Write the group algebra as $\mathbb{C}[X,X^{-1}]\oplus \mathbb{C}[X,X^{-1}]Y$, with $Y^2=1$ and $YX=X^{-1}Y$. Commuting $f(X)+g(X)Y$ with $X$ forces $g=0$, and commuting with $Y$ forces $f(X)=f(X^{-1})$. Every symmetric Laurent polynomial is a polynomial in $z=X+X^{-1}$, using the recursion $X^{m+1}+X^{-m-1}=z(X^m+X^{-m})-(X^{m-1}+X^{1-m})$. The algebra is free of rank four over this center, with basis $1,X,Y,XY$; the relation $X^2-zX+1=0$ proves spanning, and applying $X\mapsto X^{-1}$ proves independence of $1,X$.

The matrix-block sizes in the Artin–Wedderburn theorem for $\mathbb{C}[G]$ recover the irreducible character degrees with their multiplicities. They determine the degree-square sum and the number $|G:G^{\prime }|$ of linear characters. Isaacs' character-degree criterion for p-nilpotency therefore detects each normal p-complement and applying it for every prime also detects nilpotent groups. The set of distinct degrees without multiplicities does not contain the same information.

Charlie measures his qubit of a GHZ state in the $X$ basis. The two equally likely outcomes leave Alice and Bob in the Bell states ${\Phi }^{+}$ or ${\Phi }^{-}$, respectively. Charlie sends the outcome bit $c$ to Bob, who applies the Pauli operator $X{Z}^c$, giving ${\Psi }^{+}$ in either case. This is deterministic LOCC entanglement localization. If Charlie's record is ignored, Alice and Bob instead have the separable mixture $(|00⟩⟨00|+|11⟩⟨11|)/2$. The classical outcome is necessary to turn the conditional entanglement into a known usable Bell pair.

If $f$ is constant on the boundary and $g$ is a regular harmonic function throughout a volume, the divergence theorem applied to $f\nabla g$ gives ${\int }_V\nabla f\cdot \nabla g={\int }_{\partial V}f{\partial }_{n}g$. The constant boundary value factors out; a second application of the divergence theorem makes the remaining flux ${\int }_V\Delta g=0$. This requires enough regularity to use Green's first identity over the entire volume.

In a proper partial graph coloring, let $R_v$ count coloured neighbours and $D_v$ count their distinct colours. The saving $S_v=R_v-D_v$ leaves $K-D_v$ available colours for only $d(v)-R_v$ initially uncoloured neighbours. If the displayed inequality holds at every vertex, ordinary greedy completion succeeds. Each additionally coloured neighbour removes at most one available colour while also reducing the number of uncoloured neighbours by one.

For $N\ge 1$ great circles on a sphere, with no coincident circles or triple intersection, the number of open regions is $R_N=N(N-1)+2$. The first circle gives two regions. Each subsequent circle meets the previous $n-1$ circles at $2(n-1)$ distinct points and is divided into that many arcs, each splitting a previous region. Summing the increments proves the formula. If the circles are $a_i\cdot x=0$, each region has one fixed sign vector $(\mathrm{sgn}(a_i\cdot x){)}_i$. Conversely a realized strict sign vector has a connected region: its cone is an intersection of open half-spaces, and radial projection of line segments connects any two points.

For $h_t={\partial }_x(-A{h}^3+B{h}^3{h}_x)$, $A,B>0$, a front $h=F(x-ct)$ joining $h_0$ behind it to zero at its tip has $c=A{h}_0^2$. Integrating once gives $F^{\prime }=(A/B)(1-h_0^2/F^2)$ andNear the tip, $F\sim (-3c\xi /B{)}^{1/3}$. Its unbounded slope signals the failure of the small-slope lubrication approximation sufficiently close to the contact line.

For a film of thickness $H$, axial length $L\gg H$, and velocity $U$, the ratio of convective inertia to transverse viscous stress in lubrication theory is $U{H}^2/(\nu L)$. On a small incline of slope $\theta$, lubrication gravity-current flux gives $U\sim g\theta H^2/\nu$ when $H\ll \theta L$, and $U\sim g{H}^3/(\nu L)$ when $H\gg \theta L$. The respective criteria are $g\theta H^4/({\nu }^{2}L)\ll 1$ and $g{H}^5/({\nu }^2{L}^2)\ll 1$. Unsteady inertia also requires $H^2/(\nu T)\ll 1$ if the evolution timescale $T$ is independent of $L/U$.

A well-mixed particle-laden gravity current in a prismatic channel with $b(z)=C\sqrt{z}$ has volume $V=2C{h}^{3/2}L/3$. Negligible entrainment gives fixed $V$. Vertical particle deposition flux is $W_s\varphi b(h)L$, so $\dot{\varphi }=-3W_s\varphi /(2h)$. Close the front by $\dot{L}=F\sqrt{G\varphi h}$, where $G=g({\rho }_p-{\rho }_0)/{\rho }_0$. Eliminating time givesThe concentration tends to zero at a finite limiting runout length of a gravity current, although reaching that limit takes infinite time in this model.

In a horizontal prismatic V-shaped valley with section area $A=\lambda h^2$, a fixed-volume high-Reynolds-number gravity-current box model has $h=(W/(\lambda X){)}^{1/2}$ and front speed $\dot{X}=\mathrm{Fr}\sqrt{gh}$. Integrating gives $X=[5\mathrm{Fr}\sqrt{g}(W/\lambda {)}^{1/4}t/4{]}^{4/5}$ after the initial release scale is neglected. This is an inertial closure, distinct from the viscous V-shaped channel gravity current, whose exponent is different.

For a constant-volume unit-width particle-laden gravity current with $M=Lh$, define the particle-fluid reference reduced gravity $g_0^{\prime }=g({\rho }_p-{\rho }_0)/{\rho }_0$, so the current has $g^{\prime }=g_0^{\prime }\varphi$. A constant front Froude number and absorbing-bed hindered settling give $\dot{L}=F\sqrt{g_0^{\prime }\varphi M/L}$ and $\dot{\varphi }=-(V_{s}L/M)\varphi (1-\varphi )$. Eliminating time and integrating gives the displayed runout length of a gravity current. For large runout and dilute initial concentration, $L_{\infty }^5\sim 25F^2{M}^3{g}_0^{\prime }({\varphi }_0+2{\varphi }_0^2/3+\cdots )/V_s^2$. This $g_0^{\prime }$ is buoyancy per unit particle fraction, not the initial current reduced gravity $g_0^{\prime }{\varphi }_0$.

A gravity-current box model in a prismatic channel with width equal to height has volume $V_0=L{h}^2/2$. Closing the front speed by $\dot{L}=F\sqrt{g^{\prime }h}$ gives the displayed law and long-time spreading $L\propto t^{4/5}$. This is an integral approximation with constant volume and reduced gravity, not an exact uniform-depth solution of the local shallow water equations.

A caustic is the source-plane image of a gravitational-lensing critical curve. Crossing a generic fold caustic creates or destroys a pair of images of opposite parity. For a circular lens, a tangential critical circle maps to an aligned-source point, while a radial critical circle can map to a finite source-plane circle. Critical curves and caustics consequently occupy different planes and must not be identified as the same angular locus.

For circular reduced deflection $\alpha (r)=r\bar{\kappa }(r)$, a displacement perpendicular to the radial direction is multiplied by $1-\bar{\kappa }$. Vanishing of this eigenvalue gives a tangential gravitational-lensing critical curve. Its source position is $\beta =r-\alpha (r)=0$, so its caustic is the origin in the circular model. Near it, images are stretched tangentially into arcs or an Einstein ring.

The local and mean lensing convergence obey $2\kappa =2\bar{\kappa }+r{\bar{\kappa }}^{\prime }$. Hence the radial derivative of the lens mapping is $1-{\alpha }^{\prime }(r)=1-2\kappa +\bar{\kappa }$. Its zeros are radial gravitational-lensing critical curves, where radial stretching diverges and images merge on crossing the mapped gravitational-lensing caustic.

A collision-free positive-mass configuration satisfies the displayed equation when every gravitational acceleration points proportionally toward a common centre. With $W=-{\sum }_{i<j}G{m}_i{m}_j/|{\mathbf{a}}_i-{\mathbf{a}}_j|$, homogeneity gives $\Lambda I=-W>0$, $I={\sum }_i{m}_i|{\mathbf{a}}_i{|}^2$. Pairwise force cancellation implies ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Such a configuration generates a nonrotating homothetic gravitational motion by multiplying all position vectors by one positive scale factor.

Choose $r$ of the $n$ basis vectors and require their images in the quotient to be a basis. Normalize those columns to the identity; the remaining $r(n-r)$ entries are arbitrary affine coordinates. Between column sets $I,J$, the overlap is the principal open subset where $\det A_J$ is a unit, and the new normalized matrix is $A_J^{-1}A$. These matrix transformations satisfy the cocycle identity. Locally every surjective matrix has an invertible maximal minor, so the charts cover all rank-$r$ locally free quotients. The normalized matrices glue the universal quotient and give the natural identification with its functor of points.

// Target: ringed-space.bigb

The complex Grassmannian of $k$-planes in ${\mathbb{C}}^n$ identifies with the displayed Hermitian orthogonal projection matrices. A plane is recovered as the image of its projection. Near a plane $E$, a neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. Its projection is $W(W^{\ast }W{)}^{-1}{W}^{\ast }$ with $W=(I,Z{)}^T$. The unitary group acts transitively by conjugation, and its tangent directions are $[A,P]$ with $A^{\ast }=-A$.

For a continuous map $f:X\to Y$ with Hausdorff space $Y$, its graph of a function is closed in the product topology. If $y\ne f(x)$, choose disjoint neighborhoods of those two points in $Y$. Continuity supplies a neighborhood of $x$ whose image lies in the neighborhood of $f(x)$; its product with the other neighborhood misses the graph.

If $X$ is a compact space and $f:X\to Y$ is a continuous map, the graph of a function of $f$ is compact in the product topology on $X\times Y$. Indeed,is a continuous map onto that graph, so the result follows from the continuous image of a compact space theorem. No Hausdorff assumption on $Y$ is needed.

The coordinate projection map restricted to the graph is a continuous inverse to $G$. Thus the graph is actually homeomorphic to $X$. For instance, the graph of $x^2$ over $[0,1]$ has the same intrinsic topology as the interval, even though it is curved in the plane.