Proof. We denote by V the E-vector space underlying $\rho $ . Part (a) is [Reference Curtis and ReinerCR62, Corollary 69.8] with the ring A from there being the image of $E[H]$ in ${\mathrm {End}}_E(V)$ . Part (b) holds by [Reference Serre and ScottSer77, Proposition 22]. Part (c) follows from [Reference Curtis and ReinerCR62, Theorem 49.2]. For (d), note that $\rho $ is a subrepresentation of ${\mathrm {Ind}}_N^H{\mathrm {Res}}_N^H\rho $ , and hence ${\mathrm {Ind}}_H^G\rho $ is a subrepresentation of ${\mathrm {Ind}}_N^G{\mathrm {Res}}_N^H\rho $ . By (c) applied to the irreducible summands of $\rho $ , we see that ${\mathrm {Res}}_N^H\rho $ is semisimple. Now, the result can be found in [Reference WebbWeb16, Chapter 5, Exercise 8].

To prove Part (e), let $V'\subset {\mathrm {Ind}}_H^G \rho $ be an irreducible G-subrepresentation. Then by (b) the representation ${\mathrm {Res}}_N^G V'$ contains $\nu ^g$ for some $g\in G$ . As $V'$ is a G-representation, we deduce $ \nu ^g\subset {\mathrm {Res}}_N^G V'$ for all $g\in G$ . By hypothesis, the $\nu ^g$ , $g\in G/N$ , are pairwise nonisomorphic, and hence $\bigoplus _{g\in G/N} \nu ^g\subset {\mathrm {Res}}_H^G V'$ . By (b) the left-hand side is isomorphic to $ {\mathrm {Res}}_N^G {\mathrm {Ind}}_H^G\rho $ so that for dimension reasons we must have $V'={\mathrm {Ind}}_H^G\rho $ .

We next prove Part (f). Because the solution space of a linear system of equations has the same dimension over its field of definition and over any extension, one has

$$ \begin{align*}{\mathrm{Hom}}_{E[N]}(\rho,\rho^g)\otimes_E{{{E}^{\mathrm{alg}}}}\cong{\mathrm{Hom}}_{{{{E}^{\mathrm{alg}}}}[N]}(\rho\otimes_E{{{E}^{\mathrm{alg}}}},\rho^g\otimes_E{{{E}^{\mathrm{alg}}}}).\end{align*} $$

This allows one by base change $E\to {{E}^{\mathrm {alg}}}$ to reduce one direction of (f) to (e). For the converse, assume that ${\mathrm {Ind}}_H^G\rho $ is absolutely irreducible. Because ${\mathrm {Ind}}_H^G$ is an exact functor, $\rho $ must be absolutely irreducible and hence also $\rho ^g$ for all $g\in G$ . Because ${\mathrm {Ind}}_H^G\rho $ is absolutely irreducible, Frobenius reciprocity yields

$$ \begin{align*}E\cong {\mathrm{End}}_{E[G]}({\mathrm{Ind}}_H^G\rho)\cong {\mathrm{Hom}}_{E[G]}(\rho,{\mathrm{Res}}_H^G{\mathrm{Ind}}_H^G\rho)\stackrel{(c)}= {\mathrm{Hom}}_{E[G]}(\rho,\oplus_{g\in G/H} \rho^g).\end{align*} $$

Hence, $\rho $ is isomorphic to $\rho ^g$ if and only if $g\in H$ , and this completes the proof of (f).

We now prove Part (g). Note that by (b) the only if direction is clear. For the other direction, note first that by [Reference Curtis and ReinerCR81, Lemma 10.12] we have ${\mathrm {Ind}}_H^G \rho \cong {\mathrm {Ind}}_H^G \rho ^g $ for all $g\in G$ . Since induction and direct sum commute, we also have

$$ \begin{align*}{\mathrm{Ind}}_H^G\big( \bigoplus_{g\in G/H} \rho^g \big)= \bigoplus_{g\in G/H} \big({\mathrm{Ind}}_H^G\rho^g \big) = \big({\mathrm{Ind}}_H^G\rho \big)^{\oplus [G:H]}.\end{align*} $$

The same formula applies to $\rho '$ , and so our hypothesis gives $({\mathrm {Ind}}_H^G\rho )^{\oplus [G:H]}\cong ({\mathrm {Ind}}_H^G\rho ')^{\oplus [G:H]}$ . The Krull–Schmidt theorem (see [Reference Curtis and ReinerCR62, Theorem 14.5]) now yields ${\mathrm {Ind}}_H^G\rho \cong {\mathrm {Ind}}_H^G\rho '$ . Part (h) follows from the uniqueness of composition factors and the irreducibility of the $\rho ^g$ .

Proof. Lacking a precise reference, we give a proof. Let A be an invertible $n\times n$ -matrix over E such that

(2) $$ \begin{align} A \rho(g) A^{-1} = \chi(g) \rho(g) \mbox{ for all }g \in G. \end{align} $$

From equation (2), one deduces $ A^{m} \rho (g) A^{-{m}} = \chi ^{m}(g) \rho (g)= \rho (g)$ for all $g\in G$ . Since $\rho $ is absolutely irreducible, [Reference Curtis and ReinerCR62, (29.13)] implies that for some $\lambda \in E$ . Define $E':=E(\sqrt [{m}]{\lambda })$ . Let $A'=\sqrt [{m}]{\lambda }^{-1}A$ in ${\mathrm {GL}}_n(E')$ so that equation (2) also holds for $A'$ and also . Since ${m} \cdot 1$ is invertible in E, it follows, using the Jordan form, that $A'$ is semisimple. Moreover, $A'$ is diagonalizable over $E'$ since E contains a primitive ${m}$ -th root of unity.

After a change of basis over $E'$ , we may write A as a block diagonal matrix with diagonal blocks $A_1,\ldots ,A_{{m}}$ such that for $i=1,\ldots ,{m}$ each $A_i$ is a scalar matrix with $ n_i\ge 0$ and $\sum _{i=1,\ldots ,{m}} n_i=n$ . For all $g\in G$ and $i,j=1,\ldots ,{m}$ , we decompose $\rho (g)$ correspondingly into blocks $\rho _{i,j}(g)$ so that equation (2) turns into

(3) $$ \begin{align} \zeta^{i-j} \rho_{i,j}(g)= \chi(g) \rho_{i,j}(g). \end{align} $$

Choose $g\in G$ such that $\chi (g)=\zeta $ . Then $\rho _{i,j}(g)$ is zero unless $i-j\equiv 1\pmod {m}$ . Since $\rho (g)$ is invertible, all $\rho _{i+1,i}(g)$ and $\rho _{{m},1}(g)$ must be invertible and hence square matrices and of nonzero size. We deduce that all $n_i$ are equal, hence nonzero, and hence equal to $n/{m}$ . In particular, ${m}$ divides n, proving (a).

Next, for $h\in H$ and for all $i,j=1,\ldots ,{m}$ , equation (3) becomes $\zeta ^{i-j}\rho _{i,j}(h)= \rho _{i,j}(h)$ so that $\rho (h)=\bigoplus _{i=1}^{m}\rho _{i,i}(h)$ is a block diagonal matrix and each $\rho _{i,i}\colon H\to {\mathrm {GL}}_{n/{m}}(k)$ , $h\mapsto \rho _{i,i}(h),$ is a representation of dimension ${n}/{{m}}$ . In particular, the restriction satisfies

$$ \begin{align*}{\mathrm{Res}}^G_H\rho\otimes_EE'=\bigoplus_{i=1}^{m} \rho_{i,i}. \end{align*} $$

We choose $\rho '=\rho _{1,1}$ and consider ${\mathrm {Ind}}_H^G \rho '$ . By [Reference Curtis and ReinerCR62, (10.8) Frobenius Reciprocity Theorem], we have

$$ \begin{align*}{\mathrm{Hom}}_G({\mathrm{Ind}}_H^G\rho',\rho\otimes_EE') = {\mathrm{Hom}}_{H} (\rho',{\mathrm{Res}}_H^G\rho\otimes_EE') \neq 0. \end{align*} $$

Let $f\colon {\mathrm {Ind}}_H^G\rho '\to \rho \otimes _EE'$ be a nonzero G-homomorphism. Since $\rho $ is irreducible, it must be surjective, and because $\dim \rho =n=m\cdot n/m=\dim {\mathrm {Ind}}_H^G\rho '$ , its kernel must be zero so that f is an isomorphism. Next, note that ${\mathrm {Ind}}_H^G$ is an exact functor; see [Reference Curtis and ReinerCR81, § 10, Exercise 20]. Hence, $\rho '$ is absolutely irreducible because $\rho $ is so. This completes the proof of (b).

Part (c) follows from Lemma 2.1.4 (b) and (f). Part (d) easily follows from the continuity of ${\mathrm {Res}}_H^G\rho \otimes _EE'\cong \bigoplus _{g\in G/H}(\rho ')^g$ , using that all linear topologies on a finite-dimensional vector space over $E'$ that are compatible with the topology on $E'$ are equivalent.

Concerning (e), we only prove the first assertion; the proof of the second then follows from (d). For this, let V be the E-vector space underlying $\rho $ , and let T be an ${\cal O}_E$ -lattice in V. The stabilizer of T is an open subgroup of ${\mathrm {GL}}_n(E)$ and hence, by the continuity of $\rho $ , the latice T is fixed by an open subgroup $G'$ of G. Therefore, $G/G'$ is finite. Thus, $T':=\bigcap _{g\in G/G'} gT$ is an ${\cal O}_E$ -lattice in V, and this lattice is clearly G-stable. Choosing an ${\cal O}_E$ -basis of $T'$ , that is then also an E-basis of V, assertion (e) for $\rho $ is clear.

Proof. If $\rho \otimes _EE'\cong {\mathrm {Ind}}_H^G\rho '$ , then Lemma 2.1.3 implies

$$ \begin{align*}(\rho\otimes_EE')\otimes\chi\cong ({\mathrm{Ind}}_H^G\rho')\otimes\chi\cong {\mathrm{Ind}}_H^G(\rho'\otimes{\mathrm{Res}}^G_H\chi)\cong{\mathrm{Ind}}_H^G\rho'\cong \rho\otimes_EE', \end{align*} $$

and this implies $\rho \otimes \chi \cong \rho $ by [Reference Curtis and ReinerCR62, 29.7].

Conversely, suppose that $\rho \cong \rho \otimes \chi $ . After replacing E by a separable extension of degree at most $(n^2)!$ (see Lemma A.2.7 and Remark A.2.8) we may assume that $\rho $ is an absolutely completely reducible G-representation over $E'$ , that is, $\rho =\oplus _{j\in J}\rho ^{\prime }_j$ for absolutely irreducible representations $\rho ^{\prime }_j$ for $j\in J$ . We regroup this decomposition according to orbits under iterated twisting by $\chi $ . This gives rise to a decomposition

(4) $$ \begin{align} \rho\cong \bigoplus_{i\in I} \big( \bigoplus_{j=0}^{{m}_i-1} \rho_i\otimes \chi^j\big)^{\oplus r_i}, \end{align} $$

for integers $r_i>0$ , absolutely irreducible representations $\rho _i\colon G\to {\mathrm {GL}}_{n_i}(E')$ , and divisors ${m}_i$ of ${m}$ , for $i\in I$ so that $\rho _i\otimes \chi ^{{m}_i}\cong \rho _i$ , and no $\rho _i$ is isomorphic to $\rho _{i'}\otimes \chi ^j$ for some $j\in \{0,\ldots ,{m}_{i'}-1\}$ and $i'\in I$ . We have $G\supset H_i:={\operatorname {ker}{ \chi }}^{{m}_i}\supset H$ , $[H_i:H]={m}_i$ , and ${\mathrm {Res}}^G_{H_i}\chi $ is a character of order ${m}_i$ .

By Theorem 2.2.1, we find Kummer extensions $E^{\prime }_i$ of $E'$ of degree dividing $m_i$ and representations $\rho _i^{\prime \prime }\colon H_i\to {\mathrm {GL}}_{n_i/{m}_i}(E^{\prime }_i)$ such that ${\mathrm {Ind}}_{H_i}^G\rho _i^{\prime \prime }\cong \rho _i\otimes _{E'}E^{\prime }_i$ . Let

be the trivial representation of H on $E'$ . Then

where the second and fourth isomorphism follows from Lemma 2.1.3. Let $E"$ be the composite of the $E^{\prime }_i$ , and set $\rho ':=\bigoplus _{i\in I} ( {\mathrm {Res}}_H^{H_i} \rho _i^{\prime \prime }\otimes _{E^{\prime }_i}E")^{\oplus r_i}$ so that clearly $[E":E']<m^n$ . The first assertion of the corollary is now evident from the above and from (4). The remaining assertions follow from Lemma 2.1.4 (b) and (d).

Proof. By Lemma 2.1.4 (c), we have $V_H=\oplus _{g\in G/H^*} W^g$ for some irreducible H-module W over ${{{E}^{\mathrm {alg}}}}$ and some subgroup $H^*\subset G$ with $H\subset H^*$ . Since $G/H\cong {\mathbb {Z}}/p{\mathbb {Z}}$ , in Part (a) of the present lemma, we must have $H^*=H$ , and then the assertion follows from Lemma 2.1.4 (e).

We now prove (b). Let W be as in (1). By choosing the same underlying E vector space, we assume that ${\mathrm {Res}}_H^GW={\mathrm {Res}}_H^GV$ . Let $g\in G$ be a generator of $G/H$ , and let $A,B\in {\mathrm {Aut}}_{E}({\mathrm {Res}}_H^G V)$ be the automorphisms given by the action of g on W and V, respectively. Because ${\mathrm {Res}}_H^GV$ is absolutely irreducible, there exists a nonzero scalar $\lambda \in E$ such that $B=\lambda A$ . As $g^p\in H$ , we find $A^p=B^p=\lambda ^p A^p$ . Because ${\mathrm {Char}}\ E=p>0$ , we must have $\lambda =1$ , and so (1) is proved.

For (2), write ${\mathrm {Ind}}_H^G V_H=\bigoplus _{i\in I}W_i$ with indecomposable G-modules $W_i$ . Let $W_i^{\prime }$ be an irreducible quotient of $W_i$ as a G-submodule. From Lemma 2.1.4 (b) and (c) and the irreducibility of $V_H$ , we deduce ${\mathrm {Res}}_H^G W_i^{\prime }\cong V_H$ for all i, and by (1), we find $W_i^{\prime }\cong V\otimes _E{{{E}^{\mathrm {alg}}}}$ . The following inequality implies $\#I=1$ and the uniqueness of $W_1'$ and thus gives (2):

$$ \begin{align*} \#I &\le \dim_{{{E}^{\mathrm{alg}}}}{\mathrm{Hom}}_G( \bigoplus_i W_i,V\otimes_E{{{E}^{\mathrm{alg}}}}) \ = \ \dim_{{{E}^{\mathrm{alg}}}}{\mathrm{Hom}}_G( {\mathrm{Ind}}_H^GV_H,V\otimes_E{{{E}^{\mathrm{alg}}}}) \\ &= \dim_{{{E}^{\mathrm{alg}}}}{\mathrm{Hom}}_H( V_H,V_H) \ = \ 1. \end{align*} $$

To see (3), observe that if $V\otimes _E{{{E}^{\mathrm {alg}}}}$ was induced, then by Lemma 2.1.4 (b) then $V_H$ had to be reducible. For (4), note that we have ${\mathrm {Ind}}_H^G V_H\cong (V\otimes _E{{{E}^{\mathrm {alg}}}})\otimes _E{\mathrm {Ind}}_H^G E$ . Now, clearly the semisimplification of ${\mathrm {Ind}}_H^G E$ is the trivial module $E^p$ , and this shows (4).

Proof. (a) The continuity of $\lambda _A$ follows from that of $\rho $ . It is a homomorphism because of

$$ \begin{align*}A+\lambda_A(gh)1_n\!=\!ghAh^{-1}g^{-1}\!=\!g(A+\lambda_A(h)1_n)g^{-1}\!=\!gAg^{-1}+\lambda_A(h) 1_n\!=\!A+(\lambda_A(g)+\lambda_A(h))1_n.\end{align*} $$

To see (b) note that the image of $\lambda _A$ is p-elementary abelian because $p\cdot 1=0$ in E. By (a) $H_A={\mathop {\mathrm{Ker}}\nolimits }\, \lambda _A\supseteq \Phi (G)$ , and hence $H_\rho \supseteq \Phi (G)$ . Part (c) is clear from (a) and (b) since by assumption, G is topologically finitely generated and hence so is $G/\Phi (G)$ . For Part (d), denote by $\chi _A(T)\in E[T]$ the characteristic polynomial of A. Then equation (5) implies $\chi _A(T)=\chi _A(T+\lambda _A(g))$ for all $g\in G_K$ , and this proves (d).

For (e), one easily verifies that $\lambda _\bullet $ is the boundary map of cohomology $H^0(G,{\overline {\mathrm {End}}}_E(V))\to H^1(G,E)$ induced from the sequence (6); Part (a) shows that the target module is $ {\mathrm {Hom}}_{\operatorname {cont}}(G,(E,+))$ ; the cocyle condition is easily verified. Moreover, $\lambda _A$ is trivial if and only if $A\in {\mathrm {End}}_G(V)$ . Hence, $\overline {A}\to \lambda _{\overline {A}}$ is defined and injective. The homomorphism property is straightforward.

(f) Raising equation (5) to the power p and using ${\mathrm {Char}}(E)=p$ we find

$$ \begin{align*}\forall g\in G_K: \rho(g)A^{p}\rho(g)^{-1}=A^{p}+\lambda_A(g)^{p} 1_n.\end{align*} $$

Since ${\mathrm {End}}_G'(V)$ is clearly an E-vector space and $\lambda _\bullet $ is E-linear, Part (f) follows. To see (g), let A be in ${\mathrm {End}}_G'(V){\smallsetminus } E$ so that $\Lambda _A\subset (E,+)$ is nontrivial and finite. Let $\Lambda \subset \Lambda _A$ be a sub ${{\mathbb {F}}\,\!{}}_p$ -vector space of codimension $1$ , and let f be the p-linear polynomial $\prod _{\lambda \in \Lambda }(T-\lambda )$ . Then $\Lambda _{f(A)}$ has order p by (f), and $H_A$ has index p by its definition in (b).

In (h), the asserted commutativity is clear from equation (5); the assertion on the $V_\lambda $ and $V^{\prime }_\lambda $ is then immediate. For (i), choose an eigenvalue $\lambda \in {{{E}^{\mathrm {alg}}}}$ for which $\dim V_\lambda $ is minimal. By (c), we have $\dim V_\lambda \cdot \#\Lambda _A\le n$ with equality if and only if A is semisimple. Because A and $\rho |_{H_A}\otimes _E{{{E}^{\mathrm {alg}}}}$ commute, the action of $H_A$ preserves $V_\lambda $ . Let $V_{{{E}^{\mathrm {alg}}}}:=V\otimes _E{{{E}^{\mathrm {alg}}}}$ . Frobenius reciprocity gives a nonzero homomorphism in

$$ \begin{align*}{\mathrm{Hom}}_{G}({\mathrm{Ind}}_{H_A}^{G}V_\lambda,V_{{{E}^{\mathrm{alg}}}})\cong {\mathrm{Hom}}_{H_A}(V_\lambda,V_{{{E}^{\mathrm{alg}}}}|_{H_A}).\end{align*} $$

Since $V_{{{E}^{\mathrm {alg}}}}$ is an irreducible $G_K$ -representation, it follows that

$$ \begin{align*}n\le \dim {\mathrm{Ind}}_{H_A}^{G}V_\lambda =[G:H_A] \cdot \dim V_\lambda=\# \Lambda_A\cdot \dim V_\lambda\le n.\end{align*} $$

Hence, we must have equality and so A is semisimple. For (j), all but the last assertion follow from the proof for (i).

(k) Using equation (5), we compute for $A,B\in {\mathrm {End}}^{\prime }_{G_K}(V)$ and all $g\in G_K$ that

$$ \begin{align*}g(AB-BA)g^{-1}=(A+\lambda_A(g)1_n)(B+\lambda_B(g)1_n)-(B+\lambda_B(g)1_n)(A+\lambda_A(g)1_n)=(AB-BA). \end{align*} $$

Since $E={\mathrm {End}}_G(V)$ , we conclude that $AB-BA$ is a scalar matrix. We also know that A (and B) is semisimple. To conclude, we may work over ${{{E}^{\mathrm {alg}}}}$ so that we may assume that A has diagonal form. But then it is elementary to see that $AB-BA$ has entries $0$ along the diagonal and hence this scalar matrix must be zero. It follows that any $A,B\in {\mathrm {End}}^{\prime }_G(V)$ commute, and we conclude using (j).

(l) We need to show that for all $A,B\in {\mathrm {End}}_G'(V)$ there exist $\mu ,\nu \in E{\smallsetminus }\{0\}$ such that $H_{\mu A+\nu B}= H_A\cap H_B$ . Let W be the ${{\mathbb {F}}\,\!{}}_p$ -vector space $G/(H_A\cap H_B)$ , and regard $\lambda _A$ and $\lambda _B$ as ${{\mathbb {F}}\,\!{}}_p$ -linear maps $W\to E$ . Note that $d:=\dim _{{{\mathbb {F}}\,\!{}}_p}W\le m$ . Let $\underline {B}:=(b_1,\ldots ,b_d) $ be an ${{\mathbb {F}}\,\!{}}_p$ basis of W. Suppose also without loss of generality that $\dim _{{{\mathbb {F}}\,\!{}}_p}G_K/H_A,\dim _{{{\mathbb {F}}\,\!{}}_p}G_K/H_B< d$ since otherwise we are done.

For $\nu \in {{{E}^{\mathrm {alg}}}}$ set $C_\nu :=\lambda _A+\nu \lambda _B$ . Since the common kernel of $\lambda _A$ and $\lambda _B$ is $0\subset W$ , there exists $\nu \in {{{E}^{\mathrm {alg}}}}$ such that $C_\nu $ is injective, that is, such that the vectors $(C_\nu b_i)_{i=1,\ldots ,d}$ are ${{\mathbb {F}}\,\!{}}_p$ -linearly independent in E. This means that the Moore determinant of these vectors is nonzero. In other words, the determinant of the $d\times d$ -square matrix with $(i,j)$ -entry given by $(\lambda _A(b_i)+\nu \lambda _B(b_i))^{p^{j-1}}$ is nonzero. As a function of $\nu $ , this is a polynomial of degree at most $(p^d-1)/(p-1)$ . It is not identically zero because of its value at $\nu $ , and hence it can have at most $p^d-1<p^m-1$ zeros. It follows that for some $\nu '\in E$ it is nonzero because $\#E\ge p^m$ , and this completes the proof of (l).

Proof. Assume on the contrary that we can find $A\in {\mathrm {End}}_G'(V){\smallsetminus } E$ . We may assume that $\Lambda _A$ has order p by Lemma 2.3.2(g), and we also may assume $E={{{E}^{\mathrm {alg}}}}$ since this leaves $\dim _E{\overline {\mathrm {End}}}_G(V)$ unchanged. Then $H_A$ has index p in G and we have p distinct subspaces $V_\lambda $ of $V_{{{E}^{\mathrm {alg}}}}$ that are stabilized by $H_A$ . By hypotheses and Lemma 2.3.1, the restrictions $\rho _i|_{H_A}$ are absolutely irreducible. This already implies $p=2$ . We also find that the extension of $\rho _2$ by $\rho _1$ becomes trivial when restricted to $H_A$ , and that these restrictions must agree with the two distinct $V_\lambda $ . It follows by Frobenius reciprocity that we have a nonzero map ${\mathrm {Ind}}_{H_A}^G(\rho _2|_{H_A})\to \rho $ for $i=1,2$ . By Lemma 2.3.1(b)(4) all simple subquotients of ${\mathrm {Ind}}_{H_A}^G(\rho _2|_{H_A})$ are isomorphic to $\rho _2$ . But this is absurd, since $\rho $ is nonsplit and $\rho _2$ is not a submodule of $\rho $ . We reach a contradiction even for $p=2$ .

Proof. Part (a) is immediate from class field theory. If $\kappa $ is finite, then Part (b) is well-known; to deduce it one applies the inflation restriction sequence to $G\supset H:={\mathop {\mathrm{Ker}}\nolimits }\, {\operatorname {ad}_{{{\overline {\rho }}}}}$ . If $\kappa $ is a local field, the assertion is proved later in Corollary 3.3.6. We invite the reader to check that there is no circular reasoning involved.

Proof of Theorem 3.3.1.

If ${\mathrm {Char}}\ E=0$ , then this is [Reference KisinKis03, Proposition 9.5]. We give the proof if $p={\mathrm {Char}}\ E>0$ . It closely follows that of [Reference KisinKis03, Proposition 9.5]. We consider a commutative diagram

(8)

with $A\in {\mathcal {A}r}_E$ and $I\subset A$ is a square zero ideal, with the solid arrows given, and we seek to construct a dashed arrow g so that the two triangular subdiagrams commute. If ${{ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{E,\rho _E}}}}} $ is universal, we also have to show that the dashed arrow is unique. Note that A and I are finite-dimensional E-vector spaces. Also, the bottom arrow induces a pair of homomorphism $ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}\to A/I$ and $E\to A/I$ , where the second one is simply the E-algebra structure map.

By possibly conjugating $\hat \rho $ by some matrix in $\Gamma _n(\widehat {R})$ , we can assume that $\rho ^{\operatorname {ver}}_{\rho _E}\otimes _{{ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{E,\rho _E}}}}}\widehat {R}=\hat \rho $ . Following the proof in [Reference KisinKis03, Proposition 9.5], one shows that there exists an ${\cal O}$ -subalgebra $A^\circ $ of A such that

1. (a) $A^\circ $ is free over ${\cal O}$ of rank equal to $\dim _E A$ and $A^\circ \otimes _{\cal O} E=A$ ,

2. (b) the image of $A^\circ $ under $A\to E$ is ${\cal O}$ , and so $A^\circ \in {\widehat {\mathcal {A}}r}_{\kappa '}$

3. (c) the image of $\rho ^{\operatorname {ver}}_{\rho _E}\otimes _{ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{E,\rho _E}}}}A$ lies in ${\mathrm {GL}}_n(A^\circ )$ ,

4. (d) the homomorphism $ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}\to A/I$ factors via $A^\circ /I^\circ $ where $I^\circ =I\cap A^\circ $ .

Write $\rho _{A^\circ }$ for $\rho ^{\operatorname {ver}}_{\rho _E}\otimes _{ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{E,\rho _E}}}}A$ considered with its image in ${\mathrm {GL}}_n(A^\circ )$ . Then $\rho _{A^\circ }$ reduces to $\rho ^{\operatorname {ver}}_{{\overline {\rho }}'}\otimes _{ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}}A^\circ /I^\circ $ modulo $I^\circ $ , and thus by the versality of $ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}$ there is a homomorphism $g^\circ \colon {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}\to A^\circ $ such that $\rho ^{\operatorname {ver}}_{{\overline {\rho }}'}\otimes _{ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}}A^\circ $ is strictly equivalent to $\rho _{A^\circ }$ . Let $g\colon \widehat {R}\to A$ be the homomorphism obtained from $g^\circ \otimes {\mathrm {id}}$ under completion. It is now not difficult to see that both triangles in diagram (8) commute with this choice of g.

It remains to show the uniqueness of g if ${ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}}$ is universal. The argument in [Reference KisinKis03, Proposition 9.5] shows that there is in fact a directed system $A_n^\circ $ , $n\in {\mathbb {N}_{\geq 1}}$ , satisfying (a) – (d) such that $\bigcup _n A_n^\circ =A$ . Now if one has $g_1,g_2$ completing the diagram (8) to two commutative diagrams, there have to be homomorphisms $g_1^\circ ,g_2^\circ \colon {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}\to A_n^\circ $ for n sufficiently large that give rise to $g_1$ and $g_2$ , respectively. The corresponding deformations $G\to {\mathrm {GL}}_n(A_n^\circ )$ of $\bar \rho '$ do agree over A and hence they will agree for n sufficiently large, that is, they represent the same strict equivalence class. Because $ {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{\Lambda ',{\overline {\rho }}'}}}$ is universal, they define the same ring maps $g_1^\circ =g_2^\circ $ and hence $g_1=g_2$ .

Proof. We first show that $\psi $ is an isomorphism in the case $\kappa '=\kappa $ . Let $G=\widehat {{\mathbb {Z}}}$ be the free profinite group on one topological generator $\gamma $ , and let $\bar \rho \colon G\to {\mathrm {GL}}_1(\kappa )$ be the trivial representation given by $\gamma \mapsto 1$ . Because $H^1(G,\kappa )\cong \kappa $ and $H^2(G,\kappa )=0$ , we have ${R^{\operatorname {univ}}_{\Lambda ,{\overline {\rho }}}}=\Lambda [[T]]$ for the resulting universal ring with universal deformation $\rho ^u\colon G\to {\mathrm {GL}}_1(\Lambda [[T]])$ given by $\gamma \mapsto 1+T$ .

Let $f\colon {R^{\operatorname {univ}}_{\Lambda ,{\overline {\rho }}}}=\Lambda [[T]]\rightarrow L$ be the specialization that is given by reduction mod p composed with the injection $\iota \colon \kappa [[T]]{\hookrightarrow } L=\kappa ((s)),T\mapsto s$ . The corresponding representation at L is $\rho _L\colon \Gamma \to {\mathrm {GL}}_1(L),\gamma \mapsto 1+s$ , and for its universal ring we find ${R^{\operatorname {univ}}_{L,\rho _L}}=L[[X]]$ with universal representation

$$ \begin{align*}\rho_L^u\colon\Gamma\to {\mathrm{GL}}_1(L[[X]]),\quad\gamma\mapsto 1+s+X.\end{align*} $$

Let $\widehat {R}$ be the completion of ${R^{\operatorname {univ}}_{\Lambda ,{\overline {\rho }}}}\otimes _{\Lambda } L=\kappa [[T]]\otimes _\kappa L$ at the kernel ${\mathfrak {q}}$ of the homomorphism $f_L\colon \kappa [[T]]\otimes _\kappa L\to L$ that maps $g(T)\otimes h\in \kappa [[T]]\otimes _\kappa L$ to $g(s)h\in L$ . Under $f_L$ , the nonzero elements of $\kappa [[T]]\otimes 1$ map to $L^\times $ , and therefore $\kappa [[T]]\otimes _\kappa L\to \widehat {R}$ extends to $\kappa ((T))\otimes _\kappa L\to \widehat {R}$ , and completion gives an isomorphism $\kappa ((T))\widehat {\otimes }_\kappa L\stackrel \simeq {\longrightarrow } \widehat {R}$ . We now invoke Theorem 3.3.1. It asserts that the L-algebra map

$$ \begin{align*}L[[X]]={R^{\operatorname{univ}}_{L,\rho _L}}\stackrel{\simeq }{\longrightarrow } \widehat{R}=\kappa ((T))\widehat{\otimes } L,\end{align*} $$

that, by its very definition, sends $\rho _L^u(\gamma )=1+s+X$ to $(\rho ^u\otimes _{\Lambda } L)(\gamma )=1+T\otimes 1$ , is an isomorphism. Because s on the left is mapped to $1\otimes s$ on the right, we find that $X\mapsto T\otimes 1-1\otimes s$ . This proves the assertion on $\psi $ for $\kappa '=\kappa $ .

To complete the proof, it remains to explain the reduction of a general finite extension $\kappa '\supset \kappa $ to the case $\kappa '=\kappa $ just treated. For this observe that $L\cong \kappa ' (\!( s )\!)\cong \kappa (\!( s )\!)\otimes _\kappa \kappa '$ so that $L\otimes _\kappa L\to L$ can be written as the map

$$ \begin{align*}\kappa (\!( s )\!) \otimes_\kappa\kappa (\!( s )\!)\otimes_\kappa(\kappa'\otimes_\kappa\kappa')\to \kappa' (\!( s )\!),\quad f\otimes g\otimes \alpha\otimes\beta\mapsto fg\alpha\beta.\end{align*} $$

Since $\kappa '\supset \kappa $ is a finite Galois extension, the ring $A:=\kappa '\otimes _\kappa \kappa '$ is isomorphic to the product of fields $(\kappa ')^{[\kappa ':\kappa ]}$ , and A contains a primitive idempotent corresponding to each factor. Under the multiplication map $A\otimes A\to \kappa ',\lambda \otimes \mu \mapsto \lambda \mu $ , all but one of these map to zero. Hence, all but one of these primitive idempotents lie in $\mathfrak {q}$ , and so they vanish under completion at $\mathfrak {q}$ . One deduces $L\widehat {\otimes }_\kappa L\cong L\widehat {\otimes }_{\kappa '} L$ , and this completes the reduction to $\kappa '=\kappa $ .

Proof. Consider $R\to R_{\mathfrak {p}} \to \widehat {R}_{\mathfrak {p}}$ . Tensoring with ${\kappa ({\mathfrak {p}})}$ over $\kappa $ , it yields a diagram

where $\iota $ denotes completion and where the dashed arrows $\iota '$ and $\iota "$ will now be constructed. For the existence of $\iota '$ , we use the universal property of localization. Thus, we need to show that $R{\smallsetminus }\mathfrak {p}\otimes 1$ is mapped under $\iota $ to the units in $\widehat {R}$ . The ring $\widehat {R}$ is local with residue map induced from ${\varphi }$ , and therefore we need to show that ${\varphi }\circ \iota (R{\smallsetminus }\mathfrak {p}\otimes 1)$ lies in ${\kappa ({\mathfrak {p}})}^\times $ , but this is clear from the definitions and the inclusion $R/\mathfrak {p}{\hookrightarrow } {\kappa ({\mathfrak {p}})}$ . Regarding $\iota "$ , we first note that $\mathfrak {p}\otimes _\kappa {\kappa ({\mathfrak {p}})}$ maps to $\mathfrak {q}$ under $\iota $ and hence $\mathfrak {p}^n\otimes _\kappa {\kappa ({\mathfrak {p}})}$ to $\mathfrak {q}^n$ . Hence, the existence of $\iota '$ gives a compatible system of homomorphisms $R_{\mathfrak {p}}/R_{\mathfrak {p}} \mathfrak {p}^n \to (R\otimes _\kappa {\kappa ({\mathfrak {p}})})/\mathfrak {q}^n$ , and this provides the construction of $ \iota "$ .

Let $\pi $ denote the reduction map $\pi \colon \widehat {R}\to {\kappa ({\mathfrak {p}})}$ , set ${\varphi }'=\pi \circ \iota '$ and ${\varphi }"=\pi \circ \iota "$ and define $\mathfrak {q}'={\operatorname {ker}{{\varphi }}} '$ and $\mathfrak {q}"={\operatorname {ker}{{\varphi }}} "$ . Then the arguments just given provide a commutative diagram with canonical isomorphisms in the bottom row

where by slight abuse of notation we denote the middle and right vertical maps again $\iota '$ and $\iota "$ . Note that by the Cohen structure theorem in equal characteristic the ring $\widehat {R}_{\mathfrak {p}}$ contains ${\kappa ({\mathfrak {p}})}$ as a subfield. Focusing on the right-most arrow and using that $R_{\mathfrak {p}}$ is regular if and only if $\widehat {R}_{\mathfrak {p}}$ is so, it will suffice to prove the following assertion.

Let $\mathcal {R}$ be a complete Noetherian local ${\kappa ({\mathfrak {p}})}$ -algebra with residue field ${\kappa ({\mathfrak {p}})}$ and residue homomorphism $\pi \colon \mathcal {R}\to {\kappa ({\mathfrak {p}})}$ , let $\psi \colon \mathcal {R}\otimes _\kappa {\kappa ({\mathfrak {p}})}\to {\kappa ({\mathfrak {p}})}$ be the homomorphism $r\otimes x\mapsto \pi (r)\cdot x$ , let $\mathfrak {Q}={\operatorname {ker}{ \psi }}$ and let $\widehat {\mathcal {R}}$ be the completion of $\mathcal {R}\otimes _\kappa {\kappa ({\mathfrak {p}})}$ at $\mathfrak {Q}$ . Then we assert that $\widehat {\mathcal {R}}\cong \mathcal {R} [\![ t ]\!]$ .

To prove the assertion, note first that if $\mathcal {S}_1$ and $\mathcal {S}_2$ are ${\kappa ({\mathfrak {p}})}$ -algebras with maximal ideals $\mathfrak {P}_1$ and $\mathfrak {P}_2$ such that ${\kappa ({\mathfrak {p}})}$ is in both cases the residue field, then the completion of $\mathcal {S}:=\mathcal {S}_1\otimes _{{\kappa ({\mathfrak {p}})}}\mathcal {S}_2$ at the maximal ideal $\mathfrak {m}:=\mathfrak {P}_1\otimes _{{\kappa ({\mathfrak {p}})}}\mathcal {S}_2+\mathcal {S}_1\otimes _{{\kappa ({\mathfrak {p}})}}\mathfrak {P}_2$ is isomorphic to

If furthermore $\mathcal {S}_1$ is complete with respect to $\mathfrak {P}_1$ and if

, then the completion of $\mathcal {S}$ at $\mathfrak {m}$ is $\mathcal {S}_1 [\![ T ]\!]$ . We apply this to $\mathcal {S}_1=\mathcal {R}$ , $\mathcal {S}_2= {\kappa ({\mathfrak {p}})}\otimes _\kappa {\kappa ({\mathfrak {p}})}$ , $\mathfrak {P}_2={\operatorname {ker}{( {\kappa ({\mathfrak {p}})}\otimes _\kappa {\kappa ({\mathfrak {p}})}\to {\kappa ({\mathfrak {p}})},x\otimes y\mapsto xy)}}$ . Then by Corollary 3.3.4, we have

, and we deduce $\widehat {\mathcal {R}}\cong \mathcal {R}[\![ T ]\!]$ .

Proof. The corollary will follow from the simple fact that the rank of a coherent sheaf cannot decrease under specialization: Let ${\cal O}$ be the valuation ring of E. By possibly passing to a finite extension of E, we may assume that $E^n$ contains a $\rho (G)$ -stable ${\cal O}$ -lattice whose reduction is $\bar \rho $ . Let $R:= {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{{\cal O},{\overline {\rho }}}}}$ . We may assume that R is Noetherian, that is, that $\dim _\kappa H^1(G,{\operatorname {ad}_{{{\bar \rho }}}})$ is finite since else there is nothing to show. Let also $ \widehat {R}$ and $R':= {R{{\phantom {\overline {m}}}}^{\operatorname {ver}}_{{E,\rho }}}$ be as in Theorem 3.3.1, and denote by f a map $f\colon R\to {\cal O}$ given by the versality of R.

Denote by $\widehat {\Omega }_{R/{\cal O}}=\lim _n \Omega _{(R/{\mathfrak {m}}_R^n)/{\cal O}}$ the module of continuous Kähler differentials. Since R is a quotient of a power series ring over ${\cal O}$ in finitely many variables, as an R-module $\widehat {\Omega }_{R/{\cal O}}$ is finitely generated. By Nakayama’s lemma, we have

$$ \begin{align*}\dim_E \widehat{\Omega}_{R/{\cal O}}\otimes_RE\le \dim_\kappa \widehat{\Omega}_{R/{\cal O}}\otimes_R\kappa.\end{align*} $$

Let ${\mathfrak {p}}=({\mathop {\mathrm{Ker}}\nolimits }\, f)$ , let ${\mathfrak {m}}_E={\mathop {\mathrm{Ker}}\nolimits }(R_{\mathfrak {p}}\to E)$ , with the map induced from f, and let $\widehat {{\mathfrak {m}}}_E$ be the maximal ideal of $\widehat {R}$ . Then by [Reference Böckle, Khare and ManningBKM21, Lemma 7.3], we have $ \widehat {\Omega }_{R/{\cal O}}\otimes _RE\cong {\mathfrak {p}}/{\mathfrak {p}}^2\otimes _R E\cong {\mathfrak {m}}_E/{\mathfrak {m}}_E^2 $ , and furthermore from Lemma 3.3.5 and Theorem 3.3.1, again combined with [Reference Böckle, Khare and ManningBKM21, Lemma 7.3], we obtain ${\mathfrak {m}}_E/{\mathfrak {m}}_E^2 \oplus E\cong \widehat {{\mathfrak {m}}}_E/\widehat {{\mathfrak {m}}}_E^2 \cong \widehat {\Omega }_{\widehat {R}/E}\otimes _{\widehat {R}}E\to \widehat {\Omega }_{R'/E}\otimes _{R'}E$ , where the last map is surjective. Hence,

$$ \begin{align*}\dim_E \widehat{\Omega}_{R'/E}\otimes_{R'}E \le 1+\dim_\kappa \widehat{\Omega}_{R/{\cal O}}\otimes_R\kappa.\end{align*} $$

By [Reference MazurMaz97, § 17, § 21], the dual of $\widehat {\Omega }_{R/{\cal O}}\otimes _R\kappa $ is the mod ${\mathfrak {m}}_{\cal O}$ -tangent space of R at ${\mathfrak {m}}_R$ and the dual of $\widehat {\Omega }_{R'/E}\otimes _{R'}E$ is the tangent space of $R'$ at ${\mathfrak {m}}_{R'}$ , and the latter can be identified with $H^1(G,{\operatorname {ad}_{{{\bar \rho }}}})$ and $H^1(G,{\operatorname {ad}_{{\rho }}} )$ , respectively. This proves the corollary.

Proof. If $\kappa $ is finite, the above statement is just the usual Tate local duality. If $\kappa $ is local, let ${\cal O}$ be its valuation ring. Because $G_K$ is compact, one can find an ${\cal O}$ -lattice T in V that is stable under $G_K$ . Let $j\ge 0$ . Then [Reference NekovářNek06, Theorem 5.2.6] asserts that each $H^j(G_K,T')$ , $T'\in \{T,T^\vee (1)\}$ , is a finitely generated ${\cal O}$ -module and moreover it gives a spectral sequence

(9) $$ \begin{align} {\mathrm{Ext}}^i_{\cal O}(H^{2-j}(G_K,T^\vee(1)),{\cal O})\Longrightarrow H^{i+j}(G_K,T). \end{align} $$

Because ${\cal O}$ is regular and of dimension $1$ , the groups ${\mathrm {Ext}}^1_{\cal O}(\cdot ,{\cal O})$ are finitely generated ${\cal O}$ -torsion modules and ${\mathrm {Ext}}^i_{\cal O}(\cdot ,{\cal O})=0$ for $i\ge 2$ . After tensoring the spectral sequence (9) with $\kappa $ over ${\cal O}$ , Parts (b) and (a) are clear. Part (c) follows from [Reference NekovářNek06, Theorem 4.6.9 and 5.2.11] applied to T, again after tensoring with $\kappa $ over ${\cal O}$ .

Proof. Let $A\to B$ be a small extension in ${\mathcal {A}r}_{\Lambda }$ . Let I be its kernel so that $I^2\subset {\mathfrak {m}}_A I=0$ . For the relative smoothness, we need to show the surjectivity of

$$ \begin{align*}{\mathcal{D}}_{{\bar\rho}}(A)\,{\longrightarrow}\, {\mathcal{D}}_{{\bar\rho}}(B)\times_{{\mathcal{D}}_{{\mathrm{det}}{\bar\rho}}(B)}{\mathcal{D}}_{{\mathrm{det}}{\bar\rho}}(A). \end{align*} $$

So suppose we are given deformations $\rho _B\in {\mathcal {D}}_{{\bar \rho }}(B)$ and $\tau _A\in {\mathcal {D}}_{{\mathrm {det}}{\bar \rho }}(A)$ with ${\mathrm {det}}\rho _B=\tau _A\otimes _A B\in {\mathcal {D}}_{{\mathrm {det}}{\bar \rho }}(B)$ . We need to find a deformation $\rho _A\in {\mathcal {D}}_{{\bar \rho }}(A)$ such that $\rho _A\otimes _A B=\rho _B$ and ${\mathrm {det}}\rho _A=\tau _A$ .

Recall from [Reference MazurMaz89, p. 398] that there is a canonical obstruction class $\mathcal {O}(\rho _B)\in H^2(G_K,{\operatorname {ad}_{{\rho }}} )\otimes _\kappa I$ , which vanishes if and only if there exists a deformation of ${\bar \rho }$ to A that lifts $\rho _B$ . Because of the existence of the deformation $\tau _A$ that maps to ${\mathrm {det}}\rho _B$ , the obstruction class $\mathcal {O}({\mathrm {det}}\rho _B)\in H^2(G_K,{\operatorname {ad}_{{{\mathrm {det}}{\bar \rho }}}} )\otimes _\kappa I$ vanishes. By Theorem 3.4.1, the long exact sequence of Galois cohomology arising from the sequence (11) gives the left exact sequence

By Theorem 3.4.1 the sequence is dual to the right exact sequence

that arises from the sequence (12). By our hypothesis, the map $H^0({\mathrm {diag}}(1))$ is an isomorphism, and so by duality the same holds for $H^2({\mathrm {tr}})$ . By a short explicit computation, one sees that $\mathcal {O}(\rho _B)$ maps to $\mathcal {O}({\mathrm {det}}\rho _B)=0$ under $H^2({\mathrm {tr}})\otimes _{\kappa }{\mathrm {id}}_I$ , and this implies the vanishing of $\mathcal {O}(\rho _B)$ .

We have now proved that there exists $\rho _A'\in {\mathcal {D}}_{\bar \rho }(A)$ mapping to $\rho _B\in {\mathcal {D}}_{\bar \rho }(B)$ . However, this lift need not satisfy ${\mathrm {det}}\rho ^{\prime }_{A}=\tau _A$ . At this point, we note that our hypothesis in fact implies that $H^2(G_K,{\operatorname {ad}_{{{\bar \rho }}}^0})=0$ so that

(13) $$ \begin{align} H^1(G_K,{\operatorname{ad}_{{{\bar\rho}}}})\,{\longrightarrow}\, H^1(G_K,{\operatorname{ad}_{{{\mathrm{det}}{\bar\rho}}}} )=H^1(G_K,\kappa) \end{align} $$

is surjective. Now, ${\mathrm {det}}\rho ^{\prime }_{A}$ and $\tau _A$ are deformations of $\tau _{B}$ and the space of all such deformations is a principal homogeneous space under $H^1(G_K,\kappa )$ , that is, the tangent space of the deformation problem, by [Reference SchlessingerSch68, Remark 2.15], and likewise the deformations of $\rho _B$ form a principal homogeneous space under $H^1(G_K,{\operatorname {ad}_{{{\bar \rho }}}})$ . Since the map (13) is surjective we can thus alter $\rho ^{\prime }_A$ by a class in $H^1(G_K,{\operatorname {ad}_{{{\bar \rho }}}})$ into some other deformation $\rho _A$ of $\rho _B$ that also satisfies ${\mathrm {det}}\rho _A=\tau _A$ . This completes the proof of the formal smoothness. Note also that if , then the above two sequences are exact on both sides and hence $H^0(G_K,\overline {{\operatorname {ad}_{{{\bar \rho }}}}}(1))=0$ .

By Proposition 3.1.5, it follows that the natural map ${R^{\operatorname {univ}}_{{\mathrm {det}} {\bar {\rho }}}} \rightarrow {R^{\operatorname {univ}}_{{\bar {\rho }}}}$ is formally smooth of relative dimension $h=h^1(K,{\operatorname {ad}_{{{\bar \rho }}}})-h^1(K,{\operatorname {ad}_{{{\mathrm {det}}{\bar \rho }}}} )$ . It remains to identify h with the number in the lemma. Since the map (13) is surjective, by the long exact sequence for $H^*(G_K,\cdot )$ applied to the sequence (11), we deduce

$$ \begin{align*}h=h^1(K,{\operatorname{ad}_{{{\bar\rho}}}^0}) -h^0(K,{\operatorname{ad}_{{{\mathrm{det}}{\bar\rho}}}} ) + h^0(K,{\operatorname{ad}_{{{\bar\rho}}}}) - h^0(K,{\operatorname{ad}_{{{\bar\rho}}}^0})=h^1(K,{\operatorname{ad}_{{{\bar\rho}}}^0}) -1+1- h^0(K,{\operatorname{ad}_{{{\bar\rho}}}^0}).\end{align*} $$

Since $h^2(K,{\operatorname {ad}_{{{\bar \rho }}}^0})=0$ by hypothesis and the duality statement of Theorem 3.4.1, the Euler characteristic formula of Theorem 3.4.1 implies $h=d(n^2-1)$ .