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===Gradient of a vector field=== {{see also|Covariant derivative}} Since the [[total derivative]] of a vector field is a [[linear mapping]] from vectors to vectors, it is a [[tensor]] quantity. In rectangular coordinates, the gradient of a vector field {{math|1='''f''' = ( ''f''{{i sup|1}}, ''f''{{i sup|2}}, ''f''{{i sup|3}})}} is defined by: <math display="block">\nabla \mathbf{f}=g^{jk}\frac{\partial f^i}{\partial x^j} \mathbf{e}_i \otimes \mathbf{e}_k,</math> (where the [[Einstein summation notation]] is used and the [[tensor product]] of the vectors {{math|'''e'''<sub>''i''</sub>}} and {{math|'''e'''<sub>''k''</sub>}} is a [[dyadic tensor]] of type (2,0)). Overall, this expression equals the transpose of the Jacobian matrix: <math display="block">\frac{\partial f^i}{\partial x^j} = \frac{\partial (f^1,f^2,f^3)}{\partial (x^1,x^2,x^3)}.</math> In curvilinear coordinates, or more generally on a curved [[Riemannian manifold|manifold]], the gradient involves [[Christoffel symbols]]: <math display="block">\nabla \mathbf{f}=g^{jk}\left(\frac{\partial f^i}{\partial x^j}+{\Gamma^i}_{jl}f^l\right) \mathbf{e}_i \otimes \mathbf{e}_k,</math> where {{math|''g''{{i sup|''jk''}}}} are the components of the inverse [[metric tensor]] and the {{math|'''e'''<sub>''i''</sub>}} are the coordinate basis vectors. Expressed more invariantly, the gradient of a vector field {{math|'''f'''}} can be defined by the [[Levi-Civita connection]] and metric tensor:<ref>{{harvnb|Dubrovin|Fomenko|Novikov|1991|pages=348β349}}.</ref> <math display="block">\nabla^a f^b = g^{ac} \nabla_c f^b ,</math> where {{math|β<sub>''c''</sub>}} is the connection.
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