Time reversal and parity symmetry

I was previously under the misapprehension that time $T$ and parity $P$ symmetries in conjunction ($PT$) were a reflection in $(3+1)$-dimensional space-time, where

$$P: \vec x \to -\vec x$$ $$T: t \to -t$$

but, in fact, because it has determinant $(-1)^4=1$, $PT$ is merely a rotation (by $\pi$). Is this correct? Shouldn't the symmetry we seek be a reflection in space time, to generalize parity so that it includes time?

In other words, if $P$ is a reflection in $3$ spatial dimensions, why isn't $PT$ defined to be a reflection in $(3+1)$-dimensional space-time?

My thought is that $P$ is an intuitive symmetry in spatial dimensions. In the spirit of relativity, why don't we generalize $P$ from a reflection in space to a reflection in space-time (called, say, $PT$)? Why are $P$ and $T$ reflections considered separately? It's problematic to me because their combination ($PT$) is not a reflection. I suppose I know that $P$ and $T$ are separate because only $T$ must be anti-unitary etc...

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$PT$ is just a rotation by $\pi$ in space-time. I don't quite understand what you mean by your last sentence. –  Prahar Oct 27 '13 at 14:30
Are you suggesting we should always take PT as a whole and never separately? It doesn't sound like a neat idea, since with P and T we have more symmetry transformations to play with than PT as a whole. –  Jia Yiyang Oct 27 '13 at 16:18
@Prahar I've tried to make my question clearer, it was a bit vague - let me know if I have failed :) I agree that $PT$ is a rotation in space-time. –  innisfree Oct 27 '13 at 23:26

$PT-, T-, P-$ transformations refer to subgroup of discrete transformations of the Lorentz group. They transform connected components of the Lorentz group between each other ($PT$ transformation transforms $L^{\uparrow}_{+}$ representation to $L^{\downarrow}_{+}$). In general, they can't be represented as the special case of rotation, which refer to subgroup of continuous transformations. You can't get some other connected component from orthochronous group by acting of any continuous transtormation's matrix on your representation. So, by nature, it can't be rotation.
But $PT$ is a rotation. I don't understand your answer. –  innisfree Oct 27 '13 at 22:53
@innisfree . It can't be the 3-rotation, because it changes the sign of $0$-component of representation. I wrote about the Lorentz group, which has 3-rotation subgroups (not about some $SO(4)$-group). It has subgroups of continuous and discrete transformations. The second can transform representation from one connected component to another, the first doesn't. $PT$-transformation, for example, transforms rep from $L^{\uparrow}_{+}$ to $L^{\downarrow}_{-}$. –  Andrew McAddams Oct 27 '13 at 23:00
OK, we agree that $PT$ is a four-rotation by $\pi$? My question is, why isn't $PT$ a four-reflection? in the way that $P$ is a three-reflection. This seems to me to be the natural generalization from $\vec x$ and $P$ symmetry to $(t, \vec x)$ and $PT$. –  innisfree Oct 27 '13 at 23:15