# Infinitesimal $SO(N)$ transformations

An infinitesimal $SO(N)$ transformation matrix can be written :

$$R_{ij} = \delta_{ij}+\theta_{ij}+O(\theta^2)$$

Now it has to be shown that $\theta_{ij}$ is real and anti-symmetric.

I've started with the orthogonality condition like following:

$$R^TR=\boldsymbol 1$$ $$\implies (R^TR)_{ij}=\delta_{ij}$$ $$\implies \sum_kR^T_{ik}R_{jk}=\delta_{ij}$$ $$\implies \sum_kR_{ik}R_{jk}=\delta_{ij}$$ $$\implies \sum_kR_{ik}R_{ik}=\delta_{ii}$$ $$\implies \sum_jR_{ij}R_{ij}=1$$

Now i can stick my infinitesimal form of $R_{ij}$ into the above formula:

$$\sum_j(\delta_{ij}+\theta_{ij}+O(\theta^2))(\delta_{ij}+\theta_{ij}+O(\theta^2))=1$$ $$\implies \sum_j (\delta_{ij}\delta_{ij}+2\theta_{ij}\delta_{ij}+O(\theta^2))=1$$

As you can easily see my calculations are going nowhere.

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• You are on the right track. Note that ${}^t \theta_{ij} = \theta_{ji}$ and please be careful with (dummy) indices. You should never write anything like $\delta_{ii}$ in a covariant calculation. – Vibert Jun 15 '13 at 19:47

$$R^{T}_{ij}=\delta_{ij}+\theta_{ij}^{T}=\delta_{ij}+\theta_{ji}$$
$$\sum_j(\delta_{ij}+\theta_{ij}+O(\theta^2))(\delta_{ij}+\theta_{ji}+O(\theta^2))=\sum_j(\delta_{ij}+\theta_{ji})(\delta_{ij}+\theta_{ij})=\sum_j(\delta_{ij}\delta_{ij}+\delta_{ij}(\theta_{ij}+\theta_{ji})+\theta_{ij}\theta_{ji})=1$$
This is true, if $\theta_{ij}=-\theta_{ji}$. To prove that $\theta_{ij}$ is real, evaluate this $\mathrm{Im}[(R^{T}R)_{ij}]$. You will find that $\mathrm{Im}[(R^{T}R)_{ij}]=0$.
• I don't fully agree with the calculation. Why repeat the index $i$ twice but sum over $j$ explicitly? It's much cleaner to write $$(R \cdot {}^tR)_{ik} = (\delta_{ij} + \theta_{ij})(\delta_{jk} + \theta_{kj}) = \delta_{ik} + (\theta_{ik} + \theta_{ki}) + \mathrm{O}(\theta^2)$$ to reach the desired conclusion. – Vibert Jun 15 '13 at 21:08