$\mathbb{Z}_{p}[t]/(P(t))$ is a DVR if $P(t)$ is irreducible in $\mathbb{F}_{p}[t]$

Let $p$ be a prime and $P(t) \in \mathbb{Z}_{p}[t]$ a monic polynomial whose image in $\mathbb{F}_{p}$ modulo $p$ (which we also denote by $P(t)$ if there is no ambiguity) is irreducible. In this posting, we give a proof of the following fact, which we learned from some lectures notes by Andrew Sutherland (Lemma 10.13 and Example 10.16).

Theorem. Keeping the notation as above, the quotient ring $R = \mathbb{Z}_{p}[t]/(P(t))$ is a DVR whose maximal ideal is generated by $p$.

Proof. We first show that $R$ is a local ring whose maximal ideal is $(p)$. First, note that $$R/(p) \simeq \mathbb{F}_{p}[t]/(P(t))$$ is a field because we assumed that $P(t)$ is irreducible in $\mathbb{F}_{p}[t].$ Hence $(p)$ is indeed a maximal ideal of $R.$ Let $\mathfrak{m}$ be any maximal ideal of $R.$ We want to show that $p \in \mathfrak{m}.$

For contradiction, suppose that $p \notin \mathfrak{m}.$ Then since $\mathfrak{m}$ is a maximal ideal, we must have $\mathfrak{m} + (p) = R.$ Since $R$ is Noetherian, we have $$\mathfrak{m} = (r_{1}, \dots, r_{l}) = Rr_{1} + \cdots + Rr_{l}$$ for some $r_{1}, \dots, r_{l} \in R.$ We have $$R = \mathbb{Z}_{p} \oplus \bar{t} \mathbb{Z}_{p} \oplus \cdots \oplus \bar{t}^{d-1} \mathbb{Z}_{p},$$ where $d = \deg(P),$ so $R$ is finite over $\mathbb{Z}_{p}.$ Since $\mathfrak{m}$ is finite over $R,$ we see that $\mathfrak{m}$ is finite over $\mathbb{Z}_{p}.$ That is, we have some $z_{1}, \dots, z_{n} \in \mathfrak{m}$ such that $$\mathfrak{m} = \mathbb{Z}_{p}z_{1} + \cdots + \mathbb{Z}_{p}z_{n}.$$ Since $\mathfrak{m} + (p) = R,$ it follows that the images $\bar{z}_{1}, \dots, \bar{z}_{n}$ of $z_{1}, \dots, z_{n} \in R$ generate $R/(p)$ as an $\mathbb{F}_{p}$-vector space. Since $\mathbb{Z}_{p}$ is a local ring with the maximal ideal $(p),$ by Nakayama's lemma, this implies that $R$ is generated by $z_{1}, \dots, z_{n}$ as a $\mathbb{Z}_{p}$-module, but then this implies that $$R = \mathbb{Z}_{p}z_{1} + \cdots + \mathbb{Z}_{p}z_{n} \subset Rz_{1} + \cdots + Rz_{n} = \mathfrak{m},$$ which is a contradiction. Thus, we conclude that $p \in \mathfrak{m},$ showing that $R$ is a local ring with the maximal ideal $(p).$ Since $R$ is a Noetherian local domain that is not a field, whose maximal ideal is principal, we conclude that $R$ is a DVR. $\Box$

Sandpile groups of random graphs: 2. Erdős–Rényi model

First, let's consider the following question:

Question. What is the distribution of the number $N_{\Gamma}$ of the spanning trees of a random graph $\Gamma$?

Of course, we need to make sense of what "random" means in the above question. One way to generate random graphs is to use the Erdős–Rényi model. That is, given $n \in \mathbb{Z}_{\geq 2}$ and a real number $0 < q < 1$, we first fix $n$ vertices and choose a graph with these vertices at random by requiring that for each pair of vertices, the probability that the two chosen vertices are connected by an edge is $q.$ We write $G(n, q)$ to mean the set of all graphs with $n$ vertices without self-loops and double edges with the probability measure given by the Erdős–Rényi model. We note that $G(n, q)$ is a finite set of graphs, each of which has the $n$ given vertices with

• no self-loops;
• no double edges;
• and $M$ edges ($0 \leq M \leq {n \choose 2}$),

and each graph in $G(n, q)$ with $M$ edges occurs with the probability $q^{M}(1 - q)^{{n \choose 2} - M}.$ Hence, the probability that the number of edges of an Erdős–Rényi random graph is equal to $M$ is ${{n \choose 2} \choose M}q^{M}(1 - q)^{{n \choose 2} - M}.$ That is, the distribution of the number of edges of a random $\Gamma \in G(n, q)$ follows the binomial distribution with parameters $q$ and ${n \choose 2}.$ In particular, the average number of the edges in a graph randomly chosen in $G(n, q)$ is ${n \choose 2}q.$

Cayley's formula says that the total number of spanning trees on given $n$ vertices is $n^{n-2},$ and the probability that each spanning tree occurs in a random graph in $G(n, q)$ is $q^{n-1}.$ Hence, given $0 \leq m \leq n^{n-2},$ we have $$\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(N_{\Gamma} = m) = {n^{n-2} \choose m} (q^{n-1})^{m}(1 - q^{n-1})^{n^{n-2} - m}.$$ That is, the number $N_{\Gamma}$ of a random graph $\Gamma \in G(n, q)$ follows the binomial distribution with parameters $q$ and $n^{n-2}.$ In particular, the average number of spanning trees of $\Gamma$ is $n^{n-2}q^{n-1}.$

Now, we know completely what the distribution of the number of spanning trees of $\Gamma$ randomly chosen from $G(n, q).$ In the last posting, we discussed how Kirchhoff's theorem tells us that the number $N_{\Gamma}$ of spanning trees of a graph $\Gamma$ is the product of the non-negative invariant factors of its sandpile group $S_{\Gamma}.$

Question. What is the distribution of the sandpile group $S_{\Gamma}$ of $\Gamma$ randomly chosen in $G(n, q)$?

It turns out from a paper of Wood (Corollary 9.3) that for any finite abelian group $A,$ we have $$\lim_{n \rightarrow \infty}\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma} \simeq A) = 0.$$ We already know that the expected number $$n^{n-2}q^{n-1} = \frac{(nq)^{n-1}}{n}$$ of $N_{\Gamma}$ is large (which is in particular nonzero) when $n$ is large, so we expect $S_{\Gamma}$ to be a finite group. More precisely, we have $$\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma} \text{ is infinite}) = \underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(N_{\Gamma} = 0) = (1 - q^{n-1})^{n^{n-2}}.$$ If $0 < q < 1$ is fixed, then this implies that $$\lim_{n \rightarrow \infty}\underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma} \text{ is infinite}) = 0.$$ Hence, the fact that the probability that $S_{\Gamma}$ is isomorphic to a specific finite abelian group is small for large $n$ has to do with that the events we are considering are too fine.

Remark. This is a strange phenomenon. Heuristically, one may say that there are too many candidates for $S_{\Gamma}$ when we take a random $\Gamma \in G(n, q),$ which makes it less likely if we try to specify the isomorphism class of $S_{\Gamma},$ but there are other similar instances, where we choose from many finitely abelian groups but the probability does not converge to $0.$ For example, see this paper of Nguyen and Wood (Theorem 1.1 with $u = 1$).

It also turns out that given a prime $p,$ the $p$-part $S_{\Gamma}[p^{\infty}]$ of the sanpile group of a random graph $\Gamma \in G(n, q)$ has a nontrivial distribution. The following theorem is from a paper of Wood (Theorem 1.1):

Theorem (Wood). Let $p$ be a prime and $G$ a finite abelian $p$-group so that $$G \simeq \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\lambda_{2}}\mathbb{Z} \times \cdots \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}$$ with $\lambda_{1} \geq \lambda_{2} \geq \cdots \geq \lambda_{l} \geq 0.$ Let $(\mu_{1}, \mu_{2}, \dots)$ be the transpose of the partition $(\lambda_{1}, \lambda_{2}, \dots, \lambda_{l}).$ Then $$\begin{align*}\lim_{n \rightarrow \infty} & \underset{\Gamma \in G(n, q)}{\mathrm{Prob}}(S_{\Gamma}[p^{\infty}] \simeq G) \\ &= \frac{p^{-\sum_{r \geq 1} \mu_{r}(\mu_{r}+1)/2}\prod_{i=1}^{\lambda_{1}}\prod_{j=1}^{\lfloor (\mu_{i} - \mu_{i+1})/2 \rfloor} (1 - p^{-2j})^{-1}}{|G||\mathrm{Aut}(G)|}\prod_{k \geq 0} (1 - p^{-2k-1}).\end{align*}$$

Sandpile groups of random graphs -- 1 Preliminaries

In this posting, we start discussing the topic of random sandpile groups.

1 Preliminaries

Let $n \in \mathbb{Z}_{\geq 1}$ and $\Gamma$ a graph (undirected) with verticies $1, 2, \dots, n.$ Given $1 \leq i \neq j \leq n,$ define $\deg(i, j)$ to be the number of edges between vertex $i$ and vertex $j.$ We define $\deg(i)$ to be the number of edges between vertex $i$ and any other vertices. Define the adjacency matrix $A(\Gamma)$ of $\Gamma$ to be the $n \times n$ matrix whose $(i, j)$ entry is $\deg(i, j).$ We define the Laplacian of $\Gamma$ as

$$L(\Gamma) := \mathrm{diag}(\deg(1), \deg(2), \dots, \deg(n)) - A(\Gamma).$$

Remark. Note that $L(\Gamma)$ cannot tell whether there are self-loops on any vertex of the graph $\Gamma.$

Because of the way we defined the degree of each vertex, it is always the case that

• each row of $L(\Gamma)$ sums to $0$ and
• each column of $L(\Gamma)$ sums to $0.$

Denote by $L_{ij}$ the $(i, j)$-entry of $L = L(\Gamma).$ Then

$$L\begin{bmatrix} k_{1} \\ \vdots \\ k_{n} \end{bmatrix} = \begin{bmatrix} L_{11}k_{1} + \cdots + L_{1n}k_{n} \\ \vdots \\ L_{n1}k_{1} + \cdots + L_{nn}k_{n} \end{bmatrix},$$

so if we add all the entries of the last $n \times 1$ matrix, we have

$$(L_{11} + \cdots + L_{n1})k_{1} + \cdots + (L_{1n} + \cdots + L_{nn})k_{n} = 0,$$

since each column of $L = L(\Gamma)$ sums to $0.$ That is, the image $\mathrm{im}(L)$ of $L = L(\Gamma)$ always sits inside

$$(\mathbb{Z}^{n})_{0} := \{(r_{1}, \dots, r_{n}) \in \mathbb{Z}^{n} : r_{1} + \cdots + r_{n} = 0\}.$$

The sandpile group of $\Gamma$ is defined as $S_{\Gamma} := (\mathbb{Z}^{n})_{0}/\mathrm{im}(L).$

Question. When we choose $\Gamma$ at random, how are the sandpile groups $S_{\Gamma}$ distributed?

Of course, this question needs to be polished a bit. 

Lemma. We have

$$\ker(L(\Gamma)) = \bigoplus\mathbb{Z}(e_{i_{1}} + \cdots + e_{i_{k}}),$$

where the direct sum is over the subsets $\{i_{1}, \dots, i_{k}\}$ of the vertex set $\{1, 2, \dots, n\}$ of $\Gamma$, each of which consists of vertices from a connected component of $\Gamma.$ In particular, we see that $\ker(L(\Gamma))$ is a free $\mathbb{Z}$-submodule of $\mathbb{Z}^{n}$ with rank equal to the number of connected components of $\Gamma.$

Remark. It follows from Lemma that $\mathrm{im}(L(\Gamma)) \simeq \mathbb{Z}^{n}/\ker(L(\Gamma))$ is a free $\mathbb{Z}$-module with rank $n - |\pi_{0}(\Gamma)|,$ where we denote by $\pi_{0}(\Gamma)$ the set of connected components of $\Gamma.$ Thus, we note that the sandpile group $S_{\Gamma} = (\mathbb{Z}^{n})_{0}/\mathrm{im}(L(\Gamma))$ is a free $\mathbb{Z}$-module with rank $|\pi_{0}(\Gamma)| - 1.$ In particular, we have

Corollary. The sandpile group $S_{\Gamma}$ of a graph $\Gamma$ is finite if and only if $\Gamma$ is connected.

Proof of Lemma. Note that the elements $e_{i_{1}} + \cdots + e_{i_{k}}$ we consider are $\mathbb{Z}$-linearly independent because the subsets $\{i_{1},  \dots, i_{k}\}$ of $\{1, 2, \dots, n\}$ we consider are disjoint.

We write $L = L(\Gamma)$ for convenience. Let $i_{1} < \cdots < i_{k}$ be the vertices of a connected component of $\Gamma.$ We have 

$$L(e_{i_{1}} + \cdots + e_{i_{k}}) = L(e_{i_{1}}) + \cdots + L(e_{i_{k}}) = \begin{bmatrix} L_{1,i_{1}} + \cdots + L_{1,i_{k}} \\ L_{2,i_{1}} + \cdots + L_{2,i_{k}} \\ \vdots \\ L_{n,i_{1}} + \cdots + L_{n,i_{k}}\end{bmatrix}.$$

Now, for each $1 \leq i \leq n,$ we know that if $i$ does not belong to the connected component, then $L_{i,i_{1}} = \cdots = L_{i,i_{k}} = 0.$ Otherwise (if $i$ belongs to the connected component), we have $i = i_{j}$ for some $1 \leq j \leq k$ and $$L_{i_{j},i_{1}} + \cdots + L_{i_{j},i_{k}} = L_{i_{j},1} + \cdots + L_{i_{j},n} = 0.$$ Hence, we must have that $L(e_{i_{1}} + \cdots + e_{i_{k}}) = 0$ in $\mathbb{Z}^{n},$ so $$e_{i_{1}} + \cdots + e_{i_{k}} \in \ker(L).$$ This shows that

$$\ker(L) \supset \bigoplus\mathbb{Z}(e_{i_{1}} + \cdots + e_{i_{k}}),$$

so it remains to show the opposite inclusion.

Let $x = \begin{bmatrix}x_{1} \\ \vdots \\ x_{n}\end{bmatrix} \in \ker(L) \subset \mathbb{Z}^{n}.$ We have $$0 = x^{T}Lx = \sum_{1 \leq i, j \leq n}L_{ij}x_{i}x_{j} = \sum (x_{i} - x_{j})^{2},$$ where the last sum is over $1 \leq i, j \leq n$ such that $v_{i}$ and $v_{j}$ are connected by an edge. (This is by definition of $L = L(\Gamma).$) This implies that for any connected component $\{i_{1}, \dots, i_{k}\},$ we have $x_{i_{1}} = \cdots = x_{i_{k}}$, and these are only relations among $x_{1}, \dots, x_{n}.$ Hence, we see that $x$ is a $\mathbb{Z}$-linear combination of elements of the form $e_{i_{1}} + \cdots + e_{i_{k}}$ in $\mathbb{Z}^{n},$ where $\{i_{1}, \dots, i_{k}\}$ vary in connected components of $\Gamma.$ This finishes the proof. $\Box$

Lemma. Let $n \in \mathbb{Z}_{\geq 2}.$ Given any graph $\Gamma$ with $n$ vertices, its sandpile group $S_{\Gamma}$ is isomorphic to the cokernel of any $(n - 1) \times (n - 1)$ submatrix of $L(\Gamma).$ In particular, the cokernels of $(n - 1) \times (n - 1)$ submatrice of $L(\Gamma)$ are isomorphic to each other.

Proof. We write $L = L(\Gamma).$ Since all the entries of any row or column sum to $0,$ given any $1 \leq i, j \leq n,$ we may add all the other rows to the $i$-th row and then all the other columns to the $j$-th column and replace all the entries of the $i$-th row and the $j$-th column into $0$'s. Since these operations are elementary row operations over $\mathbb{Z},$ this does not change the cokernel of $L$ up to isomorphisms. Denote by $L^{(i,j)},$ the new $n \times n$ matrix. Then we have $$\mathrm{cok}(L) \simeq \mathbb{Z}^{n}/\mathrm{im}(L^{(i,j)}).$$ Note that $$\mathrm{im}(L^{(i,j)}) \subset \mathbb{Z}e_{1} \oplus \cdots \oplus \mathbb{Z}e_{j-1} \oplus \mathbb{Z}e_{j+1} \oplus \cdots \oplus \mathbb{Z}e_{n}.$$ This implies that $$S_{\Gamma} = (\mathbb{Z}^{n})_{0}/\mathrm{im}(L) \simeq (\mathbb{Z}e_{1} \oplus \cdots \oplus \mathbb{Z}e_{j-1} \oplus \mathbb{Z}e_{j+1} \oplus \cdots \oplus \mathbb{Z}e_{n})/\mathrm{im}(L^{(i,j)}).$$ Let $L_{(i,j)}$ be the $(n-1) \times (n-1)$ matrix given by erasing the $i$-th row and the $j$-th column of $L.$ Then we have $$\begin{align*}(\mathbb{Z}^{n})_{0}/\mathrm{im}(L) &\simeq (\mathbb{Z}e_{1} \oplus \cdots \oplus \mathbb{Z}e_{j-1} \oplus \mathbb{Z}e_{j+1} \oplus \cdots \oplus \mathbb{Z}e_{n})/\mathrm{im}(L^{(i,j)}) \\ &\simeq \mathbb{Z}^{n-1}/\mathrm{im}(L_{(i,j)}) \\ &= \mathrm{cok}(L_{(i,j)}),\end{align*}$$ so $S_{\Gamma} \simeq \mathrm{cok}(L_{(i,j)}),$ as desired. $\Box$

Theorem (Kirchhoff). Let $n \in \mathbb{Z}_{\geq 2}.$ The number of spanning trees of a graph $\Gamma$ with $n$ vertices is equal to the determinant of any $(n-1) \times (n-1)$ submatrix of $L(\Gamma).$ In particular, if $\Gamma$ is connected, then this number is equal to the size of the sandpile group $S_{\Gamma}.$

Remark. Let $L$ be the Laplacian of a graph $\Gamma$ with $n$ vertices with $n \geq 2,$ and $L'$ the $(n-1) \times (n-1)$ submatrix of $L$ given by erasing the last row and the last columns. Then there are always $P, Q \in \mathrm{GL}_{n-1}(\mathbb{Z})$ such that $$PL'Q = \mathrm{diag}(\lambda_{1}, \dots, \lambda_{n-1})$$ with $\lambda_{1} \geq \cdots \geq \lambda_{n-1} \geq 0$ in $\mathbb{Z}.$ Hence, we have $$S_{\Gamma} \simeq \mathrm{cok}(L') \simeq \mathbb{Z}/\lambda_{1}\mathbb{Z} \times \cdots \times \mathbb{Z}/\lambda_{n-1}\mathbb{Z},$$ and Kirchohoff's theorem tells us that $$\det(L') = \lambda_{1} \cdots \lambda_{n-1}$$ is the number of spanning trees of $\Gamma.$ That is, Kirchohoff's theorem says that the number $N_{\Gamma}$ of spanning trees of $\Gamma$ is equal to the product of (non-negative) invariant factors of its sandpile group $S_{\Gamma}.$

A vague question. If we choose $\Gamma$ at random, what does the distribution of $N_{\Gamma}$ look like? More generally, what does the distribution of $S_{\Gamma}$ look like?

Surjection moment method

In a paper by M.M.Wood (Theorem 3.1), we find the following theorem:

Theorem (Wood). Let 

• $(X_{n})_{n \geq 1}$ and $(Y_{n})_{n \geq 1}$ be sequences of random finitely generated abelian groups and
• $\mathcal{A}_{m}$ be the set of isomorphism classes of finite abelian groups with exponent dividing $m \in \mathbb{Z}_{\geq 1}$ (i.e., $mG = 0$ for all $G \in \mathcal{A}_{m}$).

Suppose that for every $G \in \mathcal{A}_{m},$ the limit  $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)).$$ exists and $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) \leq |\wedge^{2}G|.$$

Then for every $H \in \mathcal{A}_{m},$ the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists, and $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)).$$

If we also have $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(Y_{n}, G))$$ for all $G \in \mathcal{A}_{m},$ then $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) = \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H).$$

Remarks. 

1. Note that it is implicit in the statement of the theorem that if we denote by $\mathcal{S}$ the probability space of finitely generated abelian groups for $X_{n},$ then for any $H \in \mathcal{A}_{m},$ the subset $$\{X \in \mathcal{S} : X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H\}$$ is measurable, and similarly for $Y_{n}.$
2. It is also implicit in the statement that the expected value $\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$ exists for every $n \geq 1,$ and similarly for $Y_{n}.$
3. In the references we follow, it is not mentioned that the abelian groups appearing as representatives of classes in $\mathcal{A}_{m}$ are of finite size. It seems that the proof of Theorem requires it, so we assume this. (This does not affect the main results in the references.)

Rest of this posting. We cover the proof of the above theorem, which we learn from another paper by Wood (Theorem 8.3).

Step 1. We first assume that for every $H \in \mathcal{A}_{m}$ the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists and deduce $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z})\#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$$ from this assumption.

Claim. For every $G \in \mathcal{A}_{m},$ there exists $G' \in \mathcal{A}_{m}$ such that the infinite sum $$\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')}$$ converges.

Proof of Claim. If $m = 1,$ then the only isomorphism class in $\mathcal{A}_{m}$ is that of the trivial group, so the claim immediately holds. Hence, assume that $m > 1.$ Consider the prime factorization $m = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}},$ and write $$H = H_{1} \oplus \cdots \oplus H_{k}$$ with $p_{i}^{e_{i}}H_{i} = 0$ for $H \in \mathcal{A}_{m},$ which implies that $$\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')} = \sum_{H_{1} \in \mathcal{A}_{p_{1}^{e_{1}}}}\frac{\#\mathrm{Hom}(H_{1}, G_{1})}{\#\mathrm{Hom}(H_{1}, G'_{1})} \cdots \sum_{H_{k} \in \mathcal{A}_{p_{k}^{e_{k}}}}\frac{\#\mathrm{Hom}(H_{k}, G_{k})}{\#\mathrm{Hom}(H_{k}, G'_{k})}$$ for any $G' \in \mathcal{A}_{m}.$ This reduces the problem to the case where $m = p^{e}$ where $p$ is a prime and $e \in \mathbb{Z}_{\geq 1}.$ Without loss of generality, assume that $G$ is nontrivial. In this case, we have $$G \simeq \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\lambda_{2}}\mathbb{Z} \cdots \times \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}$$ with $e \geq \lambda_{1} \geq \cdots \geq \lambda_{l} \geq 1,$ so if we let $\lambda = (\lambda_{1}, \dots, \lambda_{l})$ be the corresponding partition, then we choose $$G' = \mathbb{Z}/p^{\mu_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\mu_{2}}\mathbb{Z} \cdots \times \mathbb{Z}/p^{\mu_{l'}}\mathbb{Z},$$ where $\mu_{i} = 2\lambda'_{i} + 1$ for $1 \leq i \leq e.$ (Here, we have $\lambda'_{i} = 0$ for $\lambda_{1} < i \leq e.$) By using a standard formula that counts the number of group maps from a finite abelian $p$-groups to another (Lemma 7.1 of Wood), we can compute that $$\sum_{H \in \mathcal{A}_{p^{e}}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')} = \sum_{c_{1} \geq \cdots \geq c_{e} \geq 0}p^{-c_{1}(\lambda_{1} + 1) \cdots - c_{e}(\lambda_{e} + 1)},$$ which is convergent as it is bounded by a convergent geometric series. This proves the claim. $\Box$

Back to Step 1, we have $$\begin{align*}\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') &= \mathbb{E}( \#\mathrm{Hom}(X_{n}, G')) \\ &= \sum_{H \subset G'}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H)),\end{align*}$$ where $H$ ranges over subgroups of (a representative of) fixed $G' \in \mathcal{A}_{m}.$ The second identity is evident, but the first identity needs an explanation. By definition, we have $$\begin{align*}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G')) &= \int_{X \in \mathcal{S}} \#\mathrm{Hom}(X, G') \mu_{n}(X), \end{align*}$$ where $\mu_{n}$ means the given probability measure on the probability space $\mathcal{S}$ of finitely generated abelian groups, where $X_{n}$ takes place. Since $mG' = 0,$ we have $$\#\mathrm{Hom}(X, G') = \#\mathrm{Hom}(X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}, G')$$ for any $X \in \mathcal{S}$ so that $$\begin{align*}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G')) &= \int_{X \in \mathcal{S}} \#\mathrm{Hom}(X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}, G') \mu_{n}(X) \\ &= \sum_{B \in \mathcal{A}_{m}}\int_{\substack{X \in \mathcal{S}: \\ (X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}) \simeq B}}\#\mathrm{Hom}(B, G')\mu_{n}(X) \\ &= \sum_{B \in \mathcal{A}_{m}}\int_{\substack{X \in \mathcal{S}: \\ (X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}) \simeq B}}\mu_{n}(X) \#\mathrm{Hom}(B, G') \\ &= \sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G'),\end{align*}$$ as desired.

By hypothesis, the limit $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H))$$ exists for each subgroup $H$ of $G'$, so $$\lim_{n \rightarrow \infty}\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') = \sum_{H \subset G'}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H))$$ exists as well. In particular, there exists a bound $D_{G'} > 0$ independent of $n$ such that $$\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') \leq D_{G'}.$$ Thus, for any fixed $H \in \mathcal{A}_{m},$ we have $$\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G') \leq D_{G'}.$$ This implies that $$\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq D_{G'}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')},$$ so $$\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq D_{G'}\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')},$$ and by Claim the right-hand side converges as $n \rightarrow \infty.$ Thus, by the Lebesgue dominated convergence theorem, the left-hand side also converges and $$\begin{align*}&\lim_{n \rightarrow \infty}\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G),\end{align*}$$ where we used our assumption that the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists (which will be proven below). Since $$\#\mathrm{Hom}(H, G) = \sum_{B \subset G}\#\mathrm{Sur}(H, B)$$ where the sum is over subgroups $B$ of $G,$ subtracting the corresponding quantities for all subgroups of $G,$ we obtain $$\begin{align*}&\lim_{n \rightarrow \infty}\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G).\end{align*}$$ A similar computation as above gives $$\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G),$$ so $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$$ by the hypothesis. This finishes Step 1.

Step 2. We assume that $\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ and $\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ both converge as $n \rightarrow \infty$ for every $H \in \mathcal{A}_{m}.$  and $$\begin{align*}&\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G)\end{align*}$$ for every $G \in \mathcal{A}_{m}.$ By Step 1, both sides are equal to $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(Y_{n}, G)).$$

Then we show that $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) = \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$

Using a similar argument as in Step 1, we have $$\begin{align*}&\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \\ &= \sum_{B \subset G}\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{B \subset G} \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)),\end{align*}$$ where $B$ ranges over subgroups of $G.$  We can also deduce a similar identity for $(Y_{n})_{n \geq 1}.$

We need two lemmas to finish Step 2, and this is where we need $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) \leq |\wedge^{2}G|.$$

Lemma 1. Let $\lambda = (\lambda_{1} \geq \cdots \geq \lambda_{l})$ be a partition and $$G = \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \cdots \times \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}.$$ Then $$\sum_{B \subset G}|\textstyle\bigwedge^{2}G| \leq F^{\lambda_{1}}p^{\sum_{i=1}^{\infty}\lambda'_{i}(\lambda'_{i}-1)/2},$$ where $B$ ranges over subgroups of $G$ and $$F = \frac{2}{1 - 2^{-1/8}} \prod_{i=1}^{\infty}\frac{1}{1 - 2^{-i}}.$$

Remark. For a later use, we note that the number of parts of $\lambda$ in Lemma 1 is at most $\lambda_{1}.$

For the next lemma, we need more notation. Fix any distinct primes $p_{1}, \dots, p_{s}$ and integers $e_{1}, \dots, e_{k} \geq 1.$ Let $M_{j}$ be the set of partitions with at most $e_{j}$ parts, and set $M_{(e_{1}, \dots, e_{k})} = M := M_{1} \times \cdots \times M_{k}.$ Given any $\mu \in M,$ we may write $$\mu = (\mu^{(1)}, \dots, \mu^{(k)}).$$ Since $\mu^{(j)} \in M_{j}$, we write $\mu^{(j)} = (\mu^{(j)}_{1} \geq \cdots \geq \mu^{(j)}_{e_{j}}),$ where $\mu^{(j)}_{i} \geq 0.$

Lemma 2. Assume the above notation. For each $\lambda \in M,$ suppose that we have $x_{\lambda}, y_{\lambda} \in \mathbb{R}_{\geq 0}$ such that $$\sum_{\mu \in M}x_{\mu}\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} = \sum_{\mu \in M}y_{\mu}\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} \leq \prod_{j=1}^{s}F^{e_{j}}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}(\lambda_{i}^{(j)}-1)/2},$$ for some constant $F > 0$ that does not depend on $\lambda.$ Then we have $x_{\lambda} = y_{\lambda}$ for all $\lambda \in M.$

Going back to Step 2, prime factorize $m = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$ with $e_{j} \geq 1.$ Then there is a bijection between $M_{(e_{1}, \dots, e_{k})} = M$ and $\mathcal{A}_{m}$ by mapping each $$\lambda = (\lambda^{(1)}, \dots, \lambda^{(k)}) \in M$$ to a finite abelian group, whose $p_{j}$-part corresponds to the conjugate of the partition $\lambda^{(j)}$ for $1 \leq j \leq k.$ (This is where we need abelian groups in $\mathcal{A}_{m}$ to be finite.) What's important here is that if $G, H \in \mathcal{A}_{m}$ correspond to $\lambda, \mu \in M$ respectively under this bijection, then $$\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} = \#\mathrm{Hom}(H, G).$$ Now, if we let $$x_{\lambda} := \lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ and $$y_{\lambda} := \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H),$$ then to finish Step 2 with the help of Lemma 2, we only need to show that $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq \prod_{j=1}^{s}F^{e_{j}}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}(\lambda_{i}^{(j)}-1)/2},$$ for some constant $F > 0$ that does not depend on $G \in \mathcal{A}_{m}.$ The left-hand side is $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G)) = \sum_{B \subset G}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)),$$ where $B$ varies over subgroups of $G,$ and we now use the assumption that $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)) \leq |\wedge^{2} B|,$$ which gives us $$\begin{align*}\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) &= \sum_{B \subset G}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)) \\ &\leq \sum_{B \subset G} |\wedge^{2}B| \\ &\leq \prod_{j=1}^{k} F^{e_{j}}p^{\sum_{i=1}^{\infty}\lambda'_{i}(\lambda'_{i}-1)/2},\end{align*}$$ where $$F = \frac{2}{1 - 2^{-1/8}} \prod_{i=1}^{\infty}\frac{1}{1 - 2^{-i}}.$$ We used Lemma 1 along with how direct sums work with wedge products. This finishes Step 2.

Remark. Step 2 is the most nontrivial step, and Lemma 1 and Lemma 2 are where interesting bounds are analyzed. Lemma 2 seems to be where the nontriviality lies. We will cover these two lemmas in a different post, and think about whether the bound given here can be improved.

 Step 3. We show that $\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ converges as $n \rightarrow \infty.$

Since the sequence in question is bounded in $[0, 1],$ by the Bolzano--Weierstrass theorem, there must be a subsequence $X_{n_{1}}, X_{n_{2}}, X_{n_{3}}, \dots$ such that the limit $$L = \lim_{i \rightarrow \infty}\mathrm{Prob}(X_{n_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists for any $H \in \mathcal{A}_{m}.$

For contradiction, suppose that the sequence in question does not converge for some $H \in \mathcal{A}_{m}.$ This means that there exists $\epsilon > 0$ and a subsequence $X_{m_{1}}, X_{m_{2}}, X_{m_{3}}, \dots$ such that $$|\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) - L| \geq \epsilon.$$ We may refine this subsequence by the Bolzano--Weirstrass theorem to assume that the limit $$\lim_{i \rightarrow \infty}\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists. The contradiction comes from the fact that $$\lim_{i \rightarrow \infty}\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \neq L,$$ because if we go through Step 1 and Step 2 with these two subsequences, Step 2 tells us that the two limits must be equal. This finishes the proof of Step 3.

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.

Abelian varieties: 3. Abelian schemes are commutative group schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. We also follow Mumford's book.

Convention. When we discuss an abelian scheme over a base scheme $S,$ we will assume that $S$ is nonempty.

We have discussed how an abelian variety $A$ over $\mathbb{C}$ is a commutative group scheme using analytic techniques. More specifically, we have gone through the following two steps:

Step 1. We have studied the conjugation action of each element of $A(\mathbb{C})$ on the tangent space $T_{e}A(\mathbb{C})$ at the identity $e$.

Step 2. We have made use of the exponential map $T_{e}A(\mathbb{C}) \rightarrow A(\mathbb{C}).$

We are going to show that an abelian variety over any field is a commutative group scheme. Step 1 works in algebraic setting, but Step 2 does not.

Remark. In characteristic $p > 0,$ it is known that there are examples of automorphisms that act trivially on the tangent space but not on the variety. (We might add some examples later.)

Let $A$ be an abelian variety over a field $k.$ The strategy we take is that we are going to consider the conjugations $c_{t} : A(k) \rightarrow A(k)$ as a flat family parametrized by $t \in A(k).$ We will show that $c_{t}$ is constant in $t.$

Our specific goal. Given an abelian scheme $A$ over $S,$ for $a \in A(S)$ (i.e., an $S$-scheme map $a : S \rightarrow A$), we define the left-translation by $a$ as the map $$l_{a} := m_{A} \circ (a \circ \pi_{A}, \mathrm{id}_{A}) : A \rightarrow A \times_{S} A \rightarrow A,$$ where $m_{A}$ is the multiplication map of the group $S$-scheme $A,$ and $\pi_{A} : A \rightarrow S$ is the structure map. Even though the above definition for $l_{a}$ is simple, it is not the best definition to work with. Another way to describe it is that given any $S$-scheme $T,$ the map $l_{a} : A \rightarrow A$ gives $l_{a, T} : A(T) \rightarrow A(T)$ defined by $x \mapsto \pi_{T}^{*}(a) \cdot x = (a \circ \pi_{T}) \cdot x,$ where $\pi_{T} : T \rightarrow S$ is the structure map. This allows us to show that $l_{e} = \mathrm{id}_{G}$ and $l_{g \cdot g'} = l_{g} \circ l_{g'}$ for all $g, g' \in G(S).$ In particular, we have $l_{g \cdot g^{-1}} = \mathrm{id}_{G} = l_{g^{-1} \cdot g}.$ This discussion works for any group scheme $G$ over $S.$ Details can be found in this posting.

Theorem. Let $A, B$ be abelian $S$-schemes with identities $e_{A}, e_{B}.$ If $S$ is a Noetherian scheme, given any $S$-scheme map $f : A \rightarrow B,$ there is a factorization $f = l_{f_{S}(e_{A})} \circ h$ for some map $h : A \rightarrow B$ of $S$-groups, where $f_{S}(e_{A}) = f \circ e_{A} : S \rightarrow A \rightarrow B$ as usual.

Corollary. If $S$ is a Noetherian scheme, any abelian $S$-scheme $A$ is a commutative group scheme.

Proof. Denote by $i_{A} : A \rightarrow A$ the inversion map. Since $i_{A}(e_{A}) = e_{A} : S \rightarrow A,$ we have $l_{i_{A}(e_{A})} = l_{e_{A}} = m_{A} \circ (e_{A} \circ \pi, \mathrm{id}_{A}) = \mathrm{id}_{A},$ where the last equality follows from a group scheme axiom for $A.$ Thus, applying Theorem for $f = i_{A},$ it shows that $i_{A}$ is a group scheme map. From here, it follows that $A$ is a commutative group scheme. $\Box$

Hence, it remains to show Theorem. First, we note that it is enough to show that $l_{f_{S}(e_{A})^{-1}} \circ f : A \rightarrow B$ is a map of $S$-groups. Since $$\begin{align*}(l_{f_{S}(e_{A})^{-1}} \circ f)_{S}(e_{A}) &= l_{f_{S}(e_{A})^{-1}} \circ f \circ e_{A} \\ &= l_{f_{S}(e_{A})^{-1}} \circ f_{S}(e_{A}) \\ &= l_{f_{S}(e_{A})^{-1},S}(f_{S}(e_{A})) \\ &= f_{S}(e_{A})^{-1} \cdot f_{S}(e_{A}) \\ &= e_{B}\end{align*}$$ in $B(S)$ because the $S$-scheme structure map for $S$ is $\mathrm{id}_{S}.$ Thus, this reduces the problem to the case where $f_{S}(e_{A}) = e_{B},$ and we desire to prove that $f : A \rightarrow B$ is an $S$-group map. This means that given any $S$-scheme $T$ and $a, a' \in A(T),$ we have $$f_{T}(m_{A,T}(a, a')) = f_{T}(a \cdot a') = f_{T}(a) \cdot f_{T}(a') = m_{B,T}(f_{T}(a), f_{T}(a')).$$ In terms of $S$-scheme map language, this precisely states that $$f \circ m_{A} = m_{B} \circ (f \times_{S} f) : A \times_{S} A\rightarrow B,$$ and thus this is what we need to prove. For each $S$-scheme $T,$ we may instead try to prove $$m_{B,T}(f_{T}(m_{A,T}(a, a')), i_{B}(m_{B,T}(f_{T}(a), f_{T}(a')))) = e_{B,T} \in B(T)$$ for all $a, a' \in A(T).$ In the $S$-scheme map language, this means that we want to prove $$m_{B} \circ (f \circ m_{A}, i_{B} \circ m_{B} \circ (f \times_{S} f)) = e_{B} \circ \pi_{A \times_{S} A} : A \times_{S} A \rightarrow B,$$ where again $e_{B} = e_{B,S} : S \rightarrow B$ is the identity element of the group $B(S)$ and $\pi_{A \times_{S} A} : A \times_{S} A \rightarrow S$ is the $S$-scheme structure map. The map on the left-hand side is the composition $$A \times_{S} A \rightarrow B \times_{S} B \rightarrow B$$ of the relevant maps.

Reduction of the problem. We write $$g := m_{B} \circ (f \circ m_{A}, i_{B} \circ m_{B} \circ (f \times_{S} f)) : A \times_{S} A \rightarrow B.$$ We want to show that $g = e_{B} \circ \pi_{A \times_{S} A}.$ Note that given any $S$-scheme $T,$ the map $g_{T} : A(T) \times A(T) \rightarrow B(T)$ is given by $(a, a') \mapsto f_{T}(a \cdot a') \cdot (f_{T}(a) \cdot f_{T}(a'))^{-1}.$ Hence, we have $$\begin{align*}g \circ (\mathrm{id}_{A}, e_{A} \circ \pi_{A}) \circ a &= g_{T}(a, e_{A,T}) \\ &= f_{T}(a \cdot e_{A,T}) \cdot (f_{T}(a) \cdot f_{T}(e_{A,T}))^{-1} \\ &= f_{T}(a \cdot e_{A,T}) \cdot (f_{T}(a) \cdot e_{B,T})^{-1} \\ &= e_{B,T} \\ &= e_{B} \circ \pi_{T}\end{align*}$$ because, using $f_{S}(e_{A}) = e_{B},$ we have $$\begin{align*}f_{T}(e_{A,T}) &= f \circ e_{A} \circ \pi_{T}\\ &= f_{S}(e_{A}) \circ \pi_{T}\\ &= e_{B} \circ \pi_{T} \\&= e_{B,T}.\end{align*}$$ Hence, if we prove that $g : A \times_{S} A \rightarrow B$ is a constant map, then it will follow that for every $a, a' \in A(T),$ we have $g_{T}(a, a') = e_{B, T}.$ The only $S$-scheme map $A \times_{S} A \rightarrow B$ such that $A(T) \times A(T) \rightarrow B(T)$ is giving the constant value of $e_{B,T}$ is $e_{B} \circ \pi_{A \times_{S} A},$ so this will finish the proof. Thus, it remains to show that $g : A \times_{S} A \rightarrow B$ is constant.

To finish the proof of Theorem, we use the following "rigidity" lemma:

Lemma (Rigidity). Fix any Noetherian scheme $S.$ Let  $\pi_{X} : X \rightarrow S$ be a proper flat $S$-scheme such that $\dim_{\kappa(s)}(H^{0}(X_{s}, \mathscr{O}_{X_{s}})) = 1$ for all $s \in S,$ where $$X_{s} = X \times_{S} \mathrm{Spec}(\kappa(s)),$$ the fiber at $s.$ If $\phi : X \rightarrow Y$ is any $S$-scheme map such that there exists any $s \in S$ such that the restriction $X_{s} \rightarrow Y_{s}$ of $\phi$ is constant, then $\phi$ is constant on the connected component of $s$ in $S.$

Remark. An $S$-scheme map $X \rightarrow Y$ is said to be constant if it factors through the structure map of $X.$ In particular, in such situation, for any $S$-scheme $T,$ the induced map $X(T) \rightarrow Y(T)$ is constant in set-theoretic sense. Given a field $k,$ a constant map of $k$-schemes $X \rightarrow Y$ necessarily give a constant map on the underlying topological spaces, as it factors as $X \rightarrow \mathrm{Spec}(k) \rightarrow Y.$ I don't think the converse is true: there are probably many examples where a $k$-scheme map is topologically constant but not a constant map according to the definition we are using. However, I have not thought deeply about such examples. (I may add them later if there are such though.)

Proof of Theorem. We use the notations given before Lemma. Write $$p_{1}^{-1}(e_{A}) := S \times_{A \times_{S} A} A,$$ with respect to $e_{A} : S \rightarrow A$ and $p_{1} : A \times_{S} A \rightarrow A.$ Given any $S$-scheme $T,$ the $S$-scheme map $$p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A$$ induces the set map $$(p_{1}^{-1}(e_{A}))(T) \rightarrow A(T) \times A(T),$$ which necessarily maps to elements of the form $(a, e_{A,T})$ where $a \in A(T)$ may vary. We have checked before that $g_{T}(a, e_{A,T}) = e_{B,T}$ constantly regardless of the choice of $a,$ so this implies that the composition $p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A \xrightarrow{g} B$ is constant. This implies that the following composition is also constant: $$p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A \xrightarrow{(p_{1}, g)} A \times_{S} B.$$ Then now note that $(p_{1}, g)$ is an $A$-scheme map that is constant on the fiber at any point in $A$ in the image of $e_{A} : S \rightarrow A.$ Now note that

• $A$ is connected,
• both $A \times_{S} A$ and $A \times_{S} B$ are proper and smooth over $A$ (base change) and hence also flat over $A$ (e.g., 25.2.2. (iii) in Vakil), and
• for $s \in A,$ we have $(A \times_{S} A)_{s} = p_{1}^{-1}(s) \simeq \mathrm{Spec}(\kappa(s)) \times_{A} A \times_{S} A \simeq A_{s},$ which is geometrically connected, so $\dim_{\kappa(s)} (A \times_{S} A)_{s} = 1.$

Thus, we may apply Rigidity Lemma to conclude that that $(p_{1}, g)$ is constant. Since $g$ is the composition $$A \times_{S} A \xrightarrow{(p_{1}, g)} A \times_{S} B \rightarrow B,$$ where the latter one is the projection map onto $B,$ the fact that $(p_{1}, g)$ is constant implies that $g$ is constant. This finishes the proof $\Box$