Fermionic generators of course don't act just geometrically on a bosonic space; all differential operators acting on bosonic coordinates are bosonic.

At most, you could consider a superspace extension of $AdS_5 \times S^5$ but superspaces are not too useful if there are too many supercharges (they have too many components). So it's a kind of misguided approach to ask about the action of the supercharges on the spacetime only; one should learn what is the action of the supergroup at the Hilbert space – the whole actual theory – and it is pretty straightforward if you define the $N=4$ gauge theory.

Witten – when he mentions that the group acting on the AdS-space times the sphere is the supergroup – really wants to say that $PSU(2,2|4)$ is the (or "a") maximal supergroup of symmetries that a theory defined on $AdS_5\times S^5$ may have. But he surely doesn't mean that all the generators - the fermionic ones in particular - may be defined as differential operators acting on the 10 bosonic coordinates of this spacetime only.

For the terminology of superalgebras (and the same supergroups), look at page 58 of this Kac's review (PDF):

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1103900590

Finally, the extra $P$ in $PSU$ means that one eliminates a block-diagonal "hypercharge-like" generator from $SU$. It's similar to the embedding of $SU(3)\times SU(2)\times U(1)$ in $SU(5)$ in grand unification; in the superalgebra case, one can consistently eliminate the $U(1)$ here, at least if the number of bosonic entries and fermionic entries (dimensions of the fundamental representation) are equal. And it's equal, both are 4 for $PSU(2,2|4)$. If you check a SE question about an $SU(5)$ decomposition here

Introduction to Physical Content from Adjoint Representations

the equal dimensions allow us to set the "hypercharges" of the off-block-diagonal entries (all the fermionic generators) which were $\pm 5/6$ above to zero and eliminate the "hypercharge" $U(1)$ which was just shown to become a center (generator commuting with all others) completely.

Without the $P$ which stands for "projective", the bosonic subgroup of $SU(2,2|4)$ would really be $SU(2,2)\times SU(4)\times U(1)$ with the extra last factor that actually gets eliminated in $PSU$.

This post imported from StackExchange Physics at 2014-03-22 17:21 (UCT), posted by SE-user Luboš Motl