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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.

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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. –  Reid Barton Nov 1 '09 at 20:06
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}? –  David Jordan Nov 1 '09 at 21:10
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? –  Eric Wofsey Nov 1 '09 at 22:28
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. –  David Jordan Nov 2 '09 at 2:27
ps - sorry about the terse/crammed nature of the reply. character limits! =] –  David Jordan Nov 2 '09 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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A is not cocommutative, which was one of the assumptions. –  Reid Barton Nov 17 '09 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.

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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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Eric must have meant "A-modules", not "A-algebras"; then the claim is true (in this case at least). –  Reid Barton Nov 7 '09 at 16:49
Yes, that was a typo. Sorry about that. –  Eric Wofsey Nov 7 '09 at 19:07

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