Let $f_1,\ldots,f_r \in \mathbb{R}[x_1,\ldots,x_n]$ be $r$ homogeneous polynomials of the same odd degree $d$, where $d \in \{3,5,7,\ldots\}$.

For which values of $r,n,d$ there exists a real solution $a=(a_1,\ldots,a_n) \in \mathbb{R}^n-\{(0,\ldots,0)\}$ to the system of $r$ inequalities $f_i \geq 0$?

For $r=2$ the answer is positive, namely, there exists such $a$, by the following argument that can be found in the answer to this question: By continuity there exists $t \in \mathbb{R}^n$ with $|t| = 1$ and $f_1(t) = f_1(-t) = 0$. Clearly, at least one of $\{f_2(t),f_2(-t)\}$ is $\geq 0$. Then for $s=t$ or $s=-t$, we have: $f_1(s)=0$ and $f_2(s) \geq 0$.

Actually, I am interested in the case $n=4$; I really apologize that I am not explaining why (it is quite complicated; hopefully later I will be able to explain).


(1) I wonder if the following idea may yield an answer to my question: For example, $n=4,d=3$, and order the monomials: $y_1:=x_1^3,y_2:=x_1^2x_2,y_3:=x_1^2x_3,y_4:=x_1^2x_4,x_1x_2x_1,\ldots,y_m:=x_4^3$. Then the system of $r$ inequalities $f_i \geq 0$ becomes a linear system in $m$ new variables $y_1,\ldots,y_m$. Now, $0=(0,\ldots,0) \in \mathbb{R}^m$ is a solution, and if the associated matrix is invertible, then $0$ is a unique solution, but this does not imply that the original system has $0 \in \mathbb{R}^n$ as a unique solution, so there is hope to get non-trivial solutions. On the other hand, if the associated matrix is not invertible, then there are non-trivial solutions, but I am not sure if it is possible to obtain from one of them a solution to the original system, probably not necessarily.

(2) This paper seems relevant.

(3) I have already asked the above question here, but have not received any comments, neither to it nor to another similar question.

Any hints and comments are welcome!

Edit: I have now found this question which almost answers my question, provided that I understood it correctly and if we take $k=0$ in that question notations, then we get that my question, if we replace $\mathbb{R}$ by $\mathbb{C}$, has a positive answer (= a common zero for all the $f_i$'s) for $n > r$.


1 Answer 1


I cannot say much about $\geq0$. For $=0$ one can prove the following statement: if $r\leq n-1$ then then there is an $a\in \mathbf R^n\setminus\{0\}$ such that $f_i(a)=0$ for all $i$.

Indeed, each polynomial $f_i$ defines a vanishing set $Z(f_i)$ in real projective $(n-1)$-space $\mathbf P^{n-1}(\mathbf R)$. Since $f_i$ is of odd degree, each vanishing set $Z(f_i)$ is homologically nontrivial, i.e., represents the nonzero homology class of $H_{n-2}(\mathbf P^{n-1}(\mathbf R),\mathbf Z/2)$. Since $r\leq n-1$, the intersection product of the vanishing sets is homologically nontrivial. In particular, the intersection $\bigcap_{i=1}^r Z(f_i)$ is nonempty, i.e., there is an $a\in\mathbf R^n\setminus\{0\}$ such that $f_i(a)=0$ for all $i$. This proves the statement above.

By your argument for the case $r=2$, you can then deduce that there is an $a\in\mathbf R^n\setminus\{0\}$ such that $f_i(a)\geq0$ for all $i$, if $r\leq n$.


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