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What is the exterior derivative of a function?

The exterior derivative is defined to be the unique ℝ -linear mapping from k -forms to (k + 1) -forms satisfying the following properties: df is the differential of f , for 0 -forms ( smooth functions) f . d(df ) = 0 for any 0 -form (smooth function) f .

What is the exterior derivative of a smooth 1-form?

Proposition 1 (Exterior Derivative of a 1-Form). For any smooth 1-form ω and smooth vector fields X and Y , (1) dω(X,Y)=X(ω(Y))−Y(ω(X))−ω([X,Y]). Proof. Since any smooth 1-form can be expressed locally as a sum of terms of the form udv for smooth functions u and v, it suffices to consider that case. Suppose ω = udv, and X, Y are smooth vector fields.

How do you find the exterior derivative of a differential k-form?

Thinking of a function as a zero-form, the exterior derivative extends linearly to all differential k -forms using the formula when is a -form and where is the wedge product . The exterior derivative of a -form is a -form. For example, for a differential k -form where is the permutation tensor . It is always the case that .

How do you find the exterior derivative of a parallelogram?

The above depicts the exterior derivative of a 1-form d φ ( v, w), which is the sum of φ along the boundary of the completed parallelogram defined by v and w. So if in the diagram ε = 1, we have d φ ( v, w) = ( 2 − 1) − ( 0 − 0) + 3 = 4.


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