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Feb 4, 2016 at 12:54 comment added Bulois Michael Sorry, no idea for your particular equation. But generally, I would say that it is indeed usually possible to describe a quotient in either of the two following cases: -a relatively small closed subvariety where nilpotents do not behave very badly (e.g. commuting variety in $\mathfrak g\times \mathfrak g$) -an open locus which keeps only the very general points
Feb 4, 2016 at 12:02 comment added Ehud Meir what if we restrict somehow the variety? that is: if instead of taking the entire $End(V\otimes V)$ we just take some constructible subset? Are there any nice quotients which one can describe? One of the things I am interested in is the subvariety of all solutions for the Young-Baxter Equation: $(T\otimes 1) (1\otimes T)(T\otimes 1) = (1\otimes T) (T\otimes 1) (1\otimes T)$
Feb 4, 2016 at 8:26 history edited Bulois Michael CC BY-SA 3.0
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Feb 4, 2016 at 8:23 comment added Bulois Michael It smells like a wild problem to me. Indeed, just in $(\mathbb K\otimes\mathfrak g)\oplus(\mathfrak g\otimes \mathbb K)\cong\mathfrak g \oplus\mathfrak g$, you have some closed orbits arising from nilpotent elements. E.g. in $\mathfrak{sl}_2\times\mathfrak{sl}_2$, $(e,f)$ has a closed orbit. The same phenomenon should also arise in $\mathfrak g\otimes\mathfrak g$
Feb 3, 2016 at 16:35 comment added Ehud Meir We can use another work of Procesi, to describe explicitly all the invariants: They can be described in the following way: if $T:V\otimes V\rightarrow V\otimes V$, then take $Tr(\sigma T^{\otimes n}):V^{\otimes 2n}\rightarrow V^{\otimes 2n}$, where $n$ is some number and $\sigma\in S_{2n}$ acts by permuting the tensor factors. By considering the partial traces $T_1,T_2:V\rightarrow V$ I am also convinced now that the generic stabilizer is finite. However, I will be happy to have a better description of some possible quotient spaces.
Feb 3, 2016 at 16:29 history answered Bulois Michael CC BY-SA 3.0