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A map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ is differentiable at $x$ if there is a linear map $L$ such thatthen $Df{|}_x=L$. The inverse function theorem says that if $f$ is continuously differentiable near $x$ and $Df{|}_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.

Since $F(A)=A^{T}A$ is polynomial,Thus $\ker DF{|}_I=T$, the space of skew-symmetric matrices.

DefineThen $D\Phi {|}_I(H)=2H$, so the inverse function theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi :U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^{T}A-I$ and its skew part is $A-A^T$. Consequentlyand therefore $\Phi (U\cap \mathcal{O})=V\cap T$.

For any $R\in \mathcal{O}$, left multiplication $A\mapsto R^{T}A$ is a homeomorphism preserving $\mathcal{O}$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal{O}$ homeomorphic to an open subset of $T$.

Take $x$ vertically downward and $y$ outward from the wall, with the film occupying $0<y<h$. The ambient pressure isAt $y=h$, normal stress continuity gives $p=p_a$ and zero ambient shear gives $\mu U^{\prime }(h)=0$; at the wall, no slip gives $U(0)=0$.

For steady parallel flow $u=(U(y),0)$, incompressibility is automatic and Navier–Stokes reduces toSince $p_x={\rho }_{a}g$, integration yields

Position and momentum areDifferentiating the expectation, substituting Schrödinger's equation, and integrating by parts gives Ehrenfest's relationThus the mean momentum changes according to the classical mean force $-kx$.

Insert the transform definitions and use Fubini:For the indicator of $(-1,1)$,Putting $g=f$ and $x=0$ in the convolution theorem givesand hence

The Euler–Lagrange equations areFor $\omega \ne 0$ their general solution iswhile for $\omega =0$, $x=c_{1}t+c_2$.

The action has continuous translation symmetry in $x$; it is also invariant up to endpoint terms under adding homogeneous Jacobi solutions $at+b$ to $x$ and $a\cos t+b\sin t$ to $y$. For $\omega =0$ it additionally has continuous time-translation symmetry; for nonzero $\omega$ only the corresponding discrete period remains.

A complete set of four independent first integrals isIn particular $({\dot{y}}^2+y^2)/2=(C_3^2+C_4^2)/2$ is conserved; when $\omega =0$, the usual total energy is conserved as well.

For the Möbius action on the upper half-plane, a noncentral element is elliptic if it has one fixed point in the interior and none on the boundary, parabolic if it has one boundary fixed point, and hyperbolic if it has two boundary fixed points. Examples areFixed points satisfy a real quadratic. Its discriminant is $(\mathrm{tr}g{)}^2-4$, so the three mutually exclusive cases $|\mathrm{tr}g|<2$, $=2$, and $>2$ give precisely the three types.

If $g^n=I$ and $g\ne \pm I$, its eigenvalues are nonreal reciprocal roots of unity. Thus $|\mathrm{tr}g|<2$ and $g$ is elliptic; no such element is parabolic or hyperbolic.

If a finite group is not simple, choose a maximal proper normal subgroup and repeat inside it. Orders strictly decrease, so the process terminates; maximality makes every quotient simple.

For $S_4$ one composition series isThe successive quotients have orders $2,3,2,2$, hence are simple. As subgroups of $S_4$, the groups $S_4,A_4,V_4$, and $1$ are normal. The selected order-two subgroup is not normal, since conjugation permutes the three double transpositions.