Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.


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 $f \in \mathfrak g \subseteq k[x, y]$.

share|cite|improve this answer
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
Lie algebras don't have a $1$. – Jim Feb 4 '13 at 4:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.