Checking the antipode for the dual of the Group Hopf Algebra

Question

Consider the vector space $\mathbb{C}G\,=\left\{ f:\,G\longrightarrow\mathbb{C}\right\}$ , and define the following

• multiplication: $\mu\left(f\otimes g\right)\left(x\right):=f\left(x\right)g\left(x\right)$,
• co-multiplication: $\triangle\left(f\right)\left(x\otimes y\right)\,:=f\left(xy\right)$,
• unit: $\eta\left(x\right):=1_{G}$,
• co-unit: $\epsilon\left(f\right)\,:=f\left(1_{G}\right)$,
• antipode: $S\left(f\right)\left(x\right):=f\left(x^{-1}\right)$.

It is well known it is a Hopf Algebra (look at the second example on Wikipedia article). Anyway I'm trying to demonstrate that is valid the relation for the antipode $$\mu\circ\left(S\otimes id\right)\circ\triangle=\eta\circ\varepsilon=\mu\circ\left(id\otimes S\right)\circ\triangle,$$ and I'm getting quite confused on how should I proceed since it result to me that $$\left(\mu\circ\left(S\otimes id\right)\circ\triangle\left(f\right)\right)\left(x\otimes y\right)=f\left(x^{-1}y\right)\neq f\left(xy^{-1}\right)$$ Can anybody write che proof explicitly or at least give me an hint on how to proceed?

Answer 1

I write $f = \sum \lambda_g \delta_g$, and compute $\eta \circ \varepsilon (f) = \eta(\lambda_1) = \lambda_1 (\sum_g \delta_g)$.

Now $\Delta(f) = \sum_{g,h} \lambda_g \delta_h \otimes \delta_{h^{-1}g}$ so $((S \otimes id) \circ \Delta )(f) = \sum_{g,h} \lambda_g \delta_{h^{-1}} \otimes \delta_{h^{-1}g}$.

Finally $\mu \circ (S \otimes id) \circ \Delta(f) = \sum_{g,h} \lambda_g \delta_{h^{-1}}\delta_{h^{-1}g} = \lambda_1 (\sum_g \delta_g)$

Answer 2

Write $A$ for your Hopf algebra, and $k={\mathbb C}$, and $e$ for the identity element of $G$.

For clarity, let's view $A$ only using the RHS of your definition - namely, as $k$-valued functions on $G$, and denote by $\delta\in A$ for the function defined by $$\delta (g ) = 1\in k\text{, for all $g\in G$.}$$

We have

• $\varepsilon \colon A \to k$ is evaluation at $e$: $$\varepsilon a = a( e).$$
• $\eta \colon k \to A$, with $\eta (z ) = z\delta$, for $z\in k$.
• Therefore, one has $\eta \circ \varepsilon \colon A \to A$, with $$\left[(\eta\circ \varepsilon) (a )\right] (g) = a(e) \delta (g) =a(e).$$

On the other hand,

• $\Delta\colon A \to A\otimes A$. The algebra $A\otimes A$ is the set of $k$-valued functions on $G\times G$, and $(\Delta a) (g, h) = a ( gh ).$ (Your evaluation of the tensor product $x\otimes y$ didn't make sense, and I think that is part of the confusion in your attempt to prove the equality.)
  • $\mu (a \otimes b ) (g) = a ( g ) b ( g)$. In other words, $\mu$ is the dual to the diagonal map $G \to G \times G$.

Your equality (to be proved) $$ \mu\circ\left(S\otimes id\right)\circ\triangle=\eta\circ\varepsilon=\mu\circ\left(id\otimes S\right)\circ\triangle$$ is an equality of $A\to A$ maps - i.e., we should evaluate both sides on an arbitrary element $a \in A$, and see that the result is the same function on $G$.

So, if $a\in A$, and $g\in G$, you want to test the value (in $k$) of $$ \left[(\mu\circ\left(S\otimes id\right)\circ\triangle) (a) \right] (g) $$

Running through the definitions, we get that that is equal to

$$ \left[(( S \otimes id )\circ \triangle ) (a )\right] ( g, g ) =\triangle (a) ( g^{-1},g ) = a ( g^{-1} g) = a(e).$$

Edit/Addendum

As requested, here is a clarification (I hope!) of 'Running through the definitions':

First of all, it might be worthwhile to repeat the above 'dual to diagonal' (maybe 'transpose' would have been a better word than 'dual') statement for $\mu$. Identify $A\otimes A$ with the $k$-valued functions on $G \times G$. Then, if $c\in A\otimes A$, we have, by duality/transpose, i.e., by definition after the identification of $A\otimes A$ with functions on $G \times G$, that $$\mu (c ) (g) = c (g,g ).$$

Next, set $\phi= \left(S\otimes id\right)\circ\triangle $.

Then $$ (\mu\circ\left(S\otimes id\right)\circ\triangle) (a) = (\mu \circ \phi ) ( a ) = \mu ( \phi ( a ) ).$$

Therefore, by the previous observation, $$ \mu (\phi ( a ) ) ( g ) = \phi (a ) ( g, g).$$ (Note that $\phi : A \to A\otimes A$, so the evaluation of $\phi(a)$ on $(g,g)$ makes sense.)

Similarly, $$\phi ( a ) = (\ \left(S\otimes id\right)\circ\triangle\ ) (a ) = \left(S\otimes id\right) ( \triangle (a ) ).$$ Therefore $$ \phi(a) (g,g) = (\triangle (a ) ) (g^{-1}, g).$$ This last is equal to $ a ( g^{-1}g ) = a (e ) $.