# How to decompose a divergence operator

I am reading a paper, and I see someone decompose a divergence operator as follows. Could someone judge and see if it is correct?

$$\nabla \cdot {\bf{v}} = \left( {{\bf{n}} \cdot \nabla } \right){v_n} + {\nabla _\parallel } \cdot {{\bf{v}}_t}$$

Notation: $\bf{v}$ is velocity vector field, and $n$ means normal, $t$ is tangent component.

I do not understand why ${\bf{v}}_t$ is a vector and $v_n$ is not.

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Well, you can keep $\mathbf v_n$ as a vector and decompose $\mathbf v$ into its normal and tangential components as $\mathbf v = \mathbf v_n + \mathbf v_t$. But $\mathbf v_n$ has only one degree of freedom, so it's easier to write $\mathbf v_n = v_n \mathbf n$ and deal with only a scalar $v_n$. The tangential component on the other hand has two degrees of freedom in three dimensions, so you can't do this... Unless you introduce a basis, say vectors $\mathbf a, \mathbf b$, on the tangent plane, and then write $\mathbf v_t = v_a \mathbf a + v_b \mathbf b$ and $\nabla_\parallel\cdot\mathbf v_t = (\mathbf a\cdot\nabla)v_a + (\mathbf b\cdot\nabla)v_b$.