most recent 30 from http://mathoverflow.net 2013-05-23T01:31:56Z http://mathoverflow.net/feeds/question/120127 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120127/representation-theory-for-the-exterior-algebra Oscar Guajardo 2013-01-28T16:59:50Z 2013-01-28T21:49:21Z

<p>Hello everyone.</p> <p>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 $\Phi^{-1}$). </p> <p>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. </p> <p>The question can be rephrased as: what are the irreducible modules over the exterior algebra?</p>

http://mathoverflow.net/questions/120127/representation-theory-for-the-exterior-algebra/120149#120149 MTS 2013-01-28T21:49:21Z 2013-01-28T21:49:21Z

<p>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 $0$.</p> <p>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.</p> <p>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.</p>