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I am currently learning symmetries/group theory and I learnt that the fundamental representation and the anti-fundamental representation of $SL(2,\mathbb{C})$ are not equivalent. This means that no similarity transformation can map one of them to the other.

My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.

Can anyone please explain the explanation of my professor, or perhaps give another explanation?

The First StyleBender
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    Two representations are equivalent iff there is an intertwiner. In this case that means that you need to find a matrix $S$ such that $U=S\cdot \overline{U}\cdot S^{-1}$, where $U$ is a transformation matrix of $(1/2,0)$ spinors and $\overline{U}$ is the corresponding matrix for $(0,1/2)$ matrices. If you can find such an $S$, you prove your professor wrong, but I'm afraid you won't succeed. ;-) –  Nov 25 '18 at 02:27
  • @marmot Is there a rigorous way to show that it isn't possible to find a matrix S? – The First StyleBender Nov 25 '18 at 02:34
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    Yes, of course. Hint: establish that $S$ is unity by just looking at rotations (in your professor's conventions) and then show that this does not work for boosts. –  Nov 25 '18 at 02:36
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    I hope someone else walks you through. My problem is that I do not know what you know. Do you know Schur's lemma? If yes, it is simple, if not, I'd have to explain it here. (If you don't, I'd recommend to just believe you professor for the time being and attend a lecture on group theory, if possible. If you are interested in theoretical physics, this will be a very useful course.) –  Nov 25 '18 at 03:21
  • I voted to migrate this to Math.SE. – AccidentalFourierTransform Nov 25 '18 at 03:47
  • Math mods: Please merge. – Qmechanic Dec 01 '18 at 12:44

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