Proof. Write $(\mathfrak {m}_{\mathrm {max}})^{\mathrm {ram}} = \Delta _1+\dots +\Delta _r$ and $(\mathfrak {m}^{\mathrm {max}})^{\mathrm {ram}} = \Delta _{r+1} + \dots + \Delta _t$ . Then $\mathfrak {m}^{\mathrm {ram}} = \Delta _1+\dots +\Delta _t$ by Proposition 2.7. Hence we have

$$ \begin{align*} \lambda_{\mathfrak{m}} &= \sum_{i=1}^t (0,\dots,0,\underbrace{1, \dots, 1}_{l(\Delta_i)}) \\&= \sum_{i=1}^r (0,\dots,0,\underbrace{1, \dots, 1}_{l(\Delta_i)}) + \sum_{i=r+1}^t (0,\dots,0,\underbrace{1, \dots, 1}_{l(\Delta_i)}) \\&= \lambda_{\mathfrak{m}_{\mathrm{max}}} + \lambda_{\mathfrak{m}^{\mathrm{max}}}. \end{align*} $$

This completes the proof.

Proof. For a segment $[a,b]_{\rho }$ , we denote by $\Delta [a,b]_{\rho }$ the generalised Steinberg representation: that is, the unique irreducible quotient of

$$\begin{align*}\rho|\cdot|^a \times \rho|\cdot|^{a+1} \times \dots \times \rho|\cdot|^b. \end{align*}$$

As in the Langlands classification, we write $\pi = L([a_1,b_1]_{\rho _1} + \dots + [a_r,b_r]_{\rho _r})$ if $\pi $ is the unique irreducible subrepresentation of

$$\begin{align*}\Delta[a_1,b_1]_{\rho_1} \times \dots \times \Delta[a_r,b_r]_{\rho_r} \end{align*}$$

with $\rho _i$ unitary and $a_1+b_1 \leq \dots \leq a_r+b_r$ . Then the Zelevinsky dual $\pi ^{\sharp }$ of $\pi $ is given by

$$\begin{align*}\pi^{\sharp} = Z([a_1,b_1]_{\rho_1} + \dots + [a_r,b_r]_{\rho_r}). \end{align*}$$

By [Reference Zelevinsky42, 8.1 Theorem], the highest derivative $(\pi ^{\sharp })^-$ of $\pi ^{\sharp }$ is

$$\begin{align*}(\pi^{\sharp})^- = Z([a_1,b_1-1]_{\rho_1} + \dots + [a_r,b_r-1]_{\rho_r}). \end{align*}$$

Here, if $a_i = b_i$ , we ignore $[a_i,b_i-1]_{\rho _i}$ . Hence

$$\begin{align*}\pi^{\mathrm{ram}} = ((\pi^{\sharp})^-)^{\sharp} = L([a_1,b_1-1]_{\rho_1} + \dots + [a_r,b_r-1]_{\rho_r}). \end{align*}$$

Therefore, the map $\pi \mapsto \pi ^{\mathrm {ram}}$ corresponds to $V \mapsto V/\mathrm {Ker}\, N$ (see, e.g., [Reference Rodier37]). Since $\pi $ is unipotent, the corresponding V satisfies that $V = V^{I_F}$ so that $V^{\mathrm {ram}} = V/\mathrm {Ker}\, N$ .

Proof. The ‘only if’ part is trivial. We prove the ‘if’ part. Assume the two conditions. Let us choose an isomorphism

$$\begin{align*}f_1 \colon (\mathrm{Image}\, N,N|_{\mathrm{Image}\, N}) \xrightarrow{\cong} (\mathrm{Image}\, N',N'|_{\mathrm{Image}\, N'}) \end{align*}$$

of $VN$ -pairs. Let us also choose homogeneous elements $v_1,\ldots ,v_r \in \mathrm {Image}\, N$ whose images in $\mathrm {Image}\, N/\mathrm {Image}\, N^2$ form a basis of this space. For $i=1,\ldots ,r$ , let us choose homogeneous elements $e_1,\ldots ,e_r \in V$ and $e^{\prime }_1,\ldots ,e^{\prime }_r \in V'$ in such a way that we have $N(e_i) = v_i$ and $N'(e^{\prime }_i) = f_1 (v_i)$ for $i=1,\ldots ,r$ . Let W (respectively, $W'$ ) denote the graded vector subspace of V (respectively, $V'$ ) generated by $\mathrm {Image}\, N$ and $e_1,\ldots ,e_r$ (respectively, $\mathrm {Image}\, N'$ and $e^{\prime }_1,\ldots ,e^{\prime }_r$ ).

Let $\overline {N}\colon V/\mathrm {Image}\, N \to \mathrm {Image}\, N/\mathrm {Image}\, N^2$ denote the homomorphism induced by N. It follows from the construction of W that the restriction of $\overline {N}$ to $W/\mathrm {Image}\, N$ gives an isomorphism $W/\mathrm {Image}\, N \xrightarrow {\cong } \mathrm {Image}\, N/\mathrm {Image}\, N^2$ . Hence we have $V/\mathrm {Image}\, N = \mathrm {Ker}\, \overline {N} \oplus (W/\mathrm {Image}\, N)$ . By applying the snake lemma to the commutative diagram

we see that the homomorphism $\alpha \colon \mathrm {Ker}\, N \to \mathrm {Ker}\, \overline {N}$ induced by the quotient map $V \twoheadrightarrow V/\mathrm {Image}\, N$ is surjective. Let us choose a graded vector subspace $U \subset \mathrm {Ker}\, N$ such that the restriction of $\alpha $ to U gives an isomorphism $U \xrightarrow {\cong } \mathrm {Ker}\, \overline {N}$ . Since $V/\mathrm {Image}\, N = \mathrm {Ker}\, \overline {N} \oplus (W/\mathrm {Image}\, N)$ , we have $V = U \oplus W$ .

A similar argument shows that there exists a graded vector subspace $U' \subset \mathrm {Ker}\, N'$ such that $V' = U' \oplus W'$ . Since V and $V'$ are isomorphic as graded vector spaces, U and $U'$ are isomorphic as graded vector spaces. Let us choose an isomorphism $f_2 \colon U \to U'$ of graded vector spaces.

Let $f \colon V \to V'$ denote the homomorphism defined as follows: $f(v)=f_1(v)$ for $v \in \mathrm {Image}\, N$ , $f(e_i) = e^{\prime }_i$ for $i=1,\ldots ,r$ and $f(u)=f_2(u)$ for $u \in U$ . Then f is an isomorphism of $VN$ -pairs from $(V,N)$ to $(V',N')$ . This completes the proof.

Proof. We only give a proof of assertion (1). We can prove assertion (2) in a similar manner.

Let us choose homogeneous, linearly independent elements $v_1, \ldots , v_m \in V$ such that V is a direct sum of $\mathrm {Image}\, N$ and the subspace of V generated by $v_1, \ldots , v_m$ . For $i=1,\ldots ,m$ , we let $d_i$ denote the degree of $v_i$ . Given $L' \in S(\mathrm {Image}\, N, N|_{\mathrm {Image}\, N})$ , choose a homogeneous element $w_i \in V$ of degree $d_i-1$ that satisfies $L'(N(v_i)) = N(w_i)$ for each $i=1, \ldots , m$ . Let L denote the unique $\mathbb {C}$ -linear map $V \to V$ such that $L(v) = L'(v)$ for $v \in \mathrm {Image}\, N$ and that $L(v_i) = w_i$ for $i=1,\ldots ,m$ . It is then straightforward to check that $L \in S(V,N)$ . It follows from the construction of L that $L|_{\mathrm {Image}\, N} = L'$ . Hence the claim follows.

Proof. Easy.

Proof. It is easy to see that there exists a Zariski open subset $S^{\theta }(V,N) \subset S(V,N)$ such that for $L \in S(V,N)$ , both $(V,N,L)$ and $(\mathrm {Image}\, L, N|_{\mathrm {Image}\, L}, L|_{\mathrm {Image}\, L})$ are admissible triples if and only if $L \in S^{\theta }(V,N)$ .

It remains to show that $S^{\theta }(V,N)$ is dense in $S(V,N)$ . Since $S(V,N)$ is irreducible as an algebraic variety over $\mathbb {C}$ , it suffices to show that $S^{\theta }(V,N)$ is nonempty. Let us choose $L \in S^o(V,N)$ . Since the morphism $S(V,L) \to S(\mathrm {Image}\, L, L|_{\mathrm {Image}\, L})$ is surjective by Lemma 3.2, there exists $N' \in S(V,L)$ such that both $(V,N',L)$ and $(\mathrm {Image}\, L, N'|_{\mathrm {Image}\, L}, L|_{\mathrm {Image}\, L})$ are admissible triples. Then $(V,N)$ and $(V,N')$ are isomorphic since both correspond to the same multisegment. Hence $(V,N',L)$ is isomorphic to $(V,N,L')$ for some $L' \in S(V,N)$ . Since $L'$ belongs to $S^{\theta }(V,N)$ , it follows that $S^{\theta }(V,N)$ is nonempty, as desired.

Proof of Proposition 2.4

Let $\pi = Z(\mathfrak {m})$ be an irreducible representation of $G_n$ . We decompose $\mathfrak {m}$ as

$$\begin{align*}\mathfrak{m} = \mathfrak{m}' + \mathfrak{m}_1+\dots+\mathfrak{m}_{t}, \end{align*}$$

where

• • each segment in $\mathfrak {m}'$ is not unipotent;

• • each $\mathfrak {m}_i$ is of type $\chi _i$ for some unramified character $\chi _i$ of $F^{\times }$ for $1 \leq i \leq t$ ;

• • if $i \not = j$ , then $\chi _i\chi _j^{-1}$ is not of the form $|\cdot |^a$ for any $a \in \mathbb {Z}$ .

Set $\pi ' = Z(\mathfrak {m}')$ and $\pi _i = Z(\mathfrak {m}_i)$ . Then $\pi $ is isomorphic to the parabolic induction $\pi ' \times \pi _1 \times \cdots \times \pi _{t}$ .

For $\Pi = \pi , \pi ', \pi _1, \ldots , \pi _{t}$ , let $\Pi ^{(0)} = \Pi $ and $\Pi ^{(i)}$ denote the highest derivative of $\Pi ^{(i-1)}$ for $i \geq 1$ . Then we have $\pi ^{(i)} = \pi ^{\prime (i)} \times \pi _1^{(i)} \times \cdots \times \pi _{t}^{(i)}$ for any integer $i \ge 0$ . Thus, to prove the claim, we may assume that $\mathfrak {m} = \mathfrak {m}'$ or $\mathfrak {m} = \mathfrak {m}_1$ .

First, we consider the case where $\mathfrak {m} = \mathfrak {m}'$ . Let us write $\pi = Z(\mathfrak {m})$ and $\mathfrak {m} = [a_1,b_1]_{\rho _1} + \cdots +[a_r,b_r]_{\rho _r}$ . Then $\rho _1, \ldots , \rho _r$ are ramified cuspidal representations. For $i=1,\ldots ,r$ , let $c_i = c_{\rho _i}$ denote the conductor of $\rho _i$ . Then for $j \geq 0$ , we have $\pi ^{(j)} = Z(\mathfrak {m}^{(j)})$ , where

$$\begin{align*}\mathfrak{m}^{(j)} = \sum_{1 \le i \le r \atop b_i - a_i \ge j} [a_i, b_i -j]_{\rho_i}. \end{align*}$$

This shows that the conductor of $\pi ^{(j)}$ is equal to

$$\begin{align*}c^{(j)} = \sum_{1 \le i \le r \atop b_i - a_i \ge j} (b_i -a_i + 1 -j) c_{i}. \end{align*}$$

Hence we have

$$\begin{align*}c^{(j)} - c^{(j+1)} = \sum_{1 \le i \le r \atop b_i - a_i \ge j} c_{i}. \end{align*}$$

From this, one can easily see that

$$\begin{align*}\lambda_{\pi} = \sum_{i=1}^r (0,\ldots,0, \underbrace{c_i,\ldots,c_i}_{b_i-a_i+1}) = \lambda_{\mathfrak{m}}, \end{align*}$$

as desired.

Now we consider the case where $\pi = Z(\mathfrak {m})$ is of type $\chi $ for an unramified character $\chi $ of $F^{\times }$ . Let us consider the $VN$ -pair $(V,N)$ corresponding to $\mathfrak {m}$ . For $i \ge 0$ , let us write $\pi ^{(i)} = Z(\mathfrak {m}^{(i)})$ . As we remarked at the beginning of Section 2.3, we have $\pi ^{(1)} = Z(\mathfrak {m}^{-})$ . Hence $\mathfrak {m}^{(i)}$ is obtained from $\mathfrak {m}$ by the i-fold iteration of the operation $(\ )^{-}$ . Therefore, it follows from Lemma 3.3 that $\mathfrak {m}^{(i)}$ corresponds to the $VN$ -pair $(\mathrm {Image}\, N^i, N|_{\mathrm {Image}\, N^i})(-i)$ . Let us choose $L \in S^{\theta }(V,N)$ such that $L|_{\mathrm {Image}\, N^i}$ belongs to $S^o(\mathrm {Image}\, N^i, N|_{\mathrm {Image}\, N^i})$ for any integer $i \ge 0$ . By Lemma 3.2, such an L exists. Then the conductor of $\pi ^{(i)}$ is equal to the dimension of $\mathrm {Image}\, L \circ N^i$ . Hence if we write $\lambda _{\pi } = (\lambda _1, \dots , \lambda _n)$ and $d_i = \dim \mathrm {Image}\, L \circ N^i$ for $i \geq 0$ , then we have

$$\begin{align*}\lambda_k = d_{n-k} - d_{n-k+1} \end{align*}$$

for $k = 1, \dots , n$ . Let us write $\pi ^{\mathrm {ram}} = Z(\mathfrak {m}^{\mathrm {ram}})$ with $\mathfrak {m}^{\mathrm {ram}} = \Delta _1+\dots +\Delta _r$ and $\Delta _i = [a_i,b_i]_{\chi }$ for $1 \leq i \leq r$ . Then $\lambda _{\mathfrak {m}} = (\lambda ^{\prime }_1,\dots ,\lambda ^{\prime }_n)$ , with

$$\begin{align*}\lambda^{\prime}_k = \sum_{\substack{1 \leq i \leq r \\ b_i-a_i \geq n-k}}1 \end{align*}$$

for $k = 1,\dots ,n$ . By Lemmas 3.3 and 3.4, $\mathfrak {m}^{\mathrm {ram}}$ corresponds to the $VN$ -pair $(\mathrm {Image}\, L, N|_{\mathrm {Image}\, L})$ . Since L and N commute, we have $\dim \mathrm {Image}\, N^i \circ L = d_i$ for $i \geq 0$ . Hence we have

$$\begin{align*}d_i-d_{i+1} = \sum_{\substack{1 \leq i \leq r \\ b_i-a_i \geq i}}1 \end{align*}$$

for $i = 0,\dots ,n-1$ . Therefore, we have

$$\begin{align*}\lambda^{\prime}_k = d_{n-k} - d_{n-k+1} = \lambda_k \end{align*}$$

for $k = 1,\dots ,n$ . This completes the proof.

Proof. Let $(V,N)$ be the $VN$ -pair corresponding to the multisegment $\mathfrak {m}$ . If we choose a sufficiently general $L \in S(V,N)$ , then $(\mathfrak {m}^{-})^{\mathrm {ram}}$ and $(\mathfrak {m}^{\mathrm {ram}})^{-}$ correspond to the pairs $(\mathrm {Image}\, L\circ N, N|_{\mathrm {Image}\, L \circ N})(-1)$ and $(\mathrm {Image}\, N\circ L, N|_{\mathrm {Image}\, N \circ L})(-1)$ , respectively. Since $L\circ N=N \circ L$ , the claim follows.

Proof. By replacing $\chi $ with $\chi |\cdot |^c$ for some integer c, we may and will assume that there exists an integer r such that any segment in $\mathfrak {m}$ is contained in $[1,r]_{\chi }$ .

For two integers $a,b$ with $1 \leq a \leq b \leq r$ , let $d_{a,b}=d_{a,b}(\mathfrak {m})$ denote the multiplicity of the segment $[a,b]_{\chi }$ in $\mathfrak {m}$ . When $a>b$ , we set $d_{a,b}=0$ . Then it follows from the result of Knight–Zelevinsky [Reference Knight and Zelevinsky16, Theorem 1.2] that $\mathrm {Card}(\mathfrak {m}^{\sharp }_{(a)})$ is equal to the right-hand side of the equality (1.6) in [Reference Knight and Zelevinsky16] for $(i,j)=(a,a+1)$ .

For two integers $x,y$ with $x \leq y$ , let $[x,y]$ denote the set of integers c satisfying $x \le c \le y$ . Let $a \in [1,r-1]$ . We rewrite the right-hand side of (1.6) in [Reference Knight and Zelevinsky16] for $(i,j)=(a,a+1)$ . Let us first recall some notation in [Reference Knight and Zelevinsky16]. They fix a positive integer r and consider the set S of pairs of integers $(i,j)$ such that $1\leq i \leq j \leq r$ . For $ 1 \leq i \leq j \leq r$ , they consider the set $T_{i,j}$ of functions $\nu \colon [1,i] \times [j,r] \to [i,j]$ such that $\nu (k,l) \leq \nu (k', l')$ whenever $k\leq k', l \leq l'$ .

Let $1 \leq a \leq r$ . We only use the case $i=a, j=a+1$ and consider $T_{a, a+1}$ . In this case, any function $\nu \in T_{a, a+1}$ takes one of two values $a, a+1$ . We express this using Figure 1. The rectangle depicts the set $[1,a] \times [a+1, r]$ . The upper-left corner is $(1,a+1)$ , the lower-left corner is $(a,a+1)$ , the upper-right corner is $(1,r)$ and the lower-right corner is $(a, r)$ . Because of the condition on $\nu $ , there exists a bold line as in the picture such that $\nu $ takes the value a on the left (call the region L) and the value $a+1$ on the right (call the region R).

Figure 1

We look at the sum from $(1.6)$ [Reference Knight and Zelevinsky16]:

$$\begin{align*}\sum_{(k,l) \in [1,a]\times [a+1, r]} d_{\nu(k,l)+k-a, \nu(k,l)+l-a-1}. \end{align*}$$

This equals

$$\begin{align*}\sum_{(k,l) \in L} d_{k,l-1} + \sum_{(k,l) \in R} d_{k+1, l}. \end{align*}$$

Now consider Figure 2. The rectangle depicts the set $U=[1,a+1] \times [a, r]$ . Let $L'$ be the region L moved to the left by 1 and $R'$ be the region R moved down by 1. These are subsets of U, and the complement $V_{\nu }=U\setminus (L' \cup R')$ is shown in blue in the picture.

Figure 2

A path from $(a+1,a)$ to $(1,r)$ is a map $p\colon [0,r] \to \mathbb {Z} \times \mathbb {Z}$ satisfying the following conditions:

1. (1) $p(0)=(a+1,a)$ .

2. (2) For $i=1, \ldots , r$ , the element $p(i) \in \mathbb {Z} \times \mathbb {Z}$ is equal to $p(i-1)-(1,0)$ or $p(i-1)+(0,1)$ .

3. (3) $p(r)=(1,r)$ .

Then $V_{\nu }$ is equal to the image of a path from $[a+1, a]$ to $[1,r]$ . By sending $\nu $ to this path, we obtain a bijection from $T_{a, a+1}$ to the set $A_a$ of paths from $(a+1,a)$ to $(1,r)$ .

Notice now that the sum above is equal to

$$\begin{align*}\sum_{(k,l) \in L'} d_{k,l} +\sum_{(k,l) \in R'} d_{k,l} = \sum_{(k,l) \in U} d_{k,l} - \sum_{(k,l) \in V_{\nu}} d_{k,l}. \end{align*}$$

Conversely, given a path from $[a+1, a]$ to $[a,1]$ , we obtain a function $\nu \in T_{a, a+1}$ such that $V_{\nu }$ is the image of the given path. Thus, the right-hand side of (1.6) of [Reference Knight and Zelevinsky16] is equal to

$$\begin{align*}\sum_{(k,l) \in U} d_{k,l} -\max_{p \in A_a} \sum_{i=0}^{r} d_{p(i)}. \end{align*}$$

Notice that $\mathrm {Card}(\mathfrak {m}_a) = \sum _{(k,l) \in U}d_{k,l}$ . From this, we see that $\mathrm {Card}(\mathfrak {m}^{\sharp }_{(a)})$ is equal to

$$\begin{align*}\mathrm{Card}(\mathfrak{m}_a) - E_a(\mathfrak{m}), \end{align*}$$

where

$$\begin{align*}E_a(\mathfrak{m}) = \max_{p \in A_a} \sum_{i=0}^{r} d_{p(i)}. \end{align*}$$

For $p \in A_a$ , let $\mathfrak {m}_{a,p}$ denote the full-sub-multisegment of $\mathfrak {m}_a$ that consists of the segments $[a',b']_{\chi }$ in $\mathfrak {m}_a$ of the form $(a',b') = p(i)$ for some integer $i \in [0,r]$ . Then $\mathfrak {m}_{a,p}$ is totally ordered, and we have $\sum _{i=0}^{r} d_{p(i)} = \mathrm {Card}(\mathfrak {m}_{a,p})$ . By sending p to $\mathfrak {m}_{a,p}$ , we obtain a map from $A_a$ to the set $T_a$ of totally ordered full-sub-multisegments of $\mathfrak {m}_a$ . In general, this map is neither injective nor surjective. However, for any totally ordered full-sub-multisegment $\mathfrak {m}'$ of $\mathfrak {m}_a$ , there exists a path $p \in A_a$ such that $\mathfrak {m}'$ is a full-sub-multisegment of $\mathfrak {m}_{a,p}$ . In particular $\mathrm {Card}(\mathfrak {m}') \le \mathrm {Card}(\mathfrak {m}_{a,p})$ for this p.

Thus, we obtain an equality

$$\begin{align*}\max_{\mathfrak{m}' \in T_a} \mathrm{Card}(\mathfrak{m}') = \max_{p \in A_a} \mathrm{Card}(\mathfrak{m}_{a,p}) = E_a(\mathfrak{m}), \end{align*}$$

which completes the proof.

Proof of Proposition 2.7

We prove the claim by induction on $l(\mathfrak {m})$ .

Let $(V,N)$ , $(V_1,N_1)$ and $(V_2,N_2)$ be the $VN$ -pairs corresponding to the multisegments $\mathfrak {m}$ , $\mathfrak {m}_{\mathrm {max}}$ and $\mathfrak {m}^{\mathrm {max}}$ , respectively. Let us consider the sets $S(V,N)$ , $S(V_1,N_1)$ and $S(V_2,N_2)$ introduced in Section 3.1. Let us choose sufficiently general $L \in S(V,N)$ , $L_1 \in S(V_1,N_1)$ and $L_2 \in S(V_2,N_2)$ . Then the six multisegments $\mathfrak {m}^{\mathrm {ram}}$ , $(\mathfrak {m}^{-})^{\mathrm {ram}}$ , $(\mathfrak {m}_{\mathrm {max}})^{\mathrm {ram}}$ , $((\mathfrak {m}_{\mathrm {max}})^{-})^{\mathrm {ram}}$ , $(\mathfrak {m}^{\mathrm {max}})^{\mathrm {ram}}$ and $((\mathfrak {m}^{\mathrm {max}})^{-})^{\mathrm {ram}}$ correspond to the pairs $(\mathrm {Image}\, L, N|_{\mathrm {Image}\, L})$ , $(\mathrm {Image}\, L\circ N, N|_{\mathrm {Image}\, L \circ N})(-1)$ , $(\mathrm {Image}\, L_1, N_1|_{\mathrm {Image}\, L_1})$ , $(\mathrm {Image}\, L_1\circ N_1, N_1|_{\mathrm {Image}\, L_1 \circ N_1})(-1)$ , $(\mathrm {Image}\, L_2, N_2|_{\mathrm {Image}\, L_2})$ and $(\mathrm {Image}\, L_2\circ N_2, N_2|_{\mathrm {Image}\, L_2 \circ N_2})(-1)$ , respectively.

To prove the claim, it suffices to show that the pair $(\mathrm {Image}\, L, N|_{\mathrm {Image}\, L})$ is isomorphic to the pair $(\mathrm {Image}\, L_1 \oplus \mathrm {Image}\, L_2, N_1|_{\mathrm {Image}\, L_1} \oplus N_2|_{\mathrm {Image}\, L_2})$ . By the inductive hypothesis, the claim is true for the multisegment $\mathfrak {m}^{-}$ . Hence it follows from Lemma 3.6 that

$$\begin{align*}(\mathrm{Image}\, L\circ N, N|_{\mathrm{Image}\, L\circ N}) \cong (\mathrm{Image}\, L_1\circ N_1 \oplus \mathrm{Image}\, L_2 \circ N_2, N_1|_{\mathrm{Image}\, L_1 \circ N_1} \oplus N_2|_{\mathrm{Image}\, L_2 \circ N_2}). \end{align*}$$

Hence by Lemma 3.1, it suffices to show that the graded vector space $\mathrm {Image}\, L$ is isomorphic to the graded vector space $\mathrm {Image}\, L_1 \oplus \mathrm {Image}\, L_2$ .

Let $\mathfrak {m}_a$ and $\mathfrak {m}^{\sharp }_{(a)}$ be as in Proposition 3.7. Note that the dimension of the degree-a-part of $\mathrm {Image}\, L$ is equal to $\mathrm {Card}(\mathfrak {m}^{\sharp }_{(a)})$ . Let $\mathfrak {m}'$ be a totally ordered full-sub-multisegment of $\mathfrak {m}_a$ with the maximum cardinality. When $\mathfrak {m}_a$ is nonempty, the maximal segment $\Delta _1$ of $\mathfrak {m}'$ must belong to $\mathfrak {m}_{\mathrm {max}}$ , since otherwise one can find a totally ordered full-sub-multisegment of $\mathfrak {m}_a$ that strictly contains $\mathfrak {m}'$ by adding to $\mathfrak {m}'$ a segment of $\mathfrak {m}_{\mathrm {max}}$ that contains $\Delta _1$ , which is a contradiction. It is then easy to see that

• • when $\mathfrak {m}_a$ is nonempty, $\mathfrak {m}' - \Delta _1$ is a totally ordered full-sub-multisegment of $(\mathfrak {m}^{\mathrm {max}})_a$ with the maximum cardinality; and

• • $\Delta _1$ , which is regarded as a multisegment with $\mathrm {Card}(\Delta _1) = 1$ , is a totally ordered full-sub-multisegment of $(\mathfrak {m}_{\mathrm {max}})_a$ with the maximum cardinality.

Thus, it follows from Proposition 3.7 that the dimension of the degree-a-part of $\mathrm {Image}\, L$ is equal to the sum of those of $\mathrm {Image}\, L_1$ and $\mathrm {Image}\, L_2$ , as desired.

Proof. We can easily see that the lexicographic order on $\Lambda _n$ is compatible with the monoid structure on $\Lambda _n$ given by $+$ . Hence the claim follows.

Proof. We may assume $M= \mathfrak {o}/\mathfrak {p}^{\lambda _1} \oplus \cdots \oplus \mathfrak {o}/\mathfrak {p}^{\lambda _n}$ . Since $(\ )^{\vee }$ commutes with finite direct sums, we may further assume that $M = \mathfrak {o}/\mathfrak {p}^{\lambda }$ . Then we have $M^{\vee } \cong \mathfrak {p}^{-\lambda }/\mathfrak {o}$ . Hence by choosing a uniformiser $\varpi \in \mathfrak {p}$ , we obtain a desired isomorphism $M \cong M^{\vee }$ .

Proof. Since an injective homomorphism $M' \hookrightarrow M$ induces a surjective homomorphism $M^{\vee } \twoheadrightarrow {M'}^{\vee }$ , claim (1) follows from claim (2) and Lemma 4.2. Let us prove claim (2) below.

Let $M, M'$ be $\mathfrak {o}$ -modules of finite length, and suppose that there exists a surjective homomorphism $M \twoheadrightarrow M'$ . Let us take an integer $n \geq 1$ such that both $[M]$ and $[M']$ belong to $|\mathcal {C}^n|$ . We prove the claim by induction on n. If $n=1$ , then the claim is clear. We assume $n>1$ . Let us write $\mathrm {seq}_n([M])=(\lambda _1,\ldots ,\lambda _n)$ and $\mathrm {seq}_n([M'])=(\lambda ^{\prime }_1,\ldots ,\lambda ^{\prime }_n)$ .

First, suppose that $\lambda _1> \lambda ^{\prime }_1$ . Then we have $[M'] < [M]$ as claimed. Next, suppose that $\lambda _1 = \lambda ^{\prime }_1$ . Then both $M/\mathfrak {p}^{\lambda _1} M$ and $M'/\mathfrak {p}^{\lambda _1} M'$ are isomorphic to $(\mathfrak {o}/\mathfrak {p}^{\lambda _1})^{\oplus n}$ , and we have $[M] = [\mathfrak {p}^{\lambda _1} M] \vee [(\mathfrak {o}/\mathfrak {p}^{\lambda _1})^{\oplus n}]$ and $[M'] = [\mathfrak {p}^{\lambda _1} M'] \vee [(\mathfrak {o}/\mathfrak {p}^{\lambda _1})^{\oplus n}]$ . Note that the surjective homomorphism $M \twoheadrightarrow M'$ induces a surjective homomorphism $\mathfrak {p}^{\lambda _1} M \twoheadrightarrow \mathfrak {p}^{\lambda _1} M'$ . Hence by Lemma 4.1, we are reduced to proving claim (2) for $\mathfrak {p}^{\lambda _1} M$ and $\mathfrak {p}^{\lambda _1} M'$ . Since both $[\mathfrak {p}^{\lambda _1} M]$ and $[\mathfrak {p}^{\lambda _1} M']$ belong to $|\mathcal {C}^{n-1}|$ , the inductive hypothesis proves the claim in the case where $\lambda _1 = \lambda ^{\prime }_1$ .

Finally, suppose that $\lambda _1 < \lambda ^{\prime }_1$ . Again in this case, the surjective homomorphism $M \twoheadrightarrow M'$ induces a surjective homomorphism $\mathfrak {p}^{\lambda _1} M \twoheadrightarrow \mathfrak {p}^{\lambda _1} M'$ . Note that $\mathfrak {p}^{\lambda _1} M$ is generated by fewer than n elements, whereas the minimum number of generators of $\mathfrak {p}^{\lambda _1} M'$ is equal to n. This leads to a contradiction.

Proof. Let n, $n'$ and $n"$ denote the minimal numbers of generators of the $\mathfrak {o}$ -modules M, $M'$ and $M"$ , respectively. We prove the claim by induction on $n'+n"$ .

If $n'+n"=0$ , then we have $M'=M"=0$ , and the claim is clear. Since $M \to M"$ and $M^{\vee } \to M^{\prime \vee }$ are surjective, we have $n \geq n"$ and $n \geq n'$ . If $n> \max \{n',n"\}$ , then the claim is obvious. Hence we may assume that $n=\max \{n',n"\}$ . By considering the short exact sequence

$$\begin{align*}0 \to M^{\prime\prime\vee} \to M^{\vee} \to M^{\prime\vee} \to 0 \end{align*}$$

instead of equation (4.1), if necessary, we may further assume that $n=n"$ . Let us write $\mathrm {seq}_n(M")=(\lambda ^{\prime \prime }_1,\ldots ,\lambda ^{\prime \prime }_n)$ and $I = \mathfrak {p}^{\lambda ^{\prime \prime }_1}$ . Then both $M/I M$ and $M"/I M"$ are isomorphic to $(\mathfrak {o}/I)^{\oplus n}$ , and we have $[M]=[I M] \vee [(\mathfrak {o}/I)^{\oplus n}]$ and $[M"]=[I M"] \vee [(\mathfrak {o}/I)^{\oplus n}]$ . Moreover, equation (4.1) induces a short exact sequence

$$\begin{align*}0 \to M' \to IM \to IM" \to 0. \end{align*}$$

Since $\mathrm {seq}_n(IM")=(0,\lambda ^{\prime \prime }_2-\lambda ^{\prime \prime }_1,\ldots ,\lambda ^{\prime \prime }_n-\lambda ^{\prime \prime }_1)$ , the minimal number of generators of $IM"$ is strictly smaller than $n"$ . Hence, by induction, we have $[IM] \geq [M'] \vee [IM"]$ . By adding $[(\mathfrak {o}/I)^{\oplus n}]$ to both sides and using Lemma 4.1, we obtain the desired inequality.

Proof. Let us write $I = \mathfrak {p}^{\lambda }$ . Let us choose a positive integer n such that $[M] \in |\mathcal {C}^n|$ . Let us write $\mathrm {seq}_n([M]) = (\lambda _1,\ldots ,\lambda _n)$ . For $i=1,\ldots ,n$ , set $\lambda ^{\prime \prime }_i = \min \{\lambda ,\lambda _i\}$ . Then $M/IM$ is isomorphic to $\bigoplus _{i=1}^n\mathfrak {o}/\mathfrak {p}^{\lambda ^{\prime \prime }_i}$ and $IM$ is isomorphic to $\bigoplus _{i=1}^n \mathfrak {p}^{\lambda ^{\prime \prime }_i}/\mathfrak {p}^{\lambda _i}$ . Thus, we have $[M]=[IM]\vee [M/IM]$ , as desired.

Proof. First, we prove the existence and uniqueness of N. Let n, $n'$ and $n"$ denote the minimal numbers of generators of the $\mathfrak {o}$ -modules M, $M'$ and $M"$ , respectively. We prove the claim by induction on $n'+n"$ .

If $n'+n"=0$ , then we have $M=M'=M"=0$ , and the claim is obvious. The relation $[M]=[M']\vee [M"]$ implies that $n=\max \{n',n"\}$ . By considering $M^{\vee }$ instead of M if necessary, we may assume that $n=n"$ . Let us write $\mathrm {seq}_n(M")=(\lambda ^{\prime \prime }_1,\ldots ,\lambda ^{\prime \prime }_n)$ and $I = \mathfrak {p}^{\lambda ^{\prime \prime }_1}$ . Then for any $\mathfrak {o}$ -submodule $N \subset M$ satisfying $[M/N]=[M"]$ , we have $N \subset IM$ . Since $\mathrm {seq}_n(IM")=(0,\lambda ^{\prime \prime }_2-\lambda ^{\prime \prime }_1,\ldots ,\lambda ^{\prime \prime }_n-\lambda ^{\prime \prime }_1)$ , the minimal number of generators of $IM"$ is strictly smaller than $n"$ . Hence, by induction, there exists a unique $\mathfrak {o}$ -submodule $N \subset IM$ satisfying $[N] = [M']$ and $[IM/N]=[IM"]$ . Since N is contained in $IM$ , the $\mathfrak {o}$ -module $I(M/N)$ is isomorphic to $IM/N$ , and hence $(M/N)/I(M/N)$ is isomorphic to $M/IM$ . It follows from Lemma 4.5 that we have $[M/N] = [IM/N] \vee [M/IM] = [M"]$ . Hence the claim follows.

Finally, let us prove the last assertion of the proposition. Let $N' \subset M$ be an $\mathfrak {o}$ -submodule other than N. Suppose that $[N'] \geq [M']$ and $[M/N'] \geq [M"]$ . Since $N' \neq N$ , we have either $[N']>[M']$ or $[M/N']> [M"]$ . Hence it follows from Lemma 4.1 and Proposition 4.4 that

$$\begin{align*}[M] \geq [N'] \vee [M/N']> [M'] + [M"] = [M], \end{align*}$$

which is a contradiction. Hence we have either $[N'] < [M']$ or $[M/N'] < [M"]$ . This completes the proof.

Proof. We prove the existence and uniqueness of $\mathrm {F}^0_{\bullet }M$ by induction on r. If $r = 1$ , it is obvious. If $r> 1$ , set $[M'] = [M_1]$ and $[M"] = [M_2] \vee \dots \vee [M_{r}]$ . By Proposition 4.6, there exists a unique $\mathfrak {o}$ -submodule N of M such that $[N] = [M_1]$ and $[M/N] = [M_2] \vee \dots \vee [M_{r}]$ . By the inductive hypothesis, we have a unique filtration $\mathrm {F}^0_{\bullet }(M/N)$ satisfying the conditions with respect to $[M/N] = [M_2] \vee \dots \vee [M_{r}]$ . By setting $\mathrm {F}^0_{i+1}M$ to be the inverse image of $\mathrm {F}^0_i(M/N)$ for $1 \leq i \leq r-1$ , and $\mathrm {F}^0_{1}M = N$ , we obtain $\mathrm {F}^0_{\bullet }M$ .

The last assertion follows from the same argument as in the proof of Proposition 4.6.

Proof. Let us take an arbitrary lift $\widetilde {y}' \in M$ of y and set $x'=f(\widetilde {y}')$ . Since the images of x and $x'$ coincide in $N/f(M')$ , there exists $z \in M'$ satisfying $x-x' = f(z)$ . Then the element $\widetilde {y} = \widetilde {y'} + z \in M$ has the desired property.

Proof. Since N is of finite length over a noetherian local ring $\mathfrak {o}$ , one can take a projective cover $\beta \colon P \to N$ of N (see [Reference Anderson and Fuller1, 17.16 Examples (3)]). Then there exist homomorphisms $\gamma \colon L \to P$ and $\gamma ' \colon L' \to P$ satisfying $f = \beta \circ \gamma $ and $f' = \beta \circ \gamma '$ . Since projective covers are essential surjections, the homomorphisms $\gamma $ and $\gamma '$ are surjective. Hence by the projectivity of P, one can choose a right inverse s and $s'$ of $\gamma $ and $\gamma '$ , respectively. Since $\mathrm {Ker}\, \gamma $ and $\mathrm {Ker}\, \gamma '$ are free $\mathfrak {o}$ -modules of the same rank, there exists an isomorphism $\alpha ' \colon \mathrm {Ker}\, \gamma \xrightarrow {\cong } \mathrm {Ker}\, \gamma '$ of $\mathfrak {o}$ -modules. By taking the direct sum of $\alpha '$ and the isomorphism $s(P) \xrightarrow {\cong } s'(P)$ given by $s' \circ \gamma $ , we obtain a desired isomorphism $\alpha \colon L \to L'$ .

Proof. Let $L' = \mathfrak {o}^{\oplus n}$ , and let $f' \colon L' \twoheadrightarrow N$ denote the surjection that sends the standard basis of $L'$ to the elements $x_1,\ldots ,x_n$ . By applying Lemma 4.9, we obtain an isomorphism $\alpha \colon L \xrightarrow {\cong } L'$ satisfying $f=f'\circ \alpha $ . Then the image under $\alpha ^{-1}$ of the standard basis of $L'$ gives a desired basis of L.