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rephrase 1st sentence
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Ryan Budney
Ryan Budney
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I suppose there's a natural way to quantify angive a type of global quantative answer to this question. A vector bundle is a family of vector spaces over a base space, $f : E \to B$. $f$ is a continuous function, $B$ is a topological space and $f^{-1}(b)$ is a vector space for all $b\in B$. Moreover it is a continuous family of vector spaces in the sense that vector addition $E \oplus E \to E$ and scalar multiplication $\mathbb R \times E \to E$ are continuous.

If vector spaces typically had natural basis, vector bundles would typically be trivial. i.e. $E \simeq V \times B$ and under that homeomorphism, $f$ would be conjugate to projection $\pi : V \times B \to B$, $\pi(v,b) = b$, since choosing such a conjugation is equivalent to choosing (continuously) a basis for each vector space $f^{-1}(b)$. But this generally can't be done. The Moebius band being the first interesting counter-example. The non-triviality of the Moebius band from this perspective would be a reflection of the difficulty choosing a basis for 1-dimensional vector spaces.

I suppose there's a natural way to quantify an answer to this question. A vector bundle is a family of vector spaces over a base space, $f : E \to B$. $f$ is a continuous function, $B$ is a topological space and $f^{-1}(b)$ is a vector space for all $b\in B$. Moreover it is a continuous family of vector spaces in the sense that vector addition $E \oplus E \to E$ and scalar multiplication $\mathbb R \times E \to E$ are continuous.

If vector spaces typically had natural basis, vector bundles would typically be trivial. i.e. $E \simeq V \times B$ and under that homeomorphism, $f$ would be conjugate to projection $\pi : V \times B \to B$, $\pi(v,b) = b$, since choosing such a conjugation is equivalent to choosing (continuously) a basis for each vector space $f^{-1}(b)$. But this generally can't be done. The Moebius band being the first interesting counter-example. The non-triviality of the Moebius band from this perspective would be a reflection of the difficulty choosing a basis for 1-dimensional vector spaces.

I suppose there's a natural way to give a type of global quantative answer to this question. A vector bundle is a family of vector spaces over a base space, $f : E \to B$. $f$ is a continuous function, $B$ is a topological space and $f^{-1}(b)$ is a vector space for all $b\in B$. Moreover it is a continuous family of vector spaces in the sense that vector addition $E \oplus E \to E$ and scalar multiplication $\mathbb R \times E \to E$ are continuous.

If vector spaces typically had natural basis, vector bundles would typically be trivial. i.e. $E \simeq V \times B$ and under that homeomorphism, $f$ would be conjugate to projection $\pi : V \times B \to B$, $\pi(v,b) = b$, since choosing such a conjugation is equivalent to choosing (continuously) a basis for each vector space $f^{-1}(b)$. But this generally can't be done. The Moebius band being the first interesting counter-example. The non-triviality of the Moebius band from this perspective would be a reflection of the difficulty choosing a basis for 1-dimensional vector spaces.

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Ryan Budney
Ryan Budney
  • 44.4k
  • 2
  • 139
  • 245

I suppose there's a natural way to quantify an answer to this question. A vector bundle is a family of vector spaces over a base space, $f : E \to B$. $f$ is a continuous function, $B$ is a topological space and $f^{-1}(b)$ is a vector space for all $b\in B$. Moreover it is a continuous family of vector spaces in the sense that vector addition $E \oplus E \to E$ and scalar multiplication $\mathbb R \times E \to E$ are continuous.

If vector spaces typically had natural basis, vector bundles would typically be trivial. i.e. $E \simeq V \times B$ and under that homeomorphism, $f$ would be conjugate to projection $\pi : V \times B \to B$, $\pi(v,b) = b$, since choosing such a conjugation is equivalent to choosing (continuously) a basis for each vector space $f^{-1}(b)$. But this generally can't be done. The Moebius band being the first interesting counter-example. The non-triviality of the Moebius band from this perspective would be a reflection of the difficulty choosing a basis for 1-dimensional vector spaces.