Peng Zhou

This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong.
====== Difference between Kleisili and Eilenberg-Moore reconstruction ======

===== 1. The question =====

Suppose $F:C\to D$ is a conservative functor admitting a left adjoint $F^L:D\to C$. We obtain a monad
$$
T=FF^L:D\to D.
$$

Under the Barr--Beck hypotheses, we have
$$
C\simeq \operatorname{Alg}_T(D).
$$

On the other hand, we can construct the Kleisli category $K=\operatorname{Kl}(T)$, whose objects are those of $D$, with
$$
\operatorname{Hom}_K(X,Y)=\operatorname{Hom}_D(X,TY).
$$

The Kleisli category embeds fully faithfully into $\operatorname{Alg}_T(D)$ as the category of free $T$-algebras.

A natural question is:

**If $D$ is already perfect, i.e. $D\simeq\operatorname{Perf}(D)$, do we necessarily have**
$$
\boxed{\operatorname{Perf}(K)\simeq\operatorname{Alg}_T(D)\,?}
$$

The answer is no, even in a very elementary example.

===== 2. Example: the dual numbers =====

Let $k$ be a field, and consider the algebra
$$
A=k[\varepsilon]/(\varepsilon^2).
$$

Take
$$
D=\operatorname{Perf}(k),
$$
the category of bounded complexes of finite-dimensional $k$-vector spaces.

In particular, $D$ is already stable and idempotent-complete:
$$
D\simeq\operatorname{Perf}(D).
$$

Define the monad
$$
T(V)=A\otimes_k V.
$$

Its unit and multiplication are induced by the unit and multiplication of $A$.

The Eilenberg--Moore category is
$$
\boxed{
\operatorname{Alg}_T(D)
\simeq D^b(\operatorname{mod}_{fd}A).
}
$$

Indeed, a $T$-algebra is an $A$-module whose underlying $k$-complex is perfect, or equivalently, an $A$-complex with bounded finite-dimensional cohomology.

On the other hand, the Kleisli category consists of free $A$-modules of the form $A\otimes_k V$, for $V\in\operatorname{Perf}(k)$.

Taking its perfect envelope gives
$$
\boxed{
\operatorname{Perf}(\operatorname{Kl}(T))
\simeq\operatorname{Perf}(A)
=K^b(\operatorname{proj}A).
}
$$

Thus the question becomes whether
$$
\operatorname{Perf}(A)
\overset{?}{=}
D^b(\operatorname{mod}_{fd}A).
$$

===== 3. A module which is not perfect =====

Consider the simple $A$-module
$$
M=k=A/(\varepsilon),
$$
on which $\varepsilon$ acts by zero.

Clearly $M$ is a finite-dimensional $k$-vector space, so it is an object of $\operatorname{Alg}_T(D)$.

However, $M$ is not a perfect $A$-module.

To see this, consider its free resolution:
$$
\cdots
\xrightarrow{\varepsilon} A
\xrightarrow{\varepsilon} A
\xrightarrow{\varepsilon} A
\xrightarrow{\varepsilon} A
\longrightarrow k
\longrightarrow 0.
$$

This complex is exact because
$$
\ker(\varepsilon:A\to A)
=(\varepsilon)
=\operatorname{im}(\varepsilon:A\to A).
$$

Thus $k$ has an infinite periodic free resolution.

In fact, applying $\operatorname{Hom}_A(-,k)$ gives
$$
0\longrightarrow k
\xrightarrow{0}k
\xrightarrow{0}k
\xrightarrow{0}k
\xrightarrow{0}\cdots,
$$
because $\varepsilon$ acts trivially on $k$.

Consequently,
$$
\boxed{
\operatorname{Ext}^n_A(k,k)\cong k
\qquad\text{for every }n\geq 0.
}
$$

If $k$ were perfect over $A$, it would admit a bounded complex of finitely generated projective $A$-modules as a resolution. This would imply
$$
\operatorname{Ext}^n_A(k,k)=0
\qquad\text{for }n\gg0,
$$
a contradiction.

Therefore
$$
k\notin\operatorname{Perf}(A).
$$

===== 4. Conclusion =====

We have exhibited an object
$$
k\in\operatorname{Alg}_T(D)
$$
which does not belong to the thick subcategory generated by free $T$-algebras.

Hence
$$
\boxed{
\operatorname{Perf}(\operatorname{Kl}(T))
\subsetneq
\operatorname{Alg}_T(D).
}
$$

This happens despite the fact that $D=\operatorname{Perf}(k)$ is already perfect, and the forgetful functor
$$
F:D^b(\operatorname{mod}_{fd}A)\to\operatorname{Perf}(k)
$$
is conservative and monadic.

**The point:** Being perfect after forgetting the $T$-algebra structure does not imply being perfect as a $T$-algebra.

The simple module $k$ is perfect as a $k$-complex, but has infinite projective dimension as an $A$-module.

===== 5. When does the equality hold? =====

A useful sufficient condition is that $A$ has finite global dimension.

Indeed, if
$$
\operatorname{gldim}(A)<\infty,
$$
then every finite-dimensional $A$-module admits a finite projective resolution, so
$$
D^b(\operatorname{mod}_{fd}A)
\simeq\operatorname{Perf}(A).
$$

Thus, for a finite-dimensional $k$-algebra $A$,
$$
\boxed{
\operatorname{Perf}(\operatorname{Kl}(A\otimes_k-))
\simeq
\operatorname{Alg}_{A\otimes_k-}(\operatorname{Perf}(k))
}
$$
whenever $A$ has finite global dimension.

The dual numbers provide a counterexample precisely because
$$
\operatorname{gldim}(k[\varepsilon]/(\varepsilon^2))=\infty.
$$

More generally, the correct condition is that every $T$-algebra can be obtained from free $T$-algebras using finitely many cones, shifts, and retracts.