Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith homogeneous polynomials $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2{|}_l$, $B=Q_3{|}_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a morphism of algebraic varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese residual plane conics are exactly the fibres of a morphism of $f$, giving the conic bundle from a line on a smooth cubic surface. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.

Write $q_{s,t}=z^{\mathsf{T}}M(s,t)z$ with $z=(u,v,w{)}^{\mathsf{T}}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ areThus the conic discriminant $\Delta (s,t)=\det M(s,t)$ is a homogeneous polynomial of degree five.

Smoothness also shows that every zero of $\Delta$ is simple and has matrix rank two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ has a zero over the algebraically closed field; that point would be singular in the total family. This proves the reduced singular fibres of a smooth conic bundle assertion and also rules out $\Delta$ vanishing identically.

The discriminant quintic of a cubic surface conic bundle consequently has five distinct zeros in ${\mathbb{P}}^1$. At each, a rank-two plane conic splits into two distinct projective lines $l_i,l_i^{\prime }$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original plane section is therefore $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.

Every unitary matrix is unitarily diagonalizable. Part (ii) therefore proves the stronger matrix statement $\varphi (A(t))=I$, not merely a statement about its eigenvalues. The kernel of a group homomorphism is a normal subgroup, and it contains every upper unipotent. By part (iii) their normal closure is the entire group. ThusThis proves the triviality of finite-dimensional unitary representations of SL2R. In fact the argument used only the algebraic homomorphism property and finite-dimensional unitarity; it did not need continuity.

Let $N$ be the normal closure of the upper unipotents $U(t)=A(t)$. Conjugation by $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ givesThus $N$ also contains every lower unipotent $L(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. We prove explicitly that these elementary unipotent generators of SL2R generate the full group.

For $a\ne 0$, multiplication givesSo all determinant-one diagonal matrices lie in $N$. If $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ has $a\ne 0$ and $ad-bc=1$, thenbecause its lower-right entry is $(1+bc)/a=d$. If $a=0$, then $c\ne 0$ and left multiplication by $U(1)$ makes the upper-left entry equal to $c$. The resulting matrix lies in $N$ by the preceding factorization, and $U(1{)}^{-1}\in N$ restores $g$. Therefore

Put $U=\varphi (A(t))$. Applying the group homomorphism $\varphi$ to the identity just proved shows that $U$ is conjugate to $U^{m^2}$ for every positive integer $m$. Thus the multiset of its eigenvalues is unchanged by taking their $m^2$th powers.

Take $m=2$. The fourth-power operation permutes this finite multiset. For each eigenvalue $z$, iterating its permutation cycle gives $z^{4^r}=z$ for some positive integer $r$. Since $U$ is unitary, $z\ne 0$, so $z^{4^r-1}=1$: every eigenvalue is a root of unity. Choose a positive integer $d$ divisible by all their orders. Every eigenvalue of $U^{d^2}$ is then one. Because its eigenvalue multiset is also the original one, we obtainThis is square-power rigidity of a finite nonzero spectrum. Finiteness of the dimension is essential to the permutation-cycle argument.

Let $E_{12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, so $A(t)=I+t{E}_{12}$ and $E_{12}^2=0$. Multiplication givesThe nilpotence also gives $(I+t{E}_{12}{)}^k=I+kt{E}_{12}$ for every nonnegative integer $k$, since all higher binomial terms vanish. Hence, for each positive integer $m$,These are unipotent matrices in SL2R. The integer is positive, since the displayed diagonal matrix is undefined at $m=0$.

For short and long simple roots $\alpha ,\beta$ of $G_2$, the long roots form an $A_2$ root system with simple roots ${\delta }_1=\beta$ and ${\delta }_2=3\alpha +\beta$. If ${\omega }_1,{\omega }_2$ are the corresponding $A_2$ fundamental weights, then $\alpha ={\omega }_2-{\omega }_1$ and ${\alpha }^{\vee }={\delta }_2^{\vee }-{\delta }_1^{\vee }$.

For $E:y^2=x^3-x-1$ over ${\mathbb{F}}_3$, the Trace of Frobenius is three andIn particular, $\#E({\mathbb{F}}_{3^r})=3^r+1$ exactly when $r\equiv 3(\mathrm{mod}6)$.

The effective conductance between two terminals is the reciprocal of their effective resistance. Under a unit voltage difference it equals the total current flowing from one terminal to the other.

The sheaf ${\mathcal{K}}_X$ assigns to an open subset of a complex manifold its meromorphic functions. Its subsheaf ${\mathcal{K}}_X^{\ast }$ consists of the nonzero meromorphic functions, while ${\mathcal{O}}_X^{\ast }$ consists of the nowhere-zero holomorphic functions.

The dual root system of $\Phi$ is ${\Phi }^{\vee }=\{{\alpha }^{\vee }:\alpha \in \Phi \}$. It has the transposed Cartan matrix, so its Dynkin diagram reverses every multiple-edge arrow; in particular, $B_n$ and $C_n$ are dual.

For a generator $a\in H^1(L(p);{\mathbb{F}}_p)$, Poincare duality and the Bockstein isomorphism for a three-dimensional lens space make $t(a)$ nonzero. Replacing $a$ by $na$ multiplies $t(a)$ by $n^2$. An orientation-reversing homotopy equivalence multiplies the evaluation by $-1$, so its existence forces $n^2=-1$ in ${\mathbb{F}}_p$ for some $n$.

For a finite-dimensional irreducible highest-weight representation with highest weight $\lambda$, every weight in the Weyl group orbit $W\lambda$ is extremal and has weight multiplicity one.

The self-dual and anti-self-dual combinations of the skew generators form two commuting copies of ${\mathfrak{sl}}_2(\mathbb{C})$. Consequentlyand its Adjoint representation decomposes as $(3,1)\oplus (1,3)$.

The roots of maximal length in a root system form a root subsystem. A chamber for the original system lies in a unique chamber for this long-root subsystem, because its interior avoids every long-root reflecting hyperplane.

For $t$ in a Cartan subalgebra $\mathfrak{t}$, the root-space decomposition givesThus $t$ is regular exactly when $\alpha (t)\ne 0$ for every root $\alpha$.

The kernel pair of a morphism $f:A\to B$ is the pullback in a category $A{\times }_{B}A$ with its two projections to $A$. It is the universal parallel pair equalized by $f$.

Every invertible complex matrix has the unique polar decompositionwhere $P$ is a positive-definite Hermitian matrix and $U$ is a unitary matrix. Writing $P=e^B$ gives $B=\frac{1}{2}\log (M{M}^{\ast })$, which is Hermitian.

A set belongs to the Pinsker sigma-algebra exactly when, modulo a null set, it belongs to $\mathcal{T}(\xi )$ for some finite measurable partition $\xi$. For a zero-entropy binary partition one may take that partition itself: zero conditional entropy makes its present atom measurable from every remote future.

Choose the common dominating probability measure $\nu =(\mu +{\mu }^{\prime })/2$ and let $p=d\mu /d\nu$, $q=d{\mu }^{\prime }/d\nu$ be the Radon-Nikodym derivatives. The assumed second moments imply first-moment integrability by the Cauchy-Schwarz inequality. Since $Y$ is a separable Banach space, the measurable $f$ is a strongly measurable function, and both expected values exist as Bochner integrals.

The difference of these Bochner integrals satisfiesThe first inequality is the norm bound for a Bochner integral, and the second is the scalar Cauchy-Schwarz inequality. Since $(\sqrt{p}+\sqrt{q}{)}^2\le 2(p+q)$, the first factor is at mostWith the Hellinger distance convention of the original PDF, the second factor is $\sqrt{2}{d}_{\mathrm{Hell}}(\mu ,{\mu }^{\prime })$. Multiplication proves the Hellinger bound for differences of expectations:The proof uses only the two laws and their common dominating measure; no absolute continuity of measures assumption between $\mu$ and ${\mu }^{\prime }$ is needed.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes' theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.