# Spanning polynomials with powers

So, here a cute little algebraic question that I came across thinking about a finite groups question Noah asked me (he or I might blog about that later):
Choose your favorite set of m linear polynomials in n variables (if you don’t like coordinates, your favorite m element subset of your favorite n dimensional subspace). When do the the jth powers of these polynomials span all polynomials of degree j?

Of course, for j=1, this is easy. It isn’t immediately clear what happens even for j=2; I think the answer is something like you must contain a subset of $\binom j 2$ polynomials such that no k dimensional subspace contains more than $\binom k2$ of them. Any thoughts? (Also, special bonus bonus points to whoever figures out what representation theory question this is supposed to solve).

## 14 thoughts on “Spanning polynomials with powers”

1. I apologize for asking you what is “representation theory”. Googling about it seems to me that it is a very active research topic. A short info would be appreciated.

2. David Speyer says:

Assume that the characteristic is greater than j.

Think of your set as a collection of $m$ points in $\mathbb{P}^{n-1}$. Isn’t the condition just that these points do not lie on any hypersurface of degree $j$?

To see this, notice that your condition is that the images of these points in $\mathbb{P}(\mathrm{Sym}^j K^n)$ do not lie on any hyperplane. Your map from $\mathbb{P}(K^n) \to \mathbb{P}(\mathrm{Sym}^j K^n)$ is the standard $j$-uple embedding except that some coefficients are multiplied by binomial coefficients. Given the condition on the characteristic, none of these coefficients is zero, so your map is the $j$-uple embedding up to a linear automorphism of the target. So, the question is how to find points whose images under the $j$-uple embedding do not lie on a hyperplane; this is the same as finding points who do not lie on a degree $j$ hypersuface.

Of course, with my luck, your representation theory question will be something that’s only interesting in characteristic $2$.

3. Considering linear functionals on the space of j-dimensional polynomials shows that this is the same as finding a set of points of projective (n-1)-space on which no nontrivial degree j polynomial vanishes. For example, when j=2 and n=3, this is the same as finding a set of (at least) 6 points in P^2 which are not contained in any (possibly singular) conic.

Unfortunately (fortunately?), away from characteristic 2 this is definitely not expressible in terms of linear subspaces. I’m a little unclear what your intended condition is, but for example the six points [cos(k*pi/3): sin(k*pi/3): 1] are solutions of x^2 + y^2 = z^2, but any three are linearly independent.

Perhaps an algebraic geometer can help.

4. David Speyer says:

By the way, in low characteristic, I think the following is the answer: Write $j = \sum a_i p^i$, with $0 \leq a_i \leq p-1$. Then the condition should be that your points do not lie on any hypersurface of degree $\sum a_i$.

5. You’re in luck, I only really cared about characteristic zero.

6. Ah, good. My claim at 4 was false. (Doesn’t work for $n=2$, $j=p+1$.) I’m not sure that there is a nice answer in low characteristic.

7. David-

I think small characteristic breaks the rest of our argument anyways, for roughly the same reasons (the representation theory of the symmetric group goes all pear shaped), so no worries.

8. Let me stick to characteristic zero.

In dimension two ( $n=2$ ) the vector space $V$ generated
by the $j$th powers of any set of $k$ linear polynomials has
dimension $\min( k , j+1)$.

In dimension greater than two ($n > 2$), the optimal
lower bound for the dimension of $V$ is $\min(k, j(n-1) +1)$.
Moreover, if $k \ge 2n +1$ and the equality holds, then there exists a rational normal curve in $\mathbb P(L)$, $L$ is the
space of linear polynomials in $n$ variables, passing through
the $k$ points determined by the linear polynomials.

Recall that a rational normal curve in $\mathbb P^n$, in suitable coordinates, is just the image of the natural morphism $\mathbb P( K^2) \to \mathbb P( Sym^{n} K)$

This result is usually called “Castelnuovo’s Lemma”. It is used
to bound the arithmetical genus of irreducible non-degenerate projective curves.

There are variants of this result saying that if the dimension
of $V$ is nearly minimal and $k$ is big enough then the points
determined by the linear polynomials lie in rational normal scrolls.
For details see for instance “Curves of almost maximal genus” by
Eisenbud and Harris, available at Eisenbud’s web page.

9. Ben,
Thanks!
I do realize how silly my comment was.
Américo

10. I forgot to say that I am considering sets of $k$ linear polynomials
in general position: no linear relation exists among $n$
or less of these polynomials.

11. James G says:

I must have misunderstood this question, but clearing it up for me might clear it up for other people.

Let V be a vector space over char 0 field C and SV the free unital commutative algebra. Now is your question: given k elements in V+C (call their span W) when do have W^j = F_jSV? Here F_j denotes the filtration with F_jSV consisting of the polynomials of degree less than or equal to j.

If so, then homogenise the linear polynomials (replace 1 with a new variable) , so W^j lives in the jth homogeneous graded piece of S(V+C). Now you just want to know whether W^j is the whole homogeneous piece, but now you’ve asked the question in this way it’s clear.

That is, when your k linear polynomials span the whole of V+C.

Sorry if that’s nonsensical. It’s been a long day.

James

12. Now is your question: given k elements in V+C (call their span W) when do have W^j = F_jSV?

No. You’re making this too hard. I have some homogenous linear polynomials. I take the jth power of each of them. Does that span homogeneous degree j polynomials or not? (i’ll note, it’s already been answered above. I’m just clearifying what was answered).