For $n\ge 2$, contraction with the invariant symmetric bilinear form splits off the trivial line in the symmetric square of the defining special orthogonal group representation. Its null space is the traceless symmetric square. It contains a highest-weight vector of weight $2e_1$, and the Weyl dimension formula gives $(n+1)(2n-1)$, exactly its dimension. Complete reducibility of compact-group representations therefore makes that null space irreducible. The entire symmetric square is reducible because of the invariant line.

For the defining $n$-dimensional special linear Lie algebra module, with $n\ge 2$, write $T_i(p)=e_i\otimes p(y_1,\dots ,y_n)$. The equivariant contraction is $C(T_i(p))={\partial }_{y_i}p$. The map $\iota (f)={\sum }_i{e}_i\otimes y_{i}f$ is also equivariant, since ${\sum }_i{e}_i\otimes y_i$ is an invariant identity tensor. Direct differentiation gives $C\iota =(n+1)I$, so the displayed direct sum follows. Its trace-free part is an irreducible representation of highest weight ${\omega }_1+2{\omega }_{n-1}$. Its highest-weight vector is $e_1\otimes y_n^2$; the Weyl dimension formula gives dimension $n(n-1)(n+2)/2$, exactly $\dim \ker C$.

// Target: semisimple-lie-algebra.bigb

In a nontrivial symmetric balanced incomplete block design, $\det (N{)}^2=r^2(r-\lambda {)}^{t-1}$. If $t$ is even, $t-1$ is odd. For every prime, the exponent in the left side is even and that in $r^2$ is even, so its exponent in $r-\lambda$ must be even. Thus $r-\lambda$ is a perfect square. This supplies an elementary necessary existence condition without claiming it is sufficient.

For an axisymmetric random velocity dispersion tensor with ${\Pi }_{11}={\Pi }_{22}=M{\sigma }_0^2$ and ${\Pi }_{33}=M{\sigma }_0^2(1-\delta )$, contracting with the line-of-sight unit vector $(\sin i,0,\cos i)$ gives the displayed mass-weighted velocity dispersion. A major-axis rotation amplitude projects as $V_r=V\sin i$. By contrast, an azimuthally averaged line-of-sight streaming root-mean-square is $V\sin i/\sqrt{2}$ when $V^2$ is the full in-plane streaming mean square. Aperture weights and the definition of the measured rotation statistic must match the intrinsic statistic.

In normalization $Q=T_3+Y$, one Standard Model generation has left quark and lepton doublet hypercharges $1/6$ and $-1/2$, and right singlet hypercharges $2/3,-1/3,-1$. Including colour multiplicity, ${\mathrm{tr}}_{L}Y=6/6-2/2=0$. On every left doublet $T_3^2=I/4$, and all right fields have $T_3=0$, so the mixed weak-hypercharge gauge anomaly vanishes. The cubic left trace is $6(1/6{)}^3+2(-1/2{)}^3=-2/9$, and the right trace is $3(2/3{)}^3+3(-1/3{)}^3+(-1{)}^3=-2/9$. Their difference is zero, proving generation-by-generation cancellation of these two anomalies.

Start the symmetric stochastic attrition process at $(N,\lfloor \alpha N\rfloor )$, with $0<\alpha <1$, and let $X^N$ be the final first-population size divided by $N$. The bound holds for all sufficiently large $N$, depending on $\alpha ,\delta >0$. At absorption, one coordinate is zero. If the first survives, the terminal squared difference is $N^2(X^N{)}^2$; if it is extinguished, that difference is nonpositive. In either case the indicated deviation forces an error of at least $\sqrt{1-{\alpha }^2}\delta N^2$ from the limiting squared invariant. The initial rounding error and total conditional drift are only $O(N)$. Apply Azuma-Hoeffding inequality to the centered squared-population near-invariant for stochastic attrition, using its $2N$ increments bounded by $4N$, to obtain the displayed exponent.

A spinor representation is a representation of a Spin group, or infinitesimally of a Special orthogonal Lie algebra, constructed from a Clifford algebra. In even dimensions it can split into two chiral spinor representations.

A fixed volume form makes wedge pairing on ${\Lambda }^2{\mathbb{R}}^4$ a symmetric form of signature $(3,3)$. The exterior-square action of special linear group matrices preserves it. Its kernel is $\{\pm I\}$: a transformation acting trivially on every bivector preserves every two-plane and hence every line, so is scalar, with scalar square one. The derivative is injective and both groups have dimension fifteen, so the connected image is the full identity component. This is a concrete real spin-group cover.

Congruence $X\mapsto AX{A}^T$ on real symmetric two-by-two matrices preserves the quadratic form $-\det X$ of signature $(2,1)$ when $\det A=1$. Its kernel is $\{\pm I\}$ and its derivative is injective: $aX+X{a}^T=0$ for every symmetric $X$ implies $a=0$. Both Lie groups have dimension three, so the connected image is the full identity component. Also $A^{T}JA=(\det A)J$ identifies $Sp(2,\mathbb{R})$ with $SL(2,\mathbb{R})$.

On two-by-two Hermitian matrices, the determinant of $X=tI+x\cdot \sigma$ is $t^2-|x{|}^2$. The action $X\mapsto AX{A}^{†}$ preserves it for $A\in SL(2,\mathbb{C})$. A kernel element is unitary and commutes with every Hermitian matrix, hence is $\pm I$. The derivative is injective and both real dimensions are six, giving the full proper orthochronous Lorentz group as the connected image.

Fix a volume form on a four-dimensional real symplectic vector space. Wedge pairing on its exterior square has signature $(3,3)$. The symplectic group fixes the bivector inverse to the symplectic form. Its orthogonal complement is five-dimensional with signature $(3,2)$ after choosing the appropriate overall sign. The induced representation has kernel $\{\pm I\}$: identity action on the entire exterior square preserves every two-plane and thus every line, forcing a scalar transformation with square one. Its derivative is injective and both group dimensions are ten, so its image is the identity component of $SO(3,2)$.

At fixed angular coordinates, differentiating the Cartesian coordinates with respect to radial distance gives this unit vector. It is the local direction of increasing radius in the spherical coordinate system.

A finite family of pairwise commuting normal operators on a finite-dimensional complex inner product space, commuting also with each other's adjoints, generates a commutative unital C-star algebra. Its finite maximal ideal space identifies the algebra with functions on finitely many points. The characteristic functions of those points pull back under the Gelfand transform to mutually orthogonal self-adjoint linear projections summing to the identity. Their ranges are the simultaneous eigenspaces, giving simultaneous diagonalization without choosing separate incompatible eigenbases.

A Specht filtration is a finite chain of submodules whose successive quotients are Specht modules. In positive characteristic it need not split and is not a composition series, because its Specht factors may themselves be reducible. Dimensions and ordinary lifted characters can still be added along the filtration.

For a row composition $b$ and a weakly decreasing marking $a$ with $0\le a_i\le b_i$, mark the first $a_i$ cells of each row. Antisymmetrize within the marked columns and span the resulting polytabloids in the Young permutation module $M^b$. This gives $S^{a,b}$; $S^{0,b}=M^b$ and $S^{b,b}=S^b$ for a proper partition $b$. Its characteristic-independent Specht filtration is obtained by successively marking a cell or raising an unmarked row tail, with column-cancellation maps and the good-letter counting recursion.

An ordinary Specht module of non-self-conjugate shape stays irreducible on the alternating group, and conjugate shapes give the same restriction. A self-conjugate shape splits into two irreducibles of equal degree, interchanged by an odd permutation. The index-two inner-product identity is $⟨{\chi }^{\lambda }\downarrow ,{\chi }^{\mu }\downarrow ⟩={\delta }_{\lambda \mu }+{\delta }_{{\lambda }^{\prime }\mu }$, and Frobenius reciprocity proves exhaustion. The statement applies for $n\ge 2$; $A_1$ is trivial.

For a partition of an integer $\lambda$, let ${\psi }_{i,v}:M^{\lambda }\to M^{\lambda (i,v)}$ sum all tabloids obtained by retaining $v$ elements of row $i+1$ and moving its other ${\lambda }_{i+1}-v$ elements into row $i$, leaving the other row sets unchanged. Here $0\le v<{\lambda }_{i+1}$ and $\lambda (i,v)$ replaces $({\lambda }_i,{\lambda }_{i+1})$ by $({\lambda }_i+{\lambda }_{i+1}-v,v)$; Young permutation modules are defined for compositions too. Then $S^{\lambda }={\bigcap }_{i,v}\ker {\psi }_{i,v}$ over every field. Pairwise column cancellation gives inclusion of the Specht module in each kernel. The reverse inclusion follows by straightening the tabloid coefficients using these adjacent-row relations: the remaining free leading coefficients are exactly those of the standard polytabloid basis.

A filling $T$ of shape $\lambda$ and content $\alpha$ defines a module homomorphism ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: given a numbered tableau, sum over distinct assignments obtained by permuting $T$ within its rows, and put each numbered entry into output row $j$ when its assigned label is $j$. Restricting to $S^{\lambda }$ gives ${\hat{\Theta }}_T$. For Semistandard Young tableaux $T$, these restrictions are linearly independent over every field. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ if the characteristic of a field is not two, or if $\lambda$ is a two-regular partition of an integer. For two-singular shapes in characteristic two, they need not span. Column antisymmetrization and triangular leading tabloids establish independence; straightening the images of a cyclic polytabloid establishes spanning under the stated hypotheses.

The polytabloids $e_t$ indexed by standard Young tableaux of shape $\lambda$ form a basis of $S^{\lambda }$ over every field, and a basis of its integral lattice over $\mathbb{Z}$. Column antisymmetry and Garnir relations straighten any polytabloid into this set. Ordering leading tabloids by the rows containing successive entries makes the corresponding coefficient matrix triangular with diagonal entries one, proving independence. In particular the dimension is the standard-tableau count in every characteristic.

When spatial and spin factors separate, a spin orbital combines a spatial orbital ${\varphi }_a$ with a normalized spin state ${\chi }_{\sigma }$. For an electron, choose spin one-half along an axis with $\sigma =\uparrow ,\downarrow$. The two spin orbitals $u_{a\uparrow }$ and $u_{a\downarrow }$ are orthogonal even though their spatial orbital is identical, since their inner product contains $⟨{\chi }_{\uparrow }|{\chi }_{\downarrow }⟩=0$. When spatial and spin variables do not separate, the complete one-particle state is instead a multicomponent wavefunction $u(\mathbf{r},s)$; exclusion still applies to that complete quantum state.