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Let $R_{\theta}$ be the rotation by an angle $\theta$.

Is it then true that for multi-indices $\alpha$ of fixed order $j$ and any smooth function $f$ we have

$$\sum_{\vert \alpha \vert=j}(R_{\theta}z)^{\alpha} \partial^{\alpha}f(x) = \sum_{\vert \alpha \vert=j}(z)^{\alpha} (R_{-\theta}\partial)^{\alpha}f(x)$$

It is true for $j=1$, which just follows since $R_{-\theta}$ is the adjoint map of $R_{\theta}.$

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  • $\begingroup$ Can you clarify? "z" versus "x"? And if you're in $\mathbb R^n$ or $\mathbb C^n$, is the "rotate by angle $\theta$ diagonal across every coordinate of $\mathbb C^n$, or... what? And is $(R_\theta z)^\alpha$ a multiplication operator by the monomial? More detail, please? :) $\endgroup$ Jul 25, 2021 at 4:25
  • $\begingroup$ @paulgarrett indeed, they are different arguments, $z$ and $x$, no connection between them. Both of them are arguments in $\mathbb R^2$ let's say and we rotate $z$ as a vector. $(R_{\theta}z)$ is then a vector and by indices $\alpha$ we address the individual entries, so it is a multiplication by monomials. $\endgroup$
    – Dreifuss
    Jul 25, 2021 at 7:27

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If I am understanding your notation correctly (which admittedly, I may not be), I believe this is already false for $j = 2$. Let me write: $$ R_{\theta} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \qquad f = f(x,y) \qquad z = \begin{bmatrix} u \\ v \end{bmatrix} $$ Then for $j = 2$, the LHS is: \begin{align*} \text{LHS} &= (u\cos\theta - v\sin\theta)^2 \frac{\partial^2 f}{\partial x^2} \\ &\hspace{1cm} + (u\cos\theta - v\sin\theta) (u\sin\theta + v\cos\theta) \frac{\partial^2 f}{\partial x \partial y} \\ &\hspace{1cm} + (u\sin\theta + v\cos\theta)^2 \frac{\partial^2 f}{\partial y^2} \end{align*} and the RHS is: \begin{align*} \text{RHS} &= u^2 \left( \cos^2\theta \frac{\partial^2 f}{\partial x^2} + 2\sin\theta\cos\theta \frac{\partial^2 f}{\partial x \partial y} + \sin^2\theta \frac{\partial^2 f}{\partial y^2} \right) \\ &\hspace{1cm} + uv \left( -\sin\theta \cos\theta \frac{\partial^2 f}{\partial x^2} + (\cos^2\theta - \sin^2\theta) \frac{\partial^2 f}{\partial x \partial y} +\sin\theta \cos\theta \frac{\partial^2 f}{\partial y^2} \right) \\ &\hspace{1cm} + v^2 \left( \sin^2\theta \frac{\partial^2 f}{\partial x^2} - 2\sin\theta \cos\theta \frac{\partial^2 f}{\partial x \partial y} + \cos^2\theta \frac{\partial^2 f}{\partial y^2} \right) \end{align*} If you fully expand out the products, you will find that these two quantities differ by: $$ \text{LHS} - \text{RHS} = \sin\theta\cos\theta \left( (v^2-u^2) \frac{\partial^2 f}{\partial x \partial y} - u v \left(\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2}\right) \right) $$ This is certainly not identically zero, so $\text{LHS} \ne \text{RHS}$ in general.

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