Posted by: Alexander Ellis | April 27, 2008

## Shoulda Series 1: Choosing Bases

Inspired by Tim Gowers’s illuminating informal discussions of mathematical topics, and in particular this one on vector spaces, I am starting my own “Shoulda Series.” That is, a series of notes on things about which I might have said: “Someone should have told me this a long time ago!” In particular, I don’t want to use this series to “give away” the sorts of things you should really figure out for yourself.

Question: Why do we avoid choosing bases for vector spaces whenever possible? In particular, why is an isomorphism defined independent of bases “better” than one which uses bases?

One of thousands of possible answers: Since I’m a geometer at heart, I think the following is a great reason. In short, the moral is: basis-free isomorphisms generalize to vector bundles and basis-dependent ones usually do not. This is because “choosing a basis” of a vector space is the point-wise analogue of the “choosing a local trivialization” of a vector bundle. To do anything globally, one generally needs to use several different trivializations; a bona fide isomorphism of vector bundles requires a family of vector space isomorphisms compatible with the transition functions between the various trivializations. (For those who know what this means, the precise check is that the Cech 1-cocycles corresponding to the bundles in question differ by a Cech 1-coboundary, where the coefficient sheaf is the constant sheaf associated to $\text{GL}(n)$.)

For a wealth of examples, consider the following. Any two finite-dimensional vector spaces (over the same field) are isomorphic, but not naturally so. But there are usually several different isomorphism classes of (real or complex) vector bundles of a given dimensionon on a given space.

A more specific example will also bring out a slight improvement of our moral above. Let $V$ be a real vector space and $E$ be a real vector bundle. Recall that the isomorphism between $V$ and its dual $V^*\cong\text{Hom}(V,\mathbb{R})$ is not natural. Hence $E$ is not always isomorphic to its dual $E^*=\text{Hom}(E,\mathbb{R})$. (Here, $\mathbb{R}$ is the trivial vector bundle $M\times\mathbb{R}\to\mathbb{R}$.) But if $V$ is given an inner product $\langle\cdot,\cdot\rangle$, the isomorphism $V\cong V^*$ becomes natural: the map is simply $v\mapsto (w\mapsto\langle v,w\rangle)$. So if a vector bundle $E$ admits an inner product (in the sense of vector bundles), then we should expect an isomorphism $E\cong E^*$. Indeed this is the case, and in fact this is a common phenomenon: using a partition of unity, you can check that every real vector bundle on a paracompact base space admits an inner product.

However, the situation is different with complex vector bundles. Let $F$ be a complex vector bundle. While every complex vector bundle paracompact base space admits a Hermitian inner product, this product induces an isomorphism $\overline{F}\cong F^*$ between the dual bundle $F^*$ and the complex conjugate bundle $\overline{F}$; neither of these need be isomorphic to $F$ itself.

For example, consider the tangent bundle $TS^2$ to the Riemann sphere and its dual, $T^*S^2\cong\text{Hom}(TS^2,\mathbb{R})$. Considered as real 2-plane bundles, these are isomorphic since $S^2$ is paracompact. But they are non-isomorphic as complex line bundles. For readers familiar with characteristic classes, there is an easy way to see this algebraically. The Chern classes of a complex vector bundle take values in the integral cohomology of the base space. They obey the relation $c_i(F^*)=(-1)^ic_i(F)$, so if $c_i(F)\neq0$ for any odd $i$, then $F$ is not isomorphic to $F^*$. The Stiefel-Whitney classes $w_i(E)$ of a real vector bundle obey the same law. But since they take values in $\mathbb{Z}/2\mathbb{Z}$ cohomology, $1=-1$ and the relation is an equality!

Aside: Unoriented phenomena (e.g. Stiefel-Whitney classes) tend to use $\mathbb{Z}/2\mathbb{Z}$ relations, while oriented phenomena tend to use $\mathbb{Z}$ relations. Complex vector spaces have a canonical orientation on their underlying real vector bundles, so complex phenomena (e.g. Chern classes) fall under oriented phenomena. Examples are mod 2 versus oriented intersections (see Guillemin & Pollack, Differential Topology) and the oriented versus unoriented versions of Poincaré duality (see Hatcher, Algebraic Topology, freely available online). While I’m on the subject of references, two great references for vector bundles and characteristic classes are Milnor & Stasheff’s Characteristic Classes and Hatcher’s Vector Bundles and K-Theory (the latter is unfinished and freely available online).

We conclude with our improved moral statement.

Improved moral: A natural isomorphism of vector spaces generalizes to vector bundles. An isomorphism of vector spaces making use of a structure which “globalizes” well (e.g. inner products, when the base space is paracompact) will also generalize to vector bundles. Other isomorphisms often will not.

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## Responses

1. Your argument seems to be a higher level version of the “basis-free algebra is geometry”, that somehow coordinates were introduced to better handle the geometry, but we should be able to introduce the coordinates that better suit every case, and therefore geometric arguments should not depend on what basis we choose. Since Vector Bundles, LCSs, Affine Varieties are somehow geometrical extensions and variations of vector spaces (loosely speaking), an argument about the geometry of spaces independent of basis should be easier to translate into this other cases.

Love the idea of the shoulda series!

2. […] may or not really count as part of the “Shoulda Series” (1), since I’m pretty sure someone did tell me this at some point. But either way, I had to […]