My last few posts have been about zero sections of vector bundles, and how the homology class of a zero section is governed by a cohomology class associated with the vector bundle, known as the Euler class. I worked through some examples where we can compute the Euler class of a vector bundle using another mysterious cohomology class called the Thom class, but never explained where the Thom class comes from. In this post, I will motivate these constructions using intersection theory and Poincaré duality, where cohomology classes are used to determined how different subsets of a manifold may intersect. Proofs of the theorems discussed here can be found, e.g. in these notes by Michael Hutchings.

Zero sets as intersections

If we have a vector bundle \(\RR^d \to E \to B\) over an \(n\)-dimensional base space \(B\), then a section \(\sigma\) of \(E\) can be viewed as an \(n\)-dimensional submanifold of that \((n+d)\)-dimensional manifold \(E\), which is essentially the graph of \(\sigma\). Then the zero set of \(\sigma\) can be thought of as the intersection of this submanifold with the submanifold arising from the zero section of \(E\).

This intersection picture immediately explains one mysterious fact: why must the zero sets of all sections live in the same homology class in \(B\)? Well, the graphs of any two sections are homotopic to each other (since we can linearly interpolate between the sections in each fiber). And the homology class of the intersection of two submanifolds does not change when you apply a homotopy. So the zero set of any section always resides in some fixed homology class, determined by the vector bundle \(E\).

And if we do a little more work, we can figure out what that homology class must be. Under Poincaré duality, we can characterize intersections of submanifolds using products of their dual cohomology classes. In particular, suppose let \(M\) be a compact \(n\)-dimensional manifold without boundary, and let \(A\) and \(B\) be submanifolds of dimension \(i\) and \(j\) respectively. Then \(A\) and \(B\) are dual to cohomology classes \(\alpha \in H^{n-i}(M)\) and \(\beta \in H^{n-j}(M)\), where dual class \(\alpha\) is characterized by the property that for any \((n-i)\)-dimensional submanifold \(S\), the integral \(\int_S \alpha\) of \(\alpha\) along \(S\) should equal the number of signed intersections between \(S\) and \(A\) (and similarly for \(\beta\). The product of these cohomology classes \(\alpha \smile \beta \in H^{n - (i + j - n)}(M)\) is then dual to the \((i+j-n)\)-dimensional intersection \(A \cap B\). If \(M\) has a boundary \(\partial M\), then everything works out basically the same way as long as \(\alpha\) and \(\beta\) are taken to be relative cohomology classes in \(H^\bullet(M, \partial M)\).

So if we had a cohomology class \(\gamma \in H^d(E)\) which was Poincaré dual to the zero section of \(E\), then we could use it to find the homology class of the zero set of any other section. Furthermore, since all sections are homotopic to each other, this cohomology class will actually be dual to all sections of the vector bundle! And this special cohomology class is precisely the Thom class.

Finding the Thom class

Now, one problem here is that our vector bundle \(E\) not compact, so the version of Poincaré duality which we've been considering here does not apply directly. But, there's a simple fix: rather than considering the entire vector bundle, we can instead build a disk bundle \(D(E)\) where we consider only a compact disk \(D^d_x \subset \RR^d_x\) from each fiber. So long as our base space \(B\) is compact, the graph of any section \(\sigma\) will be a compact subset of \(E\), so it will be contained in such a disk bundle. And of course, every disk bundle contains the zero section.

With that technical detail taken care of, we can look for a cohomology class in our disk bundle which is the Poincaré dual of the zero section. Of course, since our disk bundle is a manifold with boundary, this class should lie in the relative cohomology group \(H^d(D(E), \partial D(E))\). And the boundary of our disk bundle is precisely the sphere bundle \(S(E)\) over \(B\). So the Poincaré dual of the zero section should be a cohomology class \(c \in H^d(D(E), S(E))\). It turns out that this class is precisely the Thom class of our vector bundle.

The Thom class for an oriented vector bundle \(E\) is usually defined as a cohomology class \(c \in H^d(D(E), S(E))\) such that for any point in the base space \(x \in B\), the restriction of \(c\) to the fiber \((D^d_x, S^{d-1}_x)\) yields the generator of \(H^d(D^d_x, S^{d-1}_x)\) which agrees with the orientation of the fiber \(\RR^d_x\). Essentially, this means that if we evaluate \(c\) on a positively-oriented fiber, we should get 1. But this is exactly what the Poincaré dual gives on each fiber, since there is 1 intersection between each fiber and the zero section.

Restricting to the Euler class

So now we have a nice picture of the Thom class \(c \in H^d(D(E),S(E))\): it is the Poincaré dual of the zero section of the disk bundle \(D(E)\) (and hence also the dual to all other sections as well), and provides information about where the zeros of the graph of our section lie in \(D(E)\). And for a more direct description of the zeros, we can restrict \(c\) to the base space \(B\) (viewed as the zero section of \(E\)). This restriction yields the Euler class \(e \in H^d(B)\). For any \(d\)-dimensional submanifold \(S \subseteq B\), the value \(e(S)\) of the Euler class evaluated on \(S\) is equal to the value of the Thom class evaluated on \(S\), viewed as a submanifold of the disk bundle, which is precisely the number of intersections between \(S\) and any section of \(E\)!

The Thom isomorphism theorem

Finally, we can use Poincaré duality to understand the Thom isomorphism theorem. If we let \(p : D(E) \to B\) denote the projection from our disk bundle onto the base space \(B\), then the which states that the mapping Thom isomorphism theorem gives an isomorphism \[\begin{aligned} \Phi : H^k(B) &\to H^{k+d}(D(E), S(E))\\ x &\mapsto p^*(x) \smile c \end{aligned}\] (As always, \(d\) is the rank of our vector bundle \(E\), and \(c \in H^d(D(E), S(E))\) is the Thom class).

In de Rham cohomology, where our cohomology classes are represented by differential forms, you can view the inverse mapping as partial integration over the fibers of \(D(E)\): taking a \((k+d)\)-form on \(D(E)\) and integrating it over the \(d\)-dimensional fibers yeilds a \(k\)-dimensional form on the base space \(B\).

Abstractly, the fact that these cohomology groups are isomorphic is not hard to establish. By Poincaré duality, we have isomorphisms \(H^k(B) \cong H_{n-k}(B)\) and \(H^{k+d}(D(E),S(E)) \cong H_{n-k}(D(E))\), and these two homology groups are isomorphic because \(D(E)\) deformation retracts onto \(B\).

But why do all of the isomorphisms take this form using the Thom class? First, we can look at the case of \(H^0(B) \cong H_n(B)\). So long as \(B\) is connected and orientable, \(H^0(B)\) is generated by an element dual to the fundamental class \([B] \in H_n(B)\). If we want to map this generator into \(H^{d}(D(E), S(E))\) as outlined above, then we should take this fundamental class \([B]\), include it into the disk bundle \(D(E)\) as the zero section, and then take the Poincaré dual of the zero section. But this is precisely the Thom class \(c\)! Since this isomorphism maps the generator of \(H^0(B)\) to \(c\), it maps \(n \in \ZZ \cong H^0(B)\) to \(nc \in H^d(D(E), S(E))\), which can indeed be written as \(x \mapsto p^*(x) \smile c\).

How about the higher-dimensional cohomology groups? Given a cohomology class \(\eta \in H^k(B)\), we first look for an \((n-k)\)-dimensional submanifold \(\Sigma\) dual to \(\eta\). We then want to include \(\Sigma\) into \(D(E)\) using the zero section and take its dual. If we want to lift \(\Sigma\) up into \(D(E)\), we might try taking the pullback of its dual \(p^*(\eta)\). However, this will generally introduce unwanted contributions in different parts of the fibers of \(D(E)\). To ensure that this pullback only lifts \(\Sigma\) along the zero section, we can explicitly intersect the result with the zero section—which just means taking the product \(p^*(\eta) \smile c\).

On an orientable \(n\)-dimensional manifold \(M\), the following four spaces are all isomorphic: \[H_i(M;\RR), H_{n-i}(M;\RR), H^i(M;\RR), H^{n-i}(M;\RR).\] This fact really boils down to two distinct isomorphisms: \(H_i(M;\RR) \simeq H^i(M;\RR)\) and \(H_i(M;\RR) \simeq H^{n-i}(M;\RR)\). The first (known as the universal coefficients theorem), behaves surprisingly differently from the second (known as Poincaré duality) . This can be especially confusing on two-dimensional surfaces where \(H^1(M;\RR)\) is literally the same space as \(H^{2-1}(M;\RR)\). But, even then the two isomorphisms are quite different! The first isomorphism follows fairly directly from the definition of cohomology, whereas the second is a deep result about the structure of manifolds.

The first isomorphism

Recall that in (simplicial) homology, our fundamental objects of study are simplicial chains, i.e. formal linear combinations of simplices. We let \(C_k(\RR)\) denote the set of \(k\)-chains (i.e. combinations of \(k\)-dimensional simplices with real coefficients). The boundary operator \(\partial_k : C_k(\RR) \to C_{k-1}\) takes a \(k\) dimensional simplex to its \(k-1\)-dimensional boundary. Together, all of these fit together into a chain complex: \[ \ldots \xrightarrow{\partial_{k+2}}C_{k+1}(\RR)\xrightarrow{\partial_{k+1}} C_k(\RR) \xrightarrow{\partial_k} C_{k-1}(\RR) \xrightarrow{\partial_{k-1}} \ldots\] The \(i\)th homology group \(H_i(M;\RR)\) is defined to be \(\ker \partial_k / \im \partial_{k+1}\). Concretely, an element of \(H_i(M;\RR)\) may be represented by a closed chain \(c_k\), and two such representatives are homologous if they differ by a boundary \(\partial_{k+1}A\).

To define cohomology, we dualize the whole picture. We let \(C^k(\RR)\) denote the space of \(k\)-cochains, i.e. the dual space of \(C_k(\RR)\). Concretely, a \(k\)-cochain still looks like an assignment of a number to each \(k\)-simplex, but it is helpful to distinguish the two. We define \(d_k : C^k(\RR) \to C^{k+1}(\RR)\) as the adjoint of \(\partial_{k+1}\). These fit into a chain complex going the other way: \[ \ldots \xleftarrow{d_{k+1}}C^{k+1}(\RR)\xleftarrow{d_{k}} C^k(\RR) \xleftarrow{d_{k-1}} C^{k-1}(\RR) \xleftarrow{d_{k-2}} \ldots\] Now, we define the \(i\)th cohomology group \(H^i(M;\RR)\) as \(\ker d_k / \im d_{k-1}\). Concretely, an element of \(H^i(M;\RR)\) may be represented by a closed cochain \(\gamma_k\), and two such representatives are cohomologous if they differ by a coboundary \(d_{k-1}\alpha\).

Since we dualized at the level of chains, rather than directly dualizing the homology groups, it's not immediately obvious that \(H^i\) should be dual to \(H_i\). But it follows very quickly from the definitions. We just need to check that the pairing between \(C_k(\RR)\) and \(C^k(\RR)\) descends to (co)homology classes. Suppose we have a closed chain \(c\) and closed cochain \(\gamma\). Then \[ \begin{aligned} \pair{c}{\gamma + d\alpha} &= \pair c \gamma + \pair{c}{d\alpha},\\ &= \pair c \gamma + \pair{\partial c}{\alpha},\\ &= \pair c \gamma, \end{aligned} \] using the fact that \(d\) is the adjoint of \(\partial\) and \(c\) is closed. Hence, the pairing is well-defined on cohomology classes. An analogous calculation shows that it is well-defined on homology classes as well.

Furthermore, it turns out that (on nice space, using real coefficients) this pairing is nondegenerate. The existence of a nondegenerate pairing identifies \(H^i(M;\RR)\) as the dual space of \(H_i(M;\RR)\). Since these are finite-dimensional vector spaces, we can conclude that they are isomorphic. Note, though, that this isomorphism is generally not canonical (i.e. the two spaces are not naturally isomorphic). One must pick a basis of \(H_i(M;\RR)\) to map between the two, and the resulting map depends entirely on the chosen basis.

So far, we haven't used the manifold structure of \(M\) at all! And indeed, the universal coefficient theorem applies to all topological space, although things get slightly more complicated if one wants to use integer coefficients rather than the real coefficients that we've been using.

Poincaré Duality

Our second isomorphism, though, is deeply entwined with the manifold structure of \(M\). Indeed, even stating that \(H_i(M;\RR) \simeq H^{n-i}(M;\RR)\) requires the dimension of \(M\), which doesn't necessarily make sense on general topological spaces.

When working with real coefficients, it can be illuminating to write the isomorphism as \((H^i(M;\RR))^* \simeq H^{n-i}(M;\RR)\) instead (which is equivalent thanks to our first isomorphsim). In this form, Poincaré duality asserts the existence of a nondegenerate pairing between \(H^i(M;\RR)\) and \(H^{n-i}(M;\RR)\).

In de Rham cohomology this pairing is given by the wedge product: \(\pair{[\alpha]}{[\beta]} \mapsto \int_M \alpha \wedge \beta\). The more traditional Poincaré duality map \(H_i(M;\RR) \to H^{n-i}(M;\RR)\) essentially pulls back this pairing along the isomorphism \(H_i(M;\RR) \to (H^i(M;\RR))^*\). Concretely, given a homology class \([\Gamma] \in H_i(M;\RR)\), we can view integration over \([\Gamma]\) as a linear functional on \(H_i(M;\RR)\). Since the wedge product pairing is nondegenerate, we can represent this using the wedge product (or cup product) against some form \(\gamma\): \[ \int_\Gamma \omega = \int_M \gamma \wedge \omega\;\forall \omega. \] Poincaré duality is then the mapping \(H_i(M;\RR) \to H^{n-i}(M;\RR)\) given by \([\Gamma] \mapsto [\gamma]\). Note that this mapping is canonical; it does not depend on any arbitrary choices, or even a metric on \(M\).

But if you do have a metric, you can consider another nondegenerate pairing of differential forms: the inner product. The inner product of two \(k\)-forms can be written as \[ \langle\langle\alpha, \beta\rangle\rangle = \int_M \alpha \wedge \star \beta. \] The Hodge star operator gives us a mapping from \(k\) forms to \(n-k\)-forms which we can interpret abstractly as using the metric to identify \(H^k(M;\RR)\) with its dual, and then using the Poincaré duality pairing to identify its dual with \(H^{n-k}(M;\RR)\).

So far, we've looked at Poincaré duality as a pairing on cohomology classes, and we've pulled it back along one component to get a mapping from homology classes to cohomology classes. If we pull back both components, then we get a pairing on homology classes: the intersection pairing. If \(N_1\) and \(N_2\) are submanifolds of \(M\) of complementary dimension, then the intersection pairing of \([N_1]\) and \([N_2]\) counts signed intersections between the submanfolds.

Duality on surfaces

On surfaces, both the universal coefficient theorem and Poincaré duality give maps from \(H_1(M;\RR)\) to \(H^1(M;\RR)\). But these maps are quite different! To obtain the universal coefficient theorem mapping, we first fix a basis for \(H_1(M;\RR)\). Then, we map each basis loop to a 1-form which integrates to 1 along that loop and zero along all others. This gives us a well-defined duality mapping, but it depends on the choice of basis. On the other hand, Poincaré duality maps any loop to a 1-form whose integral counts intersections with that loop. This gives a canonical mapping, not requiring a choice of basis.

My last post discussed zero sections of vector bundles, and showed some examples where the zero sections are determined by a particular cohomology class called the Euler class. In this post, I'll go through two constructions of the Euler class and work through calculations of Euler classes using both constructions. The first construction—using Chern classes—is geometric and a bit more concrete, but obscures the fact that the Euler class depends only on the topology of our vector bundle. The second construction—using the Thom class—is purely topological but fairly abstract.

Geometric construction via the first Chern class

For complex line bundles, the Euler class is equal to the first Chern class (and on general complex vector bundles, it is equal to the top nonzero Chern class). This relation allows us to take a more geometric view of the Euler class.

Although the Chern class is a topological invariant of a complex vector bundle, independent of any choice of geometry, it is convenient to define the Chern class using a connection on the vector bundle. It turns out that the resulting Chern class is well-defined, independent of the particular connection used in this definition. So for now, suppose that we have a complex vector bundle \(\CC^d \to E \to B\). This notation means that we have a vector bundle \(E\) over a base space \(B\), where above every point of \(x \in B\) we have a \(d\)-dimensional complex vector space \(\CC_x^d\). Suppose also that we have a connection \(\nabla\) on \(E\) which allows us to parallel transport vectors between the different space \(\CC_x^d\). The curvature of this connection is a matrix-valued 2-form \(\Omega\). The \(d \times d\) matrix \(\Omega(X, Y)\) tells us how a vector in our line bundle gets rotated if we use the connection to parallel transport the vector around an infinitesimal loop in the \(X, Y\) plane. The first Chern class is then defined to be the 2-form \(\tr \Omega\), where \(\tr \Omega(X, Y)\) gives the trace of the matrix \(\Omega(X, Y)\). (In general, the \(k\)th Chern class is given by \(\tr \Omega^{\wedge k}\), the trace of the \(k\)th wedge power of \(\Omega\)). On a complex line bundle, \(\Omega\) is a \(1 \times 1\) matrix, so the first Chern class is equal to the 2-form \(\Omega\) itself.

Since Chern classes only exist for complex vector bundles, we cannot use this approach to understand the simpler vector bundles like the cylinder and Möbius strip which I talked about last time. But we can compute the Euler class for the tangent bundle of the sphere \(TS^2\). And the computation is a lot quicker than the Thom space construction which we will see next: if we take \(\nabla\) to be the Levi-Civita connection, then the usual Gaussian curvature 2-form \(\Omega\) is our first Chern class, and hence also our Euler class.

Since \(H^2(S^2) \cong \ZZ\), \(\Omega\) is cohomologous to a scalar multiple of the normalized area form. And to compute the scalar, we can simply integrate \(\Omega\) over our domain \(S^2\). By the Gauss-Bonnet theorem, \(\int_S^2 \Omega = \chi(S^2) = 2\). Hence, we conclude again that our Euler class is given by the element \(2 \in \ZZ \cong H^2(S^2)\), and we see that this value of \(2\) comes directly from the Euler characteristic \(\chi(S^2)\).

Topological construction via the Thom class

The Euler class is often described by constructing an equally-mysterious cohomology class called the Thom class and pulling it pack through a sequence of maps. In this section, we'll try to unpack the definitions and work through some examples. Formally, the Thom class is a certain relative cohomology class in the unit disk bundle associated to our vector bundle, and the Euler class is the restriction of the Thom class to our base space, viewed as the zero section of the unit disk bundle. We will stat by reviewing some facts about relative homology and cohomology.

Relative homology

Given a topological space \(X\), the \(k\)th absolute homology group \(H_k(X)\) consists of \(k\)-dimensional boundary-free subsets which are not themselves the boundary of any \((k+1)\)-dimensional subset. For any subset \(A \subset X\), the \(k\)th homology group relative to \(A\), denoted \(H_k(X, A)\), relaxes the boundary-free condition and instead consists of \(k\)-dimensional subsets whose boundaries lie in \(A\), and do not form the boundary of any \((k+1)\)-dimensional subset even when combined with any portion of \(A\).

One common example which helps to motivate these definitions is the homology group \(H_1(M, \partial M)\) of a surface \(M\) relative to its boundary \(\partial M\). If we have a connected surface of nontrivial topology, a basis for \(H_1(M)\) provides a maximal collection of loops which we can cut along without disconnecting the surface.

However, this property no longer holds for surfaces with boundary. Even for an annulus, the loops in the first homology group now disconnect our surface.

There are two problems here: (1) nontrivial loops in the first homology group may disconnect the surface. Even though they cut the surface into two pieces, they are not technically equal to the boundary of either piece, since the pieces may also have boundary components arising from the global boundary \(\partial M\). (2) there are cuts which we can make without disconnecting the surface—cuts in the radial direction. But they are not included in the first homology group since they are not loops.

The definition of relative homology groups solves both problems: in \(H_1(M, \partial M)\), we say that a set is the boundary of a region if it can be combined with part of \(\partial M\) to form the boundary of a region. And rather than restricting the elements of our homology group to be loops, we allow any subset whose boundary is entirely contained in \(\partial M\).

Under mild technical conditions on the subset \(A\), the relative homology group \(H_k(X, A)\) is equal to the absolute homology group of the quotient space \(H_k(X / A)\) where we collapse all of \(A\) to a single point.

Long exact sequence of homology

If we know the homology groups of \(X\) and \(A\), we can use them to calculate the relative homology groups of the pair \((X, A)\). The key tool is a long exact sequence involving all three sets of homology groups: \[ \cdots \xrightarrow{} H_{n+1}(X, A) \xrightarrow{\partial} H_n(A) \xrightarrow{i_*} H_n(X) \xrightarrow{j_*} H_n(X, A) \xrightarrow{\partial} H_{n-1}(A) \xrightarrow{} \cdots\] Here the function \(i_*\) is the map on homology induced by the inclusion \(i : A \to X\), the function \(j_*\) is the map on homology induced by the quotient map \(j : X \to X/A\), and the function \(\partial\) is the standard boundary map (which maps from \(H_n(X, A) \to H_{n-1}(A)\) since the boundary of a relative homology class must always lie in \(A\)).

Relative cohomology

The whole relative story can be dualized to relative cohomology. Although we will mostly need cohomology with \(\ZZ\) or \(\ZZ_2\) coefficients later on, I will briefly discuss relative de Rham cohomology here, since I find that it helps to highlight about the differences between homology and cohomology.

On a manifold \(X\) without boundary, the elements of the de Rham cohomology group \(H^k(X)\) are differential \(k\)-forms \(\eta\) whose exterior derivative \(d\eta\) vanishes. Two differential forms \(\eta, \omega\) are said to be equal if their difference \(\eta - \omega\) is itself the exterior derivative of some \((k-1)\) form. On a manifold with boundary, the \(k\)-forms in \(H^k(X)\) are constrained to lie tangent to the boundary \(\partial X\). On the other hand, the elements of the relative cohomology group \(H^k(X, \partial X)\) are given by \(k\)-forms whose restrictions to the boundary are zero.

Returning to the annulus example, the absolute cohomology group \(H^1(A)\) is generated by harmonic the 1-form \(\eta\) which circulates around the annulus counterclockwise, whereas the relative cohomology group \(H^1(A, \partial A)\) is generated by the harmonic 1-form \(\rho\) which flows in the radial direction from the inner boundary to the outer boundary.

In de Rham cohomology for manifolds without boundary, a cohomology class \([\eta] \in H^{n-k}(X)\) is dual to some homology class \([\gamma] \in H_k(X)\) if the integral of \(\eta\) over each \((n-k)\)-dimensional subset is equal to the number of intersections between \(\gamma\) and that subset. On manifolds with boundary, one should pair absolute homology classes in \(H_k(X)\) with relative cohomology classes in \(H^{n-k}(X, \partial X)\) and vice versa. Note, for instance, how the generator \(\eta\) of \(H^1(A)\) integrates to 1 the generator of \(H_1(A)\), even though two concentric curves in the annulus may not intersect each other! On the other hand, the generator \(\rho\) of \(H^1(A, \partial A)\) integrates to 1 along radial segments, which always intersect the generator loop of \(H_1(A)\) once.

Long exact sequence of cohomology

Just as for homology, the relative cohomology groups of a pair \((X, A)\) fit into a long exact sequence alongside the cohomology groups of \(X\) and \(A\): \[ \cdots \xrightarrow{} H^{n-1}(A) \xrightarrow{\delta} H^n(X, A) \xrightarrow{j^*} H^n(X) \xrightarrow{i^*} H^n(A) \xrightarrow{\delta} H^{n+1}(X, A) \xrightarrow{} \cdots\] The maps here are the duals of the maps in the long exact sequence for homology: the function \(i^*\) is the map on cohomology induced by the inclusion \(i : A \to X\), the function \(j^*\) is the map on cohomology induced by the quotient map \(j : X \to X/A\), and the function \(\delta\) is the codifferential (the dual of the boundary map \(\partial : H^{n+1}(X, A) \to H^n(A)\)).

The Thom space

To define the Thom class and Thom space, we start with an oriented vector bundle \(\RR^d \to E \to B\) over a base space \(B\). An orientation of \(E\) amounts to a continuous choice of orientation of the vector spaces \(\RR^d_x\) at each point \(x\). We then build two associated bundles. First, we build the disk bundle \(D^d \to D(E) \to B\), whose fiber at a point \(x \in B\) is the unit disk in our vector space \(\RR^n_x\) above \(x\). Then we build the sphere bundle \(S^{d-1} \to S(E) \to B\) whose fiber at a point \(x \in B\) is the boundary of the disk above \(x\) in \(D(E)\).

The Thom class is a cohomology class \(c \in H^d(D(E), S(E))\) whose restriction to any fiber \((D^d_x, S^{d-1}_x)\) yields the positively-oriented generator of \(H^d(D^d_x, S^{d-1}_x) \cong \ZZ\). And the Euler class is a cohomology class \(e \in H^d(B)\) obtained by restricting \(c\) to the zero section of \(E\), which we can identify with \(B\).

The Thom class \(c\) provides us with an isomorphism \(\Phi^k : H^k(B) \to H^{k+d}(D(E), S(E))\) given by taking the cup product with \(c\). Explicitly, the mapping is given by \(\Phi^k(x) = p^*(x) \smile c\), where \(p : E \to B\) denotes the canonical projection onto the base space. (In de Rham cohomology, we can equivalently write \(\Phi^k(\eta) = p^* \eta \wedge c\)).

Thom classes for bundles on \(S^1\)

We'll start by computing the Thom classes for two simple line bundles: the cylinder and the Möbius trip.

The Thom class of the cylinder

First, let's look at how the Thom class is constructed on the cylinder \(S^1 \times \RR\). In this case, our disk bundle \(D(E) = S^1 \times I\) is simply the annulus \(A\), and our sphere bundle \(S(E) = S^1 \times S^0\) is the pair of disjoint circles lying on the boundary \(\partial A\) of \(D(E)\). Hence, the Thom class lies in the relative cohomology group \(H^1(A, \partial A)\) which we investigated earlier. In particular, the Thom class is precisely the generator \(\rho\) which we identified above. Since \(\rho\) is orthogonal to the centerline of the annulus, the Euler class \(e\) is zero.

The Thom class of the Möbius strip

Now, what about the Möbius strip? We can still build the Thom space in exactly the same way: our disk bundle \(D(E)\) is a Möbius strip of unit width, and the sphere bundle \(S(E)\) is the boundary of \(D(E)\), which is topologically a circle. However, the Möbius strip is not orientable, so we run into trouble when we try to define the Thom class. The solution is to switch from considering standard cohomology with integer coefficients to use \(\ZZ_2\) coefficients instead. Working mod 2 essentially ignores signs, and allows us to treat all surfaces as if they were oriented. In this case, the Thom class is a cohomology class \(c \in H^d(D(E), S(E);\ZZ_2)\) whose restriction to any fiber yields the nontrivial element of \(H^d(D_x^d, S_x^{d-1};\ZZ_2)\cong \ZZ_2\).

In the case of the Möbius strip, then, we are interested in the relative cohomology group \(H^1(M, \partial M; \ZZ_2)\), where \(M\) is the unit-width Möbius strip. Now that we're using \(\ZZ_2\) coefficients, we cannot use de Rham cohomology any more, so we will do some explicit calculations on a particular cell decomposition of the Möbius strip, which includes the zero section in its 1-skeleton.

Our cell decomposition has three vertices, five edges, and two faces, so \(C^0 = \ZZ_2^3\), \(C^1 = \ZZ_2^5\) and \(C^2 = \ZZ_2^2\). The coboundary maps \(\delta_0 : C^0 \to C^1\) and \(\delta_1 : C^1 \to C^2\) are given by \[ \delta_0 = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}, \quad \text{and} \quad \delta_1 = \begin{pmatrix} 1 & 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 & 1 \end{pmatrix}. \] Hence, our absolute cohomology groups are \[\begin{aligned} H^0(M;\ZZ_2) &= \ker \delta_0 = \langle(1, 1, 1)\rangle \cong \ZZ_2,\\ H^1(M;\ZZ_2) &= \ker \delta_1 / \im \delta_0 = \langle(0, 1, 0, 1, 0), (1, 1, 0, 0, 0), (1, 0, 1, 0, 1)\rangle / \im \delta_0 = \ZZ_2,\\ H^2(M;\ZZ_2) &= C^2 / \im \delta_1 = 0. \end{aligned}\]

The relative cohomology is a little easier, since, we quotient each space of cochains by the space of boundary cochains. So all of our spaces get a little smaller: \(C^0(M, \partial M) = \ZZ_2^3 / \ZZ_2^2 = \ZZ_2\), \(C^1(M, \partial M) = \ZZ_2^5 / \ZZ_2^2 = \ZZ_2^3\), and \(C^2(M, \partial M) = \ZZ_2^2\). And we obtain relative boundary maps \[\delta_0' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \quad \text{and} \quad \delta_1' = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}.\] Our resulting relative cohomology groups are \[\begin{aligned} H^0(M, \partial M;\ZZ_2) &= \ker \delta_0' = 0,\\ H^1(M, \partial M;\ZZ_2) &= \ker \delta_1' / \im \delta_0' = \langle(1, 1, 0), (1, 0, 1)\rangle / \langle (1, 1, 0)\rangle \cong \ZZ_2,\\ H^2(M, \partial M;\ZZ_2) &= C^2 / \im \delta_1' = \ZZ_2. \end{aligned}\] The most important result for us is that the first relative cohomology group \(H^1(M, \partial M;\ZZ_2)\) has one nontrivial element \(c\), which is represented by the relative cochain with a 1 on edges \(e_1\) and \(e_4\) and zero everywhere else. This element \(c\) is the Thom class of the Möbius strip.

Since the Möbius strip is nonorientable, it does not technically have an Euler class. But restricting \(c\) to the centerline of the Möbius strip pulls \(c\) back onto the base space \(S^1\), yielding a sort of "mod 2" Euler class \(e \in H^1(S^1; \ZZ_2)\). In this case, \(e\) is the single nontrivial class in \(H^1(S^1;\ZZ_2)\). In particular, \(e\) is nonzero, and thus distinct from the Euler class for the cylinder which we computed earlier. (Formally, this "mod 2" Euler class is known as the top Stiefel-Whitney class, and is equal to the Euler class taken mod 2 on orientable vector bundles where the Euler class is defined)

The Thom class for \(TS^2\)

Now, let's get a little more ambitious and try to compute the Thom class for the tangent bundle \(TS^2\) of the two-dimensional sphere. For this example, we will have to use more long exact sequences and other algebraic machinery, since I do not know how to compute the relevant cohomology groups directly.

The sphere bundle \(S(TS^2)\) is the set of all unit tangent vectors on the sphere, and can be identified with the group \(SO(3)\) of \(3 \times 3\) rotation matrices: any point \(x \in S^2\) is a unit vector in \(\RR^3\), and a unit tangent vector at \(x\) provides a second orthogonal vector. We can take these two vectors to be the first two columns of a rotation matrix, and they uniquely determine the last column, which must be their cross product. The group \(SO(3)\) can also be identified with the real projective space \(\RP^3\), as both can be expressed as the quotient of the set of unit quaternions modulo \(\pm 1\). If we look up the cohomology groups of \(\RP^3\), we find that they are \[ H^0(\RP^3) = \ZZ, \quad H^1(\RP^3) = 0, \quad H^2(\RP^3) = \ZZ_2, \quad H^3(\RP^3) = \ZZ.\]

As far as I know, the disk bundle \(D(TS^2)\) does not have such a nice description, but its topology is quite simple: since each disk is contractible, \(D(TS^2)\) deformation retracts onto the sphere \(S^2\), and in particular its cohomology groups are all isomorphic to the corresponding cohomology groups of \(S^2\).

Since we know the cohomology groups for \(S(TS^2) \cong \RP^3\) and \(D(TS^2)\), we can use the long exact sequence for cohomology to work out the relative cohomology group \(H^2(D(TS^2), S(TS^2))\) which the Thom class \(c\) lives in. In particular, we have get an exact sequence \[ H^{1}(\RP^3) \xrightarrow{\delta} H^2(D(TS^2), \RP^3) \xrightarrow{j^*} H^2(D(TS^2)) \xrightarrow{i^*} H^2(\RP^3) \xrightarrow{\delta} H^{3}(D(TS^2), \RP^3) \] The first term \(H^1(\RP^3)\) is equal to zero. The last term is also equal to zero, since the Thom isomorphism tells us that \(H^3(D(TS^2), \RP^3) \cong H^1(S^2) = 0.\) So we obtain a short exact sequence \[ 0 \xrightarrow{} H^2(D(TS^2), \RP^3) \xrightarrow{j^*} H^2(D(TS^2)) \xrightarrow{i^*} H^2(\RP^3) \xrightarrow{} 0 \] Since \(D(TS^2)\) deformation retracts onto \(S^2\), the central term \(H^2(D(TS^2))\) is isomorphic to \(H^2(S^2) \cong \ZZ\). And the right-hand term \(H^2(\RP^2)\) is equal to \(\ZZ_2\). Hence, our short exact sequence is really \[ 0 \xrightarrow{} H^2(D(TS^2), \RP^3) \xrightarrow{j^*} \ZZ \xrightarrow{i^*} \ZZ_2 \xrightarrow{} 0 \] and thus we must have \(H^2(D(TS^2), S(TS^2)) \cong \ZZ\), and its generator is the Thom class of \(TS^2\). Furthermore, our short exact sequence shows that this generator maps to the element \(2 \in \ZZ \cong H^2(S^2)\). Hence, the Euler class of the sphere is twice the generator of \(H^2(S^2)\), reflecting the fact that the Euler characteristic \(\chi(S^2) = 2\).

The Thom class for a twisted line bundle over the torus

In my last post, I discussed a twisted line bundle over the torus. We built a real line bundle by specifying that as you go around one generator loop, the fibers sweep out an annulus, whereas if you go around the other generator loop the fibers sweep out a Möbius strip. Let us denote this bundle by \(\RR \to E \to T^2\). We will compute the Thom class of \(E\).

We start by drawing the torus as a square with opposite edges identified. To make the disk bundle associated to our line bundle, we can thicken the square to obtain a cube. Now we just need to identify the opposite sides of the cube properly.

Note that our bundle has annuli sitting above the loops which run parallel to edge \(e_1\), and has Möbius strips sitting above the loops which run parallel to edge \(e_2\). We have relative cochain groups are \(C^0(E, \partial E) = \ZZ_2^1, C^1(E, \partial E) = \ZZ_2^4, C^2(E, \partial E) = \ZZ_2^5, C^3 = \ZZ_2^2\). The coboundary maps are given by \[ \delta_0 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix},\quad \delta_1 = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 0 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \end{pmatrix}, \quad \delta_2 = \begin{pmatrix} 1 & 0 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 & 1 \end{pmatrix} \] Our resulting relative cohomology groups are \[\begin{aligned} H^0(E, \partial E;\ZZ_2) &= \ker \delta_0 = 0,\\ H^1(E, \partial E;\ZZ_2) &= \ker \delta_1 / \im \delta_0 = \frac{\langle(0, 1, 1, 0), (0, 0, 1, 1)\rangle}{\langle (0, 0, 1, 1)\rangle} \cong \ZZ_2,\\ H^2(E, \partial E;\ZZ_2) &= \ker \delta_2 / \im \delta_1 = \frac{\langle(1, 0, 0, 0, 1), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 1)\rangle}{\langle (0, 0, 0, 1, 1), (0, 1, 1, 0, 0)\rangle} \cong \ZZ_2^2,\\ H^3(E, \partial E;\ZZ_2) &= C^3 / \im \partial_2 \cong \ZZ_2 \end{aligned}\] The Thom class of \(E\) is the nontrivial element of \(H^1(E, \partial E; \ZZ_2)\), which is represented by a cochain supported on edges \(e_2\) and \(e_3\). Restricting to the torus \(T^2\) lying in the center, we find that our mod-2 Euler class is represented by the cocycle supported on edge \(e_2\)—the edge whose fiber was twisted into a Möbius strip.

This post walks through some interesting results on real and complex line bundles, leading up to a discussion of the Euler class for complex line bundles, which characterizes their zero sections. Briefly, a complex line bundle is a topological object which generalizes the space of complex-valued functions defined on some domain. The "sections" of the complex line bundle locally behave like complex functions, but their global structure must obey certain constraints. In particular, their zero sets must belong to a particular topological class, prescribed by the line bundle's Euler class. Here, I detail some concrete, low-dimensional examples that I found useful for thinking about these topics.

Prelude: vector fields on the sphere

The hairy ball theorem famously states that any continuous tangent vector field on the sphere must have a zero, i.e. there must be some point on the sphere where the vector field is zero. Indeed, this result can be strengthened to tell us about the required number of zeros: the sum of the indices of the zeros must equal the Euler characteristic of the domain. In particular, since the sphere has Euler characteristic 2, any vector field on the sphere must have at least one zero.

On the other hand, it's very easy to define an \(\RR^2\)-valued function on the sphere with no zeros: even the constant function \(f(x) = (1, 0)\) works. Although tangent vector fields and \(\RR^2\)-valued functions sound very similar when you first learn about them—and indeed, any tangent vector field can locally be written as an \(\RR^2\)-valued function—their global behavior is quite different. And in particular, the zero sets of tangent vector fields must obey delicate constraints which vector-valued functions are free to ignore. To get a better understanding of what's going on here, it may help to step down a dimension, and look at real-valued functions on the one dimensional sphere, i.e. the circle.

The circle

Let's start by looking at the circle \(S^1 := \{ x \in \RR^2 \;:\; \|x\| = 1\}\), and considering real-valued functions \(f : S^1 \to \RR^1\). We can identify these functions with functions \(f : [0, 1] \to \RR^1\) defined on the unit interval subject to the constraint that \(f(0) = f(1)\). And the nice thing about moving down to one dimension is that these functions are very easy to visualize. We can graph these functions over the unit interval, and we can even draw these graphs on the cylinder to emphasize the constraint that \(f(0) = f(1)\).

In one dimension, though, tangent vector fields are a lot simpler. It's easy to construct non-vanishing vector fields on the circle—for instance, you can take the vector field that always points counterclockwise.

Indeed, tangent vector fields on the circle are in one-to-one correspondence with real-valued functions on the circle. So they don't really help us understand what's going on with the hairy ball theorem in 2D. We will have to look elsewhere.

Twisted functions on the circle

The key is to consider the picture of graphs drawn on the cylinder. We can introduce a twist and consider "graphs" which are drawn on the Möbius strip instead.

While this construction may sound kind of arbitrary, but this notion of generalized graphs ends up being surprisingly useful throughout math and physics. Although these graphs no longer represent continuous functions defined on the circle, they can still be written using functions \(f : [0, 1] \to \RR^1\) defined on the unit interval. The difference is that thanks to the twist in the Möbius strip, we have a new compatibility condition: this time, we require that \(f(0) = - f(1)\). For now, I will call these functions "twisted functions," since their graphs are continuous on the twisted Möbius strip¹. And importantly, our new compatibility condition forces all of our twisted functions to have at least one zero! They must pass zero as they go from negative to positive (or vice versa). Using this Möbius strip example, we can now return to two-dimensional surfaces and find some more interesting examples of twisted functions.

Twisted functions on the torus

Now we will move back to the setting of two-dimensional surfaces, but we will continue to think about real-valued functions to start with. Consider the torus \(T^2\), and a real-valued function \(f : T^2 \to \RR\). We can draw the graph of \(f\) as a two-dimensional surfaces living inside of the three-dimensional thickened torus \(T^2 \times I\).

Any loop \(S^1 \subset T^2\) in the torus becomes a topological cylinder (or equivalently, an annulus) in the thickened cylinder. What happens if we replace some of those cylinders with Möbius strips instead?

For instance, suppose we replace the cylinder over each of the longitudinal loops with a Möbius strip. The figure below shows the resulting twisted thickened torus, with several of the component Möbius strips highlighted.

We saw above that a twisted function drawn on the Möbius strip must have a zero somewhere. This torus has a Möbius strip on each of its lateral loops, so a twisted function defined on the torus must have a zero on each lateral loop. That means that the zero set of the twisted function must be a curve which wraps one or more times around the torus!

And of course, the direction that the zero set wraps around the torus was determined by the fact that we chose to put Möbius strips over the longitudinal loops. If we twisted the thickened torus in other ways, then we could force the zero sets of our twisted functions to follow different paths along the torus.

The homological perspective

In general, if we start with an oriented \(n\)-dimensional manifold \(M\), and we look at the zero set \(S\) of a single real function, we expect \(S\) to be an \((n-1)\)-dimensional submanifold. In this case, the zero set defines an \((n-1)\)-dimensional homology class \([S] \in H_{n-1}(M)\). By Poincaré duality, we can identify homology classes \(S \in H_{n-1}(M)\) with one-dimensional cohomology classes \([\eta] \in H^1(M)\). In the language of differential forms, the Poincaré dual of a homology class \(S\) is a harmonic 1-form \(\eta\) with flux 1 through \(S\) and zero flux through all other homology classes of submanifolds.

All this is pretty abstract, so it may help to revisit our torus example. In that case, we saw that our zero set \(S\) wrapped horizontally around the torus. The dual 1-form \(\eta\) should thus wind around the torus in the longitudinal direction—exactly the way that we glued our Möbius strips.

Unfortunately, this twisting trick doesn't let us prescribe the exact homology class \([S] \in H_{n-1}(M)\) of the zero set. Going back to the circle example, we can start from a cylinder and add a single twist to ensure that our zero set contains at least one point. But we cannot then add a second twist to ensure that the zero set contains at least two points: topologically speaking, adding that second twist just brings us back from the Möbius strip to the cylinder again! In general, adding these twists only allows us to control the mod-2 homology class of the zero set, \([S] \in H_{n-1}(M;\mathbb{Z}/2\mathbb{Z})\). But amazingly, this restriction disappears in the complex case!

Mathematical terminology: vector bundles

But before proceeding, it will help to lay out a little more mathematical terminology that I've been avoiding so far. When we have a domain \(M\) and a vector-valued function \(f : M \to \RR^d\), we can draw its graph in the product space \(M \times \RR^d\). For each point \(x \in M\) this product space contains a separate copy of \(\RR^d\), which we denote by \(\RR^d_x\). And all of these copies of \(\RR^d\) fit together in a continuous way just as we built a cylinder earlier by gluing together a bunch of lines over a circle.

The product space \(M \times \RR^d\) is an example of a vector bundle, a collection of vector spaces "bundled up" over the points of some base space. Locally, vector bundles look like simple product spaces, and in particular, they still contain a separate copy of \(\RR_x^d\) for each point \(x\) in the base space. But globally they can behave quite differently. For instance, any small patch of the Möbius strip looks identically to a small patch on the cylinder, but we already saw that the Möbius strip has different global properties.

Another classic example of a vector bundle is the tangent bundle \(TM\) of a manifold \(M\). If \(M\) is an \(n\)-dimensional manifold, then \(TM\) contains an \(n\)-dimensional vector space \(T_xM\) associated to each point \(x \in M\). In general, the tangent space \(TM\) is not equal to the simple product space \(M \times \RR^n\), leading to results like the hairy ball theorem for the two-dimensional sphere which we saw earlier.

Vector bundles are often used to define spaces of generalized functions, like the graphs that we drew on the Möbius strip, or tangent vector fields on a manifold. Whereas an ordinary vector-valued function takes in a point \(x\) in the domain and returns a vector \(f(x) \in \RR^d\), our twisted functions on a vector bundle will instead return a vector \(f(x) \in \RR^d_x\) living in the special vector space associated to point \(x\). These twisted functions are usually referred to as sections (or cross-sections) of the vector bundle.

The complex case

If we look at the zero set \(S\) of a single complex function on an \(n\)-dimensional manifold \(M\), then we generically expect \(S\) to be an \((n-2)\)-dimensional submanifold, defining a homology class \(S \in H_{n-2}(M)\). And again, Poincaré duality allows us to identify \(S\) with its dual cohomology class \(\eta \in H^2(M)\). But this time, unlike in the real case, we can use special twists to force \(S\) to lie in any homology class of \(H_{n-2}(M)\) that we like: picking twists gives us complete control over the topology of the zero sets.

Another way of saying this is that the cohomology class \(\eta\) is determined by the topological structure of the vector bundle that we build over \(M\). Since our vector spaces are always the complex line \(\CC^1\), these vector bundles are referred to as complex line bundles. And the cohomology class \(\eta \in H^2(M)\) is known as the first Chern class, or the Euler class, of the line bundle.

These sorts of facts about complex line bundles, sections, and Chern classes are usually proved using pretty abstract arguments involving the classifying space \(\CC P^\infty\), which provides an alternative representation of complex line bundles. Instead of taking that path, I will work through some details of the complex analogue of the Möbius strip. We won't prove the general versions of these results, but I found the example helpful for building intuition.

Unfortunately, even the simplest complex line bundles are quite complicated. In the real case, we started with a one-dimensional circle, and built a real line bundle by gluing on one-dimensional lines to obtain a two-dimensional Möbius strip. In the complex case, we will start with the two-dimensional sphere \(S^2\) and build a complex line bundle by gluing on two-dimensional complex planes to obtain an object with four real dimensions (or two complex dimensions if you want to go all in with complex thinking). So it will be a lot harder to draw pictures.

Starting with the sphere

My discussion here follows the treatment in Chapter 1 of Hatcher's book on vector bundles. First, Hatcher notes that we can build a complex line bundle on the sphere by the following construction: we think of the sphere \(S^2\) as the union of the northern hemisphere \(D^+\) and the southern hemisphere \(D^-\), which intersect exactly along the equator \(D^+ \cap D^- = S^1\).

On each hemisphere, we build our complex line bundle as the product \(D^\pm \times \CC\), and we glue together these two halves using a "clutching map" \(f : S^1 \to \CC \setminus 0\) which tells us how the vector spaces at each point of the equator in northern hemisphere should be rotated and scaled before gluing them to the corresponding vector spaces in the southern hemisphere. This construction is still pretty opaque, so we can start by considering how it applies to the real case of the Möbius strip again.

Clutching maps for the Möbius strip

We start by decomposing the circle into northern and southern intervals \(I^\pm\), which intersect at the equator \(S^0\), consisting of a pair of opposite points. Once we thicken each hemisphere separately, we get a pair of strips \(I^\pm \times I\), and the only remaining question is how to glue them together again.

This gluing is determined by a clutching map \(f : S^0 \to \RR \setminus 0\). The map \(f\) gives a nonzero number for the left and right points of the equator; for each side, we glue the ends of our strips directly if the number is positive and glue them with a twist if the number is negative.

The fact that we only need the sign of \(f\), rather than the precise value, reflects the fact the vector bundle we get from this clutching map construction depends only on the homotopy class of the clutching map. Furthermore, we can restrict our attention to basepoint-preserving maps (which always send the left endpoint of \(S^0\) to \(1\)), yielding only two homotopy classes of basepoint-preserving maps: those which send the left point to a positive number, and those which send the left point to a negative number.

So the two real line bundles over the circle that we saw earlier—the cylinder and the Möbius strip—are the only two possibilities, and correspond exactly to the two homotopy classes of basepoint-preserving maps from \(S^0 \to \RR\setminus 0\).

Complex line bundles on the sphere

The situation with complex line bundles on the two-dimensional sphere \(S^2\) is exactly analogous. Except this time, our clutching functions are maps \(f : S^1 \to \CC \setminus 0\). Any such map is homotopic to a map \(g : S^1 \to S^1\) (e.g. the normalization of \(f\)), so we get a complex line bundle on the sphere for each homotopy class of maps from \(S^1\) to itself. The homotopy classes of maps from \(S^1\) to \(S^1\) are precisely the loops in \(\pi_1(S^1) \cong \ZZ\), so we have one complex line bundle over the sphere for each integer.

That's a good sign: we wanted to rig up our complex line bundles to force the zero sets to lie in a particular homology class in \(H_0(S^2) \cong \ZZ\). So at least we have the right number of complex line bundles. Now we will look more closely at how the clutching map construction determines the zeros of a section of our complex line bundle.

Suppose we start with a complex line bundle on the sphere, and a section \(\Gamma\). We can think of the hemisphere decomposition that we've been using as a pair of charts for the vector bundle, with the clutching map acting as our transition function between the charts. But, of course, it's not important that our hemispheres cut the sphere exactly in half. Topologically speaking, we can pick any two disks that cover the sphere and meet along a circle. So we're free to make the hemisphere \(D^+\) huge, covering almost all of the sphere, and the other hemisphere \(D^-\) tiny.

If we put \(D^-\) in a place where our section \(\Gamma\) is nonzero, then we can change \(\Gamma\) by a small homotopy to make it constant in \(D^-\) without changing its zero set. Now, all of the interesting behavior of \(\Gamma\) occurs over \(D^+\), where our charts allow us to write \(\Gamma\) as an honest complex function \(\Gamma : D^+ \to \CC\). However, \(\Gamma\) cannot be any arbitrary complex function: on the boundary of \(D^+\), it must agree with the value from \(D^-\). To be precise, for each point \(x \in \partial D^+\), the value \(\Gamma(x)\) (written in the \(D^+\) chart) must equal \(f(x)\) times the constant value of \(\Gamma\) on \(D^-\) (written in the \(D^-\) chart).

Up to multiplication by a nonzero complex constant (which does not change the zeros of \(\Gamma\)), this means that we have prescribed nonzero vectors for \(\Gamma(x) \in \RR^2\) on \(S^1 = \partial D^+\), and \(\Gamma\) continuously interpolates those values into the interior. The homotopy class of our clutching map \(f\) determines the turning number of \(\Gamma\) along the boundary \(\partial D^+\), i.e. the total number of rotations that \(\Gamma\) makes as you walk along the boundary. Now, the standard Poincaré-Hopf theorem tells us that the sum of the indices of the zero set of our vector field inside the disk is precisely equal to this turning number!

The tangent bundle of the sphere

Finally, we'll take a look at how everything works out for the familiar tangent bundle \(TS^2\) of the two-dimensional sphere. This calculation is taken from Example 1.9 of Hatcher's book on vector bundles.

To construct a clutching map for \(TS^2\), we first decompose \(S^2\) into its northern and southern hemispheres in the usual way, with the equator \(S^1\) running around the middle of the sphere. We define our chart on the northern hemisphere by picking an arbitrary vector at the north pole and parallel transporting it along lines of longitude to all other points in the northern hemisphere. This defines the \(x\)-axis of our tangent basis at each point, and we get \(y\)-axes by rotating these vectors 90 degrees counterclockwise about the normal vector to each point.

This choice of bases allows us to write any vector field in the northern hemisphere as an element of \(D^+ \times \RR^2\). We can similarly define tangent bases for points in the southern hemisphere by parallel transporting a vector in the same direction form the south pole. Now, we just need to work out the clutching map for this choice of charts on the northern and southern hemispheres. The clutching map should tell us the relative angle between the two \(x\)-axis vectors which we obtain at points on the equator.

Note that if we start off with the vector \((1, 0, 0)\) at the north and south poles, then our vectors at the equator point in the same direction at the \(\pm y\) ends of the sphere, and point in opposite directions at the \(\pm x\) points of the sphere.

In general, if we parameterize the points on the equator by their angle \(\theta\) with the \(+y\) point \((0, 1, 0)\), the angle at which our vectors meet is given by \(f(\theta) = 2\theta\). Hence, the homotopy class \([f]\) of our clutching map corresponds to the element \(2 \in \pi_1(S^1) \simeq \ZZ\). As we saw earlier, this means that the sum of the indices of the zeros of a section must be \(2\). Which is exactly the result guaranteed by the usual Poincaré-Hopf theorem.

Footnotes

1. Formally, these "twisted functions" are called "sections of a vector bundle," but we won't need this general language until later.

Suppose we have a 2\(-d\) quadratic form \(Q : \mathbb{R}^2 \to \mathbb{R}\). How can we compute its determinant?

This problem is a little under-specified. In fact, we can always pick a basis for \(\RR^2\) so that our quadratic form is diagonal with diagonal entries \(0\) or \(\pm 1\). So there is always some basis where our quadratic form has determinant \(0\) or \(\pm 1\).

But usually, we have a choice of inner product on \(\mathbb{R}^2\), and if we restrict to orthonormal bases with respect to this inner product then \(Q\) does indeed have well-defined eigenvectors, and a well-defined trace and determinant. And once we have the eigenvectors, the trace and determinant are easy to evaluate. If \(e_1, e_2\) are the eigenvectors with eigenvalues \(\lambda_1, \lambda_2\), then the trace of \(Q\) is given by \[ \tr Q = \lambda_1 + \lambda_2 = Q(e_1) + Q(e_2),\] and the determinant of \(Q\) is given by \[ \det Q = \lambda_1 \lambda_2 = Q(e_1)Q(e_2).\]

All this eigenvector business is actually not needed for computing the trace: we can pick any orthonormal vectors \(X, Y \in \mathbb{R}^2\), and the trace is given by \(Q(X) + Q(Y)\).

However, the determinant is trickier. In general, \(\det Q\) is not equal to \(Q(X)Q(Y)\) for a general pair of orthonormal vectors \(X\) and \(Y\). To see what goes wrong, we can write out \(Q\) as a matrix in the \(X, Y\) basis. The diagonal entries of this matrix are simply \(Q(X)\) and \(Q(Y)\). To obtain the off-diagonal entries, we can apply the polariziation identity (essentially a fancy name for the fact that \(xy = \tfrac{1}{2}(x+y)^2 - \tfrac{1}{2}x^2 - \tfrac{1}{2}y^2\)). Now that we have the matrix entries for \(Q\), we can apply the usual formula that \(\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = a d - b c\) and we find that \[\det Q = Q(X) Q(Y) - \tfrac{1}{4}\left(Q(X+Y) - Q(X) - Q(Y)\right)^2.\]

This formula simplifies a little if we define a new unit vector \(Z := \tfrac{1}{\sqrt{2}}\left(X + Y\right)\) which lies half way in between \(X\) and \(Y\). Then we can do a little bit of algebra to find that \[\begin{aligned} \det Q &= Q(X) Q(Y) - \left(Q(Z) - \tfrac{1}{2}\left(Q(X) + Q(Y)\right)\right)^2\\ &= Q(X)Q(Y) - Q(Z)^2 - \tfrac{1}{4}\left(Q(X)+Q(Y)\right)^2 + Q(Z)\left(Q(X) + Q(Y)\right)\\ &= \tfrac{1}{4}\left(Q(X) - Q(Y)\right)^2 + Q(Z) \left(Q(X) + Q(Y) - Q(Z)\right). \end{aligned}\]

We can simplify this formula even more by introducing the complementary unit vector \(W := \tfrac{1}{\sqrt{2}}\left(X - Y\right)\). Note that \(Z, W\) also form an orthonormal basis for \(\RR^2\). Since we can use the any orthonormal basis of \(\RR^2\) to compute the trace of \(Q\), we know that \(Q(X) + Q(Y) = \tr Q = Q(Z) + Q(W)\), and therefore the expression \(Q(X) + Q(Y) - Q(Z)\) in our formula is simply equal to \(Q(W)\). So in the end, the determinant of \(Q\) is given by \[\det Q = \tfrac{1}{4}\left(Q(X) - Q(Y)\right)^2 + Q(Z) Q(W),\] where \(X\) and \(Y\) form one orthonormal basis, and \(Z\) and \(W\) form another orthonormal basis rotated by \(45^\circ\).

As one last sanity check, note that if we take \(Z\) and \(W\) to be the eigenvectors \(e_1\) and \(e_2\), then \(Q(X) = Q(Y)\), and we so find again that the determinant is given by \(\det Q = Q(e_1) Q(e_2)\). But now we see that this is really a special case, and in general we need to add a correction term involving a pair of vectors rotated by \(45^\circ\) from our starting vectors.

Today, I want to write about a pretty simple problem: finding the line where two planes intersect. At the end of the day, it boils down to a simple formula that's easy to find elsewhere online. However, I find the derivation interesting, and it serves as a nice introduction to some powerful techniques that can be used on harder problems like intersecting conic sections.

But first, in case you just want the formula: the planes \(\langle n_1, x\rangle + d_1 = 0\) and \(\langle n_2, x \rangle + d_2 = 0\) intersect along the line \(r_o + t r_d\), where \[\begin{aligned} r_d &:= n_1 \times n_2,\\ r_o &:= \frac {r_d \times(d_2n_1 - d_1n_2)} {\|r_d\|^2}. \end{aligned}\]

Prelude: intersecting two lines in the plane

When solving these sorts of problems, it is helpful to work in homogeneous coordinates. So we represent a point \((x, y) \in \mathbb{R}^2\) by the vector \((x, y, 1) \in \mathbb{R}^3\) (or any scalar multiple of this vector). And similarly, we represent the line \(ax + by + c = 0\) using the vector \((a, b, c) \in \mathbb{R}^3\) (or any scalar multiple of this vector). It's no coincidence that both points and lines share the same representation here, but that's a discussion for another day.

Suppose now that we want to find the intersection between two lines \(\ell_1\) and \(\ell_2\) (which are represented as vectors in \(\mathbb{R}^3\). A point \(p\) lies on lines \(\ell_1\) if \(\langle p, \ell_1\rangle = 0\), and similarly for \(\ell_2\), so their intersection must be a vector \(p \in \mathbb{R}^3\) which is simultaneously orthogonal to both \(\ell_1\) and \(\ell_2\). We can construct such a vector easily by taking the cross product \(\ell_1 \times \ell_2\).

Visually, each of our lines in \(\mathbb{R}^2\) becomes a plane passing through the origin when represented in homogeneous coordinates in \(\mathbb{R}^3\) and the point in which the lines intersect in \(\mathbb{R}^2\) becomes the line passing through the origin in \(\mathbb{R}^3\). This line can easily be found by taking the cross product of the planes' normal vectors, and yields homogeneous coordinates for the intersection point in \(\mathbb{R}^2\). Computing the intersection of these planes is particularly simple since both planes pass through the origin; computing the intersection between general planes will take us a bit more work later on.

Before we move on, though, I'll mention that this exact same construction also works to compute the line that connects two points. If we have points \(p_1, p_2\), then the line passing through these two points must be given by a vector \(\ell \in \mathbb{R}^3\) which is simultaneously orthogonal to \(p_2\) and \(p_2\), i.e. \(\ell = p_1 \times p_2\).

Lines in 3D

Now let's move to 3D. Since we've moved up a dimensions, points and planes are represented by vectors in \(\mathbb{R}^4\). But what about lines? One natural representation of lines is given by Plücker coordinates. If we consider the line traced out by \(r(t) = r_o + t r_d \in \mathbb{R^3}\), its Plücker coordinates are given by the vector \[(r_o \times r_d, r_d) \in \mathbb{R}^6.\] This formula looks a little strange at first, but it has many nice properties: for instance, if we had picked some other point on the line as the base point, say \(r_o + \lambda r_d\), then we would still get the same Plücker coordinates since \((r_o + \lambda r_d) \times r_d = r_o \times r_d\). Similarly, if we scale the line direction \(r_d\) by a constant \(\lambda\), then our Plücker coordinates also get scaled by \(\lambda\), but still represent the same line. And moreover, we can recover an equation from the line by its Plücker coordinates. We can recover the direction from the last 3 coordinates, and we can find a point on the line by applying the identity that \(r_d \times (r_o \times r_d) = \|r_d\|^2 r_o - \langle r_o, r_d\rangle r_d\), yielding the point on our line which is closest to the origin.

What if we're given two points \(p_1, p_2\) and we want the line \(\ell\) connecting them? To start with let's represent out points using ordinary coordinate vectors in \(x_1, x_2 \in \mathbb{R}^3\). We can think of our line as starting at \(x_1\), and proceeding in direction \(x_2 - x_1\), so we obtain Plücker coordinates \[\ell = ((x_2 - x_1) \times x_1, x_2-x_1) = (x_2 \times x_1, x_2 - x_1).\] But if you stare at this formula for a minute, and are familiar with wedge products, you may notice that this formula can be simplified to \(\ell = (x_2, 1) \wedge (x_1, 1).\) Indeed, if we write each \(p_i\) using homogeneous coordinates in \(\mathbb{R}^4\), the equation becomes \[\ell = p_2 \wedge p_1.\]

Planes in 3D

What about the plane \(P\) passing through three points \(p_1, p_2, p_3\)? We can construct its homogeneous coordinates as \(P = *(p_1 \wedge p_2 \wedge p_3)\), noting that \[\langle P, p_i \rangle = p_1 \wedge p_2 \wedge p_3 \wedge p_i = 0,\] for each of our points \(p_i\). Similarly, the plane passing through a point \(p\) and a line \(\ell\) is given by \(P = *(\ell \wedge p)\).

This formula will help us determine when a line \(\ell\) is contained in a plane \(P\): the line \(\ell\) lies in \(P\) if there is some point \(p\) such that \(P = *(\ell \wedge p)\). Using another wedge product identity, we can write this equation as \(P = -\iota_{p^\flat} *\ell\), which in turn is true if and only if \(P \wedge *\ell = 0\).

Intersecting planes in 3D

Now we can finally find the formula to intersect two planes in 3D. If we are given two planes \(P_1\) and \(P_2\), the line \(\ell\) contained in their intersection must satisfy the equations \[\begin{aligned}P_1 \wedge *\ell &= 0,\\P_2 \wedge *\ell &= 0.\end{aligned}\] It turns out that this is precisely the system of equations that we solved above to find the intersection of lines in 2D, just written in exterior algebra! And, it has the same solution (written in exterior algebra): \(\ell = *(P_1 \wedge P_2)\).

All that remains is to unpack all of the exterior algebra a formula in terms of more familiar vector algebra operations. Let's denote our planes as \(P_i = (n_i, d_i)\). Then, as we saw above, their wedge product is given by \[P_1 \wedge P_2 = (n_1 \times n_2, d_2 n_1 - d_1 n_2).\] Then the Hodge star simply swaps the components of this vectors, yielding \[\ell = *(P_1 \wedge P_2) = (d_2 n_1 - d_1 n_2, n_1 \times n_2).\] Finally, we can use the formula from earlier to convert these Plücker coordinates into a point on our line and a direction. Putting everything together, the planes \(\langle n_1, x\rangle + d_1 = 0\) and \(\langle n_2, x \rangle + d_2 = 0\) intersect along the line \(r_o + t r_d\), where \[\begin{aligned} r_d &:= n_1 \times n_2,\\ r_o &:= \frac {r_d \times(d_2n_1 - d_1n_2)} {\|r_d\|^2}. \end{aligned}\]

A more symmetric form

I'll quickly note at the end that everything I covered here is arguably cleaner if you use exterior algebra from the beginning. Rather than representing both points and hyperplanes using vectors in \(\mathbb{R}^{n+1}\), we can represent points as vectors in \(p \in \mathbb{R}^{n+1}\) and hyperplanes as vectors in the dual space \(P \in \Lambda^n \mathbb{R}^{n+1}\). And, rather than looking at the inner product between \(p\) and \(P\) (which is no longer well-defined), we can consider the primal-dual pairing \(\langle P, p\rangle = *(p \wedge P).\) One nice aspect of version of the theory is that then our formula for the line \(\ell\) between two points \(p_1, p_2\) is always given by \(\ell = p_1 \wedge p_2\), regardless of whether we are considering \(\mathbb{R}^2\), \(\mathbb{R}^2\), or \(\mathbb{R}^n\).

Suppose we have two conic sections and we want to find their intersections. If the conics are circles, then this is easy, and can be done in closed form. However, if we have, say, a pair of hyperbolas then things are harder. For instance, a pair of hyperbolas can intersect in four points, unlike a pair of circles which generally intersect in two points.

However, there is still a pretty slick algorithm for computing the intersection points between any two conic sections, which I'll discuss in this post. A detailed description can be found in Perspectives on Projective Geometry by Jürgen Richter-Gebert.

Matrix representation of conic sections

It turns out to be much easier to do these kind of calculations on conics if we represent them as symmetric matrices via homogeneous coordinates. Explicitly, a conic can be written as the set of points \(x, y\) satisfying an equation of the form \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.\] This equation can be rewritten in matrix notation as \[\begin{pmatrix}x & y & 1\end{pmatrix} \begin{pmatrix}A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F\end{pmatrix} \begin{pmatrix}x\\y\\1\end{pmatrix}=0,\] and we can identify our conic with the matrix in the middle of this equation. Writing conics as matrices allows us to work with them algebraically: for instance, we can add together two conics by adding together their matrices. This kind of algebraic manifpulation of conics is the key to the algorithm for computing intersections that I describe below.

Computing the intersections

Suppose we have two conics given by symmetric matrices \(Q_1, Q_2 \in \mathbb{R}^{3 \times 3}\) with intersection points \(p_1, \ldots, p_4\). Explicitly, this means that for each \(p_i\), we have \[p_i^T Q_1 p_i = p_i^T Q_2 p_i = 0.\]

But by linearity, this means that for any coefficients \(\lambda, \mu \in \mathbb{R}\), we also have \[p_i^T (\lambda Q_1 + \mu Q_2) p_i = 0,\] defining a whole family of conic sections that also pass through these four points. If we plot some of the other conics generated from the two hyperbolas shown above, we see that we find some hyperbolas, some ellipses, and even a pair of diagonal lines all passing through our four intersection points.

Those two diagonal lines turn out to be the key to locating the intersection points \(p_i\). After all, it's pretty easy to solve for the intersection between a conic and a line – you just have to solve a quadratic equation.

Formally, the two diagonal lines make up a degenerate hyperbola. A matrix represents a degenerate conic if its determinant is zero, so we can find this degenerate hyperbola by solving \(\det(\lambda Q_1 + \mu Q_2) = 0\), which is a cubic equation (since \(Q_1\) and \(Q_2\) are 3 \(\times\) 3 matrices). Then we simply need to identify the two lines which compose this degenerate hyperbola and intersect them with one of the original hyperbolas. In summary, the algorithm is as follows:

1. Find \(\lambda, \mu \neq 0\) such that \(\det(\lambda Q_1 + \mu Q_2) = 0\).
2. Decompose the degenerate hyperbola \(\lambda Q_1 + \mu Q_2\) into a pair of lines \(g, h\).
3. Intersect \(g\) and \(h\) with \(Q_1\) to identify all four intersection points.

Below, I'll discuss each of the steps in some more detail.

Finding the degenerate hyperbola

Richter-Gebert gives a detailed algorithm for finding the roots of a cubic equation in homogeneous coordinates, but if you have access to a linear algebra library, you can find the roots more easily. First, if \(Q_2\) is itself degenerate (i.e. \(\det(Q_2) = 0\)), then taking \(\lambda = 0, \mu = 1\) gives us the solution that we desire. On the other hand, if \(Q_2\) is nondegenerate (i.e. invertible), then we can set \(\lambda = 1\) and multiply through by \(Q_2^{-1}\) and instead solve \[\det\left(\mu I - \left(-Q_1 Q_2^{-1}\right)\right)=0.\] Now, this problem may not obviously look easier, but there is a big advantage: the solutions \(\mu\) are precisely the eigenvalues of the matrix \(-Q_1Q_2^{-1}\), which can easily be computed by standard linear algebra libraries!

Decomposing the degenerate hyperbola into lines

For now, suppose we have some degenerate hyperbola represented by a symmetric matrix \(A \in \mathbb{R}^{3\times3}\). This degenerate hyperbola is really just a pair of lines, which can be represented by two vectors \(g, h \in \mathbb{R}^3\). Explicitly, a point \(x\) lies on the first line if \(g^Tx = 0\), and similarly for \(h\). Hence, \(x\) lies on the union of the two lines whenever \(x^T gh^T x = 0\). So if our symmetric matrix \(A\) represents this pair of lines, then (up to scale), \(A\) must equal the symmetrized matrix \(gh^T + hg^T\). Richter-Gebert gives a neat algorithm for computing \(g\) and \(h\) from \(A\). First, he notes that the cofactor matrix of \(A\) is given by \(A^\triangle = -(g \times h)(g\times h)^T\). And furthermore, a direct calculation shows that \(gh^T - hg^T\) is simply the cross product matrix \([g\times h]_\times\), and hence \(2gh^T = A + [g\times h]_\times\).

Concretely, then, \(g\) and \(h\) may be obtained form \(A\) as follows:

1. \(B \gets A^\triangle\)
2. \(i \gets\) the index of a nonzero diagonal entry of \(B\)
3. \(\beta \gets \sqrt{-B(i,i)}\)
4. \(p \gets B(:, i) / \beta\)
5. \(C \gets A + [p]_\times\)
6. \(i, j \gets\) the index of a nonzero entry of \(C\)
7. \(g = C(i, :), h = C(:, j)\)

Intersecting a line with a conic

Suppose we wish to intersect the conic \(A \in \mathbb{R}^{3\times 3}\) with the line \(g \in \mathbb{R}^3\). Perhaps unsurprisingly, Richter-Gebert also has a neat algorithm for intersecting a conic with a line in homogeneous coordinates. I find the motivation for this procedure to be the most mysterious, although the steps themselves are fairly simple. First, we assume that \(g(2) \neq 0\), reindexing the entries if necessary. Then, we do the following:

1. \(B \gets [g]_\times^T A [g]_\times\)
2. \(\alpha \gets \frac 1 {g(2)} \sqrt{B(0, 1)^2 - B(0, 0) B(1, 1)}\)
3. \(C \gets B + \alpha [g]_\times\)
4. \(i, j \gets\) the index of a nonzero entry of \(C\)
5. \(g = C(i, :), h = C(:, j)\)

This post lists some useful formulas for evaluating the distortion of maps between triangle meshes. The distortions considered here are intrinsic, meaning they only depend on the change in triangle edge lengths. Conveniently, they also happen to have fairly simple formulas in terms of the triangle edge lengths.

Background

There are many ways to measure the distortion of a map \(\phi : M \to N\) between surfaces. The general strategy is to compute the Jacobian \(J_\phi\), which is a \(2\times 2\) matrix, and consider different functions of its singular values \(\sigma_1, \sigma_2\). For instance, the product \(\sigma_1\sigma_2\) measures the area distortion of \(\phi\) (as it is simply the determinant \(\det J_\phi\)), while the ratio \(\sigma_1 / \sigma_2\) measures the amount of anisotropic stretching induced by \(\phi\). (See e.g. section 2 of Khodakovsky et al. [ 2003; free version] for more details.)

A piecewise-linear map between triangle meshes, deforms each triangle by a linear map, so its distortion is constant on each triangle. Below, I give some formulas for computing such per-triangle distortions intrinsically, that is, computing the distortions using only the lengths of the initial and deformed triangles. In each formula, I refer to the triangle's vertices as \(i, j, k\). The initial edge lenths are denoted \(\ell_{ij}, \ell_{jk}, \ell_{ki}\), and the corner angles are denoted \(\alpha_i, \alpha_j, \alpha_k\). Quantities measured after deformation are denoted \(\tilde \ell_{ij}\), etc.

Area Distortion

The area distortion \(\sigma_1\sigma_2\) is simply given by the ratio of the deformed triangle's area to the original triangle's area. Using Heron's formula, one can show that the area of a triangle is given by \[\text{area}_{ijk} := \tfrac{1}{2\sqrt2}\sqrt{\left(\ell^2\right)^T \!\!\! A \ell^2},\] where \(\ell^2\) denotes the vector of squared edge lengths $(\ell_{ij}^2, \ell_{jk}^2, \ell_{ki}^2)^T$, and \(A\) is the matrix \[ A = \frac 12 \begin{pmatrix}-1 & 1 & 1 \\ 1 & -1 & 1\\ 1 & 1 & -1\end{pmatrix}.\] Hence, the area distortion is given by \[\sigma_1\sigma_2 = \sqrt{\frac{(\tilde \ell^2)^T A \tilde \ell^2}{(\ell^2)^T A \ell^2}}.\]

Symmetrized Anisotropic Distortion

Before considering the anisotropic distortion \(\sigma_1/\sigma_2\) itself, we begin with a symmetrized version \(\frac{\sigma_1}{\sigma_2} + \frac{\sigma_2}{\sigma_1}\), which is given by a similar formula: \[ \frac{\sigma_1}{\sigma_2} + \frac{\sigma_2}{\sigma_1} = \frac{\left(\ell^2\right)^T \!\!\! A \tilde \ell^2}{\sqrt{\left(\ell^2\right)^T \!\!\! A\, \ell^2}\sqrt{(\tilde \ell^2)^T \! A \tilde \ell^2}}. \] This formula can also be written directly in terms of the angles \(\alpha_i\) as \[ \frac{\sigma_1}{\sigma_2} + \frac{\sigma_2}{\sigma_1} = \begin{pmatrix} \cot \alpha_i\\\cot\alpha_j\\\cot\alpha_k\end{pmatrix}^T A^{-1} \begin{pmatrix} \cot \tilde\alpha_i\\\cot\tilde\alpha_j\\\cot\tilde\alpha_k\end{pmatrix}, \] where \(A^{-1}\) is given by \[ A^{-1} = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}.\] The equivalence of these two expressions follows from the identity \[ \begin{pmatrix} \cot \alpha_i \\ \cot \alpha_j \\ \cot\alpha_k \end{pmatrix} = \frac {A^{-1}\ell^2} {\sqrt{\left(\tilde \ell^2\right)^T \!\!\! A \tilde \ell^2}}, \] (which can itself be derived using the law of cosines, the sine formula for area, and the definition that $\cot \alpha = \tfrac{\cos\alpha}{\sin\alpha}$.)

A derivation of the cotan formula for distortion, courtesy of Boris Springborn, can be found here, and connections to hyperbolic geometry are discussed, e.g., in p.11 of Joshua Bowman's PhD thesis with John H. Hubbard.

Anisotropic Distortion

The anisotropic distortion can easily be computed from the symmetrized anisotropic distortion by solving the quadratic equation \(y = x + \frac 1x\). Concretely, if the symmetrized distortion is given by \(d_s\), then the ordinary anisotropic distortion is given by \[ \frac{\sigma_1}{\sigma_2} = \frac 12 \left(d_s + \sqrt{d_s^2-4}\right) \]

Other distortions

Once you have computed the distortions \(\sigma_1\sigma_2\) and \(\sigma_1 / \sigma_2\), then you can find the values of the two singular values \(\sigma_1\) and \(\sigma_2\) by multiplying and dividing the distortion values respectively. These singular values can then be used to evaluate many other measurements of distortion.

For discussion of intrinsic calculations for distortion in the context of elasticity, see Appendix B of Sassen et al. [2020; arxiv version]

Everyone who discusses Möbius transformations mentions that they map circles to circles, but it can be hard to find concrete equations describing exactly how a given circle is changed by a given Möbius transformation.

I recently had to derive a formula for the curvature of a circle after applying a certain Möbius transformation, and the resulting formula was surprisingly simple. Suppose we start with a circle centered at \(x \in \mathbb{R}^2\) with radius \(r > 0\). If we apply a Möbius transformation which fixes the unit circle and sends some point \(z\) inside the unit disk to the origin, then the circle's curvature becomes \[ \tilde r = \frac{1-\|z\|^2}{1 + 2\langle x, z \rangle + (\|x\|^2 - r^2)\|z\|^2}r. \]

In particular, I was interested in Möbius transformations fixing the unit circle. If we ignore rotations, which don't affect the curvature of circles anyway, such a Möbius transformations is determined entirely by the point \(z\) which is sent to zero: \[f_z(p) := \frac{p-z}{1-\bar zp}.\] (See e.g. the wikipedia article.) Note that the inverse of \(f_z\) is given by \(f_{-z}\), since \(f_z(f_{-z}(p)) = p\).

Suppose we start with a circle of radius \(r\) centered at a point \(x \in \mathbb{R}^2\). We can encode our circle as the zero level set of a quadratic form \[Q(p) := a \|p\|^2 - \langle b, p \rangle + c = 0,\] with coefficients \(a = 1\), \(b = 2x\), and \(c = \|x\|^2 - r^2\). We can recover the center from \(Q\) as \(x = \tfrac c {2a}\), and the radius as \(r = \tfrac 1 {2a} \sqrt{\|b\|^2 - 4 ac}\).

Now we can determine how the Möbius transformation \(f_{-z}\) affects our circle by finding the zero set of the transformed map \(Q(f_{z}(p))\). The calculation is easier to do if we express \(Q(p)\) using complex numbers as \(Q(p) := a \bar p p - \Re(\bar b p) + c\). If we substitute in the definition of \(f_z(p)\) and do some nasty algebra, we find that \[\begin{aligned}\|1-\bar zp\|^2Q(f_z(p)) &= \left((a + \Re(\bar b z) + c \|z\|^2\right) \|p\|^2\\ &\quad- \Re\left((2 a \bar z + \bar b + b \bar z^2 + 2 c z)p\right)\\ &\quad+ (a \|z\|^2 + \Re(\bar b z) + c) \end{aligned}.\]

That is, up to a scalar multiple, \(Q(f_z(p))\) is itself a quadratic form with coefficients \[\begin{aligned} \tilde a &:= a + \Re(\bar b z) + c \|z\|^2\\ \tilde b &:= 2az + b + \bar b z^2 + 2 c \bar z\\ \tilde c &:= a \|z\|^2 + \Re(\bar b z) + c. \end{aligned}\]

We can now extract the transformed radius from \(Q(f_z(p))\). Another gnarly calculation shows that \[ \|\tilde b\|^2 - 4 \tilde a \tilde c = (\|b\|^2 - 4 a c)\left(1-\|z\|^2\right)^2 = 4a^2 r^2 \left(1-\|a\|^2\right)^2. \] Therefore, the radius of the transformed circle is given by \[ \tilde r = \frac{1-\|z\|^2}{1 + \tfrac 1a\langle b, z \rangle + \tfrac ca \|z\|^2}r, \] as promised above. (The sign on the \(\langle b z\rangle\) term is flipped in the earlier expression above since there we applied \(f_z\) to the circle, rather than \(f_{-z}\).)

The \(1-\|z\|^2\) term suggests that there may be a slick proof of this formula using hyperbolic geometry: Möbius transformations which fix the unit circle are precisely the isometries of the Poincaré disk model of the hyperbolic plane, whose metric tensor is given by: \[ds^2 = \frac{4 dz^2}{\left(1-\|z\|^2\right)^2}.\] But I'm happy with this concrete calculation for now.