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The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once one establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, $$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), one obtains Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theoremOn the generalized Gauss-Bonnet theorem.

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once one establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, $$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), one obtains Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once one establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, $$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), one obtains Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

corrected the construction by introducing the normalized tangent vector field
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Robert Bryant
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The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once you knowone establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $X:M_\epsilon\to \mathsf{S}(M)$$U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $X^*\Omega = X^*(\mathrm{d}\Pi)$$U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, we have $$ \int_{M_{\epsilon}}X^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} X^*\Pi. $$$$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $X^*\Pi$$U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $X$$U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), we haveone obtains Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once you know this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $X:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $X^*\Omega = X^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, we have $$ \int_{M_{\epsilon}}X^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} X^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $X^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $X$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), we have Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once one establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, $$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), one obtains Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

Fixed an error and expanded the explanation of the index calculation
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Robert Bryant
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The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once you know this, theChern's proof of the Chern-GaussGauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $X:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $X^*\Omega = X^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, we have $$ \int_{M_{\epsilon}}X^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} X^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ andwhile the $i$-th summand on the right-hand side converges to the integral over the unit sphere $\mathsf{S}_{z_i}(M)$index of $X$ at $z_i$. (This is because $X^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $X$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$. (Of course, 'converges' is a little silly in this case because these integrals are, in fact, constant, independent of $\epsilon$ because they are integers.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), we have Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at theChern's definition of $\Omega$$\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once you know this, the proof of the Chern-Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $X:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $X^*\Omega = X^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, we have $$ \int_{M_{\epsilon}}X^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} X^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ and the $i$-th summand on the right-hand side converges to the integral over the unit sphere $\mathsf{S}_{z_i}(M)$ of the pullback of the volume form of $\mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $X$, whose degree is, by definition, the index of $X$ at $z_i$. (Of course, 'converges' is a little silly in this case because these integrals are, in fact, constant, independent of $\epsilon$ because they are integers.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), we have Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at the definition of $\Omega$, which uses the Pfaffian. This comes out of the computation that Chern does, and it is essentially a geometric fact. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once you know this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $X:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $X^*\Omega = X^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, we have $$ \int_{M_{\epsilon}}X^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} X^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $X^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $X$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), we have Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question On the generalized Gauss-Bonnet theorem.

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Robert Bryant
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