Vector spaces without natural bases - MathOverflow

Vector spaces without natural bases

2010-07-18T20:34:25Z

Does anyone know any nice examples of vector spaces without a basis that is in some sense "natural".

To clarify what I mean, suppose we look at $\mathbb{R}^2$. We define $\mathbb{R}^2$ as pairs of real numbers. In some sense, what we are doing is expressing vectors in terms of a natural basis : (1,0) and (0,1). This is not what I want.

An example that I thought of is a tangent space to a manifold. When one picks a tangent space to a manifold, there is no natural basis that one can pick.

Are there other nice examples?

Answer by Anon for Vector spaces without natural bases

2010-07-18T20:40:58Z

The obvious example is $\mathbb{R}$, as a vector space over $\mathbb{Q}$; the existence of such a basis requires the axiom of choice.

Answer by Nick Salter for Vector spaces without natural bases

2010-07-18T20:45:06Z

To expand on Anon's answer, I'd like to discuss one way in which the lack of a "natural" basis has some utility. A Hamel basis is a basis for $\mathbb{R}$ over $\mathbb{Q}$. Hamel bases are quite useful, due to their interactions with Cauchy functions (real-valued functions that satisfy an "additive" functional equation $f(x+y) = f(x) + f(y)$. This functional equation is equivalent to being linear over $\mathbb{Q}$. Examples of the utility of Cauchy functions abound. One approach to proving that the cube and the tetrahedron are not equidecomposable (Hilbert's 3rd problem) is to pick the $\mathbb{Q}-$linearly independent set ${1, \pi}$ and, by the magic of AC, this extends to a Hamel basis. Setting up the right Cauchy function then resolves the problem. For more on this, see "Conjecture and Proof" by Miklós Laczkovich.

Answer by José Figueroa-O'Farrill for Vector spaces without natural bases

2010-07-18T20:45:40Z

Most vector spaces I've met don't have a natural basis. However this is question that comes up when teaching linear algebra. You want to motivate abstract vector spaces instead of working with $\mathbb{R}^n$ (or your favourite field in place of $\mathbb{R}$). One simple example, is this.

Consider $\mathbb{R}^n$ ($n>2$) as a euclidean space relative to the "dot" product and let $v = (1,1,\dots,1)$. Then the subspace $V \subset \mathbb{R}^n$ of vectors orthogonal to $v$ does not have a natural basis. If you don't like introducing an inner product, then take $V$ to be the annihilator of $v$ in the dual of $\mathbb{R}^n$. This actually comes up when discussing the root space of $\mathfrak{su}(n)$, say.

Answer by Willie Wong for Vector spaces without natural bases

2010-07-18T21:04:29Z

Another example is most function spaces defined over $\mathbb{R}$. The space of square integrable functions $L^2(\mathbb{R})$ doesn't have a natural basis. You would like one in the trigonometric functions $e^{2\pi i n x}$ in view of Plancherel's theorem and the Fourier transform, but they are not actually in $L^2$. (Compare the case on a torus, where the "natural" basis exists.)

Answer by Harald Hanche-Olsen for Vector spaces without natural bases

2010-07-18T21:20:11Z

The solution space of a homogeneous (ordinary or partial) linear differerential equation has no natural basis.

Answer by Akhil Mathew for Vector spaces without natural bases

2010-07-18T21:46:34Z

Hilbert spaces don't generally have nice bases in the sense of linear algebra. Neither does the ring of formal power series $k[[X]]$ over a field $k$. (These have "bases" with "infinite linear combinations" that only make sense because of completeness.)

Answer by Qiaochu Yuan for Vector spaces without natural bases

2010-07-18T22:54:03Z

The vector space of polynomials (possibly of some fixed degree). This is a case where many students, I think, are tempted to privilege the basis ${ 1, x, x^2, ... }$, but to do so is to 1) privilege evaluation at $0$ over evaluation at other points, and 2) miss out on the utility of other bases like ${ 1, x, {x \choose 2}, ... }$.

Answer by Pete L. Clark for Vector spaces without natural bases

2010-07-19T00:26:47Z

Let $K$ be a field, let $S$ be a set, and consider the $K$-vector space $\operatorname{Map}(S,K)$ of all functions from $S$ to $K$.

When $S$ is finite, $\operatorname{Map}(S,K)$ has a natural basis: for each $x \in S$, let $\delta_x$ be the function which takes $1$ at $x$ and $0$ otherwise. However, when $S$ is infinite, these "Dirac" functions span only the set of finitely nonzero functions. In this case, the idea that there is no "natural basis" can probably be stated and proven in categorical language. (If you wish to do so as an addendum to this answer, please feel free!)

Note that one may also look at this construction in terms of the distinction between direct products and direct sums.

Answer by Ryan Budney for Vector spaces without natural bases

2010-07-19T00:39:39Z

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.

Answer by Martin Brandenburg for Vector spaces without natural bases

2010-07-19T08:01:44Z

This example generalizes some of the others already mentioned: Take an infinite family of vector spaces $(V_i)_{i \in I}$. Now what about $\prod_{i \in I} V_i$ , can you write down a basis?

Also, it is easy to construct an infinite multilinear tensor product $\bigotimes_{i \in I} V_i$. However, writing down a basis is equivalent to find a set of representatives of $\prod_{i \in I} V_i \setminus \{0\} / \sim$, where $\sim$ identifies families of elements, which differ only at finitely many indices. And this cannot be done explicitely.

Answer by Benoît Kloeckner for Vector spaces without natural bases

2010-07-19T08:44:51Z

For teaching purposes, the most simple example (which I use frequently in a first course in linear algebra) is a generic sub-vector space of $\mathbb{R}^n$. Any vector plane in the $3$-space that is not cardinal works.

Answer by Vectornaut for Vector spaces without natural bases

2010-08-03T19:31:50Z

As a physicist, I would say the most obvious example is $n$-dimensional Euclidean space, with $n > 1$. Since a few people have mentioned casually that Euclidean spaces do have natural bases, I should explain myself...

Informally, a Euclidean space is supposed to be an idealization of something like a giant sheet of paper with an origin marked in pencil, or interstellar space with an origin marked by a certain star. If you're in the habit of carrying around a tape measure, a space like this has a natural metric, and you can turn it into a vector space in the obvious way (using the metric to define scalar multiplication and the parallelogram rule to define addition).

From this point of view, Euclidean space clearly has no natural basis, because if you're stranded on a giant sheet of paper, or floating in interstellar space, there's no natural set of "special" directions.

Unfortunately, I don't know offhand how to formalize this argument. My guess is that you would start with Hilbert's axioms for Euclidean $n$-space, and choose an arbitrary point to be the origin. Hartshorne mentions in Geometry: Euclid and Beyond that in Hilbert's framework, the congruence classes of line segments naturally become the positive elements of an ordered field, which is of course isomorphic to $\mathbb{R}$. Choosing an arbitrary congruence class of line segments to be the "unit segments," you get a metric on your space. You can then turn the set of points into a vector space, using the metric to define scalar multiplication and the parallelogram rule to define addition (just like before, but now rigorously). It seems obvious to me that this vector space will have no natural basis.

Answer by Anweshi for Vector spaces without natural bases

2010-08-03T20:06:39Z

The vector space $\mathbb C / \mathbb R$ does not have a preferred basis. Among the two bases ${1, i}$ and ${1, -i}$, there is no reason to prefer one over the other. The choice of one of these amounts to a choice of an orientation for the plane.

Answer by Bruce Westbury for Vector spaces without natural bases

2010-08-03T20:25:10Z

Cohomology with coefficients in $\mathbb{Q}$.