Recently, I saw a construction of topological invariant for $\pi_3(U(n))$ with $n\geq 2$: $$ N=\frac{1}{24\pi^2}\int_{S^3} d^3x\ \epsilon^{ijk} Tr[(U^{-1}\partial_{x_i}U)(U^{-1}\partial_{x_j}U)(U^{-1}\partial_{x_k}U)]\ , $$ where $U\in U(n)$ depends on $\boldsymbol{x}=(x_1,x_2,x_3)\in S^3$, $\epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$, and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why?
Update 02/02/2019
I think I got an argument for $n=2$. In this case, $U=e^{i \varphi} q$ with $q\in SU(2)$. Due to the trace and the Levi-Civita symbol in $N$, $\varphi$ does not contribute to $N$. As $Tr[q^{\dagger}\partial_i q]=0$ and $(q^{\dagger}\partial_i q)^{\dagger}=-q^{\dagger}\partial_i q$, $q^{\dagger}\partial_i q$ in geneeral has the form $q^{\dagger}\partial_i q=i(A_i \sigma_x +B_i \sigma_y+C_i \sigma_z )$ with $A_i,B_i,C_i$ real and $\sigma$'s being the Pauli matrices. As a result, we have $$ N=\frac{1}{2\pi^2}\int_{S^3} d^3x\ \epsilon^{ijk} A_i B_i C_i\ . $$ Furthermore, $SU(2)$ is diffeomorphic to $S^3$, and thus $q$ can be parametrized as the three angles $\boldsymbol{\Omega}=(\psi,\theta,\phi)$ of $S^3$. By choosing the right convention, $N$ can be further written as $$ N=\frac{1}{2\pi^2}\int_{S^3} d^3x \left|\frac{\partial\boldsymbol{\Omega}}{\partial\boldsymbol{x}}\right| \sin(\theta)\sin^2(\phi)\ , $$ where $\left|\frac{\partial\boldsymbol{\Omega}}{\partial\boldsymbol{x}}\right|$ is the determinant of the Jacobian matrix. Clearly, $N$ indicates how many times the $\boldsymbol{x}$-$S^3$ warps around the $\boldsymbol{\Omega}$-$S^3$.
For $n>2$, I may guess that some $SU(2)$ subgroup of $U(n)$ accounts for $N$, but I can be wrong.