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According to Loday and Prashvili's paper, to defined universal enveloping algebra of Leibniz algebras we need $g^{r}$ and $g^{l}$ as two copies of the Leibniz algebra $g$. What does it mean by copies of $g$?

Another question is that, we know that the opposite algebra of left Leibniz algebra is right Leibniz algebra, then is there any way to describe universal enveloping algebra via opposite algebra?

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  • $\begingroup$ I know nothing about Leibniz algebras, but.. "copy of" = "algebra isomorphic to" sound sensible $\endgroup$
    – Qfwfq
    Commented Jun 4, 2016 at 20:25

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For reference the paper mentioned in the question is here. As Qfwfq points out in the comments we are taking two Leibniz algebras $\mathfrak g^l$ and $\mathfrak g^r$ both isomorphic to our original Leibniz algebra $\mathfrak g$. Note the superscript is not saying anything about being a left or right Leibniz algebra, but rather is referring to the left or right factor in the direct sum used the the definition of the universal enveloping algebra. In fact, if we follow the paper take $\mathfrak g$ to a right Leibniz algebra, the both $\mathfrak g^l$ and $\mathfrak g^r$ are right Leibniz algebras since they are isomorphic to $\mathfrak g$. We can define the universal enveloping algebra without the superscripts as $UL(\mathfrak g) = T(\mathfrak g \otimes \mathfrak g)/I$ where $I$ is defined by the following relations:

$$(0,x) \otimes (0,y) - (0,y) \otimes (0,x) = (0,[x,y])$$ $$(x,0) \otimes (0,y) - (0,y) \otimes (x,0) = ([x,y],0)$$ $$(y,y) \otimes (x,0) = 0$$

Perhaps you prefer to see it written this way.

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