Wisconsin Bourbaki Seminar

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 24 March 2014

Notes and References

This is an excerpt from notes of the Wisconsin Bourbaki Seminar, Fall 1993. The notes were written by Oliver Eng, Susan Hollingsworth, Mark Logan, Arun Ram and Louis Solomon.

§2

Let K be a commutative ring with unit, let E be a K-module, and let E* be the dual of E. Let ϕ denote the canonical homomorphism from E⊗E* to End(E).
1. A pseudo-reflection of $E$ is a non-identity element of $End (E)$ of the form $s_{x, y^{*}} = 1 - ϕ (x \otimes y^{*}),$ where $x \in E$ and $y^{*} \in E^{*} .$ A pseudo-reflection $s$ is called a reflection if one may choose $x, y^{*}$ such that $⟨ x, y^{*} ⟩ = 2;$ show that then one has that $s^{2} = 1$ and that $s (x) = - x .$
2. Let $x \in E,$ $y^{*} \in E^{*}$ such that $⟨ x, y^{*} ⟩ = 1$ and let $s$ be the reflection corresponding to the pair $(2 x, y^{*}) .$ Show that $E$ is the direct sum of the submodule $K x$ generated by $x$ and the orthogonal $H$ of $y^{*} .$ Show that $K x$ is free with base $x,$ and that $s$ is equal to $1$ on $H$ and to $- 1$ on $K x .$
Solution.
1. The map $ϕ$ is given by $\begin{matrix} ϕ : & E \otimes E^{*} & ⟶ & End (E) \\ x \otimes y^{*} & ⟼ & x ⟨ \cdot, y^{*} ⟩ . \end{matrix}$ Since $⟨, ⟩$ is bilinear $\begin{matrix} s_{x, y^{*}}^{2} & = & (1 - ϕ (x \otimes y^{*})) (1 - ϕ (x \otimes y^{*})) \\ = & 1 - 2 ϕ (x \otimes y^{*}) + ϕ (x \otimes y^{*}) (ϕ (x \otimes y^{*})) \\ = & 1 - 2 ϕ (x \otimes y^{*}) + x ⟨ x ⟨ \cdot, y^{*} ⟩, y^{*} ⟩ \\ = & 1 - 2 ϕ (x \otimes y^{*}) + x ⟨ \cdot, y^{*} ⟩ ⟨ x, y^{*} ⟩ \\ = & 1 - 2 ϕ (x \otimes y^{*}) + ϕ (x \otimes y) ⟨ x, y^{*} ⟩ \\ = & 1 - ϕ (x \otimes y^{*}) (⟨ x, y^{*} ⟩ - 2) \end{matrix}$ or any pseudo-reflection $s_{x, y^{*}} .$ Thus, with the assumption that $⟨ x, y^{*} ⟩ = 2,$ we have that $s_{x, y^{*}}^{2} = 1 .$
  From the definition of pseudo-reflection we have that $s_{x, y^{*}} (x) = [1 - ϕ (x \otimes y^{*})] (x) = x - x ⟨ x, y^{*} ⟩ .$ Thus, if $⟨ x, y^{*} ⟩ = 2,$ we have that $s_{x, y^{*}} (x) = - x .$
2. Let H={h∈E | ⟨h,y*⟩=0} and Kx={kx | k∈K}.
  We show that if ⟨x,y*⟩=1 then
  1. $K x + H = E .$
  2. $K x \cap H = 0 .$
  3. $K x$ is free with basis $x .$
  4. $s_{2 x, y^{*}} (k x) = - k x$ for any $k \in K .$
  5. $s_{2 x, y^{*}} (h) = h$ for any $h \in H .$
  1. Let $e \in E .$ Then $e = ⟨ e, y^{*} ⟩ x + (e - ⟨ e, y^{*} ⟩ x) .$ Since $\begin{matrix} ⟨ e - ⟨ e, y^{*} ⟩ x, y^{*} ⟩ & = & ⟨ e, y^{*} ⟩ - ⟨ e, y^{*} ⟩ ⟨ x, y^{*} ⟩ \\ = & ⟨ e, y^{*} ⟩ (1 - ⟨ x, y^{*} ⟩) \\ = & 0, \end{matrix}$ we have that $e - ⟨ e, y^{*} ⟩ x \in H .$ It follows that $e \in K x + H .$
  2. Let $e \in K x \cap H .$ Write $e = k x$ for some $k \in K .$ Then $0 = ⟨ e, y^{*} ⟩ = ⟨ k x, y^{*} ⟩ = k 1 = k .$ So $k = 0$ and $e = k x = 0 .$
  3. To show that $K x$ is free with basis $x$ we must show that every element can be written uniquely in the form $k x$ for some $k \in K .$ To see this, suppose that $k x = k' x$ for some $k, k' \in K .$ Then $0 = ⟨ 0, y^{*} ⟩ = ⟨ (k - k') x, y^{*} ⟩ = (k - k') ⟨ x, y^{*} ⟩ = k - k' .$ So $k = k' .$
  4. Let $k x \in K x .$ Then, since $⟨ x, y^{*} ⟩ = 1,$ $s_{2 x, y^{*}} (k x) = k x - ⟨ k x, y^{*} ⟩ 2 x = k x - k 2 x = - k x .$
  5. Let $h \in H .$ Then $s_{2 x, y^{*}} (h) = h - ⟨ h, y^{*} ⟩ 2 x = h - 0 = h . ◻$
With the notations as in exercise 1, show that $det (s_{x, y^{*}}) = 1 - ⟨ x, y^{*} ⟩$ if $E$ is a free $K -module$ of finite type.

Solution. Let $e_{1}, e_{2}, \dots, e_{n}$ be a basis of $E .$ Then, for any $u \in End (E),$ $\begin{matrix} u (e_{1}) \land u (e_{2}) \land \dots \land u (e_{n}) = det (u) (e_{1} \land e_{2} \land \dots \land e_{n}) . & (*) \end{matrix}$ We have that $\begin{matrix} s_{x, y^{*}} (e_{1} \land e_{2} \land \dots \land e_{n}) & = & (e_{1} - ⟨ e_{1}, y^{*} ⟩ x) \land (e_{2} - ⟨ e_{2}, y^{*} ⟩ x) \land \dots \land (e_{n} - ⟨ e_{n}, y^{*} ⟩ x) \\ = & e_{1} \land \dots \land e_{n} - \sum_{i} e_{1} \land \dots \land e_{i - 1} \land ⟨ e_{i}, y^{*} ⟩ x \land e_{i + 1} \land \dots \land e_{n} + 0 . \end{matrix}$ Substituting $x = c_{1} e_{1} + c_{2} e_{2} + \dots + c_{n} e_{n}$ gives that $\begin{matrix} s_{x, y^{*}} (e_{1} \land e_{2} \land \dots \land e_{n}) & = & e_{1} \land \dots \land e_{n} \\ + \sum_{i} ⟨ e_{i}, y^{*} ⟩ e_{1} \land \dots \land e_{i - 1} \land (c_{1} e_{1} + c_{2} e_{2} + \dots + c_{n} e_{n}) \land e_{i + 1} \land \dots \land e_{n} \\ = & e_{1} \land \dots \land e_{n} - \sum_{i} ⟨ e_{i}, y^{*} ⟩ e_{1} \land \dots \land e_{i - 1} \land c_{i} e_{i} \land e_{i + 1} \land \dots \land e_{n} \\ = & e_{1} \land \dots \land e_{n} - \sum_{i} ⟨ c_{i} e_{i}, y^{*} ⟩ e_{1} \land \dots \land e_{i - 1} \land e_{i} \land e_{i + 1} \land \dots \land e_{n} \\ = & e_{1} \land \dots \land e_{n} - ⟨ x, y^{*} ⟩ e_{1} \land \dots \land e_{n} \\ = & (1 - ⟨ x, y^{*} ⟩) e_{1} \land e_{2} \land \dots \land e_{n} . \end{matrix}$ Thus $det (s_{x, y^{*}}) = 1 - ⟨ x, y^{*} ⟩ . ◻$
Let V be a complex Hilbert space with basis e1,…,eℓ. For 1≤i≤ℓ, let si be a unitary pseudo-reflection such that si(ei)=ciei where ci≠1. An element of V is invariant under s if and only if it is orthogonal to ei. Let W be the subgroup of GL(V) generated by the si.
1. Let $i$ be an integer $\geq 1 .$ Show that every element of $⋀^{i} V$ invariant under $W$ is zero.
2. Suppose that $W$ is finite. Show that for all endomorphisms $A$ of $V$ one has: $\begin{matrix} \sum_{w \in W} det (A - w) = Card (W) det (A), \\ \sum_{w \in W} det (1 - A w) = Card (W) . \end{matrix}$ Deduce that, for all $A \in End (V)$ there exists $w \in W$ such that $A w$ does not have any nonzero fixed point.
3. Let $Γ$ be the graph with vertices in the set $[1, ℓ]$ and edges $(i, j)$ determined by the pairs $e_{i}, e_{j}$ such that $e_{i}$ and $e_{j}$ are not orthogonal. Show that $V$ is a simple $W -module$ if and only if $Γ$ is connnected and nonempty.
4. Suppose that $V$ is a simple $W -module.$ Show that the $W -modules$ $⋀^{i} V$ $(0 \leq i \leq ℓ)$ are simple. Show that these modules are pairwise nonisomorphic.
Solution. Let us first see that an element of V is invariant under si if and only if it is orthogonal to ei. Suppose v∈V is such that si(v)=v. Then, since si is unitary, ⟨v,ei⟩= ⟨si(v),si(ei)⟩= ⟨v,ciei⟩= c‾i⟨v,ei⟩. Since ci≠1 it follows that ⟨v,ei⟩=0. So v is orthogonal to ei. To prove the converse, suppose that v is orthogonal to ei. Then, since si is a pseudoreflection, (1-si)v= c(1-si)ei= c(1-ci)ei for some c∈ℂ. Thus ⟨(1-si)v,ei⟩ =c(1-ci) ⟨ei,ei⟩. On the other hand, ⟨(1-si)v,ei⟩ = ⟨v,ei⟩- ⟨siv,ei⟩ = 0-ci-1‾ ⟨siv,siei⟩ = ci-1‾ ⟨v,ei⟩ = 0. Since ci≠1 and ⟨ei,ei⟩≠0 it follows that c=0. So (1-si)v=0 and thus siv=v.
1. The proof is by induction on $ℓ = dim V .$ When $dim V = 1$ then $e_{1}$ is a basis of $V .$ Suppose that $w c e_{1} = c e_{1},$ for all $w \in W .$ Then $s_{1} c e_{1} = c e_{1} \Rightarrow c_{1} c e_{1} \Rightarrow c (1 - c_{1}) e_{1} = 0 .$ Since $c_{1} \neq 0$ we have that $c = 0 .$ Now suppose that $dim V = ℓ .$ Let $V'$ be the subspace spanned by ${e_{1}, e_{2}, \dots, e_{ℓ - 1}}$ and let $e \in {(V')}^{⊥} .$ Let $W'$ be the group generated by $s_{1}, s_{2}, \dots, s_{ℓ - 1} .$ If $v \in ⋀^{i} V$ then $v = a + b \land e, where a \in ⋀^{i} V' and b \in ⋀^{i} V .$ Suppose that $v \in ⋀^{i} V$ and that $w v = v$ for all $w \in W .$ If $w \in W'$ then $\begin{matrix} w v & = & w (a + b \land e) \\ = & w a + w b \land w e \\ = & w a + w b \land e since e \in {(V')}^{⊥} \\ = & a + b \land e since w v = v . \end{matrix}$ So $w a = a$ and $w b = b$ for all $w \in W' .$ By induction we have that $a = 0$ and $b = 0 .$ So $v = a + b \land e = 0 .$
2. We use the determinant relation given in $(*) .$ $det (A - w) = (A - w) (e_{1} \land \dots \land e_{ℓ}) = \sum_{I \subseteq [1, ℓ]} \pm (A e_{i_{1}} \land \dots \land A e_{i_{k}} \land w e_{j_{1}} \land \dots \land w e_{j_{m}}),$ where the sum is over all subsets $I = {i_{1}, i_{2}, \dots, i_{k}}$ of $[1, ℓ] = {1, 2, \dots, ℓ}$ and $I^{c} = {j_{1}, j_{2}, \dots, j_{m}}$ is the complement of $I$ in $[1, ℓ] .$ Then $\begin{matrix} \sum_{w \in W} det (A - w) & = & \sum_{w \in W} \sum_{I \subseteq [1, ℓ]} \pm (A e_{i_{1}} \land \dots \land A e_{i_{k}} \land w e_{j_{1}} \land \dots \land w e_{j_{m}}) \\ = & \sum_{I \subseteq [1, ℓ]} \pm (A e_{i_{1}} \land \dots \land A e_{i_{k}} \land (\sum_{w \in W} w e_{j_{1}} \land \dots \land w e_{j_{m}})) . \end{matrix}$ Then, by part a), provided $m \geq 1,$ $\sum_{w \in W} w e_{j_{1}} \land \dots \land w e_{j_{m}} = 0$ since the left hand side is an invariant element of $⋀^{m} V .$ So $\begin{matrix} \sum_{w \in W} det (A - w) & = & \sum_{w \in W} A e_{1} \land \dots \land A e_{ℓ} + \sum_{\underset{I \neq [1, ℓ]}{I \subset [1, ℓ]}} \pm A e_{i_{1}} \land \dots \land A e_{i_{k}} \land 0 \\ = & \sum_{w \in W} A e_{1} \land \dots \land A e_{ℓ} \\ = & (det A) Card (W) e_{1} \land \dots \land e_{ℓ} . \end{matrix}$ This proves the identity $\sum_{w \in W} det (A - w) = Card (W) det A .$ The identity $\sum_{w \in W} det (1 - A w) = Card (W)$ is proved in exactly the same fashion. It remains to show that for all $A \in End (V)$ there exists $w \in W$ such that $A w$ does not have any nonzero fixed point. Suppose that there does not exist $w \in W$ such that $A w$ has no nonzero fixed point. Then for each $w \in W$ there is a nonzero vector $v \in V$ such that $(1 - A w) v = 0 .$ So, for each $w \in W,$ $det (1 - A w) = 0 .$ So $\sum_{w \in W} det (1 - A w) = 0 .$ This is a contradiction to the identity above.
3. $\Rightarrow :$ Suppose that $Γ$ is connected. Assume that $V'$ is a nonzero $W -submodule$ of $V .$ Let $x \in V',$ $x \neq 0 .$ Then $s_{i} x \neq x$ for some $i \in [1, ℓ] = {1, 2, \dots, ℓ}$ since $⟨ x, e_{i} ⟩ = 0$ for all $1 \leq i \leq ℓ$ is impossible. Let $i \in [1, ℓ]$ be such that $s_{i} x \neq x .$ Since $s_{i}$ is a pseudo reflection we have that $(1 - s_{i}) x = c e_{i} \in V' .$ Let $k$ be any other element of $[1, ℓ] .$ Since $Γ$ is connected there is a path $e_{i} = e_{i_{1}} \to e_{i_{2}} \to \dots \to e_{i_{m}} = e_{k}$ in $Γ$ connecting $e_{i}$ and $e_{k} .$ Assume $e_{i_{j}} \in V' .$ Since $s_{i_{j + 1}}$ is a pseudoreflection, $(1 - s_{i_{j + 1}}) e_{i_{j}} = c e_{i_{j + 1}} \in V' .$ Since $e_{i_{j + 1}}$ and $e_{i_{j}}$ are connected in the graph $Γ$ they are not orthogonal and thus $s_{i_{j + 1}} e_{i_{j}} \neq e_{i_{j}}$ and so $c \neq 0 .$ So $e_{i_{j + 1}} \in V' .$ In this way we show that $e_{k} \in V' .$ Thus $e_{k} \in V'$ for all $k \in [1, ℓ] .$ So $V' = V .$ Thus $V$ is a simple $W -module.$
  $\Leftarrow :$ Suppose that $Γ$ is not connected and let $Γ_{1}$ be a connected component of $Γ .$ Let $e_{i_{1}}, e_{i_{2}}, \dots, e_{i_{k}}$ be the vertices in $Γ_{1} .$ Then let $V_{1}$ be the span of the basis elements $e_{i_{j}}$ corresponding to the vertices in $Γ_{1} .$ Let $k \in [1, ℓ]$ and let $e_{i} \in V_{1}$ Then $e_{i} \in V_{1}$ and if $e_{k}$ is a vertex in $Γ_{1},$ then, since $(1 - s_{k}) e_{i} = c e_{k} \in V_{1},$ we have that $s_{k} e_{i} \in V_{1} .$ If $e_{k}$ is not a vertex in $Γ_{1}$ then $⟨ e_{k}, e_{i} ⟩ = 0$ and it follows that $s_{k} e_{i} = e_{i} \in V_{1} .$ Thus, $V_{1}$ is stable under the action of $W .$ So $V_{1}$ is a proper submodule of $V .$ So $V$ is not simple.
4. We will do this problem by completing the following steps.
  1. Let $s_{j} \in S .$ Then the rank of $s_{j}$ on $⋀^{p} V$ is $(\binom{n - 1}{i - 1}) .$
  2. The modules $⋀^{i} V$ and $⋀^{j} V$ are nonisomorphic if $i \neq j .$
  Let j be a vertex in the graph Γ such that Γ-{j} is connected. Let V′=span {ei | i≠j} and let e∈(V′)⊥ so that V=V′⊕ℂe.
  1. $s_{j} e = e + c e_{j},$ where $c \in ℂ$ and $c \neq 0 .$
  2. $s_{j} e = v' + d e,$ where $v' \in V',$ $v' \neq 0$ and $d \in ℂ .$
  It is clear that ⋀0V and ⋀nV are irreducible since they are 1-dimensional. Let 0<p<n. Then ⋀pV=U1⊕U2, where U1=⋀pV′ and U2=(⋀p-1V′)∧e. By induction on dim V we know that U1 and U2 are irreducible under the action of W′.
  1. $U_{1}$ and $U_{2}$ are not invariant under the action of $W .$
  2. If $E$ is a nonzero submodule of $⋀^{p} V$ then $E = ⋀^{p} V .$
  1. Note that if $v_{1}, \dots, v_{p} \in V$ then, since $s_{j}$ is a pseudo-reflection we have that for every $i,$ $s_{j} v_{i} = v_{i} + d_{i} e_{j}$ for some $d_{i} \in ℂ .$ Thus $\begin{matrix} (s_{j} - 1) v_{1} \land \dots \land v_{p} & = & s_{j} v_{1} \land \dots \land s_{j} v_{p} - v_{1} \land \dots \land v_{p} \\ = & (v_{1} + d_{1} e_{j}) \land \dots \land (v_{p} + d_{p} e_{j}) - v_{1} \land \dots \land v_{p} \\ = & \sum_{k = 1}^{p} d_{k} v a_{1} \land \dots \land e_{j} \land \dots \land v_{p}, \end{matrix}$ and also $\begin{matrix} s_{j} (e_{j} \land v_{1} \land \dots \land v_{p - 1}) & = & c_{j} e_{j} \land (v_{1} + d_{1} e_{j}) \land \dots \land (v_{p - 1} + d_{p - 1} e_{j}) - e_{j} \land v_{1} \land \dots \land v_{p - 1} \\ = & (c_{j} - 1) e_{j} \land v_{1} \land \dots \land v_{p - 1} . \end{matrix}$ The first equality shows that the action of $s_{j}$ on $⋀^{p} V$ is a map onto $e_{j} \land (⋀^{p - 1} V) .$ The second equality shows that $(s_{j} - 1)$ acts by a nonzero constant on $e_{j} \land (⋀^{p - 1} V) .$ It follows that the rank of $s_{j} - 1$ acting on $⋀^{p} V$ is $dim (e_{j} \land (⋀^{p - 1} V)) = (\binom{n - 1}{p - 1}) .$
  2. Assume that $i \neq j$ and that $⋀^{i} V$ and $⋀^{j} V$ are isomorphic. Then we have that $(\binom{n}{i}) = dim (\overset{i}{⋀} V) = dim (\overset{j}{⋀} V) = (\binom{n}{n - j}),$ giving that $i = n - j .$ We also must have that the rank of $s_{j} - 1$ on $⋀^{i} V$ is the same as the rank of $s_{j} - 1$ on $⋀^{j} V .$ It follows that $(\binom{n - 1}{i - 1}) = (\binom{n - 1}{n - 1 - (j - 1)}),$ from which we get that $i - 1 = n - 1 - j + 1$ or $i = n - j + 1 .$ This is a contradiction to $i = n - j .$ Thus $⋀^{i} V$ and $⋀^{j} V$ are nonisomorphic.
  3. Since $s_{j}$ is a pseudo-reflection, $(1 - s_{j}) e = c e_{j} .$ Since $e_{j}$ is not in $V'$ we know that $e NOTPERP e_{j} .$ So $s_{j} e \neq e .$ So $c \neq 0 .$
  4. We know $s_{j} e = v' + d e$ for some $v' \in V'$ and some $d \in ℂ .$ If $v' = 0$ then $s_{j} e = d e .$ In view of c) we see that this implies that $e_{j} = \frac{d - 1}{c} e,$ which is a contradiction since $e_{j} NOTPERP V' .$ So $v' \neq 0 .$
  5. We have assumed that $0 < p < n .$ So $p - 1 \leq n - 2$ and thus we can choose vectors $v_{1}^{'}, v_{2}^{'}, \dots, v_{p - 1}^{'} \in V'$ such that $v_{i}^{'}$ is orthogonal to $e_{j}$ for all $i .$ Let $u_{0} = v_{1}^{'} \land \dots \land v_{p - 1}^{'} .$ Then, by d), $s_{j} (u_{0} \land e) = s_{j} u_{0} \land s_{j} e = u_{0} \land (v' + d e) = u_{0} \land v' + d u_{0} \land e,$ where $u_{0} \land v' \in U_{1}$ and is nonzero. This argument shows that $U_{2}$ is not invariant under the action of $W .$ Since $U_{1} = U_{2}^{⊥}$ in $⋀^{p} V$ and we know that the inner product is $W -invariant,$ we get also that $U_{1}$ is not invariant under the action of $W .$
  6. Let $E \subseteq ⋀^{p} V$ be a nonzero submodule. Suppose that $E \cup U_{1} = 0 .$ Then let $u \in E,$ $u \neq 0,$ and write $u = u_{1} + u_{2}$ where $u_{i} \in U_{i} .$ Since $E \cup U_{1} = 0,$ $u_{2} \neq 0 .$ By the induction hypothesis applied to $W'$ we know that $U_{1}$ and $U_{2}$ are irreducible and inequivalent. Thus there exists an element $a \in ℂ [W']$ such that $a u_{1} = 0$ and $a u_{2} \neq 0 .$ So $a u = a u_{2} \in E .$ So $E \cup U_{2} \neq 0 .$ Since $U_{2}$ is irreducible as a $W' -module,$ $U_{2} \subseteq E .$ Then, by d) we know that there is an element $u \in E$ such that $u = u_{1} + u_{2}$ and $u_{1} \neq 0 .$ So $u_{1} = u_{2} - u \in E .$ Again, by the irreducibility of $U_{1}$ as a $W' -module,$ $U_{1} \subseteq E .$ So $E = ⋀^{p} V .$ A similar argument holds in the case that $E \cup U_{2} = 0 .$ Thus $⋀^{p} V$ is irreducible. $◻$
Notes. Bourbaki notes that this exercise is due to R. Steinberg. The solution to part d) is also due to R. Steinberg (unpublished notes).
Concerning part b): Let F be a field of arbitrary characteristic, let G be a finite group, and let U be a finite dimensional G-module. For g∈G let gU denote the corresponding endomorphism of U. Let UG be the submodule of G-invariants. A projection onto UG means an idempotent F-linear transformation with range UG.

Lemma: Let $π$ be any projection of $U$ onto $U^{G} .$ Suppose $b \in End (U)$ satisfies $b U^{G} \subseteq U^{G} .$ Then $\sum_{g \in G} trace (b g_{U}) = | G | trace (b π) .$

Corollary: If $U^{G} = 0$ and $b \in End (U)$ then $\sum_{g \in G} trace (b g_{U}) = 0 .$

Proof of Lemma: Choose an $F -subspace$ $U'$ of $U$ such that $U = U^{G} \oplus U'$ and use matrices relative to this decomposition. If $a \in End (U)$ let $[a]$ denote the corresponding matrix. Then $| G | [π] = (\begin{matrix} | G | I & C_{12} \\ 0 & 0 \end{matrix}) \sum_{g \in G} [g_{U}] = (\begin{matrix} | G | I & D_{12} \\ 0 & 0 \end{matrix})$ where $I$ is the identity matrix of size $dim U$ and $C_{12}, D_{12}$ are matrices of appropriate size. Since $b U^{G} \subseteq U^{G}$ we have $[b] = (\begin{matrix} B_{11} & B_{12} \\ 0 & B_{22} \end{matrix}) .$ Thus $| G | [b] [π] = (\begin{matrix} | G | B_{11} I & B_{11} C_{12} \\ 0 & 0 \end{matrix}) and [b] \sum_{g \in G} [g_{U}] = (\begin{matrix} | G | B_{11} I & B_{11} D_{12} \\ 0 & 0 \end{matrix}) .$ The assertion follows. $◻$

Remark: Note that the argument shows that $[b] (| g | [π] - \sum_{g \in G} [g_{U}])$ is nilpotent.

Now let $G$ be a finite group and let $V$ be a $G -module.$ Assume as in the above exercise that ${(⋀^{p} V)}^{G} = 0$ for $1 \leq p \leq n$ but make no hypothesis on the characteristic of the ground field $F .$ We may view $⋀^{p} V$ as a module for the semigroup $End (V) \supset G .$ If $a \in End (V)$ let $\land^{p} a$ be the corresponding endomorphism of $⋀^{p} V .$ Thus $\land^{p} (a b) = \land^{p} a \cdot \land^{p} b .$ For $a \in End (V)$ we have $det (1 - a) = \sum_{p = 0}^{n} {(- 1)}^{p} trace (\land^{p} a) .$ Let’s take this as known. Replace $a$ by $a g .$ Then $det (1 - a g) = \sum_{p = 0}^{n} trace (\land^{p} a \cdot \land^{p} g) .$ If $p > 0,$ use the hypothesis ${(⋀^{p} V)}^{G} = 0$ and apply the Corollary with $U = ⋀^{p} V$ and $b = \land^{p} a .$ This gives $\sum_{g \in G} trace (\land^{p} a \cdot \land^{p} g) = 0$ so $\sum_{g \in G} det (1 - a g) = \sum_{g \in G} trace (\land^{0} a \land^{0} g) = | G | .$

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