Recall that $U_q(sl_2)$ is the quotient of the free associative $\mathbb{Q}(q)$-algebra on generators $E$, $F$, $K^{\pm 1}$ such that $KE = q^2 EK$, $KF = q^{-2} FK$, and $[E,F] = \frac{K - K^{-1}}{q - q^{-1}}$. It turns out that there is a Hopf algebra structure on $U_q(sl_2)$ such that $\Delta(K) = K \otimes K$, $\Delta(E) = E \otimes K + 1 \otimes E$, and $\Delta(F) = F \otimes 1 + K^{-1} \otimes F$. My question is the following: suppose I only declare that the coproduct must satisfy $\Delta(K) = K \otimes K$. Does this uniquely determine the above formulas for the Hopf algebra structure on $U_q(sl_2)$? In other words, does specifying $\Delta(K)$ uniquely specify $\Delta(E)$ and $\Delta(F)$?
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1$\begingroup$ It certainly can't uniquely do so in the (boring) sense that you can flip the order of the factors in all tensor products $\endgroup$– Jules LamersCommented Aug 31 at 14:38
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$\begingroup$ From my own (physics) experience with $U_q(sl_2)$, I choose the same coproduct as your $\Delta(K)$ but different coproducts on $E$ and $F$. In particular, I would choose my coproducts on $E$ and $F$ to maintain that $E$ and $K$ are Hermitian conjugates in matrix representations of $U_q(sl_2)$. That is, a more natural coproduct (to me) is given in eq. 4 of this paper. $\endgroup$– user196574Commented Aug 31 at 14:41
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$\begingroup$ Are you asking whether $U_q(sl_2)$ could have another coalgebra structure making it a Hopf algebra with $K$ being grouplike? $\endgroup$– Christian LompCommented Oct 7 at 14:54
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