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](http://www.jstor.org/stable/1969302) (also available [here](http://www.maths.ed.ac.uk/~aar/papers/chern7.pdf)). 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](http://www.jstor.org/stable/1971013). 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?
> 
> 
> 



Gauss-Bonnet invariant Ω: explicit intrinsic expression for Π in Ω=dΠ?
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.


