Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 7 / 3

Suppose first that the bounded Fredholm operator $T:H\to H$ has Fredholm index zero. Put $N=\ker T$ and $C=(\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$. These finite-dimensional linear subspaces have the same dimension. The restricted map $T:N^{\perp }\to \mathrm{ran}T$ is a bounded bijection, with bounded inverse by the bounded inverse theorem. Choose an isomorphism $J:N\to C$ and extend it by zero on $N^{\perp }$, writing the resulting finite-rank operator as $K$.

Relative to the orthogonal decompositions $H=N\oplus N^{\perp }$ and $H=C\oplus \mathrm{ran}T$, the operator $T+K$ is the direct sum of $J$ and the invertible restricted map. Thus $T+K$ is invertible, and $T=(T+K)-K$ is the sum of an invertible operator and a compact operator.

Conversely write $T=U+K$ with $U$ invertible and $K$ compact. It suffices to treat $I+C$ with $C=U^{-1}K$. A compact operator on a Hilbert space admits arbitrary operator norm approximation by finite-rank operators: take a finite epsilon-net of its unit-ball image and project onto the span of the net. Choose $F$ of finite rank with $\Vert C-F\Vert <1$, put $R=C-F$, and factorThe first factor is invertible by the Neumann series. Set $G=(I+R{)}^{-1}F$ and $E=\mathrm{ran}G$, a finite-dimensional linear subspace. On $E\oplus E^{\perp }$, the second factor has block formLeft multiplication by the invertible block matrix $\left(\begin{array}{cc}I& -B\\ 0& I\end{array}\right)$ reduces it to $A\oplus I$. Its range is closed, and its kernel and cokernel have equal dimension by finite-dimensional rank-nullity theorem for $A:E\to E$. Invertible factors preserve all these properties. Hence $T$ is a Fredholm operator of Fredholm index zero, proving both directions:

The Clifford algebra relation impliesIndeed move $C(v)$ past $C(b)$ and then past $C(a)$ using the two anticommutator identities; the three-generator products cancel.

Since all $C(w)$ are self-adjoint,Each scalar vanishes because $T$ is skew-adjoint. This proves $\boxed{\pi (T{)}^{\ast }=-\pi (T)}$.

For the requested commutator, the first identity givesThe first sum is $C(Tv)$ by the orthonormal basis expansion of $v$. For the second, the skew-adjoint property gives ${\sum }_i(T{e}_i,v)e_i=-Tv$. By real linearity of $C$, the difference is thereforeThis calculation explains the factor $1/4$: it makes infinitesimal rotations of $V$ act correctly on its Clifford generators.

Products below combine matrix multiplication with the wedge product of differential forms of differential forms. Differentiate the projection identity $p^2=p$ to getMultiplying by $p$ on both sides also gives $pdpp=0$; applying the same argument to $1-p$ gives $(1-p)dp(1-p)=0$. Thus at each point $dp$ is off-diagonal relative to $\mathrm{im}p\oplus \ker p$, whereas $(dp{)}^2$ is diagonal and commutes with $p$. For $\omega =p(dp{)}^2$,because $d(dp)=0$.

The graded Leibniz rule and cyclicity of the matrix trace give, since $\omega$ has even degree,Here $(d\omega ){\omega }^{k-1}$ is off-diagonal: $(dp{)}^3$ is off-diagonal and every power of $\omega$ is diagonal. Its matrix trace is zero. Hence $\boxed{d\mathrm{tr}({\omega }^k)=0}$ for every $k\ge 1$; for $k=0$ the trace is the constant matrix size.

For the determinant, work over the commutative algebra of even differential forms. If $f(0)=1$, write $f(\omega )=I+N$, where $N$ has strictly positive degree and is nilpotent for dimensional reasons. The finite power series identityholds over this algebra, as follows either from the formal determinant identity over characteristic-zero commutative rings or by differentiating $\det (I+sN)$ in the scalar variable $s$. Each $N^j$ is a linear combination of positive powers of $\omega$. The matrix trace of the logarithm is therefore a closed differential form, and its finite exponential is closed by the graded Leibniz rule. Thus $\boxed{d\det f(\omega )=0}$. The projection connection interprets $\omega$ as the curvature of $pd$ on $\mathrm{im}p$, but the proof above directly establishes both closedness claims.

The positive sign printed in the exponential uses the skew-adjoint convention for the Dirac operator: $D^{\ast }=-D$, so $D_{-}=-D_{+}^{\ast }$. Put $A=D_{+}$ and define the nonnegative self-adjoint operatorsFor $t>0$, the assumed compact elliptic spectral theorem for these elliptic differential operators gives discrete eigenvalues, finite multiplicities and trace-class operators $e^{-tP}$, $e^{-tQ}$.

If $Pu=\lambda u$ with $\lambda >0$, then $Q(Au)=A(Pu)=\lambda Au$ and $Au\ne 0$. Conversely $A^{\ast }$ sends the $\lambda$ eigenspace of $Q$ to that of $P$. On these eigenspaces the inverse of $A$ is ${\lambda }^{-1}{A}^{\ast }$. Thus the positive spectra agree with multiplicity, and absolute convergence of the operator traces permits cancellation term by term:The last expression is the Fredholm index of $A$, since its closed range has orthogonal complement $\ker A^{\ast }$. ConsequentlyThis is the supertrace cancellation used in Question 1. If the Dirac operator is instead defined to be self-adjoint, $D_{-}=(D_{+}{)}^{\ast }$ and the heat exponentials must have a minus sign. With that convention the positive exponentials in the printed formula are generally unbounded and have no finite operator trace.