Timeline for Faithful representations and tensor powers
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 16, 2010 at 22:05 | comment | added | Jim Humphreys |
Notational detail: usually people write either $GL(n,\mathbb{C}) or more functorially, $GL_n(\mathbb{C})$. The original post uses a highly nonstandard form.
|
|
| Mar 15, 2010 at 11:03 | comment | added | darij grinberg | ... and the one that doesn't obviously vanish is k_k * PROD_{i in J\{k}} (1 - it). Hence, k_k * PROD_{i in J\{k}} (1 - it) = 0. But PROD_{i in J\{k}} (1 - it) is not 0 for t = 1 / k (because the elements of J are distinct), so we must have k_k = 0. (Sorry for my notation!) This must hold for every index k. So the linear combination has zero coefficients. | |
| Mar 15, 2010 at 11:02 | comment | added | darij grinberg | So let J be a finite set of rationals, and let k_j be a rational number for every j in J (yes, rationals parametrized by rationals). Assume that SUM_{j in J} k_j * 1 / (1 - jt) = 0. Multiplying this equation by PROD_{i in J} (1 - it), we get SUM_{j in J} (k_j * PROD_{i in J\{j}} (1 - it)) = 0. This is a polynomial identity. Now, evaluate it at t = 1 / k for some k in J (if k = 0, then this means taking the formal limit t -> infinity, i. e. considering the highest monomials). This t satisfies 1 - kt = 0, so all but one terms of the sum vanish ... | |
| Mar 15, 2010 at 0:06 | comment | added | Steven Blömski | I understand this and how it completes the proof. I just don't see why they are linearly independent. | |
| Mar 14, 2010 at 23:55 | comment | added | darij grinberg | Well, the functions 1 / (1-at) are linearly independent, so that any linear combination of them with not all coefficients zero must be a nontrivial rational function. Or what exactly don't you get? | |
| Mar 14, 2010 at 23:29 | comment | added | Steven Blömski | @darij grinberg: I don't get the contradiction for the proof of linear independence.. | |
| Mar 14, 2010 at 23:00 | comment | added | Jim Humphreys | Characters only work well in characteristic 0, but their values lie in fairly small number fields. So you are not involved with the entire field of complex numbers. To work in characteristic p requires a real change of methods and usually a reformulation of the questions asked. | |
| Mar 14, 2010 at 22:33 | comment | added | darij grinberg | Yes, but here we're back to the point where we need analysis (we need it to see the eigenvalues as complex numbers, not just as elements in a formal algebraic extension of the rationals). | |
| Mar 14, 2010 at 22:17 | comment | added | Jim Humphreys | Concerning Brauer's proof, which uses standard character theory over fields of characteristic 0, the test for an element of a group to lie in the kernel of an arbitrary representation is that its character value should agree with the degree of the representation (the value of the character at 1). Here the given representation is faithful. The estimates here of character values rely on these being sums of roots of unity. I suppose Brauer wanted to emphasize the brevity of his proof, besides which he knew characters inside out. | |
| Mar 14, 2010 at 21:32 | comment | added | darij grinberg | Because: (1) this power series is a linear combination of functions of the form 1 / (1-at) with rational a; (2) at least one of these functions occurs with nonzero coefficient in this linear combination (namely, the function 1 / (1-at) for a = dim V, because chi(C) = dim V only for C = [e]); (3) the functions 1 / (1-at) for distinct rational a are linearly independent (in fact, assume that they are not, and multiply with their common denominator!). The complex (in every meaning of this word) thing here is to prove that chi(C) = dim V only for C = [e]. | |
| Mar 14, 2010 at 21:02 | comment | added | Steven Blömski | In Fulton-Harris, page 517, they conclude with "the right-hand side is a nontrivial rational function". How can this be proven? | |
| Mar 14, 2010 at 20:59 | comment | added | darij grinberg | How does Brauer conclude this: "Since X is faithful, one of the A_j consists only of the unit element"? | |
| Mar 14, 2010 at 20:01 | comment | added | Jim Humphreys | There are two additional short proofs in Proc. AMS just after Curtis-Reiner first appeared, avoiding Burnside's use of complex numbers. (These back issues are to appear on the AMS journal page after scanning but haven't yet. I can access them through the UMass library or JSTOR, but these are restricted.) The two papers appear in the collected papers of the two authors: R. Steinberg, Complete sets of representations of algebras, Proc. Amer. Math. Soc. 13 (1962), 746-747 R. Brauer, A note on theorems of Burnside and Blichfeldt, Proc. Amer. Math. Soc. 15 (1964), 31-34 | |
| Mar 14, 2010 at 19:10 | comment | added | darij grinberg | I THINK you mean the same proof as in Fulton-Harris. Alas, I don't see how to show that the pole is indeed a pole (i. e., not cancelled by another term) without the use of complex numbers. | |
| Mar 14, 2010 at 18:59 | comment | added | Torsten Ekedahl | Another proof is in Curtis-Reiner: Representation theory of finite groups and associative algebras, Thm 32.9 in the first edition. To my mind the proof is very cute, it considers the generating series $\sum_k a_k t^k$, where $a_k$ is the number of times the irreducible representation occurs in the $k$'th tensor power of $\rho$. Using the orthogonality formulas for characters they sum this up as a rational function which, because $\rho$ is faithful, has a simple pole at $t=1/n$ and hence many of the $a_k$ are non-zero. | |
| Mar 14, 2010 at 18:46 | history | edited | darij grinberg | CC BY-SA 2.5 |
added 291 characters in body; added 160 characters in body
|
| Mar 14, 2010 at 18:41 | history | answered | darij grinberg | CC BY-SA 2.5 |