# Orthogonal Part Operator

I have some problems in acting with this operator algebrically. If I define: $$\boldsymbol p(\boldsymbol x)\in \mathbb R^3 \ \ ,\ \ P_{\alpha\beta}:=\delta_{\alpha\beta}-p_\alpha p_\beta$$ how does it affect for instance this kind of terms?

$$\boldsymbol{p}\cdot\nabla\boldsymbol{p}\ \\ \nabla^2\boldsymbol{p}\\ \boldsymbol{p}(\nabla\cdot\boldsymbol{p})$$

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Sorry, can you clarify the domains and how should we consider this $\boldsymbol{p}$? Is it a variable? Is it the argument of $P$? I assume $\boldsymbol{p}\in\Bbb R^n$ is a culomn vector. How do you exactly mean the $P$ operator? Is its matrix $I-\boldsymbol{p}^T\cdot\boldsymbol{p}$, or does $P$ act on the tensor space? – Berci Feb 21 '13 at 12:52
Yes, I need to be clearer. $\boldsymbol{p}$ is a column vector of $\mathbb R^3$, function of $\boldsymbol{x}$, respect to which the differentiation in $\nabla$ is carried. The $P$ operator is exactly the matrix $P_{\alpha\beta}=\delta_{\alpha\beta}-p_\alpha p_\beta$ – usumdelphini Feb 21 '13 at 12:57
Yes, now it's much clearer.. and how do you mean 'How does it affect' the listed terms? Do you mean $P{\boldsymbol p}\cdot \nabla(P{\boldsymbol p})$? – Berci Feb 21 '13 at 15:57
No, I simply meant the application of $P$ to those terms: $P(\boldsymbol{p}\cdot\nabla\boldsymbol{p})$ – usumdelphini Feb 21 '13 at 16:02
But isn't the latter one a scalar? How do you 'plug it' into a matrix? – Berci Feb 21 '13 at 16:51