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$\newcommand{\Irr}{\mathrm{Irr}}$ 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?

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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 $H = k$.

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  • $\begingroup$ 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 $m_U = \dim U$. $\endgroup$ – Jo Be May 15 at 7:35
  • $\begingroup$ 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 $m_U = \text{dim}(k^{m_U}) = \text{dim}(U)$. $\endgroup$ – Matthias Klupsch May 15 at 8:43

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