Wedderburn theorem version for superalgebras

Question

I am looking for an example of usage of wedderburn theorem version for superalgebras (which is a $\mathbb{Z}_2-$graded algebra). The theorem states that if $ A$ is finite dimensional $\mathbb{Z}_2-$graded simple associative algebra (it can be assumed it is over a field) then $ A=M\otimes D$ where $D$ is a division superalgebra (every homogeneous element is invertible) and $M$ is a matrix superalgebra.
I know that for example the theorem imples that $ \mathbb{C}=\mathbb{R}\otimes\mathbb{C},$ but it is a trivial result and I'm looking for an example that is not trivial (but not too complicated).
Thank you.

Answer 1

I've been reading John Baez's "Tenfold Way" blog post series from the $n$-Category Cafe (how division superalgebras relate to ten types of matter, compact symmetric spaces, and Bott periodicity for stable homotopy groups), and this question gives me an opportunity to test my understanding.

Super Abstract Algebra. Some things in algebra have "super" ($\mathbb{Z}_2$-graded) versions:

• super vector spaces $V=V_0\oplus V_1$,
• direct sums, with $(V\oplus W)_i=V_i\oplus W_i$
• tensor products, with $(V\otimes W)_i=\bigoplus_{j+j'=i}V_j\otimes W_{j'}$
• superalgebras $A=A_0\oplus A_1$, with $A_iA_j\subseteq A_{i+j}$,
• graded ideals $I\trianglelefteq A$, expressible as $I=I_0\oplus I_1$ with $I_i\le A_i$
• supermodules $M=M_0\oplus M_1$ over superalgebras, with $A_iM_j\subseteq M_{i+j}$
• direct sums and tensor products of supermodules, generalizing vector spaces
• supercommutator, with $[a,b]=ab-(-1)^{ij}ba$ if $a\in A_i,b\in A_j$, extended linearly
• supercommutative / supercentral defined using supercommutator
• simple (no nontrivial two-sided graded ideals)
• central simple (super simple, super center is base field)
• tensor product $A\otimes B$ of superalgebras, where (for $b_1\in B_i$, $a_2\in A_j$) $$(a_1\otimes b_1)(a_2\otimes b_2)=(-1)^{ij}(a_1a_2\otimes b_1b_2)$$

It's good to get acquainted with some examples of superalgebras.

Any plain old algebra $A$ can be considered a "purely even" superalgebra $A\oplus 0$. It is possible for superalgebras to be isomorphic as vanilla algebras but not as superalgebras: consider $\mathbb{C}\oplus0$ versus $\mathbb{R}\oplus\mathbb{R}i$ as two possible graded algebra structures on the complex numbers.

Note, because of the homomorphism $\mathbb{Z}\to\mathbb{Z}_2$, any $\mathbb{Z}$-graded algebra becomes $\mathbb{Z}_2$-graded by collecting the even and odd parts together. For example, the polynomial ring $k[x]$ and Laurent polynomial ring $k[x,x^{-1}]$ over a field $k$ are $\mathbb{Z}$-graded by degree, so are $\mathbb{Z}_2$-graded by parity.

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Matrix Superalgebras. $A(p\!\mid\!q)$ consists of $(p+q)\times(p+q)$ block matrices over $A$: $$ A(p\!\mid\!q)=\begin{bmatrix} A_0^{p\times p} & A_1^{p\times q} \\ A_1^{q\times p} & A_0^{q\times q}\end{bmatrix} \oplus \begin{bmatrix} A_1^{p\times p} & A_0^{p\times q} \\ A_0^{q\times p} & A_1^{q\times q}\end{bmatrix} $$ (Notation varies.) So, being in an off-diagonal block flips parity.

• Exercise. Verify $A(p\!\mid\!q)$ is indeed a super algebra and $A(q\!\mid\!p)\cong A(p\!\mid\!q)$.

• Exercise. Verify $A^{p\mid q}=(A_0^p\times A_1^q)\oplus(A_1^p\times A_0^q)$ is an $A(p\!\mid\!q)$-supermodule.

• Exercise. Verify $A(p\!\mid\!q)\cong A\otimes k(p\!\mid\!q)$ are superalgebra isomorphic (using the super tensor product $\otimes$, assuming $A$ is a $k$-algebra where the scalar field $k$ is considered purely even).

• Exercise. Verify $A(p\!\mid\!q)(r\!\mid\!s)\cong A(pr+qs\!\mid\! ps+qr)$ are superalgebra isomorphic. (Notice how it's kind of like multiplying $(p+q\tau)(r+s\tau)$ in the split complex numbers where $\tau^2=+1$.)

Notice how $k(p\!\mid\!q)$ puts a graded structure on the linear operators on $k^{p\mid q}$. More generally, if $M,N$ are supermodules, then the hom-space is a super vector space (at least) with grading compatible with tensor-hom adjunction: $\hom(M,N)=\bigoplus_{j+j'=i}\hom(M_j,N_{j'})$.

Super Wedderburn Theorem: every central simple superalgebra is isomorphic to a matrix superalgebra over a division superalgebra. Supposedly the usual proof of vanilla Wedderburn translates into super.

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Division Superalgebras. A superalgebra $A$ is division if its homogeneous elements are invertible. There are ten real division superalgebras, eight central simple over $\mathbb{R}$ and two over $\mathbb{C}$:

• $D_0=\mathbb{R}\oplus 0$
• $D_1=\mathbb{R}\oplus\mathbb{R}e$ with $e^2=-1$
• $D_2=\mathbb{C}\oplus\mathbb{C}e$ with $e^2=-1$, $ei=-ie$ ("antilinear")
• $D_3=\mathbb{H}\oplus\mathbb{H}f$ with $f^2=+1$, $f$ ungraded central
• $D_4=\mathbb{H}\oplus 0$
• $D_5=\mathbb{H}\oplus\mathbb{H}e$ with $e^2=-1$, $e$ ungraded central
• $D_6=\mathbb{C}\oplus\mathbb{C}f$ with $f^2=+1$, $fi=-if$ ("antilinear")
• $D_7=\mathbb{R}\oplus\mathbb{R}f$ with $f^2=+1$
• $C_0=\mathbb{C}\oplus 0$
• $C_1=\mathbb{C}\oplus\mathbb{C}g$ with $g^2=\pm1$, $g$ ungraded central

Assuming Frobenius' theorem for real division algebras, proving these are the real division superalgebras (up to isomorphism) isn't that hard. It's done in the blog posts I mentioned at the outset.

Context: If the categories of modules over two rings are equivalent, the rings are called Morita equivalent. There is an equivalent formulation using the $2$-category with rings as objects, bimodules as morphisms, and bimodule maps as $2$-morphisms. Morita equivalence classes of central simple algebras form a group under tensor products called the Brauer group, for example $\mathrm{Br}(\mathbb{R})=\{[\mathbb{R}],[\mathbb{H}]\}\cong\mathbb{Z}_2$. All of this translates into the super setting, where the super Brauer monoid is $$\mathrm{BW}(\mathbb{R})=\mathrm{Bw}(\mathbb{R})\sqcup\mathrm{Bw}(\mathbb{C})=\{[D_0],\cdots,[D_7]\}\sqcup\{[C_0],[C_1]\}\cong\mathbb{Z}_8\sqcup\mathbb{Z}_2.$$

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Exterior Algebra. $\Lambda V=\bigoplus_{i\ge0}\Lambda^i V$ is $\mathbb{Z}$-graded, hence $\mathbb{Z}_2$-graded. As a superalgebra, it is supercommutative: even elements commute with both even and odd elements, odd elements commute with even elements and anticommute with odd elements. (Indeed, assuming char $\ne2$, if $A$ is supercommutative then any $a\in A_1$ satisfies $a^2=0$.)

Clifford Superalgebras. Considered a "quantization" or "deformation" of the exterior algebra, $C\ell(p,q)$ is freely generated by $p$ square roots of $-1$ and $q$ square roots of $+1$, all independent, odd, and anticommuting. In other words, $C\ell(p,q)=D_1^{\otimes p}\otimes D_{-1}^{\otimes q}$. (Interpret the index of $D$ mod $8$. Note this uses super tensor products; most treatments of Clifford algebras ignore super stuff and consider only vanilla tensor products. Many sources use the opposite sign convention, especially those that call them "geometric algebras.")

Exercise. Verify the following (I did so case-by-case): $$ D_1\otimes D_n \cong \begin{cases} D_{n+1} & \dim D_{n+1}>\dim D_n \\ D_{n+1}(1\!\mid\!1) & \dim D_{n+1}<\dim D_n \end{cases} $$ $$ D_{-1}\otimes D_n\cong \begin{cases} D_{n-1} & \dim D_{n-1}>\dim D_n \\ D_{n-1}(1\!\mid\!1) & \dim D_{n-1}<\dim D_n \end{cases} $$

Imagine "superdimensions" $(p\!\mid\!q)$ can be added and multiplied like split complex numbers (see earlier exercise). Then the "power" $(1\!\mid\!1)^k$ is $(1\!\mid\!0)$ if $k=0$ or is $(2^{k-1}\!\mid\!2^{k-1})$ otherwise.

Exercise. Prove $C\ell(p,q)\cong D_{p-q}(1\!\mid\!1)^k$ where $k=p+q-\lg\dim D_{p-q}$ (hint: induction).

I haven't determined the matrix algebra "superranks" of $D_p\otimes D_q$ for all $p,q<8$, nor have I determined precisely when $D_k(p\!\mid\!q)\cong D_k(r\!\mid\!s)$ yet. (For example, $D_{-1}(2\!\mid\!0)\cong D_{-1}(1\!\mid\!1)$ I believe.)

Answer 2

After thinking $M_2(\mathbb{R})$ is a super algebra with the decomposition of $ \mathbb{R}\langle\begin{pmatrix} 1 & 0\\ 0 & 0\\ \end{pmatrix},\begin{pmatrix} 0 & 0\\ 0 & 1\\ \end{pmatrix}\rangle \oplus \mathbb{R}\langle\begin{pmatrix} 0 & 1\\ 0 & 0\\ \end{pmatrix},\begin{pmatrix} 0 & 0\\ 1 & 0\\ \end{pmatrix}\rangle.$ A matrix algera can be $ GL(\mathbb{C})$ where $ \mathbb{C}$ is an $ \mathbb{R}$ vector space (and a $ \mathbb{Z}_2-$graded space). I am not sure that I know how to decompose it to even and odd parts (suppose to be with even and odd maps, i.e. preserving parity and reversing parity). By the thorem $M_2(\mathbb{R})=GL(\mathbb{C})\otimes D $ where $ D$ can be the centralizer of $ GL(\mathbb{C})$ (graded commute). As a non trivial conclusion $ \dim GL(\mathbb{C})=2.$ I hope I have no mistakes. $GL(\mathbb{C})$ is actually all the $ 2\times 2$ invertible matrices. It is all based on (Theorem 2.33) https://www.researchgate.net/publication/228954911_Fundamental_Theorem_of_Finite_Dimensional_Z_2-Graded_Associative_Algebras