# Inversion symmetry on surface and spin

Let us assume you have a 3D bulk periodic crystal which has inversion symmetry e.g. $$r\rightarrow -r$$. Assume we are considering spinful operators with $$S=1/2$$.

Now let us imagine cutting a surface of this, lets say the surface $$(x, y, 0)$$. This surface should inherit the inversion symmetry of the bulk, e.g. it should be invariant under $$(x,y,0)\rightarrow (-x,-y,0)$$.

"In a two-dimensional plane, a simultaneous flip of all coordinates in sign is not a parity transformation; it is the same as a 180°-rotation. "

I can believe that this is not parity as e.g. this question points out that 2D parity would imply that spin is odd under parity. However, I cannot quite justify that this should be a $$180°$$ rotation.

The surface symmetry $$(x,y,0)\rightarrow (-x,-y,0)$$ looks a lot like a $$180°$$-rotation, except that a $$180°$$-rotation would also act on spins as $$e^{i\pi S}$$, e.g. it would give a factor of $$+i$$ for spin up and a factor $$-i$$ for spin down.

However, I would expect this factor to be absent on the surface, as there is no such factor in the bulk symmetry which the surface symmetry is inherited from (e.g. the bulk parity does not act on spin, so the surface symmetry should equally not act on spin).

So what is this effective symmetry? Is it a spinless $$180°$$-rotation in an otherwise spinful system? And if so, does anyone have an intuitive physical picture why this rotation does not give the usual factors for the spins?

Or should this symmetry actually act on spin?

Any help is greatly appreciated!

To make thing more formal recall that from group structure : $$i = r_z(180) \times \sigma_z$$
On 2D spatial coordinates, the action of the reflection with respect to the $$z$$ axis is identity ($$\sigma_z \sim e$$), so for the operators we have $$R_{z, space} (180) = \Pi_{space}$$ whereas the action on the spin degree freedom of the inversion operator is identity ($$i \sim e$$) because spin is an angular momentum. So $$R_{z, spin} (180) = \sigma^{-1}_{z,spin}$$