Faithful representations and tensor powers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:06:25Z http://mathoverflow.net/feeds/question/18194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers Faithful representations and tensor powers Steven Blömski 2010-03-14T18:28:42Z 2013-04-21T21:44:19Z <p>The following result was mentionned earlier in <a href="http://mathoverflow.net/questions/18132/faithful-characters-of-finite-groups" rel="nofollow">this thread</a>, I searched a bit in the related threads and couldn't find a proof. I would really like to see a proof of it:</p> <p>Let $G$ be a finite group and $\rho : G \rightarrow GL(\mathbb{C}, n)$ a faithful representation of $G$. Then every irreducible representation of $G$ is contained in some tensor power of $\rho$.</p> http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers/18196#18196 Answer by darij grinberg for Faithful representations and tensor powers darij grinberg 2010-03-14T18:41:44Z 2010-03-14T18:46:49Z <p>See problem 3.26 in <a href="http://math.mit.edu/~etingof/replect.pdf" rel="nofollow">Etingof's "Introduction to representation theory"</a>. If you have troubles with understanding the hint, feel free to ask me. (The first sentence uses the fact that if a vector space over an infinite field is the union of finitely many subspaces, then one of these subspaces is the whole vector space. The surjectivity of the map $SV\to F\left(G,\mathbb C\right)$ is because a polynomial can take any arbitrary finite set of values at some given distinct points. In order to conclude from this, note that this map $SV\to F\left(G,\mathbb C\right)$ is a homomorphism of representations of $G$.)</p> <p>This proof works over any algebraically closed field of characteristic $0$. This can't quite be said about the proof in Fulton-Harris, if I remember it right.</p> http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers/63034#63034 Answer by Geoff Robinson for Faithful representations and tensor powers Geoff Robinson 2011-04-26T14:09:03Z 2011-04-26T14:09:03Z <p>As I have said elsewhere on Mathoverflow, in dealing with related questions, I think the simplest and the best proof of this result is due to Blichfeldt. It is simpler than both the power series type argument which appears in Burnside's book and the Vandermonde determinant argument of Brauer. Since we are dealing with characteristic zero representations of finite groups, we need only deal with characters. Let $\chi$ be the character afforded by $\rho$, and let $\alpha_{1},\alpha_{2},\ldots,\alpha_{m}$ be all the distinct values taken by $\chi$ on non-identity elements of $G$. Note that the class function $\chi \prod_{i=1}^{m}(\chi - \alpha_{i}1)$ vanishes on all non-identity elements of $G$, but does not vanish at $1_{G}$. Hence this class function is a non-zero multiple of the regular character (in fact a rational algebraic integer multiple). On the other hand, it may be written in the form $\sum_{j=1}^{m+1} a_{j}\chi^{j}$ for certain rational integers $a_{i}$ (note that ${\alpha_{1},\ldots \alpha_{m} }$ is a set of algebraic integers closed under algebraic conjugation). Since any irreducible character $\mu$ of $G$ has non-zero inner product with the regular character, we see that $\langle \chi^{j},\mu \rangle \neq 0$ for some $j$ with $1 \leq j \leq m+1$. (It is not necessary to use the factor $\chi$ if any $\alpha_{i}$ is already zero and, in any case, the factor $\chi$ is only used to make sure that we use strictly positive tensor powers of $\rho$, avoiding the issue of the trivial module). Note that Blichfeldt's argument yields that $\prod_{i=1}^{m} (\chi(1) - \alpha_{i})$ is an integer multiple of $|G|$.</p> http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers/63043#63043 Answer by David Speyer for Faithful representations and tensor powers David Speyer 2011-04-26T14:35:50Z 2011-04-26T14:35:50Z <p>Another proof (not really that different from Geoff's, but appealing to a somewhat different intuition): Let $W$ be the representation corresponding to $\rho$, let $\underline{1}$ be the trivial representation, and let $V$ be the representation which we want to appear in some representation of $W^{\otimes N}$. I will show instead that $V$ appears in some representation of $(W \oplus \underline{1})^{\otimes N}$; this is equivalent because $(W \oplus \underline{1})^{\otimes N} = \bigoplus_{k=0}^N \binom{N}{k} W^{\otimes k}$. </p> <p>Let $\chi$ be the character of $V$ and let $\psi$ be the character of $W$. Then <code>$$\dim \mathrm{Hom}_G(V, (W \oplus \underline{1})^{\otimes N}) = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} (\psi(g)+1)^N. \quad (*)$$</code> We want to show that this Hom space is nontrivial for large $N$. </p> <p>We have $|\psi(g)| \leq \dim W$ for all $g \in G$ and, since $W$ is faithful, $\psi(g)$ is $\dim W$ if and only if $g=e$. So $|\psi(g)+1| \leq \dim W + 1$, with equality precisely for $g=e$. So the right hand side of <code>$(*)$</code> is a finite sum of exponentials, and the term $(\dim V) (\dim W + 1)^N$ has a larger base than any of the others. So the right hand side is positive for large $N$, and we see that the irrep $V$ appears in $(\underline{1} \oplus W)^{\otimes N}$ for sufficiently large $N$.</p> <p>I explain how to modify this for compact Lie groups in <a href="http://mathoverflow.net/questions/58633/does-every-irreducible-representation-of-a-compact-group-occur-in-tensor-products/58644#58644" rel="nofollow">this answer</a>.</p> http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers/128090#128090 Answer by Marat for Faithful representations and tensor powers Marat 2013-04-19T14:15:37Z 2013-04-19T14:15:37Z <p>By Satz 90, the fraction field of the symmetric algebra of a faithful representation of a finite group contains all irreducible representation. It remains to get rid of the denominators, just multiplying by the product of their conjugates. </p> http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers/128293#128293 Answer by Marat for Faithful representations and tensor powers Marat 2013-04-21T21:44:19Z 2013-04-21T21:44:19Z <p>Any semilinear Galois representation is trivial (the skew product of a field $K$ with a finite group $G$ of its automorphisms is the endomorphism algebra of $K$ as vector space over the fixed subfield), so extension to $K$ of coefficients of any irreducible representation of $G$ is isomorphic to a direct sum of copies of $K$. </p>