Pullback and exterior derivative

  • Thread starter Silviu
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Then you can use the Cartan formula to show that ##f^* \omega = f^* \varphi \wedge df^*X + f^* d\varphi \wedge df^*X##. Finally you can use the induction hypothesis and the fact that ##df^*X = f^*dX##.In summary, the problem shows that for a given vector field ##\omega## in a manifold ##N## and a function ##f## from another manifold ##M## to ##N##, the exterior derivative of the pullback of ##\omega## is equal to the pullback of the exterior derivative of ##\omega##. This is shown by using the Cartan formula and
  • #1
Silviu
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Homework Statement


Let ##\omega \in \Omega^r(N)## and let ##f:M \to N##. Show that ##d(f^*\omega)=f^*(d\omega)##

Homework Equations


##\Omega^r(N)## is the vector field of r-form at a given point in the manifold N, ##f^*## is the pullback function and ##d## is the exterior derivative

##(f^*\omega)(X_1, . . ., X_r) = \omega(f_* X_1, ..., f_* X_r)##, for r vectors ##X_1## and ##f_*## is the differential map

The Attempt at a Solution


I am not sure how to express the pullback in general, without making it act on some vectors, and I am not sure how to act on vectors when I have the exterior derivative, too. For example in ##d(f^*\omega)## do I first act on r vectors with ##f^*\omega## and then apply ##d##, or I apply ##d## and act with everything on r+1 vectors? Any help would be appreciated. Thank you!
 
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  • #2
Formally you can proceed by induction on ##r## and write ##\omega = \varphi \wedge dX##.
 

Related to Pullback and exterior derivative

1. What is the pullback of a differential form?

The pullback of a differential form is a mathematical operation that allows us to transform a differential form defined on one manifold to another manifold using a smooth map between them. It is denoted by the symbol * and is often used in the study of differential geometry and topology.

2. How is the pullback related to the exterior derivative?

The exterior derivative of a differential form, denoted by d, is a linear map that assigns a new form to the given form by taking the partial derivative of each coefficient with respect to each coordinate. The pullback and exterior derivative are related in that d and * commute with each other, meaning that the pullback of a form followed by the exterior derivative is the same as the exterior derivative of the pullback form.

3. What is the geometric interpretation of the exterior derivative?

The exterior derivative can be viewed as measuring the infinitesimal change in a differential form as we move along a given direction in the manifold. It is also seen as a generalization of the gradient operator in multivariable calculus, where the exterior derivative measures the rate of change of a function in different directions.

4. How is the exterior derivative used in integration?

In integration, the exterior derivative is used to define the integral of a form over a manifold. This integral, known as the de Rham integral, is a generalization of the Riemann integral in calculus and plays a crucial role in the study of differential geometry and topology. It allows us to integrate forms of different degrees over different manifolds, providing a powerful tool for solving a variety of mathematical problems.

5. Are there any applications of pullback and exterior derivative in real-life?

Yes, the concepts of pullback and exterior derivative have many real-life applications. They are used in fields such as physics, engineering, and computer science to model and analyze physical systems, design algorithms, and solve optimization problems. For example, in physics, the exterior derivative is used to describe electromagnetic fields and in computer science, it is used in computer graphics and computer vision.

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