A two-sided line plume whose axial velocity and density anomaly both have triangular transverse profiles. For half-width $b$, ${\int }_{-b}^{b}f(x/b{)}^{n}dx=2b/(n+1)$. These profile moments set the different coefficients of stored mass, momentum, and their fluxes. It differs from a wall plume because both edges entrain ambient fluid.

Conjugation by every element of $G$ preserves a normal Lie subgroup $H$. Its differential therefore preserves $\mathfrak{h}=T_{e}H$. Differentiate that invariance along $\exp (tX)$ to obtain $\mathrm{ad}(X)\mathfrak{h}\subseteq \mathfrak{h}$, proving that $\mathfrak{h}$ is an ideal of a Lie algebra. Closed subgroups have a canonical Lie-subgroup structure; an arbitrary abstract subgroup need not.

The special linear group acts on determinant-one real symmetric matrices by matrix congruence. Symmetry and determinant are preserved, and signatures are preserved by Sylvester's law of inertia. The determinant-one set is a regular level set of dimension $n(n+1)/2-1$, with tangent directions $C=C^T$ satisfying $\mathrm{tr}(B^{-1}C)=0$. Its infinitesimal left-action fields are $XB+B{X}^T$; converting them to a Lie algebra representation requires the infinitesimal left-action sign convention.

In a local exponential chart, the mixed second differential of the group commutator defines a bilinear map on the tangent space at the identity. Inversion after swapping the two group elements proves antisymmetry. Differentiating conjugation gives $[X,Y]=\mathrm{ad}(X)Y$; applying naturality of the Lie bracket to the Adjoint representation of a Lie group then proves the Jacobi identity. For a matrix group this recovers $XY-YX$ by expanding through the mixed term.

The relation $[x,y]=-[y,x]$ for a bilinear map proposed as a Lie bracket. An Alternating Lie bracket always satisfies it. Conversely, setting $y=x$ gives $2[x,x]=0$, so antisymmetry implies alternation over a field of characteristic different from two.

In characteristic two, the sign disappears and this relation just says $[x,y]=[y,x]$. It does not force $[x,x]=0$, even when the Jacobi identity holds. For example, on a two-dimensional vector space with basis $u,v$, set $[u,u]=v$ and every other basis bracket to zero, then extend bilinearly. Every nested bracket vanishes, so the Jacobi identity holds, and the product is symmetric. Nevertheless it is not a Lie algebra, because $[u,u]\ne 0$.

The real Lie algebra of the unitary group is $\mathfrak{u}(N)=\{X:X^{†}=-X\}$, of real dimension $N^2$. Differentiating unitarity gives necessity; exponentiating any skew-Hermitian matrix gives a curve in the group, proving sufficiency.

For a smooth homomorphism of Lie groups, naturality gives $f(\exp X)=\exp (d{f}_{e}X)$. Therefore $\log (f(g))=d{f}_e(\log g)$ locally. Apply this to the group commutator and differentiate its two curve parameters to obtain the displayed identity for the Lie bracket from local group commutators. In particular the differential of a group representation is a Lie algebra representation.

For the current with $(1-{\gamma }^5)$ rather than $P_L$, the spin-summed symmetric leptonic tensor is $8(q_{1\mu }{q}_{2\nu }+q_{1\nu }{q}_{2\mu }-g_{\mu \nu }{q}_1\cdot q_2)$. Integrating the standard two-body Lorentz-invariant phase space of two massless particles givesIts antisymmetric Levi-Civita symbol term integrates to zero. The normalization changes by a factor of four if the current uses $P_L$ instead. The result separates the leptonic integration from hadronic pseudoscalar-to-pseudoscalar form factors in a three-body decay rate.

Complementary pivoting follows adjacent bases while maintaining all but one label of a complementarity system. The Lemke-Howson algorithm for a bimatrix game follows the duplicated label between two tableaux until it restores the dropped label at a nonzero completely labelled pair. Normalization then gives a Nash equilibrium.

An uncountable subset $S$ of the real numbers whose intersection with every null set for Lebesgue measure is countable. Testing only Borel sets suffices because every Lebesgue-null set has a Borel null superset. No measurability of $S$ is assumed.

If $0<\theta <1$ and nonnegative measurable functions satisfy $h((1-\theta )x+\theta y)\ge f(x{)}^{1-\theta }g(y{)}^{\theta }$ for all $x,y\in {\mathbb{R}}^n$, then their Lebesgue integrals obeyThe quantile derivative identity and monotone substitution inequality prove the one-dimensional case; Tonelli theorem extends it by induction. It implies the Brunn–Minkowski inequality.

For increasing concave $F$, write $q=F^{-1}(1/2)$. Under the smallest-maximizer convention, the leader wins with probability greater than $1/2$ exactly when $v>1/F^{\prime }(q)$. The sign of $v{F}^{\prime }(q)-1$ places the maximizer before or after $q$. At equality, concavity alone permits a flat interval of optima; strict concavity or an explicit selection is needed for the strict comparison at that type.

If a concave leader objective $vF(b)-b$ is constant over an interval, all efforts there are best responses, but they can have different winning probabilities. A flat positive-density interval of $F$ produces this at the type $v=1/F^{\prime }$. Thus strict threshold statements must distinguish strict concavity from a selection such as the smallest maximizing effort. For an atom-free type distribution, the specified threshold type has zero ex ante probability.

For a simple symmetric random walk, geometric blocking and the Borel-Cantelli first lemma yield the upper estimate ${\mathrm{limsup}}_n{S}_n/\sqrt{2n\log \log n}\le 1$ almost surely. Equality requires a separate lower-bound argument.

For a full Euclidean lattice and a positive definite real symmetric matrix $A$, the Gaussian theta sum obeysUse the Fourier transform kernel $e^{-2\pi ix\cdot z}$, the identity $\widehat{e^{-\pi x^{T}Ax}}(z)=(\det A{)}^{-1/2}{e}^{-\pi z^T{A}^{-1}z}$, and the Poisson summation formula for a Euclidean lattice. No integrality or self-duality of the Euclidean lattice is assumed.

Starting Landweber iteration at zero with fixed step $\tau$, the full data-to-solution coefficient on a singular value $\sigma >0$ isThis follows by summing the scalar geometric series in a singular system of a compact operator. For fixed $n$, $g_n(\sigma )=n\tau \sigma +O({\sigma }^3)$ near zero and extends with $g_n(0)=0$. Its dimensionless damping factor is $\sigma g_n(\sigma )$, so the two common filter conventions must not be confused. Writing $\alpha =1/n$ uses a discrete parameter sequence; if negative multipliers occur, noninteger real powers are not a definition of the same iteration.

In a singular system of a compact operator, Landweber iteration with a Tikhonov initial value has coefficients $g_{\alpha ,n}({\sigma }_i)(y,u_i)$. Summing the finite geometric series gives the displayed filter. For fixed $n$, $g_{\alpha ,n}(\sigma )=(n+1)\sigma /\alpha +O({\sigma }^3)$ at zero, so each finite iterate is bounded. Under $\alpha >\Vert A{\Vert }^2/2$ the filter approaches $1/\sigma$ on every positive singular mode; noisy data require early stopping of Landweber iteration.

Choose a reference path between Lagrangian submanifolds $L_0,L_1$ in a symplectic manifold. With a connecting homotopy strip oriented by $ds\wedge dt$, define $\mathcal{A}(\gamma )=-\int u^{\ast }\omega$. If the path space is connected and $H_2(M,L_0\cup L_1)=0$, relative symplectic-area independence makes this single-valued up to a global constant. Its derivative is $d\mathcal{A}(\xi )=\int \omega (\dot{\gamma },\xi )dt$, so its critical points are the constant paths in $L_0\cap L_1$. For a compatible almost complex structure, its formal downward $L^2$ flow is the J-holomorphic curve equation $u_s+J{u}_t=0$.

For binary knapsack optimization, $U(\lambda )$ is an upper bound for every $\lambda \ge 0$. Its minimum is computable from the finitely many profit-to-weight breakpoints and agrees with the fractional knapsack problem optimum. Zero-weight positive-profit terms are added independently.

For perturbing acceleration $R\hat{\mathbf{r}}+T\hat{\mathbf{\theta }}$ on an osculating orbital element system with fixed central parameter $\mu$,To prove it, differentiate the specific orbital energy $-\mu /(2a)$, obtaining $\dot{a}=(2a^2/\mu )(R\dot{r}+Th/r)$. Substitute $\dot{r}=\mu e\sin f/h$, $h/r=\mu (1+e\cos f)/h$ and $h^2=\mu a(1-e^2)$. Here $f$ is the true anomaly; a normal perturbing acceleration does no instantaneous work.