For $U=e^{2i\pi/f_\pi}$, the <Maurer-Cartan form> expands as
$$
U^\dagger\partial_\mu U
=\frac{2i}{f_\pi}\partial_\mu\pi
+\frac12\left[\frac{-2i\pi}{f_\pi},\frac{2i\partial_\mu\pi}{f_\pi}\right]+\cdots.
$$
Substituting, using cyclicity of the trace, and collecting even powers gives
$$
\boxed{\mathcal L_\pi=operatorname{tr}(\partial_\mu\pi)^2
-\frac{2}{3f_\pi^2}\operatorname{tr}\!\left(\pi^2(\partial_\mu\pi)^2-(\pi\partial_\mu\pi)^2\right)+O(\pi^6)}.
$$
Odd terms cancel because the sigma-model metric is invariant under $U\leftrightarrow U^\dagger$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 305 / 2 / iii / Solution

For a matrix <Lie group>, the left-invariant <Maurer-Cartan form> is
$$
\boxed{\rho=g^{-1}dg}.
$$
Differentiating $g^{-1}g=I$ gives $d(g^{-1})=-g^{-1}(dg)g^{-1}$. Hence
$$
d\rho=d(g^{-1})\wedge dg
=-g^{-1}dg\wedge g^{-1}dg
=-\rho\wedge\rho,
$$
and therefore
$$
\boxed{d\rho+\rho\wedge\rho=0}.
$$

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 3 / a / Solution

Represent the <real affine group> by
$$
g(a,b)=
\begin{pmatrix}
e^a&b\\0&1
\end{pmatrix}.
$$
Matrix multiplication reproduces
$$
(a,b)(a',b')=(a+a',b+e^ab').
$$
The <Maurer-Cartan form> is
$$
g^{-1}dg=
\begin{pmatrix}
da&e^{-a}db\\0&0
\end{pmatrix},
$$
so a basis of left-invariant one-forms is
$$
\boxed{\sigma^1=da,
\qquad \sigma^2=e^{-a}db}.
$$
The dual left-invariant vector fields are
$$
\boxed{E_1=\partial_a,
\qquad E_2=e^a\partial_b}.
$$
Indeed $\sigma^i(E_j)=\delta^i_j$, and left translation preserves the one-forms and vector fields.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 3 / c / Solution

{c}
{wiki=Maurer–Cartan_form}

For a matrix Lie group, the left-invariant Maurer-Cartan form is $\rho=g^{-1}dg$. It identifies each tangent space with the Lie algebra and obeys $d\rho+\rho\wedge\rho=0$.


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