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Over a field of characteristic different from two, the rank of a quadratic form is the rank of a linear map associated with its symmetric bilinear form, equivalently the matrix rank of a symmetric coefficient matrix. For an $n$-dimensional vector space this is $n-\dim \mathrm{rad}B$. A change of basis acts by matrix congruence, which preserves matrix rank.

For example, $q(x,y,z)=x^2-y^2$ has rank two, although it vanishes at the nonzero vector $(1,1,0)$. Its radical of a bilinear form is the $z$-axis; an isotropic vector need not lie in that radical. In characteristic two, the polar bilinear form of a quadratic form can lose square terms: $q(x)=x^2$ is nonzero but its polar form is zero. The rank of that polar form alone therefore does not record all quadratic information; conventions for quadratic rank in characteristic two require separate care.

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.

A simple bivector can be expressed as one exterior product of two vectors. Changing the pair need not change its area element: $u\wedge (v+cu)=u\wedge v$ for every scalar $c$, by bilinearity and $u\wedge u=0$. This operation shears the represented parallelogram while preserving its oriented area.

Every bivector in a real vector space of dimension at most three is simple, but this fails in dimension four. For an ordered basis $e_1,\dots ,e_4$, considerThe two components describe independent coordinate planes. Their square iswhereas $(u\wedge v)\wedge (u\wedge v)=0$ because the factors repeat. Thus this $B$ cannot be one simple bivector. Over the real numbers in dimension four, the converse also holds: a nonzero bivector with $B\wedge B=0$ is simple. The nonzero factor two is important; that criterion cannot be applied unchanged in characteristic two.

Over a field $k$, the trace-zero $2\times 2$ matrices form the special linear Lie algebra ${\mathfrak{sl}}_2(k)$. The matrix trace of a commutator is zero, so this is a Lie subalgebra of the general linear Lie algebra. A basis isIn fact $\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)=aH+bE+cF$, uniquely. The individual products areSubtracting in the stated order givesFor characteristic different from two, this is the standard sl2 triple; see Etingof, Section 11.4. The products remain valid in characteristic two, but there $H=I$ is central and $2E=2F=0$, so the familiar characteristic zero structure changes.

The following table gives all nine basis brackets. Each entry is $[\text{row},\text{column}]$; the diagonal vanishes by alternation, and exchanging row and column reverses the sign.

Bilinearity now computes any commutator, for exampleThe three structure constants of a Lie algebra relations encode all of this multiplication. They also show why the sl2 triple is useful in Lie algebra representations: the actions of $E$ and $F$ move between eigenvalues of the action of $H$ in steps of two.

For $q=2^{2a+1}\ge 8$, the Suzuki group $Sz(q)={}^2{B}_2(q)$ is a twisted finite group of Lie type. In characteristic two, the algebraic symplectic group over a field $S{p}_4$ has an exceptional endomorphism $\sigma$ satisfying ${\sigma }^2=F_2$. Fixed points of ${\sigma }^{2a+1}$ give $Sz(q)$; its natural matrix representation has degree four.^[ref]

It is simple, of order $q^2(q^2+1)(q-1)$, and acts two-transitively on the $q^2+1$ points of a Suzuki ovoid. A point stabilizer has order $q^2(q-1)$ and contains a subgroup of order $q^2$ acting regularly on the remaining points. At $q=8$ the degree is $65$ and the order is $29120$. The parameter $q=2$ instead gives a nonsimple group of order $20$.

Over a field of characteristic different from two, every square matrix decomposes uniquely asHere $S$ is a symmetric matrix and $K$ is a skew-symmetric matrix. Transposing proves those properties. For uniqueness, a matrix in both classes satisfies $M=M^T=-M$, hence $2M=0$ and $M=0$. This is a direct sum decomposition of the vector space of square matrices. In characteristic two, division by two is unavailable and the two classes coincide, so the same decomposition statement fails.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

The symmetric square of a group representation on $V$ is the quotient vector spacewith the induced tensor product representation. If the characteristic of a field is different from two, $vw\mapsto (v\otimes w+w\otimes v)/2$ identifies it with the invariant subspace fixed by the factor-exchange map $\tau$. Over the complex numbers, the projection operator $(I+\tau )/2$ givesIn characteristic two, the quotient definition still works, but this averaging projection operator does not exist. Real coordinate-quadratic perturbations of a nonmagnetic electronic multiplet transform in this symmetric square.

Over a field of characteristic two, the linear map $T:M_n(F)\to M_n(F)$, $A\mapsto A+A^T$, has kernel of a linear map the symmetric matrices, of dimension $n(n+1)/2$. Its image of a linear map is the linear subspace of symmetric matrices with zero diagonal, of dimension $n(n-1)/2$. Indeed, $T(A)=0$ means $A=A^T$, and every symmetric zero-diagonal matrix is the image of the matrix obtained by retaining just its entries above the diagonal. Thus the two dimensions still add to $n^2$, as required by the rank-nullity theorem.