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If $\mathfrak{g}$ is a Lie algebra (over $k$ ), then we can construct its universal enveloping algebra $U(\mathfrak{g})$. We can define $\Delta:U(\mathfrak{g})\rightarrow U(\mathfrak{g})\otimes U(\mathfrak{g}) $ as a natural extension of $\mathfrak{g}\ni x\mapsto 1\otimes x + x\otimes 1\in U(\mathfrak{g})\otimes U(\mathfrak{g})$. We can also define $\varepsilon:U(\mathfrak{g})\rightarrow k$ by $\varepsilon(x)=0, \mathfrak{g}\ni x\neq 1, \ \varepsilon(1)=1$. There is theorem that $U(\mathfrak{g})$ with $\Delta,\varepsilon$ form a bialgebra. $\Delta(1)$ is from definition equal to $1\otimes 1 + 1\otimes 1$, but if $U(\mathfrak{g})$ forms a bialgebra it should be equal $1\otimes 1$, because in bialgebra unit element should be group-like. Where is the mistake ?

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The mistake is that the definition you give only works for $x\in\mathfrak{g}$. A general element is not of this form. The coproduct will be the unital algebra homomorphism having this form for each element of the Lie algebra, which is the generating set, and the coproduct of a general element is a sum of products of these.

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  • $\begingroup$ In other words, you misunderstood what "the natural extension" of the map $x\ni\mathfrak g\mapsto x\otimes1+1\otimes x\in U(\mathfrak g)\otimes U(\mathfrak g)$ to the whole of $U(\mathfrak g)$ is. $\endgroup$ – Mariano Suárez-Álvarez Jul 11 '15 at 18:00
  • $\begingroup$ Thanks ! Is there any simple method to conclude that $\Delta(1)=1\otimes 1$ without checking that $U(\mathfrak{g})$ form bialgebra ? $\endgroup$ – mikis Jul 11 '15 at 18:20
  • $\begingroup$ @mikis $\Delta(1)=1\otimes 1$ simply by definition. Otherwise it could not be a unital algebra homomorphism. $\endgroup$ – Matt Samuel Jul 11 '15 at 18:22
  • $\begingroup$ Right. It's was obvious. I should go to sleep, it's too late for me ;) $\endgroup$ – mikis Jul 11 '15 at 18:29

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