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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?

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Dear Sergeib, Some of the most common occurences are when one forms $Hom$ spaces. E.g. if $V$ and $W$ are two reps. of a group $G$ over a field $k$, then $Hom_{k[G]}(V,W)$ is naturally a $k$-vector space, but (in general) has no natural basis. More generally, if one applies linear-algebra or multilinear-algebra type contsructions (Homs, tensor products, etc.) to objects with a vector spaces structure, then one will obtain vector spaces that typically have no natural basis. –  Emerton Jul 18 '10 at 20:44
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Another example (related to my first) is the formation of subspaces: e.g. if $l:V \to k$ is a non-zero linear functional on a $k$-vector space $V$, then (even if $V$ has some specified basis) the kernel of $V$ typically has no preferred basis. –  Emerton Jul 18 '10 at 20:45
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I'm having real trouble seeing the point of this question. Do you think having a supply of vector spaces without obvious bases will illuminate some point of linear algebra for you? –  Ben Webster Jul 18 '10 at 20:47
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This really should be community wiki. –  Willie Wong Jul 18 '10 at 20:57
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Another good way to motivate basis free constructons is that there are many vector spaces with more than one natural basis. –  Noah Snyder Jul 18 '10 at 21:38

16 Answers 16

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.

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I have just taught a linear algebra course where the textbook (David Lay) offers an alternative route to those who don't want to teach abstract vector spaces: a lot of the same material (spans, linear independence, bases, coordinates) presented for subspaces of $\mathbb{R}^n.$ I found it a very convincing motivation for more abstract approach and, not unexpectedly, the students, in turn, found it just as difficult. –  Victor Protsak Jul 19 '10 at 7:32
    
I had exactly the same issue with exactly the same book this spring! –  Daniel Larsson Jul 19 '10 at 9:18

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

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I think you need a little bit more than this: Defining said basis and proving that your definition actually works are different things. Hypothetically, you could define a very natural basis for some vector space V, and then only use AC at the very end to show V was actually all of R. –  Michael Burge Jul 19 '10 at 3:33
    
The preceding remark was given a long time ago, but I'd still ask: who said anything about "defining a basis"? The OP asks for "any nice examples of vector spaces without a basis that is in some sense 'natural'." And this answer seems to give exactly that. –  Todd Trimble Jul 14 at 16:31

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.)

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

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I guess one could argue that here it is natural to look for a basis of eigenvectors with respect to differentiation. –  Qiaochu Yuan Jul 18 '10 at 22:51
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@Yuan: Even then, if your eigenspaces aren't 1-dimensional you're sunk. –  Ryan Budney Jul 19 '10 at 0:29
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I never said constant coefficients. –  Harald Hanche-Olsen Jul 19 '10 at 7:14

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.

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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.

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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 $\{ 1, x, {x \choose 2}, ... \}$.

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Well the same could be said about any other "natural basis". It always comes down to privileging one basis over all other basis. From the viewpoint of pure linear algebra, there are no natural basis at all (excluding the empty set as the only and hence natural basis of the zero vector space). All basis are equal if one doesn't specify a point of view that comes from outside linear algebra. To decide whether some item is "natural" always requires knowledge on what you want to do with it. If I only want to test equality of explicitly given polynoms the basis (x^k) could be called natural. –  Johannes Hahn Jul 18 '10 at 23:44
    
Very good point. Perhaps I should have said "the vector space of regular functions on the affine line." Then the point I'm trying to make with comment 1) is that the affine line is homogeneous with respect to its automorphism group. –  Qiaochu Yuan Jul 18 '10 at 23:57
    
Once again,better mathematical living through The Axiom of Choice. –  Andrew L Jul 20 '10 at 9:42

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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Nature could be evil and provide us with natural basis which do not depend continuously on the basis :) –  Mariano Suárez-Alvarez Jul 19 '10 at 2:20
    
on the basis of your map $f$, that is. –  Mariano Suárez-Alvarez Jul 19 '10 at 2:21

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.

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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.)

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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.

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Probably the simplest example --- $\{\ (x_1,...,x_n)\ |\ x_1+...+x_n=0\ \}$. Any choice of a basis distorts some symmetry...

As pointed out by Joe Silverman, this was in fact already mentioned in one of the previous answers. Let me still leave it, I think in this form it is sort of a "bare bones" example...

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This is exactly José Figueroa-O'Farrill's answer, i.e., the set of points in $\mathbb{R}^n$ that are orthogonal to $(1,1,\ldots,1)$. –  Joe Silverman Jul 14 at 14:46
    
@JoeSilverman Oops :( Did not notice, it is quite down there. Thanks for pointing –  მამუკა ჯიბლაძე Jul 14 at 14:48

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.

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Indeed, every Desarguesian projective geometry (i.e. a set with an incidence relation satisifying the axioms for a projective space together with Desargues' theorem) is generated by a vector space over a skew field. (This skew field is a field if Pappus' theorem holds) Furthermore there is a isomorphism that turns any given basis of that vector space into another basis, so all bases are equally natural, provided that you do not distinguish isomorphic geometries. –  Gabriel Ebner Aug 4 '10 at 1:06
    
@Gabriel Ebner: Cool! What kind of textbook would cover this stuff? How could I figure out which module generates, say, the real projective plane? –  Vectornaut Aug 5 '10 at 16:58
    
@Vectornaut I learned it from Beutelspacher's Projective Geometry (German; English translation available). To actually construct the vector space you need to pick a hyperplane of points at infinity, and an origin. The scalings centered at the origin correspond to elements of the skew field; the translations form the underlying vector space of the affine space (the complement of the hyperplane). (E.g. you can define a translation as an automorphism that has all points at infinity as fixed points, and is invariant on all lines through a point at infinity (the direction of the translation)). –  Gabriel Ebner Aug 6 '10 at 16:31

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.

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

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I'll use this question as an occasion to advertise to mathematicians a interesting piece of terminology commonly used in condensed matter physics: ``degenerate ground state manifold''. In mathematical terms, this translates to ``eigenspace for the lowest eigenvalue of the Hamiltonian, whose dimension is $\ge2$, and that does not have a preferred basis of eigenvectors''.


Let me analyze the individual terms of the phrase:

State: Here, a ``state'' is an eigenvector of the Hamiltonian.

ground: A state is a ``ground state'' if its corresponding eigenvalue (=energy) is the lowest.

degenerate: Generically, the eigenspaces corresponding to the various eigenvalues of the Hamiltonian will be one-dimensional . When that doesn't happen, an energy level is called degenerate. The word ``degeneracy'' is then used to refer to its dimension.

Manifold: Here, physicists use the term ``manifold'' here because the lowest energy eigenspace does not have a natural basis (or has more than one natural basis that one could write down). This is somewhat similar to the use of the term ``manifold'' by mathematicians: a manifold is a space on which one does not have preferred choices of coordinates (or one has multiple choices of coordinate systems).

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