Let $A$ be a (finite-dimensional graded cocommutative) Hopf algebra over a field $k$, $E$ be a Hopf subalgebra, and $R=A \otimes_E k$. Then the comultiplication on $A$ induces a coalgebra structure on $R$. Furthermore, $R$ is a coalgebra in the monoidal category of $A$-modules, with $A$ acting on $R \otimes R$ diagonally via the comultiplication. Define an internal $R$-comodule to be an object $M$ which is simultaneously an $A$-module and an $R$-comodule such that the structure map $M \to R \otimes M$ is a map of $A$-modules, for the diagonal $A$-module structure on the tensor product.

$A$ itself is naturally an internal $R$-comodule, via the comultiplication $A \to A \otimes A \to R \otimes A$. For any $E$-module $N$, $A \otimes_E N$ then inherits an internal $R$-comodule structure from $A$. Conversely, if $M$ is an internal $R$-comodule, $N={m:d(m)=1 \otimes m}$ is an $E$-module, where $d:M \to R \otimes M$ is the structure map.

Is it true (possibly under some reasonable niceness hypotheses) that these two functors between E-modules and internal $R$-comodules are inverse? In particular, I'd like to interpret this in terms of faithfully flat descent: $A$ is faithfully flat over $E$, and I want to say that for an $A$-module $M$, there is a natural bijection between descent data that allows us to identify $M=A \otimes_E N$ for an $E$-module $N$ and internal $R$-comodule structures $M \to R \otimes M$.

Sorry if I'm getting some things wrong about what hypotheses are needed for this to make sense; I'm trying to understand this in a specific example and don't know much of the general theory.

  • $\begingroup$ I think I once checked by hand that this is true in the case E = k. I don't know what A//E means in general, though. $\endgroup$ Nov 1, 2009 at 20:06
  • $\begingroup$ To clarify, does A//E mean that you take the E-invariants, A^E= {x in A s.t. ad_e(x)=eps(e) x}, and then quotient this by the ideal generated by {e-eps(e): e in E}? $\endgroup$ Nov 1, 2009 at 21:10
  • $\begingroup$ A//E is supposed to mean A \otimes_E k, i.e. A modulo the ideal generated by the augmentation ideal of E. What is ad_e in your definition of A^E? $\endgroup$ Nov 1, 2009 at 22:28
  • $\begingroup$ a Hopf algebra H and in particular it's sub algebra E acts on the vector space H in three ways: left multiplication, h.x=hx, right multiplication, h.x = xS(h) (S is the antipode, here used to make right multiplication a left action), and the adjoint h.x = h_1xS(h_2) (\Delta(h)=h_1 ot h_2 is Sweedler's notation. The latter has the pleasant feature that the multiplication of H is equivariant for this action, h.xy=h1xyS(h2)=h1xS(h2)h3y S(h4). So I confused the "//" symbol for "quantum hamiltonian reduction, which you can do in this context, and has formula like I gave. $\endgroup$ Nov 2, 2009 at 2:27
  • $\begingroup$ ps - sorry about the terse/crammed nature of the reply. character limits! =] $\endgroup$ Nov 2, 2009 at 2:31

3 Answers 3


A very small example where the answer is no:

Suppose $k$ has characteristic not two and $A=k\langle x,y:x^2=1, y^2=0\rangle$ with $\Delta(x)=x\otimes x$, $\Delta(y)=y\otimes 1+x\otimes y$, $\varepsilon(x)=1$ and $\varepsilon(y)=0$; this is the Sweedler Hopf algebra. Let $E$ be the subHopf algebra generated by $x$, which has $\{1,x\}$ as a basis. Then $R=k\otimes_EA$ has $\{\overline 1=1\otimes 1,\overline y=1\otimes y\}$ as a basis, and its coalgebra structure is given by $\Delta(\overline 1)=\overline 1\otimes\overline 1$, $\Delta(\overline y)=\overline y\otimes\overline1+\overline1\otimes\overline y$, $\varepsilon(\overline1)=1$ and $\varepsilon(\overline y)=0$.

Since $E\cong k\times k$ as an algebra, the category $\mathrm{Mod}_E$ is semisimple.

On the other hand, suppose $M\in\mathrm{Mod}_A^R$. One can check that the right $R$-comodule structure $\rho$ of $M$ is determined by a linear map $\phi:M\to M$ such that $\phi^2=0$ by the equation $$\rho(m)=m\otimes\overline 1+\phi(m)\otimes\overline y.$$ Similarly, the $A$-module structure on $M$ is easily seen to be such that $m\cdot y=0$ for all $m\in M$ and $\phi(m\cdot x)=\phi(m)\cdot x$ for all $m\in M$. It follows that one can identify an object $M$ of $\mathrm{Mod}_A^R$ with a $4$-tuple $(M^+,M^-,\phi^+,\phi^-)$ such that $M=M^+\oplus M^-$ is the decomposition of $M$ as direct sum of the eigenspaces of right multiplication by $x$ (the only possible eigenvalues are $1$ and $-1$, and it is diagonalizable) and $\phi^{\pm}:M^\pm\to M^\pm$ are the restrictions of the map $\phi$ to $M^+$ and $M^-$ (so in particular they square to zero). Moreover, morphisms in $\mathrm{Mod}_A^R$ have the obvious description in terms of these $4$-tuples.

Now, it is very easy to see using this description that $\mathrm{Mod}_A^R$ is not semisimple: for example, the object $(k^2,0,\left(\begin{array}{cc}0&1\\\\0&0\end{array}\right),0)$ is not semisimple (in fact, the category is the direct sum of two copies of the module category over the quiver $\bullet\to\bullet$). It follows that $\mathrm{Mod}_E$ and $\mathrm{Mod}_A^R$ are not equivalent in this case.

(The answer is yes, though, in the two extreme cases where (i) $E=k$ or (ii) $E=A$ (the first one is the «fundamental theorem of Hopf algebras», the second one is trivial)

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    $\begingroup$ A is not cocommutative, which was one of the assumptions. $\endgroup$ Nov 17, 2009 at 1:53

OK, this question is still bothering me, and I still don't know the answer. Truth to tell, I suspect it is false.

I write to point out that your two functors are adjoint. More precisely, suppose we have a map of R-comodules from

A tensor_E M --> N

where M is an E-module. Then we get an induced E-module map on the primitives

P(A tensor_E M) --> PN

There is an obvious map M --> P(A tensor_E M)

that takes m to 1 tensor m. Thus we get an E-module map M --> PN.

Conversely, if we have an E-module map M --> PN, then we get an R-comodule map

A tensor_E M --> A tensor_E PN

then the multiplication map A tensor_E PN --> N is an R-comodule map, so we get

an R-comodule map A tensor_E M --> N, and this makes the functors adjoint.


I have a suggestion for you. Try it when $A=k[G]$ for a finite group $G$ and $E=k[H]$ for a subgroup $H$. Then $R$ should be $k[G/H]$, which of course will only be a coalgebra and not a Hopf algebra if $H$ is not normal. This example leads me to doubt your claim that $R$ is a coalgebra in the category of $A$-algebras, since I don't think $R$ is an $A$-algebra unless E is normal.

Anyway, your desired result should be something about induction and restriction in this case. Indeed, an E-module N is just a representation of $H$. A tensor over $E$ with $N$ is just the induced G-representation.

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    $\begingroup$ Eric must have meant "A-modules", not "A-algebras"; then the claim is true (in this case at least). $\endgroup$ Nov 7, 2009 at 16:49
  • $\begingroup$ Yes, that was a typo. Sorry about that. $\endgroup$ Nov 7, 2009 at 19:07

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