About this edition

This is an independent English edition of Holger Brenner’s course Bündel, Garben und Kohomologie (Osnabrück 2019–2020). Its scope is exactly 30 lectures and 30 worksheets, Units 1–30, in source order. All 495 exercises are retained. The translation scope includes the twenty-five publicly available solutions. For the other 470 exercises, the edition preserves documented negative search results; it does not invent solutions absent from the source.

Translation, checking and edition production are assisted by OpenAI Codex gpt-5.6-sol, Ultra. The work is carried out on the user’s instructions. This model-provenance statement does not replace credits to the source author or human contributors. The edition is not an official publication of Holger Brenner, the University of Osnabrück, Wikiversity, the Wikimedia Foundation or OpenAI, and does not imply their endorsement.

Frozen source authority

The textual authority consists of the frozen semantic Wikiversity revisions. The course root has pageid 108997, revision 1052895 dated 2025-08-27T20:02:50Z, and MediaWiki SHA-1 e881e7c53531765849865454d4d0c643c2066c6d. The frozen course index records 30 lecture identities and 30 worksheet identities. The course manifest, authority/wikiversity-bgk/course/COURSE_AUTHORITY_MANIFEST.json, is 34,279 bytes, with SHA-256 ea0bf346e261db8ed80b7565f7746e95c79e0c376d25d9fbce5d96879dff7dd8. Official PDFs serve as visual and numbering witnesses, not as replacements for the more recent semantic authority.

Each of Units 1–30 has an authority manifest, official unit PDFs and component-rights records. The edition’s scope covers the lecture, worksheet and available source solutions of every unit. Source closure records file identities, exercise order, public solutions and negative candidate results. English reader checks, reproducible-build reports and release-file identities are separate evidence from these frozen source records; Indonesian build evidence is not presented as proof of an English build.

Structure and study routes

The following order is that of the source lectures and worksheets; these routes do not authorise renumbering. Each range forms a coherent reading route.

Exercises and solutions

Public solutions occur in Unit 2 (Exercise 2.4), Unit 3 (3.1), Unit 5 (5.5), Unit 7 (7.14), Unit 8 (8.3, 8.4 and 8.11), Unit 11 (11.9, 11.13 and 11.14), Unit 12 (12.5 and 12.10), Unit 16 (16.12), Unit 18 (18.6, 18.17 and 18.18), Unit 19 (19.10), Unit 21 (21.3, 21.9 and 21.10), Unit 22 (22.19), Unit 25 (25.1), Unit 28 (28.6), and Unit 29 (29.5 and 29.12).

No public solutions were found at the source boundary for Units 1, 4, 6, 9, 10, 13, 14, 15, 17, 20, 23, 24, 26, 27 or 30. In units with some public solutions, every other exercise also has its own negative result. Thus 470 is a count of documented negative search results, not blank solution pages and not proof that solutions have never existed elsewhere.

Media and accessibility

Eight reader-media positions with available component binaries occur in Units 1, 2, 4 and 29. Each asset retains its credit, creator, source, alternative text and component licence. Unit 8 has one media position declaring File:Spektrum_von_Z._xcf, but that source binary is missing. The edition preserves the creator’s name and the source’s inline licence label, the meaning of the caption, alternative text and the fact of the missing binary. It does not claim recovery or fabricate a replacement source asset.

Units 3, 5–7, 9–28 and 30 have no reader-media positions. The official unit and course PDFs are authority witnesses, not reader illustrations, so they are not counted as illustration assets. The absence of media positions in those units is not missing content.

Licence, attribution and component rights

The frozen semantic course text and this translation are shared under Creative Commons Attribution-ShareAlike 4.0 International. The frozen Commons metadata for the official course PDF states CC BY-SA 4.0, whereas the visible notice on page 265 of that PDF states CC BY-SA 3.0. The edition preserves both records and makes no blanket relicensing claim.

Media and other components remain subject to their respective rights notices; the text licence does not automatically relicense them all. Attribution, ShareAlike obligations, component notices and the non-endorsement statement must be retained when this edition is copied, adapted or distributed.

How to use this reader

Read each lecture before attempting the worksheet with the same number. Follow the routes in order: the language of bundles and sheaves; sheaf operations and spectra; affine and projective schemes; modules and differentials; tangent bundles, Picard groups and divisor classes; homological and cohomological tools; then projective-curve geometry through Riemann–Roch. Read edition notes where they occur to distinguish source content from verifiable corrections, translation decisions, notation notes and media provenance. Do not interpret a negative solution search as missing translated content. The terminology guide that follows helps identify corresponding terms across units without obscuring mathematical distinctions.

Terminology guide

This guide connects German terms with their use in Units 1–30. It adapts the independent Indonesian edition’s connective glossary for English readers. Synonymous translations do not denote new mathematical objects: read them together with the definitions, hypotheses and notation in the relevant unit. The preferred forms here support cross-unit searching without changing formulae or wording inside source quotations.

Objects that must be distinguished

Corresponding terms across units

Terms in the later units

Notation and origin of this guide

Prose terminology does not change source notation. For example, an image may still be written with the source operator bild, and the word spectrum does not require replacing Spek by Spec inside a formula. Differences in indices, equality signs, isomorphisms and hypotheses are not vocabulary variants.

This connective guide is based on Holger Brenner’s course and the terminology decisions of the Indonesian edition, adapted to English. AI-assisted preparation: OpenAI Codex gpt-5.6-sol, Ultra. This connective text is licensed under CC BY-SA 4.0; every source component retains its own credits and licence. This is not an official guide from the source author or institutions and does not imply their endorsement.

Lecture 1: Parameter-Dependent Systems of Linear Equations and Vector Bundles

Parameter-dependent systems of linear equations

Consider the real linear equation

$$7u-5v+2w=0.$$

Its solution set

$$L=\left\{(u,v,w)\in\mathbb R^3\mid 7u-5v+2w=0\right\} \subset\mathbb R^3$$

is a two-dimensional real vector subspace of $\mathbb R^3$. Solving such a linear equation means, among other things, finding a basis for $L$. In this case, for example,

$$L=\left\langle \begin{pmatrix}5\\7\\0\end{pmatrix}, \begin{pmatrix}2\\0\\-7\end{pmatrix} \right\rangle.$$

The methods of solution are largely independent of the particular coefficients of the linear equation, although we shall see below the limitations of this statement. If we replace specific numbers by coefficients depending functionally on parameters, we may ask how the solution space varies with those parameters. For example, consider the linear equation depending on a parameter $s$,

$$7u-5v+(s^2-3s-10)w=0.$$

For each $s$, the solution space $L_s$ depends on $s$, but remains a two-dimensional subspace,

$$L_s\subset\mathbb R^3.$$

In other words, the solution space is a plane moving through space as $s$ varies. We may ask for which values of $s$ the vector

$$\begin{pmatrix}5\\-3\\8\end{pmatrix}$$

is a solution, that is, belongs to $L_s$. We may also ask whether there are distinct parameters $s,t$ for which

$$L_s=L_t$$

as subspaces of $\mathbb R^3$; whether the solution space always has a basis of the form

$$\left\langle \begin{pmatrix}a\\b\\0\end{pmatrix}, \begin{pmatrix}c\\0\\d\end{pmatrix} \right\rangle;$$

or whether there is always a solution vector of the form

$$\begin{pmatrix}1\\0\\e\end{pmatrix}.$$

Recall that the algorithm for solving systems of linear equations, Gaussian elimination, branches when certain coefficients are $0$ or become $0$ during the algorithm. The equation

$$7u-5v+0w=0$$

has solution space

$$\left\langle \begin{pmatrix}5\\7\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix} \right\rangle$$

and contains no vector of the form $\begin{pmatrix}1&0&e\end{pmatrix}^{\mathsf T}$. Since $s=-2$ and $s=5$ are the roots of the quadratic polynomial $s^2-3s-10$, at these two parameter values the parametrised equation above becomes

$$7u-5v+0w=0.$$

Thus, for these two values, $L_s$ contains no vector of the form $\begin{pmatrix}1&0&e\end{pmatrix}^{\mathsf T}$. For all other parameter values, the solution space contains the vector

$$\begin{pmatrix} 1\\[2pt]0\\[2pt]-\dfrac{7}{s^2-3s-10} \end{pmatrix}.$$

A certain aspect of the solution space therefore itself depends functionally on the parameter.

It is natural to study the dependence of a linear equation or a system of linear equations on parameters in two stages. In the first stage, the coefficients of the equations themselves are treated as variables, or universal parameters, and we study how the solution spaces vary with them. In particular, we want to understand qualitative jumps in the behaviour of the solution spaces. In the second stage, we impose additional, more or less restrictive conditions on the universal parameters, or allow them to depend functionally on other parameters.

Example 1.1: one equation in two variables

Consider the general real linear equation

$$su+tv=0$$

in the variables $u,v$ and parameters $s,t$, which serve as indeterminate coefficients. We want to understand the solution space

$$L_{(s,t)}=\left\{(u,v)\mid su+tv=0\right\}\subseteq\mathbb R^2$$

as a function of the parameters $(s,t)$. An extreme case occurs at $(s,t)=(0,0)$: every $(u,v)$ satisfies the equation, so the solution space is the whole two-dimensional space $\mathbb R^2$. If $(s,t)\ne(0,0)$, the solution space is one-dimensional, and a basis vector for this solution line is

$$\begin{pmatrix}t\\-s\end{pmatrix}.$$

Thus, over the parameter space $\mathbb R^2\setminus\{(0,0)\}$, the solution space has the uniform description

$$L_{(s,t)}= \left\{c\begin{pmatrix}t\\-s\end{pmatrix}\mathrel{\Big|}c\in\mathbb R\right\}.$$

A more compact interpretation is obtained by considering the total solution space

$$L=\left\{(s,t,u,v)\mid su+tv=0\right\}\subseteq\mathbb R^4.$$

Note that $L$ is not a linear subspace of $\mathbb R^4$. The solution space for a particular parameter value $(s,t)$ is obtained by intersecting $L$ with the affine plane $(s,t)\times\mathbb R^2$. Under the total projection

$$L\longrightarrow\mathbb R^2\times\mathbb R^2 \stackrel{p_{s,t}}{\longrightarrow}\mathbb R^2, \qquad (s,t,u,v)\longmapsto(s,t),$$

$L_{(s,t)}$ is the fibre over $(s,t)$. The total solution space displays both the variation of the solution lines with the parameter and their degeneration into a solution plane over the origin. The behaviour away from the parameter origin is described by the restriction

$$L'=L\setminus\bigl(\{(0,0)\}\times\mathbb R^2\bigr) =p^{-1}\!\left(\mathbb R^2\setminus\{(0,0)\}\right) \longrightarrow\mathbb R^2\setminus\{(0,0)\}.$$

Each fibre of this restricted projection is a one-dimensional solution space. Moreover, there is a bijection

$$\begin{aligned} \left(\mathbb R^2\setminus\{(0,0)\}\right)\times\mathbb R &\longrightarrow L',\\ (s,t;c)&\longmapsto(s,t,ct,-cs), \end{aligned}$$

which is linear for each parameter $(s,t)$. On the left is the direct product of the base space $\mathbb R^2\setminus\{(0,0)\}$ and the fibre $\mathbb R$, which is independent of the base point. On the right is a family of varying lines in $\mathbb R^2$, but the bijection translates one description into the other.

Editorial note - order of factors and base space. In the definition of $L'$, the source prints $\mathbb R^2\times(0,0)$, whereas equality with $p^{-1}(\mathbb R^2\setminus\{(0,0)\})$ requires removing the fibre over the origin, namely $\{(0,0)\}\times\mathbb R^2$. The source prose subsequently prints $\mathbb R^2\times(0,0)$ again as the base space, whereas the domain of the immediately preceding bijection is $\mathbb R^2\setminus\{(0,0)\}$. This edition follows the two displayed maps and explicitly records the discrepancy.

Example 1.2: one equation in three variables

Consider the general real linear equation

$$ru+sv+tw=0$$

in the variables $u,v,w$ and parameters $r,s,t$, which serve as indeterminate coefficients. We want to understand the solution space

$$L_{(r,s,t)}= \left\{(u,v,w)\mid ru+sv+tw=0\right\}\subseteq\mathbb R^3$$

as a function of the parameters $(r,s,t)$. At $(r,s,t)=(0,0,0)$, the solution space is the whole of $\mathbb R^3$. If $(r,s,t)\ne(0,0,0)$, it is two-dimensional. We exclude the origin from the parameter space and consider the total solution space

$$\begin{aligned} L={}&\left\{(r,s,t,u,v,w)\mathrel{\Big|} ru+sv+tw=0, (r,s,t)\ne(0,0,0)\right\}\\ &\subseteq\left(\mathbb R^3\setminus\{(0,0,0)\}\right)\times\mathbb R^3, \end{aligned}$$

together with the projection $p$ to $\mathbb R^3\setminus\{(0,0,0)\}$. The fibre of $p$ over a particular parameter $(r,s,t)$ is the solution space $L_{(r,s,t)}$ of the equation determined by that parameter tuple.

Editorial note - notation for the origin and scope of the product. In this inclusion the source prints $\mathbb R^3\setminus\{0,0,0\}\times\mathbb R^3$, without writing the origin as a tuple or using parentheses to separate the base space from the fibre factor. The projection in the next sentence uniquely determines the intended meaning. This edition writes $(\mathbb R^3\setminus\{(0,0,0)\})\times\mathbb R^3$.

Can we give a basis for each solution space that depends on the parameters in an explicit computational and algebraic way? Since we have removed the origin,

$$\mathbb R^3\setminus\{(0,0,0)\} =\{(r,s,t)\mid r\ne0\}\cup \{(r,s,t)\mid s\ne0\}\cup \{(r,s,t)\mid t\ne0\}.$$

The base space can therefore be written as a union of three open sets. Over the open set $r\ne0$, for example, a basis is given by

$$(s,-r,0)\quad\text{and}\quad(t,0,-r).$$

The condition $r\ne0$ ensures that the two vectors are linearly independent. Indeed, the two vectors are well-defined solutions everywhere, but when $r=0$ they lose their linear independence and hence do not form a basis everywhere. In any case, the map

$$\begin{aligned} \{(r,s,t)\mid r\ne0\}\times\mathbb R^2 &\longrightarrow L|_{\{(r,s,t)\mid r\ne0\}},\\ (r,s,t;c,d)&\longmapsto\bigl(r,s,t;\,c(s,-r,0)+d(t,0,-r)\bigr) \end{aligned}$$

is a computationally simple bijection between the product of the base space with $\mathbb R^2$ and the solution space over $\{(r,s,t)\mid r\ne0\}$.

Editorial note - base coordinates. The source writes only the fibre vector on the right-hand side of this map. Since its codomain is the total space $L$, the unchanged base coordinates $(r,s,t)$ are displayed here as well.

We now ask whether it is possible to give, globally on all of $\mathbb R^3\setminus\{(0,0,0)\}$, a basis of the solution space that varies with the base point. The question is whether there exist two functions $u(r,s,t)$ and $v(r,s,t)$ with values in $\mathbb R^3$ that always form a basis of the corresponding fibre, and in particular belong to it. With no further conditions on $u$ and $v$, this is possible by a case-by-case definition. However, it is no longer possible if both functions are required to be continuous. By continuity, the global functions $u$ and $v$ are already determined by their values on the dense open set

$$U=\{(r,s,t)\mid r\ne0\} \subseteq\mathbb R^3\setminus\{(0,0,0)\}.$$

Using the basis over $U$ given above, we can write

$$u=\alpha(r,s,t)\begin{pmatrix}s\\-r\\0\end{pmatrix} +\beta(r,s,t)\begin{pmatrix}t\\0\\-r\end{pmatrix}$$

and

$$v=\gamma(r,s,t)\begin{pmatrix}s\\-r\\0\end{pmatrix} +\delta(r,s,t)\begin{pmatrix}t\\0\\-r\end{pmatrix},$$

where $\alpha,\beta,\gamma,\delta$ are continuous real-valued functions on $U$. We cannot expect these coefficient functions to be defined on all of $\mathbb R^3$, so the argument in the continuous case becomes more complicated. The result will follow from Theorem 2.3; see Remark 2.4.

For now, we therefore restrict attention to rational functions whose denominators may contain a power of $r$, that is, rational functions on $U$. Consider

$$\begin{aligned} u &=\alpha\begin{pmatrix}s\\-r\\0\end{pmatrix} +\beta\begin{pmatrix}t\\0\\-r\end{pmatrix}\\ &=\frac{P}{r^m}\begin{pmatrix}s\\-r\\0\end{pmatrix} +\frac{Q}{r^n}\begin{pmatrix}t\\0\\-r\end{pmatrix}, \end{aligned}$$

where $P,Q$ are polynomials and factors of $r$ have been cancelled wherever possible. Since $u$ as a whole is defined on all of $\mathbb R^3$, the exponent $m$, and likewise $n$, is at most $1$; otherwise $u$ would have a pole. For $m=n=1$, the first component gives a polynomial equation of the form

$$rN+sP+tQ=0, \qquad N,P,Q\in\mathbb R[r,s,t].$$

In this case, the relevant idea being the Koszul resolution,

$$(N,P,Q)=A(-s,r,0)+B(t,0,-r)+C(0,t,-s)$$

for polynomials $A,B,C\in\mathbb R[r,s,t]$. Similarly, $v$ has a representation in terms of $(N',P',Q')$ and $(A',B',C')$. Write

$$X=\mathbb R^3\setminus\{(0,0,0)\}$$

and consider the map

$$\begin{aligned} \varphi:X\times\mathbb R^3&\longrightarrow L\subseteq X\times\mathbb R^3,\\ (r,s,t;a,b,c)&\longmapsto (r,s,t;\,a(-s,r,0)+b(t,0,-r)+c(0,t,-s)). \end{aligned}$$

Under this map, the polynomial tuples $(-A,-B,-C)$ and $(-A',-B',-C')$, viewed as maps $X\to X\times\mathbb R^3$, are sent to $u$ and $v$. By assumption, $u$ and $v$ form a basis of every fibre of $L$, so $(A,B,C)$ and $(A',B',C')$ are linearly independent at every point. The tuple $(t,s,-r)$ is sent by $\varphi$ to $0$ in every fibre. Therefore,

$$(A,B,C),\qquad(A',B',C'),\qquad(t,s,-r)$$

form a basis of $\mathbb R^3$ at every point: $(t,s,-r)$ cannot be a linear combination of the first two tuples, since applying $\varphi$ would then give a nontrivial relation between $u$ and $v$. However, the determinant of the matrix

$$\begin{pmatrix} A&B&C\\ A'&B'&C'\\ t&s&-r \end{pmatrix}$$

is a polynomial combination of the variables $r,s,t$, and so is not a unit in the polynomial ring. In the real case, we cannot yet conclude that this determinant has a real zero in $X$; for example, it might have the form $r^2+s^2+t^2$. However, if we replace $\mathbb R$ by $\mathbb C$, the algebraic argument is unchanged, and we can conclude that the determinant has a zero in

$$X_{\mathbb C}=\mathbb C^3\setminus\{(0,0,0)\}.$$

Thus such a global basis cannot exist at every point.

Editorial note - sign of the lifted tuples. The source says that $(A,B,C)$ and $(A',B',C')$ map to $u$ and $v$. With its displayed convention $rN+sP+tQ=0$, however, $u=-(N,P,Q)$, and similarly for $v$. Negating the two coefficient tuples gives the stated lifts. The linear-independence and determinant argument is unchanged.

Example 1.3: two equations in three variables

Consider the general real system of linear equations

$$au+bv+cw=0$$

and

$$du+ev+fw=0$$

in the variables $u,v,w$ and parameters $a,b,c,d,e,f$, which serve as indeterminate coefficients of the system. If the parameters are sufficiently general, or more precisely, if there is no linear relation between the two equations, then the solution space

$$L_{(a,b,c,d,e,f)}= \left\{(u,v,w)\mathrel{\Big|} au+bv+cw=0\ \text{and}\ du+ev+fw=0\right\} \subseteq\mathbb R^3$$

is a line. Under this condition, the parameters therefore determine a family of varying lines in $\mathbb R^3$. The relevant parameter space for this family of lines is

$$P=\left\{(a,b,c,d,e,f)\mathrel{\Big|} (a,b,c)\ \text{and}\ (d,e,f)\ \text{linearly independent}\right\}.$$

Altogether, we obtain the total solution space

$$\begin{aligned} L=\{&(a,b,c,d,e,f,u,v,w)\mid au+bv+cw=0\ \text{and}\ du+ev+fw=0\}\\ &\subseteq P\times\mathbb R^3, \end{aligned}$$

together with its projection to $P$.

Can this line, or a basis element of it, be specified globally as a function of the parameters? Viewing the two equations as orthogonality relations, we seek a nonzero vector perpendicular to both constraint vectors

$$\begin{pmatrix}a\\b\\c\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}d\\e\\f\end{pmatrix}.$$

Their cross product has this property, namely

$$\begin{pmatrix} bf-ce\\ -af+cd\\ ae-bd \end{pmatrix}.$$

For the properties of the cross product used here, the source refers to Lemma 33.3 in Lineare Algebra (Osnabrück 2024-2025).

Thus there is a bijection

$$\begin{aligned} P\times\mathbb R&\longrightarrow L,\\ (a,b,c,d,e,f;s)&\longmapsto (a,b,c,d,e,f;s(bf-ce),s(-af+cd),s(ae-bd)). \end{aligned}$$

Editorial note - two coordinate names in the source. The source displays the second constraint vector as $(e,f,g)$, although the system and parameter space define it as $(d,e,f)$. In the final map, the source also prints the middle component as $-af+ce$, whereas the cross product printed immediately before it gives $-af+cd$. This edition uses $(d,e,f)$ and $-af+cd$, which can be verified directly from the two equations, while recording both source typographical errors.

In Examples 1.1 and 1.3 there are global polynomial trivialisations: polynomial functions translate the complicated geometric object into the simple object $P\times\mathbb R$, where $P$ is the base space. By contrast, such a global trivialisation is impossible in Example 1.2, although local trivialisations exist over the three specified open sets. Geometric objects of this kind are called vector bundles.

Real vector bundles

Definition 1.4: real vector bundle

Let $X$ be a topological space and $r\in\mathbb N$. A real vector bundle of rank $r$ is a topological space $V$ together with a continuous map

$$p:V\longrightarrow X$$

such that every fibre $p^{-1}(x)$ is an $r$-dimensional real vector space, and there is an open cover

$$X=\bigcup_{i\in I}U_i$$

together with homeomorphisms over $U_i$,

$$\varphi_i:p^{-1}(U_i)\longrightarrow U_i\times\mathbb R^r,$$

which induce a linear isomorphism on each fibre,

$$(\varphi_i)_x:p^{-1}(x)\longrightarrow\mathbb R^r.$$

The space $V$ is also called the total space, and $X$ the base space of the vector bundle. The fibre over $x$ is often denoted by

$$V_x=p^{-1}(x).$$

In the examples above, $X$ is the relevant parameter space, namely the locus of parameters for which the solution spaces have minimal dimension. This dimension is the rank $r$ in the definition above: respectively, $1,2,1$. In the first and third examples, the open cover consists only of the base space itself; these two bundles have a global trivialisation. In the second example, there is a cover by three open sets over which trivialisations have been given.

In the homeomorphism

$$p^{-1}(U)\longrightarrow U\times\mathbb R^r,$$

the right-hand side carries the product topology, $\mathbb R^r$ its natural Euclidean topology, and $p^{-1}(U)$ the topology induced from $V$. Thus every fibre $V_x$ carries the natural topology of a finite-dimensional real vector space. A homeomorphism over $U$ means that the diagram

$$\begin{array}{ccc} p^{-1}(U)&\stackrel{\varphi}{\longrightarrow}&U\times\mathbb R^r\\ &\searrow p&\downarrow\operatorname{pr}_1\\ &&U \end{array}$$

commutes.

Editorial note - rank symbol in the diagram. The source diagram prints $U\times\mathbb R^n$, whereas the definition and all surrounding formulae specify the rank as $r$. This edition displays $\mathbb R^r$.

The product $X\times\mathbb R^r$ is a vector bundle called the trivial vector bundle.

Lemma 1.5: restriction of a vector bundle

Let

$$p:V\longrightarrow X$$

be a real vector bundle over a topological space $X$. For every open set $W\subseteq X$, the restriction

$$p^{-1}(W)\longrightarrow W$$

is also a vector bundle.

Proof

Simply restrict the fibrewise linear homeomorphisms

$$\varphi_i:p^{-1}(U_i)\longrightarrow U_i\times\mathbb R^r$$

to

$$\varphi_i|_{W\cap U_i}: p^{-1}(W\cap U_i)\longrightarrow(W\cap U_i)\times\mathbb R^r.$$

$\square$

The restriction of a vector bundle to each $U_i$ is trivial. Thus every vector bundle is locally trivial.

Definition 1.6: homomorphism of vector bundles

Let $E$ and $F$ be real vector bundles over a topological space $X$. A homomorphism of vector bundles

$$\varphi:E\longrightarrow F$$

is a continuous map over $X$ such that, for every $x\in X$, the induced map

$$\varphi_x:E_x\longrightarrow F_x$$

is $\mathbb R$-linear.

Definition 1.7: isomorphism of vector bundles

Let $E$ and $F$ be real vector bundles over a topological space $X$. A homomorphism of vector bundles

$$\varphi:E\longrightarrow F$$

is called an isomorphism if there is a homomorphism

$$\psi:F\longrightarrow E$$

whose composition with $\varphi$, in either order, is the identity map.

The tangent bundle of a manifold

We now discuss another particularly important vector bundle, present on every manifold: the tangent bundle.

Every point $P\in M$ of a manifold has a tangent space $T_PM$. The tangent space is an $n$-dimensional vector space, where $n$ is the dimension of the manifold. Its elements are tangent vectors, or “infinitesimal directions” at that point. Initially, tangent directions at two distinct points have nothing to do with one another: their definitions depend only on arbitrarily small open neighbourhoods of the respective points, and the Hausdorff property allows these neighbourhoods to be chosen disjoint.

The picture for an open set $V\subseteq\mathbb R^n$ is quite different. For every $Q\in V$, the tangent space $T_QV$ can be identified naturally with the ambient vector space $\mathbb R^n$. A vector $v\in\mathbb R^n$ is assigned the tangent vector determined by the linear curve $t\mapsto Q+tv$. Since this identification applies at every point, there is a direct parallelism between the tangent spaces for

$$Q\in V\subseteq\mathbb R^n.$$

A manifold is covered by open sets diffeomorphic to open subsets of Euclidean space. It is therefore natural to expect that its various tangent spaces are not completely isolated. The concept of the tangent bundle brings all the tangent spaces together and reflects their local interconnection.

Source illustration - Tangent_bundle.svg. Two visualisations of the tangent bundle of a circle. In the upper picture, the tangent space at each point $P$ of the circle is placed tangentially to the circle and realised as a one-dimensional affine subspace of $\mathbb R^2$. This embedding creates intersections that do not exist in the tangent bundle itself, since the base point $P$ must also be taken into account. In the lower picture, the tangent spaces are arranged in parallel over the points of the circle, producing a cylinder.

Definition 1.8: the tangent bundle as a disjoint union

Let $M$ be a differentiable manifold. The set

$$TM=\biguplus_{P\in M}T_PM,$$

together with the projection map

$$\begin{aligned} \pi:TM&\longrightarrow M,\\ (P,v)&\longmapsto P, \end{aligned}$$

is called the tangent bundle of $M$.

A point $u\in TM$ always has a base point $P\in M$ and is an element of the tangent space $T_PM$. It is usually written $(P,v)$ with $P\in M$ and $v\in T_PM$. For an open set $V\subseteq\mathbb R^n$,

$$TV=V\times\mathbb R^n,$$

so it is a product space. This does not hold for an arbitrary manifold. Initially, the tangent bundle merely takes the disjoint union of the various tangent spaces, without identifying different tangent spaces with one another. However, the topology we shall shortly put on the tangent bundle adds a “neighbourhood structure” between the tangent spaces.

Definition 1.9: tangent map

Let $M$ and $N$ be differentiable manifolds and

$$\varphi:M\longrightarrow N$$

a differentiable map. Let $TM$ and $TN$ be the corresponding tangent bundles. The tangent map

$$T(\varphi):TM\longrightarrow TN$$

is the disjoint union of the tangent maps at the individual points, namely

$$T(\varphi)=\biguplus_{P\in M}T_P(\varphi).$$

Example 1.10: local trivialisation from a chart

Let $M$ be a differentiable manifold and

$$\alpha:U\longrightarrow V$$

a chart, where $V\subseteq\mathbb R^n$ is open. The chart induces a natural bijection

$$\begin{aligned} T(\alpha^{-1}):TV=V\times\mathbb R^n&\longrightarrow TU,\\ (Q,v)&\longmapsto \left(\alpha^{-1}(Q),[s\mapsto\alpha^{-1}(Q+sv)]\right). \end{aligned}$$

Here $s$ ranges over a real interval $I$ chosen so that $Q+sv\in V$ (compare Lemma 77.5 in Analysis (Osnabrück 2014-2016)). Since $V\times\mathbb R^n$ is a product of topological spaces,

$$TV=V\times\mathbb R^n$$

is itself a topological space. It is natural to transfer this topology to $TU$, and then construct a topology on the whole tangent bundle $TM$.

Definition 1.11: topology of the tangent bundle

Let $M$ be a differentiable manifold of dimension $n$ and

$$TM=\biguplus_{P\in M}T_PM$$

its tangent bundle, with projection

$$\begin{aligned} \pi:TM&\longrightarrow M,\\ (P,v)&\longmapsto P. \end{aligned}$$

Equip the tangent bundle with the following topology: a subset $W\subseteq TM$ is open if and only if, for every chart

$$\alpha:U\longrightarrow V,$$

the set

$$T(\alpha)\left(W\cap\pi^{-1}(U)\right)$$

is open in $V\times\mathbb R^n$.

In particular, for every open set $U\subseteq M$, the inverse image

$$\pi^{-1}(U)=TU\subseteq TM$$

is open; in other words, the projection $\pi$ is continuous. With these conventions, the tangent bundle of a differentiable manifold is a real vector bundle. If

$$M=\bigcup_{i\in I}U_i$$

is an open cover by sets $U_i$ homeomorphic to open sets $V_i\subseteq\mathbb R^n$, then the charts

$$\alpha_i:U_i\longrightarrow V_i$$

directly provide trivialisations

$$TM|_{U_i}=TU_i \stackrel{T(\alpha_i)}{\longrightarrow} TV_i=V_i\times\mathbb R^n.$$

A remarkable number of properties of a manifold are reflected in properties of its tangent bundle. The tangent bundle may be trivial even when $M$ is not homeomorphic to an open subset of $\mathbb R^n$.

Worksheet 1: Vector Bundles and Tangent Bundles

In the following exercises, we use the notation

$$D(r)=\left\{(r,s,t)\in\mathbb R^3\mid r\ne0\right\}.$$

Exercise 1.1

For the vector bundle

$$\begin{aligned} L={}&\left\{(r,s,t,u,v,w)\mathrel{\Big|} ru+sv+tw=0,\ (r,s,t)\ne(0,0,0)\right\}\\ &\subseteq \left(\mathbb R^3\setminus\{(0,0,0)\}\right)\times\mathbb R^3 \longrightarrow\mathbb R^3\setminus\{(0,0,0)\}, \end{aligned}$$

determine linear trivialisations over $D(r)$, $D(s)$ and $D(t)$, that is, bases depending on $r,s,t$ over $D(r)$ and so on. Determine the change-of-basis maps on

$$D(rs)=D(r)\cap D(s).$$

Exercise 1.2

For the vector bundle

$$\begin{aligned} L={}&\left\{(r,s,t,u,v,w)\mathrel{\Big|} ru+sv+tw=0,\ (r,s,t)\ne(0,0,0)\right\}\\ &\subseteq \left(\mathbb R^3\setminus\{(0,0,0)\}\right)\times\mathbb R^3 \longrightarrow\mathbb R^3\setminus\{(0,0,0)\}, \end{aligned}$$

determine all parameters $(r,s,t)$ for which the vector $(3,7,4)$ belongs to the fibre $L_{(r,s,t)}$.

Exercise 1.3

Show that, in Example 1.2 and over $D(r)$, the formula

$$u(r,s,t)= \frac{t}{r}\begin{pmatrix}s\\-r\\0\end{pmatrix} -\frac{s}{r}\begin{pmatrix}t\\0\\-r\end{pmatrix}$$

defines a parameter-dependent vector in the solution space that extends polynomially to all of $\mathbb R^3$, even though the coefficient functions $t/r$ and $-s/r$ are defined only on $D(r)$ and do not extend. Is $u(r,s,t)$ part of a basis at every point?

Exercise 1.4

In Example 1.3, determine the parameters for which the solution space $L_{(a,b,c,d,e,f)}$ is one-, two- or three-dimensional. Are these parameter sets open or closed?

Exercise 1.5

Show that, over an arbitrary field $K$, for two linearly independent vectors

$$u=\begin{pmatrix}a\\b\\c\end{pmatrix} \quad\text{and}\quad v=\begin{pmatrix}d\\e\\f\end{pmatrix},$$

the family consisting of $u$, $v$ and their cross product

$$\begin{pmatrix}a\\b\\c\end{pmatrix} \times \begin{pmatrix}d\\e\\f\end{pmatrix}$$

need not form a basis of $K^3$.

Exercise 1.6

Consider the topological space

$$Y:=\left\{(s,t,u,v)\in\mathbb R^4\mid su+tv=1\right\}$$

with projection

$$\begin{aligned} p:Y&\longrightarrow\mathbb R^2\setminus\{(0,0)\}=X,\\ (s,t,u,v)&\longmapsto(s,t). \end{aligned}$$

1. Show that every fibre of $p$ is homeomorphic to a real line.

2. Show that

   $$\varphi(s,t)=(s,t,u(s,t),v(s,t)) =\left(s,t,\frac{s}{s^2+t^2},\frac{t}{s^2+t^2}\right)$$

   defines a continuous map $\varphi:X\to Y$ with

   $$p\circ\varphi=\operatorname{Id}_X.$$

3. Define a homeomorphism between $Y$ and $X\times\mathbb R$.

4. Show that there is no polynomial map $\psi:X\to Y$ with

   $$p\circ\psi=\operatorname{Id}_X.$$

Exercise 1.7

Show that a real vector bundle over a point, that is, over a one-point topological space, is the same as a finite-dimensional real vector space.

Exercise 1.8

Let $p:V\to X$ be a real vector bundle over a topological space $X$. Show that $V$ is a Hausdorff space if and only if $X$ is a Hausdorff space.

Exercise 1.9

Let $X$ be a topological space. Show that the identity map

$$\operatorname{Id}_X:X\longrightarrow X$$

can be regarded as a real vector bundle of rank $0$.

Exercise 1.10

Let $X$ be a topological space. Show that a homomorphism of trivial vector bundles

$$\varphi:X\times\mathbb R^n\longrightarrow X\times\mathbb R^m$$

is the same as an $m\times n$ matrix whose entries are continuous functions from $X$ to $\mathbb R$.

Exercise 1.11

Let $U\subseteq\mathbb R^n$ be open and $f:U\to\mathbb R^m$ a continuously differentiable map. Show that the total differential, in the form

$$\begin{aligned} U\times\mathbb R^n&\longrightarrow U\times\mathbb R^m,\\ (x,v)&\longmapsto\bigl(x,(Df)_x(v)\bigr), \end{aligned}$$

defines a homomorphism from the vector bundle $U\times\mathbb R^n$ to the vector bundle $U\times\mathbb R^m$.

Exercise 1.12

Show that the tangent bundle $TS^1$ of the $1$-sphere $S^1$ is homeomorphic to the product $S^1\times\mathbb R$.

How is this exercise related to Example 1.1?

Exercise 1.13

Give an example of a differentiable curve

$$\gamma:[0,1)\longrightarrow S^1$$

such that the limit

$$\lim_{t\to1}\gamma(t)$$

exists, but the limit

$$\lim_{t\to1}\bigl(\gamma(t),T_t(\gamma)(1)\bigr)$$

does not exist in $TS^1$.

Exercise 1.14

Show that the map

$$\begin{aligned} TS^1&\longrightarrow\mathbb R^2,\\ ((a,b),t(-b,a))&\longmapsto(a,b)+t(-b,a) \end{aligned}$$

has two preimages for every point $(x,y)\in\mathbb R^2$ outside the unit disc, one preimage for every point on the unit circle, and no preimage for any point inside the open unit disc. Interpret this geometrically.

Exercise 1.15

Give an example of an injective differentiable map

$$\varphi:M\longrightarrow N$$

between two differentiable manifolds $M$ and $N$ such that the associated tangent map

$$T(\varphi):TM\longrightarrow TN$$

is not injective.

Exercise 1.16

Give an example of a surjective differentiable map

$$\varphi:M\longrightarrow N$$

between two differentiable manifolds $M$ and $N$ such that the associated tangent map

$$T(\varphi):TM\longrightarrow TN$$

is not surjective.

Exercise 1.17

Let $M$ and $N$ be differentiable manifolds and $\varphi:M\to N$ a differentiable map. Show that the associated tangent map

$$T(\varphi):TM\longrightarrow TN$$

is continuous.

Public Solution Coverage for Worksheet 1

At the frozen revision boundary, the source provides no public solution for any of the 17 exercises on Worksheet 1. The frozen solution-candidate query and exercise map record negative results for Exercises 1.1-1.17.

No new solutions have been created for this edition. This section merely documents the scope of the source so that the absence of solutions is not mistaken for content lost during translation.

Lecture 2: Sections, the Hairy Ball Theorem, and Gluing Data

Sections

Definition 2.1: continuous section

Let $X$ and $Y$ be topological spaces, and let

$$p:Y\longrightarrow X$$

be a continuous map. A continuous section of $p$ is a continuous map

$$s:X\longrightarrow Y$$

such that

$$p\circ s=\operatorname{Id}_X.$$

For example, we may think of $Y$ as a vector bundle over $X$. A section can exist only if $p$ is surjective, as is always the case for a vector bundle. A section is sometimes identified with its image; this causes no difficulty, since every section is injective. The zero section plays a special role: to each base point $P$, it assigns the zero vector in the vector space $V_P$. Sections of tangent bundles have a name of their own.

Definition 2.2: vector field

Let $M$ be a differentiable manifold. A map

$$F:M\longrightarrow TM$$

satisfying

$$F(P)\in T_PM$$

for every point $P\in M$ is called a (time-independent) vector field.

The hairy ball theorem

Source illustration - Hairy_ball_one_pole.jpg. A continuous vector field on the $2$-sphere must have at least one zero.

Theorem 2.3: the hairy ball theorem

On the $2$-sphere, every continuous vector field

$$f:S^2\longrightarrow TS^2$$

has at least one zero.

In particular, the tangent bundle of the $2$-sphere is not trivial. There are various interpretations of this theorem. For example, it says that there is always a point on the Earth’s surface where there is no wind, when the instantaneous horizontal wind is regarded as a continuous vector field. Similarly, it is impossible to lay all the spines of a hedgehog flat against its body.

Remark 2.4: application to Example 1.2

The hairy ball theorem explains why the vector bundle $L$ from Example 1.2, over

$$\mathbb R^3\setminus\{(0,0,0)\},$$

has no continuous trivialisation. First,

$$S^2\subset\mathbb R^3\setminus\{(0,0,0)\},$$

so we can restrict $L$ to $S^2$. If $L$ itself were trivial, this restriction would also be trivial. But the restriction of $L$ to the unit sphere is the tangent bundle of the unit sphere. Indeed, the condition

$$ru+sv+tw=0$$

can be interpreted as an orthogonality relation, and the extrinsic tangent space at a point $(r,s,t)$ of the sphere is determined by this relation. If the tangent bundle were trivial, there would be two continuous vector fields $u$ and $v$ forming a basis of the tangent space at every point of the sphere. The hairy ball theorem, however, says that even a single vector field must have a zero, and $0$ cannot belong to a basis.

Gluing data for topological spaces

A vector bundle $V\to X$ is “assembled” from the trivial vector bundles $V|_{U_i}\to U_i$ for an open cover of $X$. The precise way these pieces are assembled determines the vector bundle, and can be described conveniently by gluing data. We first need gluing data for topological spaces in general.

The underlying question is: what must we know about an open cover

$$X=\bigcup_{i\in I}U_i$$

in order to reconstruct the space $X$? The short answer is that we need to know the $U_i$, the pairwise intersections $U_i\cap U_j$ as subsets both of $U_i$ and of $U_j$, how these two copies are identified, and a compatibility condition on the identifications involving each triple of sets.

Source illustration - Inclusion-exclusion.svg. Three overlapping sets illustrate the need for a compatibility condition on triple intersections.

Definition 2.5: gluing data for topological spaces

Gluing data for topological spaces consist of:

1. a family of topological spaces $(U_i)_{i\in I}$;

2. for each pair $(i,j)$, an open subset

   $$U_{ij}\subseteq U_i,$$

   with $U_{ii}=U_i$;

3. for each pair $(i,j)$, a homeomorphism

   $$\varphi_{ji}:U_{ij}\longrightarrow U_{ji},$$

   with $\varphi_{ii}=\operatorname{Id}_{U_i}$;

4. for all $i,j,k\in I$, the cocycle condition

   $$\varphi_{kj}\circ\varphi_{ji}=\varphi_{ki}$$

   holds as an equality of maps from $U_{ik}\cap U_{ij}$ to $U_k$.

Lemma 2.6: reconstructing a space from gluing data

Suppose gluing data $(U_i)_{i\in I}$ for topological spaces are given. Then there exist a uniquely determined topological space $X$, an open cover

$$X=\bigcup_{i\in I}V_i,$$

and homeomorphisms

$$\psi_i:U_i\longrightarrow V_i$$

such that

$$\psi_i(U_{ij})=V_i\cap V_j$$

and

$$\psi_i|_{U_{ij}} =\psi_j|_{U_{ji}}\circ\varphi_{ji}.$$

Proof

Let $Y$ be the disjoint union of the $U_i$. Define an equivalence relation $\sim$ on $Y$ by declaring $x_i\in U_i$ and $x_j\in U_j$ equivalent when

$$x_i\in U_{ij},\qquad x_j\in U_{ji},\qquad \varphi_{ji}(x_i)=x_j.$$

The properties of an equivalence relation are ensured by the cocycle condition; see Exercise 2.14. Set

$$X:=Y/{\sim}$$

and equip $X$ with the quotient topology. The composites

$$U_i\longrightarrow Y\longrightarrow X$$

are the maps $\psi_i$, and the $V_i$ are their images. Thus $\psi_i:U_i\to V_i$ are homeomorphisms. For $x\in U_i$,

$$\psi_i(x)\in V_j$$

if and only if $x\in U_{ij}$, since precisely in this case $x$ is identified with $\varphi_{ji}(x)$. Hence

$$\psi_i(U_{ij})=V_i\cap V_j.$$

Editorial note - order of indices in the proof. Under the convention in Definition 2.5, $\varphi_{ji}:U_{ij}\to U_{ji}$. In the preceding sentence, the source prints $\varphi_{ij}(x)$ for $x\in U_{ij}$, although the correctly typed map is $\varphi_{ji}(x)$. This edition displays the indices consistent with the domain and codomain in the proof and preserves the source form in this note.

Commutativity of the diagram

$$\begin{array}{ccc} U_{ij}&\stackrel{\varphi_{ji}}{\longrightarrow}&U_{ji}\\ &\searrow\psi_i&\downarrow\psi_j\\ &&V_i\cap V_j \end{array}$$

follows in the same way. $\square$

Lemma 2.7: gluing continuous maps

Suppose gluing data $(U_i)_{i\in I}$ for topological spaces are given. Let $Z$ be another topological space, and suppose continuous maps

$$\theta_i:U_i\longrightarrow Z$$

are given satisfying

$$\theta_i|_{U_{ij}} =\bigl(\theta_j|_{U_{ji}}\bigr)\circ\varphi_{ji}.$$

Then there is a unique continuous map

$$\theta:X\longrightarrow Z$$

such that

$$\theta|_{V_i}\circ\psi_i=\theta_i,$$

where $X$ is the topological space determined by the gluing data as in Lemma 2.6, whose notation we also use.

Editorial note - ill-typed composition identity. The source prints $(\psi_i)^{-1}\circ\theta|_{V_i}=\theta_i$, but this composition is not defined: $\theta|_{V_i}$ takes values in $Z$, whereas $(\psi_i)^{-1}$ has domain $V_i$. This edition displays the correctly typed identity $\theta|_{V_i}\circ\psi_i=\theta_i$ in the lemma and preserves the source form in this note.

Proof

See Exercise 2.18.

Gluing data for vector bundles

Definition 2.8: gluing data for real vector bundles

Gluing data for a real vector bundle of rank $r$ over a topological space $X$ consist of:

1. an open cover

   $$X=\bigcup_{i\in I}U_i;$$

2. a family of real vector bundles of rank $r$,

   $$(E_i\longrightarrow U_i)_{i\in I};$$

3. for each pair $(i,j)$, an isomorphism of vector bundles

   $$\varphi_{ji}:E_i|_{U_i\cap U_j} \longrightarrow E_j|_{U_i\cap U_j}$$

   over $U_i\cap U_j$;

4. for all $i,j,k\in I$, the cocycle condition

   $$\varphi_{kj}\circ\varphi_{ji}=\varphi_{ki}$$

   holds as an equality of maps from $E_i|_{U_i\cap U_j\cap U_k}$ to $E_k|_{U_i\cap U_j\cap U_k}$.

Remark 2.9: matrix description

Typically, the vector bundles in item (2) of Definition 2.8 are trivial bundles over $U_i$, namely

$$E_i=\mathbb R^r\times U_i.$$

The isomorphisms in item (3) are then simply bijective linear maps

$$\varphi_{ji}:\mathbb R^r\longrightarrow\mathbb R^r$$

depending continuously on the base point in $U_i\cap U_j$. They can be described compactly as continuous maps

$$\varphi_{ji}:U_i\cap U_j\longrightarrow \operatorname{GL}_r(\mathbb R)$$

into the general linear group. Thus an invertible $r\times r$ matrix is assigned continuously to each base point; continuity means that every matrix entry is a continuous function. This is called a matrix description of the bundle. The cocycle condition still applies.

Lemma 2.10: gluing vector bundles

Suppose gluing data $(E_i)_{i\in I}$ over a topological space

$$X=\bigcup_{i\in I}U_i.$$

are given. Then there exist a uniquely determined real vector bundle $E\to X$ and isomorphisms

$$\psi_i:E_i\longrightarrow E|_{U_i}$$

such that

$$\psi_i|_{E_i|_{U_i\cap U_j}} =\psi_j|_{E_j|_{U_i\cap U_j}}\circ\varphi_{ji}.$$

Proof

The existence of a topological space $E$ with these properties follows from Lemma 2.6. The open sets to be glued are

$$W_{ij}:=E_i|_{U_i\cap U_j},$$

and the existence of a continuous map to $X$ follows from Lemma 2.7. Every fibre $E_x$ has a well-defined vector space structure inherited from $E_i$ for any open neighbourhood $x\in U_i$. Independence of the choice of $i$ follows because, for $x\in U_i\cap U_j$, the hypothesis supplies an isomorphism of vector bundles

$$\varphi_{ji}:E_i|_{U_i\cap U_j} \longrightarrow E_j|_{U_i\cap U_j},$$

inducing a vector space isomorphism

$$(E_i)_x\longrightarrow(E_j)_x.$$

Editorial note - order of indices in the fibre isomorphism. Definition 2.8 specifies $\varphi_{ji}:E_i|_{U_i\cap U_j}\to E_j|_{U_i\cap U_j}$. The source prints $\varphi_{ij}$ in the sentence of the proof whose map goes from $E_i$ to $E_j$. This edition displays $\varphi_{ji}$, consistent with the domain and codomain, in the proof and preserves the source form in this note.

$\square$

Source illustration - Fiddler_crab_mobius_strip.gif. A Möbius strip, arising by reversing the fibre on one overlap component when two local trivialisations are glued.

Example 2.11: the Möbius strip from gluing data

On the one-dimensional sphere

$$S^1=\left\{(x,y)\in\mathbb R^2\mid x^2+y^2=1\right\},$$

consider the open cover

$$S^1=U\cup V,$$

with

$$U=S^1\setminus\{(0,1)\}, \qquad V=S^1\setminus\{(0,-1)\}.$$

We shall describe gluing data for a real vector bundle of rank $1$. Both open sets are homeomorphic to the real line. Their intersection is

$$\begin{aligned} U\cap V &=S^1\setminus\{(0,1),(0,-1)\}\\ &=\left\{(x,y)\in S^1\mid x\ne0\right\}. \end{aligned}$$

This set is not connected, but is homeomorphic to two disjoint open real half-lines (or, equivalently, two real lines). Set

$$L=U\times\mathbb R, \qquad M=V\times\mathbb R.$$

Define an isomorphism

$$\varphi:L|_{U\cap V}\longrightarrow M|_{U\cap V}$$