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Positive definite inner product on Lie algebra.

  1. Jul 29, 2014 #1
    While studying Yang-Mills theory, I've come across the statement that
    there exists a positive-definite inner product on the lie algebra ##\mathfrak g## iff the group ##G## is compact and simple. Why is this true, and how it is proved?
  2. jcsd
  3. Jul 31, 2014 #2
    I'm not entirely sure what you mean here. The Lie algebra [itex] \mathfrak{g} [/itex] coming from a finite dimensional manifold is just a finite dimensional vector space so of course we can always construct a positive definite inner product on it just by choosing a basis and using the standard inner product on n-tuples.

    In fact, this can even be made Ad-invariant for a compact Lie group. If we let [itex] G [/itex] be any compact Lie group, we have the representation [itex] \mathrm{Ad}:G\to GL(\mathfrak{g}) [/itex] and we also have a left Haar measure [itex] dx [/itex] on [itex] G [/itex]. Define the inner product by
    [tex] (u,v)=\int_G \langle Ad(x)u, Ad(x)v \rangle dx [/tex]
    where [itex] \langle \cdot,\cdot \rangle [/itex] is just any inner product. It is straightforward to check that this is a positive-definite, ad-invariant inner product.

    If the Lie group is also simple as well as compact, then this will be a unique Ad-invariant inner product (up to scaling by a constant I think) on the Lie algebra so perhaps the statement is meant to be there is a unique Ad-invariant inner product on [itex] \mathfrak{g} [/itex] iff the group [itex] G [/itex] is compact and simple?
  4. Aug 2, 2014 #3
    Yes, I did forget the Ad-invariant part. Perhaps, you can help me with a related question:
    when G is simple, then as you said the inner product is unique up to scale.. Let's chose one of them, say the inner product ##h: \times \mathfrak g \to \mathbb{R}##. Then in the Lagrangian
    ##L = - \frac{1}4 h_{kl}F^k_{\mu \nu} F^{l \mu \nu}##
    with ##h_{kl}## the matrix of the relevant inner product. Any other Ad-invariant inner product ##g_{kl}## is proportional to this one, say ##h_{kl} = c g_{kl}##, which means that we can redefine ##\sqrt{c}F_{\mu \nu} \to F_{\mu \nu}## rendering the lagrangian equivalent.

    But now suppose G is not simple. Then there are not one unique Ad-invariant inner product (up to scale) but several -- is it then irrelevant which one we choose in the Yang-Mill's lagrangian? As the above argument fails in this case, why is this so?
  5. Aug 2, 2014 #4
    The easiest example to see that these inner products don't need to be unique on lie algebras that are not simple is just to take a semisimple lie algebra [itex] \mathfrak{g}=\mathfrak{h_1}\oplus \cdots\oplus \mathfrak{h_n} [/itex]. Then you can define a bi-invariant inner product on each summand however now you can scale the inner products independently in each factor rather than a single overall scaling factor.

    As to how this affects the Lagrangian, I presume by equivalent you just mean it yields the same physics (ie. equations of motion?) In this case I believe the Lagrangian is not equivalent (you probably shouldn't trust me here though since I'm definitely not a physicist.) My reasoning is the Hodge star operator appears in the equation of motion, which here is the Yang-Mills equation [itex] \star d_D\star F=0[/itex], as well as in the definition of the lagrangian. Changing the inner product will change the Hodge star which usually will change the Yang-Mills equation itself. You would have to look at the functional dependence of the Yang Mills equation on the inner product to determine exactly for which class of transformations of the inner product it remains invariant.
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