Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

3 Answers

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)

share|improve this answer
    
A is not cocommutative, which was one of the assumptions. –  Reid Barton Nov 17 '09 at 1:53
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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