11
$\begingroup$

Is there a generalization of the Cauchy-Schwarz inequality along the following lines? Let $V$ be an inner product space (for simplicity of notation, let us work over the real numbers). Let $v_1, \ldots, v_n$ be in $V$. Let $G$ denote the Gram matrix of the $v_i$, namely, $G$ consists of all possible $(v_i, v_j)$, as $i,j = 1, \ldots, n$, where $(-,-)$ is the inner product in $V$. The usual Cauchy-Schwarz inequality, with $n=2$, can be written as follows, to get rid of square roots:

$$ \det(G) = (v_1,v_1)(v_2,v_2) - (v_1,v_2)^2 \geq 0, $$

with equality iff $v_1$ and $v_2$ both belong to some $1$-dimensional subspace of $V$. So in this case, for $n=2$, the LHS is a homogeneous polynomial in $G$ of degree $2$, and equality is achieved iff $v_1$ and $v_2$ both belong to some $1$-dimensional subspace.

For the general $n$ case, is there a higher degree homogeneous polynomial in $G$ which is non-negative for any $v_1, \ldots, v_n$ in $V$, and which vanishes iff the $v_i$, for $i = 1,\ldots, n$ all lie in some $1$-dimensional subspace of $V$?

(I suspect there may be such a polynomial of degree $2 \lfloor \frac{n(n+1)}{4} \rfloor$. So for instance, if $n=2$, the expected degree is $2$. If $n=3$, the expected degree is $6$, and so on.)

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Indeed the Gramian is positive semi-definite, so its determinant is always nonnegative, and is positive just when the vectors are linearly independent. See en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant $\endgroup$
    – Kevin Casto
    Commented Jul 31, 2020 at 19:35
  • $\begingroup$ @KevinCasto, yes but, the determinant of the Gramian vanishes iff the vectors are linearly dependent. What I would like is though, a polynomial which vanishes iff the vectors lie in the same $1$-dimensional subspace. $\endgroup$
    – Malkoun
    Commented Jul 31, 2020 at 19:49
  • 6
    $\begingroup$ Actually, there's an octic polynomial: $$Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i<j\le n} ((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2)^2.$$ $\endgroup$
    – Robert Bryant
    Commented Jul 31, 2020 at 19:52
  • $\begingroup$ @RobertBryant, ah yes true! The famous sum of squares trick, when working over $\mathbb{R}$. Thank you. How can one obtain all such polynomials? Can one use some form of the positivstellensatz perhaps? $\endgroup$
    – Malkoun
    Commented Jul 31, 2020 at 19:55
  • $\begingroup$ @RobertBryant, could you please write it as an answer? The answer turned out to be simple (and I should have thought about it), but it is guiding me in the right direction (for the problem I am interested in, which inspired this post). Is the post too trivial? Should I delete it? $\endgroup$
    – Malkoun
    Commented Jul 31, 2020 at 20:02

1 Answer 1

Reset to default
16
$\begingroup$

Yes, because the OP stated that the ground field is $\mathbb{R}$, one can simply take the octic polynomial $$ Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i < j\le n} \bigl((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2\bigr)^2, $$ which will do the trick.

Improve this answer
$\endgroup$
4
  • 7
    $\begingroup$ Come to think about it, are the squares necessary? We already know that the expressions inside parentheses are nonnegative. $\endgroup$
    – Malkoun
    Commented Jul 31, 2020 at 20:28
  • 1
    $\begingroup$ @Malkoun: Yes, you are absolutely correct. I didn't think of that, but, yes, simply the sum of the terms would be enough. Of course, this implies that this construction would work for any ordered field. $\endgroup$
    – Robert Bryant
    Commented Jul 31, 2020 at 21:28
  • 7
    $\begingroup$ If one removes the square, then this (now quartic) polynomial also has a geometric interpretation: it is the trace of the exterior square $\bigwedge^2(T^* T)$ of the square $T^* T$ of the linear transformation $T: {\bf R}^n \to V$ that maps the standard basis to $v_1,\dots,v_n$. (It is also the square of the Frobenius norm of $\bigwedge^2 T$.) $\endgroup$
    – Terry Tao
    Commented Aug 1, 2020 at 3:24
  • 1
    $\begingroup$ I guess another way to express what Terry is saying is that the identity $$ (v,v)(w,w) = (v,w)^2 + |v\wedge w|^2 = (v,w)^2+(v{\wedge}w,v{\wedge}w)$$ (with the natural inner product on $\Lambda^2(V)$) already shows that $(v,v)(w,w) - (v,w)^2$ is a sum of squares anyway, so the natural quartic polynomial would be $$P(v_1,\ldots,v_n) = \sum_{1\le i < j\le n} |v_i\wedge v_j|^2.$$ In particular, it is expressed as a sum of squares. $\endgroup$
    – Robert Bryant
    Commented Aug 1, 2020 at 23:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .