Invariant quadratic form

Discussion in 'Physics & Math' started by QuarkHead, Sep 5, 2009.

  1. QuarkHead Remedial Math Student Valued Senior Member

    Messages:
    1,740
    The context here is Lie algebra. So, I have text here that states that, for some such algebra (I can give the context, if it is deemed relevant) that, \(\text{Tr}(T_i,T_j) = -\frac{1}{2}\delta_{ij}\). It (my text, that is) refers to the \(T_i\) as "representation matrices". I don't understand what this means. Do they mean the adjoint rep. of group generators, or what I would prefer to call basis vectors (for the algebra)?

    And what gives with the RHS of the equality above? Half of Kronecker? Uhh?

    I can just about convince myself that \(\text{Tr}(T_i,T_j)\) is the Killing form, usually thought of as a symmetric bilinear form, here expressed rather strangely, since I was taught to write this as \(\text{Tr}(ad(X)ad(Y))\), where X and Y are basis vectors for some algebra and, say, \(ad(X)Y = [X,Y]\). Anyway, if I am right to interpret this as the Killing form, then what they call "representation matrices" turn out to be the adjoint maps. Yes, no?

    But.... my text calls this an invariant quadratic form.

    I know very well what a quadratic form is, and I equally well know there are circumstances where quadratic forms and symmetric bilinear forms coincide.

    So, here is my question: Is my text using a really far-out sort of notation, or is there something I am missing here?

    As always, my text is the Spanish one I referenced before
     
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    The Killing form is defined using the ad representation, so the generators \(T_{a}\) have matrix representations with entries \((T_{a})^{b}_{c} = f^{b}_{ac}\). If you know the commutation relations, you know the matrices. From that you have \(\kappa_{ab} = Tr(ad(T_{a})ad_{Tb}) = Tr( T_{a}\cdot T_{b}) = Tr ( (T_{a})^{i}_{j}(T_{b})^{j}_{k}) = (T_{a})^{i}_{j}(T_{b})^{j}_{i}\). The Kronecker delta definition you give means that \((T_{a})^{i}_{j}(T_{b})^{j}_{i} = -\frac{1}{2}\delta_{ab}\), ie \(\kappa_{ab} = -\frac{1}{2}\delta_{ab}\). In other words the Killing form is -1/2 times the identity matrix.

    The other bilinear forms are such that you don't pick the adjoint representation you pick something else, such as the fundamental representation. By fixing the factor (in this case -1/2) you're basically rescaling the generators in some way and often (say in the application of Lie algebras to quantum field theory) there's a nice choice of your representation.
     
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