Fibre integrals and the Thom isomorphism

The first order of business will be proving a generalization of the regular value submanifold theorem which apply to manifolds with boundary, as we will be defining fibre integrals for manifolds with and without boundary.

Lemma. If \(M\) is a smooth manifold without boundary and \(f : M \rightarrow \mathbb{R}\) is smooth, with regular values \(a\) and \(b\), then \(S = f^{-1}([a, b])\) is an embedded submanifold with boundary of \(M\) of equal dimension to \(M\). Moreover, the boundary of \(S\) is \(f^{-1}(a) \sqcup f^{-1}(b)\).

Proof. Let \(n = \dim(M)\). Consider first the subspace \(f^{-1}(a)\). Since \(a\) is a regular value, for some \(p \in f^{-1}(a)\), pick some charts \((U, \varphi)\) around \(p\) and \((V, \psi)\) around \(a\) so that \(\psi \circ f \circ \varphi^{-1}\) is projection to the last coordinate. Obviously, \(a \in V\), so without loss of generality, let \(V = (a - \varepsilon, a + \varepsilon)\). We also assume that \(U \subset f^{-1}((a - \varepsilon, a + \varepsilon))\). It then follows that

\begin{align} \varphi(U \cap f^{-1}([a, b])) &= \{ \varphi(x) \in \varphi(U) \ | \ (f \circ \varphi^{-1})(\varphi(x)) \in [a, a + \varepsilon) \} \\ &= \{(y^1, \dots, y^n) \in \varphi(U) \ | \ y^n \in \psi([a, a + \varepsilon))\} \end{align}

Since \(\psi : (a - \varepsilon, a + \varepsilon) \rightarrow \widetilde{U}\) is a homeomorphism, \(\widetilde{U} \subset \mathbb{R}\) is connected, and is therefore an open interval, \((b_0, b_1)\). We also must have \(a \pm \varepsilon\) sent to the endpoints, so \([a, a + \varepsilon)\) will be sent to either \([c, b_1)\) or \((b_0, c]\) for \(c = \psi(a)\), so \(\varphi(U \cap f^{-1}([a, b]))\) is the set of all \((y^1, \dots, y^n) \in \varphi(U)\) such that \(y^n \geq c\) or \(y^n \leq c\). In the former case, we have

\begin{equation} \varphi(U \cap f^{-1}([a, b])) = \varphi(U) \cap \{ (s^1, \dots, s^n) \in \mathbb{R}^n \ | \ s^n \geq c\} \end{equation}

so we can simply re-define \(\varphi\) to \(\widetilde{\varphi}(x) = \varphi(x) - (0, \dots, 0, c)\), which is also a valid coordinate chart, and

\begin{align} \widetilde{\varphi}(U \cap f^{-1}([a, b])) &= \left[ \varphi(U) \cap \{ (s^1, \dots, s^n) \in \mathbb{R}^n \ | \ s^n \geq c\} \right] - (0, \dots, 0, c) \\ &= \widetilde{\varphi}(U) \cap \mathbb{H}^n \end{align}

so that \(\widetilde{\varphi}\) is a valid boundary slice chart. The latter case follows similarly. It is clear that every \(p \in f^{-1}((a, b))\) is contained in a standard coordinate chart for \(M\), and we can apply a similar strategy for \(q \in f^{-1}(b)\) to get boundary slice coordinate charts here.

Remark. Clearly, we also have a variant of this lemma when \([a, b]\) is replaced with \([a, b)\), following the same proof. In this case, the boundary is simply \(f^{-1}(a)\).

Theorem. If \(f : M \rightarrow N\) is a map from smooth manifold with boundary \(M\) to smooth manifold without boundary \(N\), and if \(c \in N\) is a regular value of both \(f\) and \(f|_{\partial M}\), then \(S = f^{-1}(c)\) is a smooth submanifold with boundary, and \(\partial S = \partial M \cap S\).

Proof. To begin, note that by smoothness of \(f\), we can pick boundary chart \((U, \varphi)\) around some \(p \in \partial M \cap S\) in \(M\) (assume \(\varphi(p) = (0, \dots, 0)\)) and chart \((\psi, V)\) around \(c\) in \(N\) (assume \(\psi(c) = (0, \dots, 0)\)) such that \(\psi \circ f \circ \varphi^{-1}\) admits a smooth extension to \(\widetilde{f}\) on open \(W \subset \mathbb{R}^m\) around the origin. We can shrink \(W\) to be small enough so that \(\widetilde{f}\) is a submersion, as \(f\) is regular at \(p\), so \(\psi \circ f \circ \varphi^{-1}\) is regular at \(\varphi(p) = (0, \dots, 0)\), so \(\widetilde{f}\) is as well, and thus is regular in some neighbourhood of the origin. Additionally, we know that \(f|_{U \cap \partial M} = f|_{\partial U}\) is regular at \(p\). We also know that \(\varphi^{-1} : \varphi(U) \rightarrow U\) is a diffeomorphism and restricts to a diffeomorphism of the boundaries, so \((\psi \circ f \circ \varphi^{-1})|_{\partial \varphi(U)}\) must be regular at \(\varphi(p)\), which implies that

\begin{equation} \widetilde{f}|_{W \cap \partial \varphi(U)} = (\psi \circ f \circ \varphi^{-1})|_{W \cap \partial \varphi(U)} \end{equation}

is regular at \(\varphi(p)\), as \(W \cap \partial \varphi(U)\) is an open submanifold of \(\partial \varphi(U)\). Thus, \(\widetilde{f}|_{W \cap \partial \varphi(U)}\) is regular in some neighbourhood of \(\varphi(p)\) in \(W \cap \partial \varphi(U)\), so we can shrink \(W\) even more to assume this is the case on \(W \cap \partial \varphi(U)\) itself. Finally, we define \(\widetilde{U}\) to be \(\varphi^{-1}(W) \subset U\), so that \(\varphi(\widetilde{U}) = W \cap \mathbb{H}^n\), and \(\widetilde{f}\) and \(\widetilde{f}|_{\partial \varphi(\widetilde{U})}\) are both submersions.

We know that \(\widetilde{S} = \widetilde{f}^{-1}(0, \dots, 0) \subset \mathbb{R}^n\) is an embedded submanifold of dimension \(m - n\). Let \(\pi : W \rightarrow \mathbb{R}\) be the map which projects onto the last coordinate. Of course, \(\varphi(\widetilde{U}) = \pi^{-1}([0, \infty))\). Note that

\begin{equation} (\pi|_{\widetilde{S}})^{-1}([0, \infty)) = \pi^{-1}([0, \infty)) \cap \widetilde{f}^{-1}(0, \dots, 0) = \varphi(\widetilde{U} \cap f^{-1}(c)) = \varphi(\widetilde{U} \cap S) \end{equation}

If we can show that \(0\) is a regular value of \(\pi|_{\widetilde{S}}\), then we can apply the previous lemma to conclude that \(\varphi(\widetilde{U} \cap S)\) is an embedded submanifold with boundary \((\pi|_{\widetilde{S}})^{-1}(0) = \varphi(S \cap \partial \widetilde{U})\). It then follows that \(\widetilde{U} \cap S\) will be embedded submanifold with boundary \(S \cap \partial \widetilde{U}\). Repeating this argument at all \(p \in \partial M \cap S\), and using standard arguments for \(p \in \text{Int}(M) \cap S\) show that \(S\) is a submanifold with boundary with \(\partial S = \partial M \cap S\).

So, let's prove the final thing. Of course, \(\pi^{-1}(0)\) is a submanifold of \(W\) of dimension \(m - 1\), and \(\widetilde{f}|_{\partial \varphi(\widetilde{U})} = \widetilde{f}|_{\pi^{-1}(0)}\). Let \(j : \pi^{-1}(0) \rightarrow W\) be the inclusion map, so \(\widetilde{f}|_{\pi^{-1}(0)} = \widetilde{f} \circ j\). We will then have \(\dim \text{Im}(\widetilde{f}_{*} \circ j_{*}) = n\) everywhere on \(\pi^{-1}(0) \subset W\), by the fact that this map is a submersion. It then follows from rank-nullity that

\begin{equation} \dim \text{Ker}(\widetilde{f}_{*} \circ j_{*}) = m - n - 1 \end{equation}

If \(0\) isn't a regular value of \(\pi|_{\widetilde{S}}\), then we must be able to choose some \(q \in \widetilde{S} \cap \pi^{-1}(0)\) such that all of the tangent vectors in \(T_q \widetilde{S} = \text{Ker}(\widetilde{f}_{*, q})\) are tangent to \(\pi^{-1}(0)\). In other words, the kernel of \(\text{Ker}(\widetilde{f}_{*, q}) \subset \text{Im}(j_{*, q})\). It follows that \(j_{*, q}\) yields an isomorphism of \(\text{Ker}(\widetilde{f}_{*, q} \circ j_{*, q})\) and \(\text{Ker}(\widetilde{f}_{*, q})\), so they would have the same dimension. But we know that

\begin{equation} \dim \text{Ker}(\widetilde{f}_{*, q}) = \dim T_q \widetilde{S} = m - n \end{equation}

which is a contradiction, as the dimensions are not the same. Thus, \(0\) is, in fact, a regular value of \(\pi|_{\widetilde{S}}\), and the proof is finally complete.

---

Another basic thing I'd like to define before getting into the fibre integrals is the idea of an oriented fibre bundle.

Definition. If \(\pi : E \rightarrow B\) is a smooth fibre bundle with fibre \(F\) is said to be oriented as a fibre bundle if each submanifold \(\pi^{-1}(p) = E_p\), as well as \(F\), is endowed with an orientation, and we have a trivialization of the bundle with maps \(\Phi_U : \pi^{-1}(U) \to U \times F\) which are orientation-preserving diffeomorphisms when restricted to any fibre. In the specific case that \(E\) is a real vector bundle, we automatically equip the model fibre \(F = \mathbb{R}^n\) with the standard orientation.

Claim. A manifold \(M\) is orientable (in the usual sense) if and only if its tangent bundle \(TM\) is orientable as a fibre bundle (with fibre \(\mathbb{R}^n\)). Note that \(TM\) is alway orientable as a manifold in its own right.

Proof. If \(M\) is orientable, then it admits a continuous pointwise orientation, which is to say that each fibre \(T_p M\) admits an orientation \(\mu_p\) such that in a neighbourhood \(U\) of every point, there is a smoothly-varying frame \(Y_1, \dots, Y_n\) which gives the orientation in \(U\). It follows that at \(q\), we have orientation \([Y_{1, q}, \dots, Y_{n, q}]\) for the fibre \(T_q M\), and trivialization \(\Phi_U\) which takes tangent vector \(v_q\) to its components with respect to the frame of the \(Y_{j, q}\). Thus, the restriction of \(\Phi_U\) to a fibre \(T_q M\) is the linear map taking \(Y_{j, q}\) to \(e_j\), and hence is orientation-preserving when \(\mathbb{R}^n\) is equipped with the standard orientation.

The converse follows from the fact that the trivializations \(\Phi_U : \pi^{-1}(U) \to U \times \mathbb{R}^n\) immediately give us smoothly-varying frame consisting of \(\Phi_U^{-1}((p, e_j))\) for \(p \in U\) and \(j = 1, \dots, n\) which give the pointwise orientation on each \(T_q M\). \(\blacksquare\)

Claim. If \(\pi : E \rightarrow B\) is an oriented fibre bundle, and \(B\) is itself oriented, then there is a natural induced orientation on \(E\).

Proof. We will write down an oriented atlas for \(E\). Pick oriented atlas \((U_{\alpha}, \varphi_{\alpha})\) for \(M\) such that each \(U_{\alpha}\) is trivialized by \(\Phi_{\alpha} : \pi^{-1}(U_{\alpha}) \rightarrow U_{\alpha} \times F\) which is orientation-preserving when restricted to fibres. Let \((V_{\beta}, \psi_{\beta})\) be an oriented atlas for \(F\). Then our claim is that \((\Phi_{\alpha}^{-1}(U_{\alpha} \times V_{\beta}), (\varphi_{\alpha} \times \psi_{\beta}) \circ \Phi_{\alpha})\) is an oriented atlas for \(E\). We note that the determinant of the transition function Jacobians are of the form

\begin{align} \det D \left( (\varphi_{\alpha} \times \psi_{\beta}) \circ \Phi_{\alpha} \circ \Phi_{\alpha'}^{-1} \circ (\varphi_{\alpha'} \times \psi_{\beta'})^{-1} \right) &= \det D(\Phi_{\alpha} \circ \Phi_{\alpha'}^{-1}) \det D( (\varphi_{\alpha} \circ \varphi_{\alpha'}^{-1}) \times (\psi_{\beta} \circ \psi_{\beta'}^{-1})) \\ &= \det D(\Phi_{\alpha} \circ \Phi_{\alpha'}^{-1}) \det D(\varphi_{\alpha} \circ \varphi_{\alpha'}^{-1}) \det D(\psi_{\beta} \circ \psi_{\beta'}^{-1}) \end{align}

The final two determinants in the product are positive, as we chose oriented charts. Moreover, note that \((\Phi_{\alpha} \circ \Phi_{\alpha'}^{-1})(s, f) = (s, f')\) for some \(f'\) depending on both \(s\) and \(f\). Thus, the Jacobian is of the form

\begin{equation} D(\Phi_{\alpha} \circ \Phi_{\alpha'}^{-1}) = \begin{pmatrix} \mathbb{I} & 0 \\ A & B \end{pmatrix} \end{equation}

where \(B\), is the derivative of the transition function when we restrict to a fibre. In particular, \(\Phi_{\alpha} \circ \Phi_{\alpha'}^{-1}\) restricted to \(F\) is an orientation-preserving map from \(F\) to itself, and thus has a positive Jacobian determinant, so \(\det(B)\) is positive, and from the above formula, \(\det D(\Phi_{\alpha} \circ \Phi_{\alpha'}^{-1})\) is positive as well. \(\blacksquare\)

Remark. A fibre bundle over an oriented manifold is not necessarily orientable: just look at the example of the Mobius band (as a fibre bundle over a circle, which is orientable). We require the assumption that the fibre bundle itself is orientable (as a fibre bundle).

---

Let \(\pi : E \rightarrow B\) be a smooth fibre bundle (where \(E\) is allowed to have boundary but \(B\) is not). Of course, \(\pi\) is locally a projection to base, in the sense that for any \(p \in B\), there exists open set \(U\) around \(p\) and a diffeomorphism \(\varphi : \pi^{-1}(U) \to U \times F\) such that \(\text{proj} \circ \varphi = \pi\). Therefore, locally, we can study the projection map from \(U \times F\) to \(U\) and the restriction to boundary, projection from \(U \times \partial F\) to \(U\). Both maps are submersions, so it follows from the above theorem that the fibres \(\pi^{-1}(b)\) will be \(\dim(E) - \dim(B) = e - b = n\) dimensional submanifolds for all \(b \in B\), with boundary given by their intersection with the boundary of \(E\). Assume that \(k > n\). We assume, moreover, that \(E\) is an oriented fibre bundle. Let \(\alpha\) be a \(k\) -form on \(E\) with vertical compact support, which is to say that if \(\iota_b : \pi^{-1}(b) \to E\) is the inclusion of a fibre, then \(\iota_b^{*} \alpha\) is compact. This of course holds if \(\alpha\) just has compact support outright.

Claim. Let \(\pi : E \rightarrow B\) be a smooth submersion, let \(X\) be a smooth vector field on \(B\). Then there exists a smooth vector field \(\widetilde{X}\) on \(E\) such that \(\pi_{*, p}(\widetilde{X}_p) = X_{\pi(p)}\) for all \(p \in E\).

Proof. Around \(p \in E\), pick coordinates \((U, \varphi) = (U, x^1, \dots, x^m)\) and \((V, \psi) = (V, y^1, \dots, y^m)\) so that \(\psi \circ \pi \circ \varphi^{-1}\) is a projection. Write \(X\) in \(V\) as \(\sum_{j = 1}^{m} X^j \frac{d}{dy^j}\) and define \(\widetilde{X}|_{U}\) as \(\sum_{j = 1}^{m} X^j \frac{d}{d x^j}\). We do this for some open cover \(U_{\alpha}\) and then combine via a partition of unity subordinate to the cover \(\varphi_{\alpha}\) to obtain \(\widetilde{X}\). \(\blacksquare\)

From here, let \(v_1, \dots, v_{k - n}\) be tangent vectors at some \(b \in B\). We can extend them to vector fields and lift via the above claim to obtain vector fields in \(E\), \(\widetilde{v_j}\). Let \(\beta\) be the \(n\) -form obtained from contracting \(\alpha\) with the lifted vector fields. Restrict \(\beta\) to submanifold \(\pi^{-1}(b)\), on which it is a top-form with compact support on an oriented manifold. We then define the map \(\pi_{*} \alpha : T_b B \times \cdots \times T_b B \rightarrow \mathbb{R}\) as

\begin{equation} (\pi_{*} \alpha)_b(v_1, \dots, v_{k - n}) = \int_{\pi^{-1}(b)} \iota_b^{*} \beta \end{equation}

where \(\iota_b : \pi^{-1}(b) \rightarrow E\) is inclusion. Of course, this map is alternating multilinear. We still must show that this map is well-defined independent of our choice of lifts, and that it is smooth in \(b\). To do so, we pass to coordinates. We have a trivialization of the fibre bundle which acts as a slice/submanifold chart for the fibres. In particular, we have open neighbourhood of \(U\) and a diffeomorphism \(\Phi : \pi^{-1}(U) \rightarrow U \times F\) such that \(\pi = \text{proj} \circ \Phi\). Assume without loss of generality that \((U, \varphi) = (U, x^1, \dots, x^b)\) is also a coordinate chart around \(b\). Note that the entire fibre \(\pi^{-1}(b)\) is contained in \(\pi^{-1}(U)\), let \((V, \psi) = (V, y^1, \dots, y^n)\) be coordinates around \(f \in F\) with \(\Phi(z) = (b, f)\) for some \(z \in \pi^{-1}(b)\). Then immediately \(W = \Phi^{-1}(U \times V)\) and

\begin{equation} (\varphi \times \psi) \circ \Phi = (x^1 \circ \Phi, \dots, x^b \circ \Phi, y^1 \circ \Phi, \dots, y^n \circ \Phi) = (\widetilde{x}^1, \dots, \widetilde{x}^b, \widetilde{y}^1, \dots, \widetilde{y}^n) \end{equation}

is a coordinate chart around \(z\). We write \(\widetilde{v}_j = X^q_j \frac{d}{d\widetilde{x}^q} + Y^r_j \frac{d}{d\widetilde{y}^r}\) in \(W\) as our lifted vector fields. If \(v_j = 0\) for some \(j\), then clearly \(X_j^q(p) = 0\) for any \(p \in \pi^{-1}(b)\), for all \(q\). Thus,

\begin{equation} \beta_p = \alpha_p \left( \cdots, \widetilde{v_{1, p}}, \dots, \widetilde{v_{k - n, p}} \right) = \alpha_p \left( \cdots, \widetilde{v_{1, p}}, \dots, Y^{r}_{j}(p) \frac{d}{d\widetilde{y}^{r}}, \dots, \widetilde{v_{k - n, p}} \right) \end{equation}

If we then take any \(n\) tangent vectors at \(p \in \pi^{-1}(b)\), and push forward by inclusion into \(E\), each will be a linear combination of the tangent vectors \(\frac{d}{d\widetilde{y}^j}\) at \(p\), which when plugged into the above expression, will always yield \(0\), as we will always have a repeated index (particularly, there are \(n\) different coordinates \(\widetilde{y}^j\) and \(n + 1\) slots in the form where we are plugging in tangent vectors of the form \(\frac{d}{d\widetilde{y}^j}\)). Thus, \(\beta_p = 0\) for all \(p \in \pi^{-1}(b)\), in this case. This immediately proves that our map is well-defined, independent of the chosen extension and lift of \(v_1, \dots, v_{k - n}\), via multilinearity of the differential form.

Finally, let us prove smoothness.

Remark. This is where we will actually use the full strength of the assumption that \(E\) is an oriented fibre bundle.

The chart \((W, (\varphi \times \psi) \circ \Phi)\) around \(z\) gives rise to the chart around \(z\) for \(\pi^{-1}(b)\), \((W \cap \pi^{-1}(b), \theta) = (W \cap \pi^{-1}(b), \widetilde{y}^1 \circ \iota_b, \dots, \widetilde{y}^n \circ \iota_b)\). We can write \(\beta\), a smooth form on \(E\), as \(\beta = \sum_{I, J} C_{I, J} d\widetilde{x}^I \wedge d\widetilde{y}^J\). We can also assume that \(\Phi\) is orientation-preserving when restricted to each fibre. It then follows that

\begin{equation} (\iota_b^{*} \beta)_{\Phi^{-1}(b, f)} = \sum_{I, J} (C_{I, J} \circ \Phi^{-1})(b, f) \iota_b^{*} d\widetilde{x}^I \wedge \iota_b^{*} d\widetilde{y}^J = (C \circ \Phi^{-1})(b, f) d(\widetilde{y}^1 \circ \iota_b) \wedge \cdots \wedge d(\widetilde{y}^n \circ \iota_b) \end{equation}

where all terms with non-trivial \(d\widetilde{x}^I\) are eliminated when pulled-back by \(\iota_b\), as the functions \(\widetilde{x}^j \circ \iota_b\) are constant. We have the diffeomorphism \(\Phi : W \rightarrow \Phi(W) \subset U \times F\). Inside \(U \times F\), we have submanifold \(\{b\} \times V\) and the smooth embedding \(\Phi \circ \iota_b\) from submanifold \(W \cap \pi^{-1}(b)\) to \(\{b\} \times V\). These manifolds are of the same dimension, so this is actually a diffeomorphism, and as a result, we have a diffeomorphism of \(V\) with \(W \cap \pi^{-1}(b)\), where we take \(f\) to \(\Phi^{-1}(b, f)\). Call this diffeomorphism \(\Psi\). By assumption, this diffeomorphism is always orientation-preserving. We then have

\begin{equation} \int_{\pi^{-1}(b) \cap W} \iota_b^{*} \beta = \int_{V} (\Psi)^{*} \iota_b^{*} \beta = \int_V (C \circ \Phi^{-1})(b, \cdot) dy^1 \wedge \cdots \wedge dy^n \end{equation}

The right hand side will clearly be smooth in \(b\), as in coordinates, we can always differentiate under the integral sign. The integral over all of \(\pi^{-1}(b)\) is a sum over terms of the above form, glued together via a partition of unity. Thus, such a sum is also smooth in \(b\). We have therefore shown that \(\pi_{*} \alpha\) is a \(k - n\) form on the base space.

---

Remark. It is pretty easy to see that if forms \(\alpha\) and \(\omega\) on \(E\) agree inside \(\pi^{-1}(U)\), then their fibre integrals will agree on \(U\). This is simply due to the fact that for \(b \in U\), \(\pi^{-1}(b) \subset \pi^{-1}(U)\), which is the region over which we integrate.

---