I have [asked this question on math.se](https://math.stackexchange.com/questions/356925/a-basis-of-the-symmetric-power-consisting-of-powers), but did not get an answer - I was quite surprised because I thought that lots of people must have though about this before:


Let $V$ be a complex vector space with basis $x\_1,\ldots,x\_n\in V$. Denote by $v\_1\odot\cdots\odot v\_k$ the image of $v\_1\otimes\cdots\otimes v\_k$ in the symmetric power $\newcommand{\Sym}{\mathrm{Sym}}\Sym^k(V)$. It is well-known that the Elements $v^{\odot k}$ for $v\in V$ generate this space (see, for instance, [this answer on math.se](https://math.stackexchange.com/questions/137912/can-e-n-always-be-written-as-a-linear-combination-of-n-th-powers-of-linear-p/138411#138411)), so they must contain a basis. 


In other words, let $N=\binom{n+k-1}k$, then there must be $v\_1,\ldots,v\_N\in V$ with 
$$\mathrm{Sym}^k V = \mathbb Cv\_1^{\odot k} \oplus \cdots \oplus \mathbb C v\_N^{\odot k}.$$
I am looking for an explicit description of such a basis. Is such a description known? Is there maybe even a *"nice"* or somewhat *"natural"* choice for the $v\_i$ as linear combinations of the $x\_i$?



A basis of the symmetric power consisting of powers
I would look up a book on the calculus of finite differences in a multivariate setting.
The claim here is to show that for any multi-index $\alpha=(\alpha\_1,\ldots,\alpha\_n)$
of length $k$ one can express the multiple derivative at zero
$$
\left(\frac{\partial}{\partial t}\right)^\alpha \ 
(t\_1x\_1+\cdots+ t\_n x\_n)^k
$$
as a linear combination of finite difference analogous expressions
which should only involve the evaluation of $(t\_1x\_1+\cdots+ t\_n x\_n)^k$
at integer points $(t\_1,\ldots,t\_n)$ with nonnegative coordinates adding up
to $k$. This is the same as the above candidate basis considered by you and Peter.
I don't know if there exists a multivariate analogue of the [Newton series](http://en.wikipedia.org/wiki/Finite_difference). If so
then this would immediately imply the wanted statement.




---


Edit:
Apparently there is such a formula due to Lascoux and Schutzenberger, see Theorem 9.6.1 page 148 in the book
"Symmetric functions and combinatorial operators on polynomials" by Alain Lascoux.
Another source on the web is [here](http://phalanstere.univ-mlv.fr/~al/ARTICLES/NewtonInterp.ps.gz). It also has the required property here which is that the number of finite differences taken is the same as the degree of the multiplying Schubert polynomial.




---


Edit: @Jesko you're right it is a bit more complicated than what I said. Also, the Lascoux-Schutzenberger formula might not be the simplest to use here.


First note that expressions
$$
\prod\_{i=1}^{n-1} (x\_i-x\_n)^{\beta\_i}\ \times\ (kx\_n)^{k-|\beta|}\ ,
$$
where $\beta$ ranges over multiindices with $n-1$ components and length $|\beta|\le k$,
form a basis.
Now you get the latter as derivatives
$$
\left(\frac{\partial}{\partial t}\right)^{\beta} \ 
\left(t\_1x\_1+\cdots+ t\_{n-1} x\_{n-1}+\left(k-\sum\_{i=1}^{n-1}t\_i\right)x\_n\right)^k
$$
at $t=0$.


Call $f(t\_1,\ldots,t\_{n-1})$ the polynomial function to be hit with derivatives.
One has a multivariate Newton expansion for it:
$$
f(t)=\sum\_{m} (t-a)^m \partial^m f(a\_{11},a\_{21},\ldots,a\_{n-1,1})
$$
as follows.
Here $a$ stands for a matrix of indeterminates $(a\_{ij})$ with $1\le i\le n-1$
and $1\le j\le d$, with $d$ high enough.
Let $\partial\_{ij}$ denote the divided difference operator acting on functions of these indeterminates
as
$$
\partial\_{ij} g=\frac{1}{a\_{i,j+1}-a\_{ij}}\left(
g({\rm argument\ with\ }a\_{i,j+1}\ {\rm and}\ a\_{ij}\ {\rm exchanged})- g
\right)\ .
$$
The notation $m=(m\_1,\ldots,m\_{n-1})$ is for a multiindex with nonnegative entries.
We also write the corresponding operator
$$
\partial^m = \prod\_{i=1}^{n-1} \left(\partial\_{i, m\_i}
\cdots\partial\_{i,2}\partial\_{i,1}\right)
$$
noting that finite difference operators concerning different groups of variables commute.
Finally 
$$
(t-a)^m=\prod\_{i=1}^{n-1} \left((t\_i-a\_{i,m\_i})\cdots(t\_i-a\_{i,2})(t\_i-a\_{i,1})\right)\ .
$$
The formula basically amounts to applying Newton's univariate formula in each coordinate direction separately.
Now use this with the choice $a\_{i,j}=j-1$, then take the beta derivative in the $t$'s
and that should be it.


