# Yang-Mills equations are not elliptic [closed]

How does one prove that the Yang-Mills equations (from classical Yang-Mills theory) are not elliptic?

Alternatively, how does one calculate the principal symbol of the Yang-Mills equations?

Can someone please present a proof of nonellipticity or derive the principal symbol, or point out where this has been shown in the literature.

• Just look at the locally trivial abelian case: A connection is just a $1$-form $\alpha$, and the Yang-Mills equation for it is $\mathrm{d}^*\mathrm{d}\alpha=0$. The symbol of $\mathrm{d}^*\mathrm{d}$ is $\sigma_\xi(\mathrm{d}^*\mathrm{d}) = \pm \ast (w_\xi)\ast\circ w_\xi$, where $w_\xi(\beta) = \xi\wedge\beta$. In particular, $\sigma_\xi(\xi)=0$, hence, $\sigma_\xi$ is not an isomorphism. Now repeat for the general Lie group $G$, since the symbol is essentially the same. – Robert Bryant Jul 28 '14 at 8:57
• I posted the question at math.stackexchange.com/questions/880998/…. – mdg Jul 30 '14 at 22:27

In fact Yang-Mills equations are elliptic modulo gauge transformations. In simple terms this can be explained as follows (credit: Jonathan Evans, http://www.homepages.ucl.ac.uk/~ucahjde/yangmills.htm). For example, in magnetostatics the vector potential satisfies the equation $$\nabla\times(\nabla\times\vec{A})=\nabla(\nabla\cdot \vec{A})-\nabla^2\vec{A}=\mu_0\vec{J}. \tag{1}$$ The equation is not elliptic because of the ﬁrst term. However, using the gauge freedom $\vec{A}\to\vec{A}^\prime=\vec{A}+\nabla f$, we can find such $\vec{A}^\prime$ that solves $\nabla\cdot \vec{A}^\prime=0$. All what is needed is to solve the Poisson’s equation for $f$: $$\nabla^2 f=-\nabla\cdot \vec{A}.$$ Then (1) becomes an elliptic equation for $\vec{A}^\prime$. What such kind of "good gauge" exists in general case of Yang-Mills equations is explained in http://projecteuclid.org/euclid.cmp/1103920742 (Removable singularities in Yang-Mills fields, by K.K. Uhlenbeck).