12.5 The signature complex

12.5.1 The Hirzebruch signature

Let $M$ be a compact orientable manifold of even dimension, $m=2l$ . Let $[\omega ]$ and $[\eta ]$ be the elements of the ‘middle’ cohomology group $H^l(M;\mathbb{R})$ . We consider a bilinear form $H^l(M;\mathbb{R})\times H^l(M;\mathbb{R})\to \mathbb{R}$ defined by

$$\sigma ([\omega ],[\eta ])\equiv {\int }_M\omega \wedge \eta \text{\hspace{1em}(12.52)}$$

cf example 11.8. This definition is independent of the representatives of $[\omega ]$ and $[\eta ]$ . The form $\sigma$ is symmetric if $l$ is even ( $m\equiv 0\mathrm{mod}4$ ) and anti-symmetric if $l$ is odd ( $m\equiv 2\mathrm{mod}4$ ). Poincaré duality shows that the bilinear form $\sigma$ has the maximal rank $b^l=\dim H^l(M;\mathbb{R})$ and is, hence, non-degenerate. If $l\equiv 2k$ is even, the symmetric form $\sigma$ has real eigenvalues, $b^{+}$ of which are positive and $b^{-}$ of which are negative ( $b^{+}+b^{-}=b^l$ ). The Hirzebruch signature is defined by

$$\tau (M)\equiv b^{+}-b^{-}.\text{\hspace{1em}(12.53)}$$

If $l$ is odd, $\tau (M)$ is defined to vanish (an anti-symmetric form has pure imaginary eigenvalues). In the following, we set $l=2k$ .

The Hodge $\ast$ satisfies ${\ast }^2=1$ when acting on a $2k$ -form in a $4k$ -dimensional manifold $M$ and hence $\ast$ has eigenvalues $\pm 1$ . Let ${\mathrm{Harm}}^{2k}(M)$ be the set of harmonic $2k$ -forms on $M$ . We note that ${\mathrm{Harm}}^{2k}(M)\cong H^{2k}(M;\mathbb{R})$ and each element of $H^{2k}(M;\mathbb{R})$ has a unique harmonic representative. ${\mathrm{Harm}}^{2k}(M)$ is separated into disjoint subspaces,

$${\mathrm{Harm}}^{2k}(M)={\mathrm{Harm}}_{+}^{2k}(M)\oplus {\mathrm{Harm}}_{-}^{2k}(M)\text{\hspace{1em}(12.54)}$$

according to the eigenvalue of $\ast$ . This separation block diagonalizes the bilinear form $\sigma$ . In fact, for ${\omega }^{\pm }\in {\mathrm{Harm}}_{\pm }^{2k}(M)$ ,

$$\sigma \left({\omega }^{+},{\omega }^{+}\right)={\int }_M{\omega }^{+}\wedge {\omega }^{+}={\int }_M{\omega }^{+}\wedge \ast {\omega }^{+}=\left({\omega }^{+},{\omega }^{+}\right)>0$$

where $({\omega }^{+},{\omega }^{+})$ is the standard positive-definite inner product defined by (7.181). We also find

$$\begin{array}{rl}& \sigma ({\omega }^{-},{\omega }^{-})=-{\int }_M{\omega }^{-}\wedge \ast {\omega }^{-}=-({\omega }^{-},{\omega }^{-})<0\\ & \sigma ({\omega }^{+},{\omega }^{-})=-{\int }_M{\omega }^{+}\wedge \ast {\omega }^{-}=-{\int }_M{\omega }^{-}\wedge \ast {\omega }^{+}=-\sigma ({\omega }^{+},{\omega }^{-})=0\end{array}$$

where we have noted that $\alpha \wedge \ast \beta =\beta \wedge \ast \alpha$ for any forms $\alpha$ and $\beta$ . Hence, $\sigma$ is block diagonal with respect to ${\mathrm{Harm}}_{+}^{2k}(M)\oplus {\mathrm{Harm}}_{-}^{2k}(M)$ and, moreover, $b^{\pm }={\dim }_{\mathbb{R}}{\mathrm{Harm}}_{\pm }^{2k}(M)$ . Now $\tau (M)$ is expressed as

$$\tau (M)=\dim {\mathrm{Harm}}_{+}^{2k}(M)-\dim {\mathrm{Harm}}_{-}^{2k}(M).\text{\hspace{1em}(12.55)}$$

Exercise 12.2

Let $\dim M=4k$ . Show that

$$\tau (M)=\chi (M)\mathrm{mod}2.\text{\hspace{1em}(12.56)}$$

[Hint: Use the Poincaré duality to show that $\chi (M)=b^{2k}$ mod 2.]

12.5.2 The signature complex and the Hirzebruch signature theorem

Let $M$ be an $m$ -dimensional compact Riemannian manifold without a boundary and let $g$ be the given metric. Consider an operator

$$\mathfrak{D}\equiv \mathrm{d}+{\mathrm{d}}^{†}.\text{\hspace{1em}(12.57)}$$

$\mathfrak{D}$ is a square root of the Laplacian: ${\mathfrak{D}}^2=\mathrm{d}{\mathrm{d}}^{†}+{\mathrm{d}}^{†}\mathrm{d}=\Delta$ . To show that $\mathfrak{D}$ is elliptic, it suffices to verify that $\Delta$ is elliptic since the symbol of a product of operators is the product of symbols. Let us compute the symbol of $\Delta$ . As for $\mathrm{d}$ , we have $\sigma (\mathrm{d},\xi )\omega =\xi \wedge \omega$ . As for ${\mathrm{d}}^{†}$ , it can be shown that (Palais 1965, pp77-8)

$$\sigma \left({\mathrm{d}}^{†},\xi \right)=-i_{\xi }.\text{\hspace{1em}(12.58)}$$

Here $i_{\xi }:{\Omega }_p^r(M)\to {\Omega }_p^{r-1}(M)$ is an interior product defined by (cf. (5.79))

$$\begin{array}{rl}& i_{\xi }\left({\mathrm{dx}}^{{\mu }_1}\wedge \cdots \wedge {\mathrm{dx}}^{{\mu }_r}\right)\\ & \equiv \sum _{j=1}^r(-1{)}^{j+1}{\mathrm{g}}^{{\mu }_j\mu }{\xi }_{\mu }\mathrm{d}{x}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{\hat{x}}^{{\mu }_j}\wedge \cdots \wedge \mathrm{d}{x}^{{\mu }_r}\end{array}$$

where the one-form under $\hat{\xi }$ is omitted and we put $\xi ={\xi }_{\mu }\mathrm{d}{x}^{\mu }$ . Now the symbol of the Laplacian is obtained from (12.58) as

$$\begin{array}{rl}\sigma (\Delta ,\xi )\omega & =\sigma \left(\mathrm{d}{\mathrm{d}}^{†}+{\mathrm{d}}^{†}\mathrm{d},\xi \right)\omega =-\left[\xi \wedge i_{\xi }(\omega )+i_{\xi }(\xi \wedge \omega )\right]\\ & =-i_{\xi }(\xi )\wedge \omega =-\Vert \xi {\Vert }^2\omega \end{array}$$

where $\omega$ is an arbitrary $r$ -form and the norm $\Vert \cdot \Vert$ is taken with respect to the given Riemannian metric. Finally, we obtain

$$\sigma (\Delta ,\xi )=-\Vert \xi {\Vert }^2.\text{\hspace{1em}(12.59)}$$

Thus, the Laplacian $\Delta$ is elliptic and so is $\mathfrak{D}=\mathrm{d}+{\mathrm{d}}^{†}$ .

Since the Laplacian $\Delta ={\mathfrak{D}}^2$ is self-dual on ${\Omega }^{\ast }(M)$ , the index of $\Delta$ vanishes trivially. It is also observed that $\mathfrak{D}={\mathfrak{D}}^{†}$ on ${\Omega }^{\ast }(M)$ and, hence, $\mathrm{ind}\mathfrak{D}=0$ . To construct a non-trivial index theorem, we have to find a complex on which $\mathfrak{D}\ne {\mathfrak{D}}^{†}$ .

Exercise 12.3

Consider the restriction ${\mathfrak{D}}^{\mathrm{e}}$ of $\mathfrak{D}$ to even forms, ${\mathfrak{D}}^{\mathrm{e}}:{\Omega }^{\mathrm{e}}(M{)}^{\mathbb{C}}\to {\Omega }^{\mathrm{o}}(M{)}^{\mathbb{C}}$ where ${\Omega }^{\mathrm{e}}(M{)}^{\mathbb{C}}\equiv \oplus {\Omega }^{2i}(M{)}^{\mathbb{C}}$ and ${\Omega }^{\mathrm{o}}(M{)}^{\mathbb{C}}\equiv \oplus {\Omega }^{2i+1}(M{)}^{\mathbb{C}}$ . The adjoint of ${\mathfrak{D}}^{\mathrm{e}}$ is ${\mathfrak{D}}^{\mathrm{o}}\equiv {\mathfrak{D}}^{\mathrm{e}†}:{\Omega }^{\mathrm{o}}(M{)}^{\mathrm{C}}\to {\Omega }^{\mathrm{e}}(M{)}^{\mathrm{C}}$ . Show that

$$\mathrm{ind}{\mathfrak{D}}^{\mathrm{e}}=\dim \ker {\mathfrak{D}}^{\mathrm{e}}-\dim \ker {\mathfrak{D}}^{\mathrm{o}}=\chi (M).$$

[Hint: Prove $\ker {\mathfrak{D}}^{\mathrm{e}}=\oplus {\mathrm{Harm}}^{2i}(M)$ and $\ker {\mathfrak{D}}^{\mathrm{o}}=\oplus {\mathrm{Harm}}^{2i+1}(M)$ . This complex, although non-trivial, does not yield anything new.]

If $\dim M=m=2l$ , we have $\ast \ast \eta =(-1{)}^r\eta$ for $\eta \in {\Omega }^r(M{)}^{\mathbb{C}}$ . We define an operator $\pi :{\Omega }^r(M{)}^{\mathbb{C}}\to {\Omega }^{m-r}(M{)}^{\mathbb{C}}$ by

$$\pi \equiv {\mathrm{i}}^{r(r-1)+l}\ast \text{\hspace{1em}(12.60)}$$

Observe that $\pi$ is a ‘square root’ of $(-1{)}^r\ast \ast =1$ . In fact, for $\omega \in {\Omega }^r(M{)}^{\mathbb{C}}$ ,

$$\begin{array}{rl}{\pi }^2\omega & ={\mathrm{i}}^{r(r-1)+l}\pi (\ast \omega )={\mathrm{i}}^{r(r-1)+l+(2l-r)(2l-r-1)+l}\ast \ast \omega \\ & ={\mathrm{i}}^{2r^2}\ast \ast \omega =(-1{)}^r\ast \ast \omega =\omega \end{array}\text{\hspace{1em}(12.61)}$$

where we have noted that $r\equiv r^2\mathrm{mod}2$ . We easily verify (exercise) that

$$\{\pi ,\mathfrak{D}\}=\pi \mathfrak{D}+\mathfrak{D}\pi =0.\text{\hspace{1em}(12.62)}$$

Let $\pi$ act on ${\Omega }^{\ast }(M{)}^{\mathbb{C}}=\oplus {\Omega }^r(M{)}^{\mathbb{C}}$ . Since ${\pi }^2=1$ , the eigenvalues of $\pi$ are $\pm 1$ . Then we have a decomposition of ${\Omega }^{\ast }(M{)}^{\mathbb{C}}$ into the $\pm 1$ eigenspaces ${\Omega }^{\pm }(M)$ of $\pi$ as

$${\Omega }^{\ast }(M{)}^{\complement }={\Omega }^{+}(M)\oplus {\Omega }^{-}(M).\text{\hspace{1em}(12.63)}$$

Since $\mathfrak{D}$ anti-commutes with $\pi$ , the restriction of $\mathfrak{D}$ to ${\Omega }^{+}(M)$ defines an elliptic complex called the signature complex,

$${\mathfrak{D}}_{+}:{\Omega }^{+}(M)\to {\Omega }^{-}(M)\text{\hspace{1em}(12.64)}$$

where ${\mathfrak{D}}_{+}\equiv \mathfrak{D}{|}_{{\Omega }^{+}(M)}$ . The index of the signature complex is

$$\begin{array}{rl}\mathrm{ind}{\mathfrak{D}}_{+}& =\dim \ker {\mathfrak{D}}_{+}-\dim \ker {\mathfrak{D}}_{-}\\ & =\dim \mathrm{Harm}(M{)}^{+}-\dim \mathrm{Harm}(M{)}^{-}\end{array}\text{\hspace{1em}(12.65)}$$

where ${\mathfrak{D}}_{-}\equiv {\mathfrak{D}}_{+}^{†}:{\Omega }^{-}(M)\to {\Omega }^{+}(M)$ and $\mathrm{Harm}(M{)}^{\pm }\equiv \{\omega \in {\Omega }^{\pm }(M)|{\mathfrak{D}}_{\pm }\omega =0\}$ . On the RHS of (12.65), all the contributions except those from the harmonic $l$ -forms cancel out. To see this, we separate $\ker {\mathfrak{D}}_{+}$ and $\ker {\mathfrak{D}}_{-}$ as

$$\ker {\mathfrak{D}}_{\pm }={\mathrm{Harm}}^l(M{)}^{\pm }\oplus \sum _{0\le r<l}[{\mathrm{Harm}}^r(M{)}^{\pm }\oplus {\mathrm{Harm}}^{m-r}(M{)}^{\pm }]$$

where ${\mathrm{Harm}}^r(M{)}^{\pm }\equiv \mathrm{Harm}(M{)}^{\pm }\cap {\Omega }^r(M)$ . If $\omega \in {\mathrm{Harm}}^r(M)$ , we have $\omega \pm \pi \omega \in {\mathrm{Harm}}^r(M{)}^{\pm }\oplus {\mathrm{Harm}}^{m-r}(M{)}^{\pm }$ . Then a map $\omega +\pi \omega \to \omega -\pi \omega$ defines an isomorphism between ${\mathrm{Harm}}^r(M{)}^{+}\oplus {\mathrm{Harm}}^{m-r}(M{)}^{+}$ and ${\mathrm{Harm}}^r(M{)}^{-}\oplus {\mathrm{Harm}}^{m-r}(M{)}^{-}$ . Now the index simplifies as

$$\mathrm{ind}{\mathfrak{D}}_{+}=\dim {\mathrm{Harm}}^{2k}(M{)}^{+}-\dim {\mathrm{Harm}}^{2k}(M{)}^{-}\text{\hspace{1em}(12.66)}$$

where we put $l=2k$ as before (the index vanishes if $l$ is odd). It is important to note that ${\mathrm{Harm}}^{2k}(M{)}^{\pm }={\mathrm{Harm}}_{\pm }^{2k}(M)$ since $\pi =\ast$ in ${\mathrm{Harm}}^{2k}(M)$ , see (12.54). Now the index (12.66) reduces to the Hirzebruch signature,

$$\mathrm{ind}{\mathfrak{D}}_{+}=\tau (M).\text{\hspace{1em}(12.67)}$$

The derivation of the topological index is rather technical and we simply quote the result from Shananan (1978). Let ${\wedge }^{\pm }{T}^{\ast }{M}^{\mathbb{C}}$ be the subspace of $\wedge T^{\ast }{M}^{\mathbb{C}}$ such that ${\Omega }^{\pm }(M)=\Gamma (M,{\wedge }^{\pm }{T}^{\ast }{M}^{\mathbb{C}})$ . Then we have

$$\begin{array}{rl}\text{t o p o l o g i c a l i n d e x}& =(-1{)}^l{\int }_M\mathrm{ch}\left({\wedge }^{+}{T}^{\ast }{M}^{\mathbb{C}}-{\wedge }^{-}{T}^{\ast }{M}^{\mathbb{C}}\right)\frac{\mathrm{Td}(T{M}^{\mathbb{C}})}{e(TM)}{|}_{\text{v o l}}\\ & =2^l{\int }_M\prod _{i=1}^l\frac{x_i/2}{\tanh x_i/2}{|}_{\mathrm{vol}}={\int }_M\prod _{i=1}^l\frac{x_i}{\tanh x_i}{|}_{\mathrm{vol}}\end{array}$$

where the last equality is true only for the $2l$ -forms in the expansion and $x_i=x_i(T{M}^{\mathbb{C}})$ . Now we have obtained the Hirzebruch signature theorem

$$\tau (M)={\int }_{M}L(TM){|}_{\text{v o l}}\text{\hspace{1em}(12.68)}$$

where $L$ is the Hirzebruch $L$ -polynomial defined by (11.91). Since $L$ is even in $x_i$ , $\tau (M)$ vanishes if $m=2\mathrm{mod}4$ . For example, $\tau (M)=0$ for $m=2$ . If $m=4$ , we have

$$\tau (M)={\int }_M\frac{1}{3}{p}_1(TM)=-\frac{1}{24{\pi }^2}\int \mathrm{tr}{\mathcal{R}}^2.\text{\hspace{1em}(12.69)}$$

As in the case of the Dolbeault complex, we may twist the signature complex, see Eguchi et al (1980), for example.

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