 at 14:17 history made wiki Post Made Community Wiki by Benjamin Steinberg