## A Projective Module that is Not Free

We work over a fixed commutative unital ring $R.$ An $R$-module is projective if given any surjection $f : M \to N$ and map $g : P \to N$ there exists a (not necessarily unique) lift $h : M \to P$ such that $g = fh.$

Proposition. All free modules are projective.

Pf. We need the axiom of choice. Lift the image of each generator $f(e_i)$ to some $m_i$ if the fiber over $f(e_i).$ Q.E.D.

The converse is not true in general.

Counterexample. $R \times \{0\}$ is projective but not free (as a $R$-module).

“One can turn monads into adjunctions and adjunctions into monads, but one doesn’t always return where one started.” – John Baez

Let $F : C \to D, U : D \to C$ be an adjunction. We get a monad $T = FU : C \to C.$ This is something like a map $\{\text{adjunctions}\} \to \{\text{monads}\}$ (actually, this can be encoded by a certain $2$-functor). Conversely,  suppose we have an abstract monad $T : C \to C.$ One can construct an adjunction between $C$ and its so-called ‘Eilenberg-Moore category of algebras’ $C^T.$

Definition (Eilenberg-Moore Category).  Let $(T,\eta)$ be a monad on $C.$ Then, the category of T-algebras (or Eilenberg-Moore category$C^T$ has

• objects: an object $x \in C$ together with a morphism $h:Tx \to x$ in $C$ such that the diagramscommute.
• morphisms: a morphism $\varphi : (x,h) \to (x',h')$ of $T$-algebras is a $C$-morphism $f: x \to x'$ such that TFDC

Example. Consider the monad $(-)_\ast : Set \to Set$ which adjoins an element $\ast.$ Then, $C^T$ is the category of pointed sets.

The Eilenberg-Moore construction gives an adjunction. There is forgetful functor $U C^T \to C$ given on objects by projection onto the first factor and on morphisms by identity. It admits a left adjoint. This is the free T-algebra functor $F : C \to C^T$ where

• $F(x) = \mu_x : T^2x \to Tx$
• $F(f : x \to x') = Tf : Tx \to Tx'$

Now, suppose our monad came from an adjunction between categories $C,D.$ There is a canonical comparison functor $k : D \to C^T.$ If $k$ is an equivalence of categories, the adjunction is said to be monadic. Beck’s monadicity theorem characterizes monadic adjunctions. In general, $k$ is not an equivalence of categories.

Counterexample ([1]). Consider the forgetful functor $U : Top \to Set.$ Its left adjoint is the discrete space functor. So, $Top$ cannot possibly be the Eilenberg-Moore category. Ultimately, this is related to the fact that $U$ does not reflect isomorphisms (a hypothesis of Beck’s monadicity theorem).

References

[2.] Awodey, S. Category Theory. Oxford University Press, 2006.

## Morita Equivalence but not Equivalence

### Non-Commutative Algebra with an Example

This is technically a counterexample, it is not one in the traditional sense. Normally, a counterexample is an object for which some intuition says it should not exist. Nobody thinks all Morita equivalent algebras and equivalent. But, just as nobody thinks all exact sequences split, it is helpful to have examples of non-splitting exact sequences at one’s elbow.

Morita equivalence is a property of (associative) algebras $A$ and $B.$ It means they have the same representation theory.

Definition 1. Let $A,B$ be associative algebras over a field $k.$ Then, $A$ is said to be Morita equivalent to $B$, written $A \sim B$, if there is an equivalence of categories $A\textbf{-mod} \simeq B\textbf{-mod}.$

To paraphrase: being acted on by $A$ is essentially the same as being acted on by $B.$

Notation: Throughout, let $A,B$ denote associative algebras over a fixed field $\mathbb{k}$

For commutative algebras, Morita equivalence coincides with isomorphism.

Lemma 2. $Z(A\textbf{-mod}) \simeq Z(A)$ where $Z(A)$ denotes the center of $A$ and $Z(A\textbf{-mod})=\text{End}(\text{Id}_{A\textbf{-mod}}).$

Proof.  Let $z \in A$ be central. Then, define a natural transformation whose component $\varphi_M : M \to M$ at a left $A$-module $M$ is the action of $z$ on $M$. By centrality of $z$$\varphi_M$ is a module homomorphism. Conversely, given a natural transformation $\varphi : \text{Id}_{A\textbf{-mod}} \to \text{Id}_{A\textbf{-mod}}$ define $z:= \varphi_A(1_A) \in A.$ This $z$ is central by naturality of $\varphi.$ Q.E.D.

Proposition 3. Let $A,B$ be commutative. Then, $A \sim B$ if and only if $A$ is isomorphic to $B.$

Proof. Clearly, if $A \cong B$ then $A \sim B.$ Conversely, suppose $A \sim B.$ Then, $Z(A\textbf{-mod}) \simeq Z(B\textbf{-mod}).$ By Lemma 2, we are done. Q.E.D.

To show this result is “sharp” we construct non-isomorphic algebras that are Morita equivalent. By Proposition 3. (at least) one algebra must be non-commutative.

Theorem 4. For $n \geq 1$, $A \sim \text{Mat}_n(A).$

Proof. Let $e \in \text{Mat}_n(A)$ denote the matrix with a $1$ in the $1 \times 1$ spot and zeros everywhere else. Then, $A \simeq e \cdot \text{Mat}_n(A) \cdot e$ and $\text{Mat}_n(A) =\text{Mat}_n(A) \cdot e \cdot\text{Mat}_n(A).$ So by Morita’s Theorem [1, Corollary 2.3.2] we’re done. Q.E.D.

For $n \geq 1$, Mat$_n(A)$ is non-commutative. So, for a commutative algebra $A$ we cannot have an isomorphism between $A$ and Mat$_n(A).$

References

[1] Ginzburg, V. Lectures on Noncommutative Geometry. https://arxiv.org/pdf/math/0506603.pdf

## The Symplectic Sum: An Infuriating Construction

### Never Let Anybody Chase Your Diagrams for You

The symplectic sum construction upsets me. It is canonical, but not blessed by the categorical dialectic.

Definition 1.  A symplectic vector space is a pair $(V,\omega)$ where $V$ is a vector space and $\omega$ is a non-degenerate anti-symmetric bilinear form.

Example 2. Consider $\mathbb{R}^n \oplus (\mathbb{R}^n)^\ast.$ Then,

$\omega((v,\ell),(v',\ell')) := \ell'(v)-\ell(v')$

is symplectic.

Exercise 3. Show that a symplectic vector space must have even dimension.

Construction 4. Let $(V_1,\omega_1),(V_2,\omega_2)$ be symplectic vector spaces. Then,

$\omega((v_1,v_2),(w_1,w_2)) := \omega_1(v_1,w_1) + \omega(w_2,v_2)$

makes $(V_1 \oplus V_2, \omega)$ into a symplectic vector space.

Proof. We show non-degeneracy. Fix $(v_1,v_2) \in V_1 \oplus V_2.$ Suppose that

$\omega((v_1,v_2),(w_1,w_2)) := \omega_1(v_1,w_1) + \omega_2(v_2,w_2) =0$

for all $0 \neq (v_2,w_2) \in V_1\otimes V_2.$ Take $v_2 \neq 0$ and $w_2 = 0.$ Then, $(v_2,w_2) \neq 0.$ But, we get

$\omega_1(v_1,w_1) = -\omega_2(v_2,0) = 0$

for all $v_2 \in V_2.$ So, because $\omega_1$ is symplectic, we have $v_1=0.$ Thus, $\omega_2(v_2,w_2) = 0$ for all $\omega_2 \neq 0.$ By non-degeneracy of $\omega_2$ we have $v_2=0$ as well. This completes the proof.

Construction 4. is, in an intuitive sense, the canonical symplectic product form. Yet, $(V_1 \oplus V_2, \omega_1 \oplus \omega_2)$  is not the categorical product.

Definition 5. (The Symplectic Category) Let $(V_1,\omega_1), (V_2,\omega_2)$ be symplectic vector spaces. A linear map $T:V_1 \to V_2$ is symplectic if $\omega_2(Tv,Tw) = \omega_1(v,w).$ A symplectomorphism is an invertible symplectic map.

Exercise 6. Check that the composition of symplectic maps is symplectic and that the identity is symplectic. So, there is a category Symp of symplectic spaces and symplectic linear maps.

Exercise 7. Show that every symplectic vector space is symplectomorphic to the one given as Example 2.

Exercise 8. Show that symplectomorphisms preserve volume. (Hint: Use non-degeneracy to show that $\omega^n$ is a volume form).

“Counter”-example 9. The projection $\pi : (V_1 \oplus V_2,\omega=\omega_1 \oplus \omega_2) \to (V_1,\omega_1)$ is not symplectic.

Proof. Compute $\omega_1(\pi v,\pi w) = \omega_1(v_1,w_1) \neq \omega(v,w)$ whenever $\omega_2(v_2,w_2) \neq 0$. This completes the proof.

## Quirkiness with Kan Extensions

Given two functors $F : C \to D$ and $K : C \to D$left kan extension of $F$ along $K$ is a functor Lan$_kF : D \to E$ along with a natural transformation $\eta : F \to \text{Lan}_KF K$ universal in that for any pair $(G: D \to E, \gamma: F \to GK)$ there is a unique natural transformation $\alpha : \text{Lan}_KF \to G$ such that $\gamma = \alpha_K \eta$ as in the diagram

Our counterexample is meant to define intuition about this commutativity “up to universal natural transformation.” A properly commutating diagram may not actually be the Kan extension.  In other words, if $F$ factors through $K$ by some $H$ then $(H,1_H)$ is not necessarily the same as $(\text{Lan}_KF, \eta).$

Spoiler Alert: This is exercise 1.1.3. in [1]. If you want to find out for yourself, stop reading.

Take $1 = C, E=Set$ and $D$ arbitrary. Let $d \in D$ be an arbitrary object and let $d : 1 \to D$ denote the functor constant at $d.$ We take $F = \ast$ e.g. $F$ is the functor constant at some singleton set $\ast.$ Then, we can (by abuse of notation) write $\ast : D \to Set$ again for the functor constant at $\ast.$ Indeed we have $\ast = \ast \circ d.$ But, $\ast : D \to Set$ is not the (left) Kan extension of $\ast$ along $d.$

Claim: $(\text{Lan}_d\ast,\eta)=(\text{Hom}_D(d,-),1_D)$

Why is this true?  Because elements $x \in Fd$ are in bijection with natural transformations $x : \ast \to F.d$ By Yoneda’s lemma, these elements are in natural bijection with natural transformations $\text{Hom}_D(d,-) \to F.$ This is exactly the universal property of left Kan extensions.

References

[1] Riehl, E. Categorical Homotopy Theory. Cambridge University Press, 2014.

## A Presheaf Image that is not a Sheaf Image

### The Global Sections Functor is not Exact

Presheaf morphisms are the same as sheaf morphisms; they’re natural transformations: we get one sequence for each open set. So, it seems odd that exactness can hold globally, but not for each component. After all, an exact sequence is a bunch of objects and morphisms between them. Yet, taking sections of an exact sequence of sheaves does not, in general, yield exact sequences.

How? Because, an exact sequence is not just some objects and morphisms. An exact sequence is also the fact that the image of each morphism is the kernel of the next. The image and the kernel are objects that depend on an ambient abelian category.

Definition 1. Let $\mathcal{A}$ be an abelian category and $f : A \to B$ a morphism is $\mathcal{A}.$ Then, the kernel of $f$ is a map $i : K \to A$ universal with respect to the property that $f \circ i = 0.$

Definition 2. Let $\mathcal{A}$ be an abelian category and $f : A \to B$ a morphism is $\mathcal{A}.$ Then, the cokernel of $f$ is a map $s : B \to C$ such that $s \circ f = 0$ universally. The image of $f$ is $\text{im} f := \text{ker coker} f.$

Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of presheaves. The kernel of $\varphi$ is a presheaf given by $\text{ker}_\varphi (U) = \text{ker}\varphi_U.$ The image presheaf is defined the same way. Now, the presheaf kernel of a map of sheaves is always a sheaf. In particular, the presheaf kernel is isomorphic to the sheaf kernel. But, the presheaf image of a map of sheaves is in general not a sheaf! Images in the abelian category of sheaves are constructed by sheafifying the presheaf image.

Let $X = \mathbb{C}^\ast.$ Consider the exponential sequence of sheaves on $X$

$0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^\ast \to 0,$

with maps given by $\times 2\pi i : \mathbb{Z} \to \mathcal{O}_X$ and $\text{exp} : \mathcal{O}_X \to \mathcal{O} _X^\ast.$ On each open set, it is an exact sequence of abelian groups. So, it is an exact sequence as a sequence of presheaves of abelian groups. Moreover, one checks that exp$: \mathcal{O}_X \to \mathcal{O}_X^\ast$ is surjective on the stalks. It is a fact that exactness of a sequence of sheaves can be checked at the level of stalks. Thus, this is an exact sequence of sheaves.

Taking global sections does not yield an exact sequence because there is no global logarithm. So, the exponential is not surjective at the level of global sections.

As suggested, this is because the presheaf image of the exponential is not a sheaf. Consider the sections $t_i := \text{id}_{|U_i} \in \mathcal{O}^\ast_X(U_i).$ Using the logarithm function, there are sections $s_j \in \mathcal{O}_X(U_j)$ such that $t_i = \text{exp}(s_i).$ Since $t_i{_|U_i \cap U_j} =t_j{_|U_i \cap U_j}$ the sheaf axioms glue the exp$(s_i)$ to get a global section $t$ that restricts to $\text{id}_{|U_i}$ on each $U_i.$ But, the identity is not in the image of exp. So, im$_\text{pre}\text{exp}$ fails the gluing axiom.

## Good Quotients from Bad Actions

### Not all Good Actions are Free

In constructing examples of complex manifolds the transport of structure philosophy is useful. This principle means only doing hard work for a handful of examples and then throwing that well-earned structure around, through maps, to get new creations bearing the same structure.

For example, proving that $(\mathbb{Z},+)$ is associative is an annoying set-theoretic exercise. When proving that $(\mathbb{Z}[i], +)$ is associative we do not start from pure set theory. Rather, we borrow known associativity of $\mathbb{Z}.$ We say that addition of Gaussian integers is defined pointwise

$(n_1+m_1i) + (n_2 + m_2i) = (n_1+m_1) + i(n_2 + m_2)$

and so is associative because it is associative in each coordinate by associativity of integer addition.

The case is similar in topology. Proving that $\pi_1(S^1) \simeq \mathbb{Z}$ is kind of hard. Computing $\pi_1$ of the sphere, torus and figure eight are not as hard; we use tools like van Kampen and Mayer-Vietoris to reduce the problem to computing $\pi_1(S^1).$

Writing down examples of complex manifolds is the same game. Our basic examples are open subsets $U \subset \mathbb{C}^n$. We seek to get new manifolds by taking orbit spaces of a group action. We shall describe some common conditions upon which a quotient space is a Hausdorff complex manifold and then give an example of an action not satisfying these common conditions but whose quotient is nonetheless a Hausdorff complex manifold. This (counter)example teaches us that not all good actions are free.

$\textbf{Definition 1.}$ A group action $G \times X \to X$ is said to be free if $g \cdot x \neq x$ whenever $g \neq 1.$

$\textbf{Definition 2.}$ A group action $G \times X \to X$ is said to be proper if the map $(g,x) \mapsto (g \cdot x,x)$ is proper.

Quotients by free and proper actions are manifolds.

$\textbf{Theorem 3.}$ Let $G \times X \to X$ be a free and proper group action on a manifold $X.$ Then, $X/G$ is a (Hausdorff) complex manifold.

$\textbf{Proof.}$ For all $x,y \in X$ such that $x \notin G \cdot y$ there exist open nieghborhoods $x \in U, y \in V$ such that, for any $g \in G$$U \cap g\cdot V = \empty.$ Now, suppose we have charts $(U_i,\varphi_i)$ that witness $X$ as a complex manifold and such that $g \cdot U_i \cap U_i = \empty$ for all $i$ and for all $1 \neq g \in G.$ Then, $U_i \cong \pi(U_i)$ (where$\pi$ denotes the projection$\pi : X \to X/G.$) We get holomorphic charts

$\pi^{-1} \circ \varphi_i : \pi(U_i) \to \mathbb{C}^n.$

The properness condition ensures the quotient is Hausdorff.

Now, this is just one way to get a complex structure on a quotient. This counterexample comes from D. Huybrechts Complex Geometry p. 60: Consider a complex torus $\mathbb{C}/\Gamma.$ The group $\mathbb{Z}/n\mathbb{Z}$ acts on $\mathbb{C}/\Gamma$ by $z \mapsto -z.$ This action has four fixed points: $0, \tau_1/2,\tau_2/2, (\tau_1+\tau_2)/2$, where $\Gamma = \tau_1\mathbb{Z} + \tau_2\mathbb{Z}.$ A little dexterity with the Weierstrass function shows this quotient is isomorphic to $\mathbb{P}^1.$ This quotient is indeed a complex manifold (goodbut the action is not free (bad).