We know that for any Leibniz algebra $L$ we can associated its Lie algebra denoted by $L_{Lie}$. for example the ideal generated by $\{[x,x] | x\in L\}$ determines the non-Lie character of $L$. Is it possible to find the ideal which is the largest ideal of $L$ determining the non-Lie character of $L$? if so, how can we find that?
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Let $L$ be a Leibniz algebra and $I$ be the vector space spanned by $\{[x,x]\mid x \in L\}$. It is known to be an ideal, called the Leibniz kernel. The quotient Leibniz algebra $\mathfrak{g}_{L}=L/I$ is called liezation of $L$ and the projection $f:L\to \mathfrak{g}_{L}$ is universal in the sense that if you have a Leibniz algebra homomorphisms $\phi:L\to \mathfrak{g}$, where $\mathfrak{g}$ is a Lie algebra, then there exists a Lie algbera homomorphism $h:\mathfrak{g}_{L}\to \mathfrak{g}$ such that $h \circ f=\phi$ (any Leibniz morphism to a Lie algebra factors through the liezation).
It is not clear what you mean by non-Lie character of a Leibniz algebra. Sure, there are ideals bigger than the Leibniz kernel, say the right annihilator $Ann^r(L)$ (if you consider right Leibniz algebras), the quoitient by which is a Lie algebra $L/Ann^r(L)$ anti-isomoprhic to the inner derivations of the Leibniz algebra $L$. But if you consider the "semisimple" Leibniz algebras, then biggest among those ideals is the Leibniz kernel.
