For a monad $(T,\eta ,\mu )$, the map ${\eta }_{TA}:TA\to TTA$ is a split equalizer of ${\eta }_{TTA},T{\eta }_{TA}$ with retraction ${\mu }_A$ and splitting $T{\mu }_A$. The map $T{\eta }_A:TA\to TTA$ is a split equalizer of $TT{\eta }_A,T{\eta }_{TA}$ with retraction ${\mu }_A$ and splitting ${\mu }_{TA}$. Each assertion follows from the two unit laws and naturality of the unit or multiplication. No assumption that $T$ preserves arbitrary equalizers is needed.

Every self-adjoint operator on a separable complex Hilbert space is diagonal modulo a compact operator. For a bounded operator the compact self-adjoint error can have arbitrarily small Hilbert-Schmidt norm. In a cyclic spectral representation, partition a bounded real interval into $N$ intervals of width at most $\delta$ and successively bisect. Orthogonal step-function differences give basis vectors supported in the parent intervals. A diagonal value chosen in each support interval gives total squared column error at most $3N{\delta }^2$. Since $N=O({\delta }^{-1})$, this tends to zero. Apply the construction to countably many cyclic summands with summable squared error bounds. For an unbounded operator first split into bounded spectral intervals, obtaining a bounded compact error on the same domain.

For induction of $S^{\mu }⊠S^{\lambda }$ to the symmetric group on $|\mu |+|\lambda |$ letters, the Specht filtration multiplicity of $S^{\nu }$ is the number of Semistandard Young tableaux of skew shape $\nu /\mu$, content $\lambda$, and lattice word read from right to left in each row, taking rows from top to bottom. In characteristic zero these are irreducible constituents; in positive characteristic they remain Specht factors rather than simple constituents. Content $(1)$ gives the add-one-cell induction rule.

A positive-rank holomorphic vector bundle on a compact Riemann surface is semistable when every nonzero proper holomorphic subbundle has slope at most its own. Equivalently one tests coherent subsheaves: saturation on a smooth curve preserves rank, increases degree, and gives a subbundle with torsion-free quotient. Strict inequality defines stability, which is stronger.

Fix a Hermitian metric on a holomorphic vector bundle $E$ over a compact Riemann surface and a positive area form $\omega$. Write $i{F}_E=K\omega$ for its Chern curvature. If every eigenvalue of the Hermitian endomorphism $K$ is at most $C$, the curvature formula for a holomorphic subbundle gives $\mathrm{deg}F\le \mathrm{rank}(F)C{\int }_X\omega /(2\pi )$. The second-fundamental-form term is nonpositive. Compactness supplies a finite $C$ independent of the subbundle. If $F_E=0$, every subbundle has nonpositive degree.

Let $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, and let $f_i$ be the dual basis. The equivariant contraction sends $t_{ij}\otimes f_k$ to ${\delta }_{ik}{e}_j+{\delta }_{jk}{e}_i$. Its kernel has a weight basis consisting of $t_{ii}\otimes f_k$ for $i\ne k$, $t_{ij}\otimes f_k$ for three distinct indices with $i<j$, and $t_{ij}\otimes f_j-\frac{1}{2}{t}_{ii}\otimes f_i$ for $i\ne j$. The three families have sizes six, three and six. This is the irreducible fifteen-dimensional highest-weight representation with highest weight $2L_1-L_3$.

For a smooth algebraic variety of dimension $r$ over a field, the sheaf of Kähler differentials over a field is locally free of rank $r$ and defines its algebraic cotangent bundle. Its dual is the tangent sheaf of a scheme. For a smooth hypersurface of degree $d$ in ${\mathbb{P}}^n$, the Conormal exact sequence for Kähler differentials is the short exact sequence $0\to {\mathcal{O}}_X(-d)\to {\Omega }_{{\mathbb{P}}^n}^1{|}_X\to {\Omega }_X^1\to 0$.

At the self-dual circle, left and right dimension-one currents generate $SU(2{)}_L\times SU(2{)}_R$. The compact-coordinate vertex $\partial Y_L\bar{\partial }{Y}_R$ is a Cartan–Cartan component of a scalar field in the $(3,3)$ representation. Its expectation value changes the circle radius and leaves $U(1{)}_L\times U(1{)}_R$ unbroken. The four extra charged vectors acquire the displayed mass, and four scalar field polarizations become their longitudinal modes. This realizes the Higgs mechanism geometrically. In a bosonic string the ever-present tachyon and additional scalar field instabilities prevent interpreting the whole family as a stable vacuum.

At $R=\sqrt{{\alpha }^{\prime }}$, a chiral compact boson has operator product expansion $Y_L(z)Y_L(w)\sim -{\alpha }^{\prime }\log (z-w)/2$. The currents $J^3=i\partial Y_L/\sqrt{{\alpha }^{\prime }}$ and $J^{\pm }=:e^{\pm 2i{Y}_L/\sqrt{{\alpha }^{\prime }}}:$ have conformal weight one. Contractions give $J^3(z)J^3(w)\sim 1/[2(z-w{)}^2]$, $J^3(z)J^{\pm }(w)\sim \pm J^{\pm }(w)/(z-w)$ and $J^{+}(z)J^{-}(w)\sim 1/(z-w{)}^2+2J^3(w)/(z-w)$. These are the level-one SU(2) Lie algebra currents. The right-moving copy supplies a second factor. Combining a current with a noncompact coordinate derivative gives a massless string vertex operator for each of the six vectors; four are additional to the generic two Cartan vectors.

For the unitary group $U(n)$ with $n\ge 3$, the third tensor power of its defining representation decomposes as displayed. The first two summands have highest weights $(3,0,\dots )$ and $(1,1,1,0,\dots )$ and dimensions $\left(\genfrac{}{}{0ex}{}{n+2}{3}\right)$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$. The mixed summand has highest weight $(2,1,0,\dots )$ and dimension $n(n^2-1)/3$. In the three-dimensional weight space spanned by permutations of $e_1\otimes e_1\otimes e_2$, the vectors killed by all positive root vectors have coefficient sum zero, giving multiplicity two. For $n=2$, the exterior cube vanishes and $V_{(2,1)}=\det \otimes {\mathbb{C}}^2$.

The Verma module $M(0)$ has basis $v_n=f^n{v}_0$, with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ is $M(-2)$, and the quotient is the trivial one-dimensional module. A complementary submodule would be trivial and hence lie in the zero-weight space $\mathbb{C}{v}_0$, but $f{v}_0=v_1\ne 0$. Therefore the displayed extension does not split. This explains why the finite-dimensional hypothesis in the Weyl complete reducibility theorem is essential.

For $N=q-1$, write a diagonal-torus character as $\theta (\mathrm{diag}(a,b,(ab{)}^{-1}))=\chi (a{)}^r\chi (b{)}^s$, with $\chi$ a generator of the character group of ${\mathbb{F}}_q^{\times }$. Its Weyl stabilizer is trivial exactly when $r,s,r-s$ are all nonzero modulo $N$ and, if $3\mid N$, the unordered pair $\{r,s\}$ is not $\{N/3,2N/3\}$. The first conditions exclude transpositions; the exceptional pair excludes cyclic permutation of three characters modulo a common twist.

For a nonzero primitive class $v\in H^{p,q}$ of degree $k=p+q$ on a compact Kähler manifold of complex dimension $d$, the displayed expression is positive. Together with orthogonality of complementary Hodge bidegrees, this makes primitive cohomology a polarized Hodge structure when the Kähler class is rational, after clearing denominators. Positivity on the full cohomology requires the signs prescribed by its primitive decomposition.

By positive unfolding of a cusp-form square, $\sum |a_n{|}^2{n}^{-\beta }<\infty$ for $\beta >k$. Choose $k<\beta <2\sigma -1$. The Cauchy-Schwarz inequality bounds the absolute sum for the L-function of a cusp form by

On a compact Kähler manifold $X$, if $P$ is a positive holomorphic line bundle, then $H^i(X,K_X\otimes P)=0$ for $i>0$. The canonical-bundle twist is part of the hypothesis and conclusion; positivity alone does not remove it on a general manifold. For a complex torus, the canonical bundle is trivial, so higher cohomology of a positive line bundle vanishes without an additional twist.

With zero degree assigned to the zero element of the homomorphism group of elliptic curves, the degree parallelogram law makes this symmetric pairing additive in both arguments. Its diagonal is the degree of an isogeny; it is positive definite on every finite-dimensional rational span after extension to the real numbers. The integral polarization $2B(\varphi ,\psi )$ is integer-valued, while $B$ can be half-integral. For instance, if an elliptic curve has an automorphism $\omega$ with ${\omega }^2+\omega +1=0$, then $1+\omega =-{\omega }^2$ has degree one, so $B(1,\omega )=-1/2$.

If nonnegative integer-valued random variables $X_n$ have all fixed factorial moments converging to ${\lambda }^k$, then their convergence in distribution is to the Poisson distribution with mean $\lambda$. The first-moment bound gives uniform tightness, and bounds on higher moments give uniform integrability of each fixed lower moment. Every subsequential limit therefore has the indicated factorial moments. For such a limit $X$ and $t\ge 0$, the nonnegative binomial expansion and the monotone convergence theorem give $\mathbb{E}(1+t{)}^X={\sum }_{k\ge 0}{t}^k{\lambda }^k/k!=e^{\lambda t}$. This identifies its moment-generating function as that of the Poisson distribution, uniquely determining the limit.

A degree-$g$ line bundle with vanishing first sheaf cohomology has exactly one section up to scale, by Riemann-Roch theorem. Its zero divisor is unique and varies in families by cohomology and base change for line bundles on a curve. For a fixed divisor $B$ of degree $N-g$, the open subset of $C^{(g)}$ where $\mathcal{O}(E)$ has vanishing first cohomology gives a chart in degree $N$, by $E\mapsto \mathcal{O}(B+E)$. Different charts are glued by the unique residual divisor of $\mathcal{O}(B+E-B^{\prime })$ wherever that line bundle is again nonspecial. Normalizing the universal bundles along a fixed rational point gives their canonical compatible gluing and a representing scheme for the Picard functor of a curve.

We give a quantitative construction for the whole open set, including the case of infinitely many component intervals. Pointwise divergence at a single point is insufficient here. Write $z=e^{2\pi it}$ and use normalized measure on $\mathbb{T}$.

First let $K$ be a finite union of closed arcs with $|K|\le 3\epsilon$, where $\epsilon$ is small. Choose a smooth function nonnegative function $u$, equal to one on $K$, with $u\le 1$ and $\eta :=\int u\le 6\epsilon$. This follows by making the transition regions around the finitely many arc endpoints sufficiently short. Define the analytic Herglotz integralIts real part is $(\eta +P[u](z))/(2\eta )>0$, where $P[u]$ is the disk Poisson integral, and $G(0)=1$. Smoothness of $u$ makes $G$ continuous and smooth function up to the circle; on $K$ its boundary real part is at least $1/(2\eta )$. The branch $H=\log G$ therefore has $H(0)=0$ andThe complex logarithm has no branch problem because $G$ lies in the right half-plane. Its smooth function boundary values give absolutely summable Taylor coefficients. Choose an analytic polynomial $A$, with zero constant coefficient, uniformly within $1/4$ of $H$ on the closed disk. Put $P(t)=\mathrm{Im}A(e^{2\pi it})/\pi$. Then $\Vert P{\Vert }_{\infty }\le 1/2+1/(4\pi )<1$.

If $D=\mathrm{deg}A$ and $m>D$, the modulated trigonometric polynomial $Q(t)=e^{2\pi imt}P(t)$ has spectrum in $[m-D,m+D]$. A sharp Fourier partial sum at $m$ selects exactly the negative-frequency half of $P$, since $A(0)=0$:Consequently, for every $t\in K$,The carrier frequency $m$ can be arbitrarily large. This compact-set Fourier amplification lemma uses a bounded imaginary part to obtain a large one-sided Fourier multiplier; no assertion about pointwise divergence is used.

Now let $E$ be any open set of measure $\delta$. DefineIf this is zero take $f=0$. Otherwise choose a decreasing sequence ${\delta }_j\downarrow 0$, with ${\delta }_1=\delta$, so small for $j\ge 2$ thatFor $j=1$ this inequality also holds: $B(\delta )>4$ and $B(\delta )/2\ge B(\delta )/4+1$. Let $L_j$ be an increasing compact exhaustion of $E$.

Using compact batching of a small open set, we recursively choose finite closed-arc unions $K_j\subset E$, of measure at most $3{\delta }_j$, such that $L_j$ is covered by the interiors of $K_1,\dots ,K_j$ andHere is the measure-theoretic justification. At step $j$, the uncovered portion has measure at most ${\delta }_j$. Its intersection with $L_j$ after removing the previously covered interiors is compact, and can be covered by finitely many closed arcs lying in $E$ with total length at most ${\delta }_j$ plus an arbitrarily small excess. By inner regularity of Lebesgue measure, a further compact subset of the uncovered open portion captures all but ${\delta }_{j+1}$ of its measure; cover it similarly. The combined finite arc union has length at most $3{\delta }_j$. This both covers the outstanding part of $L_j$ and ensures the next remainder estimate. In particular $E\subset {\bigcup }_j{K}_j$.

Apply the finite-arc construction to $K_j$ to get $P_j$ of degree $D_j$. Choose carriers $m_j$ recursively so that the bands $[m_j-D_j,m_j+D_j]$ are positive and strictly ordered with no overlap. SetThe frequency-separated Fourier block series and the Weierstrass M-test give a continuous complex-valued function with $\Vert f{\Vert }_{\infty }\le {\sum }_j{2}^{-j}=1$. At the cutoff $m_j$, all earlier bands have been included in full and every later band is absent. Thus on $K_j$,Every point of $E$ lies in some $K_j$, soValues of $\psi$ outside $0<\delta \le 1$ may be defined arbitrarily. The construction handles an arbitrary open set, not just one point or a finite collection of intervals. It also prevents cancellation explicitly: earlier bands contribute a bounded complete sum, and later bands contribute nothing at the chosen cutoff.

Set $p=t-\lambda r$ using part (a). At every node,If $\lambda =0$, the error vanishes. Otherwise any degree-at-most-$n$ polynomial $q$ with strictly smaller error would make $q-p$ have alternating nonzero signs at the increasingly ordered nodes (reversed globally if $\lambda <0$). It would then have at least $n+1$ distinct roots by the intermediate value theorem, impossible. Hence