# 数学代写|霍普夫代数代写Hopf algebra代考|Math6329

## 数学代写|霍普夫代数代写Hopf algebra代考|Hopf algebras

Let $(C, \Delta, \varepsilon)$ be a coalgebra, and $(A, M, u)$ an algebra. We define on the set $\operatorname{Hom}(C, A)$ an algebra structure in which the multiplication, denoted by $*$ is given as follows: if $f, g \in \operatorname{Hom}(C, A)$, then
$$(f * g)(c)=\sum f\left(c_1\right) g\left(c_2\right)$$
for any $c \in C$. The multiplication defined above is associative, since for $f, g, h \in \operatorname{Hom}(C, A)$ and $c \in C$ we have
\begin{aligned} ((f * g) * h)(c) &=\sum(f * \cdot g)\left(c_1\right) h\left(c_2\right) \ &=\sum f\left(c_1\right) g\left(c_2\right) h\left(c_3\right) \ &=\sum f\left(c_1\right)(g * h)\left(c_2\right) \ &=(f *(g * h))(c) \end{aligned}
The identity element of the algebra $\operatorname{Hom}(C, A)$ is $u \varepsilon \in \operatorname{Hom}(C, A)$, since
$$(f *(u \varepsilon))(c)=\sum f\left(c_1\right)(u \varepsilon)\left(c_2\right)=\sum f\left(c_1\right) \varepsilon\left(c_2\right) 1=f(c)$$
hence $f *(u \varepsilon)=f$. Similarly, $(u \varepsilon) * f=f$.
Let us note that if $A=k$, then $*$ is the convolution product defined on the dual algebra of the coalgebra $C$. This is why in the case $A$ is an arbitrary algebra we will also call * the convolution product.

Let us consider a special case of the above construction. Let $H$ be a bialgebra. We denote by $H^c$ the underlying coalgebra. $H$, and by $H^a$ the underlying algebra of $H$. Then we can define as above an algebra structure on $\operatorname{Hom}\left(H^c, H^a\right)$, in which the multiplication is defined by $(f * g)(h)=$ $\sum f\left(h_1\right) g\left(h_2\right)$ for any $f, g \in H o m\left(H^c, H^a\right)$ and $h \in H$, and the identity element is $u \varepsilon$. We remark that the identity map $I: H \rightarrow H$ is an element of $\operatorname{Hom}\left(H^c, H^a\right)$

## 数学代写|霍普夫代数代写Hopf algebra代考|Examples of Hopf algebras

In this section we give some relevant examples of Hopf algebras.
1) The group algebra. Let $G$ be a (multiplicative) group, and $k G$ the associated group algebra. This is a $k$-vector space with basis ${g \mid g \in G}$, so its elements are of the form $\sum_{g \in G} \alpha_g g$ with $\left(\alpha_g\right)_{g \in G}$ a family of elements from $k$ having only a finite number of non-zero elements. The multiplication is defined by the relation
$$(\alpha g)(\beta h)=(\alpha \beta)(g h)$$
for any $\alpha, \beta \in k, g, h \in G$, and extended by linearity.
On the group algebra $k G$ we also have a coalgebra structure as in Example 1.1.4 1), in which $\Delta(g)=g \otimes g$ and $\varepsilon(g)=1$ for any $g \in G$. We already know that the group algebra becomes in this way a bialgebra. We note that until now we only used the fact that $G$ is a monoid. The existence of the antipode is directly related to the fact that the elements of $G$ are invertible.

Indeed, the map $S: k G \rightarrow k G$, defined by $S(g)=g^{-1}$ for any $g \in G$, and then extended linearly, is an antipode of the bialgebra $k G$, since
$$\sum S\left(g_1\right) g_2=S(g) g=g^{-1} g=1=\varepsilon(g) 1$$
and similarly, $\sum g_1 S\left(g_2\right)=\varepsilon(g) 1$ for any $g \in G$.
It is clear that if $G$ is a monoid which is not a group, then the bialgebra $k G$ is not a Hopf algebra.
If $G$ is a finite group, then Proposition 4.2.11 shows that on $(k G)^$ we also have a Hopf algebra structure, which is dual to the one on $k G$. We recall that the algebra $(k G)^$ has a complete system of orthogonal idempotents $\left(p_g\right){g \in G}$, where $p_g \in(k G)^$ is defined by $p_g(h)=\delta{g, h}$ for any $g, h \in G$. Therefore,
$$p_g^2=p_g, p_g p_h=0 \text { for } g \neq h, \sum_{g \in G} p_g=1_{(k G)^}$$
The coalgebra structure of $(k G)^$ can be described using Remark 1.3.10, and is given by $$\Delta\left(p_g\right)=\sum_{x \in G} p_x \otimes p_{x^{-1} g}, \varepsilon\left(p_g\right)=\delta_{1, g}$$ The antipode of $(k G)^$ is defined by $S\left(p_g\right)=p_{g^{-1}}$ for any $g \in G$.

# 霍普夫代数代考

## 数学代写|霍普夫代数代写Hopf algebra代考|Hopf algebras

$$(f * g)(c)=\sum f\left(c_1\right) g\left(c_2\right)$$

$$((f * g) * h)(c)=\sum(f * \cdot g)\left(c_1\right) h\left(c_2\right)=\sum f\left(c_1\right) g\left(c_2\right) h\left(c_3\right)=\sum f\left(c_1\right)(g * h)\left(c_2\right)$$
$$(f *(u \varepsilon))(c)=\sum f\left(c_1\right)(u \varepsilon)\left(c_2\right)=\sum f\left(c_1\right) \varepsilon\left(c_2\right) 1=f(c)$$

\sum S\left(g_1\right) g_2=S(g) g=g^{-1} g=1=\varepsilon(g) 1
$$同样， \sum g_1 S\left(g_2\right)=\varepsilon(g) 1 对于任何 g \in G. 很明显，如果 G 是一个不是群的么半群，那么双代数 k G 不是 Hopf 代数。 于任何 g, h \in G. 所以，$$
p_{-} g^{\wedge} 2=p_{-} g, p_{-} g p_{-} h=0 \backslash t \text { text }{\text { for }} g \backslash \text { neq h, } \backslash \text { sum__g }{\text { in } G} p_{-} g=1_{-}\left{(k G)^{\wedge}\right}
$$的余代数结构 \left[(k \mathrm{G})^{\wedge}\right. 可以使用备注 1.3 .10 进行描述，并由下式给出$$
\Delta\left(p_g\right)=\sum_{x \in G} p_x \otimes p_{x^{-1} g}, \varepsilon\left(p_g\right)=\delta_{1, g}


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