Proof.

1. 1. Let M be an $A*G$ -module. Then we define an isomorphism

   $$ \begin{align*} \alpha_M:I_GR_G(M)=\operatorname{\mathrm{k}} G\otimes_{A|}M\rightarrow M\otimes \operatorname{\mathrm{k}} G, g\otimes m\mapsto gm\otimes g. \end{align*} $$
   It is easy to check that this is an isomorphism of $A*G$ -modules and that $\alpha =(\alpha _M)_M$ defines a natural isomorphism.

2. 2. Let M be an A-module. Then we define an isomorphism

   $$ \begin{align*} \beta_M:R_GI_G(M)=_{A|}\operatorname{\mathrm{k}} G\otimes M\rightarrow \bigoplus_{g\in G}gM, g\otimes m\mapsto (\delta_{gg'}m)_{g'\in G}, \end{align*} $$
   where $\delta _{gg'}$ is the Kronecker delta of g and $g'$ . It is easy to check that this is an isomorphism of A-modules and that $\beta =(\beta _M)_M$ defines a natural isomorphism.

3. 3. This is obvious.

4. 4. $R_G$ is a restriction functor and thus exact. Moreover, since tensor products over $\operatorname {\mathrm {k}}$ are exact, $I_G$ and $-\otimes V$ are exact.

5. 5. Since A is an $A*G$ -module via $g\cdot a=g(a)$ ,

   $$ \begin{align*}I_G(A)\cong A\otimes \operatorname{\mathrm{k}} G=A*G.\end{align*} $$
   Since the projective modules in $\operatorname {\mathrm {mod}} A$ are exactly the modules isomorphic to direct sums of direct summands of A, and the projective modules in $\operatorname {\mathrm {mod}} A*G$ are exactly the modules isomorphic to direct sums of direct summands of $A*G$ , this implies that $I_G$ preserves projectives.

   On the other hand, if M is an A-module such that $I_G(M)$ is projective, then so is $R_GI_G(M)\cong \bigoplus _{g\in G}gM$ , and since $M=eM$ is a direct summand of $\bigoplus _{g\in G}gM$ , this implies that M is projective. Similarly,

   $$ \begin{align*}R_G(A*G)\cong \bigoplus_{g\in G}gA\cong |G|A.\end{align*} $$
   Since the projective modules in $\operatorname {\mathrm {mod}} A$ are exactly the modules isomorphic to direct sums of direct summands of A, and the projective modules in $\operatorname {\mathrm {mod}} A*G$ are exactly the modules isomorphic to direct sums of direct summands of $A*G$ , this implies that $R_G$ preserves projectives. Moreover, if M is an $A*G$ -module such that $R_GM$ is projective, then so is $I_GR_G(M)\cong M\otimes \operatorname {\mathrm {k}} G$ and since $\operatorname {\mathrm {k}} |\operatorname {\mathrm {k}} G$ ,
   $$ \begin{align*}M\cong M\otimes \operatorname{\mathrm{k}} |M\otimes \operatorname{\mathrm{k}} G.\end{align*} $$
   This implies that M is projective.

   Finally, since V is an indecomposable projective $\operatorname {\mathrm {k}} G$ -module, it is isomorphic to a direct summand of $\operatorname {\mathrm {k}} G$ , so that

   $$ \begin{align*}-\otimes V|-\otimes \operatorname{\mathrm{k}} G\cong I_GR_G,\end{align*} $$
   and thus $-\otimes V$ preserves and reflects projectives.

6. 6. Since $\operatorname {\mathrm {D}} \operatorname {\mathrm {k}} G\cong \operatorname {\mathrm {k}} G$ as a $\operatorname {\mathrm {k}} G$ -module, this is analogous to the previous statement replacing A by $\operatorname {\mathrm {D}} A$ and $A*G$ by $\operatorname {\mathrm {D}} A*G$

7. 7. Let S be a semisimple A-module. Then $\operatorname {\mathrm {rad}}(A)S=(0)$ and hence

   $$ \begin{align*} \operatorname{\mathrm{rad}}(A*G)I_G(S)=(\operatorname{\mathrm{rad}}(A)\otimes \operatorname{\mathrm{k}} G)\operatorname{\mathrm{k}} G\otimes S =\operatorname{\mathrm{rad}}(A)(\operatorname{\mathrm{k}} G\otimes S)\\ =\sum_{g\in G}\operatorname{\mathrm{k}} G\otimes (g(\operatorname{\mathrm{rad}}(A))S) =\sum_{g\in G}\operatorname{\mathrm{k}} G\otimes (\operatorname{\mathrm{rad}}(A)S) =(0), \end{align*} $$
   where the first equality follows from Lemma 2.11 and $g(\operatorname {\mathrm {rad}}(A))=\operatorname {\mathrm {rad}}(A)$ since G acts on A via algebra automorphisms. Thus, $I_G(S)$ is semisimple.

   On the other hand, if S is a simple $A*G$ module, then $\operatorname {\mathrm {rad}}(A*G)S=(0)$ and hence

   $$ \begin{align*} \operatorname{\mathrm{rad}}(A)R_G(S)=\operatorname{\mathrm{rad}}(A)S \subseteq \operatorname{\mathrm{rad}}(A*G)S =(0) \end{align*} $$
   so that $R_G(S)$ is semisimple. Thus, $R_G$ preserves and reflects semisimple modules.

   Since for every $A*G$ -module M we have

   $$ \begin{align*} R_G(M\otimes V)\cong R_G(M)\otimes V\cong \dim_{\operatorname{\mathrm{k}}}(V) R_G(M), \end{align*} $$
   this implies that $-\otimes V$ also preserves and reflects semisimple modules.

   Moreover, if S is an A-module such that $I_G(S)$ is semisimple, then so is $R_GI_G(S)$ and hence S, since S is a direct summand of $R_GI_G(S)\cong \bigoplus _{g\in G}gS$ ; and if S is an $A*G$ -module such that $R_G(S)$ is semisimple, then so is $I_GR_G(S)$ and hence S, since S is a direct summand of $I_GR_G(S)\cong S\otimes \operatorname {\mathrm {k}} G$ .

8. 8. Let M be an A-module. Then, by Lemma 2.11,

   $$ \begin{align*} \operatorname{\mathrm{rad}}(I_GM)=\operatorname{\mathrm{rad}}(A*G)I_GM=(\operatorname{\mathrm{rad}}(A)\otimes \operatorname{\mathrm{k}} G)(\operatorname{\mathrm{k}} G\otimes M). \end{align*} $$
   Since G acts on A via algebra automorphisms, we have $g(\operatorname {\mathrm {rad}}(A))=\operatorname {\mathrm {rad}}(A)$ for every $g\in G$ . Thus,
   $$ \begin{align*} (\operatorname{\mathrm{rad}}(A)\otimes \operatorname{\mathrm{k}} G)(\operatorname{\mathrm{k}} G\otimes M)=\operatorname{\mathrm{k}} G\otimes \operatorname{\mathrm{rad}}(A)M=\operatorname{\mathrm{k}} G\otimes \operatorname{\mathrm{rad}}(M)=I_G(\operatorname{\mathrm{rad}}(M)). \end{align*} $$

   Since $I_G$ is exact, we thus have $I_G\circ \operatorname {\mathrm {top}}\cong \operatorname {\mathrm {top}}\circ I_G$ .

   On the other hand, let M be an $A*G$ -module. Then

   $$ \begin{align*} R_G(\operatorname{\mathrm{rad}}(M))=R_G((\operatorname{\mathrm{rad}}(A)\otimes G)M)=R_G(\operatorname{\mathrm{rad}}(A)M)=\operatorname{\mathrm{rad}}(A)R_G(M), \end{align*} $$

   so that, since $R_G$ is exact, we have $R_G\circ \operatorname {\mathrm {top}}\cong \operatorname {\mathrm {top}}\circ R_G$ .

   Finally, let M be an $A*G$ -module. Then, by Lemma 2.11, we have

   $$ \begin{align*} \operatorname{\mathrm{rad}}(M\otimes V)=\operatorname{\mathrm{rad}}(A*G) (M\otimes V)=(\operatorname{\mathrm{rad}}(A)\otimes \operatorname{\mathrm{k}} G)(M\otimes V)\\ =(\operatorname{\mathrm{rad}}(A*G)M)\otimes V=\operatorname{\mathrm{rad}}(M)\otimes V. \end{align*} $$

   Thus, since $-\otimes V$ is exact, we have $(-\otimes V)\circ \operatorname {\mathrm {top}}\cong \operatorname {\mathrm {top}}\circ (-\otimes V)$ .

Proof.

1. 1.–3. These are obvious.

2. 4. $R_{G/H}$ is a restriction functor and thus exact. Moreover, $\operatorname {\mathrm {k}} H$ is semisimple, so that tensoring over $\operatorname {\mathrm {k}} H$ is exact.

3. 5.–7. Since

   $$ \begin{align*} I_{G/H}| I_{G/H}\circ (-\otimes \operatorname{\mathrm{k}} H)\cong I_{G/H}\circ I_H\circ R_H\cong I_G\circ R_H \end{align*} $$
   and
   $$ \begin{align*} R_{G/H}| (-\otimes \operatorname{\mathrm{k}} H)\circ R_{G/H}\cong I_H\circ R_H\circ R_{G/H}\cong I_{H}\circ R_G, \end{align*} $$
   this follows from Proposition 2.12(5–7) for H and G.

4. 8. This follows analogously to 2. in Proposition 2.12.

Proof. Let $L'$ be a simple $A*G$ -module. Then $R_GL'\cong \bigoplus _{L\in \operatorname {\mathrm {Sim}}(A)}[R_GL':L]L$ is semisimple and $L'$ is a summand of

$$ \begin{align*} L'\otimes \operatorname{\mathrm{k}} G\cong I_GR_G L'\cong \bigoplus_{L\in \operatorname{\mathrm{Sim}}(A)}[R_GL':L]I_GL. \end{align*} $$

Since $L'$ is simple, this implies that there is some $L\in \operatorname {\mathrm {Sim}}(A)$ such that $L'|I_GL$ .

The second statement is analogous.

Proof.

1. 1. By definition of a composition factor, there is a factor module $M"$ of M such that we have an embedding $\iota :L'\rightarrow M"$ .

   Now let N be a maximal submodule of $M"$ subject to $N\cap \operatorname {\mathrm {im}}(\iota )=(0)$ . Then ${M':=M"/N}$ is also a factor module of M. Denote by

   $$ \begin{align*}\pi:M"\rightarrow M'\end{align*} $$
   the canonical projection. Then $\pi \circ \iota :L'\rightarrow M'$ is injective, so that $L'|\operatorname {\mathrm {soc}}(M')$ . Write $\operatorname {\mathrm {soc}}(M')=\pi \circ \iota (L')\oplus S$ . Then $N\subseteq \pi ^{-1}(S)$ and $\iota (L')\cap \pi ^{-1}(S)=(0)$ . Hence, by maximality of N, $\pi ^{-1}(S)=N$ , so that $S=(0)$ .

2. 2. This is dual to 1.

Proof. Suppose $\leq $ is adapted, and let M be a module with a composition factor $L'$ such that $L'$ is incomparable to every summand L of its top $\operatorname {\mathrm {top}}(M)$ .

By Lemma 3.3, M has a factor module $M'$ with simple socle $L'$ . Let L be any summand of $\operatorname {\mathrm {top}}(M')$ . Then, since $M'$ is a factor module of M, we have $\operatorname {\mathrm {top}}(M')|\operatorname {\mathrm {top}}(M)$ . Hence, L is also a summand of $\operatorname {\mathrm {top}}(M)$ , and thus in particular incomparable to $L'$ . Moreover, we can again apply Lemma 3.3 to obtain a submodule $M"$ of $M'$ with simple top $\operatorname {\mathrm {top}}(M")\cong L$ . As $M"$ is a submodule of $M'$ , $\operatorname {\mathrm {soc}}(M")|\operatorname {\mathrm {soc}}(M')\cong L'$ , so that $M"$ has simple socle isomorphic to $L'$ . Since L and $L'$ are incomparable, $M"$ thus has a composition factor $L">L, L'$ , and since $L"$ is a composition factor of $M"$ , which is a submodule of a factor module of M, $L"$ is also a composition factor of M. Hence, this proves the first implication.

On the other hand, suppose that every module M which has a composition factor $L'$ such that $L'$ is incomparable to every summand of its top has a composition factor $L"$ and a composition factor $L|\operatorname {\mathrm {top}}(M)$ such that $L">L$ , and let M be a module with simple top L and simple socle $L'$ such that L and $L'$ are incomparable.

Then, by assumption, M has a composition factor $L"$ such that $L">L$ . Without loss of generality, we can choose $L"$ maximal with respect to $L">L$ . Then, by Lemma 3.3, M has a submodule $M'$ with simple top $\operatorname {\mathrm {top}}(M')\cong L"$ . Since M has simple socle $L'$ , $M'$ also has simple socle $L'$ . Now, if $L'$ and $L"$ were incomparable, then by assumption $M'$ would have a composition factor $L"'>L"$ , which is a contradiction to the maximality of $L"$ . Hence, $L'$ and $L"$ are comparable. Since $L">L$ and $L'$ and L are incomparable, this implies that $L'<L"$ .

Proof.

1. 1. We have $L'\nleq L$ for every summand $L'$ of $\operatorname {\mathrm {top}}(\ker (\pi _L))$ by definition. So let $f:P_L\rightarrow M$ be an epimorphism such that $L'\nleq L$ for every summand $L'$ of $\operatorname {\mathrm {top}}(\ker (f))$ . Then we have a projection $\pi :\bigoplus _{L'\nleq L}n_{L'}P_{L'}\rightarrow \ker (f)$ for some $n_{L'}\in \mathbb {N}_0$ . Composing with the embedding yields that

   $$ \begin{align*} \ker(f)\subseteq \sum_{L'\nleq L, \varphi\in \operatorname{\mathrm{Hom}}_A(P_{L'}, P_L)}\operatorname{\mathrm{im}}(\varphi)=\ker(\pi_L). \end{align*} $$
   Hence, $\pi _L$ factors through f.

2. 2. By definition, $L'\leq L$ for every composition factor $L'$ of $\Delta _L$ . So let $f:P_L\rightarrow M$ such that $L'\nleq L$ for every composition factor $L'$ of M. Let $L'\nleq L$ and let $\varphi :P_{L'}\rightarrow P_L$ . Then, since all composition factors of M are less than or equal to L, $f\circ \varphi =0$ . Hence, $\operatorname {\mathrm {im}}(\varphi )\subseteq \ker (f)$ , so that $\ker (\pi _L)\subseteq \ker (f)$ and thus f factors through $\pi _L$ .

3. 3. This is analogous to 1.

4. 4. This is analogous to 2.

Proof. By definition, the module $\widehat {\Delta }_{L'}$ has a projective presentation

for some integers $n_{L"}\in \mathbb {N}_0$ . Suppose $\operatorname {\mathrm {Ext}}^1_A(\widehat {\Delta }_{L'}, \Delta _{L})\neq (0)$ . Then

$$ \begin{align*} \operatorname{\mathrm{Hom}}_A({\bigoplus_{L">L'}n_{L"}P_{L"}}, \Delta_L)\neq (0), \end{align*} $$

so that for some $L">L'$ ,

$$ \begin{align*} \operatorname{\mathrm{Hom}}_A({P_{L"}}, \Delta_L)\neq (0). \end{align*} $$

Thus, $L"$ is a composition factor of $\Delta _L$ . Now since every composition factor of $\Delta _{L}$ is less than or equal to L, this implies $L' < L"\leq L$ .

Proof.

1. 1 ⇒ 2 Suppose $\Delta _L=\widehat {\Delta }_L$ for every $L\in \operatorname {\mathrm {Sim}}(A)$ , and let M be an A-module with simple top L and socle $L'$ . Suppose that no composition factor $L"$ of M is bigger than L. Since $L=\operatorname {\mathrm {top}}(M)$ , there is an epimorphism $\pi _M:P_L\rightarrow M$ . Now since no composition factor $L"$ of M is bigger than L, Lemma 3.6 yields a homomorphism $g:\widehat {\Delta }_L\rightarrow M$ such that $\pi _M=g\circ \widehat {\pi }_L$ . In particular, g is an epimorphism, so that, since every composition factor of $\widehat {\Delta }_L=\Delta _L$ is less than or equal to L, every composition factor of M is less than or equal to L. In particular, $L'$ and L are comparable.

2. 2 ⇒ 3 This holds by Remark 3.9.

3. 3 ⇒ 4 This holds by definition.

4. 4 ⇒ 1 Suppose that there is some $L\in \operatorname {\mathrm {Sim}}(A)$ such that $\Delta _L\neq \widehat {\Delta }_L$ . Then $\widehat {\Delta }_L$ has some composition factor $L'\nleq L$ . Let $L'$ be a maximal such composition factor. Then there is a nonzero homomorphism $f:P_{L'}\rightarrow \widehat {\Delta }_L$ . Moreover, since $L'$ is a maximal composition factor of $\widehat {\Delta }_L$ , we have $L"\ngtr L'$ for every composition factor $L"$ of $\widehat {\Delta }_L$ . Thus, Lemma 3.6 yields a homomorphism $g: \widehat {\Delta }_{L'}\rightarrow \widehat {\Delta }_L$ such that $f=g\circ \widehat {\pi }_{L'}$ . In particular,

   $$ \begin{align*} \operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_{L'}, \widehat{\Delta}_L)\neq (0).\\[-39pt] \end{align*} $$

Proof.

1. 1 ⇒ 2 Recall that if $\leq $ is adapted, then $(\Delta _L)_L=(\widehat {\Delta }_L)_L$ . Let $L'$ be a composition factor of $\operatorname {\mathrm {rad}}(\widehat {\Delta }_{L})=\operatorname {\mathrm {rad}}(\Delta _L)$ . Then $L'\leq L$ . If $L=L'$ , then this induces a nonzero homomorphism $P_{L}\rightarrow \operatorname {\mathrm {rad}}(\Delta _{L})$ and thus, by Lemma 3.6, an endomorphism $\Delta _{L}\rightarrow \operatorname {\mathrm {rad}}(\Delta _{L})\rightarrow \Delta _{L}$ which is neither zero nor invertible. This is a contradiction. Thus, $L'<L$ .

2. 2 ⇒ 3 Suppose $\operatorname {\mathrm {Hom}}_A(\widehat {\Delta }_{L'}, \operatorname {\mathrm {rad}}(\widehat {\Delta }_{L}))\neq (0)$ . Then, since $\widehat {\Delta }_{L'}$ has simple top $L'$ , $L'$ is a composition factor of $\operatorname {\mathrm {rad}}(\widehat {\Delta }_{L})$ and thus $L'<L$ .

3. 3 ⇒ 1 If $L\neq L'$ , then

   $$ \begin{align*} \operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_{L'}, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{L}))= (0)\Leftrightarrow \operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_{L'}, \widehat{\Delta}_{L})=(0). \end{align*} $$

   Thus,

   $$ \begin{align*} \operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_{L'}, \widehat{\Delta}_{L})\neq (0)\Rightarrow L'\leq L, \end{align*} $$

   so that $\leq $ is adapted, $\widehat {\Delta }_L=\Delta _L$ , and $(\Delta _L)_L$ is a standardizable set by Lemma 3.10.

   In particular, $\operatorname {\mathrm {Ext}}^1(\Delta _L, \Delta _L)=(0).$

   Moreover, $\operatorname {\mathrm {Hom}}_A(\widehat {\Delta }_L, \operatorname {\mathrm {rad}}(\widehat {\Delta }_L))=(0)$ , so that any endomorphism of $\widehat {\Delta }_L$ is either surjective or zero. Thus, $\operatorname {\mathrm {End}}_A(\Delta _L)=\operatorname {\mathrm {End}}_A(\widehat {\Delta }_L)\cong \operatorname {\mathrm {k}}.$

Proof. Let $L^{(A)}:=A/\operatorname {\mathrm {rad}}(A)$ and $L^{(B)}:=B/\operatorname {\mathrm {rad}}(B)$ . By Remark 3.18, $L^{(A)}$ and $L^{(B)}$ are up to isomorphism the unique maximal semisimple subalgebras of A and B, respectively.

In particular, a semisimple subalgebra of A is a maximal semisimple subalgebra if and only if it has the same vector space dimension as $L^{(A)}$ . Since any semisimple subalgebra of B is also a semisimple subalgebra of A, this means that B contains a maximal semisimple subalgebra of A if and only if $\dim _{\operatorname {\mathrm {k}}}L^{(A)}=\dim _{\operatorname {\mathrm {k}}}L^{(B)}$ .

Thus, B is a strong exact Borel subalgebra of A if and only if $\dim _{\operatorname {\mathrm {k}}}L^{(B)}=\dim _{\operatorname {\mathrm {k}}}L^{(A)}$ .

Let $L_1^B, \dots , L_n^B$ be a set of representative of the simple B-modules. Write $\Delta _i^A:=A\otimes _B L_i^B$ and $L_i^A:=\operatorname {\mathrm {top}}(A\otimes _B L_i^B)$ . Since B is an exact Borel subalgebra of A, $L_1^A, \dots , L_n^A$ is a set of representatives of the simple A-modules, and $\Delta _1^A, \dots , \Delta _n^A$ are the corresponding standard modules.

Let $X:=A\operatorname {\mathrm {rad}}(B)/(\operatorname {\mathrm {rad}}(A)\cap A\operatorname {\mathrm {rad}}(B))$ . Then, as A-modules,

$$ \begin{align*} L^{(A)}= A/\operatorname{\mathrm{rad}}(A) &\cong ((\operatorname{\mathrm{rad}}(A)+A\operatorname{\mathrm{rad}}(B))/\operatorname{\mathrm{rad}}(A))\oplus A/(\operatorname{\mathrm{rad}}(A)+A\operatorname{\mathrm{rad}}(B))\\ &\cong A\operatorname{\mathrm{rad}}(B)/(\operatorname{\mathrm{rad}}(A)\cap A\operatorname{\mathrm{rad}}(B))\oplus \operatorname{\mathrm{top}}(A/A\operatorname{\mathrm{rad}}(B))\\ &\cong X\oplus \operatorname{\mathrm{top}}\left(A\otimes_B B/\operatorname{\mathrm{rad}}(B)\right)\\ &\cong X\oplus \operatorname{\mathrm{top}}\left(\bigoplus_{i}[\operatorname{\mathrm{top}} B: L_i^B]\Delta_{L_i^A}\right)\\ &\cong X\oplus \bigoplus_{i}[L^{(B)}: L_i^B]L_i^A. \end{align*} $$

Recall that for any $n\in \mathbb {N}$ we have $\dim _{\operatorname {\mathrm {k}}}\operatorname {\mathrm {k}}^n=n=[\operatorname {\mathrm {Mat}}_n(\operatorname {\mathrm {k}}):\operatorname {\mathrm {k}}^n]$ . Moreover, by the Wedderburn–Artin theorem, any finite-dimensional semisimple algebra over $\operatorname {\mathrm {k}}$ is isomorphic to a direct sum of matrix rings, and for any two finite-dimensional algebras R and $R'$ , $\operatorname {\mathrm {Sim}}(R\oplus R')$ is the disjoint union of $\operatorname {\mathrm {Sim}}(R)$ and $\operatorname {\mathrm {Sim}}(R')$ where R acts on $L'\in \operatorname {\mathrm {Sim}}(R')$ via zero and analogously for $L\in \operatorname {\mathrm {Sim}}(R)$ . Hence, the equation above generalizes to any finite-dimensional semisimple algebra over $\operatorname {\mathrm {k}}$ , so that we have

$$ \begin{align*} \dim_{\operatorname{\mathrm{k}}}L_i^A&=[L^{(A)}:L_i^A] \end{align*} $$

for every $1\leq i\leq n$ . Thus,

$$ \begin{align*} \dim_{\operatorname{\mathrm{k}}}L_i^A&=[L^{(A)}:L_i^A]=[X:L_i^A]+[L^{(B)}: L_i^B]=[X:L_i^A]+\dim_{\operatorname{\mathrm{k}}}L_i^B, \end{align*} $$

so that

$$ \begin{align*} \dim_{\operatorname{\mathrm{k}}} L^{(A)}&=\dim_{\operatorname{\mathrm{k}}} X+\sum_i[L^{(B)}: L_i^B]\dim_{\operatorname{\mathrm{k}}} L_i^A\\ &= \dim_{\operatorname{\mathrm{k}}} X+\sum_i[L^{(B)}: L_i^B]([X:L_i^A]+\dim_{\operatorname{\mathrm{k}}}(L_i^B))\\ &=\dim_{\operatorname{\mathrm{k}}} X+\sum_{i}[X:L_i^A][L^{(B)}:L_i^B]+ \dim_{\operatorname{\mathrm{k}}}L^B. \end{align*} $$

Hence, B is a strong exact Borel subalgebra of A if and only if $X=(0)$ , that is, if $A\operatorname {\mathrm {rad}}(B)\subseteq B$ .

Proof.

1. 1. Clearly, $\leq _G$ is G-stable, and

   $$ \begin{align*} L<L'\Leftrightarrow \operatorname{\mathrm{k}} G\otimes L<_{G}\operatorname{\mathrm{k}} G\otimes L'. \end{align*} $$
   Now suppose that $\leq '$ is another G-stable partial order on $\operatorname {\mathrm {Sim}}(A*G)$ such that
   $$ \begin{align*} L<L'\Leftrightarrow \operatorname{\mathrm{k}} G\otimes L<'\operatorname{\mathrm{k}} G\otimes L'. \end{align*} $$
   Let $S, S'\in \operatorname {\mathrm {Sim}}(A*G)$ . Then there are $L, L'\in \operatorname {\mathrm {Sim}}(A)$ such that $L|_{A|}S$ , $L'|_{A|}S'$ . Thus, $\operatorname {\mathrm {k}} G\otimes L| \operatorname {\mathrm {k}} G\otimes {}_{A|}S\cong S\otimes \operatorname {\mathrm {k}} G$ and $\operatorname {\mathrm {k}} G\otimes L'| \operatorname {\mathrm {k}} G\otimes {}_{A|}S'\cong S'\otimes \operatorname {\mathrm {k}} G$ .

   Moreover, by Corollary 2.14, there are $L", L"'\in \operatorname {\mathrm {Sim}}(A)$ such that $S|\operatorname {\mathrm {k}} G\otimes L", S'|\operatorname {\mathrm {k}} G\otimes L"'$ . In particular, $L|{}_{A|}\operatorname {\mathrm {k}} G\otimes L"$ and $L'|{}_{A|}\operatorname {\mathrm {k}} G\otimes L"'$ , so that $L\cong gL"$ and $L'\cong hL"'$ for some $g, h\in G$ . Hence, $\operatorname {\mathrm {k}} G\otimes L\cong \operatorname {\mathrm {k}} G\otimes L"$ and $\operatorname {\mathrm {k}} G\otimes L'\cong \operatorname {\mathrm {k}} G\otimes L"'$ . Thus,

   $$ \begin{align*} S<'S'\Leftrightarrow S\otimes \operatorname{\mathrm{k}} G<'S'\otimes \operatorname{\mathrm{k}} G &\Rightarrow \operatorname{\mathrm{k}} G\otimes L<'\operatorname{\mathrm{k}} G\otimes L'\\ &\Leftrightarrow L<L' \Leftrightarrow \operatorname{\mathrm{k}} G\otimes L<_G\operatorname{\mathrm{k}} G\otimes L' \Rightarrow S<_{G}S', \end{align*} $$
   and analogously $S<_GS'\Rightarrow S<' S'.$

2. 2. Clearly, $<^{\prime }_{|G}$ is G-equivariant and

   $$ \begin{align*} {}_{A|}S<^{\prime}_{|G}{}_{A|}S'&\Leftrightarrow \operatorname{\mathrm{k}} G\otimes _{A|}S<'\operatorname{\mathrm{k}} G\otimes _{A|}S'\\ &\Leftrightarrow S\otimes \operatorname{\mathrm{k}} G<'S'\otimes \operatorname{\mathrm{k}} G\Leftrightarrow S<S'. \end{align*} $$
   Moreover, if $\leq $ is another G-equivariant partial order on $\operatorname {\mathrm {Sim}}(A)$ such that
   $$ \begin{align*} S<'S'\Leftrightarrow _{A|}S<_{A|}S', \end{align*} $$
   then
   $$ \begin{align*} L<L'&\Leftrightarrow \bigoplus_{g\in G}gL<\bigoplus_{g\in G}gL' \Leftrightarrow _{A|}\operatorname{\mathrm{k}} G\otimes L<_{A|}\operatorname{\mathrm{k}} G\otimes L'\\ &\Leftrightarrow \operatorname{\mathrm{k}} G\otimes L<'\operatorname{\mathrm{k}} G\otimes L' \Leftrightarrow L<^{\prime}_G L' \end{align*} $$
   for all $L, L'\in \operatorname {\mathrm {Sim}}(A)$ .

Proof. This is a direct consequence of the fact that g induces an order-preserving automorphism of $\operatorname {\mathrm {mod}} A$ .

Proof. Since $\operatorname {\mathrm {k}} G\otimes P_{L}$ is projective with top

$$ \begin{align*}\operatorname{\mathrm{top}}(\operatorname{\mathrm{k}} G\otimes P_{L})\cong \operatorname{\mathrm{k}} G\otimes \operatorname{\mathrm{top}}(P_{L})=\operatorname{\mathrm{k}} G\otimes L\end{align*} $$

by Proposition 2.12(8), there is an isomorphism

$$ \begin{align*} \varphi_L': \bigoplus_{S\in \operatorname{\mathrm{Sim}}(A*G)}[\operatorname{\mathrm{k}} G\otimes L:S]P_{S}\rightarrow \operatorname{\mathrm{k}} G\otimes P_{L}. \end{align*} $$

For the sake of notation, fix a set $\mathcal {S}_L$ of simple $A*G$ -modules such that for any $S\in \operatorname {\mathrm {Sim}}(A*G)$ there are exactly $[\operatorname {\mathrm {k}} G\otimes L:S]$ modules in $\mathcal {S}_L$ which are isomorphic to S and consider instead the isomorphism

$$ \begin{align*} \varphi_L: \bigoplus_{S\in \mathcal{S}_L}P_{S}\rightarrow \operatorname{\mathrm{k}} G\otimes P_{L}. \end{align*} $$

For every $S\in \mathcal {S}_L$ , consider the map

$$ \begin{align*} f_S:=(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)\circ \varphi_L\circ \iota_{P_{S}}:P_{S}\rightarrow \operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_{L}, \end{align*} $$

where $\iota _{P_{S}}:P_{S}\rightarrow \bigoplus _{S'\in \mathcal {S}_L}P_{S'}$ is the canonical embedding.

Then since $\widehat {\Delta }_{L}$ has a filtration by $L'\ngtr _A L$ and $\operatorname {\mathrm {k}} G\otimes -$ is exact, $\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }_{L}$ has a filtration by $\operatorname {\mathrm {k}} G\otimes L'$ such that $L'\ngtr _A L$ . Since for every $L'\ngtr _A L$ any two simple summands $S|\operatorname {\mathrm {k}} G\otimes L$ and $S'|\operatorname {\mathrm {k}} G\otimes L'$ fulfill $S'\ngtr _A S$ , this means that no composition factor $S'$ of $\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }_{L}$ is greater than any summand S of $\operatorname {\mathrm {k}} G\otimes L$ .

Hence, Lemma 3.6 implies that for any $S\in \mathcal {S}_L$ there is a homomorphism $\gamma _S:\widehat {\Delta }_{S}\rightarrow \operatorname {\mathrm {k}} G\otimes \widehat {\Delta }_{L}$ such that $f_S=\gamma _S\circ \widehat {\pi }_S$ . Thus, we obtain a homomorphism

$$ \begin{align*} \gamma:\bigoplus_{S\in \mathcal{S}_L}\widehat{\Delta}_{S}\rightarrow \operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_{L}, (x_S)_S\mapsto \sum_S \gamma_S(x_S) \end{align*} $$

such that

$$ \begin{align*} \gamma\circ (\widehat{\pi}_S)_S((x_S)_S)&=\sum_S \gamma_S\circ \widehat{\pi}_S(x_S) =\sum_S f_S(x_S)\\ &=\sum_S(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)\circ \varphi_L\circ \iota_{P_S}(x_S) =(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)\circ \varphi_L((x_S)_S). \end{align*} $$

In other words, the diagram

commutes. On the other hand, note that by Proposition 2.12(2) and Lemma 4.5 we have a commutative diagram

where the horizontal arrows are isomorphisms. Since $R_G(\widehat {\Delta }_S)$ has a filtration by $R_G(S')$ where $S'\in \operatorname {\mathrm {Sim}}(A*G)$ such that $S'\ngtr _{A*G} S$ and any simple summand $L'$ of $R_G(S')$ fulfills $S'|\operatorname {\mathrm {k}} G\otimes L'$ , we have for any composition factor $L'$ of $R_G(\widehat {\Delta }_S)$ that $\operatorname {\mathrm {k}} G\otimes L'\ngtr S$ . Thus, for $S|\operatorname {\mathrm {k}} G\otimes L$ , $R_G(\widehat {\Delta }_S)$ has no composition factor $L'$ such that $\operatorname {\mathrm {k}} G \otimes L'$ is greater than $\operatorname {\mathrm {k}} G\otimes L$ and thus no composition factor $L'$ which is greater than L.

Hence, analogously to the construction of $\gamma $ , we can construct an A-linear map $\gamma ':R_G(\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }_{L})\rightarrow R_G(\bigoplus _S \widehat {\Delta }_S)$ such that the diagram

commutes.

We obtain a diagram

where both squares commute, that is,

$$ \begin{align*} \gamma'\circ R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)&=R_G((\widehat{\pi}_S)_S)\circ R_G(\varphi_L)^{-1}\\ \text{and } R_G(\gamma)\circ R_G((\widehat{\pi}_S)_S)&= R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)\circ R_G(\varphi_L). \end{align*} $$

In particular,

$$ \begin{align*} R_G(\gamma)\circ \gamma'\circ R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)&=R_G(\gamma)\circ R_G((\widehat{\pi}_S)_S)\circ R_G(\varphi_L)^{-1}\\ &= R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)\circ R_G(\varphi_L)\circ R_G(\varphi_L)^{-1}\\ &=R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L) \end{align*} $$

and

$$ \begin{align*} \gamma'\circ R_G(\gamma)\circ R_G((\widehat{\pi}_S)_S)&=\gamma'\circ R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\pi}_L)\circ R_G(\varphi_L)\\ &=R_G((\widehat{\pi}_S)_S)\circ R_G(\varphi_L)^{-1}\circ R_G(\varphi_L)\\ &=R_G((\widehat{\pi}_S)_S). \end{align*} $$

Since $R_G(\operatorname {\mathrm {k}} G\otimes \widehat {\pi }_L)$ and $R_G((\widehat {\pi }_S)_S)$ are epimorphisms, this implies that $\gamma '=R_G(\gamma )^{-1}$ . In particular, $\gamma $ is bijective and hence an isomorphism.

Thus,

$$ \begin{align*} \operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_L\cong \bigoplus_{S\in \mathcal{S}_L}\widehat{\Delta}_S\cong \bigoplus_{S\in \operatorname{\mathrm{Sim}}(A*G)}[\operatorname{\mathrm{k}} G\otimes L:S]\widehat{\Delta}_S. \\[-53pt]\end{align*} $$

Proof. By Corollary 2.14, there is a simple A-module L such that $[\operatorname {\mathrm {k}} G\otimes L:S]\neq 0$ . By Proposition 4.6, this implies that $\widehat {\Delta }_S$ is a direct summand of $\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }_L$ . Hence, $ {}_{A|}\widehat {\Delta }_{S}$ is a direct summand of

$$ \begin{align*} R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_L)\cong \bigoplus_{g\in G}g\widehat{\Delta}_L\cong \bigoplus_{g\in G}\widehat{\Delta}_{gL}, \end{align*} $$

where the last isomorphism follows from Lemma 4.5. In particular, $ {}_{A|}\widehat {\Delta }_{S}$ is a direct sum of pseudostandard modules. Moreover, by Proposition 2.12(8),

$$ \begin{align*} \operatorname{\mathrm{top}}({}_{A|}\widehat{\Delta}_{S})\cong {}_{A|}\operatorname{\mathrm{top}}(\widehat{\Delta}_{S})\cong {}_{A|}S \end{align*} $$

so that, since every pseudostandard module $\widehat {\Delta }_L$ has simple top L, we obtain

$$ \begin{align*} {}_{A|}\widehat{\Delta}_{S}\cong\bigoplus_{L\in \operatorname{\mathrm{Sim}}(A)}[_{A|}S:L]\widehat{\Delta}_{L}.\\[-53pt] \end{align*} $$

Proof. The first claim follows directly from Proposition 4.6 and Corollary 2.14, while the second claim follows from Corollary 4.7 and Corollary 2.14.

Proof. By Proposition 4.6 and Corollary 2.13(1),

$$ \begin{align*} \bigoplus_{S\in \operatorname{\mathrm{Sim}}(A*G)}[\operatorname{\mathrm{k}} G\otimes L:S]\widehat{\Delta}_{S}\cong \operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_{L} &\cong \operatorname{\mathrm{k}} G\otimes_{\operatorname{\mathrm{k}} H_L} \operatorname{\mathrm{k}} H_L\otimes \widehat{\Delta}_{L}\\ &\cong \operatorname{\mathrm{k}} G\otimes_{\operatorname{\mathrm{k}} H_L} (\widehat{\Delta}_{L}\otimes \operatorname{\mathrm{k}} H_L)\\ &\cong \bigoplus_{V\in \operatorname{\mathrm{Sim}}(\operatorname{\mathrm{k}} H_L)}[\operatorname{\mathrm{k}} H_L:V]\operatorname{\mathrm{k}} G\otimes_{\operatorname{\mathrm{k}} H_L}(\widehat{\Delta}_L\otimes V). \end{align*} $$

Moreover, $\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_L} (\widehat {\Delta }_L\otimes V)$ has simple top $\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_L} (L\otimes V)$ and is in particular indecomposable.

Thus, it is isomorphic to $\widehat {\Delta }_{\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_L} (L\otimes V)}$ .

Moreover, $\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_L}(\widehat {\Delta }_L\otimes V)\cong \widehat {\Delta }_{\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_L} (L\otimes V)}$ and $\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_{L'}}( \widehat {\Delta }_{L'}\otimes W)\cong \widehat {\Delta }_{\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_{L'}} (L'\otimes W)}$ are isomorphic if and only if $\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_L} (L\otimes V)$ and $\operatorname {\mathrm {k}} G\otimes _{\operatorname {\mathrm {k}} H_{L'}} (L'\otimes W)$ are isomorphic, which, by Proposition 2.16, is the case if and only if there is a $g\in G$ such that $gL\cong L'$ and $gV\cong W$ .

Proof. By Lemma 3.13 and Proposition 4.3, $\leq _A$ is adapted and $(\Delta _L^A)_L$ is exceptional if and only if

$$ \begin{align*} \operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_L, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{L'}))\neq (0)\Rightarrow L<_AL' \end{align*} $$

and $\leq _{A*G}$ is adapted and $(\Delta _S^{A*G})_S$ is exceptional if and only if

$$ \begin{align*} &\operatorname{\mathrm{Hom}}_{A*G}(\widehat{\Delta}_S, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{S'}))\neq (0)\\ \Rightarrow &L<_AL'\text{ for some (equivalently all) }L, L' \text{ such that }S|\operatorname{\mathrm{k}} G\otimes L\text{ and } S'|\operatorname{\mathrm{k}} G\otimes L'. \end{align*} $$

Now if $\leq _A$ is adapted and $(\Delta _L^A)_L$ exceptional, then for all $S, S'\in \operatorname {\mathrm {Sim}}(A*G)$ and $L, L'\in \operatorname {\mathrm {Sim}}(A)$ such that $S|\operatorname {\mathrm {k}} G\otimes L, S'|\operatorname {\mathrm {k}} G\otimes L'$ we have

$$ \begin{align*} &\operatorname{\mathrm{Hom}}_{A*G}(\widehat{\Delta}_S, \operatorname{\mathrm{rad}}(\widehat{\Delta}_S')) \subseteq \operatorname{\mathrm{Hom}}_{A*G}(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_L, \operatorname{\mathrm{rad}}(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_{L'}))\\ \subseteq & \operatorname{\mathrm{Hom}}_{A}(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_L, \operatorname{\mathrm{rad}}(\operatorname{\mathrm{k}} G\otimes\widehat{\Delta}_{L'})) \cong \bigoplus_{g, g'\in G}\operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_{g(L)}, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{g'(L')})). \end{align*} $$

Thus, if $\operatorname {\mathrm {Hom}}_{A*G}(\widehat {\Delta }_S, \operatorname {\mathrm {rad}}(\widehat {\Delta }_{S'}))\neq (0)$ , there exist $g, g'\in G$ such that

$$ \begin{align*} \operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_{g(L)}, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{g'(L')}))\neq (0). \end{align*} $$

By assumption, this implies $gL<_A g'L$ , and hence, since $\leq _A$ is G-equivariant, $L<_A L'$ . Thus, $\operatorname {\mathrm {k}} G\otimes L<_{A*G} \operatorname {\mathrm {k}} G\otimes L'$ , so that $S<_{A*G}S'$ .

On the other hand, if $\leq _{A*G}$ is adapted and $(\Delta _S^{A*G})$ exceptional, then

$$ \begin{align*} &\operatorname{\mathrm{Hom}}_A(\widehat{\Delta}_L, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{L'}))\neq (0)\\ \Rightarrow &\operatorname{\mathrm{Hom}}_{A*G}(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_L, \operatorname{\mathrm{rad}}(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_{L'}) )\neq (0)\\ \Rightarrow &\bigoplus_{S|\operatorname{\mathrm{k}} G\otimes L, S'|\operatorname{\mathrm{k}} G\otimes L'}\operatorname{\mathrm{Hom}}_{A*G}(\widehat{\Delta}_S, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{S'})) \neq (0)\\ \Rightarrow &\exists S|\operatorname{\mathrm{k}} G\otimes L, S'|\operatorname{\mathrm{k}} G\otimes L': \operatorname{\mathrm{Hom}}_{A*G}(\widehat{\Delta}_S, \operatorname{\mathrm{rad}}(\widehat{\Delta}_{S'}))\neq (0)\\ \Rightarrow &\exists S|\operatorname{\mathrm{k}} G\otimes L, S'|\operatorname{\mathrm{k}} G\otimes L':S<_{A*G}S'\\ \Leftrightarrow &L<_AL'.\\[-40pt] \end{align*} $$

Proof.

1. 1. Suppose $(A, \leq _A)$ is quasi-hereditary. Then $\leq _A$ is adapted and $(\Delta _L^A)$ is exceptional. So, by Lemma 4.12, $\leq _{A*G}$ is adapted and $(\Delta _S^{A*G})$ is exceptional. Moreover, A has a filtration by the $\Delta _L=\widehat {\Delta }_L$ . Hence, $A*G\cong \operatorname {\mathrm {k}} G\otimes A$ has a filtration by $\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }_L$ . Since these decompose into a direct sum of $\widehat {\Delta }_S$ , where $S|\operatorname {\mathrm {k}} G\otimes L$ , by Proposition 4.6 and $\Delta _S=\widehat {\Delta }_S$ by Lemma 3.10, this implies that $A*G$ has a filtration by standard modules $\Delta _S=\widehat {\Delta }_S$ . Hence, $A*G\in F(\Delta ^{A*G})$ .

   On the other hand, suppose $A*G$ is quasi-hereditary. Then $\leq _{A*G}$ is adapted and $(\Delta _S^{A*G})$ is exceptional, so that $\leq _A$ is adapted and $(\Delta _L^A)$ is exceptional. Moreover, $A*G$ has a filtration as an $A*G$ -module by standard modules. Hence, as an A-module $A*G\cong \bigoplus _{g\in G}gA\cong |G|A$ has a filtration by the $R_G(\Delta _S)$ , and by Proposition 2.12(2), Lemma 3.10, and Lemma 4.5,

   $$ \begin{align*} R_G(\Delta_S)=R_G(\widehat{\Delta}_S)|R_G(\operatorname{\mathrm{k}} G\otimes \widehat{\Delta}_{L})\cong \bigoplus_{g\in G}\widehat{\Delta}_{g(L)}=\bigoplus_{g\in G}\Delta_{g(L)} \end{align*} $$
   for $S|\operatorname {\mathrm {k}} G\otimes L'$ , so that, since the standard modules of A are indecomposable, $R_G(\Delta _S)$ is a direct sum of standard modules and hence $|G|A\in \mathcal {F}(\Delta )$ . Since by Proposition 3.15, $\mathcal {F}(\Delta )$ is closed under direct summands, $A\in \mathcal {F}(\Delta )$ .

2. 2. By 1., $(A, \leq _A)$ is quasi-hereditary if and only if $(A*G, \leq _{A*G})$ is quasi-hereditary. Moreover, if the projective dimension of $\Delta ^A$ is less than or equal to 1, so is the projective dimension of $\Delta ^{A*G}=\widehat {\Delta }^{A*G}|\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }^A=\operatorname {\mathrm {k}} G\otimes \Delta ^A$ , and if the projective dimension of $\Delta ^{A*G}$ is less than or equal to 1, so is the projective dimension of $\Delta ^A=\widehat {\Delta }^A|R_G(\widehat {\Delta }^{A*G})=R_G(\Delta ^{A*G})$ .

3. 3. By 1., $(A, \leq _A)$ is quasi-hereditary if and only if $(A*G, \leq _{A*G})$ is quasi-hereditary. Moreover, if $\Delta ^A$ is semisimple, so is $\Delta ^{A*G}=\widehat {\Delta }^{A*G}|\operatorname {\mathrm {k}} G\otimes \widehat {\Delta }^A=\operatorname {\mathrm {k}} G\otimes \Delta ^A$ , and if $\Delta _{A*G}$ is semisimple, so is $\Delta ^A=\widehat {\Delta }^A|R_G(\widehat {\Delta }^{A*G})=R_G(\Delta ^{A*G})$ .

Proof. It is easy to see that the action given by Remark 2.4 is an action of algebra automorphisms. Let $f\otimes g\in \operatorname {\mathrm {End}}_A(R_GM)*G$ . Then, for any $h,h'\in G$ , $a\in A$ , $m\in M$ , we have

$$ \begin{align*} (\theta_M(f\otimes g))((a\otimes h')(h\otimes m))&=(\theta_M(f\otimes g))(h'h\otimes (hh')^{-1}(a)m)\\ &=h'hg^{-1}\otimes f(g(hh')^{-1}(a)m)\\ &=h'hg^{-1}\otimes ((hh'g^{-1})^{-1}(a)f(gm))\\ &=(a\otimes h')(hg^{-1}\otimes f(gm))\\ &=(a\otimes h')\theta_M(f\otimes g)(h\otimes m). \end{align*} $$

Thus, $\theta _M(f\otimes g)$ is $A*G$ -linear, so that $\theta _M$ is well defined. Moreover, for any $f, f'\in \operatorname {\mathrm {End}}_A(R_GM)$ , $g,g',h\in G$ , $m\in M$ , we have

$$ \begin{align*} \theta_M(f\otimes g)\circ\theta_M(f'\otimes g')(h\otimes m)&=\theta_M(f\otimes g)(h(g')^{-1}\otimes f'(g'm))\\ &=h(g')^{-1}g^{-1}\otimes f(gf'(g'm))\\ &=h(gg')^{-1}\otimes f\circ (g\cdot f')(gg'm)\\ &=\theta_M(f\circ (g\cdot f')\otimes gg')(h\otimes m)\\ &=\theta_M((f\otimes g)(f'\otimes g'))(h\otimes m). \end{align*} $$

Thus, $\theta _M$ is an algebra homomorphism. Let $t\in \operatorname {\mathrm {End}}_{A*G}((\operatorname {\mathrm {k}} G\otimes M))$ . Then we can identify $R_Gt\in \operatorname {\mathrm {End}}_{A}(\bigoplus _{g\in G}gM)$ with a matrix $R_Gt=(t_{g,h})_{g, h\in G}$ where

$$ \begin{align*} t_{g,h}:gM\rightarrow hM \end{align*} $$

and since t is $A*G$ -linear we have

$$ \begin{align*} t_{g,h}=t_{g'g,g'h} \end{align*} $$

for all $g,g',h\in G$ . For $g\in G$ , let

$$ \begin{align*} f_g(t):M\rightarrow M, f_g(t):=t_{g,e}\circ (\operatorname{\mathrm{tr}}_g^M)^{-1} \end{align*} $$

and set

$$ \begin{align*} \tau_M(t):=\sum_{g\in G}f_g\otimes g. \end{align*} $$

Then, for all $m\in M, h\in G$ , we have

$$ \begin{align*} \theta_M(\tau_M(t))(h\otimes m)&=\sum_{g\in G} \theta_M(f_g(t)\otimes g)(h\otimes m)\\ =\sum_{g\in G}hg^{-1}\otimes f_g(t)(gm)&=\sum_{g\in G}hg^{-1}=\sum_{g\in G} hg^{-1}\otimes t_{g,e}\circ (\operatorname{\mathrm{tr}}_g^M)^{-1}(gm)\\ =\sum_{g\in G} hg^{-1}\otimes t_{g,e}(m)&=\sum_{g\in G} hg^{-1}\otimes t_{h,hg^{-1}}(m)\\ =\sum_{g\in G}g\otimes t_{h,g}(m)&=t(h\otimes m). \end{align*} $$

On the other hand, for $f\otimes g\in \operatorname {\mathrm {End}}_A(R_GM)*G$ , we have

$$ \begin{align*} f_h(\theta_M(f\otimes g))=0 \end{align*} $$

unless $h=g$ and

$$ \begin{align*} f_g(\theta_M(f\otimes g))=f \end{align*} $$

Thus, $\tau _M(\theta _M(f\otimes g))=f\otimes g$ , so that $\tau _M=\theta _M^{-1}$ , and $\theta _M$ is an isomorphism.

Proof. First of all, note that $B*G$ is a well-defined subalgebra of $A*G$ , since $g(B)\subseteq B$ for all $g\in G$ , and that $A*G$ is quasi-hereditary by Theorem 4.14.

Let $(L_i^B)_{1\leq i\leq n}$ be a set of representatives of the isomorphism classes of simple B-modules, and let $(L_j^{B*G})_{1\leq i\leq m}$ be a set of representatives of the isomorphism classes of simple $B*G$ -modules.

Suppose first that $(B, \leq _B)$ is an exact Borel subalgebra of $(A, \leq _A)$ .

Recall that $R_G$ both preserves and reflects short exact sequences, and note that we have a natural isomorphism

(4.1) $$ \begin{align} (A\otimes_B -)\circ R_G\rightarrow R_G\circ ((A*G)\otimes_{B*G}-) \end{align} $$

given by

$$ \begin{align*} A\otimes_B R_GM\rightarrow R_G((A*G)\otimes_{B*G}M), a\otimes m\mapsto (a\otimes 1_G)\otimes m \end{align*} $$

with inverse given by

$$ \begin{align*} R_G((A*G)\otimes_{B*G}M)\rightarrow A\otimes_B R_GM, (a\otimes g)\otimes m\mapsto a\otimes gm. \end{align*} $$

Thus, if $A\otimes _B-$ is exact, so is $(A*G)\otimes _{B*G}-$ .

By assumption, we have a bijection

$$ \begin{align*} \operatorname{\mathrm{Sim}}(B)\rightarrow \operatorname{\mathrm{Sim}}(A), L_i^B\mapsto L_i^A:=\operatorname{\mathrm{top}}(A\otimes_B L_i^B) \end{align*} $$

such that $L_i^B\leq _BL_j^B$ if and only if $L_i^A\leq _AL_j^A$ . In particular, since

$$ \begin{align*} g(\operatorname{\mathrm{top}}(A\otimes_B L_i^B))=\operatorname{\mathrm{top}}(g(A\otimes_B L_B))\cong \operatorname{\mathrm{top}}(A\otimes_B gL_B)) \end{align*} $$

and $\leq _A$ is G-invariant, so is $\leq _B$ . Thus, it induces a partial order $\leq _{B*G}$ on $\operatorname {\mathrm {Sim}}(B*G)$ according to Proposition 4.3.

Next, we want to show that there is a bijection between isomorphism classes of simple modules given by

$$ \begin{align*} \operatorname{\mathrm{Sim}}(B*G)\rightarrow \operatorname{\mathrm{Sim}}(A*G), L^{B*G}_j\mapsto L^{A*G}_j:=\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_j^{B*G}). \end{align*} $$

Note that for this, it suffices to show that for any semisimple $B*G$ -module S, the induced map

$$ \begin{align*} s_2:\operatorname{\mathrm{End}}_{B*G}(S)\rightarrow \operatorname{\mathrm{End}}_{A*G}(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}S)), f\mapsto \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_{A*G}\otimes f) \end{align*} $$

is a bijection. Recall from Proposition 2.5 that $\operatorname {\mathrm {End}}_{B*G}(S)=\operatorname {\mathrm {End}}_B(R_GS)^G$ and $\operatorname {\mathrm {End}}_{A*G}(\operatorname {\mathrm {top}}((A*G)\otimes _{B*G}S))=\operatorname {\mathrm {End}}_{A}(R_G\operatorname {\mathrm {top}}((A*G)\otimes _{B*G}S))^G$ where the G-action is given as in Remark 2.6. Moreover, note that by Proposition 2.12(8) and 4.1, we have a natural isomorphism

$$ \begin{align*} \alpha: R_G\circ \operatorname{\mathrm{top}}\circ ((A*G)\otimes_{B*G}-)\rightarrow \operatorname{\mathrm{top}}\circ (A\otimes_B-)\circ R_G,\\ \alpha_M: R_G\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}M)\rightarrow \operatorname{\mathrm{top}}(A\otimes_B R_GM),\\ (a\otimes g)\otimes m+\operatorname{\mathrm{rad}}(A*G)\otimes M\mapsto a\otimes gm+\operatorname{\mathrm{rad}}(A)\otimes M \end{align*} $$

with inverse given by

$$ \begin{align*} \alpha_M^{-1}(a\otimes m+\operatorname{\mathrm{rad}}(A)\otimes M)=(a\otimes 1_G)\otimes m+\operatorname{\mathrm{rad}}(A*G)\otimes M. \end{align*} $$

Transporting the $A*G$ -module structure of $R_G\operatorname {\mathrm {top}}((A*G)\otimes _{B*G}S)$ along $\alpha _S$ gives rise to an $A*G$ -module structure on $\operatorname {\mathrm {top}}(A\otimes _B R_GS)$ , which is given by

$$ \begin{align*} g( a\otimes s+\operatorname{\mathrm{rad}}(A)\otimes S):=g(a)\otimes gs+\operatorname{\mathrm{rad}}(A)\otimes S. \end{align*} $$

According to Remark 2.4, this induces a G-action on $\operatorname {\mathrm {End}}_A(\operatorname {\mathrm {top}}(A\otimes _B R_GS))$ . With this G-action, the homomorphism

$$ \begin{align*} s_1: \operatorname{\mathrm{End}}_B(R_GS)\rightarrow \operatorname{\mathrm{End}}_A(\operatorname{\mathrm{top}}(A\otimes_B R_GS)),\\ f\mapsto \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f) \end{align*} $$

becomes G-equivariant, since we have

$$ \begin{align*} \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes (g \cdot f))(a\otimes s+\operatorname{\mathrm{rad}}(A)\otimes S)=a\otimes (g\cdot f)(s)+\operatorname{\mathrm{rad}}(A)\otimes S\\ =a\otimes g(f(g^{-1}s))+\operatorname{\mathrm{rad}}(A)\otimes S =g(\operatorname{\mathrm{id}}_A(g^{-1}(a)))\otimes g(f(g^{-1}s))+\operatorname{\mathrm{rad}}(A)\otimes S\\ =g(\operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f)(g^{-1}(s\otimes a+\operatorname{\mathrm{rad}}(A)\otimes S)))= g\cdot (\operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f))(a\otimes s+\operatorname{\mathrm{rad}}(A)\otimes S). \end{align*} $$

Additionally, since B is an exact Borel subalgebra of A, $s_1$ is an isomorphism. Hence, it restricts to an isomorphism

$$ \begin{align*} s_1^G:\operatorname{\mathrm{End}}_B(R_GS)^G\rightarrow \operatorname{\mathrm{End}}_A(\operatorname{\mathrm{top}}(A\otimes_B R_GS))^G,\\ f\mapsto \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f). \end{align*} $$

Moreover, for any $f\in \operatorname {\mathrm {End}}_B(R_GS)^G$ , we have

$$ \begin{align*} &\alpha_S^{-1} \circ\operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f)\circ \alpha_S((a\otimes g)\otimes s+\operatorname{\mathrm{rad}}(A*G)\otimes S)\\ =\ &\alpha_S^{-1} \circ\operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f)(a\otimes g(s)+\operatorname{\mathrm{rad}}(A)\otimes S)\\ =\ &\alpha_S^{-1}(a\otimes f(g(s))+\operatorname{\mathrm{rad}}(A)\otimes S)=\alpha_S^{-1}(a\otimes g(f(s))+\operatorname{\mathrm{rad}}(A)\otimes S)\\ =\ &(a\otimes 1)\otimes g(f(s))+\operatorname{\mathrm{rad}}(A*G)\otimes S=(a\otimes g)\otimes f(s)+\operatorname{\mathrm{rad}}(A*G)\otimes S\\ =\ &\operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_{A*G}\otimes f)((a\otimes g)\otimes s+\operatorname{\mathrm{rad}}(A*G)\otimes S). \end{align*} $$

Hence, conjugation of $s_1^G$ by $\alpha _S$ gives the desired isomorphism $s_2$ . Thus, we have a bijection

$$ \begin{align*} \operatorname{\mathrm{Sim}}(B*G)\rightarrow \operatorname{\mathrm{Sim}}(A*G)\\ L_j^{B*G}\mapsto L_j^{A*G}=\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_j^{B*G}). \end{align*} $$

Moreover, for every $L_i^{B*G}, L_j^{B*G}\in \operatorname {\mathrm {Sim}}(B*G)$ , we have, using Definition 4.2, Proposition 4.3, and the fact that B is an exact Borel subalgebra of A, that

$$ \begin{align*} L_i^{B*G}<_{B*G}L_j^{B*G}\Leftrightarrow R_G L_i^{B*G}<_{B}R_GL_j^{B*G}\Leftrightarrow \operatorname{\mathrm{top}}(A\otimes_B R_GL_i^{B*G})<_A \operatorname{\mathrm{top}}(A\otimes_B R_GL_j^{B*G})\\ \Leftrightarrow R_G(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_i^{B*G}))<_A R_G(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_j^{B*G}))\\ \Leftrightarrow \operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_i^{B*G})<_{A*G}\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_j^{B*G})\Leftrightarrow L_i^{A*G}<L_j^{A*G}. \end{align*} $$

Finally, by Proposition 4.6,

$$ \begin{align*} A*G\otimes_{B*G}(\operatorname{\mathrm{k}} G \otimes L^B)&\cong \operatorname{\mathrm{k}} G\otimes (A\otimes_B L^B)\\ &\cong \operatorname{\mathrm{k}} G\otimes \bigoplus_{L_i^A\in \operatorname{\mathrm{Sim}}(A)}\Delta_{L_i^A}\\ &\cong \bigoplus_{L_i^A\in \operatorname{\mathrm{Sim}}(A)}\bigoplus_{L_j^{A*G}\in \operatorname{\mathrm{Sim}}(A*G)}[\operatorname{\mathrm{k}} G\otimes L_i^A:L_j^{A*G}]\Delta_{L_j^{A*G}}. \end{align*} $$

Since for every simple $B*G$ -module $L_j^{B*G}$ we have that $A*G\otimes _{B*G}L_j^{B*G}$ is a summand of $A*G\otimes _{B*G}\operatorname {\mathrm {k}} G \otimes L^B$ , the module $A*G\otimes _{B*G}L_j^{B*G}$ is isomorphic to a direct sum of standard modules. Since it has indecomposable top $L_j^{A*G}$ , this implies that $A*G\otimes _{B*G}L_j^{B*G}\cong \Delta _{L_j^A*G}$ .

Thus, $(B*G, \leq _{B*G})$ is an exact Borel subalgebra of $(A*G, \leq _{A*G})$ .

On the other hand, suppose that $(B*G, \leq _{B*G})$ is an exact Borel subalgebra of $(A*G, \leq _{A*G})$ . Recall that $I_G$ both preserves and reflects short exact sequences, and note that we have a natural isomorphism

(4.2) $$ \begin{align} ((A*G)\otimes_{B*G} -)\circ I_G\rightarrow I_G\circ (A\otimes_{B}-), \end{align} $$

given by

$$ \begin{align*} (A*G)\otimes_{B*G} \operatorname{\mathrm{k}} G\otimes M\rightarrow \operatorname{\mathrm{k}} G\otimes A\otimes_{B} M, (a\otimes g)\otimes h\otimes m\mapsto gh\otimes gh(a)\otimes m \end{align*} $$

with inverse given by

$$ \begin{align*} \operatorname{\mathrm{k}} G\otimes A\otimes_{B} M \rightarrow (A*G)\otimes_{B*G} \operatorname{\mathrm{k}} G\otimes M, g\otimes a\otimes m\mapsto (g^{-1}(a)\otimes 1)\otimes g\otimes m. \end{align*} $$

Thus, if $(A*G)\otimes _{B*G}-$ is exact, so is $A\otimes _{B}-$ .

By assumption, we have a bijection

$$ \begin{align*} \operatorname{\mathrm{Sim}}(B*G)\rightarrow \operatorname{\mathrm{Sim}}(A*G), L_j^{B*G}\mapsto L_j^{A*G}:=\operatorname{\mathrm{top}}((A*G)\otimes_{B*G} L_j^{B*G}) \end{align*} $$

such that $L_i^{B*G}\leq _{B*G}L_j^{B*G}$ if and only if $L_i^{A*G}\leq _{A*G}L_j^{A*G}$ . In particular, since

$$ \begin{align*} (A*G)\otimes_{B*G}(L_j^{B*G}\otimes V)\cong (((A*G)\otimes_{B*G}L_j^{B*G})\otimes V) \end{align*} $$

for every irreducible representation V of G, and since $\leq _{A*G}$ is G-stable, so is $\leq _{B*G}$ . Thus, it induces a partial order $\leq _B$ on $\operatorname {\mathrm {Sim}}(B)$ according to Proposition 4.3.

Next, we want to show that there is a bijection between isomorphism classes of simple modules given by

$$ \begin{align*} \operatorname{\mathrm{Sim}}(B)\rightarrow \operatorname{\mathrm{Sim}}(A), L^{B}_i\mapsto L^{A}_i:=\operatorname{\mathrm{top}}(A\otimes_{B}L_i^{B}). \end{align*} $$

Note that for this, it suffices to show that for any semisimple B-module S, the induced map

(4.3) $$ \begin{align} s_1:\operatorname{\mathrm{End}}_{B}(S)\rightarrow \operatorname{\mathrm{End}}_{A}(\operatorname{\mathrm{top}}(A\otimes_{B}S)), f\mapsto \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_{A}\otimes f) \end{align} $$

is a bijection, and, in fact, this holds as soon as it holds for some semisimple B-module S such that $[S:L_i^B]\neq 0$ for all $L_i^B\in \operatorname {\mathrm {Sim}}(B)$ . By Corollary 2.14, it thus suffices to consider the case $S=R_GL^{B*G}$ , where $L^{B*G}=\bigoplus _{L_j^{B*G}\in \operatorname {\mathrm {Sim}}(B*G)}L_j^{B*G}$ . By Lemma 4.16, we have an isomorphism

$$ \begin{align*} \theta_{L^{B*G}}:\operatorname{\mathrm{End}}_B(R_GL^{B*G})*G\rightarrow \operatorname{\mathrm{End}}_{B*G}(I_GR_GL^{B*G}), f\otimes g\mapsto (h\otimes x\mapsto hg^{-1}\otimes f(gx)) \end{align*} $$

as well as an isomorphism

$$ \begin{align*} \theta_{(A*G)\otimes_{B*G}L^{B*G}}:\operatorname{\mathrm{End}}_B(R_G(A*G)\otimes_{B*G}L^{B*G})*G\rightarrow \operatorname{\mathrm{End}}_{A*G}(I_GR_G(A*G)\otimes_{B*G}L^{B*G}),\\ f\otimes g\mapsto (h\otimes (a\otimes g)\otimes m\mapsto hg^{-1}\otimes f(g\cdot ((a\otimes g)\otimes m))). \end{align*} $$

Moreover, since $B*G$ is a regular exact Borel subalgebra of $A*G$ , we have an isomorphism

$$ \begin{align*} s_2:\operatorname{\mathrm{End}}_{B*G}(I_GR_GL^{B*G})\rightarrow \operatorname{\mathrm{End}}_{A*G}(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}I_GR_GL^{B*G})) f\mapsto \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_{A*G}\otimes f). \end{align*} $$

Additionally, by Proposition 2.12(8) and (4.2), there are natural isomorphisms

$$ \begin{align*} \phi_1:\operatorname{\mathrm{top}}\circ (A\otimes_B-)\circ R_G&\rightarrow R_G\circ \operatorname{\mathrm{top}}\circ ((A*G)\otimes_{B*G}-),\\ \phi_1^M:\operatorname{\mathrm{top}}(A\otimes_B R_GM)&\rightarrow R_G(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}M)),\\ a\otimes m +\operatorname{\mathrm{rad}}(A)\otimes R_GM&\mapsto (a\otimes 1_G)\otimes m+\operatorname{\mathrm{rad}}(A*G)\otimes M \end{align*} $$

and

$$ \begin{align*} \phi_2:I_G\circ R_G\circ \operatorname{\mathrm{top}}\circ ((A*G)\otimes_{B*G}-)&\rightarrow \operatorname{\mathrm{top}}\circ ((A*G)\otimes_{B*G}-)\circ I_G\circ R_G\\ \phi_2^M:I_GR_G\operatorname{\mathrm{top}}(((A*G)\otimes_{B*G}M))&\rightarrow \operatorname{\mathrm{top}}(A*G)\otimes_{B*G}I_GR_GM,\\ h\otimes ((a\otimes g)\otimes m+I_GR_G\operatorname{\mathrm{rad}}(A*G)&\otimes M)\\ \mapsto & (h(a)\otimes h)\otimes (1_G\otimes g(m))+\operatorname{\mathrm{rad}}(A*G)\otimes I_GR_GM, \end{align*} $$

which give rise to isomorphisms

$$ \begin{align*} \varphi_1:\operatorname{\mathrm{End}}_A(\operatorname{\mathrm{top}}(A\otimes_B R_GL^{B*G}))\rightarrow \operatorname{\mathrm{End}}_A(R_G(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L^{B*G}))) \end{align*} $$

and

$$ \begin{align*} \varphi_2:\operatorname{\mathrm{End}}_{A*G}(\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}I_GR_GL^{B*G}))\rightarrow \operatorname{\mathrm{End}}_{A*G}(I_GR_G\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L^{B*G})). \end{align*} $$

Now consider the diagram

For $f\in \operatorname {\mathrm {End}}_B(R_GL^{B*G})$ , $g, h\in G$ , $a\in A$ , and $x\in L^{B*G}$ , we have

$$ \begin{align*} &\varphi_2\circ s_2\circ \theta_{R_GL^{B*G}}(f\otimes g)(1_G\otimes ((1_A\otimes 1_G)\otimes x+\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G}))\\ &=(\phi_2^{L^{B*G}})^{-1}\circ (s_2\circ \theta_{R_GL^{B*G}})(f\otimes g)\circ \phi_2^{L^{B*G}}(1_G\otimes ((1_A\otimes 1_G)\otimes x+\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G}))\\ &=(\phi_2^{L^{B*G}})^{-1}\circ (s_2\circ \theta_{R_GL^{B*G}})(f\otimes g)((1_A\otimes 1_G)\otimes 1_G\otimes x+\operatorname{\mathrm{rad}}(A*G)\otimes I_GR_GL^{B*G})\\ &=(\phi_2^{L^{B*G}})^{-1} \circ (\operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_{A*G}\otimes \theta_{R_GL^{B*G}}(f\otimes g)))((1_A\otimes 1_G)\otimes 1_G\otimes x+\operatorname{\mathrm{rad}}(A*G)\otimes I_GR_GL^{B*G})\\ &=(\phi_2^{L^{B*G}})^{-1}((1_A\otimes 1_G)\otimes \theta_{R_GL^{B*G}}(f\otimes g)(1_G\otimes x)+\operatorname{\mathrm{rad}}(A*G)\otimes I_GR_GL^{B*G})\\ &=(\phi_2^{L^{B*G}})^{-1}((1_A\otimes 1_G)\otimes g^{-1}\otimes f(gx)+\operatorname{\mathrm{rad}}(A*G)\otimes I_GR_GL^{B*G})\\ &=(\phi_2^{L^{B*G}})^{-1}((g^{-1}(1_A)\otimes g^{-1})\otimes 1_G\otimes f(gx)+\operatorname{\mathrm{rad}}(A*G)\otimes I_GR_GL^{B*G})\\ &=g^{-1}\otimes ((1_A\otimes 1_G)\otimes f(g(x)+\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G})) \end{align*} $$

and

$$ \begin{align*} &\theta_{\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L^{B*G})}\circ (\varphi_1\otimes\operatorname{\mathrm{id}}_{\operatorname{\mathrm{k}} G})\circ (s_1\otimes\operatorname{\mathrm{id}}_{\operatorname{\mathrm{k}} G})(f\otimes g)(1_G\otimes (1_A\otimes 1_G)\otimes x\\ &\hspace{3cm} +I_GR_G\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G})\\ &=\theta_{\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L^{B*G})}((\varphi_1\circ s_1)(f)\otimes g)(1_G\otimes ((1_A\otimes 1_G)\otimes x+I_GR_G\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G}))\\ &=g^{-1}\otimes ((\varphi_1\circ s_1)(f))(g((1_A\otimes 1_G)\otimes x+\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G}))\\ &=g^{-1}\otimes \phi_1^{L^{B*G}}\circ s_1(f)\circ (\phi_1^{L^{B*G}})^{-1}(g((1_A\otimes 1_G)\otimes x+\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G}))\\ &=g^{-1}\otimes \phi_1^{L^{B*G}}\circ s_1(f)(1_A\otimes gx+\operatorname{\mathrm{rad}}(A)\otimes R_GL^{B*G})\\ &=g^{-1}\otimes \phi_1^{L^{B*G}}\circ \operatorname{\mathrm{top}}(\operatorname{\mathrm{id}}_A\otimes f)(1_A\otimes gx+\operatorname{\mathrm{rad}}(A)\otimes R_GL^{B*G})\\ &=g^{-1}\otimes \phi_1^{L^{B*G}}(1_A\otimes f(gx)+\operatorname{\mathrm{rad}}(A)\otimes R_GL^{B*G})\\ &=g^{-1}\otimes ((1_A\otimes 1_G)\otimes f(gx)+\operatorname{\mathrm{rad}}(A*G)\otimes L^{B*G}). \end{align*} $$

As all maps are $A*G$ -linear, and elements of the form $1_G\otimes ((1_A\otimes 1_G)\otimes x+I_GR_G\operatorname {\mathrm {rad}}(A*G)\otimes L^{B*G})$ generate $I_GR_G\operatorname {\mathrm {top}}((A*G)\otimes _{B*G}L^{B*G})$ as an $A*G$ -module, this proves that the diagram commutes. Since we know all maps except $s_1\otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}$ to be isomorphisms, we can conclude that $s_1\otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}$ , and thus $s_1$ is an isomorphism. This shows that we have a bijection between the isomorphism classes of simple modules given by

$$ \begin{align*} \operatorname{\mathrm{Sim}}(B)\rightarrow \operatorname{\mathrm{Sim}}(A), L^{B}_i\mapsto L^{A}_i:=\operatorname{\mathrm{top}}(A\otimes_{B}L_i^{B}). \end{align*} $$

Moreover, using Definition 4.2, Proposition 4.3, and the fact that $B*G$ is an exact Borel subalgebra of $A*G$ , we have that

$$ \begin{align*} L_i^B<_B L_j^B\Leftrightarrow &I_G L_i^B<_{B*G} I_GL_j^B \\ \Leftrightarrow &\operatorname{\mathrm{top}}((A*G)\otimes_{B*G}I_G L_i^B)<_{A*G} \operatorname{\mathrm{top}}((A*G)\otimes_{B*G}L_j^B)\\ \Leftrightarrow &I_G (\operatorname{\mathrm{top}}(A\otimes_B L_i^B))<_{A*G} I_G(\operatorname{\mathrm{top}}(A\otimes_B L_j^B)) \\ \Leftrightarrow &\operatorname{\mathrm{top}}(A\otimes_B L_i^B)<_A \operatorname{\mathrm{top}}(A\otimes_B L_j^B)\Leftrightarrow L_i^A<_A L_j^A. \end{align*} $$

Finally, by Corollary 4.7,

$$ \begin{align*} A\otimes_B R_GL^{B*G}\cong R_G((A*G)\otimes_{B*G}L^{B*G})\cong R_G(\Delta^{A*G})\cong\bigoplus_{i=1}^n [L^{A*G}:L_i^A]\Delta^A_{L_i^A}. \end{align*} $$

Since for every simple B-module $L_i^{B}$ we have that $A\otimes _B L_i^B$ is a summand of $A\otimes _B R_GL^{B*G}$ , the module $A\otimes _B L_i^B$ is isomorphic to a direct sum of standard modules. Since it has indecomposable top $L_i^{A}$ , this implies that $A\otimes _{B}L_i^{B}\cong \Delta _{L_i^A}$ .

Thus, $(B, \leq _{B})$ is an exact Borel subalgebra of $(A, \leq _{A})$ .

Proof.

1. 1. Suppose B is a strong exact Borel subalgebra. Then, by Lemma 3.19, $A\operatorname {\mathrm {rad}}(B)\subseteq \operatorname {\mathrm {rad}}(A)$ . Hence, $A*G\operatorname {\mathrm {rad}}(B*G)\subseteq \operatorname {\mathrm {rad}}(A*G)$ by Lemma 2.11, so that again by Lemma 3.19 $B*G$ is a strong Borel subalgebra of $A*G$ . On the other hand, suppose that $B*G$ is a strong Borel subalgebra of $A*G$ . Then $A*G\operatorname {\mathrm {rad}}(B*G)\subseteq \operatorname {\mathrm {rad}}(A*G)$ , so that again by Lemma 2.11

   $$ \begin{align*} A\operatorname{\mathrm{rad}}(B)\subseteq (A*G\operatorname{\mathrm{rad}}(B*G))\cap A\subseteq \operatorname{\mathrm{rad}}(A*G)\cap A=\operatorname{\mathrm{rad}}(A). \end{align*} $$

2. 2. Suppose that B is normal. Then the inclusion $\iota :B\rightarrow A$ has a splitting $\pi :A\rightarrow B$ as right B-modules whose kernel is a right ideal of A. Since tensoring over $\operatorname {\mathrm {k}}$ is exact, the inclusion $\iota \otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}$ has the splitting $\pi \otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}$ which is a right $B*G$ -module homomorphism whose kernel is a right ideal of $A*G$ .

   On the other hand, suppose that $B*G$ is normal. Then $\iota \otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}$ has a splitting $\pi ':A*G\rightarrow B*G$ of right $B*G$ -modules whose kernel is a right ideal in $A*G$ . Since the fixed point functor $-^G$ for the G-action given by left multiplication is exact by [Reference Martínez Villa21, Lem. 3], $(\pi ')^G$ is a splitting of the embedding $(\iota \otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G})^G$ as right $(B*G)^G$ -modules such that its kernel is a right ideal in $(A*G)^G$ . Now since the upward arrows in the commutative diagram

   
   are isomorphisms of algebras, $(\pi ')^G$ induces a splitting of $\iota $ as right B-modules such that its kernel is a right ideal in A.

3. 3.+4. Assume that B is homological (resp. regular). In the latter case, we have already seen that $B*G$ is normal. Let $P^B$ be a projective resolution of $L^B$ . Then $A\otimes _B P^B$ is a projective resolution of $\Delta ^A$ , $\operatorname {\mathrm {k}} G\otimes P^B$ is a projective resolution of $\operatorname {\mathrm {k}} G\otimes L^B$ , $\operatorname {\mathrm {k}} G\otimes (A\otimes _B P^B)$ is a projective resolution of $\operatorname {\mathrm {k}} G\otimes (A\otimes _B L^B)$ , and similarly for the restriction of the induction. Now G acts on $\operatorname {\mathrm {End}}_B(R_G(\operatorname {\mathrm {k}} G\otimes P^B))$ and on $\operatorname {\mathrm {End}}_A(R_G(\operatorname {\mathrm {k}} G\otimes (A\otimes _B P^B)))$ , and since $R_G(\operatorname {\mathrm {k}} G\otimes L^B)$ is semisimple, the map

   $$ \begin{align*} \operatorname{\mathrm{End}}_B(R_G(\operatorname{\mathrm{k}} G\otimes P^B))\rightarrow \operatorname{\mathrm{End}}_A(A\otimes_B R_G(\operatorname{\mathrm{k}} G\otimes P^B)), f\mapsto \operatorname{\mathrm{id}}_A\otimes f \end{align*} $$
   is an epimorphism in homology of degree 1 and an isomorphism in homology of degree strictly greater than 1 (resp. an isomorphism in homology of degree strictly greater than 0).As in Theorem 4.17 conjugating with the isomorphism (4.1)
   $$ \begin{align*} A\otimes_B R_G(\operatorname{\mathrm{k}} G\otimes P^B)\cong R_G((A*G)\otimes_{B*G}(\operatorname{\mathrm{k}} G\otimes P^B)) \end{align*} $$
   yields an isomorphism
   $$ \begin{align*} \operatorname{\mathrm{End}}_B(R_G(\operatorname{\mathrm{k}} G\otimes P^B))\rightarrow \operatorname{\mathrm{End}}_A(R_G(A*G)\otimes_{B*G}(\operatorname{\mathrm{k}} G\otimes P^B)), f\mapsto \operatorname{\mathrm{id}}_{A*G}\otimes f, \end{align*} $$
   which is, as before, G-equivariant, so that it induces a homomorphism
   $$ \begin{align*} &\operatorname{\mathrm{End}}_{B*G}(\operatorname{\mathrm{k}} G\otimes P^B)=\operatorname{\mathrm{End}}_B(R_G(\operatorname{\mathrm{k}} G\otimes P^B))^G\\ \rightarrow &\operatorname{\mathrm{End}}_{A*G}((A*G)\otimes_{B*G}(\operatorname{\mathrm{k}} G\otimes P^B))=\operatorname{\mathrm{End}}_A(R_G(A*G)\otimes_{B*G}(\operatorname{\mathrm{k}} G\otimes P^B))^G,\\ f&\mapsto \operatorname{\mathrm{id}}_{A*G}\otimes f. \end{align*} $$
   Since the fixed point functor $-^G$ is exact, this is an epimorphism in homology of degree 1 and an isomorphism in homology of degree strictly greater than 1 (resp. an isomorphism in homology of degree strictly greater than 0).Thus, we obtain an epimorphism in degree 1 and an isomorphism in degree strictly greater than 1 (resp. an isomorphism in degree strictly greater than 0)
   $$ \begin{align*} \operatorname{\mathrm{Ext}}_{B*G}^{*}(\operatorname{\mathrm{k}} G\otimes L^B, \operatorname{\mathrm{k}} G\otimes L^B)&\rightarrow \operatorname{\mathrm{Ext}}_{A*G}^{*}((A*G)\otimes_{B*G}L^B, (A*G)\otimes_{B*G}L^B),\\ [f]&\mapsto [\operatorname{\mathrm{id}}_{A*G}\otimes f]. \end{align*} $$
   The result now follows from Corollary 2.14.

   On the other hand, suppose $B*G$ is homological (resp. regular). In the latter case, we have already seen that B is normal. Let $P^{B*G}$ be a projective resolution of $L^{B*G}$ . Then $(A*G)\otimes _{B*G} P^{B*G}$ is a projective resolution of $\Delta _{A*G}$ , $R_GP^{B*G}$ is a projective resolution of $R_G(L^{B*G})$ , and $R_G((A*G)\otimes _{B*G} P^{B*G})\cong A\otimes _B R_G(P^{B*G})$ is a projective resolution of $R_G(A*G)\otimes _{B*G} L^{B*G})\cong A\otimes _B R_G(L^{B*G})$ and similarly for the induction of the restriction. Since $L^{B*G}\otimes \operatorname {\mathrm {k}} G$ is semisimple, we have that by assumption

   $$ \begin{align*} \operatorname{\mathrm{Ext}}^*_{B*G}(L^{B*G}\otimes \operatorname{\mathrm{k}} G, L^{B*G}\otimes \operatorname{\mathrm{k}} G)\rightarrow \operatorname{\mathrm{Ext}}^*_{A*G}(\Delta^{A*G}\otimes \operatorname{\mathrm{k}} G, \Delta^{A*G}\otimes \operatorname{\mathrm{k}} G) \end{align*} $$
   is an epimorphism in degree 1 and an isomorphism in degree strictly greater than 1 (resp. an isomorphism in degree strictly greater than 0).Moreover, G acts on $\operatorname {\mathrm {End}}_{B}(R_G(P^{B*G}))$ and on $\operatorname {\mathrm {End}}_A(R_G((A*G)\otimes _{B*G}P^{B*G}))$ via $g\cdot f=gf(g^{-1}-)$ and, arguing as in Theorem 4.17, we obtain a commutative diagram
   
   where $\theta _{P^{B*G}}$ and $\theta _{P^{B*G}}$ and $\theta _{(A*G)\otimes _{B*G}P^{B*G}}$ are the isomorphisms from Lemma 4.16,
   $$ \begin{align*} s_1:\operatorname{\mathrm{End}}_{B}(R_GP^{B*G})\rightarrow \operatorname{\mathrm{End}}_A(A\otimes_B P^{B*G}), f\mapsto \operatorname{\mathrm{id}}_A\otimes f\\ s_2:\operatorname{\mathrm{End}}_{B*G}(I_GR_GP^{B*G})\rightarrow \operatorname{\mathrm{End}}_{A*G}((A*G)\otimes_{B*G} I_GR_GP^{B*G}), f\mapsto \operatorname{\mathrm{id}}_{A*G}\otimes f \end{align*} $$
   are the maps obtained from the induction functor, and $\varphi _1$ and $\varphi _2$ are the isomorphisms arising from the natural isomorphisms (4.1) and (4.2).

   Note that all maps except $s_1\otimes \operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}$ and $s_2$ are isomorphisms and that by assumption $s_2$ is an epimorphism in homology degree 1 and an isomorphism in homology in degree strictly greater than 1 (resp. an isomorphism in homology in degree strictly greater than 0).Thus, $\operatorname {\mathrm {id}}_{\operatorname {\mathrm {k}} G}\otimes s_1$ is an epimorphism in homology degree 1 and an isomorphism in homology in degree strictly greater than 1 (resp. an isomorphism in homology in degree strictly greater than 0).This implies that $s_1$ is in homology degree 1 and an isomorphism in homology in degree strictly greater than 1 (resp. an isomorphism in homology in degree strictly greater than 0).

   We obtain that the vector space homomorphism

   $$ \begin{align*} H^*(s_1):\operatorname{\mathrm{Ext}}_{B}^*(R_GL^{B*G}, R_GL^{B*G})\rightarrow \operatorname{\mathrm{Ext}}_A^*(A\otimes_B L^{B*G}, A\otimes_B L^{B*G}), [f]\mapsto [\operatorname{\mathrm{id}}_A\otimes f] \end{align*} $$
   is an epimorphism in degree 1 and an isomorphism in degree strictly greater than 1 (resp. an isomorphism in degree strictly greater than 0).

   Moreover, since $A\otimes _B L_i^B\cong \Delta _{L_i^A}$ for every $L_i^B\in \operatorname {\mathrm {Sim}}(B)$ and

   $$ \begin{align*}R_G(L^{B*G})\cong \bigoplus_{L_i^B\in \operatorname{\mathrm{Sim}}(B)}[R_G(L^{B*G}):L_i^B]L_i^B\end{align*} $$
   by Proposition 2.12[7], this induces for every $L_i^B, L_j^B\in \operatorname {\mathrm {Sim}}(B)$ a vector space homomorphism
   $$ \begin{align*} [R_G(L^{B*G}):L_i^B][R_G(L^{B*G}):L_j^B]&\operatorname{\mathrm{Ext}}^*_B(L_i^B, L_j^B)\\\rightarrow &[R_G(L^{B*G}):L_i^B][R_G(L^{B*G}):L_j^B]\operatorname{\mathrm{Ext}}^*_A(\Delta_{L_i^A}, \Delta_{L_j^A}),\\ f\mapsto &\operatorname{\mathrm{id}}_A\otimes f, \end{align*} $$
   which is an epimorphism in degree 1 and an isomorphisms in degree strictly greater than 1 (resp. an isomorphism in degree strictly greater than 0). Since by Corollary 2.14 $[R_G(L^{B*G}):L_i^B]\neq 0$ for all i, this implies that $(B,\leq _B)$ is homological resp. regular.