Timeline for Multiple factors of the character of a representation
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2010 at 20:22 | comment | added | darij grinberg | It was "obvious" to me in the sense that it was my first idea. I don't remember having seen it anywhere else, but this doesn't mean I haven't seen it anywhere else. | |
| Nov 18, 2010 at 14:46 | answer | added | Jim Humphreys | timeline score: 3 | |
| Nov 18, 2010 at 14:24 | comment | added | Denis Serre | @Darij. Yes, I want to work with characters, not with representation. Nevertheless, I found your use of Amitsur-Levitski's theorem. I never saw an application of this beautiful statement. Did you borrow it from elsewhere or was is obvious to you ? | |
| Nov 18, 2010 at 13:53 | comment | added | darij grinberg | Oh, and I'm working over an algebraically closed field of characteristic $0$ all the time. Over a non-algebraically closed one, I don't think we can do it algorithmically. In fact, consider a cyclic group. Whether some irreps of degree $1$ over $\overline K$ are actually defined over $K$ or not depends on whether $K$ has some roots of unity, which we cannot know per se. But if we allow a roots-of-unity oracle, then I suspect that we can do pretty much all of representation theory constructively, including computing all the irreps. | |
| Nov 18, 2010 at 13:47 | comment | added | darij grinberg | ... split $V_{k-1}$ away from $V_k$ by using Maschke's theorem (totally constructive) and get it. Now, of course, this is not what you are asking for because you want to do it with characters rather than representations. Is there a way to construct a representation from its character constructively, without decomposing it into irreducibles? | |
| Nov 18, 2010 at 13:45 | comment | added | darij grinberg | ... $\left(ABCD–BACD−ABDC+BADC−ACBD+CABD+ACDB\right.$ $\left. –CADB+ADBC–DABC−ADCB+DACB+CDAB−CDBA−DCAB+DCBA \right.$ $\left.–BDAC+BDCA+DBAC–DBCA+BCAD−BCDA−CBAD+CBDA\right)v=0$ for all $A,B,C,D\in \mathrm{End}_G\left(V\right)$. What I used here is the Amitsur-Levitzki identity, or, more precisely, the fact that it is an identity for $M_2$ but not for $M_3$. Similarly we can find, for each $k\in \mathbb N$, the direct sum of all irreps which occur with multiplicity $\leq k$ in $V$. This is not exactly the direct sum of all irreps with multiplicity $= k$ in $V$, but now you can | |
| Nov 18, 2010 at 13:43 | comment | added | darij grinberg | ... irreps of multiplicity $1$ - namely, this is the subset $V_1$ of $V$ consisting of all $v\in V$ such that $\left(AB-BA\right)v=0$ for all $A,B\in\mathrm{End}_G\left(V\right)$. Why? Because it is the subset of $V$ on which all $M_{n_i}\left(\mathbb C\right)$ with $n_i\geq 2$ act as zero. Similarly we can obtain the direct sum of all irreps of multiplicity $\leq 2$ in $V$ - it is the subset $V_2$ of $V$ consisting of all $v\in V$ such that ... | |
| Nov 18, 2010 at 13:40 | comment | added | darij grinberg | At least we can do this with REPRESENTATIONS. In fact, if $V$ is a representation of $G$, then $\mathrm{End}_G\left(V\right)$ is a direct product of matrix rings $M_{n_i}\left(\mathbb C\right)$ (with $n_i$ being the multiplicities of the irreps in $V$), and $V$ can be seen as a direct sum of their standard representations $\mathbb C^{n_i}$. Now, while it is hard (if possible at all) to actually (algorithmically) decompose the ring $\mathrm{End}_G\left(V\right)$ into the direct product $M_{n_i}\left(\mathbb C\right)$, we can still find our which part of $V$ is the direct sum of all ... | |
| Nov 18, 2010 at 12:52 | history | asked | Denis Serre | CC BY-SA 2.5 |