Proof. Note that ${V = H_s \bigoplus \mathbb {F} \alpha _s}$ . Any vector v can be written in the form ${v = v_s + c_v \alpha _s}$ where $v_s \in H_s$ and $c_v \in \mathbb {F}$ . Then

$$ \begin{align*}s \cdot v = v_s + \lambda_s c_v \alpha_s = v + (\lambda_s - 1) c_v \alpha_s.\end{align*} $$

The linear function $f: v \mapsto (\lambda _s - 1) c_v$ is the desired function.

Proof. Note that $\{\alpha _s, v_1, \dots , v_{n-1}\}$ is a basis of V. Denote by $B^+$ and $B^-$ the two sets of vectors in (3.2) and (3.3), respectively. Then the disjoint union $B^+ \cup B^-$ is a basis of $\bigwedge ^d V$ by Lemma 3.1. Clearly, $B^+ \subseteq V_{d,s}^+$ and $B^- \subseteq V_{d,s}^-$ . Therefore, ${\bigwedge ^d V = V_{d,s}^+ \bigoplus V_{d,s}^-}$ and the result follows.

Proof. Point (1) in Corollary 3.3 is directly derived from Lemma 3.2. For (2), suppose ${s \cdot \alpha _i = \alpha _i + c_i \alpha _s}$ ( $c_i \in \mathbb {F}$ ) for $i = 2, \dots , n$ (see Lemma 2.2). Then,

$$ \begin{align*} s \cdot (\alpha_s \wedge \alpha_{i_1} \wedge \dots \wedge \alpha_{i_{d-1}}) & = (\lambda_s \alpha_s) \wedge (\alpha_{i_1} + c_{i_1} \alpha_s) \wedge \dots \wedge (\alpha_{i_{d-1}} + c_{i_{d-1}} \alpha_s) \\ & = \lambda_s \alpha_s \wedge \alpha_{i_1} \wedge \dots \wedge \alpha_{i_{d-1}}. \end{align*} $$

Thus, $\alpha _s \wedge \alpha _{i_1} \wedge \dots \wedge \alpha _{i_{d-1}} \in V_{d,s}^-$ . Note that by Lemma 3.1, the $\binom {n-1}{d-1}$ vectors in (3.4) are linearly independent, and that $\dim V_{d,s}^- = \binom {n-1}{d-1}$ by Lemma 3.2. Thus, the vectors in (3.4) form a basis of $V_{d,s}^-$ .

Proof. By Corollary 3.3(2), $V_{d,s_i}^-$ ( $1 \le i \le k$ ) has a basis

$$ \begin{align*}B_i := \{\alpha_{j_1} \wedge \dots \wedge \alpha_{j_d} \mid 1 \le j_1 < \dots < j_d \le n, \text{ and}\ j_l = i\ \text{for some}\ l\}.\end{align*} $$

Note that by Lemma 3.1, the ambient space $\bigwedge ^d V$ has a basis

$$ \begin{align*}B : = \{\alpha_{j_1} \wedge \dots \wedge \alpha_{j_d} \mid 1 \le j_1 < \dots < j_d \le n\}\end{align*} $$

and $B_i \subseteq B$ , for all $i = 1, \dots , k$ . By Lemma 3.4,

$$ \begin{align*}\bigcap_{i=1}^{k} V_{d,s_i}^- = \bigcap_{i=1}^{k} \langle B_i \rangle = \bigg\langle \bigcap_{i=1}^k B_i \bigg\rangle.\end{align*} $$

If $0 \le d < k$ , then $\bigcap _{1 \le i \le k} B_i = \emptyset $ and $\bigcap _{1 \le i\le k} V_{d, s_i}^- = 0$ . If $k \le d \le n$ , then

$$ \begin{align*}\bigcap_{i=1}^k B_i = \{\alpha_1 \wedge \dots \wedge \alpha_k \wedge \alpha_{j_{k+1}} \wedge \dots \wedge \alpha_{j_d} \mid k+1 \le j_{k+1} < \dots < j_d \le n\}\end{align*} $$

and $\bigcap _{1 \le i \le k} B_i$ is a basis of $\bigcap _{1 \le i\le k} V_{d, s_i}^-$ .

Proof. If $\bigwedge ^k V \simeq \bigwedge ^l V$ , then $\dim \bigwedge ^k V = \dim \bigwedge ^l V$ and $\dim V_{k,s}^+ = \dim V_{l,s}^+$ . By Lemmas 3.1 and 3.2, this is equivalent to

$$ \begin{align*} \binom{n}{k} = \binom{n}{l}, \quad \binom{n-1}{k} = \binom{n-1}{l}. \end{align*} $$

The two equalities force $k = l$ .

Proof. We do induction downwards on $\lvert I \cap J \rvert $ . If $\lvert I \cap J \rvert =d$ , then $I=J$ and there is nothing to prove.

If $I \ne J$ , then there exist vertices $r \in I \setminus J$ and $t \in J \setminus I$ . Since G is connected, there is a path connecting r and t, say,

$$ \begin{align*}(r = r_0, r_1,r_2, \dots, r_l = t).\end{align*} $$

Let $0 = i_0 < i_1 < \dots < i_k < l$ be the indices such that $\{r_{i_0}, r_{i_1}, \dots , r_{i_k} \}$ is the set of vertices in I on this path, that is, $\{r_{i_0}, \dots , r_{i_k} \} = \{r_i \mid 0 \le i < l, r_i \in I\}$ .

Clearly, $r_{i_k+1}, r_{i_k+2}, \dots , r_l \notin I$ . So beginning with I, we can move $r_{i_k}$ to $r_{i_k+1}$ , then to ${r_{i_k+2}}$ , and finally to t. Therefore, the set $I_1 := (I \setminus \{r_{i_k}\}) \cup \{t\}$ can be obtained from I by finite steps of moves. Similarly, from $I_1$ , we can move $r_{i_{k-1}}$ to $r_{i_k}$ so that we obtain

$$ \begin{align*}I_2 : = (I_1 \setminus \{r_{i_{k-1}}\}) \cup \{r_{i_k}\} = (I \setminus \{r_{i_{k-1}}\}) \cup \{t\}.\end{align*} $$

Do this recursively, and finally we get $I_{k+1} := (I \setminus \{r_{i_{0}}\}) \cup \{t\} = (I \setminus \{r\}) \cup \{t\}$ from I by finite steps of moves.

Moreover, we have $\lvert I_{k+1} \cap J \rvert = \lvert I \cap J \rvert + 1$ . By the induction hypothesis, J can be obtained from $I_{k+1}$ by finite steps of moves. It follows that J can be obtained from I by finite steps of moves.

Proof. For two vertices $r,t \in S$ , define the distance $\operatorname {d}(r,t)$ in G to be

$$ \begin{align*} \operatorname{d}(r,t) := \min \{m \in \mathbb{N} & \mid \exists r_1, \dots, r_{m-1}\in S \text{ such that } (r, r_1, \dots, r_{m-1}, t) \text{ is a path in } G\}. \end{align*} $$

Let $m = \max \{\operatorname {d}(r,t) \mid r,t \in I\}$ , and suppose $s, s^\prime \in I$ such that $\operatorname {d}(s, s^\prime ) = m$ .

We claim first that

(4.1) $$ \begin{align} \text{for any}\ r \in I \setminus\{s\}, r\ \text{is connected to}\ s^\prime\ \text{in the subgraph}\ G(S \setminus \{s\}). \end{align} $$

Otherwise, any path in G connecting r and $s^\prime $ (note that G is connected) must pass through s. It follows that $\operatorname {d}(r, s^\prime )> \operatorname {d}(s, s^\prime ) = m$ , which contradicts our choice of m.

Next we claim that

(4.2) $$ \begin{align} \text{for any}\ t \in S \setminus \{s\},\ t\ \text{is connected to}\ s^\prime\ \text{in the subgraph}\ G(S \setminus \{s\}), \end{align} $$

and therefore, $G(S \setminus \{s\})$ is connected.

In (4.2), the case where $t \in I \setminus \{s\}$ has been settled in the claim (4.1). Thus, we may assume $t \in S \setminus I$ . Let $(t = t_0, t_1, \dots , t_p = s)$ be a path of minimal length in G connecting t and s. Then, $t_1, \dots , t_{p-1} \in S \setminus \{s\}$ . In particular, t is connected to $t_{p-1}$ in $G(S \setminus \{s\})$ . If $t_{p-1} \in I$ , then by the claim (4.1), $t_{p-1}$ is connected to $s^\prime $ in $G(S \setminus \{s\})$ . Thus, t is connected to $s^\prime $ in $G(S \setminus \{s\})$ as desired. If $t_{p-1} \notin I$ , then, since $\{t_{p-1}, s\} \in E$ , there is another vertex $r \in I \setminus \{s\}$ such that $\{t_{p-1},r\} \in E$ . By the claim (4.1) again, r is connected to $s^\prime $ in ${G(S \setminus \{s\})}$ , and so is t. See Figure 1 for an illustration.

Figure 1 Illustration for the proof of Lemma 4.4.

Proof. Otherwise, suppose $S = I \sqcup J$ such that for any $i \in I$ and $j \in J$ , $\{i, j\}$ is never an edge, that is, $s_j \cdot \alpha _i = \alpha _i$ . If $V = V_I :=\langle \alpha _i \mid i \in I \rangle $ , then for any $j \in J$ , $\alpha _j$ is a linear combination of $\{\alpha _i \mid i \in I\}$ . It follows that $s_j \cdot \alpha _j = \alpha _j$ , which is absurd. Therefore, ${V_I \ne V}$ .

By Lemma 2.2, $s_i \cdot V_I \subseteq V_I$ for any $i \in I$ . However, for any $j \in J$ , $s_j$ acts trivially on $V_I$ . Since W is generated by $\{s_1, \dots , s_k\}$ , $V_I$ is closed under the action of W, that is, $V_I$ is a proper submodule. This contradicts the assumption that V is a simple W-module.

Proof. Let $U = \langle \alpha _1, \dots , \alpha _k \rangle \subseteq V$ . By Lemma 2.2, $s_i \cdot U \subseteq U$ for any $i \in S$ . Thus, U is a W-submodule. However, V is a simple W-module. So $U = V$ .

Proof. Suppose we have found a subset $J \subseteq S$ such that:

1. (a) V is spanned by $\{\alpha _i \mid i \in J\}$ ;

2. (b) the subgraph $G(J)$ is connected.

For example, S itself is such a subset by Claims 5.1 and 5.2. If the vectors ${\{\alpha _i \mid i\in J\}}$ are linearly independent, then we are done.

Now suppose $\{\alpha _i \mid i\in J\}$ are linearly dependent. By a permutation of indices, we may assume $J = \{1, \dots , h\}$ , $h \le k$ , and

(5.2) $$ \begin{align} c_1 \alpha_1 + \dots + c_l \alpha_l = 0 \quad\text{for some}\ c_1, \dots, c_l \in \mathbb{F}^\times, l \le h. \end{align} $$

If there exists $j \in J$ such that $j \ge l+1$ and $s_j \cdot \alpha _i \ne \alpha _i$ for some $i \le l$ , then

$$ \begin{align*}s_j \cdot (c_1 \alpha_1 + \dots + \widehat{c_i \alpha_i} + \dots + c_l \alpha_l) \ne c_1 \alpha_1 + \dots + \widehat{c_i \alpha_i} + \dots + c_l \alpha_l.\end{align*} $$

Here, $\widehat {c_i \alpha _i}$ means this term is omitted. Thus, there is an index $i^\prime $ with $i^\prime \le l$ and $i^\prime \ne i$ such that $s_j \cdot \alpha _{i^\prime } \ne \alpha _{i^\prime }$ . In other words, if $l+1 \le j \le h$ , then one of the following is satisfied:

1. (1) for any $i \le l$ , $\{i,j\}$ is never an edge;

2. (2) there exist at least two indices $i, i^\prime \le l$ such that $\{i,j\}, \{i^\prime , j\} \in E$ .

Applying Lemma 4.4 to the subset $\{1, \dots , l\} \subseteq J$ , we see that there is an index $i_0 \le l$ such that the subgraph $G(J \setminus \{i_0\})$ is connected. Moreover, V is spanned by ${\{\alpha _i \mid i \in J \setminus \{i_0\}\}}$ by our assumption (5.2). Thus, $J \setminus \{i_0\}$ satisfies the conditions (a) and (b), and $J \setminus \{i_0\}$ has a smaller cardinality than J.

Apply the arguments above recursively. Finally, we will obtain a subset $I \subseteq S$ as claimed.

Proof. Apply Proposition 3.5(2) to $s_{i_1}, \dots , s_{i_d} \in W$ .

Proof. To simplify notation, we assume $I_1 = \{1, \dots , d\}$ , $I_2 = (I_1 \setminus \{d\}) \cup \{d+1\}$ and $\{d, d+1\} \in E$ is an edge. In view of Lemma 2.2, for $i = 1, \dots , d$ , we assume

$$ \begin{align*}s_{d+1} \cdot \alpha_i = \alpha_i + c_i \alpha_{d+1}, \quad c_i \in \mathbb{F}. \end{align*} $$

Then, $c_d \ne 0$ . We have

$$ \begin{align*} s_{d+1} \cdot (\alpha_1 \wedge \dots \wedge \alpha_d) & = (\alpha_1 + c_1 \alpha_{d+1}) \wedge \dots \wedge (\alpha_d + c_d \alpha_{d+1})\\ & = \alpha_1 \wedge \dots \wedge \alpha_d + \sum_{i=1}^{d} (-1)^{d-i} c_i \cdot \alpha_1 \wedge \dots \wedge \widehat{\alpha}_i \wedge \dots \wedge \alpha_d \wedge \alpha_{d+1}. \end{align*} $$

Hence,

$$ \begin{align*} \varphi \bigl(s_{d+1} & \cdot (\alpha_1 \wedge \dots \wedge \alpha_d)\bigr) = \varphi \bigl(\alpha_1 \wedge \dots \wedge \alpha_d + \sum_{i=1}^{d} (-1)^{d-i} c_i \cdot \alpha_1 \wedge \dots \wedge \widehat{\alpha}_i \wedge \dots \wedge \alpha_{d+1}\bigr) \\ &\hspace{-10.2pt} = \gamma_{1,\dots, d} \cdot \alpha_1 \wedge \dots \wedge \alpha_d + \sum_{i=1}^{d} (-1)^{d-i} c_i \gamma_{1,\dots,\widehat{i},\dots, d+1} \cdot \alpha_1 \wedge \dots \wedge \widehat{\alpha}_i \wedge \dots \wedge \alpha_{d+1} \end{align*} $$

and also equals

$$ \begin{align*} s_{d+1} &\cdot \varphi(\alpha_1 \wedge \dots \wedge \alpha_d) = \gamma_{1,\dots, d} s_{d+1} \cdot (\alpha_1 \wedge \dots \wedge \alpha_d) \\ & = \gamma_{1,\dots, d} \cdot \alpha_1 \wedge \dots \wedge \alpha_d + \sum_{i=1}^{d} (-1)^{d-i} c_i \gamma_{1,\dots, d} \cdot \alpha_1 \wedge \dots \wedge \widehat{\alpha}_i \wedge \dots \wedge \alpha_{d+1}.\qquad\quad\qquad \end{align*} $$

Note that $c_d \ne 0$ , and that the vectors involved in the equation above are linearly independent. Thus, we have $\gamma _{1,\dots , d} = \gamma _{1,\dots , d-1, d+1}$ which is what we want.

Proof. We do induction on k and begin with the case $k = 2$ . Let $I_0$ be a basis of $H_1 \cap H_2$ . Extend $I_0$ to a basis of $H_1$ , say, $I_0 \sqcup I_1$ , and to a basis of $H_2$ , say, $I_0 \sqcup I_2$ . Then $I_0 \sqcup I_1 \sqcup I_2$ is a basis of $H_1 + H_2$ . Further, extend $I_0 \sqcup I_1 \sqcup I_2$ to a basis of V, say, $I_0 \sqcup I_1 \sqcup I_2 \sqcup I_3$ .

We define a total order $\le $ on the set of vectors $I_0 \sqcup I_1 \sqcup I_2 \sqcup I_3$ , and we write $v_1 < v_2$ if $v_1 \le v_2$ and $v_1 \ne v_2$ . By Lemma 3.1, B, $B_1$ , $B_2$ , $B_0$ are bases of $\bigwedge ^d V$ , $\bigwedge ^d H_1$ , $\bigwedge ^d H_2$ , $\bigwedge ^d (H_1 \cap H_2)$ , respectively, where

$$ \begin{align*} B & := \{v_1 \wedge \dots \wedge v_d \mid v_1, \dots, v_d \in I_0 \sqcup I_1 \sqcup I_2 \sqcup I_3, \text{ and } v_1 < \dots < v_d\}, \\ B_1 & := \{v_1 \wedge \dots \wedge v_d \mid v_1, \dots, v_d \in I_0 \sqcup I_1, \text{ and } v_1 < \dots < v_d\}, \\ B_2 & := \{v_1 \wedge \dots \wedge v_d \mid v_1, \dots, v_d \in I_0 \sqcup I_2, \text{ and } v_1 < \dots < v_d\}, \\ B_0 & := \{v_1 \wedge \dots \wedge v_d \mid v_1, \dots, v_d \in I_0, \text{ and } v_1 < \dots < v_d\}. \end{align*} $$

(The sets $B_1, B_2, B_0$ may be empty.) Moreover, $B_1, B_2 \subseteq B$ , and $B_0 = B_1 \cap B_2$ . Apply Lemma 3.4 to the vector space $\bigwedge ^d V$ . We obtain

$$ \begin{align*}\bigg(\bigwedge^d H_1\bigg) \cap \bigg(\bigwedge^d H_2\bigg) = \langle B_1 \rangle \cap \langle B_2 \rangle = \langle B_1 \cap B_2 \rangle = \langle B_0 \rangle = \bigwedge^d (H_1 \cap H_2).\end{align*} $$

For $k \ge 3$ , by the induction hypothesis,

$$ \begin{align*} \bigcap_{i=1}^{k} \bigg(\bigwedge^d H_i\bigg) = \bigg( \bigcap_{i=1}^{k-1} \bigg(\bigwedge^d H_i\bigg) \bigg) \cap \bigg(\bigwedge^d H_k\bigg) & = \bigg( \bigwedge^d \bigg(\bigcap_{i=1}^{k-1} H_i\bigg) \bigg) \cap \bigg(\bigwedge^d H_k\bigg) \\ & = \bigwedge^d \bigg( \bigg(\bigcap_{i=1}^{k-1} H_i\bigg) \cap H_k \bigg) = \bigwedge^d \bigg(\bigcap_{i=1}^{k} H_i\bigg) \end{align*} $$

as desired.

Proof. By Corollary 3.3(1), $V_{d,s_i}^+ = \bigwedge ^d H_i$ for all i. Therefore,

$$ \begin{align*} \bigg\{v \in \bigwedge^d V \mid s_i \cdot v = v \mbox{ for all } i\bigg\} = \bigcap_{i=1}^{k} V_{d,s_i}^+ = \bigcap_{i=1}^{k} \bigg(\bigwedge^d H_i\bigg) = \bigwedge^d \bigg(\bigcap_{i=1}^{k} H_i\bigg). \end{align*} $$

The last equality follows from Lemma 6.1.

Proof. Fix an identification of linear spaces $\bigwedge ^n V \simeq \mathbb {F}$ . For any $d = 0, 1, \dots , n$ , we define a bilinear map

$$ \begin{align*} f: \bigg(\bigwedge^{n-d} V\bigg) \times \bigg(\bigwedge^{d} V\bigg) & \to \bigwedge^{n} V \simeq \mathbb{F}, \\ (u,v) & \mapsto u \wedge v. \end{align*} $$

Clearly, this induces an isomorphism of linear spaces

$$ \begin{align*} \varphi: \bigwedge^{n-d} V & \xrightarrow{\sim} \bigg(\bigwedge^{d} V\bigg)^*, \\ u & \mapsto f(u,-). \end{align*} $$

Note that for any $w \in W$ , $u \in \bigwedge ^{n-d} V$ , $v \in \bigwedge ^{d} V$ , we have

$$ \begin{align*}f(w \cdot u, w \cdot v) = (w\cdot u) \wedge (w \cdot v) = w \cdot (u \wedge v) = \det(\rho(w)) u \wedge v = \det(\rho(w)) f(u,v).\end{align*} $$

Therefore, for $w \in W$ , $u \in \bigwedge ^{n-d} V$ ,

$$ \begin{align*} \varphi(w \cdot u) = f(w\cdot u, -) = f(w\cdot u, (w w ^{-1} \cdot -)) = \det (\rho(w))f(u, (w^{-1} \cdot -)). \end{align*} $$

This implies $u \mapsto f(u,-) \otimes 1$ is an isomorphism $\bigwedge ^{n-d} V \xrightarrow {\sim } (\bigwedge ^d V)^* \bigotimes (\det \circ \rho )$ of W-modules.

Question 7.1. Can we remove the technical condition (4) in the statement of Theorem 1.2?

Question 7.2. Is it possible to find two nonisomorphic simple W-modules $V_1, V_2$ satisfying the conditions of Theorem 1.2, and two integers $d_1, d_2$ with $0 < d_i < \dim V_i$ , such that $\bigwedge ^{d_1} V_1 \simeq \bigwedge ^{d_2} V_2$ as W-modules?

Question 7.3. What kinds of simple W-modules V have the property that the modules $\bigwedge ^d V$ , $0 \le d \le \dim V$ , are simple and pairwise nonisomorphic? Can we formulate any other sufficient conditions (in addition to Theorem 1.2) or any necessary conditions?