Elliptic operator with real coefficients on $\mathbb{R}^2$?

Question

Supposedly, an elliptic first order differential operator on $\mathbb{R}^2$ with real coefficients does not exist. But $$p_m(\xi_1,-a_1 \xi_1/ a_2)=-a_1 i \xi_1 +a_1 i /xi_1=0$$ for any real $\xi_1$, so any such operator is elliptic. Why is this wrong?

Answer 1

Why it is wrong?

Because it's a bunch of unexplained notation. It is wrong to write such a thing.

The symbol of a first order operator in $\mathbb R^2$ is a linear form in $\xi_1,\xi_2$: that is, $$p(\xi_1,\xi_2) = a_1\xi_1 + a_2 \xi_2,\quad (\xi_1,\xi_2)\in \mathbb R^2$$ The coefficients $a_1,a_2$ are real numbers (which may depend on a point $x\in\mathbb R^2$, but this does not matter). No matter what $a_1,a_2$ are, the map $$\xi\mapsto a_1\xi_1 + a_2 \xi_2$$ has nontrivial kernel, since it's a map from $\mathbb R^2$ into $\mathbb R$. Thus, the operator is not elliptic