# Higher order generalization of Cauchy-Schwarz?

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.)

• 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 – Kevin Casto Jul 31 at 19:35
• @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. – Malkoun Jul 31 at 19:49
• Actually, there's an octic polynomial: $$Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i<j\le n} ((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2)^2.$$ – Robert Bryant Jul 31 at 19:52
• @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? – Malkoun Jul 31 at 19:55
• @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? – Malkoun Jul 31 at 20:02

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.
• 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 $\bigwedge^2 T$.) – Terry Tao Aug 1 at 3:24
• 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. – Robert Bryant Aug 1 at 23:20