# Covariant splittings of Hopf algebra projections

What is an example of a pair of Hopf algebras $$(A,B)$$ with a surjective Hopf algebra map $$\phi:A \to B$$ such that $$\phi$$ does not admit a $$B$$-bi-comodule splitting $$s:B \to A$$? To be clear, the right $$B$$-comodule structure on $$A$$ is given by $$(\textrm{id} \otimes \phi) \circ \Delta_A: A \to A \otimes B,$$ where $$\Delta_A$$ is the coproduct of $$A$$, and the left coaction is defined similarily.

• Did you tried with group algebras? (And a surjective, non split, group homomorphism ) Sep 22, 2020 at 23:26
• At a glance, it looks like group algebras might actually not yield the example; if you identify the quotient group $Q = G/N$ with a subset $\tilde Q$ of $G$ via an arbitrary lifiting, the inherited coalgebra structure on the lift $k \cdot \tilde Q$ is extended linearly from $\Delta(q) = q \otimes q$, so there's a coalgebraic splitting, just not an algebraic one. Or am I missing something?
– jdc
Sep 26, 2020 at 5:36
• It's unlikely to be what you want, but in case it's not an accident that you haven't specified a fixed base ring, there are examples that fail to admit even group-theoretic splittings, e.g., $\mathbb Z[x] \to \mathbb F_2[x]$.
– jdc
Sep 26, 2020 at 5:42

According to results of Borel from 1954, the mod-2 homology Hopf algebra $$H_9 = H_* (\mathrm{Spin}(9);\mathbb F_2)$$ is the exterior algebra on one generator each of degrees 3, 5, 6, 7, and 15. The standard inclusion $$i$$ of $$\mathrm{Spin}(9)$$ in $$\mathrm{Spin}(10)$$ preserves this exterior algebra structure, but $$H_{10} = H_*(\mathrm{Spin}(10);\mathbb F_2)$$ has a new generator $$u_{9}$$ of degree 9 such that $$H_{10}$$ is a free module of rank two over $$i_* H_9$$ with basis $$\{1,u_9\}$$, following from the collapse of the Serre spectral sequence of the fiber bundle $$\mathrm{Spin}(9) \to \mathrm{Spin}(10) \to S^9$$. The new generator $$u_9$$ doesn't anticommute with the old ones as one might expect: giving the other generators the obvious names, one has
$$u_6 u_9 + u_9 u_6 = u_{15}$$
in $$H_{10}$$.
Particularly, the injection of left $$H_9$$-modules $$H_9 \to H_{10}$$ does not split. For degree reasons, $$u_9$$ would have to go to $$u_3 u_6$$ or zero under the splitting, but the product $$H_9 \otimes H_{10} \to H_{10}$$ sends $$u_6 \otimes u_9 \mapsto u_{15} + u_9 u_6,$$ while the product $$H_9 \otimes H_9 \to H_9$$ sends $$u_6 \otimes 0 \mapsto 0 \neq u_{15} + 0 u_6$$ and also $$u_6 \otimes u_3 u_6 \mapsto 0 \neq u_{15} + (u_3 u_6) u_6.$$
A $$H^*(\mathrm{Spin}(9);\mathbb F_2)$$-comodule splitting of the cohomological Hopf algebra map $$i^*\colon H^*(\mathrm{Spin}(10);\mathbb F_2) \to H^*(\mathrm{Spin}(9);\mathbb F_2)$$ would lead on dualization to a forbidden module splitting of the sort we ruled out in the previous paragraph, so $$i^*$$ is an example of the type you wanted.