Tensor product of simple representations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:12:44Z http://mathoverflow.net/feeds/question/57997 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57997/tensor-product-of-simple-representations Tensor product of simple representations shenghao 2011-03-09T21:06:39Z 2011-03-10T00:23:33Z <p>Let $G$ be a linear algebraic group over some field, and let $V$ and $W$ be two simple rational representations of $G.$ Is $V\otimes W$ semi-simple?</p> <p>I was trying to convince myself that if $G$ has a faithful semi-simple representation, then $G$ is linearly reductive, and was reduced to the question above. The problem I have in mind is over characteristic 0, but answers addressing char. $p$ is equally appreciated too!</p> http://mathoverflow.net/questions/57997/tensor-product-of-simple-representations/57999#57999 Answer by Bruce Westbury for Tensor product of simple representations Bruce Westbury 2011-03-09T21:31:13Z 2011-03-09T21:31:13Z <p>Let $G=SL_2(F_p)$. Put $V_k$ the $k+1$-dimensional representation. Then $V_k$ is simple for $0\le k\le p-1$. Take $0\le r,s\le p-1$ with $r+s>p$. Then $V_r\otimes V_s$ is not semisimple.</p> http://mathoverflow.net/questions/57997/tensor-product-of-simple-representations/58016#58016 Answer by fherzig for Tensor product of simple representations fherzig 2011-03-09T23:59:12Z 2011-03-09T23:59:12Z <p>If the characteristic is $p$ then, by a theorem of Serre, it's true provided $dim(V) + dim(W) &lt; p+2$. To be safe, let me assume that the base field is algebraically closed. (One should be safe over a perfect field though.) The example given by Bruce Westbury shows that the above condition is (in some sense) best possible. In fact, Serre showed this result is even true for <em>arbitrary</em> groups (not even algebraic)! Serre's paper is extremely beautiful and well worth reading. It in fact reduces the general case to the case of algebraic groups, and in that situation uses some ideas of Jantzen. The paper is available here for free:</p> <p><a href="http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN356556735_0116&amp;DMDID=dmdlog35" rel="nofollow">http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN356556735_0116&amp;DMDID=dmdlog35</a></p> <p>As Jim Humphreys wrote in the mathscinet review (MR1253203):</p> <p><em>There are not many interesting theorems of the form: "If $G$ is any group, then $\ldots$…''. But an old theorem of Chevalley and a new theorem proved in this paper certainly qualify.</em></p> http://mathoverflow.net/questions/57997/tensor-product-of-simple-representations/58018#58018 Answer by George McNinch for Tensor product of simple representations George McNinch 2011-03-10T00:23:33Z 2011-03-10T00:23:33Z <p>If $G$ is a(ny) group, if $k$ is a field of characteristic 0, and if $V$ and $W$ are semisimple finite dimensional $kG$ modules, then $V \otimes_k W$ is indeed semisimple as a $kG$-module. This is due to Chevalley, and (I think I'm not off-base in saying this) inspired the characteristic $p>0$ result of Serre mentioned in other answers/comments.</p> <p>The argument goes as follows: it is enough to prove the result after replacing $k$ by an algebraic closure. Now replace $G$ by the Zariski closure of its image in $GL(V) \times GL(W)$ -- this Zariski closure leaves invariant the same subspaces of $V \otimes_k W$ as does $G$, so we may suppose $G$ to be a linear algebraic group over $k$.</p> <p>Since representations of finite groups in char. 0 are semisimple, a $G$-representation is semisimple just in case that is true upon restriction to the connected component $G^0$. Thus we may and will suppose $G$ to be connected.</p> <p>Finally, note that $G$ has a faithful semisimple representation, namely $V \oplus W$. Thus the unipotent radical of $G$ is trivial so that $G$ is a connected and reductive group over $k$. Now the semisimplicity of $V \otimes W$ follows (<em>every</em> finite dimensional rational representation of $G$ is semisimple).</p>