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Finally, apply ɛ to the equality<br />

ɛ(x)1 = S(x (1) ) ∙ x (2)<br />

to get<br />

ɛ(x) = ɛ(x)ɛ(1) = ɛ(S(x (1) )ɛ(x (2) ) = ɛ ◦ S(x) for all x ∈ H .<br />

✷<br />

Proposition 2.5.6.<br />

Let H be a Hopf algebra. Then the following identities are equivalent:<br />

(a) S 2 = id H<br />

(b) ∑ x S(x (2))x (1) = ɛ(x)1 H for all x ∈ H.<br />

(c) ∑ x x (2)S(x (1) ) = ɛ(x)1 H for all x ∈ H.<br />

Proof.<br />

We show (b) ⇒ (a) by first showing from (b) that S ∗ S 2 is the identity for the convolution<br />

product. In graphical notation, (b) reads<br />

S<br />

=<br />

Thus<br />

S<br />

S ∗ S 2 = S =<br />

S 2<br />

S<br />

=<br />

(b)<br />

S<br />

= = η ◦ ɛ<br />

For comparison, we also compute in equations:<br />

S ∗ S 2 (x) = ∑ ( ∑ )<br />

S(x (1) )S 2 (x (2) ) = S S(x (2) )x (1)<br />

(x)<br />

(x)<br />

(b)<br />

= S(ɛ(x)1) = ɛ(x)S(1) = ɛ(x)1 .<br />

Then conclude S 2 = id by the uniqueness of the inverse of S with respect to the convolution<br />

product.<br />

Conversely, assume S 2 = id H<br />

S<br />

S<br />

=<br />

S<br />

S<br />

S<br />

S 2 = id<br />

S S S S<br />

= = = = η ◦ ɛ<br />

where we used S 2 = id, the fact that S is an anticoalgebra map, again S 2 = id and then the<br />

defining property of the antipode S. The equivalence of (c) and (a) is proven in complete<br />

35

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