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

Let $k$ be a field. Let $C$ be an abelian $k$-linear category with a symmetric tensor product $\otimes$ and internal homomorphisms, such that $\mathrm{End}(1)=k$. Let $M$ be another $k$-linear abelian category, and let $$h:M\to C$$ be an exact, faithful, $k$-linear functor. If $C$ has arbitrary limits and $M$ is small, we can understand the endomorphism ring of $h$ as a ring object (a monoid) in $C$, as follows. It is the equaliser $$R := \ker \left( \prod_{m\in M}\mathrm{End}(h(m)) \xrightarrow{\quad u \quad} \prod_{m\to n}\mathrm{Hom}(h(m),h(n))\right)$$ where $u$ is "precomposition minus postcomposition". The given functor $h$ factors then canonically over the category of $R$-modules in $C$ $$M \xrightarrow{\quad\widetilde h \quad} R\mathrm{-Mod}_C \xrightarrow{\quad f \quad} C$$ where $f$ is the forgetful functor. My question is:

Is the functor $\widetilde h$ fully faithful?

For example, if $C$ is the category of vector spaces over $k$, then the answer is Yes. For this, $k$ does not even have to be a field. On the other hand, $C$ could be much larger, for example, the category of sheaves of $k$-vector spaces on a connected topological space. This is the case i am ultimately interested in.

Remark: The existence of limits in $C$ was just for convenience, without them, one can still view $R$ as a pro-ring object and consider $R$-modules. The hypothesis $\mathrm{End}(1)=k$ is essential if one wishes for a positive answer, because the category of modules $R\mathrm{-Mod}_C$ is always $\mathrm{End}(1)$-linear.

Remark: Also, the whole story smells like it was some variant of the Freyd-Mitchell theorem, but i don't see a concrete way to link it to that.

Added: The proof that in the case $C = \mathrm{Vect}(k)$ the answer to my question is Yes goes roughly as follows: In this special case, we can view $M$, or at least the $\mathrm{Ind}M$, as a $C$-module, that is, for every vector space $V$ and object $m \in M$ we can give a sense to $\mathrm{Hom}(V,m)$ and $V\otimes m$ as objects in $M$. At this point, $\widetilde h$ is then even an equivalence. Define an object $$X := \ker \left( \prod_{m\in M}\mathrm{Hom}(h(m),m) \xrightarrow{\quad u \quad} \prod_{m\to n}\mathrm{Hom}(h(m), n)\right)$$ in $M$. Then $h(X)$ is $R$ as a right $R$-module, and given a left $R$-module $V$, we can define an object $Y$ of $M$ by $$Y = X \otimes_R V = \mathrm{coker}(X \otimes R \otimes V \xrightarrow{\quad v \quad} X \otimes V)$$ with $v$ = "left action on $V$ minus right action $X$". But then, $\widetilde h(Y)$ is isomorphic to the given $R$-module $V$, and one can treat morphisms in the same way: $V\to V'$ yields $X\otimes_RV \to X\otimes_RV'$. Note that if we only consider finite dimensional spaces, then the passage to ind-objects is superfluous, and $\widetilde h$ is an equivalence of categories.

The proof generalises for as far as $M$ is a $C$-module. But what is there to do if, for example, $M$ is "graded local systems" on a topological space and $C$="sheaves" and $h$=forget?

share|improve this question
    
Interesting question! Note that $R = \int^{m \in M} \underline{\mathrm{Hom}}(h(m),h(m))$. Can you say something about the proof when $C=\mathsf{Vect}(k)$ (in particular why it doesn't generalize to sheaves)? –  Martin Brandenburg Oct 4 '13 at 13:23
2  
The whole story is more like a variant of Grothendieck's Galois theory since you've provided the "fiber functor" $h$ as part of the data. In Freyd-Mitchell the whole point is to cook up a suitable fiber functor. –  Qiaochu Yuan Oct 4 '13 at 18:13
add comment

1 Answer

No. What follows appears to be a counterexample for $C = \text{Vect}$ (I don't understand where in your argument you prove fullness).

Let $M = \text{Vect}^{op}, C = \text{Vect}$, and let $h : \text{Vect}^{op} \to \text{Vect}$ be the contravariant functor $V \mapsto V^{\ast} \cong \text{Hom}(V, k)$. If smallness is important to you pretend that the first $\text{Vect}$ means vector spaces of at most countable dimension. The endomorphism ring of $h$ is $k$, so the lift $\tilde{h}$ is just $h$ again.

I claim that $h$ is not full. To see this, if $V$ is a countable-dimensional vector space, regarded as an object in $\text{Vect}^{op}$, then the induced map

$$\text{End}(V, V) \to \text{End}(V^{\ast}, V^{\ast})$$

has the property that its source has dimension $\aleph_0^{\aleph_0} = 2^{\aleph_0}$, but its target has dimension at least $\left( 2^{\aleph_0} \right)^{2^{\aleph_0}}$.

(Taking the opposites of familiar abelian categories seems to be my new favorite trick! Note that vector space duality establishes that $\text{FinVect}$ is equivalent to its opposite. $\text{Vect}$ itself is the ind-completion of $\text{FinVect}$, so $\text{Vect}^{op}$ is the pro-completion; in other words, it's the category of profinite vector spaces. This category is also known as the category of linearly compact vector spaces: see this MO question and this n-cafe post for details. This gives some intuition for why $h$ is not full: it's the forgetful functor and it doesn't see the topology.

In the special case that $k = \mathbb{F}_p$ we can be a little more explicit: $\text{Vect}^{op}$ in this case is the full subcategory of profinite groups consisting of the ones whose finite quotients are all elementary abelian $p$-groups.)

share|improve this answer
    
+1, especially "it's the forgetful functor and it doesn't see the topology" was very enlightening! –  Martin Brandenburg Oct 4 '13 at 22:27
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.