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Collection of sketches / project ideas for RepLAB
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Do we need to harmonize isotypic components to solve SDPs? #8

Open denisrosset opened 4 years ago

denisrosset commented 4 years ago

Basically, given irreducible representations, say rep1 and rep2, from the same isotypic component, do we need to have rep1 and rep2 in the same basis to solve the block-diagonalized SDP?

My intuition is that we only need one basis vector from each to construct the 2x2 block that is finally solved.

If that's true, that'd remove a source of numerical errors (the basis harmonization step) in the use of RepLAB on SDPs, and would have an impact on the data structures in isotypic components.

jdbancal commented 4 years ago

Good question. Say we have indeed two copies of an identical representations. If we can write the structure of the commutant in some basis as M \otimes Id_2, then the positivity condition can be expressed as simply M>=0. What would be the form of the commutant if the bases of both representations are not identical? Would we have M_1 \oplus M_2? Or something more precise? Would you have a (small) concrete example to better understand what is at stake exactly here?

denisrosset commented 4 years ago

First, we are not speaking of the commutant, but the "Hermitian invariant" space, as the SDP matrix transforms as V * X * V', not V * X * inv(V) where V is the representation image.

I guess that space would have the structure M (x) A where A is a positive definite matrix; and M (x) A being SDP is the same as M being SDP; in essence, we don't need all the rows and columns of M (x) A, just a submatrix of it.