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.

  • 1
    $\begingroup$ Did you tried with group algebras? (And a surjective, non split, group homomorphism ) $\endgroup$ – Marco Farinati Sep 22 '20 at 23:26
  • $\begingroup$ 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? $\endgroup$ – jdc Sep 26 '20 at 5:36
  • $\begingroup$ 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., $\mathbb Z[x] \to \mathbb F_2[x]$. $\endgroup$ – jdc Sep 26 '20 at 5:42

I'll give an example "occuring in nature." It's not the simplest possible, but you can get a simpler one by removing the generators of degrees 3, 5, and 7, which don't feature in the argument.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.