# 数学代写|霍普夫代数代写Hopf algebra代考|CO739

## 数学代写|霍普夫代数代写Hopf algebra代考|Solutions to exercises

Exercise 4.1.9 Let $k$ be a field and $n \geq 2$ a positive integer. Show that there is no bialgebra structure on $M_n(k)$ such that the underlying algebra structure is the matrix algebra.
Solution: The argument is similar to the one that was used in Example 1.4.17. Suppose there is a bialgebra structure on $M_n(k)$, then the counit $\varepsilon: M_n(k) \rightarrow k$ is an algebra morphism. Then the kernel of $\varepsilon$ is a two-sided ideal of $M_n(k)$, so it is either 0 or the whole of $M_n(k)$. Since $\varepsilon(1)=1$, we have $\operatorname{Ker}(\varepsilon)=0$ and we obtain a contradiction since $\operatorname{dim}\left(M_n(k)\right)>$ $\operatorname{dim}(k)$

Exercise 4.2.16 Let $H$ be a finite dimensional Hopf algebra over a field $k$ of characteristic zero. Show that if $x \in H$ is a primitive element, i.e. $\Delta(x)=x \otimes 1+1 \otimes x$, then $x=0$.
Solution: If there were a non-zero primitive element $x$, we prove by induction that the $1, x, \ldots, x^n$ are linearly independent for any positive integer $n$, and this will provide a contradiction, due to the finite dimension of $H$. The claim is clear for $n=1$, since $a 1+b x=0$ implies by applying $\varepsilon$ that $a=0$, and then, since $x \neq 0$, that $b=0$. Assume the assertion true for $n-1$ (where $n \geq 2$ ), and let $\sum_{p=0, n} a_p x^p=0$ for some scalars $a_0, \ldots, a_p$. Then by applying $\Delta$ we find that
$$\sum_{p=0, n} \sum_{i=0, p} a_p\left(\begin{array}{c} p \ i \end{array}\right) x^i \otimes x^{p-i}=0$$
Choose some $1 \leq i, j \leq n-1$ such that $i+j=n$, and let $h_i^, h_j^ \in H^$ such that $h_i^\left(x^t\right)=\delta_{i, t}$ for any $0 \leq t \leq n-1$ and $h_j^\left(x^t\right)=\delta_{j, t}$ for any $0 \leq t \leq n-1$ (this is possible since $1, x, \ldots, x^{n-1}$ are linearly independent). Then by applying $h_i^ \otimes h_j^*$ to the above relation we obtain that $a_n\left(\begin{array}{l}n \ i\end{array}\right)=0$, and since $k$ has characteristic zero we have $a_n=0$. Then again by the induction hypothesis we must have $a_0, \ldots, a_{n-1}=0$.

## 数学代写|霍普夫代数代写Hopf algebra代考|The definition of integrals for a bialgebra

Let. $H$ be a bialgebra. Then $H^$ has an algebra structure which is dual to the coalgebra structure on $H$. The multiplication is given by the convolution product. To simplify notation, if $h^, g^* \in H^$ we will denote the product of $h^$ and $g^$ in $H^$ by $h^* g^$ instead of $h^ * g^*$.

Definition 5.1.1 A map $T \in H^$ is called a left integral of the bialgebra $H$ if $h^ T=h^(1) T$ for any $h^ \in H^*$. The set of left integrals of $H$ is denoted by $\int_l$. Left integrals of $H^{\text {cop }}$ are called right integrals for $H$, and their set is denoted by $\int_r$.

Remark 5.1.2 It is clear that $T \in H^*$ is a left integral if and only if $\sum T\left(x_2\right) x_1=T(x) 1 \forall x \in H$, and it is a right integral if and only if $\sum T\left(x_1\right) x_2=T(x) 1 \forall x \in H$

We discuss briefly the name given to the above notion. Let $G$ be a compact group. A Haar integral on $G$ is a linear functional $\lambda$ on the space of continuous functions from $G$ to $\mathbf{R}$, which is translation invariant, i.e.
$$\lambda(x f)=\lambda(f)$$
for any continuous $f: G \rightarrow \mathbf{R}$, and any $x \in G$. Then the restriction of $\lambda$ to the Hopf algebra $\tilde{R}{\mathbf{R}}(G)$ of continuous representative functions on $G$ is an integral in the sense of the above definition. Indeed, \begin{aligned} \lambda(x f) &=\lambda(f), \quad \forall f \in \tilde{R}{\mathbf{R}}(G), x \in G \Leftrightarrow \ \lambda\left(\sum f_2(x) f_1\right) &=\lambda(f) 1, \quad \forall f \in \tilde{R}_{\mathbf{R}}(G), x \in G \Leftrightarrow \end{aligned} \begin{aligned} \sum \lambda\left(f_1\right) f_2(x) &=\lambda(f) \mathbf{1}(x), \forall f \in \tilde{R}{\mathbf{R}}(G), x \in G \Leftrightarrow \ \sum \lambda\left(f_1\right) f_2 &=\lambda(f) \mathbf{1}, \forall f \in \tilde{R}{\mathbf{R}}(G) \Leftrightarrow \ \sum \lambda\left(f_1\right) \mu\left(f_2\right) &=\lambda(f) \mu(\mathbf{1}), \forall f \in \tilde{R}{\mathbf{R}}(G), \mu \in \tilde{R}{\mathbf{R}}(G)^* \Leftrightarrow \ (\lambda \mu)(f) &=(\mu(\mathbf{1}) \lambda)(f), \forall f \in \tilde{R}{\mathbf{R}}(G), \mu \in \tilde{R}{\mathbf{R}}(G)^* \Leftrightarrow \ \lambda \mu &=\mu(\mathbf{1}) \lambda, \forall \mu \in \tilde{R}_{\mathbf{R}}(G)^* \end{aligned}
and this explains the use of the name “integral” for bialgebras.

# 霍普夫代数代考

## 数学代写|霍普夫代数代写Hopf algebra代考|Solutions to exercises

，然后，因为 $x \neq 0$ ，那 $b=0$. 假设断言为真 $n-1$ （在哪里 $n \geq 2$ )，然后让 $\sum_{p=0, n} a_p x^p=0$ 对于

$$\sum_{p=0, n} \sum_{i=0, p} a_p(p i) x^i \otimes x^{p-i}=0$$

## 数学代写|霍普夫代数代写Hopf algebra代考|The definition of integrals for a bialgebra

$$\lambda(x f)=\lambda(f)$$

\begin{aligned} &\lambda(x f)=\lambda(f), \quad \forall f \in \tilde{R} \mathbf{R}(G), x \in G \Leftrightarrow \lambda\left(\sum f_2(x) f_1\right) \quad=\lambda(f) 1, \quad \forall f \in \tilde{R}_{\mathbf{R}}(G), x \in G \Leftrightarrow \ &\sum \lambda\left(f_1\right) f_2(x)=\lambda(f) \mathbf{1}(x), \forall f \in \tilde{R} \mathbf{R}(G), x \in G \Leftrightarrow \sum \lambda\left(f_1\right) f_2 \quad=\lambda(f) \mathbf{1}, \forall f \in \tilde{R} \mathbf{R}(G) \Leftrightarrow \end{aligned}

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