# Yang Mills Lagrangian positive?

I have often seen the notation $F_{ij} F^{ij} = F^2$, for the term in the Yang-Mills lagrangian, and I was wondering about the justification of this. Is this a positive quantity?

For concreteness, let $A = A_i dx^i$ be a $U(1)$ potential on some spacetime $(M,g)$. Then $$F = dA = F_{ij}dx^i \wedge dx^j$$ The coefficient $F_{ij}$ is a smooth function on $M$, and the quantity $$F_{ij} F^{ij} = g^{im} g^{jn} F_{ij} F_{mn}$$ is also a smooth function on $M$. Is this smooth function positive?

• Is $g$ assumed to be $diag(1, -1, -1, -1)$ or $diag(-1,1,1,1)$? – Darkseid Sep 25 '17 at 3:42
• Neither - $g$ is assumed to be a generic Lorentzian metric (in fact, for the situation that I care about the metric is actually Riemannian). – Mark B Sep 25 '17 at 4:29
• I can rephrase the question. Is $p^2$ (where $p$ is a time-like four-vector) negative or positive? – Darkseid Sep 25 '17 at 4:33
• For the cases I am interested in, the signature of the spacetime is $(-,+,...,+)$ or $(+,+,...,+)$. I am not necessarily working in four dimensions, but I will take whatever answers I can get. – Mark B Sep 25 '17 at 4:38
• In Lorentzian space, note that the YM Lagrangian couldn't possibly be positive-definite since then all $D$ components of the gauge field would be physically propagating degrees of freedom. As we well know, only $D-2$ d.o.f. are physical. Therefore, there must be ghost degrees of freedom which do not propagate in order to restore unitary and Lorentz invariance of the theory. – Prahar Sep 25 '17 at 12:06

1. The YM Lagrangian density $${\cal L}_E~=~\frac{1}{4}{\rm Tr}(F_{\mu\nu}F^{\mu\nu})\tag{1}$$ is non-negative if we assume Euclidean ($E$) signature $(+,+,+,+)$, non-negative Killing metric, and real-valued gauge fields.

2. In Minkowski ($M$) signature $(\mp,\pm,\pm,\pm)$, the YM Lagrangian density $${\cal L}_M~=~-\frac{1}{4}{\rm Tr}(F_{\mu\nu}F^{\mu\nu}) \tag{2}$$ is neither non-positive nor non-negative. This is already evident in E&M (=abelian YM), where $${\cal L}_M~=~\frac{1}{2}(\vec{E}^2-\vec{B}^2).\tag{3}$$ In contrast, the Hamiltonian density ${\cal H}_M\geq 0$ is non-negative.

• Is this still true in Euclidean signature? In that case, wouldn't one have $\mathcal{L} = \frac{1}{2} (\vec{E}^2 + \vec{B}^2)$? – Mark B Sep 26 '17 at 0:48
• $\uparrow$ Yes! I updated the answer. – Qmechanic Sep 26 '17 at 8:02

$F_{ij} F^{ij}$ is a scalar and thus independent of coordinate system. A metric can be locally diagonalized by selecting appropriate coordinates, in diagonalized metric this product takes following form:

$$\sum_{ij} F_{ij} F_{ij} g^i_i g^j_j$$

Which appears to be dependent on the metric signature in general, and might take any sign.

• Just to nitpick a little: A metric can be diagonalized by a local Lorentz transformation, but not necessarily by a choice of coordinates! In 4 dimensions, the metric has 6 degrees of freedom (10 independent components, minus 4 coordinate redefinitions). But a diagonal metric has only 4 degrees of freedom, so there must be 4d metrics which cannot be diagonalized by a coordinate transformation (the Kerr metric is one example). – Ben Niehoff Sep 26 '17 at 8:41