Chiral Lagrangian Field

Question

In the chiral model $SU(N)_R × SU(N)_L$ with gauged Left-handed $SU(N)$, we take as the field the $SU(N)_L$-valued $\Sigma (x)$, defined as $$\Sigma(x) = \exp\big( \frac{2i}{v} \chi^a(x)T^a\big).$$

What is the reason for which, in the chiral lagrangian, we take the exponential field $\Sigma (x)$ defined this way, instead of a "Yang Mills"-like, $$\Sigma(x) = c \chi^a (x)T^a ,$$ with $c$ a suitable constant.

Is there a meaning in taking this field as an exponential of the generators?

Also, I don't understand which representation is taken into account for $T^a$ when we have the term, $$\mathcal{L} = \frac{ v^2}{4}\operatorname{Tr}\big[\big(\partial_\mu\Sigma \big)^{\dagger}\big(\partial_\mu\Sigma \big)\big],$$ since it doesn't act on any given representation of the group.

Answer 1

Is there a meaning to take this field as an exponential function of the generators?

From your basic Lie group theory you presumably appreciate the logic of the chiral model of the astounding Gürsey: in short,

• The projective coordinate ("pion") fields $\chi^a$ are in the Lie algebra, in representation $T^a$, often the fundamental.

• The exponentials $\Sigma$ you are asking about are in the Lie group generated by said algebra $T$, and not in the Lie algebra. Expanding in inverse v you see they include the identity matrix, so they are not expandable as a linear combination of generators, as you propose. If the T are hermitian, the $\Sigma$ are unitary. They transform in the group representation the $\chi$s transform in the algebra representation of. Specifically, $\Sigma \to L\Sigma R$, for L and R in the mutually commuting left- and right- SU(N)s, respectively.

• The left-invariant $\Sigma ^\dagger \partial_\mu \Sigma$ and right-invariant $\Sigma \partial_\mu \Sigma ^\dagger$ currents, nevertheless, are in the Lie algebra; this should be at the heart of your group theory and chiral model course. They transform in the adjoint of the R and L groups respectively!

The lagrangian you wrote is "magically" invariant under both the L and the R groups, since, by the cyclicity of the trace and $\Sigma ^\dagger \partial \Sigma + \partial \Sigma ^\dagger \Sigma =\partial (\Sigma ^\dagger \Sigma )=0$, $$ \operatorname{Tr} \partial_\mu \Sigma^\dagger \partial ^\mu \Sigma = \operatorname{Tr} \partial_\mu \Sigma^\dagger ~ ( \Sigma \Sigma^\dagger) ~ \partial ^\mu \Sigma = -\operatorname{Tr} (\Sigma^\dagger\partial_\mu \Sigma) ~~ (\Sigma^\dagger \partial ^\mu \Sigma) \\ = \operatorname{Tr} \Sigma \partial_\mu \Sigma^\dagger \partial ^\mu \Sigma \Sigma^\dagger = -\operatorname{Tr} (\Sigma\partial_\mu \Sigma^\dagger) ~~ (\Sigma \partial ^\mu \Sigma^\dagger). $$

You must verify all of the above statements for N=2, as your instructor has made you do, mindful of the algebra su(2) to group SU(2) map you must have mastered, reminiscent of Euler's identity, $$ \exp (i\theta \hat n\cdot \vec \sigma)= \cos\theta ~~1\!\! 1 + i\sin\theta ~ ~ \hat n\cdot \vec \sigma. $$