Editing Laplace operator (section)

===Generalization===
The Laplacian of any [[tensor field]] <math>\mathbf{T}</math> ("tensor" includes scalar and vector) is defined as the [[divergence]] of the [[gradient]] of the tensor:
<math display="block">\nabla ^2\mathbf{T} = (\nabla \cdot \nabla) \mathbf{T}.</math>

For the special case where <math>\mathbf{T}</math> is a [[scalar (mathematics)|scalar]] (a tensor of degree zero), the [[Laplacian]] takes on the familiar form.

If <math>\mathbf{T}</math> is a vector (a tensor of first degree), the gradient is a [[covariant derivative]] which results in a tensor of second degree, and the divergence of this is again a vector. The formula for the vector Laplacian above may be used to avoid tensor math and may be shown to be equivalent to the divergence of the [[Jacobian matrix]] shown below for the gradient of a vector:
<math display="block">\nabla \mathbf{T}= (\nabla T_x, \nabla T_y, \nabla T_z) = \begin{bmatrix}
T_{xx} & T_{xy} & T_{xz} \\
T_{yx} & T_{yy} & T_{yz} \\
T_{zx} & T_{zy} & T_{zz}
\end{bmatrix} ,
\text{ where } T_{uv} \equiv \frac{\partial T_u}{\partial v}.</math>

And, in the same manner, a [[dot product]], which evaluates to a vector, of a vector by the gradient of another vector (a tensor of 2nd degree) can be seen as a product of matrices:
<math display="block"> \mathbf{A} \cdot \nabla \mathbf{B}
= \begin{bmatrix} A_x & A_y & A_z \end{bmatrix} \nabla \mathbf{B}
= \begin{bmatrix} \mathbf{A} \cdot \nabla B_x & \mathbf{A} \cdot \nabla B_y & \mathbf{A} \cdot \nabla B_z \end{bmatrix}.</math>
This identity is a coordinate dependent result, and is not general.