Why does a cutoff break gauge invariance? It's been stated repeatedly that introducing a sharp momentum cutoff $\Lambda$ into a gauge theory breaks gauge invariance. Apparently, this is because momentum modes directly at the cutoff cannot be subjected to a gauge transformation.
Is this the only problem and could someone point to some resource where this calculation is done explicitly?
 A: A part of the problem with this question is what is meant by gauge invariance. I think the issue with gauge fields is covariance. The momentum operator is made gauge covariant by $\hat P~=~\hat p~+~ie\hat A$ then for $\hat A~\rightarrow~\hat A~+~\partial\chi$ these transformations are incomplete if we demand that $|\hat P|~<~\Lambda$. The whole set of possible transformations on the principal bundle are not permitted. 
This carries over to the fields as well.  The transforms gauge connections in general are 
$$
A'~=~g^{-1}Ag~+~g^{-1}\partial g.
$$
The gauge transformation here is seen from letting $g~=~e^{\chi}$ $=\simeq~1~+~\chi$ for $\chi$ small so that 
$$
A'~=~A~+~[A,~\chi]~+~\partial\chi~+~O(\chi^2),
$$
where the commutator is zero in the caseabove. The fields can be seen from the covariant momentum according to differential forms
$$
{\bf F}~=~{\bf P}\wedge {\bf P}~=~d\wedge {\bf A}~+~{\bf A}\wedge {\bf A}.
$$
The transformed fields are then $F'~=~g^{-1}Fg$ that gives
$$
F'~=~F~+~[F,~\chi].
$$
A cut off again means that the range of transformations is restricted.
What is invariant is the Lagrangian. For ${\cal L}~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$, with indices restored, or ${\cal L}~=~-\frac{1}{4}F\cdot F$, is such that ${\cal L}'~=~-\frac{1}{4}g^{-1}Fg\cdot g^{-1}Fg$ is just a transformation of a scalar so ${\cal L}'~=~{\cal L}$. Consequently the cut-off does not appear to influence the gauge invariance of the Lagrangian.
