3 A possible extension

Pushing the above formal analogy a little forward, and taking into account a possible gauge field type formulation of string dynamics [6], one would expect to find a similar relation between the Nambu-Goto action and a non-Abelian Born-Infeld type action

$$\sqrt{\,\left\{\, X^\mu\ , X^\nu\,\right\}_{\mathrm{PB}} \... ...{\, \mathrm{Tr}\mathbf{F}_{\mu\nu}\, \mathbf{F}^{\mu\nu}\,}$$ (14)

While being suggestive, relation (14) suffers from various problems not present in (13), e.g. the very definition of the non-Abelian version of the Born-Infeld action is ambiguous [7] . As the relation (13) can be obtained through several, non-trivial steps, including ``quenching'', large-$N$ expansion, Wigner-Weyl-Moyal quantization, it is the purpose of this communication to investigate how this approach can be, eventually, extended to the square root type gauge action in (14).
The non-perturbative aspects of the Yang-Mills models are better described by transforming the original gauge field theory into a Matrix Quantum Mechanics. Such a transition is realized through dimensional reduction and quenching. The technical steps which allows to ``get rid of'' the internal, non-Abelian, indices $i$, $j$ and replace the spacetime coordinates $x^\mu$ with two continuous coordinates $\left(\, \sigma^0\ ,\sigma^1\,\right)$, are described in some detail elsewhere [9], and will not be repeated here. For the reader convenience we shall give only sketch the main steps. The general procedure can be summarized as follows.
Take the large-$N$ limit, i.e. let the row and column labels $i$, $j$ to range over arbitrarily large values. Thus, $SU( N )\to U( N )$ and the group of spacetime translations fits into the diagonal part of $U( \infty )$. By neglecting off-diagonal components, spacetime dependent dynamical variables can be shifted to the origin by means of a translation operator $\mathbf{U}(x)$: since the translation group is Abelian one can choose the matrix $\mathbf{U}(x)$ to be a plane wave diagonal matrix [8]

$$\mathbf{U}_{ab}(x)=\delta_{ab}\exp\left( i q^a{}_\mu x^\mu \right) \ ,$$ (15)

where $q^a {}_\mu$ are the eigenvalues of the four-momentum $\mathbf{q }_\mu$. Then

$$\mathbf{A}_\mu(x)=\exp\left(-i\mathbf{q }_\mu x^\mu \right) ... ...equiv \mathbf{U}^\dagger(x) \mathbf{A}_\mu^{(0)} \mathbf{U}(x)$$

and in view of the equality

$$\mathbf{D}_ \mu \mathbf{A}_\nu = i \mathbf{U}^\dagger(x) \l... ...u+\mathbf{A}_\mu^{(0)} , \mathbf{A}_\nu \right] \mathbf{U}(x),$$

which when antisymmetrized yields

$$\mathbf{D}_{[ \mu} \mathbf{A}_{\nu ]} = i \mathbf{U}^\dagger... ...m{q})} , \mathbf{A}_\nu^{(\mathrm{q})} \right] \mathbf{U}(x) ,$$

we can see that the translation is compatible with the covariant differentiation, so that

$$\mathbf{F}_{\mu\nu}(x)= \exp\left(-i\mathbf{q }_\mu x^\mu \r... ...thbf{U}^\dagger(x) \mathbf{F}_{\mu\nu}^{(0)} \mathbf{U}(x) \ .$$