# Vector spaces with natural bases

Actually, I would claim (apparently in accord with many comments and answers to Sergeib's question ) that this is the default situation and that it is a rare event if a naturally occurring vector space does have a natural basis or even if it has any basis that can be described explicitly: the notorious $\mathbb Q \; -$ vector space $\mathbb R$ being the foremost example of this impossibility of exhibiting an explicit basis . Of course tautological examples like polynomial rings don't contradict this thesis, since they are defined as free vector spaces on some set!

In fact, the only example I can see without thinking twice of a natural vector space with a big natural non-tautological basis is the rational function field $k(X)$ seen as a $k$ - vector space .

Namely, consider a field $k$ and the set of monic polynomials $f_i(X) \; (i \in I)$ irreducible over $k$. Then $k(X)$ has the following $k$ - basis:

the mononomials $X^k (k\in \mathbb N)$ and the rational fractions $\frac{X^m}{f_i(X)^s}$ (with $i\in I,\; s>0$ and $m< deg f_i$ )

In particular this natural basis is non denumerable if $k$ is non denumerable..

Question Which other vector spaces do you know for which some (preferably big) explicit basis can be given, but which are not clearly constructed as free vector spaces over a set?

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Community wiki? –  Qiaochu Yuan Jul 18 '10 at 23:43
I would note that one can look at (multi-variate) polynomial rings as $\text{Sym}(V^*)$, and such a ring has a "natural" basis iff $V$ does (where $V$ is a finite-dimensional vector space). –  Charles Staats Jul 19 '10 at 0:19
@Qiaochu: done ! –  Georges Elencwajg Jul 19 '10 at 0:40

I'll echo Ben Webster's comment from the other thread, but in the other direction: find some natural commuting operators whose simultaneous eigenspaces are all one-dimensional. This is how, for example, one gets the root space decomposition in Lie theory and the decomposition of spaces of modular forms into eigenforms. (Admittedly, the former is only unique up to a choice of Cartan subalgebra.)

I guess that to get an actual basis instead of a basis-up-to-scalar-multiplication one needs a little more. For cusp forms one usually takes the normalization where the coefficient of $q$ in the Fourier expansion is $1$, whereas for the Lie algebra I guess we take a normalization where the structure constants are as nice as possible.

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Qiaochu, the key property you need is "multiplicity one", but note that, in general, a decomposition into 1-dim subspaces doesn't yield a basis, not even a "signed basis". In the case of the root space decomposition of a semisimple Lie algebra $\mathfrak{g},$ it holds for the orthogonal complement of the Cartan subalgebra $\mathfrak{h}\subset \mathfrak{g}$, but not for $\mathfrak{h}$ itself, so you don't even get a decomposition into 1-dim subspaces, and there are at least 2 choices of "normalization where the structure constants are as nice as possible" differing by signs. –  Victor Protsak Jul 19 '10 at 7:55
In the case of modular forms, the multiplicity one property is really deep. There are more elementary examples of a full eigenvector basis of a max commuting algebra where everything works, but the signed basis issue is typically unavoidable. –  Victor Protsak Jul 19 '10 at 7:58
I somewhat disagree with the premise of the question: I want to point out that after imposing more structure, this also becomes a default situation. The general setting I have in mind is commutative algebra and there are various techniques to construct a "preferred" SAGBI basis in the coordinate ring or homogeneous coordinate ring of a variety. Particularly useful in representation theory are standard monomial bases in $K[G/U],$ where $G$ is a semisimple algebraic group and $U$ is the unipotent radical of a parabolic subgroup.
$H^*(\mathbb{CP}^{\infty},k) = k[x]$ has basis $\{x^i : i \geq 0\}$. Of course, every other nontrivial computation of (co)homology groups fits in here.