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.
answered Jul 11, 2015 at 17:47
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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. $\endgroup$ Commented Jul 11, 2015 at 18:00
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$\begingroup$ Thanks ! Is there any simple method to conclude that $\Delta(1)=1\otimes 1$ without checking that $U(\mathfrak{g})$ form bialgebra ? $\endgroup$– mikisCommented Jul 11, 2015 at 18:20
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1$\begingroup$ @mikis $\Delta(1)=1\otimes 1$ simply by definition. Otherwise it could not be a unital algebra homomorphism. $\endgroup$ Commented Jul 11, 2015 at 18:22
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1$\begingroup$ Right. It's was obvious. I should go to sleep, it's too late for me ;) $\endgroup$– mikisCommented Jul 11, 2015 at 18:29