[Nakahara GTP] 11.4 Pontrjagin and Euler classes

 This article is one of Manifold, Differential Geometry, Fibre Bundle.

11.3.1 Pontrjagin classes

Let $E$ be a real vector bundle over an $m$-dimensional manifold $M$ with $\dim_\mathbb{R}E=k$. If $E$ is endowed with the fibre metric, we may introduce orthonormal frames at each fibre. The structure group may be reduced to $O(k)$ from $GL(k,\mathbb{R})$. Since the generators of $\mathfrak{o}(k)$ are skew symmetric, the field strength $\mathcal{F}$ of $E$ is also skew symmetric. A skew-symmetry matrix $A$ is not diagonalizable by element of a subgroup of $GL(k,\mathbb{R})$. It is, however, reducible to block diagonal form as $$\begin{align}A\rightarrow \begin{pmatrix}0&\lambda_1& & & 0\\ -\lambda_1 &0& & & \\ & & 0 & \lambda_2 & \\ & & -\lambda_2&0& \\ 0 & & & & \ddots\end{pmatrix}\rightarrow \begin{pmatrix}-\lambda_1& & & & \\ &-i\lambda_1& &0& \\ & & i\lambda_2& & \\ & 0& & -i\lambda_2& \\ &  & & & \ddots\end{pmatrix}\end{align}$$ where the second diagonalization is achieved only by an element of $GL(k,\mathbb{C})$. If $k$ is odd, the last diagonal element is set to zero. For example, the generator of $\mathfrak{o}(3)=\mathfrak{s}\mathfrak{o}(3)$ generating rotations around the $z$-axis is $$T_z=\begin{pmatrix}0&1&0\\ -1&0&0\\ 0&0&0\end{pmatrix}.$$

The total Pontrjagin class is defined by $$\begin{align}p(\mathcal{F})\equiv \det \left( I+\frac{\mathcal{F}}{2\pi}\right).\end{align}$$ From the skew symmetry $\mathcal{F}^t=-\mathcal{F}$, it follows that $$\det \left( I+\frac{\mathcal{F}}{2\pi}\right)=\det \left( I+\frac{\mathcal{F}^t}{2\pi}\right)=\det\left( I-\frac{\mathcal{F}}{2\pi}\right).$$ Therefore, $p(\mathcal{F})$ is an even function in $\mathcal{F}$. The expansion of $p(\mathcal{F})$ is $$\begin{align}p(\mathcal{F})=1+p_1(\mathcal{F})+p_2(\mathcal{F})+\cdots\end{align}$$ where $p_j(\mathcal{F})$ is a polynomial of order $2j$ and is an element of $H^{4j}(M;\mathbb{R})$. We note that $p_j(\mathcal{F})=0$ for either $2j>k=\dim E$ or $4j>\dim M$.

Let us diagonalize $\mathcal{F}/2\pi$ as $$\begin{align}\frac{\mathcal{F}}{2\pi}\rightarrow A\equiv \begin{pmatrix}-ix_1& & & & \\ &ix_1& &0& \\ & & -ix_2& & \\ & 0& & -ix_2& \\ &  & & & \ddots\end{pmatrix}\end{align}$$ where $x_k\equiv -\lambda_k/2\pi$, $\lambda_k$ being the eigenvalues of $\mathcal{F}$. The sign has been chosen to simplify the Euler calss defined here. The generating function of $p(\mathcal{F})$ is given by $$\begin{align}p(\mathcal{F})=\det(I+A)=\prod^{[k/2]}_{i=1}(1+x^2_i)\end{align}$$ where $$[k/2]=\rightarrow \begin{cases}k/2& \mbox{if }k\mbox{ is even}\\ (k-1)/2& \mbox{if }k\mbox{ is odd.}\end{cases}$$ In (5) only even powers appear, reflecting the skew symmetry. Each Pontrjagin class is computed from (5) as $$\begin{align}p_j(\mathcal{F})=\sum^{[k/2]}_{i_1<i_2<\cdots<i_j}x^2_{i_1}x^2_{i_2}\cdots x^2_{i_j}.\end{align}$$ To write $p_j(\mathcal{F})$ in terms of the curvature two-form $\mathcal{F}/2\pi$, we first note that $$\mbox{tr}\left(\frac{\mathcal{F}}{2\pi}\right)^{2j}=\mbox{tr}A^{2j}=2(-1)^j\sum^{[k/2]}_{i=1}x^{2j}_i.$$ It then follows that $$\begin{align}p_1(\mathcal{F})=\sum_ix^2_i=-\frac{1}{2}\left(\frac{1}{2\pi}\right)^2\mbox{tr}\mathcal{F}^2\\ p_2(\mathcal{F})=\sum_{i<j}x^2_ix^2_j=\frac{1}{2}\left[ \left(\sum_ix^2_i\right)^2-\sum_ix^4_i\right]\nonumber\\ =\frac{1}{8}\left(\frac{1}{2\pi}\right)^4[(\mbox{tr}\mathcal{F}^2)^2-2\mbox{tr}\mathcal{F}^4]\\ p_2(\mathcal{F})=\sum_{i<j<k}x^2_ix^2_jx_k^2 =\frac{1}{48}\left(\frac{1}{2\pi}\right)^6[-(\mbox{tr}\mathcal{F}^2)^3+6\mbox{tr}\mathcal{F}^2\mbox{tr}\mathcal{F}^4-8\mbox{tr}\mathcal{F}^6]\\ \vdots\nonumber\\ p_{[k/2]}(\mathcal{F})=x_1^2x_2^2\cdots x_{[k/2]}^2=\left(\frac{1}{2\pi}\right)^2\det \mathcal{F}.\end{align}$$

It is easy to guess that the Pontrjagin classes are written in terms of Chern classes. Since Chern classes are defined only for complex vector bundles, we must complexify the fibre of $E$ so the complex numbers make sense. The resulting vector bundle is denoted by $E^\mathbb{C}$. Let $A$ be a skew-symmetric real matrix. We find that $$\det (I+iA)=\det \begin{pmatrix}1+x_1& & & & \\ &1-x_1& &0& \\ & & 1+x_2& & \\ & 0& & 1-x_2& \\ &  & & & \ddots\end{pmatrix}\\ =\prod^{[k/2]}_{i=1}(1-x^2_i)=1-p_1(A)+p_2(A)-\cdots$$ from which it follows that $$\begin{align}p_j(E)=(-1)^jc_{2j}(E^\mathbb{C}).\end{align}$$

Example 11.5. Let $M$ be a four-dimensional Riemannian manifold. When the orthonormal frame $\{\hat{e}_\alpha\}$ is employed, the structure group of the tangent bundle $TM$ may be reduced to $O(4)$. Let $\mathcal{R}=\frac{1}{2}\mathcal{R}_{\alpha\beta}\theta^\alpha\wedge\theta^\beta$ be the curvature two-form. For the tangent bundle, it is common to write $p(M)$ instead of $p(\mathcal{R})$. We have $$\begin{align}\det\left( I+\frac{\mathcal{R}}{2\pi}\right)=1-\frac{1}{8\pi^2}\mbox{tr}\mathcal{R}^2+\frac{1}{128\pi^4}[(\mbox{tr}\mathcal{R}^2)^2-2\mbox{tr}\mathcal{R}^4].\end{align}$$ Each Pontrjagin class is given by $$\begin{align}p_(M)=1\\ p_1(M)=-\frac{1}{8\pi^2}\mbox{tr}\mathcal{R}^2=-\frac{1}{8\pi^2}\mathcal{R}_{\alpha\beta}\mathcal{R}_{\beta\alpha}\\ p_2(M)=\frac{1}{128\pi^4}[(\mbox{tr}\mathcal{R}^2)^2-2\mbox{tr}\mathcal{R}^4]=\left( \frac{1}{2\pi}\right)^4\det \mathcal{R}.\end{align}$$ Although $p_2(M)$ vanishes as a differential form, we need in the next subsection to compute the Euler class.

11.4.2 Euler classes

Let $M$ be a $2l$-dimensional orientable Riemannian manifold and let $TM$ be the tangent bundle of $M$. We denote the curvature by $\mathcal{R}$. It is always possible to reduce the structure group of $TM$ down to $SO(2l)$ by employing an orthonormal frame. The Euler class $e$ of $M$ is defined by the square root of the $4l$-form $p_l$, $$\begin{align}e(A)e(A)=p_l(A).\end{align}$$ Both sides should be understood as functions of a $2l\times 2l$ matrix $A$ and not of the curvature $\mathcal{R}$, since $p_1(\mathcal{R})$ vanishes identically. However, $e(M)\equiv e(\mathcal{R})$ thus defined is a $2l$-form and, indeed, gives a volume element of $M$. If $M$ is an odd-dimensional manifold we define $e(M)=0$, see later.

Example 11.6. Let $M=S^2$ and consider the tangent bundle $TS^2$. From example 7.14, we find the curvature two-form, $$\mathcal{R}_{\theta\phi}=-\mathcal{R}_{\phi\theta}=\sin^2\theta\frac{d\theta\wedge d\phi}{\sin \theta}=\sin \theta d\theta\wedge d\phi$$ where we have noted that $g_{\theta\theta}=\sin^2\theta$. Although $p_1(S^2)=0$ as a differential form, we compute it to find the Euler form. We have $$p_1(S^2)=-\frac{1}{8\pi^2}\mbox{tr}\mathcal{R}^2=-\frac{1}{8\pi^2}[\mathcal{R}_{\theta\phi}\mathcal{R}_{\phi\theta}+\mathcal{R}_{\phi\theta}\mathcal{R}_{\theta\phi}]\\ =\left(\frac{1}{2\pi}\sin\theta d\theta\wedge d\phi\right)^2$$ from which we read off $$\begin{align}e(S^2)=\frac{1}{2\pi}\sin\theta d\theta \wedge d\phi.\end{align}$$

It is interesting to note that $$\begin{align}\int_{S^2}e(S^2)=\frac{1}{2\pi}\int^{2\pi}_0d\phi\int^\pi_0d\theta\sin\theta=2\end{align}$$ which is the Euler characteristic of $S^2$. This is not just a coincidence. Let us take another convincing example, a torus $T^2$. Since $T^2$ admits a flat connection, the curvature vanishes identically. It the follows that $e(T^2)\equiv 0$ and $\chi(T^2)=0$. These are special cases of the Gauss-Bonnet theorem, $$\begin{align}\int_M e(M)=\chi(M)\end{align}$$ for a compact orientable manifold $M$. If $M$ is odd dimensional both $e$ and $\chi$ vanish.

In general, the determinant of a $2l\times 2l$ skew-symmetric matrix $A$ is a square of a polynomial called the Pfaffian $\mbox{Pf}(A)$, $$\begin{align}\det A=\mbox{Pf}(A)^2.\end{align}$$ We show that the Pfaffian is given by $$\begin{align}\mbox{Pf}(A)=\frac{(-1)^l}{2^ll!}\sum_p\mbox{sgn}(P)A_{P(1)P(2)}A_{P(3)P(4)}\cdots A_{P(2l-1)P(2l)}\end{align}$$ where the phase has been chosen for later convenience. We first note that a skew-symmetric matrix $A$ can be block diagonalized by an element of $O(2l)$ as $$\begin{align}S^\dagger AS=\Lambda=\begin{pmatrix}0&\lambda_1& & & & & \\-\lambda_1 & 0& & & & 0& \\ & & 0& \lambda_2& & & \\ & &-\lambda_2&0& & & \\ & & & & \ddots& & \\ & 0& & & & 0& \lambda_l\\ & & & & & -\lambda_l& 0\end{pmatrix}.\end{align}$$ It is easy to see that $$\det A=\det \Lambda =\prod^l_{i=1}\lambda^2_i.$$ To compute $\mbox{Pf}(\Lambda)$, we note that the non-vanishing terms in (21) are of form $A_{12}A_{34}\cdots A_{2l-1,2l}$. Moreover, there are $2^l$ ways of changing the suffices as $A_{ij}\rightarrow A_{ji}$, such as $$A_{12}A_{34}\cdots A_{2l-1,2l}\rightarrow A_{21}A_{34}\cdots A_{2l-1,2l}$$ and $l!$ permutations of the pairs of indices, for example, $$A_{12}A_{34}\cdots A_{2l-1,2l}\rightarrow A_{34}A_{12}\cdots A_{2l-1,2l}.$$ Hence, we have $$\mbox{Pf}(\Lambda)=(-1)^lA_{12}A_{34}\cdots A_{2l-1,2l}=(-1)^l\prod^l_{i=1}\lambda_i.$$ Thus, we conclude that a block diagonal matrix $\Lambda$ satisfies $$\det \Lambda=\mbox{Pf}(\Lambda)^2.$$ To show that (20) is ture for any skew-symmetric matrices (not necessarily block diagonal) we use the following lemma, $$\begin{align}\mbox{Pf}(X^tAX)=\mbox{Pf}(A)\det X.\end{align}$$ If $S^t AS=\Lambda$ for $S\in O(2l)$, we have $A=S\Lambda S^t$, hence $$\mbox{Pf}(S\Lambda S^t)=\mbox{Pf}(\Lambda)\det S=(-1)^l\prod^l_{i=1}\lambda_i \det S.$$ We finally find $\det A=\mbox{Pf}(A)^2$ for a skew-symmetric matrix $A$.

Note that $\mbox{Pf}(A)$ is $SO(2l)$ invariant but changes sign under an improper rotation $S$ ($\det S=-1$) of $O(2l)$.

Exercise 11.3. Show that the determinant of an odd-dimensional skew-symmetric matric vanishes. This is why we put $e(M)=0$ for an odd-dimensional manifold.

The Euler class is defined in terms of the curvature $\mathcal{R}$ as $$\begin{align}e(M)=\mbox{Pf}(\mathcal{R}/2\pi)\nonumber\\ =\frac{(-1)^l}{(4\pi)^ll!}\sum_P \mbox{sgn}(P)\mathcal{R}_{P(1)P(2)}\cdots \mathcal{R}_{P(2l-1)P(2l)}.\end{align}$$ The generating function is obtained by taking $x_j=-\lambda_i/2\pi$, $$\begin{align}e(x)=x_1x_2\cdots x_l=\prod^l_{i=1}x_i.\end{align}$$ The phase $(-1)^l$ has been chosen to simplify the RHS.

Example 11.7. Let $M$ be a four-dimensional orientable manifold. The structure group of $TM$ is $SO(4)$, see example 11.5. The Euler class is obtained from (24) as $$\begin{align}e(M)=\frac{1}{2(4\pi)^2}\epsilon^{ijkl}\mathcal{R}_{ij}\wedge\mathcal{R}_{kl}.\end{align}$$ This is in agreement with the result of example 11.5. The relevent Pontrjagin class is $$p_2(M)=\frac{1}{128\pi^4}[(\mbox{tr}\mathcal{R}^2)^2-2\mbox{tr}\mathcal{R}^4]=x^2_1x^2_2.$$ Since $e(M)=x_1x_2$, we have $p_2(M)=e(M)\wedge e(M)$. This is written as a matrix identity, $$\frac{1}{128\pi^4}[(\mbox{tr}A^2)^2-2\mbox{tr}A^4]=\left( \frac{1}{2(4\pi)^4}\epsilon^{ijkl}A_{ij}A_{kl}\right)^2.$$

11.4.3 Hirzebruch $L$-polynomial and $\hat{A}$-genus

The Hizebruch $L$-polynomial is defined by $$\begin{align}L(x)=\prod^k_{j=1}\frac{x_j}{\tanh x_j}\nonumber\\ =\prod^k_{j=1}\left( 1+\sum_{n\ge 1}(-1)^{n-1}\frac{2^{n2}}{(2n)!}B_nx^{2n}_j\right)\end{align}$$ where the $B_n$ are Bernoulli numbers, see (11.3.3). The function $L(x)$ is even in $x_j$ and can be written in terms of the Pontrjagin classes, $$\begin{align}L(\mathcal{F})=1+\frac{1}{3}p_1+\frac{1}{45}(-p^2_1+7p_2)+\frac{1}{945}(2p^3_1-13p_1p_2+62p_3)+\cdots\end{align}$$ where $p_j$ stands for $p_j(\mathcal{F})$. From the splitting principle, we find that $$\begin{align}L(E\oplus F)=L(E)\wedge L(F).\end{align}$$

The $\hat{A}$ ($A$-roof) genus $\hat{A}(\mathcal{F})$ is defined by $$\begin{align}\hat{A}(\mathcal{F})=\prod^k_{j=1}\frac{x_j/2}{\sinh (x_j/2)}\nonumber\ \prod^k_{j=1}\left( 1+\sum_{n\ge 1}(-1)^n\frac{(2^{2n}-2)}{(2n)!}B_nx^{2n}_j\right).\end{align}$$ This is an even function of $x_j$ and can be expanded in $p_j$. $\hat{A}$ is also called the Dirac genus by physicists. It satisfies $$\begin{align}\hat{A}(E\oplus F)=\hat{A}(E)\wedge \hat{A}(F).\end{align}$$ $\hat{A}$ is written in terms of the Pontrjagin classes as $$\begin{align}\hat{A}(\mathcal{F})=1-\frac{1}{24}p_1+\frac{1}{5760}(7p_1^2-4p_2)\nonumber\\ +\frac{1}{967680}(-31p_1^3+44p_1p_2-16p_3)+\cdots.\end{align}$$

Example 11.8. Let $M$ be a compact connected and orientable four-dimensional manifodl. Let us consider the symmetric bilinear form $\sigma:H^2(M;\mathbb{R})\times H^2(M;\mathbb{R})\rightarrow \mathbb{R}$ defined by $$\begin{align}\sigma([\alpha],[\beta])=\int_M\alpha\wedge\beta.\end{align}$$ $\sigma$ is a $b^2\times b^2$ symmetric matrix where $b^2=\dim H^2(M;\mathbb{R})$ is the Betti number. Clearly $\sigma$ is non-degenerate since $\sigma([\alpha],[\beta])=0$ for any $[\alpha]\in H^2(M;\mathbb{R})$ implies $[\beta]=0$. Let $p$ ($q$) be the number of positive (negative) eigenvalues of $\sigma$. The Hirzebruch signature of $M$ is $$\tau (M)\equiv p-q.$$ According to the Hirzebruch signature theorem, this number is also given in terms of the $L$-polynomial as $$\begin{align}\tau(M)=\int_ML_1(M)=\frac{1}{3}\int_Mp_1(M).\end{align}$$