2
$\begingroup$

This question is motivated by Why do we need the local triviality condition when working with vector bundles?

A general bundle is a triple $(E,p,B)$, where $E,B$ are topological spaces and $p : E \to B$ is a map. This concept can be found for example in [1]. A $k$-dimensional pre-vector bundle is defined as a bundle such that each fiber $E_b = p^{-1}(b)$, $b \in B$, has the structure of a $k$-dimensional topological vector space over $\mathbb K$. Atiyah [2] denotes this as a family of vector spaces. A vector bundle is then a locally trivial pre-vector bundle. Most standard constructions for vector bundles also work for pre-vector bundles (direct sum, ...).

In the question quoted above the OP states "If the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$, then the local triviality condition shouldn't be necessary." But what should a continuous parameterization be?

In the general case I do not see an obvious definition. So let us look at the following special case. Let $\pi : B \times \mathbb K^n \to B$ be the projection. To each $b \in B$ assign a $k$-dimensional subspace $f(b) \subset \mathbb K^n$. Then $E(f) = \bigcup_{b \in B} \{b\} \times f(b)$ is a subspace of $E$ and $\pi$ restricts to $p : E(f) \to B$. This gives us a $k$-dimensional pre-vector bundle $\xi(f)$ over $B$ which may look completely erratic. A classification of these bundles up to isomorphism is practically impossible, and there is no manifest connection with the topology on $B$.

However, we can regard the above assigment as a function $f : B \to G_k(\mathbb K^n)$ into the Grassmann variety which has a nice metric. It makes sense to regard the function $f$ as the parameterization of $\xi(f)$. Then the parameterization can be defined to be continuous if $f$ is continuous.

Question: What is the relation between "continuously parameterized" and "locally trivial"? Do these concepts agree?

[1] Husemoller, Dale. Fibre bundles. Vol. 5. New York: McGraw-Hill, 1966. See Chapter 2.

[2] Atiyah, Michael. K-theory. CRC Press, 2018.

$\endgroup$
7
  • $\begingroup$ See also math.stackexchange.com/q/2308321. $\endgroup$
    – Paul Frost
    Commented Oct 24, 2019 at 14:06
  • 1
    $\begingroup$ In some sense the "continuous" part of the continuous parametrization means that at small scales, the bundle doesn't twist wildly, (in the same sense that continuous functions means at small scales there's no wild behaviour), and so over this small set, the bundle is trivial, so we have local triviality. $\endgroup$ Commented Oct 24, 2019 at 14:07
  • $\begingroup$ I might guess that continuously parametrized bundles are equivalent to spaces with a free and transitive general linear action, so to find an example of one of these that isn’t a vector bundle you could find a nasty action of the general linear group on your space so the quotient map isn’t locally trivial. $\endgroup$ Commented Oct 24, 2019 at 16:45
  • 1
    $\begingroup$ It seems that your new definition is the same as the pull-back of the tautological bundle from the Grassmannian. Then it is a vector bundle in the traditional sense. $\endgroup$ Commented Oct 24, 2019 at 23:02
  • 1
    $\begingroup$ @MoisheKohan You are of course right, I shoud have seen that myself. I extended your comment to an answer. $\endgroup$
    – Paul Frost
    Commented Oct 25, 2019 at 10:36

1 Answer 1

1
$\begingroup$

Moishe Kohan's comment contains the answer of the question. This community wiki elaborates it a little.

Let us first observe that if $\xi = (E,p,B)$ is a pre-vector bundle and $f : X \to B$ is a (not necessarily continuous) function on a space $X$, we get a pullback pre-vector bundle $$f^*(\xi) = (f^*(E), p^*,X)$$ where $$f^*(E) = \bigcup_{x \in X} \{x \} \times p^{-1}(f(x)) \subset X \times E$$ and $p^*$ is the restriction of the projection $X \times E \to X$.

Now we generalize the construction of the question (see [1]). Define $$\mathbb K^\infty = \{(x_i)_{i \in \mathbb N} \mid x_i \in \mathbb K, x_i = 0 \text{ for almost all } i \} .$$ This is a vector space with an obvious inner product and we may regard each $\mathbb K^n$ as a genuine subspace of $\mathbb K^\infty$. Doing so, we have $\mathbb K^\infty = \bigcup_{n \in \mathbb N} \mathbb K^n$.

For $0 \le m \le \infty$ and $k \in \mathbb N$ let $G_k(\mathbb K^m)$ denote the set of all $k$-dimensional linear subspaces of $\mathbb K^m$. For $m < \infty$ these are the well-known Grassmann varieties. Each $G_k(\mathbb K^n)$ is a genuine subspace of $G_k(\mathbb K^{n+1})$, and we define $G_k(\mathbb K^\infty) = \bigcup_{n \ge k} G_k(\mathbb K^n)$ as a set, and $U \subset G_k(\mathbb K^\infty)$ to be open iff $U \cap G_k(\mathbb K^n)$ is open in $G_k(\mathbb K^n)$ for all $n$. Thus $G_k(\mathbb K^\infty)$ is the direct limit of the sequence of spaces $G_k(\mathbb K^n)$ bonded by inclusions.

The tautological (or canonical bundle) $\gamma^m_k$ over $G_k(\mathbb K^m)$ has as total space $$E^m_k = \bigcup_{V \in G_k(\mathbb K^m)} \{V\} \times V \subset G_k(\mathbb K^m) \times \mathbb K^m$$ with obvious projection $\pi$ onto the base. The fiber over $V \in G_k(\mathbb K^m)$ is nothing else than $\{V\} \times V$, i.e. a copy of $V \subset \mathbb K^m$. Again we have $E_k^\infty = \bigcup_{n \ge k} E_k^n$. It is well-known that $\gamma^m_k$ is locally trivial.

Now let $f : B \to G_k(\mathbb K^m)$ be any (not necessarily continuous) function. The pullback pre-vector bundle $f^*(\gamma^m_k)$ over $B$ has as total space $$f^*(E_k^m) = \bigcup_{b \in B} \{b \} \times \pi^{-1}(f(b)) = \bigcup_{b \in B} \{b \} \times \{f(b)\} \times f(b) \subset B \times G_k(\mathbb K^m) \times \mathbb K^m .$$ In general this is completely erratic. Let us say that $f^*(\gamma^m_k)$ is continuously parameterized if $f$ is continuous.

Note, however, that $f^*(\gamma^m_k)$ is not the same pre-vector bundle as $\xi(f)$ which was defined in the question. But $f^*(\gamma^m_k)$ is continuously parameterized if and and only $\xi(f)$ is. It seems that we now habe an adequate interpretation of "the spirit of a vector bundle is to continuously parametrize a family of vector spaces by $B$", at least for pre-vector bundles of the form $f^*(\gamma^m_k)$ and $\xi(f)$.

Fact 1. If $f^*(\gamma^m_k)$ is continuously parameterized, then it is locally trivial.

This is well-known. Pullbacks of vector bundles along continuous maps are always vector bundles. This shows that being continuously parameterized is even stronger than locally trivial.

Fact 2. If $f^*(\gamma^m_k)$ is locally trivial, then it is continuously parameterized.

Let $s_0 : B \to f^*(E_k^m) \subset B \times G_k(\mathbb K^m) \times \mathbb K^m$ be the zero-section which is given $s_0(b) = (b,f(b),0)$. Each $b \in B$ has an open neighborhood $U \subset B$ such that the restriction of $f^*(\gamma^m_k)$ to $U$ is trivial. This implies that $s_0 \mid_U$ is continuous. Since the projection $p_2 :B \times G_k(\mathbb K^m) \times \mathbb K^m \to G_k(\mathbb K^m)$ is continuous, we see that $f \mid_U = p_2 \circ s_0 \mid_U$ is continuous. Thus $f$ is continuous.

Fact 3. If $f$ is continuous, then $f^*(\gamma^m_k)$ and $\xi(f)$ are isomorphic. In particular, $\xi(f)$ is locally trivial.

To see this, define $\phi_f : B \times \mathbb K^m \to B \times G_k(\mathbb K^m) \times \mathbb K^m, \phi(b,v) = (b,f(b),v)$. We have $\phi_f(E(f)) = f^*(E_k^m)$ so that we get an induced $\phi'_f : E(f) \to f^*(E_k^m)$ which is bijection which maps the fiber over $b$ in $E(f)$ by a linear isomorphism to the fiber over $b$ in $f^*(E_k^m)$. It is continuous iff $f$ is continuous. Next define $\psi : B \times G_k(\mathbb K^m) \times \mathbb K^m \to B \times \mathbb K^m$ as the projection. Clearly $\psi$ is continuous and $\psi(f^*(E_k^m)) = E(f)$. Hence the restriction $\psi_f : f^*(E_k^m) \to E(f)$ is a morphism of pre-vector bundles which is continuous and fiberwise linearly isomorphic. It is an isomorphism of pre-vector bundles iff $f$ is continuous.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .