# Positive 4-form

Denote by $$W$$ the space of all symmetric bilinear forms on $$\mathbb{R}^n$$.

Let $$Q$$ be a quadratic form on $$W$$. Suppose that $$Q(b)\geqslant 0$$ for any $$b\in W$$ such that $$b(X,Y)=\ell(X)\cdot\ell(Y)$$ for some linear function $$\ell$$.

Is it possible to find a nonnegative quadratic form $$\tilde Q$$ such that $$\tilde Q(b)=Q(b)$$ for any $$b$$ as above?

• The forms $$b$$ as above (up to sign) can be described by quadratic equations $$b(X,Y)\cdot b(Z,W)=b(X,Z)\cdot b(Y,W).$$ This leads to a more general version of this question, but it has a negative answer (thanks to Terry Tao).

• I know that the answer is yes for $$n=2$$.

(6 October 2023) I'll leave the original argument below because it seems that many people liked it, but, in fact, it wanders around and introduces a lot of unnecessary information, which obscures the essential point, so here is a cleaned up version:

The answer 'yes' when $$n\le 3$$ but 'no' when $$n\ge4$$: Let $$V = \mathbb{R}^n$$, so $$W = S^2(V^*)$$ and the quadratic forms on $$W$$ are $$S^2(W^*) = S^2\bigl(S^2(V^*)^*\bigr) = S^2(S^2(V))$$ (since $$S^2(V^*)$$ is canonically isomorphic to $$S^2(V)^*$$). The (linear) 'multiplication map' $$\mu:S^2(S^2(V))\to S^4(V)$$ is surjective. Thus, each quadratic form on $$W$$ defines a quartic function on $$V^*$$, and every quartic function on $$V^*$$ is of the form $$\mu(Q)$$ for some $$Q\in S^2(W^*)$$.

The question is whether every non-negative quartic function on $$V^*$$ is of the form $$\mu(Q)$$ for some non-negative quadratic form $$Q \in S^2(W^*)$$. This is equivalent to asking whether every non-negative quartic function on $$V^*$$ is a sum of squares of quadratic functions on $$V^*$$.

However, it is known that for $$n\ge4$$, there exist non-negative quartic polynomials $$P$$ on $$V^*$$ that cannot be written as a sum of squares of quadratic polynomials on $$V^*$$, while, for $$n\le 3$$, any non-negative quartic polynomial $$P$$ on $$V^*$$ can be written as a sum of squares of quadratic polynomials.

Original Argument:

The answer 'yes' when $$n\le 3$$ but 'no' when $$n\ge4$$. Here is why:

Let $$V = \mathbb{R}^n$$, so $$W = S^2(V^*)$$ and the quadratic forms on $$W$$ are $$S^2(W^*) = S^2\bigl(S^2(V^*)^*\bigr)$$. It is known that there is a canonical $$\mathrm{GL}(V)$$-invariant exact sequence $$0\longrightarrow \Lambda^4(V)\longrightarrow S^2\bigl(\Lambda^2(V)\bigr)\longrightarrow S^2\bigl(S^2(V^*)^*\bigr)\longrightarrow S^4(V)\longrightarrow 0,$$ thus, there is a $$\mathrm{GL}(V)$$-module $$K(V)$$ such that $$S^2\bigl(\Lambda^2(V)\bigr) = \Lambda^4(V)\oplus K(V) \qquad\text{and}\qquad S^2\bigl(S^2(V^*)^*\bigr) = S^4(V)\oplus K(V).$$ The module $$K(V)$$ is $$\mathrm{GL}(V)$$-irreducible and of dimension $$n^2(n^2{-}1)/12$$. It is exactly the set of quadratic forms $$Q$$ on $$W$$ that vanish on all of the symmetric bilinear forms on $$V$$ of the form $$B(v,w) = \ell(v)\ell(w)$$ for some $$\ell\in V^*$$. (The quadratic forms with this latter property must be some $$\mathrm{GL}(V)$$-invariant submodule of $$S^2\bigl(S^2(V^*)^*\bigr)$$, and it clearly does not contain $$S^4(V)$$.)

Suppose that $$Q$$ be nonnegative on all of the rank 1 elements of $$W$$. This is equivalent to the condition that, when we write $$Q = Q' + Q''$$, with $$Q'\in S^4(V)$$ and $$Q''\in K(V)$$, then $$Q'$$, when considered as a quartic polynomial on $$V^*$$, be a nonnegative quartic polynomial.

If $$\tilde Q$$ were a non-negative quadratic form on $$W$$, then it would be a sum of squares of linear functions on $$W$$, and hence, if $$\tilde Q - Q$$ lay in $$K(V)$$, it would follow that $$Q'$$ would be a sum of squares of quadratic functions on $$V^*$$.

However, it is known that for $$n\ge4$$, there exist non-negative quartic polynomials $$Q'$$ on $$V^*$$ that cannot be written as a sum of squares of quadratic polynomials on $$V^*$$.

Thus, for $$n\ge 4$$, the answer to the OP's question is 'no'.

Meanwhile, when $$n\le 3$$, every non-negative quartic polynomial in $$n$$ variables is a sum of squares of quadratic polynomials in those $$n$$ variables, so the above analysis shows that, when $$Q = Q' + Q''$$ where $$Q'\in S^4(V)$$ is a nonnegative function on $$V^*$$ and $$Q''\in K(V)$$, there does exist a nonnegative quadratic $$\tilde Q\in S^2\bigl(S^2(V^*)^*\bigr)$$ such that $$\tilde Q - Q$$ lies in $$K(V)$$.

• Thank you very much. I guess you wanted to say, but did not say that $K(V)$ is the space of curvature tensors on $V$. Oct 1 at 0:02
• @AntonPetrunin: Yes, of course that's true. It's interesting that $K(V)$ is irreducible as a $\mathrm{GL}(V)$-module, whereas it's reducible as an $\mathrm{SO}(V)$-module (which is the way we usually think of it). Oct 1 at 0:24
• Also, for another way that this $\mathrm{GL}(V)$ modules shows up, see my answer at mathoverflow.net/q/100372. Oct 1 at 0:36
• @PeterLeFanuLumsdaine: There's not that much to say. The usual multiplicity formulae in $\mathrm{GL}(V)$-representation theory show that $S^2(\Lambda^2(V))\simeq \Lambda^4(V)\oplus K(V)$ while $S^2(S^2(V))\simeq S^4(V)\oplus K(V)$, where $K(V)$ is irreducible. The center map of the exact sequence on decomposable elements is $$(v_1\wedge v_2)\circ(v_3\wedge v_4)\longmapsto (v_1\circ v_3)\circ(v_2\circ v_4)- (v_1\circ v_4)\circ(v_2\circ v_3).$$ It's clearly nonzero, so it has to match the two copies of $K(V)$, making $\Lambda^4(V)$ be the kernel and $S^4(V)$ be the cokernel. Oct 2 at 14:06
• @PeterLeFanuLumsdaine: Oh, and I should have pointed out that, since $S^2(V^*)$ is canonically isomorphic to $\bigl(S^2(V)\bigr)^*$, we also have $S^2\bigl(S^2(V^*)^*\bigr)$ is canonically isomorphic to $S^2(S^2(V))$. Oct 2 at 14:10