Let $P_{1},...,P_{k}$ be first order differential operators on $\mathbb{R}^{n}$. I want to show that $P_{1}^{2}+\cdots+P^{2}_{k}$ is never elliptic if $k<n$.

We have

$$ P_{1}^{2}+\cdots+P^{2}_{k}=\sum_{i=1}^{k}P^{2}_{i}=\sum_{i=1}^{k}\sum_{|\alpha|\le 1}a_{\alpha}\left(\frac{\partial^{2\alpha}}{\partial x_{i}^{2\alpha}}\right). $$ Hence $$ \begin{aligned} \mathcal{F}\left(\sum_{i=1}^{k}P^{2}_{i}u\right)(\xi)&=\sum_{n=1}^{k}\sum_{|\alpha|\le 1}a_{\alpha}\xi_{i}^{2\alpha}\hat{u}(\xi). \end{aligned} $$ Thus $\sum_{i=1}^{k}P_{i}$ is elliptic if $$ p_{1}(\xi)=\sum_{i=1}^{k}\sum_{|\alpha|=1}a_{\alpha}\xi_{i}^{2\alpha} $$ has no real zeroes except $\xi=0$. I don't really see it though.


The symbol of a first order differential operator is a linear form; its zero set is a hyperplane of codimension $1$. Squaring does not change the zero set. For $k<n$ such operators, the intersection of zero sets is a hyperplane of codimension at most $k$, because codimension is subadditive under intersection. Hence, the dimension of the zero set is $n-k>0$.

(If it helps, one can translate all statements about codimension into statements about the dimension of orthogonal complement.)

  • $\begingroup$ Unfortunately, I am not familiar with the notion of codimension. $\endgroup$ – Jason Born Jan 26 '17 at 13:39

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