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Let $C$ be a coalgebra, and take $c\in C$. Then $c$ is group-like if $\Delta c=c\otimes c$ and $\epsilon(c)=1_k$, and the set of group-like elements is denoted $G(C)$.

For $g,h\in G(C)$, $c$ is $g,h$-primitive if $\Delta c=c\otimes g+h\otimes c$; the set of such elements is denoted $P_{g,h}(C)$.

If $C$ is a bialgebra and $g=h=1$, then elements of $P(C):=P_{1,1}(C)$ are simply called primitive elements of $C$.

So let $\mathfrak{g}$ be a Lie algebra over a field $k$ (with char$=0$) and let $B=U(\mathfrak{g})$ be its universal enveloping algebra. Then $B$ is a bialgebra via $\Delta x=x\otimes 1+1\otimes x$ and $\epsilon(x)=0_k$ for all $x\in\mathfrak{g}$.

The claim then is that $P(B)=\mathfrak{g}$. For this to be true, we need $1\in G(B)$, meaning we need $\Delta 1=1\otimes 1$ and $\epsilon(1)=1_k$. But neither of these is true!

What's going on here? What am I missing?

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1 Answer 1

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The unit $1 \in U(\mathfrak g)$ is indeed grouplike. Both $\Delta(1) = 1 \otimes 1$ and $\varepsilon(1) = 1$ hold because both $\Delta$ and $\varepsilon$ are homomorphisms of $k$-algebras.

Edit:

Here is an example. Let $k$ be a field and let $\mathfrak g = kx \oplus ky$ be a two dimensional abelian lie algebra. So $\{x, y\}$ is a basis of $\mathfrak g$ and $[x, y] = 0$.

Now the universal enveloping algebra of $\mathfrak g$ is the symmetric algebra $S^\ast(\mathfrak g)$ modulo the commutator relations. Well, the symmetric algebra $S^\ast(\mathfrak g)$ can be identified with $k\langle x, y\rangle$, the ring of non-commuting polynomials in $x$ and $y$. The commutator relations say exactly that $x$ and $y$ commute. So when we take the quotient we get that the enveloping algebra is a polynomial ring:

$U(\mathfrak g) = k[x, y]$

The inclusion $\mathfrak g \subseteq k[x, y]$ is the obvious, it's the span of $x$ and $y$ in $k[x, y]$ so the map

$\Delta\colon k[x, y] \to k[x, y] \otimes_k k[x, y]$

is an algebra map defined by $x \mapsto x \otimes 1 + 1 \otimes x$ and $y \mapsto y \otimes 1 + 1 \otimes y$.

This does not mean that those formulas hold for all elements of $k[x, y]$.

For example $\Delta(xy) \neq xy \otimes 1 + 1 \otimes xy$. Instead we compute $\Delta(xy)$ as:

$\begin{align*} \Delta(xy) &= \Delta(x)\Delta(y) \\ &= (x \otimes 1 + 1 \otimes x)(y \otimes 1 + 1 \otimes y) \\ &= xy \otimes 1 + x \otimes y + y \otimes x + 1 \otimes xy \end{align*}$

Now if you think about the degree's of the polynomials you'll see that if $\Delta$ is defined this way then a polynomial $f$ satisfies $\Delta(f) = f \otimes 1 + 1 \otimes f$ if and only if $f$ is homogeneous of degree $1$, i.e., if $f \in \mathfrak g \subseteq k[x, y]$.

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But I don't see how. By definition, $\Delta(1)=1\otimes 1+1\otimes 1$ and $\epsilon(1)=0$. –  Bey Feb 2 '13 at 19:56
    
No, by definition $\Delta(x) = x \otimes 1 + 1 \otimes x$ and $\varepsilon(x) = 0$ **for all $x in \mathfrak g$**. The algebra $U(\mathfrak g)$ is generated by $\mathfrak g \subseteq U(g)$ so to specify an algebra map you need only specify it's value on the generators. I will edit my answer with an example which might make things clearer. –  Jim Feb 2 '13 at 20:44
    
Ah I see. I was mistakenly thinking that the $1$ in $\mathfrak{g}$ is the same as the $1$ in $U(\mathfrak{g})$. –  Bey Feb 2 '13 at 21:08
1  
Lie algebras don't have a $1$. –  Jim Feb 4 '13 at 4:42

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