I've started reading Milnor, Stasheff - Characteristic Classes and at page $18$ they proved that $\mathbb{R}^n$-bundle $\xi$ is trivial if and only if $\xi$ admits $n$ cross sections $s_1, \dots , s_n$ which are everywhere independent.

Inside the proof they define the operator $$f: B \times \mathbb{R}^n \to E $$ $$ f(b,x) = x_1 s_1(b)+ \cdots + x_n s_n(b) $$

and they claim it is continuous.

But my doubt here is: which topology $E$ is equipped with?

Looking at the definition of vector bundle $E \to B$ I have:

1) $E$ is a topological space (so I don't know what topology I have)

2) a continuous map $\pi : E \to B$ called the projection (so at least I have all the open of the form $\pi^{-1}(U)$ where $U$ is open in $B$ (which has a fixed topology).

3) each fiber is a vector space

and then the conditions of local triviality hold: for each $b \in B$ there exist a neighborhood $U \subset B$, a natural number $n$ and a homeomorphism $$ h: U \times \mathbb{R}^n \to \pi^{-1}(U)$$ so that, for each $b \in B$, the correspondence $x \mapsto h(b,x)$ is an isomorphism between $\mathbb{R}^n$ and the fiber over $b$.he

this last property gives me some information about the subspace topology of each $\pi^{-1}(U)$. I think it's not a problem considering $U$ open and so I have a whole new class of open subset in $E$ : every $V$ such that $V=h(A \times B)$ where $A$ and $B$ are respectively open subsets of $B$ and $\mathbb{R}^n$.

So is this classification of the topology over E exhaustive? in other terms, these and only these are the open subsets in $E$?

other two doubts related to the main question:

A) Is the property (in relation to $E$) of being an Hausdorff space dependent on the fact that $B$ is Hausdorff ? (and so without any new information over $B$ we can't say anything?)

B) Can someone shows me how to prove (a kind of step by step) the continuity of the function $f$ defined above? so I can understand the basic mechanics of this kind of reasoning.

ADDENDUM I write down this kind of proof of the continuity. I can't judge if it is correct or no because I'm not very convicted of it, so please give it a check:

Consider $W \subset E$ an open set. So by the discussion above its intersections with open of the form $\pi^{-1}(U)$ with $U$ open in $B$ are homeomorphic to open set of $B \times \mathbb{R}^n$. Consider one of this intersection $\bar{W}$. Then $f^{-1}(\bar{W})$ (I'm using local coordinates) contains elements of the form $(b,v)$ where $b \in B$ and $v$ is the image of the linear map which assign at every n-uple $v_1, \dots , v_n$ $v_1s_1(b)+ \cdots + v_ns_n(b)$ and a linear mapping in a finite dimensional vector space is continuous. So it's pre image is an open set of $\mathbb{R}^n$ and by definition of product topology the pre image of $\bar{W}$ throughout $f$ is an open.

It is clear that I'm a novice in this field, (I've started to read this topic after attending a basic course about classical differential geometry which ended with an extremely rapid view over the concepts of vector bundle and vector fields) so I need exhaustive answers without implicit passages (I need to learn those passages, so I must see them at least ) :) thanks in advance

  • $\begingroup$ By "nowhere independent" you probably meant "everywhere independent" or "nowhere dependent". $\endgroup$
    – Michael
    Commented Mar 2, 2014 at 23:13
  • $\begingroup$ nowhere dependent I mean :) edit right now! thanks :) $\endgroup$
    – Riccardo
    Commented Mar 2, 2014 at 23:15

1 Answer 1


Here you only need to care about the topology coming from the neighborhood basis. Let $B$ be a topological space. Assuming we can find local trivializations $$ f_{U}:U_{E}\cong U\times \mathbb{R}^{n} $$ such that on $U_{E}$ this is given by product topology of $U\times \mathbb{R}^{n}$. Then you "glue" two such neighborhoods together using $$g_{UV}:f_{V}f_{U}^{-1}:(U\cap V)\times \mathbb{R}^{n}\cong (U\cap V)_{E}\rightarrow (U\cap V)_{E}$$ And the quotient topology you get by choosing an atlas $\bigcup U_{i}=B$ and $g_{U_{i},U_{j}}$ is the desired topology on $E$.

If you have $n$ nowhere vanishing independent sections, then you can use them to construct a global map $$E\cong B\times \mathbb{R}^{n}$$so $E$ has the product topology in this case. For continuity, consider the $n=1$ case. If we know the bundle is $1$ dimensional, then the map $$B\times \mathbb{R}:(v,x)\rightarrow f(v)x$$must be continuous because multiplication by $x$ (a scalar) is continuous (part of definition of $E$), and $f$ is continuous by definition. Now for $n>1$ it follows from addition of finitely many continuous maps is continuous.

The only step to be verified is multiplication by scalar is indeed continuous. We consider two approaches. Some people define the map $F_{x}\rightarrow \mathbb{R}^{n}$ to be an vector space isomorphism, then there is no question about scalar multiplication's continuity. Some people define $U_{E}\cong U\times \mathbb{R}^{n}$ directly, so $F_{x}\cong \mathbb{R}^{n}$ as topological vector spaces. Then scalar multiplication's continuity follows from that of the TVS.

  • $\begingroup$ mmh the concept of "gluing" is somewhat obscure to me. In fact i would expect an union, ora a quotient, but I cannot see how you glue $V_E$ and $U_E$. thank for the answer anyway! $\endgroup$
    – Riccardo
    Commented Mar 2, 2014 at 23:22
  • $\begingroup$ in the definition of your last map I think x and v are swapped. $v$ is a scalar and $f$ is a section so it takes values over $B$. What do you mean by "part of definition of $E$"? $\endgroup$
    – Riccardo
    Commented Mar 2, 2014 at 23:29
  • $\begingroup$ "Gluing" is essentially taking the union of two disjoint sets $U$ and $V$ as if they were not disjoint, by identifying a subset of $U$ with a subset of $V$ and calling it "$U\cap V$." $\endgroup$
    – bradhd
    Commented Mar 3, 2014 at 0:19
  • $\begingroup$ Sorry for the typos. I mean you can identify $(v,a)\in v\times \mathbb{R}^{n}$ with $(v,b)\in v\times \mathbb{R}^{n}$ by the map $g_{UV}$'s action on $a$, which give us $b$. You can think about $E$ as $\bigcup U_{i}\times \mathbb{R}^{n}$ quotient out in such a process. The topology would not change as locally it is homeomorphic to $U_{i}\times \mathbb{R}^{n}$. $\endgroup$ Commented Mar 3, 2014 at 5:03
  • 2
    $\begingroup$ This is implicit in the definition, because if you identify $F_{x}\cong_{f} \mathbb{R}^{n}$ by a linear map, then since multiplication by $x\in \mathbb{R}$ is continuous in $\mathbb{R}^{n}$. I think you require $f$ to be a vector space isomorphism(see Hatcher's definition). However, even if you do not require $f$ to be a vector space isomorphism, there is only one topology available in a finite dimensional vector space (see terrytao.wordpress.com/2011/05/24/…). So multiplication in $F_{x}$ is continuous as it is continuous in $\mathbb{R}^{n}$. $\endgroup$ Commented Mar 3, 2014 at 8:23

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