I was reading the paper "Algebras graded by groups" by Knus. I want to test and further my understanding of the paper by asking several questions.
Since the paper is not readily available I will detail everything necessary to understand my questions. Please do not be intimidated by the length of this post, the prelimenaries are very easy to read and are included in case of confusion.
Prelimenaries
Let $K$ be a field and let $G$ be an abelian group, a pairing (or bicharacter) of $G$ is a map $\phi:G\times G\rightarrow K^*$ such that
- $\phi(g_1+g_2,h)=\phi(g_1,h)\phi(g_2,h)$,
- $\phi(g,h_1+h_2)=\phi(g,h_1)\phi(g,h_2)$.
A pairing $\phi$ is symmetric if $\phi(g,h)=\phi(h,g)$. A pairing $\phi$ is called non-degenerate if $(\forall h\in G:\phi(g,h)=1)\Rightarrow g=e_G$.
A unital finite-dimensional $K$-algebra $A$ is called $G$-graded if $A=\bigoplus_{g\in G}A_g$ and $A_gA_h\subset A_{gh}$, the unit lives in degree zero. A subspace $I\subset A$ is graded if it is the direct sum of the intersections $I\cap A_g$. We call the graded algebra $A$ simple if there are no proper graded two-sided ideals.
We call $A$ central if the only homogeneous elements $x\in A$ such that $xa=\phi(x,a)ax$ or $ax=\phi(a,x)xa$ for all homogeneous $a\in A$, are in $K$.
Let $H^2(G,K^*)$ be the second cohomology group of $G$ with coefficients in the trivial $G$-module $K^*$. Let $f$ be a nonmalized $2$-cocycle on $G$ and denote by $KG$ the group algebra on $G$. We define the twisted group algebra $K_fG$ by twisting the multiplication on $KG$, i.e. $g\cdot h:=f(g,h)gh$. This turns $K_fG$ into an associative algebra.
The Brauer group $\text{Br}(K,G)$
Let $\phi$ be a fixed pairing of an abelian group $G$. Two $G$-graded central simple algebras $A$ and $B$ are called equivalent if there exist two $G$-graded vector spaces $V_1$ and $V_2$ such that $$A\otimes \text{End}_K(V_1)\cong B\otimes \text{End}_K(V_2).$$ Let $\text{Br}(K,G)$ denote the set of equivalence classes, the tensor product induces a group structure on $B(K,G)$. Here $A\otimes B$ is given the structure of an algebra by defining $$(a\otimes b)(a'\otimes b'):=\phi(b,a')(aa'\otimes bb').$$ One can show that the tensor product of two graded central simple algebras as above is again graded central simple.
Main theorem
Let $G$ be an abelian group and $\phi$ a symmetric pairing on $G$ which is non-degenerate on any finite subgroup of $G$. Then a graded central simple algebra $A$ such that $\text{char}(K)$ does not divide $\dim_K(A)$ is isomorphic to $$K_f(H)\otimes M_n(D,H').$$ Here $H$ and $H'$ are subgroups of $G$ such that $G=H\times H'$, $f$ is an abelian class in $H^2(H,K^*)$. $M_n(D,H')$ is a matrix algebra over a division ring $D$ graded by $H'$.
Questions
- What exactly is meant by $M_n(D,H')$ is a matrix algebra over a division ring $D$ graded by $H'$? Similarly what is the precise meaning of $\text{End}_k(V)$ if $V$ is $G$-graded? Is this the usual endomorphism ring or a ring of graded morphisms?
- Suppose $G=\left\{e_G\right\}$, then any $G$-graded central simple $K$-algebra $A$ is simply a central simple $K$-algebra. Thus by the main theorem $A\cong K_f(G)\otimes M_n(D,G)$. Clearly $K_f(G)\cong K$, hence $A\cong M_n(D,G)$. Whatever the answer to the previous question is, a trivial grading should give that $A\cong M_n(D)$. Clearly the size of the Brauer group $\text{Br}(K,G)$ depends on the number of division rings over $K$.
- Suppose $K=\mathbb{R}$, by a result of Frobenius the only division rings over $\mathbb{R}$ are $\mathbb{R},\mathbb{C}$ and the quaternions $\mathbb{H}$. Hence any central simple algebra over $\mathbb{R}$ is isomorphic to either $M_n(\mathbb{R}), M_n(\mathbb{C})$ or $M_n(\mathbb{H})$. But $M_n(\mathbb{C})$ is not central as the center is two-dimensional. Does it follow that the Brauer group $\text{Br}(K,\left\{e_G\right\})$ contains at most two elements? To be more precise, a priori it's possible that $M_n(\mathbb{H})$ live in different classes for different $n$.
- Let $B$ be an ordinary central simple algebra over $K$, be declaring this thing to live in degree zero we can view any central simple algebra as a $G$-graded central simple algebra. Can we view the usual Brauer group $\text{Br}(K)$ as a subgroup of $\text{Br}(K,G)$?
- Consider $K=\mathbb{R}$ and $G=\mathbb{Z}_2\times \mathbb{Z}_2$ and let $\phi$ be the trivial pairing. We know that the quaternions $\mathbb{H}$ are a real division ring. On the other hand we could view the quaternions $\mathbb{H}$ as a $G$-graded algebra by declaring $\deg(i)=(1,0), \deg(j)=(0,1)$ and $\deg(k)=(1,1)$. Let me denote the graded version by $\mathbb{H}_G$. Notice that $\mathbb{H}_G$ is a graded central simple algebra. In light of the main theorem, $\mathbb{H}_G\cong K_f(H)\otimes M_n(D,H')$, can we specify $H,H',f'$ and $n$? Depeding on the answer to $1$, a natural answer could be $\mathbb{H}_G\cong M_1(\mathbb{H}, G)$.
- Last but not least: In general one can define the Brauer group of a braided monoidal category. Now let $G$ be a group and $\phi$ a pairing. One can consider the Hopf algebra $KG$ and using $\phi$ one can define coquasi-triangular structure on $KG$. Denote by $H$ the resulting coquasi-triangular Hopf algebra. Is the Brauer group $\text{Br}(K,G)$ equal to $\text{Br}(\text{Mod}^H)$, the Brauer group of the braided monoidal category of $H$-comodules?
