Return to Answer

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a continuous local section (that is, a continuous map $s\colon\thinspace U_i\to E$ such that $p\circ s =\operatorname{incl}\colon\thinspace U_i\hookrightarrow B$).

Then the minimum number of sets in a trivializing cover for the vector bundle $E\to B$ equals the sectional category of the frame bundle of $F(E)\to B$.

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be contractible themselves, or even connected (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a continuous local section (that is, a continuous map $s\colon\thinspace U_i\to E$ such that $p\circ s =\operatorname{incl}\colon\thinspace U_i\hookrightarrow B$).

Then the minimum number of sets in a trivializing cover for the vector bundle $E\to B$ equals the sectional category of the frame bundle $F(E)\to B$.

Addendum: One can show using obstruction theory that for a $r$-connected manifold $B$, $$\operatorname{cat}(B)< \frac{\dim(B)+1}{r+1}+1.$$

So if, say, your manifold is simply-connected, you can find a trivializing cover with roughly half as many sets.

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a local section. Then the minimum number of sets in a trivializing cover for $E\to B$ is the sectional category of the frame bundle of $F(E)\to B$.

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a continuous local section (that is, a continuous map $s\colon\thinspace U_i\to E$ such that $p\circ s =\operatorname{incl}\colon\thinspace U_i\hookrightarrow B$). 

Then the minimum number of sets in a trivializing cover for the vector bundle $E\to B$ equals the sectional category of the frame bundle of $F(E)\to B$.

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ are not necessarily connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater then or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of open sets needed to cover $B$ such that your bundle $\xi$ is trivial on each, you can do better, using the notion of sectional category of a bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a local section. Then the minimum number of sets in a trivializing cover for $\xi$ is the sectional category of the associated frame bundle of $\xi$.

This should be a comment to the answer of Andreas Blass, but was too long.

The Lusternik-Schnirelmann category $\operatorname{cat}(X)$ of a space $X$ is the smallest number $k$ such that $X$ has a cover by open sets $U_1,\ldots , U_k$ which are contractible in $X$. This means the inclusions $U_i\hookrightarrow X$ are null-homotopic. Note that the $U_i$ need not be connected themselves (although each $U_i$ should be contained within a connected component $X_i$). Note also that $\operatorname{cat}(X)$ is greater than or equal to the minimum number of open sets needed to trivialize any vector bundle on $X$, by the bundle creeping lemma and the fact that any bundle over a point is trivial.

One of the first theorems about LS-category is that if $X$ is paracompact, then $\operatorname{cat}(X)\le \operatorname{dim}(X)+1$, where $\operatorname{dim}$ denotes the Lebesgue covering dimension (which for manifolds agrees with the usual dimension). So Andreas Blass is correct.

If you want the minimum number of sets in a trivializing cover for your bundle $E$, you can do better, using the notion of sectional category of a fibre bundle (also known as the Schwarz genus). The sectional category $\operatorname{secat}(p)$ of a fibre bundle $p\colon\thinspace E\to B$ is the smallest number $k$ such that $B$ has a cover by open sets $U_1,\ldots , U_k$ on each of which $p$ admits a local section. Then the minimum number of sets in a trivializing cover for $E\to B$ is the sectional category of the frame bundle of $F(E)\to B$.