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The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

A function $f:U\to {\mathbb{R}}^n$ has a Frechet derivative at $a\in U$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe remainder condition is a limit in all directions at once, not merely the existence of partial derivatives.

For the stated scalar two-variable result, fix $a=(a_1,a_2)$ and choose a small rectangle contained in the open set $U$. Split the increment into two coordinate segments. Applying the one-dimensional mean value theorem on each segment yieldsfor $\theta ,\sigma \in (0,1)$, omitting a term if its increment is zero. Subtract $h{f}_x(a)+k{f}_y(a)$. By continuity of the partial derivatives, for any $\epsilon >0$ the remainder has modulus at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ for sufficiently small increments. This proves differentiability, with Jacobian matrix $(f_x(a),f_y(a))$.

Away from the origin the specified rational function has a nonzero denominator and is smooth. At the origin, $f(t,0)=t$ and $f(0,t)=2t^2$, so the partial derivatives are $f_x(0,0)=1$, $f_y(0,0)=0$. Any Frechet derivative there would therefore be $A(h,k)=h$. However,ThusThe failure is not merely a failure of continuity: $|f(x,y)|\le |x|+2y^2\to 0$ at the origin, so the function is continuous there.

The chain rule for Frechet derivatives states that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, thenThe composition on the right is a composition of linear maps, equivalently a product of Jacobian matrices in coordinates.

Apply the chain rule to the affine function $x\mapsto (x,c-x)$. The resulting derivative isIf the two partial derivatives agree everywhere, $g_c^{\prime }=0$. The mean value theorem makes $g_c$ constant on $\mathbb{R}$. Define $h(c)=f(c,0)$, which is itself differentiable by the chain rule. Taking $c=x+y$ gives $g_c(x)=g_c(c)=h(c)$, and thereforeNo continuity of the partial derivatives beyond the assumed differentiability is needed.

Let $P$ be the averaging linear mapFor a unitary representation of a finite group, $P$ is the orthogonal projection onto the invariant subspace $V^G$, and $P\rho (h)=\rho (h)P=P$. Applying this to $\rho (g)v=\lambda v$ gives $Pv=\lambda Pv$. Since $\lambda \ne 1$, $Pv=0$. The vector $v$ is a cyclic vector for a group representation, so its translates span $V$, and every translate also has projection zero. Thus $P=0$ andThis is the cyclic eigenvector obstruction to invariant vectors. Unitarity also gives a direct proof by pairing $v$ with an invariant vector and then pairing every translate; the averaging proof makes the role of cyclicity especially transparent.

Use the PDF's covector-first convention: its space ${\mathcal{T}}_l^k$ consists of tensor fields inThis reverses the order in which some texts list tensor type. For a finite-dimensional real vector space $V$, define the tensor contraction first on decomposable tensors:The hats mean omission, with the other factors left in their original order. The formula is a multilinear map of its individual factors, so the universal property of a tensor product gives a unique linear map with this formula.

Apply it with $V=T_{p}M$ at every $p$. Evaluation of a covector on a vector is basis independent: under a change of frame, one factor transforms by a matrix and the other by its inverse transpose, and the matrices cancel in the pairing. Thus the fiber maps agree on overlapping vector bundle trivializations. In a local dual basis, the coefficients of the contracted tensor are finite sums of coefficients with the selected covariant and contravariant indices set equal. They remain smooth. Henceis a globally defined smooth contraction for $k,l\ge 1$. It is also linear over $C^{\infty }(M)$.

Pullback of a differential form commutes with the exterior derivative, so $f^{\ast }$ gives a linear map on de Rham cohomology. Functoriality and $f\circ f=\mathrm{id}$ imply $(f^{\ast }{)}^2=I$. Over the real numbers, define the complementary projectionsThey obey $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Every cohomology class therefore decomposes asThe first term is fixed and the second negated by $f^{\ast }$. Their intersection is zero, since a class in both equals its own negative. ConsequentlyThis is the eigenspace decomposition of a linear involution. The factor $1/2$ uses the real coefficient field; a corresponding argument over characteristic two would fail. At the level of differential forms, the same averaging projections give representatives of either parity whenever the class has that parity.

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.

For a formal associative deformation, a completion convention is necessary. The usual star product lives on the formal power series moduleThis is the adic completion of a module applied to the ordinary tensor product, rather than literally the ordinary $A{\otimes }_{k}k[[t]]$ when $A$ is infinite-dimensional. For example, ${\sum }_{n\ge 0}{X}^n{t}^n$ lies in $k[X][[t]]$ but not in the ordinary tensor product, whose coefficient spaces have finite-dimensional span. The two agree when $A$ is finite-dimensional. We interpret the printed notation in this standard completed sense; the infinite iteration below requires that interpretation.

A star product is a $k[[t]]$-bilinear, unital product continuous for the adic topology satisfying associativity of the formIt is a trivial formal deformation if a $k[[t]]$-linear automorphism continuous for the adic topology $T=\mathrm{id}+{\sum }_{r\ge 1}{t}^r{T}_r$, with $T(1)=1$, satisfies $T(a\ast b)=T(a)T(b)$.

If $\mathrm{Dim}(A)\le 1$, then $H{H}^2(A,A)=0$. Suppose changes of coordinates have removed all coefficients below order $r$. The order-$r$ part of associativity then says $\delta {\mu }_r=0$. Hence ${\mu }_r=\delta g_r$ for a $k$-linear map $g_r:A\to A$. Its normalization gives $g_r(1)=0$, since $(\delta g_r)(1,1)=g_r(1)$. Transport the product by $T_r=\mathrm{id}+t^r{g}_r$:The order-$r$ coefficient becomes ${\mu }_r-\delta g_r=0$, and lower coefficients remain zero. Repeating constructs compatible changes of coordinates modulo every $t^N$. They converge in the adic topology to an invertible $T$ fixing $1$, with inverse obtained coefficient by coefficient. The limit product is ordinary multiplication. Therefore every star product is trivial under the completed formal-series convention. In fact, the argument only needs the vanishing of $H{H}^2(A,A)$, not all of Hochschild cohomological dimension at most one.

On continuous functions on the p-adic integers, define the forward difference operator and the Mahler coefficients byWriting $Sf(x)=f(x+1)$, the explicit iterate follows from ${\Delta }^n=(S-I{)}^n$. The ultrametric inequality and integral binomial coefficients imply $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$ and $|a_n(f){|}_p\le \Vert f{\Vert }_{\infty }$, where the supremum norm is taken over ${\mathbb{Z}}_p$.

The Mahler theorem states that every such $f$ has a unique expansion with uniform convergenceConversely, every sequence in ${\mathbb{Q}}_p$ tending to zero yields a continuous function by this expansion. The binomial polynomials form an orthonormal expansion in the non-Archimedean sense: $\Vert f{\Vert }_{\infty }={\sup }_n|a_n(f){|}_p$.

Here is the requested proof under the permitted coefficient-decay assumption. For $n\ge 1$, the binomial polynomial $B_n(x)=x(x-1)\cdots (x-n+1)/n!$ is a continuous function on ${\mathbb{Z}}_p$. Its values on nonnegative integers are integral, and those integers are dense in the p-adic integers. Since ${\mathbb{Z}}_p$ is a closed set in ${\mathbb{Q}}_p$, $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$. Also $B_n(n)=1$, so $\Vert B_n{\Vert }_{\infty }=1$, including $B_0=1$.

If $a_n\to 0$, the ultrametric inequality gives the uniform tail boundCompleteness of ${\mathbb{Q}}_p$ gives a uniform limit $F$, and the uniform limit theorem makes $F$ continuous. At any nonnegative integer $m$, all $B_n(m)$ with $n>m$ vanish. Finite binomial inversion givessince the inner sum is $(1-1{)}^{m-j}$. Hence $F=f$ on a dense subset and therefore on all of ${\mathbb{Z}}_p$. The values at $m=0,1,2,\dots$ recover each coefficient recursively because $B_m(m)=1$; this proves uniqueness. The expansion bounds $\Vert f{\Vert }_{\infty }$ by $\sup |a_n{|}_p$, and the earlier coefficient inequality proves equality. This also proves the converse statement.

Although the problem allows us to assume decay, it can be established independently. By compactness and uniform continuity, approximate $f$ uniformly by $h$ constant on residue classes modulo $p^r$. On this finite-dimensional space, $S^{p^r}=I$ andThe matrix of ${\Delta }^{p^r}$ has entries divisible by $p$, so its operator norm is at most $p^{-1}$; consequently $\Vert {\Delta }^{j{p}^r+s}h{\Vert }_{\infty }\le p^{-j}\Vert h{\Vert }_{\infty }$ for $0\le s<p^r$. Thus $a_n(h)\to 0$. Since $|a_n(f)-a_n(h){|}_p\le \Vert f-h{\Vert }_{\infty }$, arbitrary uniform approximation proves automatic decay of Mahler coefficients.

For the last claim, construct the discrete antidifferentiation on the p-adic integersIts coefficient sequence is $0,a_0,a_1,\dots$, still tending to zero, so it is continuous. By Pascal's identity, $\Delta B_{n+1}=B_n$. The boundedness of $\Delta$ permits applying it to the uniform limit, giving $\Delta G=f$. For the stated linear map, invariance under translation by one now givesThus translation-invariant linear forms on p-adic continuous functions vanish:No continuity of $T$ has been assumed or used. In particular, one must not justify this by applying $T$ termwise to an infinite Mahler expansion; it is the existence of a continuous discrete antiderivative that makes the conclusion valid.

Use the natural cubic spline convention appropriate to the interior design knots: the spline is linear on $[a,x_1]$ and $[x_n,b]$. We also need $n\ge 2$. These points will be important for uniqueness.

A cubic spline is a function in $C^2[a,b]$ whose restriction to each of $[a,x_1]$, $[x_1,x_2]$, through $[x_n,b]$ is a polynomial of degree at most three. A spline knot is a junction between consecutive pieces. For a natural cubic spline, the two exterior pieces are affine functions; equivalently, one uses the natural endpoint conditions $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$ on the spline over $[x_1,x_n]$ and then extends linearly with matching derivatives. Merely prescribing $g^{\prime \prime }(a)=g^{\prime \prime }(b)=0$ on otherwise arbitrary exterior cubic pieces is a different condition and does not provide the unique interpolant invoked in this question.

The natural cubic spline interpolant $N\mathbf{v}$ to values $\mathbf{v}=(v_1,\dots ,v_n{)}^T$ is the natural cubic spline satisfying $(N\mathbf{v})(x_i)=v_i$. Its existence and uniqueness for $n\ge 2$ are the interpolant fact allowed in the PDF. The interpolation map $N$ is a linear map: a linear combination of interpolants satisfies the same spline, boundary and value conditions, so uniqueness identifies it with the interpolant to the corresponding linear combination of value vectors.

Let $g=N\mathbf{v}$, let $\tilde{g}\in C^2[a,b]$ have the same values, and put $r=\tilde{g}-g$. Then $r(x_i)=0$. We prove the orthogonalityApply integration by parts twice on each polynomial interval. On an interior interval, $g^{\prime \prime \prime \prime }=0$, andThe $g^{\prime \prime \prime }r$ terms vanish because $r$ vanishes at the knots. The $g^{\prime \prime }{r}^{\prime }$ terms cancel at internal knots because $g^{\prime \prime }$ and $r^{\prime }$ are continuous, and vanish at $x_1,x_n$ because $g^{\prime \prime }$ is zero there. On the exterior intervals, $g^{\prime \prime }=0$ identically. Summing therefore gives the claimed orthogonality without assuming that $g^{\prime \prime \prime }$ is continuous across knots.

Expanding the square now gives the minimum roughness property of the natural cubic spline interpolantEquality forces $r^{\prime \prime }=0$ everywhere, since $r^{\prime \prime }$ is continuous. Thus $r$ is an affine function. It vanishes at two distinct design points, so $r=0$, proving equality if and only if $\tilde{g}=g$.

For the penalized fit, take the allowed spline roughness penalty matrix $\Gamma$ to be a symmetric matrix with ${\mathbf{z}}^T\Gamma \mathbf{z}\ge 0$; thus it is a positive semidefinite matrix. If its supplied representative were not symmetric, its symmetric part defines the same quadratic form and is all that is needed. For any $\tilde{g}$, let $\mathbf{v}=(\tilde{g}(x_i){)}_{i=1}^n$. Replacing it by $N\mathbf{v}$ preserves every fitted value and can only decrease the second derivative roughness penalty. Therefore it is enough to minimizeover ${\mathbb{R}}^n$. The matrix $A_{\lambda }=I_n+\lambda \Gamma$ is a positive-definite matrix, sinceCompleting the quadratic or differentiating gives the unique solutionThis is the cubic smoothing spline. More explicitly, the normal equations are $(I_n+\lambda \Gamma ){\hat{\mathbf{v}}}_{\lambda }=\mathbf{Y}$, andFor an arbitrary function with fitted vector $\mathbf{v}$, the earlier orthogonality also givesBoth terms are nonnegative. Equality forces $\mathbf{v}={\hat{\mathbf{v}}}_{\lambda }$ and, by the proved interpolation equality case, $\tilde{g}=N\mathbf{v}={\hat{g}}_{\lambda }$. This proves both existence in $C^2[a,b]$ and uniqueness over the entire stated function class, not just over splines.

The argument is deterministic for each observed $\mathbf{Y}$. The fixed-design nonparametric regression and homoscedasticity assumptions motivate the squared-error loss, but Gaussian errors are not required.

At least two distinct design points are required for uniqueness. Question 4 does not explicitly restate $n\ge 2$, and this hypothesis is necessary. With one design point, every function $\tilde{g}(x)=Y_1+c(x-x_1)$ has zero residual and zero second derivative roughness penalty, for any real $c$. Thus the printed unrestricted uniqueness conclusion would be false for $n=1$. The corrected theorem assumes at least two distinct knots and the stated linear-tail convention. At $\lambda =0$, uniqueness over all of $C^2[a,b]$ also fails; the stipulated $\lambda >0$ eliminates that degeneracy.

Put $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$. The invertible linear map $x\mapsto u=Dx$ takes the ellipsoid to the unit sphere $S^n$. Write $u=(v,s)$ with $v\in {\mathbb{R}}^n$, and let $N=(0,1)$ and $S=(0,-1)$ be its two poles. The following two stereographic projections define manifold charts on the ellipsoid:Their ranges are ${\mathbb{R}}^n$. With $r^2=|w{|}^2$, their inverses, followed by $D^{-1}$, areThese are continuous and smooth functions, and directly satisfy the defining constraint. On the overlap, the smooth transition map isIt is its own inverse and is smooth away from zero. The two domains are open in the subspace topology and cover the ellipsoid. Its subspace topology is inherited from Euclidean space, so the ellipsoid is Hausdorff and has a second-countable space topology. Thus these charts form a smooth atlas and make it an $n$-dimensional smooth manifold, with an explicit diffeomorphism to $S^n$. For the degenerate dimension $n=0$, each chart consists of a single point with range ${\mathbb{R}}^0$, and their overlap is empty; the same conclusion holds.

Write $n=2m-1$. On ${\mathbb{R}}^{2m}$ define the orthogonal linear mapIt satisfies $J^T=-J$ and $J^{T}J=I$. Hence $⟨x,Jx⟩=0$ and $\Vert Jx\Vert =\Vert x\Vert$. Restricting to the unit sphere givesThis continuous unit tangent field is available exactly because the ambient dimension is even.

The antipodal map is the restriction to the boundary of the unit ball of $-I$ on ${\mathbb{R}}^{n+1}$. The induced orientation on the sphere is reversed or preserved according to the determinant of that ambient linear map, namely $(-1{)}^{n+1}$. Its degree is therefore $(-1{)}^{n+1}$. With integral coefficients, for $n\ge 1$ the induced top-homology map isAt $n=0$, ordinary $H_0(S^0)={\mathbb{Z}}^2$ instead has its two generators exchanged; multiplication by $-1$ is the statement on reduced $H_0(S^0)\cong \mathbb{Z}$. This distinguishes the small-dimension exception from the usual top-degree formula.

View $X$ as the matrix of a linear map $T:F^m\to F^n$ of matrix rank $r$. Choose vectors $u_1,\dots ,u_r$ whose images form a basis of $\mathrm{im}T$, extend them by a basis of $\ker T$ to a basis of $F^m$, and extend $T{u}_1,\dots ,T{u}_r$ to a basis of $F^n$. In these two bases the matrix of $T$ is the rank normal formThe two changes of basis give invertible $P,Q$ with the displayed matrix equal to $Q^{-1}XP$.

For the block upper triangular matrix $A=\left(\begin{array}{cc}B& D\\ 0& C\end{array}\right)$, every nonzero term in the Leibniz formula for determinants sends the rows belonging to $C$ into the columns belonging to $C$; bijectivity then sends the remaining rows into the $B$ columns. The permutation sum consequently factors into the determinant sums for $B$ and $C$, proving

Finally suppose $L(X)=0$, so $AX=XB$: the map $X:{\mathbb{C}}^m\to {\mathbb{C}}^n$ is an intertwining operator. If $v$ lies in the generalized eigenspace of $B$ for $\lambda$, then for some $k$,Because $A$ and $B$ have no common eigenvalue, $A-\lambda I$ is invertible, hence $Xv=0$. The generalized eigenspaces of $B$ span ${\mathbb{C}}^m$, so $X=0$. Thus the Sylvester equation operator $L$ is injective.

Use the decomposition into Lagrangian subspaces $V=V_{+}\oplus V_{-}$ from part (a). The bilinear form $\Omega$ pairs $V_{+}$ and $V_{-}$ nondegenerately: a vector in either summand annihilating the other already annihilates its own summand and hence all of $V$, so is zero.

Choose a basis $e_1,\dots ,e_n$ of $V_{+}$. The linear mapis an linear isomorphism. Let $f_j$ correspond to the dual basis of the $e_i$. Then $\Omega (e_i,f_j)={\delta }_{ij}$, and the pairings within each summand vanish. Thus

A map $f:A\to {\mathbb{R}}^m$ is differentiable at $p\in A$ when there is a linear map $Df(p):{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe linear map $Df(p)$ is the Frechet derivative of $f$ at $p$.

The multivariable chain rule states that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, thenIndeed, write $f(p+h)=f(p)+Ah+r(h)$ and $g(f(p)+k)=g(f(p))+Bk+s(k)$, where $r(h)=o(|h|)$ and $s(k)=o(|k|)$. Since $k=Ah+r(h)=O(|h|)$,and the last two terms are $o(|h|)$. This proves the formula.

For matrix inversion near the identity matrix,The last term is $O(\Vert h{\Vert }^2)$ because inversion is bounded near $I$. Therefore the inversion map is differentiable at $I$ and

A Cartan subalgebra $\mathfrak{h}$ of a finite-dimensional complex semisimple Lie algebra is a maximal abelian subalgebra consisting of semisimple elements. The root-space decomposition isand the nonzero functionals $\alpha$ are the roots. A Cartan-Weyl basis consists of a basis $H_i$ of $\mathfrak{h}$ and root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$. Its brackets have the formand $[E_{\alpha },E_{\beta }]=N_{\alpha \beta }{E}_{\alpha +\beta }$ when $\alpha +\beta$ is a root, and zero when $\alpha +\beta$ is neither a root nor zero.

For the complexified so4 Lie algebra, take $H^1=T^{(12)}$ and $H^2=T^{(34)}$. WriteDirect use of the stated commutation relations givesThe simultaneous eigenvectors, hence the step generators, may be chosen asThus the roots relative to $(H^1,H^2)$ are $(\pm i,\pm i)$. Replacing $H^a$ by $-i{H}^a$ gives the usual real coordinates $(\pm 1,\pm 1)$. The only nonzero brackets between step generators, apart from those obtained by antisymmetry, are

An isomorphism of Lie algebras is a bijective linear map preserving the Lie bracket. DefineThenThe two spans are commuting copies of the complexified ${\mathfrak{su}}_2$, and together contain all six basis elements of ${\mathfrak{so}}_4$. Chiral decomposition of the complexified so4 Lie algebra therefore givesUnder this isomorphism the Adjoint representation is the direct sum of the adjoint representations of the two factors:

Write $B$ for the open unit ball. Because $T$ is a linear map, the hypothesis scales toFix $y\in B$. Choose numbers $r$ and $\theta$ such thatthen $r/(1-\theta )<1$.

Set $z_0=y$. Inductively, if $\Vert z_{j-1}\Vert <r{\theta }^{j-1}$, the scaled density statement lets us chooseand we put $z_j=z_{j-1}-T{x}_j$. The geometric series givesBecause $X$ is a Banach space, $x={\sum }_{j\ge 1}{x}_j$ exists and lies in $B$. Since $T$ is a bounded linear operator, it is continuous, whileTherefore $Tx=y$. As $y\in B$ was arbitrary,This is the dense open-unit-ball image criterion.