Orthogonality of inner product of generators

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1. Jul 20, 2015

PineApple2

Hi, this is a rather mathematical question. The inner product between generators of a Lie algebra is commonly defined as $\mathrm{Tr}[T^a T^b]=k \delta^{ab}$. However, I don't understand why this trace is orthogonal, i.e. why the trace of a multiplication of two different generators is always zero.

2. Jul 20, 2015

The_Duck

This is a choice we make for convenience. The generators form a basis for vector space that is the Lie algebra, and it is convenient to choose an orthogonal basis. We can always choose a basis that is orthogonal.

3. Jul 20, 2015

Orodruin

Staff Emeritus
The only thing you should worry about is whether the trace tr(AB), where A and B are elements of the Lie algebra, is an inner product or not. If you have an inner product, you can always select an orthogonal basis.

4. Jul 20, 2015

vanhees71

This is the great thing with semi-simple compact Lie groups. Their generators can always be chosen such as they are "orthogonal" in the sense you wrote. This implies that the Lie group, as a differentiable manifold (with the group operations providing differentiable mappings), has a "natural" metric (Riemann-space) structure and you can thus easily derive things like the invariant measure for integrations over the group (Haar measure), using Weyl's unitarity trick (proving that all finite-dimensional representations are equivalent to a unitary one) etc. Last but not least, all semi-simple compact Lie groups are identified ("Cartan catalogue"). For details, see Weinberg, Quantum Theory of Fields, Vol. 2.

5. Jul 20, 2015

PineApple2

I see, that makes sense. Thank you all for answering