In the paper in quantum fields theory by Gribov,V.; (1978) "Quantization of non-Abelian gauge theories". Nuclear Physics B. 139: 1–19; in Section 3 the author makes the following claim from PDE and operators theory without any explanation which I would like to understand.

Let $\frak{g}$ be a Lie algebra of a compact Lie group. (If you feel uncomfortable with general Lie algebras you may think of a special case $\frak{g}=\mathbb{R}^3$ with the operation of Lie bracket $[\cdot,\cdot]$ equal to the vector product $\times $.) For $\mu=1,\dots,4$ let $$A_\mu\colon \mathbb{R}^4\to \frak{g}$$ be fixed smooth functions with compact support and satisfying $\partial_\mu A_\mu=0$ (where there is a summation convention in repeated indexes).

Consider the differential operator $L$ on $\frak{g}$-valued functions $$L\alpha =-\Delta \alpha +[A_\mu,\partial_\mu \alpha],$$ where $\Delta$ is the ordinary Laplacian acting component-wise, $[\cdot,\cdot]$ is the Lie bracket. Clearly this is a symmetric operator.

As far as I understand, Gribov claims that $L$ has no negative discrete spectrum provided $A_\mu$ are very small (in some sense). Why??

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    $\begingroup$ Consider the corresponding quadratic form $\int |\nabla \alpha|^2 + A_\mu \partial_\mu \alpha \cdot \alpha$. In $\mathbb{R}^4$ by Sobolev $W^{1,2} \to L^4$ you can bound $\int A_\mu \partial_\mu \alpha \cdot \alpha$ by $C \|A_\mu\|_{L^4} \|\partial \alpha\|_{L^2}^2$. So provided that $\|A_\mu\|_{L^4} < C^{-1}$ you have that the quadratic form is positive semidefinite. $\endgroup$ – Willie Wong May 29 '19 at 2:19
  • $\begingroup$ @WillieWong: I am not sure how you use Sobolev theorem and get rid of the term $\alpha$ without derivatives. Are you using also some version of Poincaré inequality in $\mathbb{R}^n$? $\endgroup$ – MKO May 29 '19 at 17:59
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    $\begingroup$ Gagliardo-Nirenberg-Sobolev on $\mathbb{R}^4$ states that $\| u\|_{L^4} \leq C \|\partial u\|_{L^2}$ (among other things). en.wikipedia.org/wiki/… $\endgroup$ – Willie Wong May 29 '19 at 21:18
  • $\begingroup$ @WillieWong : Many thanks. Seems to work. $\endgroup$ – MKO May 29 '19 at 22:32

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