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Let $(H, m, u, \Delta, \epsilon, S)$ be a Hopf algebra and $h\in H$. Do we have the equality

$$u\epsilon(h)\otimes h = h$$

Or maybe put $1\otimes h$ on the RHS or something else "isomorphic"?

EDIT: To give the context: this is what the exercise asks me to prove:

$$\sum_{(h)} h_{(1)}S(h_{(2)})\otimes h_{(3)} = h$$

My solution is: $$\sum_{(h)} h_{(1)}S(h_{(2)})\otimes h_{(3)} \\= ((m \circ(I\otimes S))\otimes I)(\Delta \otimes I) (h) \\= (m\circ (I\otimes S)\circ \Delta) \otimes I (h) \\= u\epsilon (h) \otimes h $$

Now the step I would need to make is the one in the title. Or have I already made a mistake in my calculations?

EDIT2: I see I made the mistake: $\Delta_2$ should be $(I\otimes\Delta)\Delta$ so what we are actually left with is

$$(u\epsilon \otimes I) \Delta(h)$$

But how does this simplify to $h$ (or $h\otimes 1 \otimes 1$ or something like that)?

EDIT3 Ok, now I see it:

$$(u\epsilon \otimes I) \Delta(h) = \sum_{(h)} \epsilon(h_{(1)})h_{(2)} = h$$ by the counital property as mentioned. (Or maybe you put the tensor product in there or not; there's the natural isomorphism $\mathbb{K}\otimes H \simeq H$, but I guess that goes away when you multiply: $1\otimes h = 1h = h$).

Thank you very much!

ploosu2
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    We have $\sum_{(h)} \varepsilon(h_{(1)}) h_{(2)} = h = \sum_{(h)} h_{(1)} \varepsilon(h_{(2)}) = h$ by the antipode condition … You may want to be more specific in what you’re looking for. – Jendrik Stelzner May 29 '19 at 16:42
  • @JendrikStelzner I edited. Does it make any sense now? – ploosu2 May 29 '19 at 17:18
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    @ploosu2 think for a moment on the outcome of $(\epsilon\otimes I)\circ\Delta$ and you'll find your answer – Ender Wiggins May 29 '19 at 17:21
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    @JendrikStelzner actually that's the counital condition :) , The antipode Condition is $\sum_{(h)}h_{(1)}S(h_{(2)}) = u\epsilon(h) =\sum_{(h)}S(h_{(1)})h_{(2)} $ – Ender Wiggins May 29 '19 at 17:23
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    The element $\sum_{(h)} h_{(1)} S(h_{(2)}) \otimes h_{(3)}$ lives in $H \otimes H$ and thus cannot be $h$. I think you want $\sum_{(h)} h_{(1)} S(h_{(2)}) h_{(3)} = h$ instead, which should be true. @EnderWiggins: You’re right, silly me. (I noticed it too late and couldn’t change it anymore.) – Jendrik Stelzner May 29 '19 at 17:29

1 Answers1

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Let me try to put some order, just of the sake of clarity. The antipode condition says that $$\label{antipode}\tag{$\dagger$} \sum_{(h)}h_{(1)}S(h_{(2)}) = \varepsilon(h)1_H = \sum_{(h)}S(h_{(1)})h_{(2)} $$ for every $h\in H$, that is to say, $$\label{ant2}\tag{$\ddagger$} I*S=m\circ(I\otimes S)\circ \Delta = u \circ \varepsilon = m\circ (S\otimes I)\circ \Delta=S*I. $$ You want to prove that $$\sum_{(h)}h_{(1)}S(h_{(2)})\otimes h_{(3)} = 1_H\otimes h$$ (which seems to me the only reasonable thing to prove, by reading your post, and it is one of the exercises on Sweedler's book, whence it seems to me a reasonable exercise that could be given to a student).

Solution 1 (elementwise) \begin{align} \sum_{(h)}h_{(1)}S(h_{(2)})\otimes h_{(3)} & = \sum_{(h)}h_{(1)_{(1)}}S(h_{(1)_{(2)}})\otimes h_{(2)} \stackrel{\eqref{antipode}}{=} \sum_{(h)}\varepsilon(h_{(1)})1_H\otimes h_{(2)} = 1_H\otimes h. \end{align} Solution 2 (commutative diagrams) \begin{align} (m\otimes I)\circ(I\otimes S\otimes I)\circ (\Delta\otimes I)\circ \Delta \stackrel{\eqref{ant2}}{=} (u\otimes I)\circ (\varepsilon\otimes I)\circ\Delta = u\otimes I. \end{align}

Ender Wiggins
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