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PutSince $x,y,z$ are unit vectors,The feasible inner products with two unit vectors satisfyThe desired pair $p=q=1/2$ is feasible exactly when $c\ge -1/2$. In that case every maximizer has these inner products. If $c<-1/2$, symmetry and strict convexity force $p=q$, and the nearest feasible diagonal point isUsing the vector triple product,Hence

Rotations preserve dot products and cross-product lengths, while $z^{\prime }=2z$. ThusLetThe target pair is feasible precisely whenorOn this interval,

Outside this interval, let $\lambda >0$ be the unique number for whichand write $r=(p,q{)}^T$. These equations give the unique weighted projection of $t$ onto the feasible ellipse, and therefore determine the maximizing inner products solely from $c=y\cdot z$. In the remaining cases,Neither answer depends on ${\theta }_1$ or ${\theta }_2$.

Write $B(x,y)=⟨\alpha x,y⟩$. The hypothesis says $B(x,x)=0$. Expanding at $x+y$ giveswhereas expanding at $x+iy$ givesTherefore $B(x,y)=0$ for every $x,y$, soThis is the complex polarization argument for a vanishing quadratic form.

The conclusion is false over a real inner product space. On ${\mathbb{R}}^2$, the nonzero operatorsatisfies $⟨Jx,x⟩=0$ for every $x$.

For every $x$,If $\alpha$ is normal, these quantities are equal. Conversely, equality of the norms for all $x$ givesThe operator in parentheses is self-adjoint, so polarization makes it zero. ThereforeThe same equivalence holds over a real inner product space: for a self-adjoint operator $T$, the real polarization identityshows that a vanishing quadratic form forces $T=0$.

Let $J_0$ be the standard complex structure on ${\mathbb{R}}^{2n}$, let ${\omega }_0$ be the standard symplectic form, and let $g_0$ be the Euclidean inner product. They satisfyThe two-out-of-three property for unitary structures says that a real linear map preserving any two of these structures preserves the third. In terms of the corresponding matrix groups,

For example, if $A$ preserves $J_0$ and ${\omega }_0$, thenso $A$ preserves $g_0$. If it preserves $J_0$ and $g_0$, the second displayed identity shows that it preserves ${\omega }_0$. Finally, $J_0$ is uniquely determined by $g_0(J_{0}u,v)={\omega }_0(u,v)$, so preservation of $g_0$ and ${\omega }_0$ implies $A{J}_0=J_{0}A$. The common intersection is therefore the unitary group.

Any symplectic form on $S^2$ orients its tangent bundle. Choose a compatible complex structure and inner product; this reduces the structure group to $SO(2)$. Since every line in an oriented symplectic plane is Lagrangian,is an oriented circle bundle.

If a transition function of $T{S}^2$ rotates vectors through an angle $\alpha$, its action on unoriented lines rotates the coordinate $e^{2i\theta }$ through $2\alpha$. The Euler class of the Lagrangian-line bundle of an oriented plane bundle therefore givesBy the Poincaré-Hopf theorem,for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal{L}Gr(T{S}^2)$ is $\pm 4$, which is nonzero. A smoothly trivial oriented circle bundle has zero Euler class, so $\mathcal{L}Gr(T{S}^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.

When all masses are equal, the free energy depends on the real vector $\mathbf{\varphi }$ only through inner products, so its symmetry is the orthogonal group $O(N)$. The mean-field potential isFor ${\mu }_0^2>0$, its unique minimum is $\mathbf{\varphi }=0$, which preserves $O(N)$. For ${\mu }_0^2<0$, the minima form the sphereChoosing one minimum leaves the subgroup $O(N-1)$ that fixes its direction, so spontaneous symmetry breaking gives $\boxed{O(N)\to O(N-1)}$.

The oriented normal vectors to the planet and dust-belt planes areTheir inner product is the cosine of the mutual inclination, soIf ${\varphi }_d=\Omega$, this reduces to $\cos I_m=\cos (I-I_d)$; equal inclinations then give $I_m=0$. If both planes are face-on, it likewise gives $I_m=0$, independently of their undefined nodal longitudes. Reversing an observationally unidentified ascending node by $\pi$ produces the familiar orbital-plane orientation degeneracy for an axisymmetric belt.

Set $c=⟨\psi |\varphi ⟩$. The off-diagonal Hermiticity identity holds automatically, whilemust be real. Thusis necessary and sufficient. For $g\ne 0$, put $r=gc/|g|\in \mathbb{R}$. Since the two normalized spanning vectors are linearly independent, $|r|<1$.

The normalized eigenvectors and eigenvalues areDirect application of $H$ verifies the eigenvalue equations. Their unnormalized inner product isso the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If $g=0$, $H=0$ and every vector is an eigenvector.

The operator $F$ maps the computational orthonormal basis $\{|j⟩\}$ to the orthonormal basis $\{|{\varphi }_j⟩\}$. It therefore preserves inner products and is surjective, so $F^{†}F=F{F}^{†}=I$ and $F$ is unitary.

For every tangent vector $v\perp x_0$, stationarity of $Q$ on the sphere gives $2v^{T}A{x}_0=0$. Hence $A{x}_0$ is parallel to $x_0$, say $A{x}_0=E{x}_0$. Taking the inner product with $x_0$ gives $E=Q(x_0)$. The spectral theorem shows that this is the largest eigenvalue.

For $A=\left(\begin{array}{cc}1& t\\ t& 1\end{array}\right)$ the eigenvalues are $1\pm t$, soIts graph is a V translated upward and is convex.

A root system in the real inner product space $V$ is a finite spanning set $\Phi \subset V\setminus \{0\}$ such thatwhere the orthogonal reflectionFor a crystallographic root system one additionally requires $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$; that condition is not needed for general finite reflection groups.

A fundamental system of a root system is a basis $\Delta \subset \Phi$ such that every root has either all nonnegative or all nonpositive coordinates in this basis. Its associated positive system of a root system isand $\Phi =\Pi \bigsqcup (-\Pi )$.

Write $k=\mu (1-\beta )$. Taking the cross product of the equation of motion with $\mathbf{r}$ givesSince $\mathbf{r}\cdot \mathbf{h}=0$ at every time, the orbit lies in the fixed plane perpendicular to the conserved specific angular momentum $\mathbf{h}$. Taking the inner product with $\dot{\mathbf{r}}$ gives the conserved specific orbital energyFor $\beta >1$, the effective inverse-square force is repulsive and the potential term is positive.

The restricted trace map is continuous, and $V$ is its kernel, so $V$ is a closed vector subspace of the Hilbert space $H^1(U)$. It is therefore complete in the $H^1$ norm. Part a shows that the gradient normis equivalent to that norm, so it is complete as well. It comes from the inner productConsequently $(V,\Vert \cdot {\Vert }_V)$ is a Hilbert space.

An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

Mackey and induced representation sum over $H\backslash G/K$. Coprimality makes every $H\cap gK{g}^{-1}$ trivial, each double coset has size $|H||K|$, and each local inner product is $(\dim V)(\dim W)$. Therefore

The prescribed form is a sum of products of real polynomial evaluations. It is a symmetric bilinear form, since each evaluation is a linear map. AlsoEquality forces $f(-1)=f(0)=f(1)=0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, by the root bound for a polynomial. Thus equality holds only for the zero polynomial, establishing positive definiteness. Therefore the form is a real inner product, and $V$ is an inner product space. This is a discrete sampling inner product on polynomial spaces.

Use $f,g$ from i. Their inner products with $x^2$ areThus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, soBy ii the three polynomials are orthogonal vectors, and normalizing $h$ givesThis orthonormal set is a basis of the three-dimensional vector space $V$. It is the result of the Gram-Schmidt process applied to $1,x,x^2$.

The relevant inner products areHence $1$ and $x$ are orthogonal vectors. Dividing by their norms gives the orthonormal basisof their linear span $U$.

$A$ is not an orthogonal matrix. Its first column has squared Euclidean normso $A^{T}A\ne I$. The reason this is compatible with ii is that $1,x,x^2$ is not an orthonormal basis for the given inner product. Its Gram matrix isThe correct condition on this matrix representation of a linear map iswhich follows either by direct multiplication or by the inner product preservation just proved. This is the rule for an orthogonal transformation in a nonorthonormal basis.

If $T$ is an orthogonal transformation and $e_1,e_2,e_3$ is an orthonormal basis, thenThe images form an orthonormal set, hence are linearly independent and form an orthonormal basis in dimension three.

Conversely, suppose the images $T{e}_1,T{e}_2,T{e}_3$ of one chosen orthonormal basis form an orthonormal basis. Write $u={\sum }_i{a}_i{e}_i$ and $v={\sum }_j{b}_j{e}_j$. By linearity of $T$ and bilinearity,Thus mapping one orthonormal basis to an orthonormal basis is equivalent to preserving every inner product. In the converse, a single chosen orthonormal basis suffices.