Case 3

Consider the case $d - j = -n$. Then we get $h_0(x, z) = - u_0(x) \log\vert z\vert$, after possibly subtracting a term depending only on $x$. We have proved the following result.

To compute $u_0(x)$, the coefficient of logarithmic divergence, we change coordinates by a local diffeomorphism $\psi(x)$. Note that

$$\log \vert\psi(x) - \psi(y)\vert \sim \log\vert\psi'(x)\cdot(x - y)\vert \sim \log\vert x - y\vert \as y \to x,$$

while $k_P(x,y) \mapsto k_P(\psi(x), \psi(y)) L(x,y)$, where $L(x, y) \to \vert\det\psi'(x)\vert$ as $y \to x$, by the change of variables formula for $\vert d^ny\vert$. (We use a 1-density, not an oriented volume form, to do integration; however, if we agree to fix an orientation on $M$ and use only coordinate changes that preserve the orientation, for which $\det\psi'(x) > 0$ at each $x$, then we need not make this distinction). Thus the $\log$-divergent term transforms as follows:

$$-u_0(x)\log\vert x - y\vert \mapsto -u_0(\psi(x))\vert\det\psi'(x)\vert \log\vert x - y\vert.$$

For the case of scalar pseudodifferential operators, this is all we need. In the general case of operators acting on sections of a vector bundle $E \to M$, we replace $u_0(x) \in \End E_x$ by its matrix trace $\tr u_0(x) \in \bC$. The previous formula then says that the 1-density $\tr u_0(x) \vert d^n x\vert$ is invariant under local coordinate changes.

Now, when we regularize $p_{-n}(x,\xi)$ to obtain this 1-density after applying $\sF_2^{-1}$, we can first subtract the homogeneous ``principal part'', at each $x \in U$, since this will not change the coefficient of logarithmic divergence. This subtraction is done by replacing $p_{-n}(x,\xi)$ by its average over the sphere $\vert\xi\vert = 1$ in the cotangent space $T_x^* M$. That is to say, we get the same $u_0(x)$ if we replace $p_{-n}(x,\xi)$ by $\Om_n^{-1} \vert\xi\vert^{-n} \int_{\vert\om\vert=1} p_{-n}(x,\om) \sg$. On applying () (with $C = 0$) at each $x$, we conclude that

$$\tr u_0(x) = \int_{\vert\om\vert=1} \tr p_{-n}(x,\om) \sg.$$

We shall now show that $\Wres$ is a trace on the algebra of classical pseudodifferential operators on $M$ acting on a given vector bundle.

We begin with another important property of homogneous functions on $\bR^n \setminus \{0\}$. We shall make use of the Euler vector field on this space:

$$R = \sum_{j=1}^n \xi_j \dd{\xi_j} = r \dd{r}.$$

Notice that $h$ is $\la$-homogeneous if and only if $Rh = \la h$, since $Rh(r\om) = r\dd{r}(r^\la h(\om)) = \la r^\la h(\om) = \la h(r\om)$.

To show that the trace is unique (up to constants) when $n > 1$, let $T$ be any trace on the algebra of classical pseudodifferential operators. Again we suppose that all symbols are supported in a coordinate chart $U \subset M$, and we note that the formulas for composition of symbols give the commutation relations

$$[x^j, f]= i \pd{f}{\xi_j}, \qquad [\xi_j, f] = -i \pd{f}{x^j}.$$

By Lemmas and , $T(P)$ thus depends only on the homogeneous term $\tr p_{-n}(x,\xi)$, of degree $-n$, and moreover $T(P) = 0$ if $\int_{\vert\xi\vert=1} \tr p_{-n}(x,\xi) \sg = 0$. We can replace $\tr p_{-n}(x,\xi)$ with $\vert\xi\vert^{-n} \int_{\vert\xi\vert=1} \tr p_{-n}(x,\xi) \sg$, without changing $T(P)$. Now $f \mapsto T(f(x) \vert\xi\vert^{-n})$ is a linear functional on $\Coo_c(U)$ that kills derivatives with respect to each $x^j$, so it is a multiple of the Lebesgue integral:

$$T(f) = C \int_U f(x) \vert d^n x\vert \words{for some} C \in \bC.$$

Therefore,

$$T(P) = C \int_U \int_{\vert\xi\vert=1} \tr p_{-n}(x,\xi) \sg \vert d^n x\vert = C \Wres(P).$$

What we have gained? We no longer need the full spectrum of the Dirac operator: its principal symbol is enough to give the Wodzicki residue.