Normalising Generators of a Lie Algebra

Question

Ok, so I'm asking this in physics because I'm currently working through part of Srednicki's text on QFT, even though it's really a maths question.

In Srednicki's chapter on non-Abelian gauge theory, he introduces the generators of a Lie group. At the moment we're only analysing $SU(N)$, which is defined by $M M^\dagger = 1$ and $\det(M) = 1$ for all $M \in SU(N)$

And the corresponding conditions on the generators of the group are $T = T^\dagger$ and ${\rm Tr}(T) = 0$ for all $T \in \mathfrak{su}(N)$

Then what I don't understand is that Srednicki tells me that we should normalise our generators so that $${\rm Tr}(T^i T^j) = \frac{1}{2}\delta^{ij}.$$

So presumably this arises because our set of $N^2-1$ generators is a basis for the tangent space of $SU(N)$ at the identity, and we choose it to be orthogonal and then need a condition to normalise the lengths of all of the basis vectors? Why did the condition Srednicki gave do that? And where did we input that the vectors are orthogonal?

Answer 1

Just a guess... The purpose is to reproduce the nice features of $SU(2)$. With that convention, the generators of $SU(2)$ are, in terms of Pauli matrices $$T^i = \frac{1}{2}\sigma^i$$ So a transformation with parameters $\theta_i$ is given by $$U=\exp\left(-i\frac{1}{2}\theta_i\sigma^i\right)$$ Things get interesting when you realize that the elements of $SU(2)$ are related to usual rotations $SO(3)$ (namely, $SU(2)$ is the double cover of $SO(3)$), and the parameters $\theta_i$ are equal to the angles of rotation around the axes. If we would have chosen another convention, extra factors would appear, and the parameters woul be proportional (but not equal) to the angles. Not a big problem, but a bit uglier.

Once you have chosen a convention for $SU(2)$, it seems natural to generalise it to $SU(N)$.

Addendum: This convention is quite common, but not universal. For example, Elvang and Huang (arxiv:1308.1697) choose $\textrm{Tr}(T^a T^b) = \delta^{ab}$ with structure constants $[T^a, T^b]=i \tilde{f}^{abc}T^c$. They are related to the "usual" structure constants by $\tilde{f}^{abc} = \sqrt{2}f^{abc}$. In this way, they get rid of some $\sqrt{2}$ factors.

Answer 2

The Lie algebra $\mathfrak{su}(N)$, viewed as a vector space of matrices, can be equipped with the following standard inner product: \begin{align} \langle X,Y\rangle = \mathrm{tr}(X^\dagger Y), \end{align} where $X^\dagger Y$ is the matrix product of $X^\dagger$ and $Y$, and $\mathrm{tr}$ is the trace. Since $X^\dagger = X$ for all $X\in\mathfrak{su}(N)$, the right hand side reduces to $\mathrm{tr}(XY)$. Thus, the condition Srednicki writes expresses orthogonality with respect to this standard inner product, and Srednicki chooses to normalize the generators to have norm-squared $1/2$.