Coadjoint action

February 13th, 2008 ~ Posted in: Mathematics

I’ve been looking at Lie algebras/groups for CDS202 tonight: I have to compute the coadjoint action of SO(3) on the dual of its Lie algebra, and find the coadjoint orbits. Exactly what all of that means, and why I care, has yet to be determined. In fact, I suspect that I don’t care.

So far, I’ve found the Lie algebra– the set of skew-symmetric matrices– but what exactly is the dual space? I can’t think of any other way to get a concrete representation of the dual space than to use an inner product, but can I just introduce one? I’m going to do that, and see what happens. Hopefully something conforming to intuition (i.e. geometrically meaningful) will pop out.

Here’s an interesting calculation that came up in my reading. Let \hat : v=(v_1, v_2, v_3) \in \R^3 \rightarrow \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & - v_1 \\ -v_2 & v_1 & 0 \end{pmatrix} \in \mathfrak{o} be a mapping from the Lie algebra (\R^3, \times) to the Lie algebra the set of skew-symmetric matrices with the usual bracket [A,B] = AB - BA. You don’t have to know what all those terms mean (I sure as hell don’t have a firm grasp on this myself) to consider the question:

Show that for all A \in \mathfrak{o} and v \in \R^3, it’s true that \widehat{Av} = A\hat{v}A^{-1}.

Mathematics ~

3 Responses to “Coadjoint action”

  • 1. ObsessiveMathsFreak
    February 14th, 2008 at 8:07 pm

    So far, I’ve found the Lie algebra– the set of skew-symmetric matrices– but what exactly is the dual space? I can’t think of any other way to get a concrete representation of the dual space than to use an inner product, but can I just introduce one?

    Do you mean the Dual space for Lie Algebras, or Dual spaces in general? I’ll try to answer the second one.

    The dual space of a vector space is the set of linear functions that map vectors to real numbers(Yawn). Yes, as you have noticed, a concrete way to get a representation of elements in the dual space is to use the inner product.

    Now for the great, dark and terrible secret of the dual space which no one ever points out.

    …..

    An inner product is the only way to represent an element of the dual space!!

    Or at least, every element can be transformed so that it is represented in this way. Basically, there’s no element \phi in the dual space that can’t be represented as \phi(\mathbf{v})=\left for some vector \mathbf{w} in the vector space. \mathbf{w} is unique for each \phi.

    Hopefully that makes sense. Basically, the dual space elements are represented by an inner product, or at the very least, you can define an inner product using them.

  • 2. Alex
    February 15th, 2008 at 10:47 pm

    Thanks OMF, I ended up working with the inner product representation of the dual space. It’s just offputting to have to say things like, if f is an element of the dual space, the coadjoint map evaluated at f is \langle g \alpha_f g^{-1}, \cdot \rangle where \alpha_f is the element in the primal space that corresponds to f via the Riesz representation. Seems unnaturally roundabout.

    In fact, a lot of my problems with the way this book approaches differential geometry stem from the fact that we make and use identifications, all the time, in ways that aren’t really obvious to me. As an example, if the book did the above example, it would from the beginning say that an element in the dual space is an element in the primal space. In this case, that wouldn’t be too confusing, but once you start making layers of arguments where you skirt around issues of the identifications you used, things get really confusing.

  • 3. ObsesiveMathsFreak
    February 16th, 2008 at 9:11 pm

    So that’s what the Riesz representation theorem was about. I often heard the name, read proofs of the theorem, but have not made the connection till now that it was a proof of the fact that dual elements were the left hand side of inner products.

    The proofs I was reading dealt with infinite dimensional Hilbert spaces (or was it Banach spaces?). In retrospect I see now why the proof was so opaque.

    I sometimes wonder do the people who write these textbooks actually understand what their theorems are about? If they do, they don’t carry their knowladge across in the text.

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