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The transpose is a map m ∗ x : H ∗ → H ∗ , for each x ∈ H. One checks graphically that it
defines the structure of a left H-module on H ∗ . We write
h ⇀ h ∗
∈ H ∗
for the image of h ∗ ∈ H ∗ under the left action of h ∈ H. Thus
〈h ⇀ h ∗ , g〉 = 〈h ∗ , gh〉 for all g ∈ H .
One can perform this construction quite generally for an algebra in a rigid monoidal
category. In this case, one has to take the left dual for this construction to work. This is
again immediately obvious from the graphical proof.
3. In the same vein, the transpose of left multiplication defines a right action of H on H ∗ .
We write
h ∗ ↼ h ∈ H ∗
for the image of h ∗ ∈ H ∗ under the right action of h ∈ H. Thus
〈h ∗ ↼ h, g〉 = 〈h ∗ , hg〉 for all g ∈ H .
One can perform this construction quite generally for an algebra in a rigid monoidal
category. In this case, one has to take the right dual for this construction to work. This
is again immediately obvious from the graphical proof.
4. Since the antipode is an antialgebra morphism, we can use it to turn left actions into
right actions and vice versa.
In this way, we get a left action of H on H ∗ by
It obeys
(h ⇁ h ∗ ) := (h ∗ ↼ S(h))
〈h ⇁ h ∗ , g〉 = 〈h ∗ , S(h)g〉 for all g ∈ H .
Similarly, we get a right action of H on H ∗ by
(h ∗ ↽ h) := (S(h) ⇀ h ∗ )
with
〈h ∗ ↽ h, g〉 = 〈h ∗ , gS(h)〉 for all g ∈ H .
The following Lemma will be important:
Lemma 3.1.11.
Let H be a finite-dimensional Hopf algebra. Then H ∗ with right H action ↽ and right coaction
ρ is a Hopf module.
Proof.
We have to show that
ρ(f ↽ h) = (f (0) ↽ h (1) ) ⊗ (f (1) ↽ h (2) )
for all f ∈ H ∗ and h ∈ H. By the definition of the coaction ρ, this amounts to showing for all
p ∈ H ∗ and x ∈ H:
〈p, x (1) 〉〈f ↽ h, x (2) 〉 = 〈f (0) ↽ h (1) , x〉〈p, f (1) ∙ h (2) 〉 .
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