I'm reading the book "Symmetry and the Standard Model" by Matthew Robinson. On page 80 of the book the author asserts that given a representation $D(g)$ of a group $g$, we can expand around the identity element as:

$$D_n(g(0 + \delta \alpha_i)) = \mathbb{I} + \delta \alpha_i \frac{\partial D_n (g(\alpha_i))}{\partial \alpha_i}\Bigr|_{\alpha_i=0} + ...$$

And we can define the generators of this representations as

$$X_i = -i \frac{\partial D_n}{\partial \alpha_i}\Bigr|_{\alpha_i=0}$$

Then the author derives the generators for $SO(2)$ arguing that rotations in the plane leave the scalar product $v^Tv$ invariant and for some generator of SO(2) we have that $$v \rightarrow R(\theta) v = \exp(i \theta X) v$$

So expanding up to first order we get:

$$v^T e^{i \theta X^T}e^{i \theta X}v = v^T(1 + i\theta X^T + i \theta X) v = v^Tv + v^Ti \theta(X+X^T)v$$

So we conclude that $X = - X^T$ for the dot product to vanish, asserting that $X$ is given by:

$$X= \frac{1}{i} \left[ \begin {array}{cc} 0&1\\ -1&0\end {array} \right] $$

But why does the generator assumes this form? In particular, why are the digonal elements zero? Why can't they any other number? Since this will still product an antisymmetric 2x2 matrix.

  • 2
    $\begingroup$ I don't understand your last sentence. A matrix with non-zero diagonal is not anti-symmetric, since $A_{ii} = -A_{ii}$ does not hold for any $A_{ii}\neq 0$. $\endgroup$
    – ACuriousMind
    Feb 24, 2021 at 16:49
  • $\begingroup$ Right, I was just being silly. In my mind I had a completely different definition of antisymmetric (that is all the off diagonal elements reflect and change sign). $\endgroup$
    – mathripper
    Feb 24, 2021 at 20:26

1 Answer 1


Why are the digonal elements zero?

You said it yourself that $X = -X^T$. Along the diagonal elements this means that

$$ X_{ii} = -X_{ii} \qquad \text{(no sum on $i$)} $$

So the only solution for the diagonal entries is that they are zero.


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