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Free Leibniz algebras are defined as follows:

Let $X$ be a set and $F(X)$ be a non associative algebra and on that let $I$ be two sided ideal generated by $[a,[b,c]]-[[a,b],c]-[[a,c],b]$ for $a,b,c \in F(X)$. Then $L(X)=F(X)/I$ is called free Leibniz algebra.

In the case of free Lie algebras we have an obvious map from free Lie algebra to a Lie algebra. Is a a map $\phi \colon L(X) \to L$, where $L$ is an arbitrary Leibniz algebra definable?

Jendrik Stelzner
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Nil
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  • There a free non-associative algebra $F(X)$, a free Leibniz algebra $Le(X)$, a free Lie algebra $L(X)$, and canonical surjective homomorphisms $F(X)\to Le(X)\to L(X)$. – YCor Jul 05 '16 at 22:36
  • Thank you very much. I have one one more enquiry in the case of Leibniz algeebra: Does this canonical homomorphis map induce a map $\theta Le(X) / Le_{i} \to L(X)/L_{i} $ , ?(where $Le_{i}$ and $L_{i} $ are the i-th term of lower central series.) – Nil Jul 05 '16 at 22:57
  • $\theta: Le(X)/Le(X){i} \to L(X)/L(X){i}$ – Nil Jul 05 '16 at 23:04
  • If one defines the lower series in an algebra $A$ as: $A_i$ is generated by all possible products involving $i$ elements, then clearly every homomorphism $A\to B$ maps $A_i$ into $B_i$, hence induces $A/A_i\to B/B_i$. – YCor Jul 06 '16 at 23:17

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The composite $X \to F(X) \to F(X)/I = L(X)$ is called the canonical map from $X$ to $L(X)$. The free Leibniz algebra satisfies the following universal property:

Let $X$ be a set, let $L(X)$ be the free Leibniz algebra on $X$, and let $i \colon X \to L(X)$ be the canonical map. For every Leibniz algebra $L$ and every map $f \colon X \to L$, there exists a unique homomorphism of Leibniz algebras $φ \colon L(X) \to L$ with $φ ∘ i = f$.

This is basically the same universal property as for free Lie algebras, but with every instance of “Lie algebra” replaced by “Leibniz algebra”.

Jendrik Stelzner
  • 16,401