Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 2

Use ${\mathbb{T}}^n={\mathbb{R}}^n/(2\pi \mathbb{Z}{)}^n$ and the periodic Sobolev space conventionThe Cauchy-Schwarz inequality givesTo check the last sum, partition the lattice into $|k|<1$ and dyadic shells $2^j\le |k|<2^{j+1}$. A shell contains at most $C_n{2}^{jn}$ points and contributes at most $C_n{2}^{j(n-2s)}$. This is a convergent geometric series exactly when $s>n/2$.

Thus the Fourier series ${\sum }_k\hat{f}(k)e^{ik\cdot x}$ converges absolutely and uniformly on the torus. Its sum is a continuous function and has the original Fourier coefficients, so it represents the original periodic distribution. In particular the representative is unique andThis proves the requested inclusion, with its continuous Sobolev embedding estimate.

Let $V$ be a finite-dimensional real or complex inner product space, and let $A\subseteq \mathrm{End}(V)$ be a unital subalgebra stable under the adjoint operator. Define its commutant of an operator algebra by $A^{\prime }=\{T:Ta=aT\text{ for all }a\in A\}$. The double commutant theorem for star-algebras asserts $\boxed{A^{\prime \prime }=A}$. The inclusion $A\subseteq A^{\prime \prime }$ follows immediately from the definition; we prove the reverse.

Write $N=\dim V$, fix a basis $v_1,\dots ,v_N$, and act diagonally on $V^N$. The linear subspaceis invariant under every diagonal $a$. Since $A$ is closed under adjoints, its orthogonal complement $W^{\perp }$ is invariant too: $⟨aw,z⟩=⟨w,a^{\ast }z⟩=0$ for $w\in W^{\perp }$, $z\in W$. Hence the orthogonal projection $P$ onto $W$ commutes with every diagonal $a$.

Writing $P$ as an $N$ by $N$ block matrix $(P_{ij})$, this says $P_{ij}a=a{P}_{ij}$, so each $P_{ij}\in A^{\prime }$. If $b\in A^{\prime \prime }$, it commutes with every $P_{ij}$, and its diagonal action therefore commutes with $P$ and preserves $W$. Because $1\in A$, $(v_1,\dots ,v_N)\in W$. Therefore there is one $a\in A$ such thatEquality on the basis gives $b=a$, proving the theorem over either scalar field.

The identity hypothesis is necessary. For a possibly nonunital adjoint-stable subalgebra, adjoining the ambient identity does not change its commutant of an operator algebra, so the exact statement is $A^{\prime \prime }=A+\mathbb{F}I$. For example $A=\{0\}$ has $A^{\prime \prime }=\mathbb{F}I$ on a nonzero $V$. Thus the usual exam formulation takes the algebra to contain the identity.

At a fixed $t$, write $A=f(t)$ and $B=f^{\prime }(t)$. Differentiating the matrix exponential in the direction $B$ givesThe order of factors matters because matrix multiplication need not be commutative.

To prove the derivative of the matrix exponential formula, expand its power series. The coefficient linear in $h$ in $(A+hB{)}^k$ is ${\sum }_{j=0}^{k-1}{A}^{j}B{A}^{k-1-j}$. On bounded sets of matrices, the exponential series and its directional derivative series converge uniformly: the latter is dominated in operator norm by $\Vert B\Vert {\sum }_{k\ge 1}k\Vert A{\Vert }^{k-1}/k!$. Terms with two or more $hB$ factors have sum $O(h^2)$ locally. Hence the derivative isOn the other hand, the integral in the box expands absolutely asThe integral identity follows by repeated integration by parts. Grouping by $a+b=k-1$ gives the same derivative. Finally differentiability means $f(t+h)=A+hB+o(h)$; the locally bounded derivative of the exponential sends the $o(h)$ error to $o(h)$, proving the chain rule even when $f^{\prime }$ is not continuous. If $[f,f^{\prime }]=0$, the formula simplifies to $e^f{f}^{\prime }$.

Let $e(\omega )$ denote exterior multiplication, $e(\omega )\alpha =\omega \wedge \alpha$. For a $q$-form $\alpha$, evaluation of the proposed operator on vector fields $Y_0,\dots ,Y_q$ givessince ${\sum }_i{\omega }_i(Y_a)X_i=Y_a$. This expression is independent of the chosen local basis.

Expand each covariant derivative into the derivative of $\alpha (Y_0,\dots ,{\hat{Y}}_a,\dots ,Y_q)$ minus the terms obtained by differentiating each argument. For every pair $a<b$, those argument terms combine intoThe Levi-Civita connection is a torsion-free connection, so the vector in this expression is $[Y_a,Y_b]$. The result is exactly the defining evaluation formula for the exterior derivative:Therefore $\boxed{d=\sum _{i}e({\omega }_i){\nabla }_{X_i}}$. No orthonormality or compactness is needed for this identity; the dual-frame property and vanishing torsion tensor suffice.