Timeline for Orbit of the identity matrix under Lie group algebra actions
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2012 at 12:51 | comment | added | Claudio Gorodski | John: it is a pleasure to see applications of Lie groups going on. | |
| Mar 25, 2012 at 20:59 | vote | accept | John Jiang | ||
| Mar 25, 2012 at 20:58 | comment | added | John Jiang | @Claudio:thanks a lot for the clarification! I now appreciate semi-simplicity more. | |
| Mar 25, 2012 at 20:58 | comment | added | Claudio Gorodski | Sorry, in my comment above I meant that $\mathbf R^4$ splits into $2$ irreducible $T^2$-components, which are $\mathfrak{so}(2)$ and the traceless symmetric $2\times2$ matrices. | |
| Mar 25, 2012 at 20:46 | comment | added | Claudio Gorodski | In your case, the tensor product representation $\mathbf R^n\otimes\mathbf R^n$ is equivalent to acting by left and right multiplication on $M(n,\mathbf R)$. | |
| Mar 25, 2012 at 20:44 | comment | added | Claudio Gorodski | On the other hand, for $n\geq3$, $SO(n)$ is compact semisimple. If $G$, $H$ are compact semisimple and act irreducibly on $U$, $V$, resp., then $G\times H$ acts irreducibly on $U\otimes V$ by the tensor product representation. | |
| Mar 25, 2012 at 20:40 | comment | added | Claudio Gorodski | For $n=2$ you have a $2$-torus $T^2$ acting on $\mathbf R^4$, and any real irreducible representation of a torus is $2$-dimensional, so $\mathbf R^4$ splits as $2$ irreducible $T^2$-components $\mathbf R.I\oplus \mathbf R J$. Here I am talking about real representations and real dimensions. | |
| Mar 25, 2012 at 20:13 | comment | added | John Jiang | Why doesn't this work for $n=2$? Is real irreducibility not good enough? | |
| Mar 25, 2012 at 20:06 | history | answered | Claudio Gorodski | CC BY-SA 3.0 |