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Let $\mathfrak g$ be a semisimple Lie algebra, $\mathfrak h\subset\mathfrakg$ a reductive subalgebra such that $\mathfrak h^\perp$ is a complementary$\mathfrak h$-submodule of $\mathfrak g$. In 1983, Bogoyavlenski claimed thatone obtains a Poisson commutative subalgebra of the symmetric algebra${\mathcal S}(\mathfrak g)$ by taking the subalgebra ${\mathcal Z}$ generatedby the bi-homogeneous components of all $H\in{\mathcal S}(\mathfrakg)^{\mathfrak g}$. But this is false, and we present a counterexample. We alsoprovide a criterion for the Poisson commutativity of such subalgebras${\mathcal Z}$. As a by-product, we prove that ${\mathcal Z}$ is Poissoncommutative if $\mathfrak h$ is abelian and describe ${\mathcal Z}$ in thespecial case when $\mathfrak h$ is a Cartan subalgebra. In this case,${\mathcal Z}$ appears to be polynomial and has the maximal transcendencedegree $(\mathrm{dim}\,\mathfrak g+\mathrm{rk}\,\mathfrak g)/2$.

Reductive subalgebras of semisimple Lie algebras and Poisson commutativity