# Normalising Generators of a Lie Algebra

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?

• The Lie algebra $\mathfrak{su}(N)$ is just a vector space. It has a basis, which we can normalize if we want. That's all there is to this, mathematically. As for the particular normalization Srednicki chooses: I'm not sure why one would want this. – Danu May 1 '15 at 18:58
• Check Killing form of the Lie Algebra. Anyway that choice is the standard one in physics. – Rexcirus May 1 '15 at 21:38

## 2 Answers

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 $1/2$.

• Ok that's helpful, thanks. So is this inner product only standard for $\mathfrak{su}(N)$, or in general for any Lie algebra? Because if it's only standard for $\mathfrak{su}(N)$, why is it defined with the Hermitian conjugate if you're then going to get rid of that anyway? And why is it natural to define this to be our inner product? Also presumably when you say that the vectors have norm $\frac{1}{2}$, you mean $\frac{1}{\sqrt{2}}$? – Joe May 3 '15 at 0:42
• @Joe The definition above gives an inner product on any real or complex vector space of square matrices. It is commonly called the Frobenius or Hilbert-Schmidt inner product. Naturalness comes from thinking of matrices as rearranged column vectors and then applying the standard inner product on $\mathbb C^{n^2}$. See en.wikipedia.org/wiki/Matrix_multiplication#Frobenius_product . Note that I wrote norm-squared $1/2$. – joshphysics May 3 '15 at 17:06
• Oh so you did, yes that makes more sense then. Ok thanks I will have a read of that – Joe May 3 '15 at 18:07

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.

• Welcome to Physics.SE; what a nice first answer! – rob May 1 '15 at 21:38