## Yang Mills

To define the Yang-Mills Lagrangian, we need to define the ‘Trace’ of an End(E) valued form. Recall that the Trace of a matrix is the sum of its diagonal enTries. The Trace is independent of the choice of basis – an invariant notion that is independent of the choice of basis. A definition of the Trace that mkes this clear is as follows. Consider ${End(V) \simeq V \otimes V^*}$ – an isomorphism that does not depend on any choice of basis – so the pairing between V and ${V^*}$ defines a linear map $Tr: End(V) \rightarrow \mathbb{R}$ $v \otimes f \mapsto f(v)$ To see that this v is really a Trace, pick ${e_i}$ of V and let ${\epsilon^j}$ be a dual basis of ${V^*}$. Writing ${T \;\in\; End(V)}$ as $T = T^i_j e_i \otimes \epsilon^j$ We have $Tr(T) = T^i_je_i(\epsilon^j) = T^i_j \delta_i^j = T^i_i$ which is of course the sum of the diagonal enTries. \newline This implies that if we have a section T of ${End(E),}$ we can define a funciton Tr(T) on the base manifold M whose value at ${p \in M}$ is the Trace of the endomorphism T(p) of the fiber ${E_p}$: $Tr(T)(p) = Tr(T(p))$ If ${T \in \Gamma(End(E))}$ and ${\omega \in \Omega^p(M)}$ we define $\displaystyle Tr(T \otimes \omega) = Tr(T)\omega$

Now we can write down the Yang-Mills Lagrangian: If D is a connection on E, this is the n-form given by $\displaystyle \mathcal{L}_{YM} = \frac{1}{2} Tr(F \wedge \textasteriskcentered F) \ \ \ \ \ (1)$

where F is the curvature of D. Note that by the defintion of the hodge star operator (also in this collection of notes), we can write this in local co-ordinates as $\displaystyle \mathcal{L}_{YM} = \frac{1}{4} Tr (F_{\mu \nu}F^{\mu \nu})vol \ \ \ \ \ (2)$

If we integrate ${\mathcal{L}_{YM}}$ over M we get the Yang-Mills action $\displaystyle S_YM = \frac{1}{2} \int_M Tr (F \wedge \textasteriskcentered F) \ \ \ \ \ (3)$

This needs some elaboration. So let us explain these formulas better. We choose the physics convention ${F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu -ig[A_\mu,A_\nu]}$ where the generators of the Lie algebra are Hermitian. $\displaystyle \mathcal{L}_{YM} = \mathcal{I} = - \int Tr(F \wedge \textasteriskcentered F) \ \ \ \ \ (4)$

by another convention (there is a lot of ambiguity of signs in this subject). The first thing to note is that F has vector and Lie algebra indices. The Trace is over the Lie algebra, not over the vector indices. The vector indices are just those of the field sTrength in QED. In Yang-Mills the curvature form is Lie Algebra valued. \newline In this case ${F_{\mu \nu} = F^{a}_{\mu \nu}T^a}$ where the summation convention is used,and where ${T^a}$ are the generators of ${\mathfrak{su}(n)}$. To be explicit, F has not only tensor components but maTrix components $(F_{\mu \nu})_{ij} = F^{a}_{\mu \nu}T^a T^a_{ij}$ The inner product of F with itself ${ = \int F \wedge \textasteriskcentered F}$ where \textasteriskcentered is the Hodge \textasteriskcentered – operator. Thus we are calculating ${\mathcal{I} = -Tr}$. It is a standard exercise to find the exterior product of two r-forms. We find $F \wedge \textasteriskcentered F = \frac{1}{2!}F_{\mu \nu}F^{\mu \nu} dx^1 \wedge \cdots \wedge dx^4$ Note that the differential forms don’t ‘know’ the Lie algebra. The algebra hasn’t come into the calcuation yet. ${Tr(F_{\mu \nu}F^{\mu \nu})=Tr(F^{a}_{\mu \nu}T^aF^{\mu \nu b}T^b}$ ${= Tr(T^aT^b)F^{a}_{\mu \nu}F^{\mu \nu b}}$ ${= \frac{1}{2}\delta^{ab}F^{a}_{\mu \nu}F^{\mu \nu b}}$ ${=\frac{1}{2}F^{a}_{\mu \nu}F^{\mu \nu a}}$ where e have used the standard normalization convention for the ${T^a}$, ${Tr T^a T^b = \frac{1}{2}\delta^{ab}}$. (This comes from the fact we want ${\mathfrak{su}(2)}$ to live in ${\mathfrak{su}(n)}$, and the generators of ${\mathfrak{su}(2)}$ are taken to be ${T^a = \frac{\sigma^a}{2}}$ where ${\sigma^a}$ are the Pauli matrices.) Thus, we find $\mathcal{I} = - Tr (F \wedge \textasteriskcentered F)$ which can be written as $\frac{1}{4} \int d^4 x F^{a}_{\mu \nu}F^{a \mu \nu}$

## 2 thoughts on “Yang Mills”

1. Jay R. Yablon says:

You may want to look at what I just posted on my Blog. It shows an exact solution for the Yang-Mills Largrangian which quantizes the theory and yields an exact propagator.

2. Jay R. Yablon says: