fa.functional analysis – Density of a graded algebra

I’m trying to prove the following proposition:

If $ v in V $ and $ Y in mathfrak{so} (V) $ then $(dotmu(Y), B(v)) = B(Yv)$.

By definition
$(dotmu(Y), B(v)) = dotmu(Y)B(v) – B(v)dotmu(Y).$ I arrive at
= B(v),dotmu(Y),f_S + B(Yv),f_S ,
iff B(Yv), f_S = dotmu(Y),B(v),f_S – B(v),dotmu(Y),f_S,.$$

The set $ {f_S: g cdot S : : mathrm{exists}} $
generates a dense subspace of $ mathcal F ^ + (V) $, the even subalgebra of $ mathcal F (V) $. But this doesn’t imply that ${f_S: g cdot S : : mathrm{exists}}$ is dense in $mathcal F (V)$ since density is not commutative. ¿Any idea on how to conclude this?

Note: Here $mathcal F (V)$ is the Fermionic Fock space: $bigoplus_{k=0}^m V^{wedge k}$, where $V$ is a (complex) vector space and $mathfrak{so} (V)$ is the Lie algebra of the special ortogonal group.

Thank you :).