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I am trying to find three independent infinitesimal transformations that leave my lagrangian invariant.

$$\mathcal{L}=-\frac{1}{2}\partial_{\mu}\phi_I\partial^{\mu}\phi^I-\frac{1}{2}m^2\phi_I\phi^I $$ The indecis $I=1,2,3$ represent three different scalar fields. I have assumed that the transformation is $$\phi^I\rightarrow (\mathbb{1}-i\epsilon_AT^{IJ}_A)\phi^J$$ where A is a label $A=1,2,3$ for the three different transformations, $\epsilon$ is infinitesimal and $T^{IJ}$ is a $3\times 3$ matrix.

I have assumed that I can drop higher order $\epsilon$ so after insertion I get that the variation of the lagrangian is $$\delta\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_I\partial^{\mu}i\epsilon T_A^{IJ}\phi^J+\frac{1}{2}\partial_{\mu}i\epsilon T_A^{IJ}\phi^I\partial^{\mu}\phi_J $$ which I want to be equal to zero. Now I have matrix equation but it is here that I run into trouble. I think I am confusing myself in trying to solve this matrix equation. I would appreciate some help in solving it and some conformation that my analysis/method is correct.

EDIT

To clarify my problem: we have a matrix equation with the matrix sizes $1\times2\cdot2\times2\cdot2\times1+2\times2\cdot2\times1\cdot1\times2=1\times1+2\times2=0$ and I do not see how to solve it.

user405158
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  • Any $SO(3)$ rotation of the fields leaves the Lagrangian invariant. You could take $T_A^{IJ}$ to be three generators of the Lie algebra of $SO(3)$, i.e. take them to be the spin-1 Pauli matrices. – Kenny Wong Apr 04 '17 at 10:40
  • Are you looking for a transformation for $\mathfrak{so}(3)$ internal symmetry? –  Apr 04 '17 at 10:41
  • That was what I assumed was the correct answer but I could not see how the matrix multiplication was satisfied then. The Pauli matrices are 2x2 and I have that $\phi^I$ are 3x1, how could that work? – user405158 Apr 04 '17 at 13:55
  • Okej, I understand now which matrices you mean. But I still do not see how it leaves the Lagrangian invariant. Is my matrix equation wrong? – user405158 Apr 04 '17 at 15:50

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