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With the mostly-plus metric signature $(-,+,+,+)$ and coordinates $X^{\mu }=(ct,x,y,z)$, the Minkowski metric is the nondegenerate bilinear formIt is a symmetric bilinear form, but not a positive-definite bilinear form. For example, $U=(1,1,0,0)$ is nonzero yet $U\cdot U=0$, making it a null vector. This does not make the Minkowski metric degenerate: some other vectors have nonzero pairing with $U$.

Applied to infinitesimal displacements, the Minkowski metric givesFor a timelike curve, a clock records $d\tau =\sqrt{-d{s}^2}/c$. For a null curve, $d{s}^2=0$. For spacelike separation, the positive Minkowski interval measures a different kind of separation. A Lorentz transformation preserves this form, expressed in matrix notation by ${\Lambda }^T\eta \Lambda =\eta$, although it need not preserve ordinary Euclidean distance in the four-component coordinate list.

No. On the real vector space $V={\mathbb{R}}^2$, use the symmetric bilinear formIts matrix of a bilinear form is $\mathrm{diag}(1,-1)$, whose determinant is $-1\ne 0$, so it is a nondegenerate bilinear form. LetA vector $(s,t)$ belongs to the bilinear orthogonal complement $U^{\perp }$ exactly when $s-t=0$. HenceIn particular, the claimed direct sum fails. The isotropic vector $(1,1)$ has zero self-pairing, although it pairs nontrivially with other vectors of $V$.

The dimension identity in part (i) counts both vector subspaces but does not force their intersection of vector subspaces to be zero. For a symmetric bilinear form, $U\cap U^{\perp }$ is the radical of a bilinear form restricted to $U$. Thus the precise criterion isIndeed, trivial intersection of vector subspaces and part (i) give a sum of vector subspaces of dimension $\dim V$. A positive-definite bilinear form would guarantee this for every $U$; symmetry alone does not.

The invariant positive-definite bilinear form makes every $\rho (g)$ orthogonal in an orthonormal real basis. Every element of $\mathrm{SU}(2)$ has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$, so this group is homeomorphic to $S^3$ and connected. The continuous function $\det \rho (g)$ takes values in $\{-1,1\}$ and equals one at the identity. It is therefore identically one. Hence $\boxed{\rho (\mathrm{SU}(2))\subseteq \mathrm{SO}(3)}$.

The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

A real symmetric bilinear form is positive-definite when $B(v,v)>0$ for every nonzero vector. It defines an inner product and the associated norm $\sqrt{B(v,v)}$. Symmetry is part of this definition: positive diagonal values by themselves do not imply symmetry.^[ref]

The identity matrix $A=\mathrm{diag}(1,1)$ gives $B(v,w)=v_1{w}_1+v_2{w}_2$ and $B(v,v)=v_1^2+v_2^2$. For $v=(3,4)$ this is $25$, giving norm $5$. A weighted example, $A=\mathrm{diag}(4,1)$, gives norm $\sqrt{4x^2+y^2}$: the first coordinate costs twice as much in length. Such inner products let geometry assign different weights to different directions.

Every positive-definite bilinear form is nondegenerate: a vector in its radical of a bilinear form has $B(v,v)=0$, so positivity forces $v=0$. The converse fails, as $\mathrm{diag}(1,-1)$ shows. Even positivity of the associated quadratic form does not force a general bilinear form to be symmetric: $A=\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$ has $v^{T}Av=x^2+y^2$ but $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$.

The complex Killing form is not positive-definite on the entire complex Cartan subalgebra; its restriction to the Euclidean subspace of a Cartan subalgebra is a real positive-definite form.

A bilinear form $B$ is symmetric when $B(v,w)=B(w,v)$ for every pair of vectors $v,w$. Equivalently, its matrix of a bilinear form in any basis is a symmetric matrix, since $A_{ij}=B(e_i,e_j)=B(e_j,e_i)=A_{ji}$. Symmetry makes the relation $B(v,w)=0$ reciprocal, giving the bilinear orthogonal complement a geometric interpretation.

Over the real numbers, set $q(v)=B(v,v)$. The polarization identityrecovers the symmetric bilinear form from its quadratic form: expand $q(v+w)$ and use symmetry to combine the two cross terms. For $q(x,y)=x^2-y^2$, take $v=(2,1)$ and $w=(1,3)$. Then $q(v+w)=-7$, $q(v)=3$, and $q(w)=-8$, so $B(v,w)=(-7-3+8)/2=-1=2\cdot 1-1\cdot 3$. The same formula works whenever the characteristic is not two.^[ref]

Three real diagonal matrices show how much variety symmetry allows. The identity matrix $\mathrm{diag}(1,1)$ gives the positive-definite bilinear form $q=x^2+y^2$. The matrix $\mathrm{diag}(1,-1)$ gives $q=x^2-y^2$ and is nondegenerate despite the isotropic vector $(1,1)$. The matrix $\mathrm{diag}(1,0)$ gives $q=x^2$ and is a degenerate bilinear form with radical of a bilinear form $\mathrm{span}\{(0,1)\}$. Their level sets make the distinction visible: a circle, two branches of a hyperbola, and two parallel lines for $q=1$.

The inertia of a bilinear form records the numbers of positive, negative, and zero directions after matrix congruence diagonalization. By Sylvester's law of inertia, these counts survive change of basis; the three examples have inertia of a bilinear form $(2,0,0)$, $(1,1,0)$, and $(1,0,1)$. Historically, Sylvester stated this invariance in an 1852 paper on reducing real quadratic forms to sums of positive and negative squares. Ghys and Ranicki reproduce his statement and distinguish it from his subsequent 1853 proof.^[ref]