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Let $H=(H, \mathbb{K}, m, \eta, \Delta, \epsilon, S)$ be a Hopf algebra. Then there is a (convolution) product on $\operatorname{End}_{\mathbb{K}}(H)$ given by $f\star g=m\circ (f\otimes g)\circ \Delta$, making it into a group.

  1. Is it correct to argue that this product is commutative? I suppose the answer is no, but an answer to this question claims otherwise.

  2. How does this product interact with the tensor product (of linear operations)? I tried a couple of things and tried manipulating defining identities for this question, but no success yet.

Bumblebee
  • 18,974

2 Answers2

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Then there is a (convolution) product on End() given by ⋆=∘(⊗)∘Δ, making it into a group.

In general, $End_K(H)$ is not a group with this structure. The neutral element is $\epsilon$ and if you take a linear map $f:H\to H$, that sends $1$ to zero, then there cannot be a a linear map $g:H\to H$ such that $f*g=\epsilon$, since $(f*g)(1) = f(1)g(1) = 0 \neq 1 = \epsilon(1)$. However, the set of $K$-linear algebra homomorphisms $f:H\to H$ forms a group under the convolution product, where the inverse of an algebra homomorphism $f$ is $f\circ S$.

Is it correct to argue that this product is commutative?

In general not. Take for instance the $4$-dimensional Sweedler algebra $H=K\langle g,x : g^2=1, x^2=0, xg=-gx\rangle$ with comultiplication $\Delta(g)=g\otimes g$ and $\Delta(x)=x\otimes 1 + g \otimes x$. Then $H$ has basis $1,g,x,gx$. Consider the dual basis $p_1, p_g, p_x, p_{gx}$ as $K$-linear maps from $H$ to $H$, i.e. $p_x(x)=1$, while $p_x(g)=p_x(gx)=p_x(1)=0$. Then $$p_x*p_1 = p_x \neq 0 = p_1*p_x,$$ since $(p_x*p_1)(x)=p_x(x)p_1(1) + p_x(g)p_1(x) = 1$ and $(p_1*p_x)(x)=p_1(x)p_x(1)+p_1(g)p_x(x)=0$. Thus the convolution product is not commutative.

Christian Lomp
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I think I figured out an answer to the second part of the question. First notice that $$f\star g=(H\xrightarrow{\Delta}H\otimes H\xrightarrow{f\otimes g}H\otimes H\xrightarrow{m}H).$$ And secondly, composition (of composable linear transformations) and tensor product commutes: $$(f\otimes g)\circ(F\otimes G)=(f\circ F)\otimes(g\circ G).$$

Hence the diagram below commutes: $\require{AMScd}$ \begin{CD} H\otimes H @>\Delta\otimes \operatorname{id}>> H\otimes H\otimes H\\ @V (f\star g)\otimes h V V @VV (f\otimes g)\otimes h V\\ H\otimes H @<m\otimes \operatorname{id}<< H\otimes H\otimes H \end{CD}

For a positive answer to the first part of the question, I suppose the Hopf algebra must be commutative. But I have not figured out the details yet.

Bumblebee
  • 18,974