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In pg. 81 of the Physics from Symmetry book, the author deduced that the $(\frac{1}{2}, \frac{1}{2})$ representation of the Lorentz $SO(1,3)$ group acts on $2\times2$ Hermitain matrices using the following arguments:

  • First assume that the $(\frac{1}{2}, \frac{1}{2})$ representation acts on general $2\times2$ matrices.

  • Any $2\times2$ matrix can be written as a sum of a Hermitian and an anti-Hermitian matrix.

  • Transformations in the $(\frac{1}{2}, \frac{1}{2})$ representation always transform a $2\times2$ Hermitian matrix into another $2\times 2 $ Hermitian matrix and equivalently an anti-Hermitian matrix into another anti-Hermitian matrix.

  • The above point tells us that the if $(\frac{1}{2}, \frac{1}{2})$ representation acts on general $2\times2$ matrices, it is a reducible representation. But this contradicts the fact that the $(\frac{1}{2}, \frac{1}{2})$ representation is irreducible.

  • Thus the $(\frac{1}{2}, \frac{1}{2})$ representation acts on $2\times 2$ Hermitian matrices.

The leap in logic from the 4th to 5th point confuses me. Why can't the $(\frac{1}{2}, \frac{1}{2})$ representation act on $2\times2$ anti-Hermitian matrices?

Qmechanic
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TaeNyFan
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    Possible duplicates: https://physics.stackexchange.com/q/160794/2451 , https://physics.stackexchange.com/q/28505/2451 and links therein. – Qmechanic Nov 30 '20 at 16:26
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    The point is antihermitean matrices A map to hermitian ones H by $ H\equiv i A$, so $A^\dagger = -A \leftrightarrow H^\dagger = H, $ so you didn't miss anything by ignoring the antihermitean block. Recall a real number multiplying a complex number! – Cosmas Zachos Nov 30 '20 at 16:44

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