Is there some sort of intuition for why I should care about this? (Sorry to be a dumb computer scientist)
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Replying to @Meaningness @alicemazzy
It lets you automagically decompose a space/algebra/module of dimension n^2 into two pieces of dimension (n^2 + n)/2 and (n^2 - n)/2. Assume you have an algorithm that scales as dim^4 or 2^dim...
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It also lets you decompose higher-order powers V^(\tensor 2n) into correspondingly more pieces. Also those pieces tend to have interesting properties, eg https://en.wikipedia.org/wiki/Symplectic_geometry … uses exterior powers on the cotangent space of a manifold to do something or other
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Replying to @St_Rev @alicemazzy
OK, I was looking at that same Wiki article a couple of months ago for an unrelated reason (coming at it from Hamiltonian mechanics), so maybe someday I need to understand this!
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Replying to @Meaningness @alicemazzy
The little wedge thingies are (secretly) exterior product symbols, ie x \wedge y = x \tensor y - y \tensor x in some tensor space.
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This might help, or not at all https://en.wikipedia.org/wiki/Exterior_algebra …
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Replying to @St_Rev @alicemazzy
Ah, ok, the R^2 and R^3 examples are clear, so that gives me a (probably mistaken) idea that I have some feel for this
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Replying to @Meaningness @alicemazzy
The horrible mind-searing truth is that you can decompose an arbitrary tensor power V^{\tensor n} into pieces corresponding to the representations of the symmetric group S_n. But it's so horrible that no one wants to write out how to do it.
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"Ja, vell, obviously you chust use ze central idempotents e_i of kS_n" "Which are?" "Eh..."
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Note that I wanted to do all of this in prime characteristic, which is WAY HARDER STILL. Obviously I made enough progress to finish my dissertation but grinding through this kind of thing is not my strong point.
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Do I remember (super-vaguely, from years ago) that you use some sort of software to do the computations?
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