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Let $H$ be a Hopf Algebra (over a field $K$), with comultiplication $\Delta$, counit $\varepsilon$, and antipode $S$.

A $K$-subspace V is said to be:

  • A right ideal if $VH \subseteq V$
  • A right coideal if $\Delta (V) \subseteq V \otimes H$ (some call this a left coideal...)

Suppose that there is a nonzero right ideal $J$ in $H$, such that $\dim_K (J)$ is finite. Then $\dim_K (H)$ is finite.

The proof breaks in two parts:

  1. If $V$ is a nonzero $K$-subspace of a Hopf algebra $H$ that is both a right ideal and a right coideal, then $V=H$.
  2. If $J\subseteq H$ is a finite-dimensional, nonzero right ideal of $H$, then $N:= H^* J = \{ f\cdot x = \sum x_{(1)}f(x_{(2)}) : f\in H^* , x\in J\}$ is finite dimensional, nonzero, and it is both a right ideal and a right coideal (thus $H=N$ is finite dimenisonal)

Now, the first part is easy (you prove that there is $x\in V$ such that $\varepsilon (x) = 1$, then $1_H = \varepsilon(x)1_H = \sum x_{(1)}S(x_{(2)}) \in V$).

My problem is the second part. I can prove that $N$ is a finite dimensional subspace: since $J$ is finite dimensional, we can assume it is contained in a finite dimensional subcoalgebra $D$ of $H$; then $N\subseteq D $.

Does anyone have suggestions on how to prove that $N$ is a right ideal and a right coideal?

de wa nai
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2 Answers2

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I'll use Sweedler-notation without the sum symbol.

For $h\in{H},f\in{H^*},x\in{J}$ you have:

$m_{(1)}f(m_{(2)})h=m_{(1)}h_{(1)}f(m_{(2)}h_{(2)}S(h_{(3)}))=m_{(1)}h_{(1)}(S(h_{(3)})\rightharpoonup f)(m_{(2)}h_{(2)})=\\(mh_{(1)})_{(1)} (S(h_{(2)}))\rightharpoonup f)(mh_{(1)})_{(2)}\in{N},$

where I used the notation $(a\rightharpoonup g)(b)=g(ba)$.

This shows that $N$ is indeed a right ideal. The proof that $N$ is also a right-coideal should be similiar, maybe this calculation helps you a bit. In fact, $N$ is the right-coideal generated by $J$.

Janik
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A long time passed but I think that this question is still missing an exhaustive answer.

In fact, why $N=H^*\cdot J$ is a right $H$-coideal has not been clarified (even if this is trivial, once you see the trick). All the other parts are treated in detail in the paper Integrals for Hopf Algebras by Sweedler.

Consider $H$ as a right $H$-comodule via its own comultiplication $\Delta$: this makes of it a rational left $H^*$-module (essentially by definition) with action $f\cdot h=h_1f(h_2)$. Take a (non-zero) right ideal $J\subseteq H$. Then $H^*\cdot J$ is the left $H^*$-submodule of $H$ generated by $J$ and, as such, it is rational (see Theorem 2.1.3, part a, in Sweelder's book Hopf Algebras).

For the sake of completeness, I will just sketch a direct proof of this claim. Let $C$ be any coalgebra and $N\subseteq M$ a $C^*$-submodule of a rational $C^*$-module $M$. For every $n\in N$ we have that

  1. $f\cdot n\in N$ for every $f\in C^*$ because $N$ is a $C^*$-submodule;
  2. there exists an element $\sum_{i}m_i\otimes c_i\in M\otimes C$ such that $f\cdot n=\sum_{i}m_if(c_i)$ for every $f\in C^*$ because $n\in M$, which is rational;
  3. in 2 we may assume that the $c_i$'s are linearly independent (whence we may consider elements $c_i^*\in C^*$ such that $c_j^*(c_i)=\delta_{i,j}$, the Kronecker's delta).

Now, we may collect together these observations to conclude that for every $j$ $$m_j=\sum_{i}m_ic_j^*(c_i)=c_j^*\cdot n\in N$$ and hence $\sum_{i}m_i\otimes c_i\in N\otimes C$, proving that $N$ is rational as well.

Ender Wiggins
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