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Let the Gauss-Bonnet form be $\Omega\propto\text{Pf}(\Omega^i{}_j)$ with $\Omega^i{}_j$ the curvature 2-form of an even-dimensional manifold with dim=$n$. The Gauss-Bonnet form is exact, as shown in the explicit construction in Chern 1944 (also available here). We may write $\Omega = d\Pi$ with $\Pi$ a rank $n-1$ form. Chern gave a construction for $\Pi$ in terms of an arbitrary vector field.

Compare this to the Chern-Simons form, as developed in Chern and Simons 1974. The Chern-Simons form is also exact, and Chern and Simons 1974 give an explicit construction without any arbitrary vector field—just in terms of the connection 1-form and the curvature 2-form.

Question: is there a modern version of Chern 1944 which, similar to Chern and Simons 1974, gives an explicit construction for $\Pi$ without reference to an arbitrary choice of vector field, but rather is only in terms of the connection 1-form and curvature 2-form?

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  • $\begingroup$ For surfaces, $\Pi$ is the connection 1-form. $\endgroup$
    – Ben McKay
    Feb 3, 2015 at 14:56
  • $\begingroup$ What about for dimension 4 or higher? $\endgroup$ Feb 3, 2015 at 15:27
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    $\begingroup$ Chern's 1944 construction of $\Pi$ does not depend on a choice of vector field. He defines $\Pi$ as a canonical $n{-}1$ form on $M^{2n-1}=\mathsf{S}(R)$, the unit sphere bundle of the oriented Riemannian $n$-manifold $R^n$ (where $n$ is even) and shows that $\mathrm{d}\Pi=\pi^*\Omega$, where $\Omega$ is the Gauss-Bonnet form on $R^n$ and $\pi:M\to R$ is the base-point mapping. The choice of a vector field only comes into play at the end, when he uses it to get a section $V$ of the bundle $\pi:M\to R$ away from a single point, where he then uses other properties of $\Pi$ to prove the theorem. $\endgroup$ Feb 3, 2015 at 16:02
  • $\begingroup$ @RobertBryant yes, I did not say it 'depends on', just that it is 'in terms of'—but it indeed must be independent. Since it is independent, what is the explicit expression without reference to the vector? $\endgroup$ Feb 3, 2015 at 16:12
  • $\begingroup$ @duetosymmetry: Chern 1944 gives the formula for $\Pi$ explicitly as a sum of $n/2$ terms, and it involves factorial coefficients and factors of $\pi^{n/2}$. It doesn't make sense for me to type it out here, because you have Chern's paper in front of you. See equation $(24)$ and look back at $(15)$ for the definition of the individual $\Phi_m$. One could, of course, pull $\Pi$ back to the $\mathrm{SO}(m)$ frame bundle, where it would become a polynomial with constant coefficients in the connection and curvature forms, but Chern doesn't do it that way; he wants it on the sphere bundle. $\endgroup$ Feb 3, 2015 at 16:26

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I see that the OP may not be entirely convinced by my comments, so let me try this, which may help. It's understandable that reading the older literature can be confusing; the classical language is often quite different from ours. Here is a slightly different interpretation of Chern's construction of the transgressed form $\Pi$ that may make it clearer what is going on.

Let $n=2p>0$ be an even integer and let $(R^n,g)$ be an oriented Riemannian $n$-manifold. Let $\pi:B\to R$ be the principal right $\mathrm{SO}(n)$-bundle consisting of the oriented $g$-orthonormal frames on $R$, i.e., each $e\in B$ with $\pi(e)=x\in R$ is of the form $e = (e_1,\ldots e_n)$ where $(e_1,\ldots,e_n)$ form an oriented, $g$-orthonormal basis of $T_xR$.

As usual, one has the tautological $1$-forms $\omega_i$ defined by $\omega_i(v) = e_i\cdot \pi'(v)$ for $v\in T_eB$. There exist unique connection $1$-forms $\omega_{ij}=-\omega_{ji}$ on $B$ satisfying the first structure equations, $$\mathrm{d}\omega_i = - \omega_{ij}\wedge\omega_j.$$ (NB: My $\omega_{ij}$ are the negatives of Chern's $\omega_{ij}$. I'm sorry about that, but I can't switch back to Chern's conventions without getting confused. On the bright side, the first structure equations actually play no role in the sequel, which is why the Gauss-Bonnet formula for the Euler class works for any $\mathrm{SO}(n)$-connection on any oriented orthogonal bundle over $R$, not just the tangent bundle.)

The curvature $2$-forms are defined by the second structure equations $$\Omega_{ij} = \mathrm{d}\omega_{ij} + \omega_{ik}\wedge\omega_{kj} = \tfrac12 R_{ijkl}\,\omega^k\wedge\omega^l, $$ and they satisfy the Bianchi identities $\mathrm{d}\Omega_{ij} = \Omega_{ik}\wedge\omega_{kj}-\omega_{ik}\wedge\Omega_{kj}$.

The Gauss-Bonnet $n$-form on $B$ is $$ \tilde\Omega = \frac1{2^{n}\pi^{n/2}(n/2)!}\ \epsilon_{i_1i_2\cdots i_n} \Omega_{i_1i_2}\wedge\Omega_{i_3i_4}\wedge\cdots\wedge \Omega_{i_{n-1}i_n}\ , $$ where the sum is over all permutations in $S_n$. The Bianchi identities imply that $\mathrm{d}\tilde\Omega=0$, and, since $\tilde\Omega$ is a multiple of $\omega_1\wedge\cdots\wedge\omega_n$, it follows that there exists a unique $n$-form $\bar\Omega$ on $M$ such that $\pi^*\bar\Omega = \tilde\Omega$.

All this is standard, except that, for clarity, I am distinguishing $\tilde\Omega$ and $\bar\Omega$. (Chern uses the same letter $\Omega$ for both.) Similarly, below, I will try to distinguish forms that Chern identifies when they differ via pullback under a submersion with connected fibers.

What Chern does next, though, is what seems to be causing the confusion about 'arbitrary vector fields'. He lets $u = (u_i):S^{n-1}\to \mathbb{R}^n$ denote the inclusion of the $(n{-}1)$-sphere into $\mathbb{R}^n$, and he defines a submersion $\bar\pi: B\times S^{n-1}\to M^{2n-1}$, where $M^{2n-1}\subset TR$ is the unit sphere bundle, by $\bar\pi(e,u) = u_ie_i$. He then constructs an $(n{-}1)$-form $\tilde\Pi$ on $B\times S^{n-1}$ that has the property that $\mathrm{d}\tilde\Pi = \tilde\Omega$ and has the property that there exists an $(n{-}1)$-form $\Pi$ on $M$ (not $R$) satisfying $\bar\pi^*\Pi = \tilde\Pi$. It is important to recognize that Chern's $u_i$ are not the components of a vector field on anything, arbitrary or otherwise.

N.B.: Actually, Chern says that he is going to work locally on an open subset (say) $O\subset R$ and choose a section of $B$ over $U$, i.e., an 'oriented orthonormal frame field' on $O$, he identifies the part of $M$ that lies over $O$ with $O\times S^{n-1}$ and constructs $\Pi$ on $O\times S^{n-1}$, and then he says that the resulting $\Pi$ doesn't depend on the local choice of frame field. However, he never actually uses the local section in any of his calculations, so they are perfectly valid on $B\times S^{n-1}$.

Now, there's a way to avoid introducing the $u_i$ at all, and you may like this better. Consider the submersion $\pi_1:B\to M$ defined by $\pi_1(e) = e_1$. The fibers of this map are (connected) $\mathrm{SO}(n{-}1)$-orbits that are the leaves of the system $$ \omega_1=\omega_2=\cdots=\omega_n=\omega_{12}=\omega_{13}=\cdots=\omega_{1n}=0. $$ Then one constructs an $(n{-}1)$-form $\tilde\Pi$ directly on $B$ that is a polynomial with constant coefficients in the forms $$ \{\omega_{12},\omega_{13},\cdots,\omega_{1n}\}\cup \bigl\{\ \Omega_{ij}\ \bigl|\ 1<i,j\le n\ \bigr\} $$ and that will satisfy $\mathrm{d}\tilde\Pi = \tilde\Omega$. This $\tilde \Pi$ will be the $\pi_1$-pullback of a unique form $\Pi$ on $M$ that does the job.

For example, as Ben remarked, when $n=2$, you can take $$\tilde\Pi = \frac{\omega_{12}}{2\pi},$$ since, in that case, we have $$ \mathrm{d}\left(\frac{\omega_{12}}{2\pi}\right) = \frac{\Omega_{12}}{2\pi} = \frac{K\ dA}{2\pi} $$ When $n=4$, you can take $$ \tilde\Pi = \frac{2\,\omega_{12}\wedge\omega_{13}\wedge\omega_{14} +\bigl(\omega_{12}\wedge\Omega_{34}+\omega_{13}\wedge\Omega_{42} +\omega_{14}\wedge\Omega_{23}\bigr)}{(2\pi)^2} $$ and verify, using the structure equations and the Bianchi identities, that $$ \mathrm{d}\tilde\Pi = \frac{ \bigl(\Omega_{12}\wedge\Omega_{34}+\Omega_{13}\wedge\Omega_{42} +\Omega_{14}\wedge\Omega_{23}\bigr)}{(2\pi)^2} = \tilde\Omega $$

Of course, Chern computes the explicit formula for all $n$. To put Chern's general formula in the above form, you can just set $u_1=1$, $u_2=u_3=\cdots=u_n=0$ in Chern's formulae (and switch the signs appropriately because my $\omega_{ij}$ are the negatives of Chern's $\omega_{ij}$. That's all there is to it.

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  • $\begingroup$ Thanks, Robert, for putting in the effort to write this didactic answer. It really is helpful, and gets almost all the way where I wanted to go. I know fully well that none of the expressions depend on an arbitrary choice of coordinate system. This can be easily seen for the SO(n) invariance of the frame indices in $\tilde{\Omega}$. However, in the examples, this is not transparent. For $n=2$ it's clear that we can take $\tilde{\Pi}\propto\epsilon^{ij}\omega_{ij}$. For $n=4$, it's also clear that we can write the rightmost term as $\tilde{\Pi}\supset\epsilon^{ijkl}\omega_{ij}\wedge\Omega_{ij}$ $\endgroup$ Feb 8, 2015 at 20:17
  • $\begingroup$ This latter expression clearly has a frame SO(n) invariance. However, I can't for the life of me deduce what frame-independent expression leads to the first term (which contains $\omega\wedge\omega\wedge\omega$). Do you have any hits for how to proceed to find such an expression for the first term? $\endgroup$ Feb 8, 2015 at 20:22
  • $\begingroup$ @duetosymmetry: (Responding to your first comment) Actually, $\mathrm{SO}(n)$-invariance does not hold for any of the terms in $\tilde \Pi$. For example, note that, when $n=4$, that second term only involves the $\omega_{1k}$ for $k=2,3,4$, and not all $6$ of the $\omega_{ij}$, as it would have to if it were fully $\mathrm{SO}(4)$-invariant. In fact, the terms in $\tilde \Pi$ are only $\mathrm{SO}(n{-}1)$-invariant, which makes sense because the fibers of $\pi_1:B\to M$ are $\mathrm{SO}(n{-}1)$-orbits. If you are still doubtful, when I have time, I'll put in a more explicit explanation. $\endgroup$ Feb 9, 2015 at 12:42
  • $\begingroup$ this is exactly what confuses me. Shouldn't there exist an SO($n$) invariant expression, like there does for the Chern-Simons form? $\endgroup$ Feb 9, 2015 at 15:28
  • $\begingroup$ @duetosymmetry: Ah, I see what's bothering you. Yes, of course, there is an $\mathrm{SO}(n)$-invariant expression on $B$, say $\tilde\Phi$ whose exterior derivative is $\tilde\Omega$, but it is useless for proving the Gauss-Bonnet theorem via Chern's method because $\tilde\Phi$ does not descend to $M$. Chern's point is that $\tilde \Pi$ does descend to $M$, which provides the key link between the Gauss-Bonnet integral and the sum of indices of a vector field on $R$ with isolated zeros because $\tilde \Pi$ pulls back to each fiber of $M\to R$ to be the volume form on that fiber. $\endgroup$ Feb 9, 2015 at 17:04

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