Do we need to harmonize isotypic components to solve SDPs?

[denisrosset]

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]

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]

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.