Let B be a bialgebra. Show that the following are equivalent

Question

Let B be a bialgebra. Show that the following are equivalent:

1. B is a Hopf algebra.
2. The maps $T_1$, $T_2$: $B \otimes B \to B \otimes B$ determined by $T_1(a \otimes b) = \sum a_{(1)} \otimes a_{(2)} b$ and $T_2(a \otimes b) = \sum a b_{(1)} \otimes b_{(2)}$ are bijective maps.

So the Sweedler notation throws me off here. In trying to prove 1 implies 2 for injectivity on $T_1$, I write let $ \sum a_{(1)} \otimes a_{(2)} b$ = $\sum a'_{(1)} \otimes a'_{(2)} b'$. I don't think I'm allowed to assume $a_{(1)} = a'_{(1)}$, or am I wrong about that? I ask because there is a problem before this one that does something analogous with monoids. The maps are (a, b) to (a, ab) and (a,b) to (ab, b), and I'm asked to show that these two maps being bijective is equivalent to the monoid being a group. I did that; I'm just wondering if it's the same proof technique? If what I wrote above for the first component in Sweedler notation is true, then showing 1 implies 2 is easy by using the antipode. However, I am also stuck on 2 implies 1. Again, I'm thinking the process is analogous to the monoid problem, but I'm not sure (again, the notation makes me question my ideas).

Can someone help me to get going on this?

Answer 1

First of all, notice that writing $a_{(1)} = a_{(1)}'$ does not properly make sense. The Heyneman-Sweedler sigma notation $\Delta(a) = \sum a_{(1)} \otimes a_{(2)}$ is a easy-to-handle way of denoting the more appropriate $\Delta(a) = \sum_{i=1}^{n_a} b_i \otimes c_i$.

Despite of this, there is an easier way to show that 1 implies 2, which is that of exhibiting an explicit inverse for $T_1$ and $T_2$. I would suggest you to try with $$T_1': B \otimes B \to B \otimes B, \qquad a \otimes b \mapsto \sum a_{(1)} \otimes S\left(a_{(2)}\right)b$$ for example, where $S : B \to B$ is the antipode.

Once you proved that 1 implies 2, proving that 2 implies 1 becomes much easier. Call, for example, $T_1'$ the inverse of $T_1$. Since $T_1$ is right $B$-linear with respect to the regular right $B$-action on both the domain and the codomain, $T_1'$ has to satisfy the same property and hence $$T_1'(a \otimes bc) = T_1'(a \otimes b)\cdot (1 \otimes c).$$ By looking at the inverse of $T_1$ from the previous implication, it comes natural to define $$S := (\varepsilon \otimes B) \circ T_1' \circ (B \otimes u): B \to B$$ where $u : \Bbbk \to B$ is the unit of $B$ as a $\Bbbk$-algebra. It follows that $$ \begin{aligned} \sum S(a_{(1)})a_{(2)} & = \sum (\varepsilon \otimes B)\big(T_1'(a_{(1)}\otimes 1)\big)a_{(2)} \\ & = \sum (\varepsilon \otimes B)\big(T_1'(a_{(1)}\otimes 1)\cdot(1 \otimes a_{(2)})\big) \\ & = (\varepsilon \otimes B)\left(T_1'\left(\sum a_{(1)}\otimes a_{(2)}\right)\right) \\ & = (\varepsilon \otimes B)\left(T_1'\left(T_1(a \otimes 1)\right)\right) = \varepsilon(a)1, \end{aligned} $$ which is one of the two identities required to have an antipode. I am sure that now you may imagine how to use the inverse of $T_2$ to complete the proof.

Therefore, let me conclude by performing a Hopf-algebraic virtuosity.

In fact, the invertibility of $T_1$ alone already suffices to prove that $S$ defined above is an antipode. The reason is that $T_1$ is also colinear with respect to the regular left coactions on both domain and codomain, whence $$(\Delta \otimes B)\big(T_1'(a \otimes b)\big) = \sum a_{(1)} \otimes T_1'(a_{(2)} \otimes b).$$ If now we apply $B \otimes \varepsilon \otimes B$ to both sides, we find that $$T_1'(a \otimes b) = \sum a_{(1)} \otimes (\varepsilon \otimes B) \left(T_1'(a_{(2)} \otimes b)\right)$$ and so $$a \otimes 1 = T_1T_1'(a \otimes 1) = \sum a_{(1)} \otimes a_{(2)}(\varepsilon \otimes B) \left(T_1'(a_{(3)} \otimes 1)\right) = \sum a_{(1)} \otimes a_{(2)}S\left(a_{(3)}\right).$$ A further application of $\varepsilon \otimes B$ entails that $$\varepsilon(a)1 = \sum a_{(1)}S\left(a_{(2)}\right).$$