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For smooth surfaces, Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. Consequently any smooth local isometry $f:M\to N$ satisfies $K_N(f(p))=K_M(p)$. The surprise is that $K$ is defined using the shape operator, which describes bending in space, yet its value can be recovered from measurements made entirely within the surface.

In his 1827 paper, Carl Friedrich Gauss expressed Gaussian curvature through the metric coefficients and their partial derivatives in Article 11, then stated its invariance under development in Article 12. Article 13 explicitly compares a plane with surfaces developable onto it, including cylindrical surfaces. The paper was presented on 8 October 1827 and printed in 1828; the Morehead–Hiltebeitel English translation appeared in 1902. ^[ref]

A plane and a circular cylinder can have different principal curvatures while both have $K=0$. Their matching first fundamental forms exhibit the actual local isometry. In contrast, no nonempty open patch of a sphere of radius $R$ can be flattened smoothly into a plane while preserving its first fundamental form: $1/R^2$ cannot become zero. Making the patch small reduces practical distortion, but never produces an exact local isometry. This is a geometric obstruction to a perfectly distance-preserving flat map of a spherical world, independent of where one places a seam.

The following proof outline connects the two descriptions of Gaussian curvature. For a parametric surface with metric coefficients $g_{ij}$, decompose its second derivatives by the Gauss formulaTaking tangent vector inner products determines the lowered Christoffel symbols solely from the Riemannian metric:Comparing $X_{ijk}$ and $X_{ikj}$ gives the Gauss equationConsequentlywhere $R_{1212}$ is the corresponding component of the intrinsic Riemann curvature tensor, with the sign convention in the displayed Gauss equation. The final expression uses only $g_{ij}$ and its first two derivatives.

For equal lengths, end masses $m$, middle mass $M$, and equal springs between adjacent bobs, reflection symmetry gives mode vectors $(1,1,1)$, $(1,0,-1)$ and $(1,-2m/M,1)$. Their squared frequencies are displayed above. These vectors are orthogonal in the inner product weighted by $\mathrm{diag}(m,M,m)$. Equal end masses, rather than equality of all three masses, suffice for this symmetry reduction.

Over a field of characteristic different from two, three linear projections summing to the identity annihilate one another in both orders. Each pair sum is the complementary projection to the third. The sum of two linear projections criterion therefore proves zero products. Every vector is the sum of its three images; applying each projection to a zero sum proves that these images form a direct sum. Choosing a basis in each image gives a simultaneous eigenbasis, with triples of eigenvalues $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. This is algebraic orthogonality, and does not assume that the projections are orthogonal projections for a specified inner product.

The dual of the trace-class operators, and also of the normed space of finite-rank operators with the trace norm, is isometrically the space of bounded operators. The pairing is $F_B(T)=\mathrm{Tr}(BT)$. For $R_{x,y}z=⟨z,y⟩x$, one has $\Vert R_{x,y}{\Vert }_1=\Vert x\Vert \Vert y\Vert$ and $F_B(R_{x,y})=⟨Bx,y⟩$, using an inner product linear in its first argument. A bounded functional on these rank-one operators gives a bounded sesquilinear form and hence a unique $B$ by the Riesz representation theorem. The singular value decomposition proves $\Vert F_B\Vert =\Vert B\Vert$ and the triangle inequality for the trace norm.

For a nonzero fractional ideal $\mathfrak{b}$ of a number field, its trace-dual lattice is the fractional idealThe equality follows from the definition of the inverse different and invertibility of fractional ideals. Under the positive metric on the Minkowski embedding of a number field, the Euclidean dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$: coordinatewise complex conjugation converts the positive inner product into the trace pairing. This assertion does not require that conjugation preserve the number field itself.

With the defining two-dimensional matrix trace and inner product $⟨X,Y⟩=-\mathrm{tr}(XY)$, the displayed Pauli matrix basis is orthonormal. The Pauli matrix multiplication law gives the indicated structure constants of a Lie algebra. With the alternative form $-2\mathrm{tr}(XY)$, the orthonormal basis is $-i{\sigma }_a/2$ and the structure constant is one. The bracket and inner-product normalizations must be kept consistent in projected gauge-field formulas.

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces:

If the state-preparation unitary sends $|0⟩$ to $|\psi ⟩=\sin \chi |g⟩+\cos \chi |b⟩$, use the orthonormal good/bad basis. The marked-space phase is diagonal there, and conjugating a phase on $|0⟩$ gives the rank-one phase in the displayed formula. Their product is unitary. An arbitrary preparation unitary not sending zero to this state need not preserve this particular two-dimensional subspace. Matrices in unnormalized good/bad vectors must not be described as unitary in the ordinary Euclidean inner product.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

On ${\mathbb{R}}^{2n}$ let $J_0$ be the standard complex structure, let ${\omega }_0$ be the standard symplectic form, and let $g_0(u,v)={\omega }_0(u,J_{0}v)$ be the Euclidean inner product. A linear map preserving any two of $J_0$, ${\omega }_0$, and $g_0$ preserves the third. Equivalently,

For self-adjoint operators $A,B$ and a normalized quantum state in their relevant product operator domains, put $u=A\psi$, $v=B\psi$, $z=⟨u,v⟩$. With the inner product linear in its second entry, $⟨C⟩=2\mathrm{Im}z$. The Cauchy-Schwarz inequality givesEquality holds exactly when $u,v$ are linearly dependent and $\mathrm{Re}z=0$. This includes either zero vector; if both vectors are nonzero it is equivalent to $B\psi =i\lambda A\psi$ for real nonzero $\lambda$. Applying this inequality to the centred observables $A-⟨A⟩I$ and $B-⟨B⟩I$ gives the Robertson uncertainty principle.

Now take one equation for two unknowns:The linear system $x_1+2x_2=3$ has an entire affine line of exact solutions. Since $A$ has full row rank, its Moore-Penrose inverse isThe kernel is $\mathrm{span}\{(-2,1{)}^T\}$, and all exact solutions areBecause $x_{\ast }$ is orthogonal to $(-2,1{)}^T$,The Euclidean norm is therefore smallest exactly at $t=0$. The zero residual alone cannot select a solution; the minimum-norm rule supplies the additional criterion.

This selection depends on the specified inner product. Rescaling one unknown changes what counts as a small Euclidean norm, so units and weighting matter in an application. A minimum-norm answer is mathematically canonical for the chosen geometry, rather than an automatic guarantee of physical plausibility.

For Hermitian $J_3$ and adjoint angular momentum ladder operators $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space, a highest-weight vector of weight $j$ obeys $J_{+}|j,j⟩=0$. The commutation relations give $J_{+}(J_{-}{)}^{\ell }|j,j⟩=\ell (2j-\ell +1)(J_{-}{)}^{\ell -1}|j,j⟩$. Taking inner products forces $2j$ to be a nonnegative integer: otherwise the first negative factor would give a negative squared norm. The norm first vanishes at $\ell =2j+1$, so the generated unitary representation has dimension $2j+1$. Without positivity an abstract highest-weight module can remain infinite-dimensional even at an integral highest weight.

A unitary highest-weight Virasoro module is generated by a normalized vector satisfying the displayed relations, with a positive-definite inner product and $L_n^{†}=L_{-n}$. Its Virasoro descendants have $L_0$ eigenvalues $h+N$ for nonnegative integral levels $N$. The commutator gives $\Vert L_{-1}|h⟩{\Vert }^2=2h$, so unitarity requires $h\ge 0$. Such a module is a quotient of a Virasoro Verma module; its Unitary Virasoro discrete series describes the cases with central charge below one.

A unitary operator $U$ on a complex vector space with an inner product satisfies $U^{\ast }U=U{U}^{\ast }=I$ and therefore preserves inner products.

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.

For $n\ge 2$, the Gram matrix of $n$ unit vectors with common distinct-pair inner product $t$ is $G=(1-t)I+t11^T$. Its eigenvalues are $1+(n-1)t$ on the all-ones direction and $1-t$ on its orthogonal complement. It is a positive-definite symmetric matrix exactly when $-1/(n-1)<t<1$, permitting a linearly independent realization in ${\mathbb{R}}^n$. An explicit construction takes the columns ofThe two orthogonal projections have product zero, so $B^{T}B=G$; positive eigenvalues also make $B$ invertible. At either endpoint the corresponding eigenspace becomes a linear dependence.

A vector is an element of a vector space. Its defining operations are vector addition and scalar multiplication. An arrow in a Euclidean space is one example; a polynomial or a function can also be a vector when the chosen vector space uses the corresponding addition and scalar multiplication.

For instance, the polynomials $p(x)=1+x$ and $q(x)=x-x^2$ belong to the real vector space of polynomials of degree at most two. Their vector addition gives $(p+q)(x)=1+2x-x^2$, and scalar multiplication gives $(2p)(x)=2+2x$. In the monomial basis $(1,x,x^2)$ these are the coordinate vectors $(1,1,0)$, $(0,1,-1)$, $(1,2,-1)$ and $(2,2,0)$ respectively. The vectors are the polynomials; these tuples are their representations in a chosen basis.

A general vector space does not itself prescribe lengths or angles. A norm supplies a size, while an inner product also supplies angle information and orthogonality. The familiar arrow geometry below belongs to Euclidean vectors, where this extra structure is available.

A symmetric positive-definite matrix $M$ defines an inner product on real coordinate vectors byFor diagonal $M$, positive diagonal entries weight the squared coordinate errors. This is the finite-coordinate version of a weighted inner product; the positive-definite matrix example appears in Francis Narcowich, Inner Products and Norms, standard inner products, example 8.

For $M=\mathrm{diag}(4,1)$, the squared norm of $(a,b{)}^T$ is $4a^2+b^2$. A first-coordinate error of one costs as much as a second-coordinate error of two. The unit sphere is the ellipse $4a^2+b^2=1$, and orthogonality means $4u_1{v}_1+u_2{v}_2=0$. Thus “closest” and “perpendicular” depend on the chosen inner product. The orthogonal projection residual example shows how this changes a best approximation.

For a positive weight $w$, the weighted inner product of functions is $⟨f,g{⟩}_w=\int \overline{f}gw$. It is the usual inner product in the L2 space for the measure $w(x)dx$. A Sturm-Liouville eigenfunction expansion uses this weight to make its eigenfunctions orthogonal.

The weight specifies how different locations contribute to a squared error: $\Vert f-g{\Vert }_w^2=\int |f-g{|}^{2}w$. More weight makes an error in that region more costly. The weighted Euclidean inner product makes the same choice for a finite list of coordinates.

For an explicit real-function example on $[0,1]$, let $w(x)=1+x$ and approximate $f(x)=x$ by a constant function. The orthogonal projection onto a finite-dimensional subspace spanned by the function $1$ givesThe best unweighted constant is $1/2$. Giving greater weight to larger $x$ moves the best constant upward, so a weighting choice changes the approximation even though the allowed family of functions is unchanged.