Determining the grouplike elements of a Hopf algebra

Question

Here is the question I am trying to solve:

For any Lie algebra $L$ determine the group of grouplike elements of the Hopf algebra $U(L).$

Some definitions:

1-To any Lie Algebra $L$ we assign an (associative) algebra $U(L),$ called the enveloping algebra of $L,$ and a morphism of Lie algebras $i_L: L \to L(U(L)).$ We define the enveloping algebra as follows. Let $I(L)$ be the two-sided ideal of the tensor algebra $T(L)$ generated by all elements of the form $xy - yx - [x,y]$ where $x,y$ are elements of $L.$ We define $$U(L) = T(L)/ I(L)$$

Where $U(L)$ is filtered as a quotient algebra of $T(L).$

2- Let $V$ be a vector space. Define $T^0 (V) = k, T^1(V) = V$ and $T^n(V) = V ^{\otimes n}$ (the tensor product of $n$ copies of $V$) if $n > 1.$ The canonical isomorphisms $$T^n (V) \otimes T^m(V) \cong T^{n + m}(V)$$ induce an associative product on the vector space $T(V) = \otimes_{n \geq 0} T^n (V).$ Equipped with this algebra structure, $T(V)$ is called the tensor algebra of $V.$ The product in $T(V)$ is explicitly defined by $$(x_1 \otimes \dots \otimes x_n)(x_{n + 1} \otimes \dots \otimes x_{n+m}) = x_1 \otimes \dots \otimes x_n \otimes x_{n + 1} \otimes \dots \otimes x_{n+m} \quad \quad \quad (5.1)$$

where $x_1, \dots , x_{n+1}, \dots , x_{n+m}$ are elements of $V.$ The unit for this product is the image of the unit element $1$ in $k = T^0(V).$ Let $i_V$ be the canonical embedding of $V = T^1 (V)$ into $T(V).$ By $(5.1)$ we have $$x_1 \otimes \dots \otimes x_n = i_V(x_1) \dots i_V(x_n)$$ which allows us to set $x_1 \dots x_n = x_1 \otimes \dots \otimes x_n$ whenever $x_1, \dots , x_n$ are elements of $V.$

3- A grouplike element of a coalgebra $(H, \Delta, \epsilon),$ i.e., an element $x \neq 0$ such that $$\Delta (x) = x \otimes x.$$

4- I also know that the set of grouplike elements of $H$ is a group.

Still I do not know how to answer it. Could someone show me some details please?

EDIT:

5- I also know the universal property of $U(L).$

EDIT-2:

Here is a theorem I have also but I do not know if it may help in my case or not:

Answer 1

Grouplike elements in a Hopf algebra are invertible, and an immediate consequence of the PBW theorem is that the only invertible elements in an enveloping algebras are the scalar multiples of the identity.

It follows that a grouplike element is of the form $x1_H$ for some scalar $x$, and from there it is easy to check that $x$ has to be $1$.

Answer 2

Let us recall the PBW theorem. Let $\pi : T(L) \rightarrow U(L)=T(L)/I(L) $ be the canonical projection. Write $T_m = T^0(L) \oplus T^1(L) \oplus \cdots T^m(L)$ and $U_m = \pi(T_m)$. Set $U_{-1} = 0$ and $G^m = U_m/U_{m-1}$. The PBW theorem states that the graded unital associative algebra $G = \bigoplus_{m=0}^{\infty} G^m$ is isomorphic (as graded algebras) to the symmetric algebra $S(L) = \bigoplus_{m=0}^{\infty}S^m(L)$.

Let $x \in U(L)$ be a nonzero element and $x_0$ be the homogeneous component of $x$ with highest degree. In other words, $x = x_0$ mod $U_{m-1}$ where $x_0 \neq 0$ and $\deg(x_0) = m$. We will show $x^2$ has a nonzero homogeneous part of degree $2m$. According to the PBW theorem, we can consider the isomorphism of graded algebras $\psi: G= \bigoplus_{i=0}^{\infty} G^i \rightarrow S(L) = \bigoplus_{i=0}^{\infty} S^i(L)$. Here $0 \neq \overline{x} \in G^m=U_m/U_{m-1}$ is mapped to $0 \neq \psi(\overline{x}) \in S^m(L)$. Since the symmetric algebra $S(L)$ is an integral domain, $\psi(\overline{x})^2 = \psi(\overline{x^2}) \in S^{2m}(L)$ is also nonzero. Thus $\overline{x^2} \in G^{2m}$ cannot be zero.

Denote the multiplication map of $U(L)$ by $m: U(L) \otimes U(L) \rightarrow U(L)$. For every $x \in U_m$, we know $m\Delta(x)$ lies in $U_m$. This can be proved directly, but you may use the proposition in your question. However, if $x \in U_m \setminus U_{m-1}$ is a grouplike element, then $m\Delta(x)= m(x \otimes x) = x^2$ has a nonzero homogeneous part of degree $2m$. If follows that $m \geq 2m$, or $x$ is a scalar. Now $\Delta(x) = x \Delta(1)=x (1 \otimes 1)$ should be equal to $x\otimes x = x^2 (1 \otimes 1)$, whence $x = 1$.

---

(Added)

If $x \neq 0 $ satisfies $\Delta(x) = x \otimes x$, then $x = \mbox{id}(x) = (\epsilon \otimes \mbox{id})(\Delta x) = \epsilon(x)x$ and thus $\epsilon(x) = 1$. By the axiom of antipode $1 = xS(x) = S(x)x$, whence $x$ is invertible. From here, Mariano's answer proves the statement.