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Hello everyone.

Let $V$ and $W$ be finite dimensional vector spaces over some field $K$. Consider $\rho:Sym(V)\to End(W)$ a homomorphism of algebras with unit (i.e., a representation of $Sym(V)$ on $W$). Fixing a basis $\{v_1,\ldots,v_m\}$ of $V$ there is associated an isomorphism $\Phi:K[x_1,\ldots,x_m]\to Sym(V)$ given by the extension of the mapping $\Phi(x_j)=v_j$ via the universal property of the symmetric algebra. This isomorphism makes the theory of representations of $Sym(V)$ very manageable because knowing where a basis element $v_j$ goes is tantamount to knowing where the monomial $x_j$ goes into $End(W)$ (using $\Phi^{-1}$).

My question is: are there any analgous results for the exterior algebra $\Lambda(V)$? This algebra can be thought of as the algebra of antisymmetric polynomials in $V$. As a $\mathbb{Z}_2$-graded algebra, the representation homomorphism $\rho:\Lambda(V)\to End(W)$ would have to be zero on the odd part, but since $V$ is odd in this algebra (it is the space of generators) any analogue of the property above for the symmetric algebra would only hold for the trivial representation (1 maps to the identity mapping and everything else goes to 0). That's quite boring. Perhaps I'm utterly wrong in regarding $End(W)$ as a purely even superalgebra but that's the setting I'm mostly interested in.

The question can be rephrased as: what are the irreducible modules over the exterior algebra?

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You want irreducible modules of the exterior algebra in what sense? Are you looking at the exterior algebra as a graded algebra (and over what group?) and you want the modules to be graded? –  Mariano Suárez-Alvarez Jan 28 '13 at 17:07
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The $\mathbb Z_2$-graded algebra $\Lamda V$ is generated by odd elements, so there are no non-trivial maps $\mathbb Z_2$-graded algebra maps $\Lambda V\to A$ to a purely even algebra $A$. –  Mariano Suárez-Alvarez Jan 28 '13 at 17:39
    
I apologise. Yes, I meant $\mathbb{Z}_2$-graded modules. So I guess the appropriate setting would be that of supervector spaces. Do you know if there are any references on that topic? –  Oscar Guajardo Jan 29 '13 at 20:54

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up vote 2 down vote accepted

I'm not sure if this answers your question, but if $V$ is finite-dimensional and you are just asking for representations of the algebra structure, then the only irreducible representation of $\Lambda(V)$ is the trivial one-dimensional representation, in which $V$ acts as $0$.

Indeed, let $W$ be an irreducible representation of $\Lambda(V)$. Let $0 \neq x \in V$, and consider $\mathrm{ker}(x) \subseteq W$ and $\mathrm{im}(x) \subseteq W$. Since $x$ anticommutes with all generators of $\Lambda(V)$, both of these subspaces of $W$ are actually $\Lambda(V)$-submodules.

We are trying to show that $x$ acts as zero. If not, then since $W$ is irreducible, we must have that $\mathrm{ker}(x) = (0)$. Since the image is nonzero, again by irreducibility we conclude that $\mathrm{im}(x) = W$, so $x$ is an automorphism on $W$. But then $x \wedge x = 0$ in $\Lambda(V)$, so the square of this automorphism of $W$ is zero, which is a contradiction.

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(With a little technology, you can say this as follows: the algebra is artinian local, and the radical is the positive part, so simple modules are modules over $\Lambda V/\Lambda^+V=K$, that is, vector spaces) –  Mariano Suárez-Alvarez Jan 28 '13 at 22:10

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