Every finite-dimensional Hopf algebra is isomorphic to a dual Hopf algebra

abstract-algebrahopf-algebrasrepresentation-theory

1. Context

Let $H$ be a Hopf algebra over a field $\mathbb{k}$. Denote by $I_l(H)$, $I_r(H)$ its space of left integrals/right integrals respectively.
I am studying a proof of the following proposition:

If $H$ is finite dimensional, then $\dim(I_l(H)) = \dim(I_r(H)) = 1$.

The proof seems to use the following lemma:

For any finite dimensional Hopf algebra $H$ there exists a Hopf algebra $M$ such that $M^*$ and $H$ are isomorphic as Hopf algebras.

2. Question

Why does the lemma hold?

3. A bit more context, if needed

The proof continues as follows:

Due to that lemma it suffices to show that for any finite-dimensional Hopf algebra $M$ its dual Hopf algebra $M^*$ satisfies $\dim(I_l(M^*)) = \dim(I_r(M^*))=1$.

One endows the vector space $M^*$ with the structure of a (right) $H$-Hopf-module (with the lower harpoon action, and a certain coaction $\Delta_{M^*}$). Using the Fundamental theorem of Hopf modules as well as the fact that $M^*$ is finite dimensional, one obtains $\dim((M^*)^{\mathrm{coH}})=1$. Here, $(M^*)^{\mathrm{coH}} = \{x \in M^* : \Delta_{M^*}(x) = x \otimes 1\}$ denotes the space of right coinvariants. Using finite-dimensionality again, a calculation shows that $(M^*)^{\mathrm{coH}} = I_l(M^*)$. Analogously, one proves that $\dim(I_r(M^*)) = 1$.

Best Answer

We have for every vector space $V$ a natural homomorphism $$ φ_V \colon V \longrightarrow V^{**} \, \quad v \longmapsto [v^* \mapsto v^*(v)] \,. $$ If $V$ is finite-dimensional, then $φ_V$ is an isomorphism. If $H$ is a finite-dimensional Hopf algebra, then $φ_H$ is not just an isomorphism of vector spaces, but an isomorphism of Hopf algebras. Therefore, $$ H ≅ H^{**} $$ as Hopf algebras.

Let us check the above claim that $φ_H$ is an isomorphism of Hopf algebras. It remains to show that $φ_H$ is a homomorphism of Hopf-algebras. We recall how the dual of a Hopf algebra is constructed:

Let $H$ be a finite-dimensional Hopf algebra. The multiplication of $H^*$ is defined by $$ (h^*_1 h^*_2)(h) = \sum_{(h)} h^*_1( h_{(1)} ) \, h^*_2( h_{(2)} ) \,. $$ The unit of $H^*$ is precisely the counit of $H$ (which is a linear map from $H$ to $𝕜$, and therefore an element of $H^*$), i.e., $1_{H^*} = ε_H$. The comultiplication of $H^*$ is implicitly determined by $$ h^*(h_1 h_2) = \sum_{(h^*)} h^*_{(1)}(h_1) \, h^*_{(2)}(h_2) \,. $$ The counit of $H^*$ is given by evalution at the unit of $H$, i.e., $ε_{H^*}(h^*) = h^*(1_H)$. The antipode of $H^*$ is given by the dual map of the antipode of $H$, i.e., $S_{H^*} = S_H^*$.

Let us now show that $φ$ is a homomorphism of bialgebras:

  • The chain of equalities \begin{align*} φ(h_1 h_2)(h^*) = h^*(h_1 h_2) = \sum_{(h^*)} h^*_{(1)}(h_1) \, h^*_{(2)}(h_2) &= \sum_{(h^*)} φ(h_1)(h^*_{(1)}) \, φ(h_2)(h^*_{(2)}) \\ &= (φ(h_1) φ(h_2))(h^*) \end{align*} shows that $φ$ is multiplicative.

  • That $φ$ preserves units follows from $φ(1_H)(h^*) = h^*(1_H) = ε_{H^*}(h^*) = 1_{H^{**}}(h^*)$.

  • The chain of equalities $$ φ(h)(h^*_1 h^*_2) = (h^*_1 h^*_2)(h) = \sum_{(h)} h^*_1( h_{(1)} ) \, h^*_2( h_{(2)} ) = \sum_{(h)} φ(h_{(1)})(h^*_1) \, φ(h_{(2)})(h^*_2) $$ shows that the element $\sum_{(h)} φ(h_{(1)}) ⊗ φ(h_{(2)})$ satisfies the defining equality of the element $Δ_{H^{**}}(φ(h))$. In other words, $φ$ is comultiplicative.

  • The equalities $ε_{H^{**}}(φ(h)) = φ(h)(1_{H^*}) = 1_{H^*}(h) = ε_H(h)$ show that $φ$ preserves counits.

This shows that $φ$ is a homomorphism of bialgebras, and therefore (automatically) a homomorphism of Hopf algebras.

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