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For a finite-dimensional complex Lie algebra $\mathfrak{g}$, the following three conditions are equivalent descriptions of a semisimple Lie algebra.

First, its solvable radical is zero. An ideal of a Lie algebra is a linear subspace $I$ satisfying $[\mathfrak{g},I]\subseteq I$. A solvable Lie algebra is one whose derived series, defined by $I^{(0)}=I$ and $I^{(r+1)}=[I^{(r)},I^{(r)}]$, eventually becomes zero. The solvable radical is the largest solvable ideal of a Lie algebra; its vanishing is equivalent to the absence of nonzero solvable ideals.

Second, the Killing formis a nondegenerate bilinear form: if $\kappa (u,v)=0$ for every $v$, then $u=0$. Here $\mathrm{ad}u(v)=[u,v]$ is the Adjoint representation, and $\mathrm{tr}$ is the trace. This equivalence is the Cartan criterion for semisimplicity.

Third, $\mathfrak{g}$ is a direct sum of Lie algebras that are nonabelian simple Lie algebras. A simple Lie algebra is nonabelian and has no ideals of a Lie algebra other than zero and itself. Thus the three equivalent criteria are

Let $\dim V=2n$ and let $V_{+}=\ker (S-I)$ and $V_{-}=\ker (S+I)$ be the two eigenspaces. The involution identity gives a direct sum, sinceFor $u,v$ in either one of these eigenspaces, the anti-symplectic involution identity impliesso both are isotropic subspaces of a symplectic vector space. Such a subspace has dimension at most $n$: $W\subset W^{\Omega }$ and the nondegenerate bilinear form $\Omega$ gives $\dim W^{\Omega }=2n-\dim W$, where $W^{\Omega }$ is the symplectic orthogonal complement. As their dimensions add to $2n$, both have dimension $n$. Therefore

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

The mapis a nonzero linear functional: otherwise $v$ would lie in the radical of a bilinear form, contradicting nondegeneracy. Its kernel is $v^{\perp }$, so the rank-nullity theorem gives $\dim v^{\perp }=n-1$. Antisymmetry gives $⟨v,v⟩=-⟨v,v⟩=0$, hence $v\in v^{\perp }$.

Suppose $w\in W$ is orthogonal to every vector of $W$. Since $W\subseteq v^{\perp }$, it is also orthogonal to $v$, and therefore toFor a nondegenerate bilinear form, $(v^{\perp }{)}^{\perp }=\mathbb{R}v$: both sides have dimension one and the latter is contained in the former. Thus $w\in W\cap \mathbb{R}v=\{0\}$, proving that the restriction to $W$ is nondegenerate.

The space $W$ has dimension $n-2$ and again carries a nondegenerate antisymmetric form. Induction, starting from the zero-dimensional space, shows that $\dim W$ is even. Therefore $n=\dim W+2$ is even. Equivalently, every finite-dimensional symplectic vector space has even dimension.

For an arbitrary finite-dimensional complex Lie algebra $L$, a Cartan subalgebra is a nilpotent Lie subalgebra $H$ equal to its own normalizer. When $L$ is semisimple, this is equivalently a maximal abelian subalgebra consisting of elements that act semisimply in the Adjoint representation.

Choose such an $H$. Simultaneous diagonalization gives the root-space decompositionThe nonzero weights $\alpha \in H^{\ast }$ are the roots. The restriction of the Killing form $B$ to $H$ is a nondegenerate bilinear form, so each $\alpha$ corresponds to a unique $t_{\alpha }\in H$ with $\alpha (h)=B(t_{\alpha },h)$. On the real span of these $t_{\alpha }$, the restriction of $B$ supplies a positive-definite inner product after choosing the standard real form. The sl2 subalgebra associated with a root givesand shows that these reflections preserve the finite set $\Phi$. Thus the roots form a finite reduced crystallographic root system, whose Weyl group is generated by these reflections.

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

Because ${\alpha }^{2^d-1}=1$,so the period divides $2^d-1$.

Conversely, suppose $r$ is an eventual period. Then for every sufficiently large $n$,Any block of $2^d-1$ consecutive powers of $\alpha$ runs through every element of $K^{\times }$. Hencefor every $y\in K$, including $y=0$. The assumed nondegenerate bilinear form forces ${\alpha }^r-1=0$. Since $\alpha$ generates the cyclic group $K^{\times }$ of order $2^d-1$, this means $2^d-1\mid r$. The least positive period is thereforeas in a trace-generated maximal-period linear-feedback sequence.

The trace form of a number field isIt is bilinear because the field trace is linear. If $x\ne 0$, choose $y=x^{-1}$. ThenThus no nonzero $x$ lies in the radical, and $B$ is a nondegenerate bilinear form. This is the nondegeneracy of the trace form of a number field.

The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

An invariant bilinear form $B$ defines a linear mapInvariance givesso ${\Phi }_B$ is an $G$-homomorphism. Its kernel is a $G$-subrepresentation of irreducible $V$. Hence either ${\Phi }_B=0$, so $B=0$, or ${\Phi }_B$ is injective and therefore a linear isomorphism $V\to V^{\ast }$. In the latter case $B$ is a nondegenerate.

If $B$ and $C$ are two nonzero invariant forms, thenBy Schur lemma, this endomorphism is scalar, so $B=\lambda C$. Cases involving the zero form are immediate. Thus any two invariant bilinear forms are proportional.

All edges in the Type E(p,q,k) Coxeter graph are unlabelled, so the factor $a^2$ in part c is one. Separate the arm of length $p$ from the central vertex. The remaining two arms form a type $A_{q+k+1}$ chain, while deleting the central vertex leaves the disjoint type $A_q$ and type $A_k$ chains. Using $\det G(A_r)=r+1$ in the formula from part c yieldsThe associated bilinear form is degenerate exactly whenThe positive-integer solutions of $1/a+1/b+1/c=1$, up to permutation, are $(3,3,3)$, $(2,4,4)$, and $(2,3,6)$. Consequently the degenerate arm-length triples are the permutations ofFor every other allowed $(p,q,k)$ the determinant is nonzero, so the form is nondegenerate.

For finite-dimensional vector spaces $V,W$ over a field $F$, a bilinear pairing $B:V\times W\to F$ is perfect when the induced linear maps $V\to W^{\ast }$ and $W\to V^{\ast }$ are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case $V=W$.

A pseudo-Riemannian manifold is a smooth manifold with a smooth metric tensor that is a nondegenerate bilinear form on each tangent space. Positivity is not required. A Riemannian manifold is the positive-definite case, while a Lorentzian manifold has exactly one temporal direction in its metric signature.

If $B_1$ is a nondegenerate bilinear form on a finite-dimensional vector space and $B_2$ is any bilinear form, there is a unique linear map $\alpha$ such thatThe right radical of $B_2$ is $\ker \alpha$.

A symplectic manifold is an even-dimensional smooth manifold $M$ equipped with a closed nondegenerate differential two-form $\omega$.

The symplectic orthogonal complement of a subspace $W$ is $W^{\omega }=\{v:\omega (v,w)=0\text{ for every }w\in W\}$. The nondegenerate bilinear form gives $\dim W^{\omega }=\dim V-\dim W$. An isotropic subspace of a symplectic vector space satisfies $W\subset W^{\omega }$; a Lagrangian subspace satisfies equality.

A subspace $E$ is totally isotropic for $B$ when $B(u,v)=0$ for all $u,v\in E$. For a real nondegenerate bilinear form of inertia $(p,q)$, its dimension is at most $\min (p,q)$.

Let $K={\mathbb{F}}_{2^d}$, let $\alpha$ generate $K^{\times }$, and let $T:K\to {\mathbb{F}}_2$ be nonzero linear with nondegenerate bilinear form $(x,y)\mapsto T(xy)$. If the minimal polynomial of $\alpha$ is $P(X)=X^d+{\sum }_{j<d}{c}_j{X}^j$, then $x_n=T({\alpha }^n)$ satisfiesso it is produced by an LFSR of length at most $d$. Its period is exactly $2^d-1$: any period $r$ would imply $T(({\alpha }^r-1)y)=0$ for every $y\in K$, hence ${\alpha }^r=1$ by nondegeneracy.