# A certain decomposition of a semisimple Hopf algebra

$$\newcommand{\Irr}{\mathrm{Irr}}$$ Let $$H$$ be a finite-dimensional semisimple Hopf algebra over an algebraically closed field $$k$$, and let $$\Irr(H)$$ be the set of (choices of representatives of) isomorphism classes of simple $$H$$-modules. I was told that \begin{align}\label{eq:question} H \cong \bigoplus_{U\in\Irr(H)} U\otimes U^* \end{align} is a consequence of Artin-Wedderburn, but I am not well-read in basic facts, I think.

Artin-Wedderburn tells me that \begin{align} H \cong \bigoplus_{U\in\Irr(H)} k^{m_U\times m_U}, \end{align} where $$m_U$$ is the multiplicity of $$U$$ in $$H$$. The right hand side of the equation in question is the same as \begin{align} \bigoplus_{U\in\Irr(H)} k^{\dim U\times \dim U}. \end{align} Why are the last two expressions now the same?

Note that Artin-Wedderburn says that $$H \to \bigoplus_{U \in \text{Irr}(H)} \text{End}_k(U), h \mapsto (u \mapsto h u)_{U}$$ is an isomorphism of $$k$$-algebras. This is either what you show when proving Artin-Wedderburn or it follows from the isomorphism $$H \cong \bigoplus_{U} k^{m_U \times m_U}$$ as it clearly holds for the algebra on the right.

Also note that we have an isomorphism $$U \otimes U^* \to \text{End}_k(U), v \otimes f \mapsto (u \mapsto f(u)v)$$ of $$k$$-vector spaces.

This gives $$H \cong \bigoplus_{U} U \otimes U^*$$ as $$k$$-vector spaces. Note that $$\text{End}_k(U)$$ becomes a $$H$$-module via $$(h\varphi)(u) = h\varphi(u)$$ for $$h \in H$$, $$u \in U$$ and $$\varphi \in \text{End}_k(U)$$ and $$U \otimes U^*$$ is an $$H$$-module via $$h (u \otimes f)) = (h u) \otimes f$$.

Given these module structures, the above isomorphisms actually become isomorphisms of $$H$$-modules and so $$H \cong \bigoplus_{U} U \otimes U^*$$ as $$H$$-modules given the above module structure of $$U \otimes U^*$$.

However, there is a huge danger here and that is that these module structures are not the ones we would expect in the context of Hopf-algebras. The endomorphism ring $$\text{End}_{k}(U)$$ usually becomes an $$H$$-module via $$(h\varphi)(u) = \sum h_1 \varphi(S(h_2)u)$$ for $$h \in H$$, $$u \in U$$ and $$\varphi \in \text{End}_k(U)$$ and $$U \otimes U^*$$ becomes an $$H$$-module via $$h(u \otimes f) = \sum (h_1u) \otimes (h_2f)$$

Note that with these module structures, the isomorphism $$U \otimes U^* \to \text{End}_k(U), v \otimes f \mapsto (u \mapsto f(u)v)$$ is still an isomorphism of $$H$$-modules, however,

$$H \to \text{End}_k(U), h \mapsto (u \mapsto h u)$$ is not an $$H$$-module homomorphism in this case.

In fact, we do not have $$H \cong \bigoplus_{U} U \otimes U^*$$ for a semisimple Hopf algebra $$H$$ with this second $$H$$-module structure unless $$H = k$$. In fact, the map $$U \otimes U^* \to k, u \otimes f \mapsto f(u)$$ is an $$H$$-module homomorphism to the trivial $$H$$-module. If we had $$H \cong \bigoplus_{U} U \otimes U^*$$ then the trivial module would have multiplicity at least $$|\text{Irr}(U)|$$ in $$H$$. By Artin-Wedderburn, the multiplicity of the trivial module in $$H$$ is its dimension and so we have to have $$|\text{Irr}(U)| = 1$$ and thus $$H = k$$.

• Thanks, the only nontrivial fact for me was that understanding why Wedderburn says "Multiplicity of simple module in regular = dimension of simple" (in the algebraically closed case). Also, I actually want $H$ not to be the regular module, but rather the adjoint representation - so I guess my question was really about a an iso of vector spaces, which I couldn't see without knowing $m_U = \dim U$. – Jo Be May 15 at 7:35
• Oh, if you are interested in the adjoint representation, then you are actually in a better shape, because then $H \to \text{End}_k(U)$ is actually an $H$-module morphism given the usual action on $\text{End}_k(U)$ and then the isomorphism $H \cong \bigoplus_{U} U \otimes U^*$ is an isomorphism of $H$-modules. Concerning the multiplicity/dimension issue, this can be considered a part of Artin-Wedderburn or deduced from it by noticing that the $H$-module $U$ corresponds to the simple $k^{m_U \times m_U}$-module and this is $k^{m_U}$, so $m_U = \text{dim}(k^{m_U}) = \text{dim}(U)$. – Matthias Klupsch May 15 at 8:43