2
$\begingroup$

I would like an explicit description of $\mathbb{R} SO(n) I_n$, i.e., the image of the identity under the action of the group algebra of $SO(n)$ by left multiplication. Equivalently, what is an explicit description of the vector space $\{\sum_i c_i A_i: A_i \in SO(n)\}$? Presumably this is well-known and in the context of all compact simple Lie groups? I need this to understand some continuous state space Markov chain of particle systems.

Edit: to those who vote to close, please state reason. If you have a one-liner answer, why not give it a shot? I will close it myself when I see a satisfactory response.

Improve this question
$\endgroup$
9
  • $\begingroup$ The question isn't clear to me. What is the precise action here, and are you just referring to the group algebra of the abstract group? $\endgroup$
    – Jim Humphreys
    Commented Mar 25, 2012 at 19:46
  • $\begingroup$ @Jim: yes just the usual left multiplication action. So the question is really asking for the vector space spanned by $\sum_i c_i A_i$, where $A_i \in SO(n)$. I will clarify in the text. $\endgroup$
    – John Jiang
    Commented Mar 25, 2012 at 19:47
  • $\begingroup$ @John: Unless I'm missing something subtle, the "natural" group action on the underlying vector space is irreducible and thus the resulting matrices should span the whole space. Is there more going on? $\endgroup$
    – Jim Humphreys
    Commented Mar 25, 2012 at 19:53
  • 1
    $\begingroup$ It is the sum of $n$ copies of the natural actio n, so it is not irreducible. $\endgroup$
    – Claudio Gorodski
    Commented Mar 25, 2012 at 19:57
  • 2
    $\begingroup$ @John: Sorry, I didn't understand what you meant by "the action" here, since I took it to be just left multiplication in the matrix algebra. $\endgroup$
    – Jim Humphreys
    Commented Mar 25, 2012 at 22:15

1 Answer 1

Reset to default
3
$\begingroup$

For $n\geq3$, it is the full space of $n\times n$ real matrices. The reason is that the $SO(n)$-orbit through the identity is the same as the $SO(n)\otimes SO(n)$-orbit by left and right multiplication, $(g,h)\cdot X=gXh^{-1}$, and this representation is irreducible for $n\geq3$. (Note that the span of any orbit is an invariant subspace, in general).

Improve this answer
$\endgroup$
7
  • $\begingroup$ Why doesn't this work for $n=2$? Is real irreducibility not good enough? $\endgroup$
    – John Jiang
    Commented Mar 25, 2012 at 20:13
  • $\begingroup$ 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. $\endgroup$
    – Claudio Gorodski
    Commented Mar 25, 2012 at 20:40
  • 2
    $\begingroup$ 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. $\endgroup$
    – Claudio Gorodski
    Commented Mar 25, 2012 at 20:44
  • $\begingroup$ In your case, the tensor product representation $\mathbf R^n\otimes\mathbf R^n$ is equivalent to acting by left and right multiplication on $M(n,\mathbf R)$. $\endgroup$
    – Claudio Gorodski
    Commented Mar 25, 2012 at 20:46
  • $\begingroup$ 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. $\endgroup$
    – Claudio Gorodski
    Commented Mar 25, 2012 at 20:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .