$\pi$, $\sigma$ - atomic transitions with respect to the magnetic field axis I am confused about the atomic transition with different polarized lights.
  I post the pictures as follows. 
  
There are four cases.
  In case 1, the right-handed circular polarized light propagates along the $\mathbf{B}$ axis. So it induces the $\delta m=+1$ transition(i.e.,$\sigma$+ transition). 
In case 2, the left-handed circular polarized light propagates along the $\mathbf{B}$  axis. So it induces the $\delta m=-1$ transition(i.e.,$\sigma$- transition). 
In case 3, the light travels normal to the $\mathbf{B}$  axis and it's linear polarized. So it induces the $\delta m=0$ transition(i.e.,$\pi$ transition). 
Question a: In case 4, what kind of transition will happen?
Question b: The linear polarized light can be decomposed into $\sigma$+ and $\sigma$- polarized light. If the linear-polarized light travels along the $\mathbf{B}$  axis, there will be $\sigma$+ and $\sigma$- transitions meanwhile. Is it right?
 A: 
a: In case 4, what kind of transition will happen?

In this case, the radiation can excite both $\sigma$ and $\pi$ transitions, although you wouldn't really call them that. More specifically, it can excite transitions which keep $L_z$ constant, and it can also excite transitions which change it by $\Delta L_z=\pm \hbar$.
To see why, simply decompose the circular polarization into a superposition of linear polarizations: one along $z$, and one along (say) $x$. The polarization along $z$ can excite the $\Delta L_z=0$ transitions. The linear polarization along $x$, on the other hand, is itself a superposition of circular polarizations with a $\vec k$ vector along $z$, and each of these excites one $|\Delta L_z|=1$ transition. (Why am I allowed to switch the wave propagation direction? This is all done inside the dipole approximation, which takes the polarization to be uniform, so the atom doesn't actually 'talk' to the propagation direction.)
To do this slightly more rigorously, the transition operator for case 4 is $\hat z+i\hat x$, and it can be decomposed as
$$
\hat z+i\hat x=\hat z+i\left(\frac{\hat x+i\hat y}{2}+\frac{\hat x-i\hat y}{2}\right).
$$
The overall matrix element for the transition is therefore
$$
\left\langle f\middle |\hat z+i\hat x\middle |i\right\rangle
=
\left\langle f\middle |\hat z\middle |i\right\rangle
+i\left\langle f\middle |\frac{\hat x+i\hat y}{2}\middle | i\right\rangle
+i\left\langle f\middle |\frac{\hat x-i\hat y}{2}\middle| i\right\rangle.
$$
Here $\pi$ transitions are those for which $\left\langle f\middle |\hat z\middle |i\right\rangle$ is nonzero, and $\sigma_\pm$ are those for which $\left\langle f\middle |\frac{\hat x\pm i\hat y}{2}\middle | i\right\rangle$ is nonzero. (Don't trust me on the sign conventions, though.)
This then answers your second question pretty directly,

b: The linear polarized light can be decomposed into $\sigma$+ and $\sigma$- polarized light. If the linear-polarized light travels along the $\mathbf{B}$  axis, there will be $\sigma$+ and $\sigma$- transitions meanwhile. Is it right?

From the above simply take 
$$
\left\langle f\middle |\hat x\middle |i\right\rangle
=
\left\langle f\middle |\frac{\hat x+i\hat y}{2}\middle | i\right\rangle
+\left\langle f\middle |\frac{\hat x-i\hat y}{2}\middle| i\right\rangle.
$$
and you get your claim.
