2nd isomorphism theorem in an abelian category

In this post, we will discuss about the "2nd isomorphism theorem" in an abelian category. This is a well-known phenomenon, but I find it difficult to find a standard source. I follow Chapter 8 of Mac Lane's book for the definition of abelian categories.

1st isomorphism theorem. Given a morphism $f : A \rightarrow B$ of abelian groups, the first isomorphism theorem states that the canonical map $A/\ker(f) \rightarrow f(B)$ is an isomorphism. This holds in an abelian category. That is, let $f : A \rightarrow B$ be a morphism in an abelian category. We define

• $\mathrm{im}(f) := \ker(\mathrm{cok}(f))$ and
• $\mathrm{coim}(f) := \mathrm{cok}(\ker(f)) = A/\ker(f).$

More precisely, we have $m : \mathrm{im}(f) \hookrightarrow B$ and $e : A \twoheadrightarrow \mathrm{coim}(f).$ From either universal property, one has a canonical map $\mathrm{coim}(f) \rightarrow \mathrm{im}(f),$ and this is necessarily an isomorphism (e.g., see Mac Lane, p.199,  Proposition 1).

Remark. It seems that the 1st isomorphism can be taken as part of axioms of a category being abelian: see Proposition 2.3 of this nLab page.

2nd isomoprhism theorem. Given a subgroups $X, Y$ of an abelian group $A,$ the 2nd isomorphism theorem states that the canonical map $$Y/(X \cap Y) \rightarrow (X + Y)/X$$ is an isomorphism. How do we get this in an abelian category? A priori, this is just the 1st isomorphism theorem applied to the map $Y \twoheadrightarrow \pi_{X}(Y),$ where $\pi_{X} : A \twoheadrightarrow A/X$ is the projection map. We similarly define $\pi_{Y} : A \twoheadrightarrow Y.$ We also denote by $\iota_{X}$ and $\iota_{Y}$ the maps $X \hookrightarrow A$ and $Y \hookrightarrow A,$ respectively.

Taking intersection in an abelian category. Note that

• $X \cap Y = \ker(\iota_{X} \circ \pi_{Y} : X \hookrightarrow A \twoheadrightarrow A/Y),$

so we define $X \cap Y := \ker(\iota_{X} \circ \pi_{Y}).$ Note that we have $$\ker(\iota_{X} \circ \pi_{Y}) \hookrightarrow \ker(\pi_{Y}) = Y$$ and if there is a common subobject $S$ of $X$ and $Y$ (with all the maps into $A$ commute), then, there is a unique map $S \hookrightarrow X \cap Y$ that commutes with the maps from $S$ and $X \cap Y$ into $A$ induced by the universal property. This means that we get a unique isomorphism $$X \cap Y \simeq Y \cap X$$ commuting with maps into $A$ (and hence those into $X$ or $Y$).

Taking sum in an abelian category. Denoting $p_{X} : X \oplus Y \rightarrow X$ and $p_{Y} : X \oplus Y \rightarrow Y$ for the projections that come with the direct product (which comes with the direct sum structure in any abelian category), we define $$X + Y := \mathrm{im}(\iota_{X}p_{X} + \iota_{Y}p_{Y}).$$ By the universal property, we can observe that $X, Y$ are subobjects of $X + Y,$ and their maps to $A$ are all commutative. Now, let $T$ be any subobject of $A$ that take $X, Y$ as subobjects (with all the maps into $A$ commute). Then there is a unique map $$X + Y \simeq \mathrm{coim}(\iota_{X}p_{X} + \iota_{Y}p_{Y}) \rightarrow T,$$ that commutes with maps from $X \oplus Y.$ The induced map is therefore (e.g., by applying Mac Lane, p.199,  Proposition 1) commutes with the maps into $A,$ so it is in particular a monomorphism. Hence, it follows that $X + Y$ is the smallest subobject of $A$ that contain $X$ and $Y.$

Proof of 2nd isomorphism theorem. By definition, we may already note that the kernel of the composition $Y \hookrightarrow A \twoheadrightarrow A/X$ is $X \cap Y.$ It remains to show that its image is given by $(X + Y)/X.$ First, it is not difficult to observe that the cokernel of this composition is given by $A/X \twoheadrightarrow A/(X + Y).$ One may then show that the kernel of this map is precisely $(X+Y)/X,$ which finishes the proof.

Some thoughts on random matrices

This post is to start writing down some thoughts on extending results in my paper with Yifeng Huang (to be referred as CH) about Haar random matrices over $\mathbb{Z}_{p}.$

Surjection moments. One thing we have not tried in our paper is to use the following strategy, popular in the literature. For example, given an odd prime $p,$ let $\mathcal{F}_{p}$ be the set of all isomorphism classes of finite abelian $p$-groups (or equivalently finite $\mathbb{Z}_{p}$-modules). Equip a structure of the probability space on $\mathcal{F}_{p},$ with the sample space consisting of all subsets of it, with a probability measure $$\mu : \mathcal{F}_{p} \rightarrow [0, 1].$$ The statement goes, if $$\mathbb{E}_{[H] \in \mathcal{F}_{p}}(|\mathrm{Sur}(H, G)|) = 1$$ for any finite abelian $p$-group $G,$ then $\mu$ is the Cohen-Lenstra measure, meaning we have $$\mu(G) := \mu(\{[G]\}) = \dfrac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{\infty}(1 - p^{\infty}),$$ for any finite abelian $p$-group $G.$ (A simple proof can be found in the proof of Lemma 8.2 of this paper by Ellenberg, Venkatesh, and Westerland.) Here, we used the following notations:

• $\mathbb{E}$ meant the expected value;
• $\mathrm{Sur}(H, G)$ meant the set of surjective $\mathbb{Z}$-linear maps from $H$ to $G$;
• $\mathrm{Aut}(G)$ meant the set of $\mathbb{Z}$-linear automorphisms of $G.$

Surjection moments for Haar $p$-adic matrices. Now, consider the set $\mathrm{Mat}_{n}(\mathbb{Z}_{p})$ of $n \times n$ matrices over $\mathbb{Z}_{p},$ the ring of $p$-adic integers. The $p$-adic topology given by $\mathbb{Z}_{p}$ makes $\mathrm{Mat}_{n}(\mathbb{Z}_{p})$ a compact Hausdorff topological group with respect to its addition, so we can consider its Haar measure, which is a unique probability measure translation-invariant under addition. Can we compute $$\mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|),$$ for a given finite abelian $p$-group $G$?

We have $$\begin{align*}\mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}|\mathrm{Sur}(\mathrm{cok}(X), G)| d\mu_{n}(X) \\ &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}|\mathrm{Sur}(\mathbb{Z}_{p}^{n}/ \mathrm{im}(X), G)| d\mu_{n}(X) \\ &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}\sum_{\bar{F} \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}/ \mathrm{im}(X), G)}1 d\mu_{n}(X) \\ &= \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}\sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)}\boldsymbol{1}_{F(-) = 0}(X) d\mu_{n}(X) \\ &= \sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)} \int_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}\boldsymbol{1}_{F(-) = 0}(X) d\mu_{n}(X) \\ &= \sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)} \mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(FX = 0),\end{align*}$$ 

• where $\mu_{n}$ denotes the Haar measure (although it could be any probability measure);
• $\boldsymbol{1}_{F(-) = 0} : \mathrm{Mat}_{n}(\mathbb{Z}_{p}) \rightarrow \{0, 1\}$ is the indicator function such that $\boldsymbol{1}_{F(-) = 0}(X) = 0$ if and only if $FX = 0.$

Note that we have $FX = 0$ if and only if $\mathrm{im}(X) \subset \ker(F),$ which is the same as saying that all the columns of $X$ sits inside $\ker(F).$ When $\mu_{n}$ is the Haar measure, each column of $X$, which is an element of $\mathbb{Z}_{p}^{n}$ sits inside $\ker(F),$ with probability $$\frac{1}{|\mathbb{Z}_{p}^{n}/\ker(F)|} = \frac{1}{|G|}.$$ The event that a column $X$ sits inside $\ker(F)$ is independent to those that other columns sit inside $\ker(F)$ (according to the Haar measure), so we have

$$\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(FX = 0) = \frac{1}{|G|^{n}} = \frac{1}{|\mathrm{Hom}(\mathbb{Z}_{p}^{n}, G)|}.$$ Thus, we have $$\begin{align*}\mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) &= \sum_{F \in \mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)} \mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(FX = 0) \\ &= \frac{|\mathrm{Sur}(\mathbb{Z}_{p}^{n}, G)|}{|\mathrm{Hom}(\mathbb{Z}_{p}^{n}, G)|} \\ &= \frac{|\mathrm{Sur}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, G/pG)|}{|\mathrm{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, G/pG)|} \\ &= \frac{(p^{n}-1)(p^{n}-p) \cdots (p^{n}-p^{r_{p}(G)-1})}{p^{n r_{p}(G)}} \\ &= \left(1 - \frac{1}{p^{n-r_{p}(G)+1}}\right) \left(1 - \frac{1}{p^{n-r_{p}(G)+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right) \\ &= \prod_{j=n-r_{p}(G)+1}^{n}(1 - p^{-j})\end{align*}$$ if $r_{p}(G) := \dim_{\mathbb{F}_{p}}(G/pG) \leq n$ and $0$ otherwise. Here, subscript $\mathbb{F}_{p}$ is to indicate linear maps over $\mathbb{F}_{p},$ and reduction over this field is possible due to Nakayama's lemma.

In particular, we have $$\lim_{n \rightarrow \infty} \mathbb{E}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) = 1.$$

Question. Does the above computation guarantee that $$\lim_{n \rightarrow \infty}\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G) = \frac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{\infty}(1 - p^{-i}) \ ?$$

First, we need to know whether if $$\mathbb{E}_{H \in \mathcal{F}_{p}}(|\mathrm{Sur}(H, G)|) = 1$$ for any finite abelian $p$-group $G$ when $\mathcal{F}_{p}$ is equipped with the Cohen-Lenstra measure. This is known to be true, and one place to find a proof is Theorem 5.3 of this paper by Fulman and Kaplan.

Next, we need to know that if this is enough to answer the question. That is, we need a version of Lévy's theorem in this setting. This also turned out to be okay. (For example, see Theorem 3.1 of this paper by Wood--the proof of it seems to be in another paper of her mentioned in the reference.)

Friedman--Washington, Cheong--Huang, and further questions. Long ago, Friedman and Washington already proved that for $n \geq r_{p}(G),$ we have $$\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G) = \frac{1}{|\mathrm{Aut}(G)|}\prod_{i=1}^{n}(1 - p^{-i}) \prod_{j=n-r_{p}(G)+1}^{n}(1 - p^{-j})$$ even without considering the surjection moments. Moreover, the probability is $0$ when $n < r_{p}(G),$ so from our previous computation, we have

$$\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G) = \frac{\mathbb{E}(|\mathrm{Sur}(\mathrm{cok}(X), G)|) \prod_{i=1}^{n}(1 - p^{-i})}{|\mathrm{Aut}(G)| }.$$

Moreover, from another result of Friedman and Washington, it turns out that

$$\begin{align*}\lim_{n \rightarrow \infty}&\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) = 0, \mathrm{cok}(X - I_{n}) \simeq G) \\ &= \lim_{n \rightarrow \infty}\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G, \mathrm{cok}(X - I_{n}) = 0) \\ &= \frac{1}{|\mathrm{Aut}(G)|}\left(\prod_{i=1}^{n}(1 - p^{-i})\right)^{2},\end{align*}$$ where $I_{n}$ is the $n \times n$ identity matrix.

Hence, the following conjecture from CH is reasonable:

Conjecture (Cheong and Huang). Given any finite abelian $p$-groups $G_{1}$ and $G_{2},$ we must have $$\lim_{n \rightarrow \infty}\mathrm{Prob}_{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}(\mathrm{cok}(X) \simeq G_{1}, \mathrm{cok}(X - I_{n}) \simeq G_{2}) = \frac{\left(\prod_{i=1}^{n}(1 - p^{-i})\right)^{2}}{|\mathrm{Aut}(G_{1})||\mathrm{Aut}(G_{2})|}.$$

Remark. We cannot find any possible use of surjection moments to attack this conjecture at the moment.

A special case of Conjecture. Conjecture implies that $$\begin{align*}&\sum_{m={1}}^{\infty}\lim_{n \rightarrow \infty}\underset{X \in \mathrm{Mat}_{n}(\mathbb{Z}_{p})}{\mathrm{Prob}}\begin{pmatrix}\mathrm{cok}(X) \simeq \mathbb{Z}/(p), \\ \mathrm{cok}(X - I_{n}) \simeq \mathbb{Z}/(p^{m}) \end{pmatrix} \\ &= \left(\frac{1}{(p-1)^{2}}\right)\left(\prod_{i=1}^{\infty}(1 - p^{-i})\right)^{2}(1 + p^{-1} + p^{-2} + \cdots) \\ &= \left(\frac{p}{(p-1)^{3}}\right)\left(\prod_{i=1}^{\infty}(1 - p^{-i})\right)^{2},\end{align*}$$ and we can compute this WITHOUT referring to Conjecture, which gives more evidence to the validity of the conjecture.

What can we do further? We believe that the trick we have in mind can compute many more sums like this, which may give more evidence to a more general conjecture given as Conjecture 2.3 in CH.

 

Update (6/6/2022). Nathan Kaplan and I proved the conjecture in our recent paper. Jungin Lee generalized this even further in his recent preprint.

Critical group of a graph

My postdoc mentor suggested me to think about some problems regarding the critical graph of a random graph, so I decided to sit down and read some basic words about this. I might write more on this if I want to store more summaries.

Critical group of a graph. When we consider a graph, we always mean a finite graph. Let $G$ be a graph with $n$ verticies. Given a vertex $v,$ the degree $\deg(v)$ is the number of paths ending at $v$ in an arbitrarily small neighborhood of $v.$ Order the vertices $v_{1}, \dots, v_{n}$ of $G.$ The diagonal matrix $D = D(G)$ of $G$ is the $n \times n$ diagonal matrix whose diagonal entries are given by $\deg(v_{1}), \dots, \deg(v_{n}).$ The adjacent matrix $A = A(G)$ of $G$ is the $n \times n$ matrix whose entries are given by defining its $(i,j)$ entry to be the number of edges between $v_{i}$ and $v_{j}.$ The Laplacian $L = L(G)$ of $G$ is defined to be $D - A.$

Convention. We will only deal with graphs whose vertices don't have self nodes. Moreover, we will assume that every pair of vertices have at max $1$ edge in between. With this convention, note that

• $A(G)$ is a $(0, 1)$ symmetric matrix with $0$'s on the diagonal, and
• The entries of any row or column of $L(G) = D(G) - A(G)$ add up to $0.$

Theorem 1. If $G$ is connected, then $\ker(L(G))$ is a cyclic group generated by $(1, 1, \dots, 1).$

Proof. Since the entries of any row of $L = L(G)$ add up to $0,$ we have $(1, 1, \dots, 1) \in \ker(L).$ Conversely, say $x = (x_{1}, \dots, x_{n}) \in \ker(L) \subset \mathbb{Z}^{n}.$ Even without connectedness hypothesis, we have $$0 = x^{T}Lx = \sum (x_{i} - x_{j})^{2},$$ where the sum is over $1 \leq i, j \leq n$ such that $v_{i}$ and $v_{j}$ are connected by an edge. Since $G$ is connected, all the distinct pairs of indices $(i, j)$ must appear in the sum, and since we work over $\mathbb{Z},$ we must have $x_{i} = x_{j}.$ Thus, we have $m := x_{1} = \cdots = x_{n},$ so $(x_{1}, \dots, x_{n}) = m (1, 1, \dots, 1) \in \mathbb{Z}(1, 1, \dots, 1),$ as desired. $\Box$

Remark. Note that if $G$ is not connected, a similar proof to Theorem 1 shows that $\ker(G)$ is generated by tuples that consist of $0$'s and $1$'s, where $1$'s are located in the coordinates corresponding to vertices that belong to each connected component.

Corollary. The rank of the cokernel $\mathrm{cok}(L(G))$ of $L(G)$ is $1$ so that $$\mathrm{cok}(L(G)) \simeq \mathbb{Z} \oplus K(G),$$ where $K(G)$ is a finite abelian group.

We call $K(G)$ (up to an isomorphism) the critical group of $G.$

A divisor of $G$ is an element of $\mathbb{Z}^{|V(G)|},$ where $V(G)$ is the set of vertices of $G.$ We write $\mathrm{Div}(G) := \mathbb{Z}^{|V(G)|}.$ Writing $n := |V(G)|,$ the degree of a divisor $D = (m_{1}, \dots, m_{n})$ is defined to be $\deg(D) := m_{1} + \cdots + m_{n}.$

Remark. It seems that the dual graph of an algebraic curve (whose vertices correspond to the irreducible components) contains quite a bit of information about the curve, which could be a motivation for studying these graph invariants for algebraic geometors. I would not try to make any analogy before I understand this story a bit better. (For example, see p.12 of this article; here, there are also graphs we do not consider in this post.)

The subgroup of divisors of degree $0$ is denoted as $\mathrm{Div}^{0}(G).$ Given a divisor $(x_{1}, \dots, x_{n})$ of $G,$ there are two kinds of chip-firing moves at each vertex $v_{i}$:

1. Fire at $v_{i}$: takes $x_{i} \mapsto x_{i} - \deg(v_{i})$ and then add $1$ to any adjacent vertex.
2. Borrow to $v_{i}$: takes $x_{i} \mapsto x_{i} + \deg(v_{i})$ and then subtract $1$ to any adjacent vertex.

Any two divisors of $G$ that can be achieved from one to another by a sequence of chip-firing moves are said to be equivalent, and note that this indeed defines an equivalence relation to all divisors. Note that chip-firing moves do not change the degree of a divisor, so this also defines an equivalence relation on the degree $0$ divisors.

Theorem 2. If $G$ is connected, then $K(G) \simeq \mathrm{Div}^{0}(G)/\sim.$

Proof. Since columns of $L(G)$ sum up to $0,$ we have $\mathrm{im}(L(G)) \subset \mathrm{Div}^{0}(G),$ so $$\mathrm{Div}^{0}(G)/\mathrm{im}(L(G)) \hookrightarrow \mathbb{Z}^{n}/\mathrm{im}(L(G)) \simeq \mathbb{Z} \oplus K(G).$$ Since $\mathrm{Div}^{0}(G)/\mathrm{im}(L(G))$ is of rank $0,$ it must sit inside the subgroup of $\mathbb{Z}^{n}/\mathrm{im}(L(G))$ corresponding to $K(G),$ which we may also write as $K(G).$ This implies that $$\mathbb{Z} \simeq \mathbb{Z}^{n}/\mathrm{Div}^{0}(G) \simeq \frac{\mathbb{Z}^{n}/\mathrm{im}(L(G))}{\mathrm{Div}^{0}(G)/\mathrm{im}(L(G))} \simeq \mathbb{Z} \oplus  \frac{K(G)}{\mathrm{Div}^{0}(G)/\mathrm{im}(L(G))}.$$ Thus, we have $$\frac{K(G)}{\mathrm{Div}^{0}(G)/\mathrm{im}(L(G))} = 0,$$ so $$\mathrm{Div}^{0}(G)/\mathrm{im}(L(G)) = K(G).$$ Hence, to finish the proof, it suffices to show that $\mathrm{im}(L(G))$ is equal to the set of degree $0$ divisors of $G$ that are equivalent to $(0, 0, \dots, 0).$ This immediately follows from observing that firing at $v_{i}$ for an $n$-tuple (i.e., a divisor) is identical to subtracting the $i$-th column of $L(G)$ from it. This finishes the proof. $\Box$

Smith normal forms. Recall that given an $n \times n$ matrix $L$ over $\mathbb{Z},$ there always exists $U, V \in \mathrm{GL}_{n}(\mathbb{Z})$ such that $ULV = \mathrm{diag}(s_{1}, \dots, s_{r}, 0, 0, \dots, 0)$ with nonzero positive integers $s_{1}, \dots, s_{r}$ satisfying $s_{1} | \cdots | s_{r}.$ This implies that $$\mathrm{cok}(L) \simeq \mathrm{cok}(ULV) \simeq \mathbb{Z}^{n - r} \oplus \mathbb{Z}/(s_{1}) \oplus \cdots \oplus \mathbb{Z}/(s_{r}),$$ which shows the uniqueness of $s_{1}, \dots, s_{n}$. We call the diagonal matrix with the entries $s_{1}, \dots, s_{r}, 0, 0, \dots, 0$ the Smith normal form of $L.$ One may note that $r$ is the rank of $L$ and recall that $U$ is given by the row operations on $L$ while $V$ is given by the column operations on $L$ over $\mathbb{Z}$ when it comes to achieving the diagonal matrix form. 

Thus, if $L = L(G)$ for some connected graph $G,$ we have $$K(G) \simeq \mathbb{Z}/(s_{1}) \oplus \cdots \oplus \mathbb{Z}/(s_{r}).$$

Lemma. Keeping the notations as above, for any $1 \leq i \leq r,$ we have $$s_{1}s_{2} \cdots s_{i} = \gcd(i \times i \text{ minors of } L).$$

Proof. This immediately follows from noticing that the gcd of $i \times i$ minors of $L$ is unchanged after any row or column operation. To see the invariance under the row operations, denote by $[R_{1} | \cdots | R_{i}]$ be an $i \times i$ submatrix of $L,$ where $R_{1}, \dots, R_{i}$ are the rows. A row operation turns this matrix into a matrix of the form $$[R_{1} | \cdots | R_{j} + cR'_{j} | \cdots | R_{i}],$$ and thus the corresponding minor is equal to $$\det[R_{1} | \cdots | R_{j} | \cdots | R_{i}] + c\det[R_{1} | \cdots | R'_{j} | \cdots | R_{i}].$$ The proof ends by noticing that $\det[R_{1} | \cdots | R'_{j} | \cdots | R_{i}]$ is one of the $i \times i$ minors before we perform the row operation. The invariance under the column operations can be similarly observed. $\Box$

Example. Let $K_{n}$ be the complete graph on $n$ vertices. To keep up with our convention, we assume that $n \geq 3.$ We have $$L(K_{n}) = \begin{bmatrix} n-1 & -1 & -1 & \cdots & -1 \\ -1 & n-1 & -1 & \cdots & -1 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ -1 & -1 & -1 & \cdots & n-1 \end{bmatrix}.$$ Using Lemma, it is immediate that $s_{1} = 1.$ The only $2 \times 2$ minors are $0, n, n(n-2),$ so Lemma implies that $s_{2} = s_{1}s_{2} = n.$ We already know $r = n-1$ and since $s_{1} | \cdots | s_{n-1},$ we have $$n^{n-2} = s_{2}^{n-2} | s_{1}s_{2} \cdots s_{n-1}.$$ By induction, we can compute that once we delete the first row and the last column of $L(K_{n-1}),$ the determinant of the resulting $(n - 1) \times (n - 1)$ matrix is $\pm n^{n-2}.$ Lemma guarantees that this number is divisible by $s_{1} \cdots s_{n-1},$ so we have $$n^{n-2} | s_{1}s_{2} \cdots s_{n-1} | n^{n-2}.$$ Thus, we have $s_{2} \cdots s_{n-1} = s_{1}s_{2} \cdots s_{n-1} = n^{n-2}.$ Since $s_{2} | \cdots | s_{n-1},$ this implies that $$s_{2} = s_{3} = \cdots = s_{n-1} = n.$$ Thus, we have $$K(K_{n}) \simeq (\mathbb{Z}/(n))^{n-2}.$$

Theorem 3. Given a connected graph $G$ with $n$ vertices, the critical group can be computed by taking the cokernel of the $(n - 1) \times (n - 1)$ matrix $L_{ij}(G)$ obtained by deleting the $i$-th row and the $j$-th column of $L(G)$ for any $1 \leq i, j \leq n.$ In particular, we have $\mathrm{cok}(L_{ij}(G)) = |K(G)|.$

Proof. This only uses the fact that the entries of any row or column of $L(G)$ add up to $0.$ If we add all the rows other then the $i$-th row to the $i$-th row of $L(G),$ the resulting matrix will only have $0$'s on its $i$-th row. This is because the entries of each column add up to $0.$ Then if we add all the columns other than the $j$-th column to the $j$-th column of this matrix, the resulting matrix will only have $0$'s on its $j$-th column. This is because the entries of each row add up to $0.$ This finishes the proof because we already know the rank of $L(G)$ is $n-1.$ $\Box$

Theorem 4 (Kirchhoff). If $G$ is connected, then $|K(G)|$ is the number of spanning trees of $G.$

Corollary (Cayley). The number of spanning trees of $K_{n}$ is $n^{n-2}.$

Cross ratio

I remember that in a complex analysis course and a combinatorics course, my teachers both discussed about cross ratio. I recently realized that Vakil's notes (19.9) contains a nice exposition about this, which I choose to regurgitate here with a bit more elementary explanations.

Let us work over a field $k.$ We can identify $$\mathrm{Aut}(\mathbb{P}^{n}) = \mathrm{PGL}_{n+1}(k) = \mathrm{GL}_{n+1}(k)/k^{\times}$$ because giving an automorphism $\pi : \mathbb{P}^{n} \rightarrow \mathbb{P}^{n}$ is the same as choosing a $k$-linear automorphism of $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) = kx_{0} + \cdots + kx_{n}$$ up to nonzero global sections of $\mathbb{P}^{n},$ which consists of precisely $k^{\times}.$ If we consider this $k$-linear automorphism as $A \in \mathrm{PGL}_{n+1}(k),$ one can check that on $\mathbb{P}^{n}(k),$ the map $\pi$ is given by $x = [x_{0} : \cdots : x_{n}] \mapsto Ax.$ Here, we note that $x$ is seen as an $(n+1) \times 1$ column vector.

One can check that given any distinct $w, x, y \in \mathbb{P}^{1}(k),$ there is precisely one $A \in \mathrm{PGL}_{2}(k)$ such that $A : (w, x, y) \mapsto (0, 1, \infty),$ where we wrote $$a = [a : 1]$$ for any $a \in k = \mathbb{A}^{1}(k)$ and $\infty = [1:0].$ Even though it is a bit convoluted-looking, one can even write down explicitly what $A$ is. (Of course, it will be only unique up to a multiple of an element in $k^{\times}.$) The exact expression is the following:

$$A = \begin{bmatrix}x_{2}(w_{1}y_{2} - y_{1}w_{2}) & -x_{1}(w_{1}y_{2} - y_{1}w_{2}) \\ -w_{2}(y_{1}x_{2} - x_{1}y_{2}) & w_{1}(y_{1}x_{2} - x_{1}y_{2})\end{bmatrix},$$

where $w = [w_{1} : w_{2}]$ and similarly for $x$ and $y.$ Given any other $k$-point $z = [z_{1} : z_{2}]$ of the projective line, the cross ratio of the four points $w, x, y, z$ is defined as $Az.$

Motivation for the nomenclature. We have 5 cases.

Case 1. Consider the case where $w, x, y, z \neq \infty$ so that we can write $w = [w : 1]$ and similarly for $x, y, z.$ Then the cross ratio of $w, x, y, z$ is precisely

$$\frac{(w -y)(z - x)}{(w - z)(y - x)}.$$

Case 2. Let $w = \infty = [1 : 0].$ Then the cross ratio is

$$\frac{z - x}{y - x}.$$

Case 3. Let $x = \infty = [1 : 0].$ Then the cross ratio is

$$\frac{w - y}{w - z}.$$

Case 4. Let $y = \infty = [1 : 0].$ Then the cross ratio is

$$\frac{x - z}{w - z}.$$

Case 5. Let $z = \infty = [1 : 0].$ Then the cross ratio is

$$\frac{w - y}{x - y}.$$

Remark. Note that the cross ratios from Case 2, 3, 4, and 5 may us think that we took each letter to infinity, but we didn't! In any case, this is great because we can just remember the expression from Case 1 as it recovers every other case.

Writing $$\lambda = \lambda(w,x,y,z) = \frac{(w -y)(z - x)}{(w - z)(y - x)}$$ to mean the cross ratio of $w, x, y, z,$ note that

1. $\lambda(w,y,x,z) = 1 - \lambda,$
2. $\lambda(w,x,z,y) = 1/\lambda,$
3. $\lambda(w,z,x,y) = 1 - 1/\lambda,$
4. $\lambda(w,y,z,x) = 1/(1-\lambda),$
5. $\lambda(w,z,y,x) = \lambda/(1-\lambda).$

Maps to projective spaces -- Part 2

Again, let us consider a variety $X$ over a field $k.$ (One can relax this condition about $X$ in various places throughout the discussion, but we won't bother.) Last time, we have discussed that when we have scheme maps $f_{0}, \dots, f_{n} : X \rightarrow \mathbb{A}^{1}$ that do not vanish all together at any single point in $X,$ we can induce a $k$-scheme map $$\pi := [f_{0} : \cdots : f_{n}] : X \rightarrow \mathbb{P}^{n}$$ such that when $X$ is locally of finite type over $k,$ for any $x \in X(k)$ such that $f_{0}(x), \dots, f_{n}(x) \in \mathbb{A}^{1}(k) = k,$ we have $\pi(x) = [f_{0}(x) : \cdots f_{n}(x)] \in \mathbb{P}^{n}(k).$ We also recall that defined this map by the associated $k$-algebra maps $$k[x_{0}/x_{i}, \dots, x_{n}/x_{i}] \rightarrow H^{0}(X_{f_{i}}, \mathscr{O}_{X})$$ such that $(x_{0}/x_{i}, \dots, x_{n}/x_{i}) \mapsto (f_{0}/f_{i}, \dots, f_{n}/f_{i}).$

We now classify all $k$-scheme maps $\pi : X \rightarrow \mathbb{P}^{n}.$ We can consider the standard open cover $$\mathbb{P}^{n} = U_{0} \cup U_{1} \cup \cdots \cup U_{n},$$ where $U_{i} = D_{\mathbb{P}^{n}}(x_{i}).$ Writing $X_{i} := \pi^{-1}(U_{i}),$ the map $\pi$ is given by gluing the $k$-algebra maps $k[x_{0}/x_{i}, \dots, x_{n}/x_{i}] \rightarrow H^{0}(X_{i}, \mathscr{O}_{X_{i}}),$ for $i = 0, 1, \dots, n,$ each of which is determined by where we send $x_{0}/x_{i}, \dots, x_{n}/x_{i}.$ Recall that we can identify $$\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}} = \mathscr{O}_{U_{i}}$$ so that we have $$H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1)) = k[x_{0}/x_{i}, \dots, x_{n}/x_{i}].$$ 

Review on $\mathscr{O}_{\mathbb{P}^{n}}(1)$. We have transition maps $$k[x_{0}/x_{i}, \dots, x_{n}/x_{i}]_{x_{j}/x_{i}} \rightarrow k[x_{0}/x_{j}, \dots, x_{n}/x_{j}]_{x_{i}/x_{j}}$$ given by the multiplication of $x_{i}/x_{j}.$ An element $s$ of $H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1))$ is given by $$s = (f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}))_{0 \leq i \leq n}$$ such that $$x_{i}f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = x_{j}f_{j}(x_{0}/x_{j}, \dots, x_{n}/x_{j})$$ for $0 \leq i, j \leq n.$ This implies that $$f_{i}(x_{0}/x_{i}, \dots, x_{n}/x_{i}) = a_{i,0}x_{0}/x_{i} + \cdots + a_{i,n}x_{n}/x_{i},$$ and $(a_{i,0}, \dots, a_{n,i}) = (a_{j,0}, \dots, a_{j,n})$ for $0 \leq i, j \leq n$ so that we can write $(a_{0}, \dots, a_{n}) = (a_{i,0}, \dots, a_{n,i})$ for $0 \leq i \leq n.$ Therefore, we have $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) \simeq k x_{0} + \cdots + k x_{n}$$ given by $(a_{0}, \dots, a_{n}) \mapsto a_{0}x_{0} + \cdots + a_{n}x_{n},$ and we shall use the identification $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) = k x_{0} + \cdots + k x_{n}.$$ With respect to this identification, the restriction map $$H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) \rightarrow H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1))$$ is given by $a_{0}x_{0} + \cdots + a_{n}x_{n} \mapsto a_{0}x_{0}/x_{i} + \cdots + a_{n}x_{n}/x_{i}.$

Review on pullbacks of quasi-coherent sheaves. Let $\pi : X \rightarrow Y$ be a scheme map and $\mathscr{G}$ a quasi-coherent sheaf on $Y.$ Then the pullback $\pi^{*}\mathscr{G}$ is a quasi-coherent sheaf on $X$ and on the restriction $\mathrm{Spec}(A) \rightarrow \mathrm{Spec}(B)$ of $\pi$ on affine opens, we have $$H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}) = H^{0}(\mathrm{Spec}(B), \mathscr{G}) \otimes_{B} A.$$ Given a map $\phi : \mathscr{G}_{1} \rightarrow \mathscr{G}_{2}$ of quasi-coherent sheaves on $Y,$ we have the induced map $\pi^{*}\phi : \pi^{*}\mathscr{G}_{1} \rightarrow \pi^{*}\mathscr{G}_{2}$ of quasi-coherent sheaves on $X$ such that the restriction $$H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}_{1}) \rightarrow H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}_{2})$$ is given by $$H^{0}(\mathrm{Spec}(B), \mathscr{G}_{1}) \otimes_{B} A \rightarrow H^{0}(\mathrm{Spec}(B), \mathscr{G}_{2}) \otimes_{B} A$$ with $s \otimes a \mapsto \phi(s) \otimes a.$

We can also have $\pi^{*} : H^{0}(Y, \mathscr{G}) \rightarrow H^{0}(X, \pi^{*}\mathscr{G})$ as follows (although this also depends on $\mathscr{G}$). Given $s \in H^{0}(Y, \mathscr{G}),$ we may consider the corresponding $\mathscr{O}_{Y} \rightarrow \mathscr{G}$, which we also call $s,$ given by $H^{0}(V, \mathscr{O}_{Y}) \rightarrow H^{0}(V, \mathscr{G})$ such that $1 \mapsto s|_{V}.$ Then we can consider its pullback $$\pi^{*}s : \mathscr{O}_{X} \simeq \pi^{*}\mathscr{O}_{Y} \rightarrow \pi^{*}\mathscr{G},$$ which corresponds to an element of $H^{0}(X, \pi^{*}\mathscr{G}).$ Restricting to the affine opens we fixed before, this gives rise to $$A \simeq B \otimes_{B} A \rightarrow H^{0}(\mathrm{Spec}(B), \mathscr{G})_{\otimes B} A = H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G})$$ given by $$b \otimes a \mapsto bs \otimes a.$$ Thus, the map $$\pi^{*} : H^{0}(Y, \mathscr{G}) \rightarrow H^{0}(X, \pi^{*}\mathscr{G})$$ must be given by gluing $$H^{0}(\mathrm{Spec}(B), \mathscr{G}) \rightarrow H^{0}(\mathrm{Spec}(A), \pi^{*}\mathscr{G}) = H^{0}(\mathrm{Spec}(B), \pi^{*}\mathscr{G}) \otimes_{B} A$$ defined by $s \mapsto s \otimes 1.$ In particular, for any $s \in H^{0}(Y, \mathscr{G})$ and open $V \subset Y,$ we have $\pi^{*}|_{V} : H^{0}(V, \mathscr{G}) \rightarrow H^{0}(\pi^{-1}(V), \pi^{*}\mathscr{G})$ such that $s|_{V} \mapsto \pi^{*}(s)|_{\pi^{-1}(V)}.$ 

Going back to the previous discussion, considering the pullback map $$\pi^{*} : H^{0}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(1)) \rightarrow H^{0}(X, \pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1)),$$ since $\mathscr{O}_{\mathbb{P}^{n}}(1)$ is locally free, one can check that $$D_{X}(\pi^{*}x_{i}) = \pi^{-1}(D_{\mathbb{P}^{n}}(x_{i})) = \pi^{-1}(U_{i}) = X_{i}.$$ The global section $s_{i} := \pi^{*}x_{i}$ does not vanish anywhere in the open subset $X_{i}$ of $X,$ so the restriction of the line bundle $\pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1)$ of $X$ on $X_{i}$ is trivial, but we will not use this trivialization yet. (We will use this later.)

Remark. What's important for us is the notion of the "inverse" of $s \in H^{0}(X, \mathscr{L})$ when $s$ is nowhere vanishing. Let $X = \bigcup_{\alpha \in I}U_{\alpha}$ be any trivializing open cover for $\mathscr{L}.$ Fix $p \in X,$ and pick any $U_{\alpha} \ni p.$ We have $\mathscr{O}_{U_{\alpha}} \simeq \mathscr{L}|_{U_{\alpha}},$ which gives $\mathscr{O}_{X,p} \simeq \mathscr{L}_{p}.$ Denote by $s'_{p}$ the element of $\mathscr{O}_{X,p}$ corresponding to $s_{p} \in \mathscr{L}_{p},$ the image of $s \in H^{0}(X, \mathscr{L}).$ We have $s_{p} \notin \mathfrak{m}_{X,p}\mathscr{L}_{p}$ (i.e., $s$ does not vanish at $p$), so $s'_{p} \in \mathfrak{m}_{X,p},$ which means that $s'_{p}$ is a unit in $\mathscr{O}_{X,p}.$ Thus, refining our trivializing open cover if necessary, we may assume that there is $w_{\alpha} \in H^{0}(U_{\alpha}, \mathscr{O}_{X})$ such that $s'|_{U_{\alpha}}w_{\alpha} = 1 \in H^{0}(U_{\alpha}, \mathscr{O}_{X}).$ These $w_{\alpha}$'s agree on the intersections of $U_{\alpha}$'s, so we may glue them to construct $w \in H^{0}(X, \mathscr{O}_{X}),$ which we refer to as the inverse of $s.$ (I am not sure whether this is a standard terminology.) Note that even though we needed a trivialization of $\mathscr{L}$ to construct this element, it does not depend on the choice of such a trivialization. 

Moreover, note that any element (not necessarily nonvanishing one) of $H^{0}(X, \mathscr{L})$ has a unique corresponding element in $H^{0}(X, \mathscr{O}_{X}),$ given by gluing the ones defined on open subsets in a trivialization of $\mathscr{L}.$ (Note that this element does not depend on the choice of a trivialization.)

Going back to our discussion, now, it makes sense for us to consider the elements $s_{0}/s_{i}, \dots, s_{n}/s_{i} \in H^{0}(X_{i}, \mathscr{O}_{X}),$ namely the elements corresponding to $s_{0}|_{X_{i}}, \dots, s_{n}|_{X_{i}} \in H^{0}(X_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1))$ multiplied by the inverse of $s_{i}|_{X_{i}} \in H^{0}(X_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1)).$ Note that $\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}} \simeq \mathscr{O}_{U_{i}},$ so $$\pi^{*}(\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}}) \simeq \pi^{*}\mathscr{O}_{U_{i}} \simeq \mathscr{O}_{X_{i}}.$$ Now, the upshot is that (one can check that) the map $$H^{0}(U_{i}, \mathscr{O}_{U_{i}}) \rightarrow H^{0}(X_{i}, \mathscr{O}_{X_{i}})$$ given by $\pi : X \rightarrow \mathbb{P}^{n}$ coincides with the one corresponding to the pullback map $$H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}}) \rightarrow H^{0}(X_{i}, \pi^{*}(\mathscr{O}_{\mathbb{P}^{n}}(1)|_{U_{i}})).$$ That is, such a map is given by $$(x_{0}/x_{i}, \dots, x_{n}/x_{i}) \mapsto (s_{0}/s_{i}, \dots, s_{n}/s_{i}).$$ So far, we have shown the following:

Theorem 1. Any $k$-scheme map $\pi : X \rightarrow \mathbb{P}^{n}$ is given by $k$-algebra maps $$H^{0}(U_{i}, \mathscr{O}_{\mathbb{P}^{n}}) \rightarrow H^{0}(\pi^{-1}(U_{i}), \mathscr{O}_{X})$$ such that $x_{0}/x_{i}, \dots, x_{n}/x_{i} \mapsto s_{0}/s_{i}, \dots, s_{n}/s_{i},$ where $s_{i}$ corresponds to the restriction of $\pi^{*}x_{i}$ on $\pi^{-1}(U_{i}).$

More on construction. If $\mathscr{L}$ is any line bundle on $X$ with global sections $s_{0}, s_{1}, \dots, s_{n} \in H^{0}(X, \mathscr{L})$ that do not vanish at a single point in $X$ altogether, then we have an open cover $X = X_{0} \cup \cdots \cup X_{n},$ where $X_{i} := D_{X}(s_{i}).$ Refining these open sets further and gluing back, we can construct a $k$-scheme map $\pi : X \rightarrow \mathbb{P}^{n}$ given by $$k[x_{0}/x_{i}, \dots, x_{n}/x_{i}] \rightarrow H^{0}(X_{i}, \mathscr{O}_{X})$$ such that $(x_{0}/x_{i}, \dots, x_{n}/x_{i}) \mapsto (s_{0}/s_{i}, \dots, s_{n}/s_{i})$ for $i = 0, 1, \dots, n.$

Notation. We write $\pi := [s_{0} : \cdots : s_{n}].$

Theorem 2. We have $\mathscr{L} \simeq \pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1)$ such that $s_{i} \leftrightarrow \pi^{*}x_{i}$ on $X.$

Remark. For any line bundle $\mathscr{L}$ on a scheme $X.$ If $s \in H^{0}(X, \mathscr{L})$ is nowhere vanishing, then we have $\mathscr{L} \simeq \mathscr{O}_{X}$ given as follows. Consider the map $\mathscr{O}_{X} \rightarrow \mathscr{L}$ given by $$H^{0}(U, \mathscr{O}_{X}) \rightarrow H^{0}(U, \mathscr{L})$$ such that $1 \mapsto s|_{U}.$ This defines an isomorphism of $\mathscr{O}_{X}$-modules for the following reason. Fixing any $p \in X,$ if we consider the map $\mathscr{O}_{X,p} \rightarrow \mathscr{L}_{p}$ on stalks at $p,$ it is an $\mathscr{O}_{X,p}$-linear isomorphism because $s_{p} \notin \mathfrak{m}_{X,p}\mathscr{L}_{p},$ which ensures that $s_{p}$ corresponds to a unit in the local ring $\mathscr{O}_{X,p}$ under the $\mathscr{O}_{X,p}$-linear isomorphism $\mathscr{L}_{p} \simeq \mathscr{O}_{X,p}$ given by the hypothesis that $\mathscr{L}$ is a line bundle.

Trivialization of a line bundle from its sections. Given any line bundle $\mathscr{L}$ on $X$ and $s_{0}, s_{1}, \dots, s_{n} \in H^{0}(X, \mathscr{L})$ that do not vanish altogether at any point in $X,$ the maps $\phi_{i} : \mathscr{O}_{X_{i}} \overset{\sim}{\longrightarrow} \mathscr{L}|_{X_{i}}$ given by $$H^{0}(U, \mathscr{O}_{X_{i}}) \simeq H^{0}(U, \mathscr{L}|_{X_{i}})$$ such that $1 \mapsto s_{i}|_{U}$ is a trivialization of the line bundle $\mathscr{L}.$ That is, any other trivialization of $\mathscr{L}$ is compatible with this one.

Proof of Theorem 2. We have $\phi_{i} : \mathscr{O}_{X_{i}} \overset{\sim}{\longrightarrow} \mathscr{L}|_{X_{i}}$ given by $$H^{0}(U, \mathscr{O}_{X_{i}}) \simeq H^{0}(U, \mathscr{L}|_{X_{i}})$$ such that $1 \mapsto s_{i}|_{U}.$ We can consider $\mathscr{L}$ as a line bundle defined by gluing $\mathscr{O}_{X_{i}}$'s with the transition functions $$\phi_{ij} := \phi_{j}^{-1}\phi_{i} : \mathscr{O}_{X_{ij}} \rightarrow \mathscr{L}|_{X_{ij}} \rightarrow \mathscr{O}_{X_{ij}},$$ where $X_{ij} = X_{i} \cap X_{j}.$ Note that these transition functions are given by $1 \mapsto s_{i} \mapsto s_{i}/s_{j}.$ Now, note that $\pi^{-1}(U_{i}) = D_{X}(s_{i}) = X_{i}$ and for any open $U \subset X_{i},$ we have $$H^{0}(U, \mathscr{L}) \simeq H^{0}(U, \pi^{*}\mathscr{O}_{\mathbb{P}^{n}}(1))$$ given by $s_{i} \mapsto \pi^{*}(x_{i}).$ This map takes $s_{i}/s_{j}$ to $\pi^{*}x_{i}/\pi^{*}x_{j},$ which finishes the proof. $\Box$

Reference. Hartshorne's book and Vakil's notes

Maps to projective spaces -- Part 1

Let $X$ be a variety over a field $k.$ Given any set-theoretic functions $f_{0}, \dots, f_{n} : X(k) \rightarrow k$ that has no common zeros, we may define a set map $X(k) \rightarrow \mathbb{P}^{n}(k)$ given by $$x \mapsto [f_{0}(x) : \cdots : f_{n}(x)].$$

Let's think about a scheme-theoretic version of this map. First, we assume that $f_{0}, \dots, f_{n}$ are elements of $H^{0}(X, \mathscr{O}_{X}).$ Note that an element $f \in H^{0}(X, \mathscr{O}_{X})$ corresponds with a $k$-scheme map $X \rightarrow \mathbb{A}^{1}$ that corresponds to the $k$-algebra map $k[t] \rightarrow H^{0}(X, \mathscr{O}_{X})$ defined by $t \mapsto f,$ so we are fixing $k$-scheme maps $f_{0}, \dots, f_{n} : X \rightarrow \mathbb{A}^{1},$ analogous to the previous situation.

A zero of $f$ is a point of $X$ that is sent to the maximal ideal $(t)$ in $\mathbb{A}^{1} = \mathrm{Spec}(k[t]).$ One can easily check that for any $x \in X,$ the following are equivalent:

1. $[f] \in \mathscr{O}_{X,x}$ sits inside $\mathfrak{m}_{X,x}$;
2. $x \in X$ is zero of $f.$

Example. Consider the case when $X = \mathrm{Spec}(k[x_{1}, \dots, x_{m}])$ and $x = (x_{1} - a_{1}, \dots, x_{m} - a_{m}).$ In this case, we can observe that $x$ is a zero of $f \in k[x_{1}, \dots, x_{m}]$ if and only if $f(a_{1}, \dots, a_{m}) = 0.$

Hence, we may now require that $f_{0}, \dots, f_{n} : X \rightarrow \mathbb{A}^{1}$ have no common zeros. (The data can be thought as $f_{0}, \dots, f_{n} \in H^{0}(X, \mathscr{O}_{X})$ as well.)

Special case: $n = 1$. We first consider the case $n = 1,$ where we see how $f_{0}, f_{1} \in H^{0}(X, \mathscr{O}_{X})$ that have no common zeros may induce a $k$-scheme map $X \rightarrow \mathbb{P}^{1}.$ Due to our hypothesis, we have the following open cover: $X = X_{f_{0}} \cup X_{f_{1}},$ where $$X_{f} := \{x \in X : f \text{ does not vanish at } x\}.$$ It is important to note that (the restriction of) $f$ is invertible in $H^{0}(X_{f}, \mathscr{O}_{X})$ because for any affine open $U = \mathrm{Spec}(R) \subset X,$ we have $$X_{f} \cap U = D_{R}(f|_{U}) \simeq \mathrm{Spec}(R_{f|_{U}}).$$

We now construct two $k$-scheme maps $X_{f_{i}} \rightarrow U_{i} = \mathrm{Spec}(k[x_{0}/x_{i}, x_{1}/x_{i}])$ for $i = 0, 1$ and glue them together. For the construction for each $i,$ it is enough to construct a $k$-algebra map $k[x_{0}/x_{i}, x_{1}/x_{i}] \rightarrow H^{0}(X_{f_{i}}, \mathscr{O}_{X}).$ We do this by the following assignments:

• $x_{0}/x_{1} \mapsto f_{0}/f_{1} := f_{0}f_{1}^{-1}$;
• $x_{1}/x_{0} \mapsto f_{1}/f_{0} := f_{1}f_{0}^{-1}.$

The $k$-scheme maps $X_{f_{0}} \rightarrow U_{0}$ and $X_{f_{1}} \rightarrow U_{1}$ agree on the intersections (which can be checked with their $k$-algebra maps), so they glue to give a $k$-scheme map $X \rightarrow \mathbb{P}^{1}.$ Let $x \in X$ be a $k$-point, and suppose that $f_{0}(x), f_{1}(x)$ are $k$-points of $\mathbb{A}^{1}$ (which always happens when $k$ is algebraically closed).

Claim. If $X$ is locally of finite type over $k,$ then the map $\pi : X \rightarrow \mathbb{P}^{1}$ we constructed sends $x$ to $[f_{0}(x) : f_{1}(x)] \in \mathbb{P}^{1}(k).$

Proof. We have two cases: either $x \in X_{f_{0}}$ or $x \in X_{f_{1}}.$ Suppose we are in the first case. We may take an affine open neighborhood $U \subset X_{f_{0}}$ of $x$ such that $W = \mathrm{Spec}(k[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{r}))$ and write $$x = (y_{1} - b_{1}, \dots, y_{m} - b_{m})$$ in $k[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{r})$ for some $b_{1}, \dots, b_{m} \in k.$ Consider the following restriction of $\pi$: $$W \hookrightarrow X_{f_{0}} \rightarrow U_{0} = \mathrm{Spec}(k[x_{1}/x_{0}]).$$ We have $x \in W,$ and we would like to compute what $\pi(x) \in W$ is. The $k$-algebra map corresponding to the above composition is $$\phi : k[x_{1}/x_{0}] \rightarrow H^{0}(X_{f_{0}}, \mathscr{O}_{X}) \rightarrow H^{0}(W, \mathscr{O}_{X}) = \frac{k[y_{1}, \dots, y_{m}]}{(g_{1}, \dots, g_{r})},$$ which is defined by $x_{1}/x_{0} \mapsto f_{1}/f_{0},$ where now we can think of $f_{0}, f_{1}$ with polynomial expressions: $f_{i} = f_{i}(y_{1}, \dots, y_{m}).$ With this notation, we have $$\begin{align*}\pi(x) &= \phi^{-1}((y_{1}-b_{1}, \dots, y_{m}-b_{m})) \\ &= \{h(x_{1}/x_{0}) \in k[x_{1}/x_{0}] : h(f_{1}/f_{0}) \in x\} \\ &= \{h(x_{1}/x_{0}) \in k[x_{1}/x_{0}] : h(f_{1}(b_{1}, \dots, b_{m})/f_{0}(b_{1}, \dots, b_{m}) = 0\} \\ &= (x_{1}/x_{0} - f_{1}(b_{1}, \dots, b_{m})/f_{0}(b_{1}, \dots, b_{m})) \\ &= [f_{0}(b_{1}, \dots, b_{m})) : f_{1}(b_{1}, \dots, b_{m})] \\ &= [f_{0}(x) : f_{1}(x)],\end{align*}$$ as desired. The other case where $x \in X_{f_{1}}$ can be proven by an almost identical argument. $\Box$

General case. One may notice that the arguments for the special case $n = 1$ works in general without any difficulty. We shall now write $\pi := [f_{0} : \cdots : f_{n}]$ for the reasons described above.