# How can parity be meaningful in an affine space?

I've recently begun a course in QFT (within a Physics Master's), and despite (admittedly limited) reading I can't get my head round the idea of parity.

Here's what I think I understand: the Minkowski spacetime of (usual) QFTs is an affine, rather than a vector, space. The Lagrangians of QFT have the continuous symmetries of the (proper, orthochronous) Lorentz group $SO^+(1,3)$, so there's no problem expressing QFT using position vectors (i.e. - we have translational invariance).

However, parity is usually expressed as 'changing the sign of the spatial components' - reflecting through an origin. Not only is it a feature of the theory - the chiral spinors $\psi_\pm$ are exchanged by parity, etc. - but it has testable predictions.

What I don't understand is how and where a parity transformation can be made without reference to the origin (and thereby working in a vector space, unlike the theory - and seemingly the Universe). Is there some generalised operator acting within the space of the theory? Is writing $\psi(\vec{x}) \mapsto \psi(-\vec{x})$ merely a notational convenience that expresses the transformation in spinor space of $\psi_{\pm} \mapsto \psi_{\mp}$ (which could be otherwise represented without reference to a vector space)? Does this problem evaporate when considering that the Lorentz group acts on the tangent space to a point, and therefore the space on which the parity transformation acts predicated on a specific point?

These are the kind of ideas I am unable to resolve. Maybe I'm seeing a contradiction where there isn't one, and this can be safely ignored thanks to isomorphisms between the relevant affine/vector spaces or some such feature - but hopefully someone understands my confusion and can point me in the right direction. Thanks!

• Just take "parity" to mean the behaviour under the parity transformation in the full Lorentz group $\mathrm{O}(1,3)$, no need to talk about origins. – ACuriousMind Nov 5 '15 at 17:04
• I think that's what I was trying to get at in my second question (in the fourth, long paragraph) - so the meaning of $\psi_+(\vec{x}) = \psi_-(\vec{x})$ is that the action of $\hat{P}$ on the object $\psi_-(\vec{x})$ is equivalently either of those two expressions? That it's only when acting with $\hat{P}$ on the position vector $\vec x$ that we need to refer to an origin, which is OK because $\vec x$ itself specifies that origin? I appreciate your response! – T. Lawson Nov 5 '15 at 17:24

1. In this answer let us focus on spatial point reflections and the spatial part of Minkowski spacetime. (There is, by the way, a similar issue with time reversal symmetry $T$.)
2. A choice of point reflection $P_0$ in affine 3-space inevitable has a fixed point ${\bf r}_0$.
3. (We can use the restricted Lorentz group to trade a point reflection $P$ to a reflection in a 2-plane, but that would have a fixed 2-plane instead.)
4. We can use the translation symmetry to trade a point reflection $P_0$ with fixed point ${\bf r}_0$ to a point reflection $P_1$ with fixed point ${\bf r}_1$.
5. So while a specific choice of point reflection $P_0$ singles out a specific fixed point ${\bf r}_0$ and hence breaks manifest translation symmetry in affine 3-space, the full Poincare group treats the affine 3-space covariantly.