Counting degrees of freedom of gauge bosons Gauge bosons are represented by $A_{\mu}$, where $\mu = 0,1,2,3$. So in general there are 4 degrees of freedom. But in reality, a photon (gauge boson) has two degrees of freedom (two polarization states). 
So, when someone asks about on-shell and off-shell degrees of freedom, I thought they are 2 and 4. But I read that the off-shell d.of. are 3. 
And my question is how to see this?
 A: One way to see this is the fact that if you write out the Lagragian in terms of the $A^0$ and $A^i$ explicitly, you will find there is no time derivatives for the $A^0$. The only candidate term $F^{00} = \partial^0 A^0 - \partial^0 A^0$ vanishes. This means we shouldn't count $A^0$ as a degree of freedom, and in fact, we should integrate it out. Now the resulting Lagrangian explicitly only has 3 DOF. If you now take this Lagrangian and vary it you will get the EOM which puts one constraint on the 3 DOF which knocks you down to the 2 DOF on-shell. 
A: On-shell dof:
Consider the constraints of "Equation of Motion" plus "Gauge Condition"(Lorenz gauge, Coulomb gauge,...), #dof=4-2.
Off-Shell dof:
Consider the constraints of "Gauge Condition" (Lorentz gauge, Coulomb gauge,...) only, #dof=4-1.
A: The four degrees of freedom *A*0,*A*1,... will reduce to three after
taking into account the Lorenz gauge condition.
By an on-shell photon we mean a massless particle, So the k^2=0 condition
will give an extra constraint on Lorenz condition such that the number of DOF
will be reduced to 2. To see it explicitly, it would be better to work with Fourier modes 
of A.  
