Quartic forms with orthogonal decomposition

Question

Let's ${\bf x}=(x_1,x_2,\ldots,x_n)^T$ and ${\bf m}$ a symmetric $n\times n$ matrix with orthogonal eigenvectors ${\bf q_i}$ and corresponding eigenvalues $\lambda_i$. Then the quadratic form of ${\bf x}$ can be expressed by projections on an orthogonal space.

$${\bf x}^T{\bf m}{\bf x}=\sum_{i=1}^n \lambda_i \Vert {\bf q_i}^T{\bf x}\Vert^2\tag{1}$$

Simple example: $${x_1} {x_2}-{x_2} {x_3} =\left(x_1,x_2,x_3\right)\begin{pmatrix}0&\frac{1}{2}&0\\\frac{1}{2}&0&-\frac{1}{2}\\0&-\frac{1}{2}&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\\= \frac{1}{\sqrt{2}}\left(-\frac{{x_1}}{2}-\frac{{x_2}}{\sqrt{2}}+\frac{{x_3}}{2}\right)^2-\frac{1}{\sqrt{2}}\left(-\frac{{x_1}}{2}+\frac{{x_2}}{\sqrt{2}}+\frac{{x_3}}{2}\right)^2\tag{2}$$ The two summands on the rhs of eq.$(2)$ result from projections on an orthogonal space.

Now I am looking for an analogous orthogonal decomposition for quartic forms with degree $4$, e.g.:

$${x_1} {x_2}x_3x_4-{x_2} {x_3}x_5x_5=\sum_{i=1}^k d_i\left(\sum_{j=1}^5c_{ij}x_j\right)^4\tag{3}$$

where $d_i,c_{ij}$ are constants to be determined. The $k$ summands on the rhs of eq.$(3)$ shall have the property that they result from projection on an orthogonal space similar as in eq.$(1)$. However, I have no idea if such concept exists beyond degree $2$ and if tensor calculus could help.

In characteristic not $2$ the quadratic form is $q(x)=b(x,x)$ with $b$ bilinear so for any $v$ such that $q(v)\ne 0$ we can look at $W={ w, b(v,w)=0}$ which is a $n-1$-dimensional subspace and $q(cv+w)=q(cv)+q(w)$, then we can repeat with $q$ restricted to $W$. It won't work for higher degree forms. — reuns, Feb 14 '22 at 23:03
@RodrigodeAzevedo This is the sum of squares decomposition. I am looking for a sum of fourths decomposition. — granular_bastard, Feb 18 '22 at 10:39
@granularbastard You are looking for a rather particular sum of quartics. Can you adapt Parrilo's method? — Rodrigo de Azevedo, Feb 18 '22 at 10:43
@RodrigodeAzevedo No, as there we get the form $(a x_1 x_2+b x_2 x_2 )^2+\ldots$ but need the form $(a x_1+b x_2 )^4+\ldots$ — granular_bastard, Feb 18 '22 at 11:14
@granularbastard Then find a way to make $a x_1 x_2 + b x_2^2$ a square! — Rodrigo de Azevedo, Feb 19 '22 at 12:53

Chris Sanders · Answer 1 · 2022-02-16T12:38:07.287

1

Firstly, in your example, $k\leq5$. If you have $(n+1)$ non-zero vectors in $\mathbb{R}^n$, they can't all be orthogonal to each other.

Let me use an even simpler example than yours. No matter which change-of-basis coefficients $c_{ij}$ you choose (by which I mean that the coefficients form an invertible matrix) you will always have

${x_1} {x_2}x_3x_4\neq \sum_{i=1}^n d_i\left(\sum_{j=1}^n c_{ij}x_j\right)^4\tag{4}$

edited Feb 16 '22 at 12:38

answered Feb 15 '22 at 06:37

Chris Sanders

7,592

That not right, e.g. $x_1 x_2 x_3 x_4 = \frac{1}{192} ({x_1}+{x_2}+{x_3}+{x_4})^4+\frac{1}{384} \left(({x_1}+{x_2}-{x_3}-{x_4})^4+({x_1}-{x_2}+{x_3}-{x_4})^4+(-{x_1}+{x_2}+{x_3}-{x_4})^4+({x_1}-{x_2}-{x_3}+{x_4})^4+(-{x_1}+{x_2}-{x_3}+{x_4})^4+(-{x_1}-{x_2}+{x_3}+{x_4})^4\right)+\frac{1}{192} \left(-({x_1}+{x_2}+{x_3}-{x_4})^4-({x_1}+{x_2}-{x_3}+{x_4})^4-({x_1}-{x_2}+{x_3}+{x_4})^4-(-{x_1}+{x_2}+{x_3}+{x_4})^4\right) $ – granular_bastard Feb 15 '22 at 08:46
Sorry, I should have doubly emphasised that the coefficients $c_{ij}$ are a change of basis, i.e. forming an invertible matrix. Your equation wouldn't then be a counterexample. – Chris Sanders Feb 16 '22 at 12:36

Quartic forms with orthogonal decomposition

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