TITLE: Vector spaces without natural bases
QUESTION [15 upvotes]: 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?

REPLY [6 votes]: One example I like to use is the $1$-dimensional vector space of multiples of some physical unit (length, time, mass): for example, the meter is a basis of the $1$-dimensional vector space of lengths, and the light-year is also a basis of it, but there is no natural basis of this vector space.
This example can also be used to illustrate multilinear algebra constructs on $1$-dimensional vector spaces: the space of speeds is the (still $1$-dimensional) space of linear maps between the $1$-dimensional vector space of time spans and the $1$-dimensional vector space of lengths (it turns out that, in special relativity, but not in classical mechanics, there is a canonical isomorphism between these spaces, i.e., a canonical basis for the space of speeds).  The space of areas it the tensor square of the space of lengths, and the space of volumes is its tensor cube.  And so on.
This kind of example makes it clear why for $1$-dimensional vector spaces $V$, the tensor product of $V$ with its dual is canonically isomorphic to the base field, so such spaces can be called "invertible" (as in "invertible sheaf").