Timeline for Is there any way to compute the restriction of morphism onto irreducible components
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2011 at 13:05 | comment | added | darij grinberg | @Alex: Ah, I see. For some reason I thought of $V$ and $W$ being the same $G$-module. | |
| Apr 25, 2011 at 11:17 | comment | added | Alex B. | In other words, the identification between the vector space Hom$_G(V,W)$ and the one-dimensional vector space $\mathbb{k}$ is non-canonical and depends on the choice of basis on the former. | |
| Apr 25, 2011 at 11:14 | comment | added | Alex B. | @darij Even if the representation is multiplicity-free, the scalars are not canonical, unless there is a canonical way of fixing an isomorphisms between the isomorphic summands. If I tell you that $V$ and $W$ are abstractly isomorphic, then the question "what scalar does the map $f:V\rightarrow W$ correspond to" doesn't make sense until you fix bases on $V$ and $W$ in such a way that one basis is a scalar multiple of f(other basis). Once you have done that, you can talk about the scalar, but the answer will depend on the bases you have chosen. | |
| Apr 25, 2011 at 10:51 | comment | added | darij grinberg | ... as elements of the group algebra) and take the trace. | |
| Apr 25, 2011 at 10:51 | comment | added | darij grinberg | However, there are a lot of cases in reality (as far as some part of algebra can be called reality) where a representation is multiplicity-free, i. e. every (absolutely) irreducible representation occurs only once in it, and the scalars therefore become canonical. For example, the restriction of an irreducible representation of $S_n$ to $S_{n-1}$ is multiplicity-free. In this case we should be able to compute the scalars e. g. by pre-composing or post-composing our morphism with the projection on the irreducible component that we want (these usually can be nicely computed ... | |
| Apr 25, 2011 at 8:54 | comment | added | zroslav | Ah, I understood. | |
| Apr 25, 2011 at 2:28 | history | answered | Alex B. | CC BY-SA 3.0 |