# Inner product on spinor space

1. Feb 11, 2016

### gentsagree

I would like to gain a more formal mathematical understanding of a construct relating to spinors.

When I write down Dirac spinors in the Weyl basis, I see why if I multiply the adjoint (conjugate transpose) of a spinor with the original spinor I don't get a SL(2,C) scalar. It just doesn't work if we stick with the conventions of multiplication between (row and column) vectors. To clarify, here I am thinking of column vectors in spinor space as elements of a complex vector space carrying the fundamental representations of the complexified Clifford algebra, Cl(C). Rows are the complex conjugate, transposed object.

Thus one sees that one needs to "rotate" the components of the Dirac spinor (say with $\gamma^{0}$), as well as complex transpose them, to be able to multiply properly into SL(2,C) scalars.

My question is the following. From the perspective of vector spaces, what goes wrong with defining the usual inner product $x^{\dagger}x$ on this complex vector space (i.e. the spinor space)? Does this construct have a particular name in the maths literature?

Thanks!

2. Feb 12, 2016

### vanhees71

The reason is that the Lorentz group is not compact and thus there are no non-trivial unitary finite-dimensional representations. For some details about these group-theoretical aspects see Appendix B of

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

3. Feb 12, 2016

### gentsagree

I can see now that all my question boils down to is that the generators of the spinor representation of Lorentz cannot be all made to be hermitian. Say we make rotations anti-hermitian and boosts hermitian. This implies the weird inner product I mention above.

Does the fact that the Lorentz group is non-compact imply in an obvious way that the generators can't be made all hermitian?

Thanks

4. Feb 12, 2016

### dextercioby

But you misunderstood the impact of non-compactness: you CAN make all its generators hermitean, but they won't be finite dimensional matrices anymore. That's the trick: you want spinors, you have to drop hermiticity for all six at the same time.

To go even further, vanhees's statement is not quite exact. The theorem involved in the representation theory of the proper Lorentz group reads:

That simple added there is important. It means the group shouldn't have proper invariant subgroups (which is the case for the proper Lorentz group). If the group is not simple, then the theorem won't hold.

Last edited: Feb 12, 2016
5. Feb 13, 2016

### vanhees71

True, I plead guilty as charged. Where is the quote of the theorem from?

Indeed, it's important to stress that there are unitary (ray) representations of the proper orthochronous Poincare group, because that implies that there exists a quantum theory in the usual sense within special relativity. You come quite far in developing it by investigating all the possible unitary representations, which goes back to Wigner's famous paper on the subject.

6. Feb 13, 2016

### dextercioby

It's the only source for this much used theorem which I could find: 2nd volume of Cornwell's book on group theory, chapter 15.

7. Feb 13, 2016

### vanhees71

Thanks!

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